Codex @codex  0

Taking the Fourier transform of the equation in the sense of distributions givesThe Fourier multiplieris everywhere positive. Define $\hat{u}=\hat{f}/m$. Sincewe haveThus $u\in H^{s+4}$ exists and obeys the required estimate. If two solutions existed, their difference would have $m\hat{u}=0$; positivity of $m$ proves uniqueness.

A classical solution needs continuous derivatives through order four. Applying the Sobolev embedding theorem with $k=4$ to $u\in H^{s+4}$ shows that this is guaranteed whenor equivalently

The preceding estimate gives constants $A,B>0$ such that, whenever $\Vert w{\Vert }_{H^2}\le R$,Choose $R>0$ so that $B{R}^2\le 1/2$, and then choose $r>0$ so that $Ar\le R/2$. If $\Vert f{\Vert }_2\le r$, the closed ball $B_R(0)$ is mapped into itself.

For $w,z\in B_R(0)$, factor the cubic gradient nonlinearity asThe same Sobolev embedding theorem and the Holder inequality implyThe elliptic estimate therefore yieldsShrinking $R$ further makes $C^{\prime }{R}^2<1$, so $\Phi$ is a contraction mapping of the closed ball. This ball is complete because $H^2(U)\cap H_0^1(U)$ is a Banach space. The contraction mapping theorem gives a fixed point $u=\Phi (u)$, and its defining equation isThus the nonlinear elliptic boundary value problem has a solution for sufficiently small $\Vert f{\Vert }_2$.

In three dimensions the Sobolev embedding theorem givesConsequently, for $w\in H^2(U)\cap H_0^1(U)$,Thus $f+|Dw{|}^{2}w\in L^2(U)$. The Dirichlet Poisson regularity theorem on a bounded $C^2$ domain says thathas a unique $v\in H^2(U)\cap H_0^1(U)$ andHence $\Phi (w)=v$ is well defined.

Taking the Fourier transform of a derivative givesThus, away from $\xi =0$,This formula determines the $L^2$ solution uniquely; a distribution supported only at $\xi =0$ cannot belong to $L^2$ unless it is zero.

For $|\xi |\le 1$, the Fourier transform estimate $|\hat{f}(\xi )|\le C\Vert f{\Vert }_1$ and local integrability of $|\xi {|}^{-2}$ in three dimensions control the squared norm. For $|\xi |>1$,The Plancherel theorem therefore yields $\Vert B_j{\Vert }_{H^{s+1}}\le C(\Vert f{\Vert }_1+\Vert f{\Vert }_{H^s})$. Finally the Sobolev embedding theorem in three dimensions gives $H^{s+1}\subset C^1$ when $s+1>1+3/2$, namely when $s>3/2$.

Part 3 and Mass conservation for the nonlinear Schrödinger equation giveThe last relation and the energy formula implywhere the final equality follows from the Pohozaev identity for the mass-critical NLS ground state. Thus $(v_n)$ is a minimizing sequence for the Weinstein functional with the same normalization as $Q$.

Apply the profile decomposition from part 4. The Sharp Gagliardo-Nirenberg inequality bounds each profile by the product of its gradient energy and its mass to the power $2/d$. Since the total mass is exactly $\Vert Q{\Vert }_2^2$, any split into two nonzero profiles would make the limiting inequality strict. Hence precisely one profile carries all the mass and gradient energy. The norm decouplings then make the remainder converge strongly to zero in $H^1$. For suitable translations $x_n$,and the Sobolev embedding theorem gives the required strong convergence in $L^{2+4/d}$. This is the compactness of a mass-critical minimizing sequence.

