Fractional Laplacian equation on a ball and explicit solutions Let us consider
\begin{align*}
(-\Delta)^s u &= 0 && x \in B_r(0) \\
u&=0 && x \in \mathbb R^N \setminus B_r(0),
\end{align*}
where $$
(-\Delta)^s u(x) = \int_{\mathbb{R}^N} \frac{u(x)-u(y)}{|x-y|^{N+2s}} dy,
$$  ($0<s<1$)
is the fractional Laplacian. How can I find solutions of this problem in the form $v(r)\omega$, with $\omega \in S^{N-1}$?
In a comment on Solution of the fractional Laplace equation on a ball, it was said that the solution in $B_r(0) \setminus \{0\}$ is
$$v(\rho) = C_1\rho(1-\rho)^{s-1} + C_2 \rho g(\rho),$$ where $g(\rho)$ is the radial profile of the Green's function $G(x,0)$ in dimension $N+2$ and that to have a solution of the problem on $B_r(0)$, we also need $C_2 \equiv 0$. Is this correct? How do you prove it?
 A: Here are some additional details to what I wrote under the previous question. Our key tool is the following result.
Bochner's relation: Let $V(x)$ be a solid harmonic polynomial of degree $\ell$: a homogeneous polynomial of degree $\ell$ such that $\Delta V = 0$. Then:

*

*the $N$-dimensional fractional Laplace operator $L$ applied to $V(x) f(|x|)$ is of the form $V(x) g(|x|)$

*with the same $f$ and $g$, the $N + 2 \ell$-dimensional fractional Laplace operator $L$ applied to $f(|y|)$ is equal to $g(|y|)$.

(In fact the above result does not depend on any particular properties of the fractional Laplace operator: $L$ can be any operator with prescribed radial Fourier symbol.)
Note: this is essentially a pointwise result, so it applies equally well to global solutions and solutions in spherically symmetric domains. This is discussed in some detail in my paper with Bartłomiej Dyda and Alexey Kuznetsov:

*

*B. Dyda, A. Kuznetsov, M. Kwaśnicki, Fractional Laplace operator and Meijer G-function, Constructive Approx. 45(3) (2017): 427–448, DOI:10.1007/s00365-016-9336-4.


If I understand the question correctly, we are asking for a vector-valued solution $u(x)$ of
$$ L u(x) = 0 \qquad \text{for } x \in B$$
of the form $u(x) = v(|x|) \tfrac{x}{|x|}$. In this case the $j$th coordinate of $u$ is equal to $v(|x|) \tfrac{x_j}{|x|} = f(|x|) x_j$, where $f(r) = v(r) / r$. Since $V(x) = x_j$ is a solid harmonic polynomial of degree $\ell = 1$, we can find $f$ by solving a radial problem
$$ L[f(|x|)] = 0 \qquad \text{for } x \in B$$
in dimension $N + 2$. But then it is known that
$$ f(r) = C (1 - r^2)^{s-1} , $$
and therefore
$$ v(r) = C r (1 - r^2)^{s - 1} . $$
If we allow for a singularity at $0$, then there is an additional solution given by the $N+2$-dimensional Green function of a ball:
$$ f(r) = C_1 (1 - r^2)^{s-1} + C_2 r^{2s - (N + 2)} \int_0^{r^{-2} - 1} \frac{t^{s - 1}}{(t + 1)^{(N + 2)/2}} \, dt . $$
Thus,
$$ v(r) = C_1 r (1 - r^2)^{s-1} + C_2 r^{2s - N - 1} \int_0^{r^{-2} - 1} \frac{t^{s - 1}}{(t + 1)^{1 + N/2}} \, dt . $$
Now of course proving the above rigorously requires some care. This seems doable, but my time is limited now, so I stop here.
