r"""
This module contains :py:meth:`~sympy.solvers.ode.dsolve` and different helper
functions that it uses.
:py:meth:`~sympy.solvers.ode.dsolve` solves ordinary differential equations.
See the docstring on the various functions for their uses. Note that partial
differential equations support is in ``pde.py``. Note that hint functions
have docstrings describing their various methods, but they are intended for
internal use. Use ``dsolve(ode, func, hint=hint)`` to solve an ODE using a
specific hint. See also the docstring on
:py:meth:`~sympy.solvers.ode.dsolve`.
**Functions in this module**
These are the user functions in this module:
- :py:meth:`~sympy.solvers.ode.dsolve` - Solves ODEs.
- :py:meth:`~sympy.solvers.ode.classify_ode` - Classifies ODEs into
possible hints for :py:meth:`~sympy.solvers.ode.dsolve`.
- :py:meth:`~sympy.solvers.ode.checkodesol` - Checks if an equation is the
solution to an ODE.
- :py:meth:`~sympy.solvers.ode.homogeneous_order` - Returns the
homogeneous order of an expression.
- :py:meth:`~sympy.solvers.ode.infinitesimals` - Returns the infinitesimals
of the Lie group of point transformations of an ODE, such that it is
invariant.
- :py:meth:`~sympy.solvers.ode_checkinfsol` - Checks if the given infinitesimals
are the actual infinitesimals of a first order ODE.
These are the non-solver helper functions that are for internal use. The
user should use the various options to
:py:meth:`~sympy.solvers.ode.dsolve` to obtain the functionality provided
by these functions:
- :py:meth:`~sympy.solvers.ode.odesimp` - Does all forms of ODE
simplification.
- :py:meth:`~sympy.solvers.ode.ode_sol_simplicity` - A key function for
comparing solutions by simplicity.
- :py:meth:`~sympy.solvers.ode.constantsimp` - Simplifies arbitrary
constants.
- :py:meth:`~sympy.solvers.ode.constant_renumber` - Renumber arbitrary
constants.
- :py:meth:`~sympy.solvers.ode._handle_Integral` - Evaluate unevaluated
Integrals.
See also the docstrings of these functions.
**Currently implemented solver methods**
The following methods are implemented for solving ordinary differential
equations. See the docstrings of the various hint functions for more
information on each (run ``help(ode)``):
- 1st order separable differential equations.
- 1st order differential equations whose coefficients or `dx` and `dy` are
functions homogeneous of the same order.
- 1st order exact differential equations.
- 1st order linear differential equations.
- 1st order Bernoulli differential equations.
- Power series solutions for first order differential equations.
- Lie Group method of solving first order differential equations.
- 2nd order Liouville differential equations.
- Power series solutions for second order differential equations
at ordinary and regular singular points.
- `n`\th order linear homogeneous differential equation with constant
coefficients.
- `n`\th order linear inhomogeneous differential equation with constant
coefficients using the method of undetermined coefficients.
- `n`\th order linear inhomogeneous differential equation with constant
coefficients using the method of variation of parameters.
**Philosophy behind this module**
This module is designed to make it easy to add new ODE solving methods without
having to mess with the solving code for other methods. The idea is that
there is a :py:meth:`~sympy.solvers.ode.classify_ode` function, which takes in
an ODE and tells you what hints, if any, will solve the ODE. It does this
without attempting to solve the ODE, so it is fast. Each solving method is a
hint, and it has its own function, named ``ode_<hint>``. That function takes
in the ODE and any match expression gathered by
:py:meth:`~sympy.solvers.ode.classify_ode` and returns a solved result. If
this result has any integrals in it, the hint function will return an
unevaluated :py:class:`~sympy.integrals.Integral` class.
:py:meth:`~sympy.solvers.ode.dsolve`, which is the user wrapper function
around all of this, will then call :py:meth:`~sympy.solvers.ode.odesimp` on
the result, which, among other things, will attempt to solve the equation for
the dependent variable (the function we are solving for), simplify the
arbitrary constants in the expression, and evaluate any integrals, if the hint
allows it.
**How to add new solution methods**
If you have an ODE that you want :py:meth:`~sympy.solvers.ode.dsolve` to be
able to solve, try to avoid adding special case code here. Instead, try
finding a general method that will solve your ODE, as well as others. This
way, the :py:mod:`~sympy.solvers.ode` module will become more robust, and
unhindered by special case hacks. WolphramAlpha and Maple's
DETools[odeadvisor] function are two resources you can use to classify a
specific ODE. It is also better for a method to work with an `n`\th order ODE
instead of only with specific orders, if possible.
To add a new method, there are a few things that you need to do. First, you
need a hint name for your method. Try to name your hint so that it is
unambiguous with all other methods, including ones that may not be implemented
yet. If your method uses integrals, also include a ``hint_Integral`` hint.
If there is more than one way to solve ODEs with your method, include a hint
for each one, as well as a ``<hint>_best`` hint. Your ``ode_<hint>_best()``
function should choose the best using min with ``ode_sol_simplicity`` as the
key argument. See
:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best`, for example.
The function that uses your method will be called ``ode_<hint>()``, so the
hint must only use characters that are allowed in a Python function name
(alphanumeric characters and the underscore '``_``' character). Include a
function for every hint, except for ``_Integral`` hints
(:py:meth:`~sympy.solvers.ode.dsolve` takes care of those automatically).
Hint names should be all lowercase, unless a word is commonly capitalized
(such as Integral or Bernoulli). If you have a hint that you do not want to
run with ``all_Integral`` that doesn't have an ``_Integral`` counterpart (such
as a best hint that would defeat the purpose of ``all_Integral``), you will
need to remove it manually in the :py:meth:`~sympy.solvers.ode.dsolve` code.
See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for
guidelines on writing a hint name.
Determine *in general* how the solutions returned by your method compare with
other methods that can potentially solve the same ODEs. Then, put your hints
in the :py:data:`~sympy.solvers.ode.allhints` tuple in the order that they
should be called. The ordering of this tuple determines which hints are
default. Note that exceptions are ok, because it is easy for the user to
choose individual hints with :py:meth:`~sympy.solvers.ode.dsolve`. In
general, ``_Integral`` variants should go at the end of the list, and
``_best`` variants should go before the various hints they apply to. For
example, the ``undetermined_coefficients`` hint comes before the
``variation_of_parameters`` hint because, even though variation of parameters
is more general than undetermined coefficients, undetermined coefficients
generally returns cleaner results for the ODEs that it can solve than
variation of parameters does, and it does not require integration, so it is
much faster.
Next, you need to have a match expression or a function that matches the type
of the ODE, which you should put in :py:meth:`~sympy.solvers.ode.classify_ode`
(if the match function is more than just a few lines, like
:py:meth:`~sympy.solvers.ode._undetermined_coefficients_match`, it should go
outside of :py:meth:`~sympy.solvers.ode.classify_ode`). It should match the
ODE without solving for it as much as possible, so that
:py:meth:`~sympy.solvers.ode.classify_ode` remains fast and is not hindered by
bugs in solving code. Be sure to consider corner cases. For example, if your
solution method involves dividing by something, make sure you exclude the case
where that division will be 0.
In most cases, the matching of the ODE will also give you the various parts
that you need to solve it. You should put that in a dictionary (``.match()``
will do this for you), and add that as ``matching_hints['hint'] = matchdict``
in the relevant part of :py:meth:`~sympy.solvers.ode.classify_ode`.
:py:meth:`~sympy.solvers.ode.classify_ode` will then send this to
:py:meth:`~sympy.solvers.ode.dsolve`, which will send it to your function as
the ``match`` argument. Your function should be named ``ode_<hint>(eq, func,
order, match)`. If you need to send more information, put it in the ``match``
dictionary. For example, if you had to substitute in a dummy variable in
:py:meth:`~sympy.solvers.ode.classify_ode` to match the ODE, you will need to
pass it to your function using the `match` dict to access it. You can access
the independent variable using ``func.args[0]``, and the dependent variable
(the function you are trying to solve for) as ``func.func``. If, while trying
to solve the ODE, you find that you cannot, raise ``NotImplementedError``.
:py:meth:`~sympy.solvers.ode.dsolve` will catch this error with the ``all``
meta-hint, rather than causing the whole routine to fail.
Add a docstring to your function that describes the method employed. Like
with anything else in SymPy, you will need to add a doctest to the docstring,
in addition to real tests in ``test_ode.py``. Try to maintain consistency
with the other hint functions' docstrings. Add your method to the list at the
top of this docstring. Also, add your method to ``ode.rst`` in the
``docs/src`` directory, so that the Sphinx docs will pull its docstring into
the main SymPy documentation. Be sure to make the Sphinx documentation by
running ``make html`` from within the doc directory to verify that the
docstring formats correctly.
If your solution method involves integrating, use :py:meth:`Integral()
<sympy.integrals.integrals.Integral>` instead of
:py:meth:`~sympy.core.expr.Expr.integrate`. This allows the user to bypass
hard/slow integration by using the ``_Integral`` variant of your hint. In
most cases, calling :py:meth:`sympy.core.basic.Basic.doit` will integrate your
solution. If this is not the case, you will need to write special code in
:py:meth:`~sympy.solvers.ode._handle_Integral`. Arbitrary constants should be
symbols named ``C1``, ``C2``, and so on. All solution methods should return
an equality instance. If you need an arbitrary number of arbitrary constants,
you can use ``constants = numbered_symbols(prefix='C', cls=Symbol, start=1)``.
If it is possible to solve for the dependent function in a general way, do so.
Otherwise, do as best as you can, but do not call solve in your
``ode_<hint>()`` function. :py:meth:`~sympy.solvers.ode.odesimp` will attempt
to solve the solution for you, so you do not need to do that. Lastly, if your
ODE has a common simplification that can be applied to your solutions, you can
add a special case in :py:meth:`~sympy.solvers.ode.odesimp` for it. For
example, solutions returned from the ``1st_homogeneous_coeff`` hints often
have many :py:meth:`~sympy.functions.log` terms, so
:py:meth:`~sympy.solvers.ode.odesimp` calls
:py:meth:`~sympy.simplify.simplify.logcombine` on them (it also helps to write
the arbitrary constant as ``log(C1)`` instead of ``C1`` in this case). Also
consider common ways that you can rearrange your solution to have
:py:meth:`~sympy.solvers.ode.constantsimp` take better advantage of it. It is
better to put simplification in :py:meth:`~sympy.solvers.ode.odesimp` than in
your method, because it can then be turned off with the simplify flag in
:py:meth:`~sympy.solvers.ode.dsolve`. If you have any extraneous
simplification in your function, be sure to only run it using ``if
match.get('simplify', True):``, especially if it can be slow or if it can
reduce the domain of the solution.
Finally, as with every contribution to SymPy, your method will need to be
tested. Add a test for each method in ``test_ode.py``. Follow the
conventions there, i.e., test the solver using ``dsolve(eq, f(x),
hint=your_hint)``, and also test the solution using
:py:meth:`~sympy.solvers.ode.checkodesol` (you can put these in a separate
tests and skip/XFAIL if it runs too slow/doesn't work). Be sure to call your
hint specifically in :py:meth:`~sympy.solvers.ode.dsolve`, that way the test
won't be broken simply by the introduction of another matching hint. If your
method works for higher order (>1) ODEs, you will need to run ``sol =
constant_renumber(sol, 'C', 1, order)`` for each solution, where ``order`` is
the order of the ODE. This is because ``constant_renumber`` renumbers the
arbitrary constants by printing order, which is platform dependent. Try to
test every corner case of your solver, including a range of orders if it is a
`n`\th order solver, but if your solver is slow, such as if it involves hard
integration, try to keep the test run time down.
Feel free to refactor existing hints to avoid duplicating code or creating
inconsistencies. If you can show that your method exactly duplicates an
existing method, including in the simplicity and speed of obtaining the
solutions, then you can remove the old, less general method. The existing
code is tested extensively in ``test_ode.py``, so if anything is broken, one
of those tests will surely fail.
"""
from __future__ import print_function, division
from collections import defaultdict
from itertools import islice
from sympy.core import Add, S, Mul, Pow, oo
from sympy.core.compatibility import ordered, iterable, is_sequence, range
from sympy.core.containers import Tuple
from sympy.core.exprtools import factor_terms
from sympy.core.expr import AtomicExpr, Expr
from sympy.core.function import (Function, Derivative, AppliedUndef, diff,
expand, expand_mul, Subs, _mexpand)
from sympy.core.multidimensional import vectorize
from sympy.core.numbers import NaN, zoo, I, Number
from sympy.core.relational import Equality, Eq
from sympy.core.symbol import Symbol, Wild, Dummy, symbols
from sympy.core.sympify import sympify
from sympy.logic.boolalg import BooleanAtom
from sympy.functions import cos, exp, im, log, re, sin, tan, sqrt, \
atan2, conjugate
from sympy.functions.combinatorial.factorials import factorial
from sympy.integrals.integrals import Integral, integrate
from sympy.matrices import wronskian, Matrix, eye, zeros
from sympy.polys import (Poly, RootOf, rootof, terms_gcd,
PolynomialError, lcm)
from sympy.polys.polyroots import roots_quartic
from sympy.polys.polytools import cancel, degree, div
from sympy.series import Order
from sympy.series.series import series
from sympy.simplify import collect, logcombine, powsimp, separatevars, \
simplify, trigsimp, denom, posify, cse
from sympy.simplify.powsimp import powdenest
from sympy.simplify.radsimp import collect_const
from sympy.solvers import solve
from sympy.solvers.pde import pdsolve
from sympy.utilities import numbered_symbols, default_sort_key, sift
from sympy.solvers.deutils import _preprocess, ode_order, _desolve
#: This is a list of hints in the order that they should be preferred by
#: :py:meth:`~sympy.solvers.ode.classify_ode`. In general, hints earlier in the
#: list should produce simpler solutions than those later in the list (for
#: ODEs that fit both). For now, the order of this list is based on empirical
#: observations by the developers of SymPy.
#:
#: The hint used by :py:meth:`~sympy.solvers.ode.dsolve` for a specific ODE
#: can be overridden (see the docstring).
#:
#: In general, ``_Integral`` hints are grouped at the end of the list, unless
#: there is a method that returns an unevaluable integral most of the time
#: (which go near the end of the list anyway). ``default``, ``all``,
#: ``best``, and ``all_Integral`` meta-hints should not be included in this
#: list, but ``_best`` and ``_Integral`` hints should be included.
allhints = (
"separable",
"1st_exact",
"1st_linear",
"Bernoulli",
"Riccati_special_minus2",
"1st_homogeneous_coeff_best",
"1st_homogeneous_coeff_subs_indep_div_dep",
"1st_homogeneous_coeff_subs_dep_div_indep",
"almost_linear",
"linear_coefficients",
"separable_reduced",
"1st_power_series",
"lie_group",
"nth_linear_constant_coeff_homogeneous",
"nth_linear_euler_eq_homogeneous",
"nth_linear_constant_coeff_undetermined_coefficients",
"nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients",
"nth_linear_constant_coeff_variation_of_parameters",
"nth_linear_euler_eq_nonhomogeneous_variation_of_parameters",
"Liouville",
"2nd_power_series_ordinary",
"2nd_power_series_regular",
"separable_Integral",
"1st_exact_Integral",
"1st_linear_Integral",
"Bernoulli_Integral",
"1st_homogeneous_coeff_subs_indep_div_dep_Integral",
"1st_homogeneous_coeff_subs_dep_div_indep_Integral",
"almost_linear_Integral",
"linear_coefficients_Integral",
"separable_reduced_Integral",
"nth_linear_constant_coeff_variation_of_parameters_Integral",
"nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral",
"Liouville_Integral",
)
lie_heuristics = (
"abaco1_simple",
"abaco1_product",
"abaco2_similar",
"abaco2_unique_unknown",
"abaco2_unique_general",
"linear",
"function_sum",
"bivariate",
"chi"
)
def sub_func_doit(eq, func, new):
r"""
When replacing the func with something else, we usually want the
derivative evaluated, so this function helps in making that happen.
To keep subs from having to look through all derivatives, we mask them off
with dummy variables, do the func sub, and then replace masked-off
derivatives with their doit values.
Examples
========
>>> from sympy import Derivative, symbols, Function
>>> from sympy.solvers.ode import sub_func_doit
>>> x, z = symbols('x, z')
>>> y = Function('y')
>>> sub_func_doit(3*Derivative(y(x), x) - 1, y(x), x)
2
>>> sub_func_doit(x*Derivative(y(x), x) - y(x)**2 + y(x), y(x),
... 1/(x*(z + 1/x)))
x*(-1/(x**2*(z + 1/x)) + 1/(x**3*(z + 1/x)**2)) + 1/(x*(z + 1/x))
...- 1/(x**2*(z + 1/x)**2)
"""
reps = {}
repu = {}
for d in eq.atoms(Derivative):
u = Dummy('u')
repu[u] = d.subs(func, new).doit()
reps[d] = u
return eq.subs(reps).subs(func, new).subs(repu)
def get_numbered_constants(eq, num=1, start=1, prefix='C'):
"""
Returns a list of constants that do not occur
in eq already.
"""
if isinstance(eq, Expr):
eq = [eq]
elif not iterable(eq):
raise ValueError("Expected Expr or iterable but got %s" % eq)
atom_set = set().union(*[i.free_symbols for i in eq])
ncs = numbered_symbols(start=start, prefix=prefix, exclude=atom_set)
Cs = [next(ncs) for i in range(num)]
return (Cs[0] if num == 1 else tuple(Cs))
[docs]def dsolve(eq, func=None, hint="default", simplify=True,
ics= None, xi=None, eta=None, x0=0, n=6, **kwargs):
r"""
Solves any (supported) kind of ordinary differential equation and
system of ordinary differential equations.
For single ordinary differential equation
=========================================
It is classified under this when number of equation in ``eq`` is one.
**Usage**
``dsolve(eq, f(x), hint)`` -> Solve ordinary differential equation
``eq`` for function ``f(x)``, using method ``hint``.
**Details**
``eq`` can be any supported ordinary differential equation (see the
:py:mod:`~sympy.solvers.ode` docstring for supported methods).
This can either be an :py:class:`~sympy.core.relational.Equality`,
or an expression, which is assumed to be equal to ``0``.
``f(x)`` is a function of one variable whose derivatives in that
variable make up the ordinary differential equation ``eq``. In
many cases it is not necessary to provide this; it will be
autodetected (and an error raised if it couldn't be detected).
``hint`` is the solving method that you want dsolve to use. Use
``classify_ode(eq, f(x))`` to get all of the possible hints for an
ODE. The default hint, ``default``, will use whatever hint is
returned first by :py:meth:`~sympy.solvers.ode.classify_ode`. See
Hints below for more options that you can use for hint.
``simplify`` enables simplification by
:py:meth:`~sympy.solvers.ode.odesimp`. See its docstring for more
information. Turn this off, for example, to disable solving of
solutions for ``func`` or simplification of arbitrary constants.
It will still integrate with this hint. Note that the solution may
contain more arbitrary constants than the order of the ODE with
this option enabled.
``xi`` and ``eta`` are the infinitesimal functions of an ordinary
differential equation. They are the infinitesimals of the Lie group
of point transformations for which the differential equation is
invariant. The user can specify values for the infinitesimals. If
nothing is specified, ``xi`` and ``eta`` are calculated using
:py:meth:`~sympy.solvers.ode.infinitesimals` with the help of various
heuristics.
``ics`` is the set of boundary conditions for the differential equation.
It should be given in the form of ``{f(x0): x1, f(x).diff(x).subs(x, x2):
x3}`` and so on. For now initial conditions are implemented only for
power series solutions of first-order differential equations which should
be given in the form of ``{f(x0): x1}`` (See issue 4720). If nothing is
specified for this case ``f(0)`` is assumed to be ``C0`` and the power
series solution is calculated about 0.
``x0`` is the point about which the power series solution of a differential
equation is to be evaluated.
``n`` gives the exponent of the dependent variable up to which the power series
solution of a differential equation is to be evaluated.
**Hints**
Aside from the various solving methods, there are also some meta-hints
that you can pass to :py:meth:`~sympy.solvers.ode.dsolve`:
``default``:
This uses whatever hint is returned first by
:py:meth:`~sympy.solvers.ode.classify_ode`. This is the
default argument to :py:meth:`~sympy.solvers.ode.dsolve`.
``all``:
To make :py:meth:`~sympy.solvers.ode.dsolve` apply all
relevant classification hints, use ``dsolve(ODE, func,
hint="all")``. This will return a dictionary of
``hint:solution`` terms. If a hint causes dsolve to raise the
``NotImplementedError``, value of that hint's key will be the
exception object raised. The dictionary will also include
some special keys:
- ``order``: The order of the ODE. See also
:py:meth:`~sympy.solvers.deutils.ode_order` in
``deutils.py``.
- ``best``: The simplest hint; what would be returned by
``best`` below.
- ``best_hint``: The hint that would produce the solution
given by ``best``. If more than one hint produces the best
solution, the first one in the tuple returned by
:py:meth:`~sympy.solvers.ode.classify_ode` is chosen.
- ``default``: The solution that would be returned by default.
This is the one produced by the hint that appears first in
the tuple returned by
:py:meth:`~sympy.solvers.ode.classify_ode`.
``all_Integral``:
This is the same as ``all``, except if a hint also has a
corresponding ``_Integral`` hint, it only returns the
``_Integral`` hint. This is useful if ``all`` causes
:py:meth:`~sympy.solvers.ode.dsolve` to hang because of a
difficult or impossible integral. This meta-hint will also be
much faster than ``all``, because
:py:meth:`~sympy.core.expr.Expr.integrate` is an expensive
routine.
``best``:
To have :py:meth:`~sympy.solvers.ode.dsolve` try all methods
and return the simplest one. This takes into account whether
the solution is solvable in the function, whether it contains
any Integral classes (i.e. unevaluatable integrals), and
which one is the shortest in size.
See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for
more info on hints, and the :py:mod:`~sympy.solvers.ode` docstring for
a list of all supported hints.
**Tips**
- You can declare the derivative of an unknown function this way:
>>> from sympy import Function, Derivative
>>> from sympy.abc import x # x is the independent variable
>>> f = Function("f")(x) # f is a function of x
>>> # f_ will be the derivative of f with respect to x
>>> f_ = Derivative(f, x)
- See ``test_ode.py`` for many tests, which serves also as a set of
examples for how to use :py:meth:`~sympy.solvers.ode.dsolve`.
- :py:meth:`~sympy.solvers.ode.dsolve` always returns an
:py:class:`~sympy.core.relational.Equality` class (except for the
case when the hint is ``all`` or ``all_Integral``). If possible, it
solves the solution explicitly for the function being solved for.
Otherwise, it returns an implicit solution.
- Arbitrary constants are symbols named ``C1``, ``C2``, and so on.
- Because all solutions should be mathematically equivalent, some
hints may return the exact same result for an ODE. Often, though,
two different hints will return the same solution formatted
differently. The two should be equivalent. Also note that sometimes
the values of the arbitrary constants in two different solutions may
not be the same, because one constant may have "absorbed" other
constants into it.
- Do ``help(ode.ode_<hintname>)`` to get help more information on a
specific hint, where ``<hintname>`` is the name of a hint without
``_Integral``.
For system of ordinary differential equations
=============================================
**Usage**
``dsolve(eq, func)`` -> Solve a system of ordinary differential
equations ``eq`` for ``func`` being list of functions including
`x(t)`, `y(t)`, `z(t)` where number of functions in the list depends
upon the number of equations provided in ``eq``.
**Details**
``eq`` can be any supported system of ordinary differential equations
This can either be an :py:class:`~sympy.core.relational.Equality`,
or an expression, which is assumed to be equal to ``0``.
``func`` holds ``x(t)`` and ``y(t)`` being functions of one variable which
together with some of their derivatives make up the system of ordinary
differential equation ``eq``. It is not necessary to provide this; it
will be autodetected (and an error raised if it couldn't be detected).
**Hints**
The hints are formed by parameters returned by classify_sysode, combining
them give hints name used later for forming method name.
Examples
========
>>> from sympy import Function, dsolve, Eq, Derivative, sin, cos, symbols
>>> from sympy.abc import x
>>> f = Function('f')
>>> dsolve(Derivative(f(x), x, x) + 9*f(x), f(x))
Eq(f(x), C1*sin(3*x) + C2*cos(3*x))
>>> eq = sin(x)*cos(f(x)) + cos(x)*sin(f(x))*f(x).diff(x)
>>> dsolve(eq, hint='1st_exact')
[Eq(f(x), -acos(C1/cos(x)) + 2*pi), Eq(f(x), acos(C1/cos(x)))]
>>> dsolve(eq, hint='almost_linear')
[Eq(f(x), -acos(C1/sqrt(-cos(x)**2)) + 2*pi), Eq(f(x), acos(C1/sqrt(-cos(x)**2)))]
>>> t = symbols('t')
>>> x, y = symbols('x, y', function=True)
>>> eq = (Eq(Derivative(x(t),t), 12*t*x(t) + 8*y(t)), Eq(Derivative(y(t),t), 21*x(t) + 7*t*y(t)))
>>> dsolve(eq)
[Eq(x(t), C1*x0 + C2*x0*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0**2, t)),
Eq(y(t), C1*y0 + C2(y0*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0**2, t) +
exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0))]
>>> eq = (Eq(Derivative(x(t),t),x(t)*y(t)*sin(t)), Eq(Derivative(y(t),t),y(t)**2*sin(t)))
>>> dsolve(eq)
set([Eq(x(t), -exp(C1)/(C2*exp(C1) - cos(t))), Eq(y(t), -1/(C1 - cos(t)))])
"""
if iterable(eq):
match = classify_sysode(eq, func)
eq = match['eq']
order = match['order']
func = match['func']
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
# keep highest order term coefficient positive
for i in range(len(eq)):
for func_ in func:
if isinstance(func_, list):
pass
else:
if eq[i].coeff(diff(func[i],t,ode_order(eq[i], func[i]))).is_negative:
eq[i] = -eq[i]
match['eq'] = eq
if len(set(order.values()))!=1:
raise ValueError("It solves only those systems of equations whose orders are equal")
match['order'] = list(order.values())[0]
def recur_len(l):
return sum(recur_len(item) if isinstance(item,list) else 1 for item in l)
if recur_len(func) != len(eq):
raise ValueError("dsolve() and classify_sysode() work with "
"number of functions being equal to number of equations")
if match['type_of_equation'] is None:
raise NotImplementedError
else:
if match['is_linear'] == True:
if match['no_of_equation'] > 3:
solvefunc = globals()['sysode_linear_neq_order%(order)s' % match]
else:
solvefunc = globals()['sysode_linear_%(no_of_equation)seq_order%(order)s' % match]
else:
solvefunc = globals()['sysode_nonlinear_%(no_of_equation)seq_order%(order)s' % match]
sols = solvefunc(match)
return sols
else:
given_hint = hint # hint given by the user
# See the docstring of _desolve for more details.
hints = _desolve(eq, func=func,
hint=hint, simplify=True, xi=xi, eta=eta, type='ode', ics=ics,
x0=x0, n=n, **kwargs)
eq = hints.pop('eq', eq)
all_ = hints.pop('all', False)
if all_:
retdict = {}
failed_hints = {}
gethints = classify_ode(eq, dict=True)
orderedhints = gethints['ordered_hints']
for hint in hints:
try:
rv = _helper_simplify(eq, hint, hints[hint], simplify)
except NotImplementedError as detail:
failed_hints[hint] = detail
else:
retdict[hint] = rv
func = hints[hint]['func']
retdict['best'] = min(list(retdict.values()), key=lambda x:
ode_sol_simplicity(x, func, trysolving=not simplify))
if given_hint == 'best':
return retdict['best']
for i in orderedhints:
if retdict['best'] == retdict.get(i, None):
retdict['best_hint'] = i
break
retdict['default'] = gethints['default']
retdict['order'] = gethints['order']
retdict.update(failed_hints)
return retdict
else:
# The key 'hint' stores the hint needed to be solved for.
hint = hints['hint']
return _helper_simplify(eq, hint, hints, simplify)
def _helper_simplify(eq, hint, match, simplify=True, **kwargs):
r"""
Helper function of dsolve that calls the respective
:py:mod:`~sympy.solvers.ode` functions to solve for the ordinary
differential equations. This minimises the computation in calling
:py:meth:`~sympy.solvers.deutils._desolve` multiple times.
"""
r = match
if hint.endswith('_Integral'):
solvefunc = globals()['ode_' + hint[:-len('_Integral')]]
else:
solvefunc = globals()['ode_' + hint]
func = r['func']
order = r['order']
match = r[hint]
if simplify:
# odesimp() will attempt to integrate, if necessary, apply constantsimp(),
# attempt to solve for func, and apply any other hint specific
# simplifications
sols = solvefunc(eq, func, order, match)
free = eq.free_symbols
cons = lambda s: s.free_symbols.difference(free)
if isinstance(sols, Expr):
return odesimp(sols, func, order, cons(sols), hint)
return [odesimp(s, func, order, cons(s), hint) for s in sols]
else:
# We still want to integrate (you can disable it separately with the hint)
match['simplify'] = False # Some hints can take advantage of this option
rv = _handle_Integral(solvefunc(eq, func, order, match),
func, order, hint)
return rv
[docs]def classify_ode(eq, func=None, dict=False, ics=None, **kwargs):
r"""
Returns a tuple of possible :py:meth:`~sympy.solvers.ode.dsolve`
classifications for an ODE.
The tuple is ordered so that first item is the classification that
:py:meth:`~sympy.solvers.ode.dsolve` uses to solve the ODE by default. In
general, classifications at the near the beginning of the list will
produce better solutions faster than those near the end, thought there are
always exceptions. To make :py:meth:`~sympy.solvers.ode.dsolve` use a
different classification, use ``dsolve(ODE, func,
hint=<classification>)``. See also the
:py:meth:`~sympy.solvers.ode.dsolve` docstring for different meta-hints
you can use.
If ``dict`` is true, :py:meth:`~sympy.solvers.ode.classify_ode` will
return a dictionary of ``hint:match`` expression terms. This is intended
for internal use by :py:meth:`~sympy.solvers.ode.dsolve`. Note that
because dictionaries are ordered arbitrarily, this will most likely not be
in the same order as the tuple.
You can get help on different hints by executing
``help(ode.ode_hintname)``, where ``hintname`` is the name of the hint
without ``_Integral``.
See :py:data:`~sympy.solvers.ode.allhints` or the
:py:mod:`~sympy.solvers.ode` docstring for a list of all supported hints
that can be returned from :py:meth:`~sympy.solvers.ode.classify_ode`.
Notes
=====
These are remarks on hint names.
``_Integral``
If a classification has ``_Integral`` at the end, it will return the
expression with an unevaluated :py:class:`~sympy.integrals.Integral`
class in it. Note that a hint may do this anyway if
:py:meth:`~sympy.core.expr.Expr.integrate` cannot do the integral,
though just using an ``_Integral`` will do so much faster. Indeed, an
``_Integral`` hint will always be faster than its corresponding hint
without ``_Integral`` because
:py:meth:`~sympy.core.expr.Expr.integrate` is an expensive routine.
If :py:meth:`~sympy.solvers.ode.dsolve` hangs, it is probably because
:py:meth:`~sympy.core.expr.Expr.integrate` is hanging on a tough or
impossible integral. Try using an ``_Integral`` hint or
``all_Integral`` to get it return something.
Note that some hints do not have ``_Integral`` counterparts. This is
because :py:meth:`~sympy.solvers.ode.integrate` is not used in solving
the ODE for those method. For example, `n`\th order linear homogeneous
ODEs with constant coefficients do not require integration to solve,
so there is no ``nth_linear_homogeneous_constant_coeff_Integrate``
hint. You can easily evaluate any unevaluated
:py:class:`~sympy.integrals.Integral`\s in an expression by doing
``expr.doit()``.
Ordinals
Some hints contain an ordinal such as ``1st_linear``. This is to help
differentiate them from other hints, as well as from other methods
that may not be implemented yet. If a hint has ``nth`` in it, such as
the ``nth_linear`` hints, this means that the method used to applies
to ODEs of any order.
``indep`` and ``dep``
Some hints contain the words ``indep`` or ``dep``. These reference
the independent variable and the dependent function, respectively. For
example, if an ODE is in terms of `f(x)`, then ``indep`` will refer to
`x` and ``dep`` will refer to `f`.
``subs``
If a hints has the word ``subs`` in it, it means the the ODE is solved
by substituting the expression given after the word ``subs`` for a
single dummy variable. This is usually in terms of ``indep`` and
``dep`` as above. The substituted expression will be written only in
characters allowed for names of Python objects, meaning operators will
be spelled out. For example, ``indep``/``dep`` will be written as
``indep_div_dep``.
``coeff``
The word ``coeff`` in a hint refers to the coefficients of something
in the ODE, usually of the derivative terms. See the docstring for
the individual methods for more info (``help(ode)``). This is
contrast to ``coefficients``, as in ``undetermined_coefficients``,
which refers to the common name of a method.
``_best``
Methods that have more than one fundamental way to solve will have a
hint for each sub-method and a ``_best`` meta-classification. This
will evaluate all hints and return the best, using the same
considerations as the normal ``best`` meta-hint.
Examples
========
>>> from sympy import Function, classify_ode, Eq
>>> from sympy.abc import x
>>> f = Function('f')
>>> classify_ode(Eq(f(x).diff(x), 0), f(x))
('separable', '1st_linear', '1st_homogeneous_coeff_best',
'1st_homogeneous_coeff_subs_indep_div_dep',
'1st_homogeneous_coeff_subs_dep_div_indep',
'1st_power_series', 'lie_group',
'nth_linear_constant_coeff_homogeneous',
'separable_Integral', '1st_linear_Integral',
'1st_homogeneous_coeff_subs_indep_div_dep_Integral',
'1st_homogeneous_coeff_subs_dep_div_indep_Integral')
>>> classify_ode(f(x).diff(x, 2) + 3*f(x).diff(x) + 2*f(x) - 4)
('nth_linear_constant_coeff_undetermined_coefficients',
'nth_linear_constant_coeff_variation_of_parameters',
'nth_linear_constant_coeff_variation_of_parameters_Integral')
"""
prep = kwargs.pop('prep', True)
if func and len(func.args) != 1:
raise ValueError("dsolve() and classify_ode() only "
"work with functions of one variable, not %s" % func)
if prep or func is None:
eq, func_ = _preprocess(eq, func)
if func is None:
func = func_
x = func.args[0]
f = func.func
y = Dummy('y')
xi = kwargs.get('xi')
eta = kwargs.get('eta')
terms = kwargs.get('n')
if isinstance(eq, Equality):
if eq.rhs != 0:
return classify_ode(eq.lhs - eq.rhs, func, ics=ics, xi=xi,
n=terms, eta=eta, prep=False)
eq = eq.lhs
order = ode_order(eq, f(x))
# hint:matchdict or hint:(tuple of matchdicts)
# Also will contain "default":<default hint> and "order":order items.
matching_hints = {"order": order}
if not order:
if dict:
matching_hints["default"] = None
return matching_hints
else:
return ()
df = f(x).diff(x)
a = Wild('a', exclude=[f(x)])
b = Wild('b', exclude=[f(x)])
c = Wild('c', exclude=[f(x)])
d = Wild('d', exclude=[df, f(x).diff(x, 2)])
e = Wild('e', exclude=[df])
k = Wild('k', exclude=[df])
n = Wild('n', exclude=[f(x)])
c1 = Wild('c1', exclude=[x])
a2 = Wild('a2', exclude=[x, f(x), df])
b2 = Wild('b2', exclude=[x, f(x), df])
c2 = Wild('c2', exclude=[x, f(x), df])
d2 = Wild('d2', exclude=[x, f(x), df])
a3 = Wild('a3', exclude=[f(x), df, f(x).diff(x, 2)])
b3 = Wild('b3', exclude=[f(x), df, f(x).diff(x, 2)])
c3 = Wild('c3', exclude=[f(x), df, f(x).diff(x, 2)])
r3 = {'xi': xi, 'eta': eta} # Used for the lie_group hint
boundary = {} # Used to extract initial conditions
C1 = Symbol("C1")
eq = expand(eq)
# Preprocessing to get the initial conditions out
if ics is not None:
for funcarg in ics:
# Separating derivatives
if isinstance(funcarg, Subs):
deriv = funcarg.expr
old = funcarg.variables[0]
new = funcarg.point[0]
if isinstance(deriv, Derivative) and isinstance(deriv.args[0],
AppliedUndef) and deriv.args[0].func == f and old == x and not new.has(x):
dorder = ode_order(deriv, x)
temp = 'f' + str(dorder)
boundary.update({temp: new, temp + 'val': ics[funcarg]})
else:
raise ValueError("Enter valid boundary conditions for Derivatives")
# Separating functions
elif isinstance(funcarg, AppliedUndef):
if funcarg.func == f and len(funcarg.args) == 1 and \
not funcarg.args[0].has(x):
boundary.update({'f0': funcarg.args[0], 'f0val': ics[funcarg]})
else:
raise ValueError("Enter valid boundary conditions for Function")
else:
raise ValueError("Enter boundary conditions of the form ics "
" = {f(point}: value, f(point).diff(point, order).subs(arg, point) "
":value")
# Precondition to try remove f(x) from highest order derivative
reduced_eq = None
if eq.is_Add:
deriv_coef = eq.coeff(f(x).diff(x, order))
if deriv_coef not in (1, 0):
r = deriv_coef.match(a*f(x)**c1)
if r and r[c1]:
den = f(x)**r[c1]
reduced_eq = Add(*[arg/den for arg in eq.args])
if not reduced_eq:
reduced_eq = eq
if order == 1:
## Linear case: a(x)*y'+b(x)*y+c(x) == 0
if eq.is_Add:
ind, dep = reduced_eq.as_independent(f)
else:
u = Dummy('u')
ind, dep = (reduced_eq + u).as_independent(f)
ind, dep = [tmp.subs(u, 0) for tmp in [ind, dep]]
r = {a: dep.coeff(df),
b: dep.coeff(f(x)),
c: ind}
# double check f[a] since the preconditioning may have failed
if not r[a].has(f) and not r[b].has(f) and (
r[a]*df + r[b]*f(x) + r[c]).expand() - reduced_eq == 0:
r['a'] = a
r['b'] = b
r['c'] = c
matching_hints["1st_linear"] = r
matching_hints["1st_linear_Integral"] = r
## Bernoulli case: a(x)*y'+b(x)*y+c(x)*y**n == 0
r = collect(
reduced_eq, f(x), exact=True).match(a*df + b*f(x) + c*f(x)**n)
if r and r[c] != 0 and r[n] != 1: # See issue 4676
r['a'] = a
r['b'] = b
r['c'] = c
r['n'] = n
matching_hints["Bernoulli"] = r
matching_hints["Bernoulli_Integral"] = r
## Riccati special n == -2 case: a2*y'+b2*y**2+c2*y/x+d2/x**2 == 0
r = collect(reduced_eq,
f(x), exact=True).match(a2*df + b2*f(x)**2 + c2*f(x)/x + d2/x**2)
if r and r[b2] != 0 and (r[c2] != 0 or r[d2] != 0):
r['a2'] = a2
r['b2'] = b2
r['c2'] = c2
r['d2'] = d2
matching_hints["Riccati_special_minus2"] = r
# NON-REDUCED FORM OF EQUATION matches
r = collect(eq, df, exact=True).match(d + e * df)
if r:
r['d'] = d
r['e'] = e
r['y'] = y
r[d] = r[d].subs(f(x), y)
r[e] = r[e].subs(f(x), y)
