def _get_eq_preprocessed(self) -> Expr:
if self.prep:
process_eq, process_func = _preprocess(self.eq, self.func)
if process_func != self.func:
raise ValueError
else:
process_eq = self.eq
return process_eq
def classify_pde(eq, func=None, dict=False, **kwargs):
"""
Returns a tuple of possible pdsolve() classifications for a PDE.
The tuple is ordered so that first item is the classification that
pdsolve() uses to solve the PDE 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 pdsolve use a different classification,
use pdsolve(PDE, func, hint=<classification>). See also the pdsolve()
docstring for different meta-hints you can use.
If ``dict`` is true, classify_pde() will return a dictionary of
hint:match expression terms. This is intended for internal use by
pdsolve(). 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 doing help(pde.pde_hintname),
where hintname is the name of the hint without "_Integral".
See sympy.pde.allhints or the sympy.pde docstring for a list of all
supported hints that can be returned from classify_pde.
Examples
========
>>> from sympy.solvers.pde import classify_pde
>>> from sympy import Function, diff, Eq
>>> from sympy.abc import x, y
>>> f = Function('f')
>>> u = f(x, y)
>>> ux = u.diff(x)
>>> uy = u.diff(y)
>>> eq = Eq(1 + (2*(ux/u)) + (3*(uy/u)))
>>> classify_pde(eq)
('1st_linear_constant_coeff_homogeneous',)
"""
prep = kwargs.pop('prep', True)
if func and len(func.args) != 2:
raise NotImplementedError("Right now only partial "
"differential equations of two variables are supported")
if prep or func is None:
prep, func_ = _preprocess(eq, func)
if func is None:
func = func_
if isinstance(eq, Equality):
if eq.rhs != 0:
return classify_pde(eq.lhs - eq.rhs, func)
eq = eq.lhs
f = func.func
x = func.args[0]
y = func.args[1]
fx = f(x,y).diff(x)
fy = f(x,y).diff(y)
# TODO : For now pde.py uses support offered by the ode_order function
# to find the order with respect to a multi-variable function. An
# improvement could be to classify the order of the PDE on the basis of
# individual variables.
order = ode_order(eq, f(x,y))
# hint:matchdict or hint:(tuple of matchdicts)
# Also will contain "default":<default hint> and "order":order items.
matching_hints = {'order': order}
if not order:
if dict:
matching_hints["default"] = None
return matching_hints
else:
return ()
eq = expand(eq)
a = Wild('a', exclude = [f(x,y)])
b = Wild('b', exclude = [f(x,y), fx, fy, x, y])
c = Wild('c', exclude = [f(x,y), fx, fy, x, y])
d = Wild('d', exclude = [f(x,y), fx, fy, x, y])
e = Wild('e', exclude = [f(x,y), fx, fy])
n = Wild('n', exclude = [x, y])
# Try removing the smallest power of f(x,y)
# from the highest partial derivatives of f(x,y)
reduced_eq = None
if eq.is_Add:
var = set(combinations_with_replacement((x,y), order))
dummyvar = deepcopy(var)
power = None
for i in var:
coeff = eq.coeff(f(x,y).diff(*i))
if coeff != 1:
match = coeff.match(a*f(x,y)**n)
if match and match[a]:
power = match[n]
dummyvar.remove(i)
break
dummyvar.remove(i)
for i in dummyvar:
coeff = eq.coeff(f(x,y).diff(*i))
if coeff != 1:
match = coeff.match(a*f(x,y)**n)
if match and match[a] and match[n] < power:
power = match[n]
if power:
den = f(x,y)**power
reduced_eq = Add(*[arg/den for arg in eq.args])
if not reduced_eq:
reduced_eq = eq
if order == 1:
reduced_eq = collect(reduced_eq, f(x, y))
r = reduced_eq.match(b*fx + c*fy + d*f(x,y) + e)
if r:
if not r[e]:
## Linear first-order homogeneous partial-differential
## equation with constant coefficients
r.update({'b': b, 'c': c, 'd': d})
matching_hints["1st_linear_constant_coeff_homogeneous"] = r
else:
if r[b]**2 + r[c]**2 != 0:
## Linear first-order general partial-differential
## equation with constant coefficients
r.update({'b': b, 'c': c, 'd': d, 'e': e})
matching_hints["1st_linear_constant_coeff"] = r
matching_hints[
"1st_linear_constant_coeff_Integral"] = r
else:
b = Wild('b', exclude=[f(x, y), fx, fy])
c = Wild('c', exclude=[f(x, y), fx, fy])
d = Wild('d', exclude=[f(x, y), fx, fy])
r = reduced_eq.match(b*fx + c*fy + d*f(x,y) + e)
if r:
r.update({'b': b, 'c': c, 'd': d, 'e': e})
