h (\frac{3}{4}-\frac{1}{4}\gamma\left(\frac x\ell\right))} & \text{on $\omega_4$}, \\
{(1+\alpha)y-\alpha h} & \text{on $\omega_5$}.
\end{cases}$$ Computing $\partial_1 v_2$, we easily see that it is of order $\alpha
h/\ell$ in $\omega_2$ and $\omega_4$. In particular, if we would simply set $v_1(x,y)=x$, we would obtain $$\label{eqdval}
  |Dv-A_2|\sim \frac{\alpha h}{\ell} \text{  in  $\omega_2$ and $\omega_4$}\,,$$ and therefore an energy at least of order $\alpha^2h^3/\ell$ in $\omega$.

It is, however, possible, via a more careful construction of $v_1$, to reduce the energy further. The idea is to use rotations, in a way similar to what was done in the study of folding pattern in blistered thin films [@B9_BCDM2000; @B9_BCDM2002]. To understand the key idea it is helpful to first consider the linearized model, in which $\operatorname{dist}(Dv, \operatorname{SO}(2) A_2)$ is replaced by $|(Dv+Dv^T)/2-A_2|$. The easier choice for $v_1$, namely, $v_1(x,y)=x$, would make the first diagonal entry in the gradient, $\partial_1v_1$, vanish. The other natural option, namely, to make the off-diagonal entries (which are equal) vanish, leads to the possibility of defining $v_1$ by solving $\partial_2 v_1+\partial_1v_2=0$. By the same estimate as in (\[eqdval\]), the term $\partial_1v_2$ is of order $\alpha h/\ell$, hence integrating in $y$ we would obtain a component $v_1$ of order $\alpha
h^2/\ell$. The diagonal 