A flat convex boundary with two characteristic continuations

A smooth convex obstacle can have infinitely many transverse reflections in a finite travel time. At their accumulation point the ordinary ray has contact of every order with the boundary. The reflected continuation and the boundary gliding continuation then pass through the same full characteristic state.

We construct the boundary, the polygon, its exact reflection law and both Hamilton lifts. The construction complements the variable-coefficient example in Tangency, gliding and generalized characteristic motion, Section 10. Its inputs are the contact-order calculation in that lesson's Section 2, its generalized-arc definition in Section 3, and ordinary smooth calculus and convergent series in the stationary-phase foundations. The exact earlier proofs are identified in the proof map.

Hörmander III, Springer 2007 edition, ISBN 978-3-540-49938-1, Example 24.3.11, printed 438, gives the convex-boundary nonuniqueness phenomenon. The formulas and interpolation below are independently chosen.

1. Prescribe chords and their reflecting tangents

D0. A polygon accumulating at the origin. For integers $j\ge1$ , set

x_j=\frac1j,\qquad d_j=x_j-x_{j+1}=\frac1{j(j+1)},\qquad m_j=2^{-j},\qquad y_j=\sum_{\ell=j}^{\infty}d_\ell m_\ell . \tag{CB1}

The series converges absolutely. Its tail satisfies

0<y_j\le m_jx_j

. Hence

P_j=(x_j,y_j)

tends to the origin, and the segment from

P_{j+1}

P_j

has slope exactly

m_j

. In increasing

x

, the polygon's slopes increase. Put also

m_0=1

, and define

\alpha_j=\arctan m_j,\qquad \beta_j=\frac{\alpha_{j-1}+\alpha_j}{2},\qquad b_j=\tan\beta_j . \tag{CB2}

If a convex smooth graph through

P_j

has derivative

b_j

there, its tangent bisects the incoming and outgoing chord directions. The equal-angle reflection law will therefore hold exactly, rather than just asymptotically.

D1. A uniformly interior interpolation parameter. Set

\theta_j=\frac{m_j-b_{j+1}}{b_j-b_{j+1}},\qquad c_j=1-\theta_j . \tag{CB3}

The denominator is positive. In fact

b_{j+1}<m_j<b_j

, and

1/9\le\theta_j\le2/3

. Here are bounds independent of

j

. Write

m=m_j\le1/2

. On

[0,1]

the derivative of

\arctan

lies between

1/2

and

1

; on the corresponding angles the derivative of

\tan

lies between

1

and

2

. Applying these bounds to the two half-angle differences gives

\frac m4\le b_j-m\le m,\qquad \frac m8\le m-b_{j+1}\le\frac m2 . \tag{CB4}

Taking the ratio of the second gap to the sum proves the assertion. In particular

1/3\le c_j\le8/9

. This uniform margin is what permits one smooth interpolation profile at every shrinking interval.

2. Construct the actual smooth convex graph

I0. An exact moment profile. Choose a nonnegative even smooth function $\kappa$ , positive on $(-a,a)$ , supported in $[-a,a]$ , with integral one, where $a=1/32$ . For this construction, take it to be the normalization of $\exp[-1/(1-(u/a)^2)]$ for $|u|<a$ , extended by zero. Every derivative at an endpoint is zero: differentiation produces a rational power of the endpoint distance times the same exponential, and $v^{-N}e^{-1/v}\to0$ . The integral is positive and finite. Define

K(v)=\int_{-\infty}^v\kappa(u)\,du,\qquad L(v)=\int_{-\infty}^v K(u)\,du . \tag{CB5}

For

0\le s\le1

, put

H_j(s)=K(s-c_j)

. It is zero near zero, one near one, and nondecreasing. Its exact mean is

\int_0^1 H_j(s)\,ds =\int_{-a}^{a}(1-c_j-u)\kappa(u)\,du =1-c_j=\theta_j . \tag{CB6}

The exchange of integrals can equally be checked by integration by parts on the compact rectangle; evenness gives the zero first moment. All derivatives of

H_j

have bounds depending only on their order, since these are translates of one fixed profile.

