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Solving a PDE using multiple transforms

  1. Aug 8, 2011 #1

    hunt_mat

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    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.
     
  2. jcsd
  3. Aug 9, 2011 #2
    Perhaps consider the behaviour of your solution at large x?
     
  4. Aug 9, 2011 #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.
     
  5. Aug 10, 2011 #4
    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
     
  6. Aug 10, 2011 #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
     
  7. Aug 10, 2011 #6
    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
     
  8. Aug 10, 2011 #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?
     
  9. Aug 10, 2011 #8
    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!
     
  10. Aug 10, 2011 #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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