Manuscript Hawk04-D@1 · Hawk04-D-v1.tex · theorem · 39b16f67

Theorem: Mass-law robustness

Stage 3: energy, erasure cost, and a conditional mass law › Stability, adiabaticity, and finite-step errors · lines 1475–1518 · proof included or linked

Hawk04-D@1#thm:mass-stability, Hawk04-D@1#eq:lifetime-bracket, Hawk04-D@1#eq:Gronwall

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Suppose an absolutely continuous trajectory remains in $[M_{c}, M_{0}]$ until its first cutoff crossing and satisfies

(1 - ρ) \frac{κ}{M ^{2}} \leq - M ˙ \leq (1 + ρ) \frac{κ}{M ^{2}}, 0 \leq ρ < 1.

Then it reaches the cutoff and

\frac{M _{0} ^{3} - M _{c} ^{3}}{3 κ ( 1 + ρ )} \leq t_{c} \leq \frac{M _{0} ^{3} - M _{c} ^{3}}{3 κ ( 1 - ρ )} .

More generally, if

M ˙ = - P (M), M ˙ = - P (M) + r (t),

and $P$ is Lipschitz with constant $L$ on a common validity interval, then, while both trajectories stay there,

∣ M (t) - M (t) ∣ \leq e^{L t} ∣ M (0) - M (0) ∣ + \int_{0}^{t} e^{L (t - s)} ∣ r (s) ∣ d s .

Proof. Multiply the first differential inequalities by $3 M^{2}$ . The derivative of $M^{3}$ lies between $- 3 κ (1 + ρ)$ and $- 3 κ (1 - ρ)$ . Integration gives finite cutoff crossing and the time bracket. For the second claim, the difference satisfies the integral inequality

∣ z (t) ∣ \leq ∣ z (0) ∣ + \int_{0}^{t} L ∣ z (s) ∣ d s + \int_{0}^{t} ∣ r (s) ∣ d s .

Gronwall's inequality gives (eq:Gronwall).

Displayed formulas in this passage

Hawk04-D@1#eq:lifetime-bracket · line 1486
$\frac{M _{0} ^{3} - M _{c} ^{3}}{3 κ ( 1 + ρ )} \leq t_{c} \leq \frac{M _{0} ^{3} - M _{c} ^{3}}{3 κ ( 1 - ρ )} .$
Hawk04-D@1#eq:Gronwall · line 1500
$∣ M (t) - M (t) ∣ \leq e^{L t} ∣ M (0) - M (0) ∣ + \int_{0}^{t} e^{L (t - s)} ∣ r (s) ∣ d s .$
unlabeled · line 1479
$(1 - ρ) \frac{κ}{M ^{2}} \leq - M ˙ \leq (1 + ρ) \frac{κ}{M ^{2}}, 0 \leq ρ < 1.$
unlabeled · line 1494
$M ˙ = - P (M), M ˙ = - P (M) + r (t),$
unlabeled · line 1514
$∣ z (t) ∣ \leq ∣ z (0) ∣ + \int_{0}^{t} L ∣ z (s) ∣ d s + \int_{0}^{t} ∣ r (s) ∣ d s .$

Theorem: Mass-law robustness

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