Introduction
The maximum diversity of a compact metric space is the largest diversity of a probability measure on it, for the similarity kernel e − d ( x , y ) e^{-d(x,y)} e − d ( x , y ) [1 ] . On a nonempty compact set E ⊂ R N E \subset\mathbb{R}^{N} E ⊂ R N
D max ( E ) = ( min μ ≥ 0 , μ ( E ) = 1 supp μ ⊂ E ∬ e − ∣ x − y ∣ d μ ( x ) d μ ( y ) ) − 1 . (1) D_{\max}(E)=\left(\min_{\substack{\mu\ge0,\ \mu(E)=1\\ \operatorname{supp}\mu\subset E}}\iint e^{-|x-y|}\,d\mu(x)\,d\mu(y)\right)^{-1}. \tag*{(1)} D m a x ( E ) = μ ≥ 0 , μ ( E ) = 1 supp μ ⊂ E min ∬ e − ∣ x − y ∣ d μ ( x ) d μ ( y ) − 1 . (1) The reciprocal of the double integral is the diversity of order two of μ \mu μ q ∈ [ 0 , ∞ ] q\in[0,\infty] q ∈ [ 0 , ∞ ] q q q [1 Theorem 7.1 and Corollary 7.3] , as Leinster and Meckes first proved for finite spaces [2 ] . Its logarithm is the maximum entropy of E E E [1 Definition 7.2] .
Rotational symmetry on a ball leaves the distribution of mass over the radii to be found. Leinster and Roff ask for the maximizing measure on a ball and conjecture that its support is a finite union of concentric spheres, their number depending on the radius [1 §10, Q (2)] .1 [3 ] . Leinster reiterated the problem in 2024, as a possible application of spectral geometry and semiclassical analysis, reporting Gimperlein’s suggestion that spectral geometry could advance the study of magnitude [4 ] . In this paper we determine the maximizing measure for every radius in dimensions two and three using a potential-theoretic approach.
Theorem 1.1 Let N ∈ { 2 , 3 } N\in\{2,3\} N ∈ { 2 , 3 } μ R \mu_R μ R e − ∣ x − y ∣ e^{-|x-y|} e − ∣ x − y ∣ B R ‾ ⊂ R N \overline{B_R}\subset\mathbb{R}^{N} B R ⊂ R N σ R \sigma_R σ R ∂ B R \partial B_R ∂ B R R N R_N R N R > 0 R>0 R > 0 σ R \sigma_R σ R
e − R = ∫ e − ∣ R e 1 − y ∣ d σ R ( y ) . e^{-R}=\int e^{-|Re_1-y|}\,d\sigma_R(y). e − R = ∫ e − ∣ R e 1 − y ∣ d σ R ( y ) . Thus R 2 ≈ 1.364 R_2\approx1.364 R 2 ≈ 1.364 R 3 ≈ 2.729 R_3\approx2.729 R 3 ≈ 2.729 0 < R ≤ R N 0<R\le R_N 0 < R ≤ R N μ R = σ R \mu_R=\sigma_R μ R = σ R D max ( B R ‾ ) D_{\max}(\overline{B_R}) D m a x ( B R ) R > R N R>R_N R > R N a ∈ ( 0 , R ) a\in(0,R) a ∈ ( 0 , R )
supp μ R = B a ‾ ∪ ∂ B R . \operatorname{supp}\mu_R=\overline{B_a}\cup\partial B_R. supp μ R = B a ∪ ∂ B R . The measure has positive density throughout B a B_a B a ∂ B R \partial B_R ∂ B R ∂ B a \partial B_a ∂ B a a < ∣ x ∣ < R a<|x|<R a < ∣ x ∣ < R e − ∣ ⋅ ∣ ∗ μ R e^{-|\cdot|}*\mu_R e − ∣ ⋅ ∣ ∗ μ R 1 / D max ( B R ‾ ) 1/D_{\max}(\overline{B_R}) 1/ D m a x ( B R )
Positive density on the central ball rules out a finite union of supporting spheres for R > R N R>R_N R > R N § 3 and the pointwise positivity proved in Theorem 5.8 .
Up to the threshold the maximum diversity of the ball is therefore that of its boundary sphere, which is the magnitude of the sphere [1 Example 8.5] . Willerton computed this magnitude for the two-sphere with the Euclidean metric [5 Theorem 12] .
Above the threshold in dimension three, Theorem 4.2 expresses the measure and maximum diversity in terms of the unique root of a scalar equation for the annulus width. For large R R R R 3 / 6 R^3/6 R 3 /6 R 2 R^2 R 2 (34) is approximately 0.928. Barceló and Carbery’s magnitude formula is R 3 / 6 + R 2 + 2 R + 1 R^3/6 + R^2 + 2R + 1 R 3 /6 + R 2 + 2 R + 1 [6 Theorem 2] .
In the plane the interior density is real analytic, and Theorem 5.8 determines it through integral equations. The density diverges like the inverse square root of the distance to its edge, and the inner radius grows linearly from zero at the threshold. At the threshold, the maximum diversity departs from the magnitude of the circle ∂ B R \partial B_R ∂ B R
For N = 4 , 5 N = 4, 5 N = 4 , 5
For a probability measure μ \mu μ E E E U μ = e − ∣ ⋅ ∣ ∗ μ U_\mu= e^{-|\cdot|} * \mu U μ = e − ∣ ⋅ ∣ ∗ μ I ( μ ) = ∫ U μ d μ I(\mu) = \int U_\mu\,d\mu I ( μ ) = ∫ U μ d μ I I I [1 Proposition 8.8] . Its equilibrium level γ = I ( μ ) \gamma= I(\mu) γ = I ( μ ) 1 / D max ( E ) 1/D_{\max}(E) 1/ D m a x ( E ) U μ U_\mu U μ [1 Definition 3.4] . A maximizing measure is balanced, with U μ = γ U_\mu= \gamma U μ = γ [1 Proposition 6.3] , and U μ ≥ γ U_\mu\ge\gamma U μ ≥ γ E E E [1 Lemma 7.7(i)] . We call the latter the equilibrium inequality, and { x ∈ E : U μ ( x ) = γ } \{x \in E : U_\mu(x) = \gamma\} { x ∈ E : U μ ( x ) = γ }
The support proof is the same for both ball problems. For the equilibrium measure μ \mu μ U = U μ U = U_\mu U = U μ J = ( 1 − Δ ) U = ( N − 1 ) ( e − ∣ ⋅ ∣ / ∣ ⋅ ∣ ) ∗ μ J = (1 - \Delta)U = (N - 1)(e^{-|\cdot|}/|\cdot|) * \mu J = ( 1 − Δ ) U = ( N − 1 ) ( e − ∣ ⋅ ∣ /∣ ⋅ ∣ ) ∗ μ U U U Δ U ≥ 0 \Delta U \ge0 Δ U ≥ 0 J = U − Δ U ≤ γ J = U - \Delta U \le\gamma J = U − Δ U ≤ γ Lemma 3.6 . In dimensions two and three J J J J J J U − γ U - \gamma U − γ Proposition 3.11 (Figure 1 ).
Figure 1. Support of the maximizing measure for R ≤ R N R \le R_N R ≤ R N R > R N R > R_N R > R N R = R N R = R_N R = R N
To recover the density we invert convolution with e − ∣ ⋅ ∣ e^{-|\cdot|} e − ∣ ⋅ ∣ R 3 \mathbb{R}^3 R 3 ( 1 − Δ ) 2 U = 8 π μ (1 - \Delta)^2 U = 8\pi\mu ( 1 − Δ ) 2 U = 8 π μ R 2 \mathbb{R}^2 R 2 ( 1 − Δ ) 3 / 2 (1 - \Delta)^{3/2} ( 1 − Δ ) 3/2 § 5 obtains it through a screened charged disk problem.
Meckes’s variational formula for the magnitude Mag ( E ) \operatorname{Mag}(E) Mag ( E ) ∣ E ∣ |E| ∣ E ∣ D max ( E ) ≤ Mag ( E ) D_{\max}(E) \leq\operatorname{Mag}(E) D m a x ( E ) ≤ Mag ( E ) [7 ] , [1 (14)] . Meckes also expresses maximum diversity as a constant multiple of the Bessel capacity of order ( N + 1 ) / 2 (N+1)/2 ( N + 1 ) /2 [8 ] . In that form
D max ( E ) = 1 c N inf { ∥ ( 1 − Δ ) ( N + 1 ) / 4 f ∥ L 2 2 : f ∈ H ( N + 1 ) / 2 ( R N ) , f ≥ 1 on E } . D_{\max}(E)=\frac{1}{c_N}\inf\left\{\left\|(1-\Delta)^{(N+1)/4}f\right\|_{L^2}^2:f\in H^{(N+1)/2}(\mathbb{R}^N),\ f\geq1\text{ on }E\right\}. D m a x ( E ) = c N 1 inf { ( 1 − Δ ) ( N + 1 ) /4 f L 2 2 : f ∈ H ( N + 1 ) /2 ( R N ) , f ≥ 1 on E } . with c N c_N c N (5) , N ! N! N ! [6 §10] . Its minimizer is U μ / γ U_\mu/\gamma U μ / γ
Apart from finite spaces, the maximizing measures Leinster and Roff compute come from positive weight measures [1 Lemma 8.2] . Three-dimensional Euclidean balls have no weight measure [9 ] , so their maximizing measure misses part of the ball [1 §8] . For R > R N R>R_N R > R N μ R / γ \mu_R/\gamma μ R / γ B a ‾ ∪ ∂ B R \overline{B_a}\cup\partial B_R B a ∪ ∂ B R [1 Lemma 5.6] . By their Corollary 7.4(ii), D max ( B R ‾ ) D_{\max}(\overline{B_R}) D m a x ( B R ) [1 §8] . Leinster and Roff’s uniform measure on a ball is normalized Lebesgue measure [1 Theorem 9.9] . In three dimensions the density is constant on the central ball, whose radius is R − O ( 1 ) R-O(1) R − O ( 1 ) O ( 1 / R ) O(1/R) O ( 1/ R ) O ( 1 / R ) O(1/R) O ( 1/ R )
Spherical to volume transitions also occur for Riesz energies with radial external fields [10 , 11 ] . The volume part in [11 ] reaches the sphere; the exponential equilibrium on a ball has an intervening mass-free annulus between these components.
