Why is exponential integrator method giving poor results? - matlab

Question

//u' + Au = g(t,u) can be solved by exponential integrators also
//Following snippet is for exp INtegrators
A = -full(Strang(11))
A[end,1]=1;A[1,end]=1;
g(t,u) = 2-u
u0 = zeros(11);u0[6]=1
nsteps = 1000
tmax = 10.0
h = tmax/nsteps
u = u0
t = 0
for k in 1:nsteps
u = expm(-h*A)*u + h*((expm(-h*A)-1)\(-h*A))*g(t,u)
t = k*h
end
//this is for euler's method
for k in 1:nsteps
u += h*(A*u + h*g(t,u))
t = k*h
end
Why are they giving poor results?
The method is exploding very badly, it should converge to [1.99]*11 , or something like that?
Is there any mistake while implementing Exp Integrator?

Answer 1

The test problem is a singular matrix. A better test is the setup:
using SpecialMatrices
A = -full(Strang(11))
g(t,u) = 2-u
u = zeros(11);u[6]=1
nsteps = 10000
tmax = 1.0
h = tmax/nsteps
t = 0
Using this, fix the h in the Euler to get (notice there's an extra h, my bad:
u = zeros(11);u[6]=1
for k in 1:nsteps
u += h*(A*u + g(t,u))
t = k*h
end
#show u
u = [0.93573,1.19361,1.26091,1.29627,1.34313,1.37767,1.34313,1.29627,1.26091,1.19361,0.93573]
But to find out what's wrong, start looking at numbers. What happens A=0? Well, we know that phi(z) = (e^z - 1)/z. By L'Hopital's rule, phi(z) -> 1 as z->0. Therefore, in order for our implementation to have the same behavior, we have to have that same result. But let's check what happens:
expm(zeros(5,5))
5×5 Array{Float64,2}:
1.0 0.0 0.0 0.0 0.0
0.0 1.0 0.0 0.0 0.0
0.0 0.0 1.0 0.0 0.0
0.0 0.0 0.0 1.0 0.0
0.0 0.0 0.0 0.0 1.0
Notice that this gives the identity matrix. So think about the limit: if the bottom is going to zero... how can this be constant? We must have that the top is going to zero... so the top is going to I.
And that's the moment of clarity: the author meant 1 in the field that you're in. So for a matrix input, 1=I. When you realize that, you fix the code:
# Norsett-Euler
u = zeros(11);u[6]=1
for k in 1:nsteps
u = expm(h*A)*u + ((expm(h*A)-I)/A)*g(t,u)
t = k*h
end
#show u
u = [0.935722,1.1936,1.26091,1.29627,1.34313,1.37768,1.34313,1.29627,1.26091,1.1936,0.935722]
Moral of the story: for programming mathematics, you also have to debug your math.
Edit
Get a more efficient form one step at a time. First, try and force another varphi term:
# Norsett-Euler
u = zeros(11);u[6]=1
for k in 1:nsteps
u = (I + A*(expm(h*A)-I)/A)*u + ((expm(h*A)-I)/A)*g(t,u)
t = k*h
end
#show u
Now gather:
# Norsett-Euler
u = zeros(11);u[6]=1
for k in 1:nsteps
u = u + ((expm(h*A)-I)/A)*(A*u + g(t,u))
t = k*h
end
#show u
This is the efficient form of the method you were trying to write. Then you can cache the operator since A is constant:
# Norsett-Euler
u = zeros(11);u[6]=1
phi1 = ((expm(h*A)-I)/A)
for k in 1:nsteps
u = u + phi1*(A*u + g(t,u))
t = k*h
end
#show u

Answer 2

The exponential Euler-Rosenbrock step should be for u'=Lu+g(t,u)
u = expm(h*L)*u + ((expm(h*L)-1)/L)*g(t,u)
Note that the inverted matrix is L (in Matlab A/B == A*inv(B) and A\B == inv(A)*B), and there is no naked h in the formula. In your case, L = -A.
An alternative method to split off the exponential part has is as follows. Set u(t) = exp(-t*A)*v(t) which translates into the differential equation
exp(-t*A)*v'(t) = g(t,exp(-t*A)*v((t))
Now apply the forward Euler step for v
v(t+h) = v(t) + h * exp(t*A)*g(t,exp(-t*A)*v((t))
which translated back into terms in u gives
u(t+h) = exp(-(t+h)*A)*v(t+h)
= exp(-h*A) * ( u(t) + h * g(t,u(t)) )
This should equally result in a first order method.