Author Topic: FE6  (Read 761 times)

Victor Ivrii

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FE6
« on: December 13, 2016, 08:00:15 PM »
Solve as $t>0$
\begin{align}
&u_{tt}-\Delta u  =0, \label{6-1}
\end{align}
with initial conditions
\begin{align}
&u(x,y,z,0)=\left\{\begin{aligned} &1\quad &&r:=\sqrt{x^2+y^2+z^2}<1,\\
&0 &&r\ge 1,\end{aligned}\right.\qquad u_t(x,y,z,0)=0\label{6-2}
\end{align}
and solve by a separation of variables.

Hint. Use spherical coordinates, observe that solution must be spherically symmetric: $u=u(r,t)$ (explain why).
Also, use equality
\begin{equation}
r  u_{rr}+2 u_r= (r u)_{rr}.
\label{6-3}
\end{equation}
« Last Edit: December 13, 2016, 08:36:29 PM by Victor Ivrii »

Victor Ivrii

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Re: FE6
« Reply #1 on: December 18, 2016, 10:10:06 AM »
Wrong solution removed. Please post a correct one

Shaghayegh A

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Re: FE6
« Reply #2 on: December 18, 2016, 02:18:34 PM »
Okay, I just figured this out. I believe this is what you do: we know solution is spherically symmetric by the hint and because the boundary conditions show that the solution has no $\theta$ or $\phi$ dependence. So converting to spherical coordinates, the problem looks like:
$$u_{tt}-u_{rr}-\frac{2}{r} u_r  =0$$
We separate variables: $u(x,t)= T(t) X(x)$. So

$$\frac{T_{tt}}{T}-\frac{R_{rr}}{R}-\frac{2}{r}\frac{R_r}{R}=0 \implies \\
\frac{T_{tt}}{T}=-\lambda \\
\frac{R_{rr}}{R}+\frac{2}{r}\frac{R_r}{R}+\lambda = 0$$where $\lambda \in R$

We know $$T(t)=A\sin(\sqrt\lambda t) +B\cos(\sqrt\lambda t)$$ The initial condition $$u_t(r,0) = 0 \implies T_t(t) = 0 \implies T(t)=B\cos(\sqrt\lambda t)$$

We also have $$\frac{R_{rr}}{R}+\frac{2}{r}\frac{R_r}{R}+\lambda = 0$$ By the hint, we write this as $$(rR)'' = -\lambda (rR)$$ Making a change of variables with $$L(r) =rR(r)$$ we get
$$L_{rr} = -\lambda L \implies L(r)=E\sin(\sqrt\lambda r)+F\cos(\sqrt \lambda r) \implies R(r) = \frac{E\sin(\sqrt\lambda r)}{r}+\frac{F\cos(\sqrt \lambda r)}{r}$$ We rule out the cos term because we get a 1/0 as r approached infinity, which is bad. So
$$R(r) = \frac{E\sin(\sqrt\lambda r)}{r}$$

Now we combine the R(r) and T(t) solutions. We call $n = \lambda$ for the sake of notation.

$$u(r,t)=\sum_{n= 1} ^{\infty} \frac{C_n \cos(nt) \sin(nr)}{r}$$

By the initial condition (equation 2) in the problem, we have $$\begin{align} &u(r,0)=f(r)=\left\{\begin{aligned} &1\quad &&r<1,\\ &0 &&r\ge 1,\end{aligned}\right.\qquad  \end{align}$$

So we have
$$u(r,0)=\sum_{n= 1} ^{\infty} C_n \sin(nr) = r f(r)$$ This is a sine fourier series with $L=\pi$, so
$$C_n = \frac{2}{\pi} \int_0^{\pi} sin(nr) r f(r) dr = \frac{2}{\pi} \int_0^1 sin(nr) r dr $$
We use integration by parts here and get
$$C_n = \frac{2}{\pi} \frac{\sin(n)-n\cos(n)}{n^2}$$
So
$$u(r,t)=\sum_{n= 1} ^{\infty} \frac{\frac{2}{\pi} \frac{\sin(n)-n\cos(n)}{n^2} \cos(nt) \sin(nr)}{r}$$
« Last Edit: December 18, 2016, 10:11:07 PM by Shaghayegh A »

XinYu Zheng

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Re: FE6
« Reply #3 on: December 18, 2016, 08:16:51 PM »
You cannot assume that the eigenvalues are integers. By writing the general solution as such a sum, you are assuming that $u=0$ at $r=\pi$, but this is not given. Indeed, there are no boundary conditions for $r>0$.

