Author Topic: Bonus Web Problem 1  (Read 4446 times)

Victor Ivrii

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Bonus Web Problem 1
« on: October 09, 2012, 03:49:08 PM »
This is not very difficult problem but it contains one tricky point

Consider heat equation with thermo-conductivity depending on the temperature:
\begin{equation}
u_t= (u^m u_x)_x
\label{eq-1}
\end{equation}
with $m>0$ and find solution(s) $u$ which are self-similar
\begin{equation}
u_\lambda:= \lambda u(\lambda x, \lambda^\gamma t )=u(x,t)\qquad \forall \lambda>0
\label{eq-2}
\end{equation}
and
\begin{equation}
u(x,t)\to 0 \qquad\text{as  } x\to \pm \infty.
\label{eq-3}
\end{equation}
Hint: first find $\gamma$ and then plugging $\lambda =t^{1/\gamma}$ reduce $u$ to a function of one variable and PDE (\ref{eq-1}) to ODE (Follow lecture 9 with the necessary modifications)
« Last Edit: October 20, 2012, 06:19:58 AM by Victor Ivrii »

Zarak Mahmud

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Re: Bonus Web Problem
« Reply #1 on: October 13, 2012, 11:38:31 AM »
Are there any resources or examples of this method apart from what is described in the lecture notes?

I'm a bit confused. In the lecture notes we have

\begin{equation*}
u_{\alpha}(x,t)=\alpha u(\alpha x, \alpha^2 t).
\label{eq-4}
\end{equation*}


but in the question there is no $t$ term


\begin{equation*}
u_{\lambda}:= \lambda u(\lambda x, \lambda^\gamma )=u(x,\lambda)\qquad \forall \lambda>0
\end{equation*}
 
« Last Edit: October 14, 2012, 05:58:46 PM by Zarak Mahmud »

Victor Ivrii

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Re: Bonus Web Problem
« Reply #2 on: October 13, 2012, 02:54:23 PM »
Are there any resources or examples of this method apart from what is described in the lecture notes?

I'm a bit confused. In the lecture notes we have

\begin{equation*}
u_{\alpha}(x,t)=\alpha u(\alpha x, \alpha^2 t).
\end{equation*}

but in the question there is no $t$ term

$$
\begin{equation*}
u_{\lambda}:= \lambda u(\lambda x, \lambda^\gamma )=u(x,\lambda)\qquad \forall \lambda>0
\end{equation*}
 $$

Right, fixed. It was unintentional error--but you should guess how to correct it :)
« Last Edit: October 14, 2012, 06:23:08 PM by Victor Ivrii »

Aida Razi

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Re: Bonus Web Problem
« Reply #3 on: October 14, 2012, 05:55:44 PM »
I was wondering if we need to plug λ=t^(-1/2) to reduce u to a function of one variable,

Victor Ivrii

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Re: Bonus Web Problem
« Reply #4 on: October 14, 2012, 06:29:39 PM »
I was wondering if we need to plug λ=t^(-1/2) to reduce u to a function of one variable,

I don't know, probably not. Why do you think it should be $t^{-\frac{1}{2}}$? -- in the lectures we found that if $u$ satisfies ordinary heat equation then $u_\lambda=\lambda u(\lambda x, \lambda^\gamma t)$ also satisfies heat equation if $\gamma=2$. Then we looked for a solution s.t. $u(x,t)=u_\lambda(x,t)=\lambda u(\lambda x, \lambda^\gamma t)$ (remember, $\gamma=2$)  and finally setting $\lambda = t^{-\frac{1}{2}}$ we eliminated one variable.

You need to deal with suggested problem in the same manner but since equation is non-linear, $\gamma$ could be different (you need to find it out!) and choice of $\lambda$ should differ as well.

