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Suppose $\Omega\subset\mathbb{R}^3$ is a bounded domain with smooth boundary. We note by $(-\Delta)^{-1}$ the inverse Laplacian i.e. $f\mapsto u$ where $u$ is the unique solution to $$-\Delta u=f,\quad u_{|\partial\Omega}=0$$

Now consider the elliptic operator $\Delta + b\cdot\nabla(-\Delta)^{-1}$ associated with, say, homogeneous Dirichlet boundary conditions, where $b$ is a smooth vector field (you can even take $b$ to be a constant vector field if that somehow makes this question answerable).

Is spectral analysis for such an operator possible? (i.e. existence of eigenvalues, properties of eigenfunctions, etc) I ask because such an operator (or rather a slightly more complicated version of it) appears as the linearization of a nonlinear operator that I'm studying. Any help/references would be greatly appreciated.

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So that's $$ -\Delta u -b\cdot\nabla v = \lambda u $$ $$ -\Delta v + u =0 $$ which you can write, with $U=(u,v)$ $$ -\Delta U + B \nabla U + C U = \lambda Q U $$ where $\Delta$ is applied line wise, and so is the gradient, whereas $$ B=\begin{pmatrix} 0& b\cdot \\ 0 &0\end{pmatrix}, C=\begin{pmatrix} 0& 0 \\ 1 &0\end{pmatrix}\text{ and }Q=\begin{pmatrix} 1& 0 \\ 0 &0\end{pmatrix}. $$ In the form you wrote it, you see immediately that there is an associated compact operator (hence properties of the spectrum), and that the problem is not self adjoint. Whether it is a cooperative or a non cooperative system will let you prove properties on the eigensolutions. In any case, it is relatively nice, because it is weakly coupled, that is, the second order operator applies to each line separately. Mitidieri and Sweers have looked at these type of problems a while ago. More recent work probably quotes them.

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  • $\begingroup$ Thank you, this is already quite helpful. Though I'm noticing that "weakly coupled" generally refers to the fact that derivatives don't act of coupling terms. But in this case we have a gradient acting on v in the equation for u. Oh well, it's not everyday that you find exactly what you want I suppose. Thanks again. $\endgroup$
    – Fozz
    Feb 23 at 17:15
  • $\begingroup$ @Fozz You are right. Not strongly coupled, not weakly coupled, vaguely coupled : ) perhaps. I would have thought that a coupling term of first order is not that bad, since it doesn't destroy uniformly ellipticity, that is, you still have, if the RHS is called $L$, $(LU,U)> \frac{1}{2} (\nabla U, \nabla U) - C(U,U)$ , From what I recall, an idea is to first look at the case when $Q=I_d$, and then discuss what applying a bounded operator $Q$, does to everything.. $\endgroup$
    – username
    Feb 23 at 19:11

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