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Two coupled non-linear differential equations in a radial $r$-direction in the region $r \in [0, \infty)$:

$$ -a\big(\partial_r^2+\frac{\partial_r}{r}-\frac{n^2}{r^2}+c\big) U(r)+ B(r) (\partial_r-\frac{n}{r}\big) V(r)=n\; k \; U(r) $$ $$ -B(r) (\partial_r+\frac{n}{r}\big) U(r) + a\big(\partial_r^2+\frac{\partial_r}{r}-\frac{n^2}{r^2}+c\big) V(r) =n\; k \; V(r), $$ We like to solve $U(r)$ and $V(r)$ (say, better analytically and exactly). What are the exact solutions of $U(r)$ and $V(r)$? (or approximate solutions?)

This is a more general set of equations extending from the post here

Here are the conditions of $B(r),a,c,k$.

The B(r) is given such that $B(r)$ is a nice smooth differentiable function, with $$B(0)=0$$ $$\lim_{r \to 0} B(r)=0$$ $$\lim_{r \to \infty} B(r)=b=\text{constant} >0,$$ and $B(r) \geq 0$ is monotonically increasing along $r \in [0, \infty)$, also $$a=\text{constant} >0.$$ $$c=\text{constant} \geq 0.$$

$$n \in \mathbb{Z}^+=\text{integer natural number constant} >0.$$

$$k=\text{constant} >0.$$ For your convenience, here $k$ can be chosen of our convenience such that the equations can be solved more easily. My suggestion and trial finds that $$ k=\frac{b^2 }{a\; c} $$ or at least the leading order $$ k=\frac{b^2 }{a\; c} + \cdots $$ look more promising for exact solutions.

Both $a$, $b$ and $c$ 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 $J_0(r),J_1(r), ...,etc$.

I suppose that their positive-valued peaks are localized near $r=0$, but oscillating modes along $r$, where $U(0)$ and $V(0)$ are nearly in their maximum, with exponential decay tails $\lim_{r \to 0} U(r)=\lim_{r \to 0} V(r)=0.$ But I could be wrong.

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.

It will be OK, for your answer, just for focus on the simple cases that $n=1$ or $n=2$. Even though one may consider generic $n \in \mathbb{Z}^+$ in general, but it is not necessary.

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The special cases of my questions boil down to here and here.

(p.s. This is some fun trial analysis done by myself.)

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    $\begingroup$ This is similar to a problem I encounter.... there I need to at least know the asymptotic behaviors. Can you set up a bounty yourself to attract more attentions? $\endgroup$ – annie heart Nov 9 '17 at 4:49
  • $\begingroup$ @annieheart The quoted post resets it as an integral equation problem (by switching to $U+V$ and $U-V$ as suggested and putting everything without derivatives to the right), so you can get something that can allow the Picard iteration scheme at least for some values of parameters (or to determine admissible asymptotics, if you know that it exists from other considerations) Alas, since the coefficients on the right are the same rather than the opposite and the signs at $n/r$ differ as well, the system does not split any more, which means that nice formulae are, probably, out of question... $\endgroup$ – fedja Nov 13 '17 at 2:05
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Due to the lack of solutions, I thought it will still be nice to consider a particualr case --- Here are the solutions for the $n=0$ case. 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 $$ U_+(r)= c_+ \int_{0}^r dw \cdot \exp(\frac{1}{a}\int_{0}^w ds (B(s) -\frac{a}{s}) $$ $$ U_-(r)= c_- \int_{0}^r dw \cdot \exp(\frac{-1}{a}\int_{0}^w ds (B(s) +\frac{a}{s}) . $$ 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 $c_+$, $c_-$ which are some constants to fit the boundary conditions.

Maybe there is a way to deal with the general $n$ from here.

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  • $\begingroup$ Thanks. As we know, this is just an partial answer, I would really be more happier if someone can provide a better full answer. $\endgroup$ – annie heart Nov 16 '17 at 16:17

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