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2 dimensional PDE and a higher dimension problem.

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    The 2d PDE

    Assume [tex]f\in S(\mathbb{R}^2)[/tex] (Schwartz space)

    Then solve

    [tex] u_{xx}(x,y) + 2u_{yy}(x,y) + 3u_{x}(x,y) -4u(x,y) = f(x,y) ; (x,y) \in\mathbb{R}^2 [/tex]

    [tex] u_{xxxx}(x,y) - u_{yy}(x,y) + 2u(x,y) = f(x,y) ; (x,y) \in\mathbb{R}^2 [/tex]

    2. Relevant equations

    The relevant equations are a whole course. None specifically come to mind except the fourier transform which is [tex]f^h = \int_{\mathbb{R}^2} f(x) e^-^2^\pi ^i^\theta^.^x dx [/tex]


    3. The attempt at a solution
    I, by all honesty have no idea how to start I've been thinking but can't. In fact that's the one thing I need help in, I think once I pick up some momentum I'll be fine.


    1. The problem statement, all variables and given/known data

    Higher d problem.

    Assume [tex]f\in S(\mathbb{R}^8)[/tex] (Schwartz space)

    Prove That

    [tex]\int_{\mathbb{R}^5} | \int_{\mathbb{R}^8} e^-^2^\pi ^i^(^\theta^,^0^,^0^,^0^).^u f(u) du |^2 d\theta = \int_{\mathbb{R}^5} | \int_{\mathbb{R}^3} f(x,y) dy |^2 dx[/tex]

    2. Relevant equations
    Plancherel's Theorem: [tex]\int |f(x)|^2 dx = \int |f^h(\theta )|^2 dc [/tex]


    3. The attempt at a solution

    Well its easy to see that inner integral on the left hand side becomes [tex] f^h(\theta ,0,0,0) which is = f^h(\theta)[/tex]. From there we can use plancherel to take the h off. But from there I get a bit confused because I'm not exactly sure what's allowed or not.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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