Incoming phase neighborhoods for Dirichlet waves

Written by GPT-6 Astra (OpenAI). Self-checked by the writing AI. Original exposition: CC0.

We prove propagation of singularities for the spectral Dirichlet wave, including strict diffraction, gliding, arbitrarily degenerate contacts and accumulating reflections. The proof first constructs positive phase weights and uniform Sobolev estimates, then builds a generalized curve inside the singular set and connects it to the initial data. The final sections establish joint regularity of the actual wave kernels, including source points approaching the wall and smooth coefficient parameters, on a region with no returning ray.

For the phase neighborhood and localized energy estimate, see Victor Ivrii, Microlocal Analysis, Sharp Spectral Asymptotics and Applications, author version of 9 July 2023, printed pp. 259–266, especially (3.4.42) and (3.4.45).

Further reading: Lars Hörmander, The Analysis of Linear Partial Differential Operators III, 2007 reprint, Definition 24.3.7 and Lemma 24.3.15, pp. 434–435 and 441–442, for generalized curves; Lemmas 24.4.3–24.4.4, pp. 445–446, for elliptic quadratic completion; pp. 448–451 for the two-root region and Sobolev regularization; Lemmas 24.4.8–24.4.9, p. 452, for flat divided square roots; p. 454 for recovery from two multipliers; and Proposition 24.5.1, Lemma 24.5.2 and Theorem 24.5.3, pp. 455–459, for general glancing propagation.

The prerequisites are the smooth-flow proof (G13), coordinate inverses, smooth quadratic division (C1)–(C17), and the flat cutoff and its derivative bounds (C18)–(C19) in the same reading. The Fourier definition of wavefront and its localization estimates are proved in the phase reading. We retain a compact semiclassical frequency patch away from the zero section; no conic homogeneity of the auxiliary cutoff is needed. The exact operator identity is proved in Dirichlet commutators and diffraction; the spectral solution, inverse-power regularization and initial-data wavefront equalities are proved in Distributional Dirichlet waves.

1. A normal-frequency coordinate along the flow

Use the gauged wave principal symbol $$p(x,y,s,\eta)=s^2+a_0(x,y,\eta),\qquad x\ge0, \tag{T1}$$ where $x$ is inward normal distance and $y$ includes physical time. At the point $z_*=(0,y_*,0,\eta_*)$ assume $$a_0(0,y_*,\eta_*)=0,\qquad -\partial_xa_0(z_*)>0. \tag{T2}$$ The latter is strict diffraction in this convention: $H_p^2x=-2\partial_xa_0>0$. Extend the smooth coefficients across $x=0$ locally. Shrink a full phase neighborhood so that $\kappa\le c=-\partial_xa_0\le K$, with positive constants. Then $H_ps=c>0$. The smooth vector field $$V=c^{-1}H_p,\qquad Vs=1,\qquad Vp=0 \tag{T3}$$ has a local smooth flow by (G13). Start its trajectories on the section $s=0$, coordinatized by $I=(x,y,\eta)-(0,y_*,\eta_*)$. Its flow map, denoted $\mathcal F(I,s)$, is a coordinate map: at $s=0$ the section directions together with $V$ are independent, since $Vs=1$. The parameter inverse theorem supplies the inverse after shrinking. Its invariant coordinate functions obey $H_pI=0$, and $\mathcal F(0,0)=z_*$.

The reference trajectory $\Gamma(s)=\mathcal F(0,s)$ stays characteristic. Its normal position satisfies $$\begin{aligned} &x(\Gamma(s))=\int_0^s\frac{2u}{c(\Gamma(u))}\,du,\\ &\frac{s^2}{K}\le x(\Gamma(s))\le\frac{s^2}{\kappa}. \end{aligned} \tag{T4}$$ For negative $s$, reverse the integral limits and the sign of $u$ to obtain the same bounds. Thus each punctured leg lies in the interior, with an explicit lower separation from the wall on every closed subsegment. This calculation does not assume a boundary propagation theorem.

Choose an orientation $\nu\in\{1,-1\}$ and a small $\sigma>0$. Let $\mathcal U$ be any prescribed open full-phase neighborhood of the compact leg $\{\Gamma(s):\sigma\le\nu s\le2\sigma\}$, contained in $x>0$. In the later propagation application it is a neighborhood where the solution is already microlocally smooth. Both choices of $\nu$ are allowed; which leg is physically incoming is determined by the time covector. Choose $\sigma$ so that a slightly larger flow box exists. Compactness and continuity allow a number $r_0>0$ such that $$\begin{gathered} \mathcal F(I,s)\in\mathcal U,\quad x(\mathcal F(I,s))>\frac{\sigma^2}{2K}\\ \text{if }|I|\le r_0,\quad \sigma\le\nu s\le2\sigma. \end{gathered} \tag{T5}$$ Indeed the compact reference segment has positive distance from the closed complement of a smaller neighborhood in $\mathcal U$, and the flow map is uniformly continuous on a compact box. Shrink $r_0$ also so that the closed box $|I|\le r_0$, $|s|\le2\sigma$ lies strictly inside the coordinate chart and its frequency patch.

