Characteristic squares and diffractive commutants

AN-04 proof supplement. Written by GPT-6 Astra (OpenAI). Independent exposition and reproducible figures are CC0-1.0; linked components retain their own terms.

At strict diffraction the normal characteristic equation has two roots which merge at the boundary. A useful commutant must have a nonpositive normal coefficient there, while its Hamilton derivative controls two independent combinations of the solution and its normal derivative. This supplement constructs those symbols, including their smooth extensions through the region where there are no real normal roots.

The mathematical antecedent is Hörmander, The Analysis of Linear Partial Differential Operators III, Springer 2007 edition, ISBN 978-3-540-49938-1, Section 24.4, printed 443–454 (PDF 458–469), particularly Lemmas 24.4.3–24.4.5, 24.4.8–24.4.9 and the symbol construction concluding Theorem 24.4.1. The proofs below use the current programme's explicit even/odd descent and extension, Sections 3–5, rather than an external preparation theorem. The proof map gives exact earlier proofs.

These are complete symbol statements. The analytic strict-diffraction estimate for the full weak form, both boundary domains and natural-dual forcing uses the weak equation in addition to these symbol inequalities. None of the symbol inequalities below tests the weak equation or establishes a solution wavefront theorem.

1. Smooth powers at a flat zero

S1. Power-stable cutoffs. Let $h\geq0$ be smooth on an open Euclidean set. Suppose, on each compact subset, that for every multi-index $\alpha$ and every $0<k<1$,

Proof. On $\{h>0\}$, repeated product and chain rules express any derivative of $h^\theta$ as a finite sum of terms

Here is the gluing argument. If a continuous function is zero on a closed set, is differentiable off it, and its candidate first derivatives extend continuously by zero, then it is differentiable at a point of that set with derivative zero. Along the segment from that point to a nearby point outside the set, take the last point of the closed set. The fundamental theorem of calculus on the remaining interval bounds the change by the segment length times the supremum of the candidate gradient. That supremum tends to zero. Apply this argument successively to all the candidate derivatives. It proves smoothness and the claimed estimates, including at zeros.

For a product use the finite Leibniz sum and apply (DS1) to both factors with the same $k$. For an average $H=\int h_\omega\,d\mu(\omega)$, differentiation and the triangle inequality give

The particular cutoff

2. The divided difference has a smooth square

S2. A square through the merging roots. Suppose $g(a,z)$ is smooth and $\partial_a g\geq0$ satisfies (DS1), with all parameter derivatives included. Locally there is a smooth real $B(r,a,z)$, defined for positive and negative $r$, such that

Proof. The exact integral

A factor $c(z)^2$, independent of $s$, may be inserted in (DS5) by replacing $B$ with $cB$. No derivative condition like (DS1) is needed for this separate factor. This observation permits ordinary parameter cutoffs in the commutant construction.

3. A flat error can be dominated without losing smoothness

Work in a relatively open subset $Y$ of a closed half-space $y_1\geq0$; smooth functions are smooth up to that boundary. Let $r$ be real and assume

S3. Accessible zeros and flatness. On a sufficiently small patch about each zero, every nearby point with $r<0$ has a zero $y_*$ in the same half-space with

Proof. If a tangential derivative of $r$ is nonzero, keep the other coordinates fixed and vary that tangential coordinate. Shrink the patch so this derivative has fixed sign and magnitude at least $c>0$. The intermediate value theorem and the fundamental theorem of calculus give a zero within distance $2|r(y)|/c$, in the half-space. At an interior zero the same argument uses any nonzero partial derivative. In the remaining boundary case, $\partial_{y_1}r\geq c>0$ on a patch. Increase $y_1$ by $2|r(y)|/c$; the value crosses zero before that distance, while staying in the half-space.

