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Poisson's Equation (I think)

  1. Mar 28, 2016 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    ##f_{xx}+f_{yy} = c## where ##c \neq 0## is a constant.

    2. Relevant equations
    None

    3. The attempt at a solution
    I tried separating but obviously this fails since the constant term has mixed variables after separating. I can solve the homogenous equation where ##c=0## but was wondering what happens if ##c \neq 0##.

    I should say this isn't homework but I didn't know where else to put it. I didn't give any boundary conditions because I'm just curious of the technique to solve.
     
  2. jcsd
  3. Mar 28, 2016 #2

    LCKurtz

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    Here's the idea. Let's say you have ##f_{xx}+f_{yy} = h(x)## and at couple of NH boundary conditions ##f(0,y) = a,~f(L,y) = b## (a little more complicated than your example). You would start by letting ##f(x,y) = u(x,y)+\Psi(x)##. Plugging that into the DE and boundary conditions gives$$
    u_{xx} +\Psi''(x) + u_{yy} = h(x)$$ $$
    u(0,y) + \Psi(0) = a,~u(L,y)+\Psi(L) = b$$
    Now, if you can find ##\Psi(x)## satisfying ##\Psi''(x) = h(x),~\Psi(0)=a,~\Psi(L) = b## you would be left with a homogeneous system in ##u##:$$
    u_{xx} + u_{yy} = 0, ~u(0,y) = u(L,y)=0$$And the ##\Psi(x)## equation can be solved by integrating twice giving two constants to take care of the ##a## and ##b##. Then you solve the homogeneous ##u## system and your anwer is ##f(x,y)=u(x,y)+\Psi(x)##.
     
  4. Mar 28, 2016 #3

    joshmccraney

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    Perfect, just what I was looking for!
     
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