characteristics

Question

I want to be able to simulate a hyperbolic equation on characteristic curves (lines). I will start with a basic one. u_{t}+2u_{x}=u^{2} with initial data u(x,0)=cos(x). The solution is u(x,t)=cos(x-2t)/(1-t*cos(x-2t)) where the characteristic curve is x=2*t+x_{0}. So the solution is defined on characteristics (method of characteristics).

x=zeros(10,5);
u=zeros(10,5);
x0=linspace(0,10,10);
t=linspace(0,5,5);
for i=1:length(x0)
    for j=1:length(t)
        x(i,j)=2*t(j)+x0(i);
        if t(j)*cos(x(i,j)-2*t(j))==1
            u(i,j)=0;
        else
            u(i,j)=cos(x(i,j)-2*t(j))/(1-t(j)*cos(x(i,j)-2*t(j)));
        end
    end
end

I'll be thankful if anyone can see my mistake.

Answer 1

I am not certain exactly if this solves your problem but you are accessing elements of your t matrix using the i and j indices. For example:

  (1-t(i)*cos(x(i)-2*t(j))

Since i can take values beyond the range of your t array, it is possible that you are mixing up your indices. At the very least you could be getting errors from accessing an element of t that is beyond its range. It is hard to say as you don't specify an error. I am just pointing out a potential problem.

It seems to me that you should use i to access elements of x0 and j to access elements of t since these are within the limits you specify in your loops. Maybe

for i=1:length(x0)
    for j=1:length(t)
        x(i,j)=2*t(j)+x0(i);
        if t(j)*cos(x(i)-2*t(j))==1
            u(i,j)=0;
        else
            u(i,j)=cos(x(i)-2*t(j))/(1-t(j)*cos(x(i)-2*t(j)));
        end
    end
end

is what you are looking for.