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.