393 lines
16 KiB
Python
393 lines
16 KiB
Python
from sympy.core import S, Pow
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from sympy.core.function import (Derivative, AppliedUndef, diff)
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from sympy.core.relational import Equality, Eq
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from sympy.core.symbol import Dummy
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from sympy.core.sympify import sympify
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from sympy.logic.boolalg import BooleanAtom
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from sympy.functions import exp
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from sympy.series import Order
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from sympy.simplify.simplify import simplify, posify, besselsimp
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from sympy.simplify.trigsimp import trigsimp
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from sympy.simplify.sqrtdenest import sqrtdenest
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from sympy.solvers import solve
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from sympy.solvers.deutils import _preprocess, ode_order
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from sympy.utilities.iterables import iterable, is_sequence
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def sub_func_doit(eq, func, new):
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r"""
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When replacing the func with something else, we usually want the
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derivative evaluated, so this function helps in making that happen.
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Examples
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========
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>>> from sympy import Derivative, symbols, Function
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>>> from sympy.solvers.ode.subscheck import sub_func_doit
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>>> x, z = symbols('x, z')
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>>> y = Function('y')
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>>> sub_func_doit(3*Derivative(y(x), x) - 1, y(x), x)
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2
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>>> sub_func_doit(x*Derivative(y(x), x) - y(x)**2 + y(x), y(x),
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... 1/(x*(z + 1/x)))
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x*(-1/(x**2*(z + 1/x)) + 1/(x**3*(z + 1/x)**2)) + 1/(x*(z + 1/x))
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...- 1/(x**2*(z + 1/x)**2)
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"""
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reps= {func: new}
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for d in eq.atoms(Derivative):
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if d.expr == func:
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reps[d] = new.diff(*d.variable_count)
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else:
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reps[d] = d.xreplace({func: new}).doit(deep=False)
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return eq.xreplace(reps)
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def checkodesol(ode, sol, func=None, order='auto', solve_for_func=True):
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r"""
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Substitutes ``sol`` into ``ode`` and checks that the result is ``0``.
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This works when ``func`` is one function, like `f(x)` or a list of
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functions like `[f(x), g(x)]` when `ode` is a system of ODEs. ``sol`` can
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be a single solution or a list of solutions. Each solution may be an
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:py:class:`~sympy.core.relational.Equality` that the solution satisfies,
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e.g. ``Eq(f(x), C1), Eq(f(x) + C1, 0)``; or simply an
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:py:class:`~sympy.core.expr.Expr`, e.g. ``f(x) - C1``. In most cases it
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will not be necessary to explicitly identify the function, but if the
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function cannot be inferred from the original equation it can be supplied
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through the ``func`` argument.
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If a sequence of solutions is passed, the same sort of container will be
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used to return the result for each solution.
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It tries the following methods, in order, until it finds zero equivalence:
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1. Substitute the solution for `f` in the original equation. This only
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works if ``ode`` is solved for `f`. It will attempt to solve it first
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unless ``solve_for_func == False``.
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2. Take `n` derivatives of the solution, where `n` is the order of
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``ode``, and check to see if that is equal to the solution. This only
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works on exact ODEs.
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3. Take the 1st, 2nd, ..., `n`\th derivatives of the solution, each time
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solving for the derivative of `f` of that order (this will always be
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possible because `f` is a linear operator). Then back substitute each
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derivative into ``ode`` in reverse order.
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This function returns a tuple. The first item in the tuple is ``True`` if
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the substitution results in ``0``, and ``False`` otherwise. The second
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item in the tuple is what the substitution results in. It should always
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be ``0`` if the first item is ``True``. Sometimes this function will
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return ``False`` even when an expression is identically equal to ``0``.
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This happens when :py:meth:`~sympy.simplify.simplify.simplify` does not
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reduce the expression to ``0``. If an expression returned by this
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function vanishes identically, then ``sol`` really is a solution to
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the ``ode``.
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If this function seems to hang, it is probably because of a hard
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simplification.
