Results 1 to 4 of 4

Math Help - Help with this PDE

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    3

    Help with this PDE

    Hello,

    I'm currently taking a course in partial differential equations. We are currently solving boundary value problems using the method of separation of variables. I'm attemping to find the seperated solutions and have gotten stuck.

    Here is the problem:

    Solve the BVP for the Laplace equation Uxx+Uyy=0, 0<x<pi, 0<y<pi, with boundary conditions u(0,y)=sin(y)+2sin(3y), u(pi,y)=0, u(x,0)=0, and u(x,pi)=0.

    Here is my attempt at solving the problem:

    Assume u(x,y)=X(x)*Y(y)
    Plug into PDE:
    Uxx+Uyy=0 --> X''Y + XY''=0 --> Y''/Y = -X''/X = lamda (I'll refer to lambda from now on as L)
    Thus Y''=LY and X''=-LX
    Plug into BC:
    u(pi,y)=0=X(pi)Y(y) --> X(pi)=0
    u(x,0)=0=X(x)Y(0) --> Y(0)=0
    u(x,pi)=0=X(x)Y(pi) --> Y(pi)=0

    Thus, we have an eigenvalue problem: Y''=LY, Y(0)=0, Y(pi)=0
    The solution to the eigenvalue problem with Dirichlet B.C. in this case is:
    Eigenvalues: Ln=-n^2, n=1,2,3,...
    Eigenfunctions: Yn(y)=sin(ny), n=1,2,3,...

    Now Xn''=-LnXn --> Xn''=n^2*Xn
    The solution to this eigenvalue problem is:
    Xn(x) = A*cosh(nx) + B*sinh(nx), n=1,2,3,...

    Now here comes the trouble... I have to plug the boundary condition of X(pi)=0 into this
    Xn(pi) = 0 = A*cosh(pi*n) + B*sinh(pi*n) , n=1,2,3,...

    In all the other problems I've done, when you solve this type of equation, either A or B ends up equally zero and drops out, but there appears to be an infinite number of solutions to this linear equation, since there is only one equation with two unknowns. Any help is appreciated. Thanks!
    Last edited by policebox; October 1st 2009 at 11:02 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,340
    Thanks
    21
    Quote Originally Posted by policebox View Post
    Hello,

    I'm currently taking a course in partial differential equations. We are currently solving boundary value problems using the method of separation of variables. I'm attemping to find the seperated solutions and have gotten stuck.

    Here is the problem:

    Solve the BVP for the Laplace equation Uxx+Uyy=0, 0<x<pi, 0<y<pi, with boundary conditions u(0,y)=sin(y)+2sin(3y), u(pi,y)=0, u(x,0)=0, and u(x,pi)=0.

    Here is my attempt at solving the problem:

    Assume u(x,y)=X(x)*Y(y)
    Plug into PDE:
    Uxx+Uyy=0 --> X''Y + XY''=0 --> Y''/Y = -X''/X = lamda (I'll refer to lambda from now on as L)
    Thus Y''=LY and X''=-LX
    Plug into BC:
    u(pi,y)=0=X(pi)Y(y) --> X(pi)=0
    u(x,0)=0=X(x)Y(0) --> Y(0)=0
    u(x,pi)=0=X(x)Y(pi) --> Y(pi)=0

    Thus, we have an eigenvalue problem: Y''=LY, Y(0)=0, Y(pi)=0
    The solution to the eigenvalue problem with Dirichlet B.C. in this case is:
    Eigenvalues: Ln=-n^2, n=1,2,3,...
    Eigenfunctions: Yn(y)=sin(ny), n=1,2,3,...

    Now Xn''=-LnXn --> Xn''=n^2*Xn
    The solution to this eigenvalue problem is:
    Xn(x) = A*cosh(nx) + B*sinh(nx), n=1,2,3,...

    Now here comes the trouble... I have to plug the boundary condition of X(pi)=0 into this
    Xn(pi) = 0 = A*cosh(pi*n) + B*sinh(pi*n) , n=1,2,3,...

    In all the other problems I've done, when you solve this type of equation, either A or B ends up equally zero and drops out, but there appears to be an infinite number of solutions to this linear equation, since there is only one equation with two unknowns. Any help is appreciated. Thanks!
    So this gives A = -\frac{B \sinh \pi n}{\cosh \pi n}. Therefoer the solution is

    u(x,t) = B \sinh n x -\frac{B \sinh \pi n}{\cosh \pi n} \cosh n x = \frac{B}{\cosh \pi n} \sinh n (\pi - x).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    3
    I'm not familiar with the identity you used to simplify this expression into one term. Can you provide me with more details of it please?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,340
    Thanks
    21
    Quote Originally Posted by policebox View Post
    I'm not familiar with the identity you used to simplify this expression into one term. Can you provide me with more details of it please?

    Thanks.
    Try this

    Nerd Paradise : Hyperbolic Identities
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum