Exact Schwarzschild waves concentrated in a photon-sphere annulus

Coordinates and angular functions. Put $r_0=3s\mathrm{/}2$ and define $logy:={\int }_1^{y}d\rho \mathrm{/}\rho$ for $y>0$; its difference quotient gives $(\log y{)}^{\mathrm{\prime }}=1\mathrm{/}y$. On each dyadic interval $[2^j,2^{j+1}]$ and $[2^{-j-1},2^{-j}]$, the integral of $1\mathrm{/}\rho$ is at least $1\mathrm{/}2$, so $\log y\to +\mathrm{\infty }$ as $y\to \mathrm{\infty }$ and $\log y\to -\mathrm{\infty }$ as $y\downarrow 0$. Define the length offset $C_{\ast }:=r_0+s\log (r_0\mathrm{/}s-1)$ and set $x(r)=r+s\log (r\mathrm{/}s-1)-C_{\ast }$, so $x(r_0)=0$. Then $x^{\mathrm{\prime }}(r)=1+s\mathrm{/}(r-s)=1\mathrm{/}f(r)>0$ and the endpoint limits are $-\mathrm{\infty },+\mathrm{\infty }$, so $x:(s,\mathrm{\infty })\to \mathbb{R}$ is bijective. Its inverse $r(x)$ is continuous by strict monotonicity; taking difference quotients in $x(r(x+h))-x(r(x))=h$ gives $r_x=1\mathrm{/}{x}^{\mathrm{\prime }}(r)=f(r)$. Thus $r$ is $C^1$, and the identity $r_x=f(r)$ inductively makes $r$ $C^k$ for every $k$, since $f$ is smooth. In polar coordinates for the unit round sphere define ${\mathrm{\Delta }}_{S^2}v:=(\sin \vartheta {)}^{-1}{\mathrm{\partial }}_{\vartheta }(\sin \vartheta {\mathrm{\partial }}_{\vartheta }v)+(\sin \vartheta {)}^{-2}{\mathrm{\partial }}_{\phi }^{2}v$ and $\mathrm{\mid }{\mathrm{\nabla }}_{S^2}v{\mathrm{\mid }}^2:=\mathrm{\mid }{\mathrm{\partial }}_{\vartheta }v{\mathrm{\mid }}^2+(\sin \vartheta {)}^{-2}\mathrm{\mid }{\mathrm{\partial }}_{\phi }v{\mathrm{\mid }}^2$; these are the scalar Laplace--Beltrami operator and the squared metric norm of the gradient for $d\mathrm{\Omega }=\sin \vartheta d\vartheta d\phi$. For an integer $l\ge 1$ set $I_n:={\int }_0^{\pi }{\sin }^n\vartheta d\vartheta$, $c_l:=(2\pi I_{2l+1}{)}^{-1\mathrm{/}2}>0$, and $Y_l:=c_l({\omega }_1+i{\omega }_2{)}^l$. The polynomial expression makes $Y_l$ smooth at both poles; in polar coordinates $Y_l=c_l{\sin }^l\vartheta e^{il\phi }$. Direct differentiation gives ${\mathrm{\Delta }}_{S^2}{Y}_l=-l(l+1)Y_l$. Integration in $\phi$ gives ${\int }_{S^2}\mathrm{\mid }{Y}_l{\mathrm{\mid }}^{2}d\mathrm{\Omega }=1$, ${\int }_{S^2}{Y}_l^{2}d\mathrm{\Omega }=0$, and ${\int }_{S^2}{\mathrm{\nabla }}_{S^2}{Y}_l\cdot {\mathrm{\nabla }}_{S^2}{Y}_{l}d\mathrm{\Omega }=0$, the latter two because their integrands have the factor $e^{2il\phi }$. Also $\mathrm{\mid }{\mathrm{\nabla }}_{S^2}{Y}_l{\mathrm{\mid }}^2=c_l^2{l}^2{\sin }^{2l-2}\vartheta (1+{\cos }^2\vartheta )$. Integration by parts gives $(2l+1)I_{2l+1}=2l{I}_{2l-1}$, and hence ${\int }_{S^2}\mathrm{\mid }{\mathrm{\nabla }}_{S^2}{Y}_l{\mathrm{\mid }}^{2}d\mathrm{\Omega }=2\pi c_l^2{l}^2(2I_{2l-1}-I_{2l+1})=l(l+1)$.

