# Prove that a map is continuous

Let $r:[0,1]\to\mathbb R$ be a continuous function and let $u_\lambda$ be the unique solution of the Cauchy Problem:

$$\begin{cases}u''(t)+\lambda r(t)u(t)=0,\quad\forall t\in [0,1],\\ u(0)=0,\quad u'(0)=1.\end{cases}$$

It is well known that $(u_{\lambda_n})$ converges uniformly on $[0,1]$ to $u_\lambda$ whenever $\lambda_n\to\lambda.$ Define the map $\tau:\mathbb R\to\mathbb R$ by setting

$$\tau(\lambda):=\inf\{t\in(0,1]\mid u_\lambda(t)=0\},$$ with the convention $\tau(\lambda)=1$ if $u_\lambda(t)\neq 0$ for every $t\in (0,1]$. Prove then that $\tau$ is continuous.

Edit:
ok so i've been trying to work on Robert hint. I am trying to prove that if $\tau(\lambda_0)=\tau_0<1$ then for some $\varepsilon>0$ small i must have $u_{\lambda_0}(\tau_0+\varepsilon)<0$ , but i seem to go nowhere farther from some silly tryings using mean value theorem or stuff like that.. Where am i missing the key point?

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Please consult math.stackexchange.com/faq#bounty for what to do when you do not receive an answer that satisfies you. Editing the question body (and thereby "bump"ing it) is preferable to posting a non-answer. – Willie Wong Sep 12 '11 at 15:16
@willie.. unfortunately i can't start a bounty yet :( – uforoboa Sep 12 '11 at 15:19
done... i posted them in a comment below but ok.. nov i've inserted them in the question as edits – uforoboa Sep 12 '11 at 15:28
Right, the point is that comments won't bump the question, but edits to the body of the question will. So if you want to gather more attention, it is better to make edits. :-) – Willie Wong Sep 12 '11 at 15:29
there's always something to learn :-).. thx willie – uforoboa Sep 12 '11 at 15:35

A few more hints:

1. Using the uniqueness of the solution to the Cauchy problem, observe that if $u_\lambda(\tau(\lambda)) = 0$, then there exists some $\epsilon > 0$ such that $u_\lambda|_{(\tau(\lambda)-\epsilon,\tau)} > 0$ and $u_\lambda|_{(\tau,\tau+\epsilon)} < 0$. (Why can't $u$ locally have the same sign on the two sides of the zero?)
2. If $u$ is a continuous function such that $u(\tau-\epsilon) > 0$ and $u(\tau+\epsilon) < 0$, what can you say about $u$?
3. Since $r(t)$ is continuous, what can you say about $u_\lambda'(t)$, for $t$ sufficiently close to 0? What does this tell you about $u_\lambda$ in, say, $(0,\epsilon)$?
4. If $u_{\lambda}|_{[a,b]} > 0$, what can you say about $u_{\eta}|_{[a,b]}$ for $\eta$ sufficiently close to $\lambda$?

To put every thing together, let $\eta$ be a small perturbation of $\lambda$, and fix a very small $\epsilon$ depending on $\lambda$ and $r$. Use 4 to show that $u_\eta$ cannot vanish on $[\epsilon,\tau(\lambda)-\epsilon]$. Use 3 to show that $u_\eta$ cannot vanish on $(0,\epsilon)$. And use 1 & 2 to show that $u_\eta$ must vanish between $(\tau(\lambda)-\epsilon, \tau(\lambda)+\epsilon)$.

Okay, more on point 1 by request.

1. The function $u \equiv 0$ is a solution to the differential equation (ignoring boundary conditions). So by uniqueness of solutions, if $\exists t_0$ such that $u_\lambda(t_0) = u_\lambda'(t_0) = 0$, $u_\lambda \equiv 0$ and cannot satisfy the boundary condition $u_\lambda'(0) = 1$. Hence for a solution to the ODE with the requisite boundary conditions, at any point where $u_\lambda(t_0) = 0$ we must have that $u_\lambda'(t_0) \neq 0$.
2. The usual regularity theory for ODEs guarantee that $u_\lambda'$ is continuous.
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@willie... It is exactly on point 1 that i can't convince myself... For the rest i know exactly how to finish, but for point 1... i'm absolutely out of ideas.. May i ask you how to proceed? – uforoboa Sep 12 '11 at 17:35
I added more hints on how to do point 1. Let me know if there are more questions. – Willie Wong Sep 12 '11 at 18:18
ok... gotcha... thanks a lot willie – uforoboa Sep 12 '11 at 18:40

Hint: if $\tau(\lambda_0) = \tau_0 < 1$ and $\epsilon > 0$ is small, there is $\delta > 0$ such that $u_\lambda(t) > \delta$ if $\epsilon \le t \le \tau_0 - \epsilon$, while $u_\lambda(\tau_0 + \epsilon) < 0$.

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ok so i've been trying to work on your hint. I am trying to prove that if $\tau(\lambda_0 )=\tau_0<1$ then for some $\varepsilon>0$ small i must have $u_{\lambda_0}(\tau_0+\varepsilon)$ , but i seem to go nowhere further from some silly tryings using mean value theorem or stuff like that.. Where am i missing the key point? – uforoboa Sep 11 '11 at 13:42
Please, can anybody give me a more specific hint, or show me the way to pick in solving this problem? I'me somewhat temptated to give up on one hand, on the other i would like to see a solution before desisting.. – uforoboa Sep 12 '11 at 14:33