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r""" This module contains :py:meth:`~sympy.solvers.ode.dsolve` and different helper functions that it uses.
:py:meth:`~sympy.solvers.ode.dsolve` solves ordinary differential equations. See the docstring on the various functions for their uses. Note that partial differential equations support is in ``pde.py``. Note that hint functions have docstrings describing their various methods, but they are intended for internal use. Use ``dsolve(ode, func, hint=hint)`` to solve an ODE using a specific hint. See also the docstring on :py:meth:`~sympy.solvers.ode.dsolve`.
**Functions in this module**
These are the user functions in this module:
- :py:meth:`~sympy.solvers.ode.dsolve` - Solves ODEs. - :py:meth:`~sympy.solvers.ode.classify_ode` - Classifies ODEs into possible hints for :py:meth:`~sympy.solvers.ode.dsolve`. - :py:meth:`~sympy.solvers.ode.checkodesol` - Checks if an equation is the solution to an ODE. - :py:meth:`~sympy.solvers.ode.homogeneous_order` - Returns the homogeneous order of an expression. - :py:meth:`~sympy.solvers.ode.infinitesimals` - Returns the infinitesimals of the Lie group of point transformations of an ODE, such that it is invariant. - :py:meth:`~sympy.solvers.ode_checkinfsol` - Checks if the given infinitesimals are the actual infinitesimals of a first order ODE.
These are the non-solver helper functions that are for internal use. The user should use the various options to :py:meth:`~sympy.solvers.ode.dsolve` to obtain the functionality provided by these functions:
- :py:meth:`~sympy.solvers.ode.odesimp` - Does all forms of ODE simplification. - :py:meth:`~sympy.solvers.ode.ode_sol_simplicity` - A key function for comparing solutions by simplicity. - :py:meth:`~sympy.solvers.ode.constantsimp` - Simplifies arbitrary constants. - :py:meth:`~sympy.solvers.ode.constant_renumber` - Renumber arbitrary constants. - :py:meth:`~sympy.solvers.ode._handle_Integral` - Evaluate unevaluated Integrals.
See also the docstrings of these functions.
**Currently implemented solver methods**
The following methods are implemented for solving ordinary differential equations. See the docstrings of the various hint functions for more information on each (run ``help(ode)``):
- 1st order separable differential equations. - 1st order differential equations whose coefficients or `dx` and `dy` are functions homogeneous of the same order. - 1st order exact differential equations. - 1st order linear differential equations. - 1st order Bernoulli differential equations. - Power series solutions for first order differential equations. - Lie Group method of solving first order differential equations. - 2nd order Liouville differential equations. - Power series solutions for second order differential equations at ordinary and regular singular points. - `n`\th order linear homogeneous differential equation with constant coefficients. - `n`\th order linear inhomogeneous differential equation with constant coefficients using the method of undetermined coefficients. - `n`\th order linear inhomogeneous differential equation with constant coefficients using the method of variation of parameters.
**Philosophy behind this module**
This module is designed to make it easy to add new ODE solving methods without having to mess with the solving code for other methods. The idea is that there is a :py:meth:`~sympy.solvers.ode.classify_ode` function, which takes in an ODE and tells you what hints, if any, will solve the ODE. It does this without attempting to solve the ODE, so it is fast. Each solving method is a hint, and it has its own function, named ``ode_<hint>``. That function takes in the ODE and any match expression gathered by :py:meth:`~sympy.solvers.ode.classify_ode` and returns a solved result. If this result has any integrals in it, the hint function will return an unevaluated :py:class:`~sympy.integrals.Integral` class. :py:meth:`~sympy.solvers.ode.dsolve`, which is the user wrapper function around all of this, will then call :py:meth:`~sympy.solvers.ode.odesimp` on the result, which, among other things, will attempt to solve the equation for the dependent variable (the function we are solving for), simplify the arbitrary constants in the expression, and evaluate any integrals, if the hint allows it.
**How to add new solution methods**
If you have an ODE that you want :py:meth:`~sympy.solvers.ode.dsolve` to be able to solve, try to avoid adding special case code here. Instead, try finding a general method that will solve your ODE, as well as others. This way, the :py:mod:`~sympy.solvers.ode` module will become more robust, and unhindered by special case hacks. WolphramAlpha and Maple's DETools[odeadvisor] function are two resources you can use to classify a specific ODE. It is also better for a method to work with an `n`\th order ODE instead of only with specific orders, if possible.
To add a new method, there are a few things that you need to do. First, you need a hint name for your method. Try to name your hint so that it is unambiguous with all other methods, including ones that may not be implemented yet. If your method uses integrals, also include a ``hint_Integral`` hint. If there is more than one way to solve ODEs with your method, include a hint for each one, as well as a ``<hint>_best`` hint. Your ``ode_<hint>_best()`` function should choose the best using min with ``ode_sol_simplicity`` as the key argument. See :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best`, for example. The function that uses your method will be called ``ode_<hint>()``, so the hint must only use characters that are allowed in a Python function name (alphanumeric characters and the underscore '``_``' character). Include a function for every hint, except for ``_Integral`` hints (:py:meth:`~sympy.solvers.ode.dsolve` takes care of those automatically). Hint names should be all lowercase, unless a word is commonly capitalized (such as Integral or Bernoulli). If you have a hint that you do not want to run with ``all_Integral`` that doesn't have an ``_Integral`` counterpart (such as a best hint that would defeat the purpose of ``all_Integral``), you will need to remove it manually in the :py:meth:`~sympy.solvers.ode.dsolve` code. See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for guidelines on writing a hint name.
Determine *in general* how the solutions returned by your method compare with other methods that can potentially solve the same ODEs. Then, put your hints in the :py:data:`~sympy.solvers.ode.allhints` tuple in the order that they should be called. The ordering of this tuple determines which hints are default. Note that exceptions are ok, because it is easy for the user to choose individual hints with :py:meth:`~sympy.solvers.ode.dsolve`. In general, ``_Integral`` variants should go at the end of the list, and ``_best`` variants should go before the various hints they apply to. For example, the ``undetermined_coefficients`` hint comes before the ``variation_of_parameters`` hint because, even though variation of parameters is more general than undetermined coefficients, undetermined coefficients generally returns cleaner results for the ODEs that it can solve than variation of parameters does, and it does not require integration, so it is much faster.
Next, you need to have a match expression or a function that matches the type of the ODE, which you should put in :py:meth:`~sympy.solvers.ode.classify_ode` (if the match function is more than just a few lines, like :py:meth:`~sympy.solvers.ode._undetermined_coefficients_match`, it should go outside of :py:meth:`~sympy.solvers.ode.classify_ode`). It should match the ODE without solving for it as much as possible, so that :py:meth:`~sympy.solvers.ode.classify_ode` remains fast and is not hindered by bugs in solving code. Be sure to consider corner cases. For example, if your solution method involves dividing by something, make sure you exclude the case where that division will be 0.
In most cases, the matching of the ODE will also give you the various parts that you need to solve it. You should put that in a dictionary (``.match()`` will do this for you), and add that as ``matching_hints['hint'] = matchdict`` in the relevant part of :py:meth:`~sympy.solvers.ode.classify_ode`. :py:meth:`~sympy.solvers.ode.classify_ode` will then send this to :py:meth:`~sympy.solvers.ode.dsolve`, which will send it to your function as the ``match`` argument. Your function should be named ``ode_<hint>(eq, func, order, match)`. If you need to send more information, put it in the ``match`` dictionary. For example, if you had to substitute in a dummy variable in :py:meth:`~sympy.solvers.ode.classify_ode` to match the ODE, you will need to pass it to your function using the `match` dict to access it. You can access the independent variable using ``func.args[0]``, and the dependent variable (the function you are trying to solve for) as ``func.func``. If, while trying to solve the ODE, you find that you cannot, raise ``NotImplementedError``. :py:meth:`~sympy.solvers.ode.dsolve` will catch this error with the ``all`` meta-hint, rather than causing the whole routine to fail.
Add a docstring to your function that describes the method employed. Like with anything else in SymPy, you will need to add a doctest to the docstring, in addition to real tests in ``test_ode.py``. Try to maintain consistency with the other hint functions' docstrings. Add your method to the list at the top of this docstring. Also, add your method to ``ode.rst`` in the ``docs/src`` directory, so that the Sphinx docs will pull its docstring into the main SymPy documentation. Be sure to make the Sphinx documentation by running ``make html`` from within the doc directory to verify that the docstring formats correctly.
If your solution method involves integrating, use :py:meth:`Integral() <sympy.integrals.integrals.Integral>` instead of :py:meth:`~sympy.core.expr.Expr.integrate`. This allows the user to bypass hard/slow integration by using the ``_Integral`` variant of your hint. In most cases, calling :py:meth:`sympy.core.basic.Basic.doit` will integrate your solution. If this is not the case, you will need to write special code in :py:meth:`~sympy.solvers.ode._handle_Integral`. Arbitrary constants should be symbols named ``C1``, ``C2``, and so on. All solution methods should return an equality instance. If you need an arbitrary number of arbitrary constants, you can use ``constants = numbered_symbols(prefix='C', cls=Symbol, start=1)``. If it is possible to solve for the dependent function in a general way, do so. Otherwise, do as best as you can, but do not call solve in your ``ode_<hint>()`` function. :py:meth:`~sympy.solvers.ode.odesimp` will attempt to solve the solution for you, so you do not need to do that. Lastly, if your ODE has a common simplification that can be applied to your solutions, you can add a special case in :py:meth:`~sympy.solvers.ode.odesimp` for it. For example, solutions returned from the ``1st_homogeneous_coeff`` hints often have many :py:meth:`~sympy.functions.log` terms, so :py:meth:`~sympy.solvers.ode.odesimp` calls :py:meth:`~sympy.simplify.simplify.logcombine` on them (it also helps to write the arbitrary constant as ``log(C1)`` instead of ``C1`` in this case). Also consider common ways that you can rearrange your solution to have :py:meth:`~sympy.solvers.ode.constantsimp` take better advantage of it. It is better to put simplification in :py:meth:`~sympy.solvers.ode.odesimp` than in your method, because it can then be turned off with the simplify flag in :py:meth:`~sympy.solvers.ode.dsolve`. If you have any extraneous simplification in your function, be sure to only run it using ``if match.get('simplify', True):``, especially if it can be slow or if it can reduce the domain of the solution.
Finally, as with every contribution to SymPy, your method will need to be tested. Add a test for each method in ``test_ode.py``. Follow the conventions there, i.e., test the solver using ``dsolve(eq, f(x), hint=your_hint)``, and also test the solution using :py:meth:`~sympy.solvers.ode.checkodesol` (you can put these in a separate tests and skip/XFAIL if it runs too slow/doesn't work). Be sure to call your hint specifically in :py:meth:`~sympy.solvers.ode.dsolve`, that way the test won't be broken simply by the introduction of another matching hint. If your method works for higher order (>1) ODEs, you will need to run ``sol = constant_renumber(sol, 'C', 1, order)`` for each solution, where ``order`` is the order of the ODE. This is because ``constant_renumber`` renumbers the arbitrary constants by printing order, which is platform dependent. Try to test every corner case of your solver, including a range of orders if it is a `n`\th order solver, but if your solver is slow, such as if it involves hard integration, try to keep the test run time down.
Feel free to refactor existing hints to avoid duplicating code or creating inconsistencies. If you can show that your method exactly duplicates an existing method, including in the simplicity and speed of obtaining the solutions, then you can remove the old, less general method. The existing code is tested extensively in ``test_ode.py``, so if anything is broken, one of those tests will surely fail.
"""
expand, expand_mul, Subs, _mexpand)
atan2, conjugate simplify, trigsimp, denom, posify, cse
#: 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. "separable", "1st_exact", "1st_linear", "Bernoulli", "Riccati_special_minus2", "1st_homogeneous_coeff_best", "1st_homogeneous_coeff_subs_indep_div_dep", "1st_homogeneous_coeff_subs_dep_div_indep", "almost_linear", "linear_coefficients", "separable_reduced", "1st_power_series", "lie_group", "nth_linear_constant_coeff_homogeneous", "nth_linear_euler_eq_homogeneous", "nth_linear_constant_coeff_undetermined_coefficients", "nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients", "nth_linear_constant_coeff_variation_of_parameters", "nth_linear_euler_eq_nonhomogeneous_variation_of_parameters", "Liouville", "2nd_power_series_ordinary", "2nd_power_series_regular", "separable_Integral", "1st_exact_Integral", "1st_linear_Integral", "Bernoulli_Integral", "1st_homogeneous_coeff_subs_indep_div_dep_Integral", "1st_homogeneous_coeff_subs_dep_div_indep_Integral", "almost_linear_Integral", "linear_coefficients_Integral", "separable_reduced_Integral", "nth_linear_constant_coeff_variation_of_parameters_Integral", "nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral", "Liouville_Integral", )
"abaco1_simple", "abaco1_product", "abaco2_similar", "abaco2_unique_unknown", "abaco2_unique_general", "linear", "function_sum", "bivariate", "chi" )
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) """
""" Returns a list of constants that do not occur in eq already. """
elif not iterable(eq): raise ValueError("Expected Expr or iterable but got %s" % eq)
ics= None, xi=None, eta=None, x0=0, n=6, **kwargs): r""" Solves any (supported) kind of ordinary differential equation and system of ordinary differential equations.
For single ordinary differential equation =========================================
It is classified under this when number of equation in ``eq`` is one. **Usage**
``dsolve(eq, f(x), hint)`` -> Solve ordinary differential equation ``eq`` for function ``f(x)``, using method ``hint``.
**Details**
``eq`` can be any supported ordinary differential equation (see the :py:mod:`~sympy.solvers.ode` docstring for supported methods). This can either be an :py:class:`~sympy.core.relational.Equality`, or an expression, which is assumed to be equal to ``0``.
``f(x)`` is a function of one variable whose derivatives in that variable make up the ordinary differential equation ``eq``. In many cases it is not necessary to provide this; it will be autodetected (and an error raised if it couldn't be detected).
``hint`` is the solving method that you want dsolve to use. Use ``classify_ode(eq, f(x))`` to get all of the possible hints for an ODE. The default hint, ``default``, will use whatever hint is returned first by :py:meth:`~sympy.solvers.ode.classify_ode`. See Hints below for more options that you can use for hint.
``simplify`` enables simplification by :py:meth:`~sympy.solvers.ode.odesimp`. See its docstring for more information. Turn this off, for example, to disable solving of solutions for ``func`` or simplification of arbitrary constants. It will still integrate with this hint. Note that the solution may contain more arbitrary constants than the order of the ODE with this option enabled.
``xi`` and ``eta`` are the infinitesimal functions of an ordinary differential equation. They are the infinitesimals of the Lie group of point transformations for which the differential equation is invariant. The user can specify values for the infinitesimals. If nothing is specified, ``xi`` and ``eta`` are calculated using :py:meth:`~sympy.solvers.ode.infinitesimals` with the help of various heuristics.
``ics`` is the set of boundary conditions for the differential equation. It should be given in the form of ``{f(x0): x1, f(x).diff(x).subs(x, x2): x3}`` and so on. For now initial conditions are implemented only for power series solutions of first-order differential equations which should be given in the form of ``{f(x0): x1}`` (See issue 4720). If nothing is specified for this case ``f(0)`` is assumed to be ``C0`` and the power series solution is calculated about 0.
``x0`` is the point about which the power series solution of a differential equation is to be evaluated.
``n`` gives the exponent of the dependent variable up to which the power series solution of a differential equation is to be evaluated.
**Hints**
Aside from the various solving methods, there are also some meta-hints that you can pass to :py:meth:`~sympy.solvers.ode.dsolve`:
``default``: This uses whatever hint is returned first by :py:meth:`~sympy.solvers.ode.classify_ode`. This is the default argument to :py:meth:`~sympy.solvers.ode.dsolve`.
``all``: To make :py:meth:`~sympy.solvers.ode.dsolve` apply all relevant classification hints, use ``dsolve(ODE, func, hint="all")``. This will return a dictionary of ``hint:solution`` terms. If a hint causes dsolve to raise the ``NotImplementedError``, value of that hint's key will be the exception object raised. The dictionary will also include some special keys:
- ``order``: The order of the ODE. See also :py:meth:`~sympy.solvers.deutils.ode_order` in ``deutils.py``. - ``best``: The simplest hint; what would be returned by ``best`` below. - ``best_hint``: The hint that would produce the solution given by ``best``. If more than one hint produces the best solution, the first one in the tuple returned by :py:meth:`~sympy.solvers.ode.classify_ode` is chosen. - ``default``: The solution that would be returned by default. This is the one produced by the hint that appears first in the tuple returned by :py:meth:`~sympy.solvers.ode.classify_ode`.
``all_Integral``: This is the same as ``all``, except if a hint also has a corresponding ``_Integral`` hint, it only returns the ``_Integral`` hint. This is useful if ``all`` causes :py:meth:`~sympy.solvers.ode.dsolve` to hang because of a difficult or impossible integral. This meta-hint will also be much faster than ``all``, because :py:meth:`~sympy.core.expr.Expr.integrate` is an expensive routine.
``best``: To have :py:meth:`~sympy.solvers.ode.dsolve` try all methods and return the simplest one. This takes into account whether the solution is solvable in the function, whether it contains any Integral classes (i.e. unevaluatable integrals), and which one is the shortest in size.
See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for more info on hints, and the :py:mod:`~sympy.solvers.ode` docstring for a list of all supported hints.
**Tips**
- You can declare the derivative of an unknown function this way:
>>> from sympy import Function, Derivative >>> from sympy.abc import x # x is the independent variable >>> f = Function("f")(x) # f is a function of x >>> # f_ will be the derivative of f with respect to x >>> f_ = Derivative(f, x)
- See ``test_ode.py`` for many tests, which serves also as a set of examples for how to use :py:meth:`~sympy.solvers.ode.dsolve`. - :py:meth:`~sympy.solvers.ode.dsolve` always returns an :py:class:`~sympy.core.relational.Equality` class (except for the case when the hint is ``all`` or ``all_Integral``). If possible, it solves the solution explicitly for the function being solved for. Otherwise, it returns an implicit solution. - Arbitrary constants are symbols named ``C1``, ``C2``, and so on. - Because all solutions should be mathematically equivalent, some hints may return the exact same result for an ODE. Often, though, two different hints will return the same solution formatted differently. The two should be equivalent. Also note that sometimes the values of the arbitrary constants in two different solutions may not be the same, because one constant may have "absorbed" other constants into it. - Do ``help(ode.ode_<hintname>)`` to get help more information on a specific hint, where ``<hintname>`` is the name of a hint without ``_Integral``.
For system of ordinary differential equations =============================================
**Usage** ``dsolve(eq, func)`` -> Solve a system of ordinary differential equations ``eq`` for ``func`` being list of functions including `x(t)`, `y(t)`, `z(t)` where number of functions in the list depends upon the number of equations provided in ``eq``.
**Details**
``eq`` can be any supported system of ordinary differential equations This can either be an :py:class:`~sympy.core.relational.Equality`, or an expression, which is assumed to be equal to ``0``.
``func`` holds ``x(t)`` and ``y(t)`` being functions of one variable which together with some of their derivatives make up the system of ordinary differential equation ``eq``. It is not necessary to provide this; it will be autodetected (and an error raised if it couldn't be detected).
**Hints**
The hints are formed by parameters returned by classify_sysode, combining them give hints name used later for forming method name.
Examples ========
>>> from sympy import Function, dsolve, Eq, Derivative, sin, cos, symbols >>> from sympy.abc import x >>> f = Function('f') >>> dsolve(Derivative(f(x), x, x) + 9*f(x), f(x)) Eq(f(x), C1*sin(3*x) + C2*cos(3*x))
>>> eq = sin(x)*cos(f(x)) + cos(x)*sin(f(x))*f(x).diff(x) >>> dsolve(eq, hint='1st_exact') [Eq(f(x), -acos(C1/cos(x)) + 2*pi), Eq(f(x), acos(C1/cos(x)))] >>> dsolve(eq, hint='almost_linear') [Eq(f(x), -acos(C1/sqrt(-cos(x)**2)) + 2*pi), Eq(f(x), acos(C1/sqrt(-cos(x)**2)))] >>> t = symbols('t') >>> x, y = symbols('x, y', function=True) >>> eq = (Eq(Derivative(x(t),t), 12*t*x(t) + 8*y(t)), Eq(Derivative(y(t),t), 21*x(t) + 7*t*y(t))) >>> dsolve(eq) [Eq(x(t), C1*x0 + C2*x0*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0**2, t)), Eq(y(t), C1*y0 + C2(y0*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0**2, t) + exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0))] >>> eq = (Eq(Derivative(x(t),t),x(t)*y(t)*sin(t)), Eq(Derivative(y(t),t),y(t)**2*sin(t))) >>> dsolve(eq) set([Eq(x(t), -exp(C1)/(C2*exp(C1) - cos(t))), Eq(y(t), -1/(C1 - cos(t)))]) """
# keep highest order term coefficient positive else: eq[i] = -eq[i] raise ValueError("It solves only those systems of equations whose orders are equal") raise ValueError("dsolve() and classify_sysode() work with " "number of functions being equal to number of equations") raise NotImplementedError else: solvefunc = globals()['sysode_linear_neq_order%(order)s' % match] else: else: else:
# See the docstring of _desolve for more details. hint=hint, simplify=True, xi=xi, eta=eta, type='ode', ics=ics, x0=x0, n=n, **kwargs)
retdict = {} failed_hints = {} gethints = classify_ode(eq, dict=True) orderedhints = gethints['ordered_hints'] for hint in hints: try: rv = _helper_simplify(eq, hint, hints[hint], simplify) except NotImplementedError as detail: failed_hints[hint] = detail else: retdict[hint] = rv func = hints[hint]['func']
retdict['best'] = min(list(retdict.values()), key=lambda x: ode_sol_simplicity(x, func, trysolving=not simplify)) if given_hint == 'best': return retdict['best'] for i in orderedhints: if retdict['best'] == retdict.get(i, None): retdict['best_hint'] = i break retdict['default'] = gethints['default'] retdict['order'] = gethints['order'] retdict.update(failed_hints) return retdict
else: # The key 'hint' stores the hint needed to be solved for.
