Manuscript VII-F@1 · VII-F.tex · lemma · 2d81c03f

Lemma: Obstruction for the axis-aligned vacuum maximum

The marked cone and a restricted obstruction › The literal-transplant obstruction · lines 314–357 · proof included or linked

VII-F@1#lem:transplant

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Consider

F_{s}^{vac} (x, y) = max {p ∣ x ∣ + a s, q ∣ y ∣ + b s}, a = 1 - p^{2}, b = 1 - q^{2},

on the axis-aligned wedge with seam $(r, 0) \sim (0, - r)$ . If $p \neq = q$ , this function does not descend continuously to the cone at $s = 0$ .

In particular, the choice $p = 3/5$ , $q = 5/13$ does not define a particle-supported slice on this seam. A positive upper cap does not repair the defect near the apex. If $p = q$ , the displayed family has no rate-driven incidence event.

Proof. For every $r > 0$ ,

F_{0}^{vac} (r, 0) = p r, F_{0}^{vac} (0, - r) = q r .

These values must agree at identified points. They do not agree when $p \neq = q$ .

A fixed positive upper cap is inactive for sufficiently small $r$ , so the same mismatch persists near the apex. If $p = q$ , then $a = b$ and

F_{s}^{vac} (x, y) = a s + p max {∣ x ∣, ∣ y ∣} .

The graph changes only by time translation; its incidence structure is independent of $s$ .

Displayed formulas in this passage

unlabeled · line 317
$F_{s}^{vac} (x, y) = max {p ∣ x ∣ + a s, q ∣ y ∣ + b s}, a = 1 - p^{2}, b = 1 - q^{2},$
unlabeled · line 339
$F_{0}^{vac} (r, 0) = p r, F_{0}^{vac} (0, - r) = q r .$
unlabeled · line 350
$F_{s}^{vac} (x, y) = a s + p max {∣ x ∣, ∣ y ∣} .$

Nearby in this section

section: The literal-transplant obstruction