Hyperbolic Equations

Hyperbolic equations have an unknown characterized by a second derivative with
respect to time, such as the wave equation:

$$\frac{{\partial }^{2}u}{\partial t^2}=a^2\frac{{\partial }^{2}u}{\partial x^2}$$

where:

• $u$ is the wave amplitude
• $x$ is position
• $t$ is time
• $a$ is a constant. For the 1D wave equation, $a$ is replaced with $c$ and is the wave speed

The 1D wave equation is:

$$\begin{array}{rl}\frac{{\partial }^{2}u}{\partial t^2}& =c^2\frac{{\partial }^{2}u}{\partial x^2}\\ \frac{{\partial }^{2}u}{\partial t^2}-c^2\frac{{\partial }^{2}u}{\partial x^2}& =0\end{array}$$

With initial conditions:

• Shape of wave: $u(x,0)=f(x);a\le x\le b$
• Velocity of wave: $\frac{\partial u}{\partial t}(x,0)=g(x);a\le x\le b$

Boundary conditions which represent the wave amplitudes at $x$ boundaries:

• $u(a,t)=l(t);t\ge 0$
• $u(b,t)=r(t);t\ge 0$

Notation

Subscript $i$ is for $x$ (displacement) h is the step size in the spatial coordinate
Subscript $j$ is for $t$ (time), $k$ is the step size in time
This is a 2D problem with grid points $x_i$ and $t$.

We start by replacing the second derivative with the centered difference. Recall that:

$$f^{\prime \prime }(x_i)=\frac{f(x_{i+1})-2f(x_i+f(x_{i-1}))}{h^2},O(h^2)$$

So the substitution is:

$$\frac{u_{i,j+1}-2u_{i,j}+u_{i,j-1}}{k^2}-c^2\frac{u_{i-1}-2u_{i,j}+u_{i+1},j}{h^2}=0$$

We let $\sigma =\frac{ck}{h}$. Rearranging for u at the next time point ($j+1$):

$$u_{i,j+1}=(2-2{\sigma }^2)u_{i,j}+{\sigma }^2{u}_{i-1,j}+{\sigma }^2{u}_{i+1,j}-u_{i,j-1}$$

However, we require values at two prior time steps: $j$ and$j-1$, to solve for $j+1$.

To address this challenge, we can approximate the first partial derivative of $u$ with respect to time (subscript $j$) as:

$$\frac{\partial u}{\partial t}\approx \frac{u_{i,j+1}-u_{i,j-1}}{2k}$$

And using our initial condition:

$$\frac{\partial u}{\partial t}(x_i,t_0)=g(x_i)\approx \frac{u_{i,1}-u_{i,-1}}{2k},j=0\text{ for IC}$$

Rearranging:

$$u_{i,-1}\approx u_{i,1}-2kg(x_i)$$

Substituting the above into the original finite difference formulation:

$$u_{i,1}=(1-{\sigma }^2)u_{i,0}+kg(x_i)+\frac{{\sigma }^2}{2}(u_{i-1,0}+u_{i+1,0})$$

This formula is for the first time step ($j=1$) and includes the initial velocity g. For all later time steps, we use:

$$u_{i,j+1}=(2-2{\sigma }^2)u_{i,j}+{\sigma }^2{u}_{i-1,j}+{\sigma }^2{u}_{i+1,j}-u_{i,j-1}$$

Since we have used 2nd order formulas for both space and time derivatives, the accuracy of the calculation will be $O(h^2)+O(k^2)$.

We use the finite difference method. Define:

$$A=\left[\begin{array}{ccccc}2-2{\sigma }^2& {\sigma }^2& 0& \dots & 0\\ {\sigma }^2& 2-2{\sigma }^2& {\sigma }^2& \ddots & ⋮\\ 0& {\sigma }^2& 2-2{\sigma }^2& \ddots & 0\\ ⋮& \ddots & \ddots & \ddots & {\sigma }^2\\ 0& \dots & 0& {\sigma }^2& 2-2{\sigma }^2\end{array}\right]$$

Then we can write the initial equation for the next time step $j=1$:

$$u_{i,1}=(1-{\sigma }^2)u_{i,0}+kg(x_i)+\frac{{\sigma }^2}{2}(u_{i-1,0}+u_{i+1,0})$$

in matrix form as:

$$\left[\begin{array}{c}{w}_{1,1}\\ ⋮\\ w_{m,1}\end{array}\right]=\frac{1}{2}A\left[\begin{array}{c}{w}_{1,0}\\ ⋮\\ w_{m,0}\end{array}\right]+k\left[\begin{array}{c}g(x_1)\\ ⋮\\ g(x_m)\end{array}\right]+\frac{1}{2}{\sigma }^2\left[\begin{array}{c}{w}_{0,0}\\ 0\\ ⋮\\ 0\\ w_{m+1,0}\end{array}\right]$$

Note that $i$ goes from $1$ to $m$, based on the spatial step size $h$. This vector is $m$ elements long, all zeros except for the first and last elements.

The subsequent time steps are:

$$\left[\begin{array}{c}{w}_{1,j+1}\\ ⋮\\ w_{m,j+1}\end{array}\right]=A\left[\begin{array}{c}{w}_{1,j}\\ ⋮\\ w_{m,j}\end{array}\right]-\left[\begin{array}{c}{w}_{1,j-1}\\ ⋮\\ g(x_m)\end{array}\right]+{\sigma }^2\left[\begin{array}{c}{w}_{0,j}\\ 0\\ ⋮\\ 0\\ w_{m+1,j}\end{array}\right]$$

Including initial boundary conditions:

$$\left[\begin{array}{c}{w}_{1,1}\\ ⋮\\ w_{m,1}\end{array}\right]=\frac{1}{2}A\left[\begin{array}{c}f(x_1)\\ ⋮\\ f(x_m)\end{array}\right]+k\left[\begin{array}{c}g(x_1)\\ ⋮\\ g(x_m)\end{array}\right]+\frac{1}{2}{\sigma }^2\left[\begin{array}{c}l(t_0)\\ 0\\ ⋮\\ 0\\ r(t_0)\end{array}\right]$$

Subsequent time steps are:

$$\left[\begin{array}{c}{w}_{1,j+1}\\ ⋮\\ w_{m,j+1}\end{array}\right]=A\left[\begin{array}{c}{w}_{1,j}\\ ⋮\\ w_{m,j}\end{array}\right]-\left[\begin{array}{c}{w}_{1,j-1}\\ ⋮\\ g(x_m)\end{array}\right]+{\sigma }^2\left[\begin{array}{c}l(t_j)\\ 0\\ ⋮\\ 0\\ r(t_j)\end{array}\right]$$