def ufuncify(args, expr, language=None, backend='numpy', tempdir=None,
flags=None, verbose=False, helpers=None):
"""Generates a binary function that supports broadcasting on numpy arrays.
Parameters
----------
args : iterable
Either a Symbol or an iterable of symbols. Specifies the argument
sequence for the function.
expr
A SymPy expression that defines the element wise operation.
language : string, optional
If supplied, (options: 'C' or 'F95'), specifies the language of the
generated code. If ``None`` [default], the language is inferred based
upon the specified backend.
backend : string, optional
Backend used to wrap the generated code. Either 'numpy' [default],
'cython', or 'f2py'.
tempdir : string, optional
Path to directory for temporary files. If this argument is supplied,
the generated code and the wrapper input files are left intact in the
specified path.
flags : iterable, optional
Additional option flags that will be passed to the backend
verbose : bool, optional
If True, autowrap will not mute the command line backends. This can be
helpful for debugging.
helpers : iterable, optional
Used to define auxillary expressions needed for the main expr. If the
main expression needs to call a specialized function it should be put
in the ``helpers`` iterable. Autowrap will then make sure that the
compiled main expression can link to the helper routine. Items should
be tuples with (<funtion_name>, <sympy_expression>, <arguments>). It
is mandatory to supply an argument sequence to helper routines.
Note
----
The default backend ('numpy') will create actual instances of
``numpy.ufunc``. These support ndimensional broadcasting, and implicit type
conversion. Use of the other backends will result in a "ufunc-like"
function, which requires equal length 1-dimensional arrays for all
arguments, and will not perform any type conversions.
References
----------
[1] http://docs.scipy.org/doc/numpy/reference/ufuncs.html
Examples
--------
>>> from sympy.utilities.autowrap import ufuncify
>>> from sympy.abc import x, y
>>> import numpy as np
>>> f = ufuncify((x, y), y + x**2)
>>> type(f)
numpy.ufunc
>>> f([1, 2, 3], 2)
array([ 3., 6., 11.])
>>> f(np.arange(5), 3)
array([ 3., 4., 7., 12., 19.])
For the F2Py and Cython backends, inputs are required to be equal length
1-dimensional arrays. The F2Py backend will perform type conversion, but
the Cython backend will error if the inputs are not of the expected type.
>>> f_fortran = ufuncify((x, y), y + x**2, backend='F2Py')
>>> f_fortran(1, 2)
3
>>> f_fortran(numpy.array([1, 2, 3]), numpy.array([1.0, 2.0, 3.0]))
array([2., 6., 12.])
>>> f_cython = ufuncify((x, y), y + x**2, backend='Cython')
>>> f_cython(1, 2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: Argument '_x' has incorrect type (expected numpy.ndarray, got int)
>>> f_cython(numpy.array([1.0]), numpy.array([2.0]))
array([ 3.])
"""
if isinstance(args, Symbol):
args = (args,)
else:
args = tuple(args)
if language:
_validate_backend_language(backend, language)
else:
language = _infer_language(backend)
helpers = helpers if helpers else ()
flags = flags if flags else ()
if backend.upper() == 'NUMPY':
routine = make_routine('autofunc', expr, args)
helps = []
for name, expr, args in helpers:
helps.append(make_routine(name, expr, args))
code_wrapper = UfuncifyCodeWrapper(CCodeGen("ufuncify"), tempdir,
flags, verbose)
return code_wrapper.wrap_code(routine, helpers=helps)
else:
# Dummies are used for all added expressions to prevent name clashes
# within the original expression.
y = IndexedBase(Dummy())
m = Dummy(integer=True)
i = Idx(Dummy(integer=True), m)
f = implemented_function(Dummy().name, Lambda(args, expr))
# For each of the args create an indexed version.
indexed_args = [IndexedBase(Dummy(str(a))) for a in args]
# Order the arguments (out, args, dim)
args = [y] + indexed_args + [m]
args_with_indices = [a[i] for a in indexed_args]
return autowrap(Eq(y[i], f(*args_with_indices)), language, backend,
tempdir, args, flags, verbose, helpers)
def as_relational(self, symbol):
"""Rewrite a FiniteSet in terms of equalities and logic operators. """
from sympy.core.relational import Eq
return Or(*[Eq(symbol, elem) for elem in self])
def eval_sum_symbolic(f, limits):
from sympy.functions import harmonic, bernoulli
f_orig = f
(i, a, b) = limits
if not f.has(i):
return f * (b - a + 1)
# Linearity
if f.is_Mul:
L, R = f.as_two_terms()
if not L.has(i):
sR = eval_sum_symbolic(R, (i, a, b))
if sR:
return L * sR
if not R.has(i):
sL = eval_sum_symbolic(L, (i, a, b))
if sL:
return R * sL
try:
f = apart(f, i) # see if it becomes an Add
except PolynomialError:
pass
if f.is_Add:
L, R = f.as_two_terms()
lrsum = telescopic(L, R, (i, a, b))
if lrsum:
return lrsum
lsum = eval_sum_symbolic(L, (i, a, b))
rsum = eval_sum_symbolic(R, (i, a, b))
if None not in (lsum, rsum):
r = lsum + rsum
if not r is S.NaN:
return r
# Polynomial terms with Faulhaber's formula
n = Wild('n')
result = f.match(i**n)
if result is not None:
n = result[n]
if n.is_Integer:
if n >= 0:
if (b is S.Infinity and not a is S.NegativeInfinity) or \
(a is S.NegativeInfinity and not b is S.Infinity):
return S.Infinity
return ((bernoulli(n + 1, b + 1) - bernoulli(n + 1, a)) /
(n + 1)).expand()
elif a.is_Integer and a >= 1:
if n == -1:
return harmonic(b) - harmonic(a - 1)
else:
return harmonic(b, abs(n)) - harmonic(a - 1, abs(n))
if not (a.has(S.Infinity, S.NegativeInfinity)
or b.has(S.Infinity, S.NegativeInfinity)):
# Geometric terms
c1 = Wild('c1', exclude=[i])
c2 = Wild('c2', exclude=[i])
c3 = Wild('c3', exclude=[i])
e = f.match(c1**(c2 * i + c3))
if e is not None:
p = (c1**c3).subs(e)
q = (c1**c2).subs(e)
r = p * (q**a - q**(b + 1)) / (1 - q)
l = p * (b - a + 1)
return Piecewise((l, Eq(q, S.One)), (r, True))
r = gosper_sum(f, (i, a, b))
if not r in (None, S.NaN):
return r
return eval_sum_hyper(f_orig, (i, a, b))
def test_dsolve_all_hint():
eq = f(x).diff(x)
output = dsolve(eq, hint='all')
# Match the Dummy variables:
sol1 = output['separable_Integral']
_y = sol1.lhs.args[1][0]
sol1 = output['1st_homogeneous_coeff_subs_dep_div_indep_Integral']
_u1 = sol1.rhs.args[1].args[1][0]
expected = {'Bernoulli_Integral': Eq(f(x), C1 + Integral(0, x)),
'1st_homogeneous_coeff_best': Eq(f(x), C1),
'Bernoulli': Eq(f(x), C1),
'nth_algebraic': Eq(f(x), C1),
'nth_linear_euler_eq_homogeneous': Eq(f(x), C1),
'nth_linear_constant_coeff_homogeneous': Eq(f(x), C1),
'separable': Eq(f(x), C1),
'1st_homogeneous_coeff_subs_indep_div_dep': Eq(f(x), C1),
'nth_algebraic_Integral': Eq(f(x), C1),
'1st_linear': Eq(f(x), C1),
'1st_linear_Integral': Eq(f(x), C1 + Integral(0, x)),
'1st_exact': Eq(f(x), C1),
'1st_exact_Integral': Eq(Subs(Integral(0, x) + Integral(1, _y), _y, f(x)), C1),
'lie_group': Eq(f(x), C1),
'1st_homogeneous_coeff_subs_dep_div_indep': Eq(f(x), C1),
'1st_homogeneous_coeff_subs_dep_div_indep_Integral': Eq(log(x), C1 + Integral(-1/_u1, (_u1, f(x)/x))),
'1st_power_series': Eq(f(x), C1),
'separable_Integral': Eq(Integral(1, (_y, f(x))), C1 + Integral(0, x)),
'1st_homogeneous_coeff_subs_indep_div_dep_Integral': Eq(f(x), C1),
'best': Eq(f(x), C1),
'best_hint': 'nth_algebraic',
'default': 'nth_algebraic',
'order': 1}
assert output == expected
assert dsolve(eq, hint='best') == Eq(f(x), C1)
def reduce_inequalities(inequalities, symbols=[]):
"""Reduce a system of inequalities with rational coefficients.
