both points, say $\gamma(x)=10x^3-15x^4+6x^5$. We subdivide $\omega$ into the five sets $$ \begin{split}
  \omega_1 &= \left\{ (x,y)\in\omega: \ 0< y <
    \frac{h}{8}+\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\
  \omega_2 &= \left\{ (x,y)\in\omega: \
    \frac{h}{8}+\frac{h}{8}\gamma\left(\frac
      x\ell\right)<y<\frac{3h}{8}+\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\ 
  \omega_3 &= \left\{ (x,y)\in\omega: \
    \frac{3h}{8}+\frac{h}{8}\gamma\left(\frac
      x\ell\right)<y<\frac{5h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\ 
  \omega_4 &= \left\{ (x,y)\in\omega:  \ \frac{5h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)<y<\frac{7h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\
  \omega_5 &= \left\{ (x,y)\in\omega: \ \frac{7h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)<y<h\right\},
 \end{split}$$ see Figure \[fig:2\]. In the domains $\omega_1$, $\omega_3$ and $\omega_5$ we shall construct $v$ so that $Dv$ is close to $B_2=\diag(1,1+\alpha)$, in the other two so that it is close to $A_2=\diag(1,1-\alpha)$. Since the boundaries are curved, there will necessarily be deviations from $A_2$ and $B_2$, and in particular the $x$-derivative will not be identically $e_1$.

![Left panel: subdivision of the domain $\omega$ used in Lemma \[eqconstronerecttheo2\]. Right panel: subdivision of the domain used in Lemma \[eqconstronerecttheo2bl\].[]{data-label="fig:2"}](fig2-crop){width="10cm"}

The basic idea is to set $v_2(\cdot,0)=0$, according to the boundary conditions, and then to construct $v_2(x,y)$ by integrating $\partial_2 v_2(x,\cdot)$ from $0$ to $y$, with $\partial_2
v_2=1+\alpha$ in $\omega_1$, $\omega_3$, $\omega_5$, and $\partial_2
v_2=1-\alpha$ on the other two sets. This leads to $$v_2(x,y)=
\begin{cases}
{(1+\alpha)y} & \text{on $\omega_1$}, \\
{(1-\alpha)y+\alpha h(\frac{1}{4}+\frac{1}{4}\gamma\left(\frac x\ell\right))} &
\text{on $\omega_2$}, \\ 
{(1+\alpha)y-\frac{\alpha h}{2}} & \text{on $\omega_3$}, \\
{(1-\alpha)y+\alpha 