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. """ else:
# odesimp() will attempt to integrate, if necessary, apply constantsimp(), # attempt to solve for func, and apply any other hint specific # simplifications return [odesimp(s, func, order, cons(s), hint) for s in sols] else: # We still want to integrate (you can disable it separately with the hint) func, order, hint)
r""" Returns a tuple of possible :py:meth:`~sympy.solvers.ode.dsolve` classifications for an ODE.
The tuple is ordered so that first item is the classification that :py:meth:`~sympy.solvers.ode.dsolve` uses to solve the ODE by default. In general, classifications at the near the beginning of the list will produce better solutions faster than those near the end, thought there are always exceptions. To make :py:meth:`~sympy.solvers.ode.dsolve` use a different classification, use ``dsolve(ODE, func, hint=<classification>)``. See also the :py:meth:`~sympy.solvers.ode.dsolve` docstring for different meta-hints you can use.
If ``dict`` is true, :py:meth:`~sympy.solvers.ode.classify_ode` will return a dictionary of ``hint:match`` expression terms. This is intended for internal use by :py:meth:`~sympy.solvers.ode.dsolve`. Note that because dictionaries are ordered arbitrarily, this will most likely not be in the same order as the tuple.
You can get help on different hints by executing ``help(ode.ode_hintname)``, where ``hintname`` is the name of the hint without ``_Integral``.
See :py:data:`~sympy.solvers.ode.allhints` or the :py:mod:`~sympy.solvers.ode` docstring for a list of all supported hints that can be returned from :py:meth:`~sympy.solvers.ode.classify_ode`.
Notes =====
These are remarks on hint names.
``_Integral``
If a classification has ``_Integral`` at the end, it will return the expression with an unevaluated :py:class:`~sympy.integrals.Integral` class in it. Note that a hint may do this anyway if :py:meth:`~sympy.core.expr.Expr.integrate` cannot do the integral, though just using an ``_Integral`` will do so much faster. Indeed, an ``_Integral`` hint will always be faster than its corresponding hint without ``_Integral`` because :py:meth:`~sympy.core.expr.Expr.integrate` is an expensive routine. If :py:meth:`~sympy.solvers.ode.dsolve` hangs, it is probably because :py:meth:`~sympy.core.expr.Expr.integrate` is hanging on a tough or impossible integral. Try using an ``_Integral`` hint or ``all_Integral`` to get it return something.
Note that some hints do not have ``_Integral`` counterparts. This is because :py:meth:`~sympy.solvers.ode.integrate` is not used in solving the ODE for those method. For example, `n`\th order linear homogeneous ODEs with constant coefficients do not require integration to solve, so there is no ``nth_linear_homogeneous_constant_coeff_Integrate`` hint. You can easily evaluate any unevaluated :py:class:`~sympy.integrals.Integral`\s in an expression by doing ``expr.doit()``.
Ordinals
Some hints contain an ordinal such as ``1st_linear``. This is to help differentiate them from other hints, as well as from other methods that may not be implemented yet. If a hint has ``nth`` in it, such as the ``nth_linear`` hints, this means that the method used to applies to ODEs of any order.
``indep`` and ``dep``
Some hints contain the words ``indep`` or ``dep``. These reference the independent variable and the dependent function, respectively. For example, if an ODE is in terms of `f(x)`, then ``indep`` will refer to `x` and ``dep`` will refer to `f`.
``subs``
If a hints has the word ``subs`` in it, it means the the ODE is solved by substituting the expression given after the word ``subs`` for a single dummy variable. This is usually in terms of ``indep`` and ``dep`` as above. The substituted expression will be written only in characters allowed for names of Python objects, meaning operators will be spelled out. For example, ``indep``/``dep`` will be written as ``indep_div_dep``.
``coeff``
The word ``coeff`` in a hint refers to the coefficients of something in the ODE, usually of the derivative terms. See the docstring for the individual methods for more info (``help(ode)``). This is contrast to ``coefficients``, as in ``undetermined_coefficients``, which refers to the common name of a method.
``_best``
Methods that have more than one fundamental way to solve will have a hint for each sub-method and a ``_best`` meta-classification. This will evaluate all hints and return the best, using the same considerations as the normal ``best`` meta-hint.
Examples ========
>>> from sympy import Function, classify_ode, Eq >>> from sympy.abc import x >>> f = Function('f') >>> classify_ode(Eq(f(x).diff(x), 0), f(x)) ('separable', '1st_linear', '1st_homogeneous_coeff_best', '1st_homogeneous_coeff_subs_indep_div_dep', '1st_homogeneous_coeff_subs_dep_div_indep', '1st_power_series', 'lie_group', 'nth_linear_constant_coeff_homogeneous', 'separable_Integral', '1st_linear_Integral', '1st_homogeneous_coeff_subs_indep_div_dep_Integral', '1st_homogeneous_coeff_subs_dep_div_indep_Integral') >>> classify_ode(f(x).diff(x, 2) + 3*f(x).diff(x) + 2*f(x) - 4) ('nth_linear_constant_coeff_undetermined_coefficients', 'nth_linear_constant_coeff_variation_of_parameters', 'nth_linear_constant_coeff_variation_of_parameters_Integral')
"""
"work with functions of one variable, not %s" % func)
n=terms, eta=eta, prep=False) # hint:matchdict or hint:(tuple of matchdicts) # Also will contain "default":<default hint> and "order":order items.
else:
# Preprocessing to get the initial conditions out # Separating derivatives deriv = funcarg.expr old = funcarg.variables[0] new = funcarg.point[0] if isinstance(deriv, Derivative) and isinstance(deriv.args[0], AppliedUndef) and deriv.args[0].func == f and old == x and not new.has(x): dorder = ode_order(deriv, x) temp = 'f' + str(dorder) boundary.update({temp: new, temp + 'val': ics[funcarg]}) else: raise ValueError("Enter valid boundary conditions for Derivatives")
# Separating functions not funcarg.args[0].has(x): else: raise ValueError("Enter valid boundary conditions for Function")
else: raise ValueError("Enter boundary conditions of the form ics " " = {f(point}: value, f(point).diff(point, order).subs(arg, point) " ":value")
# Precondition to try remove f(x) from highest order derivative
## Linear case: a(x)*y'+b(x)*y+c(x) == 0 else: b: dep.coeff(f(x)), c: ind} # double check f[a] since the preconditioning may have failed r[a]*df + r[b]*f(x) + r[c]).expand() - reduced_eq == 0:
## Bernoulli case: a(x)*y'+b(x)*y+c(x)*y**n == 0 reduced_eq, f(x), exact=True).match(a*df + b*f(x) + c*f(x)**n)
## Riccati special n == -2 case: a2*y'+b2*y**2+c2*y/x+d2/x**2 == 0 f(x), exact=True).match(a2*df + b2*f(x)**2 + c2*f(x)/x + d2/x**2)
# NON-REDUCED FORM OF EQUATION matches
# FIRST ORDER POWER SERIES WHICH NEEDS INITIAL CONDITIONS # TODO: Hint first order series should match only if d/e is analytic. # For now, only d/e and (d/e).diff(arg) is checked for existence at # at a given point. # This is currently done internally in ode_1st_power_series. not check1.has(NaN) and not check1.has(-oo): not check2.has(NaN) and not check2.has(-oo):
## Exact Differential Equation: P(x, y) + Q(x, y)*y' = 0 where # dP/dy == dQ/dx # 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 (dP/dy - dQ/dx) / Q = f(x) # then exp(integral(f(x))*equation becomes exact else: # If (dP/dy - dQ/dx) / -P = f(y) # then exp(integral(f(y))*equation becomes exact else:
except NotImplementedError: # Differentiating the coefficients might fail because of things # like f(2*x).diff(x). See issue 4624 and issue 4719. pass
# Any first order ODE can be ideally solved by the Lie Group # method
# This match is used for several cases below; we now collect on # f(x) so the matching works. # Using r[d] and r[e] without any modification for hints # linear-coefficients and separable-reduced.
## Separable Case: y' == P(y)*Q(x) # m1[coeff]*m1[x]*m1[y] + m2[coeff]*m2[x]*m2[y]*y'
## First order equation with homogeneous coefficients: # dy/dx == F(y/x) or dy/dx == F(x/y) # u1=y/x and u2=x/y
## 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. # Dummy substitution for df and f(x). # get the re-cast values for e and d # Match arguments are passed in such a way that it # is coherent with the already existing homogeneous # functions. 'd': d, 'e': e, 'y': y})
## Equation of the form y' + (y/x)*H(x^n*y) = 0 # that can be reduced to separable form
# 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)
## Almost-linear equation of the form f(x)*g(y)*y' + k(x)*l(y) + m(x) = 0 # Separate the terms having f(x) to r[d] and # remaining to r[c]
# 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
else:
# Homogeneous second order differential equation of the form # a3*f(x).diff(x, 2) + b3*f(x).diff(x) + c3, where # for simplicity, a3, b3 and c3 are assumed to be polynomials. # It has a definite power series solution at point x0 if, b3/a3 and c3/a3 # are analytic at x0. [f(x).diff(x, 2), f(x).diff(x), f(x)]).match(deq) not check.has(zoo) and not check.has(-oo): not check.has(zoo) and not check.has(-oo):
# Checking if the differential equation has a regular singular point # at x0. It has a regular singular point at x0, if (b3/a3)*(x - x0) # and (c3/a3)*((x - x0)**2) are analytic at x0. not check.has(zoo) and not check.has(-oo): not check.has(zoo) and not check.has(-oo):
# nth order linear ODE # a_n(x)y^(n) + ... + a_1(x)y' + a_0(x)y = F(x) = b
# Constant coefficient case (a_i is constant for all i) # Inhomogeneous case: F(x) is not identically 0 "nth_linear_constant_coeff_undetermined_coefficients" ] = r
# Homogeneous case: F(x) is identically 0 else:
# nth order Euler equation a_n*x**n*y^(n) + ... + a_1*x*y' + a_0*y = F(x) #In case of Homogeneous euler equation F(x) = 0 r""" Linear Euler ODEs have the form K*x**order*diff(y(x),x,order) = F(x), where K is independent of x and y(x), order>= 0. So we need to check that for each term, coeff == K*x**order from some K. We have a few cases, since coeff may have several different types. """ raise ValueError("order should be greater than 0") return False else:
# Order keys based on allhints.
# Dictionaries are ordered arbitrarily, so make note of which # hint would come first for dsolve(). Use an ordered dict in Py 3. else:
r""" Returns a dictionary of parameter names and values that define the system of ordinary differential equations in ``eq``. The parameters are further used in :py:meth:`~sympy.solvers.ode.dsolve` for solving that system.
The parameter names and values are:
'is_linear' (boolean), which tells whether the given system is linear. Note that "linear" here refers to the operator: terms such as ``x*diff(x,t)`` are nonlinear, whereas terms like ``sin(t)*diff(x,t)`` are still linear operators.
'func' (list) contains the :py:class:`~sympy.core.function.Function`s that appear with a derivative in the ODE, i.e. those that we are trying to solve the ODE for.
'order' (dict) with the maximum derivative for each element of the 'func' parameter.
'func_coeff' (dict) with the coefficient for each triple ``(equation number, function, order)```. The coefficients are those subexpressions that do not appear in 'func', and hence can be considered constant for purposes of ODE solving.
'eq' (list) with the equations from ``eq``, sympified and transformed into expressions (we are solving for these expressions to be zero).
'no_of_equations' (int) is the number of equations (same as ``len(eq)``).
'type_of_equation' (string) is an internal classification of the type of ODE.
References ========== -http://eqworld.ipmnet.ru/en/solutions/sysode/sode-toc1.htm -A. D. Polyanin and A. V. Manzhirov, Handbook of Mathematics for Engineers and Scientists
Examples ========
>>> from sympy import Function, Eq, symbols, diff >>> from sympy.solvers.ode import classify_sysode >>> from sympy.abc import t >>> f, x, y = symbols('f, x, y', function=True) >>> k, l, m, n = symbols('k, l, m, n', Integer=True) >>> x1 = diff(x(t), t) ; y1 = diff(y(t), t) >>> x2 = diff(x(t), t, t) ; y2 = diff(y(t), t, t) >>> eq = (Eq(5*x1, 12*x(t) - 6*y(t)), Eq(2*y1, 11*x(t) + 3*y(t))) >>> classify_sysode(eq) {'eq': [-12*x(t) + 6*y(t) + 5*Derivative(x(t), t), -11*x(t) - 3*y(t) + 2*Derivative(y(t), t)], 'func': [x(t), y(t)], 'func_coeff': {(0, x(t), 0): -12, (0, x(t), 1): 5, (0, y(t), 0): 6, (0, y(t), 1): 0, (1, x(t), 0): -11, (1, x(t), 1): 0, (1, y(t), 0): -3, (1, y(t), 1): 2}, 'is_linear': True, 'no_of_equation': 2, 'order': {x(t): 1, y(t): 1}, 'type_of_equation': 'type1'} >>> eq = (Eq(diff(x(t),t), 5*t*x(t) + t**2*y(t)), Eq(diff(y(t),t), -t**2*x(t) + 5*t*y(t))) >>> classify_sysode(eq) {'eq': [-t**2*y(t) - 5*t*x(t) + Derivative(x(t), t), t**2*x(t) - 5*t*y(t) + Derivative(y(t), t)], 'func': [x(t), y(t)], 'func_coeff': {(0, x(t), 0): -5*t, (0, x(t), 1): 1, (0, y(t), 0): -t**2, (0, y(t), 1): 0, (1, x(t), 0): t**2, (1, x(t), 1): 0, (1, y(t), 0): -5*t, (1, y(t), 1): 1}, 'is_linear': True, 'no_of_equation': 2, 'order': {x(t): 1, y(t): 1}, 'type_of_equation': 'type4'}
"""
# Sympify equations and convert iterables of equations into # a list of equations
raise ValueError("classify_sysode() works for systems of ODEs. " "For scalar ODEs, classify_ode should be used")
# find all the functions if not given raise ValueError("Number of functions given is less than number of equations %s" % funcs) else: raise ValueError("dsolve() and classify_sysode() work with " "functions of one variable only, not %s" % func) else: raise ValueError("dsolve() and classify_sysode() work with " "functions of one variable only, not %s" % func)
# find the order of all equation in system of odes
# find coefficients of terms f(t), diff(f(t),t) and higher derivatives # and similarly for other functions g(t), diff(g(t),t) in all equations. # Here j denotes the equation number, funcs[l] denotes the function about # which we are talking about and k denotes the order of function funcs[l] # whose coefficient we are calculating. else: is_linear_ = False else: else:
else:
else: type_of_equation = None
else: type_of_equation = None else: if order_eq == 1: type_of_equation = check_linear_neq_order1(eq, funcs, func_coef) else: type_of_equation = None else: else: type_of_equation = None elif matching_hints['no_of_equation'] == 3: if order_eq == 1: type_of_equation = check_nonlinear_3eq_order1(eq, funcs, func_coef) else: type_of_equation = None else: type_of_equation = None else: type_of_equation = None
# for equations Eq(a1*diff(x(t),t), b1*x(t) + c1*y(t) + d1) # and Eq(a2*diff(y(t),t), b2*x(t) + c2*y(t) + d2) # We can handle homogeneous case and simple constant forcings else: # Issue #9244: nonhomogeneous linear systems are not supported
# Conditions to check for type 6 whose equations are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and # Eq(diff(y(t),t), a*[f(t) + a*h(t)]x(t) + a*[g(t) - h(t)]*y(t)) # End of condition for type 6
# Equations for type 2 are Eq(a1*diff(x(t),t),b1*x(t)+c1*y(t)+d1) and Eq(a2*diff(y(t),t),b2*x(t)+c2*y(t)+d2) else: return None else: # Equations for type 1 are Eq(a1*diff(x(t),t),b1*x(t)+c1*y(t)) and Eq(a2*diff(y(t),t),b2*x(t)+c2*y(t)) else: # Equation for type 3 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), g(t)*x(t) + f(t)*y(t)) # Equation for type 4 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), -g(t)*x(t) + f(t)*y(t)) or (not cancel(r['b1']/r['c2']).has(t) and not cancel((r['c1']-r['b2'])/r['c2']).has(t)): # Equations for type 5 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), a*g(t)*x(t) + [f(t) + b*g(t)]*y(t) return "type6" else: # Equations for type 7 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), h(t)*x(t) + p(t)*y(t))
and r['b1']==r['c1']==r['b2']==r['c2']==0:
else: q[n] = 1 else: q[n] = 1 else: q[n] = 1
else: return None else: return None else: for k in 'a1 a2 d1 d2 e1 e2'.split()):
for k in 'a1 a2 b2 c1 d1 e2'.split()) and r['c1'] == -r['b2'] and \ r['d1'] == r['e2']:
(r['d2']/r['a2']).has(t) and not (r['e1']/r['a1']).has(t) and \ r['b1']==r['d1']==r['c2']==r['e2']==0:
(cancel(r['a1']*r['d2']/(r['a2']*r['d1']))).has(t) and not (r['d1']/r['e1']).has(t) and not \ (r['d2']/r['e2']).has(t) and r['b1'] == r['b2'] == r['c1'] == r['c2'] == 0:
cancel(r['d1']*r['a2']/(r['d2']*r['a1'])).has(t) and r['b1']==r['b2']==r['c1']==r['c2']==0:
cancel(r['b1']*r['a2']/(r['b2']*r['a1'])).has(t) and r['d1']==r['d2']==r['e1']==r['e2']==0:
cancel(r['e1']*r['a2']/(r['d2']*r['a1'])).has(t) and r['e1'].has(t) \ and r['b1']==r['d1']==r['c2']==r['e2']==0:
(r['b1']/r['c1']).has(t) and not (r['b2']/r['c2']).has(t) and \ (r['d1']/r['a1']).match(b/t**2) and (r['d2']/r['a2']).match(b/t**2) \ and not (r['d1']/r['e1']).has(t) and not (r['d2']/r['e2']).has(t):
else: return None
# We can handle homogeneous case and simple constant forcings. # Issue #9244: nonhomogeneous linear systems are not supported
and r['b1'] == r['c2'] == r['d3'] == 0: and r['d2']/r['a2'] == -r['b2']/r['a2'] and r['b3']/r['a3'] == -r['c3']/r['a3']: else: else: continue else: and all(not cancel(r[k1]/(r['b1'] - r[k])).has(t) for k in 'b1 c2 d3'.split() if r['b1']!=r[k]): else: break return None
return None return 'type1'
or r2[g].subs(diff(x(t),t),u).subs(diff(y(t),t),v).has(t)): else: r2[g].subs(x(t),u).subs(y(t),v).has(t)): return None
return None
x = func[0].func y = func[1].func z = func[2].func fc = func_coef t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] u, v, w = symbols('u, v, w', cls=Dummy) a = Wild('a', exclude=[x(t), y(t), z(t), t]) b = Wild('b', exclude=[x(t), y(t), z(t), t]) c = Wild('c', exclude=[x(t), y(t), z(t), t]) f = Wild('f') F1 = Wild('F1') F2 = Wild('F2') F3 = Wild('F3') for i in range(3): eqs = 0 for terms in Add.make_args(eq[i]): eqs += terms/fc[i,func[i],1] eq[i] = eqs r1 = eq[0].match(diff(x(t),t) - a*y(t)*z(t)) r2 = eq[1].match(diff(y(t),t) - b*z(t)*x(t)) r3 = eq[2].match(diff(z(t),t) - c*x(t)*y(t)) if r1 and r2 and r3: num1, den1 = r1[a].as_numer_denom() num2, den2 = r2[b].as_numer_denom() num3, den3 = r3[c].as_numer_denom() if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]): return 'type1' r = eq[0].match(diff(x(t),t) - y(t)*z(t)*f) if r: r1 = collect_const(r[f]).match(a*f) r2 = ((diff(y(t),t) - eq[1])/r1[f]).match(b*z(t)*x(t)) r3 = ((diff(z(t),t) - eq[2])/r1[f]).match(c*x(t)*y(t)) if r1 and r2 and r3: num1, den1 = r1[a].as_numer_denom() num2, den2 = r2[b].as_numer_denom() num3, den3 = r3[c].as_numer_denom() if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]): return 'type2' r = eq[0].match(diff(x(t),t) - (F2-F3)) if r: r1 = collect_const(r[F2]).match(c*F2) r1.update(collect_const(r[F3]).match(b*F3)) if r1: if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): r1[F2], r1[F3] = r1[F3], r1[F2] r1[c], r1[b] = -r1[b], -r1[c] r2 = eq[1].match(diff(y(t),t) - a*r1[F3] + r1[c]*F1) if r2: r3 = (eq[2] == diff(z(t),t) - r1[b]*r2[F1] + r2[a]*r1[F2]) if r1 and r2 and r3: return 'type3' r = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3) if r: r1 = collect_const(r[F2]).match(c*F2) r1.update(collect_const(r[F3]).match(b*F3)) if r1: if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): r1[F2], r1[F3] = r1[F3], r1[F2] r1[c], r1[b] = -r1[b], -r1[c] r2 = (diff(y(t),t) - eq[1]).match(a*x(t)*r1[F3] - r1[c]*z(t)*F1) if r2: r3 = (diff(z(t),t) - eq[2] == r1[b]*y(t)*r2[F1] - r2[a]*x(t)*r1[F2]) if r1 and r2 and r3: return 'type4' r = (diff(x(t),t) - eq[0]).match(x(t)*(F2 - F3)) if r: r1 = collect_const(r[F2]).match(c*F2) r1.update(collect_const(r[F3]).match(b*F3)) if r1: if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): r1[F2], r1[F3] = r1[F3], r1[F2] r1[c], r1[b] = -r1[b], -r1[c] r2 = (diff(y(t),t) - eq[1]).match(y(t)*(a*r1[F3] - r1[c]*F1)) if r2: r3 = (diff(z(t),t) - eq[2] == z(t)*(r1[b]*r2[F1] - r2[a]*r1[F2])) if r1 and r2 and r3: return 'type5' return None
return None
r""" Substitutes corresponding ``sols`` for each functions into each ``eqs`` and checks that the result of substitutions for each equation is ``0``. The equations and solutions passed can be any iterable.
