Solve a kind of Emden-Fowler equation Suppose $n$ is an integer, $p>1$, we have Emden-Fowler equation $v''+\frac{n-1}{r}v'+v^p=0$ derived from radial solution to $-\Delta u = u^p$. I tried to use Emden-Fowler transform $w(t) = v(t^{-\frac{1}{n-1}})$ and find the equation comes to be $w''(t) = -\frac{1}{(n-2)^2}t^{\frac{2-2n}{n-2}})w(t)^p$, which deletes the first-order derivative. However, I dont know how to deal with this equation.
 A: Letting $v=r^ku$ and $r=e^x$, where $k-1=1+pk$,
\begin{align}
u''+[2k+n-2]u'+[k^2+(n-2)k+1]u^p=0,\quad [u'=\mathrm d_x u].
\end{align}
Now taking $u'=z(u)$ & $(2k+n-2)u=t$ you'll get an Abel equation of the second kind,
\begin{align}
zz'-z=[k^2+(n-2)k+1][2k+n-2]^{p-1}t^p,\quad [z'=\mathrm d_tz].
\end{align}
This has nice solutions only for specific values of $p$, which you can see in "Handbook of exact solutions for ordinary differential equations" 2ed. by Andrei D. Polyanin & Valentin F. Zaitsev. Those solutions are unfortunately only for values of $p\leq1$.
There is, however, a proposed general solution in the paper "Exact analytic solutions of the Abel, Emden–Fowler and generalized Emden–Fowler nonlinear ODEs" by Dimitrios E. Panayotounakosa & Dimitrios C. Kravvaritis. The solution, which is a bit messy and the authors have left me to substitute a few equations together, is
\begin{align}
\left(\frac{2z}{t+c}-\frac{1}{3}\right)^3&+\left[\frac{4(\mu-F)}{t+c}-\frac{7}{3}\right]\left(\frac{2z}{t+c}-\frac{1}{3}\right)+\frac{4\mu}{3(t+c)}-\frac{20}{27}=0,\tag{1}
\\
\\
\text{where}\quad\mu(\xi)&=\frac{[(\xi\sin(\xi)+\cos(\xi))\mathrm{Ci}(\xi)+\cos(\xi)^2][4\xi\mathrm{Ci}(\xi)+\cos(\xi)]}{16e^\xi(\xi \mathrm{Ci}(\xi))^3},\\
\xi=\ln|t+c|,\quad \mathrm{Ci}&(\xi)=-\int_\xi^\infty\frac{\cos(s)}{s}\mathrm ds,\quad\text{and}\quad F=[k^2+(n-2)k+1][2k+n-2]^{p-1}t^p,
\end{align}
and $c$ is the constant of integration.
This paper (is free and) plots solutions to Abel equations with different functions $F$ and seems to verify that this solution is indeed correct. I'm not entirely convinced as I've been unable to replicate their graphs, so buyer beware.
Suppose that the solution was correct, solving equation $(1)$ for $z$ gives
\begin{align}
z=\frac{t+c}{2}\left[\frac{1}{3}+\sqrt[3]{\sqrt{j^3+q^2}-q}+\sqrt[3]{-\sqrt{j^3+q^2}-q}\right],\\
\text{where}\quad j=\frac{4(\mu-F)}{3(t+c)}-\frac{7}{9}\quad \text{and}\quad q=\frac{2\mu}{3(t+c)}-\frac{10}{27}.
\end{align}
Note that $t$, $\mu$, and $F$ are functions of $u$, and $z=u'$, so this is a separable equation for $u(x)$:
\begin{align}
\int \frac{2}{t+c}\left[\frac{1}{3}+\sqrt[3]{\sqrt{j^3+q^2}-q}+\sqrt[3]{-\sqrt{j^3+q^2}-q}\right]^{-1}\mathrm du=\ln|r|+c_2,
\end{align}
where $t=(2k+n-2)u$, $u=r^{2/(p-1)}v$, and $j$, $q$, $\mu$, and $F$ are defined above.
This is where the road of analytical solutions comes to end. Probably not what you were hoping for.
