1D wave equation with dirac delta function as an external force.

  • #1
Hey there!
I'm faced with this problem:
http://img7.imageshack.us/img7/4381/25686658nz9.png [Broken]

It's a 1D nonhomogeneous wave equation with a "right hand side" equaling to the dirac delta function in x * a sinusoidal function in t. I have to find its general solution with the constraints:
http://img177.imageshack.us/img177/8083/38983002rq3.png [Broken]

I know that the solution, by D'Alembert's theorem, is equal to a double integral over the external function. I showed this in the original problem.

I don't have a lot of experience with the dirac delta function. I know that integrals over [a,b] of the diract delta function = 1 if 0 is an element of [a,b]. The integral is 0 otherwise.

I tried switching the order of integration. Didn't help much. I don't think that integration by parts helps, either. Can somebody point me in the right direction?

Thanks!
 
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Answers and Replies

  • #2
Just curious with your initial conditions: u(x,0)= ut(x,0) = 0.
Since the initial position and velocity are zero I presume the solution must be
u(x,t) = 0.

Can't be that simple!
 
  • #3
It's not. It's not a homogeneous equation.
 
  • #4
Yes you are right the equation is nonhomogeneous. Silly me. :shy:

Let me try to integrate that delta function.
Make the substitution [tex]\theta=\varsigma-x[/tex].

[tex] Integral = \int_{-c(t-\tau)}^{c(t-\tau)} \delta (\theta +x) d\theta = 1 \\\ if \\\\ |x| < c(t-\tau)[/tex].

Hence
[tex]\\\\\\u(x,t)=\frac{1}{\omega}(1-cos(\omega t))[/tex]. :smile:
 
  • #5
Hmm.... But that function isn't dependent on x. That goes against my geometrical intuition of the problem...
 
  • #6
So the problem is not simple ha! I give up. I thought it is only an integration problem and can be solve quite easily. I'm wrong. Really sorry for the expectation.

Good that you have physical interpretation of a solution. I should have checked my solution first.

Any possibility solving the original wave equation using the Fourier or Laplace transforms? Is your x from -inf to +inf ?
 
  • #7
Yes. x can range from -inf to +inf. But the professor specifically mentioned that no question in the entire course will require Fourier transforms.
 
  • #8
I asked the professor today, and he gave me the hints I needed to figure it out. Thanks anyway guys.
 

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