This only works when each ``sols`` have one function only, like `x(t)` or `y(t)`. For each function, ``sols`` can have a single solution or a list of solutions. In most cases it will not be necessary to explicitly identify the function, but if the function cannot be inferred from the original equation it can be supplied through the ``func`` argument.
When a sequence of equations is passed, the same sequence is used to return the result for each equation with each function substitued with corresponding solutions.
It tries the following method to find zero equivalence for each equation:
Substitute the solutions for functions, like `x(t)` and `y(t)` into the original equations containing those functions. This function returns a tuple. The first item in the tuple is ``True`` if the substitution results for each equation is ``0``, and ``False`` otherwise. The second item in the tuple is what the substitution results in. Each element of the ``list`` should always be ``0`` corresponding to each equation if the first item is ``True``. Note that sometimes this function may return ``False``, but with an expression that is identically equal to ``0``, instead of returning ``True``. This is because :py:meth:`~sympy.simplify.simplify.simplify` cannot reduce the expression to ``0``. If an expression returned by each function vanishes identically, then ``sols`` really is a solution to ``eqs``.
If this function seems to hang, it is probably because of a difficult simplification.
Examples ========
>>> from sympy import Eq, diff, symbols, sin, cos, exp, sqrt, S >>> from sympy.solvers.ode import checksysodesol >>> C1, C2 = symbols('C1:3') >>> t = symbols('t') >>> x, y = symbols('x, y', function=True) >>> eq = (Eq(diff(x(t),t), x(t) + y(t) + 17), Eq(diff(y(t),t), -2*x(t) + y(t) + 12)) >>> sol = [Eq(x(t), (C1*sin(sqrt(2)*t) + C2*cos(sqrt(2)*t))*exp(t) - S(5)/3), ... Eq(y(t), (sqrt(2)*C1*cos(sqrt(2)*t) - sqrt(2)*C2*sin(sqrt(2)*t))*exp(t) - S(46)/3)] >>> checksysodesol(eq, sol) (True, [0, 0]) >>> eq = (Eq(diff(x(t),t),x(t)*y(t)**4), Eq(diff(y(t),t),y(t)**3)) >>> sol = [Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), -sqrt(2)*sqrt(-1/(C2 + t))/2), ... Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), sqrt(2)*sqrt(-1/(C2 + t))/2)] >>> checksysodesol(eq, sol) (True, [0, 0])
""" and len(set([func.args for func in funcs]))!=1: raise ValueError("func must be a function of one variable, not %s" % func) raise ValueError("solutions should have one function only") raise ValueError("number of solutions provided does not match the number of equations") sol.rhs == sol_func and not sol.lhs.has(sol_func)): solved = solve(sol, sol_func) if not solved: raise NotImplementedError dictsol[sol_func] = solved dictsol[sol_func] = sol.lhs eq = ss.expand(force=True) else: else: return (False, checkeq)
def odesimp(eq, func, order, constants, hint): r""" Simplifies ODEs, including trying to solve for ``func`` and running :py:meth:`~sympy.solvers.ode.constantsimp`.
It may use knowledge of the type of solution that the hint returns to apply additional simplifications.
It also attempts to integrate any :py:class:`~sympy.integrals.Integral`\s in the expression, if the hint is not an ``_Integral`` hint.
This function should have no effect on expressions returned by :py:meth:`~sympy.solvers.ode.dsolve`, as :py:meth:`~sympy.solvers.ode.dsolve` already calls :py:meth:`~sympy.solvers.ode.odesimp`, but the individual hint functions do not call :py:meth:`~sympy.solvers.ode.odesimp` (because the :py:meth:`~sympy.solvers.ode.dsolve` wrapper does). Therefore, this function is designed for mainly internal use.
Examples ========
>>> from sympy import sin, symbols, dsolve, pprint, Function >>> from sympy.solvers.ode import odesimp >>> x , u2, C1= symbols('x,u2,C1') >>> f = Function('f')
>>> eq = dsolve(x*f(x).diff(x) - f(x) - x*sin(f(x)/x), f(x), ... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral', ... simplify=False) >>> pprint(eq, wrap_line=False) x ---- f(x) / | | / 1 \ | -|u2 + -------| | | /1 \| | | sin|--|| | \ \u2// log(f(x)) = log(C1) + | ---------------- d(u2) | 2 | u2 | /
>>> pprint(odesimp(eq, f(x), 1, set([C1]), ... hint='1st_homogeneous_coeff_subs_indep_div_dep' ... )) #doctest: +SKIP x --------- = C1 /f(x)\ tan|----| \2*x /
"""
# First, integrate if the hint allows it. raise TypeError("eq should be an instance of Equality")
# Second, clean up the arbitrary constants. # Right now, nth linear hints can put as many as 2*order constants in an # expression. If that number grows with another hint, the third argument # here should be raised accordingly, or constantsimp() rewritten to handle # an arbitrary number of constants.
# 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. eq = [Eq(eq.rhs, eq.lhs)]
# make sure we are working with lists of solutions in simplified form. # The solution is already solved
# special simplification of the rhs # Collect terms to make the solution look nice. # This is also necessary for constantsimp to remove unnecessary # terms from the particular solution from variation of parameters # # Collect is not behaving reliably here. The results for # some linear constant-coefficient equations with repeated # roots do not properly simplify all constants sometimes. # 'collectterms' gives different orders sometimes, and results # differ in collect based on that order. The # sort-reverse trick fixes things, but may fail in the # future. In addition, collect is splitting exponentials with # rational powers for no reason. We have to do a match # to fix this using Wilds. global collectterms except Exception: pass
# Collect is splitting exponentials with rational powers for # no reason. We call powsimp to fix.
else: # The solution is not solved, so try to solve it raise NotImplementedError else:
else:
# 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).
# special simplification of the lhs. newi = Eq(newi.lhs.args[0]/C1, C1)
# We cleaned up the constants before solving to help the solve engine with # a simpler expression, but the solved expression could have introduced # things like -C1, so rerun constantsimp() one last time before returning.
# If there is only 1 solution, return it; # otherwise return the list of solutions.
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)
""" except ValueError: funcs = [s.atoms(AppliedUndef) for s in ( sol if is_sequence(sol, set) else [sol])] funcs = set().union(*funcs) if len(funcs) != 1: raise ValueError( 'must pass func arg to checkodesol for this case.') func = funcs.pop() "func must be a function of one variable, not %s" % func)
sol.lhs == func and not sol.rhs.has(func)) and not ( sol.rhs == func and not sol.lhs.has(func)): raise NotImplementedError else: order=order, solve_for_func=False) else: order=order, solve_for_func=False)
# First pass, try substituting a solved solution directly into the # ODE. This has the highest chance of succeeding.
else: # with the new numer_denom in power.py, if we do a simple # expansion then testnum == 0 verifies all solutions. else: # 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. 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) # 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 diffsols = {0: sol.lhs} else: # 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. raise NotImplementedError else: else: # This is what the solution says df/dx should be.
# Make sure the above didn't fail. else: # Substitute it into ODE to check for self consistency. # We can only substitute f(x) if the solution was # solved for f(x).
else: # No sense in overworking simplify -- just prove that the # numerator goes to zero # since solutions are obtained using force=True we test # using the same level of assumptions ## replace function with dummy so assumptions will work ## posify the expression else:
raise NotImplementedError("Unable to test if " + str(sol) + " is a solution to " + str(ode) + ".") else:
r""" Returns an extended integer representing how simple a solution to an ODE is.
The following things are considered, in order from most simple to least:
- ``sol`` is solved for ``func``. - ``sol`` is not solved for ``func``, but can be if passed to solve (e.g., a solution returned by ``dsolve(ode, func, simplify=False``). - If ``sol`` is not solved for ``func``, then base the result on the length of ``sol``, as computed by ``len(str(sol))``. - If ``sol`` has any unevaluated :py:class:`~sympy.integrals.Integral`\s, this will automatically be considered less simple than any of the above.
This function returns an integer such that if solution A is simpler than solution B by above metric, then ``ode_sol_simplicity(sola, func) < ode_sol_simplicity(solb, func)``.
Currently, the following are the numbers returned, but if the heuristic is ever improved, this may change. Only the ordering is guaranteed.
+----------------------------------------------+-------------------+ | Simplicity | Return | +==============================================+===================+ | ``sol`` solved for ``func`` | ``-2`` | +----------------------------------------------+-------------------+ | ``sol`` not solved for ``func`` but can be | ``-1`` | +----------------------------------------------+-------------------+ | ``sol`` is not solved nor solvable for | ``len(str(sol))`` | | ``func`` | | +----------------------------------------------+-------------------+ | ``sol`` contains an | ``oo`` | | :py:class:`~sympy.integrals.Integral` | | +----------------------------------------------+-------------------+
``oo`` here means the SymPy infinity, which should compare greater than any integer.
If you already know :py:meth:`~sympy.solvers.solvers.solve` cannot solve ``sol``, you can use ``trysolving=False`` to skip that step, which is the only potentially slow step. For example, :py:meth:`~sympy.solvers.ode.dsolve` with the ``simplify=False`` flag should do this.
If ``sol`` is a list of solutions, if the worst solution in the list returns ``oo`` it returns that, otherwise it returns ``len(str(sol))``, that is, the length of the string representation of the whole list.
Examples ========
This function is designed to be passed to ``min`` as the key argument, such as ``min(listofsolutions, key=lambda i: ode_sol_simplicity(i, f(x)))``.
>>> from sympy import symbols, Function, Eq, tan, cos, sqrt, Integral >>> from sympy.solvers.ode import ode_sol_simplicity >>> x, C1, C2 = symbols('x, C1, C2') >>> f = Function('f')
>>> ode_sol_simplicity(Eq(f(x), C1*x**2), f(x)) -2 >>> ode_sol_simplicity(Eq(x**2 + f(x), C1), f(x)) -1 >>> ode_sol_simplicity(Eq(f(x), C1*Integral(2*x, x)), f(x)) oo >>> eq1 = Eq(f(x)/tan(f(x)/(2*x)), C1) >>> eq2 = Eq(f(x)/tan(f(x)/(2*x) + f(x)), C2) >>> [ode_sol_simplicity(eq, f(x)) for eq in [eq1, eq2]] [28, 35] >>> min([eq1, eq2], key=lambda i: ode_sol_simplicity(i, f(x))) Eq(f(x)/tan(f(x)/(2*x)), C1)
""" # TODO: if two solutions are solved for f(x), we still want to be # able to get the simpler of the two
# See the docstring for the coercion rules. We check easier (faster) # things here first, to save time.
# See if there are Integrals for i in sol: if ode_sol_simplicity(i, func, trysolving=trysolving) == oo: return oo
return len(str(sol))
# 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 sol.rhs == func and not sol.lhs.has(func): # 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))
else: all(len(x) == 3 for x in expr.limits):
and 0 == expr.diff(i, 2)]
# this calculation can be simplified dlhs[False] = [1] else:
def constantsimp(expr, constants): r""" Simplifies an expression with arbitrary constants in it.
This function is written specifically to work with :py:meth:`~sympy.solvers.ode.dsolve`, and is not intended for general use.
Simplification is done by "absorbing" the arbitrary constants into other arbitrary constants, numbers, and symbols that they are not independent of.
The symbols must all have the same name with numbers after it, for example, ``C1``, ``C2``, ``C3``. The ``symbolname`` here would be '``C``', the ``startnumber`` would be 1, and the ``endnumber`` would be 3. If the arbitrary constants are independent of the variable ``x``, then the independent symbol would be ``x``. There is no need to specify the dependent function, such as ``f(x)``, because it already has the independent symbol, ``x``, in it.
Because terms are "absorbed" into arbitrary constants and because constants are renumbered after simplifying, the arbitrary constants in expr are not necessarily equal to the ones of the same name in the returned result.
If two or more arbitrary constants are added, multiplied, or raised to the power of each other, they are first absorbed together into a single arbitrary constant. Then the new constant is combined into other terms if necessary.
Absorption of constants is done with limited assistance:
1. terms of :py:class:`~sympy.core.add.Add`\s are collected to try join constants so `e^x (C_1 \cos(x) + C_2 \cos(x))` will simplify to `e^x C_1 \cos(x)`;
2. powers with exponents that are :py:class:`~sympy.core.add.Add`\s are expanded so `e^{C_1 + x}` will be simplified to `C_1 e^x`.
Use :py:meth:`~sympy.solvers.ode.constant_renumber` to renumber constants after simplification or else arbitrary numbers on constants may appear, e.g. `C_1 + C_3 x`.
In rare cases, a single constant can be "simplified" into two constants. Every differential equation solution should have as many arbitrary constants as the order of the differential equation. The result here will be technically correct, but it may, for example, have `C_1` and `C_2` in an expression, when `C_1` is actually equal to `C_2`. Use your discretion in such situations, and also take advantage of the ability to use hints in :py:meth:`~sympy.solvers.ode.dsolve`.
Examples ========
>>> from sympy import symbols >>> from sympy.solvers.ode import constantsimp >>> C1, C2, C3, x, y = symbols('C1, C2, C3, x, y') >>> constantsimp(2*C1*x, set([C1, C2, C3])) C1*x >>> constantsimp(C1 + 2 + x, set([C1, C2, C3])) C1 + x >>> constantsimp(C1*C2 + 2 + C2 + C3*x, set([C1, C2, C3])) C1 + C3*x
""" # This function works recursively. The idea is that, for Mul, # Add, Pow, and Function, if the class has a constant in it, then # we can simplify it, which we do by recursing down and # simplifying up. Otherwise, we can skip that part of the # expression.
# try to perform common sub-expression elimination of constant terms else:
# we do not want to factor exponentials, so handle this separately for fi in Mul.make_args(t))
# call recursively if more simplification is possible
r""" Renumber arbitrary constants in ``expr`` to have numbers 1 through `N` where `N` is ``endnumber - startnumber + 1`` at most. In the process, this reorders expression terms in a standard way.
This is a simple function that goes through and renumbers any :py:class:`~sympy.core.symbol.Symbol` with a name in the form ``symbolname + num`` where ``num`` is in the range from ``startnumber`` to ``endnumber``.
Symbols are renumbered based on ``.sort_key()``, so they should be numbered roughly in the order that they appear in the final, printed expression. Note that this ordering is based in part on hashes, so it can produce different results on different machines.
The structure of this function is very similar to that of :py:meth:`~sympy.solvers.ode.constantsimp`.
Examples ========
>>> from sympy import symbols, Eq, pprint >>> from sympy.solvers.ode import constant_renumber >>> x, C0, C1, C2, C3, C4 = symbols('x,C:5')
Only constants in the given range (inclusive) are renumbered; the renumbering always starts from 1:
>>> constant_renumber(C1 + C3 + C4, 'C', 1, 3) C1 + C2 + C4 >>> constant_renumber(C0 + C1 + C3 + C4, 'C', 2, 4) C0 + 2*C1 + C2 >>> constant_renumber(C0 + 2*C1 + C2, 'C', 0, 1) C1 + 3*C2 >>> pprint(C2 + C1*x + C3*x**2) 2 C1*x + C2 + C3*x >>> pprint(constant_renumber(C2 + C1*x + C3*x**2, 'C', 1, 3)) 2 C1 + C2*x + C3*x
""" [constant_renumber(i, symbolname=symbolname, startnumber=startnumber, endnumber=endnumber) for i in expr] ) global newstartnumber symbolname + "%d" % t) for t in range(startnumber, endnumber + 1)]
# make a mapping to send all constantsymbols to S.One and use # that to make sure that term ordering is not dependent on # the indexed value of C
r""" We need to have an internal recursive function so that newstartnumber maintains its values throughout recursive calls.
""" global newstartnumber
_constant_renumber(expr.lhs), _constant_renumber(expr.rhs))
not expr.has(*constantsymbols): # Base case, as above. Hope there aren't constants inside # of some other class, because they won't be renumbered. *[_constant_renumber(x) for x in expr.args]) else: # Renumbering happens here
r""" Converts a solution with Integrals in it into an actual solution.
For most hints, this simply runs ``expr.doit()``.
""" global y sol = expr.subs(y, f(x)) del y else:
# 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. r""" Solves 1st order exact ordinary differential equations.
A 1st order differential equation is called exact if it is the total differential of a function. That is, the differential equation
.. math:: P(x, y) \,\partial{}x + Q(x, y) \,\partial{}y = 0
is exact if there is some function `F(x, y)` such that `P(x, y) = \partial{}F/\partial{}x` and `Q(x, y) = \partial{}F/\partial{}y`. It can be shown that a necessary and sufficient condition for a first order ODE to be exact is that `\partial{}P/\partial{}y = \partial{}Q/\partial{}x`. Then, the solution will be as given below::
>>> from sympy import Function, Eq, Integral, symbols, pprint >>> x, y, t, x0, y0, C1= symbols('x,y,t,x0,y0,C1') >>> P, Q, F= map(Function, ['P', 'Q', 'F']) >>> pprint(Eq(Eq(F(x, y), Integral(P(t, y), (t, x0, x)) + ... Integral(Q(x0, t), (t, y0, y))), C1)) x y / / | | F(x, y) = | P(t, y) dt + | Q(x0, t) dt = C1 | | / / x0 y0
Where the first partials of `P` and `Q` exist and are continuous in a simply connected region.
A note: SymPy currently has no way to represent inert substitution on an expression, so the hint ``1st_exact_Integral`` will return an integral with `dy`. This is supposed to represent the function that you are solving for.
Examples ========
>>> from sympy import Function, dsolve, cos, sin >>> from sympy.abc import x >>> f = Function('f') >>> dsolve(cos(f(x)) - (x*sin(f(x)) - f(x)**2)*f(x).diff(x), ... f(x), hint='1st_exact') Eq(x*cos(f(x)) + f(x)**3/3, C1)
References ==========
- http://en.wikipedia.org/wiki/Exact_differential_equation - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 73
# indirect doctest
""" global y # This is the only way to pass dummy y to _handle_Integral # 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.
r""" Returns the best solution to an ODE from the two hints ``1st_homogeneous_coeff_subs_dep_div_indep`` and ``1st_homogeneous_coeff_subs_indep_div_dep``.
This is as determined by :py:meth:`~sympy.solvers.ode.ode_sol_simplicity`.
See the :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep` and :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep` docstrings for more information on these hints. Note that there is no ``ode_1st_homogeneous_coeff_best_Integral`` hint.
Examples ========
>>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), ... hint='1st_homogeneous_coeff_best', simplify=False)) / 2 \ | 3*x | log|----- + 1| | 2 | \f (x) / log(f(x)) = log(C1) - -------------- 3
References ==========
- http://en.wikipedia.org/wiki/Homogeneous_differential_equation - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 59
# indirect doctest
""" # There are two substitutions that solve the equation, u1=y/x and u2=x/y # They produce different integrals, so try them both and see which # one is easier. func, order, match) func, order, match) # why is odesimp called here? Should it be at the usual spot? sol1, func, order, constants, "1st_homogeneous_coeff_subs_indep_div_dep") sol2, func, order, constants, "1st_homogeneous_coeff_subs_dep_div_indep") trysolving=not simplify))
r""" Solves a 1st order differential equation with homogeneous coefficients using the substitution `u_1 = \frac{\text{<dependent variable>}}{\text{<independent variable>}}`.
This is a differential equation
.. math:: P(x, y) + Q(x, y) dy/dx = 0
such that `P` and `Q` are homogeneous and of the same order. A function `F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`. Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`. See also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`.
If the coefficients `P` and `Q` in the differential equation above are homogeneous functions of the same order, then it can be shown that the substitution `y = u_1 x` (i.e. `u_1 = y/x`) will turn the differential equation into an equation separable in the variables `x` and `u`. If `h(u_1)` is the function that results from making the substitution `u_1 = f(x)/x` on `P(x, f(x))` and `g(u_2)` is the function that results from the substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) + Q(x, f(x)) f'(x) = 0`, then the general solution is::
>>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f, g, h = map(Function, ['f', 'g', 'h']) >>> genform = g(f(x)/x) + h(f(x)/x)*f(x).diff(x) >>> pprint(genform) /f(x)\ /f(x)\ d g|----| + h|----|*--(f(x)) \ x / \ x / dx >>> pprint(dsolve(genform, f(x), ... hint='1st_homogeneous_coeff_subs_dep_div_indep_Integral')) f(x) ---- x / | | -h(u1) log(x) = C1 + | ---------------- d(u1) | u1*h(u1) + g(u1) | /
Where `u_1 h(u_1) + g(u_1) \ne 0` and `x \ne 0`.
See also the docstrings of :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best` and :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`.
