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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.

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1 Answer 1

up vote 1 down vote accepted

Question ii) is rather easy. Being $\alpha$ a constant its derivative is zero and your equation just becomes

$$dw_{xx}=-(w+\alpha)w.$$

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Oh thank you I guess I glossed over that every time, but thank you very much. –  Steve Dec 13 '11 at 0:29
    
About question iii is it just simple integration by parts or do I need something else? –  Steve Dec 13 '11 at 1:07
    
Question iii) seems to make sense only when you multiply by $w'$ and not $w$. In this way you are able to integrate this equation exactly and get the contradiction as the homework asks. The same you can do with equation (*) just multiplying by $u'$. –  Jon Dec 13 '11 at 8:00

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