### Author Topic: HA5-P6  (Read 1531 times)

#### Yunheng Chen

• Newbie
• • Posts: 4
• Karma: 0 ##### HA5-P6
« on: October 17, 2015, 05:57:08 PM »

#### Yunheng Chen

• Newbie
• • Posts: 4
• Karma: 0 ##### Re: HA5-P6
« Reply #1 on: October 17, 2015, 06:12:08 PM »
Still working on partC and i will post it as long as i finish

#### Rong Wei

• Sr. Member
•    • Posts: 43
• Karma: 0 ##### Re: HA5-P6
« Reply #2 on: October 17, 2015, 06:26:23 PM »

#### Zaihao Zhou

• Full Member
•   • Posts: 29
• Karma: 0 ##### Re: HA5-P6
« Reply #3 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

\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.)
« Last Edit: October 19, 2015, 11:08:44 AM by Zaihao Zhou »

#### Victor Ivrii ##### Re: HA5-P6
« Reply #4 on: October 21, 2015, 05:52:20 AM »
Indeed, it fails because coefficients is not non-negative everywhere.