Let $\lambda\_k$ and $\varphi\_k$ be the $k$-th eigenvalue and a corresponding eigenfunction of the fractional Laplacian in a *bounded* domain $\Omega \subset \mathbb{R}^N$, $N \geq 2$. 
That is, $\lambda\_k$ and $\varphi\_k$ satisfy
$$
\left\{
\begin{aligned}
(-\Delta)^{\alpha/2} \varphi\_k &= \lambda\_k \varphi\_k &&\text{in}~ \Omega,\\
\varphi\_k&=0 &&\text{in}~ \mathbb{R}^N \setminus \Omega.
\end{aligned}
\right.
$$
Here the fractional Laplacian $(-\Delta)^{\alpha/2}$ is given by
$$
(-\Delta)^{\alpha/2} u = - C\lim\_{\varepsilon \to 0+} \int\_{\mathbb{R}^N \setminus B\_\varepsilon(0)} \frac{u(y)-u(x)}{|y-x|^{N+\alpha}} \, dy,
$$
where $C>0$ is a some explicit constant and $\alpha \in (0,2)$. 


If $\alpha=2$, then the fractional Laplacian becomes the classical Laplace operator and it is very well known that the number of nodal domains of $\varphi\_k$ is bounded from above by $k$. This is the content of the **Courant nodal domain theorem**.


However, in the purely fractional case $\alpha \in (0,2)$ much less seems to be known in this respect. Some upper bounds on the number of nodal domains in 1D-case and $\alpha=1$ can be found in [Bañuelos, Kulczycki, 2004].


I'm interested in the following very particular case of a weak version of the Courant nodal domain theorem:



> 
> Does any **second** eigenfunction $\varphi\_2$ have a **finite number** of nodal domains?
> 
> 
> 


Maybe the answer is well-known for experts or, at least, can be relatively easy to obtain (e.g., by using some variational properties of $\lambda\_2$)? Perhaps, the answer is known in the radially-symmetric case? 
I'll appreciate any hint.



Courant nodal domain theorem for fractional Laplacian
As far as I can tell, although it is widely believed that the Courant–Hibert estimate holds for the fractional Laplacian, even finiteness of the number of nodal domains of the second eigenfunctions is open for a general domain. The only known result is that the "harmonic extension to the upper half-space" (in the sense of Caffarelli–Silvestre for $\alpha \ne 1$) has at most two nodal domains.




---


For a ball (and – if I am not mistaken – more generally, for any domain invariant under rotations), the second eigenfunction is either radial or antisymmetric. For a ball, it is conjectured that it is always antisymmetric, but the proof is only known for low dimensions; see my paper with Bartłomiej Dyda and Alexey Kuznetsov, [DOI:10.1112/jlms.12024](https://doi.org/10.1112/jlms.12024), and a follow-up by Rui A.C. Ferreira [DOI:10.1007/s00030-019-0554-x](https://doi.org/10.1007/s00030-019-0554-x).


If the second eigenfunction is antisymmetric, a rather standard Perron–Frobenius-type argument shows that it has two nodal domains. If it happens to be radial, it could have up to three nodal domains, again by the harmonic extension argument and an application of the classical Courant–Hilbert theorem. In any case, the number of nodal domains of the second eigenfunction is at most three.


