unique continuation property for overdetermined elliptic PDE

On a closed manifold $M$, let $P(f)=0$ be a linear overdetermined elliptic system of PDE of 2nd order with smooth coefficients. By overdetermined ellipticity, I mean the principal sympbol is injective. For simplicity, let's assume it is for a real-valued function $f$. If a solution $f$ is zero on an open subset, then is $f$ zero on the whole of $M$? If this is not true in general, what would be conditions for it to be true? Also please recommend some references. Thanks a lot!

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Is there any reason you cannot use Holmgren's theorem? – Michael Renardy Aug 2 '13 at 10:49
What do you mean exactly by overdetermined ? Is $P$ vector (or complex) valued ? – BS. Aug 2 '13 at 12:23
You need to specify what you mean by overdetermined. Many but not all classes of eliptic systems satisfy the unique continuation property. – Liviu Nicolaescu Aug 2 '13 at 12:51
So you have two elliptic PDEs, or one elliptic and one something else, or more than two? I think that if you already have one elliptic PDE, you should be able to apply: Aronszajn, N. A unique continuation theorem for solutions of elliptic partial differential equations or inequalities of second order. – Ben McKay Aug 2 '13 at 13:11
I cannot use Holmgren's theorem, because it requires real-analytic coefficients. It is a system of PDE for one real-valued function, but each one of those equations is alone not elliptic. – user38064 Aug 3 '13 at 4:03

(1) The most classical result, due to Aronszajn, and later to Calder\'{o}n and (independently) to H\"ormander. Take a second order elliptic operator $P$ with Lipschitz and real coefficients in the principal part. Then if $Pu=0$ on an open set $\Omega$ and u vanishes on an open set $\omega$, this implies that $u$ vanishes on the connected component of $\omega$.
(2) The same conclusion holds if $Pu=0$ on an open set $\Omega$ and $u$ is flat at a point $m$, i.e. $$\forall \alpha,\ (\partial^\alpha u)(m)=0\quad\text{or if u is not smooth, \int_{\vert x-m\vert\le R}\vert u(x)\vert dx=O(R^\infty)}.$$ (3) You cannot use Holmgren to get this since no analyticity of the coefficients is required and also the result is true if you replace the equation $Pu=0$ by an inequality $$\vert (Pu)(x)\vert\le \vert V(x) u(x)\vert+\vert W(x) \nabla u(x)\vert,$$ with $V\in L^{n/2}, W\in L^{n+\epsilon}$. The latter result is due to Jerison \& Kenig without $W$ and to Koch \& Tataru in general.