Show that there is no minimizer for a specific functional.

Question

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 $z=u(x)$?

Answer 1

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 $\mathcal{F}(u)=1.$