How to Solve this “First-order wave equation” using Euler’s time-stepping series to obtain a graph similar to above. Solve this using Runga Kutta time-stepping series also.

% This program describes a moving 1-D wave

% using the finite difference method

clc

close all;

%%

% Initialization

Nx= 480; % x-Grid

dx= 3/480; % Step size

x(:,1)= (0:Nx-1)*dx;

f= 125.7; % Wave number

U(:,1)= -16*(x(:,1)-1/2).^2

% U(:,1) = exp(-16*(x(:,1)-1/2).^2);

U(:,1)= exp(U(:,1));

U(:,1)= U(:,1).*sin(f*x(:,1))

%U(:,1)= exp(-16*(x(:,1)-1/2).^2)%sin*(f*x(:,1))

%U(:,1)= U(:,1).*sin(f*x(:,1))

mpx= (Nx+1)/2; % Mid point of x-axis

% (Mid point of 1 to 3= 2 here)

T= 10; % Total no. of time step

f= 125.7; % Wave number

dt= 0.000625; % time-step

t(:,1)= (0:T-1)*dt;

v= 1; % wave velocity

c=v*(dt/dx); % CFL condition

s1= floor(T/f);

%%

% Initial condition

x=0:.001:3;

u=@(t)exp(-16*(x-(.75*t+.5)).^2).*sin(125.7*x);

plot(x,u(0))

figure

plot(x,u(2))

Answer 1

[SAI SRUJAN]

Hi ayushman,

I understand that you are trying to solve the first-order wave equation using two different numerical time-stepping methods: Euler's method and the Runge-Kutta method.

The Euler method is a explicit time-stepping approach. To update each grid point 'i' at the subsequent time step 'n+1', you can employ the following formula within your code:

% Euler Method
for n = 1:T-1
    for i = 2:Nx
        U(i, n+1) = U(i, n) - c * (U(i, n) - U(i-1, n));
    end
end

The Runge-Kutta method is a more accurate method than Euler's method. The second-order Runge-Kutta method (RK2), also known as the midpoint method, can be applied to the wave equation as illustrated below:

% Runge-Kutta Method (RK2)
for n = 1:T-1
    for i = 2:Nx
        k1 = -c * (U(i, n) - U(i-1, n));
        k2 = -c * (U(i, n) + 0.5 * k1 - U(i-1, n) - 0.5 * k1);
        U(i, n+1) = U(i, n) + k2;
    end
end

Please follow the provided algorithmic outlines and make necessary adjustments for the boundary conditions as you proceed with your implementation.

I hope this helps!