Author Topic: FE-6  (Read 3255 times)

Victor Ivrii

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FE-6
« on: December 18, 2015, 07:49:02 PM »
Consider spherically symmetric solutions of 3D-wave equation
\begin{align}
&&&u_{tt}-u_{rr}-\frac{2}{r} u_r=0, \qquad 0<r<1,\label{6-1}\\
&\text{with boundary conditions } \notag\\
&&&|u(0,t)|<\infty, \qquad u(1,t)=0\label{6-2}\\
&\text{and initial conditions}\notag\\
&&&u(r,0)=1,\qquad u_t(r,0)=0\label{6-3}
\end{align}
and solve by a separation of variables.

Hint. $r  R''+2 R'= (rR)''$.
« Last Edit: December 18, 2015, 09:20:50 PM by Victor Ivrii »

Vivian Tan

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Re: FE-6
« Reply #1 on: December 18, 2015, 11:28:56 PM »
$\frac{T''}{T}=\frac{R''}{R}+\frac{2}{r}\frac{R'}{R}=-\omega^2$. One can check that there are only negative eigenvalues, so I have defined it this way. Let's look at the $R$ equation first
$$rR''+2R'=(rR)''=-\omega^2 rR$$
This is a second order ODE in $rR$, with solution $rR=A\cos(\omega r)+B\sin(\omega r)$. However, for $u$ to be bounded as $r\rightarrow 0$, we must have $A=0$. This also gives some insight as to why there were only negative eigenvalues to begin with: out of all possible functions, only $\frac{\sin(r)}{r}$ does not blow up at the origin. The boundary condition implies that $\omega=n\pi$. Therefore $R=B\frac{\sin(n\pi r)}{r}$.
Then solving the $T$ equation gives $T=C\cos(n\pi t)+D\sin(n\pi t)$, but the second boundary condition forces $D$ to be $0$. The general solution is, after absorbing and redefining constants:
$$u(r,t)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$
$u(r,0)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}=1\Rightarrow\sum_{n=1}^{\infty}A_n\sin(n\pi r)=r\Rightarrow A_n=2\int_{0}^{1}r\sin(n\pi r)dr=\frac{(-1)^{n+1}}{n\pi}$. Finally:
$$u(r,t)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

Bruce Wu

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Re: FE-6
« Reply #2 on: December 18, 2015, 11:48:40 PM »
Actually $$\lim_{r\rightarrow 0}\frac{\sinh(r)}{r}=1$$
So that does not blow up at the origin either. However the problem with that function (corresponding to positive eigenvalues) is that it can never be $0$ at $r\neq 0$, so the boundary conditions can never be satisfied.

Vivian Tan

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Re: FE-6
« Reply #3 on: December 18, 2015, 11:50:18 PM »
Actually $$\lim_{r\rightarrow 0}\frac{\sinh(r)}{r}=1$$
So that does not blow up at the origin either. However the problem with that function (corresponding to positive eigenvalues) is that it can never be $0$ at $r\neq 0$, so the boundary conditions can never be satisfied.

Thank you Bruce Wu, I can't believe I missed that.

Emily Deibert

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Re: FE-6
« Reply #4 on: December 19, 2015, 12:00:58 AM »
Also, in case anyone doesn't remember, note that $$\lim_{r \rightarrow 0} \frac{\sin (r)}{r} = 1$$

Bruce Wu

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Re: FE-6
« Reply #5 on: December 19, 2015, 11:12:28 AM »
Vivian I think you forgot the factor of 2 when computing $A_n$ and so it is missing from your answer. It should be:
$$u(r,t)=\sum^{\infty}_{n=1}\frac{2(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

Victor Ivrii

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Re: FE-6
« Reply #6 on: December 21, 2015, 07:35:46 AM »
Correct.

PS. Some students had trouble to start from finding $rR(r)$ and started looking for $T(t)$ which did not allow to find eigenvalues.