2. A phase whose outer error is on the chosen leg

Choose real smooth cutoffs $\rho(I)$ and $\vartheta(s)$ with values in $[0,1]$, where $\rho=1$ for $|I|\le r_0/2$ and $\rho=0$ for $|I|\ge r_0$, while $\vartheta$ is even, equals one for $|s|\le\sigma$ and vanishes for $|s|\ge2\sigma$. Their supports may be chosen strictly inside the larger flow box. Put $\zeta=\rho(I)\vartheta(s)$. Choose a fixed $\delta>0$ so small that $$\delta\sqrt{3\sigma}<r_0/2,\qquad \frac{r_0^2}{4\delta^2}>4\sigma. \tag{T6}$$ These choices precede the small positive shift $\epsilon$. Let $0<2\epsilon<\sigma$, with additional restrictions in Section 3, and define $$\begin{aligned} \phi_\nu&=|I|^2/\delta^2-\nu s,\\ q_\nu&=\nu\zeta^2\chi(\phi_\nu-\epsilon),\\ \chi(v)&= \begin{cases}e^{1/v},&v<0,\\0,&v\ge0.\end{cases} \end{aligned} \tag{T7}$$ Extend $q_\nu$ by zero outside the flow box. It is smooth with compact phase support; flatness of $\chi$ treats its level-set boundary, and the other cutoffs have closed interior support. The sign $\nu$ on $q_\nu$ is essential when the chosen regular leg has negative $s$.

Because $I$ is invariant and $H_ps=c$, $$\begin{aligned} H_p\phi_\nu&=-\nu c,\\ H_pq_\nu &=-c\zeta^2\chi'(\phi_\nu-\epsilon)+e_\epsilon,\\ e_\epsilon&=\nu\chi(\phi_\nu-\epsilon)H_p(\zeta^2). \end{aligned} \tag{T8}$$ The first term is nonnegative and is strictly positive at $z_*$. We now prove that the closed support of $e_\epsilon$ is contained in $\mathcal U$, not in an unspecified set of cutoff edges.

First $H_p\rho(I)=0$ identically, so a derivative of the invariant cutoff contributes nothing to $H_p(\zeta^2)$. At the opposite end, $\nu s\le-\sigma$ gives $\phi_\nu\ge\sigma>\epsilon$, so $\chi$ and all its derivatives vanish. At the chosen end, $\sigma\le\nu s\le2\sigma$ and $\chi(\phi_\nu-\epsilon)\ne0$ imply $$|I|^2<\delta^2(2\sigma+\epsilon)<3\delta^2\sigma. \tag{T9}$$ Thus $|I|<r_0/2$ by (T6), and (T5) places the point in the prescribed interior neighborhood. The same strict containment holds for the closed support, since all inequalities have fixed margins and the cutoff support is compact.

There is also no hidden side edge for differentiated weights. On the transition of $\rho$, where $|I|\ge r_0/2$ and $|s|\le2\sigma$, (T6) gives $\phi_\nu>2\sigma$. Hence every derivative of $\chi(\phi_\nu-\epsilon)$ vanishes in a neighborhood of that transition. The same assertions hold with $\epsilon$ replaced by $2\epsilon$. This shifted neighborhood is needed for lower-regularity terms; it is not obtained by assuming the solution regular on the entire enclosing box.

At $I=0$ and $0\le\nu s\le\sigma$, the phase is $-\nu s$, the cutoffs are one, and the weight is positive. Thus the retained neighborhood contains $z_*$ and the segment connecting it to the known leg. The invariant radius in (T9) can be made as small as required by choosing $\delta$ after $\mathcal U$ and before $\epsilon$.

3. The divided difference has the Dirichlet sign

The phase $\phi_\nu$ need not be affine in $s$ in the original coordinates. Apply the already proved smooth quadratic division directly to the full smooth cutoff $q_\nu$. With $z=(x,y,\eta)$ and $R=-a_0(z)$ it gives $$q_\nu(z,s)=\alpha(z)+s\beta(z)+p(z,s)\mu(z,s). \tag{T10}$$ For $R\ge0$ the coefficients are fixed, rather than arbitrary extensions: $$\begin{aligned} \alpha(z)&=\frac{q_\nu(z,\sqrt R)+q_\nu(z,-\sqrt R)}2,\\ \beta(z)&=\frac{q_\nu(z,\sqrt R)-q_\nu(z,-\sqrt R)}{2\sqrt R}. \end{aligned} \tag{T11}$$ At $R=0$ the latter value is $\partial_s q_\nu(z,0)$. All coefficients and the remainder are smooth through merging roots by (C1)–(C17).

We show $\beta\ge0$ at every pair of real roots under consideration, including boundary points. If $\nu=1$ and the value at the negative root vanishes, the numerator in (T11) is nonnegative, since $q_1\ge0$. If $\nu=-1$ and the value at the positive root vanishes, the same numerator is nonnegative, since $q_{-1}\le0$. These observations also treat roots outside the compact cutoff support. The only remaining case is a nonzero value at the root of sign opposite to $\nu$.

