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image-handler/image-handler10.0/venv/lib/site-packages/sympy/solvers/ode/lie_group.py

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r"""
This module contains the implementation of the internal helper functions for the lie_group hint for
dsolve. These helper functions apply different heuristics on the given equation
and return the solution. These functions are used by :py:meth:`sympy.solvers.ode.single.LieGroup`
References
=========
- `abaco1_simple`, `function_sum` and `chi` are referenced from E.S Cheb-Terrab, L.G.S Duarte
and L.A,C.P da Mota, Computer Algebra Solving of First Order ODEs Using
Symmetry Methods, pp. 7 - pp. 8
- `abaco1_product`, `abaco2_similar`, `abaco2_unique_unknown`, `linear` and `abaco2_unique_general`
are referenced from E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 7 - pp. 12
- `bivariate` from Lie Groups and Differential Equations pp. 327 - pp. 329
"""
from itertools import islice
from sympy.core import Add, S, Mul, Pow
from sympy.core.exprtools import factor_terms
from sympy.core.function import Function, AppliedUndef, expand
from sympy.core.relational import Equality, Eq
from sympy.core.symbol import Symbol, Wild, Dummy, symbols
from sympy.functions import exp, log
from sympy.integrals.integrals import integrate
from sympy.polys import Poly
from sympy.polys.polytools import cancel, div
from sympy.simplify import (collect, powsimp, # type: ignore
separatevars, simplify)
from sympy.solvers import solve
from sympy.solvers.pde import pdsolve
from sympy.utilities import numbered_symbols
from sympy.solvers.deutils import _preprocess, ode_order
from .ode import checkinfsol
lie_heuristics = (
"abaco1_simple",
"abaco1_product",
"abaco2_similar",
"abaco2_unique_unknown",
"abaco2_unique_general",
"linear",
"function_sum",
"bivariate",
"chi"
)
def _ode_lie_group_try_heuristic(eq, heuristic, func, match, inf):
xi = Function("xi")
eta = Function("eta")
f = func.func
x = func.args[0]
y = match['y']
h = match['h']
tempsol = []
if not inf:
try:
inf = infinitesimals(eq, hint=heuristic, func=func, order=1, match=match)
except ValueError:
return None
for infsim in inf:
xiinf = (infsim[xi(x, func)]).subs(func, y)
etainf = (infsim[eta(x, func)]).subs(func, y)
# This condition creates recursion while using pdsolve.
# Since the first step while solving a PDE of form
# a*(f(x, y).diff(x)) + b*(f(x, y).diff(y)) + c = 0
# is to solve the ODE dy/dx = b/a
if simplify(etainf/xiinf) == h:
continue
rpde = f(x, y).diff(x)*xiinf + f(x, y).diff(y)*etainf
r = pdsolve(rpde, func=f(x, y)).rhs
s = pdsolve(rpde - 1, func=f(x, y)).rhs
newcoord = [_lie_group_remove(coord) for coord in [r, s]]
r = Dummy("r")
s = Dummy("s")
C1 = Symbol("C1")
rcoord = newcoord[0]
scoord = newcoord[-1]
try:
sol = solve([r - rcoord, s - scoord], x, y, dict=True)
if sol == []:
continue
except NotImplementedError:
continue
else:
sol = sol[0]
xsub = sol[x]
ysub = sol[y]
num = simplify(scoord.diff(x) + scoord.diff(y)*h)
denom = simplify(rcoord.diff(x) + rcoord.diff(y)*h)
if num and denom:
diffeq = simplify((num/denom).subs([(x, xsub), (y, ysub)]))
sep = separatevars(diffeq, symbols=[r, s], dict=True)
if sep:
# Trying to separate, r and s coordinates
deq = integrate((1/sep[s]), s) + C1 - integrate(sep['coeff']*sep[r], r)
# Substituting and reverting back to original coordinates
deq = deq.subs([(r, rcoord), (s, scoord)])
try:
sdeq = solve(deq, y)
except NotImplementedError:
tempsol.append(deq)
else:
return [Eq(f(x), sol) for sol in sdeq]
elif denom: # (ds/dr) is zero which means s is constant
return [Eq(f(x), solve(scoord - C1, y)[0])]
elif num: # (dr/ds) is zero which means r is constant
return [Eq(f(x), solve(rcoord - C1, y)[0])]
# If nothing works, return solution as it is, without solving for y
if tempsol:
return [Eq(sol.subs(y, f(x)), 0) for sol in tempsol]
return None
def _ode_lie_group( s, func, order, match):
heuristics = lie_heuristics
inf = {}
f = func.func
x = func.args[0]
df = func.diff(x)
xi = Function("xi")
eta = Function("eta")
xis = match['xi']
etas = match['eta']
y = match.pop('y', None)
if y:
h = -simplify(match[match['d']]/match[match['e']])
y = y
else:
y = Dummy("y")
h = s.subs(func, y)
if xis is not None and etas is not None:
inf = [{xi(x, f(x)): S(xis), eta(x, f(x)): S(etas)}]
if checkinfsol(Eq(df, s), inf, func=f(x), order=1)[0][0]:
heuristics = ["user_defined"] + list(heuristics)
match = {'h': h, 'y': y}
# This is done so that if any heuristic raises a ValueError
# another heuristic can be used.
sol = None
for heuristic in heuristics:
sol = _ode_lie_group_try_heuristic(Eq(df, s), heuristic, func, match, inf)
if sol:
return sol
return sol
def infinitesimals(eq, func=None, order=None, hint='default', match=None):
r"""
The infinitesimal functions of an ordinary differential equation, `\xi(x,y)`
and `\eta(x,y)`, are the infinitesimals of the Lie group of point transformations
for which the differential equation is invariant. So, the ODE `y'=f(x,y)`
would admit a Lie group `x^*=X(x,y;\varepsilon)=x+\varepsilon\xi(x,y)`,
`y^*=Y(x,y;\varepsilon)=y+\varepsilon\eta(x,y)` such that `(y^*)'=f(x^*, y^*)`.
A change of coordinates, to `r(x,y)` and `s(x,y)`, can be performed so this Lie group
becomes the translation group, `r^*=r` and `s^*=s+\varepsilon`.
They are tangents to the coordinate curves of the new system.
Consider the transformation `(x, y) \to (X, Y)` such that the
differential equation remains invariant. `\xi` and `\eta` are the tangents to
the transformed coordinates `X` and `Y`, at `\varepsilon=0`.
.. math:: \left(\frac{\partial X(x,y;\varepsilon)}{\partial\varepsilon
}\right)|_{\varepsilon=0} = \xi,
\left(\frac{\partial Y(x,y;\varepsilon)}{\partial\varepsilon
}\right)|_{\varepsilon=0} = \eta,
The infinitesimals can be found by solving the following PDE:
>>> from sympy import Function, Eq, pprint
>>> from sympy.abc import x, y
>>> xi, eta, h = map(Function, ['xi', 'eta', 'h'])
>>> h = h(x, y) # dy/dx = h
>>> eta = eta(x, y)
>>> xi = xi(x, y)
>>> genform = Eq(eta.diff(x) + (eta.diff(y) - xi.diff(x))*h
... - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)), 0)
>>> pprint(genform)
/d d \ d 2 d
|--(eta(x, y)) - --(xi(x, y))|*h(x, y) - eta(x, y)*--(h(x, y)) - h (x, y)*--(x
\dy dx / dy dy
<BLANKLINE>
d d
i(x, y)) - xi(x, y)*--(h(x, y)) + --(eta(x, y)) = 0
dx dx
Solving the above mentioned PDE is not trivial, and can be solved only by
making intelligent assumptions for `\xi` and `\eta` (heuristics). Once an
infinitesimal is found, the attempt to find more heuristics stops. This is done to
optimise the speed of solving the differential equation. If a list of all the
infinitesimals is needed, ``hint`` should be flagged as ``all``, which gives
the complete list of infinitesimals. If the infinitesimals for a particular
heuristic needs to be found, it can be passed as a flag to ``hint``.
