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I'm following the article at http://www.paykin.info/irina/project_2.jsp Finite Difference method.

How to interpret this one? How to convert this to pseudo code? $$u(i,j+1) = 2u(i,j) + \left(\frac{k}{h}\right)^2[u(i+1,j) - 2u(i,j) + u(i-1,j)] - u(i,j-1)$$

Assuming the ff given values (Im not sure if values are possible).

Initial Condition:
u(i,0)=sin(i)
du(i,0)/dt=0

Boundary Condition:
u(0,j)=0
u(95,j)=0

dt=1
dx=5

What is the result when t=5.

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1 Answer 1

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The term on LHS is the variable's value at next time step i.e. (j+1) which depends on the values of variable u at previous time steps j and (j-1).

Pseudo code could be like this:

Initialize u(i,j) and let it equal to u(i,j-1).

For (all grid points except those on boundary), use the discretized equation to advance in time.

For boundary nodes, use the boundary conditions.

If I have to write in C with i varying from 0 to N, I can write:

for (i = 0; i <= N; i++)
{
     if (i > 0 && i < N)
     {
          unew[i] = 2*u[i] + (dt/dx)*(dt/dx)*(u[i+1] - 2*u[i] + u[i-1]) - uold[i]
     }
     else
     {
         unew[i] = u[i]
     }
}

Hope this helps. If still got some query, feel free to ask.

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  • $\begingroup$ let say given initial condition u(i,0)=some function. where to initialize that one? Regarding the unew and uold, are that additional variables? With given values of dt and dx, how to get the result when t=5? $\endgroup$
    – mich
    Commented May 29, 2015 at 1:10
  • $\begingroup$ Thanks Tanmay, I added example values so that I can understand more on the example. $\endgroup$
    – mich
    Commented May 29, 2015 at 1:31
  • $\begingroup$ All the variables are defined and declared at the beginning of simulation. Suppose if you want to have a sinusoidal initial condition, you can write in C as: for (i = 0; i <= N; i++) { u[i] = sin[i]; unew[i] = sin(i); uold[i] = sin(i); } If you still have a problem regarding coding or the physics, feel free to contact me: [email protected] $\endgroup$
    – Tanmay Agrawal
    Commented May 29, 2015 at 8:14

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