# Are solutions of the Beltrami Equations necessarily smooth?

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 $\Bbb{R}^{2}$: $$u_{x} = \frac{1}{\sqrt{\Delta}} (b v_{x} + c v_{y}), \qquad u_{y} = - \frac{1}{\sqrt{\Delta}} (a v_{x} + b v_{y}).$$ Boundary conditions are not imposed.

Question: Are solutions $u$ and $v$ of this system necessarily smooth?

I have been told that the Elliptic Regularity Theorem answers my question, but the theorem applies only to even-order elliptic operators and what we have here are only first-order equations, so we do not know if solutions $u$ and $v$ have even second-order partial derivatives. Observe that we have the same scenario for the Cauchy-Riemann Equations (in PDE form).

I understand that my question may be too basic for MathOverflow, so I understand if anyone wishes to close it. Thank you!

• consider the extreme case of $a=c=1$ and $b=0$. Then $u$ is harmonic. Can maybe do something similar in the more general case and still get an elliptic second order equation?? Sep 29 '15 at 5:44
• @Math604: Hi Math604. Pardon me for being a little blunt here, but there’s a problem with your comment. If we don’t know that $u$ has second-order partial derivatives, how can we claim that $u$ is harmonic? Sep 29 '15 at 6:11
• It suffices to use weak second order partial derivatives (i.e., distributions). A weak solution to a linear elliptic PDE is necessarily smooth. Or you can work with the first order system directly, since it is itself an elliptic system. Sep 29 '15 at 6:47
• Not only the solutions are smooth, but your system rewrites $du=dv\circ J$, where $J^2=-Id$. Hence in coordinates where $J$ is $(s,t)\mapsto (-t,s)$, $u+iv$ is holomorphic in $s+it$. Moreover, Ahlfors and Bers proved that this generalizes to bounded measurable $J$ (but the solutions are no longer smooth if $J$ isn't, of course). See cimat.mx/~mmoreno/teaching/fall10/AhlforsBers.pdf
– BS.
Sep 29 '15 at 8:31
• @Deane: Hi Deane. Your comment was really helpful, as it reminded me of the fact that distributions have weak derivatives of all orders. If $u$ and $v$ are distributions corresponding to locally-integrable functions, and they satisfy the Beltrami Equations in the weak sense, then by elliptic regularity, these functions are a.e.-equal to smooth functions. Hence, continuous solutions of the Beltrami Equations, viewed in the ordinary sense, are already smooth. Oct 2 '15 at 16:19