Examples
========
>>> from sympy import Function
>>> from sympy.solvers.ode.lie_group import infinitesimals
>>> from sympy.abc import x
>>> f = Function('f')
>>> eq = f(x).diff(x) - x**2*f(x)
>>> infinitesimals(eq)
[{eta(x, f(x)): exp(x**3/3), xi(x, f(x)): 0}]
References
==========
- Solving differential equations by Symmetry Groups,
John Starrett, pp. 1 - pp. 14
"""
if isinstance(eq, Equality):
eq = eq.lhs - eq.rhs
if not func:
eq, func = _preprocess(eq)
variables = func.args
if len(variables) != 1:
raise ValueError("ODE's have only one independent variable")
else:
x = variables[0]
if not order:
order = ode_order(eq, func)
if order != 1:
raise NotImplementedError("Infinitesimals for only "
"first order ODE's have been implemented")
else:
df = func.diff(x)
# Matching differential equation of the form a*df + b
a = Wild('a', exclude = [df])
b = Wild('b', exclude = [df])
if match: # Used by lie_group hint
h = match['h']
y = match['y']
else:
match = collect(expand(eq), df).match(a*df + b)
if match:
h = -simplify(match[b]/match[a])
else:
try:
sol = solve(eq, df)
except NotImplementedError:
raise NotImplementedError("Infinitesimals for the "
"first order ODE could not be found")
else:
h = sol[0] # Find infinitesimals for one solution
y = Dummy("y")
h = h.subs(func, y)
u = Dummy("u")
hx = h.diff(x)
hy = h.diff(y)
hinv = ((1/h).subs([(x, u), (y, x)])).subs(u, y) # Inverse ODE
match = {'h': h, 'func': func, 'hx': hx, 'hy': hy, 'y': y, 'hinv': hinv}
if hint == 'all':
xieta = []
for heuristic in lie_heuristics:
function = globals()['lie_heuristic_' + heuristic]
inflist = function(match, comp=True)
if inflist:
xieta.extend([inf for inf in inflist if inf not in xieta])
if xieta:
return xieta
else:
raise NotImplementedError("Infinitesimals could not be found for "
"the given ODE")
elif hint == 'default':
for heuristic in lie_heuristics:
function = globals()['lie_heuristic_' + heuristic]
xieta = function(match, comp=False)
if xieta:
return xieta
raise NotImplementedError("Infinitesimals could not be found for"
" the given ODE")
elif hint not in lie_heuristics:
raise ValueError("Heuristic not recognized: " + hint)
else:
function = globals()['lie_heuristic_' + hint]
xieta = function(match, comp=True)
if xieta:
return xieta
else:
raise ValueError("Infinitesimals could not be found using the"
" given heuristic")
def lie_heuristic_abaco1_simple(match, comp=False):
r"""
The first heuristic uses the following four sets of
assumptions on `\xi` and `\eta`
.. math:: \xi = 0, \eta = f(x)
.. math:: \xi = 0, \eta = f(y)
.. math:: \xi = f(x), \eta = 0
.. math:: \xi = f(y), \eta = 0
The success of this heuristic is determined by algebraic factorisation.
For the first assumption `\xi = 0` and `\eta` to be a function of `x`, the PDE
.. math:: \frac{\partial \eta}{\partial x} + (\frac{\partial \eta}{\partial y}
- \frac{\partial \xi}{\partial x})*h
- \frac{\partial \xi}{\partial y}*h^{2}
- \xi*\frac{\partial h}{\partial x} - \eta*\frac{\partial h}{\partial y} = 0
reduces to `f'(x) - f\frac{\partial h}{\partial y} = 0`
If `\frac{\partial h}{\partial y}` is a function of `x`, then this can usually
be integrated easily. A similar idea is applied to the other 3 assumptions as well.
References
==========
- E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra
Solving of First Order ODEs Using Symmetry Methods, pp. 8
"""
xieta = []
y = match['y']
h = match['h']
func = match['func']
x = func.args[0]
hx = match['hx']
hy = match['hy']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
hysym = hy.free_symbols
if y not in hysym:
try:
fx = exp(integrate(hy, x))
except NotImplementedError:
pass
else:
inf = {xi: S.Zero, eta: fx}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
factor = hy/h
facsym = factor.free_symbols
if x not in facsym:
try:
fy = exp(integrate(factor, y))
except NotImplementedError:
pass
else:
inf = {xi: S.Zero, eta: fy.subs(y, func)}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
factor = -hx/h
facsym = factor.free_symbols
if y not in facsym:
try:
fx = exp(integrate(factor, x))
except NotImplementedError:
pass
else:
inf = {xi: fx, eta: S.Zero}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
factor = -hx/(h**2)
facsym = factor.free_symbols
if x not in facsym:
try:
fy = exp(integrate(factor, y))
except NotImplementedError:
pass
else:
inf = {xi: fy.subs(y, func), eta: S.Zero}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
if xieta:
return xieta
def lie_heuristic_abaco1_product(match, comp=False):
r"""
The second heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = 0, \xi = f(x)*g(y)
.. math:: \eta = f(x)*g(y), \xi = 0
The first assumption of this heuristic holds good if
`\frac{1}{h^{2}}\frac{\partial^2}{\partial x \partial y}\log(h)` is
separable in `x` and `y`, then the separated factors containing `x`
is `f(x)`, and `g(y)` is obtained by
.. math:: e^{\int f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)\,dy}
provided `f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)` is a function
of `y` only.
