Toronto Math Forum
APM3462015F => APM346Home Assignments => HA5 => Topic started by: Yunheng Chen on October 17, 2015, 05:57:08 PM

http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter3/S3.2.P.html
Problem 6

Still working on partC and i will post it as long as i finish

I add question c), but I'm not sure the answer, please correct me!

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
\begin{equation}\label{eq:1}
v_t  kv_{xx} <0
\end{equation}
and the for the imaginary inner max point $(x_0,t_0)$,
\begin{equation} \label{eq:2}
v_t(x_0,t_0)  kv_{xx}(x_0,t_0) \ge 0
\end{equation}
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.)

Indeed, it fails because coefficients is not nonnegative everywhere.