1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Harmonic function in square -PDE

  1. Apr 21, 2016 #1
    Hello, I have the solution of a problem but there's something I don't understand
    1. The problem statement, all variables and given/known data
    Find the harmonic function in the square {0<x<1, 0<y<1} with the boundary conditions

    2. Relevant equations

    3. The attempt at a solution

    We first solve the problem in the square with BC
    u(x,0)=0 ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=y²

    The solution is
    u(x,y)= ∑ An cosh(nπx) sin(nπy) for n=1,2,....

    Then solving the problem in the square with BC
    u1(x,0)=x ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=0

    By separation of variable we solve
    X''=-λX with X'(0)=X'(1)=0
    Here the solution says that λ=(nπ)² n≥0 with eigenfunctions Xo=constant and Xn=cos(nπx) for n>0

    and Y''=λY
    For λ=0, we get Y''=0, solving and applying BC Yo(y)=A(y-1)
    And for n=1,2,... Yn(y)=A(cosh(nπy)-cothan(nπy)*sinh(nπy))
    u2(x,y)=Ao(y-1)+ΣAn (cosh(nπy)-cothan(nπy)*sinh(nπy))*cos(nπx)

    What I don't understant, why in part 1, n starts at 1 and not 0 ?
    And why in part 2 it starts at n=0, I understand that the sum is not defined for n=0 (coth(0)=oo) but how can I know the values that take n when i solve for the functions X and Y ?

  2. jcsd
  3. Apr 21, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In part 2 you have an eigenvalue ##\lambda = 0## with eigenfunctions ##X_0 = 1## and ##Y_0 = y-1##. So your expansion would be$$
    \sum_{n=0}^\infty c_nX_n(x)Y_n(y) = c_0(y-1) + \sum_{n=1}^\infty c_n(...)$$I didn't work through part 1 but my guess is that you don't have an eigenvalue ##\lambda = 0##.

    As an aside, you might find it instructive working the ##Y## equation to assume a form ##Y = C\sinh(n\pi y) + D\sinh(n\pi(1-y))## instead of the ##\{\sinh ,\cosh\}## pair.
  4. Apr 21, 2016 #3


    User Avatar
    Homework Helper

    The solution of [itex]Y'' = 0[/itex] which satisfies [itex]Y(0) = 0[/itex] and [itex]Y(1) = 0[/itex] is [itex]Y(y) = 0[/itex].
  5. Apr 21, 2016 #4
    I get it, thank you !
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Harmonic function in square -PDE
  1. Harmonic Function (Replies: 4)

  2. Harmonic Function (Replies: 27)