# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment 3 => Topic started by: Vitaly Shemet on October 07, 2012, 03:32:35 PM

Title: Problem 4
Post by: Vitaly Shemet 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 $x$
Title: Re: Problem 4
Post by: Victor Ivrii on October 07, 2012, 04:45:55 PM
You need to find minima and maxima. Where they are located? -- you need to find this.
Title: Re: Problem 4
Post by: Thomas Nutz 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?
Title: Re: Problem 4
Post by: Victor Ivrii 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
Title: Re: Problem 4
Post by: Jinlong Fu on October 10, 2012, 09:37:12 PM
q4
Title: Re: Problem 4
Post by: Victor Ivrii 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.
Title: Re: Problem 4
Post by: Fanxun Zeng 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
Title: Re: Problem 4
Post by: Victor Ivrii 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  $m(T)=\min \bigl(\min_{0\le x\le l} \phi(x),0\bigr)$.