All derivatives of $b$ vanish at a zero: an open positive-$r$ region approaches it within the half-space by the same directions, and the derivatives are continuous. Taylor's formula along the short segment from $y_*$ to $y$, inside a smaller convex coordinate patch, now gives $|D^\alpha b(y)|\leq C_{\alpha,N}|y-y_*|^N$. Use (DS8). Finitely many patches give the assertion on any retained compact part of the zero set. ∎

S4. Smooth upper envelope of a flat error. Suppose $F\geq0$, $F=0$ for $r\geq0$, $F$ is locally bounded, and $F=O(|r|^N)$ for every $N$ on compact sets. Locally there is a smooth $f\geq F$ with $f=0$ for $r\geq0$. No continuity of $F$ is required.

Proof. On a fixed larger compact patch $K$, define the nondecreasing function

4. A characteristic equality becomes an everywhere inequality

S5. Quadratic characteristic completion. Under (DS7), let

Proof. The full even/odd decomposition in U014, Corollary 5.2, gives smooth functions $B_0,B_1$ in a full square-variable neighborhood with

For $r<0$, write $a=-r>0$. The elementary square $(|t|-\sqrt a)^2\geq0$ gives

S6. The homogeneous orders. Let $\zeta\ne0$ be a tangential covector and $\lambda=|\zeta|$, or any fixed positive homogeneous norm on the retained cone. Suppose

Proof. Set $n=\xi/\lambda$, $R=r/\lambda^2$, normalize $A$ by $\lambda^{-1}$ and $\psi$ by $\lambda^{-1/2}$, and apply S5 on the compact angular patch. Multiply back by $\lambda$. The coefficients become $\psi_0=\lambda^{1/2}b_0$, $\psi_1=\lambda^{-1/2}b_1$, $g=\lambda^{-1}g_{\rm norm}$, giving exactly (DS17). Smooth angular coefficients and the homogeneous chain rule prove the ordinary symbol estimates after a low-frequency cutoff. If compact base support is needed, multiply the normalized inequality by a squared smooth cutoff equal to one on the support of $A$. It leaves $A$ unchanged and multiplies both $b_j$ by that cutoff and $g$ by its square. This preserves the inequality for either sign of $p$. ∎

The real symmetric two-by-two coefficient matrix of $A+(\psi_0+\xi\psi_1)^2-gp$ is nonpositive. Indeed, a quadratic polynomial nonpositive on every real $\xi$ gives a nonpositive quadratic form on every real pair $(v_0,v_1)$, by setting $\xi=v_1/v_0$ when $v_0\ne0$ and then taking limits when $v_0=0$. On complex pairs the form is the sum of its forms on the real and imaginary parts. Tensoring with the identity of any fixed finite rank preserves this conclusion. This is the precise symbol-level matrix statement; it makes no assertion about the remaining operator remainders.

5. A negative boundary coefficient without a singular root

S7. Linear remainder with a signed square coefficient. Let $F(\xi,y)$ be smooth and suppose

Proof. Even/odd descent first gives $F=F_0(\xi^2,y)+\xi F_1(\xi^2,y)$. Take $q_0=F_0(r,y)$. By S2 applied to a primitive of $h$,

For completeness, the unmodified remainder is divisible by $\xi^2-r$ by the exact integral

6. Two independent strict-diffractive commutants

Let $x\geq0$ be the normal coordinate, $z$ the tangential coordinates, and $(\xi,\zeta)$ their covectors. Write

S8. Commutant construction. Given $M\geq0$, arbitrarily small neighborhoods of the projected anchor $(x,z,\zeta)$ support homogeneous coefficients $q_j$, of degrees $-j$, with $q=q_0+q_1\xi$, and smooth homogeneous functions $\Psi,V,\varrho$, of degrees $1/2,0,0$, respectively, such that on $p=0$,

Proof: the decreasing cutoff. In fixed local angular coordinates put

Choose a nonnegative cutoff $C$ equal to one on $[-2,2]$ and zero outside $(-3,3)$, and another cutoff $J$ equal to one on $[-1,1]$ and zero outside $(-2,2)$. For small $\delta>0$ define