matching_hints["1st_linear_variable_coeff"] = r
# Order keys based on allhints.
retlist = []
for i in allhints:
if i in matching_hints:
retlist.append(i)
if dict:
# Dictionaries are ordered arbitrarily, so make note of which
# hint would come first for pdsolve(). Use an ordered dict in Py 3.
matching_hints["default"] = None
matching_hints["ordered_hints"] = tuple(retlist)
for i in allhints:
if i in matching_hints:
matching_hints["default"] = i
break
return matching_hints
else:
return tuple(retlist)
def checkpdesol(pde, sol, func=None, solve_for_func=True):
"""
Checks if the given solution satisfies the partial differential
equation.
pde is the partial differential equation which can be given in the
form of an equation or an expression. sol is the solution for which
the pde is to be checked. This can also be given in an equation or
an expression form. If the function is not provided, the helper
function _preprocess from deutils is used to identify the function.
If a sequence of solutions is passed, the same sort of container will be
used to return the result for each solution.
The following methods are currently being implemented to check if the
solution satisfies the PDE:
1. Directly substitute the solution in the PDE and check. If the
solution hasn't been solved for f, then it will solve for f
provided solve_for_func hasn't been set to False.
If the solution satisfies the PDE, then a tuple (True, 0) is returned.
Otherwise a tuple (False, expr) where expr is the value obtained
after substituting the solution in the PDE. However if a known solution
returns False, it may be due to the inability of doit() to simplify it to zero.
Examples
========
>>> from sympy import Function, symbols, diff
>>> from sympy.solvers.pde import checkpdesol, pdsolve
>>> x, y = symbols('x y')
>>> f = Function('f')
>>> eq = 2*f(x,y) + 3*f(x,y).diff(x) + 4*f(x,y).diff(y)
>>> sol = pdsolve(eq)
>>> assert checkpdesol(eq, sol)[0]
>>> eq = x*f(x,y) + f(x,y).diff(x)
>>> checkpdesol(eq, sol)
(False, (x*F(4*x - 3*y) - 6*F(4*x - 3*y)/25 + 4*Subs(Derivative(F(_xi_1), _xi_1), (_xi_1,), (4*x - 3*y,)))*exp(-6*x/25 - 8*y/25))
"""
# Converting the pde into an equation
if not isinstance(pde, Equality):
pde = Eq(pde, 0)
# If no function is given, try finding the function present.
if func is None:
try:
_, func = _preprocess(pde.lhs)
except ValueError:
funcs = [s.atoms(AppliedUndef) for s in (
sol if is_sequence(sol, set) else [sol])]
funcs = reduce(set.union, funcs, set())
if len(funcs) != 1:
raise ValueError(
'must pass func arg to checkpdesol for this case.')
func = funcs.pop()
# If the given solution is in the form of a list or a set
# then return a list or set of tuples.
if is_sequence(sol, set):
return type(sol)(map(lambda i: checkpdesol(pde, i,
solve_for_func=solve_for_func), sol))
# Convert solution into an equation
if not isinstance(sol, Equality):
sol = Eq(func, sol)
# Try solving for the function
if solve_for_func and not (sol.lhs == func and not sol.rhs.has(func)) and not \
(sol.rhs == func and not sol.lhs.has(func)):
try:
solved = solve(sol, func)
if not solved:
raise NotImplementedError
except NotImplementedError:
pass
else:
if len(solved) == 1:
result = checkpdesol(pde, Eq(func, solved[0]),
order=order, solve_for_func=False)
else:
result = checkpdesol(pde, [Eq(func, t) for t in solved],
order=order, solve_for_func=False)
# The first method includes direct substitution of the solution in
# the PDE and simplifying.
pde = pde.lhs - pde.rhs
if sol.lhs == func:
s = pde.subs(func, sol.rhs).doit()
elif sol.rhs == func:
s = pde.subs(func, sol.lhs).doit()
if s:
ss = simplify(s)
if ss:
return False, ss
else:
return True, 0
else:
return True, 0
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)