# FIRST ORDER POWER SERIES WHICH NEEDS INITIAL CONDITIONS
# TODO: Hint first order series should match only if d/e is analytic.
# For now, only d/e and (d/e).diff(arg) is checked for existence at
# at a given point.
# This is currently done internally in ode_1st_power_series.
point = boundary.get('f0', 0)
value = boundary.get('f0val', C1)
check = cancel(r[d]/r[e])
check1 = check.subs({x: point, y: value})
if not check1.has(oo) and not check1.has(zoo) and \
not check1.has(NaN) and not check1.has(-oo):
check2 = (check1.diff(x)).subs({x: point, y: value})
if not check2.has(oo) and not check2.has(zoo) and \
not check2.has(NaN) and not check2.has(-oo):
rseries = r.copy()
rseries.update({'terms': terms, 'f0': point, 'f0val': value})
matching_hints["1st_power_series"] = rseries
r3.update(r)
## Exact Differential Equation: P(x, y) + Q(x, y)*y' = 0 where
# dP/dy == dQ/dx
try:
if r[d] != 0:
numerator = simplify(r[d].diff(y) - r[e].diff(x))
# The following few conditions try to convert a non-exact
# differential equation into an exact one.
# References : Differential equations with applications
# and historical notes - George E. Simmons
if numerator:
# If (dP/dy - dQ/dx) / Q = f(x)
# then exp(integral(f(x))*equation becomes exact
factor = simplify(numerator/r[e])
variables = factor.free_symbols
if len(variables) == 1 and x == variables.pop():
factor = exp(Integral(factor).doit())
r[d] *= factor
r[e] *= factor
matching_hints["1st_exact"] = r
matching_hints["1st_exact_Integral"] = r
else:
# If (dP/dy - dQ/dx) / -P = f(y)
# then exp(integral(f(y))*equation becomes exact
factor = simplify(-numerator/r[d])
variables = factor.free_symbols
if len(variables) == 1 and y == variables.pop():
factor = exp(Integral(factor).doit())
r[d] *= factor
r[e] *= factor
matching_hints["1st_exact"] = r
matching_hints["1st_exact_Integral"] = r
else:
matching_hints["1st_exact"] = r
matching_hints["1st_exact_Integral"] = r
except NotImplementedError:
# Differentiating the coefficients might fail because of things
# like f(2*x).diff(x). See issue 4624 and issue 4719.
pass
# Any first order ODE can be ideally solved by the Lie Group
# method
matching_hints["lie_group"] = r3
# This match is used for several cases below; we now collect on
# f(x) so the matching works.
r = collect(reduced_eq, df, exact=True).match(d + e*df)
if r:
# Using r[d] and r[e] without any modification for hints
# linear-coefficients and separable-reduced.
num, den = r[d], r[e] # ODE = d/e + df
r['d'] = d
r['e'] = e
r['y'] = y
r[d] = num.subs(f(x), y)
r[e] = den.subs(f(x), y)
## Separable Case: y' == P(y)*Q(x)
r[d] = separatevars(r[d])
r[e] = separatevars(r[e])
# m1[coeff]*m1[x]*m1[y] + m2[coeff]*m2[x]*m2[y]*y'
m1 = separatevars(r[d], dict=True, symbols=(x, y))
m2 = separatevars(r[e], dict=True, symbols=(x, y))
if m1 and m2:
r1 = {'m1': m1, 'm2': m2, 'y': y}
matching_hints["separable"] = r1
matching_hints["separable_Integral"] = r1
## First order equation with homogeneous coefficients:
# dy/dx == F(y/x) or dy/dx == F(x/y)
ordera = homogeneous_order(r[d], x, y)
if ordera is not None:
orderb = homogeneous_order(r[e], x, y)
if ordera == orderb:
# u1=y/x and u2=x/y
u1 = Dummy('u1')
u2 = Dummy('u2')
s = "1st_homogeneous_coeff_subs"
s1 = s + "_dep_div_indep"
s2 = s + "_indep_div_dep"
if simplify((r[d] + u1*r[e]).subs({x: 1, y: u1})) != 0:
matching_hints[s1] = r
matching_hints[s1 + "_Integral"] = r
if simplify((r[e] + u2*r[d]).subs({x: u2, y: 1})) != 0:
matching_hints[s2] = r
matching_hints[s2 + "_Integral"] = r
if s1 in matching_hints and s2 in matching_hints:
matching_hints["1st_homogeneous_coeff_best"] = r
## Linear coefficients of the form
# y'+ F((a*x + b*y + c)/(a'*x + b'y + c')) = 0
# that can be reduced to homogeneous form.
F = num/den
params = _linear_coeff_match(F, func)
if params:
xarg, yarg = params
u = Dummy('u')
t = Dummy('t')
# Dummy substitution for df and f(x).
dummy_eq = reduced_eq.subs(((df, t), (f(x), u)))
reps = ((x, x + xarg), (u, u + yarg), (t, df), (u, f(x)))
dummy_eq = simplify(dummy_eq.subs(reps))
# get the re-cast values for e and d
r2 = collect(expand(dummy_eq), [df, f(x)]).match(e*df + d)
if r2:
orderd = homogeneous_order(r2[d], x, f(x))
if orderd is not None:
ordere = homogeneous_order(r2[e], x, f(x))
if orderd == ordere:
# Match arguments are passed in such a way that it
# is coherent with the already existing homogeneous
# functions.
r2[d] = r2[d].subs(f(x), y)
r2[e] = r2[e].subs(f(x), y)
r2.update({'xarg': xarg, 'yarg': yarg,
'd': d, 'e': e, 'y': y})
matching_hints["linear_coefficients"] = r2
matching_hints["linear_coefficients_Integral"] = r2
## Equation of the form y' + (y/x)*H(x^n*y) = 0
# that can be reduced to separable form
factor = simplify(x/f(x)*num/den)
# Try representing factor in terms of x^n*y
# where n is lowest power of x in factor;
# first remove terms like sqrt(2)*3 from factor.atoms(Mul)
u = None
for mul in ordered(factor.atoms(Mul)):
if mul.has(x):
_, u = mul.as_independent(x, f(x))
break
if u and u.has(f(x)):
h = x**(degree(Poly(u.subs(f(x), y), gen=x)))*f(x)
p = Wild('p')
if (u/h == 1) or ((u/h).simplify().match(x**p)):
t = Dummy('t')
r2 = {'t': t}
xpart, ypart = u.as_independent(f(x))
test = factor.subs(((u, t), (1/u, 1/t)))
free = test.free_symbols
if len(free) == 1 and free.pop() == t:
r2.update({'power': xpart.as_base_exp()[1], 'u': test})
matching_hints["separable_reduced"] = r2
matching_hints["separable_reduced_Integral"] = r2
## Almost-linear equation of the form f(x)*g(y)*y' + k(x)*l(y) + m(x) = 0
r = collect(eq, [df, f(x)]).match(e*df + d)
if r:
r2 = r.copy()
r2[c] = S.Zero
if r2[d].is_Add:
# Separate the terms having f(x) to r[d] and
# remaining to r[c]
no_f, r2[d] = r2[d].as_independent(f(x))
r2[c] += no_f
factor = simplify(r2[d].diff(f(x))/r[e])
if factor and not factor.has(f(x)):
r2[d] = factor_terms(r2[d])
u = r2[d].as_independent(f(x), as_Add=False)[1]
r2.update({'a': e, 'b': d, 'c': c, 'u': u})
r2[d] /= u
r2[e] /= u.diff(f(x))
matching_hints["almost_linear"] = r2
matching_hints["almost_linear_Integral"] = r2
elif order == 2:
# Liouville ODE in the form
# f(x).diff(x, 2) + g(f(x))*(f(x).diff(x))**2 + h(x)*f(x).diff(x)
# See Goldstein and Braun, "Advanced Methods for the Solution of
# Differential Equations", pg. 98
s = d*f(x).diff(x, 2) + e*df**2 + k*df
r = reduced_eq.match(s)
if r and r[d] != 0:
y = Dummy('y')
g = simplify(r[e]/r[d]).subs(f(x), y)
h = simplify(r[k]/r[d])
if h.has(f(x)) or g.has(x):
pass
else:
r = {'g': g, 'h': h, 'y': y}
matching_hints["Liouville"] = r
matching_hints["Liouville_Integral"] = r
# Homogeneous second order differential equation of the form
# a3*f(x).diff(x, 2) + b3*f(x).diff(x) + c3, where
# for simplicity, a3, b3 and c3 are assumed to be polynomials.
# It has a definite power series solution at point x0 if, b3/a3 and c3/a3
# are analytic at x0.
deq = a3*(f(x).diff(x, 2)) + b3*df + c3*f(x)
r = collect(reduced_eq,
[f(x).diff(x, 2), f(x).diff(x), f(x)]).match(deq)
ordinary = False
if r and r[a3] != 0:
if all([r[key].is_polynomial() for key in r]):
p = cancel(r[b3]/r[a3]) # Used below
q = cancel(r[c3]/r[a3]) # Used below
point = kwargs.get('x0', 0)
check = p.subs(x, point)
if not check.has(oo) and not check.has(NaN) and \
not check.has(zoo) and not check.has(-oo):
check = q.subs(x, point)
if not check.has(oo) and not check.has(NaN) and \
not check.has(zoo) and not check.has(-oo):
ordinary = True
r.update({'a3': a3, 'b3': b3, 'c3': c3, 'x0': point, 'terms': terms})
matching_hints["2nd_power_series_ordinary"] = r
# Checking if the differential equation has a regular singular point
# at x0. It has a regular singular point at x0, if (b3/a3)*(x - x0)
# and (c3/a3)*((x - x0)**2) are analytic at x0.
if not ordinary:
p = cancel((x - point)*p)
check = p.subs(x, point)
if not check.has(oo) and not check.has(NaN) and \
not check.has(zoo) and not check.has(-oo):
q = cancel(((x - point)**2)*q)
check = q.subs(x, point)
if not check.has(oo) and not check.has(NaN) and \
not check.has(zoo) and not check.has(-oo):
coeff_dict = {'p': p, 'q': q, 'x0': point, 'terms': terms}
matching_hints["2nd_power_series_regular"] = coeff_dict
if order > 0:
# nth order linear ODE
# a_n(x)y^(n) + ... + a_1(x)y' + a_0(x)y = F(x) = b
r = _nth_linear_match(reduced_eq, func, order)
# Constant coefficient case (a_i is constant for all i)
if r and not any(r[i].has(x) for i in r if i >= 0):
# Inhomogeneous case: F(x) is not identically 0
if r[-1]:
undetcoeff = _undetermined_coefficients_match(r[-1], x)
s = "nth_linear_constant_coeff_variation_of_parameters"
matching_hints[s] = r
matching_hints[s + "_Integral"] = r
if undetcoeff['test']:
r['trialset'] = undetcoeff['trialset']
matching_hints[
"nth_linear_constant_coeff_undetermined_coefficients"
] = r
# Homogeneous case: F(x) is identically 0
else:
matching_hints["nth_linear_constant_coeff_homogeneous"] = r
# nth order Euler equation a_n*x**n*y^(n) + ... + a_1*x*y' + a_0*y = F(x)
#In case of Homogeneous euler equation F(x) = 0
def _test_term(coeff, order):
r"""
Linear Euler ODEs have the form K*x**order*diff(y(x),x,order) = F(x),
where K is independent of x and y(x), order>= 0.
So we need to check that for each term, coeff == K*x**order from
some K. We have a few cases, since coeff may have several
different types.
"""
if order < 0:
raise ValueError("order should be greater than 0")
if coeff == 0:
return True
if order == 0:
if x in coeff.free_symbols:
return False
return True
if coeff.is_Mul:
if coeff.has(f(x)):
return False
return x**order in coeff.args
elif coeff.is_Pow:
return coeff.as_base_exp() == (x, order)
elif order == 1:
return x == coeff
return False
if r and not any(not _test_term(r[i], i) for i in r if i >= 0):
if not r[-1]:
matching_hints["nth_linear_euler_eq_homogeneous"] = r
else:
matching_hints["nth_linear_euler_eq_nonhomogeneous_variation_of_parameters"] = r
matching_hints["nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral"] = r
e, re = posify(r[-1].subs(x, exp(x)))
undetcoeff = _undetermined_coefficients_match(e.subs(re), x)
if undetcoeff['test']:
r['trialset'] = undetcoeff['trialset']
matching_hints["nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients"] = r
# Order keys based on allhints.
retlist = [i for i in allhints if i in matching_hints]
if dict:
# Dictionaries are ordered arbitrarily, so make note of which
# hint would come first for dsolve(). Use an ordered dict in Py 3.
matching_hints["default"] = retlist[0] if retlist else None
matching_hints["ordered_hints"] = tuple(retlist)
return matching_hints
else:
return tuple(retlist)
def classify_sysode(eq, funcs=None, **kwargs):
r"""
Returns a dictionary of parameter names and values that define the system
of ordinary differential equations in ``eq``.
The parameters are further used in
:py:meth:`~sympy.solvers.ode.dsolve` for solving that system.
The parameter names and values are:
'is_linear' (boolean), which tells whether the given system is linear.
Note that "linear" here refers to the operator: terms such as ``x*diff(x,t)`` are
nonlinear, whereas terms like ``sin(t)*diff(x,t)`` are still linear operators.
'func' (list) contains the :py:class:`~sympy.core.function.Function`s that
appear with a derivative in the ODE, i.e. those that we are trying to solve
the ODE for.
'order' (dict) with the maximum derivative for each element of the 'func'
parameter.
'func_coeff' (dict) with the coefficient for each triple ``(equation number,
function, order)```. The coefficients are those subexpressions that do not
appear in 'func', and hence can be considered constant for purposes of ODE
solving.
'eq' (list) with the equations from ``eq``, sympified and transformed into
expressions (we are solving for these expressions to be zero).
'no_of_equations' (int) is the number of equations (same as ``len(eq)``).
'type_of_equation' (string) is an internal classification of the type of
ODE.
References
==========
-http://eqworld.ipmnet.ru/en/solutions/sysode/sode-toc1.htm
-A. D. Polyanin and A. V. Manzhirov, Handbook of Mathematics for Engineers and Scientists
Examples
========
>>> from sympy import Function, Eq, symbols, diff
>>> from sympy.solvers.ode import classify_sysode
>>> from sympy.abc import t
>>> f, x, y = symbols('f, x, y', function=True)
>>> k, l, m, n = symbols('k, l, m, n', Integer=True)
>>> x1 = diff(x(t), t) ; y1 = diff(y(t), t)
>>> x2 = diff(x(t), t, t) ; y2 = diff(y(t), t, t)
>>> eq = (Eq(5*x1, 12*x(t) - 6*y(t)), Eq(2*y1, 11*x(t) + 3*y(t)))
>>> classify_sysode(eq)
{'eq': [-12*x(t) + 6*y(t) + 5*Derivative(x(t), t), -11*x(t) - 3*y(t) + 2*Derivative(y(t), t)],
'func': [x(t), y(t)], 'func_coeff': {(0, x(t), 0): -12, (0, x(t), 1): 5, (0, y(t), 0): 6,
(0, y(t), 1): 0, (1, x(t), 0): -11, (1, x(t), 1): 0, (1, y(t), 0): -3, (1, y(t), 1): 2},
'is_linear': True, 'no_of_equation': 2, 'order': {x(t): 1, y(t): 1}, 'type_of_equation': 'type1'}
>>> eq = (Eq(diff(x(t),t), 5*t*x(t) + t**2*y(t)), Eq(diff(y(t),t), -t**2*x(t) + 5*t*y(t)))
>>> classify_sysode(eq)
{'eq': [-t**2*y(t) - 5*t*x(t) + Derivative(x(t), t), t**2*x(t) - 5*t*y(t) + Derivative(y(t), t)],
'func': [x(t), y(t)], 'func_coeff': {(0, x(t), 0): -5*t, (0, x(t), 1): 1, (0, y(t), 0): -t**2,
(0, y(t), 1): 0, (1, x(t), 0): t**2, (1, x(t), 1): 0, (1, y(t), 0): -5*t, (1, y(t), 1): 1},
'is_linear': True, 'no_of_equation': 2, 'order': {x(t): 1, y(t): 1}, 'type_of_equation': 'type4'}
"""
# Sympify equations and convert iterables of equations into
# a list of equations
def _sympify(eq):
return list(map(sympify, eq if iterable(eq) else [eq]))
eq, funcs = (_sympify(w) for w in [eq, funcs])
for i, fi in enumerate(eq):
if isinstance(fi, Equality):
eq[i] = fi.lhs - fi.rhs
matching_hints = {"no_of_equation":i+1}
matching_hints['eq'] = eq
if i==0:
raise ValueError("classify_sysode() works for systems of ODEs. "
"For scalar ODEs, classify_ode should be used")
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
# find all the functions if not given
order = dict()
if funcs==[None]:
funcs = []
for eqs in eq:
derivs = eqs.atoms(Derivative)
func = set().union(*[d.atoms(AppliedUndef) for d in derivs])
for func_ in func:
order[func_] = 0
funcs.append(func_)
funcs = list(set(funcs))
if len(funcs) < len(eq):
raise ValueError("Number of functions given is less than number of equations %s" % funcs)
func_dict = dict()
for func in funcs:
if not order[func]:
max_order = 0
for i, eqs_ in enumerate(eq):
order_ = ode_order(eqs_,func)
if max_order < order_:
max_order = order_
eq_no = i
if eq_no in func_dict:
list_func = []
list_func.append(func_dict[eq_no])
list_func.append(func)
func_dict[eq_no] = list_func
else:
func_dict[eq_no] = func
order[func] = max_order
funcs = [func_dict[i] for i in range(len(func_dict))]
matching_hints['func'] = funcs
for func in funcs:
if isinstance(func, list):
for func_elem in func:
if len(func_elem.args) != 1:
raise ValueError("dsolve() and classify_sysode() work with "
"functions of one variable only, not %s" % func)
else:
if func and len(func.args) != 1:
raise ValueError("dsolve() and classify_sysode() work with "
"functions of one variable only, not %s" % func)
# find the order of all equation in system of odes
matching_hints["order"] = order
# find coefficients of terms f(t), diff(f(t),t) and higher derivatives
# and similarly for other functions g(t), diff(g(t),t) in all equations.
# Here j denotes the equation number, funcs[l] denotes the function about
# which we are talking about and k denotes the order of function funcs[l]
# whose coefficient we are calculating.
def linearity_check(eqs, j, func, is_linear_):
for k in range(order[func]+1):
func_coef[j,func,k] = collect(eqs.expand(),[diff(func,t,k)]).coeff(diff(func,t,k))
if is_linear_ == True:
if func_coef[j,func,k]==0:
if k==0:
coef = eqs.as_independent(func)[1]
for xr in range(1, ode_order(eqs,func)+1):
coef -= eqs.as_independent(diff(func,t,xr))[1]
if coef != 0:
is_linear_ = False
else:
if eqs.as_independent(diff(func,t,k))[1]:
is_linear_ = False
else:
for func_ in funcs:
if isinstance(func_, list):
for elem_func_ in func_:
dep = func_coef[j,func,k].as_independent(elem_func_)[1]
if dep!=1 and dep!=0:
is_linear_ = False
else:
dep = func_coef[j,func,k].as_independent(func_)[1]
if dep!=1 and dep!=0:
is_linear_ = False
return is_linear_
func_coef = {}
is_linear = True
for j, eqs in enumerate(eq):
for func in funcs:
if isinstance(func, list):
for func_elem in func:
is_linear = linearity_check(eqs, j, func_elem, is_linear)
else:
is_linear = linearity_check(eqs, j, func, is_linear)
matching_hints['func_coeff'] = func_coef
matching_hints['is_linear'] = is_linear
if len(set(order.values()))==1:
order_eq = list(matching_hints['order'].values())[0]
if matching_hints['is_linear'] == True:
if matching_hints['no_of_equation'] == 2:
if order_eq == 1:
type_of_equation = check_linear_2eq_order1(eq, funcs, func_coef)
elif order_eq == 2:
type_of_equation = check_linear_2eq_order2(eq, funcs, func_coef)
else:
type_of_equation = None
elif matching_hints['no_of_equation'] == 3:
if order_eq == 1:
type_of_equation = check_linear_3eq_order1(eq, funcs, func_coef)
if type_of_equation==None:
type_of_equation = check_linear_neq_order1(eq, funcs, func_coef)
else:
type_of_equation = None
else:
if order_eq == 1:
type_of_equation = check_linear_neq_order1(eq, funcs, func_coef)
else:
type_of_equation = None
else:
if matching_hints['no_of_equation'] == 2:
if order_eq == 1:
type_of_equation = check_nonlinear_2eq_order1(eq, funcs, func_coef)
else:
type_of_equation = None
elif matching_hints['no_of_equation'] == 3:
if order_eq == 1:
type_of_equation = check_nonlinear_3eq_order1(eq, funcs, func_coef)
else:
type_of_equation = None
else:
type_of_equation = None
else:
type_of_equation = None
matching_hints['type_of_equation'] = type_of_equation
return matching_hints
def check_linear_2eq_order1(eq, func, func_coef):
x = func[0].func
y = func[1].func
fc = func_coef
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
r = dict()
# for equations Eq(a1*diff(x(t),t), b1*x(t) + c1*y(t) + d1)
# and Eq(a2*diff(y(t),t), b2*x(t) + c2*y(t) + d2)
r['a1'] = fc[0,x(t),1] ; r['a2'] = fc[1,y(t),1]
r['b1'] = -fc[0,x(t),0]/fc[0,x(t),1] ; r['b2'] = -fc[1,x(t),0]/fc[1,y(t),1]
r['c1'] = -fc[0,y(t),0]/fc[0,x(t),1] ; r['c2'] = -fc[1,y(t),0]/fc[1,y(t),1]
forcing = [S(0),S(0)]
for i in range(2):
for j in Add.make_args(eq[i]):
if not j.has(x(t), y(t)):
forcing[i] += j
if not (forcing[0].has(t) or forcing[1].has(t)):
# We can handle homogeneous case and simple constant forcings
r['d1'] = forcing[0]
r['d2'] = forcing[1]
else:
# Issue #9244: nonhomogeneous linear systems are not supported
return None
# Conditions to check for type 6 whose equations are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and
# Eq(diff(y(t),t), a*[f(t) + a*h(t)]x(t) + a*[g(t) - h(t)]*y(t))
p = 0
q = 0
p1 = cancel(r['b2']/(cancel(r['b2']/r['c2']).as_numer_denom()[0]))
p2 = cancel(r['b1']/(cancel(r['b1']/r['c1']).as_numer_denom()[0]))
for n, i in enumerate([p1, p2]):
for j in Mul.make_args(collect_const(i)):
if not j.has(t):
q = j
if q and n==0:
if ((r['b2']/j - r['b1'])/(r['c1'] - r['c2']/j)) == j:
p = 1
elif q and n==1:
if ((r['b1']/j - r['b2'])/(r['c2'] - r['c1']/j)) == j:
p = 2
# End of condition for type 6
if r['d1']!=0 or r['d2']!=0:
if not r['d1'].has(t) and not r['d2'].has(t):
if all(not r[k].has(t) for k in 'a1 a2 b1 b2 c1 c2'.split()):
# Equations for type 2 are Eq(a1*diff(x(t),t),b1*x(t)+c1*y(t)+d1) and Eq(a2*diff(y(t),t),b2*x(t)+c2*y(t)+d2)
return "type2"
else:
return None
else:
if all(not r[k].has(t) for k in 'a1 a2 b1 b2 c1 c2'.split()):
# Equations for type 1 are Eq(a1*diff(x(t),t),b1*x(t)+c1*y(t)) and Eq(a2*diff(y(t),t),b2*x(t)+c2*y(t))
return "type1"
else:
r['b1'] = r['b1']/r['a1'] ; r['b2'] = r['b2']/r['a2']
r['c1'] = r['c1']/r['a1'] ; r['c2'] = r['c2']/r['a2']
if (r['b1'] == r['c2']) and (r['c1'] == r['b2']):
# Equation for type 3 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), g(t)*x(t) + f(t)*y(t))
return "type3"
elif (r['b1'] == r['c2']) and (r['c1'] == -r['b2']) or (r['b1'] == -r['c2']) and (r['c1'] == r['b2']):
# Equation for type 4 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), -g(t)*x(t) + f(t)*y(t))
return "type4"
elif (not cancel(r['b2']/r['c1']).has(t) and not cancel((r['c2']-r['b1'])/r['c1']).has(t)) \
or (not cancel(r['b1']/r['c2']).has(t) and not cancel((r['c1']-r['b2'])/r['c2']).has(t)):
# Equations for type 5 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), a*g(t)*x(t) + [f(t) + b*g(t)]*y(t)
return "type5"
elif p:
return "type6"
else:
# Equations for type 7 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), h(t)*x(t) + p(t)*y(t))
return "type7"
def check_linear_2eq_order2(eq, func, func_coef):
x = func[0].func
y = func[1].func
fc = func_coef
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
r = dict()
a = Wild('a', exclude=[1/t])
b = Wild('b', exclude=[1/t**2])
u = Wild('u', exclude=[t, t**2])
v = Wild('v', exclude=[t, t**2])
w = Wild('w', exclude=[t, t**2])
p = Wild('p', exclude=[t, t**2])
r['a1'] = fc[0,x(t),2] ; r['a2'] = fc[1,y(t),2]
r['b1'] = fc[0,x(t),1] ; r['b2'] = fc[1,x(t),1]
r['c1'] = fc[0,y(t),1] ; r['c2'] = fc[1,y(t),1]
r['d1'] = fc[0,x(t),0] ; r['d2'] = fc[1,x(t),0]
r['e1'] = fc[0,y(t),0] ; r['e2'] = fc[1,y(t),0]
const = [S(0), S(0)]
for i in range(2):
for j in Add.make_args(eq[i]):
if not (j.has(x(t)) or j.has(y(t))):
const[i] += j
r['f1'] = const[0]
r['f2'] = const[1]
if r['f1']!=0 or r['f2']!=0:
if all(not r[k].has(t) for k in 'a1 a2 d1 d2 e1 e2 f1 f2'.split()) \
and r['b1']==r['c1']==r['b2']==r['c2']==0:
return "type2"
elif all(not r[k].has(t) for k in 'a1 a2 b1 b2 c1 c2 d1 d2 e1 e1'.split()):
p = [S(0), S(0)] ; q = [S(0), S(0)]
for n, e in enumerate([r['f1'], r['f2']]):
if e.has(t):
tpart = e.as_independent(t, Mul)[1]
for i in Mul.make_args(tpart):
if i.has(exp):
b, e = i.as_base_exp()
co = e.coeff(t)
if co and not co.has(t) and co.has(I):
p[n] = 1
else:
q[n] = 1
else:
q[n] = 1
else:
q[n] = 1
if p[0]==1 and p[1]==1 and q[0]==0 and q[1]==0:
return "type4"
else:
return None
else:
return None
else:
if r['b1']==r['b2']==r['c1']==r['c2']==0 and all(not r[k].has(t) \
for k in 'a1 a2 d1 d2 e1 e2'.split()):
return "type1"
elif r['b1']==r['e1']==r['c2']==r['d2']==0 and all(not r[k].has(t) \
for k in 'a1 a2 b2 c1 d1 e2'.split()) and r['c1'] == -r['b2'] and \
r['d1'] == r['e2']:
return "type3"
elif cancel(-r['b2']/r['d2'])==t and cancel(-r['c1']/r['e1'])==t and not \
(r['d2']/r['a2']).has(t) and not (r['e1']/r['a1']).has(t) and \
r['b1']==r['d1']==r['c2']==r['e2']==0:
return "type5"
elif ((r['a1']/r['d1']).expand()).match((p*(u*t**2+v*t+w)**2).expand()) and not \
(cancel(r['a1']*r['d2']/(r['a2']*r['d1']))).has(t) and not (r['d1']/r['e1']).has(t) and not \
(r['d2']/r['e2']).has(t) and r['b1'] == r['b2'] == r['c1'] == r['c2'] == 0:
return "type10"
elif not cancel(r['d1']/r['e1']).has(t) and not cancel(r['d2']/r['e2']).has(t) and not \
cancel(r['d1']*r['a2']/(r['d2']*r['a1'])).has(t) and r['b1']==r['b2']==r['c1']==r['c2']==0:
return "type6"
elif not cancel(r['b1']/r['c1']).has(t) and not cancel(r['b2']/r['c2']).has(t) and not \
cancel(r['b1']*r['a2']/(r['b2']*r['a1'])).has(t) and r['d1']==r['d2']==r['e1']==r['e2']==0:
return "type7"
elif cancel(-r['b2']/r['d2'])==t and cancel(-r['c1']/r['e1'])==t and not \
cancel(r['e1']*r['a2']/(r['d2']*r['a1'])).has(t) and r['e1'].has(t) \
and r['b1']==r['d1']==r['c2']==r['e2']==0:
return "type8"
elif (r['b1']/r['a1']).match(a/t) and (r['b2']/r['a2']).match(a/t) and not \
(r['b1']/r['c1']).has(t) and not (r['b2']/r['c2']).has(t) and \
(r['d1']/r['a1']).match(b/t**2) and (r['d2']/r['a2']).match(b/t**2) \
and not (r['d1']/r['e1']).has(t) and not (r['d2']/r['e2']).has(t):
return "type9"
elif -r['b1']/r['d1']==-r['c1']/r['e1']==-r['b2']/r['d2']==-r['c2']/r['e2']==t:
return "type11"
else:
return None
def check_linear_3eq_order1(eq, func, func_coef):
x = func[0].func
y = func[1].func
z = func[2].func
fc = func_coef
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
r = dict()
r['a1'] = fc[0,x(t),1]; r['a2'] = fc[1,y(t),1]; r['a3'] = fc[2,z(t),1]
r['b1'] = fc[0,x(t),0]; r['b2'] = fc[1,x(t),0]; r['b3'] = fc[2,x(t),0]
r['c1'] = fc[0,y(t),0]; r['c2'] = fc[1,y(t),0]; r['c3'] = fc[2,y(t),0]
r['d1'] = fc[0,z(t),0]; r['d2'] = fc[1,z(t),0]; r['d3'] = fc[2,z(t),0]
forcing = [S(0), S(0), S(0)]
for i in range(3):
for j in Add.make_args(eq[i]):
if not j.has(x(t), y(t), z(t)):
forcing[i] += j
if forcing[0].has(t) or forcing[1].has(t) or forcing[2].has(t):
# We can handle homogeneous case and simple constant forcings.
# Issue #9244: nonhomogeneous linear systems are not supported
return None
if all(not r[k].has(t) for k in 'a1 a2 a3 b1 b2 b3 c1 c2 c3 d1 d2 d3'.split()):
if r['c1']==r['d1']==r['d2']==0:
return 'type1'
elif r['c1'] == -r['b2'] and r['d1'] == -r['b3'] and r['d2'] == -r['c3'] \
and r['b1'] == r['c2'] == r['d3'] == 0:
return 'type2'
elif r['b1'] == r['c2'] == r['d3'] == 0 and r['c1']/r['a1'] == -r['d1']/r['a1'] \
and r['d2']/r['a2'] == -r['b2']/r['a2'] and r['b3']/r['a3'] == -r['c3']/r['a3']:
return 'type3'
else:
return None
else:
for k1 in 'c1 d1 b2 d2 b3 c3'.split():
if r[k1] == 0:
continue
else:
if all(not cancel(r[k1]/r[k]).has(t) for k in 'd1 b2 d2 b3 c3'.split() if r[k]!=0) \
and all(not cancel(r[k1]/(r['b1'] - r[k])).has(t) for k in 'b1 c2 d3'.split() if r['b1']!=r[k]):
return 'type4'
else:
break
return None
def check_linear_neq_order1(eq, func, func_coef):
x = func[0].func
y = func[1].func
z = func[2].func
fc = func_coef
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
r = dict()
n = len(eq)
for i in range(n):
for j in range(n):
if (fc[i,func[j],0]/fc[i,func[i],1]).has(t):
return None
if len(eq)==3:
return 'type6'
return 'type1'
def check_nonlinear_2eq_order1(eq, func, func_coef):
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
f = Wild('f')
g = Wild('g')
u, v = symbols('u, v', cls=Dummy)
def check_type(x, y):
r1 = eq[0].match(t*diff(x(t),t) - x(t) + f)
r2 = eq[1].match(t*diff(y(t),t) - y(t) + g)
if not (r1 and r2):
r1 = eq[0].match(diff(x(t),t) - x(t)/t + f/t)
r2 = eq[1].match(diff(y(t),t) - y(t)/t + g/t)
if not (r1 and r2):
r1 = (-eq[0]).match(t*diff(x(t),t) - x(t) + f)
r2 = (-eq[1]).match(t*diff(y(t),t) - y(t) + g)
if not (r1 and r2):
r1 = (-eq[0]).match(diff(x(t),t) - x(t)/t + f/t)
r2 = (-eq[1]).match(diff(y(t),t) - y(t)/t + g/t)
if r1 and r2 and not (r1[f].subs(diff(x(t),t),u).subs(diff(y(t),t),v).has(t) \
or r2[g].subs(diff(x(t),t),u).subs(diff(y(t),t),v).has(t)):
return 'type5'
else:
return None
for func_ in func:
if isinstance(func_, list):
x = func[0][0].func
y = func[0][1].func
eq_type = check_type(x, y)
if not eq_type:
eq_type = check_type(y, x)
return eq_type
x = func[0].func
y = func[1].func
fc = func_coef
n = Wild('n', exclude=[x(t),y(t)])
f1 = Wild('f1', exclude=[v,t])
f2 = Wild('f2', exclude=[v,t])
g1 = Wild('g1', exclude=[u,t])
g2 = Wild('g2', exclude=[u,t])
for i in range(2):
eqs = 0
for terms in Add.make_args(eq[i]):
eqs += terms/fc[i,func[i],1]
eq[i] = eqs
r = eq[0].match(diff(x(t),t) - x(t)**n*f)
if r:
g = (diff(y(t),t) - eq[1])/r[f]
if r and not (g.has(x(t)) or g.subs(y(t),v).has(t) or r[f].subs(x(t),u).subs(y(t),v).has(t)):
return 'type1'
r = eq[0].match(diff(x(t),t) - exp(n*x(t))*f)
if r:
g = (diff(y(t),t) - eq[1])/r[f]
if r and not (g.has(x(t)) or g.subs(y(t),v).has(t) or r[f].subs(x(t),u).subs(y(t),v).has(t)):
return 'type2'
g = Wild('g')
r1 = eq[0].match(diff(x(t),t) - f)
r2 = eq[1].match(diff(y(t),t) - g)
if r1 and r2 and not (r1[f].subs(x(t),u).subs(y(t),v).has(t) or \
r2[g].subs(x(t),u).subs(y(t),v).has(t)):
return 'type3'
r1 = eq[0].match(diff(x(t),t) - f)
r2 = eq[1].match(diff(y(t),t) - g)
num, den = (
(r1[f].subs(x(t),u).subs(y(t),v))/
(r2[g].subs(x(t),u).subs(y(t),v))).as_numer_denom()
R1 = num.match(f1*g1)
R2 = den.match(f2*g2)
phi = (r1[f].subs(x(t),u).subs(y(t),v))/num
if R1 and R2:
return 'type4'
return None
def check_nonlinear_2eq_order2(eq, func, func_coef):
return None
def check_nonlinear_3eq_order1(eq, func, func_coef):
x = func[0].func
y = func[1].func
z = func[2].func
fc = func_coef
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
u, v, w = symbols('u, v, w', cls=Dummy)
a = Wild('a', exclude=[x(t), y(t), z(t), t])
b = Wild('b', exclude=[x(t), y(t), z(t), t])
c = Wild('c', exclude=[x(t), y(t), z(t), t])
f = Wild('f')
F1 = Wild('F1')
F2 = Wild('F2')
F3 = Wild('F3')
for i in range(3):
eqs = 0
for terms in Add.make_args(eq[i]):
eqs += terms/fc[i,func[i],1]
eq[i] = eqs
r1 = eq[0].match(diff(x(t),t) - a*y(t)*z(t))
r2 = eq[1].match(diff(y(t),t) - b*z(t)*x(t))
r3 = eq[2].match(diff(z(t),t) - c*x(t)*y(t))
if r1 and r2 and r3:
num1, den1 = r1[a].as_numer_denom()
num2, den2 = r2[b].as_numer_denom()
num3, den3 = r3[c].as_numer_denom()
if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]):
return 'type1'
r = eq[0].match(diff(x(t),t) - y(t)*z(t)*f)
if r:
r1 = collect_const(r[f]).match(a*f)
r2 = ((diff(y(t),t) - eq[1])/r1[f]).match(b*z(t)*x(t))
r3 = ((diff(z(t),t) - eq[2])/r1[f]).match(c*x(t)*y(t))
if r1 and r2 and r3:
num1, den1 = r1[a].as_numer_denom()
num2, den2 = r2[b].as_numer_denom()
num3, den3 = r3[c].as_numer_denom()
if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]):
return 'type2'
r = eq[0].match(diff(x(t),t) - (F2-F3))
if r:
r1 = collect_const(r[F2]).match(c*F2)
r1.update(collect_const(r[F3]).match(b*F3))
if r1:
if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]):
r1[F2], r1[F3] = r1[F3], r1[F2]
r1[c], r1[b] = -r1[b], -r1[c]
r2 = eq[1].match(diff(y(t),t) - a*r1[F3] + r1[c]*F1)
if r2:
r3 = (eq[2] == diff(z(t),t) - r1[b]*r2[F1] + r2[a]*r1[F2])
if r1 and r2 and r3:
return 'type3'
r = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3)
if r:
r1 = collect_const(r[F2]).match(c*F2)
r1.update(collect_const(r[F3]).match(b*F3))
if r1:
if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]):
r1[F2], r1[F3] = r1[F3], r1[F2]
r1[c], r1[b] = -r1[b], -r1[c]
r2 = (diff(y(t),t) - eq[1]).match(a*x(t)*r1[F3] - r1[c]*z(t)*F1)
if r2:
r3 = (diff(z(t),t) - eq[2] == r1[b]*y(t)*r2[F1] - r2[a]*x(t)*r1[F2])
if r1 and r2 and r3:
return 'type4'
r = (diff(x(t),t) - eq[0]).match(x(t)*(F2 - F3))
if r:
r1 = collect_const(r[F2]).match(c*F2)
r1.update(collect_const(r[F3]).match(b*F3))
if r1:
if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]):
r1[F2], r1[F3] = r1[F3], r1[F2]
r1[c], r1[b] = -r1[b], -r1[c]
r2 = (diff(y(t),t) - eq[1]).match(y(t)*(a*r1[F3] - r1[c]*F1))
if r2:
r3 = (diff(z(t),t) - eq[2] == z(t)*(r1[b]*r2[F1] - r2[a]*r1[F2]))
if r1 and r2 and r3:
return 'type5'
return None
def check_nonlinear_3eq_order2(eq, func, func_coef):
return None
def checksysodesol(eqs, sols, func=None):
r"""
Substitutes corresponding ``sols`` for each functions into each ``eqs`` and
checks that the result of substitutions for each equation is ``0``. The
equations and solutions passed can be any iterable.
