# Source code for sympy.solvers.ode

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
: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.

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.

**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.
- 2nd order Liouville differential equations.
- 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:C.Integral()
<sympy.core.C.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.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 collections import defaultdict

from sympy.core import Add, C, S, Mul, Pow, oo
from sympy.core.compatibility import ordered, iterable, is_sequence, set_union
from sympy.utilities.exceptions import SymPyDeprecationWarning
from sympy.core.exprtools import factor_terms, gcd_terms
from sympy.core.function import (Function, Derivative, AppliedUndef, diff,
expand, expand_mul)
from sympy.core.multidimensional import vectorize
from sympy.core.numbers import Rational, NaN
from sympy.core.relational import Equality, Eq
from sympy.core.symbol import Symbol, Wild, Dummy, symbols
from sympy.core.sympify import sympify

from sympy.functions import cos, exp, im, log, re, sin, tan, sqrt, sign, Piecewise
from sympy.matrices import wronskian
from sympy.polys import Poly, RootOf, terms_gcd, PolynomialError
from sympy.polys.polytools import cancel, degree
from sympy.series import Order
from sympy.simplify import collect, logcombine, powsimp, separatevars, \
simplify, trigsimp, denom, fraction, posify
from sympy.simplify.simplify import _mexpand
from sympy.solvers import solve

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",
"nth_linear_constant_coeff_homogeneous",
"nth_linear_euler_eq_homogeneous",
"nth_linear_constant_coeff_undetermined_coefficients",
"nth_linear_constant_coeff_variation_of_parameters",
"Liouville",
"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",
"Liouville_Integral",
)

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 = C.Dummy('u')
repu[u] = d.subs(func, new).doit()
reps[d] = u

return eq.subs(reps).subs(func, new).subs(repu)

[docs]def dsolve(eq, func=None, hint="default", simplify=True, **kwargs): r""" Solves any (supported) kind of ordinary differential equation. **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. **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. Examples ======== >>> from sympy import Function, dsolve, Eq, Derivative, sin, cos >>> from sympy.abc import x >>> f = Function('f') >>> dsolve(Derivative(f(x), x, x) + 9*f(x), f(x)) 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='separable_reduced') f(x) == C1/(C2*x - 1) >>> dsolve(eq, hint='1st_exact') [f(x) == -acos(C1/cos(x)) + 2*pi, f(x) == acos(C1/cos(x))] >>> dsolve(eq, hint='almost_linear') [f(x) == -acos(-sqrt(C1/cos(x)**2)) + 2*pi, f(x) == -acos(sqrt(C1/cos(x)**2)) + 2*pi, f(x) == acos(-sqrt(C1/cos(x)**2)), f(x) == acos(sqrt(C1/cos(x)**2))] >>> dsolve(eq, hint='best') f(x) == C1/(C2*x - 1) """ 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, type='ode', **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, detail: failed_hints[hint] = detail else: retdict[hint] = rv func = hints[hint]['func'] retdict['best'] = min(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): 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 rv = odesimp(solvefunc(eq, func, order, match), func, order, hint) return rv else: # We still want to integrate (you can disable it separately with the hint) match['simplify'] = False # Some hints can take advantage of this option rv = _handle_Integral(solvefunc(eq, func, order, match), func, order, hint) return rv
[docs]def classify_ode(eq, func=None, dict=False, **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', '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) from sympy import expand 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') if isinstance(eq, Equality): if eq.rhs != 0: return classify_ode(eq.lhs - eq.rhs, func, 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]) eq = expand(eq) # 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 != 1: 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 ( 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 1577 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) ## 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(C.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(C.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 1525 and issue 1620. pass # 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)): 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 if 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 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 # Euler equation case (a_i * x**i for all i) def _test_term(coeff, order): r""" Linear Euler ODEs have the form K*x**order*diff(y(x),x,order), 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. """ assert order >= 0 if coeff == 0: return True if order == 0: if x in coeff.free_symbols: return False return f(x) not in coeff.atoms() 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 # 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)
@vectorize(0)
[docs]def odesimp(eq, func, order, 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, ... 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 = Symbol('C1') # First, integrate if the hint allows it. eq = _handle_Integral(eq, func, order, hint) assert isinstance(eq, 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, x, 2*order) # 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 global collectterms 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 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 costants 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): eqi = constantsimp(eqi, x, 2*order) eq[i] = constant_renumber(eqi, 'C', 1, 2*order) # If there is only 1 solution, return it; # otherwise return the list of solutions. if len(eq) == 1: eq = eq[0] return eq
[docs]def checkodesol(ode, sol, func=None, order='auto', solve_for_func=True): r""" Substitutes sol into ode and checks that the result is 0. This only works when func is one function, like f(x). sol can be a single solution or a list of solutions. Each solution may be an :py:class:~sympy.core.relational.Equality that the solution satisfies, e.g. Eq(f(x), C1), Eq(f(x) + C1, 0); or simply an :py:class:~sympy.core.expr.Expr, e.g. f(x) - C1. In most cases it will not be necessary to explicitly identify the function, but if the function cannot be inferred from the original equation it can be supplied through the func argument. If a sequence of solutions is passed, the same sort of container will be used to return the result for each solution. It tries the following methods, in order, until it finds zero equivalence: 1. Substitute the solution for f in the original equation. This only works if ode is solved for f. It will attempt to solve it first unless solve_for_func == False. 2. Take n derivatives of the solution, where n is the order of ode, and check to see if that is equal to the solution. This only works on exact ODEs. 3. Take the 1st, 2nd, ..., n\th derivatives of the solution, each time solving for the derivative of f of that order (this will always be possible because f is a linear operator). Then back substitute each derivative into ode in reverse order. This function returns a tuple. The first item in the tuple is True if the substitution results in 0, and False otherwise. The second item in the tuple is what the substitution results in. It should always be 0 if the first item is True. Note that sometimes this function will False, but with an expression that is identically equal to 0, instead of returning True. This is because :py:meth:~sympy.simplify.simplify.simplify cannot reduce the expression to 0. If an expression returned by this function vanishes identically, then sol really is a solution to ode. If this function seems to hang, it is probably because of a hard simplification. To use this function to test, test the first item of the tuple. Examples ======== >>> from sympy import Eq, Function, checkodesol, symbols >>> x, C1 = symbols('x,C1') >>> f = Function('f') >>> checkodesol(f(x).diff(x), Eq(f(x), C1)) (True, 0) >>> assert checkodesol(f(x).diff(x), C1)[0] >>> assert not checkodesol(f(x).diff(x), x)[0] >>> checkodesol(f(x).diff(x, 2), x**2) (False, 2) """ if not isinstance(ode, Equality): ode = Eq(ode, 0) if func is None: try: _, func = _preprocess(ode.lhs) except ValueError: funcs = [s.atoms(AppliedUndef) for s in ( sol if is_sequence(sol, set) else [sol])] funcs = reduce(set.union, funcs, set()) if len(funcs) != 1: raise ValueError( 'must pass func arg to checkodesol for this case.') func = funcs.pop() # ========== deprecation handling # After the deprecation period this handling section becomes: # ---------- # if not is_unfunc(func) or len(func.args) != 1: # raise ValueError( # "func must be a function of one variable, not %s" % func) # ---------- # assume, during deprecation, that sol and func are reversed if isinstance(sol, AppliedUndef) and len(sol.args) == 1: if isinstance(func, AppliedUndef) and len(func.args) == 1: msg = "If you really do want sol to be just %s, use Eq(%s, 0) " % \ (sol, sol) + "instead." else: msg = "" SymPyDeprecationWarning(msg, feature="The order of the " "arguments sol and func to checkodesol()", useinstead="checkodesol(ode, sol, func)", issue=3384, ).warn() sol, func = func, sol elif not (isinstance(func, AppliedUndef) and len(func.args) == 1): from sympy.utilities.misc import filldedent raise ValueError(filldedent(''' func (or sol, during deprecation) must be a function of one variable. Got sol = %s, func = %s''' % (sol, func))) # ========== end of deprecation handling if is_sequence(sol, set): return type(sol)(map(lambda i: checkodesol(ode, i, order=order, solve_for_func=solve_for_func), 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): if ode_or_bool: lhs = rhs = S.Zero else: lhs = ode_or_bool.lhs rhs = ode_or_bool.rhs # No sense in overworking simplify -- just prove that the # numerator goes to zero num = trigsimp((lhs - rhs).as_numer_denom()[0]) # since solutions are obtained using force=True we test # using the same level of assumptions ## replace function with dummy so assumptions will work _func = Dummy('func') num = num.subs(func, _func) ## posify the expression num, reps = posify(num) s = simplify(num).xreplace(reps).xreplace({_func: func}) testnum += 1 else: break if not s: return (True, s) elif s is True: # The code above never was able to change s raise NotImplementedError("Unable to test if " + str(sol) + " is a solution to " + str(ode) + ".") else: return (False, s)
[docs]def ode_sol_simplicity(sol, func, trysolving=True): r""" Returns an extended integer representing how simple a solution to an ODE is. The following things are considered, in order from most simple to least: - sol is solved for func. - sol is not solved for func, but can be if passed to solve (e.g., a solution returned by dsolve(ode, func, simplify=False). - If sol is not solved for func, then base the result on the length of sol, as computed by len(str(sol)). - If sol has any unevaluated :py:class:~sympy.integrals.Integral\s, this will automatically be considered less simple than any of the above. This function returns an integer such that if solution A is simpler than solution B by above metric, then ode_sol_simplicity(sola, func) < ode_sol_simplicity(solb, func). Currently, the following are the numbers returned, but if the heuristic is ever improved, this may change. Only the ordering is guaranteed. +----------------------------------------------+-------------------+ | Simplicity | Return | +==============================================+===================+ | sol solved for func | -2 | +----------------------------------------------+-------------------+ | sol not solved for func but can be | -1 | +----------------------------------------------+-------------------+ | sol is not solved nor solvable for | len(str(sol)) | | func | | +----------------------------------------------+-------------------+ | sol contains an | oo | | :py:class:~sympy.integrals.Integral | | +----------------------------------------------+-------------------+ oo here means the SymPy infinity, which should compare greater than any integer. If you already know :py:meth:~sympy.solvers.solvers.solve cannot solve sol, you can use trysolving=False to skip that step, which is the only potentially slow step. For example, :py:meth:~sympy.solvers.ode.dsolve with the simplify=False flag should do this. If sol is a list of solutions, if the worst solution in the list returns oo it returns that, otherwise it returns len(str(sol)), that is, the length of the string representation of the whole list. Examples ======== This function is designed to be passed to min as the key argument, such as min(listofsolutions, key=lambda i: ode_sol_simplicity(i, f(x))). >>> from sympy import symbols, Function, Eq, tan, cos, sqrt, Integral >>> from sympy.solvers.ode import ode_sol_simplicity >>> x, C1, C2 = symbols('x, C1, C2') >>> f = Function('f') >>> ode_sol_simplicity(Eq(f(x), C1*x**2), f(x)) -2 >>> ode_sol_simplicity(Eq(x**2 + f(x), C1), f(x)) -1 >>> ode_sol_simplicity(Eq(f(x), C1*Integral(2*x, x)), f(x)) oo >>> eq1 = Eq(f(x)/tan(f(x)/(2*x)), C1) >>> eq2 = Eq(f(x)/tan(f(x)/(2*x) + f(x)), C2) >>> [ode_sol_simplicity(eq, f(x)) for eq in [eq1, eq2]] [26, 33] >>> min([eq1, eq2], key=lambda i: ode_sol_simplicity(i, f(x))) 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(C.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))
@vectorize(0)
[docs]def constantsimp(expr, independentsymbol, endnumber, startnumber=1, symbolname='C'): 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 in to 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 is done with limited assistance: terms of :py:class:~sympy.core.add.Add\s are collected to try join constants and powers with exponents that are :py:class:~sympy.core.add.Add\s are expanded so e^x (C_1 \cos(x) + C_2 \cos(x)) will simplify to e^x C_1 \cos(x) and 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, x, 3) C1*x >>> constantsimp(C1 + 2 + x + y, x, 3) C1 + x >>> constantsimp(C1*C2 + 2 + x + y + C3*x, x, 3) 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. if type(symbolname) is tuple: x, endnumber, startnumber, constantsymbols = symbolname else: constantsymbols = symbols( symbolname + '%i:%i' % (startnumber, endnumber + 1)) x = independentsymbol con_set = set(constantsymbols) ARGS = None, None, None, ( x, endnumber, startnumber, constantsymbols) if isinstance(expr, Equality): # For now, only treat the special case where one side of the equation # is a constant if expr.lhs in con_set: return Eq(expr.lhs, constantsimp(expr.rhs + expr.lhs, *ARGS) - expr.lhs) # this could break if expr.lhs is absorbed into another constant, # but for now, the only solutions that return Eq's with a constant # on one side are first order. At any rate, it will still be # technically correct. The expression will just have too many # constants in it elif expr.rhs in con_set: return Eq(constantsimp(expr.lhs + expr.rhs, *ARGS) - expr.rhs, expr.rhs) else: return Eq(constantsimp(expr.lhs, *ARGS), constantsimp(expr.rhs, *ARGS)) if not hasattr(expr, 'has') or not expr.has(*constantsymbols): return expr else: # ================ pre-processing ================ def _take(i): # return the lowest numbered constant symbol that appears in i # else return i c = i.free_symbols & con_set if c: return min(c) return i if not (expr.has(x) and x in expr.free_symbols): return constantsymbols[0] # collect terms to get constants together new_expr = terms_gcd(expr, clear=False, deep=True, expand=False) if new_expr.is_Mul: # don't let C1*exp(x) + C2*exp(2*x) become exp(x)*(C1 + C2*exp(x)) 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 expr = new_expr # don't allow a number to be factored out of an expression # that has no denominator if expr.is_Mul: h, t = expr.as_coeff_Mul() if h != 1 and (t.is_Add or denom(t) == 1): args = list(Mul.make_args(t)) for i, a in enumerate(args): if a.is_Add: args[i] = h*a expr = Mul._from_args(args) break # let numbers absorb into constants of an Add, perhaps # in the base of a power, if all its terms have a constant # symbol in them, e.g. sqrt(2)*(C1 + C2*x) -> C1 + C2*x if expr.is_Mul: d = sift(expr.args, lambda m: m.is_number is True) num = d[True] other = d[False] if num: for o in other: b, e = o.as_base_exp() if b.is_Add and \ all(a.args_cnc(cset=True, warn=False)[0] & con_set for a in b.args): expr = sign(Mul(*num))*Mul._from_args(other) break if expr.is_Mul: # check again that it's still a Mul i, d = expr.as_independent(x, strict=True) newi = _take(i) if newi != i: expr = newi*d elif expr.is_Add: i, d = expr.as_independent(x, strict=True) expr = _take(i) + d if expr.is_Add: terms = {} for ai in expr.args: i, d = ai.as_independent(x, strict=True, as_Add=False) terms.setdefault(d, []).append(i) expr = Add(*[k*Add(*v) for k, v in terms.items()]) # handle powers like exp(C0 + g(x)) -> C0*exp(g(x)) pows = [p for p in expr.atoms(C.Function, C.Pow) if (p.is_Pow or p.func is exp) and p.exp.is_Add and p.exp.as_independent(x, strict=True)[1]] if pows: reps = [] for p in pows: b, e = p.as_base_exp() ei, ed = e.as_independent(x, strict=True) e = _take(ei) if e != ei or e in constantsymbols: reps.append((p, e*b**ed)) expr = expr.xreplace(dict(reps)) # a C1*C2 may have been introduced and the code below won't # handle that so handle it now: once to handle the C1*C2 # and once to handle any C0*f(x) + C0*f(x) for _ in range(2): muls = [m for m in expr.atoms(Mul) if m.has(*constantsymbols)] reps = [] for m in muls: i, d = m.as_independent(x, strict=True) newi = _take(i) if newi != i: reps.append((m, _take(i)*d)) expr = expr.xreplace(dict(reps)) # ================ end of pre-processing ================ newargs = [] hasconst = False isPowExp = False reeval = False for i in expr.args: if i not in constantsymbols: newargs.append(i) else: newconst = i hasconst = True if expr.is_Pow and i == expr.exp: isPowExp = True for i in range(len(newargs)): isimp = constantsimp(newargs[i], *ARGS) if isimp in constantsymbols: reeval = True hasconst = True newconst = isimp if expr.is_Pow and i == 1: isPowExp = True newargs[i] = isimp if hasconst: newargs = [i for i in newargs if i.has(x)] if isPowExp: newargs = newargs + [newconst] # Order matters in this case else: newargs = [newconst] + newargs if expr.is_Pow and len(newargs) == 1: newargs.append(S.One) if expr.is_Function: if (len(newargs) == 0 or hasconst and len(newargs) == 1): return newconst else: newfuncargs = [constantsimp(t, *ARGS) for t in expr.args] return expr.func(*newfuncargs) else: newexpr = expr.func(*newargs) if reeval: return constantsimp(newexpr, *ARGS) else: return newexpr
[docs]def constant_renumber(expr, symbolname, startnumber, endnumber): r""" Renumber arbitrary constants in expr to have numbers 1 through N where N is endnumber - startnumber + 1 at most. 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)(map(lambda i: constant_renumber(i, symbolname=symbolname, startnumber=startnumber, endnumber=endnumber), expr)) global newstartnumber newstartnumber = 1 def _constant_renumber(expr, symbolname, startnumber, endnumber): r""" We need to have an internal recursive function so that newstartnumber maintains its values throughout recursive calls. """ constantsymbols = [Symbol( symbolname + "%d" % t) for t in range(startnumber, endnumber + 1)] global newstartnumber if isinstance(expr, Equality): return Eq( _constant_renumber( expr.lhs, symbolname, startnumber, endnumber), _constant_renumber( expr.rhs, symbolname, startnumber, endnumber)) 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: # Renumbering happens here newconst = Symbol(symbolname + str(newstartnumber)) newstartnumber += 1 return newconst else: from sympy.core.containers import Tuple if expr.is_Function or expr.is_Pow or isinstance(expr, Tuple): return expr.func( *[_constant_renumber(x, symbolname, startnumber, endnumber) for x in expr.args]) else: sortedargs = list(expr.args) # 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] sortedargs.sort( key=lambda arg: default_sort_key(arg.subs(C_1))) return expr.func( *[_constant_renumber(x, symbolname, startnumber, endnumber) for x in sortedargs]) return _constant_renumber(expr, symbolname, startnumber, endnumber)
def _handle_Integral(expr, func, order, hint): r""" Converts a solution with Integrals in it into an actual solution. For most hints, this simply runs expr.doit(). """ global y x = func.args[0] f = func.func if hint == "1st_exact": sol = (expr.doit()).subs(y, f(x)) del y elif hint == "1st_exact_Integral": sol = expr.subs(y, f(x)) del y elif hint == "nth_linear_constant_coeff_homogeneous": sol = expr elif not hint.endswith("_Integral"): sol = expr.doit() else: sol = expr return sol # FIXME: replace the general solution in the docstring with # dsolve(equation, hint='1st_exact_Integral'). You will need to be able # to have assumptions on P and Q that dP/dy = dQ/dx.
[docs]def ode_1st_exact(eq, func, order, match): r""" Solves 1st order exact ordinary differential equations. A 1st order differential equation is called exact if it is the total differential of a function. That is, the differential equation .. math:: P(x, y) \,\partial{}x + Q(x, y) \,\partial{}y = 0 is exact if there is some function F(x, y) such that P(x, y) = \partial{}F/\partial{}x and Q(x, y) = \partial{}F/\partial{}y. It can be shown that a necessary and sufficient condition for a first order ODE to be exact is that \partial{}P/\partial{}y = \partial{}Q/\partial{}x. Then, the solution will be as given below:: >>> from sympy import Function, Eq, Integral, symbols, pprint >>> x, y, t, x0, y0, C1= symbols('x,y,t,x0,y0,C1') >>> P, Q, F= map(Function, ['P', 'Q', 'F']) >>> pprint(Eq(Eq(F(x, y), Integral(P(t, y), (t, x0, x)) + ... Integral(Q(x0, t), (t, y0, y))), C1)) x y / / | | F(x, y) = | P(t, y) dt + | Q(x0, t) dt = C1 | | / / x0 y0 Where the first partials of P and Q exist and are continuous in a simply connected region. A note: SymPy currently has no way to represent inert substitution on an expression, so the hint 1st_exact_Integral will return an integral with dy. This is supposed to represent the function that you are solving for. Examples ======== >>> from sympy import Function, dsolve, cos, sin >>> from sympy.abc import x >>> f = Function('f') >>> dsolve(cos(f(x)) - (x*sin(f(x)) - f(x)**2)*f(x).diff(x), ... f(x), hint='1st_exact') 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 = Symbol('C1') # 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 = C.Integral(d, x) + C.Integral((e - (C.Integral(d, x).diff(y))), y) return Eq(sol, C1)
[docs]def ode_1st_homogeneous_coeff_best(eq, func, order, match): r""" Returns the best solution to an ODE from the two hints 1st_homogeneous_coeff_subs_dep_div_indep and 1st_homogeneous_coeff_subs_indep_div_dep. This is as determined by :py:meth:~ode_sol_simplicity. See the :py:meth:~ode_1st_homogeneous_coeff_subs_indep_div_dep and :py:meth:~ode_1st_homogeneous_coeff_subs_dep_div_indep docstrings for more information on these hints. Note that there is no 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: sol1 = odesimp( sol1, func, order, "1st_homogeneous_coeff_subs_indep_div_dep") sol2 = odesimp( sol2, func, order, "1st_homogeneous_coeff_subs_dep_div_indep") return min([sol1, sol2], key=lambda x: ode_sol_simplicity(x, func, trysolving=not simplify))
[docs]def ode_1st_homogeneous_coeff_subs_dep_div_indep(eq, func, order, match): r""" Solves a 1st order differential equation with homogeneous coefficients using the substitution u_1 = \frac{\text{<dependent variable>}}{\text{<independent variable>}}. This is a differential equation .. math:: P(x, y) + Q(x, y) dy/dx = 0 such that P and Q are homogeneous and of the same order. A function F(x, y) is homogeneous of order n if F(x t, y t) = t^n F(x, y). Equivalently, F(x, y) can be rewritten as G(y/x) or H(x/y). See also the docstring of :py:meth:~sympy.solvers.ode.homogeneous_order. If the coefficients P and Q in the differential equation above are homogeneous functions of the same order, then it can be shown that the substitution y = u_1 x (i.e. u_1 = y/x) will turn the differential equation into an equation separable in the variables x and u. If h(u_1) is the function that results from making the substitution u_1 = f(x)/x on P(x, f(x)) and g(u_2) is the function that results from the substitution on Q(x, f(x)) in the differential equation P(x, f(x)) + Q(x, f(x)) f'(x) = 0, then the general solution is:: >>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f, g, h = map(Function, ['f', 'g', 'h']) >>> genform = g(f(x)/x) + h(f(x)/x)*f(x).diff(x) >>> pprint(genform) /f(x)\ /f(x)\ d g|----| + h|----|*--(f(x)) \ x / \ x / dx >>> pprint(dsolve(genform, f(x), ... hint='1st_homogeneous_coeff_subs_dep_div_indep_Integral')) f(x) ---- x / | | -h(u1) log(x) = C1 + | ---------------- d(u1) | u1*h(u1) + g(u1) | / Where u_1 h(u_1) + g(u_1) \ne 0 and x \ne 0. See also the docstrings of :py:meth:~ode_1st_homogeneous_coeff_best and :py:meth:~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 = Symbol('C1') xarg = match.get('xarg', 0) yarg = match.get('yarg', 0) int = C.Integral( (-r[r['e']]/(r[r['d']] + u1*r[r['e']])).subs({x: 1, r['y']: u1}), (u1, None, f(x)/x)) sol = logcombine(Eq(log(x), int + log(C1)), force=True) sol = sol.subs(f(x), u).subs(((u, u - yarg), (x, x - xarg), (u, f(x)))) return sol
[docs]def ode_1st_homogeneous_coeff_subs_indep_div_dep(eq, func, order, match): r""" Solves a 1st order differential equation with homogeneous coefficients using the substitution u_2 = \frac{\text{<independent variable>}}{\text{<dependent variable>}}. This is a differential equation .. math:: P(x, y) + Q(x, y) dy/dx = 0 such that P and Q are homogeneous and of the same order. A function F(x, y) is homogeneous of order n if F(x t, y t) = t^n F(x, y). Equivalently, F(x, y) can be rewritten as G(y/x) or H(x/y). See also the docstring of :py:meth:~sympy.solvers.ode.homogeneous_order. If the coefficients P and Q in the differential equation above are homogeneous functions of the same order, then it can be shown that the substitution x = u_2 y (i.e. u_2 = x/y) will turn the differential equation into an equation separable in the variables y and u_2. If h(u_2) is the function that results from making the substitution u_2 = x/f(x) on P(x, f(x)) and g(u_2) is the function that results from the substitution on Q(x, f(x)) in the differential equation P(x, f(x)) + Q(x, f(x)) f'(x) = 0, then the general solution is: >>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f, g, h = map(Function, ['f', 'g', 'h']) >>> genform = g(x/f(x)) + h(x/f(x))*f(x).diff(x) >>> pprint(genform) / x \ / x \ d g|----| + h|----|*--(f(x)) \f(x)/ \f(x)/ dx >>> pprint(dsolve(genform, f(x), ... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral')) x ---- f(x) / | | -g(u2) | ---------------- d(u2) | u2*g(u2) + h(u2) | / <BLANKLINE> f(x) = C1*e Where u_2 g(u_2) + h(u_2) \ne 0 and f(x) \ne 0. See also the docstrings of :py:meth:~sympy.solvers.ode.ode_1st_homogeneous_coeff_best and :py:meth:~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 = Symbol('C1') 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 = C.Integral( simplify( (-r[r['d']]/(r[r['e']] + u2*r[r['d']])).subs({x: u2, r['y']: 1})), (u2, None, x/f(x))) sol = logcombine(Eq(log(f(x)), int + log(C1)), force=True) sol = sol.subs(f(x), u).subs(((u, u - yarg), (x, x - xarg), (u, f(x)))) return sol # XXX: Should this function maybe go somewhere else?
[docs]def homogeneous_order(eq, *symbols): r""" Returns the order n if g is homogeneous and None if it is not homogeneous. Determines if a function is homogeneous and if so of what order. A function f(x, y, \cdots) is homogeneous of order n if f(t x, t y, \cdots) = t^n f(x, y, \cdots). If the function is of two variables, F(x, y), then f being homogeneous of any order is equivalent to being able to rewrite F(x, y) as G(x/y) or H(y/x). This fact is used to solve 1st order ordinary differential equations whose coefficients are homogeneous of the same order (see the docstrings of :py:meth:~solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep and :py:meth:~solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep). Symbols can be functions, but every argument of the function must be a symbol, and the arguments of the function that appear in the expression must match those given in the list of symbols. If a declared function appears with different arguments than given in the list of symbols, None is returned. Examples ======== >>> from sympy import Function, homogeneous_order, sqrt >>> from sympy.abc import x, y >>> f = Function('f') >>> homogeneous_order(f(x), f(x)) is None True >>> homogeneous_order(f(x,y), f(y, x), x, y) is None True >>> homogeneous_order(f(x), f(x), x) 1 >>> homogeneous_order(x**2*f(x)/sqrt(x**2+f(x)**2), x, f(x)) 2 >>> homogeneous_order(x**2+f(x), x, f(x)) is None True """ from sympy.simplify.simplify import separatevars 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 = dum.next() eq = eq.subs(i, dummyvar) symset.remove(i) newsyms.add(dummyvar) symset.update(newsyms) if not eq.free_symbols & symset: return None # assuming order of a nested function can only be equal to zero if isinstance(eq, Function): return None if homogeneous_order( eq.args[0], *tuple(symset)) != 0 else S.Zero # make the replacement of x with x*t and see if t can be factored out t = Dummy('t', positive=True) # It is sufficient that t > 0 eqs = separatevars(eq.subs([(i, t*i) for i in symset]), [t], dict=True)[t] if eqs is S.One: return S.Zero # there was no term with only t i, d = eqs.as_independent(t, as_Add=False) b, e = d.as_base_exp() if b == t: return e
[docs]def ode_1st_linear(eq, func, order, match): r""" Solves 1st order linear differential equations. These are differential equations of the form .. math:: dy/dx + P(x) y = Q(x)\text{.} These kinds of differential equations can be solved in a general way. The integrating factor e^{\int P(x) \,dx} will turn the equation into a separable equation. The general solution is:: >>> from sympy import Function, dsolve, Eq, pprint, diff, sin >>> from sympy.abc import x >>> f, P, Q = map(Function, ['f', 'P', 'Q']) >>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x)) >>> pprint(genform) d P(x)*f(x) + --(f(x)) = Q(x) dx >>> pprint(dsolve(genform, f(x), hint='1st_linear_Integral')) / / \ | | | | | / | / | | | | | | | | P(x) dx | - | P(x) dx | | | | | | | / | / f(x) = |C1 + | Q(x)*e dx|*e | | | \ / / Examples ======== >>> f = Function('f') >>> pprint(dsolve(Eq(x*diff(f(x), x) - f(x), x**2*sin(x)), ... f(x), '1st_linear')) f(x) = x*(C1 - cos(x)) References ========== - http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 92 # indirect doctest """ x = func.args[0] f = func.func r = match # a*diff(f(x),x) + b*f(x) + c C1 = Symbol('C1') t = exp(C.Integral(r[r['b']]/r[r['a']], x)) tt = C.Integral(t*(-r[r['c']]/r[r['a']]), x) f = match.get('u', f(x)) # take almost-linear u if present, else f(x) return Eq(f, (tt + C1)/t)
[docs]def ode_Bernoulli(eq, func, order, match): r""" Solves Bernoulli differential equations. These are equations of the form .. math:: dy/dx + P(x) y = Q(x) y^n\text{, }n \ne 1\text{.} The substitution w = 1/y^{1-n} will transform an equation of this form into one that is linear (see the docstring of :py:meth:~sympy.solvers.ode.ode_1st_linear). The general solution is:: >>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x, n >>> f, P, Q = map(Function, ['f', 'P', 'Q']) >>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)**n) >>> pprint(genform) d n P(x)*f(x) + --(f(x)) = Q(x)*f (x) dx >>> pprint(dsolve(genform, f(x), hint='Bernoulli_Integral')) #doctest: +SKIP 1 ---- 1 - n // / \ \ || | | | || | / | / | || | | | | | || | (1 - n)* | P(x) dx | (-1 + n)* | P(x) dx| || | | | | | || | / | / | f(x) = ||C1 + (-1 + n)* | -Q(x)*e dx|*e | || | | | \\ / / / Note that the equation is separable when n = 1 (see the docstring of :py:meth:~sympy.solvers.ode.ode_separable). >>> pprint(dsolve(Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)), f(x), ... hint='separable_Integral')) f(x) / | / | 1 | | - dy = C1 + | (-P(x) + Q(x)) dx | y | | / / Examples ======== >>> from sympy import Function, dsolve, Eq, pprint, log >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(Eq(x*f(x).diff(x) + f(x), log(x)*f(x)**2), ... f(x), hint='Bernoulli')) 1 f(x) = ------------------- / log(x) 1\ x*|C1 + ------ + -| \ x x/ References ========== - http://en.wikipedia.org/wiki/Bernoulli_differential_equation - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 95 # indirect doctest """ x = func.args[0] f = func.func r = match # a*diff(f(x),x) + b*f(x) + c*f(x)**n, n != 1 C1 = Symbol('C1') t = exp((1 - r[r['n']])*C.Integral(r[r['b']]/r[r['a']], x)) tt = (r[r['n']] - 1)*C.Integral(t*r[r['c']]/r[r['a']], x) return Eq(f(x), ((tt + C1)/t)**(1/(1 - r[r['n']])))
[docs]def ode_Riccati_special_minus2(eq, func, order, match): r""" The general Riccati equation has the form .. math:: dy/dx = f(x) y^2 + g(x) y + h(x)\text{.} While it does not have a general solution [1], the "special" form, dy/dx = a y^2 - b x^c, does have solutions in many cases [2]. This routine returns a solution for a(dy/dx) = b y^2 + c y/x + d/x^2 that is obtained by using a suitable change of variables to reduce it to the special form and is valid when neither a nor b are zero and either c or d is zero. >>> from sympy.abc import x, y, a, b, c, d >>> from sympy.solvers.ode import dsolve, checkodesol >>> from sympy import pprint, Function >>> f = Function('f') >>> y = f(x) >>> genform = a*y.diff(x) - (b*y**2 + c*y/x + d/x**2) >>> sol = dsolve(genform, y) >>> pprint(sol, wrap_line=False) / / __________________ \\ | __________________ | / 2 || | / 2 | \/ 4*b*d - (a + c) *log(x)|| -|a + c - \/ 4*b*d - (a + c) *tan|C1 + ----------------------------|| \ \ 2*a // f(x) = ------------------------------------------------------------------------ 2*b*x >>> checkodesol(genform, sol, order=1)[0] True References ========== 1. http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Riccati 2. http://eqworld.ipmnet.ru/en/solutions/ode/ode0106.pdf - http://eqworld.ipmnet.ru/en/solutions/ode/ode0123.pdf """ x = func.args[0] f = func.func r = match # a2*diff(f(x),x) + b2*f(x) + c2*f(x)/x + d2/x**2 a2, b2, c2, d2 = [r[r[s]] for s in 'a2 b2 c2 d2'.split()] C1 = Symbol('C1') mu = sqrt(4*d2*b2 - (a2 - c2)**2) return Eq(f(x), (a2 - c2 - mu*tan(mu/(2*a2)*log(x) + C1))/(2*b2*x))
[docs]def ode_Liouville(eq, func, order, match): r""" Solves 2nd order Liouville differential equations. The general form of a Liouville ODE is .. math:: \frac{d^2 y}{dx^2} + g(y) \left(\! \frac{dy}{dx}\!\right)^2 + h(x) \frac{dy}{dx}\text{.} The general solution is: >>> from sympy import Function, dsolve, Eq, pprint, diff >>> from sympy.abc import x >>> f, g, h = map(Function, ['f', 'g', 'h']) >>> genform = Eq(diff(f(x),x,x) + g(f(x))*diff(f(x),x)**2 + ... h(x)*diff(f(x),x), 0) >>> pprint(genform) 2 2 /d \ d d g(f(x))*|--(f(x))| + h(x)*--(f(x)) + ---(f(x)) = 0 \dx / dx 2 dx >>> pprint(dsolve(genform, f(x), hint='Liouville_Integral')) f(x) / / | | | / | / | | | | | - | h(x) dx | | g(y) dy | | | | | / | / C1 + C2* | e dx + | e dy = 0 | | / / Examples ======== >>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(diff(f(x), x, x) + diff(f(x), x)**2/f(x) + ... diff(f(x), x)/x, f(x), hint='Liouville')) ________________ ________________ [f(x) = -\/ C1 + C2*log(x) , f(x) = \/ C1 + C2*log(x) ] References ========== - Goldstein and Braun, "Advanced Methods for the Solution of Differential Equations", pp. 98 - http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Liouville # indirect doctest """ # Liouville ODE: # f(x).diff(x, 2) + g(f(x))*(f(x).diff(x, 2))**2 + h(x)*f(x).diff(x) # See Goldstein and Braun, "Advanced Methods for the Solution of # Differential Equations", pg. 98, as well as # http://www.maplesoft.com/support/help/view.aspx?path=odeadvisor/Liouville x = func.args[0] f = func.func r = match # f(x).diff(x, 2) + g*f(x).diff(x)**2 + h*f(x).diff(x) y = r['y'] C1 = Symbol('C1') C2 = Symbol('C2') int = C.Integral(exp(C.Integral(r['g'], y)), (y, None, f(x))) sol = Eq(int + C1*C.Integral(exp(-C.Integral(r['h'], x)), x) + C2, 0) return sol
def _nth_linear_match(eq, func, order): r""" Matches a differential equation to the linear form: .. math:: a_n(x) y^{(n)} + \cdots + a_1(x)y' + a_0(x) y + B(x) = 0 Returns a dict of order:coeff terms, where order is the order of the derivative on each term, and coeff is the coefficient of that derivative. The key -1 holds the function B(x). Returns None if the ODE is not linear. This function assumes that func has already been checked to be good. Examples ======== >>> from sympy import Function, cos, sin >>> from sympy.abc import x >>> from sympy.solvers.ode import _nth_linear_match >>> f = Function('f') >>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) + ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) - ... sin(x), f(x), 3) {-1: x - sin(x), 0: -1, 1: cos(x) + 2, 2: x, 3: 1} >>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) + ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) - ... sin(f(x)), f(x), 3) == None True """ x = func.args[0] one_x = set([x]) terms = dict([(i, S.Zero) for i in range(-1, order + 1)]) for i in Add.make_args(eq): if not i.has(func): terms[-1] += i else: c, f = i.as_independent(func) if not ((isinstance(f, Derivative) and set(f.variables) == one_x) \ or f == func): return None else: terms[len(f.args[1:])] += c return terms def ode_nth_linear_euler_eq_homogeneous(eq, func, order, match, returns='sol'): r""" Solves an n\th order linear homogeneous variable-coefficient Cauchy-Euler equidimensional ordinary differential equation. This is an equation with form 0 = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) \cdots. These equations can be solved in a general manner, by substituting solutions of the form f(x) = x^r, and deriving a characteristic equation for r. When there are repeated roots, we include extra terms of the form C_{r k} \ln^k(x) x^r, where C_{r k} is an arbitrary integration constant, r is a root of the characteristic equation, and k ranges over the multiplicity of r. In the cases where the roots are complex, solutions of the form C_1 x^a \sin(b \log(x)) + C_2 x^a \cos(b \log(x)) are returned, based on expansions with Eulers formula. The general solution is the sum of the terms found. If SymPy cannot find exact roots to the characteristic equation, a :py:class:~sympy.polys.rootoftools.RootOf instance will be returned instead. >>> from sympy import Function, dsolve, Eq >>> from sympy.abc import x >>> f = Function('f') >>> dsolve(4*x**2*f(x).diff(x, 2) + f(x), f(x), ... hint='nth_linear_euler_eq_homogeneous') ... # doctest: +NORMALIZE_WHITESPACE f(x) == sqrt(x)*(C1 + C2*log(x)) Note that because this method does not involve integration, there is no nth_linear_euler_eq_homogeneous_Integral hint. The following is for internal use: - returns = 'sol' returns the solution to the ODE. - returns = 'list' returns a list of linearly independent solutions, corresponding to the fundamental solution set, for use with non homogeneous solution methods like variation of parameters and undetermined coefficients. Note that, though the solutions should be linearly independent, this function does not explicitly check that. You can do assert simplify(wronskian(sollist)) != 0 to check for linear independence. Also, assert len(sollist) == order will need to pass. - returns = 'both', return a dictionary {'sol': <solution to ODE>, 'list': <list of linearly independent solutions>}. Examples ======== >>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f = Function('f') >>> eq = f(x).diff(x, 2)*x**2 - 4*f(x).diff(x)*x + 6*f(x) >>> pprint(dsolve(eq, f(x), ... hint='nth_linear_euler_eq_homogeneous')) 2 f(x) = x *(C1 + C2*x) References ========== - http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation - C. Bender & S. Orszag, "Advanced Mathematical Methods for Scientists and Engineers", Springer 1999, pp. 12 # indirect doctest """ global collectterms collectterms = [] x = func.args[0] f = func.func r = match # A generator of constants constants = numbered_symbols(prefix='C', cls=Symbol, start=1) # First, set up characteristic equation. chareq, symbol = S.Zero, Dummy('x') for i in r.keys(): if not isinstance(i, str) and i >= 0: chareq += (r[i]*diff(x**symbol, x, i)*x**-symbol).expand() chareq = Poly(chareq, symbol) chareqroots = [RootOf(chareq, k) for k in xrange(chareq.degree())] # Create a dict root: multiplicity or charroots charroots = defaultdict(int) for root in chareqroots: charroots[root] += 1 gsol = S(0) # We need keep track of terms so we can run collect() at the end. # This is necessary for constantsimp to work properly. ln = log for root, multiplicity in charroots.items(): for i in range(multiplicity): if isinstance(root, RootOf): gsol += (x**root)*constants.next() assert multiplicity == 1 collectterms = [(0, root, 0)] + collectterms elif root.is_real: gsol += ln(x)**i*(x**root)*constants.next() collectterms = [(i, root, 0)] + collectterms else: reroot = re(root) imroot = im(root) gsol += ln(x)**i*(x**reroot)*( constants.next()*sin(abs(imroot)*ln(x)) + constants.next()*cos(imroot*ln(x))) # Preserve ordering (multiplicity, real part, imaginary part) # It will be assumed implicitly when constructing # fundamental solution sets. collectterms = [(i, reroot, imroot)] + collectterms if returns == 'sol': return Eq(f(x), gsol) elif returns in ('list' 'both'): # HOW TO TEST THIS CODE? (dsolve does not pass 'returns' through) # Create a list of (hopefully) linearly independent solutions gensols = [] # Keep track of when to use sin or cos for nonzero imroot for i, reroot, imroot in collectterms: if imroot == 0: gensols.append(ln(x)**i*x**reroot) else: sin_form = ln(x)**i*x**reroot*sin(abs(imroot)*ln(x)) if sin_form in gensols: cos_form = ln(x)**i*x**reroot*cos(imroot*ln(x)) gensols.append(cos_form) else: gensols.append(sin_form) if returns == 'list': return gensols else: return {'sol': Eq(f(x), gsol), 'list': gensols} else: raise ValueError('Unknown value for key "returns".')
