It is very well known that if $A\in L^\infty(B_1;\mathbb R ^{d\times d})$ is a positive definite symmetric matrix, the eigenvalue of the self adjoint operator $H^2(B_1)\cap H^1_0(B_1)\to L^2(B_1)$ $$T:u\to\text{div}(A Du)$$ are all real, and the corresponding eigenvectors can be chosen to form an orthonormal eigenbasis of $L^2(\Omega)$ and an $A$-orthogonal basis of $H^1_0(\Omega)$.

My question is whether it is known if this extends to the case when the operator is perturbed by a lower order term, namely $$T^\prime:u\to\text{div}(A Du)+C\cdot Du$$ with $C\in L^\infty(\Omega;\mathbb R^d)$? It could have complex eigenvalues and eigenvectors, but nevertheless would the span of the real and imaginary part of such vectors still be a generating family?

Giorgio Metafune has pointed out that the question is answered positively when $A$ and $C$ are constant, which was the formulation of this question before this edit. In that case, the eigenvalues and eigenvectors of $T^\prime$ are connected to that of $T$ thanks to a transformation. Indeed, note that $$ \text{div}(A Du)+C\cdot Du = \lambda u \Leftrightarrow \text{div}(A Dv)= \lambda v $$ where $v=u\exp(-\frac12 A^{-1}C \cdot x)$. It therefore immediately follows that the answer is positive. The very same argument carries over (as Giorgio Metafune pointed out) when $A$ and $C$ are non constant, but $$ A^{-1}C =D\phi $$ for some function $\phi$. The question remains in general. Decomposing the problem by this method, I suppose the next simple case is $$ T^\prime : u\to \Delta u+ \phi \cdot Du $$ where $\text{div} \phi =0$. But if a "lower order terms therefore perturbation therefore yes" argument exist, it will avoid this case by case dissection.

  • $\begingroup$ I think that this old post should contains the answer (which is positive) mathoverflow.net/questions/348931/…. However I do not know any simple perturbation argument yielding the result and I would be very interested at it. $\endgroup$ Apr 29, 2021 at 7:11
  • $\begingroup$ By the way, if $A, C$ are constant matrices, then the proof is simple. $\endgroup$ Apr 29, 2021 at 8:19
  • $\begingroup$ @GiorgioMetafune Is it? That's a good news. Yes, the question is for constant coefficients, but you are right, the wider question is the non constant case, hopefully as a perturbation of the constant case first, for smooth coefficients, and then the rough coefficient case. But first things first: why is the constant case simple? $\endgroup$
    – username
    Apr 29, 2021 at 8:27
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    $\begingroup$ Consider $A=I$ (otherwise you make a linear change of variables) and, more generally, a first order term $\nabla \phi \cdot \nabla u$. Then you can write the operator as $e^{-\phi} div (e^{\phi} \nabla u)$ which is self-adjoint with respect to the (equivalent) measure $e^{\phi}\, dx$. $\endgroup$ Apr 29, 2021 at 8:32
  • $\begingroup$ @GiorgioMetafune Thank you! I referred to your comment in the post, and stated the question in general. I also proposed a "next case" problem, but yes, I would like to know if such a perturbation result exists, which would avoid to consider a catalog of cases one by one. $\endgroup$
    – username
    Apr 29, 2021 at 9:24

1 Answer 1


In one space dimension, the answer is yes, and the eigenvalues are real and simple (Sturm-Liouville theory).

In higher space dimension, the answer is negative, because the operator needs not be diagonalisable. If you let the data $(A,C)$ depend upon a parameter, you can pass from a situation where all the eigenvalues are real to one where there is a pair of complex conjugate eigenvalues. At the transition point, the operator is not semi-simple, admitting a double eigenvalue whose eigenspace is only one-dimensional.

An explicit construction is given in this answer. Just rewrite the operator $A:D^2$ as ${\rm div}(AD)+C\cdot D$ where $-C$ is the row-wise divergence of $A$.

  • $\begingroup$ Thank you for your answer, could you clarify: why does that imply that the span (of the real and imaginary parts) of the eigenvectors does not fill the space? $\endgroup$
    – username
    Apr 29, 2021 at 11:30
  • $\begingroup$ However, the linear span of the generalized eigenfunctions is dense. This is the content of the theorem I quoted in the previous comment. $\endgroup$ Apr 29, 2021 at 12:38

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