Examples ========
>>> from sympy import Function, dsolve >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), ... hint='1st_homogeneous_coeff_subs_dep_div_indep', simplify=False)) / 3 \ |3*f(x) f (x)| log|------ + -----| | x 3 | \ x / log(x) = log(C1) - ------------------- 3
References ==========
- http://en.wikipedia.org/wiki/Homogeneous_differential_equation - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 59
# indirect doctest
""" (-r[r['e']]/(r[r['d']] + u1*r[r['e']])).subs({x: 1, r['y']: u1}), (u1, None, f(x)/x))
r""" Solves a 1st order differential equation with homogeneous coefficients using the substitution `u_2 = \frac{\text{<independent variable>}}{\text{<dependent variable>}}`.
This is a differential equation
.. math:: P(x, y) + Q(x, y) dy/dx = 0
such that `P` and `Q` are homogeneous and of the same order. A function `F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`. Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`. See also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`.
If the coefficients `P` and `Q` in the differential equation above are homogeneous functions of the same order, then it can be shown that the substitution `x = u_2 y` (i.e. `u_2 = x/y`) will turn the differential equation into an equation separable in the variables `y` and `u_2`. If `h(u_2)` is the function that results from making the substitution `u_2 = x/f(x)` on `P(x, f(x))` and `g(u_2)` is the function that results from the substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) + Q(x, f(x)) f'(x) = 0`, then the general solution is:
>>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f, g, h = map(Function, ['f', 'g', 'h']) >>> genform = g(x/f(x)) + h(x/f(x))*f(x).diff(x) >>> pprint(genform) / x \ / x \ d g|----| + h|----|*--(f(x)) \f(x)/ \f(x)/ dx >>> pprint(dsolve(genform, f(x), ... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral')) x ---- f(x) / | | -g(u2) | ---------------- d(u2) | u2*g(u2) + h(u2) | / <BLANKLINE> f(x) = C1*e
Where `u_2 g(u_2) + h(u_2) \ne 0` and `f(x) \ne 0`.
See also the docstrings of :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best` and :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`.
Examples ========
>>> from sympy import Function, pprint, dsolve >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), ... hint='1st_homogeneous_coeff_subs_indep_div_dep', ... simplify=False)) / 2 \ | 3*x | log|----- + 1| | 2 | \f (x) / log(f(x)) = log(C1) - -------------- 3
References ==========
- http://en.wikipedia.org/wiki/Homogeneous_differential_equation - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 59
# indirect doctest
""" simplify( (-r[r['d']]/(r[r['e']] + u2*r[r['d']])).subs({x: u2, r['y']: 1})), (u2, None, x/f(x)))
# XXX: Should this function maybe go somewhere else?
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
"""
# The following are not supported
# These are all constants eq.is_NumberSymbol or eq.is_number ):
# Replace all functions with dummy variables else:
# assuming order of a nested function can only be equal to zero 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
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
"""
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
"""
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 """
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
r""" Gives a power series solution to a second order homogeneous differential equation with polynomial coefficients at an ordinary point. A homogenous differential equation is of the form
.. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0
For simplicity it is assumed that `P(x)`, `Q(x)` and `R(x)` are polynomials, it is sufficient that `\frac{Q(x)}{P(x)}` and `\frac{R(x)}{P(x)}` exists at `x_{0}`. A recurrence relation is obtained by substituting `y` as `\sum_{n=0}^\infty a_{n}x^{n}`, in the differential equation, and equating the nth term. Using this relation various terms can be generated.
Examples ========
>>> from sympy import dsolve, Function, pprint >>> from sympy.abc import x, y >>> f = Function("f") >>> eq = f(x).diff(x, 2) + f(x) >>> pprint(dsolve(eq, hint='2nd_power_series_ordinary')) / 4 2 \ / 2 \ |x x | | x | / 6\ f(x) = C2*|-- - -- + 1| + C1*x*|- -- + 1| + O\x / \24 2 / \ 6 /
References ========== - http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx - George E. Simmons, "Differential Equations with Applications and Historical Notes", p.p 176 - 184
"""
# Generating the recurrence relation which works this way # a] For the second order term the summation begins at n = 2. The coefficients # p is multiplied with an*(n - 1)*(n - 2)*x**n-2 and a substitution is made such that # the exponent of x becomes n. # For example, if p is x, then the second degree recurrence term is # an*(n - 1)*(n - 2)*x**n-1, substituting (n - 1) as n, it transforms to # an+1*n*(n - 1)*x**n. # A similar process is done with the first order and zeroth order term.
else: # Seeing if the startterm can be reduced further. # If it vanishes for n lesser than startind, it is # equal to summation from n. else: else:
# Stripping of terms so that the sum starts with the same number.
else:
# Finding the recurrence relation in terms of the largest term. else:
# Checking how many values are already present
#while tcounter < terms - 2: # Assuming c0 and c1 to be arbitrary
# Post processing x - x0)**fact)
r""" Gives a power series solution to a second order homogeneous differential equation with polynomial coefficients at a regular point. A second order homogenous differential equation is of the form
.. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0
A point is said to regular singular at `x0` if `x - x0\frac{Q(x)}{P(x)}` and `(x - x0)^{2}\frac{R(x)}{P(x)}` are analytic at `x0`. For simplicity `P(x)`, `Q(x)` and `R(x)` are assumed to be polynomials. The algorithm for finding the power series solutions is:
1. Try expressing `(x - x0)P(x)` and `((x - x0)^{2})Q(x)` as power series solutions about x0. Find `p0` and `q0` which are the constants of the power series expansions. 2. Solve the indicial equation `f(m) = m(m - 1) + m*p0 + q0`, to obtain the roots `m1` and `m2` of the indicial equation. 3. If `m1 - m2` is a non integer there exists two series solutions. If `m1 = m2`, there exists only one solution. If `m1 - m2` is an integer, then the existence of one solution is confirmed. The other solution may or may not exist.
The power series solution is of the form `x^{m}\sum_{n=0}^\infty a_{n}x^{n}`. The coefficients are determined by the following recurrence relation. `a_{n} = -\frac{\sum_{k=0}^{n-1} q_{n-k} + (m + k)p_{n-k}}{f(m + n)}`. For the case in which `m1 - m2` is an integer, it can be seen from the recurrence relation that for the lower root `m`, when `n` equals the difference of both the roots, the denominator becomes zero. So if the numerator is not equal to zero, a second series solution exists.
Examples ========
>>> from sympy import dsolve, Function, pprint >>> from sympy.abc import x, y >>> f = Function("f") >>> eq = x*(f(x).diff(x, 2)) + 2*(f(x).diff(x)) + x*f(x) >>> pprint(dsolve(eq)) / 6 4 2 \ | x x x | / 4 2 \ C1*|- --- + -- - -- + 1| | x x | \ 720 24 2 / / 6\ f(x) = C2*|--- - -- + 1| + ------------------------ + O\x / \120 6 / x
References ========== - George E. Simmons, "Differential Equations with Applications and Historical Notes", p.p 176 - 184
"""
# Generating the indicial equation else: else:
[sol.is_real for sol in sollist]): # Only one series solution exists in this case. return Eq(f(x), Order(terms))
else: # Irrespective of whether m1 - m2 is an integer or not, one # Frobenius series solution exists. # Second frobenius series solution exists. serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1) else: # Check if second frobenius series solution exists.
[C0, C1]) + Order(x**terms))
r""" Returns a dict with keys as coefficients and values as their values in terms of C0 """ # In cases where m1 - m2 is not an integer
# Order term not present ser = ser.subs(x, x + x0) # Order term present else: # Removing order # Fill in with zeros, if coefficients are zero.
# Checking for cases when m1 - m2 is an integer. If num equals zero # then a second Frobenius series solution cannot be found. If num is not zero # then set constant as zero and proceed. return False else: else:
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
""" else: or f == func): else:
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 Eq(f(x), sqrt(x)*(C1 + C2*log(x)))
Note that because this method does not involve integration, there is no ``nth_linear_euler_eq_homogeneous_Integral`` hint.
The following is for internal use:
- ``returns = 'sol'`` returns the solution to the ODE. - ``returns = 'list'`` returns a list of linearly independent solutions, corresponding to the fundamental solution set, for use with non homogeneous solution methods like variation of parameters and undetermined coefficients. Note that, though the solutions should be linearly independent, this function does not explicitly check that. You can do ``assert simplify(wronskian(sollist)) != 0`` to check for linear independence. Also, ``assert len(sollist) == order`` will need to pass. - ``returns = 'both'``, return a dictionary ``{'sol': <solution to ODE>, 'list': <list of linearly independent solutions>}``.
Examples ========
>>> from sympy import Function, dsolve, pprint >>> from sympy.abc import x >>> f = Function('f') >>> eq = f(x).diff(x, 2)*x**2 - 4*f(x).diff(x)*x + 6*f(x) >>> pprint(dsolve(eq, f(x), ... hint='nth_linear_euler_eq_homogeneous')) 2 f(x) = x *(C1 + C2*x)
References ==========
- http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation - C. Bender & S. Orszag, "Advanced Mathematical Methods for Scientists and Engineers", Springer 1999, pp. 12
# indirect doctest
""" global collectterms
# First, set up characteristic equation.
# A generator of constants
# Create a dict root: multiplicity or charroots # We need keep track of terms so we can run collect() at the end. # This is necessary for constantsimp to work properly. gsol += (x**root) * constants.pop() if multiplicity != 1: raise ValueError("Value should be 1") collectterms = [(0, root, 0)] + collectterms else: reroot = re(root) imroot = im(root) gsol += ln(x)**i * (x**reroot) * ( constants.pop() * sin(abs(imroot)*ln(x)) + constants.pop() * cos(imroot*ln(x))) # Preserve ordering (multiplicity, real part, imaginary part) # It will be assumed implicitly when constructing # fundamental solution sets. collectterms = [(i, reroot, imroot)] + collectterms # HOW TO TEST THIS CODE? (dsolve does not pass 'returns' through) # Create a list of (hopefully) linearly independent solutions # Keep track of when to use sin or cos for nonzero imroot 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) return gensols else: else: raise ValueError('Unknown value for key "returns".')
r""" Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional ordinary differential equation using undetermined coefficients.
This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) \cdots`.
These equations can be solved in a general manner, by substituting solutions of the form `x = exp(t)`, and deriving a characteristic equation of form `g(exp(t)) = b_0 f(t) + b_1 f'(t) + b_2 f''(t) \cdots` which can be then solved by nth_linear_constant_coeff_undetermined_coefficients if g(exp(t)) has finite number of lineary independent derivatives.
Functions that fit this requirement are finite sums functions of the form `a x^i e^{b x} \sin(c x + d)` or `a x^i e^{b x} \cos(c x + d)`, where `i` is a non-negative integer and `a`, `b`, `c`, and `d` are constants. For example any polynomial in `x`, functions like `x^2 e^{2 x}`, `x \sin(x)`, and `e^x \cos(x)` can all be used. Products of `\sin`'s and `\cos`'s have a finite number of derivatives, because they can be expanded into `\sin(a x)` and `\cos(b x)` terms. However, SymPy currently cannot do that expansion, so you will need to manually rewrite the expression in terms of the above to use this method. So, for example, you will need to manually convert `\sin^2(x)` into `(1 + \cos(2 x))/2` to properly apply the method of undetermined coefficients on it.
After replacement of x by exp(t), this method works by creating a trial function from the expression and all of its linear independent derivatives and substituting them into the original ODE. The coefficients for each term will be a system of linear equations, which are be solved for and substituted, giving the solution. If any of the trial functions are linearly dependent on the solution to the homogeneous equation, they are multiplied by sufficient `x` to make them linearly independent.
Examples ========
>>> from sympy import dsolve, Function, Derivative, log >>> from sympy.abc import x >>> f = Function('f') >>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - log(x) >>> dsolve(eq, f(x), ... hint='nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients').expand() Eq(f(x), C1*x + C2*x**2 + log(x)/2 + 3/4)
"""
r""" Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional ordinary differential equation using variation of parameters.
This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) \cdots`.
This method works by assuming that the particular solution takes the form
.. math:: \sum_{x=1}^{n} c_i(x) y_i(x) {a_n} {x^n} \text{,}
where `y_i` is the `i`\th solution to the homogeneous equation. The solution is then solved using Wronskian's and Cramer's Rule. The particular solution is given by multiplying eq given below with `a_n x^{n}`
.. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx \right) y_i(x) \text{,}
where `W(x)` is the Wronskian of the fundamental system (the system of `n` linearly independent solutions to the homogeneous equation), and `W_i(x)` is the Wronskian of the fundamental system with the `i`\th column replaced with `[0, 0, \cdots, 0, \frac{x^{- n}}{a_n} g{\left (x \right )}]`.
This method is general enough to solve any `n`\th order inhomogeneous linear differential equation, but sometimes SymPy cannot simplify the Wronskian well enough to integrate it. If this method hangs, try using the ``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and simplifying the integrals manually. Also, prefer using ``nth_linear_constant_coeff_undetermined_coefficients`` when it applies, because it doesn't use integration, making it faster and more reliable.
Warning, using simplify=False with 'nth_linear_constant_coeff_variation_of_parameters' in :py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will not attempt to simplify the Wronskian before integrating. It is recommended that you only use simplify=False with 'nth_linear_constant_coeff_variation_of_parameters_Integral' for this method, especially if the solution to the homogeneous equation has trigonometric functions in it.
Examples ========
>>> from sympy import Function, dsolve, Derivative >>> from sympy.abc import x >>> f = Function('f') >>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - x**4 >>> dsolve(eq, f(x), ... hint='nth_linear_euler_eq_nonhomogeneous_variation_of_parameters').expand() Eq(f(x), C1*x + C2*x**2 + x**4/6)
"""
r""" Solves an almost-linear differential equation.
The general form of an almost linear differential equation is
.. math:: f(x) g(y) y + k(x) l(y) + m(x) = 0 \text{where} l'(y) = g(y)\text{.}
This can be solved by substituting `l(y) = u(y)`. Making the given substitution reduces it to a linear differential equation of the form `u' + P(x) u + Q(x) = 0`.
The general solution is
>>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x, y, n >>> f, g, k, l = map(Function, ['f', 'g', 'k', 'l']) >>> genform = Eq(f(x)*(l(y).diff(y)) + k(x)*l(y) + g(x)) >>> pprint(genform) d f(x)*--(l(y)) + g(x) + k(x)*l(y) = 0 dy >>> pprint(dsolve(genform, hint = 'almost_linear')) / // -y*g(x) \\ | || -------- for k(x) = 0|| | || f(x) || -y*k(x) | || || -------- | || y*k(x) || f(x) l(y) = |C1 + |< ------ ||*e | || f(x) || | ||-g(x)*e || | ||-------------- otherwise || | || k(x) || \ \\ //
See Also ======== :meth:`sympy.solvers.ode.ode_1st_linear`
Examples ========
>>> from sympy import Function, Derivative, pprint >>> from sympy.solvers.ode import dsolve, classify_ode >>> from sympy.abc import x >>> f = Function('f') >>> d = f(x).diff(x) >>> eq = x*d + x*f(x) + 1 >>> dsolve(eq, f(x), hint='almost_linear') Eq(f(x), (C1 - Ei(x))*exp(-x)) >>> pprint(dsolve(eq, f(x), hint='almost_linear')) -x f(x) = (C1 - Ei(x))*e
References ==========
- Joel Moses, "Symbolic Integration - The Stormy Decade", Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558 """
# Since ode_1st_linear has already been implemented, and the # coefficients have been modified to the required form in # classify_ode, just passing eq, func, order and match to # ode_1st_linear will give the required output.
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))
""" r''' Internal function of _linear_coeff_match that returns Rationals a, b, c if eq is a*x + b*f(x) + c, else None. '''
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. '''
len(fi.args) == 1 and not fi.args[0].is_Function] or set([expr])
r""" Solves a differential equation with linear coefficients.
The general form of a differential equation with linear coefficients is
.. math:: y' + F\left(\!\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}\!\right) = 0\text{,}
where `a_1`, `b_1`, `c_1`, `a_2`, `b_2`, `c_2` are constants and `a_1 b_2 - a_2 b_1 \ne 0`.
This can be solved by substituting:
.. math:: x = x' + \frac{b_2 c_1 - b_1 c_2}{a_2 b_1 - a_1 b_2}
y = y' + \frac{a_1 c_2 - a_2 c_1}{a_2 b_1 - a_1 b_2}\text{.}
This substitution reduces the equation to a homogeneous differential equation.
See Also ======== :meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_best` :meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep` :meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`
Examples ========
>>> from sympy import Function, Derivative, pprint >>> from sympy.solvers.ode import dsolve, classify_ode >>> from sympy.abc import x >>> f = Function('f') >>> df = f(x).diff(x) >>> eq = (x + f(x) + 1)*df + (f(x) - 6*x + 1) >>> dsolve(eq, hint='linear_coefficients') [Eq(f(x), -x - sqrt(C1 + 7*x**2) - 1), Eq(f(x), -x + sqrt(C1 + 7*x**2) - 1)] >>> pprint(dsolve(eq, hint='linear_coefficients')) ___________ ___________ / 2 / 2 [f(x) = -x - \/ C1 + 7*x - 1, f(x) = -x + \/ C1 + 7*x - 1]
References ==========
- Joel Moses, "Symbolic Integration - The Stormy Decade", Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558 """
r""" Solves a differential equation that can be reduced to the separable form.
The general form of this equation is
.. math:: y' + (y/x) H(x^n y) = 0\text{}.
This can be solved by substituting `u(y) = x^n y`. The equation then reduces to the separable form `\frac{u'}{u (\mathrm{power} - H(u))} - \frac{1}{x} = 0`.
The general solution is:
>>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x, n >>> f, g = map(Function, ['f', 'g']) >>> genform = f(x).diff(x) + (f(x)/x)*g(x**n*f(x)) >>> pprint(genform) / n \ d f(x)*g\x *f(x)/ --(f(x)) + --------------- dx x >>> pprint(dsolve(genform, hint='separable_reduced')) n x *f(x) / | | 1 | ------------ dy = C1 + log(x) | y*(n - g(y)) | /
See Also ======== :meth:`sympy.solvers.ode.ode_separable`
Examples ========
>>> from sympy import Function, Derivative, pprint >>> from sympy.solvers.ode import dsolve, classify_ode >>> from sympy.abc import x >>> f = Function('f') >>> d = f(x).diff(x) >>> eq = (x - x**2*f(x))*d - f(x) >>> dsolve(eq, hint='separable_reduced') [Eq(f(x), (-sqrt(C1*x**2 + 1) + 1)/x), Eq(f(x), (sqrt(C1*x**2 + 1) + 1)/x)] >>> pprint(dsolve(eq, hint='separable_reduced')) ___________ ___________ / 2 / 2 - \/ C1*x + 1 + 1 \/ C1*x + 1 + 1 [f(x) = --------------------, f(x) = ------------------] x x
References ==========
- Joel Moses, "Symbolic Integration - The Stormy Decade", Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558 """
# Arguments are passed in a way so that they are coherent with the # ode_separable function
r""" The power series solution is a method which gives the Taylor series expansion to the solution of a differential equation.
For a first order differential equation `\frac{dy}{dx} = h(x, y)`, a power series solution exists at a point `x = x_{0}` if `h(x, y)` is analytic at `x_{0}`. The solution is given by
.. math:: y(x) = y(x_{0}) + \sum_{n = 1}^{\infty} \frac{F_{n}(x_{0},b)(x - x_{0})^n}{n!},
where `y(x_{0}) = b` is the value of y at the initial value of `x_{0}`. To compute the values of the `F_{n}(x_{0},b)` the following algorithm is followed, until the required number of terms are generated.
1. `F_1 = h(x_{0}, b)` 2. `F_{n+1} = \frac{\partial F_{n}}{\partial x} + \frac{\partial F_{n}}{\partial y}F_{1}`
Examples ========
>>> from sympy import Function, Derivative, pprint, exp >>> from sympy.solvers.ode import dsolve >>> from sympy.abc import x >>> f = Function('f') >>> eq = exp(x)*(f(x).diff(x)) - f(x) >>> pprint(dsolve(eq, hint='1st_power_series')) 3 4 5 C1*x C1*x C1*x / 6\ f(x) = C1 + C1*x - ----- + ----- + ----- + O\x / 6 24 60
References ==========
- Travis W. Walker, Analytic power series technique for solving first-order differential equations, p.p 17, 18
"""
# First term return Eq(f(x), value)
# Initialisation # Derivative does not exist, not analytic return Eq(f(x), oo)
# Same logic as above return Eq(f(x), oo)
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 Eq(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
"""
# First, set up characteristic equation.
else:
# A generator of constants
# Create a dict root: multiplicity or charroots # We need to keep track of terms so we can run collect() at the end. # This is necessary for constantsimp to work properly. global collectterms raise ValueError("Value should be 1") # This ordering is important else: gensols.append(x**i*exp(root*x)) collectterms = [(i, root, 0)] + collectterms continue # Remove this condition when re and im stop returning # circular atan2 usages. gensols.append(x**i*exp(root*x)) collectterms = [(i, root, 0)] + collectterms else:
# This ordering is important return gensols else: else: raise ValueError('Unknown value for key "returns".')
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
""" returns='both')
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']``.
""" global collectterms raise NotImplementedError("Cannot find " + str(order) + " solutions to the homogeneous equation necessary to apply" + " undetermined coefficients to " + str(eq) + " (number of terms != order)") # Alternate between sin and cos else: else:
# 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. else:
raise NotImplementedError( "Could not solve `%s` using the " "method of undetermined coefficients " "(unable to solve for coefficients)." % eq)
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}
"""
r""" Test if ``expr`` fits the proper form for undetermined coefficients. """ # Make sure that there is only one trig function in the args. # See the docstring. else: else: else: expr.exp >= 0: else: else:
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. """ r""" Returns the expression without a coefficient.
