Author Topic: TT2-P1  (Read 886 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
TT2-P1
« on: November 18, 2015, 08:36:52 PM »
Solve by Fourier method
\begin{align}
& u_{tt}-u_{xx}=0\qquad -\frac{\pi}{2}<x<\frac{\pi}{2},\label{1-1}\\
& u_x|_{x=-\pi/2}=u_x|_{x=\pi/2}=0,\label{1-2}\\
&u| _{t=0}=x^2,\qquad u_t|_{t=0}=0.\label{1-3}
\end{align}

Rong Wei

  • APM346
  • Sr. Member
  • *
  • Posts: 43
  • Karma: 0
    • View Profile
Re: TT2-P1
« Reply #1 on: November 19, 2015, 12:56:19 AM »
 :)

Bruce Wu

  • APM346
  • Sr. Member
  • *
  • Posts: 57
  • Karma: 0
    • View Profile
Re: TT2-P1
« Reply #2 on: November 19, 2015, 01:26:59 AM »
I am not sure if my solution agrees with Rong Wei's. I do not think there was a need to shift coordinates. Btw there was a hint saying to only consider eigenfunctions which are even with respect to x. I will proceed with that knowledge.

By separation of variables, we have $$\frac{X''}{X}=\frac{T''}{T}=-\lambda$$
One can check that there are only positive eigenvalues, so let $\lambda=\omega^2$. Solving the $X$ equation, and only keeping the even term, we have
$$X(x)=A\cos(\omega x)$$
The boundary conditions in $x$ imply that $$X'\left(\pm\frac{\pi}{2}\right)=\mp \omega A\sin\left(\omega\frac{\pi}{2}\right)=0\Rightarrow \omega\frac{\pi}{2}=n\pi\Rightarrow\omega=2n\Rightarrow\lambda=4n^2$$
So we have $$X_{n}(x)=A_n \cos(2nx)$$
Now solving the $T$ equation, we get $$T(t)=B\cos(2nt)+C\sin(2nt)$$
The $T'(0)=0$ boundary condition implies that $C=0$. So
$$T_n(t)=B_n \cos(2nt)$$
The general solution is, after absorbing some constants $$u(x,t)=\frac{1}{2}A_0+\sum_{n=1}^{\infty}A_n\cos(2nx)\cos(2nt)$$
$u(x,0)=x^2$ implies $$A_n=\frac{2}{\pi}\int_{-\pi/2}^{\pi/2}x^2\cos(2nx)dx=\frac{(-1)^n}{n^2}$$
This is not defined for $n=0$, so we have to calculate that term separately
$$A_0=\frac{2}{\pi}\int_{-\pi/2}^{\pi/2}x^2dx=\frac{\pi^2}{6}$$
Therefore the final solution is $$u(x,t)=\frac{\pi^2}{12}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(2nx)\cos(2nt)$$

Rong Wei

  • APM346
  • Sr. Member
  • *
  • Posts: 43
  • Karma: 0
    • View Profile
Re: TT2-P1
« Reply #3 on: November 19, 2015, 01:30:32 AM »
:)
for the last step, m should start from 0, my mistake

Emily Deibert

  • APM346
  • Elder Member
  • *
  • Posts: 100
  • Karma: 0
    • View Profile
Re: TT2-P1
« Reply #4 on: November 19, 2015, 01:35:52 AM »
I got the same answer as Fei Fan Wu, but unfortunately forgot to put the factor of $\frac{1}{2}$ during the test! Oops.

Rong Wei

  • APM346
  • Sr. Member
  • *
  • Posts: 43
  • Karma: 0
    • View Profile
Re: TT2-P1
« Reply #5 on: November 19, 2015, 01:55:36 AM »
I got the same answer as Fei Fan Wu, but unfortunately forgot to put the factor of $\frac{1}{2}$ during the test! Oops.
you sleep so late,  maybe I'm wrong, but anyway, this test is pretty hard  :'(