### Author Topic: HA3 problem 5  (Read 3259 times)

#### Victor Ivrii ##### HA3 problem 5
« on: February 05, 2015, 07:33:12 PM »
The purpose of this exercise is to show that the maximum principle is not true for the equation $u_t=xu_{xx}$ which has a variable coefficient.

a.  Verify that $u=-2xt-x^2$ is a solution.
b. Find the location of its maximum in the closed rectangle $[-2\le x\le 2, 0\le t\le 1]$.
c.  Where precisely does our proof of the maximum principle break down for this equation?

#### Yiyun Liu

• Full Member
•   • Posts: 15
• Karma: 0 ##### Re: HA3 problem 5
« Reply #1 on: February 05, 2015, 09:01:29 PM »
Part(a)
\begin{array}{l}
\mathop u\nolimits_t  =  - 2x,\\
\mathop u\nolimits_x  =  - 2t - 2x\\
\mathop u\nolimits_{xx}  =  - 2
\end{array}
plugging in the equation, so âˆ’2x = âˆ’2x. u(x,t) is the solution which satisfies the equation

part(a):
By principle of maximum theorem. If a maximum is at an interior point, then ux=ut=0,then x=t=0, which lies on the bottom boundary, for u = 0 âˆ€x âˆˆ (âˆ’2, 2).  That is there is no maximum in the interior. Now consider case of the two side edges, firstly on the left side, where {(x, t)|x = âˆ’2, 0 â‰¤ t â‰¤ 1}, u(âˆ’2, t) = 4(t âˆ’ 1) â‰¤ 0 , for any 0 â‰¤ t â‰¤ 1.  Next consider the right side, where {(x, t)|x = 2, 0 â‰¤ t â‰¤ 1}, u(2, t) = âˆ’4(t + 1) < 0, for all 0 â‰¤ t â‰¤ 1.Therefore we can conclude that the maximum value of u on the bottom and sides is 0. Consider the case of the top, where {(x, t)| âˆ’ 2 < x < 2, t = 1},
u(x, 1) = âˆ’2x âˆ’ x^2. If the maximum exists, then x âˆˆ (âˆ’2, 2) and u_x (x,1) = 0.
From part(a), u_x (x,1) = âˆ’2 âˆ’ 2x. Taking u_x=0, then x=-1, hence u_max=u(-1,1)=1

part(b):

consider these situations below:
u_t = âˆ’2x, u_t(1,1)=2>0
u_xx = âˆ’2, u_xx(-1,1)=-2<0
but u(x,t) still satisfies the PDE equation.
« Last Edit: February 06, 2015, 05:39:27 PM by Yiyun Liu »

#### Victor Ivrii ##### Re: HA3 problem 5
« Reply #2 on: February 06, 2015, 11:52:32 AM »
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