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System of second order linear coupled pde with constant coefficient

  1. Oct 11, 2012 #1
    Someone know how to uncouple this system of pde?

    Δu[itex]_{1}[/itex](x) + a u[itex]_{1}[/itex](x) + b u[itex]_{2}[/itex](x) =f(x)
    Δu[itex]_{2}[/itex](x) + c u[itex]_{1}[/itex](x) + d u[itex]_{2}[/itex](x) =g(x)

    a,b,c,d are constant.

    I would like to find a solution in one, two, three dimension, possibily in terms of Green function...someone could help me, please?
     
  2. jcsd
  3. Oct 13, 2012 #2

    Ben Niehoff

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    Let

    [tex]y_1 = u_1 + \lambda_1 u_2, \qquad y_2 = u_1 + \lambda_2 u_2[/tex]
    where [itex]\lambda_1, \lambda_2[/itex] are constants. For the right choice of constants, the equations will separate when written in terms of [itex]y_1, y_2[/itex].

    [itex]\lambda_{1,2}[/itex] will have to solve a quadratic equation that involves a, b, c, d, hence generically you will get two roots.
     
  4. Oct 14, 2012 #3

    HallsofIvy

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    Why do you refer to this as a "PDE" when you have only the single independent variable, x?
     
  5. Oct 14, 2012 #4
    Thank you so much Ben Niehoff!

    Sorry for the notation for the variable x, HallsofIvy, I intended it is a vector
    x[itex]\equiv[/itex](x,y,z)
    and
    Δ[itex]\equiv[/itex][itex]\partial_{xx}+\partial_{yy}+\partial_{zz}[/itex]
     
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