I am looking for an example of a pure second order uniformly elliptic operator $L=\sum_{i,j=1}^da_{ij}(x)D_{ij}$ in a bounded domain $\Omega$ (with Dirichlet boundary conditions, for example) having a non-real eigenvalue in $L^2(\Omega)$. Under the above assumptions, the operator $L$ has a discrete spectrum and the principal eigenvalue is real; however it is non-symmetric. Note that if $a_{ij}=a$ (in particular if $d=1$), then every eigenvalue is real since the measure $d\mu=a^{-1}(x)\, dx$ symmetrizes $L$. Does anybody know such an example?

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    $\begingroup$ Small anecdote: I remember Wolfgang Arendt mentioning the same question a few years ago during coffee. He said that he already had discussed this with several people who know quite a deal about elliptic operators, but none of them knew the answer. $\endgroup$ Jan 12, 2020 at 23:34
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    $\begingroup$ Thank you for the comment. Let me point out that the spectrum is real also for operators like $\Delta+\nabla \phi \cdot \nabla$, since it is symmetric with respect to the the measure $d\mu=e^{\phi}\, dx$. In particular the spectrum is always real in 1d and this perhaps explains why it is not easy to produce counteraxamples. $\endgroup$ Jan 13, 2020 at 19:19

1 Answer 1


Here is a construction. It elaborates from perturbation analysis of eigenvalues. However it starts from the situation of a non-simple eigenvalue.

So, let me start with the standard self-adjoint $L_0=-\Delta$. I assume that the Dirichlet problem admits an eigenvalue $\lambda_0$ of multiplicity $2$ exactly. I denote $(u,v)$ an orthonormal basis of the eigenspace. A well-known example is $\lambda_0=5$ for the domain $K=(0,\pi)\times(0,\pi)$ and $$u(x)=\frac2\pi\,\sin x_1\sin2x_2,\quad v=\frac2\pi\,\sin2x_1\sin x_2.$$

Let $M=\sum a_{ij}D_{ij}$ be given with bounded (smooth) functions $a_{ij}$, and form $L_\epsilon=L+\epsilon M$. Perturbation analysis tells us that $L_\epsilon$ admits a stable plane $\Pi_\epsilon$, which depends smoothly upon $|\epsilon|<\!<1$, and $P_0={\rm vec}(u,v)$. In addition, the restriction of $L_\epsilon$ over $\Pi_\epsilon$ is similar to a matrix $C_\epsilon$ such that on the one hand $C_0=\lambda_0I_2$, and on the other hand $$\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}C_\epsilon=\begin{pmatrix} \langle Mu,u\rangle & \langle Mu,v\rangle \\ \langle Mv,u\rangle & \langle Mv,v\rangle \end{pmatrix}=:X_0.$$ Suppose now that $M$ has been chosen so that $$(\dagger)\qquad ( \langle Mu,u\rangle - \langle Mv,v\rangle )^2+4 \langle Mu,v\rangle \langle Mv,u\rangle <0. $$ Then its eigenvalues are complex conjugated. Therefore, for $\epsilon\ne0$ small enough, $C_\epsilon$ will have non-real eigenvalues. Since these are eigenvalues of the Dirichlet problem for $L_\epsilon$, this provides an example.

There remains to find coefficients $a_{ij}$ satisfying ($\dagger$). Let me just choose in my example (on the square $K$) $M=aD_{12}$. The functions $uD_{12}u-vD_{12}v$, $uD_{12}v$ and $vD_{12}u$ are linearly independent. Thus there exists a function $a$ such that $$\int_Ka(uD_{12}u-vD_{12}v)\,dx=0,\qquad\int_KauD_{12}v\,dx=1,\qquad\int_KavD_{12}u\,dx=-1.$$ This ends the construction.

Edit. It is known that one-dimensional elliptic second-order operators have a real spectrum. Actually the eigenvalues are simple, therefore the construction described above cannot be implemented.

Redit. In the construction above, the eigenvalues depend smoothly upon $\epsilon$. There is no branching phenomenon. This happens because the unperturbed operator $L_0$ is self-adjoint, thus diagonalisable.

  • $\begingroup$ Thank you very much. $\endgroup$ Nov 9, 2020 at 18:25

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