Exact scalar reduction. In $(t,x,\omega )$ the metric is $f(-d{t}^2+d{x}^2)+r^2{\gamma }_{S^2}$. In a polar sphere chart its inverse components are $(-f^{-1},f^{-1},r^{-2},(r^2{\sin }^2\vartheta {)}^{-1})$ and $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}=f{r}^2\sin \vartheta$. The Christoffel symbols of this metric-selected connection are ${\mathrm{\Gamma }}^c{}_{ab}=\frac{1}{2}{g}^{cd}({\mathrm{\partial }}_a{g}_{bd}+{\mathrm{\partial }}_b{g}_{ad}-{\mathrm{\partial }}_d{g}_{ab})$; for a scalar $u$, ${\mathrm{\nabla }}_a{\mathrm{\nabla }}_{b}u={\mathrm{\partial }}_a{\mathrm{\partial }}_{b}u-{\mathrm{\Gamma }}^c{}_{ab}{\mathrm{\partial }}_{c}u$; substituting the diagonal metric into the Christoffel formula gives $g^{ab}{\mathrm{\Gamma }}^x{}_{ab}=-2\mathrm{/}r$, $g^{ab}{\mathrm{\Gamma }}^{\vartheta }{}_{ab}=-\cot \vartheta \mathrm{/}{r}^2$, and zero for the other two traced components. Consequently ${\mathrm{\square }}_{g}u=-f^{-1}{u}_{tt}+f^{-1}{u}_{xx}+2r^{-1}{u}_x+r^{-2}(u_{\vartheta \vartheta }+\cot \vartheta u_{\vartheta }+{\sin }^{-2}\vartheta u_{\phi \phi })=f^{-1}(-u_{tt}+r^{-2}{\mathrm{\partial }}_x(r^2{u}_x))+r^{-2}{\mathrm{\Delta }}_{S^2}u$. Substituting $u=q(t,x)Y_l(\omega )\mathrm{/}r(x)$ and $r_x=f$ yields ${\mathrm{\square }}_{g}u=\frac{Y_l}{fr}(-q_{tt}+q_{xx}-W_{l}q)$, where $W_l=f(l(l+1)\mathrm{/}{r}^2+s\mathrm{/}{r}^3)\ge 0$ and $f_x=fs\mathrm{/}{r}^2$. For each fixed $l$, $W_l$ and every $x$ derivative are bounded on $\mathbb{R}$: ${\mathrm{\partial }}_x=f{\mathrm{\partial }}_r$, the rational functions extend smoothly to $r=s$, and their derivatives decay at infinity. No uniform-in-$l$ bound is needed for existence.

Positive energy and real fields. For a compactly supported complex $q$ define $e_l[q]=\frac{1}{2}{\int }_{\mathbb{R}}(\mathrm{\mid }{q}_t{\mathrm{\mid }}^2+\mathrm{\mid }{q}_x{\mathrm{\mid }}^2+W_l\mathrm{\mid }q{\mathrm{\mid }}^2)dx$. Use the sum of the real and imaginary scalar energies for a complex field. From [F2]--[F3], its energy density in $x,\omega$ is $\frac{1}{2}(r^2\mathrm{\mid }{u}_t{\mathrm{\mid }}^2+r^2\mathrm{\mid }{u}_x{\mathrm{\mid }}^2+f\mathrm{\mid }{\mathrm{\nabla }}_{S^2}u{\mathrm{\mid }}^2)$. For $q{Y}_l\mathrm{/}r$, integration over $S^2$ gives $\frac{1}{2}(\mathrm{\mid }{q}_t{\mathrm{\mid }}^2+\mathrm{\mid }{q}_x-bq{\mathrm{\mid }}^2+fl(l+1)\mathrm{\mid }q{\mathrm{\mid }}^2\mathrm{/}{r}^2)$, $b=f\mathrm{/}r$. The identity $b^{\mathrm{\prime }}+b^2=fs\mathrm{/}{r}^3$ and integration of $-b{\mathrm{\partial }}_x\mathrm{\mid }q{\mathrm{\mid }}^2$, with zero endpoint terms, prove that its total energy equals $e_l[q]$. The zero bilinear angular integrals in step 1.1 give exactly $E_T[\mathrm{Re}(q{Y}_l\mathrm{/}r)]=e_l[q]\mathrm{/}2$. The original local energy density is nonnegative on every radial subset. The factor $1\mathrm{/}2$ for real fields also holds before the radial integration by parts, so no false boundary-term identity on a finite annulus is being used.

Global smooth radial Cauchy construction. For arbitrary compact smooth data $p,v$, solve the integral equation $q=q_0+\mathcal{K}q$, where $q_0=(p(x+t)+p(x-t))\mathrm{/}2+\frac{1}{2}{\int }_{x-t}^{x+t}v(y)dy$ and $\mathcal{K}q=-\frac{1}{2}{\int }_0^t{\int }_{x-(t-a)}^{x+(t-a)}{W}_l(y)q(a,y)dyda$. On a slab of width $d$ its operator norm on bounded continuous functions is at most $\mathrm{\parallel }{W}_l{\mathrm{\parallel }}_{\mathrm{\infty }}{d}^2\mathrm{/}2$. Choose this below one. Picard iteration from $q_0$ then has geometrically summable successive sup-norm differences, so converges uniformly to the unique bounded continuous solution. Each iterate vanishes outside the unit-speed cone of the initial compact support. Spatial difference quotients of the integral equation inductively give continuous derivatives of every order: the highest differentiated unknown is acted on by the same $\mathcal{K}$, while all lower terms involve already known derivatives and bounded derivatives of $W_l$. Uniform convergence of these difference quotients follows from that same geometric estimate; the fundamental theorem of calculus identifies their limits as the derivatives. Differentiating the integral equation in time gives the initial velocity, $q_{tt}=q_{xx}-W_{l}q$, and all higher time derivatives. Thus the solution is jointly smooth. The fixed slab width depends on $l$ but not on the data size, so restart at successive endpoints and use uniqueness on overlaps to obtain a solution for every real time. Time reversal treats negative times. The iterates and their restarted solutions retain the compact unit-speed support cone. This is an actual Cauchy construction; global hyperbolicity or an unproved scalar-wave existence theorem has not been invoked.

Explicit Gaussian envelope. Let $h=(l+1\mathrm{/}2{)}^{-1}$, $V_0=f\mathrm{/}{r}^2$, $V_1=fs\mathrm{/}{r}^3-V_0\mathrm{/}4$. Then $W_l=h^{-2}{V}_0+V_1$. At $x=0$ direct differentiation gives $V_0(0)={\mathrm{\Omega }}^2$, $V_0^{\mathrm{\prime }}(0)=0$, $V_0^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{/}2=-{\mathrm{\Omega }}^4$, where $\mathrm{\Omega }=2\mathrm{/}(3\sqrt{3}s)>0$. Indeed ${\mathrm{\partial }}_r{V}_0=(-2r+3s)\mathrm{/}{r}^4$, and ${\mathrm{\partial }}_r^2{V}_0(r_0)=-3s\mathrm{/}{r}_0^5$, with ${\mathrm{\partial }}_x=f{\mathrm{\partial }}_r$. Put $z(t)=\cosh (\mathrm{\Omega }t)+i\sinh (\mathrm{\Omega }t)$, $b(t)={\mathrm{\Omega }}^2(\cosh (\mathrm{\Omega }t)-i\sinh (\mathrm{\Omega }t))\mathrm{/}z(t)$, and $a(t,y)=z(t{)}^{-1\mathrm{/}2}{e}^{-b(t)y^2\mathrm{/}2}$, using the smooth square-root branch equal to one initially. Since $\mathrm{Re}z>0$, it never vanishes. Calculation gives $\mathrm{Re}b={\mathrm{\Omega }}^2\mathrm{/}\mathrm{\mid }z{\mathrm{\mid }}^2>0$, $b^{\mathrm{\prime }}=-i(b^2+{\mathrm{\Omega }}^4)\mathrm{/}\mathrm{\Omega }$, and $(z^{-1\mathrm{/}2}{)}^{\mathrm{\prime }}\mathrm{/}{z}^{-1\mathrm{/}2}=-ib\mathrm{/}(2\mathrm{\Omega })$. Therefore $2i\mathrm{\Omega }{a}_t=-a_{yy}-{\mathrm{\Omega }}^4{y}^{2}a$. On every fixed $\mathrm{\mid }t\mathrm{\mid }\le H$, all derivatives of $a$ are polynomials in $y$ with bounded coefficients times a Gaussian having $\mathrm{Re}b\ge c_H>0$; all required weighted $L^2$ norms are bounded uniformly there.

