APM346-2015F > Test 2

TT2-P1

**Victor Ivrii**:

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**:

:)

**Bruce Wu**:

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**:

--- Quote from: Rong Wei on November 19, 2015, 12:56:19 AM --- :)

--- End quote ---

for the last step, m should start from 0, my mistake

**Emily Deibert**:

I got the same answer as Fei Fan Wu, but unfortunately forgot to put the factor of $\frac{1}{2}$ during the test! Oops.

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