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Consider the minimizing problem of $$\mathcal{F}(u)=\int_0^1 F(x,u(x),u'(x)) \, \mathrm{d}x$$ on $W^{1,4}(I)$ with $I=(0,1)$. Do you have an idea how to show that there exists no minimizer of $\mathcal{F}(u)$ if we chose $$F(x,z,p)=(p^2-1)^2+z^4$$ where $p=u'(x)$ and $z=u(x)$?

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Here, the equations are tedious but the idea I present below will hopefully be sufficient to convince you!

Heuristically, there are two contributions of "energy'' to $F$: $((u')^2-1)^2$ and $u^4$. The energy of $F$ is minimized when $u'(x)=\pm 1$ and when $|u(x)|=0$. With these ideas in mind, consider a sequence of functions $u_n(x)$, where $u_n$ is a sawtooth function (alternating piece of $u'(x)=1$ and $u'(x)=-1$) with more and more oscillations. Then, $((u'(x))^2-1)^2 =0$ for all $x$ (where $u'(x)$ is defined).

In addition, note that by taking more and more teeth in your sawtooth, you can make $\int (u_n)^4 dx$ arbitrarily small, since $u_n$ is converging uniformly to $0$. Thus, if a minimizer to $\mathcal{F}$ were to exist, it must satisfy

$$\min \mathcal{F} \leq \lim_{n\to \infty} \mathcal{F}(u_n) =0.$$

Of course, there is no hope for $\mathcal{F}(u)=0$ since for any non-zero function $u$ we have $\int u^4>0$ and for $u\equiv 0$ we have $\mathcal{F}(u)=1.$

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    Thank you for the answer! I have 2 questions. Do you mean by $ |u(x)|=0$ that $u(x)= 0$ almost everywhere? And what argument allows me to show that the sawtooth functions belong to $W^{1,4}(I)$? – 2017-01-31
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    Hi @derthomas. You're welcome! Yes, that is correct: minimization of the functional requires, in particular, that $\int u^4 dx =0$, which happens if and only if $u=0$ almost everywhere. To see that the sawtooth functions belong to $W^{1,4}(I)$ you can use trivial bounds. Using that $|u_n(x)|\leq 1$ and $|\partial_x u_n(x)|=1$ (almost everywhere), we have $\|u_n\|_{W^{1,4}}^4 =\|u_n\|^4_{L^4} + \|\partial_x u_n\|^4_{L^4}\leq 1+1=2$ – 2017-01-31
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    That's very helpful! I have one last question: a necessary condition for a minimizer of $\mathcal{F}$ is, as you said, that (1) $u'(x)=^+_-1$ and (2) $u(x)=0$ almost everywhere. Could one argue that (1) and (2) can not be verified at the same time, hence a minimizer can not exist? – 2017-01-31
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    @derthomas yup! Exactly correct! – 2017-01-31
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    Allright, you helped me to get a good view on the problem, thanks! Also, your argumentation with the toothsaw functions is very intuitive, so that helped a lot! It would be nice if you gave me an upvote on my question! ;-) – 2017-01-31