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I have gathered in my reading that there is a difference between the behavior of the wave equation depending on whether the initial position or initial velocity is zero, but my impression is not precise enough to be very clear to me.

If we have the one dimensional wave equation $u_{tt} = u_{xx}$ with $x \in \mathbb{R}$ and $t \geq 0$, with initial conditions \begin{equation} u(x,0) = f(x) \\ u_t(x,0) = g(x) \end{equation}

As I understand it, if $f(x) = 0$ for all $x$ (and $g(x)$ not zero), then the wave will propagate in one direction, whereas if $g(x) = 0$ (and $f(x)$ not zero), then the disturbance splits in half and goes both directions. This latter case can be seen by looking at d'Alembert's Formula. Though I don't have a good reference for the former case.

Can somebody help make this more precise, confirm my understanding, or point to a reference where this particular matter is treated?

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  • $\begingroup$ see my related questions physics.stackexchange.com/q/268740 and physics.stackexchange.com/q/268740 . I am also looking for intuition into the integral term in the D'Alembert formula--and a physical explanation. There are references in both of these questions, but not a complete answer for intuition into the integral term. $\endgroup$
    – user45664
    Jul 29, 2016 at 17:46
  • $\begingroup$ I have added an answer to update this comment. $\endgroup$
    – user45664
    Mar 31, 2020 at 18:44

2 Answers 2

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This is not an answer, but an unsuccessful attempt to reconcile the intuition of d'Alembert's formula with the explanation you've given, as well as an iteration of a cry for help since in my humble opinion there are same gaps in the intuition, which your question calls upon.

The general solution is $$2u(x,t)=f(x-t)+f(x+t)+\int_{x-t}^{x+t}g(\tau)d\tau.$$ As you pointed out, if $g\equiv0$ then the initial wave $f$ splits in two and the solution goes in opposite directions (I imagine a string stretched and let go from rest). This case makes sense.

For the other situation, if $f\equiv0$, then for any $x$, $u(x,t)$ approaches the integral of $g$ as $t$ goes to infinity. This makes no intuitive sense, since straight from the formula we know then that if say, $g$ is any bump function with area 1, then for any $x$, $u(x,t)$ is eventually $1$, which is like saying "if you pluck a string with unit velocity, the entire string will eventually sit one unit in the direction that you've plucked it". This statement of course makes no sense physically...

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  • $\begingroup$ I think the solution to the problem you point out is to note that $g$ and $f$ are functions of a single variable, so if $t$ goes to infinity then you are integrating over the whole real line, and so the disturbance at any given point $x$ is infinitesimally small. $\endgroup$
    – Wapiti
    Dec 12, 2015 at 16:05
  • $\begingroup$ @charlestoncrabb see my related questions physics.stackexchange.com/q/268740 and physics.stackexchange.com/q/267899 . I am also looking for intuition into the integral term in the D'Alembert formula--and a physical explanation. – user45664 22 hours ago $\endgroup$
    – user45664
    Aug 1, 2016 at 16:59
  • $\begingroup$ @charlestoncrabb i think i have a solution to the OP's question-see my answer. Also D'Alemberts formula (the one you give) is the solution for an infinite string with initial conditions but no boundary conditions. So "the entire string will eventually sit one unit in the direction that you've plucked it" is true until the pulse reaches the end of the string and reflects--which is never. $\endgroup$
    – user45664
    Mar 31, 2020 at 18:42
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Update: see

https://www.researchgate.net/publication/340085346

This paper shows that f can be any (reasonable) function. The if g = -df/dx there will be no backward wave--the wave will propagate in one direction only.

In particular see appendix C.

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