Write $r=\sqrt R$. At that root $s=-\nu r$, nonzero weight forces $$r<\epsilon,\qquad |I(z,-\nu r)|<\delta\sqrt{\epsilon}. \tag{T12}$$ Choose a fixed small product neighborhood of $z_*$ on which $I(z,s)$ is defined along these short vertical segments, and let $M\ge1$ bound $|\partial_sI|$ there. The inverse flow map and (T12) put the entire segment $|s|\le r$ in that product neighborhood once $\epsilon$ is sufficiently small. More explicitly, its initial point approaches $z_*$ uniformly as the two bounds in (T12) tend to zero; holding $z$ fixed and changing $s$ by at most $2\epsilon$ keeps it in a preselected smaller box. On the segment, $|I(z,s)|\le\delta\sqrt{\epsilon}+2M\epsilon$. Require $$\begin{aligned} &\epsilon<\sigma,\quad \delta\sqrt{\epsilon}+2M\epsilon<r_0/2,\\ &2M\delta^{-2} (\delta\sqrt{\epsilon}+2M\epsilon)<1/2 . \end{aligned} \tag{T13}$$ All are possible after $\delta$ is fixed. The cutoffs $\rho,\vartheta$ are then identically one on this whole root interval. Furthermore $|\partial_s\phi_\nu+\nu|<1/2$ there. Since $\chi'\le0$, $\partial_s q_\nu=\nu\chi'(\phi_\nu-\epsilon)\partial_s\phi_\nu\ge0$. The fundamental theorem of calculus between $-r$ and $r$ proves the remaining sign in (T11). For a vanishing root take the continuous limit of the same formulas, or differentiate directly when $a_0=0$. At the center in particular $$\beta(z_*)=-\chi'(-\epsilon)>0 \tag{T14}$$ for both orientations. Continuity gives strict positivity on a smaller neighborhood including its elliptic side, so the exact positive boundary quantization (C24) applies there.

The real-root sign alone does not supply a smooth square root at its zeros or a nonnegative elliptic extension. Sections 5–7 below prove those additional properties for this particular flat cutoff and construct the positive boundary form on its whole support.

Applying $H_p$ to (T10), using $H_pp=0$, gives the exact identity $$\begin{aligned} H_p(\alpha+s\beta) &=-c\zeta^2\chi'(\phi_\nu-\epsilon)\\ &\quad+e_\epsilon-pH_p\mu . \end{aligned} \tag{T15}$$ Thus the normal-linear principal multiplier has the required positive term on the characteristic set and an explicitly located regular-edge term. This is a symbol identity with a retained multiple of the equation; it is not yet an exact operator factorization.

4. The actual full-phase edge operator is rapidly small

Let $u$ be a fixed distribution and suppose $\mathcal U$ lies outside its wavefront set, in the interior of the spacetime collar. Multiply $u$ by a fixed interior cutoff that equals one near the compact base projection needed below. For any smooth semiclassical symbol $b_h$ with uniformly bounded derivatives, compact support in a fixed closed subset of $\mathcal U$ and frequency support in a fixed annulus, $$\|\operatorname{Op}_h(b_h)u\|_{H^k} =O(h^N)\qquad (k,N\ge0). \tag{T16}$$ This holds also for each fixed semiclassical differential derivative of the output. We give the localization argument to specify precisely what the assertion uses.

Cover the compact phase support by finitely many product patches whose smaller frequency cones have closures inside cones of rapid Fourier decay for a spatially localized $u$. Choose a finite smooth partition of the symbol, and a compact base cutoff $\lambda$ equal to one on a neighborhood of each output support. On the retained cone, the definition of wavefront gives $|\widehat{\lambda u}(\xi)|\le C_L\langle\xi\rangle^{-L}$ for every $L$. Write $F_h(\theta)=\widehat{\lambda u}(\theta/h)$. The formula $$\begin{aligned} &\operatorname{Op}_h(b_h)(\lambda u)(w)\\ &\quad=(2\pi h)^{-D} \int e^{iw\cdot\theta/h}b_h(w,\theta)F_h(\theta)\,d\theta \end{aligned} \tag{T17}$$ uses full interior variables $w=(x,y)$, $D=\dim w$, and full frequencies $\theta=(s,\eta)$. In the annulus $|\theta|\asymp1$, the Fourier factor is $O(h^L)$. Every prescribed output derivative costs at most a fixed power of $h^{-1}$; the frequency region and output base support are compact. Taking $L$ arbitrarily large proves the corresponding uniform derivative and $H^k$ bounds.

The remaining input $(1-\lambda)u$ has base support separated from the output support. In its oscillatory kernel apply $h(w-v)\cdot\partial_\theta/(i|w-v|^2)$ repeatedly to the exponential and transfer it to the amplitude. Each application gives a power of $h$, with bounded coefficients because of separation. Pair the resulting smooth kernel with the compact distribution, whose continuity bounds it by finitely many input derivatives; those derivatives and the requested output derivatives cost only a fixed power of $h^{-1}$. Arbitrarily many integrations prove the same estimates. Summing the finite partition proves (T16). The restriction away from the physical boundary is essential to this particular full-variable argument.

By (T8)–(T9), $b_h=e_\epsilon$ satisfies these hypotheses for every fixed permitted $\epsilon$. The same is true for the shifted edge $e_{2\epsilon}$. Thus their full-phase quantizations satisfy actual rapid bounds, rather than hypothetical bounds on an arbitrary enclosing annulus.

The full-phase edge in (T16) is supported in the prescribed regular incoming neighborhood. The entire transition support of the different inner cutoff in the localized glancing estimate, (L20), need not lie there; its regularity is a separate hypothesis. We next construct an exactly positive boundary form for the whole phase weight.

