Author Topic: FE-5  (Read 2688 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2599
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
FE-5
« on: December 18, 2015, 07:46:35 PM »
Consider Laplace equation in the half-strip
\begin{align}
&&&u_{xx} +u_{yy}=0 \qquad  y>0, \ 0 <x< \pi/2 \label{5-1}\\
&\text{with the boundary conditions}\notag\\
&&&u (0,y)=0,\qquad  u(\pi/2,y)=0,\label{5-2}\\
&&&u(x,0)=g(x)\label{5-3}
\end{align}
with $g(x)=\cos(x)$\;  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.
« Last Edit: December 18, 2015, 07:50:20 PM by Victor Ivrii »

Vivian Tan

  • Full Member
  • ***
  • Posts: 15
  • Karma: 0
    • View Profile
Re: FE-5
« Reply #1 on: December 18, 2015, 10:37:20 PM »
a) $\frac{X''}{X}+\frac{Y''}{Y}=0\Rightarrow \frac{X''}{X}=-\lambda,~\frac{Y''}{Y}=\lambda$
The associated eigenvalue problem is in the $x$ component:
$$\frac{X''}{X}=-\lambda,~X(0)=X(\frac{\pi}{2})=0$$

Vivian Tan

  • Full Member
  • ***
  • Posts: 15
  • Karma: 0
    • View Profile
Re: FE-5
« Reply #2 on: December 18, 2015, 10:43:25 PM »
b) There are only strictly positive eigenvalues, so let $\lambda=\omega^2$. Then $X''=-\omega^2 X\Rightarrow X=A\cos(\omega x)+B\sin(\omega x)$. The boundary conditions imply that $A=0$, and $\omega\frac{\pi}{2}=n\pi\Rightarrow\omega=2n$. The eigenfunctions and eigenvalues are:
$$\lambda_n=-4n^2,~X_n (x)=B_n\sin(2nx)$$

Vivian Tan

  • Full Member
  • ***
  • Posts: 15
  • Karma: 0
    • View Profile
Re: FE-5
« Reply #3 on: December 18, 2015, 10:56:24 PM »
$T''=4n^2 T\Rightarrow T=Ce^{2ny}+De^{-2ny}$. However, $y>0$, so for the solution to be bounded we must have $C=0$.
Then the general solution is, after absorbing and redefining some constants:
$$u(x,y)=\sum_{n=1}^{\infty}A_n \sin(2nx)e^{-2ny}$$
$u(x,0)=\cos(x)=\sum_{n=1}^{\infty} A_n\sin(2nx)\Rightarrow A_n=\frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\cos(x)\sin(2nx)dx=\frac{8n}{\pi(4n^2 -1)}$
Therefore the final solution is:
$$u(x,y)=\sum_{n=1}^{\infty}\frac{8n}{\pi(4n^2 -1)}\sin(2nx)e^{-2ny}$$

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2599
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: FE-5
« Reply #4 on: December 21, 2015, 02:32:53 AM »
Correct!

But you don't need to deduct formula for eigenvalues and eigenfunctions, just apply a known formula!


PS. Some students (including those who performed otherwise well) made the following weird statement:
Quote
Since $\cos (x)$ and $\sin (nx)$ are orthogonal
Remember, that $\cos(x)$ and $\sin (nx)$ are orthogonal only on intervals of length $\pi$ and multiple, but not on $[0,\frac{\pi}{2}]$. Conclusion that $u=0$ obviously contradicts to $u(x,0)=\cos(x)$.
« Last Edit: December 21, 2015, 03:47:22 AM by Victor Ivrii »