# Nonhomogeneous wave equation with vanishing initial conditions

• Jul 23rd 2010, 12:21 PM
Mmmm
Nonhomogeneous wave equation with vanishing initial conditions
1. The problem statement, all variables and given/known data

Let u(x,t) be the solution of the following initial value problem for the nonhomogeneous wave equation,

$\displaystyle u_{x_1x_1}+u_{x_2x_2}+u_{x_3x_3}-u_{tt}=f(x_1,x_2,x_3,t)$

$\displaystyle u(x,0)=0$ and $\displaystyle u_t(x,0)=0$

$\displaystyle x\in\Re^3 , t>0$

Use Duhamel's Principle and Kirchoff's formula to show that

$\displaystyle u(x,t)=-\frac{1}{4\pi}\int_{\overline{B}(x,t)}\frac{f(x',t-r)}{r}dx'_1dx'_2dx'_3$

where

$\displaystyle r=\left|x-x'\right|=[(x_1-x'_1)^2+(x_2-x'_2)^2+(x_3-x'_3)^2]^\frac{1}{2}$

and $\displaystyle \overline{B}(x,t)$ is the ball in $\displaystyle \Re^3$ with center at x and radius t.

2. Relevant equations

Duhamel's Principle
Let $\displaystyle v(x,t;\tau)$ be the solution of the associated (to the above initial value problem) "pulse problem"

$\displaystyle v_{x_1x_1}+v_{x_2x_2}+v_{x_3x_3}-v_{tt}=0$

$\displaystyle v(x,\tau;\tau)=0$ and $\displaystyle v_t(x,\tau;\tau)=-f(x,\tau)$
$\displaystyle x\in\Re^3 , t>\tau$

then
$\displaystyle u(x,t)=\int^t_0v(x,t;\tau)d\tau$

Kirchoff's Formula

Suppose $\displaystyle p\in C^k(\Re^3)$ where k is any integer $\displaystyle \geq$2 Then the solution of

$\displaystyle u_{x_1x_1}+u_{x_2x_2}+u_{x_3x_3}-u_{tt}=0$

$\displaystyle u(x,0)=0$ and $\displaystyle u_t(x,0)=p(x)$

$\displaystyle x\in\Re^3 , t>0$

is given by

$\displaystyle \frac{1}{4\pi t}\int_{S(x,t)}p(x')d\sigma_t$

where S(x,t) is the surface of the sphere with radius t and centre at the point x. $\displaystyle d\sigma_t$ is the element of surface on S and x' is the variable point of integration.

3. The attempt at a solution

I should split this into two parts, one for each formula.
first I need to use Kirchoff's formula to find v and then Duhamel's principle to find u.

The problem with using Kirchoff is that the initial conditions are given at t=0 whereas our initial conditions for v are at $\displaystyle t=\tau$

So a transformation into kirchoff's formula to give the required initial conditions, $\displaystyle t'=t-\tau$

so

$\displaystyle v(x,\tau;\tau)=0$ and $\displaystyle v_t(x,\tau;\tau)=-f(x,\tau)$

becomes

$\displaystyle v(x,0;\tau)=?$ and $\displaystyle v_t(x,0;\tau)=-f(x,?)$

which doesn't give me anything because of that parameter in v. The t value and the parameter must be equal to give a known value.

or can I just use anything I like for $\displaystyle \tau$ as it doesn't affect kirchoffs formula?

$\displaystyle v(x,0;0)=0$ and $\displaystyle v_t(x,0;0)=-f(x,0)$

but I can't really see how I would get to the required answer with this, so

$\displaystyle v(x,0;t-\tau)=0$ and $\displaystyle v_t(x,0;t-\tau)=-f(x,t-\tau)$

which is a bit of a desperate attempt to match the format of the given answer.

Can anyone give me a clue as to how to get Kirchoff's formula to work with this? Or am I going in completely the wrong direction?
I'm sure that this is only the first of many sticking points in this question but I thought I'd ask one question at a time...:)