# Biharmonic Operator -Fourier Transform.

• Sep 27th 2012, 03:21 AM
bluesboy91
Biharmonic Operator -Fourier Transform.
Hi guys,

just wondering if someone out there could give me a hand with the following problem.

u(x,y) multiplied by a biharmonic operator =0.

The question is asking us to prove that this can be satisfied by S(x,k) = (S"(x,k)-k^2*S(x,k)) = 0 (which is straight forward enough)

but then to solve the function with respect to the boundary conditions:

u_x(0,y)=0 and u_y(0,y)=f(y) using Fourier Transforms.

I then have to solve the system using the convolution theorem.

if anyone is a bit of a whiz with this sort of thing, could you give me a hand?

id really appreciate it.

Thanks guys.
• Sep 27th 2012, 11:22 AM
TheEmptySet
Re: Biharmonic Operator -Fourier Transform.
Quote:

Originally Posted by bluesboy91
Hi guys,

just wondering if someone out there could give me a hand with the following problem.

u(x,y) multiplied by a biharmonic operator =0.

The question is asking us to prove that this can be satisfied by S(x,k) = (S"(x,k)-k^2*S(x,k)) = 0 (which is straight forward enough)

but then to solve the function with respect to the boundary conditions:

u_x(0,y)=0 and u_y(0,y)=f(y) using Fourier Transforms.

I then have to solve the system using the convolution theorem.

if anyone is a bit of a whiz with this sort of thing, could you give me a hand?

id really appreciate it.

Thanks guys.

Your question is really hard to understand as written. When you say multiplied to do mean acted on?

Biharmonic operator acting on $u(x,y)$ would give the PDE

$\Delta ( \Delta u(x,y))=0 \iff \frac{\partial^4 u}{\partial x^4}+2\frac{\partial^4 u}{\partial x^2 \partial y^2 }+\frac{\partial^4 u}{\partial y^4 } =0$

If you take the fourier transform with respect to $y$ you get the ODE

$\frac{\partial^4 \hat{u}}{\partial x^2}-2k^2\frac{\partial^4 \hat{u}}{\partial x^2 }+k^4\hat{ u} =0$

This is a 4th order ODE in $x$ with charaterisitic equation

$m^4-2k^2+k^4=0 \iff (m-k)^2(m+k)^2$

So the solution will have the form

$\hat{u}(x,k)=(c_1+c_2x)e^{kx}+(c_3+c_4x)e^{-kx}$

I hope this helps, If this is not what you are looking for please clarify.
• Sep 27th 2012, 02:29 PM
bluesboy91
Re: Biharmonic Operator -Fourier Transform.
Hi mate,

Thanks for the help. I'm still unsure of a few things. (that solution did clarify
An initial problem I was facing).

First is,

Using the BC/IC: u_x(0,y)=0 and u_y(0,y)=f(y) how do I transform these conditions (more specifically the second one as it depends on y which is to be transformed) and then use that to find the constants c_i(k)?

Second, after getting that solution, solving the system via the convolution theorem. ( I obviously haven't attempted this one as I can't do the first part).

Thanks heaps!