Aida Razi

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Re: Bonus Web Problem
« Reply #5 on: October 14, 2012, 08:27:43 PM »
Solution is attached up to converting PDE to ODE,

Aida Razi

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Re: Bonus Web Problem
« Reply #6 on: October 14, 2012, 09:00:42 PM »
Here is the rest of the solution:

Victor Ivrii

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Re: Bonus Web Problem
« Reply #7 on: October 14, 2012, 09:54:12 PM »
AR figured out that for $u_\lambda$ to be a solution if $u$ is, $\gamma=m+2$. Correct!

Then $\lambda = t^{-\frac{1}{\gamma}}$ results in $u(x,t)=t^{-\frac{1}{\gamma}}\phi (x {-\frac{1}{\gamma}})$. Good!

Then plugging into equation transforming it AR got the first equation on page 3 (not sure why you open parenthesis and then collapsed them again)
\begin{equation*}
-\frac{1}{m+2}(\eta \phi)'=(\phi^m \phi')'
\end{equation*}
with $\phi=\phi(\eta)$.--Good!

Integrating and then understanding that $C=0$ for interesting solution AR got
\begin{equation}
-\frac{1}{m+2}\eta \phi =\phi^m \phi'
\tag{A}
\end{equation}

Then AR wrote it as
\begin{equation}
-\frac{1}{m+2}\eta d\eta =\phi^{m-1} d \phi.
\tag{B}
\end{equation}
Looks right.

Now integrating AR got
\begin{equation}
\frac{1}{2(m+2)}(C-\eta ^2) =\frac{1}{m}\phi^m \implies \phi(\eta)=\kappa \bigl(C-\eta ^2)\bigr)^{1/m}
\tag{C}
\end{equation}
with $\kappa= \bigl(\frac{m}{2(m+2)}\bigr)^{1/m}$ which would be the few last lines on page 3. The last line in her Page 3 is definitely wrong due to improper manipulation but  we have a bigger issue here: $\phi(\eta)$ as given by (C) does not tend to $0$ as $\eta \to \infty$; in contrast $C-\eta ^2$ becomes a large negative number and a real-valued $\phi$ may even fail to exist!!!

I warned that there is a tricky spot in this problem--but there is a correct solution (it does exist)

Aida Razi

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Re: Bonus Web Problem
« Reply #8 on: October 14, 2012, 10:30:10 PM »
Thank you for checking my solution.

I am just wondering, do we need to play with m in order to get ϕ(η) as given by (C) tends to 0 as η→∞?

Victor Ivrii

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Re: Bonus Web Problem
« Reply #9 on: October 15, 2012, 04:22:19 AM »
Nobody noticed Looks right. In fact transition from (A) and (B) contains a lost solution $\phi=0$ as AR divides by $\phi$. Sure, trivial solution $\phi=0$ everywhere will be met with an yawn but $\phi=0$ on some interval is a completely different matter. So we need to sew (C) and $\phi=0$ and the only way to do so with $\phi$ decaying at infinity is
\begin{equation}
\phi(\eta)=\kappa \bigl(c^2-\eta ^2\bigr)_+^{1/m} :=
\left\{\begin{aligned}
&\kappa \bigl(c^2-\eta ^2\bigr)_+^{1/m}\qquad && |\eta|\le c,\\
&0 &&|\eta|>c
\end{aligned}\right.
\tag{D}\end{equation}
where $z_+:=z$ as $z\ge 0$ and $z_+:=0$ as $z\le 0$ ($z_+$ is the positive part of $z$). In fact $\phi$ is not just decaying at infinity, it is $0$ for large arguments.

below I plot $\phi$ as $m=2$ (upper half of ellipse) and plugging in $u$ we see that this bubble stretches (squeezes) along $x$ and squeezes (stretches) along $u$ as $t$ decays (grows--respectively).

Solution looks like bubble for any $m>0$ but the bubble looks differently as $0<m<1$ and $m=1$.

« Last Edit: October 15, 2012, 04:29:18 AM by Victor Ivrii »