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.
I am having a little bit of trouble with part ii of this question, any help is greatly appreciated. Thank you.