Solving a PDE using multiple transforms

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  • #1
hunt_mat
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Main Question or Discussion Point

Suppose I have the PDE:
[tex]
\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}=\frac{1}{c^{2}} \frac{\partial^{2}u}{\partial t^{2}}
[/tex]
with
[tex]
u(0,x,y)=\partial_{t}u(0,x,y)=0
[/tex]
along with [itex]u(t,0,y)=f(y)[/itex] With [itex]x\geqslant 0[/itex]. My initial thoughts were to take the Laplace transform in t and the Fourier transform in y to get:
[tex]
\frac{\partial^{2}\hat{u}_{L}}{\partial x^{2}}-\left( k^{2}+\frac{s^{2}}{c^{2}}\right) \hat{u}_{L}=0
[/tex]
The hat donates the Fourier transform and the subscript L denotes the Laplace transform. Treating this as a standard ODE to obtain:
[tex]
\hat{u}_{L}=Ae^{x\sqrt{k^{2}+s^{2}c^{-2}}}+Be^{-x\sqrt{k^{2}+s^{2}c^{-2}}}
[/tex]
Setting x=0 will show us that:
[tex]
A+B=\hat{f}_{L}
[/tex]
I have no other way of determining the other constant (this comes from a question in a book) and I have no idea how to get another boundary condition. I am also concerned about how to do my contour integral when I compute my inverse Laplace transform.

Any suggestions.
 

Answers and Replies

  • #2
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Any suggestions.
Perhaps consider the behaviour of your solution at large x?
 
  • #3
hunt_mat
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I thought about that but it's a wave equation without damping, so physically that doesn't make sense, however I think that as c is the wave speed that for x>ct, the value should be zero and hence the solution has another condition:
[tex]
u(t,ct,y)=0
[/tex]
I would have to take the Laplace transform of this which I am not too sure how to do.
 
  • #4
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Isn't another common procedure with such problems to consider it on the whole real line of x-values, and then notice that both the PDE and initial condition are symmetric in x? And from this conclude that the solution also has to be symmetric under the transformation x <-> -x ?

In your case this would lead to A=B, no?

Torquil
 
  • #5
hunt_mat
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Normally yes (I think) but the problem specifically says that the region is [itex]x\geqslant 0[/itex], so i don't think that we can apply that argument
 
  • #6
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I suspect that you can. Isn't that the whole point of the so-called "method of images"? Your problem would be the restriction to x >= 0 of the symmetric problem on the whole real line of x-values. I'm not ruling out that I can be wrong, though... :-)

Torquil
 
  • #7
hunt_mat
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True enough, I will have a go and see if I get the right answer. So you suggestion means that [itex]A=B=\hat{f}_{L}/2[/itex] right?
 
  • #8
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I would guess so. I haven't really "gone over" your other math leading up to the expressions with A and B, though.

Btw, I think that my guess about using images assumes that both situations have a unique solution. That is not always the case for all PDEs. I've seen some cases where a unique solution is only obtained after imposing e.g. a restriction on the growth of the solution at large x. If solutions are not unique in your case, you might need to impose such a condition, e.g. large-x behaviour would then imply A=0.

This is like saying that you are searching for solutions within some more restricted function space, which could be bounded functions, L^2 functions, etc.

I don't really remember much about the theorems on uniqueness of solutions for the different PDEs...

Anyway, good luck!
 
  • #9
hunt_mat
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Getting the unknown constants is one problem but taking the inverse transforms in another matter entirely.
 

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