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Laplace equation

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data

    I need to solve Laplace equation in the domain D= 0 < x,y < pi

    Neumann boundary conditions are given:


    2. The attempt at a solution

    first, we check that the integral of directional derivative of u on the edge of D is zero. This should be the necessary condition for the existence of a solution to the problem. the condition is satisfied (the integral is zero indeed), but why is this condition sufficient and not only necessary?

    Anyway, assuming a solution does exist, I propose a solution of the form u=X(x)Y(t), and solving the appropriate SL system I find that the solution sould be of the form:
    u(x,y)= A+sigma{Cos(nx)*[Cosh(ny)+Sinh(ny)]}

    but the solution should have the form

    u(x,y)= A+By+sigma{Cos(nx)*[Coshny+Cosh[n(y-pi)]}

    where does the By come from? and does Cosh[n(y-pi)] equal Sinhny? It doesn't make sense because for example if we have y=pi , Cosh[n(y-pi)] = 1, while Sinh(ny) gives an infinite number of answers, depending on n.

    Thanks in advance...:)
  2. jcsd
  3. Apr 21, 2008 #2


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    Science Advisor

    I don't understand this. What makes you think this condition is sufficient?

    The By is because of that "+1" on the du/dy at both y= 0 and y=[itex]\pi[/itex].
  4. Apr 21, 2008 #3
    Well, this is what they write in the answer, that here the condition is "necessary and sufficient"....
    I guess it is not merely a mistake, because this is not the first time I encounter this argument.

    any suggestions regarding the problem in the equation:
  5. Oct 12, 2008 #4
    It's to satisfy the boundary conditions at 0 and pi.
    solutions are u(x,y)=A exp(ny) + Bexp(-ny)
    or =a sinh(ny)+bcosh(ny)
    or =cosh(ny)+cosh n(y-pi) this satisfy thr BC at 0 and pi
    The laplace equation is invariant under translation
    belgium 12
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