Examples
========
>>> from sympy.abc import x, y
>>> from sympy import reduce_inequalities
>>> reduce_inequalities(0 <= x + 3, [])
(-3 <= x) & (x < oo)
>>> reduce_inequalities(0 <= x + y*2 - 1, [x])
(x < oo) & (x >= 1 - 2*y)
"""
if not iterable(inequalities):
inequalities = [inequalities]
inequalities = [sympify(i) for i in inequalities]
gens = set().union(*[i.free_symbols for i in inequalities])
if not iterable(symbols):
symbols = [symbols]
symbols = (set(symbols) or gens) & gens
if any(i.is_extended_real is False for i in symbols):
raise TypeError(
filldedent('''
inequalities cannot contain symbols that are not real.
'''))
# make vanilla symbol real
recast = {
i: Dummy(i.name, extended_real=True)
for i in gens if i.is_extended_real is None
}
inequalities = [i.xreplace(recast) for i in inequalities]
symbols = {i.xreplace(recast) for i in symbols}
# prefilter
keep = []
for i in inequalities:
if isinstance(i, Relational):
i = i.func(i.lhs.as_expr() - i.rhs.as_expr(), 0)
elif i not in (True, False):
i = Eq(i, 0)
if i == True:
continue
elif i == False:
return S.false
if i.lhs.is_number:
raise NotImplementedError("could not determine truth value of %s" %
i)
keep.append(i)
inequalities = keep
del keep
# solve system
rv = _reduce_inequalities(inequalities, symbols)
# restore original symbols and return
return rv.xreplace({v: k for k, v in recast.items()})
def test_issue_13060():
A, B = symbols("A B", cls=Function)
t = Symbol("t")
eq = [Eq(Derivative(A(t), t), A(t)*B(t)), Eq(Derivative(B(t), t), A(t)*B(t))]
sol = dsolve(eq)
assert checkodesol(eq, sol) == (True, [0, 0])
def test_classify_ode():
assert classify_ode(f(x).diff(x, 2), f(x)) == \
(
'nth_algebraic',
'nth_linear_constant_coeff_homogeneous',
'nth_linear_euler_eq_homogeneous',
'Liouville',
'2nd_power_series_ordinary',
'nth_algebraic_Integral',
'Liouville_Integral',
)
assert classify_ode(f(x), f(x)) == ('nth_algebraic', 'nth_algebraic_Integral')
assert classify_ode(Eq(f(x).diff(x), 0), f(x)) == (
'nth_algebraic',
'separable',
'1st_exact',
'1st_linear',
'Bernoulli',
'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',
'nth_linear_euler_eq_homogeneous',
'nth_algebraic_Integral',
'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')
assert classify_ode(f(x).diff(x)**2, f(x)) == ('factorable',
'nth_algebraic',
'separable',
'1st_exact',
'1st_linear',
'Bernoulli',
'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_euler_eq_homogeneous',
'nth_algebraic_Integral',
'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')
# issue 4749: f(x) should be cleared from highest derivative before classifying
a = classify_ode(Eq(f(x).diff(x) + f(x), x), f(x))
b = classify_ode(f(x).diff(x)*f(x) + f(x)*f(x) - x*f(x), f(x))
c = classify_ode(f(x).diff(x)/f(x) + f(x)/f(x) - x/f(x), f(x))
assert a == ('1st_exact',
'1st_linear',
'Bernoulli',
'almost_linear',
'1st_power_series', "lie_group",
'nth_linear_constant_coeff_undetermined_coefficients',
'nth_linear_constant_coeff_variation_of_parameters',
'1st_exact_Integral',
'1st_linear_Integral',
'Bernoulli_Integral',
'almost_linear_Integral',
'nth_linear_constant_coeff_variation_of_parameters_Integral')
assert b == ('factorable',
'1st_linear',
'Bernoulli',
'1st_power_series',
'lie_group',
'nth_linear_constant_coeff_undetermined_coefficients',
'nth_linear_constant_coeff_variation_of_parameters',
'1st_linear_Integral',
'Bernoulli_Integral',
'nth_linear_constant_coeff_variation_of_parameters_Integral')
assert c == ('factorable',
'1st_linear',
'Bernoulli',
'1st_power_series',
'lie_group',
'nth_linear_constant_coeff_undetermined_coefficients',
'nth_linear_constant_coeff_variation_of_parameters',
'1st_linear_Integral',
'Bernoulli_Integral',
'nth_linear_constant_coeff_variation_of_parameters_Integral')
assert classify_ode(
2*x*f(x)*f(x).diff(x) + (1 + x)*f(x)**2 - exp(x), f(x)
) == ('factorable', '1st_exact', 'Bernoulli', 'almost_linear', 'lie_group',
'1st_exact_Integral', 'Bernoulli_Integral', 'almost_linear_Integral')
assert 'Riccati_special_minus2' in \
classify_ode(2*f(x).diff(x) + f(x)**2 - f(x)/x + 3*x**(-2), f(x))
raises(ValueError, lambda: classify_ode(x + f(x, y).diff(x).diff(
y), f(x, y)))
# issue 5176
k = Symbol('k')
assert classify_ode(f(x).diff(x)/(k*f(x) + k*x*f(x)) + 2*f(x)/(k*f(x) +
k*x*f(x)) + x*f(x).diff(x)/(k*f(x) + k*x*f(x)) + z, f(x)) == \
('factorable', 'separable', '1st_exact', '1st_linear', 'Bernoulli',
'1st_power_series', 'lie_group', 'separable_Integral', '1st_exact_Integral',
'1st_linear_Integral', 'Bernoulli_Integral')
# preprocessing
ans = ('factorable', 'nth_algebraic', 'separable', '1st_exact', '1st_linear', 'Bernoulli',
'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_undetermined_coefficients',
'nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients',
'nth_linear_constant_coeff_variation_of_parameters',
'nth_linear_euler_eq_nonhomogeneous_variation_of_parameters',
'nth_algebraic_Integral',
'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',
'nth_linear_constant_coeff_variation_of_parameters_Integral',
'nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral')
# w/o f(x) given
assert classify_ode(diff(f(x) + x, x) + diff(f(x), x)) == ans
# w/ f(x) and prep=True
assert classify_ode(diff(f(x) + x, x) + diff(f(x), x), f(x),
prep=True) == ans
assert classify_ode(Eq(2*x**3*f(x).diff(x), 0), f(x)) == \
('factorable', 'nth_algebraic', 'separable', '1st_exact',
'1st_linear', 'Bernoulli', '1st_power_series',
'lie_group', 'nth_linear_euler_eq_homogeneous',
'nth_algebraic_Integral', 'separable_Integral', '1st_exact_Integral',
'1st_linear_Integral', 'Bernoulli_Integral')
assert classify_ode(Eq(2*f(x)**3*f(x).diff(x), 0), f(x)) == \
('factorable', 'nth_algebraic', 'separable', '1st_exact', '1st_linear',
'Bernoulli', '1st_power_series', 'lie_group', 'nth_algebraic_Integral',
'separable_Integral', '1st_exact_Integral', '1st_linear_Integral',
'Bernoulli_Integral')
# test issue 13864
assert classify_ode(Eq(diff(f(x), x) - f(x)**x, 0), f(x)) == \
('1st_power_series', 'lie_group')
assert isinstance(classify_ode(Eq(f(x), 5), f(x), dict=True), dict)
#This is for new behavior of classify_ode when called internally with default, It should
# return the first hint which matches therefore, 'ordered_hints' key will not be there.
assert sorted(classify_ode(Eq(f(x).diff(x), 0), f(x), dict=True).keys()) == \
['default', 'nth_linear_constant_coeff_homogeneous', 'order']
a = classify_ode(2*x*f(x)*f(x).diff(x) + (1 + x)*f(x)**2 - exp(x), f(x), dict=True, hint='Bernoulli')
assert sorted(a.keys()) == ['Bernoulli', 'Bernoulli_Integral', 'default', 'order', 'ordered_hints']
# test issue 22155
a = classify_ode(f(x).diff(x) - exp(f(x) - x), f(x))
assert a == ('separable',
'1st_exact', '1st_power_series',
'lie_group', 'separable_Integral',
'1st_exact_Integral')
def pde_1st_linear_constant_coeff(eq, func, order, match, solvefun):
r"""
Solves a first order linear partial differential equation
with constant coefficients.
The general form of this partial differential equation is
.. math:: a \frac{\partial f(x,y)}{\partial x}
+ b \frac{\partial f(x,y)}{\partial y}
+ c f(x,y) = G(x,y)
where `a`, `b` and `c` are constants and `G(x, y)` can be an arbitrary
function in `x` and `y`.
