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I recently encountered a paper by Protter ("Unique continuation of elliptic equations") that starts out by saying "any solution of an elliptic equation that is defined on a domain $D$ must vanish on all of $D$ if it vanishes on an open set in $D$" and calls this the unique continuation principle. The problem is that this statement is so general that I'm afraid to use it. Does anyone know the rigorous statement of the unique continuation principle? In particular I wonder what are the conditions on the elliptic equation and the solution involved.

flag	asked Mar 17 2012 at 0:38 Viktor Bundle 999●1●8

The main thing is that the equation must be a linear homogeneous elliptic PDE with sufficiently smooth coefficients. In spirit, a solution to such an equation behaves analogously to a solution to the standard Laplace equation or the Cauchy-Riemann equations, which are always real or complex analytic and therefore satisfy the unique continuation principle. But alvarezpaiva gives the right reference below. – Deane Yang Mar 17 2012 at 9:40

alvarezpaiva · Answer 1 · 2012-03-17 09:09:19Z UTC

I think this is the basic reference:

A unique continuation theorem for solutions of elliptic partial differential equations or inequalities of second order by N Aronszajn - J. Math. pur. appl., IX. Sér., 1957

Aronszajn considers second order elliptic equations AND inequalities, but his theorem is even more general. You only need that the difference of two solutions vanish at all orders at some point and already the solutions must be identical. Actually, its better than that, but I forget the details.

Misha · Answer 2 · 2012-03-17 03:28:26Z UTC

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See "Critical sets of solutions to elliptic equations", by R. Hardt, M. Hoffmann-Ostenhof, T. Hoffmann-Ostenhof, and N. Nadirashvili, J. Differential Geom. Volume 51, Number 2 (1999), 359-373.

There is also an earlier paper by Lipman Bers, "Local behaviour of solutions of general linear elliptic equations", Comm. Pure Appl. Math. 8 (1955) 473-496.

answered Mar 17 2012 at 3:28

Misha
12.6k●1●26●48

	I bet there are more general results about the set of points where a solution vanishes. How messy can a closed set in the plane be if it is the zero set of a solution? Empty interior if not whole plane, but if it's one-dimensional what can it look like locally, up to diffeomorphism? – Tom Goodwillie Mar 17 2012 at 13:38
	From what I remember (reading Bers' paper ages ago, in mid-1990s) that in dimension 2 zero level sets are always locally diffeomorphic to zero sets of $Re(q(z)dz^k)$, where $q(z)dz^k$ is a degree $k$ holomorphic quadratic differential. – Misha Mar 20 2012 at 11:06

Kelei Wang · Answer 3 · 2012-03-19 01:56:53Z UTC

Besides the Carleman type estimate in Aronszajn work, Nicola Garofalo and Fang-Hua Lin, Monotonicity properties of variational integrals, Ap weights and unique continuation, using Almgren frequency monotonicity, also proved the unique continuation result.

For a linear elliptic operator, the Lipschitz continuity of the coefficient is sufficient. And as I remember, if the coefficients are Holder continuous, there are counterexamples.

unique continuation principle

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