Fourier transform of laplace operator

[LuisVela]

Hello Everybody.
I gave a quick look onto the internet but i couldn't get anything interesting.
Heres my problem.

Im solving the differential equation given by:

(-\Delta+k^2)^2u=\delta

Where \delta 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:

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

...I do know that the FT of the Laplacian is -4\pi^2\xi^2, but when the whole parenthesis is squared, i just can follow it. I don't know how to get that result...

BTW..whats the meaning of \Delta^2?

Any ideas?

 

[torquil]

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.

 

[LuisVela]

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 can't apply Laplace over again to the result...
Is that how we interpret Laplace squared?...or maybe more like the biharmonic operator.??

 

[Dickfore]

The meaning of \Delta^{2} \, \varphi is:

<br /> \Delta(\Delta \, \varphi) =<br />
<br /> = \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)<br />
<br /> + \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)<br />
<br /> + \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) =<br />
<br /> = \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}}<br />

 

[LuisVela]

So it is the biharmonic operator indeed !
Thanks a loot