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 $p$?

  • $\begingroup$ p gets integrated out $\endgroup$ – garserdt216 Apr 30 '17 at 20:10
  • $\begingroup$ 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? $\endgroup$ – user409800 Apr 30 '17 at 20:13
  • $\begingroup$ 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 $\int_0^1 x dx$. $\endgroup$ – Ian Apr 30 '17 at 20:43

Lets assume we are talking about the 1+1D wave equation describing waves on a string (D'Alembert's original application). In what you (or the teacher) have written, some things have been left out that would help to answer your question. The propagation speed has been set to 1, instead set it to c and the integration limits make more sense. The initial conditions are not given, so lets state them. Including these changes and using different variable names to make things clearer, we can restate D'Alembert's formula:

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.



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