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My question is about differential equations. I want to prove this theorem

Theorem: Let $u(t)$ be a bounded solution in $[0,\infty[$ of the equation $u''+cu'+f(u)=0$, where c is a positive constant and $f()$ is a continuous function. Then $\lim u(t) = l$ exists and $f(l) = 0$. Plus, $\lim u'(t) = 0$.

I'll sketch the proof I have here and i'll point the five questions I have.

Lemma 1: $u'$ is bounded (This one I can prove)

Lemma 2: $\int_{0}^{\infty}{u'(s)^2 ds} < \infty$ (This one I can prove too). Observation: by multiplying u' in the equation we have $\frac{d}{dt}(\frac{u'^2}{2} + F(u)) + cu'^2 = 0$, so $\frac{u'^2}{2} + F(u)$ is decreasing in [0,$\infty$[. Note: $F' = f$.

Lemma 3: $\lim u'(t) = 0$. In the demonstration we are informed that $u''$ is bounded due to the 1st lemma and I can prove it. But I don't understand where I am able to use this information. I think it's easy to see due to lemma 2 that if $u'$ is convergent them the limit is zero, but why this limit exist? That's my first question...

Lemma 4: The limit $\lim F(u(t))$ exists. We know that $\frac{u'^2}{2} + F(u)$ is decreasing in $[0,\infty[$. But why does the limit exists?... 2nd question.

Proof (using the lemmas): Suppose that $\lim \inf u(t)$ is different from $\lim \sup u(t)$. So there exists $a < b$ such that given any number $r$ in $[a,b]$ there exists $t$ arbitrarily large such that $u(t) = r$. Due to the lemma 4, $F$ is constant in $[a,b]$. (Why??? 3rd question) So $f = 0$ in $[a,b]$. Let $r=\frac{a+b}{2}$, $\delta$>0 such that $\frac{2\delta}{c} < \frac{b-a}{4}$ and $T>0$ such that $u(T)=r$ and $|u'(t)|<\delta$ for all $t \geq T$. For $t \geq T$ and at least in the interval $[T,S[$ such that $t\in[T,S[ \Rightarrow u(t)\in]a,b[$, we have: $u'' + cu'$ = 0.

So, $u=A+Be^{-ct}, A+Be^{-cT}=r$ and $|B|ce^{-ct}<\delta$. Finally we have $$ |A+Be^{-ct}-r|\leq \frac{1}{c}|cBe^{-cT} - cBe^{-ct}| \leq \frac{2\delta}{c} < \frac{b-a}{4}, $$ what means that for all $t \ge T$ the solution is $A+Be^{-ct}$ and only takes values in [$\frac{3a+b}{4}$ , $\frac{a+3b}{4}$] (4th question: why this interval??), a contradiction.

So, lim $u(t) = l$ exists. And now the equation $u'' + cu' + f(u) = 0$ itself shows that lim $u''(t$) exists and is equal to $0$ (5th and last question: why lim $u''(t)$ exists and it's $0$???).

So, $f(l) = 0$. The proof is completed

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(1) It's not at all easy to see your actual questions (and the more difficult it is for your readers, the less likely it is that you will get a useful answer. (2) You should learn to use latex (for the same reason as in (1). – Chris Godsil Jan 22 '13 at 22:09
I'll try to help with your "first question" as to why $\lim u'(t)$ exists and equals zero (in a rough and sketchy way). You know already that $u'$ is $C^1$ and that its derivative $u''$ is bounded, and now you have that $u' \in L^2[0,\infty)$. Suppose now that $\lim u'(t)$ doesn't exist; as $u'$ is square integrable this must mean that there are spikes in $u'$ as you go further out; moreover the "width" of these spikes must be diminishing due to square integrability. But then that means that $u''$ must be unbounded, or else that the maximum height of the spikes is diminishing to zero. – A Blumenthal Jan 22 '13 at 22:56
For your second question, equipped with the answer from the first, you know that $(u')^2/2 + F \circ u(t)$ is decreasing, and hence has a limit (perhaps equal to $-\infty$; see if you can't rule that out), so that because $u' \rightarrow 0$, you must have that $F \circ u(t) \rightarrow 0$ as a consequence. – A Blumenthal Jan 22 '13 at 22:59
For the third question, you know that $F \circ u$ has a limit for long times; the idea here is that $u$ is crossing over the interval $[a,b]$ infinitely often for long times, and so if $F$ attained two distinct values on $[a,b]$, then $\lim F \circ u$ wouldn't exist, as the $\limsup$ and the $\liminf$ wouldn't coincide. – A Blumenthal Jan 22 '13 at 23:02
Thanks for all the answers! I know now how to solve my 'question 4' - it's really easy =D – Raphaël Jan 23 '13 at 0:32

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