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

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?

• $\phi=$ constant is a solution. Commented Aug 19, 2023 at 12:56
• Another remark: if $\phi$ is a non-constant solution to the PDE, then $K^2/\phi$ also is. Commented Aug 19, 2023 at 13:58
• the solutions satisfy $r\left(\phi^2-K^2-\lambda \phi +2K\phi \ln \phi \right)/R = \phi$ for integration constants $R, \ \lambda$
– Sal
Commented Aug 19, 2023 at 16:43
• @Sal, do you have any reference on how to arrive at the solution? Commented Aug 20, 2023 at 2:10
• 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. Commented Aug 20, 2023 at 5:32

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?

• 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. Commented Aug 21, 2023 at 0:39
• This was an old assignment problem, which I was revisiting. Commented Aug 21, 2023 at 0:39
• @Angela There's no problem with negative $\phi$ -- I just forgot the absolute value bars in the final answer. Nice catch! Commented Aug 21, 2023 at 0:50
• can you point me to the Wolfram page? Commented Aug 22, 2023 at 3:28
• @Angela There's no real need for Wolfram - I added the integration to my post. Commented Aug 22, 2023 at 3:35