This only works when each ``sols`` have one function only, like `x(t)` or `y(t)`.
For each function, ``sols`` can have a single solution or a list of solutions.
In most cases it will not be necessary to explicitly identify the function,
but if the function cannot be inferred from the original equation it
can be supplied through the ``func`` argument.
When a sequence of equations is passed, the same sequence is used to return
the result for each equation with each function substitued with corresponding
solutions.
It tries the following method to find zero equivalence for each equation:
Substitute the solutions for functions, like `x(t)` and `y(t)` into the
original equations containing those functions.
This function returns a tuple. The first item in the tuple is ``True`` if
the substitution results for each equation is ``0``, and ``False`` otherwise.
The second item in the tuple is what the substitution results in. Each element
of the ``list`` should always be ``0`` corresponding to each equation if the
first item is ``True``. Note that sometimes this function may return ``False``,
but with an expression that is identically equal to ``0``, instead of returning
``True``. This is because :py:meth:`~sympy.simplify.simplify.simplify` cannot
reduce the expression to ``0``. If an expression returned by each function
vanishes identically, then ``sols`` really is a solution to ``eqs``.
If this function seems to hang, it is probably because of a difficult simplification.
Examples
========
>>> from sympy import Eq, diff, symbols, sin, cos, exp, sqrt, S
>>> from sympy.solvers.ode import checksysodesol
>>> C1, C2 = symbols('C1:3')
>>> t = symbols('t')
>>> x, y = symbols('x, y', function=True)
>>> eq = (Eq(diff(x(t),t), x(t) + y(t) + 17), Eq(diff(y(t),t), -2*x(t) + y(t) + 12))
>>> sol = [Eq(x(t), (C1*sin(sqrt(2)*t) + C2*cos(sqrt(2)*t))*exp(t) - S(5)/3),
... Eq(y(t), (sqrt(2)*C1*cos(sqrt(2)*t) - sqrt(2)*C2*sin(sqrt(2)*t))*exp(t) - S(46)/3)]
>>> checksysodesol(eq, sol)
(True, [0, 0])
>>> eq = (Eq(diff(x(t),t),x(t)*y(t)**4), Eq(diff(y(t),t),y(t)**3))
>>> sol = [Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), -sqrt(2)*sqrt(-1/(C2 + t))/2),
... Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), sqrt(2)*sqrt(-1/(C2 + t))/2)]
>>> checksysodesol(eq, sol)
(True, [0, 0])
"""
def _sympify(eq):
return list(map(sympify, eq if iterable(eq) else [eq]))
eqs = _sympify(eqs)
for i in range(len(eqs)):
if isinstance(eqs[i], Equality):
eqs[i] = eqs[i].lhs - eqs[i].rhs
if func is None:
funcs = []
for eq in eqs:
derivs = eq.atoms(Derivative)
func = set().union(*[d.atoms(AppliedUndef) for d in derivs])
for func_ in func:
funcs.append(func_)
funcs = list(set(funcs))
if not all(isinstance(func, AppliedUndef) and len(func.args) == 1 for func in funcs)\
and len({func.args for func in funcs})!=1:
raise ValueError("func must be a function of one variable, not %s" % func)
for sol in sols:
if len(sol.atoms(AppliedUndef)) != 1:
raise ValueError("solutions should have one function only")
if len(funcs) != len({sol.lhs for sol in sols}):
raise ValueError("number of solutions provided does not match the number of equations")
t = funcs[0].args[0]
dictsol = dict()
for sol in sols:
func = list(sol.atoms(AppliedUndef))[0]
if sol.rhs == func:
sol = sol.reversed
solved = sol.lhs == func and not sol.rhs.has(func)
if not solved:
rhs = solve(sol, func)
if not rhs:
raise NotImplementedError
else:
rhs = sol.rhs
dictsol[func] = rhs
checkeq = []
for eq in eqs:
for func in funcs:
eq = sub_func_doit(eq, func, dictsol[func])
ss = simplify(eq)
if ss != 0:
eq = ss.expand(force=True)
else:
eq = 0
checkeq.append(eq)
if len(set(checkeq)) == 1 and list(set(checkeq))[0] == 0:
return (True, checkeq)
else:
return (False, checkeq)
@vectorize(0)
[docs]def odesimp(eq, func, order, constants, hint):
r"""
Simplifies ODEs, including trying to solve for ``func`` and running
:py:meth:`~sympy.solvers.ode.constantsimp`.
It may use knowledge of the type of solution that the hint returns to
apply additional simplifications.
It also attempts to integrate any :py:class:`~sympy.integrals.Integral`\s
in the expression, if the hint is not an ``_Integral`` hint.
This function should have no effect on expressions returned by
:py:meth:`~sympy.solvers.ode.dsolve`, as
:py:meth:`~sympy.solvers.ode.dsolve` already calls
:py:meth:`~sympy.solvers.ode.odesimp`, but the individual hint functions
do not call :py:meth:`~sympy.solvers.ode.odesimp` (because the
:py:meth:`~sympy.solvers.ode.dsolve` wrapper does). Therefore, this
function is designed for mainly internal use.
Examples
========
>>> from sympy import sin, symbols, dsolve, pprint, Function
>>> from sympy.solvers.ode import odesimp
>>> x , u2, C1= symbols('x,u2,C1')
>>> f = Function('f')
>>> eq = dsolve(x*f(x).diff(x) - f(x) - x*sin(f(x)/x), f(x),
... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral',
... simplify=False)
>>> pprint(eq, wrap_line=False)
x
----
f(x)
/
|
| / 1 \
| -|u2 + -------|
| | /1 \|
| | sin|--||
| \ \u2//
log(f(x)) = log(C1) + | ---------------- d(u2)
| 2
| u2
|
/
>>> pprint(odesimp(eq, f(x), 1, set([C1]),
... hint='1st_homogeneous_coeff_subs_indep_div_dep'
... )) #doctest: +SKIP
x
--------- = C1
/f(x)\
tan|----|
\2*x /
"""
x = func.args[0]
f = func.func
C1 = get_numbered_constants(eq, num=1)
# First, integrate if the hint allows it.
eq = _handle_Integral(eq, func, order, hint)
if hint.startswith("nth_linear_euler_eq_nonhomogeneous"):
eq = simplify(eq)
if not isinstance(eq, Equality):
raise TypeError("eq should be an instance of Equality")
# Second, clean up the arbitrary constants.
# Right now, nth linear hints can put as many as 2*order constants in an
# expression. If that number grows with another hint, the third argument
# here should be raised accordingly, or constantsimp() rewritten to handle
# an arbitrary number of constants.
eq = constantsimp(eq, constants)
# Lastly, now that we have cleaned up the expression, try solving for func.
# When CRootOf is implemented in solve(), we will want to return a CRootOf
# everytime instead of an Equality.
# Get the f(x) on the left if possible.
if eq.rhs == func and not eq.lhs.has(func):
eq = [Eq(eq.rhs, eq.lhs)]
# make sure we are working with lists of solutions in simplified form.
if eq.lhs == func and not eq.rhs.has(func):
# The solution is already solved
eq = [eq]
# special simplification of the rhs
if hint.startswith("nth_linear_constant_coeff"):
# Collect terms to make the solution look nice.
# This is also necessary for constantsimp to remove unnecessary
# terms from the particular solution from variation of parameters
#
# Collect is not behaving reliably here. The results for
# some linear constant-coefficient equations with repeated
# roots do not properly simplify all constants sometimes.
# 'collectterms' gives different orders sometimes, and results
# differ in collect based on that order. The
# sort-reverse trick fixes things, but may fail in the
# future. In addition, collect is splitting exponentials with
# rational powers for no reason. We have to do a match
# to fix this using Wilds.
global collectterms
try:
collectterms.sort(key=default_sort_key)
collectterms.reverse()
except Exception:
pass
assert len(eq) == 1 and eq[0].lhs == f(x)
sol = eq[0].rhs
sol = expand_mul(sol)
for i, reroot, imroot in collectterms:
sol = collect(sol, x**i*exp(reroot*x)*sin(abs(imroot)*x))
sol = collect(sol, x**i*exp(reroot*x)*cos(imroot*x))
for i, reroot, imroot in collectterms:
sol = collect(sol, x**i*exp(reroot*x))
del collectterms
# Collect is splitting exponentials with rational powers for
# no reason. We call powsimp to fix.
sol = powsimp(sol)
eq[0] = Eq(f(x), sol)
else:
# The solution is not solved, so try to solve it
try:
floats = any(i.is_Float for i in eq.atoms(Number))
eqsol = solve(eq, func, force=True, rational=False if floats else None)
if not eqsol:
raise NotImplementedError
except (NotImplementedError, PolynomialError):
eq = [eq]
else:
def _expand(expr):
numer, denom = expr.as_numer_denom()
if denom.is_Add:
return expr
else:
return powsimp(expr.expand(), combine='exp', deep=True)
# XXX: the rest of odesimp() expects each ``t`` to be in a
# specific normal form: rational expression with numerator
# expanded, but with combined exponential functions (at
# least in this setup all tests pass).
eq = [Eq(f(x), _expand(t)) for t in eqsol]
# special simplification of the lhs.
if hint.startswith("1st_homogeneous_coeff"):
for j, eqi in enumerate(eq):
newi = logcombine(eqi, force=True)
if newi.lhs.func is log and newi.rhs == 0:
newi = Eq(newi.lhs.args[0]/C1, C1)
eq[j] = newi
# We cleaned up the constants before solving to help the solve engine with
# a simpler expression, but the solved expression could have introduced
# things like -C1, so rerun constantsimp() one last time before returning.
for i, eqi in enumerate(eq):
eq[i] = constantsimp(eqi, constants)
eq[i] = constant_renumber(eq[i], 'C', 1, 2*order)
# If there is only 1 solution, return it;
# otherwise return the list of solutions.
if len(eq) == 1:
eq = eq[0]
return eq
[docs]def checkodesol(ode, sol, func=None, order='auto', solve_for_func=True):
r"""
Substitutes ``sol`` into ``ode`` and checks that the result is ``0``.
This only works when ``func`` is one function, like `f(x)`. ``sol`` can
be a single solution or a list of solutions. Each solution may be an
:py:class:`~sympy.core.relational.Equality` that the solution satisfies,
e.g. ``Eq(f(x), C1), Eq(f(x) + C1, 0)``; or simply an
:py:class:`~sympy.core.expr.Expr`, e.g. ``f(x) - C1``. In most cases it
will not be necessary to explicitly identify the function, but if the
function cannot be inferred from the original equation it can be supplied
through the ``func`` argument.
If a sequence of solutions is passed, the same sort of container will be
used to return the result for each solution.
It tries the following methods, in order, until it finds zero equivalence:
1. Substitute the solution for `f` in the original equation. This only
works if ``ode`` is solved for `f`. It will attempt to solve it first
unless ``solve_for_func == False``.
2. Take `n` derivatives of the solution, where `n` is the order of
``ode``, and check to see if that is equal to the solution. This only
works on exact ODEs.
3. Take the 1st, 2nd, ..., `n`\th derivatives of the solution, each time
solving for the derivative of `f` of that order (this will always be
possible because `f` is a linear operator). Then back substitute each
derivative into ``ode`` in reverse order.
This function returns a tuple. The first item in the tuple is ``True`` if
the substitution results in ``0``, and ``False`` otherwise. The second
item in the tuple is what the substitution results in. It should always
be ``0`` if the first item is ``True``. Note that sometimes this function
will ``False``, but with an expression that is identically equal to ``0``,
instead of returning ``True``. This is because
:py:meth:`~sympy.simplify.simplify.simplify` cannot reduce the expression
to ``0``. If an expression returned by this function vanishes
identically, then ``sol`` really is a solution to ``ode``.
If this function seems to hang, it is probably because of a hard
simplification.
To use this function to test, test the first item of the tuple.
Examples
========
>>> from sympy import Eq, Function, checkodesol, symbols
>>> x, C1 = symbols('x,C1')
>>> f = Function('f')
>>> checkodesol(f(x).diff(x), Eq(f(x), C1))
(True, 0)
>>> assert checkodesol(f(x).diff(x), C1)[0]
>>> assert not checkodesol(f(x).diff(x), x)[0]
>>> checkodesol(f(x).diff(x, 2), x**2)
(False, 2)
"""
if not isinstance(ode, Equality):
ode = Eq(ode, 0)
if func is None:
try:
_, func = _preprocess(ode.lhs)
except ValueError:
funcs = [s.atoms(AppliedUndef) for s in (
sol if is_sequence(sol, set) else [sol])]
funcs = set().union(*funcs)
if len(funcs) != 1:
raise ValueError(
'must pass func arg to checkodesol for this case.')
func = funcs.pop()
if not isinstance(func, AppliedUndef) or len(func.args) != 1:
raise ValueError(
"func must be a function of one variable, not %s" % func)
if is_sequence(sol, set):
return type(sol)([checkodesol(ode, i, order=order, solve_for_func=solve_for_func) for i in sol])
if not isinstance(sol, Equality):
sol = Eq(func, sol)
elif sol.rhs == func:
sol = sol.reversed
if order == 'auto':
order = ode_order(ode, func)
solved = sol.lhs == func and not sol.rhs.has(func)
if solve_for_func and not solved:
rhs = solve(sol, func)
if rhs:
eqs = [Eq(func, t) for t in rhs]
if len(rhs) == 1:
eqs = eqs[0]
return checkodesol(ode, eqs, order=order,
solve_for_func=False)
s = True
testnum = 0
x = func.args[0]
while s:
if testnum == 0:
# First pass, try substituting a solved solution directly into the
# ODE. This has the highest chance of succeeding.
ode_diff = ode.lhs - ode.rhs
if sol.lhs == func:
s = sub_func_doit(ode_diff, func, sol.rhs)
else:
testnum += 1
continue
ss = simplify(s)
if ss:
# with the new numer_denom in power.py, if we do a simple
# expansion then testnum == 0 verifies all solutions.
s = ss.expand(force=True)
else:
s = 0
testnum += 1
elif testnum == 1:
# Second pass. If we cannot substitute f, try seeing if the nth
# derivative is equal, this will only work for odes that are exact,
# by definition.
s = simplify(
trigsimp(diff(sol.lhs, x, order) - diff(sol.rhs, x, order)) -
trigsimp(ode.lhs) + trigsimp(ode.rhs))
# s2 = simplify(
# diff(sol.lhs, x, order) - diff(sol.rhs, x, order) - \
# ode.lhs + ode.rhs)
testnum += 1
elif testnum == 2:
# Third pass. Try solving for df/dx and substituting that into the
# ODE. Thanks to Chris Smith for suggesting this method. Many of
# the comments below are his, too.
# The method:
# - Take each of 1..n derivatives of the solution.
# - Solve each nth derivative for d^(n)f/dx^(n)
# (the differential of that order)
# - Back substitute into the ODE in decreasing order
# (i.e., n, n-1, ...)
# - Check the result for zero equivalence
if sol.lhs == func and not sol.rhs.has(func):
diffsols = {0: sol.rhs}
elif sol.rhs == func and not sol.lhs.has(func):
diffsols = {0: sol.lhs}
else:
diffsols = {}
sol = sol.lhs - sol.rhs
for i in range(1, order + 1):
# Differentiation is a linear operator, so there should always
# be 1 solution. Nonetheless, we test just to make sure.
# We only need to solve once. After that, we automatically
# have the solution to the differential in the order we want.
if i == 1:
ds = sol.diff(x)
try:
sdf = solve(ds, func.diff(x, i))
if not sdf:
raise NotImplementedError
except NotImplementedError:
testnum += 1
break
else:
diffsols[i] = sdf[0]
else:
# This is what the solution says df/dx should be.
diffsols[i] = diffsols[i - 1].diff(x)
# Make sure the above didn't fail.
if testnum > 2:
continue
else:
# Substitute it into ODE to check for self consistency.
lhs, rhs = ode.lhs, ode.rhs
for i in range(order, -1, -1):
if i == 0 and 0 not in diffsols:
# We can only substitute f(x) if the solution was
# solved for f(x).
break
lhs = sub_func_doit(lhs, func.diff(x, i), diffsols[i])
rhs = sub_func_doit(rhs, func.diff(x, i), diffsols[i])
ode_or_bool = Eq(lhs, rhs)
ode_or_bool = simplify(ode_or_bool)
if isinstance(ode_or_bool, (bool, BooleanAtom)):
if ode_or_bool:
lhs = rhs = S.Zero
else:
lhs = ode_or_bool.lhs
rhs = ode_or_bool.rhs
# No sense in overworking simplify -- just prove that the
# numerator goes to zero
num = trigsimp((lhs - rhs).as_numer_denom()[0])
# since solutions are obtained using force=True we test
# using the same level of assumptions
## replace function with dummy so assumptions will work
_func = Dummy('func')
num = num.subs(func, _func)
## posify the expression
num, reps = posify(num)
s = simplify(num).xreplace(reps).xreplace({_func: func})
testnum += 1
else:
break
if not s:
return (True, s)
elif s is True: # The code above never was able to change s
raise NotImplementedError("Unable to test if " + str(sol) +
" is a solution to " + str(ode) + ".")
else:
return (False, s)
[docs]def ode_sol_simplicity(sol, func, trysolving=True):
r"""
Returns an extended integer representing how simple a solution to an ODE
is.
The following things are considered, in order from most simple to least:
- ``sol`` is solved for ``func``.
- ``sol`` is not solved for ``func``, but can be if passed to solve (e.g.,
a solution returned by ``dsolve(ode, func, simplify=False``).
- If ``sol`` is not solved for ``func``, then base the result on the
length of ``sol``, as computed by ``len(str(sol))``.
- If ``sol`` has any unevaluated :py:class:`~sympy.integrals.Integral`\s,
this will automatically be considered less simple than any of the above.
This function returns an integer such that if solution A is simpler than
solution B by above metric, then ``ode_sol_simplicity(sola, func) <
ode_sol_simplicity(solb, func)``.
Currently, the following are the numbers returned, but if the heuristic is
ever improved, this may change. Only the ordering is guaranteed.
+----------------------------------------------+-------------------+
| Simplicity | Return |
+==============================================+===================+
| ``sol`` solved for ``func`` | ``-2`` |
+----------------------------------------------+-------------------+
| ``sol`` not solved for ``func`` but can be | ``-1`` |
+----------------------------------------------+-------------------+
| ``sol`` is not solved nor solvable for | ``len(str(sol))`` |
| ``func`` | |
+----------------------------------------------+-------------------+
| ``sol`` contains an | ``oo`` |
| :py:class:`~sympy.integrals.Integral` | |
+----------------------------------------------+-------------------+
``oo`` here means the SymPy infinity, which should compare greater than
any integer.
If you already know :py:meth:`~sympy.solvers.solvers.solve` cannot solve
``sol``, you can use ``trysolving=False`` to skip that step, which is the
only potentially slow step. For example,
:py:meth:`~sympy.solvers.ode.dsolve` with the ``simplify=False`` flag
should do this.
If ``sol`` is a list of solutions, if the worst solution in the list
returns ``oo`` it returns that, otherwise it returns ``len(str(sol))``,
that is, the length of the string representation of the whole list.
Examples
========
This function is designed to be passed to ``min`` as the key argument,
such as ``min(listofsolutions, key=lambda i: ode_sol_simplicity(i,
f(x)))``.
>>> from sympy import symbols, Function, Eq, tan, cos, sqrt, Integral
>>> from sympy.solvers.ode import ode_sol_simplicity
>>> x, C1, C2 = symbols('x, C1, C2')
>>> f = Function('f')
>>> ode_sol_simplicity(Eq(f(x), C1*x**2), f(x))
-2
>>> ode_sol_simplicity(Eq(x**2 + f(x), C1), f(x))
-1
>>> ode_sol_simplicity(Eq(f(x), C1*Integral(2*x, x)), f(x))
oo
>>> eq1 = Eq(f(x)/tan(f(x)/(2*x)), C1)
>>> eq2 = Eq(f(x)/tan(f(x)/(2*x) + f(x)), C2)
>>> [ode_sol_simplicity(eq, f(x)) for eq in [eq1, eq2]]
[28, 35]
>>> min([eq1, eq2], key=lambda i: ode_sol_simplicity(i, f(x)))
Eq(f(x)/tan(f(x)/(2*x)), C1)
"""
# TODO: if two solutions are solved for f(x), we still want to be
# able to get the simpler of the two
# See the docstring for the coercion rules. We check easier (faster)
# things here first, to save time.
if iterable(sol):
# See if there are Integrals
for i in sol:
if ode_sol_simplicity(i, func, trysolving=trysolving) == oo:
return oo
return len(str(sol))
if sol.has(Integral):
return oo
# Next, try to solve for func. This code will change slightly when CRootOf
# is implemented in solve(). Probably a CRootOf solution should fall
# somewhere between a normal solution and an unsolvable expression.
# First, see if they are already solved
if sol.lhs == func and not sol.rhs.has(func) or \
sol.rhs == func and not sol.lhs.has(func):
return -2
# We are not so lucky, try solving manually
if trysolving:
try:
sols = solve(sol, func)
if not sols:
raise NotImplementedError
except NotImplementedError:
pass
else:
return -1
# Finally, a naive computation based on the length of the string version
# of the expression. This may favor combined fractions because they
# will not have duplicate denominators, and may slightly favor expressions
# with fewer additions and subtractions, as those are separated by spaces
# by the printer.
# Additional ideas for simplicity heuristics are welcome, like maybe
# checking if a equation has a larger domain, or if constantsimp has
# introduced arbitrary constants numbered higher than the order of a
# given ODE that sol is a solution of.
return len(str(sol))
def _get_constant_subexpressions(expr, Cs):
Cs = set(Cs)
Ces = []
def _recursive_walk(expr):
expr_syms = expr.free_symbols
if len(expr_syms) > 0 and expr_syms.issubset(Cs):
Ces.append(expr)
else:
if expr.func == exp:
expr = expr.expand(mul=True)
if expr.func in (Add, Mul):
d = sift(expr.args, lambda i : i.free_symbols.issubset(Cs))
if len(d[True]) > 1:
x = expr.func(*d[True])
if not x.is_number:
Ces.append(x)
elif isinstance(expr, Integral):
if expr.free_symbols.issubset(Cs) and \
all(len(x) == 3 for x in expr.limits):
Ces.append(expr)
for i in expr.args:
_recursive_walk(i)
return
_recursive_walk(expr)
return Ces
def __remove_linear_redundancies(expr, Cs):
cnts = {i: expr.count(i) for i in Cs}
Cs = [i for i in Cs if cnts[i] > 0]
def _linear(expr):
if expr.func is Add:
xs = [i for i in Cs if expr.count(i)==cnts[i] \
and 0 == expr.diff(i, 2)]
d = {}
for x in xs:
y = expr.diff(x)
if y not in d:
d[y]=[]
d[y].append(x)
for y in d:
if len(d[y]) > 1:
d[y].sort(key=str)
for x in d[y][1:]:
expr = expr.subs(x, 0)
return expr
def _recursive_walk(expr):
if len(expr.args) != 0:
expr = expr.func(*[_recursive_walk(i) for i in expr.args])
expr = _linear(expr)
return expr
if expr.func is Equality:
lhs, rhs = [_recursive_walk(i) for i in expr.args]
f = lambda i: isinstance(i, Number) or i in Cs
if lhs.func is Symbol and lhs in Cs:
rhs, lhs = lhs, rhs
if lhs.func in (Add, Symbol) and rhs.func in (Add, Symbol):
dlhs = sift([lhs] if isinstance(lhs, AtomicExpr) else lhs.args, f)
drhs = sift([rhs] if isinstance(rhs, AtomicExpr) else rhs.args, f)
for i in [True, False]:
for hs in [dlhs, drhs]:
if i not in hs:
hs[i] = [0]
# this calculation can be simplified
lhs = Add(*dlhs[False]) - Add(*drhs[False])
rhs = Add(*drhs[True]) - Add(*dlhs[True])
elif lhs.func in (Mul, Symbol) and rhs.func in (Mul, Symbol):
dlhs = sift([lhs] if isinstance(lhs, AtomicExpr) else lhs.args, f)
if True in dlhs:
if False not in dlhs:
dlhs[False] = [1]
lhs = Mul(*dlhs[False])
rhs = rhs/Mul(*dlhs[True])
return Eq(lhs, rhs)
else:
return _recursive_walk(expr)
@vectorize(0)
[docs]def constantsimp(expr, constants):
r"""
Simplifies an expression with arbitrary constants in it.
This function is written specifically to work with
:py:meth:`~sympy.solvers.ode.dsolve`, and is not intended for general use.
Simplification is done by "absorbing" the arbitrary constants into other
arbitrary constants, numbers, and symbols that they are not independent
of.
The symbols must all have the same name with numbers after it, for
example, ``C1``, ``C2``, ``C3``. The ``symbolname`` here would be
'``C``', the ``startnumber`` would be 1, and the ``endnumber`` would be 3.
If the arbitrary constants are independent of the variable ``x``, then the
independent symbol would be ``x``. There is no need to specify the
dependent function, such as ``f(x)``, because it already has the
independent symbol, ``x``, in it.
Because terms are "absorbed" into arbitrary constants and because
constants are renumbered after simplifying, the arbitrary constants in
expr are not necessarily equal to the ones of the same name in the
returned result.
If two or more arbitrary constants are added, multiplied, or raised to the
power of each other, they are first absorbed together into a single
arbitrary constant. Then the new constant is combined into other terms if
necessary.
Absorption of constants is done with limited assistance:
1. terms of :py:class:`~sympy.core.add.Add`\s are collected to try join
constants so `e^x (C_1 \cos(x) + C_2 \cos(x))` will simplify to `e^x
C_1 \cos(x)`;
2. powers with exponents that are :py:class:`~sympy.core.add.Add`\s are
expanded so `e^{C_1 + x}` will be simplified to `C_1 e^x`.
Use :py:meth:`~sympy.solvers.ode.constant_renumber` to renumber constants
after simplification or else arbitrary numbers on constants may appear,
e.g. `C_1 + C_3 x`.
In rare cases, a single constant can be "simplified" into two constants.
Every differential equation solution should have as many arbitrary
constants as the order of the differential equation. The result here will
be technically correct, but it may, for example, have `C_1` and `C_2` in
an expression, when `C_1` is actually equal to `C_2`. Use your discretion
in such situations, and also take advantage of the ability to use hints in
:py:meth:`~sympy.solvers.ode.dsolve`.
Examples
========
>>> from sympy import symbols
>>> from sympy.solvers.ode import constantsimp
>>> C1, C2, C3, x, y = symbols('C1, C2, C3, x, y')
>>> constantsimp(2*C1*x, set([C1, C2, C3]))
C1*x
>>> constantsimp(C1 + 2 + x, set([C1, C2, C3]))
C1 + x
>>> constantsimp(C1*C2 + 2 + C2 + C3*x, set([C1, C2, C3]))
C1 + C3*x
"""
# This function works recursively. The idea is that, for Mul,
# Add, Pow, and Function, if the class has a constant in it, then
# we can simplify it, which we do by recursing down and
# simplifying up. Otherwise, we can skip that part of the
# expression.
Cs = constants
orig_expr = expr
constant_subexprs = _get_constant_subexpressions(expr, Cs)
for xe in constant_subexprs:
xes = list(xe.free_symbols)
if not xes:
continue
if all([expr.count(c) == xe.count(c) for c in xes]):
xes.sort(key=str)
expr = expr.subs(xe, xes[0])
# try to perform common sub-expression elimination of constant terms
try:
commons, rexpr = cse(expr)
commons.reverse()
rexpr = rexpr[0]
for s in commons:
cs = list(s[1].atoms(Symbol))
if len(cs) == 1 and cs[0] in Cs:
rexpr = rexpr.subs(s[0], cs[0])
else:
rexpr = rexpr.subs(*s)
expr = rexpr
except Exception:
pass
expr = __remove_linear_redundancies(expr, Cs)
def _conditional_term_factoring(expr):
new_expr = terms_gcd(expr, clear=False, deep=True, expand=False)
# we do not want to factor exponentials, so handle this separately
if new_expr.is_Mul:
infac = False
asfac = False
for m in new_expr.args:
if m.func is exp:
asfac = True
elif m.is_Add:
infac = any(fi.func is exp for t in m.args
for fi in Mul.make_args(t))
if asfac and infac:
new_expr = expr
break
return new_expr
expr = _conditional_term_factoring(expr)
# call recursively if more simplification is possible
if orig_expr != expr:
return constantsimp(expr, Cs)
return expr
[docs]def constant_renumber(expr, symbolname, startnumber, endnumber):
r"""
Renumber arbitrary constants in ``expr`` to have numbers 1 through `N`
where `N` is ``endnumber - startnumber + 1`` at most.
In the process, this reorders expression terms in a standard way.
This is a simple function that goes through and renumbers any
:py:class:`~sympy.core.symbol.Symbol` with a name in the form ``symbolname
+ num`` where ``num`` is in the range from ``startnumber`` to
``endnumber``.
Symbols are renumbered based on ``.sort_key()``, so they should be
numbered roughly in the order that they appear in the final, printed
expression. Note that this ordering is based in part on hashes, so it can
produce different results on different machines.
The structure of this function is very similar to that of
:py:meth:`~sympy.solvers.ode.constantsimp`.
Examples
========
>>> from sympy import symbols, Eq, pprint
>>> from sympy.solvers.ode import constant_renumber
>>> x, C0, C1, C2, C3, C4 = symbols('x,C:5')
Only constants in the given range (inclusive) are renumbered;
the renumbering always starts from 1:
>>> constant_renumber(C1 + C3 + C4, 'C', 1, 3)
C1 + C2 + C4
>>> constant_renumber(C0 + C1 + C3 + C4, 'C', 2, 4)
C0 + 2*C1 + C2
>>> constant_renumber(C0 + 2*C1 + C2, 'C', 0, 1)
C1 + 3*C2
>>> pprint(C2 + C1*x + C3*x**2)
2
C1*x + C2 + C3*x
>>> pprint(constant_renumber(C2 + C1*x + C3*x**2, 'C', 1, 3))
2
C1 + C2*x + C3*x
"""
if type(expr) in (set, list, tuple):
return type(expr)(
[constant_renumber(i, symbolname=symbolname, startnumber=startnumber, endnumber=endnumber)
for i in expr]
)
global newstartnumber
newstartnumber = 1
constants_found = [None]*(endnumber + 2)
constantsymbols = [Symbol(
symbolname + "%d" % t) for t in range(startnumber,
endnumber + 1)]
# make a mapping to send all constantsymbols to S.One and use
# that to make sure that term ordering is not dependent on
# the indexed value of C
C_1 = [(ci, S.One) for ci in constantsymbols]
sort_key=lambda arg: default_sort_key(arg.subs(C_1))
def _constant_renumber(expr):
r"""
We need to have an internal recursive function so that
newstartnumber maintains its values throughout recursive calls.
"""
global newstartnumber
if isinstance(expr, Equality):
return Eq(
_constant_renumber(expr.lhs),
_constant_renumber(expr.rhs))
if type(expr) not in (Mul, Add, Pow) and not expr.is_Function and \
not expr.has(*constantsymbols):
# Base case, as above. Hope there aren't constants inside
# of some other class, because they won't be renumbered.
return expr
elif expr.is_Piecewise:
return expr
elif expr in constantsymbols:
if expr not in constants_found:
constants_found[newstartnumber] = expr
newstartnumber += 1
return expr
elif expr.is_Function or expr.is_Pow or isinstance(expr, Tuple):
return expr.func(
*[_constant_renumber(x) for x in expr.args])
else:
sortedargs = list(expr.args)
sortedargs.sort(key=sort_key)
return expr.func(*[_constant_renumber(x) for x in sortedargs])
expr = _constant_renumber(expr)
# Renumbering happens here
newconsts = symbols('C1:%d' % newstartnumber)
expr = expr.subs(zip(constants_found[1:], newconsts), simultaneous=True)
return expr
def _handle_Integral(expr, func, order, hint):
r"""
Converts a solution with Integrals in it into an actual solution.
For most hints, this simply runs ``expr.doit()``.
"""
global y
x = func.args[0]
f = func.func
if hint == "1st_exact":
sol = (expr.doit()).subs(y, f(x))
del y
elif hint == "1st_exact_Integral":
sol = expr.subs(y, f(x))
del y
elif hint == "nth_linear_constant_coeff_homogeneous":
sol = expr
elif not hint.endswith("_Integral"):
sol = expr.doit()
else:
sol = expr
return sol
# FIXME: replace the general solution in the docstring with
# dsolve(equation, hint='1st_exact_Integral'). You will need to be able
# to have assumptions on P and Q that dP/dy = dQ/dx.
[docs]def ode_1st_exact(eq, func, order, match):
r"""
Solves 1st order exact ordinary differential equations.
A 1st order differential equation is called exact if it is the total
differential of a function. That is, the differential equation
.. math:: P(x, y) \,\partial{}x + Q(x, y) \,\partial{}y = 0
is exact if there is some function `F(x, y)` such that `P(x, y) =
\partial{}F/\partial{}x` and `Q(x, y) = \partial{}F/\partial{}y`. It can
be shown that a necessary and sufficient condition for a first order ODE
to be exact is that `\partial{}P/\partial{}y = \partial{}Q/\partial{}x`.
Then, the solution will be as given below::
>>> from sympy import Function, Eq, Integral, symbols, pprint
>>> x, y, t, x0, y0, C1= symbols('x,y,t,x0,y0,C1')
>>> P, Q, F= map(Function, ['P', 'Q', 'F'])
>>> pprint(Eq(Eq(F(x, y), Integral(P(t, y), (t, x0, x)) +
... Integral(Q(x0, t), (t, y0, y))), C1))
x y
/ /
| |
F(x, y) = | P(t, y) dt + | Q(x0, t) dt = C1
| |
/ /
x0 y0
Where the first partials of `P` and `Q` exist and are continuous in a
simply connected region.
A note: SymPy currently has no way to represent inert substitution on an
expression, so the hint ``1st_exact_Integral`` will return an integral
with `dy`. This is supposed to represent the function that you are
solving for.
Examples
========
>>> from sympy import Function, dsolve, cos, sin
>>> from sympy.abc import x
>>> f = Function('f')
>>> dsolve(cos(f(x)) - (x*sin(f(x)) - f(x)**2)*f(x).diff(x),
... f(x), hint='1st_exact')
Eq(x*cos(f(x)) + f(x)**3/3, C1)
References
==========
- http://en.wikipedia.org/wiki/Exact_differential_equation
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 73
# indirect doctest
"""
x = func.args[0]
f = func.func
r = match # d+e*diff(f(x),x)
e = r[r['e']]
d = r[r['d']]
global y # This is the only way to pass dummy y to _handle_Integral
y = r['y']
C1 = get_numbered_constants(eq, num=1)
# Refer Joel Moses, "Symbolic Integration - The Stormy Decade",
# Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558
# which gives the method to solve an exact differential equation.
sol = Integral(d, x) + Integral((e - (Integral(d, x).diff(y))), y)
return Eq(sol, C1)
[docs]def ode_1st_homogeneous_coeff_best(eq, func, order, match):
r"""
Returns the best solution to an ODE from the two hints
``1st_homogeneous_coeff_subs_dep_div_indep`` and
``1st_homogeneous_coeff_subs_indep_div_dep``.
This is as determined by :py:meth:`~sympy.solvers.ode.ode_sol_simplicity`.
See the
:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`
and
:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`
docstrings for more information on these hints. Note that there is no
``ode_1st_homogeneous_coeff_best_Integral`` hint.
Examples
========
>>> from sympy import Function, dsolve, pprint
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x),
... hint='1st_homogeneous_coeff_best', simplify=False))
/ 2 \
| 3*x |
log|----- + 1|
| 2 |
\f (x) /
log(f(x)) = log(C1) - --------------
3
References
==========
- http://en.wikipedia.org/wiki/Homogeneous_differential_equation
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 59
# indirect doctest
"""
# There are two substitutions that solve the equation, u1=y/x and u2=x/y
# They produce different integrals, so try them both and see which
# one is easier.
sol1 = ode_1st_homogeneous_coeff_subs_indep_div_dep(eq,
func, order, match)
sol2 = ode_1st_homogeneous_coeff_subs_dep_div_indep(eq,
func, order, match)
simplify = match.get('simplify', True)
if simplify:
# why is odesimp called here? Should it be at the usual spot?
constants = sol1.free_symbols.difference(eq.free_symbols)
sol1 = odesimp(
sol1, func, order, constants,
"1st_homogeneous_coeff_subs_indep_div_dep")
constants = sol2.free_symbols.difference(eq.free_symbols)
sol2 = odesimp(
sol2, func, order, constants,
"1st_homogeneous_coeff_subs_dep_div_indep")
return min([sol1, sol2], key=lambda x: ode_sol_simplicity(x, func,
trysolving=not simplify))
[docs]def ode_1st_homogeneous_coeff_subs_dep_div_indep(eq, func, order, match):
r"""
Solves a 1st order differential equation with homogeneous coefficients
using the substitution `u_1 = \frac{\text{<dependent
variable>}}{\text{<independent variable>}}`.
This is a differential equation
.. math:: P(x, y) + Q(x, y) dy/dx = 0
such that `P` and `Q` are homogeneous and of the same order. A function
`F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`.
Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`. See
also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`.
If the coefficients `P` and `Q` in the differential equation above are
homogeneous functions of the same order, then it can be shown that the
substitution `y = u_1 x` (i.e. `u_1 = y/x`) will turn the differential
equation into an equation separable in the variables `x` and `u`. If
`h(u_1)` is the function that results from making the substitution `u_1 =
f(x)/x` on `P(x, f(x))` and `g(u_2)` is the function that results from the
substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) +
Q(x, f(x)) f'(x) = 0`, then the general solution is::
>>> from sympy import Function, dsolve, pprint
>>> from sympy.abc import x
>>> f, g, h = map(Function, ['f', 'g', 'h'])
>>> genform = g(f(x)/x) + h(f(x)/x)*f(x).diff(x)
>>> pprint(genform)
/f(x)\ /f(x)\ d
g|----| + h|----|*--(f(x))
\ x / \ x / dx
>>> pprint(dsolve(genform, f(x),
... hint='1st_homogeneous_coeff_subs_dep_div_indep_Integral'))
f(x)
----
x
/
|
| -h(u1)
log(x) = C1 + | ---------------- d(u1)
| u1*h(u1) + g(u1)
|
/
Where `u_1 h(u_1) + g(u_1) \ne 0` and `x \ne 0`.
See also the docstrings of
:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best` and
:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`.
Examples
========
>>> from sympy import Function, dsolve
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x),
... hint='1st_homogeneous_coeff_subs_dep_div_indep', simplify=False))
/ 3 \
|3*f(x) f (x)|
log|------ + -----|
| x 3 |
\ x /
log(x) = log(C1) - -------------------
3
References
==========
- http://en.wikipedia.org/wiki/Homogeneous_differential_equation
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 59
# indirect doctest
"""
x = func.args[0]
f = func.func
u = Dummy('u')
u1 = Dummy('u1') # u1 == f(x)/x
r = match # d+e*diff(f(x),x)
C1 = get_numbered_constants(eq, num=1)
xarg = match.get('xarg', 0)
yarg = match.get('yarg', 0)
int = Integral(
(-r[r['e']]/(r[r['d']] + u1*r[r['e']])).subs({x: 1, r['y']: u1}),
(u1, None, f(x)/x))
sol = logcombine(Eq(log(x), int + log(C1)), force=True)
sol = sol.subs(f(x), u).subs(((u, u - yarg), (x, x - xarg), (u, f(x))))
return sol
[docs]def ode_1st_homogeneous_coeff_subs_indep_div_dep(eq, func, order, match):
r"""
Solves a 1st order differential equation with homogeneous coefficients
using the substitution `u_2 = \frac{\text{<independent
variable>}}{\text{<dependent variable>}}`.
This is a differential equation
.. math:: P(x, y) + Q(x, y) dy/dx = 0
such that `P` and `Q` are homogeneous and of the same order. A function
`F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`.
Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`. See
also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`.
If the coefficients `P` and `Q` in the differential equation above are
homogeneous functions of the same order, then it can be shown that the
substitution `x = u_2 y` (i.e. `u_2 = x/y`) will turn the differential
equation into an equation separable in the variables `y` and `u_2`. If
`h(u_2)` is the function that results from making the substitution `u_2 =
x/f(x)` on `P(x, f(x))` and `g(u_2)` is the function that results from the
substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) +
Q(x, f(x)) f'(x) = 0`, then the general solution is:
>>> from sympy import Function, dsolve, pprint
>>> from sympy.abc import x
>>> f, g, h = map(Function, ['f', 'g', 'h'])
>>> genform = g(x/f(x)) + h(x/f(x))*f(x).diff(x)
>>> pprint(genform)
/ x \ / x \ d
g|----| + h|----|*--(f(x))
\f(x)/ \f(x)/ dx
>>> pprint(dsolve(genform, f(x),
... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral'))
x
----
f(x)
/
|
| -g(u2)
| ---------------- d(u2)
| u2*g(u2) + h(u2)
|
/
<BLANKLINE>
f(x) = C1*e
Where `u_2 g(u_2) + h(u_2) \ne 0` and `f(x) \ne 0`.
See also the docstrings of
:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best` and
:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`.
Examples
========
>>> from sympy import Function, pprint, dsolve
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x),
... hint='1st_homogeneous_coeff_subs_indep_div_dep',
... simplify=False))
/ 2 \
| 3*x |
log|----- + 1|
| 2 |
\f (x) /
log(f(x)) = log(C1) - --------------
3
References
==========
- http://en.wikipedia.org/wiki/Homogeneous_differential_equation
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 59
# indirect doctest
"""
x = func.args[0]
f = func.func
u = Dummy('u')
u2 = Dummy('u2') # u2 == x/f(x)
r = match # d+e*diff(f(x),x)
C1 = get_numbered_constants(eq, num=1)
xarg = match.get('xarg', 0) # If xarg present take xarg, else zero
yarg = match.get('yarg', 0) # If yarg present take yarg, else zero
int = Integral(
simplify(
(-r[r['d']]/(r[r['e']] + u2*r[r['d']])).subs({x: u2, r['y']: 1})),
(u2, None, x/f(x)))
sol = logcombine(Eq(log(f(x)), int + log(C1)), force=True)
sol = sol.subs(f(x), u).subs(((u, u - yarg), (x, x - xarg), (u, f(x))))
return sol
# XXX: Should this function maybe go somewhere else?
[docs]def homogeneous_order(eq, *symbols):
r"""
Returns the order `n` if `g` is homogeneous and ``None`` if it is not
homogeneous.
Determines if a function is homogeneous and if so of what order. A
function `f(x, y, \cdots)` is homogeneous of order `n` if `f(t x, t y,
\cdots) = t^n f(x, y, \cdots)`.
If the function is of two variables, `F(x, y)`, then `f` being homogeneous
of any order is equivalent to being able to rewrite `F(x, y)` as `G(x/y)`
or `H(y/x)`. This fact is used to solve 1st order ordinary differential
equations whose coefficients are homogeneous of the same order (see the
docstrings of
:py:meth:`~solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep` and
:py:meth:`~solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`).
Symbols can be functions, but every argument of the function must be a
symbol, and the arguments of the function that appear in the expression
must match those given in the list of symbols. If a declared function
appears with different arguments than given in the list of symbols,
``None`` is returned.
Examples
========
>>> from sympy import Function, homogeneous_order, sqrt
>>> from sympy.abc import x, y
>>> f = Function('f')
>>> homogeneous_order(f(x), f(x)) is None
True
>>> homogeneous_order(f(x,y), f(y, x), x, y) is None
True
>>> homogeneous_order(f(x), f(x), x)
1
>>> homogeneous_order(x**2*f(x)/sqrt(x**2+f(x)**2), x, f(x))
2
>>> homogeneous_order(x**2+f(x), x, f(x)) is None
True
"""
if not symbols:
raise ValueError("homogeneous_order: no symbols were given.")
symset = set(symbols)
eq = sympify(eq)
# The following are not supported
if eq.has(Order, Derivative):
return None
# These are all constants
if (eq.is_Number or
eq.is_NumberSymbol or
eq.is_number
):
return S.Zero
# Replace all functions with dummy variables
dum = numbered_symbols(prefix='d', cls=Dummy)
newsyms = set()
for i in [j for j in symset if getattr(j, 'is_Function')]:
iargs = set(i.args)
if iargs.difference(symset):
return None
else:
dummyvar = next(dum)
eq = eq.subs(i, dummyvar)
symset.remove(i)
newsyms.add(dummyvar)
symset.update(newsyms)
if not eq.free_symbols & symset:
return None
# assuming order of a nested function can only be equal to zero
if isinstance(eq, Function):
return None if homogeneous_order(
eq.args[0], *tuple(symset)) != 0 else S.Zero
# make the replacement of x with x*t and see if t can be factored out
t = Dummy('t', positive=True) # It is sufficient that t > 0
eqs = separatevars(eq.subs([(i, t*i) for i in symset]), [t], dict=True)[t]
if eqs is S.One:
return S.Zero # there was no term with only t
i, d = eqs.as_independent(t, as_Add=False)
b, e = d.as_base_exp()
if b == t:
return e
[docs]def ode_1st_linear(eq, func, order, match):
r"""
Solves 1st order linear differential equations.
These are differential equations of the form
.. math:: dy/dx + P(x) y = Q(x)\text{.}
These kinds of differential equations can be solved in a general way. The
integrating factor `e^{\int P(x) \,dx}` will turn the equation into a
separable equation. The general solution is::
>>> from sympy import Function, dsolve, Eq, pprint, diff, sin
>>> from sympy.abc import x
>>> f, P, Q = map(Function, ['f', 'P', 'Q'])
>>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x))
>>> pprint(genform)
d
P(x)*f(x) + --(f(x)) = Q(x)
dx
>>> pprint(dsolve(genform, f(x), hint='1st_linear_Integral'))
/ / \
| | |
| | / | /
| | | | |
| | | P(x) dx | - | P(x) dx
| | | | |
| | / | /
f(x) = |C1 + | Q(x)*e dx|*e
| | |
\ / /
Examples
========
>>> f = Function('f')
>>> pprint(dsolve(Eq(x*diff(f(x), x) - f(x), x**2*sin(x)),
... f(x), '1st_linear'))
f(x) = x*(C1 - cos(x))
References
==========
- http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 92
# indirect doctest
"""
x = func.args[0]
f = func.func
r = match # a*diff(f(x),x) + b*f(x) + c
C1 = get_numbered_constants(eq, num=1)
t = exp(Integral(r[r['b']]/r[r['a']], x))
tt = Integral(t*(-r[r['c']]/r[r['a']]), x)
f = match.get('u', f(x)) # take almost-linear u if present, else f(x)
return Eq(f, (tt + C1)/t)
[docs]def ode_Bernoulli(eq, func, order, match):
r"""
Solves Bernoulli differential equations.
These are equations of the form
.. math:: dy/dx + P(x) y = Q(x) y^n\text{, }n \ne 1`\text{.}
The substitution `w = 1/y^{1-n}` will transform an equation of this form
into one that is linear (see the docstring of
:py:meth:`~sympy.solvers.ode.ode_1st_linear`). The general solution is::
>>> from sympy import Function, dsolve, Eq, pprint
>>> from sympy.abc import x, n
>>> f, P, Q = map(Function, ['f', 'P', 'Q'])
>>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)**n)
>>> pprint(genform)
d n
P(x)*f(x) + --(f(x)) = Q(x)*f (x)
dx
>>> pprint(dsolve(genform, f(x), hint='Bernoulli_Integral')) #doctest: +SKIP
1
----
1 - n
// / \ \
|| | | |
|| | / | / |
|| | | | | |
|| | (1 - n)* | P(x) dx | (-1 + n)* | P(x) dx|
|| | | | | |
|| | / | / |
f(x) = ||C1 + (-1 + n)* | -Q(x)*e dx|*e |
|| | | |
\\ / / /
Note that the equation is separable when `n = 1` (see the docstring of
:py:meth:`~sympy.solvers.ode.ode_separable`).
>>> pprint(dsolve(Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)), f(x),
... hint='separable_Integral'))
f(x)
/
| /
| 1 |
| - dy = C1 + | (-P(x) + Q(x)) dx
| y |
| /
/
Examples
========
>>> from sympy import Function, dsolve, Eq, pprint, log
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(Eq(x*f(x).diff(x) + f(x), log(x)*f(x)**2),
... f(x), hint='Bernoulli'))
1
f(x) = -------------------
/ log(x) 1\
x*|C1 + ------ + -|
\ x x/
References
==========
- http://en.wikipedia.org/wiki/Bernoulli_differential_equation
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 95
# indirect doctest
"""
x = func.args[0]
f = func.func
r = match # a*diff(f(x),x) + b*f(x) + c*f(x)**n, n != 1
C1 = get_numbered_constants(eq, num=1)
t = exp((1 - r[r['n']])*Integral(r[r['b']]/r[r['a']], x))
tt = (r[r['n']] - 1)*Integral(t*r[r['c']]/r[r['a']], x)
return Eq(f(x), ((tt + C1)/t)**(1/(1 - r[r['n']])))
[docs]def ode_Riccati_special_minus2(eq, func, order, match):
r"""
The general Riccati equation has the form
.. math:: dy/dx = f(x) y^2 + g(x) y + h(x)\text{.}
While it does not have a general solution [1], the "special" form, `dy/dx
= a y^2 - b x^c`, does have solutions in many cases [2]. This routine
returns a solution for `a(dy/dx) = b y^2 + c y/x + d/x^2` that is obtained
by using a suitable change of variables to reduce it to the special form
and is valid when neither `a` nor `b` are zero and either `c` or `d` is
zero.
>>> from sympy.abc import x, y, a, b, c, d
>>> from sympy.solvers.ode import dsolve, checkodesol
>>> from sympy import pprint, Function
>>> f = Function('f')
>>> y = f(x)
>>> genform = a*y.diff(x) - (b*y**2 + c*y/x + d/x**2)
>>> sol = dsolve(genform, y)
>>> pprint(sol, wrap_line=False)
/ / __________________ \\
| __________________ | / 2 ||
| / 2 | \/ 4*b*d - (a + c) *log(x)||
-|a + c - \/ 4*b*d - (a + c) *tan|C1 + ----------------------------||
\ \ 2*a //
f(x) = ------------------------------------------------------------------------
2*b*x
>>> checkodesol(genform, sol, order=1)[0]
True
References
==========
1. http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Riccati
2. http://eqworld.ipmnet.ru/en/solutions/ode/ode0106.pdf -
http://eqworld.ipmnet.ru/en/solutions/ode/ode0123.pdf
"""
x = func.args[0]
f = func.func
r = match # a2*diff(f(x),x) + b2*f(x) + c2*f(x)/x + d2/x**2
a2, b2, c2, d2 = [r[r[s]] for s in 'a2 b2 c2 d2'.split()]
C1 = get_numbered_constants(eq, num=1)
mu = sqrt(4*d2*b2 - (a2 - c2)**2)
return Eq(f(x), (a2 - c2 - mu*tan(mu/(2*a2)*log(x) + C1))/(2*b2*x))
[docs]def ode_Liouville(eq, func, order, match):
r"""
Solves 2nd order Liouville differential equations.
The general form of a Liouville ODE is
.. math:: \frac{d^2 y}{dx^2} + g(y) \left(\!
\frac{dy}{dx}\!\right)^2 + h(x)
\frac{dy}{dx}\text{.}
The general solution is:
>>> from sympy import Function, dsolve, Eq, pprint, diff
>>> from sympy.abc import x
>>> f, g, h = map(Function, ['f', 'g', 'h'])
>>> genform = Eq(diff(f(x),x,x) + g(f(x))*diff(f(x),x)**2 +
... h(x)*diff(f(x),x), 0)
>>> pprint(genform)
2 2
/d \ d d
g(f(x))*|--(f(x))| + h(x)*--(f(x)) + ---(f(x)) = 0
\dx / dx 2
dx
>>> pprint(dsolve(genform, f(x), hint='Liouville_Integral'))
f(x)
/ /
| |
| / | /
| | | |
| - | h(x) dx | | g(y) dy
| | | |
| / | /
C1 + C2* | e dx + | e dy = 0
| |
/ /
Examples
========
>>> from sympy import Function, dsolve, Eq, pprint
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(diff(f(x), x, x) + diff(f(x), x)**2/f(x) +
... diff(f(x), x)/x, f(x), hint='Liouville'))
________________ ________________
[f(x) = -\/ C1 + C2*log(x) , f(x) = \/ C1 + C2*log(x) ]
References
==========
- Goldstein and Braun, "Advanced Methods for the Solution of Differential
Equations", pp. 98
- http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Liouville
# indirect doctest
"""
# Liouville ODE:
# f(x).diff(x, 2) + g(f(x))*(f(x).diff(x, 2))**2 + h(x)*f(x).diff(x)
# See Goldstein and Braun, "Advanced Methods for the Solution of
# Differential Equations", pg. 98, as well as
# http://www.maplesoft.com/support/help/view.aspx?path=odeadvisor/Liouville
x = func.args[0]
f = func.func
r = match # f(x).diff(x, 2) + g*f(x).diff(x)**2 + h*f(x).diff(x)
y = r['y']
C1, C2 = get_numbered_constants(eq, num=2)
int = Integral(exp(Integral(r['g'], y)), (y, None, f(x)))
sol = Eq(int + C1*Integral(exp(-Integral(r['h'], x)), x) + C2, 0)
return sol
[docs]def ode_2nd_power_series_ordinary(eq, func, order, match):
r"""
Gives a power series solution to a second order homogeneous differential
equation with polynomial coefficients at an ordinary point. A homogenous
differential equation is of the form
.. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0
For simplicity it is assumed that `P(x)`, `Q(x)` and `R(x)` are polynomials,
it is sufficient that `\frac{Q(x)}{P(x)}` and `\frac{R(x)}{P(x)}` exists at
`x_{0}`. A recurrence relation is obtained by substituting `y` as `\sum_{n=0}^\infty a_{n}x^{n}`,
in the differential equation, and equating the nth term. Using this relation
various terms can be generated.
Examples
========
>>> from sympy import dsolve, Function, pprint
>>> from sympy.abc import x, y
>>> f = Function("f")
>>> eq = f(x).diff(x, 2) + f(x)
>>> pprint(dsolve(eq, hint='2nd_power_series_ordinary'))
/ 4 2 \ / 2 \
|x x | | x | / 6\
f(x) = C2*|-- - -- + 1| + C1*x*|- -- + 1| + O\x /
\24 2 / \ 6 /
References
==========
- http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx
- George E. Simmons, "Differential Equations with Applications and
Historical Notes", p.p 176 - 184
"""
x = func.args[0]
f = func.func
C0, C1 = get_numbered_constants(eq, num=2)
n = Dummy("n", integer=True)
s = Wild("s")
k = Wild("k", exclude=[x])
x0 = match.get('x0')
terms = match.get('terms', 5)
p = match[match['a3']]
q = match[match['b3']]
r = match[match['c3']]
seriesdict = {}
recurr = Function("r")
# Generating the recurrence relation which works this way:
# for the second order term the summation begins at n = 2. The coefficients
# p is multiplied with an*(n - 1)*(n - 2)*x**n-2 and a substitution is made such that
# the exponent of x becomes n.
# For example, if p is x, then the second degree recurrence term is
# an*(n - 1)*(n - 2)*x**n-1, substituting (n - 1) as n, it transforms to
# an+1*n*(n - 1)*x**n.
# A similar process is done with the first order and zeroth order term.
coefflist = [(recurr(n), r), (n*recurr(n), q), (n*(n - 1)*recurr(n), p)]
for index, coeff in enumerate(coefflist):
if coeff[1]:
f2 = powsimp(expand((coeff[1]*(x - x0)**(n - index)).subs(x, x + x0)))
if f2.is_Add:
addargs = f2.args
else:
addargs = [f2]
for arg in addargs:
powm = arg.match(s*x**k)
term = coeff[0]*powm[s]
if not powm[k].is_Symbol:
term = term.subs(n, n - powm[k].as_independent(n)[0])
startind = powm[k].subs(n, index)
# Seeing if the startterm can be reduced further.
# If it vanishes for n lesser than startind, it is
# equal to summation from n.
if startind:
for i in reversed(range(startind)):
if not term.subs(n, i):
seriesdict[term] = i
else:
seriesdict[term] = i + 1
break
else:
seriesdict[term] = S(0)
# Stripping of terms so that the sum starts with the same number.
teq = S(0)
suminit = seriesdict.values()
rkeys = seriesdict.keys()
req = Add(*rkeys)
if any(suminit):
maxval = max(suminit)
for term in seriesdict:
val = seriesdict[term]
if val != maxval:
for i in range(val, maxval):
teq += term.subs(n, val)
finaldict = {}
if teq:
fargs = teq.atoms(AppliedUndef)
if len(fargs) == 1:
finaldict[fargs.pop()] = 0
else:
maxf = max(fargs, key = lambda x: x.args[0])
sol = solve(teq, maxf)
if isinstance(sol, list):
sol = sol[0]
finaldict[maxf] = sol
# Finding the recurrence relation in terms of the largest term.
fargs = req.atoms(AppliedUndef)
maxf = max(fargs, key = lambda x: x.args[0])
minf = min(fargs, key = lambda x: x.args[0])
if minf.args[0].is_Symbol:
startiter = 0
else:
startiter = -minf.args[0].as_independent(n)[0]
lhs = maxf
rhs = solve(req, maxf)
if isinstance(rhs, list):
rhs = rhs[0]
# Checking how many values are already present
tcounter = len([t for t in finaldict.values() if t])
for _ in range(tcounter, terms - 3): # Assuming c0 and c1 to be arbitrary
check = rhs.subs(n, startiter)
nlhs = lhs.subs(n, startiter)
nrhs = check.subs(finaldict)
finaldict[nlhs] = nrhs
startiter += 1
# Post processing
series = C0 + C1*(x - x0)
for term in finaldict:
if finaldict[term]:
fact = term.args[0]
series += (finaldict[term].subs([(recurr(0), C0), (recurr(1), C1)])*(
x - x0)**fact)
series = collect(expand_mul(series), [C0, C1]) + Order(x**terms)
return Eq(f(x), series)
[docs]def ode_2nd_power_series_regular(eq, func, order, match):
r"""
Gives a power series solution to a second order homogeneous differential
equation with polynomial coefficients at a regular point. A second order
homogenous differential equation is of the form
.. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0
A point is said to regular singular at `x0` if `x - x0\frac{Q(x)}{P(x)}`
and `(x - x0)^{2}\frac{R(x)}{P(x)}` are analytic at `x0`. For simplicity
`P(x)`, `Q(x)` and `R(x)` are assumed to be polynomials. The algorithm for
finding the power series solutions is:
1. Try expressing `(x - x0)P(x)` and `((x - x0)^{2})Q(x)` as power series
solutions about x0. Find `p0` and `q0` which are the constants of the
power series expansions.
2. Solve the indicial equation `f(m) = m(m - 1) + m*p0 + q0`, to obtain the
roots `m1` and `m2` of the indicial equation.
3. If `m1 - m2` is a non integer there exists two series solutions. If
`m1 = m2`, there exists only one solution. If `m1 - m2` is an integer,
then the existence of one solution is confirmed. The other solution may
or may not exist.
The power series solution is of the form `x^{m}\sum_{n=0}^\infty a_{n}x^{n}`. The
coefficients are determined by the following recurrence relation.
`a_{n} = -\frac{\sum_{k=0}^{n-1} q_{n-k} + (m + k)p_{n-k}}{f(m + n)}`. For the case
in which `m1 - m2` is an integer, it can be seen from the recurrence relation
that for the lower root `m`, when `n` equals the difference of both the
roots, the denominator becomes zero. So if the numerator is not equal to zero,
a second series solution exists.
Examples
========
>>> from sympy import dsolve, Function, pprint
>>> from sympy.abc import x, y
>>> f = Function("f")
>>> eq = x*(f(x).diff(x, 2)) + 2*(f(x).diff(x)) + x*f(x)
>>> pprint(dsolve(eq))
/ 6 4 2 \
| x x x |
/ 4 2 \ C1*|- --- + -- - -- + 1|
| x x | \ 720 24 2 / / 6\
f(x) = C2*|--- - -- + 1| + ------------------------ + O\x /
\120 6 / x
References
==========
- George E. Simmons, "Differential Equations with Applications and
Historical Notes", p.p 176 - 184
"""
x = func.args[0]
f = func.func
C0, C1 = get_numbered_constants(eq, num=2)
n = Dummy("n")
m = Dummy("m") # for solving the indicial equation
s = Wild("s")
k = Wild("k", exclude=[x])
x0 = match.get('x0')
terms = match.get('terms', 5)
p = match['p']
q = match['q']
# Generating the indicial equation
indicial = []
for term in [p, q]:
if not term.has(x):
indicial.append(term)
else:
term = series(term, n=1, x0=x0)
if isinstance(term, Order):
indicial.append(S(0))
else:
for arg in term.args:
if not arg.has(x):
indicial.append(arg)
break
p0, q0 = indicial
sollist = solve(m*(m - 1) + m*p0 + q0, m)
if sollist and isinstance(sollist, list) and all(
[sol.is_real for sol in sollist]):
serdict1 = {}
serdict2 = {}
if len(sollist) == 1:
# Only one series solution exists in this case.
m1 = m2 = sollist.pop()
if terms-m1-1 <= 0:
return Eq(f(x), Order(terms))
serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0)
else:
m1 = sollist[0]
m2 = sollist[1]
if m1 < m2:
m1, m2 = m2, m1
# Irrespective of whether m1 - m2 is an integer or not, one
# Frobenius series solution exists.
serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0)
if not (m1 - m2).is_integer:
# Second frobenius series solution exists.
serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1)
else:
# Check if second frobenius series solution exists.
serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1, check=m1)
if serdict1:
finalseries1 = C0
for key in serdict1:
power = int(key.name[1:])
finalseries1 += serdict1[key]*(x - x0)**power
finalseries1 = (x - x0)**m1*finalseries1
finalseries2 = S(0)
if serdict2:
for key in serdict2:
power = int(key.name[1:])
finalseries2 += serdict2[key]*(x - x0)**power
finalseries2 += C1
finalseries2 = (x - x0)**m2*finalseries2
return Eq(f(x), collect(finalseries1 + finalseries2,
[C0, C1]) + Order(x**terms))
def _frobenius(n, m, p0, q0, p, q, x0, x, c, check=None):
r"""
Returns a dict with keys as coefficients and values as their values in terms of C0
"""
n = int(n)
# In cases where m1 - m2 is not an integer
m2 = check
d = Dummy("d")
numsyms = numbered_symbols("C", start=0)
numsyms = [next(numsyms) for i in range(n + 1)]
C0 = Symbol("C0")
serlist = []
for ser in [p, q]:
# Order term not present
if ser.is_polynomial(x) and Poly(ser, x).degree() <= n:
if x0:
ser = ser.subs(x, x + x0)
dict_ = Poly(ser, x).as_dict()
# Order term present
else:
tseries = series(ser, x=x0, n=n+1)
# Removing order
dict_ = Poly(list(ordered(tseries.args))[: -1], x).as_dict()
# Fill in with zeros, if coefficients are zero.
for i in range(n + 1):
if (i,) not in dict_:
dict_[(i,)] = S(0)
serlist.append(dict_)
pseries = serlist[0]
qseries = serlist[1]
indicial = d*(d - 1) + d*p0 + q0
frobdict = {}
for i in range(1, n + 1):
num = c*(m*pseries[(i,)] + qseries[(i,)])
for j in range(1, i):
sym = Symbol("C" + str(j))
num += frobdict[sym]*((m + j)*pseries[(i - j,)] + qseries[(i - j,)])
# Checking for cases when m1 - m2 is an integer. If num equals zero
# then a second Frobenius series solution cannot be found. If num is not zero
# then set constant as zero and proceed.
if m2 is not None and i == m2 - m:
if num:
return False
else:
frobdict[numsyms[i]] = S(0)
else:
frobdict[numsyms[i]] = -num/(indicial.subs(d, m+i))
return frobdict
def _nth_linear_match(eq, func, order):
r"""
Matches a differential equation to the linear form:
.. math:: a_n(x) y^{(n)} + \cdots + a_1(x)y' + a_0(x) y + B(x) = 0
Returns a dict of order:coeff terms, where order is the order of the
derivative on each term, and coeff is the coefficient of that derivative.
The key ``-1`` holds the function `B(x)`. Returns ``None`` if the ODE is
not linear. This function assumes that ``func`` has already been checked
to be good.
Examples
========
>>> from sympy import Function, cos, sin
>>> from sympy.abc import x
>>> from sympy.solvers.ode import _nth_linear_match
>>> f = Function('f')
>>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) +
... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) -
... sin(x), f(x), 3)
{-1: x - sin(x), 0: -1, 1: cos(x) + 2, 2: x, 3: 1}
>>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) +
... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) -
... sin(f(x)), f(x), 3) == None
True
"""
x = func.args[0]
one_x = {x}
terms = {i: S.Zero for i in range(-1, order + 1)}
for i in Add.make_args(eq):
if not i.has(func):
terms[-1] += i
else:
c, f = i.as_independent(func)
if not ((isinstance(f, Derivative) and set(f.variables) == one_x) \
or f == func):
return None
else:
terms[len(f.args[1:])] += c
return terms
def ode_nth_linear_euler_eq_homogeneous(eq, func, order, match, returns='sol'):
r"""
Solves an `n`\th order linear homogeneous variable-coefficient
Cauchy-Euler equidimensional ordinary differential equation.
This is an equation with form `0 = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x)
\cdots`.
These equations can be solved in a general manner, by substituting
solutions of the form `f(x) = x^r`, and deriving a characteristic equation
for `r`. When there are repeated roots, we include extra terms of the
form `C_{r k} \ln^k(x) x^r`, where `C_{r k}` is an arbitrary integration
constant, `r` is a root of the characteristic equation, and `k` ranges
over the multiplicity of `r`. In the cases where the roots are complex,
solutions of the form `C_1 x^a \sin(b \log(x)) + C_2 x^a \cos(b \log(x))`
are returned, based on expansions with Eulers formula. The general
solution is the sum of the terms found. If SymPy cannot find exact roots
to the characteristic equation, a
:py:class:`~sympy.polys.rootoftools.CRootOf` instance will be returned
instead.
>>> from sympy import Function, dsolve, Eq
>>> from sympy.abc import x
>>> f = Function('f')
>>> dsolve(4*x**2*f(x).diff(x, 2) + f(x), f(x),
... hint='nth_linear_euler_eq_homogeneous')
... # doctest: +NORMALIZE_WHITESPACE
Eq(f(x), sqrt(x)*(C1 + C2*log(x)))
Note that because this method does not involve integration, there is no
``nth_linear_euler_eq_homogeneous_Integral`` hint.
The following is for internal use:
- ``returns = 'sol'`` returns the solution to the ODE.
- ``returns = 'list'`` returns a list of linearly independent solutions,
corresponding to the fundamental solution set, for use with non
homogeneous solution methods like variation of parameters and
undetermined coefficients. Note that, though the solutions should be
linearly independent, this function does not explicitly check that. You
can do ``assert simplify(wronskian(sollist)) != 0`` to check for linear
independence. Also, ``assert len(sollist) == order`` will need to pass.
- ``returns = 'both'``, return a dictionary ``{'sol': <solution to ODE>,
'list': <list of linearly independent solutions>}``.
Examples
========
>>> from sympy import Function, dsolve, pprint
>>> from sympy.abc import x
>>> f = Function('f')
>>> eq = f(x).diff(x, 2)*x**2 - 4*f(x).diff(x)*x + 6*f(x)
>>> pprint(dsolve(eq, f(x),
... hint='nth_linear_euler_eq_homogeneous'))
2
f(x) = x *(C1 + C2*x)
References
==========
- http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation
- C. Bender & S. Orszag, "Advanced Mathematical Methods for Scientists and
Engineers", Springer 1999, pp. 12
# indirect doctest
"""
global collectterms
collectterms = []
x = func.args[0]
f = func.func
r = match
# First, set up characteristic equation.
chareq, symbol = S.Zero, Dummy('x')
for i in r.keys():
if not isinstance(i, str) and i >= 0:
chareq += (r[i]*diff(x**symbol, x, i)*x**-symbol).expand()
chareq = Poly(chareq, symbol)
chareqroots = [rootof(chareq, k) for k in range(chareq.degree())]
# A generator of constants
constants = list(get_numbered_constants(eq, num=chareq.degree()*2))
constants.reverse()
# Create a dict root: multiplicity or charroots
charroots = defaultdict(int)
for root in chareqroots:
charroots[root] += 1
gsol = S(0)
# We need keep track of terms so we can run collect() at the end.
# This is necessary for constantsimp to work properly.
ln = log
for root, multiplicity in charroots.items():
for i in range(multiplicity):
if isinstance(root, RootOf):
gsol += (x**root) * constants.pop()
if multiplicity != 1:
raise ValueError("Value should be 1")
collectterms = [(0, root, 0)] + collectterms
elif root.is_real:
gsol += ln(x)**i*(x**root) * constants.pop()
collectterms = [(i, root, 0)] + collectterms
else:
reroot = re(root)
imroot = im(root)
gsol += ln(x)**i * (x**reroot) * (
constants.pop() * sin(abs(imroot)*ln(x))
+ constants.pop() * cos(imroot*ln(x)))
# Preserve ordering (multiplicity, real part, imaginary part)
# It will be assumed implicitly when constructing
# fundamental solution sets.
collectterms = [(i, reroot, imroot)] + collectterms
if returns == 'sol':
return Eq(f(x), gsol)
elif returns in ('list' 'both'):
# HOW TO TEST THIS CODE? (dsolve does not pass 'returns' through)
# Create a list of (hopefully) linearly independent solutions
gensols = []
# Keep track of when to use sin or cos for nonzero imroot
for i, reroot, imroot in collectterms:
if imroot == 0:
gensols.append(ln(x)**i*x**reroot)
else:
sin_form = ln(x)**i*x**reroot*sin(abs(imroot)*ln(x))
if sin_form in gensols:
cos_form = ln(x)**i*x**reroot*cos(imroot*ln(x))
gensols.append(cos_form)
else:
gensols.append(sin_form)
if returns == 'list':
return gensols
else:
return {'sol': Eq(f(x), gsol), 'list': gensols}
else:
raise ValueError('Unknown value for key "returns".')
def ode_nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients(eq, func, order, match, returns='sol'):
r"""
Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional
ordinary differential equation using undetermined coefficients.
This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x)
\cdots`.
These equations can be solved in a general manner, by substituting
solutions of the form `x = exp(t)`, and deriving a characteristic equation
of form `g(exp(t)) = b_0 f(t) + b_1 f'(t) + b_2 f''(t) \cdots` which can
be then solved by nth_linear_constant_coeff_undetermined_coefficients if
g(exp(t)) has finite number of lineary independent derivatives.
Functions that fit this requirement are finite sums functions of the form
`a x^i e^{b x} \sin(c x + d)` or `a x^i e^{b x} \cos(c x + d)`, where `i`
is a non-negative integer and `a`, `b`, `c`, and `d` are constants. For
example any polynomial in `x`, functions like `x^2 e^{2 x}`, `x \sin(x)`,
and `e^x \cos(x)` can all be used. Products of `\sin`'s and `\cos`'s have
a finite number of derivatives, because they can be expanded into `\sin(a
x)` and `\cos(b x)` terms. However, SymPy currently cannot do that
expansion, so you will need to manually rewrite the expression in terms of
the above to use this method. So, for example, you will need to manually
convert `\sin^2(x)` into `(1 + \cos(2 x))/2` to properly apply the method
of undetermined coefficients on it.
After replacement of x by exp(t), this method works by creating a trial function
from the expression and all of its linear independent derivatives and
substituting them into the original ODE. The coefficients for each term
will be a system of linear equations, which are be solved for and
substituted, giving the solution. If any of the trial functions are linearly
dependent on the solution to the homogeneous equation, they are multiplied
by sufficient `x` to make them linearly independent.
Examples
========
>>> from sympy import dsolve, Function, Derivative, log
>>> from sympy.abc import x
>>> f = Function('f')
>>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - log(x)
>>> dsolve(eq, f(x),
... hint='nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients').expand()
Eq(f(x), C1*x + C2*x**2 + log(x)/2 + 3/4)
"""
x = func.args[0]
f = func.func
r = match
chareq, eq, symbol = S.Zero, S.Zero, Dummy('x')
for i in r.keys():
if not isinstance(i, str) and i >= 0:
chareq += (r[i]*diff(x**symbol, x, i)*x**-symbol).expand()
for i in range(1,degree(Poly(chareq, symbol))+1):
eq += chareq.coeff(symbol**i)*diff(f(x), x, i)
if chareq.as_coeff_add(symbol)[0]:
eq += chareq.as_coeff_add(symbol)[0]*f(x)
e, re = posify(r[-1].subs(x, exp(x)))
eq += e.subs(re)
match = _nth_linear_match(eq, f(x), ode_order(eq, f(x)))
match['trialset'] = r['trialset']
return ode_nth_linear_constant_coeff_undetermined_coefficients(eq, func, order, match).subs(x, log(x)).subs(f(log(x)), f(x)).expand()
def ode_nth_linear_euler_eq_nonhomogeneous_variation_of_parameters(eq, func, order, match, returns='sol'):
r"""
Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional
ordinary differential equation using variation of parameters.
This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x)
\cdots`.
This method works by assuming that the particular solution takes the form
.. math:: \sum_{x=1}^{n} c_i(x) y_i(x) {a_n} {x^n} \text{,}
where `y_i` is the `i`\th solution to the homogeneous equation. The
solution is then solved using Wronskian's and Cramer's Rule. The
particular solution is given by multiplying eq given below with `a_n x^{n}`
.. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx
\right) y_i(x) \text{,}
where `W(x)` is the Wronskian of the fundamental system (the system of `n`
linearly independent solutions to the homogeneous equation), and `W_i(x)`
is the Wronskian of the fundamental system with the `i`\th column replaced
with `[0, 0, \cdots, 0, \frac{x^{- n}}{a_n} g{\left (x \right )}]`.
This method is general enough to solve any `n`\th order inhomogeneous
linear differential equation, but sometimes SymPy cannot simplify the
Wronskian well enough to integrate it. If this method hangs, try using the
``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and
simplifying the integrals manually. Also, prefer using
``nth_linear_constant_coeff_undetermined_coefficients`` when it
applies, because it doesn't use integration, making it faster and more
reliable.
Warning, using simplify=False with
'nth_linear_constant_coeff_variation_of_parameters' in
:py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will
not attempt to simplify the Wronskian before integrating. It is
recommended that you only use simplify=False with
'nth_linear_constant_coeff_variation_of_parameters_Integral' for this
method, especially if the solution to the homogeneous equation has
trigonometric functions in it.
Examples
========
>>> from sympy import Function, dsolve, Derivative
>>> from sympy.abc import x
>>> f = Function('f')
>>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - x**4
>>> dsolve(eq, f(x),
... hint='nth_linear_euler_eq_nonhomogeneous_variation_of_parameters').expand()
Eq(f(x), C1*x + C2*x**2 + x**4/6)
"""
x = func.args[0]
f = func.func
r = match
gensol = ode_nth_linear_euler_eq_homogeneous(eq, func, order, match, returns='both')
match.update(gensol)
r[-1] = r[-1]/r[ode_order(eq, f(x))]
sol = _solve_variation_of_parameters(eq, func, order, match)
return Eq(f(x), r['sol'].rhs + (sol.rhs - r['sol'].rhs)*r[ode_order(eq, f(x))])
[docs]def ode_almost_linear(eq, func, order, match):
r"""
Solves an almost-linear differential equation.
The general form of an almost linear differential equation is
.. math:: f(x) g(y) y + k(x) l(y) + m(x) = 0
\text{where} l'(y) = g(y)\text{.}
This can be solved by substituting `l(y) = u(y)`. Making the given
substitution reduces it to a linear differential equation of the form `u'
+ P(x) u + Q(x) = 0`.
The general solution is
>>> from sympy import Function, dsolve, Eq, pprint
>>> from sympy.abc import x, y, n
>>> f, g, k, l = map(Function, ['f', 'g', 'k', 'l'])
>>> genform = Eq(f(x)*(l(y).diff(y)) + k(x)*l(y) + g(x))
>>> pprint(genform)
d
f(x)*--(l(y)) + g(x) + k(x)*l(y) = 0
dy
>>> pprint(dsolve(genform, hint = 'almost_linear'))
/ // -y*g(x) \\
| || -------- for k(x) = 0||
| || f(x) || -y*k(x)
| || || --------
| || y*k(x) || f(x)
l(y) = |C1 + |< ------ ||*e
| || f(x) ||
| ||-g(x)*e ||
| ||-------------- otherwise ||
| || k(x) ||
\ \\ //
See Also
========
:meth:`sympy.solvers.ode.ode_1st_linear`
Examples
========
>>> from sympy import Function, Derivative, pprint
>>> from sympy.solvers.ode import dsolve, classify_ode
>>> from sympy.abc import x
>>> f = Function('f')
>>> d = f(x).diff(x)
>>> eq = x*d + x*f(x) + 1
>>> dsolve(eq, f(x), hint='almost_linear')
Eq(f(x), (C1 - Ei(x))*exp(-x))
>>> pprint(dsolve(eq, f(x), hint='almost_linear'))
-x
f(x) = (C1 - Ei(x))*e
References
==========
- Joel Moses, "Symbolic Integration - The Stormy Decade", Communications
of the ACM, Volume 14, Number 8, August 1971, pp. 558
"""
# Since ode_1st_linear has already been implemented, and the
# coefficients have been modified to the required form in
# classify_ode, just passing eq, func, order and match to
# ode_1st_linear will give the required output.
return ode_1st_linear(eq, func, order, match)
def _linear_coeff_match(expr, func):
r"""
Helper function to match hint ``linear_coefficients``.
