2 Numerical solutions

The approach we will adopt is similar to that seen in Section 32.4 where we looked at parabolic equations. We use the notation

$u_j^n$

to denote an approximation to $u$ evaluated at $x=j\times \delta x$ , $t=n\times \delta t$ . Approximating the derivatives in the PDE

$u_{tt}=c^2{u}_x$

by central differences we obtain the numerical difference equation

$\frac{u_j^{n+1}-2u_j^n+u_j^{n-1}}{{\left(\delta t\right)}^2}=c^2\frac{u_{j+1}^n-2u_j^n+u_{j-1}^n}{{\left(\delta x\right)}^2}.$

Multiplying through by ${\left(\delta t\right)}^2$ this can be rearranged to give

$u_j^{n+1}=2u_j^n-u_j^{n-1}+{\mu }^2\left(u_{j+1}^n-2u_j^n+u_{j-1}^n\right)$

in which $\mu =\frac{c\delta t}{\delta x}$ is called the Courant number.

The equation above gives $u_j^{n+1}$ in terms of $u$ -approximations at earlier time-steps (that is, all the appearances of $u$ on the right-hand side have a superscript smaller than $n+1$ ).

Figure 9

Thinking of the numerical stencil graphically we have the situation shown above. We may think of the values on the right-hand side of the equation “pointing to" a new value on the left-hand side.

(We will deal with how to carry out the first time-step shortly.)

The time-stepping process has much in common with the corresponding procedure for parabolic problems. The following Example will help establish the general idea.

Example 18

Given that $u=u\left(x,t\right)$ satisfies the wave equation $u_{tt}=c^2{u}_{xx}$ in $t>0$ and $0<x<1$ with boundary conditions $u\left(0,t\right)=u\left(1,t\right)=0$ $\left(t>0\right)$ with wave speed $c=1.2$ .

The numerical method    $u_j^{n+1}=2u_j^n-u_j^{n-1}+{\mu }^2\left(u_{j+1}^n-2u_j^n+u_{j-1}^n\right)$   where $\mu =c\delta t∕\delta x$ , is implemented using $\delta x=0.25$ and $\delta t=0.1$ .

Suppose that, after 5 time-steps, the following data forms part of the numerical solution:

$\begin{array}{ccccccc}\hfill u_0^4& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill u_0^5& \hfill =\hfill & 0.0000\hfill \\ \hfill u_1^4& \hfill =\hfill & 0.9242\hfill & \hfill \hfill & \hfill u_1^5& \hfill =\hfill & 0.7110\hfill \\ \hfill u_2^4& \hfill =\hfill & -0.0020\hfill & \hfill \hfill & \hfill u_2^5& \hfill =\hfill & -0.0059\hfill \\ \hfill u_3^4& \hfill =\hfill & -0.9624\hfill & \hfill \hfill & \hfill u_3^5& \hfill =\hfill & -0.7409\hfill \\ \hfill u_4^4& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill u_4^5& \hfill =\hfill & 0.0000\hfill \end{array}$

Carry out the next time-step so as to find an approximation to $u$ at $t=6\delta t$ .

Solution

In this case $\mu =1.2\times 0.1∕0.25=0.48$ and the required time-step is carried out as follows:

to 4 decimal places and these are the approximations to $u\left(0,6\delta t\right)$ , $u\left(0.25,6\delta t\right)$ , $u\left(0.5,6\delta t\right)$ , $u\left(0.75,6\delta t\right)$ and $u\left(1,6\delta t\right)$ , respectively.

The diagram below shows the numerical results that appeared in the example above. It can be seen that the example was a (rather coarse) model of a standing wave with two antinodes.

Figure 10

Task!

Suppose that $u=u\left(x,t\right)$ satisfies the wave equation $u_{tt}=c^2{u}_{xx}$ in $t>0$ and $0<x<1$ . It is given that $u$ satisfies boundary conditions $u\left(0,t\right)=u\left(1,t\right)=0$ $\left(t>0\right)$ and initial conditions that need not be stated for the purposes of this question. The application is such that the wave speed $c=1.2$ .

The numerical method $u_j^{n+1}=2u_j^n-u_j^{n-1}+{\mu }^2\left(u_{j+1}^n-2u_j^n+u_{j-1}^n\right)$   where $\mu =c\delta t∕\delta x$ , is implemented using $\delta x=0.25$ and $\delta t=0.2$ .

Suppose that, after 8 time-steps, the following data forms part of the numerical solution:

$\begin{array}{ccccccc}\hfill u_0^7& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill u_0^8& \hfill =\hfill & 0.0000\hfill \\ \hfill u_1^7& \hfill =\hfill & 0.6423\hfill & \hfill \hfill & \hfill u_1^8& \hfill =\hfill & 0.4640\hfill \\ \hfill u_2^7& \hfill =\hfill & 0.8976\hfill & \hfill \hfill & \hfill u_2^8& \hfill =\hfill & 0.6792\hfill \\ \hfill u_3^7& \hfill =\hfill & 0.6789\hfill & \hfill \hfill & \hfill u_3^8& \hfill =\hfill & 0.4668\hfill \\ \hfill u_4^7& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill u_4^8& \hfill =\hfill & 0.0000\hfill \end{array}$

Carry out the next time-step so as to find an approximation to $u$ at $t=9\delta t$ .

Answer

The above Task concerns a stretched string oscillating in such a way that at the $9^{\text{th}}$ time-step the string is approximately flat. The motion continues with $u$ taking negative values. Figure 11 below uses data calculated above, and also data for the next two time-steps so as to show subsequent progress of the solution.

Figure 11