Fourier transform of laplace operator

  • Thread starter LuisVela
  • Start date
33
0
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:

[tex](4\pi^2\xi^2+k^2)^2\hat{u}=1[/tex]

...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:
649
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.
 
33
0
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.??
 
2,956
5
The meaning of [itex]\Delta^{2} \, \varphi[/itex] is:

[tex]
\Delta(\Delta \, \varphi) =
[/tex]
[tex]
= \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)
[/tex]
[tex]
+ \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)
[/tex]
[tex]
+ \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) =
[/tex]
[tex]
= \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}}
[/tex]
 
33
0
So it is the biharmonic operator indeed !
Thanks a loot
 

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