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

For a radial function $u=u(r)$ on ${\mathbb{R}}^d$, the radial Laplacian isThe problem therefore becomesThe equation has the scaling symmetryChoosing $\lambda =u(0{)}^{-(p-1)/2}$ normalizes any positive solution to $u_{\lambda }(0)=1$.

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PutThen $u=r^{-a}v$ and direct differentiation givesSince $ap=a+2$ and $d-2-2a=2(s_c-1)$, the Emden-Fowler transformation turns the equation intowhere

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Seek a homogeneous solution $u(r)=c{r}^{-\beta }$. Matching the powers in $\Delta u+u^p=0$ gives $\beta p=\beta +2$, henceSincematching coefficients givesThus the positive Singular homogeneous solution of the Lane-Emden equation is

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When $s_c=1$, one has $p=(d+2)/(d-2)$. The normalized Aubin-Talenti bubblesolves 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 $r\to \infty$,so $\alpha =d-2$. Moreover $u^{\prime }(r)=O(r^{-(d-1)})$ at infinity and $u^{\prime }(r)=O(r)$ near zero. Thereforeand $\nabla u\in L^2({\mathbb{R}}^d)$.

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Let $A=c_{\infty }^{p-1}$ andUsing the transformed equation,Thus $J$ is a strict Lyapunov function away from equilibria.

On ${\mathbb{R}}_{+}$, $f(0)=0$, it decreases to its unique minimum at $v=c_{\infty }$, wherethen increases, crosses zero at $v=((p+1)/2{)}^{1/(p-1)}{c}_{\infty }$, and tends to $+\infty$.

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Take the regular radial solution with $u(0)=1$. In the transformed variables it satisfiesso $J(s)\to 0$. Since $s_c>1$, the Lyapunov function immediately becomes negative. The trajectory cannot reach $v=0$, where $J={\dot{v}}^2\ge 0$, and $J<0$ confines it below the positive zero of $f$. Hence it remains positive and bounded for all $s$.

The identity $\dot{J}=-4(s_c-1){\dot{v}}^2$ and the LaSalle invariance principle force the omega-limit set to consist of equilibria. The negative limiting energy excludes $v=0$, leavingConsequentlyso $\alpha =2/(p-1)$. At infinity, $|u^{\prime }(r)|\sim a{c}_{\infty }{r}^{-a-1}$, andbecause $s_c>1$. Thus $\nabla u\notin L^2({\mathbb{R}}^d)$.

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