Kernel identities
For N ≥ 2 N\geq2 N ≥ 2 K ( x ) = e − ∣ x ∣ K(x)=e^{-|x|} K ( x ) = e − ∣ x ∣ L ( x ) = e − ∣ x ∣ / ∣ x ∣ L(x)=e^{-|x|}/|x| L ( x ) = e − ∣ x ∣ /∣ x ∣ x ≠ 0 x\ne0 x = 0 L L L + ∞ +\infty + ∞ h ^ ( ξ ) = ∫ e − i x ⋅ ξ h ( x ) d x \widehat{h}(\xi)=\int e^{-ix\cdot\xi}h(x)\,dx h ( ξ ) = ∫ e − i x ⋅ ξ h ( x ) d x
( 1 − Δ ) K = ( N − 1 ) L (2) (1-\Delta)K=(N-1)L \tag*{(2)} ( 1 − Δ ) K = ( N − 1 ) L (2) and, away from the origin,
Δ L ( x ) = e − ∣ x ∣ ∣ x ∣ 3 ( ∣ x ∣ 2 + ( 3 − N ) ∣ x ∣ + 3 − N ) . (3) \Delta L(x)=\frac{e^{-|x|}}{|x|^3}\left(|x|^2+(3-N)|x|+3-N\right). \tag*{(3)} Δ L ( x ) = ∣ x ∣ 3 e − ∣ x ∣ ( ∣ x ∣ 2 + ( 3 − N ) ∣ x ∣ + 3 − N ) . (3) The first identity holds as distributions on R N \mathbb{R}^N R N K K K ε \varepsilon ε N = 2 , 3 N=2,3 N = 2 , 3
For α , ζ > 0 \alpha,\zeta>0 α , ζ > 0
∫ 0 ∞ t − 1 / 2 e − α t − ζ / t d t = π α e − 2 α ζ , ∫ 0 ∞ t − 3 / 2 e − α t − ζ / t d t = π ζ e − 2 α ζ . (4) \int_{0}^{\infty} t^{-1/2}e^{-\alpha t-\zeta/t}\,\mathrm{d}t=\sqrt{\frac{\pi}{\alpha}}e^{-2\sqrt{\alpha\zeta}},\qquad\int_{0}^{\infty} t^{-3/2}e^{-\alpha t-\zeta/t}\,\mathrm{d}t=\sqrt{\frac{\pi}{\zeta}}e^{-2\sqrt{\alpha\zeta}}. \tag*{(4)} ∫ 0 ∞ t − 1/2 e − α t − ζ / t d t = α π e − 2 α ζ , ∫ 0 ∞ t − 3/2 e − α t − ζ / t d t = ζ π e − 2 α ζ . (4) and by monotone convergence the first holds at ζ = 0 \zeta=0 ζ = 0 α = 0 \alpha=0 α = 0 t ↦ ζ / ( α t ) t\mapsto\zeta/(\alpha t) t ↦ ζ / ( α t ) ζ / α \sqrt{\zeta/\alpha} ζ / α u = α t − ζ / t u=\sqrt{\alpha}t-\sqrt{\zeta}/t u = α t − ζ / t α \sqrt{\alpha} α ζ \sqrt{\zeta} ζ 2 e − 2 α ζ ∫ R e − u 2 d u 2e^{-2\sqrt{\alpha\zeta}}\int_{\mathbb{R}}e^{-u^2}\,\mathrm{d}u 2 e − 2 α ζ ∫ R e − u 2 d u α = ∣ x ∣ 2 \alpha=|x|^2 α = ∣ x ∣ 2 ζ = 1 / 4 \zeta=1/4 ζ = 1/4 K K K e − t ∣ x ∣ 2 e^{-t|x|^2} e − t ∣ x ∣ 2
K ^ ( ξ ) = c N ( 1 + ∣ ξ ∣ 2 ) ( N + 1 ) / 2 , c N = 2 N π ( N − 1 ) / 2 Γ ( N + 1 2 ) . (5) \widehat{K}(\xi)=\frac{c_N}{(1+|\xi|^2)^{(N+1)/2}},\qquad c_N=2^N\pi^{(N-1)/2}\Gamma\left(\frac{N+1}{2}\right). \tag*{(5)} K ( ξ ) = ( 1 + ∣ ξ ∣ 2 ) ( N + 1 ) /2 c N , c N = 2 N π ( N − 1 ) /2 Γ ( 2 N + 1 ) . (5) By Legendre’s duplication formula c N c_N c N N ! N! N ! [1 Proposition 9.7] ; in particular c 3 = 8 π c_3=8\pi c 3 = 8 π η \eta η
I ( η ) = ∬ K ( x − y ) d η ( x ) d η ( y ) = 1 ( 2 π ) N ∫ R N c N ∣ η ^ ( ξ ) ∣ 2 ( 1 + ∣ ξ ∣ 2 ) ( N + 1 ) / 2 d ξ . I(\eta)=\iint K(x-y)\,\mathrm{d}\eta(x)\,\mathrm{d}\eta(y)=\frac{1}{(2\pi)^N}\int_{\mathbb{R}^N}\frac{c_N|\widehat{\eta}(\xi)|^2}{(1+|\xi|^2)^{(N+1)/2}}\,\mathrm{d}\xi. I ( η ) = ∬ K ( x − y ) d η ( x ) d η ( y ) = ( 2 π ) N 1 ∫ R N ( 1 + ∣ ξ ∣ 2 ) ( N + 1 ) /2 c N ∣ η ( ξ ) ∣ 2 d ξ . Zero energy means that the continuous function η ^ \widehat{\eta} η η = 0 \eta=0 η = 0 μ , ν \mu,\nu μ , ν U μ = K ∗ μ U_\mu=K*\mu U μ = K ∗ μ
I ( ν ) − I ( μ ) = I ( ν − μ ) + 2 ∫ U μ d ( ν − μ ) . (6) I(\nu)-I(\mu)=I(\nu-\mu)+2\int U_\mu\,\mathrm{d}(\nu-\mu). \tag*{(6)} I ( ν ) − I ( μ ) = I ( ν − μ ) + 2 ∫ U μ d ( ν − μ ) . (6) Hence U μ ≥ c U_\mu\ge c U μ ≥ c E E E U μ = c U_\mu=c U μ = c E E E c = I ( μ ) c=I(\mu) c = I ( μ )
For N = 3 N=3 N = 3 (5) reads K ^ ( ξ ) = 8 π ( 1 + ∣ ξ ∣ 2 ) − 2 \widehat{K}(\xi)=8\pi(1+|\xi|^2)^{-2} K ( ξ ) = 8 π ( 1 + ∣ ξ ∣ 2 ) − 2
( 1 − Δ ) 2 K = 8 π δ 0 in R 3 . (7) (1-\Delta)^2K=8\pi\delta_0\quad\text{in }\mathbb{R}^3. \tag*{(7)} ( 1 − Δ ) 2 K = 8 π δ 0 in R 3 . (7) In the plane the first formula in (G), with α = ∣ x ∣ 2 \alpha=|x|^2 α = ∣ x ∣ 2 ζ = 1 / 4 \zeta=1/4 ζ = 1/4 π L \sqrt{\pi}L π L α = 0 \alpha=0 α = 0 ζ = ( 1 + ∣ ξ ∣ 2 ) / 4 \zeta=(1+|\xi|^2)/4 ζ = ( 1 + ∣ ξ ∣ 2 ) /4
L ^ ( ξ ) = 2 π 1 + ∣ ξ ∣ 2 , K ^ ( ξ ) = 2 π ( 1 + ∣ ξ ∣ 2 ) 3 / 2 . \widehat{L}(\xi)=\frac{2\pi}{\sqrt{1+|\xi|^2}},\qquad\widehat{K}(\xi)=\frac{2\pi}{(1+|\xi|^2)^{3/2}}. L ( ξ ) = 1 + ∣ ξ ∣ 2 2 π , K ( ξ ) = ( 1 + ∣ ξ ∣ 2 ) 3/2 2 π . For N = 2 N = 2 N = 2 (2) and (3) read
( 1 − Δ ) K = L , Δ L ( x ) = e − ∣ x ∣ ( 1 ∣ x ∣ + 1 ∣ x ∣ 2 + 1 ∣ x ∣ 3 ) > 0 ( x ≠ 0 ) . (8) (1-\Delta)K=L,\qquad\Delta L(x)=e^{-|x|}\left(\frac{1}{|x|}+\frac{1}{|x|^2}+\frac{1}{|x|^3}\right)>0\quad(x\ne0). \tag*{(8)} ( 1 − Δ ) K = L , Δ L ( x ) = e − ∣ x ∣ ( ∣ x ∣ 1 + ∣ x ∣ 2 1 + ∣ x ∣ 3 1 ) > 0 ( x = 0 ) . (8) Support on balls
Let μ \mu μ B R ‾ ⊂ R N \overline{B_R}\subset\mathbb{R}^N B R ⊂ R N R > 0 R>0 R > 0 N ≥ 2 N\ge2 N ≥ 2
U = K ∗ μ , γ = I ( μ ) , J = ( N − 1 ) L ∗ μ . U=K*\mu,\qquad\gamma=I(\mu),\qquad J=(N-1)L*\mu. U = K ∗ μ , γ = I ( μ ) , J = ( N − 1 ) L ∗ μ . By (2) , ( 1 − Δ ) U = J (1-\Delta)U=J ( 1 − Δ ) U = J J J J L 1 ( R N ) L^1(\mathbb{R}^N) L 1 ( R N )
U ≥ γ on B R ‾ , U = γ on supp μ . (9) U\ge\gamma\quad\text{on }\overline{B_R},\qquad U=\gamma\quad\text{on }\operatorname{supp}\mu. \tag*{(9)} U ≥ γ on B R , U = γ on supp μ . (9) By uniqueness μ \mu μ x ↦ ∣ x ∣ x\mapsto|x| x ↦ ∣ x ∣ μ r a d \mu_{\mathrm{rad}} μ rad S = supp μ r a d S=\operatorname{supp}\mu_{\mathrm{rad}} S = supp μ rad t > 0 t>0 t > 0 σ t \sigma_t σ t ∂ B t \partial B_t ∂ B t
μ = μ r a d ( { 0 } ) δ 0 + ∫ ( 0 , R ] σ t d μ r a d ( t ) , U ( r ) = γ ( r ∈ S ) . (10) \mu=\mu_{\mathrm{rad}}(\{0\})\delta_0+\int_{(0,R]}\sigma_t\,d\mu_{\mathrm{rad}}(t),\qquad U(r)=\gamma\quad(r\in S). \tag*{(10)} μ = μ rad ({ 0 }) δ 0 + ∫ ( 0 , R ] σ t d μ rad ( t ) , U ( r ) = γ ( r ∈ S ) . (10) Boundary measure in dimension three
Let σ R \sigma_R σ R ∂ B R ⊂ R 3 \partial B_R\subset\mathbb{R}^3 ∂ B R ⊂ R 3 U ∂ = K ∗ σ R U_{\partial}=K*\sigma_R U ∂ = K ∗ σ R
Υ ( R ) = 2 R 2 e − R + ( 1 + 2 R ) e − 2 R − 1. (11) \Upsilon(R)=2R^2e^{-R}+(1+2R)e^{-2R}-1. \tag*{(11)} Υ ( R ) = 2 R 2 e − R + ( 1 + 2 R ) e − 2 R − 1. (11) Lemma 3.1. For R > 0 R>0 R > 0
U ∂ ( r ) = e − R R ( ( 1 + R ) sinh r r − cosh r ) ( 0 < r ≤ R ) . (12) U_{\partial}(r)=\frac{e^{-R}}{R}\left((1+R)\frac{\sinh r}{r}-\cosh r\right)\quad(0<r\le R). \tag*{(12)} U ∂ ( r ) = R e − R ( ( 1 + R ) r sinh r − cosh r ) ( 0 < r ≤ R ) . (12) Its center value is e − R e^{-R} e − R
U ∂ ( R ) = 1 − ( 1 + 2 R ) e − 2 R 2 R 2 . (13) U_{\partial}(R)=\frac{1-(1+2R)e^{-2R}}{2R^2}. \tag*{(13)} U ∂ ( R ) = 2 R 2 1 − ( 1 + 2 R ) e − 2 R . (13) If R ≤ 2 R\le2 R ≤ 2 U ∂ U_{\partial} U ∂ [ 0 , R ] [0,R] [ 0 , R ] R > 2 R>2 R > 2 r R ∈ ( 0 , R ) r_R\in(0,R) r R ∈ ( 0 , R )
U ∂ ( r ) > min { U ∂ ( 0 ) , U ∂ ( R ) } ( 0 < r < R ) . (14) U_{\partial}(r)>\min\{U_{\partial}(0),U_{\partial}(R)\}\quad(0<r<R). \tag*{(14)} U ∂ ( r ) > min { U ∂ ( 0 ) , U ∂ ( R )} ( 0 < r < R ) . (14) Proof. For 0 < r < R 0<r<R 0 < r < R r e 1 re_1 r e 1 σ R \sigma_R σ R [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] u u u
U ∂ ( r ) = 1 2 r R ∫ R − r R + r u e − u d u , U_{\partial}(r)=\frac{1}{2rR}\int_{R-r}^{R+r}ue^{-u}\,du, U ∂ ( r ) = 2 r R 1 ∫ R − r R + r u e − u d u , which is (12) . The center value follows by continuity, and (13) is the case r = R r=R r = R
For the extrema introduce ℓ ( r ) = ( r coth r − 1 ) / r 2 \ell(r) = (r\coth r - 1)/r^2 ℓ ( r ) = ( r coth r − 1 ) / r 2 r > 0 r > 0 r > 0
U ∂ ′ ( r ) = e − R R sinh r ( ( 1 + R ) ℓ ( r ) − 1 ) , ℓ ( 0 + ) = 1 3 , U_{\partial}'(r) = \frac{e^{-R}}{R}\sinh r\bigl((1+R)\ell(r)-1\bigr), \qquad\ell(0+) = \frac{1}{3}, U ∂ ′ ( r ) = R e − R sinh r ( ( 1 + R ) ℓ ( r ) − 1 ) , ℓ ( 0 + ) = 3 1 , ℓ ′ ( r ) = − 1 4 r 3 sinh 2 r ∑ n ≥ 3 ( 2 n − 4 ) ( 2 r ) 2 n ( 2 n ) ! < 0. \ell'(r) = -\frac{1}{4r^3\sinh^2 r}\sum_{n\geq3}\frac{(2n-4)(2r)^{2n}}{(2n)!} < 0. ℓ ′ ( r ) = − 4 r 3 sinh 2 r 1 n ≥ 3 ∑ ( 2 n )! ( 2 n − 4 ) ( 2 r ) 2 n < 0. Thus ( 1 + R ) ℓ ( r ) − 1 (1+R)\ell(r)-1 ( 1 + R ) ℓ ( r ) − 1 ( R − 2 ) / 3 (R-2)/3 ( R − 2 ) /3
R 2 ( ( 1 + R ) ℓ ( R ) − 1 ) = 2 R ( 1 + R ) e 2 R − 1 − 1 < 0 , R^2\bigl((1+R)\ell(R)-1\bigr)=\frac{2R(1+R)}{e^{2R}-1}-1<0, R 2 ( ( 1 + R ) ℓ ( R ) − 1 ) = e 2 R − 1 2 R ( 1 + R ) − 1 < 0 , since e 2 R − 1 > 2 R + 2 R 2 e^{2R}-1>2R+2R^2 e 2 R − 1 > 2 R + 2 R 2 R ≤ 2 R\leq2 R ≤ 2 U ∂ ′ < 0 U_{\partial}'<0 U ∂ ′ < 0 ( 0 , R ] (0,R] ( 0 , R ] R > 2 R>2 R > 2 r R ∈ ( 0 , R ) r_R\in(0,R) r R ∈ ( 0 , R ) U ∂ ( r ) > U ∂ ( 0 ) U_{\partial}(r)>U_{\partial}(0) U ∂ ( r ) > U ∂ ( 0 ) r < R r<R r < R U ∂ ( r ) > U ∂ ( R ) U_{\partial}(r)>U_{\partial}(R) U ∂ ( r ) > U ∂ ( R ) (14) follows. □
Proposition 3.2 The function Υ \Upsilon Υ R 3 R_3 R 3 ( 0 , R 3 ) (0,R_3) ( 0 , R 3 ) ( R 3 , ∞ ) (R_3,\infty) ( R 3 , ∞ ) σ R \sigma_R σ R B R ‾ ⊂ R 3 \overline{B_R}\subset\mathbb{R}^3 B R ⊂ R 3 R ≤ R 3 R\leq R_3 R ≤ R 3 U ∂ ( R ) U_{\partial}(R) U ∂ ( R ) R < R 3 R<R_3 R < R 3 ∂ B R \partial B_R ∂ B R R = R 3 R=R_3 R = R 3 ∂ B R ∪ { 0 } \partial B_R\cup\{0\} ∂ B R ∪ { 0 } R > R 3 R>R_3 R > R 3 B R B_R B R
Proof. Differentiating,
Υ ′ ( R ) = 2 R e − R ( 2 − R − 2 e − R ) . \Upsilon'(R)=2Re^{-R}\bigl(2-R-2e^{-R}\bigr). Υ ′ ( R ) = 2 R e − R ( 2 − R − 2 e − R ) . The last factor starts at zero with positive derivative and is strictly concave, tending to − ∞ -\infty − ∞ Υ \Upsilon Υ − 1 -1 − 1 R 3 R_3 R 3 Lemma 3.1 ,
U ∂ ( 0 ) − U ∂ ( R ) = Υ ( R ) 2 R 2 . (15) U_{\partial}(0)-U_{\partial}(R)=\frac{\Upsilon(R)}{2R^2}. \tag*{(15)} U ∂ ( 0 ) − U ∂ ( R ) = 2 R 2 Υ ( R ) . (15) If R ≤ R 3 R\leq R_3 R ≤ R 3 (14) imply U ∂ ≥ U ∂ ( R ) U_{\partial}\geq U_{\partial}(R) U ∂ ≥ U ∂ ( R ) B R ∖ { 0 } B_R\setminus\{0\} B R ∖ { 0 } R = R 3 R=R_3 R = R 3 § 2 .
For R > R 3 R>R_3 R > R 3 s > 0 s>0 s > 0
I ( ( 1 − s ) σ R + s δ 0 ) − I ( σ R ) = s 2 I ( δ 0 − σ R ) + 2 s ( U ∂ ( 0 ) − U ∂ ( R ) ) < 0 I\bigl((1-s)\sigma_R+s\delta_0\bigr)-I(\sigma_R)=s^2I(\delta_0-\sigma_R)+2s\bigl(U_{\partial}(0)-U_{\partial}(R)\bigr)<0 I ( ( 1 − s ) σ R + s δ 0 ) − I ( σ R ) = s 2 I ( δ 0 − σ R ) + 2 s ( U ∂ ( 0 ) − U ∂ ( R ) ) < 0 for sufficiently small s > 0 s>0 s > 0 σ R \sigma_R σ R B R B_R B R σ R \sigma_R σ R
Boundary measure in dimension two
Here N = 2 N=2 N = 2 U ∂ = K ∗ σ R U_{\partial}=K*\sigma_R U ∂ = K ∗ σ R j R = L ∗ σ R j_R=L*\sigma_R j R = L ∗ σ R
Proposition 3.3 For every R > 0 R>0 R > 0 U ∂ ( r ) ≥ min { U ∂ ( 0 ) , U ∂ ( R ) } U_{\partial}(r)\geq\min\{U_{\partial}(0),U_{\partial}(R)\} U ∂ ( r ) ≥ min { U ∂ ( 0 ) , U ∂ ( R )} [ 0 , R ] [0,R] [ 0 , R ]
Proof. The potentials are smooth in B R B_R B R (8) , ( 1 − Δ ) U ∂ = j R (1-\Delta)U_{\partial}=j_R ( 1 − Δ ) U ∂ = j R ( r j R ′ ( r ) ) ′ = r Δ j R ( r ) > 0 (rj_R'(r))'=r\Delta j_R(r)>0 ( r j R ′ ( r ) ) ′ = r Δ j R ( r ) > 0 0 < r < R 0<r<R 0 < r < R j R ′ ( 0 ) = 0 j_R'(0)=0 j R ′ ( 0 ) = 0 j R ′ ( r ) > 0 j_R'(r)>0 j R ′ ( r ) > 0 w = U ∂ ′ w=U_{\partial}' w = U ∂ ′
w ′ ′ + 1 r w ′ − ( 1 + 1 r 2 ) w = − j R ′ < 0 , 0 < r < R , w''+\frac{1}{r}w'-\left(1+\frac{1}{r^2}\right)w=-j_R'<0,\qquad0<r<R, w ′′ + r 1 w ′ − ( 1 + r 2 1 ) w = − j R ′ < 0 , 0 < r < R , with w ( 0 ) = 0 w(0)=0 w ( 0 ) = 0 w w w { w < 0 } \{w<0\} { w < 0 } R R R [ 0 , R ) [0,R) [ 0 , R ) w w w r 0 ∈ ( 0 , R ) r_0\in(0,R) r 0 ∈ ( 0 , R ) w w w U ∂ U_{\partial} U ∂ R R R [ 0 , R ] [0,R] [ 0 , R ]
Proposition 3.4. There is a unique positive solution R 2 R_2 R 2 e − R = 2 π ∫ 0 π / 2 e − 2 R sin t d t e^{-R}=\frac{2}{\pi}\int_0^{\pi/2}e^{-2R\sin t}\,dt e − R = π 2 ∫ 0 π /2 e − 2 R s i n t d t σ R \sigma_R σ R e − ∣ x − y ∣ e^{-|x-y|} e − ∣ x − y ∣ B R ‾ ⊂ R 2 \overline{B_R}\subset\mathbb{R}^2 B R ⊂ R 2 0 < R ≤ R 2 0<R\le R_2 0 < R ≤ R 2
Proof. The potential is constant on the circle, with value U ∂ ( R ) U_{\partial}(R) U ∂ ( R )
U ∂ ( 0 ) = e − R , U ∂ ( R ) = 2 π ∫ 0 π / 2 e − 2 R sin t d t = I 0 ( 2 R ) − L 0 ( 2 R ) . U_{\partial}(0)=e^{-R},\qquad U_{\partial}(R)=\frac{2}{\pi}\int_0^{\pi/2}e^{-2R\sin t}\,dt=I_0(2R)-L_0(2R). U ∂ ( 0 ) = e − R , U ∂ ( R ) = π 2 ∫ 0 π /2 e − 2 R s i n t d t = I 0 ( 2 R ) − L 0 ( 2 R ) . Here I 0 I_0 I 0 L 0 L_0 L 0 [12 (11.2.6), (11.5.5)] . By Proposition 3.3 the equilibrium inequality is equivalent to U ∂ ( 0 ) ≥ U ∂ ( R ) U_{\partial}(0)\ge U_{\partial}(R) U ∂ ( 0 ) ≥ U ∂ ( R )
χ ( R ) = e R U ∂ ( R ) = 2 π ∫ 0 π / 2 e R ( 1 − 2 sin t ) d t . \chi(R)=e^RU_{\partial}(R)=\frac{2}{\pi}\int_0^{\pi/2}e^{R(1-2\sin t)}\,dt. χ ( R ) = e R U ∂ ( R ) = π 2 ∫ 0 π /2 e R ( 1 − 2 s i n t ) d t . Then χ ( 0 ) = 1 \chi(0)=1 χ ( 0 ) = 1 χ ′ ( 0 ) = 1 − 4 / π < 0 \chi'(0)=1-4/\pi<0 χ ′ ( 0 ) = 1 − 4/ π < 0 χ ′ ′ > 0 \chi''>0 χ ′′ > 0 [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) 0 ≤ t ≤ π / 12 0\le t\le\pi/12 0 ≤ t ≤ π /12 χ ( R ) → ∞ \chi(R)\to\infty χ ( R ) → ∞ χ = 1 \chi=1 χ = 1 χ ( R ) ≤ 1 \chi(R)\le1 χ ( R ) ≤ 1 R ∈ [ 0 , R 2 ] R\in[0,R_2] R ∈ [ 0 , R 2 ] § 2 finishes the proof. □
Interior contact estimates
Remark 3.5. The conjecture suggested examining what a finite collection of supporting spheres would leave empty. Once a central atom is excluded such a configuration leaves a mass-free ball around the origin. We sought a maximum principle argument to test whether this is compatible with equilibrium. The auxiliary potential J = ( 1 − Δ ) U J=(1-\Delta)U J = ( 1 − Δ ) U J ≤ γ J\le\gamma J ≤ γ Lemma 3.6 proves through second differences, not needing to assume the measure has a density.