The general solution of the PDE is:
.. math::
f(x, y) = \left. \left[F(\eta) + \frac{1}{a^2 + b^2}
\int\limits^{a x + b y} G\left(\frac{a \xi + b \eta}{a^2 + b^2},
\frac{- a \eta + b \xi}{a^2 + b^2} \right)
e^{\frac{c \xi}{a^2 + b^2}}\, d\xi\right]
e^{- \frac{c \xi}{a^2 + b^2}}
\right|_{\substack{\eta=- a y + b x\\ \xi=a x + b y }}\, ,
where `F(\eta)` is an arbitrary single-valued function. The solution
can be found in SymPy with ``pdsolve``::
>>> from sympy.solvers import pdsolve
>>> from sympy.abc import x, y, a, b, c
>>> from sympy import Function, pprint
>>> f = Function('f')
>>> G = Function('G')
>>> u = f(x,y)
>>> ux = u.diff(x)
>>> uy = u.diff(y)
>>> genform = a*ux + b*uy + c*u - G(x,y)
>>> pprint(genform)
d d
a*--(f(x, y)) + b*--(f(x, y)) + c*f(x, y) - G(x, y)
dx dy
>>> pprint(pdsolve(genform, hint='1st_linear_constant_coeff_Integral'))
// a*x + b*y \
|| / |
|| | |
|| | c*xi |
|| | ------- |
|| | 2 2 |
|| | /a*xi + b*eta -a*eta + b*xi\ a + b |
|| | G|------------, -------------|*e d(xi)|
|| | | 2 2 2 2 | |
|| | \ a + b a + b / |
|| | |
|| / |
|| |
f(x, y) = ||F(eta) + -------------------------------------------------------|*
|| 2 2 |
\\ a + b /
<BLANKLINE>
\|
||
||
||
||
||
||
||
||
-c*xi ||
-------||
2 2||
a + b ||
e ||
||
/|eta=-a*y + b*x, xi=a*x + b*y
Examples
========
>>> from sympy.solvers.pde import pdsolve
>>> from sympy import Function, diff, pprint, exp
>>> from sympy.abc import x,y
>>> f = Function('f')
>>> eq = -2*f(x,y).diff(x) + 4*f(x,y).diff(y) + 5*f(x,y) - exp(x + 3*y)
>>> pdsolve(eq)
Eq(f(x, y), (F(4*x + 2*y) + exp(x/2 + 4*y)/15)*exp(x/2 - y))
References
==========
- Viktor Grigoryan, "Partial Differential Equations"
Math 124A - Fall 2010, pp.7
"""
# TODO : For now homogeneous first order linear PDE's having
# two variables are implemented. Once there is support for
# solving systems of ODE's, this can be extended to n variables.
xi, eta = symbols("xi eta")
f = func.func
x = func.args[0]
y = func.args[1]
b = match[match['b']]
c = match[match['c']]
d = match[match['d']]
e = -match[match['e']]
expterm = exp(-S(d) / (b**2 + c**2) * xi)
functerm = solvefun(eta)
solvedict = solve((b * x + c * y - xi, c * x - b * y - eta), x, y)
# Integral should remain as it is in terms of xi,
# doit() should be done in _handle_Integral.
genterm = (1 / S(b**2 + c**2)) * Integral(
(1 / expterm * e).subs(solvedict), (xi, b * x + c * y))
return Eq(
f(x, y),
Subs(expterm * (functerm + genterm), (eta, xi),
(c * x - b * y, b * x + c * y)))
def test_equality_subs1():
f = Function('f')
eq = Eq(f(x)**2, x)
res = Eq(Integer(16), x)
assert eq.subs(f(x), 4) == res
def _eval_rewrite_as_gamma(self, n, **kwargs):
from sympy import gamma, Piecewise, sqrt
return 2**(n/2)*gamma(n/2 + 1) * Piecewise((1, Eq(Mod(n, 2), 0)),
(sqrt(2/pi), Eq(Mod(n, 2), 1)))
def rsolve_hypergeometric(f, x, P, Q, k, m):
"""Solves RE of hypergeometric type.
Attempts to solve RE of the form
Q(k)*a(k + m) - P(k)*a(k)
Transformations that preserve Hypergeometric type:
a. x**n*f(x): b(k + m) = R(k - n)*b(k)
b. f(A*x): b(k + m) = A**m*R(k)*b(k)
c. f(x**n): b(k + n*m) = R(k/n)*b(k)
d. f(x**(1/m)): b(k + 1) = R(k*m)*b(k)
e. f'(x): b(k + m) = ((k + m + 1)/(k + 1))*R(k + 1)*b(k)
Some of these transformations have been used to solve the RE.
Returns
=======
formula : Expr
ind : Expr
Independent terms.
order : int
Examples
========
>>> from sympy import exp, ln, S
>>> from sympy.series.formal import rsolve_hypergeometric as rh
>>> from sympy.abc import x, k
>>> rh(exp(x), x, -S.One, (k + 1), k, 1)
(Piecewise((1/factorial(k), Eq(Mod(k, 1), 0)), (0, True)), 1, 1)
>>> rh(ln(1 + x), x, k**2, k*(k + 1), k, 1)
(Piecewise(((-1)**(k - 1)*factorial(k - 1)/RisingFactorial(2, k - 1),
Eq(Mod(k, 1), 0)), (0, True)), x, 2)
References
==========
.. [1] Formal Power Series - Dominik Gruntz, Wolfram Koepf
.. [2] Power Series in Computer Algebra - Wolfram Koepf
"""
result = _rsolve_hypergeometric(f, x, P, Q, k, m)
if result is None:
return None
sol_list, ind, mp = result
sol_dict = defaultdict(lambda: S.Zero)
for res, cond in sol_list:
j, mk = cond.as_coeff_Add()
c = mk.coeff(k)
if j.is_integer is False:
res *= x**frac(j)
j = floor(j)
res = res.subs(k, (k - j) / c)
cond = Eq(k % c, j % c)
sol_dict[cond] += res # Group together formula for same conditions
sol = []
for cond, res in sol_dict.items():
sol.append((res, cond))
sol.append((S.Zero, True))
sol = Piecewise(*sol)
if mp is -oo:
s = S.Zero
elif mp.is_integer is False:
s = ceiling(mp)
else:
s = mp + 1
# save all the terms of
# form 1/x**k in ind
if s < 0:
ind += sum(sequence(sol * x**k, (k, s, -1)))
s = S.Zero
return (sol, ind, s)
def canonical_odes(eqs, funcs, t):
r"""
Function that solves for highest order derivatives in a system
Explanation
===========
This function inputs a system of ODEs and based on the system,
the dependent variables and their highest order, returns the system
in the following form:
.. math::
X'(t) = A(t) X(t) + b(t)
Here, $X(t)$ is the vector of dependent variables of lower order, $A(t)$ is
the coefficient matrix, $b(t)$ is the non-homogeneous term and $X'(t)$ is the
vector of dependent variables in their respective highest order. We use the term
canonical form to imply the system of ODEs which is of the above form.
If the system passed has a non-linear term with multiple solutions, then a list of
systems is returned in its canonical form.
Parameters
==========
eqs : List
List of the ODEs
funcs : List
List of dependent variables
t : Symbol
Independent variable
Examples
========
>>> from sympy import symbols, Function, Eq, Derivative
>>> from sympy.solvers.ode.systems import canonical_odes
>>> f, g = symbols("f g", cls=Function)
>>> x, y = symbols("x y")
>>> funcs = [f(x), g(x)]
>>> eqs = [Eq(f(x).diff(x) - 7*f(x), 12*g(x)), Eq(g(x).diff(x) + g(x), 20*f(x))]
>>> canonical_eqs = canonical_odes(eqs, funcs, x)
>>> canonical_eqs
[Eq(Derivative(f(x), x), 7*f(x) + 12*g(x)), Eq(Derivative(g(x), x), 20*f(x) - g(x))]
>>> system = [Eq(Derivative(f(x), x)**2 - 2*Derivative(f(x), x) + 1, 4), Eq(-y*f(x) + Derivative(g(x), x), 0)]
>>> canonical_system = canonical_odes(system, funcs, x)
>>> canonical_system
[[Eq(Derivative(f(x), x), -1), Eq(Derivative(g(x), x), y*f(x))], [Eq(Derivative(f(x), x), 3), Eq(Derivative(g(x), x), y*f(x))]]
Returns
=======
List
"""
from sympy.solvers.solvers import solve
order = _get_func_order(eqs, funcs)
canon_eqs = solve(eqs, *[func.diff(t, order[func]) for func in funcs], dict=True)
systems = []
for eq in canon_eqs:
system = [Eq(func.diff(t, order[func]), eq[func.diff(t, order[func])]) for func in funcs]
systems.append(system)
if len(canon_eqs) == 1:
systems = systems[0]
return systems
def _preprocess_eqs(eqs):
processed_eqs = []
for eq in eqs:
processed_eqs.append(eq if isinstance(eq, Equality) else Eq(eq, 0))
return processed_eqs
def as_relational(self, x):
from sympy.functions.elementary.integers import floor
return And(Eq(floor(x), x), -oo < x, x < oo)
def heurisch_wrapper(f,
x,
rewrite=False,
hints=None,
mappings=None,
retries=3,
degree_offset=0,
unnecessary_permutations=None,
_try_heurisch=None):
"""
A wrapper around the heurisch integration algorithm.
