Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need help with this problem. I think that I have to use the maximum principle for the heat equation, but I don't know how.

Let $u$ be a solution of $u_t =u_{xx}$ in the rectangle $S_{T}=(0,1)\times (0,T)$, continuous in the closed set $\bar S_T$. Also suppose that $u_x$ is continuous in $[0,1]\times (0,T]$. Let $0<t_0\le T$ and $u(x,t)>m$ for $x \in [0,1]$, $t \in (0, t_0]$. Also, let $u(0,t_0)=m$. Prove that $u_x(0,t_0)>0$.

share|improve this question
    
You said that $u(x,t)>m$ for $x\in\[0,1],t\in(0,t_0]$ and then you said that $u(0,t_0)=m$ - $(0,t_0)\in [0,1]\times(0,t_0]$ –  shilov May 20 '13 at 17:43
    
If you mean $u(x,t)>m$ on the set $(0,1]\times(0,t_0]$. $\forall h>0,$ $\frac{u(h,t_0)-u(0,t_0)}{h}>0$ so $u(0,t_0)_x>0$ –  shilov May 20 '13 at 17:50
    
@shilov: Take $u(x,t_0)=x^2$. Obviously, $\forall\, h>0,\;\frac{u(h,t_0)-u(0,t_0)}{h}=h>0$, but why $u_x(0,t_0)>0$? –  mkl314 May 13 at 23:27
add comment

1 Answer 1

This is known as Giraud-type theorem for parabolic equations. For the details see: abstract at http://link.springer.com/article/10.1007/BF00967266#page-1; full text at http://booksc.org/book/12187461

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.