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Fourier transform of laplace operator

  1. Jul 22, 2010 #1
    Hello Everybody.
    I gave a quick look onto the internet but i couldnt get anything interesting.
    Heres my problem.

    Im solving the differential equation given by:

    [tex](-\Delta+k^2)^2u=\delta [/tex]

    Where [tex]\delta[/tex] is the dirac delta distribuiton (and u is thought as a distribution as well)

    The first step in the book is to apply FT to both sides of the equation...

    The result is:


    ...I do know that the FT of the Laplacian is [tex]-4\pi^2\xi^2[/tex], but when the whole parenthesis is squared, i just can follow it. I dont know how to get that result....

    BTW..whats the meaning of [tex]\Delta^2[/tex]?

    Any ideas?
    Last edited: Jul 22, 2010
  2. jcsd
  3. Jul 23, 2010 #2
    Represent u(x) as a Fourier transform of \hat{u}(xi). Also, write the Dirac delta as a fourier transform integral. Then let the operator parenthesis act inside the integral sign.
  4. Jul 24, 2010 #3
    I think i solved it.
    Fourier acting on laplace squared being equal to the square of F on Laplace is actually easy to prove....
    The only thing remaining to understand is the 'meaning' of laplace squared.
    Laplace acting on a function is an operation from R3-->R1, so you cant apply Laplace over again to the result....
    Is that how we interpret Laplace squared?....or maybe more like the biharmonic operator.??
  5. Jul 24, 2010 #4
    The meaning of [itex]\Delta^{2} \, \varphi[/itex] is:

    \Delta(\Delta \, \varphi) =
    = \frac{\partial^{2}}{\partial x^{2}} \left( \frac{\partial^{2} \, \varphi}{\partial x^{2}} + \frac{\partial^{2} \, \varphi}{\partial y^{2}} + \frac{\partial^{2} \, \varphi}{\partial z^{2}}\right)
    + \frac{\partial^{2}}{\partial y^{2}} \left( \frac{\partial^{2} \, \varphi}{\partial x^{2}} + \frac{\partial^{2} \, \varphi}{\partial y^{2}} + \frac{\partial^{2} \, \varphi}{\partial z^{2}}\right)
    + \frac{\partial^{2}}{\partial z^{2}} \left( \frac{\partial^{2} \, \varphi}{\partial x^{2}} + \frac{\partial^{2} \, \varphi}{\partial y^{2}} + \frac{\partial^{2} \, \varphi}{\partial z^{2}}\right) =
    = \frac{\partial^{4} \, \varphi}{\partial x^{4}} + \frac{\partial^{4} \, \varphi}{\partial y^{4}} + \frac{\partial^{4} \, \varphi}{\partial z^{4}} + 2 \, \frac{\partial^{4} \, \varphi}{\partial x^{2} \, \partial y^{2}} + 2 \, \frac{\partial^{4} \, \varphi}{\partial y^{2} \, \partial z^{2}} + 2 \, \frac{\partial^{4} \, \varphi}{\partial z^{2} \, \partial x^{2}}
  6. Jul 24, 2010 #5
    So it is the biharmonic operator indeed !
    Thanks a loot
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