Lemma 3.6. For every N ≥ 2 N\ge2 N ≥ 2 x ∈ B R x\in B_R x ∈ B R U ( x ) = γ U(x)=\gamma U ( x ) = γ J ( x ) ≤ γ J(x)\le\gamma J ( x ) ≤ γ μ \mu μ N = 2 N=2 N = 2 μ r a d ( { t } ) = 0 \mu_{\mathrm{rad}}(\{t\})=0 μ rad ({ t }) = 0
Proof. For z ≠ 0 z \ne0 z = 0 D 2 K ( z ) D^2K(z) D 2 K ( z ) e − ∣ z ∣ e^{-|z|} e − ∣ z ∣ − e − ∣ z ∣ / ∣ z ∣ -e^{-|z|}/|z| − e − ∣ z ∣ /∣ z ∣
δ h , e 2 K ( z ) : = K ( z + h e ) + K ( z − h e ) − 2 K ( z ) ≤ h 2 (16) \delta^2_{h,e}K(z) := K(z+he)+K(z-he)-2K(z) \le h^2 \tag*{(16)} δ h , e 2 K ( z ) := K ( z + h e ) + K ( z − h e ) − 2 K ( z ) ≤ h 2 (16) for every unit vector e e e e 1 , … , e N e_1,\ldots,e_N e 1 , … , e N
A h ( z ) = N − h − 2 ∑ i = 1 N δ h , e i 2 K ( z ) ≥ 0. A_h(z)=N-h^{-2}\sum_{i=1}^{N}\delta^2_{h,e_i}K(z)\ge0. A h ( z ) = N − h − 2 i = 1 ∑ N δ h , e i 2 K ( z ) ≥ 0. At an interior contact point δ h , e i 2 U ( x ) ≥ 0 \delta^2_{h,e_i}U(x)\ge0 δ h , e i 2 U ( x ) ≥ 0 h h h ∫ A h ( x − y ) d μ ( y ) ≤ N \int A_h(x-y)\,d\mu(y)\le N ∫ A h ( x − y ) d μ ( y ) ≤ N z ≠ 0 z\ne0 z = 0
A h ( z ) ⟶ N − K ( z ) + ( N − 1 ) L ( z ) , A_h(z)\longrightarrow N-K(z)+(N-1)L(z), A h ( z ) ⟶ N − K ( z ) + ( N − 1 ) L ( z ) , and at z = 0 z=0 z = 0 + ∞ +\infty + ∞ N − U ( x ) + J ( x ) ≤ N N-U(x)+J(x)\le N N − U ( x ) + J ( x ) ≤ N
An atom at zero would put zero in the contact set and make J ( 0 ) J(0) J ( 0 ) μ \mu μ J J J 1 / ( t ∣ θ ∣ ) 1/(t|\theta|) 1/ ( t ∣ θ ∣ )
Boundary mass and mass-free regions
Lemma 3.7 For every N ≥ 2 N\ge2 N ≥ 2 U ∈ C 1 ( R N ) U\in C^1(\mathbb{R}^N) U ∈ C 1 ( R N ) R ∈ S R\in S R ∈ S U ′ ( R ) < 0 U'(R)<0 U ′ ( R ) < 0 ε > 0 \varepsilon>0 ε > 0 U ( r ) > γ U(r)>\gamma U ( r ) > γ R − ε < r < R R-\varepsilon<r<R R − ε < r < R S ∩ ( R − ε , R ) = ∅ S\cap(R-\varepsilon,R)=\varnothing S ∩ ( R − ε , R ) = ∅ q : = μ r a d ( { R } ) > 0 q:=\mu_{\mathrm{rad}}(\{R\})>0 q := μ rad ({ R }) > 0
Proof. Since K K K μ \mu μ
∇ U ( x ) = − ∫ e − ∣ x − y ∣ x − y ∣ x − y ∣ d μ ( y ) (17) \nabla U(x)=-\int e^{-|x-y|}\frac{x-y}{|x-y|}\,d\mu(y) \tag*{(17)} ∇ U ( x ) = − ∫ e − ∣ x − y ∣ ∣ x − y ∣ x − y d μ ( y ) (17) and dominated convergence once more makes it continuous. If t 0 = max S < R t_0=\max S<R t 0 = max S < R ∣ y ∣ ≤ t 0 |y|\le t_0 ∣ y ∣ ≤ t 0 U ′ ( r ) < 0 U'(r)<0 U ′ ( r ) < 0 r > t 0 r>t_0 r > t 0 U ( R ) < U ( t 0 ) = γ U(R)<U(t_0)=\gamma U ( R ) < U ( t 0 ) = γ (9) . So R ∈ S R\in S R ∈ S U ( R ) = γ U(R)=\gamma U ( R ) = γ r = R r=R r = R
U ′ ( R ) = − ∫ e − ∣ R e 1 − y ∣ R − y 1 ∣ R e 1 − y ∣ d μ ( y ) < 0. U'(R)=-\int e^{-|Re_1-y|}\frac{R-y_1}{|Re_1-y|}\,d\mu(y)<0. U ′ ( R ) = − ∫ e − ∣ R e 1 − y ∣ ∣ R e 1 − y ∣ R − y 1 d μ ( y ) < 0. The numerator is positive except at the single point R e 1 Re_1 R e 1 U ′ U' U ′ U ( r ) > γ U(r)>\gamma U ( r ) > γ R R R (9) excludes those radii from S S S R R R μ r a d \mu_{\mathrm{rad}} μ rad
For N = 2 , 3 N=2,3 N = 2 , 3 t > 0 t>0 t > 0
j ( r , t ) = ∫ L ( r e 1 − y ) d σ t ( y ) , j ( r , 0 ) = L ( r e 1 ) . j(r,t)=\int L(re_1-y)\,d\sigma_t(y),\qquad j(r,0)=L(re_1). j ( r , t ) = ∫ L ( r e 1 − y ) d σ t ( y ) , j ( r , 0 ) = L ( r e 1 ) . For 0 < r < t 0<r<t 0 < r < t ( r N − 1 ∂ r j ( r , t ) ) ′ = r N − 1 Δ j ( r , t ) > 0 (r^{N-1}\partial_rj(r,t))'=r^{N-1}\Delta j(r,t)>0 ( r N − 1 ∂ r j ( r , t ) ) ′ = r N − 1 Δ j ( r , t ) > 0 ∂ r j ( 0 , t ) = 0 \partial_rj(0,t)=0 ∂ r j ( 0 , t ) = 0 j ( r , t ) j(r,t) j ( r , t ) r < t r<t r < t r > t r>t r > t r r r j ( r , t ) j(r,t) j ( r , t ) j ( t , t ) = + ∞ j(t,t)=+\infty j ( t , t ) = + ∞
j ( r , t ) = 1 2 r t ∫ ∣ r − t ∣ r + t e − u d u = e − ∣ r − t ∣ − e − ( r + t ) 2 r t ( r , t > 0 ) . (18) j(r,t)=\frac{1}{2rt}\int_{\lvert r-t\rvert}^{r+t}e^{-u}\,\mathrm{d}u=\frac{e^{-\lvert r-t\rvert}-e^{-(r+t)}}{2rt}\qquad(r,t>0). \tag*{(18)} j ( r , t ) = 2 r t 1 ∫ ∣ r − t ∣ r + t e − u d u = 2 r t e − ∣ r − t ∣ − e − ( r + t ) ( r , t > 0 ) . (18) In particular it is continuous across r = t r=t r = t j ( t , t ) = ( 1 − e − 2 t ) / ( 2 t 2 ) j(t,t)=(1-e^{-2t})/(2t^2) j ( t , t ) = ( 1 − e − 2 t ) / ( 2 t 2 ) j ( r , t ) ≤ 1 / max { r , t } j(r,t)\le1/\max\{r,t\} j ( r , t ) ≤ 1/ max { r , t } J ( r ) = 2 ∫ j ( r , t ) d μ r a d ( t ) J(r)=2\int j(r,t)\,\mathrm{d}\mu_{\mathrm{rad}}(t) J ( r ) = 2 ∫ j ( r , t ) d μ rad ( t ) N = 2 N=2 N = 2 r = t r=t r = t
Lemma 3.8 Let N = 2 N=2 N = 2 N = 3 N=3 N = 3 0 < r 1 < r 2 < R 0<r_1<r_2<R 0 < r 1 < r 2 < R μ ( { r 1 < ∣ x ∣ < r 2 } ) = 0 \mu(\{r_1<|x|<r_2\})=0 μ ({ r 1 < ∣ x ∣ < r 2 }) = 0 J J J J ( r 1 ) , J ( r 2 ) ≤ γ J(r_1),J(r_2)\le\gamma J ( r 1 ) , J ( r 2 ) ≤ γ 0 < r 1 < R 0<r_1<R 0 < r 1 < R μ ( B r 1 ) = 0 \mu(B_{r_1})=0 μ ( B r 1 ) = 0 B r 1 B_{r_1} B r 1 J ( r 1 ) ≤ γ J(r_1)\le\gamma J ( r 1 ) ≤ γ
Proof. Either open region is locally separated from the source support, so differentiation under the integral and (3) imply smoothness and strict subharmonicity and Lemma 3.6 bounds the endpoint values.
In the annulus split the radial integral at the source sets t ≤ r 1 t\le r_1 t ≤ r 1 t ≥ r 2 t\ge r_2 t ≥ r 2 r ↓ r 1 r\downarrow r_1 r ↓ r 1 j ( r , t ) ≤ j ( r 1 , t ) j(r,t)\le j(r_1,t) j ( r , t ) ≤ j ( r 1 , t ) t ≤ r 1 t\le r_1 t ≤ r 1 J ( r 1 ) < ∞ J(r_1)<\infty J ( r 1 ) < ∞ t t t j ( r 1 , t ) j(r_1,t) j ( r 1 , t ) t = r 1 t=r_1 t = r 1 (18) when N = 3 N=3 N = 3 N = 2 N=2 N = 2 t ≥ r 2 t\ge r_2 t ≥ r 2 r r r r 1 r_1 r 1 J ( r ) → J ( r 1 ) J(r)\to J(r_1) J ( r ) → J ( r 1 )
As r ↑ r 2 r\uparrow r_2 r ↑ r 2 j ( r , t ) ≤ j ( r 2 , t ) j(r,t)\le j(r_2,t) j ( r , t ) ≤ j ( r 2 , t ) t ≥ r 2 t\ge r_2 t ≥ r 2 t = r 2 t=r_2 t = r 2 J ( r ) → J ( r 2 ) J(r)\to J(r_2) J ( r ) → J ( r 2 ) t ≥ r 1 t\ge r_1 t ≥ r 1 j ( r , t ) ≤ j ( r 1 , t ) j(r,t)\le j(r_1,t) j ( r , t ) ≤ j ( r 1 , t ) r < r 1 r<r_1 r < r 1 J ( r ) → J ( r 1 ) J(r)\to J(r_1) J ( r ) → J ( r 1 ) r ↑ r 1 r\uparrow r_1 r ↑ r 1 J J J
Analyticity of densities
Constancy of U U U J J J J J J
Take N = 2 N=2 N = 2 B = ( 1 − Δ ) 1 / 2 \mathcal{B}=(1-\Delta)^{1/2} B = ( 1 − Δ ) 1/2 ( 1 + ∣ ξ ∣ 2 ) 1 / 2 (1+|\xi|^2)^{1/2} ( 1 + ∣ ξ ∣ 2 ) 1/2 L L L
B J = 2 π μ . (19) \mathcal{B}J=2\pi\mu. \tag*{(19)} B J = 2 π μ . (19) For τ > 0 \tau>0 τ > 0
τ = 1 2 π ∫ 0 ∞ ( 1 − e − τ t ) t − 3 / 2 d t . \sqrt{\tau}=\frac{1}{2\sqrt{\pi}}\int_0^\infty(1-e^{-\tau t})t^{-3/2}\,\mathrm{d}t. τ = 2 π 1 ∫ 0 ∞ ( 1 − e − τ t ) t − 3/2 d t . Integration by parts reduces it to the first formula in (G) with ζ = 0 \zeta= 0 ζ = 0
B h = 1 2 π ∫ 0 ∞ ( h − e − t e t Δ h ) t − 3 / 2 d t , (20) Bh = \frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\left(h-e^{-t}e^{t\Delta}h\right)t^{-3/2}\,dt, \tag*{(20)} B h = 2 π 1 ∫ 0 ∞ ( h − e − t e t Δ h ) t − 3/2 d t , (20) first for Schwartz functions and then by duality where used below, and the positive off-diagonal density in the formula is
n ( z ) = 1 8 π π ∫ 0 ∞ e − t − ∣ z ∣ 2 / ( 4 t ) t − 5 / 2 d t = ( 1 + ∣ z ∣ ) e − ∣ z ∣ 2 π ∣ z ∣ 3 ( z ≠ 0 ) . (21) n(z)=\frac{1}{8\pi\sqrt{\pi}}\int_{0}^{\infty}e^{-t-|z|^{2}/(4t)}t^{-5/2}\,dt=\frac{(1+|z|)e^{-|z|}}{2\pi|z|^{3}}\qquad(z\ne0). \tag*{(21)} n ( z ) = 8 π π 1 ∫ 0 ∞ e − t − ∣ z ∣ 2 / ( 4 t ) t − 5/2 d t = 2 π ∣ z ∣ 3 ( 1 + ∣ z ∣ ) e − ∣ z ∣ ( z = 0 ) . (21) For the last equality differentiate the second formula in (G) in ζ \zeta ζ α = 1 \alpha=1 α = 1 ζ = ∣ z ∣ 2 / 4 \zeta=|z|^{2}/4 ζ = ∣ z ∣ 2 /4
Lemma 3.9 Suppose H ∈ L 1 ( R 2 ) H\in L^{1}(\mathbb{R}^{2}) H ∈ L 1 ( R 2 ) H = c H=c H = c B a B_{a} B a a > 0 a>0 a > 0 B H BH B H B a B_{a} B a
c + ∫ R 2 ∖ B a ( c − H ( y ) ) n ( x − y ) d y . (22) c+\int_{\mathbb{R}^{2}\setminus B_{a}}(c-H(y))n(x-y)\,dy. \tag*{(22)} c + ∫ R 2 ∖ B a ( c − H ( y )) n ( x − y ) d y . (22) The proof is in Appendix A.
For N = 2 N=2 N = 2 U = γ U=\gamma U = γ B a B_{a} B a ( 1 − Δ ) U = J (1-\Delta)U=J ( 1 − Δ ) U = J J = γ J=\gamma J = γ (19) the restriction of μ \mu μ B a B_{a} B a
f ( x ) = 1 2 π [ γ + 1 2 π ∫ R 2 ∖ B a ( γ − J ( y ) ) ( 1 + ∣ x − y ∣ ) e − ∣ x − y ∣ ∣ x − y ∣ 3 d y ] . (23) f(x)=\frac{1}{2\pi}\left[\gamma+\frac{1}{2\pi}\int_{\mathbb{R}^{2}\setminus B_{a}}(\gamma-J(y))\frac{(1+|x-y|)e^{-|x-y|}}{|x-y|^{3}}\,dy\right]. \tag*{(23)} f ( x ) = 2 π 1 [ γ + 2 π 1 ∫ R 2 ∖ B a ( γ − J ( y )) ∣ x − y ∣ 3 ( 1 + ∣ x − y ∣ ) e − ∣ x − y ∣ d y ] . (23) It is nonnegative because μ \mu μ
Support classification
Lemma 3.10 For t > 0 t>0 t > 0 p ( r , t ) = ( K ∗ σ t ) ( r e 1 ) p(r,t)=(K*\sigma_{t})(re_{1}) p ( r , t ) = ( K ∗ σ t ) ( r e 1 ) p ( ⋅ , t ) p(\cdot,t) p ( ⋅ , t ) C 2 C^{2} C 2 r = t r=t r = t
∂ r 3 p ( t + , t ) − ∂ r 3 p ( t − , t ) = 2 t 2 . \partial_{r}^{3}p(t+,t)-\partial_{r}^{3}p(t-,t)=\frac{2}{t^{2}}. ∂ r 3 p ( t + , t ) − ∂ r 3 p ( t − , t ) = t 2 2 . For a > 0 a>0 a > 0 1 B a d x \mathbf{1}_{B_{a}}\,dx 1 B a d x C 3 C^{3} C 3 r = a r=a r = a
Proof. The distance substitution gives
r p ( r , t ) = ( 1 + ∣ r − t ∣ ) e − ∣ r − t ∣ − ( 1 + r + t ) e − ( r + t ) 2 t . rp(r,t)=\frac{(1+|r-t|)e^{-|r-t|}-(1+r+t)e^{-(r+t)}}{2t}. r p ( r , t ) = 2 t ( 1 + ∣ r − t ∣ ) e − ∣ r − t ∣ − ( 1 + r + t ) e − ( r + t ) . The function v ↦ ( 1 + ∣ v ∣ ) e − ∣ v ∣ v\mapsto(1+|v|)e^{-|v|} v ↦ ( 1 + ∣ v ∣ ) e − ∣ v ∣ C 2 C^{2} C 2 4 4 4 r p ( r , t ) rp(r,t) r p ( r , t ) 2 / t 2/t 2/ t r = t r=t r = t p p p 2 / t 2 2/t^{2} 2/ t 2
For the volume potential, integrate p ( r , t ) p(r,t) p ( r , t ) 4 π t 2 d t 4\pi t^{2}\,dt 4 π t 2 d t 0 < t < a 0<t<a 0 < t < a r r r a a a t < a / 4 t<a/4 t < a /4 r r r t ≥ a / 4 t\ge a/4 t ≥ a /4 r r r t = r t=r t = r
Proposition 3.11 Let N = 2 N=2 N = 2 N = 3 N=3 N = 3 μ = σ R \mu=\sigma_R μ = σ R a ∈ ( 0 , R ) a\in(0,R) a ∈ ( 0 , R )
S = [ 0 , a ] ∪ { R } , supp μ = B a ‾ ∪ ∂ B R , U ≡ γ on B a ‾ . S=[0,a]\cup\{R\},\qquad\operatorname{supp}\mu=\overline{B_a}\cup\partial B_R,\qquad U\equiv\gamma\quad\text{on }\overline{B_a}. S = [ 0 , a ] ∪ { R } , supp μ = B a ∪ ∂ B R , U ≡ γ on B a . In the latter case μ ( ∂ B R ) > 0 \mu(\partial B_R)>0 μ ( ∂ B R ) > 0 U ( r ) > γ U(r)>\gamma U ( r ) > γ a < r < R a<r<R a < r < R N = 2 N=2 N = 2 B a B_a B a μ ( ∂ B a ) = 0 \mu(\partial B_a)=0 μ ( ∂ B a ) = 0 N = 3 N=3 N = 3
μ = γ 8 π 1 B a d x + m σ a + q σ R , m , q > 0. (24) \mu=\frac{\gamma}{8\pi}\mathbf{1}_{B_a}\,dx+m\sigma_a+q\sigma_R,\qquad m,q>0. \tag*{(24)} μ = 8 π γ 1 B a d x + m σ a + q σ R , m , q > 0. (24) Proof. Suppose that an open region is either a mass-free annulus between two interior contact radii or a mass-free central ball bounded by an interior contact sphere. By Lemma 3.8 , J J J γ \gamma γ J < γ J<\gamma J < γ u = U − γ u=U-\gamma u = U − γ
( 1 − Δ ) u = J − γ < 0. (25) (1-\Delta)u=J-\gamma<0. \tag*{(25)} ( 1 − Δ ) u = J − γ < 0. (25) If u u u u ≡ 0 u\equiv0 u ≡ 0
By Lemma 3.7, R R R S S S S ∖ { R } S\setminus\{R\} S ∖ { R } μ = σ R \mu=\sigma_R μ = σ R S ∖ { R } S\setminus\{R\} S ∖ { R } a a a a = 0 a=0 a = 0 ( 0 , a ) (0,a) ( 0 , a ) S S S S S S S S S S = [ 0 , a ] ∪ { R } S=[0,a]\cup\{R\} S = [ 0 , a ] ∪ { R } (9) the potential equals γ \gamma γ B a ‾ \overline{B_a} B a
If an additional contact radius r 2 r_2 r 2 ( a , R ) (a,R) ( a , R ) ( a , r 2 ) (a,r_2) ( a , r 2 ) U > γ U>\gamma U > γ a < r < R a<r<R a < r < R
For N = 2 N=2 N = 2 Lemma 3.6 excludes mass on ∂ B a \partial B_a ∂ B a (23) defines a nonnegative real analytic radial density in B a B_a B a B a B_a B a ∂ B a \partial B_a ∂ B a B a B_a B a
For N = 3 N=3 N = 3 ( 1 − Δ ) 2 U = 8 π μ (1-\Delta)^2U=8\pi\mu ( 1 − Δ ) 2 U = 8 π μ μ ∣ B a = ( γ / 8 π ) d x \mu|_{B_a}=(\gamma/8\pi)\,dx μ ∣ B a = ( γ /8 π ) d x ∂ B a \partial B_a ∂ B a ∂ B R \partial B_R ∂ B R (24) holds with m ≥ 0 m\ge0 m ≥ 0 q > 0 q>0 q > 0 s = r − a s=r-a s = r − a w ( s ) = r ( U ( r ) − γ ) / r w(s)=r(U(r)-\gamma)/r w ( s ) = r ( U ( r ) − γ ) / r Lemma 3.10 , the inner sphere potential is C 2 C^2 C 2 2 / a 2 2/a^2 2/ a 2 C 3 C^3 C 3 a a a U = γ U=\gamma U = γ
w ( 0 ) = w ′ ( 0 ) = w ′ ′ ( 0 ) = 0 , w ′ ′ ′ ( 0 ) = 2 m a γ , ( ∂ s 2 − 1 ) 2 w = − ( a + s ) . w(0)=w'(0)=w''(0)=0,\qquad w'''(0)=\frac{2m}{a\gamma},\qquad(\partial_s^2-1)^2w=-(a+s). w ( 0 ) = w ′ ( 0 ) = w ′′ ( 0 ) = 0 , w ′′′ ( 0 ) = aγ 2 m , ( ∂ s 2 − 1 ) 2 w = − ( a + s ) . If m = 0 m=0 m = 0 w ( 4 ) ( 0 ) = − a < 0 w^{(4)}(0)=-a<0 w ( 4 ) ( 0 ) = − a < 0 w ( s ) = − a s 4 / 24 + O ( s 5 ) < 0 w(s)=-as^4/24+O(s^5)<0 w ( s ) = − a s 4 /24 + O ( s 5 ) < 0 s > 0 s>0 s > 0 U ≥ γ U\ge\gamma U ≥ γ m > 0 m>0 m > 0
Remark 3.12 Let F F F supp μ R ⊆ F ⊆ B R ‾ \operatorname{supp}\mu_R \subseteq F \subseteq\overline{B_R} supp μ R ⊆ F ⊆ B R μ R \mu_R μ R F F F D max ( F ) = D max ( B R ‾ ) D_{\max}(F)=D_{\max}(\overline{B_R}) D m a x ( F ) = D m a x ( B R )
The measure μ R \mu_R μ R F F F F F F B R B_R B R F F F (1) and [1 Corollaries 7.3 and 7.5] .
