Laplace's equation on a rectangle with mixed boundary conditions

  1. 1. The problem statement, all variables and given/known data
    Solve Laplace's equation inside the rectangle [itex]0 \le x \le L[/itex], [itex]0 \le y \le H[/itex] with the following boundary conditions

    [tex] u(0,y) = g(y)\text{, } u(L,y) = 0\text{, } u_y(x,0) = 0\text{, and } u(x,H) = 0[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I know that with Dirichlet boundary conditions one can simply superpose 4 solutions to 4 other problems corresponding to one side held fixed and the others held at 0. Can the same technique be generalzed for mixed boundary conditions, like I have above? I don't think so, because when I do that the solution I get for
    [tex] u(0,y) = g(y)\text{, } u(L,y) = 0\text{, } u(x,0) = 0\text{, and } u(x,H) = 0 [/tex]
    does not satisfy [itex]u_y(x,0) = 0[/tex].

    Does anyone have a hint for how I might find solutions which simultaneously satisfy the boundary condition at [itex]u(0,y)\text{ and for }u_y(x,0)[/itex]?
     
  2. jcsd
  3. gabbagabbahey

    gabbagabbahey 5,013
    Homework Helper
    Gold Member

    Why not find the general 2D solution to Laplace's equation, using separation of variables (i.e. [tex]u(x,y) \equiv X(x)Y(y)[/tex])and then substitute your boundary conditions to find the particular solution?
     
  4. HallsofIvy

    HallsofIvy 40,673
    Staff Emeritus
    Science Advisor

    Let v(x,y)= u(x,y)- xg(y)/L

    Then [itex]\nabla^2 v= \nabla^2 u- xg"(y)/L= -xg"(y)/L[/itex] since [itex]\nabla^2 u= 0[/itex].

    The boundary conditions on v are v(0,y)= 0, v(L, y)= g(y)- g(y)= 0, vy(x, 0)= -xg'(0)/L, v(x,H)= -xg(H)/L.

    Because the boundary conditions on x are both 0, you can write v as a Fourier sine series:
    [tex]v(x,y)= \sum_{n=1}^\infty A_n(y)sin(n\pi x/L)[/tex]

    You will need to write -xg"(y)/L as a Fourier sine series in x so you can treat g"(y) as a constant.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?