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Let $\mathcal E$ be the set of all functions $u\in C^1([0,2])$ such that $u(x)\geq 0$ for every $x\in[0,2]$ and |u'(x)+u^2(x)|<1 for every $x\in [0,2]$.

Prove that the set $\mathcal F:=\{u_{|[1,2]}: u\in\mathcal E\}$ is an equicontinuous subset of $C^0([1,2]).$

The point I am stuck on is that i can't see how to combine the strange hypothesis imposed on every $u\in\mathcal E$, in particular i solved the two differential equations u'(x)=1-u^2(x),\qquad u'(x)=-1-u^2(x), which result to be the extremal case of the condition given. In particular the two solutions are $u_1(x)=\frac{ae^t-be^{-t}}{ae^t+be^{-t}},\qquad u_2(x)=\frac{a\cos(x)-b\sin(x)}{a\cos(x)+b\sin(x)}.$ I feel however i'm not ong the right path so any help is appreciated.

P.S. Those above are a big part of my efforts and thoughts on this problem so i hope they won't be completely useless :P

Edit In the first case the derivative is u'_1(x)=\frac{2ab}{(ae^t+be^{-t})}\geq 0 while for the other function we have, for $x\in[0,2],$ u'_2(x)=-\frac{\sin(2x) ab}{(a\cos(x)+b\sin(x))^2}\leq 0. Moreover $u_1(1)>u_2(1)$, since $\frac{ae-b^{e-1}}{ae+be^{-1}}>\frac{a\cos(1)-b\sin(1)}{a\sin(1)+b\cos(1)}\Leftrightarrow (a^2e+be^{-1})(\sin(1)-\cos(1)),$ and $\sin(1)>\cos(1).$ Now, all this bounds i've found are useful to solve the problem?

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    I removed my earlier comment about one minute after posting because I saw the flaw in that argument (I only see your comment now). Didn't intend to scare you...2011-09-16

1 Answers 1

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Suppose $u \in \mathcal{E}$.

It's enough to show $u(t) \le 3$ for all $t \in [1,2]$, since then we'll have -10 \le u'(t) \le 1, and any set of functions with uniformly bounded first derivatives is certainly equicontinuous. We also know that u' \le 1 on $[0,2]$, and so by the mean value theorem it suffices to show that $u(1) \le 2$. If $u(0) \le 1$ we are also done, so assume $u(0) > 1$.

Let $v$ be the solution of v'(t) = 1 - v(t)^2 with $v(0) = u(0) > 1$. This is given by your formula for $u_1$ with, say, $b=1$ and some $a < -1$. I claim $u(t) \le v(t)$. This will complete the proof, since it is easy to check that $v(1) < 2$. (We have $v(1) = \frac{ae-e^{-1}}{ae+e^{-1}}$, which is increasing in $a$; compute its value at $a=-1$.)

Set $w(t) = v(t) - u(t)$. We have $w(0)=0$ and w'(t) > u(t)^2 - v(t)^2. Suppose to the contrary there exists $s \in [0,1]$ such that $u(s) > v(s)$; let $s_0$ be the infimum of all such $s$. Then necessarily $u(s_0) = v(s_0)$, so $w(s_0)=0$ and w'(s_0) > u(s_0)^2 - v(s_0)^2 = 0. So for all small enough $\epsilon$, $w(s_0 + \epsilon) > 0$. This contradicts our choice of $s_0$ as the infimum. So in fact $u \le v$ and we are done.