# Thread: Wave equation: d'Alemberts Method

1. ## Wave equation: d'Alemberts Method

Hi all.

I have two questions.

The first question: Let us say I am given a PDE on this form:

$\displaystyle \frac{d^2}{dx^2}u+\frac{d^2}{dt^2}u=0$

for $\displaystyle 0<x<L$ and $\displaystyle t>0$. Does this mean that the solution u(x,y) is defined on the interval 0<x<L and t>0?

The second question: I wish to solve an initial- and boundary value problem for the wave equation using d'Alembert's method. The initial displacement is zero, and the initial velocity is given by:
$\displaystyle g(x)=x(\pi - x)$

Thus the solution is on the form:

$\displaystyle u(x,t)=\int_{x-ct}^{x+ct}{g^*(z)dz},$
where $\displaystyle g^*$ is the odd extension of $\displaystyle g$. What is the easiest way of solving this? Do I have to find the sine series expansion?

Thanks in advance. I really appreciate any help, since I - at the moment - cannot get help from anywhere else.

Regards,
Niles.

2. That looks like Laplace's equation for the first and I assume you mean the wave equation in 1D over a finite domain for the second right? The answer to the first is yes. For the second, I'll assume its:

$\displaystyle \text{D.E}\quad \frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial^2 u}{\partial x^2},\quad 0\leq x\leq \pi,\quad -\infty<t<\infty$

$\displaystyle \text{B.C}\quad u(0,t)=0\quad u(\pi,t)=0$

$\displaystyle \text{I.C}\quad u(x,0)=0\quad u_t(x,0)=g(x)=x(\pi-x)$

Ok then, the solution is simply:

$\displaystyle u(x,t)=\frac{1}{2a}\int_{x-at}^{x+at} \tilde{g_0}(\tau)d\tau$

where $\displaystyle \tilde{g_0}(x)$ is the odd extension of $\displaystyle g(x)$.

Can you derive the odd extension for $\displaystyle g(x)$ and then plot $\displaystyle u(x,t)$ for say 10 seconds? Tell you what, I'll use $\displaystyle g(x)=e^{x}\sin(x)$ and show what the odd extension for it between -2pi and 2pi looks like as well as the (cryptic) Mathematica code I used to plot it. Now, if I wanted to calculate u(x,t), I'd use those "conditional" expressions in the integral whenever the parameters x-at and x+at reaches each region.

Code:
f[x_] := Which[0 <= x <= Pi, Exp[x]*Sin[x], Pi <= x <= 2*Pi,
(-Exp[2*Pi - x])*Sin[2*Pi - x], -Pi <= x <= 0,
(-Exp[-x])*Sin[-x], -2*Pi <= x <= -Pi,
(-Exp[2*Pi + x])*Sin[2*Pi - x]];
Plot[f[x], {x, -2*Pi, 2*Pi}]

3. First of all, thanks for replying.

You are absolutely correct about #2. I am sorry for not writing it directly.

I took this exercise (d'Alembert) from an old exam, where the student is not allowed to use any electrical devices. In that case, is there no other way of solving it than finding the Fourier sine series?

4. Hey Niles. I don't know. Taking the odd extension in the way I suggested is the only way I know but I'm no expert in the matter.