Similar to expr.as_independent(x)[1], except it only works multiplicatively. """
else: else: # If you get stuck in this loop, then _test_term is probably # broken else:
# 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).
r""" Solves an `n`\th order linear differential equation with constant coefficients using the method of variation of parameters.
This method works on any differential equations of the form
.. math:: f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + a_0 f(x) = P(x)\text{.}
This method works by assuming that the particular solution takes the form
.. math:: \sum_{x=1}^{n} c_i(x) y_i(x)\text{,}
where `y_i` is the `i`\th solution to the homogeneous equation. The solution is then solved using Wronskian's and Cramer's Rule. The particular solution is given by
.. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx \right) y_i(x) \text{,}
where `W(x)` is the Wronskian of the fundamental system (the system of `n` linearly independent solutions to the homogeneous equation), and `W_i(x)` is the Wronskian of the fundamental system with the `i`\th column replaced with `[0, 0, \cdots, 0, P(x)]`.
This method is general enough to solve any `n`\th order inhomogeneous linear differential equation with constant coefficients, but sometimes SymPy cannot simplify the Wronskian well enough to integrate it. If this method hangs, try using the ``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and simplifying the integrals manually. Also, prefer using ``nth_linear_constant_coeff_undetermined_coefficients`` when it applies, because it doesn't use integration, making it faster and more reliable.
Warning, using simplify=False with 'nth_linear_constant_coeff_variation_of_parameters' in :py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will not attempt to simplify the Wronskian before integrating. It is recommended that you only use simplify=False with 'nth_linear_constant_coeff_variation_of_parameters_Integral' for this method, especially if the solution to the homogeneous equation has trigonometric functions in it.
Examples ========
>>> from sympy import Function, dsolve, pprint, exp, log >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(f(x).diff(x, 3) - 3*f(x).diff(x, 2) + ... 3*f(x).diff(x) - f(x) - exp(x)*log(x), f(x), ... hint='nth_linear_constant_coeff_variation_of_parameters')) / 3 \ | 2 x *(6*log(x) - 11)| x f(x) = |C1 + C2*x + C3*x + ------------------|*e \ 36 /
References ==========
- http://en.wikipedia.org/wiki/Variation_of_parameters - http://planetmath.org/VariationOfParameters - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 233
# indirect doctest
"""
gensol = ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match, returns='both') match.update(gensol) return _solve_variation_of_parameters(eq, func, order, match)
r""" Helper function for the method of variation of parameters and nonhomogeneous euler eq.
See the :py:meth:`~sympy.solvers.ode.ode_nth_linear_constant_coeff_variation_of_parameters` docstring for more information on this method.
The parameter ``match`` should be a dictionary that has the following keys:
``list`` A list of solutions to the homogeneous equation, such as the list returned by ``ode_nth_linear_constant_coeff_homogeneous(returns='list')``.
``sol`` The general solution, such as the solution returned by ``ode_nth_linear_constant_coeff_homogeneous(returns='sol')``.
"""
# some ODEs. See issue 4662, for example.
# To reduce commonly occuring sin(x)**2 + cos(x)**2 to 1 # 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)") raise NotImplementedError("Cannot find " + str(order) + " solutions to the homogeneous equation nessesary to apply " + "variation of parameters to " + str(eq) + " (number of terms != order)")
r""" Solves separable 1st order differential equations.
This is any differential equation that can be written as `P(y) \tfrac{dy}{dx} = Q(x)`. The solution can then just be found by rearranging terms and integrating: `\int P(y) \,dy = \int Q(x) \,dx`. This hint uses :py:meth:`sympy.simplify.simplify.separatevars` as its back end, so if a separable equation is not caught by this solver, it is most likely the fault of that function. :py:meth:`~sympy.simplify.simplify.separatevars` is smart enough to do most expansion and factoring necessary to convert a separable equation `F(x, y)` into the proper form `P(x)\cdot{}Q(y)`. The general solution is::
>>> from sympy import Function, dsolve, Eq, pprint >>> from sympy.abc import x >>> a, b, c, d, f = map(Function, ['a', 'b', 'c', 'd', 'f']) >>> genform = Eq(a(x)*b(f(x))*f(x).diff(x), c(x)*d(f(x))) >>> pprint(genform) d a(x)*b(f(x))*--(f(x)) = c(x)*d(f(x)) dx >>> pprint(dsolve(genform, f(x), hint='separable_Integral')) f(x) / / | | | b(y) | c(x) | ---- dy = C1 + | ---- dx | d(y) | a(x) | | / /
Examples ========
>>> from sympy import Function, dsolve, Eq >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(Eq(f(x)*f(x).diff(x) + x, 3*x*f(x)**2), f(x), ... hint='separable', simplify=False)) / 2 \ 2 log\3*f (x) - 1/ x ---------------- = C1 + -- 6 2
References ==========
- M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", Dover 1963, pp. 52
# indirect doctest
""" (r['y'], None, u)), Integral(-r['m1']['coeff']*r['m1'][x]/ r['m2'][x], x) + C1)
r""" This function is used to check if the given infinitesimals are the actual infinitesimals of the given first order differential equation. This method is specific to the Lie Group Solver of ODEs.
As of now, it simply checks, by substituting the infinitesimals in the partial differential equation.
.. math:: \frac{\partial \eta}{\partial x} + \left(\frac{\partial \eta}{\partial y} - \frac{\partial \xi}{\partial x}\right)*h - \frac{\partial \xi}{\partial y}*h^{2} - \xi\frac{\partial h}{\partial x} - \eta\frac{\partial h}{\partial y} = 0
where `\eta`, and `\xi` are the infinitesimals and `h(x,y) = \frac{dy}{dx}`
The infinitesimals should be given in the form of a list of dicts ``[{xi(x, y): inf, eta(x, y): inf}]``, corresponding to the output of the function infinitesimals. It returns a list of values of the form ``[(True/False, sol)]`` where ``sol`` is the value obtained after substituting the infinitesimals in the PDE. If it is ``True``, then ``sol`` would be 0.
""" raise ValueError("ODE's have only one independent variable") else: raise NotImplementedError("Lie groups solver has been implemented " "only for first order differential equations") else:
else: except NotImplementedError: raise NotImplementedError("Infinitesimals for the " "first order ODE could not be found") else:
(xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) eta: S(sol[deta]).subs(func, y)} else:
r""" This hint implements the Lie group method of solving first order differential equations. The aim is to convert the given differential equation from the given coordinate given system into another coordinate system where it becomes invariant under the one-parameter Lie group of translations. The converted ODE is quadrature and can be solved easily. It makes use of the :py:meth:`sympy.solvers.ode.infinitesimals` function which returns the infinitesimals of the transformation.
The coordinates `r` and `s` can be found by solving the following Partial Differential Equations.
.. math :: \xi\frac{\partial r}{\partial x} + \eta\frac{\partial r}{\partial y} = 0
.. math :: \xi\frac{\partial s}{\partial x} + \eta\frac{\partial s}{\partial y} = 1
The differential equation becomes separable in the new coordinate system
.. math :: \frac{ds}{dr} = \frac{\frac{\partial s}{\partial x} + h(x, y)\frac{\partial s}{\partial y}}{ \frac{\partial r}{\partial x} + h(x, y)\frac{\partial r}{\partial y}}
After finding the solution by integration, it is then converted back to the original coordinate system by subsituting `r` and `s` in terms of `x` and `y` again.
Examples ========
>>> from sympy import Function, dsolve, Eq, exp, pprint >>> from sympy.abc import x >>> f = Function('f') >>> pprint(dsolve(f(x).diff(x) + 2*x*f(x) - x*exp(-x**2), f(x), ... hint='lie_group')) / 2\ 2 | x | -x f(x) = |C1 + --|*e \ 2 /
References ==========
- Solving differential equations by Symmetry Groups, John Starrett, pp. 1 - pp. 14
"""
else: except NotImplementedError: raise NotImplementedError("Unable to solve the differential equation " + str(eq) + " by the lie group method") else:
" are not the infinitesimals to the given equation") else:
# This is done so that if: # a] solve raises a NotImplementedError. # b] any heuristic raises a ValueError # another heuristic can be used. except ValueError: continue else: # This condition creates recursion while using pdsolve. # Since the first step while solving a PDE of form # a*(f(x, y).diff(x)) + b*(f(x, y).diff(y)) + c = 0 # is to solve the ODE dy/dx = b/a continue except NotImplementedError: continue else: # Trying to separate, r and s coordinates # Substituting and reverting back to original coordinates except NotImplementedError: tempsol.append(deq) else: else: return [Eq(f(x), sol) for sol in sdeq]
elif num: # (dr/ds) is zero which means r is constant return Eq(f(x), solve(rcoord - C1, y)[0])
# If nothing works, return solution as it is, without solving for y if tempsol: if len(tempsol) == 1: return Eq(tempsol.pop().subs(y, f(x)), 0) else: return [Eq(sol.subs(y, f(x)), 0) for sol in tempsol]
raise NotImplementedError("The given ODE " + str(eq) + " cannot be solved by" + " the lie group method")
r""" This function is strictly meant for internal use by the Lie group ODE solving method. It replaces arbitrary functions returned by pdsolve with either 0 or 1 or the args of the arbitrary function.
The algorithm used is: 1] If coords is an instance of an Undefined Function, then the args are returned 2] If the arbitrary function is present in an Add object, it is replaced by zero. 3] If the arbitrary function is present in an Mul object, it is replaced by one. 4] If coords has no Undefined Function, it is returned as it is.
Examples ========
>>> from sympy.solvers.ode import _lie_group_remove >>> from sympy import Function >>> from sympy.abc import x, y >>> F = Function("F") >>> eq = x**2*y >>> _lie_group_remove(eq) x**2*y >>> eq = F(x**2*y) >>> _lie_group_remove(eq) x**2*y >>> eq = y**2*x + F(x**3) >>> _lie_group_remove(eq) x*y**2 >>> eq = (F(x**3) + y)*x**4 >>> _lie_group_remove(eq) x**4*y
""" elif coords.is_Pow: base, expr = coords.as_base_exp() base = _lie_group_remove(base) expr = _lie_group_remove(expr) return base**expr elif coords.is_Mul: mulargs = [] coordargs = coords.args for arg in coordargs: if not isinstance(coords, AppliedUndef): mulargs.append(_lie_group_remove(arg)) return Mul(*mulargs) return coords
r""" The infinitesimal functions of an ordinary differential equation, `\xi(x,y)` and `\eta(x,y)`, are the infinitesimals of the Lie group of point transformations for which the differential equation is invariant. So, the ODE `y'=f(x,y)` would admit a Lie group `x^*=X(x,y;\varepsilon)=x+\varepsilon\xi(x,y)`, `y^*=Y(x,y;\varepsilon)=y+\varepsilon\eta(x,y)` such that `(y^*)'=f(x^*, y^*)`. A change of coordinates, to `r(x,y)` and `s(x,y)`, can be performed so this Lie group becomes the translation group, `r^*=r` and `s^*=s+\varepsilon`. They are tangents to the coordinate curves of the new system.
Consider the transformation `(x, y) \to (X, Y)` such that the differential equation remains invariant. `\xi` and `\eta` are the tangents to the transformed coordinates `X` and `Y`, at `\varepsilon=0`.
.. math:: \left(\frac{\partial X(x,y;\varepsilon)}{\partial\varepsilon }\right)|_{\varepsilon=0} = \xi, \left(\frac{\partial Y(x,y;\varepsilon)}{\partial\varepsilon }\right)|_{\varepsilon=0} = \eta,
The infinitesimals can be found by solving the following PDE:
>>> from sympy import Function, diff, Eq, pprint >>> from sympy.abc import x, y >>> xi, eta, h = map(Function, ['xi', 'eta', 'h']) >>> h = h(x, y) # dy/dx = h >>> eta = eta(x, y) >>> xi = xi(x, y) >>> genform = Eq(eta.diff(x) + (eta.diff(y) - xi.diff(x))*h ... - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)), 0) >>> pprint(genform) /d d \ d 2 d |--(eta(x, y)) - --(xi(x, y))|*h(x, y) - eta(x, y)*--(h(x, y)) - h (x, y)*--(x \dy dx / dy dy <BLANKLINE> d d i(x, y)) - xi(x, y)*--(h(x, y)) + --(eta(x, y)) = 0 dx dx
Solving the above mentioned PDE is not trivial, and can be solved only by making intelligent assumptions for `\xi` and `\eta` (heuristics). Once an infinitesimal is found, the attempt to find more heuristics stops. This is done to optimise the speed of solving the differential equation. If a list of all the infinitesimals is needed, ``hint`` should be flagged as ``all``, which gives the complete list of infinitesimals. If the infinitesimals for a particular heuristic needs to be found, it can be passed as a flag to ``hint``.
Examples ========
>>> from sympy import Function, diff >>> from sympy.solvers.ode import infinitesimals >>> from sympy.abc import x >>> f = Function('f') >>> eq = f(x).diff(x) - x**2*f(x) >>> infinitesimals(eq) [{eta(x, f(x)): exp(x**3/3), xi(x, f(x)): 0}]
References ==========
- Solving differential equations by Symmetry Groups, John Starrett, pp. 1 - pp. 14
"""
raise ValueError("ODE's have only one independent variable") else: raise NotImplementedError("Infinitesimals for only " "first order ODE's have been implemented") else: # Matching differential equation of the form a*df + b else: else: except NotImplementedError: raise NotImplementedError("Infinitesimals for the " "first order ODE could not be found") else:
xieta = [] for heuristic in lie_heuristics: function = globals()['lie_heuristic_' + heuristic] inflist = function(match, comp=True) if inflist: xieta.extend([inf for inf in inflist if inf not in xieta]) if xieta: return xieta else: raise NotImplementedError("Infinitesimals could not be found for " "the given ODE")
for heuristic in lie_heuristics: function = globals()['lie_heuristic_' + heuristic] xieta = function(match, comp=False) if xieta: return xieta
raise NotImplementedError("Infinitesimals could not be found for" " the given ODE")
else: else: raise ValueError("Infinitesimals could not be found using the" " given heuristic")
r""" The first heuristic uses the following four sets of assumptions on `\xi` and `\eta`
.. math:: \xi = 0, \eta = f(x)
.. math:: \xi = 0, \eta = f(y)
.. math:: \xi = f(x), \eta = 0
.. math:: \xi = f(y), \eta = 0
The success of this heuristic is determined by algebraic factorisation. For the first assumption `\xi = 0` and `\eta` to be a function of `x`, the PDE
.. math:: \frac{\partial \eta}{\partial x} + (\frac{\partial \eta}{\partial y} - \frac{\partial \xi}{\partial x})*h - \frac{\partial \xi}{\partial y}*h^{2} - \xi*\frac{\partial h}{\partial x} - \eta*\frac{\partial h}{\partial y} = 0
reduces to `f'(x) - f\frac{\partial h}{\partial y} = 0` If `\frac{\partial h}{\partial y}` is a function of `x`, then this can usually be integrated easily. A similar idea is applied to the other 3 assumptions as well.
References ==========
- E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra Solving of First Order ODEs Using Symmetry Methods, pp. 8
"""
except NotImplementedError: pass else: return [inf]
except NotImplementedError: pass else: return [inf]
except NotImplementedError: pass else: return [inf]
except NotImplementedError: pass else: return [inf]
r""" The second heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = 0, \xi = f(x)*g(y)
.. math:: \eta = f(x)*g(y), \xi = 0
The first assumption of this heuristic holds good if `\frac{1}{h^{2}}\frac{\partial^2}{\partial x \partial y}\log(h)` is separable in `x` and `y`, then the separated factors containing `x` is `f(x)`, and `g(y)` is obtained by
.. math:: e^{\int f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)\,dy}
provided `f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)` is a function of `y` only.
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption satisifes. After obtaining `f(x)` and `g(y)`, the coordinates are again interchanged, to get `\eta` as `f(x)*g(y)`
References ========== - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order ODE Patterns, pp. 7 - pp. 8
"""
fx = inf[x] gy = simplify(fx*((1/(fx*h)).diff(x))) gysyms = gy.free_symbols if x not in gysyms: gy = exp(integrate(gy, y)) inf = {eta: S(0), xi: (fx*gy).subs(y, func)} if not comp: return [inf] if comp and inf not in xieta: xieta.append(inf)
return [inf]
r""" The third heuristic assumes the infinitesimals `\xi` and `\eta` to be bi-variate polynomials in `x` and `y`. The assumption made here for the logic below is that `h` is a rational function in `x` and `y` though that may not be necessary for the infinitesimals to be bivariate polynomials. The coefficients of the infinitesimals are found out by substituting them in the PDE and grouping similar terms that are polynomials and since they form a linear system, solve and check for non trivial solutions. The degree of the assumed bivariates are increased till a certain maximum value.
References ========== - Lie Groups and Differential Equations pp. 327 - pp. 329
"""
h = match['h'] hx = match['hx'] hy = match['hy'] func = match['func'] x = func.args[0] y = match['y'] xi = Function('xi')(x, func) eta = Function('eta')(x, func)
if h.is_rational_function(): # The maximum degree that the infinitesimals can take is # calculated by this technique. etax, etay, etad, xix, xiy, xid = symbols("etax etay etad xix xiy xid") ipde = etax + (etay - xix)*h - xiy*h**2 - xid*hx - etad*hy num, denom = cancel(ipde).as_numer_denom() deg = Poly(num, x, y).total_degree() deta = Function('deta')(x, y) dxi = Function('dxi')(x, y) ipde = (deta.diff(x) + (deta.diff(y) - dxi.diff(x))*h - (dxi.diff(y))*h**2 - dxi*hx - deta*hy) xieq = Symbol("xi0") etaeq = Symbol("eta0")
for i in range(deg + 1): if i: xieq += Add(*[ Symbol("xi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power) for power in range(i + 1)]) etaeq += Add(*[ Symbol("eta_" + str(power) + "_" + str(i - power))*x**power*y**(i - power) for power in range(i + 1)]) pden, denom = (ipde.subs({dxi: xieq, deta: etaeq}).doit()).as_numer_denom() pden = expand(pden)
# If the individual terms are monomials, the coefficients # are grouped if pden.is_polynomial(x, y) and pden.is_Add: polyy = Poly(pden, x, y).as_dict() if polyy: symset = xieq.free_symbols.union(etaeq.free_symbols) - 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)} return [inf]
r""" The aim of the fourth heuristic is to find the function `\chi(x, y)` that satisifies the PDE `\frac{d\chi}{dx} + h\frac{d\chi}{dx} - \frac{\partial h}{\partial y}\chi = 0`.
This assumes `\chi` to be a bivariate polynomial in `x` and `y`. By intution, `h` should be a rational function in `x` and `y`. The method used here is to substitute a general binomial for `\chi` up to a certain maximum degree is reached. The coefficients of the polynomials, are calculated by by collecting terms of the same order in `x` and `y`.
After finding `\chi`, the next step is to use `\eta = \xi*h + \chi`, to determine `\xi` and `\eta`. This can be done by dividing `\chi` by `h` which would give `-\xi` as the quotient and `\eta` as the remainder.
References ========== - E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra Solving of First Order ODEs Using Symmetry Methods, pp. 8
"""
Symbol("chi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power) for power in range(i + 1)]) soldict = soldict[0] # After finding chi, the main aim is to find out # eta, xi by the equation eta = xi*h + chi # One method to set xi, would be rearranging it to # (eta/h) - xi = (chi/h). This would mean dividing # chi by h would give -xi as the quotient and eta # as the remainder. Thanks to Sean Vig for suggesting # this method.
r""" This heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = 0, \xi = f(x) + g(y)
.. math:: \eta = f(x) + g(y), \xi = 0
The first assumption of this heuristic holds good if
.. math:: \frac{\partial}{\partial y}[(h\frac{\partial^{2}}{ \partial x^{2}}(h^{-1}))^{-1}]
is separable in `x` and `y`,
1. The separated factors containing `y` is `\frac{\partial g}{\partial y}`. From this `g(y)` can be determined. 2. The separated factors containing `x` is `f''(x)`. 3. `h\frac{\partial^{2}}{\partial x^{2}}(h^{-1})` equals `\frac{f''(x)}{f(x) + g(y)}`. From this `f(x)` can be determined.
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption satisifes. After obtaining `f(x)` and `g(y)`, the coordinates are again interchanged, to get `\eta` as `f(x) + g(y)`.
For both assumptions, the constant factors are separated among `g(y)` and `f''(x)`, such that `f''(x)` obtained from 3] is the same as that obtained from 2]. If not possible, then this heuristic fails.
