1
$\begingroup$

How to find a normalizable solution of the nonlinear differential equation below?

$$ R'' + \frac{R'}{r} - R + R^3 =0 . $$

The domain is $[0,\infty ]$ and we want the norm of the solution to be finite, i.e.,

$$2\pi \int_0^\infty R^2 r d r < \infty .$$

There is no known analytic solution of the differential equation, so we have to resort to numerics. But even with numerics, the problem seems nontrivial as the domian extends to infinity.

How about shooting method?

BTW. The solution looks like this (the original paper is here): enter image description here

$\endgroup$
2
  • $\begingroup$ One possible way is to take a look of the iteration solving $R_n'' + \frac{R_n}{r} - R_n = -R_{n-1}^3$, The starting solution $R_0 = K_0(r)$, the modified Bessel function satisfies $R_0'' + \frac{R_0}{r} - R_0 = 0$, it decays exponentially, and has a mild singularity at $r=0$. Each time you solve this iteration by the standard variation of parameter, $$R_n(r) = -K_0(r)\int^r_0 s I_0(s)R_{n-1}^3(s) ds + I_0(r) \int_0^r s K_0(s) R_{n-1}^3(s) ds + c_1 I_0(r) + c_2 K_0(r)$$ $\endgroup$
    – Yimin
    Commented May 20 at 15:07
  • $\begingroup$ This looks similar to or like a special case of math.stackexchange.com/questions/3164614/…. Only if you select the wrong, too crude boundary condition do you get "interesting" numerical results. More accurate approaches yield boring solutions as you cited. $\endgroup$
    – Lutz Lehmann
    Commented May 20 at 15:24

1 Answer 1

Reset to default
2
$\begingroup$

Origin

At $r=0$ the leading terms in the sense of a Frobenius power series expansion are the Euler-Cauchy terms $$rR''+R'=0.$$ This gives basis solutions $R\sim 1$ and $R\sim \log r$. The finite solution allows to proceed with an ordinary power series for $R(r)=a_0+a_1r+a_2r^2+...$ so that $$ [2a_2+6a_3r+12a_4r^2+..]+[a_1r^{-1}+2a_2+3a_3r+4a_4r^2+..]-[a_0+a_1r+a_2r^2+..]+[a_0^3+3a_0^2a_1r+3(a_0^2a_2+a_0a_1^2)r^2+..]=0 $$ Comparing coefficients gives $a_1=0$, $4a_2=a_0-a_0^3$, $9a_3=a_1-3a_0^3a_1=0$ etc.

This allows to move the initial point of the numerical integration away from the singularity.

Infinity

At the far end with large $r$ we expect $R\approx 0$, so that the equation is approximately $$R''-R=0.$$ This has exponential solutions $e^{\pm x}$. For a finite solution the growing term has to vanish, the expected solution has to be close to $R(r)=ce^{-r}$. So using $a=a_0$ and $c$ as parameters, one can pose 4 boundary conditions at $r_0,r_f$ as $$ R(r_0)=a+\frac a4(1-a^2)r_0^2,~~R'(r_0)=\frac a2(1-a^2)r_0\\ R(r_f)=c\exp(-r_f),~~R'(r_f)=-c\exp(r_f). $$

A straight implementation of this gives the zero-solution as answer. This is due to the situation that all equations are homogeneous, and inequalities need to be implemented via some tricks.

Getting non-zero solutions

to follow

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .