Let $ a $, $ b $ and $ c $ be real constants such that $ \Delta \stackrel{\text{df}}{=} a c - b^{2} > 0 $. The Beltrami Equations are defined as the following system of PDE’s on the domain $ \Bbb{R}^{2} $: $$ u_{x} = \frac{1}{\sqrt{\Delta}} (b v_{x} + c v_{y}), \qquad u_{y} = - \frac{1}{\sqrt{\Delta}} (a v_{x} + b v_{y}). $$
Question: Are solutions $ u $ and $ v $ of this system necessarily in $ C^{\infty} \! \left( \Bbb{R}^{2} \right) $?
My question stems from similarities I see between this system and the Cauchy-Riemann Equations in PDE form.
Thanks!
If $ (u,v) $ is a $ C^{1} $-solution pair, then it is, in fact, a $ C^{\infty} $-solution pair. To prove this, let $ F_{f} $ denote the distribution corresponding to a locally integrable function $ f $ on $ \Bbb{R}^{2} $. As $$ {\partial_{x}}(F_{u}) = F_{u_{x}}, \quad {\partial_{y}}(F_{u}) = F_{u_{y}}, \quad {\partial_{x}}(F_{v}) = F_{v_{x}} \quad \text{and} \quad {\partial_{y}}(F_{v}) = F_{v_{y}}, $$ the Beltrami Equations imply the following relations: $$ {\partial_{x}}(F_{u}) = \frac{1}{\sqrt{\Delta}} [b \cdot {\partial_{x}}(F_{v}) + c \cdot {\partial_{y}}(F_{v})] \quad \text{and} \quad {\partial_{y}}(F_{u}) = - \frac{1}{\sqrt{\Delta}} [a \cdot {\partial_{x}}(F_{v}) + b \cdot {\partial_{y}}(F_{v})]. $$ Then as $ {\partial_{xy}}(F_{u}) = {\partial_{yx}}(F_{u}) $, we obtain $$ 0 = a \cdot {\partial_{xx}}(F_{v}) + 2 b \cdot {\partial_{xy}}(F_{v}) + c \cdot {\partial_{yy}}(F_{v}). $$ The distribution $ F_{v} $ therefore belongs to the kernel of a second-order elliptic operator, so by elliptic regularity, we have $ F_{v} = F_{w} $ for some $ w \in C^{\infty} \! \left( \Bbb{R}^{2} \right) $. It now follows from continuity that $ v = w $, so we can conclude that $ u,v \in C^{\infty} \! \left( \Bbb{R}^{2} \right) $.
Note: If we only assume that $ (u,v) $ is a continuous pair with first-order partial derivatives that are not assumed to be continuous, then it is most likely that we still have the same conclusion. One has to study Section $ 15.2 $ of the reference kindly provided by Mice Elf in his comment above.