def test_polymatrix_manipulations():
M1 = PolyMatrix([[1, 2], [3, 4]], x)
assert M1.transpose() == PolyMatrix([[1, 3], [2, 4]], x)
M2 = PolyMatrix([[5, 6], [7, 8]], x)
assert M1.row_join(M2) == PolyMatrix([[1, 2, 5, 6], [3, 4, 7, 8]], x)
assert M1.col_join(M2) == PolyMatrix([[1, 2], [3, 4], [5, 6], [7, 8]], x)
assert M1.applyfunc(lambda e: 2 * e) == PolyMatrix([[2, 4], [6, 8]], x)
def prde_no_cancel_b_small(b, Q, n, DE):
"""
Parametric Poly Risch Differential Equation - No cancellation: deg(b) small enough.
Given a derivation D on k[t], n in ZZ, and b, q1, ..., qm in k[t] with
deg(b) < deg(D) - 1 and either D == d/dt or deg(D) >= 2, returns
h1, ..., hr in k[t] and a matrix A with coefficients in Const(k) such that
if c1, ..., cm in Const(k) and q in k[t] satisfy deg(q) <= n and
Dq + b*q == Sum(ci*qi, (i, 1, m)) then q = Sum(dj*hj, (j, 1, r)) where
d1, ..., dr in Const(k) and A*Matrix([[c1, ..., cm, d1, ..., dr]]).T == 0.
"""
m = len(Q)
H = [Poly(0, DE.t)]*m
for N in range(n, 0, -1): # [n, ..., 1]
for i in range(m):
si = Q[i].nth(N + DE.d.degree(DE.t) - 1)/(N*DE.d.LC())
sitn = Poly(si*DE.t**N, DE.t)
H[i] = H[i] + sitn
Q[i] = Q[i] - derivation(sitn, DE) - b*sitn
if b.degree(DE.t) > 0:
for i in range(m):
si = Poly(Q[i].nth(b.degree(DE.t))/b.LC(), DE.t)
H[i] = H[i] + si
Q[i] = Q[i] - derivation(si, DE) - b*si
if all(qi.is_zero for qi in Q):
dc = -1
M = Matrix()
else:
dc = max([qi.degree(DE.t) for qi in Q])
M = Matrix(dc + 1, m, lambda i, j: Q[j].nth(i))
A, u = constant_system(M, zeros(dc + 1, 1), DE)
c = eye(m)
A = A.row_join(zeros(A.rows, m)).col_join(c.row_join(-c))
return (H, A)
# else: b is in k, deg(qi) < deg(Dt)
t = DE.t
if DE.case != 'base':
with DecrementLevel(DE):
t0 = DE.t # k = k0(t0)
ba, bd = frac_in(b, t0, field=True)
Q0 = [frac_in(qi.TC(), t0, field=True) for qi in Q]
f, B = param_rischDE(ba, bd, Q0, DE)
# f = [f1, ..., fr] in k^r and B is a matrix with
# m + r columns and entries in Const(k) = Const(k0)
# such that Dy0 + b*y0 = Sum(ci*qi, (i, 1, m)) has
# a solution y0 in k with c1, ..., cm in Const(k)
# if and only y0 = Sum(dj*fj, (j, 1, r)) where
# d1, ..., dr ar in Const(k) and
# B*Matrix([c1, ..., cm, d1, ..., dr]) == 0.
# Transform fractions (fa, fd) in f into constant
# polynomials fa/fd in k[t].
# (Is there a better way?)
f = [Poly(fa.as_expr()/fd.as_expr(), t, field=True)
for fa, fd in f]
else:
# Base case. Dy == 0 for all y in k and b == 0.
# Dy + b*y = Sum(ci*qi) is solvable if and only if
# Sum(ci*qi) == 0 in which case the solutions are
# y = d1*f1 for f1 = 1 and any d1 in Const(k) = k.
f = [Poly(1, t, field=True)] # r = 1
B = Matrix([[qi.TC() for qi in Q] + [S(0)]])
# The condition for solvability is
# B*Matrix([c1, ..., cm, d1]) == 0
# There are no constraints on d1.
# Coefficients of t^j (j > 0) in Sum(ci*qi) must be zero.
d = max([qi.degree(DE.t) for qi in Q])
if d > 0:
M = Matrix(d, m, lambda i, j: Q[j].nth(i + 1))
A, _ = constant_system(M, zeros(d, 1), DE)
else:
# No constraints on the hj.
A = Matrix(0, m, [])
# Solutions of the original equation are
# y = Sum(dj*fj, (j, 1, r) + Sum(ei*hi, (i, 1, m)),
# where ei == ci (i = 1, ..., m), when
# A*Matrix([c1, ..., cm]) == 0 and
# B*Matrix([c1, ..., cm, d1, ..., dr]) == 0
# Build combined constraint matrix with m + r + m columns.
r = len(f)
I = eye(m)
A = A.row_join(zeros(A.rows, r + m))
B = B.row_join(zeros(B.rows, m))
C = I.row_join(zeros(m, r)).row_join(-I)
return f + H, A.col_join(B).col_join(C)
def prde_no_cancel_b_small(b, Q, n, DE):
"""
Parametric Poly Risch Differential Equation - No cancellation: deg(b) small enough.
Given a derivation D on k[t], n in ZZ, and b, q1, ..., qm in k[t] with
deg(b) < deg(D) - 1 and either D == d/dt or deg(D) >= 2, returns
h1, ..., hr in k[t] and a matrix A with coefficients in Const(k) such that
if c1, ..., cm in Const(k) and q in k[t] satisfy deg(q) <= n and
Dq + b*q == Sum(ci*qi, (i, 1, m)) then q = Sum(dj*hj, (j, 1, r)) where
d1, ..., dr in Const(k) and A*Matrix([[c1, ..., cm, d1, ..., dr]]).T == 0.
"""
m = len(Q)
H = [Poly(0, DE.t)]*m
for N in range(n, 0, -1): # [n, ..., 1]
for i in range(m):
si = Q[i].nth(N + DE.d.degree(DE.t) - 1)/(N*DE.d.LC())
sitn = Poly(si*DE.t**N, DE.t)
H[i] = H[i] + sitn
Q[i] = Q[i] - derivation(sitn, DE) - b*sitn
if b.degree(DE.t) > 0:
for i in range(m):
si = Poly(Q[i].nth(b.degree(DE.t))/b.LC(), DE.t)
H[i] = H[i] + si
Q[i] = Q[i] - derivation(si, DE) - b*si
if all(qi.is_zero for qi in Q):
dc = -1
M = Matrix()
else:
dc = max([qi.degree(DE.t) for qi in Q])
M = Matrix(dc + 1, m, lambda i, j: Q[j].nth(i))
A, u = constant_system(M, zeros(dc + 1, 1), DE)
c = eye(m)
A = A.row_join(zeros(A.rows, m)).col_join(c.row_join(-c))
return (H, A)
# else: b is in k, deg(qi) < deg(Dt)
t = DE.t
if DE.case != 'base':
with DecrementLevel(DE):
t0 = DE.t # k = k0(t0)
ba, bd = frac_in(b, t0, field=True)
Q0 = [frac_in(qi.TC(), t0, field=True) for qi in Q]
f, B = param_rischDE(ba, bd, Q0, DE)
# f = [f1, ..., fr] in k^r and B is a matrix with
# m + r columns and entries in Const(k) = Const(k0)
# such that Dy0 + b*y0 = Sum(ci*qi, (i, 1, m)) has
# a solution y0 in k with c1, ..., cm in Const(k)
# if and only y0 = Sum(dj*fj, (j, 1, r)) where
# d1, ..., dr ar in Const(k) and
# B*Matrix([c1, ..., cm, d1, ..., dr]) == 0.
# Transform fractions (fa, fd) in f into constant
# polynomials fa/fd in k[t].
# (Is there a better way?)
f = [Poly(fa.as_expr()/fd.as_expr(), t, field=True)
for fa, fd in f]
else:
# Base case. Dy == 0 for all y in k and b == 0.
# Dy + b*y = Sum(ci*qi) is solvable if and only if
# Sum(ci*qi) == 0 in which case the solutions are
# y = d1*f1 for f1 = 1 and any d1 in Const(k) = k.
f = [Poly(1, t, field=True)] # r = 1
B = Matrix([[qi.TC() for qi in Q] + [S(0)]])
# The condition for solvability is
# B*Matrix([c1, ..., cm, d1]) == 0
# There are no constraints on d1.
# Coefficients of t^j (j > 0) in Sum(ci*qi) must be zero.
d = max([qi.degree(DE.t) for qi in Q])
if d > 0:
M = Matrix(d, m, lambda i, j: Q[j].nth(i + 1))
A, _ = constant_system(M, zeros(d, 1), DE)
else:
# No constraints on the hj.
A = Matrix(0, m, [])
# Solutions of the original equation are
# y = Sum(dj*fj, (j, 1, r) + Sum(ei*hi, (i, 1, m)),
# where ei == ci (i = 1, ..., m), when
# A*Matrix([c1, ..., cm]) == 0 and
# B*Matrix([c1, ..., cm, d1, ..., dr]) == 0
# Build combined constraint matrix with m + r + m columns.
r = len(f)
I = eye(m)
A = A.row_join(zeros(A.rows, r + m))
B = B.row_join(zeros(B.rows, m))
C = I.row_join(zeros(m, r)).row_join(-I)
return f + H, A.col_join(B).col_join(C)