This method takes the result from heurisch and checks for poles in the
denominator. For each of these poles, the integral is reevaluated, and
the final integration result is given in terms of a Piecewise.
Examples
========
>>> from sympy.core import symbols
>>> from sympy.functions import cos
>>> from sympy.integrals.heurisch import heurisch, heurisch_wrapper
>>> n, x = symbols('n x')
>>> heurisch(cos(n*x), x)
sin(n*x)/n
>>> heurisch_wrapper(cos(n*x), x)
Piecewise((sin(n*x)/n, Ne(n, 0)), (x, True))
See Also
========
heurisch
"""
from sympy.solvers.solvers import solve, denoms
f = sympify(f)
if x not in f.free_symbols:
return f * x
res = heurisch(f, x, rewrite, hints, mappings, retries, degree_offset,
unnecessary_permutations, _try_heurisch)
if not isinstance(res, Basic):
return res
# We consider each denominator in the expression, and try to find
# cases where one or more symbolic denominator might be zero. The
# conditions for these cases are stored in the list slns.
slns = []
for d in denoms(res):
try:
slns += solve(d, dict=True, exclude=(x, ))
except NotImplementedError:
pass
if not slns:
return res
slns = list(uniq(slns))
# Remove the solutions corresponding to poles in the original expression.
slns0 = []
for d in denoms(f):
try:
slns0 += solve(d, dict=True, exclude=(x, ))
except NotImplementedError:
pass
slns = [s for s in slns if s not in slns0]
if not slns:
return res
if len(slns) > 1:
eqs = []
for sub_dict in slns:
eqs.extend([Eq(key, value) for key, value in sub_dict.items()])
slns = solve(eqs, dict=True, exclude=(x, )) + slns
# For each case listed in the list slns, we reevaluate the integral.
pairs = []
for sub_dict in slns:
expr = heurisch(f.subs(sub_dict), x, rewrite, hints, mappings, retries,
degree_offset, unnecessary_permutations, _try_heurisch)
cond = And(*[Eq(key, value) for key, value in sub_dict.items()])
generic = Or(*[Ne(key, value) for key, value in sub_dict.items()])
if expr is None:
expr = integrate(f.subs(sub_dict), x)
pairs.append((expr, cond))
# If there is one condition, put the generic case first. Otherwise,
# doing so may lead to longer Piecewise formulas
if len(pairs) == 1:
pairs = [(heurisch(f, x, rewrite, hints, mappings, retries,
degree_offset, unnecessary_permutations,
_try_heurisch), generic), (pairs[0][0], True)]
else:
pairs.append(
(heurisch(f, x, rewrite, hints, mappings, retries, degree_offset,
unnecessary_permutations, _try_heurisch), True))
return Piecewise(*pairs)
def test_equality_subs2():
f = Function('f')
eq = Eq(f(x)**2, 16)
assert bool(eq.subs(f(x), 3)) is False
assert bool(eq.subs(f(x), 4)) is True
def test_diophantine():
assert check_solutions((x - y) * (y - z) * (z - x))
assert check_solutions((x - y) * (x**2 + y**2 - z**2))
assert check_solutions((x - 3 * y + 7 * z) * (x**2 + y**2 - z**2))
assert check_solutions(x**2 - 3 * y**2 - 1)
assert check_solutions(y**2 + 7 * x * y)
assert check_solutions(x**2 - 3 * x * y + y**2)
assert check_solutions(z * (x**2 - y**2 - 15))
assert check_solutions(x * (2 * y - 2 * z + 5))
assert check_solutions((x**2 - 3 * y**2 - 1) * (x**2 - y**2 - 15))
assert check_solutions((x**2 - 3 * y**2 - 1) * (y - 7 * z))
assert check_solutions((x**2 + y**2 - z**2) * (x - 7 * y - 3 * z + 4 * w))
# Following test case caused problems in parametric representation
# But this can be solved by factoring out y.
# No need to use methods for ternary quadratic equations.
assert check_solutions(y**2 - 7 * x * y + 4 * y * z)
assert check_solutions(x**2 - 2 * x + 1)
assert diophantine(x - y) == diophantine(Eq(x, y))
# 18196
eq = x**4 + y**4 - 97
assert diophantine(eq, permute=True) == diophantine(-eq, permute=True)
assert diophantine(3 * x * pi - 2 * y * pi) == {(2 * t_0, 3 * t_0)}
eq = x**2 + y**2 + z**2 - 14
base_sol = {(1, 2, 3)}
assert diophantine(eq) == base_sol
complete_soln = set(signed_permutations(base_sol.pop()))
assert diophantine(eq, permute=True) == complete_soln
assert diophantine(x**2 + x * Rational(15, 14) - 3) == set()
# test issue 11049
eq = 92 * x**2 - 99 * y**2 - z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(9, 7, 51)}
assert diophantine(eq) == {
(891 * p**2 + 9 * q**2, -693 * p**2 - 102 * p * q + 7 * q**2,
5049 * p**2 - 1386 * p * q - 51 * q**2)
}
eq = 2 * x**2 + 2 * y**2 - z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(1, 1, 2)}
assert diophantine(eq) == {(2 * p**2 - q**2, -2 * p**2 + 4 * p * q - q**2,
4 * p**2 - 4 * p * q + 2 * q**2)}
eq = 411 * x**2 + 57 * y**2 - 221 * z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(2021, 2645, 3066)}
assert diophantine(eq) == \
{(115197*p**2 - 446641*q**2, -150765*p**2 + 1355172*p*q -
584545*q**2, 174762*p**2 - 301530*p*q + 677586*q**2)}
eq = 573 * x**2 + 267 * y**2 - 984 * z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(49, 233, 127)}
assert diophantine(eq) == \
{(4361*p**2 - 16072*q**2, -20737*p**2 + 83312*p*q - 76424*q**2,
11303*p**2 - 41474*p*q + 41656*q**2)}
# this produces factors during reconstruction
eq = x**2 + 3 * y**2 - 12 * z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(0, 2, 1)}
assert diophantine(eq) == \
{(24*p*q, 2*p**2 - 24*q**2, p**2 + 12*q**2)}
# solvers have not been written for every type
raises(NotImplementedError, lambda: diophantine(x * y**2 + 1))
# rational expressions
assert diophantine(1 / x) == set()
assert diophantine(1 / x + 1 / y - S.Half) == {(6, 3), (-2, 1), (4, 4),
(1, -2), (3, 6)}
assert diophantine(x**2 + y**2 +3*x- 5, permute=True) == \
{(-1, 1), (-4, -1), (1, -1), (1, 1), (-4, 1), (-1, -1), (4, 1), (4, -1)}
#test issue 18186
assert diophantine(y**4 + x**4 - 2**4 - 3**4, syms=(x, y), permute=True) == \
{(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)}
assert diophantine(y**4 + x**4 - 2**4 - 3**4, syms=(y, x), permute=True) == \
{(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)}
# issue 18122
assert check_solutions(x**2 - y)
assert check_solutions(y**2 - x)
assert diophantine((x**2 - y), t) == {(t, t**2)}
assert diophantine((y**2 - x), t) == {(t**2, -t)}
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 = set().union(funcs)
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 test_issue_22462():
for de in [
Eq(f(x).diff(x), -20*f(x)**2 - 500*f(x)/7200),
Eq(f(x).diff(x), -2*f(x)**2 - 5*f(x)/7)]:
assert 'Bernoulli' in classify_ode(de, f(x))
def pde_1st_linear_constant_coeff_homogeneous(eq, func, order, match,
solvefun):
r"""
Solves a first order linear homogeneous
partial differential equation with constant coefficients.
The general form of this partial differential equation is
.. math:: a \frac{df(x,y)}{dx} + b \frac{df(x,y)}{dy} + c f(x,y) = 0
where `a`, `b` and `c` are constants.
