Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Interior sphere barrier for the laplacian

Statement

Let n≥2, R>0, a∈Rn and α≥2n/R2. Set c=(e−αR2/4−e−αR2)−1,v(x)=c(e−α∣x−a∣2−e−αR2). On the annulus R/2<∣x−a∣<R, v is subharmonic; it equals 1 on the inner sphere and 0 on the outer sphere. At every point of the outer sphere its outward sphere derivative is strictly negative.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

A real C2 function with nonnegative Laplacian is subharmonic. (Subharmonic and superharmonic functions in rn).

[F2]

The sphere direction is (p−a)/R, and its outward derivative is the limit of (v(p)−v(p−tν))/t. (Interior sphere condition and sphere normal).

Proof

technique · direct
1.1givenalgebra

The parameter α is positive, so the denominator defining c is positive. Substitution at radii R/2 and R gives the two boundary values.

2.1F1step 1.1algebra

Writing z=x−a, Cartesian differentiation gives ∂iiv=c(4α2zi2−2α)e−α∣z∣2 and hence Δv=2cα(2α∣z∣2−n)e−α∣z∣2≥0 on the closed annulus. This is subharmonicity.

3.1F2step 1.1algebra∎

At p on the outer sphere, the supplied direction is ν=(p−a)/R. The radial derivative gives ∂νv(p)=−2cαRe−αR2<0, also equal to the one-sided quotient limit.

Remarks

Hunter’s displayed Laplacian has the sign used here. The following prose on printed p.30 says negative; that prose sign is a typo.

Depends on

Used by

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources