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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, 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))
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
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, denum = ((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 = denum.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(set([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(set([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:
sol_func = list(sol.atoms(AppliedUndef))[0]
if not (sol.lhs == sol_func and not sol.rhs.has(sol_func)) and not (\
sol.rhs == sol_func and not sol.lhs.has(sol_func)):
solved = solve(sol, sol_func)
if not solved:
raise NotImplementedError
dictsol[sol_func] = solved
if sol.lhs == sol_func:
dictsol[sol_func] = sol.rhs
if sol.rhs == sol_func:
dictsol[sol_func] = sol.lhs
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)
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 RootOf is implemented in solve(), we will want to return a RootOf
# 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:
eqsol = solve(eq, func, force=True)
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
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)
x = func.args[0]
s = True
testnum = 0
if order == 'auto':
order = ode_order(ode, func)
if solve_for_func and not (
sol.lhs == func and not sol.rhs.has(func)) and not (
sol.rhs == func and not sol.lhs.has(func)):
try:
solved = solve(sol, func)
if not solved:
raise NotImplementedError
except NotImplementedError:
pass
else:
if len(solved) == 1:
result = checkodesol(ode, Eq(func, solved[0]),
order=order, solve_for_func=False)
else:
result = checkodesol(ode, [Eq(func, t) for t in solved],
order=order, solve_for_func=False)
return result
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)
elif sol.rhs == func:
s = sub_func_doit(ode_diff, func, sol.lhs)
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)
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 RootOf
# is implemented in solve(). Probably a RootOf 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 = dict([(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)
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
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.
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)
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))
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
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?
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
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)
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