I1. Match values, slopes and every higher derivative. On $[x_{j+1},x_j]$ prescribe

\begin{split} g(x)&=b_{j+1}+(b_j-b_{j+1}) K\left(\frac{x-x_{j+1}}{d_j}-c_j\right),\\ f(x)&=y_{j+1}+\int_{x_{j+1}}^x g(v)\,dv . \end{split} \tag{CB7}

Formula (CB6) gives

f(x_j)-f(x_{j+1})=d_jm_j

, so the endpoint values are exactly

y_j,y_{j+1}

. Near either endpoint

g

is the corresponding constant

b_j

b_{j+1}

. Neighboring intervals thus have identical first derivatives and zero derivatives of every order at least two at their shared endpoint. The resulting function is smooth on

0<x\le1

, has

f'=g

, and is convex since

g'\ge0

. Extend it linearly past

x_1

with slope

b_1

I2. Verify smoothness and flatness at the accumulation point. For every integer $k\ge0$ ,

\sup_{[x_{j+1},x_j]}|g^{(k)}| \le C_k 2^{-j}[j(j+1)]^k . \tag{CB8}

For

k=0

, use

0<b_{j+1}\le g\le b_j\le2m_j

; for

k\ge1

, differentiate (CB7) and use the uniform profile bounds. The same bound for

f^{(k+1)}

tends to zero faster than every power of

x

, since

x\ge1/(j+1)

and

2^{-j}j^M\to0

for each fixed

M

. The latter limit follows, for example, from the ratio of successive terms tending to

1/2

Set $f(x)=0$ for $x\le0$ . Its value tends to zero, because $0\le f(x)\le y_j$ on the $j$ -th interval. The displayed bounds show that each right derivative tends to zero and is $o(x)$ . Inductively the difference quotient at zero gives the next derivative, also zero; on the negative side all derivatives vanish. Thus $f$ is genuinely $C^\infty$ and flat at zero. Its derivative is nondecreasing on the whole line, so its convexity persists through zero. No finite-order smoothing or formal series stands in for the actual graph.

3. Check the full billiard and wave characteristics

R0. Chords lie strictly in the permitted interior. Let

X=\{(x,y,t):y\ge f(x)\},\qquad \phi=y-f(x). \tag{CB9}

Convexity places each chord above the graph. The inequality is strict between its endpoints. Indeed the graph derivative starts below the chord slope

m_j

, ends above it, and is nondecreasing; the difference between chord and graph therefore first increases and then decreases, returning to zero only at the right endpoint. A nontrivial middle plateau cannot have zero difference, because the initial strict increase has already made it positive.

The unit velocity on chord $j$ is $v_j=(\cos\alpha_j,\sin\alpha_j)$ . At $P_j$ , reflection in the tangent line of angle $\beta_j$ takes direction $\alpha_j$ to $2\beta_j-\alpha_j=\alpha_{j-1}$ . Equivalently it takes $v_j$ to $2(v_j\cdot T_j)T_j-v_j$ , where $T_j=(\cos\beta_j,\sin\beta_j)$ . Thus it is the exact Euclidean specular reflection. The two normal components are nonzero because $\alpha_j<\beta_j<\alpha_{j-1}$ .

R1. Use finite arclength as the Hamilton parameter. Let

\ell_j=d_j\sqrt{1+m_j^2},\qquad s_j=\sum_{\ell=j}^{\infty}\ell_\ell,\qquad x_j\le s_j\le\sqrt{5/4}\,x_j . \tag{CB10}

The polygon from the origin has finite length to every vertex. On

s_{j+1}<s<s_j

, take

(x(s),y(s))=P_{j+1}+(s-s_{j+1})v_j

, and set physical time

t=s

. For the real homogeneous quadratic symbol

q(x,y,t;\xi,\eta,\tau) =\tfrac12(\xi^2+\eta^2-\tau^2),\qquad (\xi,\eta,\tau)=(v_j,-1), \tag{CB11}

we have

q=0

H_q=(\xi,\eta,-\tau;0,0,0)

, and hence precisely the stated unit-speed Hamilton pieces.