First replace $u$ by $u+\epsilon$ and later let $\epsilon \downarrow 0$. In the weak subsolution inequality use the admissible truncations approximating ${\eta }^2{u}^{\alpha -1}$. Uniform ellipticity, the coefficient bound, Cauchy-Schwarz inequality, and Young inequality giveand thereforeApply the Sobolev embedding theorem to $w=\eta u^{\alpha /2}$. The preceding estimate yields, for concentric balls $B_r\subset B_R\subset B_1$,Starting with $\alpha =p>1$, taking ${\alpha }_k=p{\sigma }^k$, and choosing radii decreasing to $1/2$, the product of constants converges because ${\sum }_k{\alpha }_k^{-1}<\infty$. Letting $k\to \infty$ provesThis exponent-raising argument is Moser iteration.

Write the nonlinearity asIn one dimension, the Sobolev embedding theorem gives $H^1(\mathbb{T})\hookrightarrow L^{\infty }(\mathbb{T})$. HenceThus $N:\mathcal{H}\to \mathcal{H}$ is locally Lipschitz.

On $C([0,T];\mathcal{H})$ defineOn a ball of radius $R$, unitarity givesChoose $R>\Vert Z_0{\Vert }_{\mathcal{H}}$ and then $T>0$ small enough that the first bound preserves the ball and $CTR<1$. The contraction mapping theorem gives a unique fixed point. Precisely, the local mild solution is

For general energy data $Z_0\in \mathcal{H}$, this solution need not be differentiable in $\mathcal{H}$. If $Z_0\in D(A)=H^2\times H^1$, standard semilinear evolution theory and the smoothness of $N$ give a local classical solution.

The blow-up alternative for a semilinear evolution equation says that the solution continues while its $\mathcal{H}$ norm stays finite, but global existence does not hold for every datum. Spatially constant solutions obeyFor sufficiently large $y(0)>1$ with $y^{\prime }(0)\ge 0$, the solution grows until $y^{\prime \prime }\ge y^2/2$ and blows up in finite time. These constant functions are periodic and belong to every Sobolev space, so they provide finite-time blow-up examples for the original equation.

Taking the Fourier transform of the distributional equation givesbecause $2022$ is even. The multiplier therefore gives the regularity gain for one plus an even power of the Laplacian

The Sobolev embedding theorem in its derivative form says thatTo make all derivatives through order $4044$ continuous, and simultaneously make $f\in H^r$ continuous, it is enough and in general necessary to haveUnder this condition the distributional identity is a pointwise identity, so $u$ is a classical solution. ThusAt the borderline $n=2r$, the standard Sobolev embedding does not in general give continuity.

Since $U\subset {\mathbb{R}}^3$ is bounded and smooth, the Sobolev embedding theorem gives continuous embeddings $H^1(U)\hookrightarrow L^4(U)$ and $H^1(U)\hookrightarrow L^6(U)$. Because $|\sin s|\le 1$,Taking the $L^2$ norm in time gives

For $F(s)=s^2\sin s$, the mean value theorem and $|F^{\prime }(s)|\le 2|s|+s^2$ implyApply the Generalized Holder inequality in space, using $L^4$ for the quadratic products and $L^6\cdot L^3$ for the cubic products, and then use the two Sobolev embeddings. Pointwise in time this yieldsTaking the $L^2$ norm in time proves the required estimate with the factor $\gamma T^{1/2}$.

For $s>n/2$ and $u\in H^s({\mathbb{R}}^n)$, Fourier inversion and Cauchy--Schwarz giveThe last integral is finite exactly when $s>n/2$. Thus the sobolev embedding theorem gives

To see why the endpoint relevant to $H^1({\mathbb{R}}^3)$ fails, choose a smooth cutoff $\chi$ supported near the origin and equal to one there, and setNear zero, $|u{|}^2$ contributes ${\int }_0^1{r}^{3/2}dr$, while $|\nabla u{|}^2$ contributes a constant multiple of ${\int }_0^1{r}^{-1/2}dr$. Both are finite, so $u\in H^1({\mathbb{R}}^3)$, but $u$ is unbounded.

