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$u_1$ and $u_2$ solve

$u_t-u_{xx}=f(u)$ for some $C^{\infty}$ function

on $(a,b)\times[0,\infty)$

On the vertical boundaries: $u_1(a,t)=u_2(a,t)=u_1(b,t)=u_2(b,t)=0$

and there exists $\tau\in[0,\infty)$ such that $u_1(x,\tau))\leq u_2(x,\tau)$ for the whole of $x\in (a,b)$

I am trying to show:

$$u_1(x,t)\geq u_2(x,t)$$

for all $(x,t)\in(a,b)\times[0,\infty)$

Attempt

Set $v=u_2-u_1$ - this is a classical solution to the heat equation and hence has a maximum and minimum on the parabolic boundary. So to show that $v(x,0)\leq 0$, is sufficient to show that $u_2(x,t)\leq u_1(x,t)$ everywhere inside $\Omega\equiv(a,b)\times[0,\infty)$.

Here's where I get stuck. The question says to assume that $\exists t_0\geq\tau$ and a $x_0 \in(a,b)$ such that $u_2$ crosses $u_1$ for the first time and show that this cannot be the case.

In terms of the function $v=u_2-u_1$ this corresponds to $v$ going through zero for the first time. Since $u_2(x,t_0)\leq u_1(x,t_0)$ this means $u_2(x,t_0)=u_1(x,t_0)$ and in between $\tau$ and $t_0$ (1 and 2 on the graph) there is are two intermediate stationary point $v_t=0$ corresponding to:

$$(u_1)_{xx}+f(u_1)=(u_2)_{xx}+f(u_2)$$

The stationary point at $\tau$ corresponds to $v_t=0, v_{xx}>0$ and $u_1=u_2$

So we have since $u_1=u_2 \Rightarrow f(u_1)=f(u_2)$, $(u_1)_{xx}=(u_2)_{xx}$ but $(u_1)_{xx}<(u_2)_{xx}$ since $v_{xx}>0$ because it is a minimum.

A graph showing $v(x,0)>0$ in red and $v(x,0)\leq 0$ in blue if $\tau = 1$ and $t_0=2$ A graph showing $v(x,0)>0$ in red and $v(x,0)\leq 0$ in blue if $\tau = 1$ and $t_0=2$

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  • $\begingroup$ I found a solution to this I will post at some point in the near future $\endgroup$ – shilov May 28 '13 at 20:59
  • $\begingroup$ Could you please post the solution? Thank you! $\endgroup$ – Hans Dec 14 '15 at 6:55

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