At a reflection the tangential covector is unchanged: for the boundary tangent $(1,f'(x),0)$ , the two velocities have the same scalar product, while $\tau=-1$ remains fixed. Their difference is a multiple of $d\phi=(-f',1,0)$ . Thus they are the two distinct characteristic roots in the same boundary fiber. Also $H_\phi^2q=1+f'^2>0$ , so the required double normal root is nondegenerate at every boundary point, and the boundary is noncharacteristic for the wave operator with symbol $q$ .

R2. Check the generalized derivative at infinite contact. On the $j$ -th chord,

|v_j-(1,0)|\le C2^{-j},\quad |y(s)|\le C2^{-j}x_j,\quad |x(s)-s|\le C\,4^{-j}x_j . \tag{CB12}

The last estimate follows by summing

0\le\ell_\ell-d_\ell\le d_\ell m_\ell^2/2

, including the current partial chord; for that partial chord use

1-\cos\alpha_j\le m_j^2/2

. Since

s

is comparable to

1/j

between successive vertices, all three errors are smaller than every fixed power of

s

. These bounds include both one-sided velocities at each reflection.

For $s\le0$ , continue by $x=t=s,y=0,\xi=1,\eta=0,\tau=-1$ . At zero the positions have derivative $(1,0,1)$ and the covectors derivative zero, including along the accumulating reflected pieces. The limiting full state is $q_0=(0,0,0;1,0,-1)$ . Along its ordinary Hamilton orbit, $\phi(s)=-f(s)$ . Flatness of $f$ proves $H_q^k\phi(q_0)=0$ for every $k$ ; thus $q_0\in G^\infty$ . The gliding field there equals $H_q$ . The resulting reflected arc therefore satisfies the actual generalized-arc definition at its accumulation time. Its reflection times are isolated individually and accumulate only at zero, which is a glancing time, not a transverse reflection.

4. Construct the different continuation through the same state

G0. The boundary arclength lift is a genuine gliding solution. Parameterize the boundary graph by its signed arclength $s$ , with $x(0)=0$ . The inverse is smooth because $ds/dx=\sqrt{1+f'(x)^2}>0$ . Set

y=f(x),\quad t=s,\quad \xi=(1+f'^2)^{-1/2},\quad \eta=f'(1+f'^2)^{-1/2},\quad \tau=-1. \tag{CB13}

It has

q=0

and

H_q\phi=0

. Direct calculation on these constraints gives

H_q^2\phi=-f''\xi^2,\qquad H_q^G=H_q-\frac{f''\xi^2}{1+f'^2}H_\phi,\qquad (\xi',\eta')=\frac{f''}{(1+f'^2)^2}(-f',1). \tag{CB14}

Differentiating (CB13) with

x'=(1+f'^2)^{-1/2}

gives exactly the last formula. Its base derivatives are

(\xi,\eta,1)

, so this is the full restricted Hamilton equation. Where

f''>0

the state is strictly gliding; where

f''=0

it is in

G^3

, and the same equation applies. This proves the gliding solution through every interpolation and flat segment.

At $s=0$ this arc has the same full state $q_0$ as the reflected arc and agrees with it for $s\le0$ . For every positive chord interval the reflected arc has $y>f(x)$ , whereas this one stays on the boundary. They are different continuations of the same smooth real constant-coefficient wave characteristic problem. Finite-contact uniqueness cannot be extended to all infinite-contact states.

5. Both specified continuations are limits of transverse broken rays

L0. Approximation of these two curves. For this example the reflected arc and the boundary arc are each uniform compressed limits of finite transverse broken rays on every compact physical-time interval in their stated domain. We do not infer this property for all nonunique generalized rays.