The general solution is of the form::
>>> from sympy.solvers import pdsolve
>>> from sympy.abc import x, y, a, b, c
>>> from sympy import Function, pprint
>>> f = Function('f')
>>> u = f(x,y)
>>> ux = u.diff(x)
>>> uy = u.diff(y)
>>> genform = a*ux + b*uy + c*u
>>> pprint(genform)
d d
a*--(f(x, y)) + b*--(f(x, y)) + c*f(x, y)
dx dy
>>> pprint(pdsolve(genform))
-c*(a*x + b*y)
---------------
2 2
a + b
f(x, y) = F(-a*y + b*x)*e
Examples
========
>>> from sympy.solvers.pde import (
... pde_1st_linear_constant_coeff_homogeneous)
>>> from sympy import pdsolve
>>> from sympy import Function, diff, pprint
>>> from sympy.abc import x,y
>>> f = Function('f')
>>> pdsolve(f(x,y) + f(x,y).diff(x) + f(x,y).diff(y))
f(x, y) == F(x - y)*exp(-x/2 - y/2)
>>> pprint(pdsolve(f(x,y) + f(x,y).diff(x) + f(x,y).diff(y)))
x y
- - - -
2 2
f(x, y) = F(x - y)*e
References
==========
- Viktor Grigoryan, "Partial Differential Equations"
Math 124A - Fall 2010, pp.7
"""
# TODO : For now homogeneous first order linear PDE's having
# two variables are implemented. Once there is support for
# solving systems of ODE's, this can be extended to n variables.
f = func.func
x = func.args[0]
y = func.args[1]
b = match[match['b']]
c = match[match['c']]
d = match[match['d']]
return Eq(
f(x, y),
exp(-S(d) / (b**2 + c**2) * (b * x + c * y)) * solvefun(c * x - b * y))
def test_solve_ics():
# Basic tests that things work from dsolve.
assert dsolve(f(x).diff(x) - 1/f(x), f(x), ics={f(1): 2}) == \
Eq(f(x), sqrt(2 * x + 2))
assert dsolve(f(x).diff(x) - f(x), f(x), ics={f(0): 1}) == Eq(f(x), exp(x))
assert dsolve(f(x).diff(x) - f(x), f(x), ics={f(x).diff(x).subs(x, 0): 1}) == Eq(f(x), exp(x))
assert dsolve(f(x).diff(x, x) + f(x), f(x), ics={f(0): 1,
f(x).diff(x).subs(x, 0): 1}) == Eq(f(x), sin(x) + cos(x))
assert dsolve([f(x).diff(x) - f(x) + g(x), g(x).diff(x) - g(x) - f(x)],
[f(x), g(x)], ics={f(0): 1, g(0): 0}) == [Eq(f(x), exp(x)*cos(x)), Eq(g(x), exp(x)*sin(x))]
# Test cases where dsolve returns two solutions.
eq = (x**2*f(x)**2 - x).diff(x)
assert dsolve(eq, f(x), ics={f(1): 0}) == [Eq(f(x),
-sqrt(x - 1)/x), Eq(f(x), sqrt(x - 1)/x)]
assert dsolve(eq, f(x), ics={f(x).diff(x).subs(x, 1): 0}) == [Eq(f(x),
-sqrt(x - S.Half)/x), Eq(f(x), sqrt(x - S.Half)/x)]
eq = cos(f(x)) - (x*sin(f(x)) - f(x)**2)*f(x).diff(x)
assert dsolve(eq, f(x),
ics={f(0):1}, hint='1st_exact', simplify=False) == Eq(x*cos(f(x)) + f(x)**3/3, Rational(1, 3))
assert dsolve(eq, f(x),
ics={f(0):1}, hint='1st_exact', simplify=True) == Eq(x*cos(f(x)) + f(x)**3/3, Rational(1, 3))
assert solve_ics([Eq(f(x), C1*exp(x))], [f(x)], [C1], {f(0): 1}) == {C1: 1}
assert solve_ics([Eq(f(x), C1*sin(x) + C2*cos(x))], [f(x)], [C1, C2],
{f(0): 1, f(pi/2): 1}) == {C1: 1, C2: 1}
assert solve_ics([Eq(f(x), C1*sin(x) + C2*cos(x))], [f(x)], [C1, C2],
{f(0): 1, f(x).diff(x).subs(x, 0): 1}) == {C1: 1, C2: 1}
assert solve_ics([Eq(f(x), C1*sin(x) + C2*cos(x))], [f(x)], [C1, C2], {f(0): 1}) == \
{C2: 1}
# Some more complicated tests Refer to PR #16098
assert set(dsolve(f(x).diff(x)*(f(x).diff(x, 2)-x), ics={f(0):0, f(x).diff(x).subs(x, 1):0})) == \
{Eq(f(x), 0), Eq(f(x), x ** 3 / 6 - x / 2)}
assert set(dsolve(f(x).diff(x)*(f(x).diff(x, 2)-x), ics={f(0):0})) == \
{Eq(f(x), 0), Eq(f(x), C2*x + x**3/6)}
K, r, f0 = symbols('K r f0')
sol = Eq(f(x), K*f0*exp(r*x)/((-K + f0)*(f0*exp(r*x)/(-K + f0) - 1)))
assert (dsolve(Eq(f(x).diff(x), r * f(x) * (1 - f(x) / K)), f(x), ics={f(0): f0})) == sol
#Order dependent issues Refer to PR #16098
assert set(dsolve(f(x).diff(x)*(f(x).diff(x, 2)-x), ics={f(x).diff(x).subs(x,0):0, f(0):0})) == \
{Eq(f(x), 0), Eq(f(x), x ** 3 / 6)}
assert set(dsolve(f(x).diff(x)*(f(x).diff(x, 2)-x), ics={f(0):0, f(x).diff(x).subs(x,0):0})) == \
{Eq(f(x), 0), Eq(f(x), x ** 3 / 6)}
# XXX: Ought to be ValueError
raises(ValueError, lambda: solve_ics([Eq(f(x), C1*sin(x) + C2*cos(x))], [f(x)], [C1, C2], {f(0): 1, f(pi): 1}))
# Degenerate case. f'(0) is identically 0.
raises(ValueError, lambda: solve_ics([Eq(f(x), sqrt(C1 - x**2))], [f(x)], [C1], {f(x).diff(x).subs(x, 0): 0}))
EI, q, L = symbols('EI q L')
# eq = Eq(EI*diff(f(x), x, 4), q)
sols = [Eq(f(x), C1 + C2*x + C3*x**2 + C4*x**3 + q*x**4/(24*EI))]
funcs = [f(x)]
constants = [C1, C2, C3, C4]
# Test both cases, Derivative (the default from f(x).diff(x).subs(x, L)),
# and Subs
ics1 = {f(0): 0,
f(x).diff(x).subs(x, 0): 0,
f(L).diff(L, 2): 0,
f(L).diff(L, 3): 0}
ics2 = {f(0): 0,
f(x).diff(x).subs(x, 0): 0,
Subs(f(x).diff(x, 2), x, L): 0,
Subs(f(x).diff(x, 3), x, L): 0}
solved_constants1 = solve_ics(sols, funcs, constants, ics1)
solved_constants2 = solve_ics(sols, funcs, constants, ics2)
assert solved_constants1 == solved_constants2 == {
C1: 0,
C2: 0,
C3: L**2*q/(4*EI),
C4: -L*q/(6*EI)}
def pde_1st_linear_constant_coeff(eq, func, order, match, solvefun):
r"""
Solves a first order linear partial differential equation
with constant coefficients.
The general form of this partial differential equation is
.. math:: a \frac{df(x,y)}{dx} + b \frac{df(x,y)}{dy} + c f(x,y) = G(x,y)
where `a`, `b` and `c` are constants and `G(x, y)` can be an arbitrary
function in `x` and `y`.