Matches the expression to the form `(a_1 x + b_1 f(x) + c_1)/(a_2 x + b_2
f(x) + c_2)` where the following conditions hold:
1. `a_1`, `b_1`, `c_1`, `a_2`, `b_2`, `c_2` are Rationals;
2. `c_1` or `c_2` are not equal to zero;
3. `a_2 b_1 - a_1 b_2` is not equal to zero.
Return ``xarg``, ``yarg`` where
1. ``xarg`` = `(b_2 c_1 - b_1 c_2)/(a_2 b_1 - a_1 b_2)`
2. ``yarg`` = `(a_1 c_2 - a_2 c_1)/(a_2 b_1 - a_1 b_2)`
Examples
========
>>> from sympy import Function
>>> from sympy.abc import x
>>> from sympy.solvers.ode import _linear_coeff_match
>>> from sympy.functions.elementary.trigonometric import sin
>>> f = Function('f')
>>> _linear_coeff_match((
... (-25*f(x) - 8*x + 62)/(4*f(x) + 11*x - 11)), f(x))
(1/9, 22/9)
>>> _linear_coeff_match(
... sin((-5*f(x) - 8*x + 6)/(4*f(x) + x - 1)), f(x))
(19/27, 2/27)
>>> _linear_coeff_match(sin(f(x)/x), f(x))
"""
f = func.func
x = func.args[0]
def abc(eq):
r'''
Internal function of _linear_coeff_match
that returns Rationals a, b, c
if eq is a*x + b*f(x) + c, else None.
'''
eq = _mexpand(eq)
c = eq.as_independent(x, f(x), as_Add=True)[0]
if not c.is_Rational:
return
a = eq.coeff(x)
if not a.is_Rational:
return
b = eq.coeff(f(x))
if not b.is_Rational:
return
if eq == a*x + b*f(x) + c:
return a, b, c
def match(arg):
r'''
Internal function of _linear_coeff_match that returns Rationals a1,
b1, c1, a2, b2, c2 and a2*b1 - a1*b2 of the expression (a1*x + b1*f(x)
+ c1)/(a2*x + b2*f(x) + c2) if one of c1 or c2 and a2*b1 - a1*b2 is
non-zero, else None.
'''
n, d = arg.together().as_numer_denom()
m = abc(n)
if m is not None:
a1, b1, c1 = m
m = abc(d)
if m is not None:
a2, b2, c2 = m
d = a2*b1 - a1*b2
if (c1 or c2) and d:
return a1, b1, c1, a2, b2, c2, d
m = [fi.args[0] for fi in expr.atoms(Function) if fi.func != f and
len(fi.args) == 1 and not fi.args[0].is_Function] or {expr}
m1 = match(m.pop())
if m1 and all(match(mi) == m1 for mi in m):
a1, b1, c1, a2, b2, c2, denom = m1
return (b2*c1 - b1*c2)/denom, (a1*c2 - a2*c1)/denom
[docs]def ode_linear_coefficients(eq, func, order, match):
r"""
Solves a differential equation with linear coefficients.
The general form of a differential equation with linear coefficients is
.. math:: y' + F\left(\!\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y +
c_2}\!\right) = 0\text{,}
where `a_1`, `b_1`, `c_1`, `a_2`, `b_2`, `c_2` are constants and `a_1 b_2
- a_2 b_1 \ne 0`.
This can be solved by substituting:
.. math:: x = x' + \frac{b_2 c_1 - b_1 c_2}{a_2 b_1 - a_1 b_2}
y = y' + \frac{a_1 c_2 - a_2 c_1}{a_2 b_1 - a_1
b_2}\text{.}
This substitution reduces the equation to a homogeneous differential
equation.
See Also
========
:meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_best`
:meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`
:meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`
Examples
========
>>> from sympy import Function, Derivative, pprint
>>> from sympy.solvers.ode import dsolve, classify_ode
>>> from sympy.abc import x
>>> f = Function('f')
>>> df = f(x).diff(x)
>>> eq = (x + f(x) + 1)*df + (f(x) - 6*x + 1)
>>> dsolve(eq, hint='linear_coefficients')
[Eq(f(x), -x - sqrt(C1 + 7*x**2) - 1), Eq(f(x), -x + sqrt(C1 + 7*x**2) - 1)]
>>> pprint(dsolve(eq, hint='linear_coefficients'))
___________ ___________
/ 2 / 2
[f(x) = -x - \/ C1 + 7*x - 1, f(x) = -x + \/ C1 + 7*x - 1]
References
==========
- Joel Moses, "Symbolic Integration - The Stormy Decade", Communications
of the ACM, Volume 14, Number 8, August 1971, pp. 558
"""
return ode_1st_homogeneous_coeff_best(eq, func, order, match)
[docs]def ode_separable_reduced(eq, func, order, match):
r"""
Solves a differential equation that can be reduced to the separable form.
The general form of this equation is
.. math:: y' + (y/x) H(x^n y) = 0\text{}.
This can be solved by substituting `u(y) = x^n y`. The equation then
reduces to the separable form `\frac{u'}{u (\mathrm{power} - H(u))} -
\frac{1}{x} = 0`.
The general solution is:
>>> from sympy import Function, dsolve, Eq, pprint
>>> from sympy.abc import x, n
>>> f, g = map(Function, ['f', 'g'])
>>> genform = f(x).diff(x) + (f(x)/x)*g(x**n*f(x))
>>> pprint(genform)
/ n \
d f(x)*g\x *f(x)/
--(f(x)) + ---------------
dx x
>>> pprint(dsolve(genform, hint='separable_reduced'))
n
x *f(x)
/
|
| 1
| ------------ dy = C1 + log(x)
| y*(n - g(y))
|
/
See Also
========
:meth:`sympy.solvers.ode.ode_separable`
Examples
========
>>> from sympy import Function, Derivative, pprint
>>> from sympy.solvers.ode import dsolve, classify_ode
>>> from sympy.abc import x
>>> f = Function('f')
>>> d = f(x).diff(x)
>>> eq = (x - x**2*f(x))*d - f(x)
>>> dsolve(eq, hint='separable_reduced')
[Eq(f(x), (-sqrt(C1*x**2 + 1) + 1)/x), Eq(f(x), (sqrt(C1*x**2 + 1) + 1)/x)]
>>> pprint(dsolve(eq, hint='separable_reduced'))
___________ ___________
/ 2 / 2
- \/ C1*x + 1 + 1 \/ C1*x + 1 + 1
[f(x) = --------------------, f(x) = ------------------]
x x
References
==========
- Joel Moses, "Symbolic Integration - The Stormy Decade", Communications
of the ACM, Volume 14, Number 8, August 1971, pp. 558
"""
# Arguments are passed in a way so that they are coherent with the
# ode_separable function
x = func.args[0]
f = func.func
y = Dummy('y')
u = match['u'].subs(match['t'], y)
ycoeff = 1/(y*(match['power'] - u))
m1 = {y: 1, x: -1/x, 'coeff': 1}
m2 = {y: ycoeff, x: 1, 'coeff': 1}
r = {'m1': m1, 'm2': m2, 'y': y, 'hint': x**match['power']*f(x)}
return ode_separable(eq, func, order, r)
[docs]def ode_1st_power_series(eq, func, order, match):
r"""
The power series solution is a method which gives the Taylor series expansion
to the solution of a differential equation.
For a first order differential equation `\frac{dy}{dx} = h(x, y)`, a power
series solution exists at a point `x = x_{0}` if `h(x, y)` is analytic at `x_{0}`.
The solution is given by
.. math:: y(x) = y(x_{0}) + \sum_{n = 1}^{\infty} \frac{F_{n}(x_{0},b)(x - x_{0})^n}{n!},
where `y(x_{0}) = b` is the value of y at the initial value of `x_{0}`.
To compute the values of the `F_{n}(x_{0},b)` the following algorithm is
followed, until the required number of terms are generated.
1. `F_1 = h(x_{0}, b)`
2. `F_{n+1} = \frac{\partial F_{n}}{\partial x} + \frac{\partial F_{n}}{\partial y}F_{1}`
Examples
========
>>> from sympy import Function, Derivative, pprint, exp
>>> from sympy.solvers.ode import dsolve
>>> from sympy.abc import x
>>> f = Function('f')
>>> eq = exp(x)*(f(x).diff(x)) - f(x)
>>> pprint(dsolve(eq, hint='1st_power_series'))
3 4 5
C1*x C1*x C1*x / 6\
f(x) = C1 + C1*x - ----- + ----- + ----- + O\x /
6 24 60
References
==========
- Travis W. Walker, Analytic power series technique for solving first-order
differential equations, p.p 17, 18
"""
x = func.args[0]
y = match['y']
f = func.func
h = -match[match['d']]/match[match['e']]
point = match.get('f0')
value = match.get('f0val')
terms = match.get('terms')
# First term
F = h
if not h:
return Eq(f(x), value)
# Initialisation
series = value
if terms > 1:
hc = h.subs({x: point, y: value})
if hc.has(oo) or hc.has(NaN) or hc.has(zoo):
# Derivative does not exist, not analytic
return Eq(f(x), oo)
elif hc:
series += hc*(x - point)
for factcount in range(2, terms):
Fnew = F.diff(x) + F.diff(y)*h
Fnewc = Fnew.subs({x: point, y: value})
# Same logic as above
if Fnewc.has(oo) or Fnewc.has(NaN) or Fnewc.has(-oo) or Fnewc.has(zoo):
return Eq(f(x), oo)
series += Fnewc*((x - point)**factcount)/factorial(factcount)
F = Fnew
series += Order(x**terms)
return Eq(f(x), series)
[docs]def ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match,
returns='sol'):
r"""
Solves an `n`\th order linear homogeneous differential equation with
constant coefficients.
This is an equation of the form
.. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x)
+ a_0 f(x) = 0\text{.}
These equations can be solved in a general manner, by taking the roots of
the characteristic equation `a_n m^n + a_{n-1} m^{n-1} + \cdots + a_1 m +
a_0 = 0`. The solution will then be the sum of `C_n x^i e^{r x}` terms,
for each where `C_n` is an arbitrary constant, `r` is a root of the
characteristic equation and `i` is one of each from 0 to the multiplicity
of the root - 1 (for example, a root 3 of multiplicity 2 would create the
terms `C_1 e^{3 x} + C_2 x e^{3 x}`). The exponential is usually expanded
for complex roots using Euler's equation `e^{I x} = \cos(x) + I \sin(x)`.
Complex roots always come in conjugate pairs in polynomials with real
coefficients, so the two roots will be represented (after simplifying the
constants) as `e^{a x} \left(C_1 \cos(b x) + C_2 \sin(b x)\right)`.
If SymPy cannot find exact roots to the characteristic equation, a
:py:class:`~sympy.polys.rootoftools.CRootOf` instance will be return
instead.
>>> from sympy import Function, dsolve, Eq
>>> from sympy.abc import x
>>> f = Function('f')
>>> dsolve(f(x).diff(x, 5) + 10*f(x).diff(x) - 2*f(x), f(x),
... hint='nth_linear_constant_coeff_homogeneous')
... # doctest: +NORMALIZE_WHITESPACE
Eq(f(x), C1*exp(x*CRootOf(_x**5 + 10*_x - 2, 0)) +
C2*exp(x*CRootOf(_x**5 + 10*_x - 2, 1)) +
C3*exp(x*CRootOf(_x**5 + 10*_x - 2, 2)) +
C4*exp(x*CRootOf(_x**5 + 10*_x - 2, 3)) +
C5*exp(x*CRootOf(_x**5 + 10*_x - 2, 4)))
Note that because this method does not involve integration, there is no
``nth_linear_constant_coeff_homogeneous_Integral`` hint.
The following is for internal use:
- ``returns = 'sol'`` returns the solution to the ODE.
- ``returns = 'list'`` returns a list of linearly independent solutions,
for use with non homogeneous solution methods like variation of
parameters and undetermined coefficients. Note that, though the
solutions should be linearly independent, this function does not
explicitly check that. You can do ``assert simplify(wronskian(sollist))
!= 0`` to check for linear independence. Also, ``assert len(sollist) ==
order`` will need to pass.
- ``returns = 'both'``, return a dictionary ``{'sol': <solution to ODE>,
'list': <list of linearly independent solutions>}``.
Examples
========
>>> from sympy import Function, dsolve, pprint
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(f(x).diff(x, 4) + 2*f(x).diff(x, 3) -
... 2*f(x).diff(x, 2) - 6*f(x).diff(x) + 5*f(x), f(x),
... hint='nth_linear_constant_coeff_homogeneous'))
x -2*x
f(x) = (C1 + C2*x)*e + (C3*sin(x) + C4*cos(x))*e
References
==========
- http://en.wikipedia.org/wiki/Linear_differential_equation section:
Nonhomogeneous_equation_with_constant_coefficients
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 211
# indirect doctest
"""
x = func.args[0]
f = func.func
r = match
# First, set up characteristic equation.
chareq, symbol = S.Zero, Dummy('x')
for i in r.keys():
if type(i) == str or i < 0:
pass
else:
chareq += r[i]*symbol**i
chareq = Poly(chareq, symbol)
chareqroots = [rootof(chareq, k) for k in range(chareq.degree())]
chareq_is_complex = not all([i.is_real for i in chareq.all_coeffs()])
# A generator of constants
constants = list(get_numbered_constants(eq, num=chareq.degree()*2))
# Create a dict root: multiplicity or charroots
charroots = defaultdict(int)
for root in chareqroots:
charroots[root] += 1
gsol = S(0)
# We need to keep track of terms so we can run collect() at the end.
# This is necessary for constantsimp to work properly.
global collectterms
collectterms = []
gensols = []
conjugate_roots = [] # used to prevent double-use of conjugate roots
for root, multiplicity in charroots.items():
for i in range(multiplicity):
if isinstance(root, RootOf):
gensols.append(exp(root*x))
if multiplicity != 1:
raise ValueError("Value should be 1")
# This ordering is important
collectterms = [(0, root, 0)] + collectterms
else:
if chareq_is_complex:
gensols.append(x**i*exp(root*x))
collectterms = [(i, root, 0)] + collectterms
continue
reroot = re(root)
imroot = im(root)
if imroot.has(atan2) and reroot.has(atan2):
# Remove this condition when re and im stop returning
# circular atan2 usages.
gensols.append(x**i*exp(root*x))
collectterms = [(i, root, 0)] + collectterms
else:
if root in conjugate_roots:
collectterms = [(i, reroot, imroot)] + collectterms
continue
if imroot == 0:
gensols.append(x**i*exp(reroot*x))
collectterms = [(i, reroot, 0)] + collectterms
continue
conjugate_roots.append(conjugate(root))
gensols.append(x**i*exp(reroot*x) * sin(abs(imroot) * x))
gensols.append(x**i*exp(reroot*x) * cos( imroot * x))
# This ordering is important
collectterms = [(i, reroot, imroot)] + collectterms
if returns == 'list':
return gensols
elif returns in ('sol' 'both'):
gsol = Add(*[i*j for (i,j) in zip(constants, gensols)])
if returns == 'sol':
return Eq(f(x), gsol)
else:
return {'sol': Eq(f(x), gsol), 'list': gensols}
else:
raise ValueError('Unknown value for key "returns".')
[docs]def ode_nth_linear_constant_coeff_undetermined_coefficients(eq, func, order, match):
r"""
Solves an `n`\th order linear differential equation with constant
coefficients using the method of undetermined coefficients.
This method works on differential equations of the form
.. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x)
+ a_0 f(x) = P(x)\text{,}
where `P(x)` is a function that has a finite number of linearly
independent derivatives.
Functions that fit this requirement are finite sums functions of the form
`a x^i e^{b x} \sin(c x + d)` or `a x^i e^{b x} \cos(c x + d)`, where `i`
is a non-negative integer and `a`, `b`, `c`, and `d` are constants. For
example any polynomial in `x`, functions like `x^2 e^{2 x}`, `x \sin(x)`,
and `e^x \cos(x)` can all be used. Products of `\sin`'s and `\cos`'s have
a finite number of derivatives, because they can be expanded into `\sin(a
x)` and `\cos(b x)` terms. However, SymPy currently cannot do that
expansion, so you will need to manually rewrite the expression in terms of
the above to use this method. So, for example, you will need to manually
convert `\sin^2(x)` into `(1 + \cos(2 x))/2` to properly apply the method
of undetermined coefficients on it.
This method works by creating a trial function from the expression and all
of its linear independent derivatives and substituting them into the
original ODE. The coefficients for each term will be a system of linear
equations, which are be solved for and substituted, giving the solution.
If any of the trial functions are linearly dependent on the solution to
the homogeneous equation, they are multiplied by sufficient `x` to make
them linearly independent.
Examples
========
>>> from sympy import Function, dsolve, pprint, exp, cos
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(f(x).diff(x, 2) + 2*f(x).diff(x) + f(x) -
... 4*exp(-x)*x**2 + cos(2*x), f(x),
... hint='nth_linear_constant_coeff_undetermined_coefficients'))
/ 4\
| x | -x 4*sin(2*x) 3*cos(2*x)
f(x) = |C1 + C2*x + --|*e - ---------- + ----------
\ 3 / 25 25
References
==========
- http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 221
# indirect doctest
"""
gensol = ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match,
returns='both')
match.update(gensol)
return _solve_undetermined_coefficients(eq, func, order, match)
def _solve_undetermined_coefficients(eq, func, order, match):
r"""
Helper function for the method of undetermined coefficients.
See the
:py:meth:`~sympy.solvers.ode.ode_nth_linear_constant_coeff_undetermined_coefficients`
docstring for more information on this method.
The parameter ``match`` should be a dictionary that has the following
keys:
``list``
A list of solutions to the homogeneous equation, such as the list
returned by
``ode_nth_linear_constant_coeff_homogeneous(returns='list')``.
``sol``
The general solution, such as the solution returned by
``ode_nth_linear_constant_coeff_homogeneous(returns='sol')``.
``trialset``
The set of trial functions as returned by
``_undetermined_coefficients_match()['trialset']``.
"""
x = func.args[0]
f = func.func
r = match
coeffs = numbered_symbols('a', cls=Dummy)
coefflist = []
gensols = r['list']
gsol = r['sol']
trialset = r['trialset']
notneedset = set([])
newtrialset = set([])
global collectterms
if len(gensols) != order:
raise NotImplementedError("Cannot find " + str(order) +
" solutions to the homogeneous equation necessary to apply" +
" undetermined coefficients to " + str(eq) +
" (number of terms != order)")
usedsin = set([])
mult = 0 # The multiplicity of the root
getmult = True
for i, reroot, imroot in collectterms:
if getmult:
mult = i + 1
getmult = False
if i == 0:
getmult = True
if imroot:
# Alternate between sin and cos
if (i, reroot) in usedsin:
check = x**i*exp(reroot*x)*cos(imroot*x)
else:
check = x**i*exp(reroot*x)*sin(abs(imroot)*x)
usedsin.add((i, reroot))
else:
check = x**i*exp(reroot*x)
if check in trialset:
# If an element of the trial function is already part of the
# homogeneous solution, we need to multiply by sufficient x to
# make it linearly independent. We also don't need to bother
# checking for the coefficients on those elements, since we
# already know it will be 0.
while True:
if check*x**mult in trialset:
mult += 1
else:
break
trialset.add(check*x**mult)
notneedset.add(check)
newtrialset = trialset - notneedset
trialfunc = 0
for i in newtrialset:
c = next(coeffs)
coefflist.append(c)
trialfunc += c*i
eqs = sub_func_doit(eq, f(x), trialfunc)
coeffsdict = dict(list(zip(trialset, [0]*(len(trialset) + 1))))
eqs = _mexpand(eqs)
for i in Add.make_args(eqs):
s = separatevars(i, dict=True, symbols=[x])
coeffsdict[s[x]] += s['coeff']
coeffvals = solve(list(coeffsdict.values()), coefflist)
if not coeffvals:
raise NotImplementedError(
"Could not solve `%s` using the "
"method of undetermined coefficients "
"(unable to solve for coefficients)." % eq)
psol = trialfunc.subs(coeffvals)
return Eq(f(x), gsol.rhs + psol)
def _undetermined_coefficients_match(expr, x):
r"""
Returns a trial function match if undetermined coefficients can be applied
to ``expr``, and ``None`` otherwise.
A trial expression can be found for an expression for use with the method
of undetermined coefficients if the expression is an
additive/multiplicative combination of constants, polynomials in `x` (the
independent variable of expr), `\sin(a x + b)`, `\cos(a x + b)`, and
`e^{a x}` terms (in other words, it has a finite number of linearly
independent derivatives).
Note that you may still need to multiply each term returned here by
sufficient `x` to make it linearly independent with the solutions to the
homogeneous equation.
This is intended for internal use by ``undetermined_coefficients`` hints.
SymPy currently has no way to convert `\sin^n(x) \cos^m(y)` into a sum of
only `\sin(a x)` and `\cos(b x)` terms, so these are not implemented. So,
for example, you will need to manually convert `\sin^2(x)` into `[1 +
\cos(2 x)]/2` to properly apply the method of undetermined coefficients on
it.
Examples
========
>>> from sympy import log, exp
>>> from sympy.solvers.ode import _undetermined_coefficients_match
>>> from sympy.abc import x
>>> _undetermined_coefficients_match(9*x*exp(x) + exp(-x), x)
{'test': True, 'trialset': set([x*exp(x), exp(-x), exp(x)])}
>>> _undetermined_coefficients_match(log(x), x)
{'test': False}
"""
a = Wild('a', exclude=[x])
b = Wild('b', exclude=[x])
expr = powsimp(expr, combine='exp') # exp(x)*exp(2*x + 1) => exp(3*x + 1)
retdict = {}
def _test_term(expr, x):
r"""
Test if ``expr`` fits the proper form for undetermined coefficients.
"""
if expr.is_Add:
return all(_test_term(i, x) for i in expr.args)
elif expr.is_Mul:
if expr.has(sin, cos):
foundtrig = False
# Make sure that there is only one trig function in the args.
# See the docstring.
for i in expr.args:
if i.has(sin, cos):
if foundtrig:
return False
else:
foundtrig = True
return all(_test_term(i, x) for i in expr.args)
elif expr.is_Function:
if expr.func in (sin, cos, exp):
if expr.args[0].match(a*x + b):
return True
else:
return False
else:
return False
elif expr.is_Pow and expr.base.is_Symbol and expr.exp.is_Integer and \
expr.exp >= 0:
return True
elif expr.is_Pow and expr.base.is_number:
if expr.exp.match(a*x + b):
return True
else:
return False
elif expr.is_Symbol or expr.is_number:
return True
else:
return False
def _get_trial_set(expr, x, exprs=set([])):
r"""
Returns a set of trial terms for undetermined coefficients.
The idea behind undetermined coefficients is that the terms expression
repeat themselves after a finite number of derivatives, except for the
coefficients (they are linearly dependent). So if we collect these,
we should have the terms of our trial function.
"""
def _remove_coefficient(expr, x):
r"""
Returns the expression without a coefficient.
Similar to expr.as_independent(x)[1], except it only works
multiplicatively.
"""
term = S.One
if expr.is_Mul:
for i in expr.args:
if i.has(x):
term *= i
elif expr.has(x):
term = expr
return term
expr = expand_mul(expr)
if expr.is_Add:
for term in expr.args:
if _remove_coefficient(term, x) in exprs:
pass
else:
exprs.add(_remove_coefficient(term, x))
exprs = exprs.union(_get_trial_set(term, x, exprs))
else:
term = _remove_coefficient(expr, x)
tmpset = exprs.union({term})
oldset = set([])
while tmpset != oldset:
# If you get stuck in this loop, then _test_term is probably
# broken
oldset = tmpset.copy()
expr = expr.diff(x)
term = _remove_coefficient(expr, x)
if term.is_Add:
tmpset = tmpset.union(_get_trial_set(term, x, tmpset))
else:
tmpset.add(term)
exprs = tmpset
return exprs
retdict['test'] = _test_term(expr, x)
if retdict['test']:
# Try to generate a list of trial solutions that will have the
# undetermined coefficients. Note that if any of these are not linearly
# independent with any of the solutions to the homogeneous equation,
# then they will need to be multiplied by sufficient x to make them so.
# This function DOES NOT do that (it doesn't even look at the
# homogeneous equation).
retdict['trialset'] = _get_trial_set(expr, x)
return retdict
[docs]def ode_nth_linear_constant_coeff_variation_of_parameters(eq, func, order, match):
r"""
Solves an `n`\th order linear differential equation with constant
coefficients using the method of variation of parameters.
This method works on any differential equations of the form
.. math:: f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + a_0
f(x) = P(x)\text{.}
This method works by assuming that the particular solution takes the form
.. math:: \sum_{x=1}^{n} c_i(x) y_i(x)\text{,}
where `y_i` is the `i`\th solution to the homogeneous equation. The
solution is then solved using Wronskian's and Cramer's Rule. The
particular solution is given by
.. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx
\right) y_i(x) \text{,}
where `W(x)` is the Wronskian of the fundamental system (the system of `n`
linearly independent solutions to the homogeneous equation), and `W_i(x)`
is the Wronskian of the fundamental system with the `i`\th column replaced
with `[0, 0, \cdots, 0, P(x)]`.
This method is general enough to solve any `n`\th order inhomogeneous
linear differential equation with constant coefficients, but sometimes
SymPy cannot simplify the Wronskian well enough to integrate it. If this
method hangs, try using the
``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and
simplifying the integrals manually. Also, prefer using
``nth_linear_constant_coeff_undetermined_coefficients`` when it
applies, because it doesn't use integration, making it faster and more
reliable.
Warning, using simplify=False with
'nth_linear_constant_coeff_variation_of_parameters' in
:py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will
not attempt to simplify the Wronskian before integrating. It is
recommended that you only use simplify=False with
'nth_linear_constant_coeff_variation_of_parameters_Integral' for this
method, especially if the solution to the homogeneous equation has
trigonometric functions in it.
Examples
========
>>> from sympy import Function, dsolve, pprint, exp, log
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(f(x).diff(x, 3) - 3*f(x).diff(x, 2) +
... 3*f(x).diff(x) - f(x) - exp(x)*log(x), f(x),
... hint='nth_linear_constant_coeff_variation_of_parameters'))
/ 3 \
| 2 x *(6*log(x) - 11)| x
f(x) = |C1 + C2*x + C3*x + ------------------|*e
\ 36 /
References
==========
- http://en.wikipedia.org/wiki/Variation_of_parameters
- http://planetmath.org/VariationOfParameters
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 233
# indirect doctest
"""
gensol = ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match,
returns='both')
match.update(gensol)
return _solve_variation_of_parameters(eq, func, order, match)
def _solve_variation_of_parameters(eq, func, order, match):
r"""
Helper function for the method of variation of parameters and nonhomogeneous euler eq.
See the
:py:meth:`~sympy.solvers.ode.ode_nth_linear_constant_coeff_variation_of_parameters`
docstring for more information on this method.
The parameter ``match`` should be a dictionary that has the following
keys:
``list``
A list of solutions to the homogeneous equation, such as the list
returned by
``ode_nth_linear_constant_coeff_homogeneous(returns='list')``.
``sol``
The general solution, such as the solution returned by
``ode_nth_linear_constant_coeff_homogeneous(returns='sol')``.
"""
x = func.args[0]
f = func.func
r = match
psol = 0
gensols = r['list']
gsol = r['sol']
wr = wronskian(gensols, x)
if r.get('simplify', True):
wr = simplify(wr) # We need much better simplification for
# some ODEs. See issue 4662, for example.
# To reduce commonly occuring sin(x)**2 + cos(x)**2 to 1
wr = trigsimp(wr, deep=True, recursive=True)
if not wr:
# The wronskian will be 0 iff the solutions are not linearly
# independent.
raise NotImplementedError("Cannot find " + str(order) +
" solutions to the homogeneous equation nessesary to apply " +
"variation of parameters to " + str(eq) + " (Wronskian == 0)")
if len(gensols) != order:
raise NotImplementedError("Cannot find " + str(order) +
" solutions to the homogeneous equation nessesary to apply " +
"variation of parameters to " +
str(eq) + " (number of terms != order)")
negoneterm = (-1)**(order)
for i in gensols:
psol += negoneterm*Integral(wronskian([sol for sol in gensols if sol != i], x)*r[-1]/wr, x)*i/r[order]
negoneterm *= -1
if r.get('simplify', True):
psol = simplify(psol)
psol = trigsimp(psol, deep=True)
return Eq(f(x), gsol.rhs + psol)
[docs]def ode_separable(eq, func, order, match):
r"""
Solves separable 1st order differential equations.
This is any differential equation that can be written as `P(y)
\tfrac{dy}{dx} = Q(x)`. The solution can then just be found by
rearranging terms and integrating: `\int P(y) \,dy = \int Q(x) \,dx`.
This hint uses :py:meth:`sympy.simplify.simplify.separatevars` as its back
end, so if a separable equation is not caught by this solver, it is most
likely the fault of that function.
:py:meth:`~sympy.simplify.simplify.separatevars` is
smart enough to do most expansion and factoring necessary to convert a
separable equation `F(x, y)` into the proper form `P(x)\cdot{}Q(y)`. The
general solution is::
>>> from sympy import Function, dsolve, Eq, pprint
>>> from sympy.abc import x
>>> a, b, c, d, f = map(Function, ['a', 'b', 'c', 'd', 'f'])
>>> genform = Eq(a(x)*b(f(x))*f(x).diff(x), c(x)*d(f(x)))
>>> pprint(genform)
d
a(x)*b(f(x))*--(f(x)) = c(x)*d(f(x))
dx
>>> pprint(dsolve(genform, f(x), hint='separable_Integral'))
f(x)
/ /
| |
| b(y) | c(x)
| ---- dy = C1 + | ---- dx
| d(y) | a(x)
| |
/ /
Examples
========
>>> from sympy import Function, dsolve, Eq
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(Eq(f(x)*f(x).diff(x) + x, 3*x*f(x)**2), f(x),
... hint='separable', simplify=False))
/ 2 \ 2
log\3*f (x) - 1/ x
---------------- = C1 + --
6 2
References
==========
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
Dover 1963, pp. 52
# indirect doctest
"""
x = func.args[0]
f = func.func
C1 = get_numbered_constants(eq, num=1)
r = match # {'m1':m1, 'm2':m2, 'y':y}
u = r.get('hint', f(x)) # get u from separable_reduced else get f(x)
return Eq(Integral(r['m2']['coeff']*r['m2'][r['y']]/r['m1'][r['y']],
(r['y'], None, u)), Integral(-r['m1']['coeff']*r['m1'][x]/
r['m2'][x], x) + C1)
[docs]def checkinfsol(eq, infinitesimals, func=None, order=None):
r"""
This function is used to check if the given infinitesimals are the
actual infinitesimals of the given first order differential equation.
This method is specific to the Lie Group Solver of ODEs.
As of now, it simply checks, by substituting the infinitesimals in the
partial differential equation.
.. math:: \frac{\partial \eta}{\partial x} + \left(\frac{\partial \eta}{\partial y}
- \frac{\partial \xi}{\partial x}\right)*h
- \frac{\partial \xi}{\partial y}*h^{2}
- \xi\frac{\partial h}{\partial x} - \eta\frac{\partial h}{\partial y} = 0
where `\eta`, and `\xi` are the infinitesimals and `h(x,y) = \frac{dy}{dx}`
The infinitesimals should be given in the form of a list of dicts
``[{xi(x, y): inf, eta(x, y): inf}]``, corresponding to the
output of the function infinitesimals. It returns a list
of values of the form ``[(True/False, sol)]`` where ``sol`` is the value
obtained after substituting the infinitesimals in the PDE. If it
is ``True``, then ``sol`` would be 0.
"""
if isinstance(eq, Equality):
eq = eq.lhs - eq.rhs
if not func:
eq, func = _preprocess(eq)
variables = func.args
if len(variables) != 1:
raise ValueError("ODE's have only one independent variable")
else:
x = variables[0]
if not order:
order = ode_order(eq, func)
if order != 1:
raise NotImplementedError("Lie groups solver has been implemented "
"only for first order differential equations")
else:
df = func.diff(x)
a = Wild('a', exclude = [df])
b = Wild('b', exclude = [df])
match = collect(expand(eq), df).match(a*df + b)
if match:
h = -simplify(match[b]/match[a])
else:
try:
sol = solve(eq, df)
except NotImplementedError:
raise NotImplementedError("Infinitesimals for the "
"first order ODE could not be found")
else:
h = sol[0] # Find infinitesimals for one solution
y = Dummy('y')
h = h.subs(func, y)
xi = Function('xi')(x, y)
eta = Function('eta')(x, y)
dxi = Function('xi')(x, func)
deta = Function('eta')(x, func)
pde = (eta.diff(x) + (eta.diff(y) - xi.diff(x))*h -
(xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)))
soltup = []
for sol in infinitesimals:
tsol = {xi: S(sol[dxi]).subs(func, y),
eta: S(sol[deta]).subs(func, y)}
sol = simplify(pde.subs(tsol).doit())
if sol:
soltup.append((False, sol.subs(y, func)))
else:
soltup.append((True, 0))
return soltup
[docs]def ode_lie_group(eq, func, order, match):
r"""
This hint implements the Lie group method of solving first order differential
equations. The aim is to convert the given differential equation from the
given coordinate given system into another coordinate system where it becomes
invariant under the one-parameter Lie group of translations. The converted ODE is
quadrature and can be solved easily. It makes use of the
:py:meth:`sympy.solvers.ode.infinitesimals` function which returns the
infinitesimals of the transformation.
The coordinates `r` and `s` can be found by solving the following Partial
Differential Equations.
.. math :: \xi\frac{\partial r}{\partial x} + \eta\frac{\partial r}{\partial y}
= 0
.. math :: \xi\frac{\partial s}{\partial x} + \eta\frac{\partial s}{\partial y}
= 1
The differential equation becomes separable in the new coordinate system
.. math :: \frac{ds}{dr} = \frac{\frac{\partial s}{\partial x} +
h(x, y)\frac{\partial s}{\partial y}}{
\frac{\partial r}{\partial x} + h(x, y)\frac{\partial r}{\partial y}}
After finding the solution by integration, it is then converted back to the original
coordinate system by subsituting `r` and `s` in terms of `x` and `y` again.
Examples
========
>>> from sympy import Function, dsolve, Eq, exp, pprint
>>> from sympy.abc import x
>>> f = Function('f')
>>> pprint(dsolve(f(x).diff(x) + 2*x*f(x) - x*exp(-x**2), f(x),
... hint='lie_group'))
/ 2\ 2
| x | -x
f(x) = |C1 + --|*e
\ 2 /
References
==========
- Solving differential equations by Symmetry Groups,
John Starrett, pp. 1 - pp. 14
"""
heuristics = lie_heuristics
inf = {}
f = func.func
x = func.args[0]
df = func.diff(x)
xi = Function("xi")
eta = Function("eta")
a = Wild('a', exclude = [df])
b = Wild('b', exclude = [df])
xis = match.pop('xi')
etas = match.pop('eta')
if match:
h = -simplify(match[match['d']]/match[match['e']])
y = match['y']
else:
try:
sol = solve(eq, df)
except NotImplementedError:
raise NotImplementedError("Unable to solve the differential equation " +
str(eq) + " by the lie group method")
else:
y = Dummy("y")
h = sol[0].subs(func, y)
if xis is not None and etas is not None:
inf = [{xi(x, f(x)): S(xis), eta(x, f(x)): S(etas)}]
if not checkinfsol(eq, inf, func=f(x), order=1)[0][0]:
raise ValueError("The given infinitesimals xi and eta"
" are not the infinitesimals to the given equation")
else:
heuristics = ["user_defined"]
match = {'h': h, 'y': y}
# This is done so that if:
# a] solve raises a NotImplementedError.
# b] any heuristic raises a ValueError
# another heuristic can be used.
tempsol = [] # Used by solve below
for heuristic in heuristics:
try:
if not inf:
inf = infinitesimals(eq, hint=heuristic, func=func, order=1, match=match)
except ValueError:
continue
else:
for infsim in inf:
xiinf = (infsim[xi(x, func)]).subs(func, y)
etainf = (infsim[eta(x, func)]).subs(func, y)
# This condition creates recursion while using pdsolve.
# Since the first step while solving a PDE of form
# a*(f(x, y).diff(x)) + b*(f(x, y).diff(y)) + c = 0
# is to solve the ODE dy/dx = b/a
if simplify(etainf/xiinf) == h:
continue
rpde = f(x, y).diff(x)*xiinf + f(x, y).diff(y)*etainf
r = pdsolve(rpde, func=f(x, y)).rhs
s = pdsolve(rpde - 1, func=f(x, y)).rhs
newcoord = [_lie_group_remove(coord) for coord in [r, s]]
r = Dummy("r")
s = Dummy("s")
C1 = Symbol("C1")
rcoord = newcoord[0]
scoord = newcoord[-1]
try:
sol = solve([r - rcoord, s - scoord], x, y, dict=True)
except NotImplementedError:
continue
else:
sol = sol[0]
xsub = sol[x]
ysub = sol[y]
num = simplify(scoord.diff(x) + scoord.diff(y)*h)
denom = simplify(rcoord.diff(x) + rcoord.diff(y)*h)
if num and denom:
diffeq = simplify((num/denom).subs([(x, xsub), (y, ysub)]))
sep = separatevars(diffeq, symbols=[r, s], dict=True)
if sep:
# Trying to separate, r and s coordinates
deq = integrate((1/sep[s]), s) + C1 - integrate(sep['coeff']*sep[r], r)
# Substituting and reverting back to original coordinates
deq = deq.subs([(r, rcoord), (s, scoord)])
try:
sdeq = solve(deq, y)
except NotImplementedError:
tempsol.append(deq)
else:
if len(sdeq) == 1:
return Eq(f(x), sdeq.pop())
else:
return [Eq(f(x), sol) for sol in sdeq]
elif denom: # (ds/dr) is zero which means s is constant
return Eq(f(x), solve(scoord - C1, y)[0])
elif num: # (dr/ds) is zero which means r is constant
return Eq(f(x), solve(rcoord - C1, y)[0])
# If nothing works, return solution as it is, without solving for y
if tempsol:
if len(tempsol) == 1:
return Eq(tempsol.pop().subs(y, f(x)), 0)
else:
return [Eq(sol.subs(y, f(x)), 0) for sol in tempsol]
raise NotImplementedError("The given ODE " + str(eq) + " cannot be solved by"
+ " the lie group method")
def _lie_group_remove(coords):
r"""
This function is strictly meant for internal use by the Lie group ODE solving
method. It replaces arbitrary functions returned by pdsolve with either 0 or 1 or the
args of the arbitrary function.
The algorithm used is:
1] If coords is an instance of an Undefined Function, then the args are returned
2] If the arbitrary function is present in an Add object, it is replaced by zero.
3] If the arbitrary function is present in an Mul object, it is replaced by one.
4] If coords has no Undefined Function, it is returned as it is.