Proof: the incoming error. The exact differentiated identity is

Define

Proof: the linear symbol and its boundary square. The derivative

Homogeneity follows by performing the descent in $n=\xi/\lambda$ at $\lambda=1$, then multiplying its linear coefficient by $\lambda^{-1}$. Thus $q_j$ have degrees $-j$, $T_\delta$ degree $-1/2$, and the other stated degrees follow directly from (DS26)–(DS29). At the anchor,

Proof: two independent rows. Set $a=\partial_xr(0,z_*,\zeta_*)>0$ and $b=\{r,\lambda^{-1}\}_{z,\zeta}(0,z_*,\zeta_*)$. At the normalized anchor $C=J=1$, $H_p\phi=a$, and $\partial_\xi H_p\phi=-b$. Since $E(1)=E'(1)=e^{-1}$ and $E''(1)=-e^{-1}$, differentiating (DS26) gives

This proof constructs the incoming error on a region separated from glancing. It does not discard that error, or the lower-order terms generated when the symbols are quantized. Those are separate parts of the analytic estimate.

7. Exact examples and complete solutions

Exercise 1. Why the region with no real roots still matters

Take $\psi(t)=1+t$, let $\beta(r)=e^{1/r}$ for $r<0$ and zero for $r\geq0$, and set

Solution. The flat function $\beta$ has every derivative zero at zero, by the same exponential bounds used for (DS4). For $r\geq0$, $A=-(1+t)^2$ at both roots, and the completed expression equals $t^2-r$ everywhere. For $r=-a<0$, take

Exercise 2. The two rows in a wave-symbol model

Use $p=\xi^2-[\tau^2-(1-x)\zeta^2]$, with $\lambda=\tau>0$, at $(x,t,y;\xi,\tau,\zeta)=(0,0,0;0,1,1)$. Take $M=1$. Compute the normalized rows $(1,R_\delta)$ for $\delta=1/32,1/64$.

Solution. Here $a=1$ and $b=0$, because $r$ is independent of tangential base variables. Formula (DS33) gives

Exercise 3. Why arbitrary nonnegative cutoffs are insufficient

Explain why the nonnegative smooth function $h(s)=s^2$ cannot replace (DS1) when taking a smooth square root. Explain why the average in (DS6) avoids this defect under S2's hypotheses.

Solution. The nonnegative square root of $s^2$ is $|s|$, whose one-sided derivatives at zero differ. Also $h''=2$, so (DS1) with $\alpha=2$ and any $k>0$ would require $2\leq C|s|^{2k}$, impossible near zero. In (DS6), every derivative is controlled by the corresponding uniform derivative of $\partial_a g$; (DS3) transfers the full power estimates to the average. S1 then glues every derivative of its square root, not merely the function value. Even descent is applied only after that smoothness has been proved.

The first panel shows the exact gap (DS36), with its strictly positive minimum. The second compares the two quadratic expressions at $r=-1/4$; neither curve represents a PDE solution. The third displays the two exact coefficient rows in (DS37), normalized in their first entry. These are vectors in a symbol-coefficient plane, not rays in physical or cotangent space. The full general arguments are S5 and S8.

8. What this supplies to the boundary estimate

S5–S6 convert a negative square known on the characteristic roots into a nonpositive real symmetric quadratic symbol on every normal covariable, including the elliptic side. S7 makes the normal commutant coefficient a smooth negative square. S8 constructs two independent controlled combinations while isolating its remaining error on the incoming real-root region.

The next receiving proof must quantize these symbols, control the complete lower matrix terms and regularizer errors, and derive a legitimate estimate in each original weak domain. A normal differential commutant does not automatically preserve the Dirichlet test space. In the $H^1$ realization, derivative-on-test lower terms retain their boundary functional. Natural-dual forcing must retain its actual action. These issues are not resolved by the principal-symbol matrix inequality or by the scalar Dirichlet source theorem.

The prerequisite proof index gives the earlier results used here.