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Suppose that $u$ is the solution to

$\begin{cases} u_t -u_{xx} =2 , &0<x<1, t>0 \\ u(x,0)=0, & 0\le x \le 1 \\ u(0,t) = u(1,t)=0, &t\ge0 \end{cases}$

Show that $u(x,t)\le x(1-x), 0\le x \le 1$

While I am studying heat equation, I think it looks trivial because, say $k(x,t)=x(1-x)$,

then

$\begin{cases} k_t -k_{xx} =2 , &0<x<1, t>0 \\ k(x,0) \ge0 = u(x,0), & 0\le x \le 1 \\ k(0,t) = k(1,t)=0, &t\ge0 \end{cases}$

So I think the inequality is satisfied, but I think it is not enough...

Am I wrong? Or any other solutions?

JAEMTO
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    This is a good start but it is not enough. You need to use the maximum principle for $v=u-k$. – MOMO Mar 27 '24 at 09:06
  • @MOMO Could you be more specific? I tried that but failed about hours... – JAEMTO Mar 27 '24 at 09:09
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    Write the Heat problem that $v$ solves. The equation (and boundary conditions) will be homogeneous therefore you could use the initial condition (which is $\leq 0$) to prove $v\leq 0$ – MOMO Mar 27 '24 at 09:12
  • @MOMO $v$ becomes heat equation $v_t -v_{xx}=0, v(x,0) \le 0 , v(0,t)=v(1,t) \le 0.$ So this satisfies $v<0$, by Maximum Principles. Right? just to verify – JAEMTO Mar 27 '24 at 09:14
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    Yes. $v\leq 0$. – MOMO Mar 27 '24 at 09:40

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