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Solving a 2nd order pde

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    I have a pde,
    16d2u/dxdy + du/dx + du/dy + au = 0 where a is constant.


    2. Relevant equations



    3. The attempt at a solution
    I have tried to solve this pde using the substitutions x=e^t and y=e^s so t=ln(x) and s=ln(y) then finding
    Du/dx= 1/x du/dt and du/dy= 1/y du/ds
    For d2u/dxdy i am unsure if my answer is correct,
    1/xy d2u/dsdt - 1/x du/dt - 1/y du/ds

    When i substitute these into the pde i get 16/xy d2u/dsdt + au = 0
    I could integrate this with respect to s and t but dont think that helps me.

    Am i using the correct method here or is there a method that is better suited to my equation

    Thank you for any help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2011 #2
    Try setting u(x,y) = s(x)t(y) and separating the equation.
     
  4. Oct 25, 2011 #3
    Sorry, not sure i follow, what do you suggest i could set them as?
     
  5. Oct 25, 2011 #4
    If you substitute u(x,y) = s(x)t(y) into your PDE, you'll obtain

    16 (ds/dx)(dt/dy) + (ds/dx) t + s (dt/dy) + a s t = 0

    Rearrange & factor:

    (dt/dy)[16 (ds/dx) + s] = -t [(ds/dx) + a s]

    Separate the variables:

    (dt/dy) / t = -[(ds/dx) + s] / [16 (ds/dx ) + a s]

    The left side is a function of x only & the right side is a function of y only. The only way that can be so is if both sides are equal and constant. So

    (dt/dy) / t = C

    -[(ds/dx) + s] / [16 (ds/dx + a s] = C

    That leaves you with two linear ODEs. C is an eigenvalue that will be determined by the PDE's boundary conditions.

    For more info, check out

    http://mathworld.wolfram.com/SeparationofVariables.html

    or

    http://en.wikipedia.org/wiki/Separation_of_variables
     
    Last edited: Oct 25, 2011
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