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
`\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption
satisfies. After obtaining `f(x)` and `g(y)`, the coordinates are again
interchanged, to get `\eta` as `f(x)*g(y)`
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 7 - pp. 8
"""
xieta = []
y = match['y']
h = match['h']
hinv = match['hinv']
func = match['func']
x = func.args[0]
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
inf = separatevars(((log(h).diff(y)).diff(x))/h**2, dict=True, symbols=[x, y])
if inf and inf['coeff']:
fx = inf[x]
gy = simplify(fx*((1/(fx*h)).diff(x)))
gysyms = gy.free_symbols
if x not in gysyms:
gy = exp(integrate(gy, y))
inf = {eta: S.Zero, xi: (fx*gy).subs(y, func)}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
u1 = Dummy("u1")
inf = separatevars(((log(hinv).diff(y)).diff(x))/hinv**2, dict=True, symbols=[x, y])
if inf and inf['coeff']:
fx = inf[x]
gy = simplify(fx*((1/(fx*hinv)).diff(x)))
gysyms = gy.free_symbols
if x not in gysyms:
gy = exp(integrate(gy, y))
etaval = fx*gy
etaval = (etaval.subs([(x, u1), (y, x)])).subs(u1, y)
inf = {eta: etaval.subs(y, func), xi: S.Zero}
if not comp:
return [inf]
if comp and inf not in xieta:
xieta.append(inf)
if xieta:
return xieta
def lie_heuristic_bivariate(match, comp=False):
r"""
The third heuristic assumes the infinitesimals `\xi` and `\eta`
to be bi-variate polynomials in `x` and `y`. The assumption made here
for the logic below is that `h` is a rational function in `x` and `y`
though that may not be necessary for the infinitesimals to be
bivariate polynomials. The coefficients of the infinitesimals
are found out by substituting them in the PDE and grouping similar terms
that are polynomials and since they form a linear system, solve and check
for non trivial solutions. The degree of the assumed bivariates
are increased till a certain maximum value.
References
==========
- Lie Groups and Differential Equations
pp. 327 - pp. 329
"""
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
if h.is_rational_function():
# The maximum degree that the infinitesimals can take is
# calculated by this technique.
etax, etay, etad, xix, xiy, xid = symbols("etax etay etad xix xiy xid")
ipde = etax + (etay - xix)*h - xiy*h**2 - xid*hx - etad*hy
num, denom = cancel(ipde).as_numer_denom()
deg = Poly(num, x, y).total_degree()
deta = Function('deta')(x, y)
dxi = Function('dxi')(x, y)
ipde = (deta.diff(x) + (deta.diff(y) - dxi.diff(x))*h - (dxi.diff(y))*h**2
- dxi*hx - deta*hy)
xieq = Symbol("xi0")
etaeq = Symbol("eta0")
for i in range(deg + 1):
if i:
xieq += Add(*[
Symbol("xi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
for power in range(i + 1)])
etaeq += Add(*[
Symbol("eta_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
for power in range(i + 1)])
pden, denom = (ipde.subs({dxi: xieq, deta: etaeq}).doit()).as_numer_denom()
pden = expand(pden)
# If the individual terms are monomials, the coefficients
# are grouped
if pden.is_polynomial(x, y) and pden.is_Add:
polyy = Poly(pden, x, y).as_dict()
if polyy:
symset = xieq.free_symbols.union(etaeq.free_symbols) - {x, y}
soldict = solve(polyy.values(), *symset)
if isinstance(soldict, list):
soldict = soldict[0]
if any(soldict.values()):
xired = xieq.subs(soldict)
etared = etaeq.subs(soldict)
# Scaling is done by substituting one for the parameters
# This can be any number except zero.
dict_ = {sym: 1 for sym in symset}
inf = {eta: etared.subs(dict_).subs(y, func),
xi: xired.subs(dict_).subs(y, func)}
return [inf]
def lie_heuristic_chi(match, comp=False):
r"""
The aim of the fourth heuristic is to find the function `\chi(x, y)`
that satisfies the PDE `\frac{d\chi}{dx} + h\frac{d\chi}{dx}
- \frac{\partial h}{\partial y}\chi = 0`.
This assumes `\chi` to be a bivariate polynomial in `x` and `y`. By intuition,
`h` should be a rational function in `x` and `y`. The method used here is
to substitute a general binomial for `\chi` up to a certain maximum degree
is reached. The coefficients of the polynomials, are calculated by by collecting
terms of the same order in `x` and `y`.
After finding `\chi`, the next step is to use `\eta = \xi*h + \chi`, to
determine `\xi` and `\eta`. This can be done by dividing `\chi` by `h`
which would give `-\xi` as the quotient and `\eta` as the remainder.
References
==========
- E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra
Solving of First Order ODEs Using Symmetry Methods, pp. 8
"""
h = match['h']
hy = match['hy']
func = match['func']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
if h.is_rational_function():
schi, schix, schiy = symbols("schi, schix, schiy")
cpde = schix + h*schiy - hy*schi
num, denom = cancel(cpde).as_numer_denom()
deg = Poly(num, x, y).total_degree()
chi = Function('chi')(x, y)
chix = chi.diff(x)
chiy = chi.diff(y)
cpde = chix + h*chiy - hy*chi
chieq = Symbol("chi")
for i in range(1, deg + 1):
chieq += Add(*[
Symbol("chi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
for power in range(i + 1)])
cnum, cden = cancel(cpde.subs({chi : chieq}).doit()).as_numer_denom()
cnum = expand(cnum)
if cnum.is_polynomial(x, y) and cnum.is_Add:
cpoly = Poly(cnum, x, y).as_dict()
if cpoly:
solsyms = chieq.free_symbols - {x, y}
soldict = solve(cpoly.values(), *solsyms)
if isinstance(soldict, list):
soldict = soldict[0]
if any(soldict.values()):
chieq = chieq.subs(soldict)
dict_ = {sym: 1 for sym in solsyms}
chieq = chieq.subs(dict_)
# After finding chi, the main aim is to find out
# eta, xi by the equation eta = xi*h + chi
# One method to set xi, would be rearranging it to
# (eta/h) - xi = (chi/h). This would mean dividing
# chi by h would give -xi as the quotient and eta
# as the remainder. Thanks to Sean Vig for suggesting
# this method.
xic, etac = div(chieq, h)
inf = {eta: etac.subs(y, func), xi: -xic.subs(y, func)}
return [inf]
def lie_heuristic_function_sum(match, comp=False):
r"""
This heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = 0, \xi = f(x) + g(y)
.. math:: \eta = f(x) + g(y), \xi = 0
The first assumption of this heuristic holds good if
.. math:: \frac{\partial}{\partial y}[(h\frac{\partial^{2}}{
\partial x^{2}}(h^{-1}))^{-1}]
is separable in `x` and `y`,
1. The separated factors containing `y` is `\frac{\partial g}{\partial y}`.
From this `g(y)` can be determined.
2. The separated factors containing `x` is `f''(x)`.
3. `h\frac{\partial^{2}}{\partial x^{2}}(h^{-1})` equals
`\frac{f''(x)}{f(x) + g(y)}`. From this `f(x)` can be determined.
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
`\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first
assumption satisfies. After obtaining `f(x)` and `g(y)`, the coordinates
are again interchanged, to get `\eta` as `f(x) + g(y)`.
