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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?

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    $\begingroup$ $\phi=$ constant is a solution. $\endgroup$
    – Gonçalo
    Commented Aug 19, 2023 at 12:56
  • $\begingroup$ Another remark: if $\phi$ is a non-constant solution to the PDE, then $K^2/\phi$ also is. $\endgroup$
    – Gonçalo
    Commented Aug 19, 2023 at 13:58
  • $\begingroup$ the solutions satisfy $r\left(\phi^2-K^2-\lambda \phi +2K\phi \ln \phi \right)/R = \phi$ for integration constants $R, \ \lambda$ $\endgroup$
    – Sal
    Commented Aug 19, 2023 at 16:43
  • $\begingroup$ @Sal, do you have any reference on how to arrive at the solution? $\endgroup$
    – Angela
    Commented Aug 20, 2023 at 2:10
  • $\begingroup$ That equation is equidimensional in $r$; i.e. invariant under a rescaling $r\rightarrow ar$. According to my copy of Bender and Orszag, the substitutions $r=e^{t}$ and $r\,\frac{d}{dr}=\frac{d}{dt}$ will reduce it to an autonomous equation. $\endgroup$
    – CW279
    Commented Aug 20, 2023 at 5:32

1 Answer 1

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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?

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    $\begingroup$ Thank you for the solution. Does the solution mean that there are no negative values of $\phi$ possible? For negative values, $ln(\phi)$ becomes complex. $\endgroup$
    – Angela
    Commented Aug 21, 2023 at 0:39
  • $\begingroup$ This was an old assignment problem, which I was revisiting. $\endgroup$
    – Angela
    Commented Aug 21, 2023 at 0:39
  • $\begingroup$ @Angela There's no problem with negative $\phi$ -- I just forgot the absolute value bars in the final answer. Nice catch! $\endgroup$
    – CW279
    Commented Aug 21, 2023 at 0:50
  • $\begingroup$ can you point me to the Wolfram page? $\endgroup$
    – Angela
    Commented Aug 22, 2023 at 3:28
  • $\begingroup$ @Angela There's no real need for Wolfram - I added the integration to my post. $\endgroup$
    – CW279
    Commented Aug 22, 2023 at 3:35

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