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PDE from a general solution

  1. Apr 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the PDE for this general solution:
    U(x,y) = Phi(x+y) + Psi(x-2y)


    2. Relevant equations



    3. The attempt at a solution
    I let my xi = x+y and my eta = x-2y and found that both roots are {-1,1/2}. From that I multiplied: (dy/dx - root1)*(dy/dx - root2) to give me the function that will give me my little a,b,c for the quadratic equation. But now I'm stuck and am not sure where to go. Any hints or ideas would be appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 19, 2007 #2

    Dick

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    I don't know what the specific technique you are trying to use is. But one PDE is clearly [tex]\frac{\partial^2 F}{\partial \xi \partial \eta}=0[/tex], right? Now just change variables to x and y.
     
  4. Apr 19, 2007 #3
    Sorry, I should clarify. The PDE that I'm trying to find from the general solution should look like 2U_xx + U_xy - U_yy = 0.
     
  5. Apr 19, 2007 #4

    Dick

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    Ok. Then you just want to factor [tex]2 \partial_x^2+\partial_x \partial_y-\partial_y^2[/tex] into linear factors, right?
     
  6. Apr 19, 2007 #5
    Since this is a hyperbolic pde (discriminant > 0) I would find the characteristics by using a form of the quadratic equation, which would give me two roots, xi and eta. But I'm working backwards; I have the roots but not I need to get back to the hyperbolic pde. Does this help?
     
  7. Apr 19, 2007 #6

    Dick

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    Not exactly. You are basically given xi and eta as functions of x and y, right? I don't see where any quadratic roots come into it. Go back to the PDE in my post #2. All you have to do is change the partial derivatives wrt xi and eta to partial derivatives wrt x and y (use the chain rule) - then you get a PDE in x and y.
     
  8. Apr 19, 2007 #7
    Thanks Dick! I managed to get a similar pde.
     
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