solution of $\nabla^2 \phi = K\phi \nabla^2 \frac{1}{\phi}$

Question

Is there any known analytical solution to the below equation? $\nabla^2 \phi = K\phi \nabla^2 \frac{1}{\phi}$, where $K$ is a constant. Assume spherical co-ordinates and spherical symmetry, i.e.,
$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r} r^2 \frac{\partial \phi}{\partial r}$. In case there is no analytical solution, is there an asymptotic solution? i.e. for large r?

Answer 1

Let $r=x$. We can write the ODE as

$$ x\,\frac{d}{dx}\left[x\cdot x\,\frac{d\phi}{dx}\right] =-K\phi\,x\,\frac{d}{dx}\left[\frac{x}{\phi^2}\cdot x\,\frac{d\phi}{dx}\right] $$

Substituting $x=e^t,x\,\frac{d}{dx}=\frac{d}{dt}$ results in, after simplification

$$ \phi''+\phi'=-K\left[\frac{\phi'\phi+\phi''\phi-2\phi'^2}{\phi^2}\right] $$

where all derivatives are taken with respect to $t$. This in an autonomous equation in $\phi$ (ie. there are no explicit appearences of the independent variable $t$). Further substituting $\phi'(t)=\xi(\phi),\phi''=\xi\xi'$ results in, after simplification

$$ \xi'-\frac{2K\xi}{\phi^2+K\phi}=-1 $$

which is a first-order linear ODE in $\xi$ with the integrating factor

$$ \exp{\left[2\ln\left(\frac{K+\phi}{\phi}\right)\right]} =\left(\frac{K+\phi}{\phi}\right)^{\!2} \equiv f(\phi) $$

Thus the linear ODE becomes

$$ \frac{d}{d\phi}\left[\xi f(\phi)\right]=-f(\phi) \implies \phi'\,f(\phi)=-\int d\phi' f(\phi') \implies \frac{f(\phi)}{\int d\phi'f(\phi')}\,d\phi=-dt $$

after replacing $\xi=\phi'$. Performing the trailing integration and changing variables back to $t=\ln{x}$ gives

$$ \int d\phi' f(\phi')=\frac{c_1}{x} $$

for some constant of integration $c_1$. The integral over $f$ is straightforward:

$$ \begin{align} \int f(\phi)\,d\phi =\int\left(\frac{K^2}{\phi^2}+\frac{2K}{\phi}+1\right)d\phi =-\frac{K^2}{\phi}+2K\ln|\phi|+\phi+c_2 \end{align} $$

Hence

$$ -\frac{K^2}{\phi}+2K\ln{|\phi|}+\phi=\frac{c_1}{x}+c_2 $$

which has the symmetry property noted by @Goncalo in the comments. Out of curiosity, where did you find this ODE? Did you just make it up, or did it arise in a physical problem?