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To use this function to test, test the first item of the tuple.
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Examples
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========
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>>> from sympy import (Eq, Function, checkodesol, symbols,
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... Derivative, exp)
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>>> x, C1, C2 = symbols('x,C1,C2')
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>>> f, g = symbols('f g', cls=Function)
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>>> checkodesol(f(x).diff(x), Eq(f(x), C1))
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(True, 0)
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>>> assert checkodesol(f(x).diff(x), C1)[0]
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>>> assert not checkodesol(f(x).diff(x), x)[0]
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>>> checkodesol(f(x).diff(x, 2), x**2)
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(False, 2)
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>>> eqs = [Eq(Derivative(f(x), x), f(x)), Eq(Derivative(g(x), x), g(x))]
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>>> sol = [Eq(f(x), C1*exp(x)), Eq(g(x), C2*exp(x))]
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>>> checkodesol(eqs, sol)
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(True, [0, 0])
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"""
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if iterable(ode):
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return checksysodesol(ode, sol, func=func)
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if not isinstance(ode, Equality):
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ode = Eq(ode, 0)
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if func is None:
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try:
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_, func = _preprocess(ode.lhs)
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except ValueError:
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funcs = [s.atoms(AppliedUndef) for s in (
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sol if is_sequence(sol, set) else [sol])]
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funcs = set().union(*funcs)
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if len(funcs) != 1:
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raise ValueError(
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'must pass func arg to checkodesol for this case.')
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func = funcs.pop()
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if not isinstance(func, AppliedUndef) or len(func.args) != 1:
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raise ValueError(
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"func must be a function of one variable, not %s" % func)
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if is_sequence(sol, set):
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return type(sol)([checkodesol(ode, i, order=order, solve_for_func=solve_for_func) for i in sol])
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if not isinstance(sol, Equality):
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sol = Eq(func, sol)
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elif sol.rhs == func:
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sol = sol.reversed
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if order == 'auto':
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order = ode_order(ode, func)
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solved = sol.lhs == func and not sol.rhs.has(func)
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if solve_for_func and not solved:
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rhs = solve(sol, func)
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if rhs:
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eqs = [Eq(func, t) for t in rhs]
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if len(rhs) == 1:
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eqs = eqs[0]
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return checkodesol(ode, eqs, order=order,
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solve_for_func=False)
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x = func.args[0]
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# Handle series solutions here
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if sol.has(Order):
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assert sol.lhs == func
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Oterm = sol.rhs.getO()
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solrhs = sol.rhs.removeO()
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Oexpr = Oterm.expr
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assert isinstance(Oexpr, Pow)
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sorder = Oexpr.exp
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assert Oterm == Order(x**sorder)
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odesubs = (ode.lhs-ode.rhs).subs(func, solrhs).doit().expand()
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neworder = Order(x**(sorder - order))
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odesubs = odesubs + neworder
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assert odesubs.getO() == neworder
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residual = odesubs.removeO()
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return (residual == 0, residual)
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s = True
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testnum = 0
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while s:
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if testnum == 0:
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# First pass, try substituting a solved solution directly into the
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# ODE. This has the highest chance of succeeding.
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ode_diff = ode.lhs - ode.rhs
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if sol.lhs == func:
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s = sub_func_doit(ode_diff, func, sol.rhs)
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s = besselsimp(s)
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else:
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testnum += 1
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continue
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ss = simplify(s.rewrite(exp))
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if ss:
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# with the new numer_denom in power.py, if we do a simple
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# expansion then testnum == 0 verifies all solutions.
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s = ss.expand(force=True)
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else:
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s = 0
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testnum += 1
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elif testnum == 1:
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# Second pass. If we cannot substitute f, try seeing if the nth
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# derivative is equal, this will only work for odes that are exact,
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# by definition.