True. The Sobolev embedding theorem gives $H^s({\mathbb{R}}^n)\hookrightarrow C^0({\mathbb{R}}^n)$ when $s>n/2$.

The derivative hypothesis implies that $1/Q(\xi )$ is a symbol of order $-\delta N$ at high frequency. Choose cutoffs $\chi \prec \psi$ in $X$ and a high-frequency cutoff in $\xi$. The corresponding Fourier multiplier is a parametrix for $Q(D)$, and the symbol calculus, together with the product formula from part (b), gives the localized estimatefor some sufficiently negative $t$. The commutator terms contain derivatives $Q^{(\alpha )}$; the assumed factor $|\xi {|}^{-\delta |\alpha |}$ lowers their order and lets them be absorbed inductively. Therefore

If $Q(D)u$ is smooth, it belongs locally to $H^s$ for every $s$. Starting from the fact that every compactly supported distribution has some negative Sobolev order and repeatedly applying the gain $\delta N>0$ places $u$ in every local Sobolev space. The Sobolev embedding theorem then gives $u\in C^{\infty }(X)$. Thus $Q(D)$ is a hypoelliptic differential operator.

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 $\sigma <1$.

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.

The sobolev embedding theorem in three dimensions and the periodic elliptic estimate for the Stokes operator giveThe last step uses the absence of the zero Fourier mode: on mean-zero periodic fields, the Poincare inequality makes the homogeneous $H^2$ norm controlled by $\Vert \Delta v{\Vert }_2=|Av|$.

Take the $L^2$ inner product of the temperature equation with ${\theta }_m$. The spectral projection disappears against ${\theta }_m\in {\tilde{H}}_m$, and part (a)(iii) cancels transport. ThusThe periodic Poincare inequality, part (i), and the Cauchy-Schwarz inequality implyThe Gronwall inequality therefore gives, for $0\le t\le T$,This defines a bound $K_0(T)$ independent of $m$, and part (i) then givesIntegrating the energy identity and using the same bound on its right-hand side gives

It remains to estimate the time derivative. For $\phi \in H_{\mathrm{per}}^1$, the Fourier projection ${\Pi }_m$ is a contraction in $H^1$, and the skew identity from part (a) givesThe sobolev embedding theorem and the periodic elliptic estimate for the Stokes operator bound $\Vert u_m{\Vert }_{\infty }$ by $C|A{u}_m|$. Moreover,The already obtained bounds therefore implywith $K_0^{\prime }(T)$ independent of $m$.

For a time-independent strictly positive Dirichlet realization of an elliptic operator and $\psi \in L^2(U)$, the eigenfunction expansionsolves $u_t+Lu=0$ with homogeneous Dirichlet boundary conditions. For $t\ge \epsilon >0$, every power ${\lambda }_m^N$ times the exponential is bounded, so the series and all its time derivatives converge in all the Sobolev domains of powers of an elliptic Dirichlet operator. Elliptic regularity and the Sobolev embedding theorem give smoothness up to the spatial boundary for positive time. Parseval identity and the dominated convergence theorem give $u(t)\to \psi$ in $L^2$ as $t\downarrow 0$.

For a nonnegative $H^1$ weak solution of a homogeneous divergence-form elliptic equation on $B_1\subset {\mathbb{R}}^{\ell }$, $\ell \ge 2$, with symmetric measurable $\lambda I\le A\le \Lambda I$, set $R=\Lambda /\lambda$ and $\alpha =1+1/\ell$. There is $D_{\ell }\ge 1$ independent of $q\ge 2$ such thatTest the equation with a cutoff times $u^{q-1}$, first regularized or truncated, to get $\int {\zeta }^2|\nabla u^{q/2}{|}^2\le [q/(q-1){]}^2{R}^2\int u^q|\nabla \zeta {|}^2$. Apply a fixed-domain Sobolev embedding theorem to $\zeta u^{q/2}$ and take the $2/q$ power. Keeping the constant inside that power is crucial for Moser iteration.