APM346-2015S > Test 1

TT1 Problem 4

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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)

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