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An outward Euclidean exit cannot return

Statement

Let $d \geq 2$ and let $Θ \subset R^{d}$ be a compact obstacle whose exterior $Ω = R^{d} ∖ Θ$ has smooth one-sided boundary. Fix $R > 0$ with $Θ \subset B_{R}$ , where $B_{R} = {z \in R^{d} : ∣ z ∣ < R}$ and ${\overline{B}}_{R} = {z \in R^{d} : ∣ z ∣ \leq R}$ ; thus the compact obstacle is contained in the open ball. Quantify over complete unit-speed generalized rays $x : R \to \overline{Ω}$ in the stationary Euclidean generalized-ray convention. In particular, each ray is defined for all forward times. If $∣ x (0) ∣ \leq R$ and $∣ x (τ) ∣ > R$ at some $τ > 0$ , then $∣ x (t) ∣ > R$ for every $t \geq τ$ and $∣ x (t) ∣ \to \infty$ as $t \to \infty$ .

More precisely, suppose there is $L > 0$ such that every complete admissible ray $y$ with $∣ y (0) ∣ \leq R$ has a time $s \in (0, L]$ satisfying $∣ y (s) ∣ > R$ . This is the common exterior-hitting bound. Then every such $y$ remains outside ${\overline{B}}_{R}$ for all $t \geq L$ and tends to spatial infinity. Every complete ray initially outside ${\overline{B}}_{R}$ also eventually remains outside ${\overline{B}}_{R}$ and tends to spatial infinity. No common time for this last assertion, measured from the initial point outside the ball, is claimed.

Statement references

Independent generalized reflected-ray and GCC convention

In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones

Facts & Assumptions

[F1]

The convention of Independent generalized reflected-ray and GCC convention supplies continuous spatial projections of complete rays on $R$ . In the Euclidean interior its Hamilton rule for $p = - τ^{2} + ∣ ξ ∣^{2}$ keeps $τ$ and $ξ$ constant, with $d t / d s = - 2 τ$ and $d x / d s = 2 ξ$ . A nonzero characteristic covector has $∣ ξ ∣ = ∣ τ ∣ > 0$ , so physical-time velocity is the constant unit vector $- ξ / τ$ . Its local rules are stationary, so $t \mapsto x (q + t)$ is again a complete admissible ray for every real $q$ . These are properties of the specified complete curves; no extension of an arbitrary ray segment is used.

[F2]

A compact subset of Euclidean space is closed by In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, since Euclidean space is Hausdorff (distinct points have disjoint balls of radius one third their distance). Hence $Ω$ is open. Every point with $∣ z ∣ \geq R$ is outside $Θ$ and has an obstacle-free neighborhood in $Ω$ . Smoothness of this exterior is assumed explicitly, as required by the generalized-ray convention.

Proof

Given: The smooth compact-obstacle setting and complete rays specified above, and the exterior-hitting bound when proving the consequence. Goal: No return after exit, divergence to infinity, and eventual escape for all initial locations.

1.1givenF1F2algebra

Fix a ray with $∣ x (0) ∣ \leq R$ and $∣ x (τ) ∣ > R$ . By continuity, the connected component of ${t > 0 : ∣ x (t) ∣ > R}$ containing $τ$ is an interval $(a, b)$ with $0 \leq a < τ < b \leq \infty$ , and $∣ x (a) ∣ = R$ . Indeed, if $a > 0$ an exterior value at $a$ would extend the component to the left, whereas a strictly interior value would contradict the exterior points immediately to its right; if $a = 0$ , continuity and $∣ x (0) ∣ \leq R$ give the same equality. On $(a, b)$ there is no obstacle boundary. The interior rule gives a single constant unit velocity $v$ on this interval: its locally constant velocities agree on overlapping interior intervals, and the interval is connected. Taking the limit at $a$ gives $x (a + s) = x (a) + s v$ for $0 < s < b - a$ . Expanding the squared norm for arbitrarily small positive $s$ shows $x (a) \cdot v \geq 0$ , since a negative value would give $∣ x (a + s) ∣^{2} = R^{2} + 2 s x (a) \cdot v + s^{2} < R^{2}$ for sufficiently small $s$ .

2.1F1step 1.1algebra

If $b < \infty$ , the ray is defined and continuous at $b$ by completeness. The line formula from step 1.1 gives $∣ x (b) ∣^{2} = R^{2} + 2 (b - a) x (a) \cdot v + (b - a)^{2} > R^{2}$ , contradicting that $b$ is the right endpoint of the exterior component: continuity would put a whole neighborhood of $b$ in the same exterior set. Thus $b = \infty$ . The same formula gives $∣ x (a + s) ∣^{2} \geq R^{2} + s^{2}$ for all $s > 0$ , proving both no return and divergence to infinity. This is an argument about an already complete ray, not a global existence or continuation theorem.

3.1givenstep 1.1step 2.1

Assume the stated common exterior-hitting bound. For any complete ray $y$ initially in ${\overline{B}}_{R}$ , take its exterior-hitting time $s \in (0, L]$ . Steps 1.1 and 2.1 imply $∣ y (t) ∣ > R$ for every $t \geq s$ , hence for every $t \geq L$ , and $∣ y (t) ∣ \to \infty$ . The bound is quantified over complete rays themselves, including every admissible branch at glancing, so no uniqueness of a ray from its initial state is needed.

4.1givenF1F2step 3.1∎

Let $x$ instead start outside ${\overline{B}}_{R}$ . If it never enters ${\overline{B}}_{R}$ at a nonnegative time, its whole forward path lies in the interior of $Ω$ , so $x (t) = x (0) + t v$ for a fixed unit vector $v$ ; the inequality $∣ x (0) + t v ∣ \geq t - ∣ x (0) ∣$ proves divergence. Otherwise let $q$ be its first entrance time into ${\overline{B}}_{R}$ . Continuity gives $q > 0$ , $q < \infty$ and $∣ x (q) ∣ = R$ . Stationarity makes $y (t) = x (q + t)$ a complete admissible ray initially in ${\overline{B}}_{R}$ . Step 3.1 applies, so $x$ remains outside for all $t \geq q + L$ and tends to infinity. This proves the asserted consequence for every initial location.

Depends on

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