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A variable-coefficient annular barrier and inward boundary derivative

Statement

Let $L = Δ_{h} - V = a^{i j} \partial_{i j} + b^{i} \partial_{i} - V$ be the geometric operator on a relatively compact smooth metric chart, with symmetric uniformly elliptic $a$ , bounded $a, b, V$ , and $V \geq 0$ . Let $B_{R} (z)$ be an interior tangent Euclidean ball at $p$ , with its closure in the chart. Suppose $u > 0$ in the ball, $u \geq 0$ on its closure, $u (p) = 0$ , $L u \leq 0$ , and $u$ is $C^{2}$ inside and $C^{1}$ at $p$ . Then the derivative in direction $(z - p) / R$ at $p$ is strictly positive. This applies to the geometric operator from the coordinate-form lemma.

Facts & Assumptions

[F1]

The smooth metric $h$ is Riemannian in the sense of Riemannian metric and riemannian manifold. The symbol $Δ_{h}$ uses the positive-divergence, metric-trace-of-Hessian convention of Laplace–Beltrami operator as the trace of the Hessian. For the general metric-chart argument, use the differential expression $Δ_{h} = a^{i j} \partial_{i j} + b^{i} \partial_{i}$ and coefficient bounds assumed in the Statement. For the smooth-exterior application, Coordinate and weak forms of the geometric Schrödinger equation supplies this differential expression for its smooth metric. Subtracting a bounded potential is a pointwise zeroth-order operation; no smoothness of the original $V$ is used below. For this application use the smooth metric on the closed ball, so the stated coefficient bounds and ellipticity hold. Fix $λ > 0$ with $a^{i j} ξ_{i} ξ_{j} \geq λ ∣ ξ ∣^{2}$ , and bounds $tr a \leq T$ , $∣ b ∣ \leq B$ and $0 \leq V \leq Q$ , where $Q \geq 0$ is a constant.

[F2]

The supplier Compact-domain comparison with a nonnegative potential gives comparison for $Δ_{h} - W$ on relatively compact smooth domains when $W \geq 0$ is smooth and bounded and the difference belongs to $C^{2} (D) \cap C^{1} (\overline{D})$ , with each component closure meeting the boundary. In its stated ZFC setting, $(Δ_{h} - W) w \geq 0$ and $w \leq 0$ on the boundary imply $w \leq 0$ inside. We use this supplier only with the smooth constant $W = Q$ and annuli whose closed outer spheres lie strictly inside $B_{R} (z)$ .

[F3]

For every integer $n \geq 1$ , Heine-Borel in ℝⁿ: with the Euclidean metric a subset of ℝⁿ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, part 2, says that a subset of Euclidean $R^{n}$ is compact if and only if it is closed and bounded. We use the Euclidean subspace metric on coordinate spheres and closed annuli. A compact coordinate set is also compact in the manifold: a manifold open cover pulls back under the inverse chart to an open cover of that coordinate set, and a finite subcover transfers back.

[F4]

By A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, a continuous real function on a nonempty compact metric space attains a minimum and a maximum. If it is strictly positive everywhere, its attained minimum is strictly positive.

[F5]

Use the half-space chart and smooth-transition conventions of Smooth charts, atlases, and structures with boundary. By Embedded smooth submanifolds with boundary, an embedded smooth submanifold with boundary is a subset with a manifold-with-boundary smooth structure for which inclusion is a smooth embedding. Step 1.2 constructs these charts and the embedding for the closed annulus; no smooth-domain assertion is imported from the comparison theorem.

Proof

Given: The hypotheses in the Statement and the suppliers and bounds in [F1]–[F5].

Goal: Prove that the inward derivative at $p$ is strictly positive.

Proof technique: direct.

1.1F1givenalgebrachoose

Put $L_{Q} = Δ_{h} - Q$ . Positivity of $u$ inside the ball gives $L_{Q} u = L u - (Q - V) u \leq 0$ . Choose $α > 0$ so large that $α^{2} λ R^{2} - 2 α T - 2 α B R - Q \geq 0$ ; this is possible because $λ R^{2} > 0$ . This choice is independent of the truncated outer radius used below.

