TT1 Problem 4

[Victor Ivrii]

Check that function $u=x^3+6xt$ satisfies diffusion equation $u_t-u_{xx}=0$ and find
\begin{align*}
&M(T)= \max _{0\le x\le L,\ 0\le t\le T} u(x,t),\\[2pt]
&m(T)= \min _{0\le x\le L,\  0\le t\le T} u(x,t).
\end{align*}

a. Where is the maximum value  $u(x,t)=M(T,L)$  achieved?
b. Where is the minimum  value $u(x,t)=m(T,L)$ achieved?
c.  Verify the  maximum and minimum principle.

[Ping Wei]

a), u(L,T) is maximum value
b), u(0,0) is minimum value

[Ping Wei]

Ut= 6x , Ux= 2x^2+6t, Uxx= 6x  so Ut - Uxx = 6x -6x =0

[Yiyun Liu]

c)
as the principle of maximum and minimum stated that the max and min must occur either on the bottom t=0 or the sides edges at x=0,L including the corners (0,T),(L,T) for some T∈t>0, obviously,  M(T) and m(t) satisfies the maximum and minimum principals.

[Victor Ivrii]

I am waiting for a), b)

[Victor Ivrii]

I am really glad that I decided to grade TT1 by myself (well, I was forced because TA has insufficient number of hours). I see by myself shortcoming of my choice of problems and of your knowledge.

Where maximum or minimum of $u(x,t)$ could be achieved? Second year calculus tells us that it could be

• In inner points of rectangle, $u_x=u_t=0$—there are no such points
• On the boundary $x=0$ or $x=L$ where $u_t=0$. We get $x=0$ and $0\le t\le T$––the whole left border. lost to majority who indicated only $(0,0)$; or on the boundary $t=0$ or $T=L$ where $u_x=0$
• In the four corner points $(0,0)$ (which is covered by what we found), $(0,T)$ (which is alsocovered by what we found), $(L,0)$ and $(L,T)$

So
a) Minimum is $0$ and it is achieved at $\{x=0, 0\le t\le T\}$
b) Maximum is $L^3+6LT$ and it is achieved at $(L,T)$.

Important: both maximum and minimum are achieved on the lateral boundary or the bottom, but neither inside nor on the upper lead (excluding its borders).