### Author Topic: FE5  (Read 303 times)

#### Victor Ivrii

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##### FE5
« on: December 13, 2016, 07:58:11 PM »
Consider Laplace equation in the half-strip
\begin{align}
&u_{xx} +u_{yy}=0 \qquad  y>0, \ 0 < x< \pi \label{5-1}
\end{align}
with the boundary conditions
\begin{align}
&u  (0,y)=u(\pi, y)=0,\label{5-2}\\
&u_y(x,0)=g(x)\label{5-3}
\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.

#### Sajjan Heerah

• Jr. Member
• Posts: 5
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##### Re: FE5
« Reply #1 on: December 14, 2016, 11:04:32 AM »
Solution attempt for 5

#### Victor Ivrii

• Elder Member
• Posts: 1332
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##### Re: FE5
« Reply #2 on: December 18, 2016, 08:25:07 AM »
Minor errors. Still I expect a correct solution to be posted.

#### Shaghayegh A

• Full Member
• Posts: 21
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##### Re: FE5
« Reply #3 on: December 18, 2016, 12:03:03 PM »
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{5-4}\\ &\frac{Y_{yy}}{Y}=\lambda \label{5-5} \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 $A_n$:

$$A_n=\frac{2}{\pi} \int_0^{\pi} \cos(x) \sin(nx) dx=\frac{2}{2 \pi} \int_0^{\pi} \Bigl(\sin(n-1)x + \sin(n+1)x \Bigr)\;dx \tag{\checkmark}\\ \\ = \frac{-1}{\pi} \left.\frac{\cos(n-1)x}{n-1} \right|_{0}^{\pi} + \left.\frac{\cos(n+1)x)}{n+1} \right|_{0}^{\pi} \implies$$

$A_n = 0$           n is odd
$A_n = \frac{4n}{\pi (n^2-1)}$    n is even

So
$C_n = 0$                  n is odd
$C_n=-\frac{4}{\pi(n^2-1)}$        n is even

So $$u(x,y) = \sum_{n = 1, \;n \;even}^{\infty} -\frac{4}{\pi (n^2-1)} \sin(nx) e^{-n y} \;for \;y>0, 0<x<\pi$$ which can be written as
$$u(x,y) = \sum_{m = 1}^{\infty} -\frac{4}{\pi \Bigl((2m)^2-1 \Bigr)} \sin(2mx) e^{-2m y}, \;\; for \;y>0, 0<x<\pi$$

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
« Last Edit: December 18, 2016, 06:29:15 PM by Victor Ivrii »

#### Ziling Lu

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• Posts: 1
• Karma: 0
##### Re: FE5
« Reply #4 on: December 18, 2016, 12:52:05 PM »
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.

#### Victor Ivrii

• Elder Member
• Posts: 1332
• Karma: 0
##### Re: FE5
« Reply #5 on: December 18, 2016, 01:14:35 PM »
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).
« Last Edit: December 18, 2016, 06:28:46 PM by Victor Ivrii »