download
raw
2.11 kB
from numpy import *
from matplotlib.pyplot import *
def solver(I, a, b, T, dt, theta):
"""
Solve u'=-a(t)*u + b(t), u(0)=I,
for t in (0,T] with steps of dt.
a and b are Python functions of t.
"""
dt = float(dt) # avoid integer division
Nt = int(round(T/dt)) # no of time intervals
T = Nt*dt # adjust T to fit time step dt
u = zeros(Nt+1) # array of u[n] values
t = linspace(0, T, Nt+1) # time mesh
u[0] = I # assign initial condition
for n in range(0, Nt): # n=0,1,...,Nt-1
u[n+1] = ((1 - dt*(1-theta)*a(t[n]))*u[n] + \
dt*(theta*b(t[n+1]) + (1-theta)*b(t[n])))/\
(1 + dt*theta*a(t[n+1]))
return u, t
def test_constant_solution():
"""
Test problem where u=u_const is the exact solution, to be
reproduced (to machine precision) by any relevant method.
"""
def exact_solution(t):
return u_const
def a(t):
return 2.5*(1+t**3) # can be arbitrary
def b(t):
return a(t)*u_const
u_const = 2.15
theta = 0.4; I = u_const; dt = 4
Nt = 4 # enough with a few steps
u, t = solver(I=I, a=a, b=b, T=Nt*dt, dt=dt, theta=theta)
print u
u_e = exact_solution(t)
difference = abs(u_e - u).max() # max deviation
tol = 1E-14
assert difference < tol
def test_linear_solution():
"""
Test problem where u=c*t+I is the exact solution, to be
reproduced (to machine precision) by any relevant method.
"""
def exact_solution(t):
return c*t + I
def a(t):
return t**0.5 # can be arbitrary
def b(t):
return c + a(t)*exact_solution(t)
theta = 0.4; I = 0.1; dt = 0.1; c = -0.5
T = 4
Nt = int(T/dt) # no of steps
u, t = solver(I=I, a=a, b=b, T=Nt*dt, dt=dt, theta=theta)
u_e = exact_solution(t)
difference = abs(u_e - u).max() # max deviation
print difference
tol = 1E-14 # depends on c!
assert difference < tol
if __name__ == '__main__':
#test_constant_solution()
test_linear_solution()

Xet Storage Details

Size:
2.11 kB
·
Xet hash:
c77e768faf8ad6e9d64f030019dea8339adc8ed5dca6146b284f9214f6f61d1c

Xet efficiently stores files, intelligently splitting them into unique chunks and accelerating uploads and downloads. More info.