Solving Linear convection equation (wave equation) by using 3rd order Runge Kutta scheme

Hi!

I am solving the wave equation by using the 3rd order RK scheme. The attached photo shows the problem.

I have managed to make the other schemes work but I think I am not implementing the 3rd order RK correctly. My code is attached below:

% MAE 5093 Homework #5- Due November 13th, 2018

%% Problem 1 C)

%% Code written by Saad Saleem

% % This MATLAB code solves linear convection equation by 3rd order RK and

% compares it with exact solution

clear all

close all

N=100; %No. of grid points

Tmax=1; % time period=1

alpha=1; %given

h=0.01; %given value of delX

delt=0.005;

maxt=Tmax/delt; %Number of time steps

c=alpha*delt/h; %c=0.5

u=zeros(N+1,maxt+1 ); %Initializing the matrix that will fill in 'u'values

x=zeros(N+1,1);%There is only 1 column but N+1 rows

% Initial condition

for i=1:N+1

x(i)=(i-1)*h ;

u(i,1)=sin(2*pi*x(i ));

end

for k=1:maxt

u0 = u(N,k ); %This is to implement the periodic BC

u(1,1)=u(1,k)-c/2*(u(2,k)-u0);

u(1,2)=3/4*u(1,k)+0.25*u(1,1)-c/8*(u(2,1)-u0);

u(1,k+1)=1/3*(u(1,k))+2/3*u(1,2)-c/3*(u(2,2)-u0);

% u(1,k+1)=u(1,k)-c/2*(u(2,k)-u0)+c^2/2*(u(2,k)-2*u(1,k)+u0 ); %Once you do the initial call, start filling in the next

for i=2:N

u(i,1)=u(i,k)-c/2*(u(i+1,k)-u(i-1,k));

u(i,2)=3/4*u(i,k)+0.25*u(i,1)-c/8*(u(i+1,1)-u(i-1,1));

u(i,k+1)=1/3*(u(i,k))+2/3*u(i,2)-c/3*(u(i+1,2)-u(i-1,2));

% u(i,k+1) =u(i,k)-c/2*(u(i+1,k)-u(i-1,k))+(c^2)/2*(u(i+1,k)-2*u(i,k)+u(i-1,k ));

end

uNp2 = u(2,k);

u(N+1,1)=u(N+1,k)-c/2*(uNp2-u(N,k));

u(N+1,2)=3/4*(u(N+1,k))+0.25*(u(N+1,1))-c/8*(uNp2-u(N,1));

u(N+1,k+1)=1/3*(u(N+1,k))+2/3*(u(N+1,2))-c/3*(uNp2-u(N,2));

end

% plot(x,u(:,10))

plot(x,u(:,10),x,u(:,20),x,u(:,30),x,u(:,40),x,u(:,50),x,u(:,60),x,u(:,70),x,u(:,100))

[Torsten]

Be careful: The notation of the Runge-Kutta method in your homework assignment is misleading.

The superscripts (1) and (2) do not indicate the time step (like the superscripts (n) and (n+1)), but intermediate stages of the Runge-Kutta-method in the computation of u^(n+1) from u^(n).

Best wishes

Torsten.

[Saad Saleem]

Torsten ,

Ah interesting - so like, how will the syntax vary then ?