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