A About ellipticity and a proof that a system of PDEs is elliptic

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Hello all,
I am dealing with a system of 2 PDEs and I think it is elliptic. I wonder what are the implications of this ellipticity. Does this mean that the solution exists, is unique and well behaved (i.e. a small perturbation in the parameters manifests itself as a small change in the solution), provided the boundary conditions are properly defined? Is there anything else deeper than this?

The 2 PDEs are ##\nabla \cdot \vec J =0## and ##\nabla \cdot \vec J_U =0## (a heat equation with 2 source terms) where ##\vec J=-\sigma \nabla V -\sigma S \nabla T##, ##\vec J_U=-\kappa \nabla T + TS\vec J + V\vec J## where ##S## is a tensor represented by a 2x2 diagonal matrix with entries ##S_{xx}## and ##S_{yy}##. The 2 unknowns are the 2 scalar fields ##T## and ##V##, ##\kappa## and ##\rho=1/\sigma## being constant positive scalars. Let ##\vec u=(T,V)## be the vector of unknowns and ##A## the operator that acts upon ##\vec u## and describes the principal part of the system of PDEs.

In order to show that the system is elliptic, I have introduced a wavevector ##\vec k = |\vec k|\vec n## where ##\vec n## is a unit vector. This way, ##\nabla V \to i\vec kV## and ##\nabla^2 V=-k^2V##.

The principal part of the heat equation is ##-\kappa(\partial_{xx}T+\partial_{yy}T) + (TS_{xx}V)\partial_x[-\sigma(\partial x V+S_{xx}\partial_x T)]+(TS_{yy}V)\partial_y[-\sigma(\partial y V+S_{yy}\partial_y T)]##.
The principal part of ##\nabla \cdot \vec J =0## is ##-\sigma(\partial_{xx}V+\partial_{yy}V)-\sigma(S_{xx}\partial_{xx}T+S_{yy}\partial_{yy}T)##.
In this case ##A=k^2\begin{bmatrix}\kappa+\sigma T (S_{xx}^2n_x^2+S_{yy}^2n_y^2) + \sigma V(S_{xx}^2n_x^2+S_{yy}^2n_y^2) & \sigma T(S_{xx}n_x^2 + S_{yy}n_y^2) +\sigma V \\ \sigma (S_{xx}n_x^2 +S_{yy}n_y^2) & \sigma \end{bmatrix}##.
We can simplify a bit this matrix noting that ##n^TSn=S_{xx}n_x^2+S_{yy}n_y^2## (where T here means the transpose, not temperature, i.e. not the unknown variable T). And ##n^TS^2n=S_{xx}^2n_x^2+S_{yy}^2n_y^2##.
For ellipticity, we need ##\det (A)>0##.
Skipping some algebraical maniulation steps, ##\det (A)=\kappa \sigma + \sigma ^2 T[(n^TS^2n)-(n^TSn)^2]##. I then used the Cauchy-Schwartz inequality with ##\vec a=\vec n## and ##\vec b =S\vec n##. This propelled me to ##n^TS^2n-(n^TSn)^2\geq 0## and so ##\det (A)\geq \kappa \sigma >0##.
 
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