5. A smooth square root at the zero weights

We now strengthen the coefficient construction on the entire cutoff, including points where $\beta=0$. Keep $z=(x,y,\eta)$ as a parameter and first regard $R$ as independent. A further smallness choice for $\epsilon$ is permitted in the construction above. Fix a small $a>0$, with $a<\sigma/2$, such that points with $|s|\le a$ and $|I|\le\delta\sqrt{2a}$ lie in the fixed product neighborhood where $|\partial_sI|\le M$, and $\delta\sqrt{2a}<r_0/2$ and $2M\delta^{-1}\sqrt{2a}<1/2$. This is possible by the inverse flow chart and then by decreasing $a$. Take $0<\epsilon<a/4$, also satisfying (T13). These choices work for both $\epsilon$ and its shift $2\epsilon$.

Where $|s|\le a$ and $\chi(\phi_\nu-\epsilon)\ne0$, we have $|I|<\delta\sqrt{2a}$. Thus both cutoffs are one there and $|\partial_s\phi_\nu+\nu|<1/2$. Consequently the function $g(z,s)=\partial_s q_\nu(z,s)$ is nonnegative on this whole small $s$ interval, for every retained $z$. On its positive set it is comparable, with fixed positive constants, to $-\chi'(\phi_\nu-\epsilon)$. Where the latter vanishes, all derivatives of $g$ vanish as well. The chain rule and (C19), applied to $-\chi'$, prove $$\begin{gathered} |D^\alpha g(z,s)| \le C_{\alpha,\gamma}g(z,s)^{1-\gamma},\\ |s|\le a,\quad 0<\gamma<1 . \end{gathered} \tag{T18}$$ Here $D^\alpha$ includes arbitrary mixed derivatives in $z,s$. Indeed on the nonzero set $g=\nu(\partial_s\phi_\nu)\chi'(\phi_\nu-\epsilon)$ with the first factor bounded away from zero in absolute value; all its derivatives and those of $\phi_\nu$ are bounded on the compact patch. Every differentiated term is therefore bounded by a constant times $(-\chi')^{1-\gamma}$. On the zero set the same inequality follows from flatness. No derivative assumption on the outer cutoffs at their zeros is needed here: they are identically one wherever this small-$s$ weight can be nonzero.

We prove an averaging fact to control the divided difference without factors singular at $R=0$. For $0\le R\le a^2$, set $$b_+(z,R)=\frac12\int_{-1}^1g(z,t\sqrt R)\,dt. \tag{T19}$$ It is smooth up to $R=0$ by the even-function proof (C1)–(C3). If $r=\sqrt R>0$, write $A=\max_{|s|\le r}g(z,s)$. The derivative estimate (T18) implies $|\partial_s g|\le C_\gamma A^{1-\gamma}$ on that interval. Starting at a maximum point, choose the direction with available interval length at least $r$. On a subinterval of length $\min(r,A^\gamma/(2C_\gamma))$ the value is at least $A/2$. Integration over $[-r,r]$ shows $$\begin{aligned} A^{1+\gamma}&\le C_\gamma b_+(z,R),\\ A&\le C_\gamma b_+(z,R)^{1/(1+\gamma)} . \end{aligned} \tag{T20}$$ To see the first inequality uniformly, divide $g$ by a fixed upper bound to make $A\le1$. Its average is at least $A\min(1,A^\gamma/(2C_\gamma r))/4$; since $r\le a$, this is at least a fixed multiple of $A^{1+\gamma}$. Restoring the upper bound only changes constants. At $r=0$ the same inequalities hold because $A=b_+(z,0)$ is bounded. The case $A=0$ is immediate.

Each $\partial_z^\alpha\partial_R^\ell b_+$ is bounded by a constant times the supremum of derivatives of $g$ through order $|\alpha|+2\ell$ on $[-r,r]$. This is the explicit repeated integral operation (C2): differentiate the even function in $r$ twice, integrate at contracted values of $r$, and repeat. The factors from differentiating $g(z,tr)$ are powers of the integration variables of absolute value at most one. There are finitely many such integrals for each order, all over fixed compact intervals with bounded total weights. Hence (T18) and (T20) give a bound by $C b_+^{(1-\gamma)/(1+\gamma)}$. Given any $0<\lambda<1$, choose $\gamma$ so small that $(1-\gamma)/(1+\gamma)>1-\lambda$ and use boundedness of $b_+$ to obtain $$|\partial_z^\alpha\partial_R^\ell b_+| \le C_{\alpha,\ell,\lambda}b_+^{1-\lambda}. \tag{T21}$$ At a zero of $b_+$ these derivatives vanish. This estimate concerns the divided coefficient itself, not just its two endpoint values.

On $\{b_+>0\}$ differentiate $S_+=\sqrt{b_+}$. A derivative of positive total order $m$ is a finite sum of products $C b_+^{1/2-j}\prod_{i=1}^j D^{\alpha_i}b_+$, where $1\le j\le m$ and each $\alpha_i$ has positive order. Formula (T21) bounds each such term by $C b_+^{1/2-j\lambda}$. Taking $\lambda<1/(4m)$ shows that every derivative through order $m$ tends to zero at the zero set: $$|D^\alpha S_+|\le C_\alpha b_+^{1/4} \quad(1\le|\alpha|\le m) \tag{T22}$$ after increasing constants on the bounded range of $b_+$. Define these derivatives to be zero on that set. They are the actual continuous derivatives. For a coordinate segment ending at a zero, split its positive portion into component intervals; the fundamental theorem bounds each change by interval length times the supremum of the proposed derivative, and at zero endpoints the function is zero. This proves differentiability there with derivative zero. Repeat for the derivative functions just obtained. Segments in $R$ at zero are taken from the right. Thus $S_+$ is smooth in all variables up to $R=0$, including arbitrary parameter derivatives and all zero weights.

This covers $r\le a$. For $r\ge a/2$, the wrong-sign root has $|s|\ge a/2>\epsilon$, so its cutoff is zero as in Section 3. The exact remaining formula is $b_+(z,R)=\zeta(z,\nu r)^2\chi(\phi_\nu(z,\nu r)-\epsilon)/(2r)$. Its nonnegative square root is $\zeta(z,\nu r)\sqrt{\chi(\phi_\nu(z,\nu r)-\epsilon)}/\sqrt{2r}$, which is smooth because $r$ is bounded away from zero and $\sqrt\chi$ is smooth and flat. The two square roots agree pointwise on the overlap, hence give a smooth square root on the full retained real-root region. The same proof applies to the shifted cutoff $2\epsilon$, since $2\epsilon<a/2$.

6. A nonnegative extension with exact division

Extend the smooth one-sided function $S_+(z,R)$ across $R=0$ using the explicit summation (C4)–(C5); call the extension $S(z,R)$. It need not be nonnegative for negative $R$. Its square is nonnegative and matches $b_+$ and all its derivatives on the real-root side. This avoids the unjustified assumption that an arbitrary extension of $b_+$ itself preserves its sign.

The extension can retain compact parameter support and can be confined to a prescribed small negative-$R$ collar. The one-sided jets of $S_+$ have compact parameter support, since $q_\nu$ does. Their summation coefficients have that same support. Choose every cutoff radius in (C5), including the zeroth, inside the desired collar, and multiply the negative-side extension by a cutoff equal to one near zero if needed. These operations leave every boundary jet unchanged. An additional parameter cutoff equal to one on the union of the jet supports also leaves every boundary jet unchanged. Gluing to the unchanged $S_+$ on $R\ge0$ remains smooth by the proved extension argument.

There is a useful more precise support choice. At $R=0$ the jets of $S_+$ vanish unless $\phi_\nu(z,0)\le\epsilon$. Indeed where $\phi_\nu(z,0)>\epsilon$ the original cutoff is identically zero in an $s$ neighborhood, and at equality all its jets, and then those of $S_+$ by (T21)–(T22), vanish. Choose the negative-side parameter cutoff equal to one on this compact set and supported in $\phi_\nu(z,0)<3\epsilon/2$, within the original compact patch. On its closure the shifted divided coefficient at $R=0$ is strictly positive, with a uniform positive lower bound: its value is $\nu\,\partial_s\phi_\nu(z,0)\chi'(\phi_\nu(z,0)-2\epsilon)$, whose two factors have the required strict signs by the smallness choices in Section 5. First construct a square-root extension for the shifted weight. By continuity it stays bounded away from zero on this compact parameter support for sufficiently small negative $R$. Then choose the extension collar for $S$ inside that region. Thus the elliptic-side adjustment for the smaller weight can be supported where the shifted boundary weight is nonvanishing. This is a support statement for the boundary coefficient, not an interior propagation estimate.

Let $b(z,R)$ be the previous, possibly signed smooth extension used in quadratic division, and put $\Delta=S^2-b$. It is identically zero for $R\ge0$ and is smooth across zero. Therefore Taylor's formula at zero gives, for every fixed derivative and every $L$, $|\partial_z^\alpha\partial_R^\ell\Delta|\le C_{\alpha,\ell,L}|R|^L$ on the negative side. The correction needed for the division remainder is $$E= \begin{cases} s\Delta(z,R)/(s^2-R),&R<0,\\ 0,&R\ge0. \end{cases} \tag{T23}$$ It is smooth. Away from $s=R=0$ the denominator is nonzero across the gluing set. Near that corner put $\omega=(s^2+|R|)^{1/2}$. The reciprocal-denominator derivative bound proved in (C16), together with the flat bounds on $\Delta$, gives $|\partial_z^\alpha\partial_s^k\partial_R^\ell E| \le C\omega^{2L-1-k-2\ell}$ after taking enough orders of flatness for the finitely many differentiated numerator terms. Since $L$ is arbitrary, every derivative tends to zero at the corner. Gluing these derivatives by the fundamental theorem proves smoothness, with zero jets there. This explicitly checks the division correction at the merging roots.

Keep the old average coefficient $a(z,R)$ and replace $b$ by $S^2$ and the old remainder $\mu$ by $\mu-E$. The division identity is still exact, because $s\Delta=(s^2-R)E$: $$\begin{aligned} q_\nu(z,s) &=a(z,R)+sS(z,R)^2\\ &\quad+(s^2-R)(\mu-E)(z,s,R). \end{aligned} \tag{T24}$$ Now substitute $R=-a_0(z)$. This gives (T10)–(T15) with $\beta(z)=S(z,-a_0(z))^2\ge0$ on both sides of the glancing set. All real-root values, the actual cutoff $q_\nu$ and the support of its full-phase edge $e_\epsilon$ are unchanged. The multiple of the equation in (T15) uses the adjusted smooth remainder and must still be retained in its later quantization.

7. Quantizing the whole boundary weight

Write $s_0(z)=S(z,-a_0(z))$ on the retained compact tangential phase patch. With an outer tangential cutoff equal to one on the coefficient support if necessary, $s_0$ is smooth and compactly supported, including at every zero of the weight. No restriction to the smaller neighborhood where $\beta$ is strictly positive is required. Set $$\begin{aligned} S_h&=\operatorname{Op}_h(s_0),\qquad B_h=S_h^*S_h,\\ A_{0,h}&=\tfrac12\bigl(\operatorname{Op}_h(\alpha) +\operatorname{Op}_h(\alpha)^*\bigr),\\ A_h&=A_{0,h}+\tfrac12(B_hd_x+d_xB_h). \end{aligned} \tag{T25}$$ The finite tangential calculus in (N19)–(N23) gives the principal symbols $\alpha,\beta$ and uniform bounds for these tangential factors and each fixed normal derivative. The lower terms are the actual product and adjoint terms; they have not been set to zero. The exact boundary commutator calculation (D5)–(D8), applied to this $A_h$, gives for a smooth Dirichlet input $u$ vanishing near the outer normal endpoint and $f=P_hu$, $$\begin{aligned} &\left(\frac ih[P_h,A_h]u,u\right) +\|S_h(0)d_xu(0)\|^2\\ &\qquad=\frac2h\operatorname{Im}(f,A_hu). \end{aligned} \tag{T26}$$ The boundary term is exactly nonnegative on the whole weight. There is no boundary error from applying a positivity theorem to left quantization. Expansion of $d_xB_h=B_hd_x-ihB_h'$ gives the same uniform bound $\|A_hu\|\le C(\|u\|+\|d_xu\|)$ as (C26). The exact bulk expansion is (C27), including $\operatorname{Re}(B_h'f,u)$. Thus every term is defined for the $H^2$ Dirichlet inputs supplied by the earlier negative-order wave reduction. The odd-mollification and trace argument following (C28) proves (T26) for that domain as well; it uses precisely the bounded factors and derivatives just verified.

This closes the smooth-square-root and full-boundary-sign steps for the incoming cutoff, for either orientation. The following sections convert the positive principal expression in (T15) into an actual bulk square estimate, keep the equation contribution and recover the full-phase edge of (T16). The last section states exactly what remains before propagation can be claimed.

8. A positive quadratic form across the elliptic side

Keep the fixed phase neighborhood and its shift $\epsilon$. Put $z=(x,y,\eta)$ and $R=-a_0(z)$. Define the smooth full-phase function $$j(z,s)=\sqrt{c(z)}\,\zeta(z,s) \sqrt{-\chi'(\phi_\nu(z,s)-\epsilon)}. \tag{T27}$$ The square root is smooth and flat by (C18)–(C19). The factor $c$ is positive on a neighborhood of the support; extension by zero outside that neighborhood is smooth. Thus $j$ has compact phase support. Apply (C1)–(C17) separately to $j$ and $e_\epsilon$, keeping real coefficients: $$\begin{aligned} j&=j_0(z)+j_1(z)s+p\mu_j(z,s),\\ e_\epsilon&=e_0(z)+e_1(z)s+p\mu_e(z,s). \end{aligned} \tag{T28}$$ The coefficient extensions have compact tangential phase support. Their negative-$R$ collars may be chosen inside a fixed product chart where $\partial_xa_0<0$, as in Section 6. We will make a particular, simpler choice for $e_0,e_1$ in Section 10.

Write the Hamilton derivative of the normal polynomial as $$\begin{aligned} F&=H_p(\alpha+\beta s)=F_0+F_1s+F_2s^2,\\ F_0&=\{a_0,\alpha\}_{\rm tan}-\beta\partial_xa_0,\\ F_1&=2\partial_x\alpha+\{a_0,\beta\}_{\rm tan},\qquad F_2=2\partial_x\beta,\\ \lambda&=F_2-j_1^2,\\ D_0&=F_0-j_0^2-e_0-\lambda a_0,\\ D_1&=F_1-2j_0j_1-e_1. \end{aligned} \tag{T29}$$ At the two real characteristic roots, (T15) and (T28) give $F=(j_0+j_1s)^2+e_0+e_1s$. Subtracting the multiple $\lambda p$ cancels the quadratic coefficient. For $R>0$, evaluating the remaining polynomial $D_0+D_1s$ at $s=\pm\sqrt R$ proves $D_0=D_1=0$. By smoothness the same holds at $R=0$. Because $t=a_0=-R$ is a coordinate, every derivative of $D_0,D_1$ is flat at $t=0$: it vanishes on $t<0$, and Taylor's formula bounds it on $t>0$ by $C_Mt^M$ for every $M$. All such bounds are uniform on the retained compact parameter set. We must account for these possibly nonzero elliptic residuals; a real-root identity alone would not do so.

There is a smooth, nonnegative, compactly supported function $k(z)$, zero for $t\le0$ and flat at zero, such that on $t>0$ $$k\ge \frac{2|D_0|}{t}+\frac{|D_1|}{\sqrt t}. \tag{T30}$$ Here is a construction with every smoothness issue retained. Use coordinates $(t,w)$ and make the negative-$R$ coefficient extensions small enough that the residuals are supported in $0<t<T/2$, inside a product chart containing $0\le t\le2T$. Choose a nonnegative parameter cutoff $\omega(w)$ equal to one on their projected support and compactly supported in that chart. Let $\psi\ge0$ be smooth, supported in $(1/4,2)$ and at least one on $[1/2,1]$. For $t_m=T2^{-m}$, $m\ge0$, let $A_m$ be the supremum of the right side of (T30) over $t\in[t_m/4,2t_m]$ and all parameters, extending the residuals by zero outside their compact parameter support. Set $a_m=A_m+2^{-m^2}$ and put $$\begin{aligned} k(t,w)&=\omega(w)\sum_{m\ge0}a_m\psi(t/t_m) &&(t>0),\\ k(t,w)&=0 &&(t\le0). \end{aligned} \tag{T31}$$ At each positive $t$ only a bounded number of summands is nonzero. On the residual support, some $t/t_m$ lies in $[1/2,1]$, so this sum dominates (T30); off that support the right side is zero. Flatness gives $A_m=O(2^{-mM})$ for every $M$. A derivative of order $l$ in $t$ costs at most $C_l2^{ml}$. Consequently all derivatives of the sum tend to zero faster than every power of $t$, uniformly in $w$. Repeated use of the fundamental theorem across zero verifies its smooth extension with zero jets. The support stays in $t<2T$ and the compact support of $\omega$. These choices concern a fixed phase neighborhood; no constant is asserted uniform as $\epsilon$ tends to zero.

Set $\ell=\lambda-k$ and $$\begin{gathered} N(z)=\begin{pmatrix} D_0+ka_0&D_1/2\\ D_1/2&k \end{pmatrix},\qquad N\ge0,\\ \begin{aligned} F={}&(j_0+j_1s)^2+(e_0+e_1s)\\ &+\ell p+(1,s)N(1,s)^{\mathsf T}. \end{aligned} \end{gathered} \tag{T32}$$ Both assertions are global on the retained coefficient patch. For $t\le0$, $N=0$. For $t>0$, (T30) gives $D_0+kt\ge kt/2$ and $|D_1|\le k\sqrt t$. If $k>0$, completing the square in the second component of a complex vector leaves the coefficient $D_0+kt-D_1^2/(4k)\ge kt/4\ge0$ in the first component. If $k=0$, both residuals vanish and the matrix is zero. This proves positivity without taking a potentially nonsmooth matrix square root. Expanding the right side of (T32) gives (T29) exactly. All coefficients in (T32) and each fixed normal derivative are smooth, compactly supported and bounded.

9. Quantizing the positive form and the equation multiple

For a real tangential coefficient $b$ denote its self-adjoint quantization by $b_h^s=(\operatorname{Op}_h(b)+\operatorname{Op}_h(b)^*)/2$. For a real normal polynomial $v=v_0+v_1s+v_2s^2$ define $$\begin{aligned} \mathcal Q_h(v;u)={}&((v_0)_h^su,u) +\operatorname{Re}((v_1)_h^sd_xu,u)\\ &+((v_2)_h^sd_xu,d_xu),\\ H_h(u)^2={}&\|u\|^2+\|d_xu\|^2,\\ J_hu={}&\operatorname{Op}_h(j_0)u +\operatorname{Op}_h(j_1)d_xu. \end{aligned} \tag{T33}$$ Inner products here include integration in the normal variable. For smooth Dirichlet inputs with compact normal support, the finite tangential products give $$\begin{aligned} &\mathcal Q_h((j_0+j_1s)^2;u)\\ &\qquad=\|J_hu\|^2+O(h)H_h(u)^2,\\ &\mathcal Q_h((1,s)N(1,s)^{\mathsf T};u)\\ &\qquad\ge-ChH_h(u)^2. \end{aligned} \tag{T34}$$ For the first line, expand the square norm. Each product of two tangential factors differs in norm by $O(h)$ from the self-adjoint quantization of its real principal product. The mixed term is bounded by $Ch\|u\|\|d_xu\|$.

For the second line, apply the proved Gaussian-packet construction to the two-component function $(u,d_xu)$ at each fixed $x$. The analysis map acts on both components. Multiplication by the positive matrix $N(y,h\xi)$ has nonnegative quadratic form pointwise, so its packet quantization is nonnegative. The kernel calculation, Gaussian second-moment estimate and left-to-Weyl correction in the Gaussian reading, Lemmas 2–3 and Theorem 4, are linear in each entry. They show that each entry differs from its self-adjoint tangential quantization by norm at most $Ch$. There are only four entries. Their errors have total form bounded by $Ch(\|u\|^2+\|d_xu\|^2)$. The constants are uniform in the compact normal interval, and integration proves the second line. This proves the matrix version actually needed here directly from the scalar kernel estimates.

For every real compact coefficient $v(z)$, let $V_h=v_h^s$. If $P_hu=f$, integration once in $x$ gives $$\begin{aligned} (V_hd_xu,d_xu) &={}\operatorname{Re}(V_hd_x^2u,u)\\ &\quad+\operatorname{Re}(-ihV_h'd_xu,u),\\ \mathcal Q_h(vp;u) &={}\operatorname{Re}(V_hf,u)\\ &\quad+O(h)H_h(u)^2. \end{aligned} \tag{T35}$$ The first line has no boundary term because $u(0)=0$ and the input vanishes near the outer endpoint. In the second line substitute $d_x^2u=f-R_hu$. The finite differential-cutoff calculation shows that the real form of $V_hR_h$ differs from $(va_0)_h^s$ by norm $O(h)$, including the full lower symbol $ha_1$. The last term in the first line is bounded by $Ch\|d_xu\|\|u\|$. This establishes the second line with the actual equation and lower terms present.

