Suppose $u:(0,1)\rar E$ solves the seconde order ODE
$$
u^\dprime(r)+br^{-1}u^\prime(r)-cr^{-2}Du(r)=0,
\quad b,c\in\C,
$$
for some linear operator $D\in\Hom(E)$. Suppose for another linear operator $A\in\Hom(E)$ the operator $r^A\colon=\exp(\log r\,A)$ is defined and $C^2$ in $r$. Then for $x\in E$ the function $u(r)=r^A x$ solves the ODE, iff
$$
\Big(A(A-1)+bA-cD\Big)r^{A-2}x=0,
\quad\mbox{i.e.}\quad
A=-\tfrac12(b-1)\pm\sqrt{\tfrac14(b-1)^2+cD}
$$
whenever this makes sense and it definitely makes sense in case $D$ is a positive, self-adjoint operator on a Hilbert space $E$.
Let $\s$ denote the normalized surface measure on the sphere $S^{d-1}$. The Poisson semigroup on $S^{d-1}$ is given by
$$
P_tf(x)
=\int_{S^{d-1}}\frac{(1-e^{-2t})}{\norm{e^{-t}x-y}^d}\,f(y)\,\s(dy)~.
$$
The generator of this semigroup is
$$
-H=\tfrac12(d-2)-\sqrt{\tfrac14(d-2)^2+\D_{S^{d-1}}}
$$
where $\D_{S^{d-1}}$ is the Laplacian on $S^{d-1}$.
The map $F:\R^+\times S^{d-1}\rar\R^d\sm\{0\}$, $F(r,x)\colon=rx$ is a coordinate transformation with inverse $z\mapsto(\norm z,z/\norm z)$. If $f:\R^d\sm\{0\}\rar\C$ is smooth and $\wt f\colon=f\circ F$, i.e. $\wt f(r,x)=f(rx)$, then the operator $\wt\D$ defined by $(\D f)(rx)=(\wt\D\wt f)(r,x)$ is called the Laplacian in spherical coordinates. We have:
\begin{equation}\label{eq1}\tag{EQ1}
-\wt\D\wt f
=\pa_r^2\wt f+(d-1)r^{-1}\pa_r\wt f-r^{-2}\D_{S^{d-1}}\wt f_r,
\quad\mbox{where}\quad
\wt f_r(x)\colon=\wt f(r,x)
\end{equation}
Let $u(r)(x)=u(r,x)$ denote the harmonic extension of $f\in C^\infty(S^{d-1})$ into its interior $B_2^d$, i.e. $u(e^{-t})(x)=P_tf(x)$, then by \eqref{eq1} we get:
$$
u^\dprime(r)+r^{-1}(d-1)u^\prime(r)-r^{-2}\D_{S^{d-1}}u(r)=0
$$
The formal solution $u(r)=r^Af$, where
$$
A=-\tfrac12(d-2)+\sqrt{\tfrac14(d-2)^2+\D_{S^{d-1}}}
$$
In contrast to the Ornstein-Uhlenbeck semigroup $A$ is not a diffusion operator.
Remark: the positive sign in front of the square root is choosen in order to get a positive operator with kernel $\ker\D_{S^{d-1}}$, i.e. the constant functions. For the other choice the operator would be strictly negative with trivial kernel. Since $P_tf(x)=e^{-tA}f$, it follows that the generator of $P_t$ is given by
$$
\tfrac12(d-2)-\sqrt{\tfrac14(d-2)^2+\D_{S^{d-1}}}
$$