Equilibrium on three-dimensional balls
Here N = 3 N=3 N = 3 μ R \mu_R μ R U = K ∗ μ R U=K*\mu_R U = K ∗ μ R γ = 1 / D max ( B R ‾ ) \gamma=1/D_{\max}(\overline{B_R}) γ = 1/ D m a x ( B R ) § 3 . By (7) , ( 1 − Δ ) 2 U = 8 π μ R (1-\Delta)^2U=8\pi\mu_R ( 1 − Δ ) 2 U = 8 π μ R Υ \Upsilon Υ (11) . For R > R 3 R>R_3 R > R 3 B a B_a B a 1 / ( 8 π ) 1/(8\pi) 1/ ( 8 π ) cosh r \cosh r cosh r sinh r / r \sinh r/r sinh r / r b = R − a b=R-a b = R − a
For b > 0 b>0 b > 0
P ( b ) = b − b 2 + b 2 e − b − sinh b 2 , Q ( b ) = b 2 + b e − b − sinh b . (26) P(b)=b-\frac{b^2+b}{2}e^{-b}-\frac{\sinh b}{2},\qquad Q(b)=b^2+be^{-b}-\sinh b. \tag*{(26)} P ( b ) = b − 2 b 2 + b e − b − 2 sinh b , Q ( b ) = b 2 + b e − b − sinh b . (26) Lemma 4.1 The function P P P β < R 3 \beta<R_3 β < R 3 β \beta β Q > 0 Q>0 Q > 0 ( 0 , R 3 ) (0,R_3) ( 0 , R 3 ) Q ( R 3 ) = 0 Q(R_3)=0 Q ( R 3 ) = 0
Proof. Expanding 4 e b P ( b ) 4e^bP(b) 4 e b P ( b )
4 e b P ( b ) = 4 b e b − e 2 b + 1 − 2 b − 2 b 2 = 2 3 b 3 − ∑ n ≥ 5 ( 2 n − 4 n ) b n n ! . 4e^bP(b)=4be^b-e^{2b}+1-2b-2b^2=\frac{2}{3}b^3-\sum_{n\geq5}\frac{(2^n-4n)b^n}{n!}. 4 e b P ( b ) = 4 b e b − e 2 b + 1 − 2 b − 2 b 2 = 3 2 b 3 − n ≥ 5 ∑ n ! ( 2 n − 4 n ) b n . After division by b 3 b^3 b 3 2 / 3 2/3 2/3 − ∞ -\infty − ∞ 2 n − 4 n > 0 2^n-4n>0 2 n − 4 n > 0 n ≥ 5 n\geq5 n ≥ 5 P P P Q Q Q Q ( b ) = e b Υ ( b ) / 2 Q(b)=e^b\Upsilon(b)/2 Q ( b ) = e b Υ ( b ) /2
Q ( b ) − 2 P ( b ) = b ( b − 2 + ( b + 2 ) e − b ) . Q(b)-2P(b)=b\left(b-2+(b+2)e^{-b}\right). Q ( b ) − 2 P ( b ) = b ( b − 2 + ( b + 2 ) e − b ) . The bracket vanishes at zero and its derivative is 1 − ( b + 1 ) e − b > 0 1-(b+1)e^{-b}>0 1 − ( b + 1 ) e − b > 0 2 P < Q 2P<Q 2 P < Q ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) Q ( R 3 ) = 0 Q(R_3)=0 Q ( R 3 ) = 0 P ( R 3 ) < 0 P(R_3)<0 P ( R 3 ) < 0 β < R 3 \beta<R_3 β < R 3
Theorem 4.2 Let R 3 R_3 R 3 Υ \Upsilon Υ
(i) If R ≤ R 3 R\leq R_3 R ≤ R 3 μ R = σ R \mu_R=\sigma_R μ R = σ R
D max ( B R ‾ ) = 2 R 2 1 − ( 1 + 2 R ) e − 2 R . (27) D_{\max}(\overline{B_R})=\frac{2R^2}{1-(1+2R)e^{-2R}}. \tag*{(27)} D m a x ( B R ) = 1 − ( 1 + 2 R ) e − 2 R 2 R 2 . (27) In particular D max ( B R 3 ‾ ) = e R 3 D_{\max}(\overline{B_{R_3}})=e^{R_3} D m a x ( B R 3 ) = e R 3 ∂ B R \partial B_R ∂ B R R < R 3 R<R_3 R < R 3 ∂ B R ∪ { 0 } \partial B_R\cup\{0\} ∂ B R ∪ { 0 } R = R 3 R=R_3 R = R 3 U > γ U>\gamma U > γ B R ‾ \overline{B_R} B R
(ii) If R > R 3 R > R_{3} R > R 3 0 < β < R 3 0 < \beta< R_{3} 0 < β < R 3
( R − b ) P ( b ) + Q ( b ) = 0 (28) (R-b)P(b)+Q(b)=0 \tag*{(28)} ( R − b ) P ( b ) + Q ( b ) = 0 (28) has a unique solution b ∈ ( β , R 3 ) b \in(\beta,R_{3}) b ∈ ( β , R 3 )
a = R − b , g = a + 1 − a + 2 b , M = a 3 6 + a g 2 + R ( a + 2 ) e b 2 b , (29) a=R-b,\qquad g=a+1-\frac{a+2}{b},\qquad M=\frac{a^{3}}{6}+\frac{ag}{2}+\frac{R(a+2)e^{b}}{2b}, \tag*{(29)} a = R − b , g = a + 1 − b a + 2 , M = 6 a 3 + 2 a g + 2 b R ( a + 2 ) e b , (29) the maximizing measure and the maximum diversity are
μ R = 1 M ( 1 8 π 1 B a d x + a g 2 σ a + R ( a + 2 ) e b 2 b σ R ) , D max ( B ‾ R ) = M . (30) \mu_{R}=\frac{1}{M}\left(\frac{1}{8\pi}\mathbf{1}_{B_{a}}\,dx+\frac{ag}{2}\sigma_{a}+\frac{R(a+2)e^{b}}{2b}\sigma_{R}\right),\qquad D_{\max}(\overline{B}_{R})=M. \tag*{(30)} μ R = M 1 ( 8 π 1 1 B a d x + 2 a g σ a + 2 b R ( a + 2 ) e b σ R ) , D m a x ( B R ) = M . (30) Both spherical masses are positive.
Proof of Theorem 4.2 (i). Proposition 3.2 determines the measure and the contact set, and its level U ∂ ( R ) U_{\partial}(R) U ∂ ( R ) (13) is 1 / D max ( B ‾ R ) 1/D_{\max}(\overline{B}_{R}) 1/ D m a x ( B R ) R = R 3 R=R_{3} R = R 3 (15) shows U ∂ ( R 3 ) = e − R 3 U_{\partial}(R_{3})=e^{-R_{3}} U ∂ ( R 3 ) = e − R 3
Supercritical equilibrium
Multiplying the potential excess by r r r
Lemma 4.3 For s > 0 s>0 s > 0
k 0 ( s ) = s cosh s − sinh s 2 , k 1 ( s ) = s sinh s − 2 cosh s + 2 2 , k_{0}(s)=\frac{s\cosh s-\sinh s}{2},\qquad k_{1}(s)=\frac{s\sinh s-2\cosh s+2}{2}, k 0 ( s ) = 2 s cosh s − sinh s , k 1 ( s ) = 2 s sinh s − 2 cosh s + 2 , k 2 ( s ) = s cosh s − 3 sinh s + 2 s 2 . k_{2}(s)=\frac{s\cosh s-3\sinh s+2s}{2}. k 2 ( s ) = 2 s cosh s − 3 sinh s + 2 s . These functions are positive, and, for every a ≥ 0 a\geq0 a ≥ 0
f a ( s ) = a k 1 ( s ) + k 2 ( s ) k 0 ( s ) f_{a}(s)=\frac{ak_{1}(s)+k_{2}(s)}{k_{0}(s)} f a ( s ) = k 0 ( s ) a k 1 ( s ) + k 2 ( s ) is strictly increasing.
Proof. Since k 0 ′ ( s ) = s sinh s / 2 > 0 k_{0}'(s)=s\sinh s/2>0 k 0 ′ ( s ) = s sinh s /2 > 0 k 0 ( 0 ) = 0 k_{0}(0)=0 k 0 ( 0 ) = 0
k 1 ( s ) = ∫ 0 s k 0 ( u ) d u , k 2 ( s ) = ∫ 0 s ( s − u ) k 0 ( u ) d u . k_{1}(s)=\int_{0}^{s}k_{0}(u)\,du,\qquad k_{2}(s)=\int_{0}^{s}(s-u)k_{0}(u)\,du. k 1 ( s ) = ∫ 0 s k 0 ( u ) d u , k 2 ( s ) = ∫ 0 s ( s − u ) k 0 ( u ) d u . The identity
k 0 ′ ( s ) 2 − k 0 ( s ) k 0 ′ ′ ( s ) = sinh 2 s − s 2 4 > 0 k_{0}'(s)^{2}-k_{0}(s)k_{0}''(s)=\frac{\sinh^{2}s-s^{2}}{4}>0 k 0 ′ ( s ) 2 − k 0 ( s ) k 0 ′′ ( s ) = 4 sinh 2 s − s 2 > 0 makes k 0 ′ / k 0 k_{0}'/k_{0} k 0 ′ / k 0 t > 0 t>0 t > 0 s > t s>t s > t
d d s log k 0 ( s − t ) k 0 ( s ) = k 0 ′ ( s − t ) k 0 ( s − t ) − k 0 ′ ( s ) k 0 ( s ) > 0. \frac{d}{ds}\log\frac{k_{0}(s-t)}{k_{0}(s)}=\frac{k_{0}'(s-t)}{k_{0}(s-t)}-\frac{k_{0}'(s)}{k_{0}(s)}>0. d s d log k 0 ( s ) k 0 ( s − t ) = k 0 ( s − t ) k 0 ′ ( s − t ) − k 0 ( s ) k 0 ′ ( s ) > 0. In each of k 1 ( s ) / k 0 ( s ) = ∫ 0 s k 0 ( s − t ) / k 0 ( s ) d t k_1(s)/k_0(s)=\int_0^s k_0(s-t)/k_0(s)\,dt k 1 ( s ) / k 0 ( s ) = ∫ 0 s k 0 ( s − t ) / k 0 ( s ) d t k 2 ( s ) / k 0 ( s ) = ∫ 0 s t k 0 ( s − t ) / k 0 ( s ) d t k_2(s)/k_0(s)=\int_0^s tk_0(s-t)/k_0(s)\,dt k 2 ( s ) / k 0 ( s ) = ∫ 0 s t k 0 ( s − t ) / k 0 ( s ) d t f a f_a f a a ≥ 0 a\geq0 a ≥ 0
Proof of Theorem 4.2(ii). The left side of (28) is positive at b = β b=\beta b = β b = R 3 b=R_3 b = R 3 a = R − b a=R-b a = R − b
b k 1 ( b ) − ( b − 1 ) k 0 ( b ) = P ( b ) , b k 2 ( b ) − ( b − 2 ) k 0 ( b ) = Q ( b ) bk_1(b)-(b-1)k_0(b)=P(b),\qquad bk_2(b)-(b-2)k_0(b)=Q(b) b k 1 ( b ) − ( b − 1 ) k 0 ( b ) = P ( b ) , b k 2 ( b ) − ( b − 2 ) k 0 ( b ) = Q ( b ) turn the root equation into
g = f a ( b ) = a + 1 − a + 2 b > 0. (31) g=f_a(b)=a+1-\frac{a+2}{b}>0. \tag*{(31)} g = f a ( b ) = a + 1 − b a + 2 > 0. (31) Define M M M (29) and the positive measure
ν = 1 8 π 1 B a d x + a g 2 σ a + R ( a + 2 ) e b 2 b σ R , \nu=\frac{1}{8\pi}\mathbf{1}_{B_a}\,dx+\frac{ag}{2}\sigma_a+\frac{R(a+2)e^b}{2b}\sigma_R, ν = 8 π 1 1 B a d x + 2 a g σ a + 2 b R ( a + 2 ) e b σ R , of total mass M M M U = K ∗ ν U=K*\nu U = K ∗ ν ν / M \nu/M ν / M U / M U/M U / M
For t > 0 t>0 t > 0 0 ≤ r ≤ t 0\leq r\leq t 0 ≤ r ≤ t
p ( r , t ) = e − t t ( ( t + 1 ) sinh r r − cosh r ) , (32) p(r,t)=\frac{e^{-t}}{t}\left((t+1)\frac{\sinh r}{r}-\cosh r\right), \tag*{(32)} p ( r , t ) = t e − t ( ( t + 1 ) r sinh r − cosh r ) , (32) with the continuous value at r = 0 r=0 r = 0 r ≥ t r\geq t r ≥ t p ( r , t ) = p ( t , r ) p(r,t)=p(t,r) p ( r , t ) = p ( t , r ) Lemma 3.1 . Since ∫ R 3 e − ∣ x ∣ d x = 8 π \int_{\mathbb{R}^3}e^{-|x|}\,dx=8\pi ∫ R 3 e − ∣ x ∣ d x = 8 π t > a t>a t > a 0 ≤ r ≤ a 0\leq r\leq a 0 ≤ r ≤ a
U v o l ( r ) = 1 − e − a 2 ( ( a 2 + 3 a + 3 ) sinh r r − ( a + 1 ) cosh r ) . U_{\mathrm{vol}}(r)=1-\frac{e^{-a}}{2}\left((a^2+3a+3)\frac{\sinh r}{r}-(a+1)\cosh r\right). U vol ( r ) = 1 − 2 e − a ( ( a 2 + 3 a + 3 ) r sinh r − ( a + 1 ) cosh r ) . Here the radial density t 2 / 2 t^2/2 t 2 /2 e − a ( a 2 + 3 a + 3 ) e^{-a}(a^2+3a+3) e − a ( a 2 + 3 a + 3 ) e − a ( a + 1 ) e^{-a}(a+1) e − a ( a + 1 ) cosh r \cosh r cosh r