For both assumptions, the constant factors are separated among `g(y)`
and `f''(x)`, such that `f''(x)` obtained from 3] is the same as that
obtained from 2]. If not possible, then this heuristic fails.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 7 - pp. 8
"""
xieta = []
h = match['h']
func = match['func']
hinv = match['hinv']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
for odefac in [h, hinv]:
factor = odefac*((1/odefac).diff(x, 2))
sep = separatevars((1/factor).diff(y), dict=True, symbols=[x, y])
if sep and sep['coeff'] and sep[x].has(x) and sep[y].has(y):
k = Dummy("k")
try:
gy = k*integrate(sep[y], y)
except NotImplementedError:
pass
else:
fdd = 1/(k*sep[x]*sep['coeff'])
fx = simplify(fdd/factor - gy)
check = simplify(fx.diff(x, 2) - fdd)
if fx:
if not check:
fx = fx.subs(k, 1)
gy = (gy/k)
else:
sol = solve(check, k)
if sol:
sol = sol[0]
fx = fx.subs(k, sol)
gy = (gy/k)*sol
else:
continue
if odefac == hinv: # Inverse ODE
fx = fx.subs(x, y)
gy = gy.subs(y, x)
etaval = factor_terms(fx + gy)
if etaval.is_Mul:
etaval = Mul(*[arg for arg in etaval.args if arg.has(x, y)])
if odefac == hinv: # Inverse ODE
inf = {eta: etaval.subs(y, func), xi : S.Zero}
else:
inf = {xi: etaval.subs(y, func), eta : S.Zero}
if not comp:
return [inf]
else:
xieta.append(inf)
if xieta:
return xieta
def lie_heuristic_abaco2_similar(match, comp=False):
r"""
This heuristic uses the following two assumptions on `\xi` and `\eta`
.. math:: \eta = g(x), \xi = f(x)
.. math:: \eta = f(y), \xi = g(y)
For the first assumption,
1. First `\frac{\frac{\partial h}{\partial y}}{\frac{\partial^{2} h}{
\partial yy}}` is calculated. Let us say this value is A
2. If this is constant, then `h` is matched to the form `A(x) + B(x)e^{
\frac{y}{C}}` then, `\frac{e^{\int \frac{A(x)}{C} \,dx}}{B(x)}` gives `f(x)`
and `A(x)*f(x)` gives `g(x)`
3. Otherwise `\frac{\frac{\partial A}{\partial X}}{\frac{\partial A}{
\partial Y}} = \gamma` is calculated. If
a] `\gamma` is a function of `x` alone
b] `\frac{\gamma\frac{\partial h}{\partial y} - \gamma'(x) - \frac{
\partial h}{\partial x}}{h + \gamma} = G` is a function of `x` alone.
then, `e^{\int G \,dx}` gives `f(x)` and `-\gamma*f(x)` gives `g(x)`
The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
`\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption
satisfies. After obtaining `f(x)` and `g(x)`, the coordinates are again
interchanged, to get `\xi` as `f(x^*)` and `\eta` as `g(y^*)`
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
hinv = match['hinv']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
factor = cancel(h.diff(y)/h.diff(y, 2))
factorx = factor.diff(x)
factory = factor.diff(y)
if not factor.has(x) and not factor.has(y):
A = Wild('A', exclude=[y])
B = Wild('B', exclude=[y])
C = Wild('C', exclude=[x, y])
match = h.match(A + B*exp(y/C))
try:
tau = exp(-integrate(match[A]/match[C]), x)/match[B]
except NotImplementedError:
pass
else:
gx = match[A]*tau
return [{xi: tau, eta: gx}]
else:
gamma = cancel(factorx/factory)
if not gamma.has(y):
tauint = cancel((gamma*hy - gamma.diff(x) - hx)/(h + gamma))
if not tauint.has(y):
try:
tau = exp(integrate(tauint, x))
except NotImplementedError:
pass
else:
gx = -tau*gamma
return [{xi: tau, eta: gx}]
factor = cancel(hinv.diff(y)/hinv.diff(y, 2))
factorx = factor.diff(x)
factory = factor.diff(y)
if not factor.has(x) and not factor.has(y):
A = Wild('A', exclude=[y])
B = Wild('B', exclude=[y])
C = Wild('C', exclude=[x, y])
match = h.match(A + B*exp(y/C))
try:
tau = exp(-integrate(match[A]/match[C]), x)/match[B]
except NotImplementedError:
pass
else:
gx = match[A]*tau
return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}]
else:
gamma = cancel(factorx/factory)
if not gamma.has(y):
tauint = cancel((gamma*hinv.diff(y) - gamma.diff(x) - hinv.diff(x))/(
hinv + gamma))
if not tauint.has(y):
try:
tau = exp(integrate(tauint, x))
except NotImplementedError:
pass
else:
gx = -tau*gamma
return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}]
def lie_heuristic_abaco2_unique_unknown(match, comp=False):
r"""
This heuristic assumes the presence of unknown functions or known functions
with non-integer powers.
