### Author Topic: Problem 4  (Read 10192 times)

#### Vitaly Shemet

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##### Problem 4
« on: October 07, 2012, 03:32:35 PM »
Is initial Gaussian centered at $0$? Considering opposite I'm getting $M(T)$ and $m(T)$  neither increasing nor decreasing, what seems suspicious to me. If it is centered then $M(T)$ is decreasing... In other words, is $u(0,0)$ or $u(l,0) = max u(x,t)$ for all $x$

#### Victor Ivrii

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##### Re: Problem 4
« Reply #1 on: October 07, 2012, 04:45:55 PM »
You need to find minima and maxima. Where they are located? -- you need to find this.

#### Thomas Nutz

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##### Re: Problem 4
« Reply #2 on: October 08, 2012, 03:53:35 PM »
I don't think that the maximum in the region $0\leq x \leq l$, $0 \leq t \leq T$ must either decreases or increase; I think it also can stay constant (e.g. iron rod that is initially very hot in the middle, then the maximum is found at t=0, x=l/2) and M(T) =const.

Am I wrong?

#### Victor Ivrii

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##### Re: Problem 4
« Reply #3 on: October 08, 2012, 04:00:48 PM »
I don't think that the maximum in the region $0\leq x \leq l$, $0 \leq t \leq T$ must either decreases or increase; I think it also can stay constant (e.g. iron rod that is initially very hot in the middle, then the maximum is found at t=0, x=l/2) and M(T) =const.

Am I wrong?

You are definitely correct. Since domain increases as $T,L$ grow, then maximum could only increase (or stay the same) and minimum could only decrease (or stay the same). The question is, what happens in the framework of the given problem

#### Jinlong Fu

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##### Re: Problem 4
« Reply #4 on: October 10, 2012, 09:37:12 PM »
q4

#### Victor Ivrii

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##### Re: Problem 4
« Reply #5 on: October 11, 2012, 04:33:52 AM »
Somehow this problem got shortened. Sure, $M(T)$ does not decrease as domain $\{0<x<l, 0<t<T\}$ increases and thus maximum over it can only increase or remain the same. Without boundary conditions we however cannot say anything more.

#### Fanxun Zeng

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##### Re: Problem 4
« Reply #6 on: December 20, 2012, 12:00:07 AM »
I just find by myself online for a well-typed clear solution for Problem 4 and share solution attached
http://www.math.uiuc.edu/~rdeville/teaching/442/hw2S.pdf

#### Victor Ivrii

Actually it contains unnecessary assumption about positivity of initial function $\phi$. Since $u=0$ on the boundary we know that $M(T)\ge 0$ and $m(T)\le 0$ anyway. Further maximum/minimum principle tells that $M(T)=\max \bigl(\max_{0\le x\le l} \phi(x),0\bigr)$ and  $m(T)=\min \bigl(\min_{0\le x\le l} \phi(x),0\bigr)$.