entry in the gradient $\partial_1 v_1$ would then be of order $\alpha (h/\ell)^2$. The additional factor $h/\ell$ in $\partial_1v_1$ with respect to $\partial_2v_1$ arises from the fact that we took an integral in the $y$ direction and a derivative in the $x$-direction. On most of our domain, $h$ will be substantially smaller than $\ell$. Therefore, this construction leads to a substantial reduction in energy.

Since we are dealing with a geometrically nonlinear model, there will be an additional error term of order $(\partial_1 v_2)^2\sim \alpha^2 (h/\ell)^2$, arising from the linearization. We also need to take care of the fact that we are not linearizing around the identity, but instead around the matrix $A_2$. Therefore, we solve $\partial_2 v_1+(1-\alpha)\partial_1v_2=0$ instead of $\partial_2
v_1+\partial_1v_2=0$, see (\[eqlinearize\]) below.

A detailed computation based on this idea motivates the definition $$v_1(x,y)=
\begin{cases}
{x} & \text{on $\omega_1$}, \\
{x+\alpha(1-\alpha)\frac{h}{4\ell}\gamma'\left(\frac
    x\ell\right)(\frac{h}{8}+\frac{h}{8}\gamma\left(\frac x\ell\right)-y)} & \text{on $\omega_2$},
\\ 
{x-\alpha(1-\alpha)\gamma'\left(\frac x\ell\right)\frac{h^2}{16\ell}}& \text{on $\omega_3$}, \\
{x-\alpha(1-\alpha)\frac{h}{4\ell}\gamma'\left(\frac
    x\ell\right)(\frac{7h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)-y)} &
\text{on $\omega_4$}, \\ 
{x} & \text{on $\omega_5$}.
\end{cases}$$ This concludes the definition of $v$. We remark that $v_1$ satisfies the boundary conditions at $x=0$ and $x=\ell$ in the regions $\omega_2$, $\omega_3$ and $\omega_4$ only if $\gamma'(0)=\gamma'(1)=0$. Therefore, in this case (at variance 