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

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

I’m going to define a new symbol: $latex 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

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

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 $latex f$, as long as $latex z$ and $latex t$ are grouped together as follows:

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

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

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

or

$latex \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 $latex f(\xi)$ is $latex f'(\xi)$, and the second derivative of $latex f(\xi)$ is $latex f”(\xi)$. In our case $latex \xi = (z – v_p t)$, so we can see that

$latex \displaystyle \frac{\partial}{\partial z} f(z – v_p t) = f'(z – v_p t)$

and, by ye olde chain rule,

$latex \displaystyle \frac{\partial}{\partial t} f(z – v_p t) = -v_p f'(z – v_p t)$

similarly

$latex \displaystyle \frac{\partial^2}{\partial z^2} f(z – v_p t) = f”(z – v_p t)$

and

$latex \displaystyle \frac{\partial^2}{\partial t^2} f(z – v_p t) = v_p^2 f”(z – v_p t)$

So we’re pretty much done proving this solution, because we can see that

$latex \displaystyle \frac{\partial^2}{\partial z^2} f(z – v_p t) = \frac{1}{v_p^2} \frac{\partial^2}{\partial t^2} f(z – v_p t) $

so

$latex \displaystyle \frac{\partial^2}{\partial z^2} f(z – v_p t) – \frac{1}{v_p^2} \frac{\partial^2}{\partial t^2} f(z – v_p t) = 0$

Hopefully you can see that there is another possible solution to the wave equation; namely,

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

In the next section we’ll explore the properties of these solutions.