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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!

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  • $\begingroup$ 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?? $\endgroup$
    – Math604
    Commented Sep 29, 2015 at 5:44
  • $\begingroup$ @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? $\endgroup$
    – Transcendental
    Commented Sep 29, 2015 at 6:11
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    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Sep 29, 2015 at 6:47
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    $\begingroup$ 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 $\endgroup$
    – BS.
    Commented Sep 29, 2015 at 8:31
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    $\begingroup$ @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. $\endgroup$
    – Transcendental
    Commented Oct 2, 2015 at 16:19

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As Deane Yang suggested you can regard the first order Beltrami system as an elliptic system, see Chapter 7.3 in the following book

Astala, Kari; Iwaniec, Tadeusz; Martin, Gaven Elliptic partial differential equations and quasiconformal mappings in the plane. Princeton Mathematical Series, 48. Princeton University Press, Princeton, NJ, 2009.

Also Chapter 5 and 6 give the answer for your question.

Since your equation is linear, you could have a representation formula for your equation, see for instance the following nice paper

Bojarski, B. V. Generalized solutions of a system of differential equations of the first order and elliptic type with discontinuous coefficients. Translated from the 1957 Russian original. With a foreword by Eero Saksman. Report. University of Jyväskylä Department of Mathematics and Statistics, 118. University of Jyväskylä, Jyväskylä, 2009.

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