1. A list of all functions and non-integer powers containing x and y
2. Loop over each element `f` in the list, find `\frac{\frac{\partial f}{\partial x}}{
\frac{\partial f}{\partial x}} = R`
If it is separable in `x` and `y`, let `X` be the factors containing `x`. Then
a] Check if `\xi = X` and `\eta = -\frac{X}{R}` satisfy the PDE. If yes, then return
`\xi` and `\eta`
b] Check if `\xi = \frac{-R}{X}` and `\eta = -\frac{1}{X}` satisfy the PDE.
If yes, then return `\xi` and `\eta`
If not, then check if
a] :math:`\xi = -R,\eta = 1`
b] :math:`\xi = 1, \eta = -\frac{1}{R}`
are solutions.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
funclist = []
for atom in h.atoms(Pow):
base, exp = atom.as_base_exp()
if base.has(x) and base.has(y):
if not exp.is_Integer:
funclist.append(atom)
for function in h.atoms(AppliedUndef):
syms = function.free_symbols
if x in syms and y in syms:
funclist.append(function)
for f in funclist:
frac = cancel(f.diff(y)/f.diff(x))
sep = separatevars(frac, dict=True, symbols=[x, y])
if sep and sep['coeff']:
xitry1 = sep[x]
etatry1 = -1/(sep[y]*sep['coeff'])
pde1 = etatry1.diff(y)*h - xitry1.diff(x)*h - xitry1*hx - etatry1*hy
if not simplify(pde1):
return [{xi: xitry1, eta: etatry1.subs(y, func)}]
xitry2 = 1/etatry1
etatry2 = 1/xitry1
pde2 = etatry2.diff(x) - (xitry2.diff(y))*h**2 - xitry2*hx - etatry2*hy
if not simplify(expand(pde2)):
return [{xi: xitry2.subs(y, func), eta: etatry2}]
else:
etatry = -1/frac
pde = etatry.diff(x) + etatry.diff(y)*h - hx - etatry*hy
if not simplify(pde):
return [{xi: S.One, eta: etatry.subs(y, func)}]
xitry = -frac
pde = -xitry.diff(x)*h -xitry.diff(y)*h**2 - xitry*hx -hy
if not simplify(expand(pde)):
return [{xi: xitry.subs(y, func), eta: S.One}]
def lie_heuristic_abaco2_unique_general(match, comp=False):
r"""
This heuristic finds if infinitesimals of the form `\eta = f(x)`, `\xi = g(y)`
without making any assumptions on `h`.
The complete sequence of steps is given in the paper mentioned below.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
hx = match['hx']
hy = match['hy']
func = match['func']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
A = hx.diff(y)
B = hy.diff(y) + hy**2
C = hx.diff(x) - hx**2
if not (A and B and C):
return
Ax = A.diff(x)
Ay = A.diff(y)
Axy = Ax.diff(y)
Axx = Ax.diff(x)
Ayy = Ay.diff(y)
D = simplify(2*Axy + hx*Ay - Ax*hy + (hx*hy + 2*A)*A)*A - 3*Ax*Ay
if not D:
E1 = simplify(3*Ax**2 + ((hx**2 + 2*C)*A - 2*Axx)*A)
if E1:
E2 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2)
if not E2:
E3 = simplify(
E1*((28*Ax + 4*hx*A)*A**3 - E1*(hy*A + Ay)) - E1.diff(x)*8*A**4)
if not E3:
etaval = cancel((4*A**3*(Ax - hx*A) + E1*(hy*A - Ay))/(S(2)*A*E1))
if x not in etaval:
try:
etaval = exp(integrate(etaval, y))
except NotImplementedError:
pass
else:
xival = -4*A**3*etaval/E1
if y not in xival:
return [{xi: xival, eta: etaval.subs(y, func)}]
else:
E1 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2)
if E1:
E2 = simplify(
4*A**3*D - D**2 + E1*((2*Axx - (hx**2 + 2*C)*A)*A - 3*Ax**2))
if not E2:
E3 = simplify(
-(A*D)*E1.diff(y) + ((E1.diff(x) - hy*D)*A + 3*Ay*D +
(A*hx - 3*Ax)*E1)*E1)
if not E3:
etaval = cancel(((A*hx - Ax)*E1 - (Ay + A*hy)*D)/(S(2)*A*D))
if x not in etaval:
try:
etaval = exp(integrate(etaval, y))
except NotImplementedError:
pass
else:
xival = -E1*etaval/D
if y not in xival:
return [{xi: xival, eta: etaval.subs(y, func)}]
def lie_heuristic_linear(match, comp=False):
r"""
This heuristic assumes
1. `\xi = ax + by + c` and
2. `\eta = fx + gy + h`
After substituting the following assumptions in the determining PDE, it
reduces to
.. math:: f + (g - a)h - bh^{2} - (ax + by + c)\frac{\partial h}{\partial x}
- (fx + gy + c)\frac{\partial h}{\partial y}
Solving the reduced PDE obtained, using the method of characteristics, becomes
impractical. The method followed is grouping similar terms and solving the system
of linear equations obtained. The difference between the bivariate heuristic is that
`h` need not be a rational function in this case.