The general solution of the PDE is::
>>> from sympy.solvers import pdsolve
>>> from sympy.abc import x, y, a, b, c
>>> from sympy import Function, pprint
>>> f = Function('f')
>>> G = Function('G')
>>> u = f(x,y)
>>> ux = u.diff(x)
>>> uy = u.diff(y)
>>> genform = a*u + b*ux + c*uy - G(x,y)
>>> pprint(genform)
d d
a*f(x, y) + b*--(f(x, y)) + c*--(f(x, y)) - G(x, y)
dx dy
>>> pprint(pdsolve(genform, hint='1st_linear_constant_coeff_Integral'))
// b*x + c*y \
|| / |
|| | |
|| | a*xi |
|| | ------- |
|| | 2 2 |
|| | /b*xi + c*eta -b*eta + c*xi\ b + c |
|| | G|------------, -------------|*e d(xi)|
|| | | 2 2 2 2 | |
|| | \ b + c b + c / |
|| | |
|| / |
|| |
f(x, y) = ||F(eta) + -------------------------------------------------------|*
|| 2 2 |
\\ b + c /
<BLANKLINE>
\|
||
||
||
||
||
||
||
||
-a*xi ||
-------||
2 2||
b + c ||
e ||
||
/|eta=-b*y + c*x, xi=b*x + c*y
Examples
========
>>> from sympy.solvers.pde import pdsolve
>>> from sympy import Function, diff, pprint, exp
>>> from sympy.abc import x,y
>>> f = Function('f')
>>> eq = -2*f(x,y).diff(x) + 4*f(x,y).diff(y) + 5*f(x,y) - exp(x + 3*y)
>>> pdsolve(eq)
f(x, y) == (F(4*x + 2*y) + exp(x/2 + 4*y)/15)*exp(x/2 - y)
References
==========
- Viktor Grigoryan, "Partial Differential Equations"
Math 124A - Fall 2010, pp.7
"""
# TODO : For now homogeneous first order linear PDE's having
# two variables are implemented. Once there is support for
# solving systems of ODE's, this can be extended to n variables.
xi, eta = symbols("xi eta")
f = func.func
x = func.args[0]
y = func.args[1]
b = match[match['b']]
c = match[match['c']]
d = match[match['d']]
e = -match[match['e']]
expterm = exp(-S(d) / (b**2 + c**2) * xi)
functerm = solvefun(eta)
solvedict = solve((b * x + c * y - xi, c * x - b * y - eta), x, y)
# Integral should remain as it is in terms of xi,
# doit() should be done in _handle_Integral.
genterm = (1 / S(b**2 + c**2)) * C.Integral(
(1 / expterm * e).subs(solvedict), (xi, b * x + c * y))
return Eq(
f(x, y),
Subs(expterm * (functerm + genterm), (eta, xi),
(c * x - b * y, b * x + c * y)))
def test_dsolve_options():
eq = x*f(x).diff(x) + f(x)
a = dsolve(eq, hint='all')
b = dsolve(eq, hint='all', simplify=False)
c = dsolve(eq, hint='all_Integral')
keys = ['1st_exact', '1st_exact_Integral', '1st_homogeneous_coeff_best',
'1st_homogeneous_coeff_subs_dep_div_indep',
'1st_homogeneous_coeff_subs_dep_div_indep_Integral',
'1st_homogeneous_coeff_subs_indep_div_dep',
'1st_homogeneous_coeff_subs_indep_div_dep_Integral', '1st_linear',
'1st_linear_Integral', 'Bernoulli', 'Bernoulli_Integral',
'almost_linear', 'almost_linear_Integral', 'best', 'best_hint',
'default', 'factorable', 'lie_group',
'nth_linear_euler_eq_homogeneous', 'order',
'separable', 'separable_Integral']
Integral_keys = ['1st_exact_Integral',
'1st_homogeneous_coeff_subs_dep_div_indep_Integral',
'1st_homogeneous_coeff_subs_indep_div_dep_Integral', '1st_linear_Integral',
'Bernoulli_Integral', 'almost_linear_Integral', 'best', 'best_hint', 'default',
'factorable', 'nth_linear_euler_eq_homogeneous',
'order', 'separable_Integral']
assert sorted(a.keys()) == keys
assert a['order'] == ode_order(eq, f(x))
assert a['best'] == Eq(f(x), C1/x)
assert dsolve(eq, hint='best') == Eq(f(x), C1/x)
assert a['default'] == 'factorable'
assert a['best_hint'] == 'factorable'
assert not a['1st_exact'].has(Integral)
assert not a['separable'].has(Integral)
assert not a['1st_homogeneous_coeff_best'].has(Integral)
assert not a['1st_homogeneous_coeff_subs_dep_div_indep'].has(Integral)
assert not a['1st_homogeneous_coeff_subs_indep_div_dep'].has(Integral)
assert not a['1st_linear'].has(Integral)
assert a['1st_linear_Integral'].has(Integral)
assert a['1st_exact_Integral'].has(Integral)
assert a['1st_homogeneous_coeff_subs_dep_div_indep_Integral'].has(Integral)
assert a['1st_homogeneous_coeff_subs_indep_div_dep_Integral'].has(Integral)
assert a['separable_Integral'].has(Integral)
assert sorted(b.keys()) == keys
assert b['order'] == ode_order(eq, f(x))
assert b['best'] == Eq(f(x), C1/x)
assert dsolve(eq, hint='best', simplify=False) == Eq(f(x), C1/x)
assert b['default'] == 'factorable'
assert b['best_hint'] == 'factorable'
assert a['separable'] != b['separable']
assert a['1st_homogeneous_coeff_subs_dep_div_indep'] != \
b['1st_homogeneous_coeff_subs_dep_div_indep']
assert a['1st_homogeneous_coeff_subs_indep_div_dep'] != \
b['1st_homogeneous_coeff_subs_indep_div_dep']
assert not b['1st_exact'].has(Integral)
assert not b['separable'].has(Integral)
assert not b['1st_homogeneous_coeff_best'].has(Integral)
assert not b['1st_homogeneous_coeff_subs_dep_div_indep'].has(Integral)
assert not b['1st_homogeneous_coeff_subs_indep_div_dep'].has(Integral)
assert not b['1st_linear'].has(Integral)
assert b['1st_linear_Integral'].has(Integral)
assert b['1st_exact_Integral'].has(Integral)
assert b['1st_homogeneous_coeff_subs_dep_div_indep_Integral'].has(Integral)
assert b['1st_homogeneous_coeff_subs_indep_div_dep_Integral'].has(Integral)
assert b['separable_Integral'].has(Integral)
assert sorted(c.keys()) == Integral_keys
raises(ValueError, lambda: dsolve(eq, hint='notarealhint'))
raises(ValueError, lambda: dsolve(eq, hint='Liouville'))
assert dsolve(f(x).diff(x) - 1/f(x)**2, hint='all')['best'] == \
dsolve(f(x).diff(x) - 1/f(x)**2, hint='best')
assert dsolve(f(x) + f(x).diff(x) + sin(x).diff(x) + 1, f(x),
hint="1st_linear_Integral") == \
Eq(f(x), (C1 + Integral((-sin(x).diff(x) - 1)*
exp(Integral(1, x)), x))*exp(-Integral(1, x)))
def pde_1st_linear_variable_coeff(eq, func, order, match, solvefun):
r"""
Solves a first order linear partial differential equation
with variable coefficients. The general form of this partial differential equation is
.. math:: a(x, y) \frac{df(x, y)}{dx} + a(x, y) \frac{df(x, y)}{dy}
+ c(x, y) f(x, y) - G(x, y)
where `a(x, y)`, `b(x, y)`, `c(x, y)` and `G(x, y)` are arbitrary functions
in `x` and `y`. This PDE is converted into an ODE by making the following transformation.
1] `\xi` as `x`
2] `\eta` as the constant in the solution to the differential equation
`\frac{dy}{dx} = -\frac{b}{a}`
Making the following substitutions reduces it to the linear ODE
.. math:: a(\xi, \eta)\frac{du}{d\xi} + c(\xi, \eta)u - d(\xi, \eta) = 0
which can be solved using dsolve.
The general form of this PDE is::
>>> from sympy.solvers.pde import pdsolve
>>> from sympy.abc import x, y
>>> from sympy import Function, pprint
>>> a, b, c, G, f= [Function(i) for i in ['a', 'b', 'c', 'G', 'f']]
>>> u = f(x,y)
>>> ux = u.diff(x)
>>> uy = u.diff(y)
>>> genform = a(x, y)*u + b(x, y)*ux + c(x, y)*uy - G(x,y)
>>> pprint(genform)
d d
-G(x, y) + a(x, y)*f(x, y) + b(x, y)*--(f(x, y)) + c(x, y)*--(f(x, y))
dx dy
Examples
========
>>> from sympy.solvers.pde import pdsolve
>>> from sympy import Function, diff, pprint, exp
>>> from sympy.abc import x,y
>>> f = Function('f')
>>> eq = x*(u.diff(x)) - y*(u.diff(y)) + y**2*u - y**2
>>> pdsolve(eq)
f(x, y) == F(x*y)*exp(y**2/2) + 1
References
==========
- Viktor Grigoryan, "Partial Differential Equations"
Math 124A - Fall 2010, pp.7
"""
from sympy.integrals.integrals import integrate
from sympy.solvers.ode import dsolve
xi, eta = symbols("xi eta")
f = func.func
x = func.args[0]
y = func.args[1]
b = match[match['b']]
c = match[match['c']]
d = match[match['d']]
e = -match[match['e']]
if not d:
# To deal with cases like b*ux = e or c*uy = e
if not (b and c):
if c:
try:
tsol = integrate(e / c, y)
except NotImplementedError:
raise NotImplementedError("Unable to find a solution"
" due to inability of integrate")
else:
return Eq(f(x, y), solvefun(x) + tsol)
if b:
try:
tsol = integrate(e / b, x)
except NotImplementedError:
raise NotImplementedError("Unable to find a solution"
" due to inability of integrate")
else:
return Eq(f(x, y), solvefun(y) + tsol)
if not c:
