- #1
Incand
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Homework Statement
Solve the problem
\begin{cases}
u_{xx} + u_{yy} = y, \; \; 0 < x < 2, \; \; 0 < y < 1\\
u(x,0)=0, \; \; u(x,1)=0\\
u(0,y)=y-y^3, \; \; y(2,y)=0.
\end{cases}
Homework Equations
N/A
The Attempt at a Solution
I start by trying to find a steady-state solution ##u_0(y)## (or I guess a solution independent of ##x## since we don't have a time variable). We have
##u_0''=y \Longrightarrow y = \frac{y^3}{6}+ay+b##. The boundary values ##u(x,0)=0, \; \; u(x,1)=0## give us that ##u_0 = \frac{y^3-y}{6}##.
Setting ##u = v+u_0## we end up with the new homogeneous problem
\begin{cases}v_{xx}+v_{yy}=0\\
v(x,0) = 0, \; \; v(x,1)=0\\
v(0,y) = \frac{7}{6}(y-y^3), \; \; v(2,y)=\frac{y-y^3}{6}\end{cases}
Separation of variables give us
##X(x) = c_1\cosh \lambda x + c_2 \sinh \lambda x## and
##Y(y) = c_3\cos \lambda y + c_3\sin \lambda y##.
The boundary conditions ##v(x,0) = 0, \; \; v(x,1)=0## give us that ##c_3=0## and ##\lambda = n\pi##. We then have solutions of the form
##v(x,y) = \sum_1^\infty \left( a_n\cosh n\pi x + b_n \sinh n\pi x \right) \sin n\pi y##.
Imposing the boundary condition ##v(0,y) = \frac{7}{6}(y-y^3)## we have that
##\frac{7}{6}(y-y^3) = \sum_1^\infty a_n \sin n\pi y##. Calculating the Fourier coefficients we have that
##a_n = \frac{7}{3}\int_0^1 (y-y^3)\sin n\pi y dy = \frac{14 (-1)^{n-1}}{\pi^3 n^3}##.
So we have
##v(x,y) = \sum_1^\infty \left( \frac{14 (-1)^{n-1}}{\pi^3 n^3}\cosh n\pi x + b_n \sinh n\pi x \right) \sin n\pi y##. Imposing the other boundary condition ##v(2,y)=\frac{y-y^3}{6}## we get
##\frac{y-y^3}{6} =\sum_1^\infty \left( \frac{14 (-1)^{n-1}}{\pi^3 n^3}\cosh n\pi 2 + b_n \sinh n\pi 2 \right) \sin n\pi y## at which point I'm stuck not knowing how to proceed.
The answer should be
##u(x,y) = \frac{1}{6}(y^3-y)+\frac{2}{\pi^3}\sum_1^\infty \frac{(-1)^{n-1}}{n^3\sinh 2n \pi} (\sinh n\pi x + 7 \sinh n\pi(2-x) )\sin n\pi y##.
I guess they choose the eigenfunction basis to the Sturm-Liouville problem as
##\sinh n\pi x## and ##\sinh n\pi (2-x)## instead which work just as well. Is there a way I can get the answer in this form or do I have to redo every calculation with these eigenfunctions?