Frequency-independent energy stability. The same integral construction with an added d'Alembert forcing term treats smooth forcing on finite slabs. For a smooth spatially compact solution of $q_{tt}-q_{xx}+W_{l}q=F$, direct differentiation and integration by parts give $e_l^{\mathrm{\prime }}[q]=\mathrm{Re}\int F\overline{q_t}dx$. For nonzero $A=\mathrm{\parallel }F{\mathrm{\parallel }}_2$ and $B=\mathrm{\parallel }{q}_t{\mathrm{\parallel }}_2$, integration of $(\mathrm{\mid }F\mathrm{\mid }\mathrm{/}A-\mathrm{\mid }{q}_t\mathrm{\mid }\mathrm{/}B{)}^2\ge 0$ gives $\int \mathrm{\mid }F{q}_t\mathrm{\mid }\le AB$; zero-norm cases give the same inequality immediately. Since $W_l\ge 0$, this bounds the energy derivative in absolute value by $\sqrt{2e_l[q]}\mathrm{\parallel }F{\mathrm{\parallel }}_2$. Apply this first to $\sqrt{e_l+\eta }$, integrate and let $\eta \downarrow 0$ to obtain $\sqrt{e_l[q](t)}\le \sqrt{e_l[q](0)}+2^{-1\mathrm{/}2}{\int }_0^t\mathrm{\parallel }F(a){\mathrm{\parallel }}_{2}da$. In particular the energy of a homogeneous wave from step 3.2 is conserved. Compact support on every finite slab justifies all integrations, even though the full spatial line is noncompact.

Compact approximation and residual. Choose a real $\chi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ from [F4], equal to one near $0$ and supported inside $\{\mathrm{\mid }r(x)-r_0\mathrm{\mid }<\delta \}$. Set $Q_h=h^{3\mathrm{/}4}{e}^{-i\mathrm{\Omega }t\mathrm{/}h}\chi (x)a(t,x\mathrm{/}\sqrt{h})$. Its support and the support of each derivative lie in the exact annulus. In $R_h=({\mathrm{\partial }}_t^2-{\mathrm{\partial }}_x^2+W_l)Q_h$, the $h^{-2}$ term cancels by $V_0(0)={\mathrm{\Omega }}^2$ and the $h^{-1}$ term cancels by the envelope equation in step 3.3. The remaining terms away from cutoff derivatives are $h^{3\mathrm{/}4}{e}^{-i\mathrm{\Omega }t\mathrm{/}h}\chi$ times $a_{tt}+V_{1}a+h^{-2}(V_0(x)-{\mathrm{\Omega }}^2+{\mathrm{\Omega }}^4{x}^2)a$. Taylor's theorem on the fixed compact support gives $\mathrm{\mid }{V}_0(x)-{\mathrm{\Omega }}^2+{\mathrm{\Omega }}^4{x}^2\mathrm{\mid }\le C\mathrm{\mid }x{\mathrm{\mid }}^3$, with $C$ independent of $h$. With $dx=\sqrt{h}dy$, the amplitude prefactor changes an envelope $L^2$ norm by $h$. Thus the first two terms are $O_H(h)$ in $L_x^2$ and the cubic term is bounded by $Ch{h}^{-2}{h}^{3\mathrm{/}2}\mathrm{\parallel }\mathrm{\mid }y{\mathrm{\mid }}^{3}a{\mathrm{\parallel }}_2=O_H(\sqrt{h})$. Cutoff commutators are $-2{\chi }^{\mathrm{\prime }}{\mathrm{\partial }}_x$ of the uncut amplitude and $-{\chi }^{\mathrm{\prime }\mathrm{\prime }}$ times that amplitude. Their supports have $\mathrm{\mid }x\mathrm{\mid }\ge d_0>0$; Gaussian decay there makes their norms $O_{H,N}(h^N)$ for every $N$, despite any fixed derivative powers of $h^{-1}$. Consequently ${\sup }_{0\le t\le H}\mathrm{\parallel }{R}_h(t){\mathrm{\parallel }}_2\le C_H\sqrt{h}$.

