Localized center modes with exponential decay tails, solved from non-linear differential equations Two coupled non-linear differential equations in a radial $r$-direction in the region $r \in [0, \infty)$:

$$-a\big(\partial_r^2+\frac{\partial_r}{r}\big) U(r)+ 
B(r) \partial_r V(r)=0,
$$
  $$
-B(r) \partial_r 
U(r)
+
a\big(\partial_r^2+\frac{\partial_r}{r}\big) V(r)
=0,
$$
  We like to solve $U(r)$ and $V(r)$. 

The B(r) is given such that $B(r)$ is a nice smooth differentiable function, with
$$B(0)=0$$
$$\lim_{r \to 0} B(r)=0$$
$$\lim_{r \to \infty} B(r)=b=constant >0,$$
and $B(r)>0$ is monotonically increasing along $r \in [0, \infty)$,
also
$$a=constant >0.$$
Both $a$ and $b$ are finite values.
I have done some analysis myself. My expected analysis find that
$U(r)$ and $V(r)$ have exponential decay tails that look like 
$$\exp[-\int_0^r B(r')^{\#} dr']$$
The ${\#}$ means some tentative power. And both $U(r)$ and $V(r)$ likely contain Bessel functions $J_0(r),J_1(r), ...,etc$.

What are the exact solutions of $U(r)$ and $V(r)$?

I suppose that they have localized center modes at $r=0$ (namely, $U(0)$ and $V(0)$ are maximum and positive) with exponential decay tails $\lim_{r \to 0} U(r)=\lim_{r \to 0} V(r)=0.$
If exact analytic solutions are NOT possible, please give arguments, and please feel free to take approximations. Personally I believe that it can be solved analytically exactly by some Bessel type functions. 
(p.s. This is not a homework problem. Just do some trial analysis done by myself.)
 A: I am assuming below that $\frac{\partial_r}{r}U$ means $\frac{1}{r}\partial_rU$, not $\partial_r(\frac{1}{r}U)$.
Let $u=\partial_rU,v=\partial_rV$; your equations become
\begin{align}
a(\partial_r+\frac{1}{r})u-B(r)v=0,\\
a(\partial_r+\frac{1}{r})v-B(r)u=0.
\end{align}
Let $y=v+u,z=v-u$.  Adding and subtracting the equations gives a decoupled system of first order linear ODEs
\begin{align}
a(\partial_r+\frac{1}{r})y-B(r)y=0,\\
a(\partial_r+\frac{1}{r})z+B(r)z=0.
\end{align}
Each of these can be solved by quadratures, which yields $u(r)$ and $v(r)$, and finally $U(r),V(r)$.
A: Here are the solutions. Define $U(r) \pm V(r)=U_\pm(r)$, and $\partial_r U_{\pm}\equiv U'_{\pm}$, the $\pm$ linear combinations give
$$
-a \partial_r (U'_+(r))+ 
(B(r) -\frac{a}{r}) (U'_+(r))=0
$$
$$
-a\partial_r (U'_-(r))- 
(B(r) +\frac{a}{r}) (U'_-(r))=0,
$$
Thus
$$
 U_+(r)= C_+\exp( c_+ \int_{0}^r dw \cdot \exp(\frac{1}{a}\int_{0}^w ds (B(s) -\frac{a}{s}) )
$$
$$
 U_-(r)=C_-\exp( c_- \int_{0}^r dw \cdot \exp(\frac{-1}{a}\int_{0}^w ds (B(s) +\frac{a}{s}) ).
$$
In the end, we can plug in to solve.
$$
U(r)=\frac{1}{2}( U_+(r)+ U_-(r)) 
$$
$$
V(r)=\frac{1}{2}( U_+(r)- U_-(r))
$$
As for appropriate boundary conditions, one can choose the values of $C_+$, $c_+$, $C_-$, $c_-$ which are some constants to fit the boundary conditions. 
(p.s. I figured very soon, and was posting the solution, when user254433 also contributes an answer. But here just showing the solutions more explicitly. )