Luyu CEN

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Re: FE6
« Reply #4 on: December 19, 2016, 10:56:31 AM »
Using spherical coodinates, equation (1) can be rewritten as
\begin{align*}
&r u_{tt} - (ru_{rr} + 2u_r) = 0\\
&(r u)_{tt} - (r u)_{rr} = 0
\end{align*}
Let
\begin{equation*}
v = ru
\end{equation*}
And we have
\begin{equation}
v_{tt} - v_{rr} = 0
\end{equation}
We translated the boundary conditions as
\begin{align} &v(r,0)=\left\{
\begin{aligned}
&r\quad &&r<1,\\ &0 &&r\ge 1,
\end{aligned}
\right.\qquad v_t(r,0)=0
\end{align}
In addition, we have a boundary condition from the continuity of u:
\begin{equation}
v|_{r=0} = 0
\end{equation}
Since the boundary conditions are spherical symmetric and only depend on $r$ and $t$, we expect the solution to be a function of $r$ and $t$ also. Thus, we separate variables by $v(r,t) = T(t)T(r)$ and we get
\begin{equation}
\frac{T''(t)}{T(t)} = \frac{R''(r)}{R(r)} = -\lambda
\end{equation}
Note this is a Dirichlet-Dirichlet boundary eigenvalue problem for R(r) with l = 1, we have
\begin{align}
&\lambda_{n} = (\pi n)^2
&R_{n} = \sin(n\pi r)
&& n = 1, 2, ...
\end{align}
Plug $\lambda$ back to (7), and since $v_t(r,0)=0$
\begin{equation}
T(t) = A_n \cos(n\pi t)
\end{equation}
Therefore
\begin{equation}
v = \sum_{n=1}^{\infty}A_n \cos(n\pi t) \sin(n\pi r)
\end{equation}
We solve the coefficients $A_n$ by
\begin{equation*}
A_n = 2\int_{0}^{1}r\sin n\pi r\,dr = -\frac{2\cos(n\pi)}{n\pi}
\end{equation*}
When $r<1$,
\begin{align*}
&v = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{2}{n\pi}\cos(n\pi t)\sin(n\pi r)\\
&u = \frac{1}{r}\sum_{n=1}^{\infty} (-1)^{n+1}\frac{2}{n\pi}\cos(n\pi t)\sin(n\pi r)
\end{align*}
When $r\geq 1$
\begin{equation*}
u = 0
\end{equation*}


XinYu Zheng

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Re: FE6
« Reply #5 on: December 19, 2016, 11:14:28 PM »
$R(r)$ does not have a two-sided Dirichlet boundary condition. Such a boundary condition would be something like $v|_{r=0}=v|_{r=1}=0$. In particular, this must be true for all $t$, while in the problem you are only given that $v|_{r=1}=0$ at $t=0$.
I think the best way to solve this problem is to just ignore separation of variables and solve IVP for $v$ using the tools of chapter 2.
« Last Edit: December 19, 2016, 11:17:39 PM by XinYu Zheng »

Victor Ivrii

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Re: FE6
« Reply #6 on: December 20, 2016, 06:22:30 AM »
Indeed, occasionally (remnant of the previous version) misleading "use separation of variables" appeared. Because everybody followed and none got a correct solution, this problem was removed and FE was normalized 90->40.

There is a correct solution:

Solution is spherically symmetric because the problem is. Then
\begin{equation}
u_{tt}- \bigl(u_{rr}+\frac{2}{r}u_r\bigr)=0\qquad r>0, t>0.
\label{6-4A}
\end{equation}
Multiplying by $r$ and using (\ref{6-3}) we arrive to the first equation below:
\begin{align}
&v_{tt}-v_{rr}=0\qquad r>0,\label{6-5}\\
&v(0,t)=0,
\label{6-6}\\
&v(r,0)=g(r)=\left\{\begin{aligned} &r
\quad &&r<1,\\
&0 &&r\ge 1,\end{aligned}\right.  && v_t(r,0)=0.
\label{6-7}
\end{align}

Continuing $g(r)$ as and odd function  $\tilde{g}(r)=\left\{\begin{aligned} &r
\quad &&|r|<1,\\
&0 &&|r|\ge 1,\end{aligned}\right.$ and solving Cauchy problem we get
\begin{equation}
v(r,t)=\frac{1}{2}\bigl( \tilde{g}(r+t)+\tilde{g}(r-t)\bigr)=\left\{\begin{aligned}
&0  &&r>t+1,\\
&\frac{1}{2}(r-t) \qquad&&1-t<r<t+1,\\
&r &&0<r<1-t,\\
&0 && 0< r<t-1
\end{aligned}\right.
\end{equation}
and finally
\begin{equation}
u(r,t)=r^{-1}v(r,t)=\left\{\begin{aligned}
&0  &&r>t+1,\\
&\frac{1}{2r}(r-t) \qquad&&1-t<r<t+1,\\
&1 &&0<r<1-t,\\
&0 && 0< r<t-1
\end{aligned}\right.
\end{equation}

« Last Edit: December 20, 2016, 06:27:00 AM by Victor Ivrii »