# Sketch graph of wave equation

## Homework Statement

A 1D wave function ψ(x,t) satisfies these initial conditions:
ψ(x,0) = 0 for all x

∂ψ/∂t (x,0) is v for -a≤x≤a
0 otherwise
Plot ψ(x,t) as a function of x at time t=a/v.

## The Attempt at a Solution

I know the 1D wave equation is given by d'Alembert's:
$\psi(x,t) = 0.5[\psi(x+vt,0) + \psi(x-vt,0)] + \frac{1}{2v} \int_{x-vt}^{x+vt} \frac{\partial \psi}{\partial t}(x,0) \mathrm dx$

But for this function because ψ(x, 0) = 0 for all x, that simplifies to
$\psi(x,t) = \frac{1}{2v} \int_{x-vt}^{x+vt} \frac{\partial \psi}{\partial t}(x,0) \mathrm dx$

I wanted to sketch ψ(x,t) first but I'm not sure how to evaluate the integral. For -a≤x≤a,
$\frac{\partial \psi}{\partial t}(x,0)$ = v,
But do I still integrate between x+vt and x-vt? Do I sub x=a or x=-a into the limits, or maybe t=0? I;m fairly sure you just integrate v between x+vt and x-vt (w.r.t x).

Then I have to plot the graph at t=a/v. And I'm not given what
$\frac{\partial \psi}{\partial t}(x,\frac{a}{v})$ is, and there's no easy relationship I can spot between the two graphs in terms of for example ψ(x,0) is ψ(x, a/v) translated or rotated etc.

In fact, I can't see how it being t=a/v instead of t=0 would affect the graph at all!

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When you do the integral, you will have a function $\psi(x,t)$ which is of x and t and is valid in [-a,a].