First check local uniqueness along the positive boundary arc, including its flat portions. Write $\sigma$ for boundary arclength, $T,N$ for its tangent and inward unit normal, and

\varkappa=\frac{f''}{(1+f'^2)^{3/2}}\ge0,\qquad F(r,\sigma)=F(0,\sigma)+rN(\sigma). \tag{CB15}

r=0

, the two coordinate derivatives are

N,T

, hence the map has a smooth local inverse. Since

N_\sigma=-\varkappa T

, its Euclidean metric is

dr^2+(1-r\varkappa)^2d\sigma^2

. The actual pulled-back wave symbol, multiplied by two, is therefore

p=\rho^2-R,\qquad R=\tau^2-\frac{\eta_\sigma^2}{(1-r\varkappa)^2},\qquad R_0=\tau^2-\eta_\sigma^2,\qquad R_1=-2\varkappa\eta_\sigma^2 . \tag{CB16}

Its boundary gliding field has constant

\eta_\sigma,\tau

, with

\sigma'=2\eta_\sigma

and

t'=-2\tau

. This factor-two Hamilton parameter agrees with the physical-time parameter of (CB11) after the stated constant rescaling.

Where $\varkappa>0$ , strict gliding uniqueness follows from the normal-energy estimate GGL:T016–T017. In an open flat segment, the higher-contact comparison has $e=0$ , so the continuation is also uniquely the boundary line there. At a nonzero endpoint of a curvature bump, the curvature is monotone just inside that bump. To verify the last assertion, on its interval (CB7) gives

f''=\frac{b_j-b_{j+1}}{d_j}\, \kappa\!\left(\frac{x-x_{j+1}}{d_j}-c_j\right),\qquad \frac{d}{dx}\log\varkappa =\frac{\kappa'(S-c_j)}{d_j\kappa(S-c_j)} -\frac{3f'f''}{1+f'^2}. \tag{CB17}

For the fixed exponential bump in I0, its logarithmic derivative tends to positive infinity at the left edge and negative infinity at the right edge. The second term is bounded on this one interval. Thus

\varkappa

increases from zero at entry and decreases to zero at exit on sufficiently small one-sided neighborhoods. The map from

x

\sigma

is strictly increasing. In the time direction entering either bump edge,

e=R_1=-2\varkappa

, on the unit gliding state, is nonincreasing from zero. GGL:T024 forces the gliding continuation; on the flat side

e=0

gives the same conclusion. Every positive boundary point therefore has local uniqueness. This uses the specific bump just chosen, rather than a claim for arbitrary flat curvature.

Every positive point of the polygon has local uniqueness as well, by ordinary flow or a transverse reflection. Neither curve has reflection accumulation at a positive time. On the negative side of zero the boundary is straight and $R_1=0$ . GGL:T019–T022, applied backward from the common state at zero, forces the unique boundary-line continuation. Hence each complete curve is the only generalized continuation through any one of its own fixed positive states on the compact interval considered: local uniqueness propagates agreement through every positive time, continuity supplies zero, and the backward zero-forcing comparison supplies negative times. This conclusion does not say that the common state at zero alone determines the positive future.

It remains to obtain approximating finite rays through nearby states. Physical time is a uniform clock for these wave characteristics near $\tau=-1$ . On a bounded physical-time interval the base velocity has magnitude one, and reflection preserves that speed and the spatial covector magnitude. Nearby normalized states therefore remain in a common compact phase region. Apply the volume argument GGL:T031–T035 on this compact region. A finite boundary-chart cover has a common normal strip estimate $O(\epsilon^3)$ ; the preceding clock grid has $O(\epsilon^{-1})$ sections. The injective finite reflected clock maps preserve the whole reduced volume. The initial states whose continuation reaches glancing before either end thus have outer measure $O(\epsilon^2)$ for every small $\epsilon$ , and form a null set. Fixed-clock wall endpoints are the null exceptions treated in GGL:A1. The same proof works forward and backward, and across the finitely many compact charts.

