3 The first time-step

In the Example and Task above we have seen how time-steps can be carried out using the numerical stencil

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

but there remains one issue which, so far, we have neglected. How do we carry out the first time-step?

3.1 Initial conditions

The initial time-step must use information from the two initial conditions

$\left.\begin{array}{ccc}\hfill u\left(x,0\right)& \hfill =\hfill & f\left(x\right)\hfill \\ \hfill u_t\left(x,0\right)& \hfill =\hfill & g\left(x\right)\hfill \end{array}\right\}0\le x\le \ell$

The first initial condition is easy enough to interpret. It gives $u_j^n$ in the case where $n=0$ . In fact

$u_j^0=f_j$

where $f_j$ is simply shorthand for $f\left(j\times \delta x\right)$ .

The second initial condition, the one involving $g$ , gives information about $u_t=\frac{\partial u}{\partial t}$ at $t=0$ . We can approximate the $t$ -derivative of $u$ at $t=0$ and $x=j\times \delta x$ by a central difference to write

$\frac{u_j^1-u_j^{-1}}{2\delta t}=g_j$

in which $g_j$ is shorthand for $g\left(j\times \delta x\right)$ .

This last expression involves $u_j^{-1}$ which, if it has a meaning at all, refers to $u$ at time $t=-\delta t$ , that is, before the initial time $t=0$ . One way to think of $u_j^{-1}$ is simply as an artificial quantity which proves useful later on. The equation above, rearranged for $u_j^{-1}$ is

$u_j^{-1}=u_j^1-2\delta t\times g_j$

3.2 The first time-step

To carry out the first time-step we put $n=0$ in the numerical stencil

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

to give

$u_j^1=2u_j^0-u_j^{-1}+{\mu }^2\left(u_{j+1}^0-2u_j^0+u_{j-1}^0\right).$

Those terms on the right-hand side with a 0 superscript are known via the function $f$ since we know that $u_j^0=f_j$ . Hence

$u_j^1=2f_j-u_j^{-1}+{\mu }^2\left(f_{j+1}-2f_j+f_{j-1}\right).$

And the $u_j^{-1}$ term is dealt with using the Key Point above to give

$u_j^1=2f_j-u_j^1+2\delta t\times g_j+{\mu }^2\left(f_{j+1}-2f_j+f_{j-1}\right).$

and therefore, moving the latest appearance of $u_j^1$ over to the left-hand side and dividing by 2,

Example 19

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 may be summarised as

$\begin{array}{ccccccc}\hfill f_0& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill g_0& \hfill =\hfill & 0.0000\hfill \\ \hfill f_1& \hfill =\hfill & 0.6000\hfill & \hfill \hfill & \hfill g_1& \hfill =\hfill & 0.1000\hfill \\ \hfill f_2& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill g_2& \hfill =\hfill & 0.2000\hfill \\ \hfill f_3& \hfill =\hfill & -0.5000\hfill & \hfill \hfill & \hfill g_3& \hfill =\hfill & 0.1000\hfill \\ \hfill f_4& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill g_4& \hfill =\hfill & 0.0000\hfill \end{array}$

The application is such that the wave speed $c=1$ .

Carry out the first two time-steps of 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 $\mu =c\delta t∕\delta x$ in which $\delta x=0.25$ and $\delta t=0.2$ .

Solution

In this case $\mu =1\times 0.2∕0.25=0.8$ and the first time-step is carried out as follows (to 4 d.p.):

The second time-step is as follows (to 4 d.p.):

Task!

Suppose that $u=u\left(x,t\right)$ satisfies the wave equation $u_{tt}=c^2{u}_{xx}$ in $t>0$ and $0<x<0.8$ . It is given that $u$ satisfies boundary conditions $u\left(0,t\right)=u\left(0.8,t\right)=0$ $\left(t>0\right)$ and initial conditions that may be summarised as

$\begin{array}{ccccccc}\hfill f_0& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill g_0& \hfill =\hfill & 0.0000\hfill \\ \hfill f_1& \hfill =\hfill & 0.1703\hfill & \hfill \hfill & \hfill g_1& \hfill =\hfill & 0.4227\hfill \\ \hfill f_2& \hfill =\hfill & 0.2364\hfill & \hfill \hfill & \hfill g_2& \hfill =\hfill & 0.5417\hfill \\ \hfill f_3& \hfill =\hfill & 0.1703\hfill & \hfill \hfill & \hfill g_3& \hfill =\hfill & 0.4227\hfill \\ \hfill f_4& \hfill =\hfill & 0.0000\hfill & \hfill \hfill & \hfill g_4& \hfill =\hfill & 0.0000\hfill \end{array}$

The application is such that the wave speed $c=1$ .

Carry out the first two time-steps of 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 $\mu =c\delta t∕\delta x$ in which $\delta x=0.2$ and $\delta t=0.11$ .

Answer