# Fourier transform of laplace operator

1. Jul 22, 2010

### LuisVela

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:

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

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 dont know how to get that result....

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

Any ideas?

Last edited: Jul 22, 2010
2. Jul 23, 2010

### 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.

3. Jul 24, 2010

### 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 cant apply Laplace over again to the result....
Is that how we interpret Laplace squared?....or maybe more like the biharmonic operator.??

4. Jul 24, 2010

### Dickfore

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

$$\Delta(\Delta \, \varphi) =$$
$$= \frac{\partial^{2}}{\partial x^{2}} \left( \frac{\partial^{2} \, \varphi}{\partial x^{2}} + \frac{\partial^{2} \, \varphi}{\partial y^{2}} + \frac{\partial^{2} \, \varphi}{\partial z^{2}}\right)$$
$$+ \frac{\partial^{2}}{\partial y^{2}} \left( \frac{\partial^{2} \, \varphi}{\partial x^{2}} + \frac{\partial^{2} \, \varphi}{\partial y^{2}} + \frac{\partial^{2} \, \varphi}{\partial z^{2}}\right)$$
$$+ \frac{\partial^{2}}{\partial z^{2}} \left( \frac{\partial^{2} \, \varphi}{\partial x^{2}} + \frac{\partial^{2} \, \varphi}{\partial y^{2}} + \frac{\partial^{2} \, \varphi}{\partial z^{2}}\right) =$$
$$= \frac{\partial^{4} \, \varphi}{\partial x^{4}} + \frac{\partial^{4} \, \varphi}{\partial y^{4}} + \frac{\partial^{4} \, \varphi}{\partial z^{4}} + 2 \, \frac{\partial^{4} \, \varphi}{\partial x^{2} \, \partial y^{2}} + 2 \, \frac{\partial^{4} \, \varphi}{\partial y^{2} \, \partial z^{2}} + 2 \, \frac{\partial^{4} \, \varphi}{\partial z^{2} \, \partial x^{2}}$$

5. Jul 24, 2010

### LuisVela

So it is the biharmonic operator indeed !
Thanks a loot