# To deal with cases when c is 0, a simpler method is used.
# The PDE reduces to b*(u.diff(x)) + d*u = e, which is a linear ODE in x
plode = f(x).diff(x) * b + d * f(x) - e
sol = dsolve(plode, f(x))
syms = sol.free_symbols - plode.free_symbols - set([x, y])
rhs = _simplify_variable_coeff(sol.rhs, syms, solvefun, y)
return Eq(f(x, y), rhs)
if not b:
# To deal with cases when b is 0, a simpler method is used.
# The PDE reduces to c*(u.diff(y)) + d*u = e, which is a linear ODE in y
plode = f(y).diff(y) * c + d * f(y) - e
sol = dsolve(plode, f(y))
syms = sol.free_symbols - plode.free_symbols - set([x, y])
rhs = _simplify_variable_coeff(sol.rhs, syms, solvefun, x)
return Eq(f(x, y), rhs)
dummy = Function('d')
h = (c / b).subs(y, dummy(x))
sol = dsolve(dummy(x).diff(x) - h, dummy(x))
if isinstance(sol, list):
sol = sol[0]
solsym = sol.free_symbols - h.free_symbols - set([x, y])
if len(solsym) == 1:
solsym = solsym.pop()
etat = (solve(sol, solsym)[0]).subs(dummy(x), y)
ysub = solve(eta - etat, y)[0]
deq = (b * (f(x).diff(x)) + d * f(x) - e).subs(y, ysub)
final = (dsolve(deq, f(x), hint='1st_linear')).rhs
if isinstance(final, list):
final = final[0]
finsyms = final.free_symbols - deq.free_symbols - set([x, y])
rhs = _simplify_variable_coeff(final, finsyms, solvefun, etat)
return Eq(f(x, y), rhs)
else:
raise NotImplementedError(
"Cannot solve the partial differential equation due"
" to inability of constantsimp")
def _solve_inequality(ie, s, linear=False):
"""Return the inequality with s isolated on the left, if possible.
If the relationship is non-linear, a solution involving And or Or
may be returned. False or True are returned if the relationship
is never True or always True, respectively.
If `linear` is True (default is False) an `s`-dependent expression
will be isolated on the left, if possible
but it will not be solved for `s` unless the expression is linear
in `s`. Furthermore, only "safe" operations which do not change the
sense of the relationship are applied: no division by an unsigned
value is attempted unless the relationship involves Eq or Ne and
no division by a value not known to be nonzero is ever attempted.
Examples
========
>>> from sympy import Eq, Symbol
>>> from sympy.solvers.inequalities import _solve_inequality as f
>>> from sympy.abc import x, y
For linear expressions, the symbol can be isolated:
>>> f(x - 2 < 0, x)
x < 2
>>> f(-x - 6 < x, x)
x > -3
Sometimes nonlinear relationships will be False
>>> f(x**2 + 4 < 0, x)
False
Or they may involve more than one region of values:
>>> f(x**2 - 4 < 0, x)
(-2 < x) & (x < 2)
To restrict the solution to a relational, set linear=True
and only the x-dependent portion will be isolated on the left:
>>> f(x**2 - 4 < 0, x, linear=True)
x**2 < 4
Division of only nonzero quantities is allowed, so x cannot
be isolated by dividing by y:
>>> y.is_nonzero is None # it is unknown whether it is 0 or not
True
>>> f(x*y < 1, x)
x*y < 1
And while an equality (or inequality) still holds after dividing by a
non-zero quantity
>>> nz = Symbol('nz', nonzero=True)
>>> f(Eq(x*nz, 1), x)
Eq(x, 1/nz)
the sign must be known for other inequalities involving > or <:
>>> f(x*nz <= 1, x)
nz*x <= 1
>>> p = Symbol('p', positive=True)
>>> f(x*p <= 1, x)
x <= 1/p
When there are denominators in the original expression that
are removed by expansion, conditions for them will be returned
as part of the result:
>>> f(x < x*(2/x - 1), x)
(x < 1) & Ne(x, 0)
"""
from sympy.solvers.solvers import denoms
if s not in ie.free_symbols:
return ie
if ie.rhs == s:
ie = ie.reversed
if ie.lhs == s and s not in ie.rhs.free_symbols:
return ie
def classify(ie, s, i):
# return True or False if ie evaluates when substituting s with
# i else None (if unevaluated) or NaN (when there is an error
# in evaluating)
try:
v = ie.subs(s, i)
if v is S.NaN:
return v
elif v not in (True, False):
return
return v
except TypeError:
return S.NaN
rv = None
oo = S.Infinity
expr = ie.lhs - ie.rhs
try:
p = Poly(expr, s)
if p.degree() == 0:
rv = ie.func(p.as_expr(), 0)
elif not linear and p.degree() > 1:
# handle in except clause
raise NotImplementedError
except (PolynomialError, NotImplementedError):
if not linear:
try:
rv = reduce_rational_inequalities([[ie]], s)
except PolynomialError:
rv = solve_univariate_inequality(ie, s)
# remove restrictions wrt +/-oo that may have been
# applied when using sets to simplify the relationship
okoo = classify(ie, s, oo)
if okoo is S.true and classify(rv, s, oo) is S.false:
rv = rv.subs(s < oo, True)
oknoo = classify(ie, s, -oo)
if (oknoo is S.true and
classify(rv, s, -oo) is S.false):
rv = rv.subs(-oo < s, True)
rv = rv.subs(s > -oo, True)
if rv is S.true:
rv = (s <= oo) if okoo is S.true else (s < oo)
if oknoo is not S.true:
rv = And(-oo < s, rv)
else:
p = Poly(expr)
conds = []
if rv is None:
e = p.as_expr() # this is in expanded form
# Do a safe inversion of e, moving non-s terms
# to the rhs and dividing by a nonzero factor if
# the relational is Eq/Ne; for other relationals
# the sign must also be positive or negative
rhs = 0
b, ax = e.as_independent(s, as_Add=True)
e -= b
rhs -= b
ef = factor_terms(e)
a, e = ef.as_independent(s, as_Add=False)
if (a.is_zero != False or # don't divide by potential 0
a.is_negative ==
a.is_positive is None and # if sign is not known then
ie.rel_op not in ('!=', '==')): # reject if not Eq/Ne
e = ef
a = S.One
rhs /= a
if a.is_positive:
rv = ie.func(e, rhs)
else:
rv = ie.reversed.func(e, rhs)
# return conditions under which the value is
# valid, too.
beginning_denoms = denoms(ie.lhs) | denoms(ie.rhs)
current_denoms = denoms(rv)
for d in beginning_denoms - current_denoms:
c = _solve_inequality(Eq(d, 0), s, linear=linear)
if isinstance(c, Eq) and c.lhs == s:
if classify(rv, s, c.rhs) is S.true:
# rv is permitting this value but it shouldn't
conds.append(~c)
for i in (-oo, oo):
if (classify(rv, s, i) is S.true and
classify(ie, s, i) is not S.true):
conds.append(s < i if i is oo else i < s)
conds.append(rv)
return And(*conds)
def pde_separate(eq, fun, sep, strategy='mul'):
"""Separate variables in partial differential equation either by additive
or multiplicative separation approach. It tries to rewrite an equation so
that one of the specified variables occurs on a different side of the
equation than the others.
:param eq: Partial differential equation
:param fun: Original function F(x, y, z)
:param sep: List of separated functions [X(x), u(y, z)]
:param strategy: Separation strategy. You can choose between additive
separation ('add') and multiplicative separation ('mul') which is
default.
