claimroutineinferred_dep

Harnack lower bound on unit circle

Harnack gives

H(T,1)\ge c\,m(b)/D

pointwise for

|T|\le1

turns band mass into a pointwise bound for the return

Claim·§6 Propagation across a thin cylinder·inside lemPropagation across a thin cylinder

Statement (paper)

There is a constant

c(M)>0

such that

H(T,1)\ge c(M)\,m(b)/D

for all

|T|\le 1

In plain English

There is a constant

c(M)>0

such that the circular mean satisfies

H(T,1)\ge c(M)\,m(b)/D

for every

|T|\le 1

, where

m(b)

is the sine-weighted mean of

H(\cdot,b)

T

Why it's here

The previous claim only gives the band mass of

H

3/4\le|\tau|\le 5/4

, at least

c\,m(b)/D

. A Harnack inequality for the subsolution mean, which is why the bound

\sup_{1/2\le s\le 3/2}V\le M

is assumed, turns this integrated mass into a pointwise lower bound at radius 1 for all

|T|\le1

. This feeds the return passage to inner radius, which lowers the bound from

s=1

H(0,r)

Classified routine:

A standard application of the Harnack inequality for nonnegative (sub/super)solutions of elliptic equations with bounded potential (as in Gilbarg–Trudinger), which turns the band-mass bound from the previous claim into a pointwise lower bound at radius 1.

Uses

claimOutward passage bounds band mass

Used by

claimReturn passage to inner radius

Figures

Two side-by-side rectangles in the $(T,s)$ plane, with $s=|\tau|$ drawn schematically. Panel (a), "Outward comparison", spans $0\le T\le3$ and $b\le s\le3$. A thick orange segment labelled "observed wave" lies on the bottom wall $s=b$ for $1\le T\le2$, and a shaded band $3/4\le s\le5/4$ is centred on a dashed line at $s=1$; below it is the factor at $s=1$, $\log 3/\log(3/b)$. Panel (b), "Inward comparison", spans $-1\le T\le1$ and $b\le s\le1$. Its thick teal top wall at $s=1$ is labelled "uniform lower bound", and dashed guides mark an orange dot at $(0,2b)$ labelled $H(0,2b)$; below it is the factor at $s=2b$, $\log 2/\log(1/b)$. Together the panels show the two logarithmic losses incurred in carrying the energy out to $s=1$ and back to the inner wall. — Figure 2schematicp. 22
The two comparisons, with radii $s=|\tau|$ drawn schematically. The boundary mass on $1\le T\le2$ bounds the mass in the shaded band from below. The mean inequality on a fixed annulus supplies a lower bound on the entire upper wall in (b). The two displayed fractions are values of the normalized radial harmonic functions. The comparison solutions with potential dominate fixed multiples of these values, giving the factor $[\log(3/b)]^{-2}$ at the dot. In the wave estimate, $b=2Kh$ : $H$ controls the wave norm on the marked segment and is controlled by the initial data at the dot.
What it shows
Two side-by-side rectangles in the $(T,s)$ plane, with $s=|\tau|$ drawn schematically. Panel (a), "Outward comparison", spans $0\le T\le3$ and $b\le s\le3$ . A thick orange segment labelled "observed wave" lies on the bottom wall $s=b$ for $1\le T\le2$ , and a shaded band $3/4\le s\le5/4$ is centred on a dashed line at $s=1$ ; below it is the factor at $s=1$ , $\log 3/\log(3/b)$ . Panel (b), "Inward comparison", spans $-1\le T\le1$ and $b\le s\le1$ . Its thick teal top wall at $s=1$ is labelled "uniform lower bound", and dashed guides mark an orange dot at $(0,2b)$ labelled $H(0,2b)$ ; below it is the factor at $s=2b$ , $\log 2/\log(1/b)$ . Together the panels show the two logarithmic losses incurred in carrying the energy out to $s=1$ and back to the inner wall.
Read from the image by claude-opus-5.5 · confidence 88%
SupportslemPropagation across a thin cylinderclaimOutward passage bounds band massclaimHarnack lower bound on unit circleclaimReturn passage to inner radiusclaimRadial solutions dominate log-affine interpolants

Sources

PaperOpen

Section: §6 Propagation across a thin cylinder
Page: —
Label: —
Line: sections/propagation.tex:79

Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

info
inferred_dep
outline: "Harnack lower bound on unit circle" rests on claim-lem-cylinder-2