Consequently finite transverse broken arcs on the entire interval have initial states dense near either selected positive state. Normalize them by their constant negative time covector and choose a sequence approaching that state. GGL:T027–T030 supplies uniformly convergent compressed subsequences, and every subsequential limit is a generalized arc through that state. The just-proved uniqueness on this interval identifies it with the chosen polygon or gliding arc. If the whole sequence did not converge, a subsequence with a fixed positive uniform error would have a further compactness limit with zero error, a contradiction. This proves the claimed uniform compressed approximation of both curves.

Convex chord interpolation and its exact equal-angle reflection

V0. What the diagram measures. The left panel shows the actual vertices $P_j$ , straight chords and the smooth convex graph, with vertical coordinates multiplied by the stated factor. The middle panel subtracts the left tangent on the interval $j=5$ and divides by $d_j(b_j-b_{j+1})$ : the chord is exactly $\theta_j S$ , the graph is $L(S-c_j)$ , and their difference is positive. The right panel uses an orthonormal frame at $P_2$ ; its two unit directions have angles $\pm(\alpha_1-\alpha_2)/2$ to the tangent. Numerical integration is used only to draw the fixed smooth profile; exact mean, convexity, flatness and reflection are proved above. The picture is a spatial section; physical time and covectors are supplied in (CB10)–(CB14).

6. Exercises with complete solutions

Exercise 1. Track the normal reflection gap

For a vertex $P_j$ , compute the incoming and outgoing values of $H_q\phi$ , including their signs. Explain why their magnitudes agree.

Solution. Write $\delta_j=(\alpha_{j-1}-\alpha_j)/2>0$ . Since $f'(x_j)=\tan\beta_j$ , $H_q\phi=\sin\alpha-\tan\beta_j\cos\alpha =\sin(\alpha-\beta_j)/\cos\beta_j$ . The incoming value is $-\sin\delta_j/\cos\beta_j$ , the outgoing value is its positive opposite. Both are nonzero and tend to zero. This is the normal-root exchange for the defining function actually used, rather than a reflection of only the base velocity without its covector.

Exercise 2. Remove a false finite-order conclusion

Show that the reflected arc approaches the tangent state faster than every algebraic power of time, although it is interior on infinitely many intervals in every positive neighborhood of zero.

Solution. On the $j$ -th interval, $s\ge1/(j+1)$ . Multiplying the bounds (CB12) by $s^{-N}$ gives at most $C_N(j+1)^N2^{-j}$ , or the still smaller analogous $4^{-j}$ bound. Each tends to zero by its successive-term ratio. The same applies to the limiting one-sided momenta at the vertices. Yet $s_j\downarrow0$ , and strict chord separation in R0 gives an open interior interval $(s_{j+1},s_j)$ for every $j$ . Infinite-order contact therefore does not imply that the reflected curve equals the boundary curve on a positive neighborhood.

Exercise 3. Why a linear derivative interpolation fails

Suppose one replaces $H_j(S)$ by $S$ . Determine the obstruction to keeping both endpoints of the same chord, and explain why the shifted smooth profile fixes it.

Solution. Linear interpolation has mean $1/2$ ; its total height is $d_j(b_j+b_{j+1})/2$ . The required height is $d_jm_j$ . These agree only if $\theta_j=1/2$ . Here, as $j\to\infty$ , $b_j/m_j\to3/2$ and $b_{j+1}/m_j\to3/4$ , by differentiability of $\tan$ and $\arctan$ at zero. Thus $\theta_j\to1/3$ , and the endpoint error is of the same scale as the height itself. Moreover linear interpolation does not match all higher derivatives at the vertices. The translated fixed profile has exact mean $\theta_j$ and is constant near both endpoints, simultaneously preserving the chord and every derivative match.

Independent exposition, proofs, exercises, figure and scripts by GPT-6.1 Sol (OpenAI), October 2026, dedicated under CC0 1.0 Universal. Linked programme materials retain their recorded terms. The source credit concerns Hörmander III, Example 24.3.11; this construction does not assert approximation of every nonunique generalized arc by transverse broken rays.