References ========== - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order ODE Patterns, pp. 7 - pp. 8
"""
except NotImplementedError: pass else: else: sol = solve(check, k) if sol: sol = sol[0] fx = fx.subs(k, sol) gy = (gy/k)*sol else: continue else: inf = {xi: etaval.subs(y, func), eta : S(0)} return [inf] else:
r""" This heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = g(x), \xi = f(x)
.. math:: \eta = f(y), \xi = g(y)
For the first assumption,
1. First `\frac{\frac{\partial h}{\partial y}}{\frac{\partial^{2} h}{ \partial yy}}` is calculated. Let us say this value is A
2. If this is constant, then `h` is matched to the form `A(x) + B(x)e^{ \frac{y}{C}}` then, `\frac{e^{\int \frac{A(x)}{C} \,dx}}{B(x)}` gives `f(x)` and `A(x)*f(x)` gives `g(x)`
3. Otherwise `\frac{\frac{\partial A}{\partial X}}{\frac{\partial A}{ \partial Y}} = \gamma` is calculated. If
a] `\gamma` is a function of `x` alone
b] `\frac{\gamma\frac{\partial h}{\partial y} - \gamma'(x) - \frac{ \partial h}{\partial x}}{h + \gamma} = G` is a function of `x` alone. then, `e^{\int G \,dx}` gives `f(x)` and `-\gamma*f(x)` gives `g(x)`
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption satisifes. After obtaining `f(x)` and `g(x)`, the coordinates are again interchanged, to get `\xi` as `f(x^*)` and `\eta` as `g(y^*)`
References ========== - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order ODE Patterns, pp. 10 - pp. 12
"""
A = Wild('A', exclude=[y]) B = Wild('B', exclude=[y]) C = Wild('C', exclude=[x, y]) match = h.match(A + B*exp(y/C)) try: tau = exp(-integrate(match[A]/match[C]), x)/match[B] except NotImplementedError: pass else: gx = match[A]*tau return [{xi: tau, eta: gx}]
else: except NotImplementedError: pass else:
A = Wild('A', exclude=[y]) B = Wild('B', exclude=[y]) C = Wild('C', exclude=[x, y]) match = h.match(A + B*exp(y/C)) try: tau = exp(-integrate(match[A]/match[C]), x)/match[B] except NotImplementedError: pass else: gx = match[A]*tau return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}]
else: hinv + gamma)) except NotImplementedError: pass else:
r""" This heuristic assumes the presence of unknown functions or known functions with non-integer powers.
1. A list of all functions and non-integer powers containing x and y 2. Loop over each element `f` in the list, find `\frac{\frac{\partial f}{\partial x}}{ \frac{\partial f}{\partial x}} = R`
If it is separable in `x` and `y`, let `X` be the factors containing `x`. Then
a] Check if `\xi = X` and `\eta = -\frac{X}{R}` satisfy the PDE. If yes, then return `\xi` and `\eta` b] Check if `\xi = \frac{-R}{X}` and `\eta = -\frac{1}{X}` satisfy the PDE. If yes, then return `\xi` and `\eta`
If not, then check if
a] :math:`\xi = -R,\eta = 1`
b] :math:`\xi = 1, \eta = -\frac{1}{R}`
are solutions.
References ========== - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order ODE Patterns, pp. 10 - pp. 12
"""
else: xitry = -frac pde = -xitry.diff(x)*h -xitry.diff(y)*h**2 - xitry*hx -hy if not simplify(expand(pde)): return [{xi: xitry.subs(y, func), eta: S(1)}]
r""" This heuristic finds if infinitesimals of the form `\eta = f(x)`, `\xi = g(y)` without making any assumptions on `h`.
The complete sequence of steps is given in the paper mentioned below.
References ========== - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order ODE Patterns, pp. 10 - pp. 12
""" xieta = [] h = match['h'] hx = match['hx'] hy = match['hy'] func = match['func'] hinv = match['hinv'] x = func.args[0] y = match['y'] xi = Function('xi')(x, func) eta = Function('eta')(x, func)
C = S(0) A = hx.diff(y) B = hy.diff(y) + hy**2 C = hx.diff(x) - hx**2
if not (A and B and C): return
Ax = A.diff(x) Ay = A.diff(y) Axy = Ax.diff(y) Axx = Ax.diff(x) Ayy = Ay.diff(y) D = simplify(2*Axy + hx*Ay - Ax*hy + (hx*hy + 2*A)*A)*A - 3*Ax*Ay if not D: E1 = simplify(3*Ax**2 + ((hx**2 + 2*C)*A - 2*Axx)*A) if E1: E2 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2) if not E2: E3 = simplify( E1*((28*Ax + 4*hx*A)*A**3 - E1*(hy*A + Ay)) - E1.diff(x)*8*A**4) if not E3: etaval = cancel((4*A**3*(Ax - hx*A) + E1*(hy*A - Ay))/(S(2)*A*E1)) if x not in etaval: try: etaval = exp(integrate(etaval, y)) except NotImplementedError: pass else: xival = -4*A**3*etaval/E1 if y not in xival: return [{xi: xival, eta: etaval.subs(y, func)}]
else: E1 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2) if E1: E2 = simplify( 4*A**3*D - D**2 + E1*((2*Axx - (hx**2 + 2*C)*A)*A - 3*Ax**2)) if not E2: E3 = simplify( -(A*D)*E1.diff(y) + ((E1.diff(x) - hy*D)*A + 3*Ay*D + (A*hx - 3*Ax)*E1)*E1) if not E3: etaval = cancel(((A*hx - Ax)*E1 - (Ay + A*hy)*D)/(S(2)*A*D)) if x not in etaval: try: etaval = exp(integrate(etaval, y)) except NotImplementedError: pass else: xival = -E1*etaval/D if y not in xival: return [{xi: xival, eta: etaval.subs(y, func)}]
r""" This heuristic assumes
1. `\xi = ax + by + c` and 2. `\eta = fx + gy + h`
After substituting the following assumptions in the determining PDE, it reduces to
.. math:: f + (g - a)h - bh^{2} - (ax + by + c)\frac{\partial h}{\partial x} - (fx + gy + c)\frac{\partial h}{\partial y}
Solving the reduced PDE obtained, using the method of characteristics, becomes impractical. The method followed is grouping similar terms and solving the system of linear equations obtained. The difference between the bivariate heuristic is that `h` need not be a rational function in this case.
References ========== - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order ODE Patterns, pp. 10 - pp. 12
"""
else: else: if term not in coeffdict: coeffdict[term] = S(1) else: coeffdict[term] += S(1)
soldict = soldict[0]
# for equations Eq(a1*diff(x(t),t), a*x(t) + b*y(t) + k1) # and Eq(a2*diff(x(t),t), c*x(t) + d*y(t) + k2) else: raise NotImplementedError("Only homogeneous problems are supported" + " (and constant inhomogeneity)")
sol = _linear_2eq_order1_type6(x, y, t, r)
r""" It is classified under system of two linear homogeneous first-order constant-coefficient ordinary differential equations.
The equations which come under this type are
.. math:: x' = ax + by,
.. math:: y' = cx + dy
The characteristics equation is written as
.. math:: \lambda^{2} + (a+d) \lambda + ad - bc = 0
and its discriminant is `D = (a-d)^{2} + 4bc`. There are several cases
1. Case when `ad - bc \neq 0`. The origin of coordinates, `x = y = 0`, is the only stationary point; it is - a node if `D = 0` - a node if `D > 0` and `ad - bc > 0` - a saddle if `D > 0` and `ad - bc < 0` - a focus if `D < 0` and `a + d \neq 0` - a centre if `D < 0` and `a + d \neq 0`.
1.1. If `D > 0`. The characteristic equation has two distinct real roots `\lambda_1` and `\lambda_ 2` . The general solution of the system in question is expressed as
.. math:: x = C_1 b e^{\lambda_1 t} + C_2 b e^{\lambda_2 t}
.. math:: y = C_1 (\lambda_1 - a) e^{\lambda_1 t} + C_2 (\lambda_2 - a) e^{\lambda_2 t}
where `C_1` and `C_2` being arbitary constants
1.2. If `D < 0`. The characteristics equation has two conjugate roots, `\lambda_1 = \sigma + i \beta` and `\lambda_2 = \sigma - i \beta`. The general solution of the system is given by
.. math:: x = b e^{\sigma t} (C_1 \sin(\beta t) + C_2 \cos(\beta t))
.. math:: y = e^{\sigma t} ([(\sigma - a) C_1 - \beta C_2] \sin(\beta t) + [\beta C_1 + (\sigma - a) C_2 \cos(\beta t)])
1.3. If `D = 0` and `a \neq d`. The characteristic equation has two equal roots, `\lambda_1 = \lambda_2`. The general solution of the system is written as
.. math:: x = 2b (C_1 + \frac{C_2}{a-d} + C_2 t) e^{\frac{a+d}{2} t}
.. math:: y = [(d - a) C_1 + C_2 + (d - a) C_2 t] e^{\frac{a+d}{2} t}
1.4. If `D = 0` and `a = d \neq 0` and `b = 0`
.. math:: x = C_1 e^{a t} , y = (c C_1 t + C_2) e^{a t}
1.5. If `D = 0` and `a = d \neq 0` and `c = 0`
.. math:: x = (b C_1 t + C_2) e^{a t} , y = C_1 e^{a t}
2. Case when `ad - bc = 0` and `a^{2} + b^{2} > 0`. The whole straight line `ax + by = 0` consists of singular points. The orginal system of differential equaitons can be rewritten as
.. math:: x' = ax + by , y' = k (ax + by)
2.1 If `a + bk \neq 0`, solution will be
.. math:: x = b C_1 + C_2 e^{(a + bk) t} , y = -a C_1 + k C_2 e^{(a + bk) t}
2.2 If `a + bk = 0`, solution will be
.. math:: x = C_1 (bk t - 1) + b C_2 t , y = k^{2} b C_1 t + (b k^{2} t + 1) C_2
""" # FIXME: at least some of these can fail to give two linearly # independent solutions e.g., because they make assumptions about # zero/nonzero of certain coefficients. I've "fixed" one and # raised NotImplementedError in another. I think this should probably # just be re-written in terms of eigenvectors...
# tempting to use this in all cases, but does not guarantee linearly independent eigenvectors else: beta = im(l1) else: raise NotImplementedError('b == 0 case not implemented') gsol1 = 2*r['b']*(C1 + C2/(r['a']-r['d'])+C2*t)*exp((r['a']+r['d'])*t/2) gsol2 = ((r['d']-r['a'])*C1+C2+(r['d']-r['a'])*C2*t)*exp((r['a']+r['d'])*t/2) else: gsol1 = C1*(r['b']*k*t-1)+r['b']*C2*t gsol2 = k**2*r['b']*C1*t+(r['b']*k**2*t+1)*C2
r""" The equations of this type are
.. math:: x' = ax + by + k1 , y' = cx + dy + k2
The general solution of this system is given by sum of its particular solution and the general solution of the corresponding homogeneous system is obtained from type1.
1. When `ad - bc \neq 0`. The particular solution will be `x = x_0` and `y = y_0` where `x_0` and `y_0` are determined by solving linear system of equations
.. math:: a x_0 + b y_0 + k1 = 0 , c x_0 + d y_0 + k2 = 0
2. When `ad - bc = 0` and `a^{2} + b^{2} > 0`. In this case, the system of equation becomes
.. math:: x' = ax + by + k_1 , y' = k (ax + by) + k_2
2.1 If `\sigma = a + bk \neq 0`, particular solution is given by
.. math:: x = b \sigma^{-1} (c_1 k - c_2) t - \sigma^{-2} (a c_1 + b c_2)
.. math:: y = kx + (c_2 - c_1 k) t
2.2 If `\sigma = a + bk = 0`, particular solution is given by
.. math:: x = \frac{1}{2} b (c_2 - c_1 k) t^{2} + c_1 t
.. math:: y = kx + (c_2 - c_1 k) t
""" else: # FIXME: a previous typo fix shows this is not covered by tests sol1 = r['b']*(r['k2']-r['k1']*k)*t**2 + r['k1']*t sol2 = k*sol1 + (r['k2']-r['k1']*k)*t
r""" The equations of this type of ode are
.. math:: x' = f(t) x + g(t) y
.. math:: y' = g(t) x + f(t) y
The solution of such equations is given by
.. math:: x = e^{F} (C_1 e^{G} + C_2 e^{-G}) , y = e^{F} (C_1 e^{G} - C_2 e^{-G})
where `C_1` and `C_2` are arbitary constants, and
.. math:: F = \int f(t) \,dt , G = \int g(t) \,dt
"""
r""" The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = -g(t) x + f(t) y
The solution is given by
.. math:: x = F (C_1 \cos(G) + C_2 \sin(G)), y = F (-C_1 \sin(G) + C_2 \cos(G))
where `C_1` and `C_2` are arbitary constants, and
.. math:: F = \int f(t) \,dt , G = \int g(t) \,dt
""" elif r['d'] == -r['a']: F = exp(Integral(r['c'], t)) G = Integral(r['d'], t) sol1 = F*(-C1*sin(G) + C2*cos(G)) sol2 = F*(C1*cos(G) + C2*sin(G))
r""" The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = a g(t) x + [f(t) + b g(t)] y
The transformation of
.. math:: x = e^{\int f(t) \,dt} u , y = e^{\int f(t) \,dt} v , T = \int g(t) \,dt
leads to a system of constant coefficient linear differential equations
.. math:: u'(T) = v , v'(T) = au + bv
""" p = cancel(r['a']/r['d']) q = cancel((r['b']-r['c'])/r['d']) sol = dsolve(Eq(diff(u(T),T), v(T)), Eq(diff(v(T),T), p*u(T)+q*v(T))) sol1 = exp(Integral(r['c'], t))*sol[1].rhs.subs(T, Integral(r['d'],t)) sol2 = exp(Integral(r['c'], t))*sol[0].rhs.subs(T, Integral(r['d'],t))
r""" The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = a [f(t) + a h(t)] x + a [g(t) - h(t)] y
This is solved by first multiplying the first equation by `-a` and adding it to the second equation to obtain
.. math:: y' - a x' = -a h(t) (y - a x)
Setting `U = y - ax` and integrating the equation we arrive at
.. math:: y - ax = C_1 e^{-a \int h(t) \,dt}
and on substituing the value of y in first equation give rise to first order ODEs. After solving for `x`, we can obtain `y` by substituting the value of `x` in second equation.
""" C1, C2, C3, C4 = symbols('C1:5') p = 0 q = 0 p1 = cancel(r['c']/cancel(r['c']/r['d']).as_numer_denom()[0]) p2 = cancel(r['a']/cancel(r['a']/r['b']).as_numer_denom()[0]) for n, i in enumerate([p1, p2]): for j in Mul.make_args(collect_const(i)): if not j.has(t): q = j if q!=0 and n==0: if ((r['c']/j - r['a'])/(r['b'] - r['d']/j)) == j: p = 1 s = j break if q!=0 and n==1: if ((r['a']/j - r['c'])/(r['d'] - r['b']/j)) == j: p = 2 s = j break if p == 1: equ = diff(x(t),t) - r['a']*x(t) - r['b']*(s*x(t) + C1*exp(-s*Integral(r['b'] - r['d']/s, t))) hint1 = classify_ode(equ)[1] sol1 = dsolve(equ, hint=hint1+'_Integral').rhs sol2 = s*sol1 + C1*exp(-s*Integral(r['b'] - r['d']/s, t)) elif p ==2: equ = diff(y(t),t) - r['c']*y(t) - r['d']*s*y(t) + C1*exp(-s*Integral(r['d'] - r['b']/s, t)) hint1 = classify_ode(equ)[1] sol2 = dsolve(equ, hint=hint1+'_Integral').rhs sol1 = s*sol2 + C1*exp(-s*Integral(r['d'] - r['b']/s, t)) return [Eq(x(t), sol1), Eq(y(t), sol2)]
r""" The equations of this type of ode are .
.. math:: x' = f(t) x + g(t) y
.. math:: y' = h(t) x + p(t) y
Differentiating the first equation and substituting the value of `y` from second equation will give a second-order linear equation
.. math:: g x'' - (fg + gp + g') x' + (fgp - g^{2} h + f g' - f' g) x = 0
This above equation can be easily integrated if following conditions are satisfied.
1. `fgp - g^{2} h + f g' - f' g = 0`
2. `fgp - g^{2} h + f g' - f' g = ag, fg + gp + g' = bg`
If first condition is satisfied then it is solved by current dsolve solver and in second case it becomes a constant cofficient differential equation which is also solved by current solver.
Otherwise if the above condition fails then, a particular solution is assumed as `x = x_0(t)` and `y = y_0(t)` Then the general solution is expressed as
.. math:: x = C_1 x_0(t) + C_2 x_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt
.. math:: y = C_1 y_0(t) + C_2 [\frac{F(t) P(t)}{x_0(t)} + y_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt]
where C1 and C2 are arbitary constants and
.. math:: F(t) = e^{\int f(t) \,dt} , P(t) = e^{\int p(t) \,dt}
""" sol1 = dsolve(r['b']*diff(x(t),t,t) - m1*diff(x(t),t)).rhs sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs sol2 = dsolve(r['c']*diff(y(t),t,t) - m2*diff(y(t),t)).rhs sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs sol1 = dsolve(diff(x(t),t,t) - (m1/r['b'])*diff(x(t),t) - (e1/r['b'])*x(t)).rhs sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs sol2 = dsolve(diff(y(t),t,t) - (m2/r['c'])*diff(y(t),t) - (e2/r['c'])*y(t)).rhs sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs else:
# for equations Eq(diff(x(t),t,t), a1*diff(x(t),t)+b1*diff(y(t),t)+c1*x(t)+d1*y(t)+e1) # and Eq(a2*diff(y(t),t,t), a2*diff(x(t),t)+b2*diff(y(t),t)+c2*x(t)+d2*y(t)+e2) sol = _linear_2eq_order2_type4(x, y, t, r) sol = _linear_2eq_order2_type10(x, y, t, r)
r""" System of two constant-coefficient second-order linear homogeneous differential equations
.. math:: x'' = ax + by
.. math:: y'' = cx + dy
The charecteristic equation for above equations
.. math:: \lambda^4 - (a + d) \lambda^2 + ad - bc = 0
whose discriminant is `D = (a - d)^2 + 4bc \neq 0`
1. When `ad - bc \neq 0`
1.1. If `D \neq 0`. The characteristic equation has four distict roots, `\lambda_1, \lambda_2, \lambda_3, \lambda_4`. The general solution of the system is
.. math:: x = C_1 b e^{\lambda_1 t} + C_2 b e^{\lambda_2 t} + C_3 b e^{\lambda_3 t} + C_4 b e^{\lambda_4 t}
.. math:: y = C_1 (\lambda_1^{2} - a) e^{\lambda_1 t} + C_2 (\lambda_2^{2} - a) e^{\lambda_2 t} + C_3 (\lambda_3^{2} - a) e^{\lambda_3 t} + C_4 (\lambda_4^{2} - a) e^{\lambda_4 t}
where `C_1,..., C_4` are arbitary constants.
