Put
$$
a=\frac2{p-1},\qquad s=\log r,
\qquad v(s)=r^au(r).
$$
Then $u=r^{-a}v$ and direct differentiation gives
$$
\Delta u=r^{-a-2}
\left(\ddot v+(d-2-2a)\dot v-a(d-2-a)v\right).
$$
Since $ap=a+2$ and $d-2-2a=2(s_c-1)$, the <Emden-Fowler transformation> turns the equation into
$$
\ddot v+2(s_c-1)\dot v-c_\infty^{p-1}v+v^p=0,
$$
where
$$
c_\infty^{p-1}=a(d-2-a)
=\frac2{p-1}\left(d-2-\frac2{p-1}\right)>0.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 154 / 1 / 1 / 2 / Solution

When $s_c=1$, one has $p=(d+2)/(d-2)$. The normalized <Aubin-Talenti bubble>
$$
u(r)=\left(1+\frac{r^2}{d(d-2)}\right)^{-(d-2)/2}
$$
solves the radial equation, has $u(0)=1$, and tends to zero. It also follows from the stated one-dimensional soliton after the <Emden-Fowler transformation>, because the damping term vanishes.

As $r\to\infty$,
$$
r^{d-2}u(r)\longrightarrow[d(d-2)]^{(d-2)/2}>0,
$$
so $\alpha=d-2$. Moreover $u'(r)=O(r^{-(d-1)})$ at infinity and $u'(r)=O(r)$ near zero. Therefore
$$
\int_{\mathbb R^d}|\nabla u|^2,dx
\asymp\int_0^\infty|u'(r)|^2r^{d-1},dr<\infty,
$$
and $\nabla u\in L^2(\mathbb R^d)$.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 154 / 1 / 1 / 4 / Solution

{c}
{wiki=Emden–Fowler_equation}

The Emden-Fowler transformation replaces a radial variable $r$ by logarithmic time $s=\log r$ and rescales the dependent variable by a power of $r$. It converts many scale-invariant radial equations into autonomous ordinary differential equations.


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 Incoming links: Emden-Fowler transformation