Toronto Math Forum
APM3462016F => APM346Tests => FE => Topic started by: Victor Ivrii on December 13, 2016, 07:58:11 PM

Consider Laplace equation in the halfstrip
\begin{align}
&u_{xx} +u_{yy}=0 \qquad y>0, \ 0 < x< \pi \label{51}
\end{align}
with the boundary conditions
\begin{align}
&u (0,y)=u(\pi, y)=0,\label{52}\\
&u_y(x,0)=g(x)\label{53}
\end{align}
with $g(x)=\cos(x)$ and condition $\max u<\infty$.
 Write the associated eigenvalue problem.
 Find all eigenvalues and corresponding eigenfunctions.
 Write the solution in the form of a series expansion.

Solution attempt for 5

Minor errors. Still I expect a correct solution to be posted.

By separation of variables, we get $u(x,y) = X(x) Y(y)$. Plugging this in $ u_{xx} +u_{yy}=0$ we get
$$\frac{X_{xx}}{X} +\frac{Y_{yy}}{Y}=0 \implies \\ \begin{align}&\frac{X_{xx}}{X}=\lambda,\label{54}\\ &\frac{Y_{yy}}{Y}=\lambda \label{55} \end{align}$$ where $\lambda \in R$.
From the boundary conditions given, we must have
$$\begin{align}&X(0) = 0\\
&X(\pi) = 0\end{align}$$ in order to have a nontrivial solution for u(x,y).
Equations (2)(4) are the associated eigenvalue problem.
We know $$X(x) = A \sin(\sqrt\lambda x)+B\cos(\sqrt\lambda x)$$ but we rule out the cosine term because of the first boundary condition on x ( equation 3). Equation 4 implies $$X(\pi) = A\sin(\sqrt\lambda \pi)= 0 \implies \sqrt\lambda \pi = n \pi, n \in N \implies \lambda = n^2 \implies X(x) = A sin(nx)$$ where $\sin(nx)$ are the X eigenfunctions and $n^2$ is the eigenvalue.
From equation (2), we get $$Y(y) = C e^{\sqrt\lambda y} +De^{\sqrt\lambda y} $$ but we rule out $C e^{\sqrt\lambda y}$ because this reaches infinity as y approaches infinity. We are left with $$Y(y)=De^{n y}$$ where $e^{ny}$ are the y eigenfunctions and $n^2$ is the eigenvalue.
So $$u(x,y) = X(x) Y(y) = \sum_{n = 1}^{\infty} C_n sin(nx) e^{n y}$$
The boundary condition $$u_y(x,0)=\cos(x) \implies cos(x) =\sum_{n = 1}^{\infty} n C_n sin(nx) $$ This is a sine fourier series with L = $\pi$, we call $A_n = n C_n$, and solve for $A_n$:
$$A_n=\frac{2}{\pi} \int_0^{\pi} \cos(x) \sin(nx) dx=\frac{2}{2 \pi} \int_0^{\pi} \Bigl(\sin(n1)x + \sin(n+1)x \Bigr)\;dx \tag{$\checkmark$}\\ \\ = \frac{1}{\pi} \left.\frac{\cos(n1)x}{n1} \right_{0}^{\pi} + \left.\frac{\cos(n+1)x)}{n+1} \right_{0}^{\pi} \implies$$
$A_n = 0$ n is odd
$A_n = \frac{4n}{\pi (n^21)}$ n is even
So
$C_n = 0$ n is odd
$C_n=\frac{4}{\pi(n^21)}$ n is even
So $$ u(x,y) = \sum_{n = 1, \;n \;even}^{\infty} \frac{4}{\pi (n^21)} \sin(nx) e^{n y} \;for \;y>0, 0<x<\pi$$ which can be written as
$$u(x,y) = \sum_{m = 1}^{\infty} \frac{4}{\pi \Bigl((2m)^21 \Bigr)} \sin(2mx) e^{2m y}, \;\; for \;y>0, 0<x<\pi$$
Edit: my final answer has remained the same, except it's written in a different form (I'm not sure what error I've made after Prof. Ivrii's checkmark, perhaps someone can point it out). I've also formatted the latex a bit differently

my picture file is too large to upload.. i have to cut it in pieces and it takes me forever to upload the pic :'( :'( :'(. but i have the same answer as Shaghayegh.

Shaghayegh,
the last correct line is the one I denoted by $(\checkmark)$; I also put ( ) where they should be. Please proceed with the correct solution after.
Ziling Also not completely correct solution
Indeed, both solutions are correct! Sorry... Shaghayegh, unwritten rule: red is mine (admin/moderator).