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.

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.

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: $T<\infty$.

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

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 $u\in H_r^1$.

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.

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 $\nabla u\notin L^2({\mathbb{R}}^d)$.

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 $+\infty$.

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 $\nabla u\in L^2({\mathbb{R}}^d)$.

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

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

Suppose $G\cong D_{10}$. For residue characteristic two, $G_1$ is a normal $2$-group, $G_0/G_1$ embeds in $k_L^{\ast }$, and $G/G_0\cong \mathrm{Gal}(k_L/{\mathbb{F}}_2)$ is cyclic. The only proper nontrivial normal subgroup of $D_{10}$ is its rotation subgroup $C_5$, and it has no nontrivial normal $2$-subgroup.

Thus either $G_0=G$ or $G_0=C_5$. In the first case $G_1=1$, so part (b) would embed the noncyclic group $D_{10}$ in the cyclic group $k_L^{\ast }$, impossible. In the second case the residue degree is $|G/G_0|=2$, so $k_L={\mathbb{F}}_4$ and $k_L^{\ast }$ has order three; it cannot contain the injected group $G_0/G_1=C_5$. Both cases contradict the ramification constraints, so no such extension exists.

The polynomial $X^3-2$ is Eisenstein over ${\mathbb{Q}}_2$, hence irreducible. Its discriminant is $-108=36(-3)$, and $-3\equiv 5(\mathrm{mod}8)$ is not a square in ${\mathbb{Q}}_2$. Therefore its splitting field has Galois group $S_3\cong D_6$.

This is always true. Every open ball in an ultrametric space is also closed: a point outside a ball has a disjoint ball of the same radius around it. Distinct points can therefore be separated by clopen sets, so every connected subset is a singleton and $K$ is totally disconnected.

This can be false. In a valued field with value group $\mathbb{Q}$, such as the Hahn series field $k((t^{\mathbb{Q}}))$, the maximal idealof the valuation ring is not principal: if $v(a)=q>0$, an element of valuation $q/2$ belongs to $\mathfrak{m}$ but not to $(a)$. Thus the valuation ring need not be a principal ideal domain.

This is always true. If, say, $|x|>|y|$, then the ultrametric inequality applied to $x=(x+y)-y$ forces $|x+y|=|x|$; applying it to $x=(x-y)+y$ similarly gives $|x-y|=|x|$.

This can be false: take $y=-x\ne 0$. Then $|x+y|=0<|x|=\min (|x|,|y|)$. The upper inequality is always the ultrametric inequality.

The HNN extension is the Baumslag-Solitar groupIts Bass-Serre tree has vertices $G/⟨u⟩$ and oriented edges $G/⟨u⟩$, with the two endpoint maps induced by the identity embedding and the index-two embedding $n\mapsto 2n$. At each vertex there is one incident edge on the identity side and two on the index-two side. The underlying unoriented tree is therefore infinite and $3$-regular, with an orientation in which every vertex has one incoming and two outgoing edges, up to reversing the convention.

WriteIts Bass-Serre tree has verticesand edges $G/C$, where $C=⟨u^2⟩=⟨v^3⟩$; the edge $gC$ joins $g⟨u⟩$ to $g⟨v⟩$. Sincethis is the infinite $(2,3)$-biregular tree: every $u$-type vertex has degree two and every $v$-type vertex has degree three.

The based Schreier coset graph has vertices $⟨a⟩g$ and an $s$-edge from $⟨a⟩g$ to $⟨a⟩gs$. At the base vertex $⟨a⟩$ there is an $a$-loop. There are no other cycles: the fundamental group of the covering is the subgroup $⟨a⟩\cong \mathbb{Z}$, and its displayed loop already generates it.

Equivalently, start with one $a$-circle at the base vertex and attach labelled trees so that every vertex has exactly one incoming and one outgoing $a$-edge and exactly one incoming and one outgoing $b$-edge. After deleting the base $a$-loop, the underlying graph is a tree. This describes the complete infinite covering and distinguishes it from its finite core, which is just the $a$-loop.