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The Problem.
Consider the initial value problem for the wave equation, $u_{tt} = c^2 u_{xx}$, $x \in \mathbb{R}$, $t > 0$, with initial conditions $u(x,0) = \phi(x), u_t(x,0) = \psi(x)$. The purpose of this exercise is to give an alternative derivation of the d'Alembert solution, based on the method characteristics. (a) Assume that $u = u(x,t)$ solves the problem and let $v = u_t + cu_x$. Show that $v_t - cv_x = 0$. (b) Find $v(x,t)$ expressed by $\phi$ and $\psi$. [Note: the problem continues for several more parts, but I'm already stuck on part (b).]

My Attempt.
I've done part (a). I have no idea what they want for part (b). I recognize that $v_t - cv_x = 0$ is a first order homogeneous PDE and has solution $\eta(x - ct)$ for $\eta$, where, by $\eta$, I mean the initial data of $v$. So I think I'm supposed to find $\eta$ in terms of the initial data for $u$, i.e. in terms of $\phi$ and $\psi$, which will be the desired answer. But, I don't know if that is the right idea, and I don't know how to do that.

Any help is greatly appreciated!

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1 Answers1

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That is almost correct. Setting $v = u_t + c u_x$, the wave equation gives us $v_t = c v_x$, where we have used the equality of mixed partials. Now, a general solution of this PDE for $v$ reads $$ v(x,t) = v_0(x+ct) , $$ where $v(x, 0) = v_0(x)$ is the initial data. Using the initial conditions for $u$, $u_t$, we note that $v_0(x) = \psi(x) + c \phi'(x)$. Thus, $$ v(x,t) = \psi(x+ct) + c\, \phi'(x+ct) . $$ You can substitute back this expression in the PDE problem for verification. FYI, the resolution of the initial-value problem via characteristics is presented in this post, which gives an alternative derivation of the d'Alembert solution.

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