Nonhomogeneous wave equation with vanishing initial conditions

[Mmmm]

Homework Statement

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

[tex]u_{x_1x_1}+u_{x_2x_2}+u_{x_3x_3}-u_{tt}=f(x_1,x_2,x_3,t)[/tex]

[tex]u(x,0)=0[/tex] and [tex]u_t(x,0)=0[/tex]

[tex]x\in\Re^3 , t>0[/tex]

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

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

where

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

and [tex]\overline{B}(x,t)[/itex] is the ball in [itex]\Re^3[/itex] with center at x and radius t.

Homework Equations

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

[tex]v_{x_1x_1}+v_{x_2x_2}+v_{x_3x_3}-v_{tt}=0[/tex]

[tex]v(x,\tau;\tau)=0[/tex] and [tex]v_t(x,\tau;\tau)=-f(x,\tau)[/tex]
[tex]x\in\Re^3 , t>\tau[/tex]

then
[tex]u(x,t)=\int^t_0v(x,t;\tau)d\tau[/tex]


Kirchoff's Formula

Suppose [itex]p\in C^k(\Re^3)[/itex] where k is any integer [itex]\geq[/itex]2 Then the solution of

[tex]u_{x_1x_1}+u_{x_2x_2}+u_{x_3x_3}-u_{tt}=0[/tex]

[tex]u(x,0)=0[/tex] and [tex]u_t(x,0)=p(x)[/tex]

[tex]x\in\Re^3 , t>0[/tex]

is given by

[tex]\frac{1}{4\pi t}\int_{S(x,t)}p(x')d\sigma_t[/tex]

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

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 [itex]t=\tau[/itex]

So a transformation into kirchhoffs formula to give the required initial conditions, [itex]t'=t-\tau[/itex]

so

[tex]v(x,\tau;\tau)=0[/tex] and [tex]v_t(x,\tau;\tau)=-f(x,\tau)[/tex]

becomes

[tex]v(x,0;\tau)=?[/tex] and [tex]v_t(x,0;\tau)=-f(x,?)[/tex]

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.

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...:)

 

[mighty2000]

Thank you for your post. It seems like you are on the right track in using Kirchoff's formula and Duhamel's principle to solve this problem. However, you are correct in noting that the initial conditions for v are not given at t=0, but rather at t=\tau. In order to use Kirchoff's formula, we need to transform the initial conditions to t=0.

One way to do this is by using the transformation t' = t - \tau, as you have mentioned. However, this transformation is not enough to get the desired initial conditions for v. We also need to consider the fact that the initial conditions for u are given at t=0, not t=\tau. So, in order to use Kirchoff's formula, we need to transform the initial conditions for v to t=0 and also transform the initial conditions for u to t=\tau.

To do this, we can use the transformation t'' = t + \tau. This transformation will give us the desired initial conditions for v at t=0, and also transform the initial conditions for u to t=\tau. So, we can rewrite the initial value problem for v as:

v_{x_1x_1}+v_{x_2x_2}+v_{x_3x_3}-v_{tt}=0

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

x\in\Re^3 , t>0

Now, we can use Kirchoff's formula to find v, and then use Duhamel's principle to find u. I hope this helps and good luck with your solution!