1.2F3F5givenconstructalgebra

For $R / 2 < s < R$ , let $r = ∣ x - z ∣$ and $D_{s} = {R / 2 < r < s}$ . Write $n$ for the chart dimension. Since $∣ p - z ∣ = R > 0$ , $n \geq 1$ . The sets $S_{R / 2} = {∣ x - z ∣ = R / 2}$ and $N_{s} = {R / 2 \leq ∣ x - z ∣ \leq s}$ are closed and bounded in $R^{n}$ , hence compact by [F3], and lie inside $B_{R} (z)$ . The sphere is nonempty, containing $z + (R / 2) e_{1}$ , where $e_{1}$ is the first coordinate unit vector. The radial inequalities show that $N_{s} = \overline{D_{s}}$ and $N_{s} ∖ D_{s}$ consists exactly of the spheres of radii $R / 2$ and $s$ . Here is an explicit smooth boundary chart near a point $y$ of either sphere. Write its radius as $r_{0}$ and choose $j$ with $y_{j} - z_{j} \neq 0$ . Set $σ = sign (y_{j} - z_{j})$ and use coordinates $({\hat{x}}_{j}, t)$ , where ${\hat{x}}_{j}$ lists all coordinates other than $x_{j}$ , with $t = r - r_{0}$ on the inner sphere and $t = r_{0} - r$ on the outer sphere. With $η = 1$ in the inner case and $η = - 1$ in the outer case, the inverse is $x_{j} = z_{j} + σ \sqrt{(r_{0} + η t)^{2} - \sum_{i \neq j} (x_{i} - z_{i})^{2}}$ , retaining the other coordinates. Near $y$ the radicand is positive and the other radial inequality is strict, so these maps are mutually inverse smooth extensions on open Euclidean neighborhoods, and $N_{s}$ corresponds exactly to $t \geq 0$ . Along with ordinary charts in $D_{s}$ , they form compatible half-space charts as in [F5]. The inclusion has the subspace topology and in each chart extends to a local Euclidean diffeomorphism, hence is a smooth embedding. Thus $N_{s}$ is a compact smooth full-dimensional closure with intrinsic boundary exactly those two spheres. Finally, for any $x$ in a component $C$ of $D_{s}$ , the radial path $z + τ (x - z) / ∣ x - z ∣$ , $∣ x - z ∣ \leq τ < s$ , lies in $D_{s}$ and in the same component as $x$ ; its limit at $τ = s$ lies in $\overline{C} \cap \partial D_{s}$ . This also covers $n = 1$ , when the annulus has two interval components and the boundary-chart list ${\hat{x}}_{j}$ is empty. All these sets lie in the chart interior, so the same smooth-domain and component-boundary assertions hold on the manifold.

2.1F1step 1.1algebra

For any $R / 2 < s < R$ , on $D_{s} = {R / 2 < r = ∣ x - z ∣ < s}$ set $ψ_{s} = e^{- α r^{2}} - e^{- α s^{2}}$ . Direct differentiation in the coordinates from [F1] gives $L_{Q} ψ_{s} = e^{- α r^{2}} (4 α^{2} a^{i j} (x - z)_{i} (x - z)_{j} - 2 α tr a - 2 α b \cdot (x - z) - Q) + Q e^{- α s^{2}}$ . Since $r \geq R / 2$ and $r \leq R$ , the parenthesis is at least $α^{2} λ R^{2} - 2 α T - 2 α B R - Q$ . Thus $L_{Q} ψ_{s} \geq 0$ for every such $s$ .

3.1F2F4step 1.1step 2.1step 1.2givenalgebrachoose

By step 1.2 the sphere $r = R / 2$ is nonempty and compactly contained in the ball. The $C^{2}$ regularity of $u$ there gives continuity, so [F4] and strict positivity give an attained minimum $μ = \min_{r = R / 2} u > 0$ . Put $M = e^{- α R^{2} / 4} > 0$ and fix $ϵ = μ / (2 M) > 0$ , independently of $s$ . On the inner sphere $0 \leq ψ_{s} \leq M$ , hence $ϵ ψ_{s} \leq u$ ; on the outer sphere $r = s$ , $ψ_{s} = 0 < u$ . The closed annulus lies inside the open ball, where $u$ is $C^{2}$ , so $w_{s} = ϵ ψ_{s} - u$ belongs to $C^{2} (D_{s}) \cap C^{1} ({\overline{D}}_{s})$ . Step 1.2 supplies its smooth relatively compact closure and proves that every component closure meets its boundary. Moreover $L_{Q} w_{s} \geq 0$ by steps 1.1 and 2.1. Apply [F2] with constant potential $Q$ to conclude $u \geq ϵ ψ_{s}$ throughout $D_{s}$ . This application uses neither regularity on $r = R$ nor smoothness of $V$ .

4.1step 3.1givenalgebra∎

Fix $x$ with $R / 2 < ∣ x - z ∣ < R$ . The comparison in step 3.1 holds for every $s \in (∣ x - z ∣, R)$ , so taking $s ↑ R$ yields $u (x) \geq ϵ ψ_{R} (x)$ , where $ψ_{R} (x) = e^{- α ∣ x - z ∣^{2}} - e^{- α R^{2}}$ . Let $v = (z - p) / R$ . For $0 < t < R / 2$ , the point $p + t v$ has radius $R - t$ and lies in this annulus. Since $u (p) = ψ_{R} (p) = 0$ , division by $t$ gives $(u (p + t v) - u (p)) / t \geq ϵ (e^{- α (R - t)^{2}} - e^{- α R^{2}}) / t$ . The assumed $C^{1}$ regularity at $p$ gives the left-hand limit $\partial_{v} u (p)$ , and the right-hand limit is $2 ϵ α R e^{- α R^{2}} > 0$ . This proves the asserted inward derivative bound with exactly the stated regularity.

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