1.2. If `D = 0` and `a \neq d`:
.. math:: x = 2 C_1 (bt + \frac{2bk}{a - d}) e^{\frac{kt}{2}} + 2 C_2 (bt + \frac{2bk}{a - d}) e^{\frac{-kt}{2}} + 2b C_3 t e^{\frac{kt}{2}} + 2b C_4 t e^{\frac{-kt}{2}}
.. math:: y = C_1 (d - a) t e^{\frac{kt}{2}} + C_2 (d - a) t e^{\frac{-kt}{2}} + C_3 [(d - a) t + 2k] e^{\frac{kt}{2}} + C_4 [(d - a) t - 2k] e^{\frac{-kt}{2}}
where `C_1,..., C_4` are arbitary constants and `k = \sqrt{2 (a + d)}`
1.3. If `D = 0` and `a = d \neq 0` and `b = 0`:
.. math:: x = 2 \sqrt{a} C_1 e^{\sqrt{a} t} + 2 \sqrt{a} C_2 e^{-\sqrt{a} t}
.. math:: y = c C_1 t e^{\sqrt{a} t} - c C_2 t e^{-\sqrt{a} t} + C_3 e^{\sqrt{a} t} + C_4 e^{-\sqrt{a} t}
1.4. If `D = 0` and `a = d \neq 0` and `c = 0`:
.. math:: x = b C_1 t e^{\sqrt{a} t} - b C_2 t e^{-\sqrt{a} t} + C_3 e^{\sqrt{a} t} + C_4 e^{-\sqrt{a} t}
.. math:: y = 2 \sqrt{a} C_1 e^{\sqrt{a} t} + 2 \sqrt{a} C_2 e^{-\sqrt{a} t}
2. When `ad - bc = 0` and `a^2 + b^2 > 0`. Then the original system becomes
.. math:: x'' = ax + by
.. math:: y'' = k (ax + by)
2.1. If `a + bk \neq 0`:
.. math:: x = C_1 e^{t \sqrt{a + bk}} + C_2 e^{-t \sqrt{a + bk}} + C_3 bt + C_4 b
.. math:: y = C_1 k e^{t \sqrt{a + bk}} + C_2 k e^{-t \sqrt{a + bk}} - C_3 at - C_4 a
2.2. If `a + bk = 0`:
.. math:: x = C_1 b t^3 + C_2 b t^2 + C_3 t + C_4
.. math:: y = kx + 6 C_1 t + 2 C_2
""" + C4*r['b']*exp(l4*t) C3*(l3**2-r['a'])*exp(l3*t) + C4*(l4**2-r['a'])*exp(l4*t) else: if r['a'] != r['d']: k = sqrt(2*(r['a']+r['d'])) mid = r['b']*t+2*r['b']*k/(r['a']-r['d']) gsol1 = 2*C1*mid*exp(k*t/2) + 2*C2*mid*exp(-k*t/2) + \ 2*r['b']*C3*t*exp(k*t/2) + 2*r['b']*C4*t*exp(-k*t/2) gsol2 = C1*(r['d']-r['a'])*t*exp(k*t/2) + C2*(r['d']-r['a'])*t*exp(-k*t/2) + \ C3*((r['d']-r['a'])*t+2*k)*exp(k*t/2) + C4*((r['d']-r['a'])*t-2*k)*exp(-k*t/2) elif r['a'] == r['d'] != 0 and r['b'] == 0: sa = sqrt(r['a']) gsol1 = 2*sa*C1*exp(sa*t) + 2*sa*C2*exp(-sa*t) gsol2 = r['c']*C1*t*exp(sa*t)-r['c']*C2*t*exp(-sa*t)+C3*exp(sa*t)+C4*exp(-sa*t) elif r['a'] == r['d'] != 0 and r['c'] == 0: sa = sqrt(r['a']) gsol1 = r['b']*C1*t*exp(sa*t)-r['b']*C2*t*exp(-sa*t)+C3*exp(sa*t)+C4*exp(-sa*t) gsol2 = 2*sa*C1*exp(sa*t) + 2*sa*C2*exp(-sa*t) elif (r['a']*r['d'] - r['b']*r['c']) == 0 and (r['a']**2 + r['b']**2) > 0: k = r['c']/r['a'] if r['a'] + r['b']*k != 0: mid = sqrt(r['a'] + r['b']*k) gsol1 = C1*exp(mid*t) + C2*exp(-mid*t) + C3*r['b']*t + C4*r['b'] gsol2 = C1*k*exp(mid*t) + C2*k*exp(-mid*t) - C3*r['a']*t - C4*r['a'] else: gsol1 = C1*r['b']*t**3 + C2*r['b']*t**2 + C3*t + C4 gsol2 = k*gsol1 + 6*C1*t + 2*C2
r""" The equations in this type are
.. math:: x'' = a_1 x + b_1 y + c_1
.. math:: y'' = a_2 x + b_2 y + c_2
The general solution of this system is given by the sum of its particular solution and the general solution of the homogeneous system. The general solution is given by the linear system of 2 equation of order 2 and type 1
1. If `a_1 b_2 - a_2 b_1 \neq 0`. A particular solution will be `x = x_0` and `y = y_0` where the constants `x_0` and `y_0` are determined by solving the linear algebraic system
.. math:: a_1 x_0 + b_1 y_0 + c_1 = 0, a_2 x_0 + b_2 y_0 + c_2 = 0
2. If `a_1 b_2 - a_2 b_1 = 0` and `a_1^2 + b_1^2 > 0`. In this case, the system in question becomes
.. math:: x'' = ax + by + c_1, y'' = k (ax + by) + c_2
2.1. If `\sigma = a + bk \neq 0`, the particular solution will be
.. math:: x = \frac{1}{2} b \sigma^{-1} (c_1 k - c_2) t^2 - \sigma^{-2} (a c_1 + b c_2)
.. math:: y = kx + \frac{1}{2} (c_2 - c_1 k) t^2
2.2. If `\sigma = a + bk = 0`, the particular solution will be
.. math:: x = \frac{1}{24} b (c_2 - c_1 k) t^4 + \frac{1}{2} c_1 t^2
.. math:: y = kx + \frac{1}{2} (c_2 - c_1 k) t^2
""" elif r['c1']*r['d2'] - r['c2']*r['d1'] == 0 and (r['c1']**2 + r['d1']**2) > 0: k = r['c2']/r['c1'] sig = r['c1'] + r['d1']*k if sig != 0: psol1 = r['d1']*sig**-1*(r['e1']*k-r['e2'])*t**2/2 - \ sig**-2*(r['c1']*r['e1']+r['d1']*r['e2']) psol2 = k*psol1 + (r['e2'] - r['e1']*k)*t**2/2 psol = [psol1, psol2] else: psol1 = r['d1']*(r['e2']-r['e1']*k)*t**4/24 + r['e1']*t**2/2 psol2 = k*psol1 + (r['e2']-r['e1']*k)*t**2/2 psol = [psol1, psol2]
r""" These type of equation is used for describing the horizontal motion of a pendulum taking into account the Earth rotation. The solution is given with `a^2 + 4b > 0`:
.. math:: x = C_1 \cos(\alpha t) + C_2 \sin(\alpha t) + C_3 \cos(\beta t) + C_4 \sin(\beta t)
.. math:: y = -C_1 \sin(\alpha t) + C_2 \cos(\alpha t) - C_3 \sin(\beta t) + C_4 \cos(\beta t)
where `C_1,...,C_4` and
.. math:: \alpha = \frac{1}{2} a + \frac{1}{2} \sqrt{a^2 + 4b}, \beta = \frac{1}{2} a - \frac{1}{2} \sqrt{a^2 + 4b}
"""
r""" These equations are found in the theory of oscillations
.. math:: x'' + a_1 x' + b_1 y' + c_1 x + d_1 y = k_1 e^{i \omega t}
.. math:: y'' + a_2 x' + b_2 y' + c_2 x + d_2 y = k_2 e^{i \omega t}
The general solution of this linear nonhomogeneous system of constant-coefficient differential equations is given by the sum of its particular solution and the general solution of the corresponding homogeneous system (with `k_1 = k_2 = 0`)
1. A particular solution is obtained by the method of undetermined coefficients:
.. math:: x = A_* e^{i \omega t}, y = B_* e^{i \omega t}
On substituting these expressions into the original system of differential equations, one arrive at a linear nonhomogeneous system of algebraic equations for the coefficients `A` and `B`.
2. The general solution of the homogeneous system of differential equations is determined by a linear combination of linearly independent particular solutions determined by the method of undetermined coefficients in the form of exponentials:
.. math:: x = A e^{\lambda t}, y = B e^{\lambda t}
On substituting these expressions into the original system and colleting the coefficients of the unknown `A` and `B`, one obtains
.. math:: (\lambda^{2} + a_1 \lambda + c_1) A + (b_1 \lambda + d_1) B = 0
.. math:: (a_2 \lambda + c_2) A + (\lambda^{2} + b_2 \lambda + d_2) B = 0
The determinant of this system must vanish for nontrivial solutions A, B to exist. This requirement results in the following characteristic equation for `\lambda`
.. math:: (\lambda^2 + a_1 \lambda + c_1) (\lambda^2 + b_2 \lambda + d_2) - (b_1 \lambda + d_1) (a_2 \lambda + c_2) = 0
If all roots `k_1,...,k_4` of this equation are distict, the general solution of the original system of the differential equations has the form
.. math:: x = C_1 (b_1 \lambda_1 + d_1) e^{\lambda_1 t} - C_2 (b_1 \lambda_2 + d_1) e^{\lambda_2 t} - C_3 (b_1 \lambda_3 + d_1) e^{\lambda_3 t} - C_4 (b_1 \lambda_4 + d_1) e^{\lambda_4 t}
.. math:: y = C_1 (\lambda_1^{2} + a_1 \lambda_1 + c_1) e^{\lambda_1 t} + C_2 (\lambda_2^{2} + a_1 \lambda_2 + c_1) e^{\lambda_2 t} + C_3 (\lambda_3^{2} + a_1 \lambda_3 + c_1) e^{\lambda_3 t} + C_4 (\lambda_4^{2} + a_1 \lambda_4 + c_1) e^{\lambda_4 t}
""" C1, C2, C3, C4 = symbols('C1:5') k = Symbol('k') Ra, Ca, Rb, Cb = symbols('Ra, Ca, Rb, Cb') a1 = r['a1'] ; a2 = r['a2'] b1 = r['b1'] ; b2 = r['b2'] c1 = r['c1'] ; c2 = r['c2'] d1 = r['d1'] ; d2 = r['d2'] k1 = r['e1'].expand().as_independent(t)[0] k2 = r['e2'].expand().as_independent(t)[0] ew1 = r['e1'].expand().as_independent(t)[1] ew2 = powdenest(ew1).as_base_exp()[1] ew3 = collect(ew2, t).coeff(t) w = cancel(ew3/I) # The particular solution is assumed to be (Ra+I*Ca)*exp(I*w*t) and # (Rb+I*Cb)*exp(I*w*t) for x(t) and y(t) respectively peq1 = (-w**2+c1)*Ra - a1*w*Ca + d1*Rb - b1*w*Cb - k1 peq2 = a1*w*Ra + (-w**2+c1)*Ca + b1*w*Rb + d1*Cb peq3 = c2*Ra - a2*w*Ca + (-w**2+d2)*Rb - b2*w*Cb - k2 peq4 = a2*w*Ra + c2*Ca + b2*w*Rb + (-w**2+d2)*Cb # FIXME: solve for what in what? Ra, Rb, etc I guess # but then psol not used for anything? psol = solve([peq1, peq2, peq3, peq4])
chareq = (k**2+a1*k+c1)*(k**2+b2*k+d2) - (b1*k+d1)*(a2*k+c2) [k1, k2, k3, k4] = roots_quartic(Poly(chareq)) sol1 = -C1*(b1*k1+d1)*exp(k1*t) - C2*(b1*k2+d1)*exp(k2*t) - \ C3*(b1*k3+d1)*exp(k3*t) - C4*(b1*k4+d1)*exp(k4*t) + (Ra+I*Ca)*exp(I*w*t)
a1_ = (a1-1) sol2 = C1*(k1**2+a1_*k1+c1)*exp(k1*t) + C2*(k2**2+a1_*k2+c1)*exp(k2*t) + \ C3*(k3**2+a1_*k3+c1)*exp(k3*t) + C4*(k4**2+a1_*k4+c1)*exp(k4*t) + (Rb+I*Cb)*exp(I*w*t) return [Eq(x(t), sol1), Eq(y(t), sol2)]
r""" The equation which come under this catagory are
.. math:: x'' = a (t y' - y)
.. math:: y'' = b (t x' - x)
The transformation
.. math:: u = t x' - x, b = t y' - y
leads to the first-order system
.. math:: u' = atv, v' = btu
The general solution of this system is given by
If `ab > 0`:
.. math:: u = C_1 a e^{\frac{1}{2} \sqrt{ab} t^2} + C_2 a e^{-\frac{1}{2} \sqrt{ab} t^2}
.. math:: v = C_1 \sqrt{ab} e^{\frac{1}{2} \sqrt{ab} t^2} - C_2 \sqrt{ab} e^{-\frac{1}{2} \sqrt{ab} t^2}
If `ab < 0`:
.. math:: u = C_1 a \cos(\frac{1}{2} \sqrt{\left|ab\right|} t^2) + C_2 a \sin(-\frac{1}{2} \sqrt{\left|ab\right|} t^2)
.. math:: v = C_1 \sqrt{\left|ab\right|} \sin(\frac{1}{2} \sqrt{\left|ab\right|} t^2) + C_2 \sqrt{\left|ab\right|} \cos(-\frac{1}{2} \sqrt{\left|ab\right|} t^2)
where `C_1` and `C_2` are arbitary constants. On substituting the value of `u` and `v` in above equations and integrating the resulting expressions, the general solution will become
.. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt, y = C_4 t + t \int \frac{u}{t^2} \,dt
where `C_3` and `C_4` are arbitrary constants.
""" else: u = C1*r['a']*cos(mul*t**2/2) + C2*r['a']*sin(mul*t**2/2) v = -C1*mul*sin(mul*t**2/2) + C2*mul*cos(mul*t**2/2)
r""" The equations are
.. math:: x'' = f(t) (a_1 x + b_1 y)
.. math:: y'' = f(t) (a_2 x + b_2 y)
If `k_1` and `k_2` are roots of the quadratic equation
.. math:: k^2 - (a_1 + b_2) k + a_1 b_2 - a_2 b_1 = 0
Then by multiplying appropriate constants and adding together original equations we obtain two independent equations:
.. math:: z_1'' = k_1 f(t) z_1, z_1 = a_2 x + (k_1 - a_1) y
.. math:: z_2'' = k_2 f(t) z_2, z_2 = a_2 x + (k_2 - a_1) y
Solving the equations will give the values of `x` and `y` after obtaining the value of `z_1` and `z_2` by solving the differential equation and substuting the result.
"""
r""" The equations are given as
.. math:: x'' = f(t) (a_1 x' + b_1 y')
.. math:: y'' = f(t) (a_2 x' + b_2 y')
If `k_1` and 'k_2` are roots of the quadratic equation
.. math:: k^2 - (a_1 + b_2) k + a_1 b_2 - a_2 b_1 = 0
Then the system can be reduced by adding together the two equations multiplied by appropriate constants give following two independent equations:
.. math:: z_1'' = k_1 f(t) z_1', z_1 = a_2 x + (k_1 - a_1) y
.. math:: z_2'' = k_2 f(t) z_2', z_2 = a_2 x + (k_2 - a_1) y
Integrating these and returning to the original variables, one arrives at a linear algebraic system for the unknowns `x` and `y`:
.. math:: a_2 x + (k_1 - a_1) y = C_1 \int e^{k_1 F(t)} \,dt + C_2
.. math:: a_2 x + (k_2 - a_1) y = C_3 \int e^{k_2 F(t)} \,dt + C_4
where `C_1,...,C_4` are arbitrary constants and `F(t) = \int f(t) \,dt`
"""
r""" The equation of this catagory are
.. math:: x'' = a f(t) (t y' - y)
.. math:: y'' = b f(t) (t x' - x)
The transformation
.. math:: u = t x' - x, v = t y' - y
leads to the system of first-order equations
.. math:: u' = a t f(t) v, v' = b t f(t) u
The general solution of this system has the form
If `ab > 0`:
.. math:: u = C_1 a e^{\sqrt{ab} \int t f(t) \,dt} + C_2 a e^{-\sqrt{ab} \int t f(t) \,dt}
.. math:: v = C_1 \sqrt{ab} e^{\sqrt{ab} \int t f(t) \,dt} - C_2 \sqrt{ab} e^{-\sqrt{ab} \int t f(t) \,dt}
If `ab < 0`:
.. math:: u = C_1 a \cos(\sqrt{\left|ab\right|} \int t f(t) \,dt) + C_2 a \sin(-\sqrt{\left|ab\right|} \int t f(t) \,dt)
.. math:: v = C_1 \sqrt{\left|ab\right|} \sin(\sqrt{\left|ab\right|} \int t f(t) \,dt) + C_2 \sqrt{\left|ab\right|} \cos(-\sqrt{\left|ab\right|} \int t f(t) \,dt)
where `C_1` and `C_2` are arbitary constants. On substituting the value of `u` and `v` in above equations and integrating the resulting expressions, the general solution will become
.. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt, y = C_4 t + t \int \frac{u}{t^2} \,dt
where `C_3` and `C_4` are arbitrary constants.
""" else: u = C1*a*cos(mul*Igral) + C2*a*sin(mul*Igral) v = -C1*mul*sin(mul*Igral) + C2*mul*cos(mul*Igral)
r""" .. math:: t^2 x'' + a_1 t x' + b_1 t y' + c_1 x + d_1 y = 0
.. math:: t^2 y'' + a_2 t x' + b_2 t y' + c_2 x + d_2 y = 0
These system of equations are euler type.
The substitution of `t = \sigma e^{\tau} (\sigma \neq 0)` leads to the system of constant coefficient linear differential equations
.. math:: x'' + (a_1 - 1) x' + b_1 y' + c_1 x + d_1 y = 0
.. math:: y'' + a_2 x' + (b_2 - 1) y' + c_2 x + d_2 y = 0
The general solution of the homogeneous system of differential equations is determined by a linear combination of linearly independent particular solutions determined by the method of undetermined coefficients in the form of exponentials
.. math:: x = A e^{\lambda t}, y = B e^{\lambda t}
On substituting these expressions into the original system and colleting the coefficients of the unknown `A` and `B`, one obtains
.. math:: (\lambda^{2} + (a_1 - 1) \lambda + c_1) A + (b_1 \lambda + d_1) B = 0
.. math:: (a_2 \lambda + c_2) A + (\lambda^{2} + (b_2 - 1) \lambda + d_2) B = 0
The determinant of this system must vanish for nontrivial solutions A, B to exist. This requirement results in the following characteristic equation for `\lambda`
.. math:: (\lambda^2 + (a_1 - 1) \lambda + c_1) (\lambda^2 + (b_2 - 1) \lambda + d_2) - (b_1 \lambda + d_1) (a_2 \lambda + c_2) = 0
If all roots `k_1,...,k_4` of this equation are distict, the general solution of the original system of the differential equations has the form
.. math:: x = C_1 (b_1 \lambda_1 + d_1) e^{\lambda_1 t} - C_2 (b_1 \lambda_2 + d_1) e^{\lambda_2 t} - C_3 (b_1 \lambda_3 + d_1) e^{\lambda_3 t} - C_4 (b_1 \lambda_4 + d_1) e^{\lambda_4 t}
.. math:: y = C_1 (\lambda_1^{2} + (a_1 - 1) \lambda_1 + c_1) e^{\lambda_1 t} + C_2 (\lambda_2^{2} + (a_1 - 1) \lambda_2 + c_1) e^{\lambda_2 t} + C_3 (\lambda_3^{2} + (a_1 - 1) \lambda_3 + c_1) e^{\lambda_3 t} + C_4 (\lambda_4^{2} + (a_1 - 1) \lambda_4 + c_1) e^{\lambda_4 t}
""" C3*(b1*k3+d1)*exp(k3*log(t)) - C4*(b1*k4+d1)*exp(k4*log(t))
+ C3*(k3**2+a1_*k3+c1)*exp(k3*log(t)) + C4*(k4**2+a1_*k4+c1)*exp(k4*log(t))
r""" The equation of this catagory are
.. math:: (\alpha t^2 + \beta t + \gamma)^{2} x'' = ax + by
.. math:: (\alpha t^2 + \beta t + \gamma)^{2} y'' = cx + dy
The transformation
.. math:: \tau = \int \frac{1}{\alpha t^2 + \beta t + \gamma} \,dt , u = \frac{x}{\sqrt{\left|\alpha t^2 + \beta t + \gamma\right|}} , v = \frac{y}{\sqrt{\left|\alpha t^2 + \beta t + \gamma\right|}}
leads to a constant coefficient linear system of equations
.. math:: u'' = (a - \alpha \gamma + \frac{1}{4} \beta^{2}) u + b v
.. math:: v'' = c u + (d - \alpha \gamma + \frac{1}{4} \beta^{2}) v
These system of equations obtained can be solved by type1 of System of two constant-coefficient second-order linear homogeneous differential equations.
""" C1, C2, C3, C4 = symbols('C1:5') u, v = symbols('u, v', function=True) T = Symbol('T') p = Wild('p', exclude=[t, t**2]) q = Wild('q', exclude=[t, t**2]) s = Wild('s', exclude=[t, t**2]) n = Wild('n', exclude=[t, t**2]) num, denum = r['c1'].as_numer_denom() dic = denum.match((n*(p*t**2+q*t+s)**2).expand()) eqz = dic[p]*t**2 + dic[q]*t + dic[s] a = num/dic[n] b = cancel(r['d1']*eqz**2) c = cancel(r['c2']*eqz**2) d = cancel(r['d2']*eqz**2) [msol1, msol2] = dsolve([Eq(diff(u(t), t, t), (a - dic[p]*dic[s] + dic[q]**2/4)*u(t) \ + b*v(t)), Eq(diff(v(t),t,t), c*u(t) + (d - dic[p]*dic[s] + dic[q]**2/4)*v(t))]) sol1 = (msol1.rhs*sqrt(abs(eqz))).subs(t, Integral(1/eqz, t)) sol2 = (msol2.rhs*sqrt(abs(eqz))).subs(t, Integral(1/eqz, t)) return [Eq(x(t), sol1), Eq(y(t), sol2)]
r""" The equations which comes under this type are
.. math:: x'' = f(t) (t x' - x) + g(t) (t y' - y)
.. math:: y'' = h(t) (t x' - x) + p(t) (t y' - y)
The transformation
.. math:: u = t x' - x, v = t y' - y
leads to the linear system of first-order equations
.. math:: u' = t f(t) u + t g(t) v, v' = t h(t) u + t p(t) v
On substituting the value of `u` and `v` in transformed equation gives value of `x` and `y` as
.. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt , y = C_4 t + t \int \frac{v}{t^2} \,dt.
where `C_3` and `C_4` are arbitrary constants.
"""
# for equations: # Eq(g1*diff(x(t),t), a1*x(t)+b1*y(t)+c1*z(t)+d1), # Eq(g2*diff(y(t),t), a2*x(t)+b2*y(t)+c2*z(t)+d2), and # Eq(g3*diff(z(t),t), a3*x(t)+b3*y(t)+c3*z(t)+d3) # We can handle homogeneous case and simple constant forcings else: # Issue #9244: nonhomogeneous linear systems are not supported raise NotImplementedError("Only homogeneous problems are supported" + " (and constant inhomogeneity)")
r""" .. math:: x' = ax
.. math:: y' = bx + cy
.. math:: z' = dx + ky + pz
Solution of such equations are forward substitution. Solving first equations gives the value of `x`, substituting it in second and third equation and solving second equation gives `y` and similarly substituting `y` in third equation give `z`.
.. math:: x = C_1 e^{at}
.. math:: y = \frac{b C_1}{a - c} e^{at} + C_2 e^{ct}
.. math:: z = \frac{C_1}{a - p} (d + \frac{bk}{a - c}) e^{at} + \frac{k C_2}{c - p} e^{ct} + C_3 e^{pt}
where `C_1, C_2` and `C_3` are arbitrary constants.
"""
r""" The equations of this type are
.. math:: x' = cy - bz
.. math:: y' = az - cx
.. math:: z' = bx - ay
1. First integral:
.. math:: ax + by + cz = A \qquad - (1)
.. math:: x^2 + y^2 + z^2 = B^2 \qquad - (2)
where `A` and `B` are arbitrary constants. It follows from these integrals that the integral lines are circles formed by the intersection of the planes `(1)` and sphere `(2)`
2. Solution:
.. math:: x = a C_0 + k C_1 \cos(kt) + (c C_2 - b C_3) \sin(kt)
.. math:: y = b C_0 + k C_2 \cos(kt) + (a C_2 - c C_3) \sin(kt)
.. math:: z = c C_0 + k C_3 \cos(kt) + (b C_2 - a C_3) \sin(kt)
where `k = \sqrt{a^2 + b^2 + c^2}` and the four constants of integration, `C_1,...,C_4` are constrained by a single relation,
.. math:: a C_1 + b C_2 + c C_3 = 0
"""
r""" Equations of this system of ODEs
.. math:: a x' = bc (y - z)
.. math:: b y' = ac (z - x)
.. math:: c z' = ab (x - y)
1. First integral:
.. math:: a^2 x + b^2 y + c^2 z = A
where A is an arbitary constant. It follows that the integral lines are plane curves.