e − a 2 ( a + 1 − g − a + 2 b ) = 0. \frac{e^{-a}}{2}\left(a+1-g-\frac{a+2}{b}\right)=0. 2 e − a ( a + 1 − g − b a + 2 ) = 0. Likewise for the coefficient of sinh r / r \sinh r/r sinh r / r
e − a 2 ( − a 2 − 3 a − 3 + g ( a + 1 ) + ( a + 2 ) ( R + 1 ) b ) = 0 , \frac{e^{-a}}{2}\left(-a^2-3a-3+g(a+1)+\frac{(a+2)(R+1)}{b}\right)=0, 2 e − a ( − a 2 − 3 a − 3 + g ( a + 1 ) + b ( a + 2 ) ( R + 1 ) ) = 0 , by the definition of g g g R = a + b R=a+b R = a + b U = 1 U=1 U = 1 B ‾ a \overline{B}_a B a
By (7) , ( 1 − Δ ) 2 U = 8 π ν (1-\Delta)^2U=8\pi\nu ( 1 − Δ ) 2 U = 8 π ν U U U r > 0 r>0 r > 0
r ( 1 − Δ ) 2 U = ( d 2 d r 2 − 1 ) 2 ( r U ) . r(1-\Delta)^2U=\left(\frac{d^2}{dr^2}-1\right)^2(rU). r ( 1 − Δ ) 2 U = ( d r 2 d 2 − 1 ) 2 ( r U ) . Lemma 3.10 gives the sphere and volume regularity at a a a s = r − a s=r-a s = r − a w ( s ) = r ( U ( r ) − 1 ) w(s)=r(U(r)-1) w ( s ) = r ( U ( r ) − 1 )
w ( 0 ) = w ′ ( 0 ) = w ′ ′ ( 0 ) = 0 , w ′ ′ ′ ( 0 ) = 2 a a g 2 = g , ( d 2 d s 2 − 1 ) 2 w = − ( a + s ) . w(0)=w'(0)=w''(0)=0,\qquad w'''(0)=\frac{2}{a}\frac{ag}{2}=g,\qquad\left(\frac{d^2}{ds^2}-1\right)^2w=-(a+s). w ( 0 ) = w ′ ( 0 ) = w ′′ ( 0 ) = 0 , w ′′′ ( 0 ) = a 2 2 a g = g , ( d s 2 d 2 − 1 ) 2 w = − ( a + s ) . The solution is
w ( s ) = g k 0 ( s ) − a k 1 ( s ) − k 2 ( s ) = k 0 ( s ) ( f a ( b ) − f a ( s ) ) . (33) w(s)=gk_0(s)-ak_1(s)-k_2(s)=k_0(s)\bigl(f_a(b)-f_a(s)\bigr). \tag*{(33)} w ( s ) = g k 0 ( s ) − a k 1 ( s ) − k 2 ( s ) = k 0 ( s ) ( f a ( b ) − f a ( s ) ) . (33) By Lemma 4.3 , w ( s ) > 0 w(s)>0 w ( s ) > 0 0 < s < b 0<s<b 0 < s < b w ( b ) = 0 w(b)=0 w ( b ) = 0 U > 1 U>1 U > 1 U = 1 U=1 U = 1
The probability measure ν / M \nu/M ν / M 1 / M 1/M 1/ M (6) it is the unique minimizer, of energy 1 / M 1/M 1/ M (28) in ( β , R 3 ) (\beta,R_3) ( β , R 3 ) B ‾ a ∪ ∂ B R \overline{B}_a\cup\partial B_R B a ∪ ∂ B R (33) it equals the contact set. □
Corollary 4.4 Let λ > 0 \lambda>0 λ > 0 (7) and scaling, κ λ ( x ) = e − λ ∣ x ∣ / ( 8 π λ ) \kappa_\lambda(x)=e^{-\lambda\lvert x\rvert}/(8\pi\lambda) κ λ ( x ) = e − λ ∣ x ∣ / ( 8 π λ ) Green kernel of ( λ 2 − Δ ) 2 (\lambda^2-\Delta)^2 ( λ 2 − Δ ) 2 R 3 \mathbb{R}^3 R 3 Matérn kernel of smoothness 1 2 \frac{1}{2} 2 1 B ‾ R \overline{B}_R B R μ λ R \mu_{\lambda R} μ λ R x ↦ x / λ x\mapsto x/\lambda x ↦ x / λ B ‾ R \overline{B}_R B R κ λ \kappa_\lambda κ λ 8 π λ D max ( B ‾ λ R ) 8\pi\lambda D_{\max}(\overline{B}_{\lambda R}) 8 π λ D m a x ( B λ R )
Proof. The dilation x ↦ λ x x\mapsto\lambda x x ↦ λ x B ‾ R \overline{B}_R B R B ‾ λ R \overline{B}_{\lambda R} B λ R κ λ \kappa_\lambda κ λ B ‾ R \overline{B}_R B R 1 / ( 8 π λ ) 1/(8\pi\lambda) 1/ ( 8 π λ )
Radius limits
Proposition 4.5 The root b = b ( R ) b=b(R) b = b ( R )
lim R ↓ R 3 b ( R ) = R 3 , lim R → ∞ b ( R ) = β . \lim_{R\downarrow R_3} b(R)=R_3,\qquad\lim_{R\to\infty} b(R)=\beta. R ↓ R 3 lim b ( R ) = R 3 , R → ∞ lim b ( R ) = β . The maximum diversity is continuous at R 3 R_3 R 3
Proof. Set G R ( t ) = ( R − t ) P ( t ) + Q ( t ) G_R(t)=(R-t)P(t)+Q(t) G R ( t ) = ( R − t ) P ( t ) + Q ( t ) (28) ; at a fixed R > R 3 R>R_3 R > R 3 β \beta β R 3 R_3 R 3 R ′ > R > R 3 R'>R>R_3 R ′ > R > R 3
G R ′ ( b ( R ) ) = ( R ′ − R ) P ( b ( R ) ) < 0 , G R ′ ( β ) = Q ( β ) > 0. G_{R'}(b(R))=(R'-R)P(b(R))<0,\qquad G_{R'}(\beta)=Q(\beta)>0. G R ′ ( b ( R )) = ( R ′ − R ) P ( b ( R )) < 0 , G R ′ ( β ) = Q ( β ) > 0. The unique root for R ′ R' R ′ b ( R ) b(R) b ( R ) R → ∞ R\to\infty R → ∞ b b b P ( b ) = − Q ( b ) / ( R − b ) = O ( 1 / R ) P(b)=-Q(b)/(R-b)=O(1/R) P ( b ) = − Q ( b ) / ( R − b ) = O ( 1/ R ) b → β b\to\beta b → β 4 e β P ( b ) / b 3 4e^{\beta}P(b)/b^3 4 e β P ( b ) / b 3 P ′ ( β ) < 0 P'(\beta)<0 P ′ ( β ) < 0 b − β = O ( 1 / R ) b-\beta=O(1/R) b − β = O ( 1/ R ) b 0 b_0 b 0 R 3 R_3 R 3 a 0 = R 3 − b 0 > 0 a_0=R_3-b_0>0 a 0 = R 3 − b 0 > 0 a 0 P ( b 0 ) + Q ( b 0 ) = 0 a_0P(b_0)+Q(b_0)=0 a 0 P ( b 0 ) + Q ( b 0 ) = 0 R = R 3 R=R_3 R = R 3 Theorem 4.2 (i). So b → R 3 b\to R_3 b → R 3 a → 0 a\to0 a → 0 (29) shows M → e R 3 M\to e^{R_3} M → e R 3 R 3 R_3 R 3
Remark 4.6 For large R R R (30) are O ( 1 / R ) O(1/R) O ( 1/ R ) R 3 R^3 R 3 R 2 R^2 R 2 M M M
D max ( B ‾ R ) = R 3 6 + 1 2 ( e β β + 1 − β − 1 β ) R 2 + O ( R ) , R → ∞ , (34) D_{\max}(\overline{B}_R)=\frac{R^3}{6}+\frac{1}{2}\left(\frac{e^\beta}{\beta}+1-\beta-\frac{1}{\beta}\right)R^2+O(R),\qquad R\to\infty, \tag*{(34)} D m a x ( B R ) = 6 R 3 + 2 1 ( β e β + 1 − β − β 1 ) R 2 + O ( R ) , R → ∞ , (34) with leading term as in [1 Proposition 9.7] . Meckes’s variational formula and [6 Theorem 2] also show
D max ( B ‾ R ) ≤ Mag ( B ‾ R ) = R 3 6 + R 2 + 2 R + 1. D_{\max}(\overline{B}_R)\leq\operatorname{Mag}(\overline{B}_R)=\frac{R^3}{6}+R^2+2R+1. D m a x ( B R ) ≤ Mag ( B R ) = 6 R 3 + R 2 + 2 R + 1. The coefficient of R 2 R^2 R 2 (34) is approximately 0.928. For R = 3 , 4 , … , 10 R = 3, 4, \ldots, 10 R = 3 , 4 , … , 10 b b b (28) and the maximum diversity from (29) , next to the magnitude.
At small radius, the formula in Theorem 4.2 (i) expands to D max ( B R ‾ ) = 1 + 4 3 R + O ( R 2 ) D_{\max}(\overline{B_R}) = 1 + \frac{4}{3}R + O(R^2) D m a x ( B R ) = 1 + 3 4 R + O ( R 2 ) Mag ( B R ‾ ) − D max ( B R ‾ ) = 2 3 R + O ( R 2 ) \operatorname{Mag}(\overline{B_R}) - D_{\max}(\overline{B_R}) = \frac{2}{3}R + O(R^2) Mag ( B R ) − D m a x ( B R ) = 3 2 R + O ( R 2 ) [13 Proposition 6.8] . That term is absent from the subthreshold maximizing probability.
Equilibrium on planar disks
In the plane K ^ = 2 π ( 1 + ∣ ξ ∣ 2 ) − 3 / 2 \widehat{K} = 2\pi(1 + |\xi|^2)^{-3/2} K = 2 π ( 1 + ∣ ξ ∣ 2 ) − 3/2 ( 1 − Δ ) 3 / 2 = ( 1 − Δ ) ( 1 − Δ ) 1 / 2 (1-\Delta)^{3/2} = (1-\Delta)(1-\Delta)^{1/2} ( 1 − Δ ) 3/2 = ( 1 − Δ ) ( 1 − Δ ) 1/2 L / ( 2 π ) L/(2\pi) L / ( 2 π ) L ^ = 2 π ( 1 + ∣ ξ ∣ 2 ) − 1 / 2 \widehat{L} = 2\pi(1 + |\xi|^2)^{-1/2} L = 2 π ( 1 + ∣ ξ ∣ 2 ) − 1/2 L L L z = 0 z = 0 z = 0 e − ∣ X ∣ / ∣ X ∣ e^{-|X|}/|X| e − ∣ X ∣ /∣ X ∣ 4 π 4\pi 4 π 1 − Δ 3 1-\Delta_3 1 − Δ 3 R 3 \mathbb{R}^3 R 3 μ \mu μ (8) , J = ( 1 − Δ ) U J = (1-\Delta)U J = ( 1 − Δ ) U L ∗ μ L * \mu L ∗ μ μ \mu μ J = γ J = \gamma J = γ U = γ U = \gamma U = γ R R R Theorem 5.10 ), and above it the planar density obeys the same edge law (Theorem 5.9 ).
Lemma 5.1 Let a > 0 a > 0 a > 0 f f f B a ⊂ R 2 B_a \subset\mathbb{R}^2 B a ⊂ R 2
∣ f ( y ) ∣ ≤ C ( a 2 − ∣ y ∣ 2 ) − 1 / 2 . |f(y)| \le C(a^2 - |y|^2)^{-1/2}. ∣ f ( y ) ∣ ≤ C ( a 2 − ∣ y ∣ 2 ) − 1/2 . Its three-dimensional Yukawa potential
∫ B a e − ∣ X − ( y , 0 ) ∣ ∣ X − ( y , 0 ) ∣ f ( y ) d y \int_{B_a} \frac{e^{-|X-(y,0)|}}{|X-(y,0)|} f(y)\,dy ∫ B a ∣ X − ( y , 0 ) ∣ e − ∣ X − ( y , 0 ) ∣ f ( y ) d y is continuous on R 3 \mathbb{R}^3 R 3 z = 0 z = 0 z = 0 L ∗ ( f d x ) L*(f\,dx) L ∗ ( f d x ) ∂ B a \partial B_a ∂ B a
Proof. Let ν \nu ν ∣ f ( y ) ∣ d y |f(y)|\,dy ∣ f ( y ) ∣ d y z = 0 z = 0 z = 0
ν ( B s ( X ) ) ≤ C a s 3 / 2 ( X ∈ R 3 , 0 < s < a / 8 ) . (35) \nu(B_s(X)) \le C_a s^{3/2} \quad(X \in\mathbb{R}^3,\ 0 < s < a/8). \tag*{(35)} ν ( B s ( X )) ≤ C a s 3/2 ( X ∈ R 3 , 0 < s < a /8 ) . (35) Projecting to the plane reduces it to planar balls. In B a / 2 B_{a/2} B a /2 O ( s 2 ) O(s^2) O ( s 2 ) a / 2 ≤ t < a a/2 \le t < a a /2 ≤ t < a t t t s < a / 8 s < a/8 s < a /8 s s s I I I 2 s 2s 2 s a 2 − t 2 ≥ a ( a − t ) a^2 - t^2 \ge a(a-t) a 2 − t 2 ≥ a ( a − t )
C a s ∫ I ( a − t ) − 1 / 2 d t ≤ C a s 3 / 2 , C_a s \int_I (a-t)^{-1/2}\,dt \le C_a s^{3/2}, C a s ∫ I ( a − t ) − 1/2 d t ≤ C a s 3/2 , uniformly over intervals I ⊂ [ a / 2 , a ] I \subset[a/2,a] I ⊂ [ a /2 , a ] 2 s 2s 2 s (35) .