Positive normalization. At $t=0$, the substitution $x=\sqrt{h}y$ and Gaussian tail bounds show $e_l[Q_h](0)\to C_0={\mathrm{\Omega }}^2\int \mathrm{\mid }a(0,y){\mathrm{\mid }}^{2}dy=\mathrm{\Omega }\sqrt{\pi }>0$. More explicitly, the time derivative's leading term $-i\mathrm{\Omega }{Q}_h\mathrm{/}h$ contributes ${\mathrm{\Omega }}^2\int \mathrm{\mid }{a}_0{\mathrm{\mid }}^2$ to the unhalved energy integral; $h^{-2}{V}_0\mathrm{\mid }{Q}_h{\mathrm{\mid }}^2$ contributes the same limit. Their cross/time-envelope terms are $O(h)$, the radial derivative energy is $O(h)$, and the $V_1$ term is $O(h^2)$. The factor $1\mathrm{/}2$ in $e_l$ therefore gives $C_0+O(h)$. In particular this energy has a positive lower bound for sufficiently large integers $l$.

Exact correction. Use step 3.2 to solve the homogeneous radial equation with exactly the smooth compact initial data $Q_h(0),{\mathrm{\partial }}_t{Q}_h(0)$, obtaining $q_h$ on all real times. Then $w_h=q_h-Q_h$ has zero initial data and forcing $-R_h$. Steps 4.1 and 4.2 give ${\sup }_{0\le t\le H}{e}_l[w_h](t)\le (H^2\mathrm{/}2)\sup \mathrm{\parallel }{R}_h(t){\mathrm{\parallel }}_2^2\le C_{H}h$. Its compact support on finite slabs follows from the support of $q_h$ and $Q_h$. Define $U_h=\mathrm{Re}(q_h{Y}_l\mathrm{/}r)$. By steps 1.1--2.1 it is a real globally smooth exact scalar wave, including at the sphere poles, with compact smooth initial data. Its conserved stationary energy is exactly $e_l[Q_h](0)\mathrm{/}2$ by steps 3.1 and 4.1.

Concentration on every required slice. Normalize ${\psi }_h=U_h\mathrm{/}\sqrt{e_l[Q_h](0)\mathrm{/}2}$; its stationary energy is exactly one on every slice. Outside $K_{\delta }$, the approximate field $\mathrm{Re}(Q_h{Y}_l\mathrm{/}r)$ and all its derivatives vanish. There the exact field equals the correction $\mathrm{Re}(w_h{Y}_l\mathrm{/}r)$. The original positive density of step 3.1 and the global energy identity consequently give $E_{T,\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}{K}_{\delta }}[{\psi }_h](t)\le e_l[w_h](t)\mathrm{/}{e}_l[Q_h](0)\le C_H^{\mathrm{\prime }}h$. The boundary spheres have measure zero, so total energy one implies $E_{T,\delta }[{\psi }_h](t)\ge 1-C_H^{\mathrm{\prime }}h$ simultaneously for all $t\in [0,H]$. Choose one sufficiently large integer $l$ with $C_H^{\mathrm{\prime }}\mathrm{/}(l+1\mathrm{/}2)<ϵ$. This proves the exact quantifiers and real-field assertion. The construction uses the full specified radial annulus and makes no assertion about a smaller angular neighborhood or a frequency uniform in $H$.