Examples
========
>>> from sympy import E, Eq, Function, pde_separate, Derivative as D
>>> from sympy.abc import x, t
>>> u, X, T = map(Function, 'uXT')
>>> eq = Eq(D(u(x, t), x), E**(u(x, t))*D(u(x, t), t))
>>> pde_separate(eq, u(x, t), [X(x), T(t)], strategy='add')
[exp(-X(x))*Derivative(X(x), x), exp(T(t))*Derivative(T(t), t)]
>>> eq = Eq(D(u(x, t), x, 2), D(u(x, t), t, 2))
>>> pde_separate(eq, u(x, t), [X(x), T(t)], strategy='mul')
[Derivative(X(x), x, x)/X(x), Derivative(T(t), t, t)/T(t)]
See Also
========
pde_separate_add, pde_separate_mul
"""
do_add = False
if strategy == 'add':
do_add = True
elif strategy == 'mul':
do_add = False
else:
assert ValueError('Unknown strategy: %s' % strategy)
if isinstance(eq, Equality):
if eq.rhs != 0:
return pde_separate(Eq(eq.lhs - eq.rhs), fun, sep, strategy)
if eq.rhs != 0:
raise ValueError("Value should be 0")
# Handle arguments
orig_args = list(fun.args)
subs_args = []
for s in sep:
for j in range(0, len(s.args)):
subs_args.append(s.args[j])
if do_add:
functions = reduce(operator.add, sep)
else:
functions = reduce(operator.mul, sep)
# Check whether variables match
if len(subs_args) != len(orig_args):
raise ValueError("Variable counts do not match")
# Check for duplicate arguments like [X(x), u(x, y)]
if has_dups(subs_args):
raise ValueError("Duplicate substitution arguments detected")
# Check whether the variables match
if set(orig_args) != set(subs_args):
raise ValueError("Arguments do not match")
# Substitute original function with separated...
result = eq.lhs.subs(fun, functions).doit()
# Divide by terms when doing multiplicative separation
if not do_add:
eq = 0
for i in result.args:
eq += i / functions
result = eq
svar = subs_args[0]
dvar = subs_args[1:]
return _separate(result, svar, dvar)
def _eval_rewrite_as_Abs(self, arg, **kwargs):
return Piecewise((0, Eq(arg, 0)), (arg / Abs(arg), True))
def roots_quartic(f):
r"""
Returns a list of roots of a quartic polynomial.
There are many references for solving quartic expressions available [1-5].
This reviewer has found that many of them require one to select from among
2 or more possible sets of solutions and that some solutions work when one
is searching for real roots but don't work when searching for complex roots
(though this is not always stated clearly). The following routine has been
tested and found to be correct for 0, 2 or 4 complex roots.
The quasisymmetric case solution [6] looks for quartics that have the form
`x**4 + A*x**3 + B*x**2 + C*x + D = 0` where `(C/A)**2 = D`.
Although no general solution that is always applicable for all
coefficients is known to this reviewer, certain conditions are tested
to determine the simplest 4 expressions that can be returned:
1) `f = c + a*(a**2/8 - b/2) == 0`
2) `g = d - a*(a*(3*a**2/256 - b/16) + c/4) = 0`
3) if `f != 0` and `g != 0` and `p = -d + a*c/4 - b**2/12` then
a) `p == 0`
b) `p != 0`
Examples
========
>>> from sympy import Poly, symbols, I
>>> from sympy.polys.polyroots import roots_quartic
>>> r = roots_quartic(Poly('x**4-6*x**3+17*x**2-26*x+20'))
>>> # 4 complex roots: 1+-I*sqrt(3), 2+-I
>>> sorted(str(tmp.evalf(n=2)) for tmp in r)
['1.0 + 1.7*I', '1.0 - 1.7*I', '2.0 + 1.0*I', '2.0 - 1.0*I']
References
==========
1. http://mathforum.org/dr.math/faq/faq.cubic.equations.html
2. https://en.wikipedia.org/wiki/Quartic_function#Summary_of_Ferrari.27s_method
3. http://planetmath.org/encyclopedia/GaloisTheoreticDerivationOfTheQuarticFormula.html
4. http://staff.bath.ac.uk/masjhd/JHD-CA.pdf
5. http://www.albmath.org/files/Math_5713.pdf
6. http://www.statemaster.com/encyclopedia/Quartic-equation
7. eqworld.ipmnet.ru/en/solutions/ae/ae0108.pdf
"""
_, a, b, c, d = f.monic().all_coeffs()
if not d:
return [S.Zero] + roots([1, a, b, c], multiple=True)
elif (c / a)**2 == d:
x, m = f.gen, c / a
g = Poly(x**2 + a * x + b - 2 * m, x)
z1, z2 = roots_quadratic(g)
h1 = Poly(x**2 - z1 * x + m, x)
h2 = Poly(x**2 - z2 * x + m, x)
r1 = roots_quadratic(h1)
r2 = roots_quadratic(h2)
return r1 + r2
else:
a2 = a**2
e = b - 3 * a2 / 8
f = _mexpand(c + a * (a2 / 8 - b / 2))
g = _mexpand(d - a * (a * (3 * a2 / 256 - b / 16) + c / 4))
aon4 = a / 4
if f is S.Zero:
y1, y2 = [sqrt(tmp) for tmp in roots([1, e, g], multiple=True)]
return [tmp - aon4 for tmp in [-y1, -y2, y1, y2]]
if g is S.Zero:
y = [S.Zero] + roots([1, 0, e, f], multiple=True)
return [tmp - aon4 for tmp in y]
else:
# Descartes-Euler method, see [7]
sols = _roots_quartic_euler(e, f, g, aon4)
if sols:
return sols
# Ferrari method, see [1, 2]
a2 = a**2
e = b - 3 * a2 / 8
f = c + a * (a2 / 8 - b / 2)
g = d - a * (a * (3 * a2 / 256 - b / 16) + c / 4)
p = -e**2 / 12 - g
q = -e**3 / 108 + e * g / 3 - f**2 / 8
TH = Rational(1, 3)
def _ans(y):
w = sqrt(e + 2 * y)
arg1 = 3 * e + 2 * y
arg2 = 2 * f / w
ans = []
for s in [-1, 1]:
root = sqrt(-(arg1 + s * arg2))
for t in [-1, 1]:
ans.append((s * w - t * root) / 2 - aon4)
return ans
# p == 0 case
y1 = e * Rational(-5, 6) - q**TH
if p.is_zero:
return _ans(y1)
# if p != 0 then u below is not 0
root = sqrt(q**2 / 4 + p**3 / 27)
r = -q / 2 + root # or -q/2 - root
u = r**TH # primary root of solve(x**3 - r, x)
y2 = e * Rational(-5, 6) + u - p / u / 3
if fuzzy_not(p.is_zero):
return _ans(y2)
# sort it out once they know the values of the coefficients
return [
Piecewise((a1, Eq(p, 0)), (a2, True))
for a1, a2 in zip(_ans(y1), _ans(y2))
]
def _solve_as_poly(f, symbol, domain=S.Complexes):
"""
Solve the equation using polynomial techniques if it already is a
polynomial equation or, with a change of variables, can be made so.
"""
result = None
if f.is_polynomial(symbol):
solns = roots(f,
symbol,
cubics=True,
quartics=True,
quintics=True,
domain='EX')
num_roots = sum(solns.values())
if degree(f, symbol) <= num_roots:
result = FiniteSet(*solns.keys())
else:
poly = Poly(f, symbol)
solns = poly.all_roots()
if poly.degree() <= len(solns):
result = FiniteSet(*solns)
else:
result = ConditionSet(symbol, Eq(f, 0), domain)
else:
poly = Poly(f)
if poly is None:
result = ConditionSet(symbol, Eq(f, 0), domain)
gens = [g for g in poly.gens if g.has(symbol)]
if len(gens) == 1:
poly = Poly(poly, gens[0])
gen = poly.gen
deg = poly.degree()
poly = Poly(poly.as_expr(), poly.gen, composite=True)
poly_solns = FiniteSet(
*roots(poly, cubics=True, quartics=True, quintics=True).keys())
if len(poly_solns) < deg:
result = ConditionSet(symbol, Eq(f, 0), domain)
if gen != symbol:
y = Dummy('y')
inverter = invert_real if domain.is_subset(
S.Reals) else invert_complex
lhs, rhs_s = inverter(gen, y, symbol)
if lhs == symbol:
result = Union(*[rhs_s.subs(y, s) for s in poly_solns])
else:
result = ConditionSet(symbol, Eq(f, 0), domain)
else:
result = ConditionSet(symbol, Eq(f, 0), domain)
if result is not None:
if isinstance(result, FiniteSet):
# this is to simplify solutions like -sqrt(-I) to sqrt(2)/2
# - sqrt(2)*I/2. We are not expanding for solution with free
# variables because that makes the solution more complicated. For
# example expand_complex(a) returns re(a) + I*im(a)
if all([
s.free_symbols == set() and not isinstance(s, RootOf)
for s in result
]):
s = Dummy('s')
result = imageset(Lambda(s, expand_complex(s)), result)
if isinstance(result, FiniteSet):
result = result.intersection(domain)
return result
else:
return ConditionSet(symbol, Eq(f, 0), domain)
def test_Piecewise():
f = Piecewise((-z + x*y, Eq(y, 0)), (-z - x*y, True))
ans = cse(f)
actual_ans = ([(x0, x*y)],
[Piecewise((x0 - z, Eq(y, 0)), (-z - x0, True))])
assert ans == actual_ans
