Solutions to the scalar wave equation

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

 

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

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

or

\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

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

and, by ye olde chain rule,

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

similarly

\displaystyle \frac{\partial^2}{\partial z^2} f(z - v_p t) = f''(z - v_p t)

and

\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

\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

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

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

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