Quantitative forced similarity estimate for real-linear Beltrami equations

Define $a=\nu \overline{\mathrm{\partial }F}\mathrm{/}\mathrm{\partial }F$ where $\mathrm{\partial }F\mathrm{\ne }0$ and $a=0$ otherwise, and define $b=\alpha \bar{F}\mathrm{/}F$ where $F\mathrm{\ne }0$ and $b=0$ otherwise. Then $a,b$ are measurable, supported in $D_1$, satisfy $\mathrm{\mid }a\mathrm{\mid }\le \kappa$, $\mathrm{\mid }b\mathrm{\mid }\le B$, and the equation becomes $\bar{\mathrm{\partial }}F=a\mathrm{\partial }F+bF+E$. The identities also hold on the zero denominator sets, because the corresponding original terms are zero there. Let $s=\mathcal{C}(I-a\mathcal{B}{)}^{-1}b$. By [F1], $\bar{\mathrm{\partial }}s-a\mathrm{\partial }s=b$, $s\to 0$ at infinity, and $\mathrm{\parallel }s{\mathrm{\parallel }}_{W^{1,r}}+\mathrm{\parallel }s{\mathrm{\parallel }}_{\mathrm{\infty }}\le CB$.

Put $G=e^{-s}F$. The weak product and composition rules follow by local smooth approximation, since both factors are continuous and bounded on compact sets and their derivatives are in $L^r$. The equation in step 1.1 gives $\bar{\mathrm{\partial }}G-a\mathrm{\partial }G=e^{-s}E$. The particular solution $G_0=\mathcal{C}(I-a\mathcal{B}{)}^{-1}(e^{-s}E)$ obeys $\mathrm{\parallel }{G}_0{\mathrm{\parallel }}_{W^{1,r}}+\mathrm{\parallel }{G}_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{e}^{CB}\mathrm{\parallel }E{\mathrm{\parallel }}_r$, by [F1]. Both $G$ and $G_0$ are continuous and tend to zero at infinity; their difference is bounded and solves the homogeneous equation. By [F2], $G=G_0$. Finally $F=e^s{G}_0$, so the product rule gives $\mathrm{\parallel }F{\mathrm{\parallel }}_{W^{1,r}}+\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{e}^{CB}(1+\mathrm{\parallel }\mathrm{\nabla }s{\mathrm{\parallel }}_r)(\mathrm{\parallel }{G}_0{\mathrm{\parallel }}_{W^{1,r}}+\mathrm{\parallel }{G}_0{\mathrm{\parallel }}_{\mathrm{\infty }})\le C{e}^{C(1+B)}\mathrm{\parallel }E{\mathrm{\parallel }}_r$. This proves the estimate.