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Consider the Bolza problem

$$ \inf\left\{F(u)=\int\limits_0^1 ((1-u'^2)^2+u^2)\, dx, u\in W^{1,4}(0,1), u(0)=0=u(1)\right\}. $$

Show that $\inf F(u)=0$, but that it does not exist an $u_0$ with $F(u_0)=0$.

Hello! To my opinion the fist step is to show that $F(u)\geq 0~\forall~u$. This is easy, I think, because the integrand is always $\geq 0$ anf therefore the integral, which means $F(u)$.

Then I have to find a sequence with $\lim\limits_{n\to\infty}F(u_n)=0$.

Can anybody help to find such a sequence? I did not have an idea yet...

How can I construct such a sequence? I don' t see that.

Thank you very much for helping!

1 Answers1

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The first part is OK. For the second part, if there is a $u \in W^{1,4}(0,1)$ with $u(0)=0=u(1)$ such that $F(u)=0$. Then $u^2=0=1-(u')^2$ a.e. in $(0,1)$, i.e. $u=0$ a.e. in $(0,1)$, but $u'\ne 0$ a.e. in $(0,1)$. This is not possible, hence there is no such a $u$.

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  • Thank you. But the first part is not finished yet! I have to find a sequence $(u_n)$ with $\lim\limits_{n\to\infty}F(u_n)=0$, haven't I? –  Jan 21 '13 at 16:28
  • @math12 Use a triangular wave: a sequence of small zigzags with slope $\pm 1$. –  Jan 21 '13 at 16:52
  • Do you mean $u_n(x)=\begin{cases}x, & x\in [0,\frac{1}{n})\-x+\frac{n}{2}, & x\in [\frac{1}{n},\frac{2}{n})\x-\frac{n}{2}, & x\in [\frac{2}{n},\frac{3}{n})\-x+\frac{n}{4}, & x\in [\frac{3}{n},\frac{4}{n})\...\-x+1, & x\in [\frac{n-1}{n},1]\end{cases}$? –  Jan 21 '13 at 17:17
  • @math12 I think the terms you add should be something like $j/n$. Or you can define $u$ in one line: $$u(x)=\int_0^x \operatorname{sign}\sin 2\pi n t,dt$$ –  Jan 21 '13 at 20:39
  • Maybe you can explain me how you came to that? –  Jan 21 '13 at 20:44
  • I have to check additionally if $F$ is weakly low semi continious$. I said that $u_n$ is a limited sequence in a reflexive space, so it has a subsequence $(u_{n_k})$ that convergences weakly against an $\overline{u}$. Because of $F(u_n)\to 0$ it is $\liminf\limits_{k\to\infty}F(u_{n_k})=\lim\limits_{k\to\infty}F(u_{n_k})=0$, too and because it does not exists an $u_0$ with $F(u_0)=0$ it is $F(\overline{u})>\limfinf\limits_{k\to\infty}F(u_{n_k})=0$ and therefore $F$ is not weak low semi continious. Right? –  Jan 21 '13 at 20:54
  • My last comment is not readable, sorry. I wanted to show that $F$ is NOT weakly low semi continious: The sequence $(u_n)$ above is limited in a reflexive space. Therefore it has a subsequence $(u_{n_k})$ that convergences weakly against an $u^$. Because of $F(u_n)\to 0$, it is again $\lim\limits_{k}F(u_{n_k})=\liminf\limits_{k\to\infty}F(u_{n_k})=0$. And because there is no $u_0$ with $F(u_0)=0$ it is to my opinion $F(u^)>\liminf\limits_{k\to\infty}F(u_{n_k})=\lim\limits_{k\to\infty}F(u_{n_k})=0$. Therefore $F$ is NOT wlsc. Right? –  Jan 21 '13 at 21:00