def get_linear_equations_solution_numerically(eqs, dt):
# Otherwise assumes it is given in functional form
n = len(eqs._diffeq_names) # number of state variables
dynamicvars = eqs._diffeq_names
# Calculate B
AB = zeros((n, 1))
d = dict.fromkeys(dynamicvars)
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
for var, i in zip(dynamicvars, count()):
AB[i] = -eqs.apply(var, d)
# Calculate A
M = zeros((n, n))
for i in range(n):
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
if isinstance(eqs._units[dynamicvars[i]], Quantity):
d[dynamicvars[i]] = Quantity.with_dimensions(1., eqs._units[dynamicvars[i]].get_dimensions())
else:
d[dynamicvars[i]] = 1.
for var, j in zip(dynamicvars, count()):
M[j, i] = eqs.apply(var, d) + AB[j]
#B=linalg.solve(M,AB)
numeulersteps = 100
deltat = dt / numeulersteps
E = eye(n) + deltat * M
C = eye(n)
D = zeros((n, 1))
for step in xrange(numeulersteps):
C, D = dot(E, C), dot(E, D) - AB * deltat
return C, D
def get_linear_equations_solution_numerically(eqs, dt):
# Otherwise assumes it is given in functional form
n = len(eqs._diffeq_names) # number of state variables
dynamicvars = eqs._diffeq_names
# Calculate B
AB = zeros((n, 1))
d = dict.fromkeys(dynamicvars)
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
for var, i in zip(dynamicvars, count()):
AB[i] = -eqs.apply(var, d)
# Calculate A
M = zeros((n, n))
for i in range(n):
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
if isinstance(eqs._units[dynamicvars[i]], Quantity):
d[dynamicvars[i]] = Quantity.with_dimensions(1., eqs._units[dynamicvars[i]].get_dimensions())
else:
d[dynamicvars[i]] = 1.
for var, j in zip(dynamicvars, count()):
M[j, i] = eqs.apply(var, d) + AB[j]
#B=linalg.solve(M,AB)
numeulersteps = 100
deltat = dt / numeulersteps
E = eye(n) + deltat * M
C = eye(n)
D = zeros((n, 1))
for step in xrange(numeulersteps):
C, D = dot(E, C), dot(E, D) - AB * deltat
return C, D
def get_linear_equations(eqs):
'''
Returns the matrices M and B for the linear model dX/dt = M(X-B),
where eqs is an Equations object.
'''
# Otherwise assumes it is given in functional form
n = len(eqs._diffeq_names) # number of state variables
dynamicvars = eqs._diffeq_names
# Calculate B
AB = zeros((n, 1))
d = dict.fromkeys(dynamicvars)
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
for var, i in zip(dynamicvars, count()):
AB[i] = -eqs.apply(var, d)
# Calculate A
M = zeros((n, n))
for i in range(n):
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
if isinstance(eqs._units[dynamicvars[i]], Quantity):
d[dynamicvars[i]] = Quantity.with_dimensions(1., eqs._units[dynamicvars[i]].get_dimensions())
else:
d[dynamicvars[i]] = 1.
for var, j in zip(dynamicvars, count()):
M[j, i] = eqs.apply(var, d) + AB[j]
#M-=eye(n)*1e-10 # quick dirty fix for problem of constant derivatives; dimension = Hz
#B=linalg.lstsq(M,AB)[0] # We use this instead of solve in case M is degenerate
B = linalg.solve(M, AB) # We use this instead of solve in case M is degenerate
return M, B
def get_linear_equations(eqs):
'''
Returns the matrices M and B for the linear model dX/dt = M(X-B),
where eqs is an Equations object.
'''
# Otherwise assumes it is given in functional form
n = len(eqs._diffeq_names) # number of state variables
dynamicvars = eqs._diffeq_names
# Calculate B
AB = zeros((n, 1))
d = dict.fromkeys(dynamicvars)
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
for var, i in zip(dynamicvars, count()):
AB[i] = -eqs.apply(var, d)
# Calculate A
M = zeros((n, n))
for i in range(n):
for j in range(n):
d[dynamicvars[j]] = 0. * eqs._units[dynamicvars[j]]
if isinstance(eqs._units[dynamicvars[i]], Quantity):
d[dynamicvars[i]] = Quantity.with_dimensions(1., eqs._units[dynamicvars[i]].get_dimensions())
else:
d[dynamicvars[i]] = 1.
for var, j in zip(dynamicvars, count()):
M[j, i] = eqs.apply(var, d) + AB[j]
#M-=eye(n)*1e-10 # quick dirty fix for problem of constant derivatives; dimension = Hz
#B=linalg.lstsq(M,AB)[0] # We use this instead of solve in case M is degenerate
B = linalg.solve(M, AB) # We use this instead of solve in case M is degenerate
return M, B