def _lambert(eq, x):
"""
Given an expression assumed to be in the form
``F(X, a..f) = a*log(b*X + c) + d*X + f = 0``
where X = g(x) and x = g^-1(X), return the Lambert solution if possible:
``x = g^-1(-c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(-f/a)))``.
"""
eq = _mexpand(expand_log(eq))
mainlog = _mostfunc(eq, log, x)
if not mainlog:
return [] # violated assumptions
other = eq.subs(mainlog, 0)
if (-other).func is log:
eq = (eq - other).subs(mainlog, mainlog.args[0])
mainlog = mainlog.args[0]
if mainlog.func is not log:
return [] # violated assumptions
other = -(-other).args[0]
eq += other
if not x in other.free_symbols:
return [] # violated assumptions
d, f, X2 = _linab(other, x)
logterm = collect(eq - other, mainlog)
a = logterm.as_coefficient(mainlog)
if a is None or x in a.free_symbols:
return [] # violated assumptions
logarg = mainlog.args[0]
b, c, X1 = _linab(logarg, x)
if X1 != X2:
return [] # violated assumptions
u = Dummy('rhs')
sol = []
# check only real solutions:
for k in [-1, 0]:
l = LambertW(d/(a*b)*exp(c*d/a/b)*exp(-f/a), k)
# if W's arg is between -1/e and 0 there is
# a -1 branch real solution, too.
if k and not l.is_real:
continue
rhs = -c/b + (a/d)*l
solns = solve(X1 - u, x)
for i, tmp in enumerate(solns):
solns[i] = tmp.subs(u, rhs)
sol.append(solns[i])
return sol
def _lambert(eq, x):
"""
Given an expression assumed to be in the form
``F(X, a..f) = a*log(b*X + c) + d*X + f = 0``
where X = g(x) and x = g^-1(X), return the Lambert solution if possible:
``x = g^-1(-c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(-f/a)))``.
"""
eq = _mexpand(expand_log(eq))
mainlog = _mostfunc(eq, log, x)
if not mainlog:
return [] # violated assumptions
other = eq.subs(mainlog, 0)
if (-other).func is log:
eq = (eq - other).subs(mainlog, mainlog.args[0])
mainlog = mainlog.args[0]
if mainlog.func is not log:
return [] # violated assumptions
other = -(-other).args[0]
eq += other
if not x in other.free_symbols:
return [] # violated assumptions
d, f, X2 = _linab(other, x)
logterm = collect(eq - other, mainlog)
a = logterm.as_coefficient(mainlog)
if a is None or x in a.free_symbols:
return [] # violated assumptions
logarg = mainlog.args[0]
b, c, X1 = _linab(logarg, x)
if X1 != X2:
return [] # violated assumptions
u = Dummy('rhs')
sol = []
# check only real solutions:
for k in [-1, 0]:
l = LambertW(d / (a * b) * exp(c * d / a / b) * exp(-f / a), k)
# if W's arg is between -1/e and 0 there is
# a -1 branch real solution, too.
if k and not l.is_real:
continue
rhs = -c / b + (a / d) * l
solns = solve(X1 - u, x)
for i, tmp in enumerate(solns):
solns[i] = tmp.subs(u, rhs)
sol.append(solns[i])
return sol
def _solve_lambert(f, symbol, gens):
"""Return solution to ``f`` if it is a Lambert-type expression
else raise NotImplementedError.
