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I have been having trouble with the following outline of a proof, any help I can get will be appreciated. Thank you in advance.

Consider a positive solution to

$(*)\qquad du_{xx}=u(\alpha-u)$ in $(a,b)$, $u(a)=0=u(b)$,

where $d$ and $\alpha$ are positive constants. Prove $u(x) \leqslant \alpha$ on $[a,b]$.

i) Suppose to the contrary that $u(x_0) > \alpha$ for some $x_0 \in (a,b)$. show there is $(a',b') \subseteq (a,b)$ so that $u(x) > \alpha$ on $(a',b')$ with $u(a')=u(b')=\alpha\,$.

ii) let $w(x) =u(x) - \alpha, x \in [a',b']$, and write $(*)$ in terms of $w$.

iii) Multiply the resulting equation by $w$, integrate from $a'$ to $b'$ and find a contradiction to the original solution.

My main problem is with the starting of it, I figure I will be able to complete the proof if I have help with part (i). Of course more help is appreciated but my main problem is with part (i). Thank you all again for your help.

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Looks like the solution to your PDE is just identically zero, unless I'm missing something.... –  Jeff Dec 8 '11 at 20:38
$du_{xx}=u(\alpha-u)$ looks more like an ODE, than a PDE... However, part (i) is just continuity: in fact, if you assume $u(x_0)>\alpha$, then you will find open intervals $I\subseteq (a,b)$ containing $x_0$ s.t. $u>\alpha$ in $I$; then take the union of all such $I$s: it is an interval, which is actually closed and s.t. in its extrema the function $u$ takes the value $\alpha$. –  Pacciu Dec 8 '11 at 21:49

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