2. Solution:
.. math:: x = C_0 + k C_1 \cos(kt) + a^{-1} bc (C_2 - C_3) \sin(kt)
.. math:: y = C_0 + k C_2 \cos(kt) + a b^{-1} c (C_3 - C_1) \sin(kt)
.. math:: z = C_0 + k C_3 \cos(kt) + ab c^{-1} (C_1 - C_2) \sin(kt)
where `k = \sqrt{a^2 + b^2 + c^2}` and the four constants of integration, `C_1,...,C_4` are constrained by a single relation
.. math:: a^2 C_1 + b^2 C_2 + c^2 C_3 = 0
"""
r""" Equations:
.. math:: x' = (a_1 f(t) + g(t)) x + a_2 f(t) y + a_3 f(t) z
.. math:: y' = b_1 f(t) x + (b_2 f(t) + g(t)) y + b_3 f(t) z
.. math:: z' = c_1 f(t) x + c_2 f(t) y + (c_3 f(t) + g(t)) z
The transformation
.. math:: x = e^{\int g(t) \,dt} u, y = e^{\int g(t) \,dt} v, z = e^{\int g(t) \,dt} w, \tau = \int f(t) \,dt
leads to the system of constant coefficient linear differential equations
.. math:: u' = a_1 u + a_2 v + a_3 w
.. math:: v' = b_1 u + b_2 v + b_3 w
.. math:: w' = c_1 u + c_2 v + c_3 w
These system of equations are solved by homogeneous linear system of constant coefficients of `n` equations of first order. Then substituting the value of `u, v` and `w` in transformed equation gives value of `x, y` and `z`.
""" b2*v(t)-b3*w(t), diff(w(t),t)-c1*u(t)-c2*v(t)-c3*w(t))
sol = _linear_neq_order1_type1(match_)
r""" System of n first-order constant-coefficient linear nonhomogeneous differential equation
.. math:: y'_k = a_{k1} y_1 + a_{k2} y_2 +...+ a_{kn} y_n; k = 1,2,...,n
or that can be written as `\vec{y'} = A . \vec{y}` where `\vec{y}` is matrix of `y_k` for `k = 1,2,...n` and `A` is a `n \times n` matrix.
Since these equations are equivalent to a first order homogeneous linear differential equation. So the general solution will contain `n` linearly independent parts and solution will consist some type of exponential functions. Assuming `y = \vec{v} e^{rt}` is a solution of the system where `\vec{v}` is a vector of coefficients of `y_1,...,y_n`. Substituting `y` and `y' = r v e^{r t}` into the equation `\vec{y'} = A . \vec{y}`, we get
.. math:: r \vec{v} e^{rt} = A \vec{v} e^{rt}
.. math:: r \vec{v} = A \vec{v}
where `r` comes out to be eigenvalue of `A` and vector `\vec{v}` is the eigenvector of `A` corresponding to `r`. There are three possiblities of eigenvalues of `A`
- `n` distinct real eigenvalues - complex conjugate eigenvalues - eigenvalues with multiplicity `k`
1. When all eigenvalues `r_1,..,r_n` are distinct with `n` different eigenvectors `v_1,...v_n` then the solution is given by
.. math:: \vec{y} = C_1 e^{r_1 t} \vec{v_1} + C_2 e^{r_2 t} \vec{v_2} +...+ C_n e^{r_n t} \vec{v_n}
where `C_1,C_2,...,C_n` are arbitrary constants.
2. When some eigenvalues are complex then in order to make the solution real, we take a llinear combination: if `r = a + bi` has an eigenvector `\vec{v} = \vec{w_1} + i \vec{w_2}` then to obtain real-valued solutions to the system, replace the complex-valued solutions `e^{rx} \vec{v}` with real-valued solution `e^{ax} (\vec{w_1} \cos(bx) - \vec{w_2} \sin(bx))` and for `r = a - bi` replace the solution `e^{-r x} \vec{v}` with `e^{ax} (\vec{w_1} \sin(bx) + \vec{w_2} \cos(bx))`
3. If some eigenvalues are repeated. Then we get fewer than `n` linearly independent eigenvectors, we miss some of the solutions and need to construct the missing ones. We do this via generalized eigenvectors, vectors which are not eigenvectors but are close enough that we can use to write down the remaining solutions. For a eigenvalue `r` with eigenvector `\vec{w}` we obtain `\vec{w_2},...,\vec{w_k}` using
.. math:: (A - r I) . \vec{w_2} = \vec{w}
.. math:: (A - r I) . \vec{w_3} = \vec{w_2}
.. math:: \vdots
.. math:: (A - r I) . \vec{w_k} = \vec{w_{k-1}}
Then the solutions to the system for the eigenspace are `e^{rt} [\vec{w}], e^{rt} [t \vec{w} + \vec{w_2}], e^{rt} [\frac{t^2}{2} \vec{w} + t \vec{w_2} + \vec{w_3}], ...,e^{rt} [\frac{t^{k-1}}{(k-1)!} \vec{w} + \frac{t^{k-2}}{(k-2)!} \vec{w_2} +...+ t \vec{w_{k-1}} + \vec{w_k}]`
So, If `\vec{y_1},...,\vec{y_n}` are `n` solution of obtained from three categories of `A`, then general solution to the system `\vec{y'} = A . \vec{y}`
.. math:: \vec{y} = C_1 \vec{y_1} + C_2 \vec{y_2} + \cdots + C_n \vec{y_n}
""" # If number of column of an eigenvector is not equal to the multiplicity # of its eigenvalue then the legt eigenvectors are calculated else:
r""" Equations:
.. math:: x' = x^n F(x,y)
.. math:: y' = g(y) F(x,y)
Solution:
.. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2
where
if `n \neq 1`
.. math:: \varphi = [C_1 + (1-n) \int \frac{1}{g(y)} \,dy]^{\frac{1}{1-n}}
if `n = 1`
.. math:: \varphi = C_1 e^{\int \frac{1}{g(y)} \,dy}
where `C_1` and `C_2` are arbitrary constants.
""" else:
r""" Equations:
.. math:: x' = e^{\lambda x} F(x,y)
.. math:: y' = g(y) F(x,y)
Solution:
.. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2
where
if `\lambda \neq 0`
.. math:: \varphi = -\frac{1}{\lambda} log(C_1 - \lambda \int \frac{1}{g(y)} \,dy)
if `\lambda = 0`
.. math:: \varphi = C_1 + \int \frac{1}{g(y)} \,dy
where `C_1` and `C_2` are arbitrary constants.
""" else: phi = C1 + Integral(1/g, v)
r""" Autonomous system of general form
.. math:: x' = F(x,y)
.. math:: y' = G(x,y)
Assuming `y = y(x, C_1)` where `C_1` is an arbitrary constant is the general solution of the first-order equation
.. math:: F(x,y) y'_x = G(x,y)
Then the general solution of the original system of equations has the form
.. math:: \int \frac{1}{F(x,y(x,C_1))} \,dx = t + C_1
"""
r""" Equation:
.. math:: x' = f_1(x) g_1(y) \phi(x,y,t)
.. math:: y' = f_2(x) g_2(y) \phi(x,y,t)
First integral:
.. math:: \int \frac{f_2(x)}{f_1(x)} \,dx - \int \frac{g_1(y)}{g_2(y)} \,dy = C
where `C` is an arbitrary constant.
On solving the first integral for `x` (resp., `y` ) and on substituting the resulting expression into either equation of the original solution, one arrives at a firs-order equation for determining `y` (resp., `x` ).
"""
r""" Clairaut system of ODEs
.. math:: x = t x' + F(x',y')
.. math:: y = t y' + G(x',y')
The following are solutions of the system
`(i)` straight lines:
.. math:: x = C_1 t + F(C_1, C_2), y = C_2 t + G(C_1, C_2)
where `C_1` and `C_2` are arbitrary constants;
`(ii)` envelopes of the above lines;
`(iii)` continuously differentiable lines made up from segments of the lines `(i)` and `(ii)`.
"""
x = match_['func'][0].func y = match_['func'][1].func z = match_['func'][2].func eq = match_['eq'] fc = match_['func_coeff'] func = match_['func'] t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] if match_['type_of_equation'] == 'type1': sol = _nonlinear_3eq_order1_type1(x, y, z, t, eq) if match_['type_of_equation'] == 'type2': sol = _nonlinear_3eq_order1_type2(x, y, z, t, eq) if match_['type_of_equation'] == 'type3': sol = _nonlinear_3eq_order1_type3(x, y, z, t, eq) if match_['type_of_equation'] == 'type4': sol = _nonlinear_3eq_order1_type4(x, y, z, t, eq) if match_['type_of_equation'] == 'type5': sol = _nonlinear_3eq_order1_type5(x, y, z, t, eq) return sol
r""" Equations:
.. math:: a x' = (b - c) y z, \enspace b y' = (c - a) z x, \enspace c z' = (a - b) x y
First Integrals:
.. math:: a x^{2} + b y^{2} + c z^{2} = C_1
.. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2
where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and `z` and on substituting the resulting expressions into the first equation of the system, we arrives at a separable first-order equation on `x`. Similarly doing that for other two equations, we will arrive at first order equation on `y` and `z` too.
References ========== -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0401.pdf
""" C1, C2 = symbols('C1:3') u, v, w = symbols('u, v, w') p = Wild('p', exclude=[x(t), y(t), z(t), t]) q = Wild('q', exclude=[x(t), y(t), z(t), t]) s = Wild('s', exclude=[x(t), y(t), z(t), t]) r = (diff(x(t),t) - eq[0]).match(p*y(t)*z(t)) r.update((diff(y(t),t) - eq[1]).match(q*z(t)*x(t))) r.update((diff(z(t),t) - eq[2]).match(s*x(t)*y(t))) n1, d1 = r[p].as_numer_denom() n2, d2 = r[q].as_numer_denom() n3, d3 = r[s].as_numer_denom() val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, d3*u-d3*v-n3*w],[u,v]) vals = [val[v], val[u]] c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1]) b = vals[0].subs(w,c) a = vals[1].subs(w,c) y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b))) z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c))) z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c))) x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a))) x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a))) y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b))) try: sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x).rhs except: sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x, hint='separable_Integral') try: sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y).rhs except: sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y, hint='separable_Integral') try: sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z).rhs except: sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z, hint='separable_Integral') return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
r""" Equations:
.. math:: a x' = (b - c) y z f(x, y, z, t)
.. math:: b y' = (c - a) z x f(x, y, z, t)
.. math:: c z' = (a - b) x y f(x, y, z, t)
First Integrals:
.. math:: a x^{2} + b y^{2} + c z^{2} = C_1
.. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2
where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and `z` and on substituting the resulting expressions into the first equation of the system, we arrives at a first-order differential equations on `x`. Similarly doing that for other two equations we will arrive at first order equation on `y` and `z`.
References ========== -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0402.pdf
""" C1, C2 = symbols('C1:3') u, v, w = symbols('u, v, w') p = Wild('p', exclude=[x(t), y(t), z(t), t]) q = Wild('q', exclude=[x(t), y(t), z(t), t]) s = Wild('s', exclude=[x(t), y(t), z(t), t]) f = Wild('f') r1 = (diff(x(t),t) - eq[0]).match(y(t)*z(t)*f) r = collect_const(r1[f]).match(p*f) r.update(((diff(y(t),t) - eq[1])/r[f]).match(q*z(t)*x(t))) r.update(((diff(z(t),t) - eq[2])/r[f]).match(s*x(t)*y(t))) n1, d1 = r[p].as_numer_denom() n2, d2 = r[q].as_numer_denom() n3, d3 = r[s].as_numer_denom() val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, -d3*u+d3*v+n3*w],[u,v]) vals = [val[v], val[u]] c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1]) a = vals[0].subs(w,c) b = vals[1].subs(w,c) y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b))) z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c))) z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c))) x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a))) x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a))) y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b))) try: sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x*r[f]).rhs except: sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x*r[f], hint='separable_Integral') try: sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y*r[f]).rhs except: sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y*r[f], hint='separable_Integral') try: sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z*r[f]).rhs except: sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z*r[f], hint='separable_Integral') return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
r""" Equations:
.. math:: x' = c F_2 - b F_3, \enspace y' = a F_3 - c F_1, \enspace z' = b F_1 - a F_2
where `F_n = F_n(x, y, z, t)`.
1. First Integral:
.. math:: a x + b y + c z = C_1,
where C is an arbitrary constant.
2. If we assume function `F_n` to be independent of `t`,i.e, `F_n` = `F_n (x, y, z)` Then, on eliminating `t` and `z` from the first two equation of the system, one arrives at the first-order equation
.. math:: \frac{dy}{dx} = \frac{a F_3 (x, y, z) - c F_1 (x, y, z)}{c F_2 (x, y, z) - b F_3 (x, y, z)}
where `z = \frac{1}{c} (C_1 - a x - b y)`
References ========== -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0404.pdf
""" C1 = symbols('C1') u, v, w = symbols('u, v, w') p = Wild('p', exclude=[x(t), y(t), z(t), t]) q = Wild('q', exclude=[x(t), y(t), z(t), t]) s = Wild('s', exclude=[x(t), y(t), z(t), t]) F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) r1 = (diff(x(t),t) - eq[0]).match(F2-F3) r = collect_const(r1[F2]).match(s*F2) r.update(collect_const(r1[F3]).match(q*F3)) if eq[1].has(r[F2]) and not eq[1].has(r[F3]): r[F2], r[F3] = r[F3], r[F2] r[s], r[q] = -r[q], -r[s] r.update((diff(y(t),t) - eq[1]).match(p*r[F3] - r[s]*F1)) a = r[p]; b = r[q]; c = r[s] F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w) F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w) F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w) z_xy = (C1-a*u-b*v)/c y_zx = (C1-a*u-c*w)/b x_yz = (C1-b*v-c*w)/a y_x = dsolve(diff(v(u),u) - ((a*F3-c*F1)/(c*F2-b*F3)).subs(w,z_xy).subs(v,v(u))).rhs z_x = dsolve(diff(w(u),u) - ((b*F1-a*F2)/(c*F2-b*F3)).subs(v,y_zx).subs(w,w(u))).rhs z_y = dsolve(diff(w(v),v) - ((b*F1-a*F2)/(a*F3-c*F1)).subs(u,x_yz).subs(w,w(v))).rhs x_y = dsolve(diff(u(v),v) - ((c*F2-b*F3)/(a*F3-c*F1)).subs(w,z_xy).subs(u,u(v))).rhs y_z = dsolve(diff(v(w),w) - ((a*F3-c*F1)/(b*F1-a*F2)).subs(u,x_yz).subs(v,v(w))).rhs x_z = dsolve(diff(u(w),w) - ((c*F2-b*F3)/(b*F1-a*F2)).subs(v,y_zx).subs(u,u(w))).rhs sol1 = dsolve(diff(u(t),t) - (c*F2 - b*F3).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs sol2 = dsolve(diff(v(t),t) - (a*F3 - c*F1).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs sol3 = dsolve(diff(w(t),t) - (b*F1 - a*F2).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
r""" Equations:
.. math:: x' = c z F_2 - b y F_3, \enspace y' = a x F_3 - c z F_1, \enspace z' = b y F_1 - a x F_2
where `F_n = F_n (x, y, z, t)`
1. First integral:
.. math:: a x^{2} + b y^{2} + c z^{2} = C_1
where `C` is an arbitrary constant.
2. Assuming the function `F_n` is independent of `t`: `F_n = F_n (x, y, z)`. Then on eliminating `t` and `z` from the first two equations of the system, one arrives at the first-order equation
.. math:: \frac{dy}{dx} = \frac{a x F_3 (x, y, z) - c z F_1 (x, y, z)} {c z F_2 (x, y, z) - b y F_3 (x, y, z)}
where `z = \pm \sqrt{\frac{1}{c} (C_1 - a x^{2} - b y^{2})}`
References ========== -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0405.pdf
""" C1 = symbols('C1') u, v, w = symbols('u, v, w') p = Wild('p', exclude=[x(t), y(t), z(t), t]) q = Wild('q', exclude=[x(t), y(t), z(t), t]) s = Wild('s', exclude=[x(t), y(t), z(t), t]) F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) r1 = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3) r = collect_const(r1[F2]).match(s*F2) r.update(collect_const(r1[F3]).match(q*F3)) if eq[1].has(r[F2]) and not eq[1].has(r[F3]): r[F2], r[F3] = r[F3], r[F2] r[s], r[q] = -r[q], -r[s] r.update((diff(y(t),t) - eq[1]).match(p*x(t)*r[F3] - r[s]*z(t)*F1)) a = r[p]; b = r[q]; c = r[s] F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w) F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w) F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w) x_yz = sqrt((C1 - b*v**2 - c*w**2)/a) y_zx = sqrt((C1 - c*w**2 - a*u**2)/b) z_xy = sqrt((C1 - a*u**2 - b*v**2)/c) y_x = dsolve(diff(v(u),u) - ((a*u*F3-c*w*F1)/(c*w*F2-b*v*F3)).subs(w,z_xy).subs(v,v(u))).rhs z_x = dsolve(diff(w(u),u) - ((b*v*F1-a*u*F2)/(c*w*F2-b*v*F3)).subs(v,y_zx).subs(w,w(u))).rhs z_y = dsolve(diff(w(v),v) - ((b*v*F1-a*u*F2)/(a*u*F3-c*w*F1)).subs(u,x_yz).subs(w,w(v))).rhs x_y = dsolve(diff(u(v),v) - ((c*w*F2-b*v*F3)/(a*u*F3-c*w*F1)).subs(w,z_xy).subs(u,u(v))).rhs y_z = dsolve(diff(v(w),w) - ((a*u*F3-c*w*F1)/(b*v*F1-a*u*F2)).subs(u,x_yz).subs(v,v(w))).rhs x_z = dsolve(diff(u(w),w) - ((c*w*F2-b*v*F3)/(b*v*F1-a*u*F2)).subs(v,y_zx).subs(u,u(w))).rhs sol1 = dsolve(diff(u(t),t) - (c*w*F2 - b*v*F3).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs sol2 = dsolve(diff(v(t),t) - (a*u*F3 - c*w*F1).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs sol3 = dsolve(diff(w(t),t) - (b*v*F1 - a*u*F2).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]
r""" .. math:: x' = x (c F_2 - b F_3), \enspace y' = y (a F_3 - c F_1), \enspace z' = z (b F_1 - a F_2)
where `F_n = F_n (x, y, z, t)` and are arbitrary functions.
First Integral:
.. math:: \left|x\right|^{a} \left|y\right|^{b} \left|z\right|^{c} = C_1
where `C` is an arbitrary constant. If the function `F_n` is independent of `t`, then, by eliminating `t` and `z` from the first two equations of the system, one arrives at a first-order equation.
References ========== -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0406.pdf
""" C1 = symbols('C1') u, v, w = symbols('u, v, w') p = Wild('p', exclude=[x(t), y(t), z(t), t]) q = Wild('q', exclude=[x(t), y(t), z(t), t]) s = Wild('s', exclude=[x(t), y(t), z(t), t]) F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) r1 = eq[0].match(diff(x(t),t) - x(t)*(F2 - F3)) r = collect_const(r1[F2]).match(s*F2) r.update(collect_const(r1[F3]).match(q*F3)) if eq[1].has(r[F2]) and not eq[1].has(r[F3]): r[F2], r[F3] = r[F3], r[F2] r[s], r[q] = -r[q], -r[s] r.update((diff(y(t),t) - eq[1]).match(y(t)*(a*r[F3] - r[c]*F1))) a = r[p]; b = r[q]; c = r[s] F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w) F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w) F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w) x_yz = (C1*v**-b*w**-c)**-a y_zx = (C1*w**-c*u**-a)**-b z_xy = (C1*u**-a*v**-b)**-c y_x = dsolve(diff(v(u),u) - ((v*(a*F3-c*F1))/(u*(c*F2-b*F3))).subs(w,z_xy).subs(v,v(u))).rhs z_x = dsolve(diff(w(u),u) - ((w*(b*F1-a*F2))/(u*(c*F2-b*F3))).subs(v,y_zx).subs(w,w(u))).rhs z_y = dsolve(diff(w(v),v) - ((w*(b*F1-a*F2))/(v*(a*F3-c*F1))).subs(u,x_yz).subs(w,w(v))).rhs x_y = dsolve(diff(u(v),v) - ((u*(c*F2-b*F3))/(v*(a*F3-c*F1))).subs(w,z_xy).subs(u,u(v))).rhs y_z = dsolve(diff(v(w),w) - ((v*(a*F3-c*F1))/(w*(b*F1-a*F2))).subs(u,x_yz).subs(v,v(w))).rhs x_z = dsolve(diff(u(w),w) - ((u*(c*F2-b*F3))/(w*(b*F1-a*F2))).subs(v,y_zx).subs(u,u(w))).rhs sol1 = dsolve(diff(u(t),t) - (u*(c*F2-b*F3)).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs sol2 = dsolve(diff(v(t),t) - (v*(a*F3-c*F1)).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs sol3 = dsolve(diff(w(t),t) - (w*(b*F1-a*F2)).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] |