# Toronto Math Forum

## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 3 => Topic started by: Zicheng Ding on February 06, 2022, 12:45:35 PM

Title: Chapter 3.2 Problem 9
Post by: Zicheng Ding 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.
Title: Re: Chapter 3.2 Problem 9
Post by: Victor Ivrii on February 06, 2022, 01:25:46 PM
You almost there. Think!
Title: Re: Chapter 3.2 Problem 9
Post by: Zicheng Ding 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.
Title: Re: Chapter 3.2 Problem 9
Post by: Victor Ivrii on February 06, 2022, 02:21:30 PM
Indeed. You got it!