Results 1 to 3 of 3

Math Help - Biharmonic Operator -Fourier Transform.

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    Australia
    Posts
    10

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Biharmonic Operator -Fourier Transform.

    Quote Originally Posted by bluesboy91 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2012
    From
    Australia
    Posts
    10

    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 10th 2012, 08:41 AM
  2. fourier transform
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 2nd 2012, 03:07 PM
  3. Laplace transform and Fourier transform what is the different?
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: December 29th 2010, 11:51 PM
  4. Replies: 0
    Last Post: April 23rd 2009, 06:44 AM
  5. from fourier transform to fourier series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2008, 07:35 AM

Search Tags


/mathhelpforum @mathhelpforum