Examples
========
>>> from sympy.solvers.ode import _lie_group_remove
>>> from sympy import Function
>>> from sympy.abc import x, y
>>> F = Function("F")
>>> eq = x**2*y
>>> _lie_group_remove(eq)
x**2*y
>>> eq = F(x**2*y)
>>> _lie_group_remove(eq)
x**2*y
>>> eq = y**2*x + F(x**3)
>>> _lie_group_remove(eq)
x*y**2
>>> eq = (F(x**3) + y)*x**4
>>> _lie_group_remove(eq)
x**4*y
"""
if isinstance(coords, AppliedUndef):
return coords.args[0]
elif coords.is_Add:
subfunc = coords.atoms(AppliedUndef)
if subfunc:
for func in subfunc:
coords = coords.subs(func, 0)
return coords
elif coords.is_Pow:
base, expr = coords.as_base_exp()
base = _lie_group_remove(base)
expr = _lie_group_remove(expr)
return base**expr
elif coords.is_Mul:
mulargs = []
coordargs = coords.args
for arg in coordargs:
if not isinstance(coords, AppliedUndef):
mulargs.append(_lie_group_remove(arg))
return Mul(*mulargs)
return coords
[docs]def infinitesimals(eq, func=None, order=None, hint='default', match=None):
r"""
The infinitesimal functions of an ordinary differential equation, `\xi(x,y)`
and `\eta(x,y)`, are the infinitesimals of the Lie group of point transformations
for which the differential equation is invariant. So, the ODE `y'=f(x,y)`
would admit a Lie group `x^*=X(x,y;\varepsilon)=x+\varepsilon\xi(x,y)`,
`y^*=Y(x,y;\varepsilon)=y+\varepsilon\eta(x,y)` such that `(y^*)'=f(x^*, y^*)`.
A change of coordinates, to `r(x,y)` and `s(x,y)`, can be performed so this Lie group
becomes the translation group, `r^*=r` and `s^*=s+\varepsilon`.
They are tangents to the coordinate curves of the new system.
Consider the transformation `(x, y) \to (X, Y)` such that the
differential equation remains invariant. `\xi` and `\eta` are the tangents to
the transformed coordinates `X` and `Y`, at `\varepsilon=0`.
.. math:: \left(\frac{\partial X(x,y;\varepsilon)}{\partial\varepsilon
}\right)|_{\varepsilon=0} = \xi,
\left(\frac{\partial Y(x,y;\varepsilon)}{\partial\varepsilon
}\right)|_{\varepsilon=0} = \eta,
The infinitesimals can be found by solving the following PDE:
>>> from sympy import Function, diff, Eq, pprint
>>> from sympy.abc import x, y
>>> xi, eta, h = map(Function, ['xi', 'eta', 'h'])
>>> h = h(x, y) # dy/dx = h
>>> eta = eta(x, y)
>>> xi = xi(x, y)
>>> genform = Eq(eta.diff(x) + (eta.diff(y) - xi.diff(x))*h
... - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)), 0)
>>> pprint(genform)
/d d \ d 2 d
|--(eta(x, y)) - --(xi(x, y))|*h(x, y) - eta(x, y)*--(h(x, y)) - h (x, y)*--(x
\dy dx / dy dy
<BLANKLINE>
d d
i(x, y)) - xi(x, y)*--(h(x, y)) + --(eta(x, y)) = 0
dx dx
Solving the above mentioned PDE is not trivial, and can be solved only by
making intelligent assumptions for `\xi` and `\eta` (heuristics). Once an
infinitesimal is found, the attempt to find more heuristics stops. This is done to
optimise the speed of solving the differential equation. If a list of all the
infinitesimals is needed, ``hint`` should be flagged as ``all``, which gives
the complete list of infinitesimals. If the infinitesimals for a particular
heuristic needs to be found, it can be passed as a flag to ``hint``.
Examples
========
>>> from sympy import Function, diff
>>> from sympy.solvers.ode import infinitesimals
>>> from sympy.abc import x
>>> f = Function('f')
>>> eq = f(x).diff(x) - x**2*f(x)
>>> infinitesimals(eq)
[{eta(x, f(x)): exp(x**3/3), xi(x, f(x)): 0}]
References
==========
- Solving differential equations by Symmetry Groups,
John Starrett, pp. 1 - pp. 14
"""
if isinstance(eq, Equality):
eq = eq.lhs - eq.rhs
if not func:
eq, func = _preprocess(eq)
variables = func.args
if len(variables) != 1:
raise ValueError("ODE's have only one independent variable")
else:
x = variables[0]
if not order:
order = ode_order(eq, func)
if order != 1:
raise NotImplementedError("Infinitesimals for only "
"first order ODE's have been implemented")
else:
df = func.diff(x)
# Matching differential equation of the form a*df + b
a = Wild('a', exclude = [df])
b = Wild('b', exclude = [df])
if match: # Used by lie_group hint
h = match['h']
y = match['y']
else:
match = collect(expand(eq), df).match(a*df + b)
if match:
h = -simplify(match[b]/match[a])
else:
try:
sol = solve(eq, df)
except NotImplementedError:
raise NotImplementedError("Infinitesimals for the "
"first order ODE could not be found")
else:
h = sol[0] # Find infinitesimals for one solution
y = Dummy("y")
h = h.subs(func, y)
u = Dummy("u")
hx = h.diff(x)
hy = h.diff(y)
hinv = ((1/h).subs([(x, u), (y, x)])).subs(u, y) # Inverse ODE
match = {'h': h, 'func': func, 'hx': hx, 'hy': hy, 'y': y, 'hinv': hinv}
if hint == 'all':
xieta = []
for heuristic in lie_heuristics:
function = globals()['lie_heuristic_' + heuristic]
inflist = function(match, comp=True)
if inflist:
xieta.extend([inf for inf in inflist if inf not in xieta])
if xieta:
return xieta
else:
raise NotImplementedError("Infinitesimals could not be found for "
"the given ODE")
elif hint == 'default':
for heuristic in lie_heuristics:
function = globals()['lie_heuristic_' + heuristic]
xieta = function(match, comp=False)
if xieta:
return xieta
raise NotImplementedError("Infinitesimals could not be found for"
" the given ODE")
elif hint not in lie_heuristics:
raise ValueError("Heuristic not recognized: " + hint)
else:
function = globals()['lie_heuristic_' + hint]
xieta = function(match, comp=True)
if xieta:
return xieta
else:
raise ValueError("Infinitesimals could not be found using the"
" given heuristic")
[docs]def lie_heuristic_abaco1_simple(match, comp=False):
r"""
The first heuristic uses the following four sets of
assumptions on `\xi` and `\eta`
.. math:: \xi = 0, \eta = f(x)
.. math:: \xi = 0, \eta = f(y)
.. math:: \xi = f(x), \eta = 0
.. math:: \xi = f(y), \eta = 0
The success of this heuristic is determined by algebraic factorisation.
For the first assumption `\xi = 0` and `\eta` to be a function of `x`, the PDE
.. math:: \frac{\partial \eta}{\partial x} + (\frac{\partial \eta}{\partial y}
- \frac{\partial \xi}{\partial x})*h
- \frac{\partial \xi}{\partial y}*h^{2}
- \xi*\frac{\partial h}{\partial x} - \eta*\frac{\partial h}{\partial y} = 0
reduces to `f'(x) - f\frac{\partial h}{\partial y} = 0`
If `\frac{\partial h}{\partial y}` is a function of `x`, then this can usually
be integrated easily. A similar idea is applied to the other 3 assumptions as well.
References
==========
- E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra
Solving of First Order ODEs Using Symmetry Methods, pp. 8
"""
xieta = []
y = match['y']
h = match['h']
func = match['func']
x = func.args[0]
hx = match['hx']
hy = match['hy']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
hysym = hy.free_symbols
if y not in hysym:
try:
fx = exp(integrate(hy, x))
except NotImplementedError:
pass
else:
inf = {xi: S(0), eta: fx}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
factor = hy/h
facsym = factor.free_symbols
if x not in facsym:
try:
fy = exp(integrate(factor, y))
except NotImplementedError:
pass
else:
inf = {xi: S(0), eta: fy.subs(y, func)}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
factor = -hx/h
facsym = factor.free_symbols
if y not in facsym:
try:
fx = exp(integrate(factor, x))
except NotImplementedError:
pass
else:
inf = {xi: fx, eta: S(0)}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
factor = -hx/(h**2)
facsym = factor.free_symbols
if x not in facsym:
try:
fy = exp(integrate(factor, y))
except NotImplementedError:
pass
else:
inf = {xi: fy.subs(y, func), eta: S(0)}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
if xieta:
return xieta
[docs]def lie_heuristic_abaco1_product(match, comp=False):
r"""
The second heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = 0, \xi = f(x)*g(y)
.. math:: \eta = f(x)*g(y), \xi = 0
The first assumption of this heuristic holds good if
`\frac{1}{h^{2}}\frac{\partial^2}{\partial x \partial y}\log(h)` is
separable in `x` and `y`, then the separated factors containing `x`
is `f(x)`, and `g(y)` is obtained by
.. math:: e^{\int f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)\,dy}
provided `f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)` is a function
of `y` only.
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
`\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption
satisifes. After obtaining `f(x)` and `g(y)`, the coordinates are again
interchanged, to get `\eta` as `f(x)*g(y)`
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 7 - pp. 8
"""
xieta = []
y = match['y']
h = match['h']
hinv = match['hinv']
func = match['func']
x = func.args[0]
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
inf = separatevars(((log(h).diff(y)).diff(x))/h**2, dict=True, symbols=[x, y])
if inf and inf['coeff']:
fx = inf[x]
gy = simplify(fx*((1/(fx*h)).diff(x)))
gysyms = gy.free_symbols
if x not in gysyms:
gy = exp(integrate(gy, y))
inf = {eta: S(0), xi: (fx*gy).subs(y, func)}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
u1 = Dummy("u1")
inf = separatevars(((log(hinv).diff(y)).diff(x))/hinv**2, dict=True, symbols=[x, y])
if inf and inf['coeff']:
fx = inf[x]
gy = simplify(fx*((1/(fx*hinv)).diff(x)))
gysyms = gy.free_symbols
if x not in gysyms:
gy = exp(integrate(gy, y))
etaval = fx*gy
etaval = (etaval.subs([(x, u1), (y, x)])).subs(u1, y)
inf = {eta: etaval.subs(y, func), xi: S(0)}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
if xieta:
return xieta
[docs]def lie_heuristic_bivariate(match, comp=False):
r"""
The third heuristic assumes the infinitesimals `\xi` and `\eta`
to be bi-variate polynomials in `x` and `y`. The assumption made here
for the logic below is that `h` is a rational function in `x` and `y`
though that may not be necessary for the infinitesimals to be
bivariate polynomials. The coefficients of the infinitesimals
are found out by substituting them in the PDE and grouping similar terms
that are polynomials and since they form a linear system, solve and check
for non trivial solutions. The degree of the assumed bivariates
are increased till a certain maximum value.
References
==========
- Lie Groups and Differential Equations
pp. 327 - pp. 329
"""
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
if h.is_rational_function():
# The maximum degree that the infinitesimals can take is
# calculated by this technique.
etax, etay, etad, xix, xiy, xid = symbols("etax etay etad xix xiy xid")
ipde = etax + (etay - xix)*h - xiy*h**2 - xid*hx - etad*hy
num, denom = cancel(ipde).as_numer_denom()
deg = Poly(num, x, y).total_degree()
deta = Function('deta')(x, y)
dxi = Function('dxi')(x, y)
ipde = (deta.diff(x) + (deta.diff(y) - dxi.diff(x))*h - (dxi.diff(y))*h**2
- dxi*hx - deta*hy)
xieq = Symbol("xi0")
etaeq = Symbol("eta0")
for i in range(deg + 1):
if i:
xieq += Add(*[
Symbol("xi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
for power in range(i + 1)])
etaeq += Add(*[
Symbol("eta_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
for power in range(i + 1)])
pden, denom = (ipde.subs({dxi: xieq, deta: etaeq}).doit()).as_numer_denom()
pden = expand(pden)
# If the individual terms are monomials, the coefficients
# are grouped
if pden.is_polynomial(x, y) and pden.is_Add:
polyy = Poly(pden, x, y).as_dict()
if polyy:
symset = xieq.free_symbols.union(etaeq.free_symbols) - {x, y}
soldict = solve(polyy.values(), *symset)
if isinstance(soldict, list):
soldict = soldict[0]
if any(x for x in soldict.values()):
xired = xieq.subs(soldict)
etared = etaeq.subs(soldict)
# Scaling is done by substituting one for the parameters
# This can be any number except zero.
dict_ = dict((sym, 1) for sym in symset)
inf = {eta: etared.subs(dict_).subs(y, func),
xi: xired.subs(dict_).subs(y, func)}
return [inf]
[docs]def lie_heuristic_chi(match, comp=False):
r"""
The aim of the fourth heuristic is to find the function `\chi(x, y)`
that satisifies the PDE `\frac{d\chi}{dx} + h\frac{d\chi}{dx}
- \frac{\partial h}{\partial y}\chi = 0`.
This assumes `\chi` to be a bivariate polynomial in `x` and `y`. By intution,
`h` should be a rational function in `x` and `y`. The method used here is
to substitute a general binomial for `\chi` up to a certain maximum degree
is reached. The coefficients of the polynomials, are calculated by by collecting
terms of the same order in `x` and `y`.
After finding `\chi`, the next step is to use `\eta = \xi*h + \chi`, to
determine `\xi` and `\eta`. This can be done by dividing `\chi` by `h`
which would give `-\xi` as the quotient and `\eta` as the remainder.
References
==========
- E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra
Solving of First Order ODEs Using Symmetry Methods, pp. 8
"""
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
if h.is_rational_function():
schi, schix, schiy = symbols("schi, schix, schiy")
cpde = schix + h*schiy - hy*schi
num, denom = cancel(cpde).as_numer_denom()
deg = Poly(num, x, y).total_degree()
chi = Function('chi')(x, y)
chix = chi.diff(x)
chiy = chi.diff(y)
cpde = chix + h*chiy - hy*chi
chieq = Symbol("chi")
for i in range(1, deg + 1):
chieq += Add(*[
Symbol("chi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
for power in range(i + 1)])
cnum, cden = cancel(cpde.subs({chi : chieq}).doit()).as_numer_denom()
cnum = expand(cnum)
if cnum.is_polynomial(x, y) and cnum.is_Add:
cpoly = Poly(cnum, x, y).as_dict()
if cpoly:
solsyms = chieq.free_symbols - {x, y}
soldict = solve(cpoly.values(), *solsyms)
if isinstance(soldict, list):
soldict = soldict[0]
if any(x for x in soldict.values()):
chieq = chieq.subs(soldict)
dict_ = dict((sym, 1) for sym in solsyms)
chieq = chieq.subs(dict_)
# After finding chi, the main aim is to find out
# eta, xi by the equation eta = xi*h + chi
# One method to set xi, would be rearranging it to
# (eta/h) - xi = (chi/h). This would mean dividing
# chi by h would give -xi as the quotient and eta
# as the remainder. Thanks to Sean Vig for suggesting
# this method.
xic, etac = div(chieq, h)
inf = {eta: etac.subs(y, func), xi: -xic.subs(y, func)}
return [inf]
[docs]def lie_heuristic_function_sum(match, comp=False):
r"""
This heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = 0, \xi = f(x) + g(y)
.. math:: \eta = f(x) + g(y), \xi = 0
The first assumption of this heuristic holds good if
.. math:: \frac{\partial}{\partial y}[(h\frac{\partial^{2}}{
\partial x^{2}}(h^{-1}))^{-1}]
is separable in `x` and `y`,
1. The separated factors containing `y` is `\frac{\partial g}{\partial y}`.
From this `g(y)` can be determined.
2. The separated factors containing `x` is `f''(x)`.
3. `h\frac{\partial^{2}}{\partial x^{2}}(h^{-1})` equals
`\frac{f''(x)}{f(x) + g(y)}`. From this `f(x)` can be determined.
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
`\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first
assumption satisifes. After obtaining `f(x)` and `g(y)`, the coordinates
are again interchanged, to get `\eta` as `f(x) + g(y)`.
For both assumptions, the constant factors are separated among `g(y)`
and `f''(x)`, such that `f''(x)` obtained from 3] is the same as that
obtained from 2]. If not possible, then this heuristic fails.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 7 - pp. 8
"""
xieta = []
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
hinv = match['hinv']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
for odefac in [h, hinv]:
factor = odefac*((1/odefac).diff(x, 2))
sep = separatevars((1/factor).diff(y), dict=True, symbols=[x, y])
if sep and sep['coeff'] and sep[x].has(x) and sep[y].has(y):
k = Dummy("k")
try:
gy = k*integrate(sep[y], y)
except NotImplementedError:
pass
else:
fdd = 1/(k*sep[x]*sep['coeff'])
fx = simplify(fdd/factor - gy)
check = simplify(fx.diff(x, 2) - fdd)
if fx:
if not check:
fx = fx.subs(k, 1)
gy = (gy/k)
else:
sol = solve(check, k)
if sol:
sol = sol[0]
fx = fx.subs(k, sol)
gy = (gy/k)*sol
else:
continue
if odefac == hinv: # Inverse ODE
fx = fx.subs(x, y)
gy = gy.subs(y, x)
etaval = factor_terms(fx + gy)
if etaval.is_Mul:
etaval = Mul(*[arg for arg in etaval.args if arg.has(x, y)])
if odefac == hinv: # Inverse ODE
inf = {eta: etaval.subs(y, func), xi : S(0)}
else:
inf = {xi: etaval.subs(y, func), eta : S(0)}
if not comp:
return [inf]
else:
xieta.append(inf)
if xieta:
return xieta
[docs]def lie_heuristic_abaco2_similar(match, comp=False):
r"""
This heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = g(x), \xi = f(x)
.. math:: \eta = f(y), \xi = g(y)
For the first assumption,
1. First `\frac{\frac{\partial h}{\partial y}}{\frac{\partial^{2} h}{
\partial yy}}` is calculated. Let us say this value is A
2. If this is constant, then `h` is matched to the form `A(x) + B(x)e^{
\frac{y}{C}}` then, `\frac{e^{\int \frac{A(x)}{C} \,dx}}{B(x)}` gives `f(x)`
and `A(x)*f(x)` gives `g(x)`
3. Otherwise `\frac{\frac{\partial A}{\partial X}}{\frac{\partial A}{
\partial Y}} = \gamma` is calculated. If
a] `\gamma` is a function of `x` alone
b] `\frac{\gamma\frac{\partial h}{\partial y} - \gamma'(x) - \frac{
\partial h}{\partial x}}{h + \gamma} = G` is a function of `x` alone.
then, `e^{\int G \,dx}` gives `f(x)` and `-\gamma*f(x)` gives `g(x)`
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
`\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption
satisifes. After obtaining `f(x)` and `g(x)`, the coordinates are again
interchanged, to get `\xi` as `f(x^*)` and `\eta` as `g(y^*)`
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
xieta = []
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
hinv = match['hinv']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
factor = cancel(h.diff(y)/h.diff(y, 2))
factorx = factor.diff(x)
factory = factor.diff(y)
if not factor.has(x) and not factor.has(y):
A = Wild('A', exclude=[y])
B = Wild('B', exclude=[y])
C = Wild('C', exclude=[x, y])
match = h.match(A + B*exp(y/C))
try:
tau = exp(-integrate(match[A]/match[C]), x)/match[B]
except NotImplementedError:
pass
else:
gx = match[A]*tau
return [{xi: tau, eta: gx}]
else:
gamma = cancel(factorx/factory)
if not gamma.has(y):
tauint = cancel((gamma*hy - gamma.diff(x) - hx)/(h + gamma))
if not tauint.has(y):
try:
tau = exp(integrate(tauint, x))
except NotImplementedError:
pass
else:
gx = -tau*gamma
return [{xi: tau, eta: gx}]
factor = cancel(hinv.diff(y)/hinv.diff(y, 2))
factorx = factor.diff(x)
factory = factor.diff(y)
if not factor.has(x) and not factor.has(y):
A = Wild('A', exclude=[y])
B = Wild('B', exclude=[y])
C = Wild('C', exclude=[x, y])
match = h.match(A + B*exp(y/C))
try:
tau = exp(-integrate(match[A]/match[C]), x)/match[B]
except NotImplementedError:
pass
else:
gx = match[A]*tau
return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}]
else:
gamma = cancel(factorx/factory)
if not gamma.has(y):
tauint = cancel((gamma*hinv.diff(y) - gamma.diff(x) - hinv.diff(x))/(
hinv + gamma))
if not tauint.has(y):
try:
tau = exp(integrate(tauint, x))
except NotImplementedError:
pass
else:
gx = -tau*gamma
return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}]
[docs]def lie_heuristic_abaco2_unique_unknown(match, comp=False):
r"""
This heuristic assumes the presence of unknown functions or known functions
with non-integer powers.
1. A list of all functions and non-integer powers containing x and y
2. Loop over each element `f` in the list, find `\frac{\frac{\partial f}{\partial x}}{
\frac{\partial f}{\partial x}} = R`
If it is separable in `x` and `y`, let `X` be the factors containing `x`. Then
a] Check if `\xi = X` and `\eta = -\frac{X}{R}` satisfy the PDE. If yes, then return
`\xi` and `\eta`
b] Check if `\xi = \frac{-R}{X}` and `\eta = -\frac{1}{X}` satisfy the PDE.
If yes, then return `\xi` and `\eta`
If not, then check if
a] :math:`\xi = -R,\eta = 1`
b] :math:`\xi = 1, \eta = -\frac{1}{R}`
are solutions.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
xieta = []
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
hinv = match['hinv']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
funclist = []
for atom in h.atoms(Pow):
base, exp = atom.as_base_exp()
if base.has(x) and base.has(y):
if not exp.is_Integer:
funclist.append(atom)
for function in h.atoms(AppliedUndef):
syms = function.free_symbols
if x in syms and y in syms:
funclist.append(function)
for f in funclist:
frac = cancel(f.diff(y)/f.diff(x))
sep = separatevars(frac, dict=True, symbols=[x, y])
if sep and sep['coeff']:
xitry1 = sep[x]
etatry1 = -1/(sep[y]*sep['coeff'])
pde1 = etatry1.diff(y)*h - xitry1.diff(x)*h - xitry1*hx - etatry1*hy
if not simplify(pde1):
return [{xi: xitry1, eta: etatry1.subs(y, func)}]
xitry2 = 1/etatry1
etatry2 = 1/xitry1
pde2 = etatry2.diff(x) - (xitry2.diff(y))*h**2 - xitry2*hx - etatry2*hy
if not simplify(expand(pde2)):
return [{xi: xitry2.subs(y, func), eta: etatry2}]
else:
etatry = -1/frac
pde = etatry.diff(x) + etatry.diff(y)*h - hx - etatry*hy
if not simplify(pde):
return [{xi: S(1), eta: etatry.subs(y, func)}]
xitry = -frac
pde = -xitry.diff(x)*h -xitry.diff(y)*h**2 - xitry*hx -hy
if not simplify(expand(pde)):
return [{xi: xitry.subs(y, func), eta: S(1)}]
[docs]def lie_heuristic_abaco2_unique_general(match, comp=False):
r"""
This heuristic finds if infinitesimals of the form `\eta = f(x)`, `\xi = g(y)`
without making any assumptions on `h`.
The complete sequence of steps is given in the paper mentioned below.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
xieta = []
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
hinv = match['hinv']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
C = S(0)
A = hx.diff(y)
B = hy.diff(y) + hy**2
C = hx.diff(x) - hx**2
if not (A and B and C):
return
Ax = A.diff(x)
Ay = A.diff(y)
Axy = Ax.diff(y)
Axx = Ax.diff(x)
Ayy = Ay.diff(y)
D = simplify(2*Axy + hx*Ay - Ax*hy + (hx*hy + 2*A)*A)*A - 3*Ax*Ay
if not D:
E1 = simplify(3*Ax**2 + ((hx**2 + 2*C)*A - 2*Axx)*A)
if E1:
E2 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2)
if not E2:
E3 = simplify(
E1*((28*Ax + 4*hx*A)*A**3 - E1*(hy*A + Ay)) - E1.diff(x)*8*A**4)
if not E3:
etaval = cancel((4*A**3*(Ax - hx*A) + E1*(hy*A - Ay))/(S(2)*A*E1))
if x not in etaval:
try:
etaval = exp(integrate(etaval, y))
except NotImplementedError:
pass
else:
xival = -4*A**3*etaval/E1
if y not in xival:
return [{xi: xival, eta: etaval.subs(y, func)}]
else:
E1 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2)
if E1:
E2 = simplify(
4*A**3*D - D**2 + E1*((2*Axx - (hx**2 + 2*C)*A)*A - 3*Ax**2))
if not E2:
E3 = simplify(
-(A*D)*E1.diff(y) + ((E1.diff(x) - hy*D)*A + 3*Ay*D +
(A*hx - 3*Ax)*E1)*E1)
if not E3:
etaval = cancel(((A*hx - Ax)*E1 - (Ay + A*hy)*D)/(S(2)*A*D))
if x not in etaval:
try:
etaval = exp(integrate(etaval, y))
except NotImplementedError:
pass
else:
xival = -E1*etaval/D
if y not in xival:
return [{xi: xival, eta: etaval.subs(y, func)}]
[docs]def lie_heuristic_linear(match, comp=False):
r"""
This heuristic assumes
1. `\xi = ax + by + c` and
2. `\eta = fx + gy + h`
After substituting the following assumptions in the determining PDE, it
reduces to
.. math:: f + (g - a)h - bh^{2} - (ax + by + c)\frac{\partial h}{\partial x}
- (fx + gy + c)\frac{\partial h}{\partial y}
Solving the reduced PDE obtained, using the method of characteristics, becomes
impractical. The method followed is grouping similar terms and solving the system
of linear equations obtained. The difference between the bivariate heuristic is that
`h` need not be a rational function in this case.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
xieta = []
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
hinv = match['hinv']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
coeffdict = {}
symbols = numbered_symbols("c", cls=Dummy)
symlist = [next(symbols) for i in islice(symbols, 6)]
C0, C1, C2, C3, C4, C5 = symlist
pde = C3 + (C4 - C0)*h -(C0*x + C1*y + C2)*hx - (C3*x + C4*y + C5)*hy - C1*h**2
pde, denom = pde.as_numer_denom()
pde = powsimp(expand(pde))
if pde.is_Add:
terms = pde.args
for term in terms:
if term.is_Mul:
rem = Mul(*[m for m in term.args if not m.has(x, y)])
xypart = term/rem
if xypart not in coeffdict:
coeffdict[xypart] = rem
else:
coeffdict[xypart] += rem
else:
if term not in coeffdict:
coeffdict[term] = S(1)
else:
coeffdict[term] += S(1)
sollist = coeffdict.values()
soldict = solve(sollist, symlist)
if soldict:
if isinstance(soldict, list):
soldict = soldict[0]
subval = soldict.values()
if any(t for t in subval):
onedict = dict(zip(symlist, [1]*6))
xival = C0*x + C1*func + C2
etaval = C3*x + C4*func + C5
xival = xival.subs(soldict)
etaval = etaval.subs(soldict)
xival = xival.subs(onedict)
etaval = etaval.subs(onedict)
return [{xi: xival, eta: etaval}]
def sysode_linear_2eq_order1(match_):
x = match_['func'][0].func
y = match_['func'][1].func
func = match_['func']
fc = match_['func_coeff']
eq = match_['eq']
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
r = dict()
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
for i in range(2):
eqs = 0
for terms in Add.make_args(eq[i]):
eqs += terms/fc[i,func[i],1]
eq[i] = eqs
# for equations Eq(a1*diff(x(t),t), a*x(t) + b*y(t) + k1)
# and Eq(a2*diff(x(t),t), c*x(t) + d*y(t) + k2)
r['a'] = -fc[0,x(t),0]/fc[0,x(t),1]
r['c'] = -fc[1,x(t),0]/fc[1,y(t),1]
r['b'] = -fc[0,y(t),0]/fc[0,x(t),1]
r['d'] = -fc[1,y(t),0]/fc[1,y(t),1]
forcing = [S(0),S(0)]
for i in range(2):
for j in Add.make_args(eq[i]):
if not j.has(x(t), y(t)):
forcing[i] += j
if not (forcing[0].has(t) or forcing[1].has(t)):
r['k1'] = forcing[0]
r['k2'] = forcing[1]
else:
raise NotImplementedError("Only homogeneous problems are supported" +
" (and constant inhomogeneity)")
if match_['type_of_equation'] == 'type1':
sol = _linear_2eq_order1_type1(x, y, t, r, eq)
if match_['type_of_equation'] == 'type2':
gsol = _linear_2eq_order1_type1(x, y, t, r, eq)
psol = _linear_2eq_order1_type2(x, y, t, r, eq)
sol = [Eq(x(t), gsol[0].rhs+psol[0]), Eq(y(t), gsol[1].rhs+psol[1])]
if match_['type_of_equation'] == 'type3':
sol = _linear_2eq_order1_type3(x, y, t, r, eq)
if match_['type_of_equation'] == 'type4':
sol = _linear_2eq_order1_type4(x, y, t, r, eq)
if match_['type_of_equation'] == 'type5':
sol = _linear_2eq_order1_type5(x, y, t, r, eq)
if match_['type_of_equation'] == 'type6':
sol = _linear_2eq_order1_type6(x, y, t, r, eq)
if match_['type_of_equation'] == 'type7':
sol = _linear_2eq_order1_type7(x, y, t, r, eq)
return sol
[docs]def _linear_2eq_order1_type1(x, y, t, r, eq):
r"""
It is classified under system of two linear homogeneous first-order constant-coefficient
ordinary differential equations.
The equations which come under this type are
.. math:: x' = ax + by,
.. math:: y' = cx + dy
The characteristics equation is written as
.. math:: \lambda^{2} + (a+d) \lambda + ad - bc = 0
and its discriminant is `D = (a-d)^{2} + 4bc`. There are several cases
1. Case when `ad - bc \neq 0`. The origin of coordinates, `x = y = 0`,
is the only stationary point; it is
- a node if `D = 0`
- a node if `D > 0` and `ad - bc > 0`
- a saddle if `D > 0` and `ad - bc < 0`
- a focus if `D < 0` and `a + d \neq 0`
- a centre if `D < 0` and `a + d \neq 0`.
1.1. If `D > 0`. The characteristic equation has two distinct real roots
`\lambda_1` and `\lambda_ 2` . The general solution of the system in question is expressed as
.. math:: x = C_1 b e^{\lambda_1 t} + C_2 b e^{\lambda_2 t}
.. math:: y = C_1 (\lambda_1 - a) e^{\lambda_1 t} + C_2 (\lambda_2 - a) e^{\lambda_2 t}
where `C_1` and `C_2` being arbitary constants
1.2. If `D < 0`. The characteristics equation has two conjugate
roots, `\lambda_1 = \sigma + i \beta` and `\lambda_2 = \sigma - i \beta`.
The general solution of the system is given by
.. math:: x = b e^{\sigma t} (C_1 \sin(\beta t) + C_2 \cos(\beta t))
.. math:: y = e^{\sigma t} ([(\sigma - a) C_1 - \beta C_2] \sin(\beta t) + [\beta C_1 + (\sigma - a) C_2 \cos(\beta t)])
1.3. If `D = 0` and `a \neq d`. The characteristic equation has
two equal roots, `\lambda_1 = \lambda_2`. The general solution of the system is written as
.. math:: x = 2b (C_1 + \frac{C_2}{a-d} + C_2 t) e^{\frac{a+d}{2} t}
.. math:: y = [(d - a) C_1 + C_2 + (d - a) C_2 t] e^{\frac{a+d}{2} t}
1.4. If `D = 0` and `a = d \neq 0` and `b = 0`
.. math:: x = C_1 e^{a t} , y = (c C_1 t + C_2) e^{a t}
1.5. If `D = 0` and `a = d \neq 0` and `c = 0`
.. math:: x = (b C_1 t + C_2) e^{a t} , y = C_1 e^{a t}
2. Case when `ad - bc = 0` and `a^{2} + b^{2} > 0`. The whole straight
line `ax + by = 0` consists of singular points. The orginal system of differential
equaitons can be rewritten as
.. math:: x' = ax + by , y' = k (ax + by)
2.1 If `a + bk \neq 0`, solution will be
.. math:: x = b C_1 + C_2 e^{(a + bk) t} , y = -a C_1 + k C_2 e^{(a + bk) t}
2.2 If `a + bk = 0`, solution will be
.. math:: x = C_1 (bk t - 1) + b C_2 t , y = k^{2} b C_1 t + (b k^{2} t + 1) C_2
"""
# FIXME: at least some of these can fail to give two linearly
# independent solutions e.g., because they make assumptions about
# zero/nonzero of certain coefficients. I've "fixed" one and
# raised NotImplementedError in another. I think this should probably
# just be re-written in terms of eigenvectors...
l = Dummy('l')
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
l1 = rootof(l**2 - (r['a']+r['d'])*l + r['a']*r['d'] - r['b']*r['c'], l, 0)
l2 = rootof(l**2 - (r['a']+r['d'])*l + r['a']*r['d'] - r['b']*r['c'], l, 1)
D = (r['a'] - r['d'])**2 + 4*r['b']*r['c']
if (r['a']*r['d'] - r['b']*r['c']) != 0:
if D > 0:
if r['b'].is_zero:
# tempting to use this in all cases, but does not guarantee linearly independent eigenvectors
gsol1 = C1*(l1 - r['d'] + r['b'])*exp(l1*t) + C2*(l2 - r['d'] + r['b'])*exp(l2*t)
gsol2 = C1*(l1 - r['a'] + r['c'])*exp(l1*t) + C2*(l2 - r['a'] + r['c'])*exp(l2*t)
else:
gsol1 = C1*r['b']*exp(l1*t) + C2*r['b']*exp(l2*t)
gsol2 = C1*(l1 - r['a'])*exp(l1*t) + C2*(l2 - r['a'])*exp(l2*t)
if D < 0:
sigma = re(l1)
if im(l1).is_positive:
beta = im(l1)
else:
beta = im(l2)
if r['b'].is_zero:
raise NotImplementedError('b == 0 case not implemented')
gsol1 = r['b']*exp(sigma*t)*(C1*sin(beta*t)+C2*cos(beta*t))
gsol2 = exp(sigma*t)*(((C1*(sigma-r['a'])-C2*beta)*sin(beta*t)+(C1*beta+(sigma-r['a'])*C2)*cos(beta*t)))
if D == 0:
if r['a']!=r['d']:
gsol1 = 2*r['b']*(C1 + C2/(r['a']-r['d'])+C2*t)*exp((r['a']+r['d'])*t/2)
gsol2 = ((r['d']-r['a'])*C1+C2+(r['d']-r['a'])*C2*t)*exp((r['a']+r['d'])*t/2)
if r['a']==r['d'] and r['a']!=0 and r['b']==0:
gsol1 = C1*exp(r['a']*t)
gsol2 = (r['c']*C1*t+C2)*exp(r['a']*t)
if r['a']==r['d'] and r['a']!=0 and r['c']==0:
gsol1 = (r['b']*C1*t+C2)*exp(r['a']*t)
gsol2 = C1*exp(r['a']*t)
elif (r['a']*r['d'] - r['b']*r['c']) == 0 and (r['a']**2+r['b']**2) > 0:
k = r['c']/r['a']
if r['a']+r['b']*k != 0:
gsol1 = r['b']*C1 + C2*exp((r['a']+r['b']*k)*t)
gsol2 = -r['a']*C1 + k*C2*exp((r['a']+r['b']*k)*t)
else:
gsol1 = C1*(r['b']*k*t-1)+r['b']*C2*t
gsol2 = k**2*r['b']*C1*t+(r['b']*k**2*t+1)*C2
return [Eq(x(t), gsol1), Eq(y(t), gsol2)]
[docs]def _linear_2eq_order1_type2(x, y, t, r, eq):
r"""
The equations of this type are
.. math:: x' = ax + by + k1 , y' = cx + dy + k2
The general solution of this system is given by sum of its particular solution and the
general solution of the corresponding homogeneous system is obtained from type1.
1. When `ad - bc \neq 0`. The particular solution will be
`x = x_0` and `y = y_0` where `x_0` and `y_0` are determined by solving linear system of equations
.. math:: a x_0 + b y_0 + k1 = 0 , c x_0 + d y_0 + k2 = 0
2. When `ad - bc = 0` and `a^{2} + b^{2} > 0`. In this case, the system of equation becomes
.. math:: x' = ax + by + k_1 , y' = k (ax + by) + k_2
2.1 If `\sigma = a + bk \neq 0`, particular solution is given by
.. math:: x = b \sigma^{-1} (c_1 k - c_2) t - \sigma^{-2} (a c_1 + b c_2)
.. math:: y = kx + (c_2 - c_1 k) t
2.2 If `\sigma = a + bk = 0`, particular solution is given by
.. math:: x = \frac{1}{2} b (c_2 - c_1 k) t^{2} + c_1 t
.. math:: y = kx + (c_2 - c_1 k) t
"""
r['k1'] = -r['k1']; r['k2'] = -r['k2']
if (r['a']*r['d'] - r['b']*r['c']) != 0:
x0, y0 = symbols('x0, y0', cls=Dummy)
sol = solve((r['a']*x0+r['b']*y0+r['k1'], r['c']*x0+r['d']*y0+r['k2']), x0, y0)
psol = [sol[x0], sol[y0]]
elif (r['a']*r['d'] - r['b']*r['c']) == 0 and (r['a']**2+r['b']**2) > 0:
k = r['c']/r['a']
sigma = r['a'] + r['b']*k
if sigma != 0:
sol1 = r['b']*sigma**-1*(r['k1']*k-r['k2'])*t - sigma**-2*(r['a']*r['k1']+r['b']*r['k2'])
sol2 = k*sol1 + (r['k2']-r['k1']*k)*t
else:
# FIXME: a previous typo fix shows this is not covered by tests
sol1 = r['b']*(r['k2']-r['k1']*k)*t**2 + r['k1']*t
sol2 = k*sol1 + (r['k2']-r['k1']*k)*t
psol = [sol1, sol2]
return psol
[docs]def _linear_2eq_order1_type3(x, y, t, r, eq):
r"""
The equations of this type of ode are
.. math:: x' = f(t) x + g(t) y
.. math:: y' = g(t) x + f(t) y
The solution of such equations is given by
.. math:: x = e^{F} (C_1 e^{G} + C_2 e^{-G}) , y = e^{F} (C_1 e^{G} - C_2 e^{-G})
where `C_1` and `C_2` are arbitary constants, and
.. math:: F = \int f(t) \,dt , G = \int g(t) \,dt
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
F = Integral(r['a'], t)
G = Integral(r['b'], t)
sol1 = exp(F)*(C1*exp(G) + C2*exp(-G))
sol2 = exp(F)*(C1*exp(G) - C2*exp(-G))
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order1_type4(x, y, t, r, eq):
r"""
The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = -g(t) x + f(t) y
The solution is given by
.. math:: x = F (C_1 \cos(G) + C_2 \sin(G)), y = F (-C_1 \sin(G) + C_2 \cos(G))
where `C_1` and `C_2` are arbitary constants, and
.. math:: F = \int f(t) \,dt , G = \int g(t) \,dt
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
if r['b'] == -r['c']:
F = exp(Integral(r['a'], t))
G = Integral(r['b'], t)
sol1 = F*(C1*cos(G) + C2*sin(G))
sol2 = F*(-C1*sin(G) + C2*cos(G))
elif r['d'] == -r['a']:
F = exp(Integral(r['c'], t))
G = Integral(r['d'], t)
sol1 = F*(-C1*sin(G) + C2*cos(G))
sol2 = F*(C1*cos(G) + C2*sin(G))
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order1_type5(x, y, t, r, eq):
r"""
The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = a g(t) x + [f(t) + b g(t)] y
The transformation of
.. math:: x = e^{\int f(t) \,dt} u , y = e^{\int f(t) \,dt} v , T = \int g(t) \,dt
leads to a system of constant coefficient linear differential equations
.. math:: u'(T) = v , v'(T) = au + bv
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
u, v = symbols('u, v', function=True)
T = Symbol('T')
if not cancel(r['c']/r['b']).has(t):
p = cancel(r['c']/r['b'])
q = cancel((r['d']-r['a'])/r['b'])
eq = (Eq(diff(u(T),T), v(T)), Eq(diff(v(T),T), p*u(T)+q*v(T)))
sol = dsolve(eq)
sol1 = exp(Integral(r['a'], t))*sol[0].rhs.subs(T, Integral(r['b'],t))
sol2 = exp(Integral(r['a'], t))*sol[1].rhs.subs(T, Integral(r['b'],t))
if not cancel(r['a']/r['d']).has(t):
p = cancel(r['a']/r['d'])
q = cancel((r['b']-r['c'])/r['d'])
sol = dsolve(Eq(diff(u(T),T), v(T)), Eq(diff(v(T),T), p*u(T)+q*v(T)))
sol1 = exp(Integral(r['c'], t))*sol[1].rhs.subs(T, Integral(r['d'],t))
sol2 = exp(Integral(r['c'], t))*sol[0].rhs.subs(T, Integral(r['d'],t))
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order1_type6(x, y, t, r, eq):
r"""
The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = a [f(t) + a h(t)] x + a [g(t) - h(t)] y
This is solved by first multiplying the first equation by `-a` and adding
it to the second equation to obtain
.. math:: y' - a x' = -a h(t) (y - a x)
Setting `U = y - ax` and integrating the equation we arrive at
.. math:: y - ax = C_1 e^{-a \int h(t) \,dt}
and on substituing the value of y in first equation give rise to first order ODEs. After solving for
`x`, we can obtain `y` by substituting the value of `x` in second equation.
