Author Topic: Chapter 3.2 Problem 9  (Read 384 times)

Zicheng Ding

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Chapter 3.2 Problem 9
« on: February 06, 2022, 12:45:35 PM »
For question 9 we have $u = - 2xt - x^2$ as a solution for $u_t = xu_{xx}$, and I found the maximum in the closed rectangle {$-2 \leq x \leq 2$, $0 \leq t \leq 1$} at $(x,t) = (-1, 1)$ on the boundary. I notice that at the maximum we have $u_t > 0$ and $u_{xx} < 0$ but since we have an $x$ in the equation, $u_t = xu_{xx}$ is still satisfied. In the proof of the maximum principle with $v = u - \varepsilon t$, the solution for this question also seems valid, so I am a little confused about where in the proof of maximum principle actually breaks down in this example.

Victor Ivrii

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Re: Chapter 3.2 Problem 9
« Reply #1 on: February 06, 2022, 01:25:46 PM »
You almost there. Think!

Zicheng Ding

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Re: Chapter 3.2 Problem 9
« Reply #2 on: February 06, 2022, 02:19:29 PM »
I am thinking that if we assume maximum is not on the boundary, then in the initially with $u_t = ku_{xx}$ we will have $u_t - ku_{xx} < 0$ as a contradiction, but in this case we will not necessarily have $u_t - xu_{xx} < 0$ since $x$ can switch signs, so we no longer have the contradiction and that breaks down the proof.

Victor Ivrii

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Re: Chapter 3.2 Problem 9
« Reply #3 on: February 06, 2022, 02:21:30 PM »
Indeed. You got it!