Wave equation: travelling solutions

Question

I know that $f(x-ct)$, $g(x+ct)$ and $f(x-ct)+g(x+ct)$ are all solutions of the wave equation $$ \frac{\partial u}{\partial t^2}=c^2\frac{\partial^2 u}{\partial x^2}. $$ This is easily shown using the chain rule. How do I interpret this as left and right travelling functions? I'm having a hard time getting the picture of what's happening.

Answer 1

Let $f:\mathbb{R} \to \mathbb{R}$ and define $F :\mathbb{R}^2 \to \mathbb{R}$ as $F(x,t)=f(x-ct)$. It will solve the equation. Pick your favourite point, for example $(0,0)$. If you watch the evolution of this point in time, you will see that its coordinates will satisfy $x-ct=0$, i.e. $x=ct$, so it will travel to the right. The argument is similar for the $x+ct$ combination.

Answer 2

if you select a certain value of $f$, say $f_*$ when $x=x_*$ and $t=t_*$and then ask for what other combinations of $x$ and $t$ does $f(x-ct)=f(x_*-ct_*)=f_*$ you'll see that that happens when $x-ct=x_*-ct_*$ ... for that to be true, the relations $t=x/c$ and $t_*=x_*/c$ must hold ... meaning that $f=f_*$ when $x=ct=x_*=ct_*$