Solve the first-order wave equation using the central difference scheme (with Matlab code)

$\frac{\partial^2u}{\partial t^2}-\frac{\partial^2u}{\partial x^2}=0,0<x<1,t>0,$
the initial boundary value condition is:
$u\left(0,x\right)=2sin\left(\pi x\right),$
$u_t\left(0,x\right)=0,0<x<1,$
$u\left(t,0\right)=u\left(t,1\right)=0,t\geq0.$
精确解： $u=sin\left(\pi\left(x-t\right)\right)+sin\left(\pi\left(x+t\right)\right)。$
(1) Takeτ $\tau$ =0.05,h=0.1 Calculate the solution of t = 0.5,1.0,1.5,2.0.
(2) Take $\tau$ =h=0.1, calculate the solution of t = 0.5,1.0,1.5,2.0.

untie:

(1) Problem analysis

Step 1: Meshing

Take the time step $\tau=\frac{T}{N}$ and space step size $h=\frac{l}{J}$ , note the rectangular area $\le t \le T, 0 \le x \le l}$ . Use two families of parallel straight lines to divide this area G into a rectangular grid. Grid nodes are represented by coordinate points $\left(t_n,x_j\right)$ means. $G_h$ Represents the set of points within the grid. $\Gamma_h=G/G_h=G-G_h$ Represents a set of boundary points. Use $u\left(t_n,x_j\right)$ represents the true solution, $u_j^n$ Expressed in $\left(t_n,x_j\right)$ is the approximate value. Define the network ratio $r^2=\frac{\tau^2}{h^2}$ 。

Step 2: Difference equation, central difference format

$u_j^{n+1}-2u_j^n+u_j^{n-1}=r^2\left(u_{j+1}^n-2u_j^n+u_{j-1}^n\right)$
$\Rightarrow u_j^{n+1}=r^2\left(u_{j-1}^n+u_{j+1}^n\right)+2\left(1-r^2\right)u_j^n-u_j^{n-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
由初值实作得： $u_j^0=sin\left(\pi x_j\right),\frac{u_j^1 -u_j^{-1}}{2\tau}=0.$
(1)式中令n=0, 得: $u_j^1=r^2\left(u_{j-1}^0+u_{j+1}^0\right)+2\left(1-r^2\right)u_j^0-u_j^{-1}，消去u_j^{-1}$ ，得:
$u_j^1=r^2\left(sin\left(\pi x_{j-1}\right)+sin\left(\pi x_{j+1}\right)\right)+\left(1-r^2\right)sin\left(\pi x_j\right).$
令 ${\ U}^n=\left(u_1^n,u_2^n,\cdots,u_{J-1}^n\right)^T,$ $A=\left[\begin{matrix}2\left(1-r^2\right)&r^2&0&\cdots&0&0&0\\r^2&2\left(1-r^2\right)&r^2&\cdots&0&0&0\\\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&0\\0&0&0&\cdots&r^2&2\left(1-r^2\right)&r^2\\0&0&0&\cdots&0&r^2&2\left(1-r^2\right)\\\end{matrix}\right]$ is a $J - 1 \times J - 1$ ,的单位矩阵，特别地， ${\ U}^0=\left(2sin\left(\pi x_1\right),2sin\left(\pi x_2\right),\cdots,2sin\left(\pi x_{J-1}\right)\right)^T,{\ U}^1=\left(u_1^1,u_2^1,\cdots,u_{J-1}^1\right)^T.$

(2) Computer program

function [U1,E]=shuangqv(J,N)
x0=0;xJ=1;h=(xJ-x0)/J;x=[x0:h:xJ]';
t0=0;tN=2;tau=(tN-t0)/N;t=[t0:tau:tN]';
r=tau^2/h^2;
u(:,1)=2*sin(pi*x(2:J,1));
for j=1:J-1
    u(j,2)=r*(sin(pi*x(j))+sin(pi*x(j+2)))+(1-r)*2*sin(pi*x(j+1));
end
U(:,1)=[u(:,2);u(:,1)];
A=diag(ones(1,J-1))*(2-2*r)+(diag(ones(1,J-2),1)+diag(ones(1,J-2),-1))*r;
E=diag(ones(1,J-1));
B=[A,-E;E,zeros(J-1,J-1)];
for n=2:N+1
    U(:,n)=B*U(:,n-1);
end
U1=U(J:2*(J-1),:);
U1=[zeros(1,N+1);U1;zeros(1,N+1)];
for j=1:J+1
    for n=1:N+1
        V(j,n)=sin(pi*(x(j)-t(n)))+sin(pi*(x(j)+t(n)));
    end
end
E=max(max(abs(U1-V)));
end
 %调试
clear all;clc;close  all;
[U1,E1]=shuangqv(10,40);
[U2,E2]=shuangqv(10,20);

(3) Question results

Insert image description here ( $\year$ =h=0.1, the images of the analytical solution and the numerical solution)
It can be seen from the image that the images of the analytical solution and the numerical solution are highly consistent. Next we look at the maximum error in the two cases:

	maximum error
$\year$ =0.05,h=0.1	0.029747384249724
$\year$ =0.1,h=0.1	1.665334536937735e-15

( $\year$ Numerical solution corresponding to the corresponding time when $τ$
Insert image description here
=0.05, h=0.1) ( $\tau$ Numerical solution corresponding to the corresponding time when $τ =0.1,h=0.1)$