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
p = 0
q = 0
p1 = cancel(r['c']/cancel(r['c']/r['d']).as_numer_denom()[0])
p2 = cancel(r['a']/cancel(r['a']/r['b']).as_numer_denom()[0])
for n, i in enumerate([p1, p2]):
for j in Mul.make_args(collect_const(i)):
if not j.has(t):
q = j
if q!=0 and n==0:
if ((r['c']/j - r['a'])/(r['b'] - r['d']/j)) == j:
p = 1
s = j
break
if q!=0 and n==1:
if ((r['a']/j - r['c'])/(r['d'] - r['b']/j)) == j:
p = 2
s = j
break
if p == 1:
equ = diff(x(t),t) - r['a']*x(t) - r['b']*(s*x(t) + C1*exp(-s*Integral(r['b'] - r['d']/s, t)))
hint1 = classify_ode(equ)[1]
sol1 = dsolve(equ, hint=hint1+'_Integral').rhs
sol2 = s*sol1 + C1*exp(-s*Integral(r['b'] - r['d']/s, t))
elif p ==2:
equ = diff(y(t),t) - r['c']*y(t) - r['d']*s*y(t) + C1*exp(-s*Integral(r['d'] - r['b']/s, t))
hint1 = classify_ode(equ)[1]
sol2 = dsolve(equ, hint=hint1+'_Integral').rhs
sol1 = s*sol2 + C1*exp(-s*Integral(r['d'] - r['b']/s, t))
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order1_type7(x, y, t, r, eq):
r"""
The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = h(t) x + p(t) y
Differentiating the first equation and substituting the value of `y`
from second equation will give a second-order linear equation
.. math:: g x'' - (fg + gp + g') x' + (fgp - g^{2} h + f g' - f' g) x = 0
This above equation can be easily integrated if following conditions are satisfied.
1. `fgp - g^{2} h + f g' - f' g = 0`
2. `fgp - g^{2} h + f g' - f' g = ag, fg + gp + g' = bg`
If first condition is satisfied then it is solved by current dsolve solver and in second case it becomes
a constant cofficient differential equation which is also solved by current solver.
Otherwise if the above condition fails then,
a particular solution is assumed as `x = x_0(t)` and `y = y_0(t)`
Then the general solution is expressed as
.. math:: x = C_1 x_0(t) + C_2 x_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt
.. math:: y = C_1 y_0(t) + C_2 [\frac{F(t) P(t)}{x_0(t)} + y_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt]
where C1 and C2 are arbitary constants and
.. math:: F(t) = e^{\int f(t) \,dt} , P(t) = e^{\int p(t) \,dt}
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
e1 = r['a']*r['b']*r['c'] - r['b']**2*r['c'] + r['a']*diff(r['b'],t) - diff(r['a'],t)*r['b']
e2 = r['a']*r['c']*r['d'] - r['b']*r['c']**2 + diff(r['c'],t)*r['d'] - r['c']*diff(r['d'],t)
m1 = r['a']*r['b'] + r['b']*r['d'] + diff(r['b'],t)
m2 = r['a']*r['c'] + r['c']*r['d'] + diff(r['c'],t)
if e1 == 0:
sol1 = dsolve(r['b']*diff(x(t),t,t) - m1*diff(x(t),t)).rhs
sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs
elif e2 == 0:
sol2 = dsolve(r['c']*diff(y(t),t,t) - m2*diff(y(t),t)).rhs
sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs
elif not (e1/r['b']).has(t) and not (m1/r['b']).has(t):
sol1 = dsolve(diff(x(t),t,t) - (m1/r['b'])*diff(x(t),t) - (e1/r['b'])*x(t)).rhs
sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs
elif not (e2/r['c']).has(t) and not (m2/r['c']).has(t):
sol2 = dsolve(diff(y(t),t,t) - (m2/r['c'])*diff(y(t),t) - (e2/r['c'])*y(t)).rhs
sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs
else:
x0, y0 = symbols('x0, y0') #x0 and y0 being particular solutions
F = exp(Integral(r['a'],t))
P = exp(Integral(r['d'],t))
sol1 = C1*x0 + C2*x0*Integral(r['b']*F*P/x0**2, t)
sol2 = C1*y0 + C2(F*P/x0 + y0*Integral(r['b']*F*P/x0**2, t))
return [Eq(x(t), sol1), Eq(y(t), sol2)]
def sysode_linear_2eq_order2(match_):
x = match_['func'][0].func
y = match_['func'][1].func
func = match_['func']
fc = match_['func_coeff']
eq = match_['eq']
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
r = dict()
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
for i in range(2):
eqs = []
for terms in Add.make_args(eq[i]):
eqs.append(terms/fc[i,func[i],2])
eq[i] = Add(*eqs)
# for equations Eq(diff(x(t),t,t), a1*diff(x(t),t)+b1*diff(y(t),t)+c1*x(t)+d1*y(t)+e1)
# and Eq(a2*diff(y(t),t,t), a2*diff(x(t),t)+b2*diff(y(t),t)+c2*x(t)+d2*y(t)+e2)
r['a1'] = -fc[0,x(t),1]/fc[0,x(t),2] ; r['a2'] = -fc[1,x(t),1]/fc[1,y(t),2]
r['b1'] = -fc[0,y(t),1]/fc[0,x(t),2] ; r['b2'] = -fc[1,y(t),1]/fc[1,y(t),2]
r['c1'] = -fc[0,x(t),0]/fc[0,x(t),2] ; r['c2'] = -fc[1,x(t),0]/fc[1,y(t),2]
r['d1'] = -fc[0,y(t),0]/fc[0,x(t),2] ; r['d2'] = -fc[1,y(t),0]/fc[1,y(t),2]
const = [S(0), S(0)]
for i in range(2):
for j in Add.make_args(eq[i]):
if not (j.has(x(t)) or j.has(y(t))):
const[i] += j
r['e1'] = -const[0]
r['e2'] = -const[1]
if match_['type_of_equation'] == 'type1':
sol = _linear_2eq_order2_type1(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type2':
gsol = _linear_2eq_order2_type1(x, y, t, r, eq)
psol = _linear_2eq_order2_type2(x, y, t, r, eq)
sol = [Eq(x(t), gsol[0].rhs+psol[0]), Eq(y(t), gsol[1].rhs+psol[1])]
elif match_['type_of_equation'] == 'type3':
sol = _linear_2eq_order2_type3(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type4':
sol = _linear_2eq_order2_type4(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type5':
sol = _linear_2eq_order2_type5(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type6':
sol = _linear_2eq_order2_type6(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type7':
sol = _linear_2eq_order2_type7(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type8':
sol = _linear_2eq_order2_type8(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type9':
sol = _linear_2eq_order2_type9(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type10':
sol = _linear_2eq_order2_type10(x, y, t, r, eq)
elif match_['type_of_equation'] == 'type11':
sol = _linear_2eq_order2_type11(x, y, t, r, eq)
return sol
[docs]def _linear_2eq_order2_type1(x, y, t, r, eq):
r"""
System of two constant-coefficient second-order linear homogeneous differential equations
.. math:: x'' = ax + by
.. math:: y'' = cx + dy
The charecteristic equation for above equations
.. math:: \lambda^4 - (a + d) \lambda^2 + ad - bc = 0
whose discriminant is `D = (a - d)^2 + 4bc \neq 0`
1. When `ad - bc \neq 0`
1.1. If `D \neq 0`. The characteristic equation has four distict roots, `\lambda_1, \lambda_2, \lambda_3, \lambda_4`.
The general solution of the system is
.. math:: x = C_1 b e^{\lambda_1 t} + C_2 b e^{\lambda_2 t} + C_3 b e^{\lambda_3 t} + C_4 b e^{\lambda_4 t}
.. math:: y = C_1 (\lambda_1^{2} - a) e^{\lambda_1 t} + C_2 (\lambda_2^{2} - a) e^{\lambda_2 t} + C_3 (\lambda_3^{2} - a) e^{\lambda_3 t} + C_4 (\lambda_4^{2} - a) e^{\lambda_4 t}
where `C_1,..., C_4` are arbitary constants.
1.2. If `D = 0` and `a \neq d`:
.. math:: x = 2 C_1 (bt + \frac{2bk}{a - d}) e^{\frac{kt}{2}} + 2 C_2 (bt + \frac{2bk}{a - d}) e^{\frac{-kt}{2}} + 2b C_3 t e^{\frac{kt}{2}} + 2b C_4 t e^{\frac{-kt}{2}}
.. math:: y = C_1 (d - a) t e^{\frac{kt}{2}} + C_2 (d - a) t e^{\frac{-kt}{2}} + C_3 [(d - a) t + 2k] e^{\frac{kt}{2}} + C_4 [(d - a) t - 2k] e^{\frac{-kt}{2}}
where `C_1,..., C_4` are arbitary constants and `k = \sqrt{2 (a + d)}`
1.3. If `D = 0` and `a = d \neq 0` and `b = 0`:
.. math:: x = 2 \sqrt{a} C_1 e^{\sqrt{a} t} + 2 \sqrt{a} C_2 e^{-\sqrt{a} t}
.. math:: y = c C_1 t e^{\sqrt{a} t} - c C_2 t e^{-\sqrt{a} t} + C_3 e^{\sqrt{a} t} + C_4 e^{-\sqrt{a} t}
1.4. If `D = 0` and `a = d \neq 0` and `c = 0`:
.. math:: x = b C_1 t e^{\sqrt{a} t} - b C_2 t e^{-\sqrt{a} t} + C_3 e^{\sqrt{a} t} + C_4 e^{-\sqrt{a} t}
.. math:: y = 2 \sqrt{a} C_1 e^{\sqrt{a} t} + 2 \sqrt{a} C_2 e^{-\sqrt{a} t}
2. When `ad - bc = 0` and `a^2 + b^2 > 0`. Then the original system becomes
.. math:: x'' = ax + by
.. math:: y'' = k (ax + by)
2.1. If `a + bk \neq 0`:
.. math:: x = C_1 e^{t \sqrt{a + bk}} + C_2 e^{-t \sqrt{a + bk}} + C_3 bt + C_4 b
.. math:: y = C_1 k e^{t \sqrt{a + bk}} + C_2 k e^{-t \sqrt{a + bk}} - C_3 at - C_4 a
2.2. If `a + bk = 0`:
.. math:: x = C_1 b t^3 + C_2 b t^2 + C_3 t + C_4
.. math:: y = kx + 6 C_1 t + 2 C_2
"""
r['a'] = r['c1']
r['b'] = r['d1']
r['c'] = r['c2']
r['d'] = r['d2']
l = Symbol('l')
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
chara_eq = l**4 - (r['a']+r['d'])*l**2 + r['a']*r['d'] - r['b']*r['c']
l1 = rootof(chara_eq, 0)
l2 = rootof(chara_eq, 1)
l3 = rootof(chara_eq, 2)
l4 = rootof(chara_eq, 3)
D = (r['a'] - r['d'])**2 + 4*r['b']*r['c']
if (r['a']*r['d'] - r['b']*r['c']) != 0:
if D != 0:
gsol1 = C1*r['b']*exp(l1*t) + C2*r['b']*exp(l2*t) + C3*r['b']*exp(l3*t) \
+ C4*r['b']*exp(l4*t)
gsol2 = C1*(l1**2-r['a'])*exp(l1*t) + C2*(l2**2-r['a'])*exp(l2*t) + \
C3*(l3**2-r['a'])*exp(l3*t) + C4*(l4**2-r['a'])*exp(l4*t)
else:
if r['a'] != r['d']:
k = sqrt(2*(r['a']+r['d']))
mid = r['b']*t+2*r['b']*k/(r['a']-r['d'])
gsol1 = 2*C1*mid*exp(k*t/2) + 2*C2*mid*exp(-k*t/2) + \
2*r['b']*C3*t*exp(k*t/2) + 2*r['b']*C4*t*exp(-k*t/2)
gsol2 = C1*(r['d']-r['a'])*t*exp(k*t/2) + C2*(r['d']-r['a'])*t*exp(-k*t/2) + \
C3*((r['d']-r['a'])*t+2*k)*exp(k*t/2) + C4*((r['d']-r['a'])*t-2*k)*exp(-k*t/2)
elif r['a'] == r['d'] != 0 and r['b'] == 0:
sa = sqrt(r['a'])
gsol1 = 2*sa*C1*exp(sa*t) + 2*sa*C2*exp(-sa*t)
gsol2 = r['c']*C1*t*exp(sa*t)-r['c']*C2*t*exp(-sa*t)+C3*exp(sa*t)+C4*exp(-sa*t)
elif r['a'] == r['d'] != 0 and r['c'] == 0:
sa = sqrt(r['a'])
gsol1 = r['b']*C1*t*exp(sa*t)-r['b']*C2*t*exp(-sa*t)+C3*exp(sa*t)+C4*exp(-sa*t)
gsol2 = 2*sa*C1*exp(sa*t) + 2*sa*C2*exp(-sa*t)
elif (r['a']*r['d'] - r['b']*r['c']) == 0 and (r['a']**2 + r['b']**2) > 0:
k = r['c']/r['a']
if r['a'] + r['b']*k != 0:
mid = sqrt(r['a'] + r['b']*k)
gsol1 = C1*exp(mid*t) + C2*exp(-mid*t) + C3*r['b']*t + C4*r['b']
gsol2 = C1*k*exp(mid*t) + C2*k*exp(-mid*t) - C3*r['a']*t - C4*r['a']
else:
gsol1 = C1*r['b']*t**3 + C2*r['b']*t**2 + C3*t + C4
gsol2 = k*gsol1 + 6*C1*t + 2*C2
return [Eq(x(t), gsol1), Eq(y(t), gsol2)]
[docs]def _linear_2eq_order2_type2(x, y, t, r, eq):
r"""
The equations in this type are
.. math:: x'' = a_1 x + b_1 y + c_1
.. math:: y'' = a_2 x + b_2 y + c_2
The general solution of this system is given by the sum of its particular solution
and the general solution of the homogeneous system. The general solution is given
by the linear system of 2 equation of order 2 and type 1
1. If `a_1 b_2 - a_2 b_1 \neq 0`. A particular solution will be `x = x_0` and `y = y_0`
where the constants `x_0` and `y_0` are determined by solving the linear algebraic system
.. math:: a_1 x_0 + b_1 y_0 + c_1 = 0, a_2 x_0 + b_2 y_0 + c_2 = 0
2. If `a_1 b_2 - a_2 b_1 = 0` and `a_1^2 + b_1^2 > 0`. In this case, the system in question becomes
.. math:: x'' = ax + by + c_1, y'' = k (ax + by) + c_2
2.1. If `\sigma = a + bk \neq 0`, the particular solution will be
.. math:: x = \frac{1}{2} b \sigma^{-1} (c_1 k - c_2) t^2 - \sigma^{-2} (a c_1 + b c_2)
.. math:: y = kx + \frac{1}{2} (c_2 - c_1 k) t^2
2.2. If `\sigma = a + bk = 0`, the particular solution will be
.. math:: x = \frac{1}{24} b (c_2 - c_1 k) t^4 + \frac{1}{2} c_1 t^2
.. math:: y = kx + \frac{1}{2} (c_2 - c_1 k) t^2
"""
x0, y0 = symbols('x0, y0')
if r['c1']*r['d2'] - r['c2']*r['d1'] != 0:
sol = solve((r['c1']*x0+r['d1']*y0+r['e1'], r['c2']*x0+r['d2']*y0+r['e2']), x0, y0)
psol = [sol[x0], sol[y0]]
elif r['c1']*r['d2'] - r['c2']*r['d1'] == 0 and (r['c1']**2 + r['d1']**2) > 0:
k = r['c2']/r['c1']
sig = r['c1'] + r['d1']*k
if sig != 0:
psol1 = r['d1']*sig**-1*(r['e1']*k-r['e2'])*t**2/2 - \
sig**-2*(r['c1']*r['e1']+r['d1']*r['e2'])
psol2 = k*psol1 + (r['e2'] - r['e1']*k)*t**2/2
psol = [psol1, psol2]
else:
psol1 = r['d1']*(r['e2']-r['e1']*k)*t**4/24 + r['e1']*t**2/2
psol2 = k*psol1 + (r['e2']-r['e1']*k)*t**2/2
psol = [psol1, psol2]
return psol
[docs]def _linear_2eq_order2_type3(x, y, t, r, eq):
r"""
These type of equation is used for describing the horizontal motion of a pendulum
taking into account the Earth rotation.
The solution is given with `a^2 + 4b > 0`:
.. math:: x = C_1 \cos(\alpha t) + C_2 \sin(\alpha t) + C_3 \cos(\beta t) + C_4 \sin(\beta t)
.. math:: y = -C_1 \sin(\alpha t) + C_2 \cos(\alpha t) - C_3 \sin(\beta t) + C_4 \cos(\beta t)
where `C_1,...,C_4` and
.. math:: \alpha = \frac{1}{2} a + \frac{1}{2} \sqrt{a^2 + 4b}, \beta = \frac{1}{2} a - \frac{1}{2} \sqrt{a^2 + 4b}
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
if r['b1']**2 - 4*r['c1'] > 0:
r['a'] = r['b1'] ; r['b'] = -r['c1']
alpha = r['a']/2 + sqrt(r['a']**2 + 4*r['b'])/2
beta = r['a']/2 - sqrt(r['a']**2 + 4*r['b'])/2
sol1 = C1*cos(alpha*t) + C2*sin(alpha*t) + C3*cos(beta*t) + C4*sin(beta*t)
sol2 = -C1*sin(alpha*t) + C2*cos(alpha*t) - C3*sin(beta*t) + C4*cos(beta*t)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type4(x, y, t, r, eq):
r"""
These equations are found in the theory of oscillations
.. math:: x'' + a_1 x' + b_1 y' + c_1 x + d_1 y = k_1 e^{i \omega t}
.. math:: y'' + a_2 x' + b_2 y' + c_2 x + d_2 y = k_2 e^{i \omega t}
The general solution of this linear nonhomogeneous system of constant-coefficient
differential equations is given by the sum of its particular solution and the
general solution of the corresponding homogeneous system (with `k_1 = k_2 = 0`)
1. A particular solution is obtained by the method of undetermined coefficients:
.. math:: x = A_* e^{i \omega t}, y = B_* e^{i \omega t}
On substituting these expressions into the original system of differential equations,
one arrive at a linear nonhomogeneous system of algebraic equations for the
coefficients `A` and `B`.
2. The general solution of the homogeneous system of differential equations is determined
by a linear combination of linearly independent particular solutions determined by
the method of undetermined coefficients in the form of exponentials:
.. math:: x = A e^{\lambda t}, y = B e^{\lambda t}
On substituting these expressions into the original system and colleting the
coefficients of the unknown `A` and `B`, one obtains
.. math:: (\lambda^{2} + a_1 \lambda + c_1) A + (b_1 \lambda + d_1) B = 0
.. math:: (a_2 \lambda + c_2) A + (\lambda^{2} + b_2 \lambda + d_2) B = 0
The determinant of this system must vanish for nontrivial solutions A, B to exist.
This requirement results in the following characteristic equation for `\lambda`
.. math:: (\lambda^2 + a_1 \lambda + c_1) (\lambda^2 + b_2 \lambda + d_2) - (b_1 \lambda + d_1) (a_2 \lambda + c_2) = 0
If all roots `k_1,...,k_4` of this equation are distict, the general solution of the original
system of the differential equations has the form
.. math:: x = C_1 (b_1 \lambda_1 + d_1) e^{\lambda_1 t} - C_2 (b_1 \lambda_2 + d_1) e^{\lambda_2 t} - C_3 (b_1 \lambda_3 + d_1) e^{\lambda_3 t} - C_4 (b_1 \lambda_4 + d_1) e^{\lambda_4 t}
.. math:: y = C_1 (\lambda_1^{2} + a_1 \lambda_1 + c_1) e^{\lambda_1 t} + C_2 (\lambda_2^{2} + a_1 \lambda_2 + c_1) e^{\lambda_2 t} + C_3 (\lambda_3^{2} + a_1 \lambda_3 + c_1) e^{\lambda_3 t} + C_4 (\lambda_4^{2} + a_1 \lambda_4 + c_1) e^{\lambda_4 t}
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
k = Symbol('k')
Ra, Ca, Rb, Cb = symbols('Ra, Ca, Rb, Cb')
a1 = r['a1'] ; a2 = r['a2']
b1 = r['b1'] ; b2 = r['b2']
c1 = r['c1'] ; c2 = r['c2']
d1 = r['d1'] ; d2 = r['d2']
k1 = r['e1'].expand().as_independent(t)[0]
k2 = r['e2'].expand().as_independent(t)[0]
ew1 = r['e1'].expand().as_independent(t)[1]
ew2 = powdenest(ew1).as_base_exp()[1]
ew3 = collect(ew2, t).coeff(t)
w = cancel(ew3/I)
# The particular solution is assumed to be (Ra+I*Ca)*exp(I*w*t) and
# (Rb+I*Cb)*exp(I*w*t) for x(t) and y(t) respectively
peq1 = (-w**2+c1)*Ra - a1*w*Ca + d1*Rb - b1*w*Cb - k1
peq2 = a1*w*Ra + (-w**2+c1)*Ca + b1*w*Rb + d1*Cb
peq3 = c2*Ra - a2*w*Ca + (-w**2+d2)*Rb - b2*w*Cb - k2
peq4 = a2*w*Ra + c2*Ca + b2*w*Rb + (-w**2+d2)*Cb
# FIXME: solve for what in what? Ra, Rb, etc I guess
# but then psol not used for anything?
psol = solve([peq1, peq2, peq3, peq4])
chareq = (k**2+a1*k+c1)*(k**2+b2*k+d2) - (b1*k+d1)*(a2*k+c2)
[k1, k2, k3, k4] = roots_quartic(Poly(chareq))
sol1 = -C1*(b1*k1+d1)*exp(k1*t) - C2*(b1*k2+d1)*exp(k2*t) - \
C3*(b1*k3+d1)*exp(k3*t) - C4*(b1*k4+d1)*exp(k4*t) + (Ra+I*Ca)*exp(I*w*t)
a1_ = (a1-1)
sol2 = C1*(k1**2+a1_*k1+c1)*exp(k1*t) + C2*(k2**2+a1_*k2+c1)*exp(k2*t) + \
C3*(k3**2+a1_*k3+c1)*exp(k3*t) + C4*(k4**2+a1_*k4+c1)*exp(k4*t) + (Rb+I*Cb)*exp(I*w*t)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type5(x, y, t, r, eq):
r"""
The equation which come under this catagory are
.. math:: x'' = a (t y' - y)
.. math:: y'' = b (t x' - x)
The transformation
.. math:: u = t x' - x, b = t y' - y
leads to the first-order system
.. math:: u' = atv, v' = btu
The general solution of this system is given by
If `ab > 0`:
.. math:: u = C_1 a e^{\frac{1}{2} \sqrt{ab} t^2} + C_2 a e^{-\frac{1}{2} \sqrt{ab} t^2}
.. math:: v = C_1 \sqrt{ab} e^{\frac{1}{2} \sqrt{ab} t^2} - C_2 \sqrt{ab} e^{-\frac{1}{2} \sqrt{ab} t^2}
If `ab < 0`:
.. math:: u = C_1 a \cos(\frac{1}{2} \sqrt{\left|ab\right|} t^2) + C_2 a \sin(-\frac{1}{2} \sqrt{\left|ab\right|} t^2)
.. math:: v = C_1 \sqrt{\left|ab\right|} \sin(\frac{1}{2} \sqrt{\left|ab\right|} t^2) + C_2 \sqrt{\left|ab\right|} \cos(-\frac{1}{2} \sqrt{\left|ab\right|} t^2)
where `C_1` and `C_2` are arbitary constants. On substituting the value of `u` and `v`
in above equations and integrating the resulting expressions, the general solution will become
.. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt, y = C_4 t + t \int \frac{u}{t^2} \,dt
where `C_3` and `C_4` are arbitrary constants.
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
r['a'] = -r['d1'] ; r['b'] = -r['c2']
mul = sqrt(abs(r['a']*r['b']))
if r['a']*r['b'] > 0:
u = C1*r['a']*exp(mul*t**2/2) + C2*r['a']*exp(-mul*t**2/2)
v = C1*mul*exp(mul*t**2/2) - C2*mul*exp(-mul*t**2/2)
else:
u = C1*r['a']*cos(mul*t**2/2) + C2*r['a']*sin(mul*t**2/2)
v = -C1*mul*sin(mul*t**2/2) + C2*mul*cos(mul*t**2/2)
sol1 = C3*t + t*Integral(u/t**2, t)
sol2 = C4*t + t*Integral(v/t**2, t)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type6(x, y, t, r, eq):
r"""
The equations are
.. math:: x'' = f(t) (a_1 x + b_1 y)
.. math:: y'' = f(t) (a_2 x + b_2 y)
If `k_1` and `k_2` are roots of the quadratic equation
.. math:: k^2 - (a_1 + b_2) k + a_1 b_2 - a_2 b_1 = 0
Then by multiplying appropriate constants and adding together original equations
we obtain two independent equations:
.. math:: z_1'' = k_1 f(t) z_1, z_1 = a_2 x + (k_1 - a_1) y
.. math:: z_2'' = k_2 f(t) z_2, z_2 = a_2 x + (k_2 - a_1) y
Solving the equations will give the values of `x` and `y` after obtaining the value
of `z_1` and `z_2` by solving the differential equation and substuting the result.
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
k = Symbol('k')
z = Function('z')
num, den = cancel(
(r['c1']*x(t) + r['d1']*y(t))/
(r['c2']*x(t) + r['d2']*y(t))).as_numer_denom()
f = r['c1']/num.coeff(x(t))
a1 = num.coeff(x(t))
b1 = num.coeff(y(t))
a2 = den.coeff(x(t))
b2 = den.coeff(y(t))
chareq = k**2 - (a1 + b2)*k + a1*b2 - a2*b1
k1, k2 = [rootof(chareq, k) for k in range(Poly(chareq).degree())]
z1 = dsolve(diff(z(t),t,t) - k1*f*z(t)).rhs
z2 = dsolve(diff(z(t),t,t) - k2*f*z(t)).rhs
sol1 = (k1*z2 - k2*z1 + a1*(z1 - z2))/(a2*(k1-k2))
sol2 = (z1 - z2)/(k1 - k2)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type7(x, y, t, r, eq):
r"""
The equations are given as
.. math:: x'' = f(t) (a_1 x' + b_1 y')
.. math:: y'' = f(t) (a_2 x' + b_2 y')
If `k_1` and 'k_2` are roots of the quadratic equation
.. math:: k^2 - (a_1 + b_2) k + a_1 b_2 - a_2 b_1 = 0
Then the system can be reduced by adding together the two equations multiplied
by appropriate constants give following two independent equations:
.. math:: z_1'' = k_1 f(t) z_1', z_1 = a_2 x + (k_1 - a_1) y
.. math:: z_2'' = k_2 f(t) z_2', z_2 = a_2 x + (k_2 - a_1) y
Integrating these and returning to the original variables, one arrives at a linear
algebraic system for the unknowns `x` and `y`:
.. math:: a_2 x + (k_1 - a_1) y = C_1 \int e^{k_1 F(t)} \,dt + C_2
.. math:: a_2 x + (k_2 - a_1) y = C_3 \int e^{k_2 F(t)} \,dt + C_4
where `C_1,...,C_4` are arbitrary constants and `F(t) = \int f(t) \,dt`
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
k = Symbol('k')
num, den = cancel(
(r['a1']*x(t) + r['b1']*y(t))/
(r['a2']*x(t) + r['b2']*y(t))).as_numer_denom()
f = r['a1']/num.coeff(x(t))
a1 = num.coeff(x(t))
b1 = num.coeff(y(t))
a2 = den.coeff(x(t))
b2 = den.coeff(y(t))
chareq = k**2 - (a1 + b2)*k + a1*b2 - a2*b1
[k1, k2] = [rootof(chareq, k) for k in range(Poly(chareq).degree())]
F = Integral(f, t)
z1 = C1*Integral(exp(k1*F), t) + C2
z2 = C3*Integral(exp(k2*F), t) + C4
sol1 = (k1*z2 - k2*z1 + a1*(z1 - z2))/(a2*(k1-k2))
sol2 = (z1 - z2)/(k1 - k2)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type8(x, y, t, r, eq):
r"""
The equation of this catagory are
.. math:: x'' = a f(t) (t y' - y)
.. math:: y'' = b f(t) (t x' - x)
The transformation
.. math:: u = t x' - x, v = t y' - y
leads to the system of first-order equations
.. math:: u' = a t f(t) v, v' = b t f(t) u
The general solution of this system has the form
If `ab > 0`:
.. math:: u = C_1 a e^{\sqrt{ab} \int t f(t) \,dt} + C_2 a e^{-\sqrt{ab} \int t f(t) \,dt}
.. math:: v = C_1 \sqrt{ab} e^{\sqrt{ab} \int t f(t) \,dt} - C_2 \sqrt{ab} e^{-\sqrt{ab} \int t f(t) \,dt}
If `ab < 0`:
.. math:: u = C_1 a \cos(\sqrt{\left|ab\right|} \int t f(t) \,dt) + C_2 a \sin(-\sqrt{\left|ab\right|} \int t f(t) \,dt)
.. math:: v = C_1 \sqrt{\left|ab\right|} \sin(\sqrt{\left|ab\right|} \int t f(t) \,dt) + C_2 \sqrt{\left|ab\right|} \cos(-\sqrt{\left|ab\right|} \int t f(t) \,dt)
where `C_1` and `C_2` are arbitary constants. On substituting the value of `u` and `v`
in above equations and integrating the resulting expressions, the general solution will become
.. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt, y = C_4 t + t \int \frac{u}{t^2} \,dt
where `C_3` and `C_4` are arbitrary constants.
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
num, den = cancel(r['d1']/r['c2']).as_numer_denom()
f = -r['d1']/num
a = num
b = den
mul = sqrt(abs(a*b))
Igral = Integral(t*f, t)
if a*b > 0:
u = C1*a*exp(mul*Igral) + C2*a*exp(-mul*Igral)
v = C1*mul*exp(mul*Igral) - C2*mul*exp(-mul*Igral)
else:
u = C1*a*cos(mul*Igral) + C2*a*sin(mul*Igral)
v = -C1*mul*sin(mul*Igral) + C2*mul*cos(mul*Igral)
sol1 = C3*t + t*Integral(u/t**2, t)
sol2 = C4*t + t*Integral(v/t**2, t)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type9(x, y, t, r, eq):
r"""
.. math:: t^2 x'' + a_1 t x' + b_1 t y' + c_1 x + d_1 y = 0
.. math:: t^2 y'' + a_2 t x' + b_2 t y' + c_2 x + d_2 y = 0
These system of equations are euler type.
The substitution of `t = \sigma e^{\tau} (\sigma \neq 0)` leads to the system of constant
coefficient linear differential equations
.. math:: x'' + (a_1 - 1) x' + b_1 y' + c_1 x + d_1 y = 0
.. math:: y'' + a_2 x' + (b_2 - 1) y' + c_2 x + d_2 y = 0
The general solution of the homogeneous system of differential equations is determined
by a linear combination of linearly independent particular solutions determined by
the method of undetermined coefficients in the form of exponentials
.. math:: x = A e^{\lambda t}, y = B e^{\lambda t}
On substituting these expressions into the original system and colleting the
coefficients of the unknown `A` and `B`, one obtains
.. math:: (\lambda^{2} + (a_1 - 1) \lambda + c_1) A + (b_1 \lambda + d_1) B = 0
.. math:: (a_2 \lambda + c_2) A + (\lambda^{2} + (b_2 - 1) \lambda + d_2) B = 0
The determinant of this system must vanish for nontrivial solutions A, B to exist.
This requirement results in the following characteristic equation for `\lambda`
.. math:: (\lambda^2 + (a_1 - 1) \lambda + c_1) (\lambda^2 + (b_2 - 1) \lambda + d_2) - (b_1 \lambda + d_1) (a_2 \lambda + c_2) = 0
If all roots `k_1,...,k_4` of this equation are distict, the general solution of the original
system of the differential equations has the form
.. math:: x = C_1 (b_1 \lambda_1 + d_1) e^{\lambda_1 t} - C_2 (b_1 \lambda_2 + d_1) e^{\lambda_2 t} - C_3 (b_1 \lambda_3 + d_1) e^{\lambda_3 t} - C_4 (b_1 \lambda_4 + d_1) e^{\lambda_4 t}
.. math:: y = C_1 (\lambda_1^{2} + (a_1 - 1) \lambda_1 + c_1) e^{\lambda_1 t} + C_2 (\lambda_2^{2} + (a_1 - 1) \lambda_2 + c_1) e^{\lambda_2 t} + C_3 (\lambda_3^{2} + (a_1 - 1) \lambda_3 + c_1) e^{\lambda_3 t} + C_4 (\lambda_4^{2} + (a_1 - 1) \lambda_4 + c_1) e^{\lambda_4 t}
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
k = Symbol('k')
a1 = -r['a1']*t; a2 = -r['a2']*t
b1 = -r['b1']*t; b2 = -r['b2']*t
c1 = -r['c1']*t**2; c2 = -r['c2']*t**2
d1 = -r['d1']*t**2; d2 = -r['d2']*t**2
eq = (k**2+(a1-1)*k+c1)*(k**2+(b2-1)*k+d2)-(b1*k+d1)*(a2*k+c2)
[k1, k2, k3, k4] = roots_quartic(Poly(eq))
sol1 = -C1*(b1*k1+d1)*exp(k1*log(t)) - C2*(b1*k2+d1)*exp(k2*log(t)) - \
C3*(b1*k3+d1)*exp(k3*log(t)) - C4*(b1*k4+d1)*exp(k4*log(t))
a1_ = (a1-1)
sol2 = C1*(k1**2+a1_*k1+c1)*exp(k1*log(t)) + C2*(k2**2+a1_*k2+c1)*exp(k2*log(t)) \
+ C3*(k3**2+a1_*k3+c1)*exp(k3*log(t)) + C4*(k4**2+a1_*k4+c1)*exp(k4*log(t))
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type10(x, y, t, r, eq):
r"""
The equation of this catagory are
.. math:: (\alpha t^2 + \beta t + \gamma)^{2} x'' = ax + by
.. math:: (\alpha t^2 + \beta t + \gamma)^{2} y'' = cx + dy
The transformation
.. math:: \tau = \int \frac{1}{\alpha t^2 + \beta t + \gamma} \,dt , u = \frac{x}{\sqrt{\left|\alpha t^2 + \beta t + \gamma\right|}} , v = \frac{y}{\sqrt{\left|\alpha t^2 + \beta t + \gamma\right|}}
leads to a constant coefficient linear system of equations
.. math:: u'' = (a - \alpha \gamma + \frac{1}{4} \beta^{2}) u + b v
.. math:: v'' = c u + (d - \alpha \gamma + \frac{1}{4} \beta^{2}) v
These system of equations obtained can be solved by type1 of System of two
constant-coefficient second-order linear homogeneous differential equations.
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
u, v = symbols('u, v', function=True)
T = Symbol('T')
p = Wild('p', exclude=[t, t**2])
q = Wild('q', exclude=[t, t**2])
s = Wild('s', exclude=[t, t**2])
n = Wild('n', exclude=[t, t**2])
num, den = r['c1'].as_numer_denom()
dic = den.match((n*(p*t**2+q*t+s)**2).expand())
eqz = dic[p]*t**2 + dic[q]*t + dic[s]
a = num/dic[n]
b = cancel(r['d1']*eqz**2)
c = cancel(r['c2']*eqz**2)
d = cancel(r['d2']*eqz**2)
[msol1, msol2] = dsolve([Eq(diff(u(t), t, t), (a - dic[p]*dic[s] + dic[q]**2/4)*u(t) \
+ b*v(t)), Eq(diff(v(t),t,t), c*u(t) + (d - dic[p]*dic[s] + dic[q]**2/4)*v(t))])
sol1 = (msol1.rhs*sqrt(abs(eqz))).subs(t, Integral(1/eqz, t))
sol2 = (msol2.rhs*sqrt(abs(eqz))).subs(t, Integral(1/eqz, t))
return [Eq(x(t), sol1), Eq(y(t), sol2)]
[docs]def _linear_2eq_order2_type11(x, y, t, r, eq):
r"""
The equations which comes under this type are
.. math:: x'' = f(t) (t x' - x) + g(t) (t y' - y)
.. math:: y'' = h(t) (t x' - x) + p(t) (t y' - y)
The transformation
.. math:: u = t x' - x, v = t y' - y
leads to the linear system of first-order equations
.. math:: u' = t f(t) u + t g(t) v, v' = t h(t) u + t p(t) v
On substituting the value of `u` and `v` in transformed equation gives value of `x` and `y` as
.. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt , y = C_4 t + t \int \frac{v}{t^2} \,dt.
where `C_3` and `C_4` are arbitrary constants.
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
u, v = symbols('u, v', function=True)
f = -r['c1'] ; g = -r['d1']
h = -r['c2'] ; p = -r['d2']
[msol1, msol2] = dsolve([Eq(diff(u(t),t), t*f*u(t) + t*g*v(t)), Eq(diff(v(t),t), t*h*u(t) + t*p*v(t))])
sol1 = C3*t + t*Integral(msol1.rhs/t**2, t)
sol2 = C4*t + t*Integral(msol2.rhs/t**2, t)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
def sysode_linear_3eq_order1(match_):
x = match_['func'][0].func
y = match_['func'][1].func
z = match_['func'][2].func
func = match_['func']
fc = match_['func_coeff']
eq = match_['eq']
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
r = dict()
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
for i in range(3):
eqs = 0
for terms in Add.make_args(eq[i]):
eqs += terms/fc[i,func[i],1]
eq[i] = eqs
# for equations:
# Eq(g1*diff(x(t),t), a1*x(t)+b1*y(t)+c1*z(t)+d1),
# Eq(g2*diff(y(t),t), a2*x(t)+b2*y(t)+c2*z(t)+d2), and
# Eq(g3*diff(z(t),t), a3*x(t)+b3*y(t)+c3*z(t)+d3)
r['a1'] = fc[0,x(t),0]/fc[0,x(t),1]; r['a2'] = fc[1,x(t),0]/fc[1,y(t),1];
r['a3'] = fc[2,x(t),0]/fc[2,z(t),1]
r['b1'] = fc[0,y(t),0]/fc[0,x(t),1]; r['b2'] = fc[1,y(t),0]/fc[1,y(t),1];
r['b3'] = fc[2,y(t),0]/fc[2,z(t),1]
r['c1'] = fc[0,z(t),0]/fc[0,x(t),1]; r['c2'] = fc[1,z(t),0]/fc[1,y(t),1];
r['c3'] = fc[2,z(t),0]/fc[2,z(t),1]
for i in range(3):
for j in Add.make_args(eq[i]):
if not j.has(x(t), y(t), z(t)):
raise NotImplementedError("Only homogeneous problems are supported, non-homogenous are not supported currently.")
if match_['type_of_equation'] == 'type1':
sol = _linear_3eq_order1_type1(x, y, z, t, r, eq)
if match_['type_of_equation'] == 'type2':
sol = _linear_3eq_order1_type2(x, y, z, t, r, eq)
if match_['type_of_equation'] == 'type3':
sol = _linear_3eq_order1_type3(x, y, z, t, r, eq)
if match_['type_of_equation'] == 'type4':
sol = _linear_3eq_order1_type4(x, y, z, t, r, eq)
if match_['type_of_equation'] == 'type6':
sol = _linear_neq_order1_type1(match_)
return sol
[docs]def _linear_3eq_order1_type1(x, y, z, t, r, eq):
r"""
.. math:: x' = ax
.. math:: y' = bx + cy
.. math:: z' = dx + ky + pz
Solution of such equations are forward substitution. Solving first equations
gives the value of `x`, substituting it in second and third equation and
solving second equation gives `y` and similarly substituting `y` in third
equation give `z`.
.. math:: x = C_1 e^{at}
.. math:: y = \frac{b C_1}{a - c} e^{at} + C_2 e^{ct}
.. math:: z = \frac{C_1}{a - p} (d + \frac{bk}{a - c}) e^{at} + \frac{k C_2}{c - p} e^{ct} + C_3 e^{pt}
where `C_1, C_2` and `C_3` are arbitrary constants.
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
a = -r['a1']; b = -r['a2']; c = -r['b2']
d = -r['a3']; k = -r['b3']; p = -r['c3']
sol1 = C1*exp(a*t)
sol2 = b*C1*exp(a*t)/(a-c) + C2*exp(c*t)
sol3 = C1*(d+b*k/(a-c))*exp(a*t)/(a-p) + k*C2*exp(c*t)/(c-p) + C3*exp(p*t)
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
[docs]def _linear_3eq_order1_type2(x, y, z, t, r, eq):
r"""
The equations of this type are
.. math:: x' = cy - bz
.. math:: y' = az - cx
.. math:: z' = bx - ay
1. First integral:
.. math:: ax + by + cz = A \qquad - (1)
.. math:: x^2 + y^2 + z^2 = B^2 \qquad - (2)
where `A` and `B` are arbitrary constants. It follows from these integrals
that the integral lines are circles formed by the intersection of the planes
`(1)` and sphere `(2)`
2. Solution:
.. math:: x = a C_0 + k C_1 \cos(kt) + (c C_2 - b C_3) \sin(kt)
.. math:: y = b C_0 + k C_2 \cos(kt) + (a C_2 - c C_3) \sin(kt)
.. math:: z = c C_0 + k C_3 \cos(kt) + (b C_2 - a C_3) \sin(kt)
where `k = \sqrt{a^2 + b^2 + c^2}` and the four constants of integration,
`C_1,...,C_4` are constrained by a single relation,
.. math:: a C_1 + b C_2 + c C_3 = 0
"""
C0, C1, C2, C3 = get_numbered_constants(eq, num=4, start=0)
a = -r['c2']; b = -r['a3']; c = -r['b1']
k = sqrt(a**2 + b**2 + c**2)
C3 = (-a*C1 - b*C2)/c
sol1 = a*C0 + k*C1*cos(k*t) + (c*C2-b*C3)*sin(k*t)
sol2 = b*C0 + k*C2*cos(k*t) + (a*C3-c*C1)*sin(k*t)
sol3 = c*C0 + k*C3*cos(k*t) + (b*C1-a*C2)*sin(k*t)
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
[docs]def _linear_3eq_order1_type3(x, y, z, t, r, eq):
r"""
Equations of this system of ODEs
.. math:: a x' = bc (y - z)
.. math:: b y' = ac (z - x)
.. math:: c z' = ab (x - y)
1. First integral:
.. math:: a^2 x + b^2 y + c^2 z = A
where A is an arbitary constant. It follows that the integral lines are plane curves.
2. Solution:
.. math:: x = C_0 + k C_1 \cos(kt) + a^{-1} bc (C_2 - C_3) \sin(kt)
.. math:: y = C_0 + k C_2 \cos(kt) + a b^{-1} c (C_3 - C_1) \sin(kt)
.. math:: z = C_0 + k C_3 \cos(kt) + ab c^{-1} (C_1 - C_2) \sin(kt)
where `k = \sqrt{a^2 + b^2 + c^2}` and the four constants of integration,
`C_1,...,C_4` are constrained by a single relation
.. math:: a^2 C_1 + b^2 C_2 + c^2 C_3 = 0
"""
C0, C1, C2, C3 = get_numbered_constants(eq, num=4, start=0)
c = sqrt(r['b1']*r['c2'])
b = sqrt(r['b1']*r['a3'])
a = sqrt(r['c2']*r['a3'])
C3 = (-a**2*C1-b**2*C2)/c**2
k = sqrt(a**2 + b**2 + c**2)
sol1 = C0 + k*C1*cos(k*t) + a**-1*b*c*(C2-C3)*sin(k*t)
sol2 = C0 + k*C2*cos(k*t) + a*b**-1*c*(C3-C1)*sin(k*t)
sol3 = C0 + k*C3*cos(k*t) + a*b*c**-1*(C1-C2)*sin(k*t)
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
[docs]def _linear_3eq_order1_type4(x, y, z, t, r, eq):
r"""
Equations:
.. math:: x' = (a_1 f(t) + g(t)) x + a_2 f(t) y + a_3 f(t) z
.. math:: y' = b_1 f(t) x + (b_2 f(t) + g(t)) y + b_3 f(t) z
.. math:: z' = c_1 f(t) x + c_2 f(t) y + (c_3 f(t) + g(t)) z
The transformation
.. math:: x = e^{\int g(t) \,dt} u, y = e^{\int g(t) \,dt} v, z = e^{\int g(t) \,dt} w, \tau = \int f(t) \,dt
leads to the system of constant coefficient linear differential equations
.. math:: u' = a_1 u + a_2 v + a_3 w
.. math:: v' = b_1 u + b_2 v + b_3 w
.. math:: w' = c_1 u + c_2 v + c_3 w
These system of equations are solved by homogeneous linear system of constant
coefficients of `n` equations of first order. Then substituting the value of
`u, v` and `w` in transformed equation gives value of `x, y` and `z`.
"""
u, v, w = symbols('u, v, w', function=True)
a2, a3 = cancel(r['b1']/r['c1']).as_numer_denom()
f = cancel(r['b1']/a2)
b1 = cancel(r['a2']/f); b3 = cancel(r['c2']/f)
c1 = cancel(r['a3']/f); c2 = cancel(r['b3']/f)
a1, g = div(r['a1'],f)
b2 = div(r['b2'],f)[0]
c3 = div(r['c3'],f)[0]
trans_eq = (diff(u(t),t)-a1*u(t)-a2*v(t)-a3*w(t), diff(v(t),t)-b1*u(t)-\
b2*v(t)-b3*w(t), diff(w(t),t)-c1*u(t)-c2*v(t)-c3*w(t))
sol = dsolve(trans_eq)
sol1 = exp(Integral(g,t))*((sol[0].rhs).subs(t, Integral(f,t)))
sol2 = exp(Integral(g,t))*((sol[1].rhs).subs(t, Integral(f,t)))
sol3 = exp(Integral(g,t))*((sol[2].rhs).subs(t, Integral(f,t)))
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
def sysode_linear_neq_order1(match_):
sol = _linear_neq_order1_type1(match_)
[docs]def _linear_neq_order1_type1(match_):
r"""
System of n first-order constant-coefficient linear nonhomogeneous differential equation
.. math:: y'_k = a_{k1} y_1 + a_{k2} y_2 +...+ a_{kn} y_n; k = 1,2,...,n
or that can be written as `\vec{y'} = A . \vec{y}`
where `\vec{y}` is matrix of `y_k` for `k = 1,2,...n` and `A` is a `n \times n` matrix.
Since these equations are equivalent to a first order homogeneous linear
differential equation. So the general solution will contain `n` linearly
independent parts and solution will consist some type of exponential
functions. Assuming `y = \vec{v} e^{rt}` is a solution of the system where
`\vec{v}` is a vector of coefficients of `y_1,...,y_n`. Substituting `y` and
`y' = r v e^{r t}` into the equation `\vec{y'} = A . \vec{y}`, we get
.. math:: r \vec{v} e^{rt} = A \vec{v} e^{rt}
.. math:: r \vec{v} = A \vec{v}
where `r` comes out to be eigenvalue of `A` and vector `\vec{v}` is the eigenvector
of `A` corresponding to `r`. There are three possiblities of eigenvalues of `A`
- `n` distinct real eigenvalues
- complex conjugate eigenvalues
- eigenvalues with multiplicity `k`
1. When all eigenvalues `r_1,..,r_n` are distinct with `n` different eigenvectors
`v_1,...v_n` then the solution is given by
.. math:: \vec{y} = C_1 e^{r_1 t} \vec{v_1} + C_2 e^{r_2 t} \vec{v_2} +...+ C_n e^{r_n t} \vec{v_n}
where `C_1,C_2,...,C_n` are arbitrary constants.
2. When some eigenvalues are complex then in order to make the solution real,
we take a llinear combination: if `r = a + bi` has an eigenvector
`\vec{v} = \vec{w_1} + i \vec{w_2}` then to obtain real-valued solutions to
the system, replace the complex-valued solutions `e^{rx} \vec{v}`
with real-valued solution `e^{ax} (\vec{w_1} \cos(bx) - \vec{w_2} \sin(bx))`
and for `r = a - bi` replace the solution `e^{-r x} \vec{v}` with
`e^{ax} (\vec{w_1} \sin(bx) + \vec{w_2} \cos(bx))`
3. If some eigenvalues are repeated. Then we get fewer than `n` linearly
independent eigenvectors, we miss some of the solutions and need to
construct the missing ones. We do this via generalized eigenvectors, vectors
which are not eigenvectors but are close enough that we can use to write
down the remaining solutions. For a eigenvalue `r` with eigenvector `\vec{w}`
we obtain `\vec{w_2},...,\vec{w_k}` using
.. math:: (A - r I) . \vec{w_2} = \vec{w}
.. math:: (A - r I) . \vec{w_3} = \vec{w_2}
.. math:: \vdots
.. math:: (A - r I) . \vec{w_k} = \vec{w_{k-1}}
Then the solutions to the system for the eigenspace are `e^{rt} [\vec{w}],
e^{rt} [t \vec{w} + \vec{w_2}], e^{rt} [\frac{t^2}{2} \vec{w} + t \vec{w_2} + \vec{w_3}],
...,e^{rt} [\frac{t^{k-1}}{(k-1)!} \vec{w} + \frac{t^{k-2}}{(k-2)!} \vec{w_2} +...+ t \vec{w_{k-1}}
+ \vec{w_k}]`
So, If `\vec{y_1},...,\vec{y_n}` are `n` solution of obtained from three
categories of `A`, then general solution to the system `\vec{y'} = A . \vec{y}`
.. math:: \vec{y} = C_1 \vec{y_1} + C_2 \vec{y_2} + \cdots + C_n \vec{y_n}
"""
eq = match_['eq']
func = match_['func']
fc = match_['func_coeff']
n = len(eq)
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
constants = numbered_symbols(prefix='C', cls=Symbol, start=1)
M = Matrix(n,n,lambda i,j:-fc[i,func[j],0])
evector = M.eigenvects(simplify=True)
def is_complex(mat, root):
return Matrix(n, 1, lambda i,j: re(mat[i])*cos(im(root)*t) - im(mat[i])*sin(im(root)*t))
def is_complex_conjugate(mat, root):
return Matrix(n, 1, lambda i,j: re(mat[i])*sin(abs(im(root))*t) + im(mat[i])*cos(im(root)*t)*abs(im(root))/im(root))
conjugate_root = []
e_vector = zeros(n,1)
for evects in evector:
if evects[0] not in conjugate_root:
# If number of column of an eigenvector is not equal to the multiplicity
# of its eigenvalue then the legt eigenvectors are calculated
if len(evects[2])!=evects[1]:
var_mat = Matrix(n, 1, lambda i,j: Symbol('x'+str(i)))
Mnew = (M - evects[0]*eye(evects[2][-1].rows))*var_mat
w = [0 for i in range(evects[1])]
w[0] = evects[2][-1]
for r in range(1, evects[1]):
w_ = Mnew - w[r-1]
sol_dict = solve(list(w_), var_mat[1:])
sol_dict[var_mat[0]] = var_mat[0]
for key, value in sol_dict.items():
sol_dict[key] = value.subs(var_mat[0],1)
w[r] = Matrix(n, 1, lambda i,j: sol_dict[var_mat[i]])
evects[2].append(w[r])
for i in range(evects[1]):
C = next(constants)
for j in range(i+1):
if evects[0].has(I):
evects[2][j] = simplify(evects[2][j])
e_vector += C*is_complex(evects[2][j], evects[0])*t**(i-j)*exp(re(evects[0])*t)/factorial(i-j)
C = next(constants)
e_vector += C*is_complex_conjugate(evects[2][j], evects[0])*t**(i-j)*exp(re(evects[0])*t)/factorial(i-j)
else:
e_vector += C*evects[2][j]*t**(i-j)*exp(evects[0]*t)/factorial(i-j)
if evects[0].has(I):
conjugate_root.append(conjugate(evects[0]))
sol = []
for i in range(len(eq)):
sol.append(Eq(func[i],e_vector[i]))
return sol
def sysode_nonlinear_2eq_order1(match_):
func = match_['func']
eq = match_['eq']
fc = match_['func_coeff']
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
if match_['type_of_equation'] == 'type5':
sol = _nonlinear_2eq_order1_type5(func, t, eq)
return sol
x = func[0].func
y = func[1].func
for i in range(2):
eqs = 0
for terms in Add.make_args(eq[i]):
eqs += terms/fc[i,func[i],1]
eq[i] = eqs
if match_['type_of_equation'] == 'type1':
sol = _nonlinear_2eq_order1_type1(x, y, t, eq)
elif match_['type_of_equation'] == 'type2':
sol = _nonlinear_2eq_order1_type2(x, y, t, eq)
elif match_['type_of_equation'] == 'type3':
sol = _nonlinear_2eq_order1_type3(x, y, t, eq)
elif match_['type_of_equation'] == 'type4':
sol = _nonlinear_2eq_order1_type4(x, y, t, eq)
return sol
[docs]def _nonlinear_2eq_order1_type1(x, y, t, eq):
r"""
Equations:
.. math:: x' = x^n F(x,y)
.. math:: y' = g(y) F(x,y)
Solution:
.. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2
where
if `n \neq 1`
.. math:: \varphi = [C_1 + (1-n) \int \frac{1}{g(y)} \,dy]^{\frac{1}{1-n}}
if `n = 1`
.. math:: \varphi = C_1 e^{\int \frac{1}{g(y)} \,dy}
where `C_1` and `C_2` are arbitrary constants.
"""
C1, C2 = get_numbered_constants(eq, num=2)
n = Wild('n', exclude=[x(t),y(t)])
f = Wild('f')
u, v, phi = symbols('u, v, phi', function=True)
r = eq[0].match(diff(x(t),t) - x(t)**n*f)
g = ((diff(y(t),t) - eq[1])/r[f]).subs(y(t),v)
F = r[f].subs(x(t),u).subs(y(t),v)
n = r[n]
if n!=1:
phi = (C1 + (1-n)*Integral(1/g, v))**(1/(1-n))
else:
phi = C1*exp(Integral(1/g, v))
phi = phi.doit()
sol2 = solve(Integral(1/(g*F.subs(u,phi)), v).doit() - t - C2, v)
sol = []
for sols in sol2:
sol.append(Eq(x(t),phi.subs(v, sols)))
sol.append(Eq(y(t), sols))
return sol
[docs]def _nonlinear_2eq_order1_type2(x, y, t, eq):
r"""
Equations:
.. math:: x' = e^{\lambda x} F(x,y)
.. math:: y' = g(y) F(x,y)
Solution:
.. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2
where
if `\lambda \neq 0`
.. math:: \varphi = -\frac{1}{\lambda} log(C_1 - \lambda \int \frac{1}{g(y)} \,dy)
if `\lambda = 0`
.. math:: \varphi = C_1 + \int \frac{1}{g(y)} \,dy
where `C_1` and `C_2` are arbitrary constants.
"""
C1, C2 = get_numbered_constants(eq, num=2)
n = Wild('n', exclude=[x(t),y(t)])
f = Wild('f')
u, v, phi = symbols('u, v, phi', function=True)
r = eq[0].match(diff(x(t),t) - exp(n*x(t))*f)
g = ((diff(y(t),t) - eq[1])/r[f]).subs(y(t),v)
F = r[f].subs(x(t),u).subs(y(t),v)
n = r[n]
if n:
phi = -1/n*log(C1 - n*Integral(1/g, v))
else:
phi = C1 + Integral(1/g, v)
phi = phi.doit()
sol2 = solve(Integral(1/(g*F.subs(u,phi)), v).doit() - t - C2, v)
sol = []
for sols in sol2:
sol.append(Eq(x(t),phi.subs(v, sols)))
sol.append(Eq(y(t), sols))
return sol
[docs]def _nonlinear_2eq_order1_type3(x, y, t, eq):
r"""
Autonomous system of general form
.. math:: x' = F(x,y)
.. math:: y' = G(x,y)
Assuming `y = y(x, C_1)` where `C_1` is an arbitrary constant is the general
solution of the first-order equation
.. math:: F(x,y) y'_x = G(x,y)
Then the general solution of the original system of equations has the form
.. math:: \int \frac{1}{F(x,y(x,C_1))} \,dx = t + C_1
"""
C1, C2, C3, C4 = get_numbered_constants(eq, num=4)
u, v = symbols('u, v', function=True)
f = Wild('f')
g = Wild('g')
r1 = eq[0].match(diff(x(t),t) - f)
r2 = eq[1].match(diff(y(t),t) - g)
F = r1[f].subs(x(t),u).subs(y(t),v)
G = r2[g].subs(x(t),u).subs(y(t),v)
sol2r = dsolve(Eq(diff(v(u),u), G.subs(v,v(u))/F.subs(v,v(u))))
for sol2s in sol2r:
sol1 = solve(Integral(1/F.subs(v, sol2s.rhs), u).doit() - t - C2, u)
sol = []
for sols in sol1:
sol.append(Eq(x(t), sols))
sol.append(Eq(y(t), (sol2s.rhs).subs(u, sols)))
return sol
[docs]def _nonlinear_2eq_order1_type4(x, y, t, eq):
r"""
Equation:
.. math:: x' = f_1(x) g_1(y) \phi(x,y,t)
.. math:: y' = f_2(x) g_2(y) \phi(x,y,t)
First integral:
.. math:: \int \frac{f_2(x)}{f_1(x)} \,dx - \int \frac{g_1(y)}{g_2(y)} \,dy = C
where `C` is an arbitrary constant.
On solving the first integral for `x` (resp., `y` ) and on substituting the
resulting expression into either equation of the original solution, one
arrives at a firs-order equation for determining `y` (resp., `x` ).
"""
C1, C2 = get_numbered_constants(eq, num=2)
u, v = symbols('u, v')
f = Wild('f')
g = Wild('g')
f1 = Wild('f1', exclude=[v,t])
f2 = Wild('f2', exclude=[v,t])
g1 = Wild('g1', exclude=[u,t])
g2 = Wild('g2', exclude=[u,t])
r1 = eq[0].match(diff(x(t),t) - f)
r2 = eq[1].match(diff(y(t),t) - g)
num, den = (
(r1[f].subs(x(t),u).subs(y(t),v))/
(r2[g].subs(x(t),u).subs(y(t),v))).as_numer_denom()
R1 = num.match(f1*g1)
R2 = den.match(f2*g2)
phi = (r1[f].subs(x(t),u).subs(y(t),v))/num
F1 = R1[f1]; F2 = R2[f2]
G1 = R1[g1]; G2 = R2[g2]
sol1r = solve(Integral(F2/F1, u).doit() - Integral(G1/G2,v).doit() - C1, u)
sol2r = solve(Integral(F2/F1, u).doit() - Integral(G1/G2,v).doit() - C1, v)
sol = []
for sols in sol1r:
sol.append(Eq(y(t), dsolve(diff(v(t),t) - F2.subs(u,sols).subs(v,v(t))*G2.subs(v,v(t))*phi.subs(u,sols).subs(v,v(t))).rhs))
for sols in sol2r:
sol.append(Eq(x(t), dsolve(diff(u(t),t) - F1.subs(u,u(t))*G1.subs(v,sols).subs(u,u(t))*phi.subs(v,sols).subs(u,u(t))).rhs))
return set(sol)
[docs]def _nonlinear_2eq_order1_type5(func, t, eq):
r"""
Clairaut system of ODEs
.. math:: x = t x' + F(x',y')
.. math:: y = t y' + G(x',y')
The following are solutions of the system
`(i)` straight lines:
.. math:: x = C_1 t + F(C_1, C_2), y = C_2 t + G(C_1, C_2)
where `C_1` and `C_2` are arbitrary constants;
`(ii)` envelopes of the above lines;
`(iii)` continuously differentiable lines made up from segments of the lines
`(i)` and `(ii)`.
"""
C1, C2 = get_numbered_constants(eq, num=2)
f = Wild('f')
g = Wild('g')
def check_type(x, y):
r1 = eq[0].match(t*diff(x(t),t) - x(t) + f)
r2 = eq[1].match(t*diff(y(t),t) - y(t) + g)
if not (r1 and r2):
r1 = eq[0].match(diff(x(t),t) - x(t)/t + f/t)
r2 = eq[1].match(diff(y(t),t) - y(t)/t + g/t)
if not (r1 and r2):
r1 = (-eq[0]).match(t*diff(x(t),t) - x(t) + f)
r2 = (-eq[1]).match(t*diff(y(t),t) - y(t) + g)
if not (r1 and r2):
r1 = (-eq[0]).match(diff(x(t),t) - x(t)/t + f/t)
r2 = (-eq[1]).match(diff(y(t),t) - y(t)/t + g/t)
return [r1, r2]
for func_ in func:
if isinstance(func_, list):
x = func[0][0].func
y = func[0][1].func
[r1, r2] = check_type(x, y)
if not (r1 and r2):
[r1, r2] = check_type(y, x)
x, y = y, x
x1 = diff(x(t),t); y1 = diff(y(t),t)
return {Eq(x(t), C1*t + r1[f].subs(x1,C1).subs(y1,C2)), Eq(y(t), C2*t + r2[g].subs(x1,C1).subs(y1,C2))}
def sysode_nonlinear_3eq_order1(match_):
x = match_['func'][0].func
y = match_['func'][1].func
z = match_['func'][2].func
eq = match_['eq']
fc = match_['func_coeff']
func = match_['func']
t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0]
if match_['type_of_equation'] == 'type1':
sol = _nonlinear_3eq_order1_type1(x, y, z, t, eq)
if match_['type_of_equation'] == 'type2':
sol = _nonlinear_3eq_order1_type2(x, y, z, t, eq)
if match_['type_of_equation'] == 'type3':
sol = _nonlinear_3eq_order1_type3(x, y, z, t, eq)
if match_['type_of_equation'] == 'type4':
sol = _nonlinear_3eq_order1_type4(x, y, z, t, eq)
if match_['type_of_equation'] == 'type5':
sol = _nonlinear_3eq_order1_type5(x, y, z, t, eq)
return sol
[docs]def _nonlinear_3eq_order1_type1(x, y, z, t, eq):
r"""
Equations:
.. math:: a x' = (b - c) y z, \enspace b y' = (c - a) z x, \enspace c z' = (a - b) x y
First Integrals:
.. math:: a x^{2} + b y^{2} + c z^{2} = C_1
.. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2
where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and
`z` and on substituting the resulting expressions into the first equation of the
system, we arrives at a separable first-order equation on `x`. Similarly doing that
for other two equations, we will arrive at first order equation on `y` and `z` too.
References
==========
-http://eqworld.ipmnet.ru/en/solutions/sysode/sode0401.pdf
"""
C1, C2 = get_numbered_constants(eq, num=2)
u, v, w = symbols('u, v, w')
p = Wild('p', exclude=[x(t), y(t), z(t), t])
q = Wild('q', exclude=[x(t), y(t), z(t), t])
s = Wild('s', exclude=[x(t), y(t), z(t), t])
r = (diff(x(t),t) - eq[0]).match(p*y(t)*z(t))
r.update((diff(y(t),t) - eq[1]).match(q*z(t)*x(t)))
r.update((diff(z(t),t) - eq[2]).match(s*x(t)*y(t)))
n1, d1 = r[p].as_numer_denom()
n2, d2 = r[q].as_numer_denom()
n3, d3 = r[s].as_numer_denom()
val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, d3*u-d3*v-n3*w],[u,v])
vals = [val[v], val[u]]
c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1])
b = vals[0].subs(w,c)
a = vals[1].subs(w,c)
y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b)))
z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c)))
z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c)))
x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a)))
x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a)))
y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b)))
try:
sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x).rhs
except:
sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x, hint='separable_Integral')
try:
sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y).rhs
except:
sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y, hint='separable_Integral')
try:
sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z).rhs
except:
sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z, hint='separable_Integral')
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
[docs]def _nonlinear_3eq_order1_type2(x, y, z, t, eq):
r"""
Equations:
.. math:: a x' = (b - c) y z f(x, y, z, t)
.. math:: b y' = (c - a) z x f(x, y, z, t)
.. math:: c z' = (a - b) x y f(x, y, z, t)
First Integrals:
.. math:: a x^{2} + b y^{2} + c z^{2} = C_1
.. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2
where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and
`z` and on substituting the resulting expressions into the first equation of the
system, we arrives at a first-order differential equations on `x`. Similarly doing
that for other two equations we will arrive at first order equation on `y` and `z`.
References
==========
-http://eqworld.ipmnet.ru/en/solutions/sysode/sode0402.pdf
"""
C1, C2 = get_numbered_constants(eq, num=2)
u, v, w = symbols('u, v, w')
p = Wild('p', exclude=[x(t), y(t), z(t), t])
q = Wild('q', exclude=[x(t), y(t), z(t), t])
s = Wild('s', exclude=[x(t), y(t), z(t), t])
f = Wild('f')
r1 = (diff(x(t),t) - eq[0]).match(y(t)*z(t)*f)
r = collect_const(r1[f]).match(p*f)
r.update(((diff(y(t),t) - eq[1])/r[f]).match(q*z(t)*x(t)))
r.update(((diff(z(t),t) - eq[2])/r[f]).match(s*x(t)*y(t)))
n1, d1 = r[p].as_numer_denom()
n2, d2 = r[q].as_numer_denom()
n3, d3 = r[s].as_numer_denom()
val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, -d3*u+d3*v+n3*w],[u,v])
vals = [val[v], val[u]]
c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1])
a = vals[0].subs(w,c)
b = vals[1].subs(w,c)
y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b)))
z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c)))
z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c)))
x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a)))
x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a)))
y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b)))
try:
sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x*r[f]).rhs
except:
sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x*r[f], hint='separable_Integral')
try:
sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y*r[f]).rhs
except:
sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y*r[f], hint='separable_Integral')
try:
sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z*r[f]).rhs
except:
sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z*r[f], hint='separable_Integral')
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
[docs]def _nonlinear_3eq_order1_type3(x, y, z, t, eq):
r"""
Equations:
.. math:: x' = c F_2 - b F_3, \enspace y' = a F_3 - c F_1, \enspace z' = b F_1 - a F_2
where `F_n = F_n(x, y, z, t)`.
1. First Integral:
.. math:: a x + b y + c z = C_1,
where C is an arbitrary constant.
2. If we assume function `F_n` to be independent of `t`,i.e, `F_n` = `F_n (x, y, z)`
Then, on eliminating `t` and `z` from the first two equation of the system, one
arrives at the first-order equation
.. math:: \frac{dy}{dx} = \frac{a F_3 (x, y, z) - c F_1 (x, y, z)}{c F_2 (x, y, z) -
b F_3 (x, y, z)}
where `z = \frac{1}{c} (C_1 - a x - b y)`
References
==========
-http://eqworld.ipmnet.ru/en/solutions/sysode/sode0404.pdf
"""
C1 = get_numbered_constants(eq, num=1)
u, v, w = symbols('u, v, w')
p = Wild('p', exclude=[x(t), y(t), z(t), t])
q = Wild('q', exclude=[x(t), y(t), z(t), t])
s = Wild('s', exclude=[x(t), y(t), z(t), t])
F1, F2, F3 = symbols('F1, F2, F3', cls=Wild)
r1 = (diff(x(t),t) - eq[0]).match(F2-F3)
r = collect_const(r1[F2]).match(s*F2)
r.update(collect_const(r1[F3]).match(q*F3))
if eq[1].has(r[F2]) and not eq[1].has(r[F3]):
r[F2], r[F3] = r[F3], r[F2]
r[s], r[q] = -r[q], -r[s]
r.update((diff(y(t),t) - eq[1]).match(p*r[F3] - r[s]*F1))
a = r[p]; b = r[q]; c = r[s]
F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w)
F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w)
F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w)
z_xy = (C1-a*u-b*v)/c
y_zx = (C1-a*u-c*w)/b
x_yz = (C1-b*v-c*w)/a
y_x = dsolve(diff(v(u),u) - ((a*F3-c*F1)/(c*F2-b*F3)).subs(w,z_xy).subs(v,v(u))).rhs
z_x = dsolve(diff(w(u),u) - ((b*F1-a*F2)/(c*F2-b*F3)).subs(v,y_zx).subs(w,w(u))).rhs
z_y = dsolve(diff(w(v),v) - ((b*F1-a*F2)/(a*F3-c*F1)).subs(u,x_yz).subs(w,w(v))).rhs
x_y = dsolve(diff(u(v),v) - ((c*F2-b*F3)/(a*F3-c*F1)).subs(w,z_xy).subs(u,u(v))).rhs
y_z = dsolve(diff(v(w),w) - ((a*F3-c*F1)/(b*F1-a*F2)).subs(u,x_yz).subs(v,v(w))).rhs
x_z = dsolve(diff(u(w),w) - ((c*F2-b*F3)/(b*F1-a*F2)).subs(v,y_zx).subs(u,u(w))).rhs
sol1 = dsolve(diff(u(t),t) - (c*F2 - b*F3).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs
sol2 = dsolve(diff(v(t),t) - (a*F3 - c*F1).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs
sol3 = dsolve(diff(w(t),t) - (b*F1 - a*F2).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
[docs]def _nonlinear_3eq_order1_type4(x, y, z, t, eq):
r"""
Equations:
.. math:: x' = c z F_2 - b y F_3, \enspace y' = a x F_3 - c z F_1, \enspace z' = b y F_1 - a x F_2
where `F_n = F_n (x, y, z, t)`
1. First integral:
.. math:: a x^{2} + b y^{2} + c z^{2} = C_1
where `C` is an arbitrary constant.
2. Assuming the function `F_n` is independent of `t`: `F_n = F_n (x, y, z)`. Then on
eliminating `t` and `z` from the first two equations of the system, one arrives at
the first-order equation
.. math:: \frac{dy}{dx} = \frac{a x F_3 (x, y, z) - c z F_1 (x, y, z)}
{c z F_2 (x, y, z) - b y F_3 (x, y, z)}
where `z = \pm \sqrt{\frac{1}{c} (C_1 - a x^{2} - b y^{2})}`
References
==========
-http://eqworld.ipmnet.ru/en/solutions/sysode/sode0405.pdf
"""
C1 = get_numbered_constants(eq, num=1)
u, v, w = symbols('u, v, w')
p = Wild('p', exclude=[x(t), y(t), z(t), t])
q = Wild('q', exclude=[x(t), y(t), z(t), t])
s = Wild('s', exclude=[x(t), y(t), z(t), t])
F1, F2, F3 = symbols('F1, F2, F3', cls=Wild)
r1 = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3)
r = collect_const(r1[F2]).match(s*F2)
r.update(collect_const(r1[F3]).match(q*F3))
if eq[1].has(r[F2]) and not eq[1].has(r[F3]):
r[F2], r[F3] = r[F3], r[F2]
r[s], r[q] = -r[q], -r[s]
r.update((diff(y(t),t) - eq[1]).match(p*x(t)*r[F3] - r[s]*z(t)*F1))
a = r[p]; b = r[q]; c = r[s]
F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w)
F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w)
F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w)
x_yz = sqrt((C1 - b*v**2 - c*w**2)/a)
y_zx = sqrt((C1 - c*w**2 - a*u**2)/b)
z_xy = sqrt((C1 - a*u**2 - b*v**2)/c)
y_x = dsolve(diff(v(u),u) - ((a*u*F3-c*w*F1)/(c*w*F2-b*v*F3)).subs(w,z_xy).subs(v,v(u))).rhs
z_x = dsolve(diff(w(u),u) - ((b*v*F1-a*u*F2)/(c*w*F2-b*v*F3)).subs(v,y_zx).subs(w,w(u))).rhs
z_y = dsolve(diff(w(v),v) - ((b*v*F1-a*u*F2)/(a*u*F3-c*w*F1)).subs(u,x_yz).subs(w,w(v))).rhs
x_y = dsolve(diff(u(v),v) - ((c*w*F2-b*v*F3)/(a*u*F3-c*w*F1)).subs(w,z_xy).subs(u,u(v))).rhs
y_z = dsolve(diff(v(w),w) - ((a*u*F3-c*w*F1)/(b*v*F1-a*u*F2)).subs(u,x_yz).subs(v,v(w))).rhs
x_z = dsolve(diff(u(w),w) - ((c*w*F2-b*v*F3)/(b*v*F1-a*u*F2)).subs(v,y_zx).subs(u,u(w))).rhs
sol1 = dsolve(diff(u(t),t) - (c*w*F2 - b*v*F3).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs
sol2 = dsolve(diff(v(t),t) - (a*u*F3 - c*w*F1).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs
sol3 = dsolve(diff(w(t),t) - (b*v*F1 - a*u*F2).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
[docs]def _nonlinear_3eq_order1_type5(x, y, t, eq):
r"""
.. math:: x' = x (c F_2 - b F_3), \enspace y' = y (a F_3 - c F_1), \enspace z' = z (b F_1 - a F_2)
where `F_n = F_n (x, y, z, t)` and are arbitrary functions.
First Integral:
.. math:: \left|x\right|^{a} \left|y\right|^{b} \left|z\right|^{c} = C_1
where `C` is an arbitrary constant. If the function `F_n` is independent of `t`,
then, by eliminating `t` and `z` from the first two equations of the system, one
arrives at a first-order equation.
References
==========
-http://eqworld.ipmnet.ru/en/solutions/sysode/sode0406.pdf
"""
C1 = get_numbered_constants(eq, num=1)
u, v, w = symbols('u, v, w')
p = Wild('p', exclude=[x(t), y(t), z(t), t])
q = Wild('q', exclude=[x(t), y(t), z(t), t])
s = Wild('s', exclude=[x(t), y(t), z(t), t])
F1, F2, F3 = symbols('F1, F2, F3', cls=Wild)
r1 = eq[0].match(diff(x(t),t) - x(t)*(F2 - F3))
r = collect_const(r1[F2]).match(s*F2)
r.update(collect_const(r1[F3]).match(q*F3))
if eq[1].has(r[F2]) and not eq[1].has(r[F3]):
r[F2], r[F3] = r[F3], r[F2]
r[s], r[q] = -r[q], -r[s]
r.update((diff(y(t),t) - eq[1]).match(y(t)*(a*r[F3] - r[c]*F1)))
a = r[p]; b = r[q]; c = r[s]
F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w)
F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w)
F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w)
x_yz = (C1*v**-b*w**-c)**-a
y_zx = (C1*w**-c*u**-a)**-b
z_xy = (C1*u**-a*v**-b)**-c
y_x = dsolve(diff(v(u),u) - ((v*(a*F3-c*F1))/(u*(c*F2-b*F3))).subs(w,z_xy).subs(v,v(u))).rhs
z_x = dsolve(diff(w(u),u) - ((w*(b*F1-a*F2))/(u*(c*F2-b*F3))).subs(v,y_zx).subs(w,w(u))).rhs
z_y = dsolve(diff(w(v),v) - ((w*(b*F1-a*F2))/(v*(a*F3-c*F1))).subs(u,x_yz).subs(w,w(v))).rhs
x_y = dsolve(diff(u(v),v) - ((u*(c*F2-b*F3))/(v*(a*F3-c*F1))).subs(w,z_xy).subs(u,u(v))).rhs
y_z = dsolve(diff(v(w),w) - ((v*(a*F3-c*F1))/(w*(b*F1-a*F2))).subs(u,x_yz).subs(v,v(w))).rhs
x_z = dsolve(diff(u(w),w) - ((u*(c*F2-b*F3))/(w*(b*F1-a*F2))).subs(v,y_zx).subs(u,u(w))).rhs
sol1 = dsolve(diff(u(t),t) - (u*(c*F2-b*F3)).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs
sol2 = dsolve(diff(v(t),t) - (v*(a*F3-c*F1)).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs
sol3 = dsolve(diff(w(t),t) - (w*(b*F1-a*F2)).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs
return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
