# What do we do with the dummy variable in d'Alembert's wave equation?

As my teacher taught, we have d'Alambert's equation to solve the wave equation, $u_{tt}-u_{xx}=0$: $$u(x,t)=\frac{1}{2}(f(x+t)+f(x-t))+\frac{1}{2}\int_{x-t}^{x+t}g(p)dp$$

What do we do with the dummy variable, $p$? The way I think about this solution, if I give an $x$ and a $t$, I should be able to plug it in. But now we have a function of $p$?

• p gets integrated out – garserdt216 Apr 30 '17 at 20:10
• I realized that, but where do we get a p in the first place? u is supposed to be a function of x and t, right? – user409800 Apr 30 '17 at 20:13
• If you have $x$ and $t$ then you can indeed plug in: for example $u(2,1)=\frac{1}{2} \left ( f(3) + f(1) + \int_1^3 g(p) dp \right )$. That integral is just represented in terms of the variable $p$ but you do not need to "choose a value" of $p$ to evaluate it. Nor do you "choose a value of $x$" to evaluate $\int_0^1 x dx$. – Ian Apr 30 '17 at 20:43

If $\phi(t,x)$ is a solution to the 1+1D wave equation and if the initial conditions $\phi(0,x)$ and $\phi_t(0,x)$ are given, then D'Alembert's Formula for the 1+1D wave equation gives
$$\phi(t,x)= \frac 12[ \phi(0,x-ct)+ \phi(0,x+ct) ]+ \frac1{2c} \int_{x-ct}^{x+ct} \phi_t(0,y)dy .$$
Now it can be seen that your $g(p)$ is actually $\phi_t(0,y)$ which is the initial velocity as a function of $y$. But by observing the limits on the integral it can be seen that $y$ is a dummy variable standing for $x$ which is distance along the string. Then the integral is of the initial velocity as a function of distance from x-ct to x+ct along the string at time zero.
So you can see that a dummy variable $y$ representing distance $x$ is needed to perform the integration, and the results of the integration will be a function of $x$ and $t$ because of the limits on the integral.