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

secondeigenfunction $\varphi_2$ have afinite numberof 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.