[docs]def ode_almost_linear(eq, func, order, match): r""" Solves an almost-linear differential equation. The general form of an almost linear differential equation is .. math:: f(x) g(y) y + k(x) l(y) + m(x) = 0 \text{where} l'(y) = g(y)\text{.} This can be solved by substituting l(y) = u(y). Making the given substitution reduces it to a linear differential equation of the form u' + P(x) u + Q(x) = 0. The general solution is >>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x, y, n >>> f, g, k, l = map(Function, ['f', 'g', 'k', 'l']) >>> genform = Eq(f(x)*(l(y).diff(y)) + k(x)*l(y) + g(x)) >>> pprint(genform) d f(x)*--(l(y)) + g(x) + k(x)*l(y) = 0 dy >>> pprint(dsolve(genform, hint = 'almost_linear')) / // -y*g(x) \\ | || -------- for k(x) = 0|| | || f(x) || -y*k(x) | || || -------- | || y*k(x) || f(x) l(y) = |C1 + |< ------ ||*e | || f(x) || | ||-g(x)*e || | ||-------------- otherwise || | || k(x) || \ \\ // See Also ======== :meth:sympy.solvers.ode.ode_1st_linear Examples ======== >>> from sympy import Function, Derivative, pprint >>> from sympy.solvers.ode import dsolve, classify_ode >>> from sympy.abc import x >>> f = Function('f') >>> d = f(x).diff(x) >>> eq = x*d + x*f(x) + 1 >>> dsolve(eq, f(x), hint='almost_linear') f(x) == (C1 - Ei(x))*exp(-x) >>> pprint(dsolve(eq, f(x), hint='almost_linear')) -x f(x) = (C1 - Ei(x))*e References ========== - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558 """ # Since ode_1st_linear has already been implemented, and the # coefficients have been modified to the required form in # classify_ode, just passing eq, func, order and match to # ode_1st_linear will give the required output. return ode_1st_linear(eq, func, order, match)
def _linear_coeff_match(expr, func): r""" Helper function to match hint linear_coefficients. Matches the expression to the form (a_1 x + b_1 f(x) + c_1)/(a_2 x + b_2 f(x) + c_2) where the following conditions hold: 1. a_1, b_1, c_1, a_2, b_2, c_2 are Rationals; 2. c_1 or c_2 are not equal to zero; 3. a_2 b_1 - a_1 b_2 is not equal to zero. Return xarg, yarg where 1. xarg = (b_2 c_1 - b_1 c_2)/(a_2 b_1 - a_1 b_2) 2. yarg = (a_1 c_2 - a_2 c_1)/(a_2 b_1 - a_1 b_2) Examples ======== >>> from sympy import Function >>> from sympy.abc import x >>> from sympy.solvers.ode import _linear_coeff_match >>> from sympy.functions.elementary.trigonometric import sin >>> f = Function('f') >>> _linear_coeff_match(( ... (-25*f(x) - 8*x + 62)/(4*f(x) + 11*x - 11)), f(x)) (1/9, 22/9) >>> _linear_coeff_match( ... sin((-5*f(x) - 8*x + 6)/(4*f(x) + x - 1)), f(x)) (19/27, 2/27) >>> _linear_coeff_match(sin(f(x)/x), f(x)) """ f = func.func x = func.args[0] def abc(eq): r''' Internal function of _linear_coeff_match that returns Rationals a, b, c if eq is a*x + b*f(x) + c, else None. ''' eq = _mexpand(eq) c = eq.as_independent(x, f(x), as_Add = True)[0] if not c.is_Rational: return a = eq.coeff(x) if not a.is_Rational: return b = eq.coeff(f(x)) if not b.is_Rational: return if eq == a*x + b*f(x) + c: return a, b, c def match(arg): r''' Internal function of _linear_coeff_match that returns Rationals a1, b1, c1, a2, b2, c2 and a2*b1 - a1*b2 of the expression (a1*x + b1*f(x) + c1)/(a2*x + b2*f(x) + c2) if one of c1 or c2 and a2*b1 - a1*b2 is non-zero, else None. ''' n, d = arg.together().as_numer_denom() m = abc(n) if m is not None: a1, b1, c1 = m m = abc(d) if m is not None: a2, b2, c2 = m d = a2*b1 - a1*b2 if (c1 or c2) and d: return a1, b1, c1, a2, b2, c2, d m = [fi.args[0] for fi in expr.atoms(Function) if fi.func != f and fi.nargs == 1 and not fi.args[0].is_Function] or set([expr]) m1 = match(m.pop()) if m1 and all(match(mi) == m1 for mi in m): a1, b1, c1, a2, b2, c2, denom = m1 return (b2*c1 - b1*c2)/denom, (a1*c2 - a2*c1)/denom
[docs]def ode_linear_coefficients(eq, func, order, match): r""" Solves a differential equation with linear coefficients. The general form of a differential equation with linear coefficients is .. math:: y' + F\left(\!\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}\!\right) = 0\text{,} where a_1, b_1, c_1, a_2, b_2, c_2 are constants and a_1 b_2 - a_2 b_1 \ne 0. This can be solved by substituting: .. math:: x = x' + \frac{b_2 c_1 - b_1 c_2}{a_2 b_1 - a_1 b_2} y = y' + \frac{a_1 c_2 - a_2 c_1}{a_2 b_1 - a_1 b_2}\text{.} This substitution reduces the equation to a homogeneous differential equation. See Also ======== :meth:sympy.solvers.ode.ode_1st_homogeneous_coeff_best :meth:sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep :meth:sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep Examples ======== >>> from sympy import Function, Derivative, pprint >>> from sympy.solvers.ode import dsolve, classify_ode >>> from sympy.abc import x >>> f = Function('f') >>> df = f(x).diff(x) >>> eq = (x + f(x) + 1)*df + (f(x) - 6*x + 1) >>> dsolve(eq, hint='linear_coefficients') [f(x) == -x - sqrt(C1 + 7*x**2) - 1, f(x) == -x + sqrt(C1 + 7*x**2) - 1] >>> pprint(dsolve(eq, hint='linear_coefficients')) ___________ ___________ / 2 / 2 [f(x) = -x - \/ C1 + 7*x - 1, f(x) = -x + \/ C1 + 7*x - 1] References ========== - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558 """ return ode_1st_homogeneous_coeff_best(eq, func, order, match)
[docs]def ode_separable_reduced(eq, func, order, match): r""" Solves a differential equation that can be reduced to the separable form. The general form of this equation is .. math:: y' + (y/x) H(x^n y) = 0\text{}. This can be solved by substituting u(y) = x^n y. The equation then reduces to the separable form \frac{u'}{u (\mathrm{power} - H(u))} - \frac{1}{x} = 0. The general solution is: >>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x, n >>> f, g = map(Function, ['f', 'g']) >>> genform = f(x).diff(x) + (f(x)/x)*g(x**n*f(x)) >>> pprint(genform) / n \ d f(x)*g\x *f(x)/ --(f(x)) + --------------- dx x >>> pprint(dsolve(genform, hint='separable_reduced')) n x *f(x) / | | 1 | ------------ dy = C1 + log(x) | y*(n - g(y)) | / See Also ======== :meth:sympy.solvers.ode.ode_separable Examples ======== >>> from sympy import Function, Derivative, pprint >>> from sympy.solvers.ode import dsolve, classify_ode >>> from sympy.abc import x >>> f = Function('f') >>> d = f(x).diff(x) >>> eq = (x - x**2*f(x))*d - f(x) >>> dsolve(eq, hint='separable_reduced') [f(x) == (-sqrt(C1*x**2 + 1) + 1)/x, f(x) == (sqrt(C1*x**2 + 1) + 1)/x] >>> pprint(dsolve(eq, hint='separable_reduced')) ___________ ___________ / 2 / 2 - \/ C1*x + 1 + 1 \/ C1*x + 1 + 1 [f(x) = --------------------, f(x) = ------------------] x x References ========== - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558 """ # Arguments are passed in a way so that they are coherent with the # ode_separable function x = func.args[0] f = func.func y = Dummy('y') u = match['u'].subs(match['t'], y) ycoeff = 1/(y*(match['power'] - u)) m1 = {y: 1, x: -1/x, 'coeff': 1} m2 = {y: ycoeff, x: 1, 'coeff': 1} r = {'m1': m1, 'm2': m2, 'y': y, 'hint': x**match['power']*f(x)} return ode_separable(eq, func, order, r)
[docs]def ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match, returns='sol'): r""" Solves an n\th order linear homogeneous differential equation with constant coefficients. This is an equation of the form .. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + a_0 f(x) = 0\text{.} These equations can be solved in a general manner, by taking the roots of the characteristic equation a_n m^n + a_{n-1} m^{n-1} + \cdots + a_1 m + a_0 = 0. The solution will then be the sum of C_n x^i e^{r x} terms, for each where C_n is an arbitrary constant, r is a root of the characteristic equation and i is one of each from 0 to the multiplicity of the root - 1 (for example, a root 3 of multiplicity 2 would create the terms C_1 e^{3 x} + C_2 x e^{3 x}). The exponential is usually expanded for complex roots using Euler's equation e{I x} = \cos(x) + I \sin(x). Complex roots always come in conjugate pairs in polynomials with real coefficients, so the two roots will be represented (after simplifying the constants) as e^{a x} \left(C_1 \cos(b x) + C_2 \sin(b x)\right). If SymPy cannot find exact roots to the characteristic equation, a :py:class:~sympy.polys.rootoftools.RootOf instance will be return instead. >>> from sympy import Function, dsolve, Eq >>> from sympy.abc import x >>> f = Function('f') >>> dsolve(f(x).diff(x, 5) + 10*f(x).diff(x) - 2*f(x), f(x), ... hint='nth_linear_constant_coeff_homogeneous') ... # doctest: +NORMALIZE_WHITESPACE f(x) == C1*exp(x*RootOf(_x**5 + 10*_x - 2, 0)) + C2*exp(x*RootOf(_x**5 + 10*_x - 2, 1)) + C3*exp(x*RootOf(_x**5 + 10*_x - 2, 2)) + C4*exp(x*RootOf(_x**5 + 10*_x - 2, 3)) + C5*exp(x*RootOf(_x**5 + 10*_x - 2, 4)) Note that because this method does not involve integration, there is no nth_linear_constant_coeff_homogeneous_Integral hint. The following is for internal use: - returns = 'sol' returns the solution to the ODE. - returns = 'list' returns a list of linearly independent solutions, for use with non homogeneous solution methods like variation of parameters and undetermined coefficients. Note that, though the solutions should be linearly independent, this function does not explicitly check that. You can do assert simplify(wronskian(sollist)) != 0 to check for linear independence. Also, assert len(sollist) == order will need to pass. - returns = 'both', return a dictionary {'sol': <solution to ODE>, 'list': <list of linearly independent solutions>}. Examples ======== >>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(f(x).diff(x, 4) + 2*f(x).diff(x, 3) - ... 2*f(x).diff(x, 2) - 6*f(x).diff(x) + 5*f(x), f(x), ... hint='nth_linear_constant_coeff_homogeneous')) x -2*x f(x) = (C1 + C2*x)*e + (C3*sin(x) + C4*cos(x))*e References ========== - http://en.wikipedia.org/wiki/Linear_differential_equation section: Nonhomogeneous_equation_with_constant_coefficients - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 211 # indirect doctest """ x = func.args[0] f = func.func r = match # A generator of constants constants = numbered_symbols(prefix='C', cls=Symbol, start=1) # First, set up characteristic equation. chareq, symbol = S.Zero, Dummy('x') for i in r.keys(): if type(i) == str or i < 0: pass else: chareq += r[i]*symbol**i chareq = Poly(chareq, symbol) chareqroots = [ RootOf(chareq, k) for k in xrange(chareq.degree()) ] # Create a dict root: multiplicity or charroots charroots = defaultdict(int) for root in chareqroots: charroots[root] += 1 gsol = S(0) # We need keep track of terms so we can run collect() at the end. # This is necessary for constantsimp to work properly. global collectterms collectterms = [] for root, multiplicity in charroots.items(): for i in range(multiplicity): if isinstance(root, RootOf): gsol += exp(root*x)*constants.next() assert multiplicity == 1 collectterms = [(0, root, 0)] + collectterms else: reroot = re(root) imroot = im(root) gsol += x**i*exp(reroot*x)*(constants.next()*sin(abs(imroot)*x) + constants.next()*cos(imroot*x)) # This ordering is important collectterms = [(i, reroot, imroot)] + collectterms if returns == 'sol': return Eq(f(x), gsol) elif returns in ('list' 'both'): # Create a list of (hopefully) linearly independent solutions gensols = [] # Keep track of when to use sin or cos for nonzero imroot for i, reroot, imroot in collectterms: if imroot == 0: gensols.append(x**i*exp(reroot*x)) else: if x**i*exp(reroot*x)*sin(abs(imroot)*x) in gensols: gensols.append(x**i*exp(reroot*x)*cos(imroot*x)) else: gensols.append(x**i*exp(reroot*x)*sin(abs(imroot)*x)) if returns == 'list': return gensols else: return {'sol': Eq(f(x), gsol), 'list': gensols} else: raise ValueError('Unknown value for key "returns".')
[docs]def ode_nth_linear_constant_coeff_undetermined_coefficients(eq, func, order, match): r""" Solves an n\th order linear differential equation with constant coefficients using the method of undetermined coefficients. This method works on differential equations of the form .. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + a_0 f(x) = P(x)\text{,} where P(x) is a function that has a finite number of linearly independent derivatives. Functions that fit this requirement are finite sums functions of the form a x^i e^{b x} \sin(c x + d) or a x^i e^{b x} \cos(c x + d), where i is a non-negative integer and a, b, c, and d are constants. For example any polynomial in x, functions like x^2 e^{2 x}, x \sin(x), and e^x \cos(x) can all be used. Products of \sin's and \cos's have a finite number of derivatives, because they can be expanded into \sin(a x) and \cos(b x) terms. However, SymPy currently cannot do that expansion, so you will need to manually rewrite the expression in terms of the above to use this method. So, for example, you will need to manually convert \sin^2(x) into (1 + \cos(2 x))/2 to properly apply the method of undetermined coefficients on it. This method works by creating a trial function from the expression and all of its linear independent derivatives and substituting them into the original ODE. The coefficients for each term will be a system of linear equations, which are be solved for and substituted, giving the solution. If any of the trial functions are linearly dependent on the solution to the homogeneous equation, they are multiplied by sufficient x to make them linearly independent. Examples ======== >>> from sympy import Function, dsolve, pprint, exp, cos >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(f(x).diff(x, 2) + 2*f(x).diff(x) + f(x) - ... 4*exp(-x)*x**2 + cos(2*x), f(x), ... hint='nth_linear_constant_coeff_undetermined_coefficients')) / 4\ | x | -x 4*sin(2*x) 3*cos(2*x) f(x) = |C1 + C2*x + --|*e - ---------- + ---------- \ 3 / 25 25 References ========== - http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 221 # indirect doctest """ gensol = ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match, returns='both') match.update(gensol) return _solve_undetermined_coefficients(eq, func, order, match)
def _solve_undetermined_coefficients(eq, func, order, match): r""" Helper function for the method of undetermined coefficients. See the :py:meth:~sympy.solvers.ode.ode_nth_linear_constant_coeff_undetermined_coefficients docstring for more information on this method. The parameter match should be a dictionary that has the following keys: list A list of solutions to the homogeneous equation, such as the list returned by ode_nth_linear_constant_coeff_homogeneous(returns='list'). sol The general solution, such as the solution returned by ode_nth_linear_constant_coeff_homogeneous(returns='sol'). trialset The set of trial functions as returned by _undetermined_coefficients_match()['trialset']. """ x = func.args[0] f = func.func r = match coeffs = numbered_symbols('a', cls=Dummy) coefflist = [] gensols = r['list'] gsol = r['sol'] trialset = r['trialset'] notneedset = set([]) newtrialset = set([]) global collectterms if len(gensols) != order: raise NotImplementedError("Cannot find " + str(order) + " solutions to the homogeneous equation nessesary to apply" + " undetermined coefficients to " + str(eq) + " (number of terms != order)") usedsin = set([]) mult = 0 # The multiplicity of the root getmult = True for i, reroot, imroot in collectterms: if getmult: mult = i + 1 getmult = False if i == 0: getmult = True if imroot: # Alternate between sin and cos if (i, reroot) in usedsin: check = x**i*exp(reroot*x)*cos(imroot*x) else: check = x**i*exp(reroot*x)*sin(abs(imroot)*x) usedsin.add((i, reroot)) else: check = x**i*exp(reroot*x) if check in trialset: # If an element of the trial function is already part of the # homogeneous solution, we need to multiply by sufficient x to # make it linearly independent. We also don't need to bother # checking for the coefficients on those elements, since we # already know it will be 0. while True: if check*x**mult in trialset: mult += 1 else: break trialset.add(check*x**mult) notneedset.add(check) newtrialset = trialset - notneedset trialfunc = 0 for i in newtrialset: c = coeffs.next() coefflist.append(c) trialfunc += c*i eqs = sub_func_doit(eq, f(x), trialfunc) coeffsdict = dict(zip(trialset, [0]*(len(trialset) + 1))) eqs = expand_mul(eqs) for i in Add.make_args(eqs): s = separatevars(i, dict=True, symbols=[x]) coeffsdict[s[x]] += s['coeff'] coeffvals = solve(coeffsdict.values(), coefflist) if not coeffvals: raise NotImplementedError( "Could not solve %s using the " "method of undetermined coefficients " "(unable to solve for coefficients)." % eq) psol = trialfunc.subs(coeffvals) return Eq(f(x), gsol.rhs + psol) def _undetermined_coefficients_match(expr, x): r""" Returns a trial function match if undetermined coefficients can be applied to expr, and None otherwise. A trial expression can be found for an expression for use with the method of undetermined coefficients if the expression is an additive/multiplicative combination of constants, polynomials in x (the independent variable of expr), \sin(a x + b), \cos(a x + b), and e^{a x} terms (in other words, it has a finite number of linearly independent derivatives). Note that you may still need to multiply each term returned here by sufficient x to make it linearly independent with the solutions to the homogeneous equation. This is intended for internal use by undetermined_coefficients hints. SymPy currently has no way to convert \sin^n(x) \cos^m(y) into a sum of only \sin(a x) and \cos(b x) terms, so these are not implemented. So, for example, you will need to manually convert \sin^2(x) into [1 + \cos(2 x)]/2 to properly apply the method of undetermined coefficients on it. Examples ======== >>> from sympy import log, exp >>> from sympy.solvers.ode import _undetermined_coefficients_match >>> from sympy.abc import x >>> _undetermined_coefficients_match(9*x*exp(x) + exp(-x), x) {'test': True, 'trialset': set([x*exp(x), exp(-x), exp(x)])} >>> _undetermined_coefficients_match(log(x), x) {'test': False} """ from sympy import S a = Wild('a', exclude=[x]) b = Wild('b', exclude=[x]) expr = powsimp(expr, combine='exp') # exp(x)*exp(2*x + 1) => exp(3*x + 1) retdict = {} def _test_term(expr, x): r""" Test if expr fits the proper form for undetermined coefficients. """ if expr.is_Add: return all(_test_term(i, x) for i in expr.args) elif expr.is_Mul: if expr.has(sin, cos): foundtrig = False # Make sure that there is only one trig function in the args. # See the docstring. for i in expr.args: if i.has(sin, cos): if foundtrig: return False else: foundtrig = True return all(_test_term(i, x) for i in expr.args) elif expr.is_Function: if expr.func in (sin, cos, exp): if expr.args[0].match(a*x + b): return True else: return False else: return False elif expr.is_Pow and expr.base.is_Symbol and expr.exp.is_Integer and \ expr.exp >= 0: return True elif expr.is_Pow and expr.base.is_number: if expr.exp.match(a*x + b): return True else: return False elif expr.is_Symbol or expr.is_Number: return True else: return False def _get_trial_set(expr, x, exprs=set([])): r""" Returns a set of trial terms for undetermined coefficients. The idea behind undetermined coefficients is that the terms expression repeat themselves after a finite number of derivatives, except for the coefficients (they are linearly dependent). So if we collect these, we should have the terms of our trial function. """ def _remove_coefficient(expr, x): r""" Returns the expression without a coefficient. Similar to expr.as_independent(x)[1], except it only works multiplicatively. """ # I was using the below match, but it doesn't always put all of the # coefficient in c. c.f. 2**x*6*exp(x)*log(2) # The below code is probably cleaner anyway. # c = Wild('c', exclude=[x]) # t = Wild('t') # r = expr.match(c*t) term = S.One if expr.is_Mul: for i in expr.args: if i.has(x): term *= i elif expr.has(x): term = expr return term expr = expand_mul(expr) if expr.is_Add: for term in expr.args: if _remove_coefficient(term, x) in exprs: pass else: exprs.add(_remove_coefficient(term, x)) exprs = exprs.union(_get_trial_set(term, x, exprs)) else: term = _remove_coefficient(expr, x) tmpset = exprs.union(set([term])) oldset = set([]) while tmpset != oldset: # If you get stuck in this loop, then _test_term is probably # broken oldset = tmpset.copy() expr = expr.diff(x) term = _remove_coefficient(expr, x) if term.is_Add: tmpset = tmpset.union(_get_trial_set(term, x, tmpset)) else: tmpset.add(term) exprs = tmpset return exprs retdict['test'] = _test_term(expr, x) if retdict['test']: # Try to generate a list of trial solutions that will have the # undetermined coefficients. Note that if any of these are not linearly # independent with any of the solutions to the homogeneous equation, # then they will need to be multiplied by sufficient x to make them so. # This function DOES NOT do that (it doesn't even look at the # homogeneous equation). retdict['trialset'] = _get_trial_set(expr, x) return retdict
[docs]def ode_nth_linear_constant_coeff_variation_of_parameters(eq, func, order, match): r""" Solves an n\th order linear differential equation with constant coefficients using the method of variation of parameters. This method works on any differential equations of the form .. math:: f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + a_0 f(x) = P(x)\text{.} This method works by assuming that the particular solution takes the form .. math:: \sum_{x=1}^{n} c_i(x) y_i(x)\text{,} where y_i is the i\th solution to the homogeneous equation. The solution is then solved using Wronskian's and Cramer's Rule. The particular solution is given by .. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx \right) y_i(x) \text{,} where W(x) is the Wronskian of the fundamental system (the system of n linearly independent solutions to the homogeneous equation), and W_i(x) is the Wronskian of the fundamental system with the i\th column replaced with [0, 0, \cdots, 0, P(x)]. This method is general enough to solve any n\th order inhomogeneous linear differential equation with constant coefficients, but sometimes SymPy cannot simplify the Wronskian well enough to integrate it. If this method hangs, try using the nth_linear_constant_coeff_variation_of_parameters_Integral hint and simplifying the integrals manually. Also, prefer using nth_linear_constant_coeff_undetermined_coefficients when it applies, because it doesn't use integration, making it faster and more reliable. Warning, using simplify=False with 'nth_linear_constant_coeff_variation_of_parameters' in :py:meth:~sympy.solvers.ode.dsolve may cause it to hang, because it will not attempt to simplify the Wronskian before integrating. It is recommended that you only use simplify=False with 'nth_linear_constant_coeff_variation_of_parameters_Integral' for this method, especially if the solution to the homogeneous equation has trigonometric functions in it. Examples ======== >>> from sympy import Function, dsolve, pprint, exp, log >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(f(x).diff(x, 3) - 3*f(x).diff(x, 2) + ... 3*f(x).diff(x) - f(x) - exp(x)*log(x), f(x), ... hint='nth_linear_constant_coeff_variation_of_parameters')) / 3 \ | 2 x *(6*log(x) - 11)| x f(x) = |C1 + C2*x + C3*x + ------------------|*e \ 36 / References ========== - http://en.wikipedia.org/wiki/Variation_of_parameters - http://planetmath.org/encyclopedia/VariationOfParameters.html - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 233 # indirect doctest """ gensol = ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match, returns='both') match.update(gensol) return _solve_variation_of_parameters(eq, func, order, match)
def _solve_variation_of_parameters(eq, func, order, match): r""" Helper function for the method of variation of parameters. See the :py:meth:~sympy.solvers.ode.ode_nth_linear_constant_coeff_variation_of_parameters docstring for more information on this method. The parameter match should be a dictionary that has the following keys: list A list of solutions to the homogeneous equation, such as the list returned by ode_nth_linear_constant_coeff_homogeneous(returns='list'). sol The general solution, such as the solution returned by ode_nth_linear_constant_coeff_homogeneous(returns='sol'). """ x = func.args[0] f = func.func r = match psol = 0 gensols = r['list'] gsol = r['sol'] wr = wronskian(gensols, x) if r.get('simplify', True): wr = simplify(wr) # We need much better simplification for # some ODEs. See issue 1563, for example. # To reduce commonly occuring sin(x)**2 + cos(x)**2 to 1 wr = trigsimp(wr, deep=True, recursive=True) if not wr: # The wronskian will be 0 iff the solutions are not linearly # independent. raise NotImplementedError("Cannot find " + str(order) + " solutions to the homogeneous equation nessesary to apply " + "variation of parameters to " + str(eq) + " (Wronskian == 0)") if len(gensols) != order: raise NotImplementedError("Cannot find " + str(order) + " solutions to the homogeneous equation nessesary to apply " + "variation of parameters to " + str(eq) + " (number of terms != order)") negoneterm = (-1)**(order) for i in gensols: psol += negoneterm*C.Integral(wronskian(filter(lambda x: x != i, gensols), x)*r[-1]/wr, x)*i/r[order] negoneterm *= -1 if r.get('simplify', True): psol = simplify(psol) psol = trigsimp(psol, deep=True) return Eq(f(x), gsol.rhs + psol)
[docs]def ode_separable(eq, func, order, match): r""" Solves separable 1st order differential equations. This is any differential equation that can be written as P(y) \tfrac{dy}{dx} = Q(x). The solution can then just be found by rearranging terms and integrating: \int P(y) \,dy = \int Q(x) \,dx. This hint uses :py:meth:sympy.simplify.separatevars as its back end, so if a separable equation is not caught by this solver, it is most likely the fault of that function. :py:meth:~sympy.simplify.separatevars is smart enough to do most expansion and factoring necessary to convert a separable equation F(x, y) into the proper form P(x)\cdot{}Q(y). The general solution is:: >>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x >>> a, b, c, d, f = map(Function, ['a', 'b', 'c', 'd', 'f']) >>> genform = Eq(a(x)*b(f(x))*f(x).diff(x), c(x)*d(f(x))) >>> pprint(genform) d a(x)*b(f(x))*--(f(x)) = c(x)*d(f(x)) dx >>> pprint(dsolve(genform, f(x), hint='separable_Integral')) f(x) / / | | | b(y) | c(x) | ---- dy = C1 + | ---- dx | d(y) | a(x) | | / / Examples ======== >>> from sympy import Function, dsolve, Eq >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(Eq(f(x)*f(x).diff(x) + x, 3*x*f(x)**2), f(x), ... hint='separable', simplify=False)) / 2 \ 2 log\3*f (x) - 1/ x ---------------- = C1 + -- 6 2 References ========== - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 52 # indirect doctest """ x = func.args[0] f = func.func C1 = Symbol('C1') r = match # {'m1':m1, 'm2':m2, 'y':y} u = r.get('hint', f(x)) # get u from separable_reduced else get f(x) return Eq(C.Integral(r['m2']['coeff']*r['m2'][r['y']]/r['m1'][r['y']], (r['y'], None, u)), C.Integral(-r['m1']['coeff']*r['m1'][x]/ r['m2'][x], x) + C1)
def checkinfsol(eq, infinitesimals, func=None, order=None): r""" This function is used to check if the given infinitesimals are the actual infinitesimals for the given first order differential equation. As of now, it simply checks, by substituting the infinitesimals in the partial differential equation. (eta.diff(x) + (eta.diff(y) - xi.diff(x))*h - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) = 0 where eta, and xi are the infinitesimals and h(x,y) = dy/dx The infinitesimals should be given in the form of a list of dicts [{xi(x, y): inf, eta(x, y): inf}], corresponding to the output of the function infinitesimals. It returns a list of values of the form [(True/False, sol)] where sol is the value obtained after substituting the infinitesimals in the PDE. If it is True, then sol would be zero. """ if isinstance(eq, Equality): eq = eq.lhs - eq.rhs if not func: eq, func = _preprocess(eq) variables = func.args if len(variables) != 1: raise ValueError("ODE's have only one independent variable") else: x = variables[0] if not order: order = ode_order(eq, func) if order != 1: raise NotImplementedError("Lie groups solver has been implemented " "only for first order differential equations") else: df = func.diff(x) a = Wild('a', exclude = [df]) b = Wild('b') match = collect(expand(eq), df).match(a*df + b) h = -match[b]/match[a] y = Symbol('y') h = h.subs(func, y) xi = Function('xi')(x, y) eta = Function('eta')(x, y) dxi = Function('xi')(x, func) deta = Function('eta')(x, func) pde = (eta.diff(x) + (eta.diff(y) - xi.diff(x))*h - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) soltup = [] for sol in infinitesimals: tsol = {xi: S(sol[dxi]).subs(func, y), eta: S(sol[deta]).subs(func, y)} sol = simplify(pde.subs(tsol).doit()) if sol: soltup.append((False, sol)) else: soltup.append((True, 0)) return soltup
[docs]def infinitesimals(eq, func=None, order=None, **kwargs): r""" The functions xi and eta, are called the infinitesimals which help in the process of finding a new co-ordinate system, in which a differential equation can be simplified. They are tangents to the coordinate curves, in which the differential equation is simplified. Consider the transformation (x, y) --> (X, Y) such that X and Y are f(x, y, lambda) and g(x, y, lambda) and such that the differential equation remains invariant. Xi and eta are the tangents to the transformed coordinates X and Y, when lambda is the identity. The infinitesimals can be found by solving the following Partial Differential Equation. >>> from sympy import Function, diff, Eq, pprint >>> from sympy.abc import x, y >>> xi, eta, h = map(Function, ['xi', 'eta', 'h']) >>> h = h(x, y) # dy/dx = h >>> eta = eta(x, y) >>> xi = xi(x, y) >>> genform = Eq(eta.diff(x) + (eta.diff(y) - xi.diff(x))*h ... - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)), 0) >>> pprint(genform) /d d \ d 2 d |--(eta(x, y)) - --(xi(x, y))|*h(x, y) - eta(x, y)*--(h(x, y)) - h (x, y)*--(x \dy dx / dy dy <BLANKLINE> d d i(x, y)) - xi(x, y)*--(h(x, y)) + --(eta(x, y)) = 0 dx dx Once the infinitesimals are found, the following partial differential equations would help us find the transformed coordinates (X, Y) 1. X.diff(x)*xi + X.diff(y)*eta = 0 2. Y.diff(x)*xi + Y.diff(y)*eta = 1 Examples ======== >>> from sympy import Function, diff >>> from sympy.solvers.ode import infinitesimals >>> from sympy.abc import x >>> f = Function('f') >>> eq = f(x).diff(x) - x**2*f(x) >>> infinitesimals(eq) [{eta(x, f(x)): exp(x**3/3), xi(x, f(x)): 0}, {eta(x, f(x)): f(x), xi(x, f(x)): 0}, {eta(x, f(x)): 0, xi(x, f(x)): x**(-2)}, {eta(x, f(x)): x**2*f(x) + f(x), xi(x, f(x)): 