Author Topic: Problem 4  (Read 5209 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

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Re: Problem 4
« Reply #7 on: December 20, 2012, 01:55:59 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

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  $m(T)=\min \bigl(\min_{0\le x\le l} \phi(x),0\bigr)$.