# Toronto Math Forum

## APM346-2015F => APM346--Home Assignments => HA5 => Topic started by: Yunheng Chen on October 17, 2015, 05:57:08 PM

Title: HA5-P6
Post by: Yunheng Chen on October 17, 2015, 05:57:08 PM
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter3/S3.2.P.html
Problem 6
Title: Re: HA5-P6
Post by: Yunheng Chen on October 17, 2015, 06:12:08 PM
Still working on partC and i will post it as long as i finish
Title: Re: HA5-P6
Post by: Rong Wei on October 17, 2015, 06:26:23 PM
Title: Re: HA5-P6
Post by: Zaihao Zhou on October 19, 2015, 11:05:51 AM
Maximum principle tells us that the maximum point will be on either t=0, x= lower limit (in this case, -2), x = higher limit (in this case 2). But from Yunheng's answer we can see the maximum point is indeed $(x,t) = (-1,1)$, not what the principle asserts.

The failure of the principle rooted in the possible negative value of $x$. A crucial step of the proof needs

\label{eq:1}
v_t - kv_{xx} <0

and the for the imaginary inner max point $(x_0,t_0)$,

\label{eq:2}
v_t(x_0,t_0) - kv_{xx}(x_0,t_0) \ge 0

to arrive at a contradiction. Where $v(x,t) = u(x,t) + \epsilon x^2$.

We can see in this example $k$ is changed to $x$, which is not a fixed positive value anymore, it has a chance of getting negative to fail both of these two equations. (Of course in the specific example t=1 is on the upper boundary, the second equation is proved differently than an inner point, but we nevertheless will arrive at (\ref{eq:2}) for contradiction purpose. Furthermore possible failure in (\ref{eq:1}) suffices.)
Title: Re: HA5-P6
Post by: Victor Ivrii on October 21, 2015, 05:52:20 AM
Indeed, it fails because coefficients is not non-negative everywhere.