Laser Photonics Solutions to the scalar wave equation

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

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

Hopefully you can see that there is another possible solution to the wave equation; namely, $\displaystyle E_x(z, t) = f(z + v_p t)$

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