Author Topic: HA3 problem 4  (Read 784 times)

Victor Ivrii

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HA3 problem 4
« on: February 05, 2015, 07:31:28 PM »
Consider a solution of the diffusion equation $u_t=u_{xx}$ in $[0\le x \le l, 0\le t <\infty]$.

Let \begin{gather*}
M(T)= \max _{[0\le x \le l, 0\le t \le T]} u(x,t),\\[4pt]
m(T)= \min _{[0\le x \le l, 0\le t \le T]} u(x,t).
\end{gather*}

a.  Does $M(T)$ increase or decrease as a function of $T$?
b. Does $m(T)$ increase or decrease as a function of $T$?

Jessica Chen

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Re: HA3 problem 4
« Reply #1 on: February 05, 2015, 09:09:54 PM »
a.$M(T)= $max $u(x,t)$, where $0\leq x\leq l, 0\leq t\leq T]$

We claim that M(T) is an non decreasing function of T.

Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1} $ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2} $.

The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus$M(T1)\leq M(T2)$

b.$m(T)=$ min $u(x,t)$, where $0\leq x\leq l, 0\leq t\leq T]$

We claim that m(T) is an non increasing function of T.

Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1} $ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2} $.

The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus  $m(T2)\leq  m(T1)$
« Last Edit: February 07, 2015, 05:09:23 PM by Jessica Chen »

Victor Ivrii

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Re: HA3 problem 4
« Reply #2 on: February 06, 2015, 11:50:46 AM »
Actually it has nothing to do with maximum principle: since domain increases maximum can only increase (or remain the same) and minimum can only decrease (or remain the same).

Maximum/minimum principle would be useful to prove that both maximum/minimum remain the same—but then one needs boundary conditions.