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s = simplify(
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trigsimp(diff(sol.lhs, x, order) - diff(sol.rhs, x, order)) -
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trigsimp(ode.lhs) + trigsimp(ode.rhs))
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# s2 = simplify(
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# diff(sol.lhs, x, order) - diff(sol.rhs, x, order) - \
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# ode.lhs + ode.rhs)
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testnum += 1
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elif testnum == 2:
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# Third pass. Try solving for df/dx and substituting that into the
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# ODE. Thanks to Chris Smith for suggesting this method. Many of
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# the comments below are his, too.
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# The method:
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# - Take each of 1..n derivatives of the solution.
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# - Solve each nth derivative for d^(n)f/dx^(n)
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# (the differential of that order)
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# - Back substitute into the ODE in decreasing order
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# (i.e., n, n-1, ...)
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# - Check the result for zero equivalence
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if sol.lhs == func and not sol.rhs.has(func):
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diffsols = {0: sol.rhs}
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elif sol.rhs == func and not sol.lhs.has(func):
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diffsols = {0: sol.lhs}
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else:
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diffsols = {}
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sol = sol.lhs - sol.rhs
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for i in range(1, order + 1):
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# Differentiation is a linear operator, so there should always
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# be 1 solution. Nonetheless, we test just to make sure.
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# We only need to solve once. After that, we automatically
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# have the solution to the differential in the order we want.
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if i == 1:
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ds = sol.diff(x)
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try:
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sdf = solve(ds, func.diff(x, i))
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if not sdf:
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raise NotImplementedError
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except NotImplementedError:
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testnum += 1
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break
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else:
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diffsols[i] = sdf[0]
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else:
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# This is what the solution says df/dx should be.
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diffsols[i] = diffsols[i - 1].diff(x)
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# Make sure the above didn't fail.
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if testnum > 2:
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continue
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else:
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# Substitute it into ODE to check for self consistency.
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lhs, rhs = ode.lhs, ode.rhs
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for i in range(order, -1, -1):
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if i == 0 and 0 not in diffsols:
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# We can only substitute f(x) if the solution was
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# solved for f(x).
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break
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lhs = sub_func_doit(lhs, func.diff(x, i), diffsols[i])
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rhs = sub_func_doit(rhs, func.diff(x, i), diffsols[i])
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ode_or_bool = Eq(lhs, rhs)
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ode_or_bool = simplify(ode_or_bool)
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if isinstance(ode_or_bool, (bool, BooleanAtom)):
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if ode_or_bool:
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lhs = rhs = S.Zero
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else:
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lhs = ode_or_bool.lhs
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rhs = ode_or_bool.rhs
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# No sense in overworking simplify -- just prove that the
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# numerator goes to zero
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num = trigsimp((lhs - rhs).as_numer_denom()[0])
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# since solutions are obtained using force=True we test
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# using the same level of assumptions
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## replace function with dummy so assumptions will work
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_func = Dummy('func')
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num = num.subs(func, _func)
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## posify the expression
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num, reps = posify(num)
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s = simplify(num).xreplace(reps).xreplace({_func: func})
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testnum += 1
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else:
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break
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if not s:
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return (True, s)
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elif s is True: # The code above never was able to change s
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raise NotImplementedError("Unable to test if " + str(sol) +
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" is a solution to " + str(ode) + ".")
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else:
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return (False, s)
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def checksysodesol(eqs, sols, func=None):
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r"""
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Substitutes corresponding ``sols`` for each functions into each ``eqs`` and
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checks that the result of substitutions for each equation is ``0``. The
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equations and solutions passed can be any iterable.
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This only works when each ``sols`` have one function only, like `x(t)` or `y(t)`.
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For each function, ``sols`` can have a single solution or a list of solutions.
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In most cases it will not be necessary to explicitly identify the function,
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but if the function cannot be inferred from the original equation it
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can be supplied through the ``func`` argument.
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When a sequence of equations is passed, the same sequence is used to return
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the result for each equation with each function substituted with corresponding
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solutions.
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It tries the following method to find zero equivalence for each equation:
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Substitute the solutions for functions, like `x(t)` and `y(t)` into the
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original equations containing those functions.