References
==========
- E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
ODE Patterns, pp. 10 - pp. 12
"""
h = match['h']
hx = match['hx']
hy = match['hy']
func = match['func']
x = func.args[0]
y = match['y']
xi = Function('xi')(x, func)
eta = Function('eta')(x, func)
coeffdict = {}
symbols = numbered_symbols("c", cls=Dummy)
symlist = [next(symbols) for _ in islice(symbols, 6)]
C0, C1, C2, C3, C4, C5 = symlist
pde = C3 + (C4 - C0)*h - (C0*x + C1*y + C2)*hx - (C3*x + C4*y + C5)*hy - C1*h**2
pde, denom = pde.as_numer_denom()
pde = powsimp(expand(pde))
if pde.is_Add:
terms = pde.args
for term in terms:
if term.is_Mul:
rem = Mul(*[m for m in term.args if not m.has(x, y)])
xypart = term/rem
if xypart not in coeffdict:
coeffdict[xypart] = rem
else:
coeffdict[xypart] += rem
else:
if term not in coeffdict:
coeffdict[term] = S.One
else:
coeffdict[term] += S.One
sollist = coeffdict.values()
soldict = solve(sollist, symlist)
if soldict:
if isinstance(soldict, list):
soldict = soldict[0]
subval = soldict.values()
if any(t for t in subval):
onedict = dict(zip(symlist, [1]*6))
xival = C0*x + C1*func + C2
etaval = C3*x + C4*func + C5
xival = xival.subs(soldict)
etaval = etaval.subs(soldict)
xival = xival.subs(onedict)
etaval = etaval.subs(onedict)
return [{xi: xival, eta: etaval}]
def _lie_group_remove(coords):
r"""
This function is strictly meant for internal use by the Lie group ODE solving
method. It replaces arbitrary functions returned by pdsolve as follows:
1] If coords is an arbitrary function, then its argument is returned.
2] An arbitrary function in an Add object is replaced by zero.
3] An arbitrary function in a Mul object is replaced by one.
4] If there is no arbitrary function coords is returned unchanged.
Examples
========
>>> from sympy.solvers.ode.lie_group import _lie_group_remove
>>> from sympy import Function
>>> from sympy.abc import x, y
>>> F = Function("F")
>>> eq = x**2*y
>>> _lie_group_remove(eq)
x**2*y
>>> eq = F(x**2*y)
>>> _lie_group_remove(eq)
x**2*y
>>> eq = x*y**2 + F(x**3)
>>> _lie_group_remove(eq)
x*y**2
>>> eq = (F(x**3) + y)*x**4
>>> _lie_group_remove(eq)
x**4*y
"""
if isinstance(coords, AppliedUndef):
return coords.args[0]
elif coords.is_Add:
subfunc = coords.atoms(AppliedUndef)
if subfunc:
for func in subfunc:
coords = coords.subs(func, 0)
return coords
elif coords.is_Pow:
base, expr = coords.as_base_exp()
base = _lie_group_remove(base)
expr = _lie_group_remove(expr)
return base**expr
elif coords.is_Mul:
mulargs = []
coordargs = coords.args
for arg in coordargs:
if not isinstance(coords, AppliedUndef):
mulargs.append(_lie_group_remove(arg))
return Mul(*mulargs)
return coords