The equality, ``f(x, a..f) = a*log(b*X + c) + d*X - f = 0`` has the
solution, `X = -c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(f/a))`. There
are a variety of forms for `f(X, a..f)` as enumerated below:
1a1)
if B**B = R for R not [0, 1] then
log(B) + log(log(B)) = log(log(R))
X = log(B), a = 1, b = 1, c = 0, d = 1, f = log(log(R))
1a2)
if B*(b*log(B) + c)**a = R then
log(B) + a*log(b*log(B) + c) = log(R)
X = log(B); d=1, f=log(R)
1b)
if a*log(b*B + c) + d*B = R then
X = B, f = R
2a)
if (b*B + c)*exp(d*B + g) = R then
log(b*B + c) + d*B + g = log(R)
a = 1, f = log(R) - g, X = B
2b)
if -b*B + g*exp(d*B + h) = c then
log(g) + d*B + h - log(b*B + c) = 0
a = -1, f = -h - log(g), X = B
3)
if d*p**(a*B + g) - b*B = c then
log(d) + (a*B + g)*log(p) - log(c + b*B) = 0
a = -1, d = a*log(p), f = -log(d) - g*log(p)
"""
nrhs, lhs = f.as_independent(symbol, as_Add=True)
rhs = -nrhs
lamcheck = [tmp for tmp in gens
if (tmp.func in [exp, log] or
(tmp.is_Pow and symbol in tmp.exp.free_symbols))]
if not lamcheck:
raise NotImplementedError()
if lhs.is_Mul:
lhs = expand_log(log(lhs))
rhs = log(rhs)
lhs = factor(lhs, deep=True)
# make sure we are inverted as completely as possible
r = Dummy()
i, lhs = _invert(lhs - r, symbol)
rhs = i.xreplace({r: rhs})
# For the first ones:
# 1a1) B**B = R != 0 (when 0, there is only a solution if the base is 0,
# but if it is, the exp is 0 and 0**0=1
# comes back as B*log(B) = log(R)
# 1a2) B*(a + b*log(B))**p = R or with monomial expanded or with whole
# thing expanded comes back unchanged
# log(B) + p*log(a + b*log(B)) = log(R)
# lhs is Mul:
# expand log of both sides to give:
# log(B) + log(log(B)) = log(log(R))
# 1b) d*log(a*B + b) + c*B = R
# lhs is Add:
# isolate c*B and expand log of both sides:
# log(c) + log(B) = log(R - d*log(a*B + b))
soln = []
if not soln:
mainlog = _mostfunc(lhs, log, symbol)
if mainlog:
if lhs.is_Mul and rhs != 0:
soln = _lambert(log(lhs) - log(rhs), symbol)
elif lhs.is_Add:
other = lhs.subs(mainlog, 0)
if other and not other.is_Add and [
tmp for tmp in other.atoms(Pow)
if symbol in tmp.free_symbols]:
if not rhs:
diff = log(other) - log(other - lhs)
else:
diff = log(lhs - other) - log(rhs - other)
soln = _lambert(expand_log(diff), symbol)
else:
#it's ready to go
soln = _lambert(lhs - rhs, symbol)
# For the next two,
# collect on main exp
# 2a) (b*B + c)*exp(d*B + g) = R
# lhs is mul:
# log to give
# log(b*B + c) + d*B = log(R) - g
# 2b) -b*B + g*exp(d*B + h) = R
# lhs is add:
# add b*B
# log and rearrange
# log(R + b*B) - d*B = log(g) + h
if not soln:
mainexp = _mostfunc(lhs, exp, symbol)
if mainexp:
lhs = collect(lhs, mainexp)
if lhs.is_Mul and rhs != 0:
soln = _lambert(expand_log(log(lhs) - log(rhs)), symbol)
elif lhs.is_Add:
# move all but mainexp-containing term to rhs
other = lhs.subs(mainexp, 0)
mainterm = lhs - other
rhs=rhs - other
if (mainterm.could_extract_minus_sign() and
rhs.could_extract_minus_sign()):
mainterm *= -1
rhs *= -1
diff = log(mainterm) - log(rhs)
soln = _lambert(expand_log(diff), symbol)
# 3) d*p**(a*B + b) + c*B = R
# collect on main pow
# log(R - c*B) - a*B*log(p) = log(d) + b*log(p)
if not soln:
mainpow = _mostfunc(lhs, Pow, symbol)
if mainpow and symbol in mainpow.exp.free_symbols:
lhs = collect(lhs, mainpow)
if lhs.is_Mul and rhs != 0:
soln = _lambert(expand_log(log(lhs) - log(rhs)), symbol)
elif lhs.is_Add:
# move all but mainpow-containing term to rhs
other = lhs.subs(mainpow, 0)
mainterm = lhs - other
rhs = rhs - other
diff = log(mainterm) - log(rhs)
soln = _lambert(expand_log(diff), symbol)
if not soln:
raise NotImplementedError('%s does not appear to have a solution in '
'terms of LambertW' % f)
return list(ordered(soln))
def _matches_commutative(self, expr, repl_dict={}, old=False):
"""
Matches Add/Mul "pattern" to an expression "expr".