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This function returns a tuple. The first item in the tuple is ``True`` if
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the substitution results for each equation is ``0``, and ``False`` otherwise.
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The second item in the tuple is what the substitution results in. Each element
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of the ``list`` should always be ``0`` corresponding to each equation if the
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first item is ``True``. Note that sometimes this function may return ``False``,
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but with an expression that is identically equal to ``0``, instead of returning
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``True``. This is because :py:meth:`~sympy.simplify.simplify.simplify` cannot
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reduce the expression to ``0``. If an expression returned by each function
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vanishes identically, then ``sols`` really is a solution to ``eqs``.
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If this function seems to hang, it is probably because of a difficult simplification.
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Examples
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========
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>>> from sympy import Eq, diff, symbols, sin, cos, exp, sqrt, S, Function
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>>> from sympy.solvers.ode.subscheck import checksysodesol
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>>> C1, C2 = symbols('C1:3')
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>>> t = symbols('t')
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>>> x, y = symbols('x, y', cls=Function)
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>>> eq = (Eq(diff(x(t),t), x(t) + y(t) + 17), Eq(diff(y(t),t), -2*x(t) + y(t) + 12))
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>>> sol = [Eq(x(t), (C1*sin(sqrt(2)*t) + C2*cos(sqrt(2)*t))*exp(t) - S(5)/3),
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... Eq(y(t), (sqrt(2)*C1*cos(sqrt(2)*t) - sqrt(2)*C2*sin(sqrt(2)*t))*exp(t) - S(46)/3)]
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>>> checksysodesol(eq, sol)
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(True, [0, 0])
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>>> eq = (Eq(diff(x(t),t),x(t)*y(t)**4), Eq(diff(y(t),t),y(t)**3))
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>>> sol = [Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), -sqrt(2)*sqrt(-1/(C2 + t))/2),
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... Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), sqrt(2)*sqrt(-1/(C2 + t))/2)]
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>>> checksysodesol(eq, sol)
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(True, [0, 0])
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"""
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def _sympify(eq):
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return list(map(sympify, eq if iterable(eq) else [eq]))
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eqs = _sympify(eqs)
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for i in range(len(eqs)):
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if isinstance(eqs[i], Equality):
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eqs[i] = eqs[i].lhs - eqs[i].rhs
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if func is None:
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funcs = []
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for eq in eqs:
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derivs = eq.atoms(Derivative)
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func = set().union(*[d.atoms(AppliedUndef) for d in derivs])
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funcs.extend(func)
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funcs = list(set(funcs))
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if not all(isinstance(func, AppliedUndef) and len(func.args) == 1 for func in funcs)\
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and len({func.args for func in funcs})!=1:
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raise ValueError("func must be a function of one variable, not %s" % func)
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for sol in sols:
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if len(sol.atoms(AppliedUndef)) != 1:
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raise ValueError("solutions should have one function only")
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if len(funcs) != len({sol.lhs for sol in sols}):
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raise ValueError("number of solutions provided does not match the number of equations")
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dictsol = {}
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for sol in sols:
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func = list(sol.atoms(AppliedUndef))[0]
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if sol.rhs == func:
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sol = sol.reversed
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solved = sol.lhs == func and not sol.rhs.has(func)
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if not solved:
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rhs = solve(sol, func)
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if not rhs:
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raise NotImplementedError
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else:
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rhs = sol.rhs
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dictsol[func] = rhs
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checkeq = []
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for eq in eqs:
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for func in funcs:
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eq = sub_func_doit(eq, func, dictsol[func])
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ss = simplify(eq)
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if ss != 0:
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eq = ss.expand(force=True)
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if eq != 0:
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eq = sqrtdenest(eq).simplify()
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else:
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eq = 0
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checkeq.append(eq)
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if len(set(checkeq)) == 1 and list(set(checkeq))[0] == 0:
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return (True, checkeq)
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else:
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return (False, checkeq)
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