Show that the functional $F(u)=\int ((u_x^2-1)^2+u_y^4)$ has zero infimum on a Sobolev space Let $\Omega=(-1,1)\times (-1,1)$. We consider the functionnal
$$F(u)=\int_{\Omega}\left[\left(\left(\frac{\partial u}{\partial x}\right)^2-1\right)^2+\left(\frac{\partial u}{\partial y}\right)^4\right]dx,$$
where $u\in W_0^{1,4}(\Omega)$. 
I am trying to prove that 
$$\inf\{F(u): u\in W_0^{1,4}(\Omega)=0\}$$
by finding a sequence $u_n\in  W_0^{1,4}(\Omega)$ such that $F(u_n)\rightarrow 0$. I chose $u_n(x)=(x^2-1/n^2)(y^2-1/n^2)$ for $(x,y)\in (-1/n,1/n)\times (-1/n,1/n)$ and $u_n(x)=0$ in $\Omega\backslash (-1/n,1/n)\times (-1/n,1/n)$. However, it did not work ! 
Can any one give some hints to find the sequence $\{u_n\}$ ?
 A: Under zero boundary condition, the $y$ derivative controls the supremum of $u$. So you need to make $|u_x^2-1|$ small while also making the supremum of $u$ small. This can only be achieved if $u_x$ quickly oscillates between $1$ and $-1$. 
As a first attempt, put $u(x,y) = d_n(x)$ where $d_n(x)=\operatorname{dist}(x,n^{-1}\mathbb{Z})$. Note that $u_x^2-1=0$ a.e. and $0\le u\le 1/(2n)$. Also, the derivative $u_y$ is identically zero. The only problem is that $u$ is not zero on top and bottom of the square. 
So, modify the constriction by $u(x,y)=d_n(x)\eta(y)$ where $\eta$ is a cut-off function, e.g. $\eta(y)=\min(1, \epsilon^{-1}(1-|y|))$. Note that $\eta(y)=1$ when $|y|<1-\epsilon$, so on most of the square this function is $1$. Now you have to worry about two things:


*

*$u_y^4 = d_n(x)^4(\eta'(y))^4$ is of the size $1/(\epsilon n)^4$. This gets integrated over a set of measure about $\epsilon$, so you need $\epsilon^{-3}n^{-4}$ to be small. 

*$(u_x^2-1)^2 = (\eta(y)-1)^2$. This terms is not a problem; it's $0$ on most of the square and at most $1$ everywhere.


So, letting $\epsilon=1/n$ does the trick.
