# wave equation

• Mar 12th 2010, 03:17 PM
sidi
wave equation
I have to find the solution to this problem:
$u_{xx}=u_{tt}+f(x,t)$
$u(x,0)=0$
$u_t(x,0)=0$

where f(x,t) is a step function:
$1$ for $|x|<= 1$ and $0< t < 1$
$0$ else

I know I have to use the duhamel's principle, but I have problem with the integration of f(x,t).

thank you
• Mar 12th 2010, 10:02 PM
TheEmptySet
Quote:

Originally Posted by sidi
I have to find the solution to this problem:
$u_{xx}=u_{tt}+f(x,t)$
$u(x,0)=0$
$u_t(x,0)=0$

where f(x,t) is a step function:
$1$ for $|x|<= 1$ and $0< t < 1$
$0$ else

I know I have to use the duhamel's principle, but I have problem with the integration of f(x,t).

thank you

So using Duhammel's principle you need to find a $\phi(x,t;s)$
Such that $\phi_{tt}=\phi_{xx}$ with initial conditions
$\phi(x,s;s)=0 \text{ and } \phi_{t}(x,s;s)=f(x,s)$

Now by D'lamberts we know that

$\phi(x,t,;s)=\frac{1}{2}\int_{x-(t-s)}^{x+(t-s)}f(z,s)dz$

Now by Duhammel's principle the solution is

$u(x,t)=\frac{1}{2}\int_{0}^{t}\int_{x-(t-s)}^{x+(t-s)}f(z,s)dz ds$

Note that the region of integration depends on the x and t varaibles. Also the forcing function is non zero in the rectangle $[-1 ,1] \times [0,1]$ (In red in the diagram). In this diagram the function will be zero in any of the shaded (blue) area's and the wave will travel in the white strips.. I hope this helps.
Attachment 15879
• Mar 13th 2010, 02:28 AM
sidi
I still don't know how to find the region of integration....
On your diagram are there the x and t axis? Where is s?