# Maximum principle for the heat equation

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 $u_x(0,t_0)>0$.

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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 '14 at 23:27