Laplace Equation Solved by Method of Separation of Variables

Homework Equations

Assume the solution has a form of:

The Attempt at a Solution

It looks like a sine Fourier series except for the 2c5 term outside of the series, so I'm not sure how to go about solving for the coefficients c5 and c10. Any idea?

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When lambda is zero, X(x) is identically zero, which means X(x)Y(y) is also zero. So there must be nothing in front of the series.

When lambda is zero, X(x) is identically zero, which means X(x)Y(y) is also zero. So there must be nothing in front of the series.
Ahhh, so for there to be a term outside the sum due to lambda=0, you must have non-zero values for both X(x) AND Y(y)? Makes sense! :D

LCKurtz
Homework Helper
Gold Member
I am puzzled by the boundary condition$$\left. \frac{\partial u}{\partial y}\right |_{y=0} = u(x,0)$$Is that supposed to be the same ##u## on both sides? Or is it just another way to say something like$$u_y(x,0) = f(x)$$some arbitrary function ##f##?

I am puzzled by the boundary condition$$\left. \frac{\partial u}{\partial y}\right |_{y=0} = u(x,0)$$Is that supposed to be the same ##u## on both sides? Or is it just another way to say something like$$u_y(x,0) = f(x)$$some arbitrary function ##f##?
It is suppose to be u on both sides; that boundary condition is stating: Y'(0)X(x)=Y(0)X(x) since it was assumed the solution to u(x,y) had the form of X(x)Y(y).

LCKurtz
OK. With that clarification for me, I would just comment about the last three lines. You already know you should have no eigenfunction for ##\lambda = 0##. Your eigenvalues are ##\lambda_n = n\pi##. Your third line from the bottom should read for the eigenfunctions ##Y_n##$$Y_n(y) = n\pi\cosh(n\pi y)+\sinh(n\pi y)$$You don't need a constant multiple in front of them and there shouldn't be an ##x## in front of the ##\cosh## term. Similarly your eigenfunctions for ##X## are$$X_n(x) = \sin(n\pi x)$$ Then you write your potential solution as$$u(x,y) =\sum_{n=1}^\infty c_nX_n(x)Y_n(y)= \sum_{n=1}^\infty c_n\sin(n\pi x)(n\pi\cosh(n\pi y)+\sinh(n\pi y))$$Now you are ready for the Fourier Series solution to the last boundary condition.