def __apply_tension(self):
"""Inspiration from:
U{http://www.ee.ucl.ac.uk/~rjames/modelling/constant-order/oned/}."""
try:
from scikits.bvp1lg import colnew
except ImportError:
warning("bvp module not found.")
raise
boundary_points = numpy.array([0, 0, 0, 1, 1, 1])
tol = 1e-6 * numpy.ones_like(boundary_points)
degrees = numpy.array([2, 2, 2])
solution = colnew.solve(boundary_points,
degrees,
self.__ode_3k,
self.__bc_nosplay,
is_linear=False,
initial_guess=self.__ic_nosplay,
tolerances=tol,
vectorized=True,
maximum_mesh_size=1000)
self.bvp_solution = solution
def __apply_tension(self):
"""Inspiration from:
U{http://www.ee.ucl.ac.uk/~rjames/modelling/constant-order/oned/}."""
try:
from scikits.bvp1lg import colnew
except ImportError:
warning("bvp module not found.")
raise
boundary_points = numpy.array([0, 0, 0, 1, 1, 1])
tol = 1e-6 * numpy.ones_like(boundary_points)
degrees = numpy.array([2, 2, 2])
solution = colnew.solve(
boundary_points, degrees, self.__ode_3k, self.__bc_nosplay,
is_linear=False, initial_guess=self.__ic_nosplay,
tolerances=tol, vectorized=True,
maximum_mesh_size=1000)
self.bvp_solution = solution
return z, fsub(t,z)
# Initial guess for the solution
N = 5
degrees = [1, 1, 1]
boundary_points = [0, S, T]
#tin = [0, .4, .8, 1.5, T]
tin = np.linspace(0, T, N)
# solve the boundary value problem
tol = [1e-5, 1e-5, 1e-5]
solution = colnew.solve(
boundary_points, degrees, fsub, gsub,
dfsub=dfsub, dgsub=dgsub,
is_linear=False, tolerances=tol, initial_guess=guess,
#extra_fixed_points=[0.800],
collocation_points=3, initial_mesh=tin,
vectorized=True, maximum_mesh_size=50, verbosity=2)
print ('grid used within colnew')
tc = solution.mesh
n = solution.nmesh
print (n)
print (tc)
# refine the grid (doubling)
t = np.zeros(2*n-1)
for i in range(n-1):
t[2*i] = tc[i]
t[2*i+1] = .5*(tc[i]+tc[i+1])
t[-1] = tc[-1]
# Initial guess for the solution
N = 5
degrees = [1, 1]
boundary_points = [0, 1]
tin = np.linspace(0, 1, N)
# solve the boundary value problem
tol = [1e-5, 1e-5]
solution = colnew.solve(boundary_points,
degrees,
fsub,
gsub,
dfsub=None,
dgsub=None,
is_linear=False,
tolerances=tol,
initial_guess=None,
collocation_points=3,
initial_mesh=tin,
vectorized=True,
maximum_mesh_size=30,
verbosity=0)
t = solution.mesh
x = solution(t)[:, 0]
p = solution(t)[:, 1]
# Calculate u(t) from x,p
u = x - p / 2.
# Calculate the cost
w = (u - x)**2
j = -2*x
z = np.array([x,p,q1,q2,j])
return z, fsub(t,z)
# Initial guess for the solution
N = 10
degrees = [1, 1, 1, 1, 1]
boundary_points = [0, 0, S[0], S[1], T]
tin = [0., 0.2, 0.4, S[0], 0.5, S[1], 1.4, T]
# solve the boundary value problem
tol = [1e-6, 1e-6, 1e-5, 1e-5, 0] # change
solution = colnew.solve(
boundary_points, degrees, fsub, gsub,
dfsub=None, dgsub=dgsub,
is_linear=False, tolerances=tol, initial_guess=guess,
collocation_points = 5, initial_mesh=tin, adaptive_mesh_selection=True,
vectorized=False, maximum_mesh_size=50, verbosity=2)
print ('grid used within colnew')
tc = solution.mesh
n = solution.nmesh
print (n)
print (tc)
# refine the grid (doubling)
t = np.zeros(2*n-1)
for i in range(n-1):
t[2*i] = tc[i]
t[2*i+1] = .5*(tc[i]+tc[i+1])
t[-1] = tc[-1]
boundary_points = np.array([0,np.pi])
tol=1.0e-8*np.ones_like(boundary_points)
def fsub(x,Z):
"""The equations"""
u,du=Z
return np.array([-u])
def gsub(Z):
"""The boundary conditions"""
u,du=Z
return np.array([u[0]-0.0,du[1]-1])
solution=colnew.solve(
boundary_points,degrees,fsub,gsub,
is_linear=True,tolerances=tol,
vectorized=True,
maximum_mesh_size=300
)
plt.ion()
x=solution.mesh
u_exact=-np.sin(x)
plt.plot(x,solution(x)[:,0],'b.')
plt.plot(x,u_exact,'g-')
plt.show()