repl_dict ... a dictionary of (wild: expression) pairs, that get
returned with the results
This function is the main workhorse for Add/Mul.
For instance:
>>> from sympy import symbols, Wild, sin
>>> a = Wild("a")
>>> b = Wild("b")
>>> c = Wild("c")
>>> x, y, z = symbols("x y z")
>>> (a+sin(b)*c)._matches_commutative(x+sin(y)*z)
{a_: x, b_: y, c_: z}
In the example above, "a+sin(b)*c" is the pattern, and "x+sin(y)*z" is
the expression.
The repl_dict contains parts that were already matched. For example
here:
>>> (x+sin(b)*c)._matches_commutative(x+sin(y)*z, repl_dict={a: x})
{a_: x, b_: y, c_: z}
the only function of the repl_dict is to return it in the
result, e.g. if you omit it:
>>> (x+sin(b)*c)._matches_commutative(x+sin(y)*z)
{b_: y, c_: z}
the "a: x" is not returned in the result, but otherwise it is
equivalent.
"""
# handle simple patterns
if self == expr:
return repl_dict
d = self._matches_simple(expr, repl_dict)
if d is not None:
return d
# eliminate exact part from pattern: (2+a+w1+w2).matches(expr) -> (w1+w2).matches(expr-a-2)
from .function import WildFunction
from .symbol import Wild
wild_part = []
exact_part = []
for p in ordered(self.args):
if p.has(Wild, WildFunction) and (not expr.has(p)):
# not all Wild should stay Wilds, for example:
# (w2+w3).matches(w1) -> (w1+w3).matches(w1) -> w3.matches(0)
wild_part.append(p)
else:
exact_part.append(p)
if exact_part:
exact = self.func(*exact_part)
free = expr.free_symbols
if free and (exact.free_symbols - free):
# there are symbols in the exact part that are not
# in the expr; but if there are no free symbols, let
# the matching continue
return None
newpattern = self.func(*wild_part)
newexpr = self._combine_inverse(expr, exact)
if not old and (expr.is_Add or expr.is_Mul):
if newexpr.count_ops() > expr.count_ops():
return None
return newpattern.matches(newexpr, repl_dict)
# now to real work ;)
i = 0
saw = set()
while expr not in saw:
saw.add(expr)
expr_list = (self.identity,) + tuple(ordered(self.make_args(expr)))
for last_op in reversed(expr_list):
for w in reversed(wild_part):
d1 = w.matches(last_op, repl_dict)
if d1 is not None:
d2 = self.xreplace(d1).matches(expr, d1)
if d2 is not None:
return d2
if i == 0:
if self.is_Mul:
# make e**i look like Mul
if expr.is_Pow and expr.exp.is_Integer:
if expr.exp > 0:
expr = C.Mul(*[expr.base, expr.base**(expr.exp - 1)], evaluate=False)
else:
expr = C.Mul(*[1/expr.base, expr.base**(expr.exp + 1)], evaluate=False)
i += 1
continue
elif self.is_Add:
# make i*e look like Add
c, e = expr.as_coeff_Mul()
if abs(c) > 1:
if c > 0:
expr = C.Add(*[e, (c - 1)*e], evaluate=False)
else:
expr = C.Add(*[-e, (c + 1)*e], evaluate=False)
i += 1
continue
# try collection on non-Wild symbols
from sympy.simplify.simplify import collect
was = expr
did = set()
for w in reversed(wild_part):
c, w = w.as_coeff_mul(Wild)
free = c.free_symbols - did
if free:
did.update(free)
expr = collect(expr, free)
if expr != was:
i += 0
continue
break # if we didn't continue, there is nothing more to do
return
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 _matches_commutative(self, expr, repl_dict={}, old=False):
"""
Matches Add/Mul "pattern" to an expression "expr".
repl_dict ... a dictionary of (wild: expression) pairs, that get
returned with the results
This function is the main workhorse for Add/Mul.
For instance:
>>> from sympy import symbols, Wild, sin
>>> a = Wild("a")
>>> b = Wild("b")
>>> c = Wild("c")
>>> x, y, z = symbols("x y z")
>>> (a+sin(b)*c)._matches_commutative(x+sin(y)*z)
{a_: x, b_: y, c_: z}
In the example above, "a+sin(b)*c" is the pattern, and "x+sin(y)*z" is
the expression.
The repl_dict contains parts that were already matched. For example
here:
>>> (x+sin(b)*c)._matches_commutative(x+sin(y)*z, repl_dict={a: x})
{a_: x, b_: y, c_: z}
the only function of the repl_dict is to return it in the
result, e.g. if you omit it:
>>> (x+sin(b)*c)._matches_commutative(x+sin(y)*z)
{b_: y, c_: z}
the "a: x" is not returned in the result, but otherwise it is
equivalent.
"""
# handle simple patterns
if self == expr:
return repl_dict
d = self._matches_simple(expr, repl_dict)
if d is not None:
return d
# eliminate exact part from pattern: (2+a+w1+w2).matches(expr) -> (w1+w2).matches(expr-a-2)
from .function import WildFunction
from .symbol import Wild
wild_part = []
exact_part = []
for p in ordered(self.args):
if p.has(Wild, WildFunction) and (not expr.has(p)):
# not all Wild should stay Wilds, for example:
# (w2+w3).matches(w1) -> (w1+w3).matches(w1) -> w3.matches(0)
wild_part.append(p)
else:
exact_part.append(p)
if exact_part:
exact = self.func(*exact_part)
free = expr.free_symbols
if free and (exact.free_symbols - free):
# there are symbols in the exact part that are not
# in the expr; but if there are no free symbols, let
# the matching continue
return None
newpattern = self.func(*wild_part)
newexpr = self._combine_inverse(expr, exact)
if not old and (expr.is_Add or expr.is_Mul):
if newexpr.count_ops() > expr.count_ops():
return None
return newpattern.matches(newexpr, repl_dict)
# now to real work ;)
i = 0
saw = set()
while expr not in saw:
saw.add(expr)
expr_list = (self.identity,) + tuple(ordered(self.make_args(expr)))
for last_op in reversed(expr_list):
for w in reversed(wild_part):
d1 = w.matches(last_op, repl_dict)
if d1 is not None:
d2 = self.xreplace(d1).matches(expr, d1)
if d2 is not None:
return d2
if i == 0:
if self.is_Mul:
# make e**i look like Mul
if expr.is_Pow and expr.exp.is_Integer:
if expr.exp > 0:
expr = C.Mul(*[expr.base, expr.base**(expr.exp - 1)], evaluate=False)
else:
expr = C.Mul(*[1/expr.base, expr.base**(expr.exp + 1)], evaluate=False)
i += 1
continue
elif self.is_Add:
# make i*e look like Add
c, e = expr.as_coeff_Mul()
if abs(c) > 1:
if c > 0:
expr = C.Add(*[e, (c - 1)*e], evaluate=False)
else:
expr = C.Add(*[-e, (c + 1)*e], evaluate=False)
i += 1
continue
# try collection on non-Wild symbols
from sympy.simplify.simplify import collect
was = expr
did = set()
for w in reversed(wild_part):
c, w = w.as_coeff_mul(Wild)
free = c.free_symbols - did
if free:
did.update(free)
expr = collect(expr, free)
if expr != was:
i += 0
continue
break # if we didn't continue, there is nothing more to do
return