Solutions to the scalar wave equation

Let’s repeat the scalar wave equation from the last section:

$\displaystyle \frac{\partial^2 E_x}{\partial z^2} - \epsilon \mu \frac{\partial^2 E_x}{\partial t^2} = 0$

I’m going to define a new symbol: $v_p = 1 / \sqrt{ \epsilon \mu }$ , which admittedly seems pretty arbitrary, but it’s my book and I can make up symbols if I want. It will make sense later. So if I use my new symbol in the scalar wave equation it becomes

$\displaystyle \frac{\partial^2 E_x}{\partial z^2} - \frac{1}{v_p^2} \frac{\partial^2 E_x}{\partial t^2} = 0$

We’re going to solve this equation using the time-honored technique of just writing down the answer and then proving that it works. The solution is any twice-differentiable function $f$ , as long as $z$ and $t$ are grouped together as follows:

$\displaystyle E_x(z, t) = f(z - v_p t)$

What I mean by ‘twice differentiable’ is that I can take the derivative of $f$ twice without getting into trouble with infinities, so in principle things like step functions are out. Other than that restriction, $f$ can be any function, so we could have (for example)

$\displaystyle f = (z - v_p t)^2$

$\displaystyle f = e^{z - v_p t}$

amongst many other choices. Now, I’ll prove that this solution actually satisfies the scalar wave equation, by substituting it into the scalar wave equation. The derivative of our function $f(\xi)$ is $f'(\xi)$ , and the second derivative of $f(\xi)$ is $f''(\xi)$ . In our case $\xi = (z - v_p t)$ , so we can see that