1}] References ========== - Solving differential equations by Symmetry Groups, John Starrett, pp. 1 - pp. 14 """ from sympy.integrals.integrals import integrate if isinstance(eq, Equality): eq = eq.lhs - eq.rhs if not func: eq, func = _preprocess(eq) variables = func.args if len(variables) != 1: raise ValueError("ODE's have only one independent variable") else: x = variables[0] if not order: order = ode_order(eq, func) if order != 1: raise NotImplementedError("Infinitesimals for only " "first order ODE's have been implemented") else: df = func.diff(x) # Matching differential equation of the form a*df + b a = Wild('a', exclude = [df]) b = Wild('b') match = kwargs.get('match', collect(expand(eq), df).match(a*df + b)) h = -simplify(match[b]/match[a]) y = Symbol('y') h = h.subs(func, y) hsyms = h.free_symbols xi = Function('xi')(x, func) eta = Function('eta')(x, func) hx = h.diff(x) hy = h.diff(y) xieta = [] # This is the PDE that has to be solved using various # heuristics. The purpose is to "intelligently" guess # the functions xi and eta. All the heuristics are cited # from the paper "Computer Algebra Solving of first order # ODE's Using Symmetry Methods" unless otherwise specified. # Here dy/dx = h # [eta.diff(x) + (eta.diff(y) - xi.diff(x))*h # - xi.diff(y)*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) = 0] # The first heuristic uses the following four sets of # assumptions on xi and eta # 1. [xi = 0, eta = f(x)] # 2. [xi = 0, eta = f(y)] # 3. [xi = f(x), eta = 0] # 4. [xi = f(y), eta = 0] # Assuming xi = 0 and eta to be a function of x, the PDE # reduces to eta.diff(x) - eta*(h.diff(y)) = 0 # If h.diff(y) is a function of x, then this can usually # be integrated easily. hysym = hy.free_symbols if y not in hysym: try: fx = exp(integrate(hy, x)) except NotImplementedError: pass else: inf = {xi: 0, eta: fx.subs(y, func)} if inf not in xieta: xieta.append(inf) # Assuming xi = 0 and eta to be a function of y, the PDE # reduces to eta.diff(y)*h - eta*(h.diff(y)) = 0 # If h.diff(y)/h is a function of y, then this can usually # be integrated easily. factor = hy/h facsym = factor.free_symbols if x not in facsym: try: fy = exp(integrate(factor, y)) except NotImplementedError: pass else: inf = {xi: 0, eta: fy.subs(y, func)} if inf not in xieta: xieta.append(inf) # Assuming eta = 0 and xi to be a function of x, the PDE # reduces to - (xi.diff(x))*h - xi*(h.diff(x)) # If -h.diff(x)/h is a function of x, then this can usually # be integrated easily. factor = -hx/h facsym = factor.free_symbols if y not in facsym: try: fx = exp(integrate(factor, x)) except NotImplementedError: pass else: inf = {xi: fx.subs(y, func), eta: 0} if inf not in xieta: xieta.append(inf) # Assuming eta = 0 and xi to be a function of y, the PDE # reduces to - xi.diff(y)*h**2 - xi*(h.diff(x)) # If -h.diff(x)/h**2 is a function of y, then this can usually # be integrated easily. factor = -hx/(h**2) facsym = factor.free_symbols if x not in facsym: try: fy = exp(integrate(factor, y)) except NotImplementedError: pass else: inf = {xi: fy.subs(y, func), eta: 0} if inf not in xieta: xieta.append(inf) # The second heuristic uses the following four sets of # assumptions on xi and eta # 1. [xi = 0, eta = f(x)*g(y)] # 2. [xi = 0, eta = f(y)*g(x)] # 3. [xi = f(x)*g(y), eta = 0] # 4. [xi = f(y)*g(x), eta = 0] # g is a function built by extracting algebraic factors from the # numerator and denominator of the ODE to be solved and f is an # unknown function to be determined by solving auxiliary ODE's # after substituting g in the PDE. # If the auxilliary ODE obtained is separable in f, it is assumed # that this heuristic works. argx = [] # Extracting factors containing x only argy = [] # Extracting factors containing y only Fx = Function('F')(x) Fy = Function('F')(y) c = Wild('c') d = Wild('d', exclude=[Fx.diff(x), Fy.diff(y)]) gcd = gcd_terms(h) if gcd.is_Pow: base, power = gcd.as_base_exp() if gcd.has(y) and not gcd.has(x) and \ gcd.is_algebraic_expr(y): argy.append(base**power) if gcd.has(x) and not gcd.has(y) and \ gcd.is_algebraic_expr(x): argx.append(base**power) elif gcd.is_Mul: factors = gcd.args for arg in factors: if arg.has(y) and not arg.has(x) and \ arg.is_algebraic_expr(y): if argy: argy.extend([arg*factory for factory in argy]) argy.append(arg) if arg.has(x) and not arg.has(y) and \ arg.is_algebraic_expr(x): if argx: argx.extend([arg*factorx for factorx in argx]) argx.append(arg) if argy: for arg in argy: # Assuming xi = 0, and eta to be f(x)*g(y), the PDE reduces to # (f(x).diff(x)*g(y) + f(x)*g(y).diff(y)*h - f(x)*g(y)*(hy) = 0) eq = (Fx.diff(x))*arg + Fx*h*(arg.diff(y)) - Fx*arg*(hy) # This means y can be successfully eliminated from eq. # The same logic applies for the four assumptions below var = separatevars(eq, [x, y], dict=True) if var: coeffx = var[x] match = coeffx.match(c*Fx.diff(x) + d) if match: match = (-match[d]/match[c]).subs(Fx, y) # This heuristic is assumed to work if the auxilliary ODE # obtained is separable in Fx and x. The same logic # is used for the cases below. red = separatevars(match, [x, y], dict=True) if red: inty = integrate(1/(red['coeff']*red[y]), y) intx = integrate(red[x], x) try: msol = solve(Eq(inty, intx), y) except NotImplementedError: pass else: for sol in msol: if not sol is S.NaN: inf = {xi: 0, eta: (sol*arg).subs(y, func)} if inf not in xieta: xieta.append(inf) # Assuming eta = 0, and xi to be f(x)*g(y), the PDE reduces to # (f(x).diff(x)*g(y)*h + f(x)*g(y).diff(y)*h**2 + # f(x)*g(y)*h.diff(x) = 0) eq = ((Fx.diff(x))*arg*h + h**2*Fx*(arg.diff(y)) + Fx*arg*(hx)) var = separatevars(eq, [x, y], dict=True) if var: coeffx = var[x] match = coeffx.match(c*Fx.diff(x) + d) if match: match = (-match[d]/match[c]).subs(Fx, y) red = separatevars(match, [x, y], dict=True) if red: inty = integrate(1/(red['coeff']*red[y]), y) intx = integrate(red[x], x) try: msol = solve(Eq(inty, intx), y) except NotImplementedError: pass else: for sol in msol: if not sol is S.NaN: inf = {xi: (sol*arg).subs(y, func), eta: 0} if inf not in xieta: xieta.append(inf) if argx: for arg in argx: # Assuming xi = 0, and eta to be g(x)*f(y), the PDE reduces to # (g(x).diff(x)*f(y) + g(x)*f(y).diff(y)*h - g(x)*f(y)*hy = 0) eq = (arg.diff(x))*Fy + arg*h*(Fy.diff(y)) - arg*hy*Fy var = separatevars(eq, [x, y], dict=True) if var: coeffy = var[y] match = coeffy.match(c*Fy.diff(y) + d) if match: match = (-match[d]/match[c]).subs(Fy, x) red = separatevars(match, [x, y], dict=True) if red: intx = integrate(1/(red['coeff']*red[x]), x) inty = integrate(red[y], y) try: msol = solve(Eq(intx, inty), x) except NotImplementedError: pass else: for sol in msol: if not sol is S.NaN: inf = {xi: 0, eta: (sol*arg).subs(y, func)} if inf not in xieta: xieta.append(inf) # Assuming eta = 0, and xi to be g(x)*f(y), the PDE reduces to # (g(x).diff(x)*f(y)*h + g(x)*f(y).diff(y)*h**2 + # g(x)*f(y)*h.diff(x) = 0) eq = ((arg.diff(x))*h*Fy + h**2*arg*(Fy.diff(y)) + arg*Fy*(h.diff(x))) var = separatevars(eq, [x, y], dict=True) if var: coeffy = var[y] match = coeffy.match(c*Fy.diff(y) + d) if match: match = (-match[d]/match[c]).subs(Fy, x) red = separatevars(match, [x, y], dict=True) if red: intx = integrate(1/(red['coeff']*red[x]), x) inty = integrate(red[y], y) try: msol = solve(Eq(intx, inty), x) except NotImplementedError: pass else: for sol in msol: if not sol is S.NaN: inf = {xi: (sol*arg).subs(y, func), eta: 0} if inf not in xieta: xieta.append(inf) # The third heuristic assumes the infinitesimals xi and eta # to be bi-variate polynomials in x and y. The assumption made here # for the logic below is that h is a rational function in x and y # though that may not be necessary for the infinitesimals to be # be bivariate polynomials. The coefficients of the infinitesimals # are found out by substituting them in the PDE and grouping terms # that are monomials in y. The degree of the assumed bivariates # are increased till a certain maximum value. if h.is_rational_function(): # The maximum degree that the infinitesimals can take is # calculated by this technique. deglist = [degree(term) for term in [h, h**2, hx, hy]] maxdeg = max(deglist) mindeg = min(deglist) if mindeg < 0: deg = maxdeg - mindeg else: deg = maxdeg deta = Function('deta')(x, y) dxi = Function('dxi')(x, y) ipde = (deta.diff(x) + (deta.diff(y) - dxi.diff(x))*h - (dxi.diff(y))*h**2 - dxi*hx - deta*hy) xieq = Symbol("xi0") etaeq = Symbol("eta0") for i in range(deg + 1): if i: xieq += Add(*[ Symbol("xi" + str(power) + str(i - power))*x**power*y**(i - power) for power in range(i + 1)]) etaeq += Add(*[ Symbol("eta" + str(power) + str(i - power))*x**power*y**(i - power) for power in range(i + 1)]) pden, denom = (ipde.subs({dxi: xieq, deta: etaeq}).doit()).as_numer_denom() pden = expand(pden) # If the individual terms are monomials, the coefficients # are grouped if pden.is_polynomial(x, y) and pden.is_Add: polyy = Poly(pden, x, y).as_dict() if polyy: symset = xieq.free_symbols.union(etaeq.free_symbols) - set([x, y]) soldict = solve(polyy.values(), *symset) if isinstance(soldict, list): soldict = soldict[0] if any(x for x in soldict.values()): xired = xieq.subs(soldict) etared = etaeq.subs(soldict) # Scaling is done by substituting one for the parameters # This can be any number except zero. dict_ = dict((sym, 1) for sym in symset) inf = {eta: etared.subs(dict_).subs(y, func), xi: xired.subs(dict_).subs(y, func)} if inf not in xieta: xieta.append(inf) break return xieta