Let $\mathcal C_h(u)=(i[P_h,A_h]u/h,u)$ and $\mathcal E_h(u)=\mathcal Q_h(e_0+e_1s;u)$. The exact formula (C27), and the principal coefficients checked in (L3)–(L6), give $$\begin{aligned} \mathcal C_h(u) ={}&\mathcal Q_h(F-\beta_xp;u)\\ &+\operatorname{Re}(B_h'f,u)\\ &+O(h)H_h(u)^2,\\ V_h={}&(\ell-\beta_x)_h^s,\\ \Lambda_h={}&V_h+B_h',\\ \mathcal C_h(u) \ge{}&\|J_hu\|^2+\mathcal E_h(u)\\ &+\operatorname{Re}(\Lambda_hf,u)\\ &-ChH_h(u)^2. \end{aligned} \tag{T36}$$ Indeed the three coefficients of $F-\beta_xp$ are $F_0-a_0\beta_x,F_1,\beta_x$, exactly the reduced bulk coefficients in (L3)–(L5). Insert (T32), apply (T34), and apply (T35) with $v=\ell-\beta_x$. This proves the last line. In particular the $B_h'f$ term has not been discarded. The family $\Lambda_h$ is uniformly bounded on tangential $L^2$, by the finite norm bounds already proved.

10. Recovering the actual interior edge operator

It remains to relate $\mathcal E_h$ to the full-phase edge whose regularity was proved in (T16). A linear normal polynomial obtained by division need not itself be supported on the incoming branch. The equation is needed for this step too.

In the present construction, $e_\epsilon$ vanishes for $|s|<\sigma$ and is supported where $x>\sigma^2/(2K)$, by (T8)–(T9). Its real-root average and divided difference therefore vanish for $0\le R<\sigma^2$. Choose $e_0=e_1=0$ for $R<0$. This is a smooth extension agreeing with every jet at zero. For $R>0$, use the exact real-root formulas. Their supports are compact and separated from the wall, since every nonzero endpoint value has that property. Define $\mu_e$ by (T28), with the smooth values at the real roots supplied by the division proof. Its output-position and tangential-frequency support is also compact and separated from the wall. Outside the projection of the support of $e_\epsilon$, all three numerator terms vanish, hence so does $\mu_e$.

For large $|s|$ the full-phase edge is zero, and $\mu_e=-(e_0+e_1s)/(s^2+a_0)$. Differentiating this rational expression, and using smoothness on the remaining compact region, gives for all fixed multi-indices $$|\partial_z^\gamma\partial_s^l\mu_e(z,s)| \le C_{\gamma l}\langle s\rangle^{-1-l}. \tag{T37}$$ In particular it is a bounded-derivative symbol in all full position and frequency variables. No isotropic order-minus-one assertion in $(s,\eta)$ is required. The finite-derivative bound and the bounded-amplitude lemma apply: replacing the ordinary frequency by $h$ times that frequency multiplies its derivatives by powers of $h\le1$ and keeps all needed seminorms uniformly bounded.

Write $\operatorname{Op}^{\rm full}_h$ for left quantization in $(x,y)$ and $(s,\eta)$. Choose a smooth normal cutoff $\theta=1$ near the normal projection of these supports, zero near both endpoints of the collar. Extend $\theta u$ by zero to the full normal line, and extend the smooth differential coefficients to have bounded derivatives on that line. For $w=\theta u$, the finite calculation gives $$\begin{aligned} \operatorname{Op}^{\rm full}_h(\mu_e)P_hw &=\operatorname{Op}^{\rm full}_h(\mu_ep)w+hZ_hw,\\ \sup_{0<h\le h_0}\|Z_h\|_{L^2\to L^2}&<\infty. \end{aligned} \tag{T38}$$ Here are the required bounds for this calculation despite the unbounded normal frequency. The normal term is exact: right composition with $d_x^2$ multiplies the left symbol by $s^2$. For the tangential differential terms, move the finitely many input derivatives onto the kernel. The leading amplitude is $\mu_e(X,\Xi)a_0(Y,\eta)$, with full positions $X,Y$ and frequency $\Xi=(s,\eta)$. Taylor's formula for each coefficient between $X$ and $Y$ replaces its difference from the value at $X$ by a position difference times an integral of a first derivative. Move that position difference onto the exponential as $h$ times a frequency derivative, and integrate by parts. The resulting amplitudes and all required derivatives are bounded: the tangential frequency is compact, (T37) bounds the normal frequency and its derivatives, and the extended differential coefficients have bounded derivatives. Terms already containing $h$, including $ha_1$, obey the same bounds. The bounded-amplitude lemma therefore proves (T38). One can first cut off $s$ at radius $L$. The bounded symbol $\mu_e$ and its needed derivatives converge in supremum norm as $L\to\infty$; the differentiated remainder amplitudes have the same property. The normal product converges on $H^2$ inputs by writing it as the bounded operator on $d_x^2w$. Thus the calculation has a well-defined limit on that domain.