Localized center modes with exponential decay tails, solved from non-linear differential equations

Question

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.)

Answer 1

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. )

Answer 2

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 $U(r),V(r)$.