Author Topic: FE-P5  (Read 1098 times)

Victor Ivrii

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FE-P5
« on: April 11, 2018, 02:43:36 PM »
 Consider Laplace equation in the half-strip
\begin{align}
&u_{xx} +u_{yy}=0 \qquad  y>0, \ 0 < x< \frac{\pi}{2} \label{5-1}\\
&u_x  (0,y)=u_x(\frac{\pi}{2}, y)=0,\label{5-2}\\
&(u_y-u)(x,0)=g(x)\label{5-3}
\end{align}
with $g(x)=1$  and condition $\max |u|<\infty$.


a Write the associated eigenvalue problem.
b  Find all  eigenvalues and corresponding eigenfunctions.
c  Write the solution in the form of  a series expansion.

George Lu

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Re: FE-P5
« Reply #1 on: April 11, 2018, 08:11:27 PM »
I shall expand upon this post later with the full list of steps I took to cancel out all my eigenfunctions and eigenvalues. For the moment, my final answer was u=-1, which is bounded and satisfies all boundary conditions (trivially, $u_x$ and $u_y$ are 0 as needed).

Update (apologies for the comparatively short response, I'm not feeling great at the moment): so we first write $u=XY$ so that the problem becomes $\frac{Y''}{Y}+\frac{X''}{X}=0$. Condition (2) implies periodicity in x, so we have associated problems $X''+\lambda X = 0$, and $Y''-\lambda Y=0$. Then $X=A_n \cos(\omega_n x) + B_n \sin(\omega_n x)$, but we can set $B_n=0$ immediately from (2). Additionally, $X'(0)=X'(\frac{\pi}{2}=0) $ so $\omega_n=2n, \lambda_n=4n^2$. Then  $Y = A_n e^{-2ny} + B_n e^{2ny} = A_n e^{-2ny} + 0$ , because u is bounded on $y>0$

So $u=A_0 + \sum_{n=1}^\infty e^{-2ny} \cos (2nx) \rightarrow (u_y-u)(x,0)=-A_0 - \sum_{n=1}^\infty (2n+1) A_n \cos (2nx) = 1 \rightarrow A_n(2n+1) = \frac{4}{\pi}\int_0^\frac{\pi}{2}\cos(2nx)$

In fact, this integral is 0 for all n. Therefore, we are just left with the condition $-A_0=1 \rightarrow u=-1$
« Last Edit: April 12, 2018, 12:20:05 PM by George Lu »

Victor Ivrii

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Re: FE-P5
« Reply #2 on: April 13, 2018, 12:20:38 PM »
General comment

This problem was included occasionally: I planned b.c.
\begin{equation}
u|_{x=0}=u|_{x=\frac{\pi}{2}}=0.
\label{5-X}
\end{equation}
 As a result the problem became really easy but also treacherous.

Typical errors:
* (The most widespread) Wrongly solve  problem for $X_n$: $X_n=\cos (2nx)$ $n=1,2,\ldots$ but forgetting $X_0=\frac{1}{2}$; as a result
$$
u(x,y)={\color{blue}{\frac{A_0}{2}}} + \sum_{n=1}^\infty A_ne^{-2ny}\cos (2nx);
$$
where highlighted term is missing. Then
$$
{\color{blue}{-\frac{A_0}{2}}}- \sum _{n=1}^\infty (2n+1)\cos(2nx)=1.
$$
This implies $A_n=0$ for $n=1,2,\ldots$ and $u(x,y)=0$ rather than correct $u(x,y)=-1$.

Remark: Basically one does not need to calculate $A_n$ here, since $g(x)=1$ is already a decomposition of $g(x)$ into $\cos$-Fourier series.

* Wrongly solved problem for $X_n$: $X_n=\sin (2nx)$...
« Last Edit: April 13, 2018, 01:19:16 PM by Victor Ivrii »

Victor Ivrii

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Re: FE-P5
« Reply #3 on: April 13, 2018, 01:18:03 PM »
Consolation prize

Solve this problem but with (\ref{5-2}) replaced by (\ref{5-X})

Andrew Hardy

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Re: FE-P5
« Reply #4 on: April 13, 2018, 04:03:03 PM »
Begin via separation of variables.
$$ X''+\lambda X = 0 $$
$$ Y''-\lambda Y=0 $$
with this the solution to the  new Dirichlet Boundary conditions gives  $ \lambda = n^2  $  $n = 1,2,... $
so
$$  X= A_n \sin(2nx) $$ where we don't have any constants that might be forgotten...
Now the solution to the Y ODE are exponential, where we discard the infinite terms so our general solution is of the form
$$ u(x,y)=\sum_{n=1}^\infty B_ne^{-2ny}\sin(2nx) $$ Then the Robin Boundary Condition gives us
$$ - \sum _{n=1}^\infty B_n (2n+1)\sin(2nx)=1. $$ This allows us to solve the traditional method that
 $$ B_n = -\int_{0}^{\frac{\pi}{2}}  (2n+1)^{-1}\sin(2nx)  \text{dx} $$ Which has the nice solution that
$$ = B_n ((2n(2n+1))^{-1}(\cos(n\pi) - 1)$$
$$ B_n = -(n(2n+1))^{-1} $$ for n odd

and therefore we have what would have been a much nicer solution of
$$ u(x,y)=-\sum_{n \text{  odd}}^\infty  (n(2n+1))^{-1} e^{-2ny}\sin(2nx)  $$

Victor Ivrii

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Re: FE-P5
« Reply #5 on: April 13, 2018, 09:13:38 PM »
OK