Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 154 (Introduction to Nonlinear Analysis)

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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The homogeneous Sobolev embedding theorem givessoNow $s_c>0$ is equivalent to $p-1>4/d$, which gives $\sigma >2/(p+1)$. Similarly $s_c<1$ is equivalent to $(d-2)(p-1)<4$, which gives $\sigma <1$.

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Interpolate between $L^2$ and ${\dot{H}}^1$ and then use the preceding Sobolev embedding theorem:Mass conservation for the nonlinear Schrödinger equation fixes $\Vert u(t){\Vert }_2=\Vert u_0{\Vert }_2$, while Energy conservation for the nonlinear Schrödinger equation and $E(u_0)\le 0$ implySince $\sigma (p+1)>2$, rearrangement givesuniformly on the maximal lifespan.

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SetSubstituting $u_t=i(\Delta u+|u{|}^{p-1}u)$ and integrating by parts gives the localized virial identityDifferentiating once more, integrating the Laplacian terms twice, and observing that the gauge-invariant nonlinearity contributes only through $\Delta \chi$, givesThese are the required formulas, with ${\chi }^{\prime \prime }|\nabla u{|}^2$ interpreted as the Hessian quadratic form for a radial weight.

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For a smooth radial function that decays at infinity,The Cauchy-Schwarz inequality therefore yields the Radial Sobolev inequalityup to the harmless common normalization of surface measure. Taking the supremum over $r\ge R$ gives the claimed estimate; density extends it from smooth radial functions to every $u\in H_r^1$.

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For ${\psi }_R(x)=R^2\psi (x/R)$, the assumptions imply $\mathrm{Hess}{\psi }_R\le I$, while on $|x|\le 2R$ one hasApply the localized virial identity and compare its interior terms withThe coefficient $d(p-1)/4$ is strictly greater than one because $s_c>0$. The resulting negative multiple of $\Vert \nabla u{\Vert }_2^2$ can be moved to the left. All errors in the nonlinear term are supported on $|x|\ge 2R$, and ${\Delta }^2{\psi }_R=O(R^{-2})$ is supported on $2R\le |x|\le 10R$. Thus

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Suppose that $T=\infty$. Since $s_c<1$ and $d\ge 3$,The Radial Sobolev inequality and Mass conservation for the nonlinear Schrödinger equation giveBecause $(p-1)/2<2$, Young inequality and a sufficiently large fixed $R$ absorb this term into the left side of the estimate from part 5. The annular term is at most $C(u_0)R^{-2}$. Using the uniform lower bound $\Vert \nabla u(t){\Vert }_2\ge c(u_0)$ from part 2 and enlarging $R$ once more yields

Since $I_{{\psi }_R}^{\prime }=2V_{{\psi }_R}$, two integrations giveThe right-hand side is negative for large $t$, contradicting $I_{{\psi }_R}(t)=\int {\psi }_R|u{|}^2\ge 0$. Hence the maximal forward lifespan is finite: $T<\infty$.

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The NLS ground state is the positive radial solution ofIt is, up to phase, translation, and scaling, the optimizer of the Sharp Gagliardo-Nirenberg inequalityConsequentlyEquality in the sharp inequality occurs precisely for the symmetry orbit of $Q$.

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For every real $a$, the gauge transform $w_a=e^{ia\psi }w$ has the same $L^2$ norm as $w$. Since $\Vert w{\Vert }_2\le \Vert Q{\Vert }_2$, the variational bound from part 1 gives $E(w_a)\ge 0$. Expanding its gradient giveswhereThis quadratic polynomial is nonnegative for every $a\in \mathbb{R}$, so its discriminant is nonpositive. Thereforewhich is the desired estimate.

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If $\Vert u_0{\Vert }_2<\Vert Q{\Vert }_2$, the sharp variational estimate and the conservation laws giveThus the $H^1$ norm remains bounded on every finite time interval. The Blowup alternative for the nonlinear Schrödinger equation then prevents a finite endpoint of the maximal lifespan, so $T=\infty$.

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Let $(u_n)$ be bounded in $H_r^1({\mathbb{R}}^2)$. After passing to a subsequence it converges weakly in $H^1$, while the Rellich-Kondrachov compactness theorem gives strong $L^4$ convergence on every bounded ball. The Radial Sobolev inequality gives uniformlyThe same estimate applies to the weak limit. Choosing $R$ large and then using local compactness proves strong convergence in $L^4({\mathbb{R}}^2)$. Henceis compact.

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By the Blowup alternative for the nonlinear Schrödinger equation, choose $t_n\uparrow T$ with $\Vert \nabla u(t_n){\Vert }_2\to \infty$, and setThe mass-critical scaling gives

The sequence is bounded and radial in $H^1$, so part 4 yields, after extraction, weak $H^1$ and strong $L^4$ convergence to some $v$. The energy identity gives $\Vert v_n{\Vert }_4^4\to 2\Vert \nabla Q{\Vert }_2^2$. The Sharp Gagliardo-Nirenberg inequality and weak lower semicontinuity force equality in every limiting norm inequality. Hence $v_n\to v$ strongly in $H^1$, and the variational characterization identifiesfor some $\gamma \in \mathbb{R}$; radiality removes translations and the gradient normalization fixes the scale.

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Choose a smooth radial cutoff ${\theta }_R$ that vanishes on $|x|\le R$, equals one on $|x|\ge 2R$, and satisfies $|\nabla {\theta }_R|\le C/R$. The first localized virial identity and the estimate from part 2, applied to $w=u(t)$ and $\psi ={\theta }_R$, givebecause the mass and energy are conserved.

For fixed $R$, the strong profile convergence from part 5 and ${\lambda }_n\to 0$ implyIntegrating the flux estimate from $t$ to $t_n$ and then letting $n\to \infty$ yieldsuniformly for $0\le t<T$. Taking $R$ sufficiently large proves the claim.

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Change variables $x={\lambda }_{n}y$ and use the profiles from part 5:Since $v_n\to Q{e}^{i\gamma }$ in $L^2$, the difference made by replacing $|v_n{|}^2$ with $|Q{|}^2$ tends to zero against the bounded function $\varphi ({\lambda }_n\cdot )$. For each fixed $y$, continuity gives $\varphi ({\lambda }_{n}y)\to \varphi (0)$, and dominated convergence then givesThus the mass measures $|u(t_n,x){|}^{2}dx$ converge weakly to the Dirac delta function $\Vert Q{\Vert }_2^2{\delta }_0$.

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