Layer-cake integration then gives, uniformly in X X X 0 < δ < a / 8 0 < \delta< a/8 0 < δ < a /8
∫ ∣ X − Y ∣ < δ d ν ( Y ) ∣ X − Y ∣ = ν ( B δ ( X ) ) δ + ∫ 0 δ ν ( B s ( X ) ) s 2 d s ≤ 3 C a δ 1 / 2 . (36) \int_{\lvert X-Y\rvert<\delta}\frac{d\nu(Y)}{\lvert X-Y\rvert}=\frac{\nu(B_\delta(X))}{\delta}+\int_0^\delta\frac{\nu(B_s(X))}{s^2}\,ds\le3Ca\delta^{1/2}. \tag*{(36)} ∫ ∣ X − Y ∣ < δ ∣ X − Y ∣ d ν ( Y ) = δ ν ( B δ ( X )) + ∫ 0 δ s 2 ν ( B s ( X )) d s ≤ 3 C a δ 1/2 . (36) Since the Yukawa kernel is bounded by the Coulomb kernel, removing its singularity inside a ball of radius δ \delta δ O ( δ 1 / 2 ) O(\delta^{1/2}) O ( δ 1/2 ) f f f
Lemma 5.2 Let 0 < a < R 0<a<R 0 < a < R ψ \psi ψ B a B_a B a ∣ ψ ( y ) ∣ ≤ C ( a 2 − ∣ y ∣ 2 ) − 1 / 2 \lvert\psi(y)\rvert\le C(a^2-\lvert y\rvert^2)^{-1/2} ∣ ψ ( y )∣ ≤ C ( a 2 − ∣ y ∣ 2 ) − 1/2 L ∗ ( ψ d x ) = j R = L ∗ σ R L*(\psi\,dx)=j_R=L*\sigma_R L ∗ ( ψ d x ) = j R = L ∗ σ R B a B_a B a z = 0 z=0 z = 0
W ( X ) = ∫ ∂ B R e − ∣ X − ( y , 0 ) ∣ ∣ X − ( y , 0 ) ∣ d σ R ( y ) − ∫ B a e − ∣ X − ( y , 0 ) ∣ ∣ X − ( y , 0 ) ∣ ψ ( y ) d y . W(X)=\int_{\partial B_R}\frac{e^{-\lvert X-(y,0)\rvert}}{\lvert X-(y,0)\rvert}\,d\sigma_R(y)-\int_{B_a}\frac{e^{-\lvert X-(y,0)\rvert}}{\lvert X-(y,0)\rvert}\,\psi(y)\,dy. W ( X ) = ∫ ∂ B R ∣ X − ( y , 0 )∣ e − ∣ X − ( y , 0 )∣ d σ R ( y ) − ∫ B a ∣ X − ( y , 0 )∣ e − ∣ X − ( y , 0 )∣ ψ ( y ) d y . Then W = 0 W=0 W = 0 W > 0 W>0 W > 0 W ( X ) → + ∞ W(X)\to+\infty W ( X ) → + ∞ X X X j R − L ∗ ( ψ d x ) j_R-L*(\psi\,dx) j R − L ∗ ( ψ d x ) B a B_a B a + ∞ +\infty + ∞ ∂ B R \partial B_R ∂ B R L 1 ( R 2 ) L^1(\mathbb{R}^2) L 1 ( R 2 ) r = R r=R r = R
Proof. The circle potential is smooth near the closed disk because a < R a<R a < R W = 0 W=0 W = 0 X = ( r , 0 , z ) X=(r,0,z) X = ( r , 0 , z ) d 2 = ( r − R ) 2 + z 2 d^2=(r-R)^2+z^2 d 2 = ( r − R ) 2 + z 2 θ \theta θ d 2 + 4 r R sin 2 ( θ / 2 ) d^2+4rR\sin^2(\theta/2) d 2 + 4 r R sin 2 ( θ /2 )
∫ − ε ε ( d 2 + s 2 ) − 1 / 2 d s , d = dist ( X , ∂ B R ) , \int_{-\varepsilon}^{\varepsilon}(d^2+s^2)^{-1/2}\,ds,\qquad d=\operatorname{dist}(X,\partial B_R), ∫ − ε ε ( d 2 + s 2 ) − 1/2 d s , d = dist ( X , ∂ B R ) , and tends to + ∞ +\infty + ∞ d ↓ 0 d\downarrow0 d ↓ 0 W W W ( 1 − Δ 3 ) W = 0 (1-\Delta_3)W=0 ( 1 − Δ 3 ) W = 0 W W W ( 1 − Δ 3 ) W < 0 (1-\Delta_3)W<0 ( 1 − Δ 3 ) W < 0 W > 0 W>0 W > 0
The asserted trace signs follow by restriction to the plane. For r r r R R R r e 1 re_1 r e 1 θ \theta θ ( r − R ) 2 + 4 r R sin 2 ( θ / 2 ) (r-R)^2+4rR\sin^2(\theta/2) ( r − R ) 2 + 4 r R sin 2 ( θ /2 ) r = R r=R r = R L ∈ L 1 ( R 2 ) L\in L^1(\mathbb{R}^2) L ∈ L 1 ( R 2 )
For the Coulomb kernel C ( x ) = ∣ x ∣ − 1 C(x)=\lvert x\rvert^{-1} C ( x ) = ∣ x ∣ − 1 E C ( η ) = ∬ C ( x − y ) d η ( x ) d η ( y ) E_C(\eta)=\iint C(x-y)\,d\eta(x)\,d\eta(y) E C ( η ) = ∬ C ( x − y ) d η ( x ) d η ( y )
Lemma 5.3 The probability measure
d μ C ( z ) = 1 { ∣ z ∣ < 1 } 2 π 1 − ∣ z ∣ 2 d z d\mu_C(z)=\frac{\mathbf{1}_{\{\lvert z\rvert<1\}}}{2\pi\sqrt{1-\lvert z\rvert^2}}\,dz d μ C ( z ) = 2 π 1 − ∣ z ∣ 2 1 {∣ z ∣ < 1 } d z has Coulomb potential π / 2 \pi/2 π /2 ∣ x − y ∣ − 1 \lvert x-y\rvert^{-1} ∣ x − y ∣ − 1
Proof For the kernel ∣ x − y ∣ − 1 \lvert x-y\rvert^{-1} ∣ x − y ∣ − 1 1 / ( π 2 1 − ∣ z ∣ 2 ) 1/(\pi^{2}\sqrt{1-\lvert z\rvert^{2}}) 1/ ( π 2 1 − ∣ z ∣ 2 ) [14 §4(i)] , and dividing by its mass 2 / π 2/\pi 2/ π μ C \mu_{\mathrm{C}} μ C π / 2 \pi/2 π /2 0 < ∣ z ∣ < 1 0<\lvert z\rvert<1 0 < ∣ z ∣ < 1 μ C \mu_{\mathrm{C}} μ C ( 2 π ) − 1 ( 1 − ∣ z ∣ ) − 1 / 2 (2\pi)^{-1}(1-\lvert z\rvert)^{-1/2} ( 2 π ) − 1 ( 1 − ∣ z ∣ ) − 1/2 (36) and the truncation argument in the proof of Lemma 5.1 make the potential continuous on R 3 \mathbb{R}^{3} R 3 π / 2 \pi/2 π /2
The Coulomb kernel is strictly positive definite on finite-energy signed measures in R 2 \mathbb{R}^{2} R 2 [15 §8.1] . A finite energy probability measure on a compact set whose potential is at least its energy throughout the set minimizes energy among probability measures on the set, and strict positive definiteness makes the minimizer unique [15 Remark 2 after Theorem 2.4] . On the closed disk the potential of μ C \mu_{\mathrm{C}} μ C π / 2 = E C ( μ C ) \pi/2=E_{\mathrm{C}}(\mu_{\mathrm{C}}) π /2 = E C ( μ C )
For the rest of this section take N = 2 N=2 N = 2 B ‾ R \overline{B}_{R} B R μ R \mu_{R} μ R U = K ∗ μ R U=K*\mu_{R} U = K ∗ μ R γ = ∫ U d μ R \gamma=\int U\,\mathrm{d}\mu_{R} γ = ∫ U d μ R J = L ∗ μ R J=L*\mu_{R} J = L ∗ μ R σ R \sigma_{R} σ R j R = L ∗ σ R j_{R}=L*\sigma_{R} j R = L ∗ σ R
For R > R 2 R>R_{2} R > R 2 Proposition 3.11 by a a a f R f_{R} f R q q q
∫ B a L ( x − y ) f R ( ∣ y ∣ ) d y = γ − q j R ( x ) for almost every x ∈ B a . \int_{B_{a}} L(x-y)f_{R}(\lvert y\rvert)\,\mathrm{d}y=\gamma-qj_{R}(x)\quad\text{for almost every }x\in B_{a}. ∫ B a L ( x − y ) f R (∣ y ∣) d y = γ − q j R ( x ) for almost every x ∈ B a . Copson solves the unscreened disk at prescribed potential by an Abel construction [14 ] , and Agra, Trizac and Bocquet reduce the screened constant-potential problem to a Fredholm equation by dual integral methods, observing the same inverse-square-root edge law by a local analysis [16 ] .
The factor 1 / π 2 1/\pi^{2} 1/ π 2 T \mathcal{T} T A a T 1 = 1 / ( π 2 a 2 − r 2 ) \mathcal{A}_{a}\mathcal{T}1=1/(\pi^{2}\sqrt{a^{2}-r^{2}}) A a T 1 = 1/ ( π 2 a 2 − r 2 ) B a B_{a} B a Lemma 5.3 .
Abel inversion
For a trial inner radius 0 < a < R 0<a<R 0 < a < R L ∗ ( f d x ) = v L*(f\,\mathrm{d}x)=v L ∗ ( f d x ) = v B a B_{a} B a B a B_{a} B a g g g g ( a ) / a 2 − r 2 g(a)/\sqrt{a^{2}-r^{2}} g ( a ) / a 2 − r 2 [17 eqs. (36)–(38)] . For a smooth radial function v v v B a B_{a} B a
( T v ) ( t ) = 1 π 2 d d t ∫ 0 t r v ( r ) t 2 − r 2 d r , h ( x ) = ∫ 0 π / 2 sin θ e − ∣ x ∣ sin θ d θ , ( P a g ) ( t ) = 1 π ∫ 0 a [ h ( t − s ) + h ( t + s ) ] g ( s ) d s . (37) \begin{aligned}
(\mathcal{T}v)(t)&=\frac{1}{\pi^{2}}\frac{\mathrm{d}}{\mathrm{d}t}\int_{0}^{t}\frac{r v(r)}{\sqrt{t^{2}-r^{2}}}\,\mathrm{d}r,\\
h(x)&=\int_{0}^{\pi/2}\sin\theta\,e^{-\lvert x\rvert\sin\theta}\,\mathrm{d}\theta,\\
(\mathcal{P}_{a}g)(t)&=\frac{1}{\pi}\int_{0}^{a}\left[h(t-s)+h(t+s)\right]g(s)\,\mathrm{d}s. \tag*{(37)}
\end{aligned} ( T v ) ( t ) h ( x ) ( P a g ) ( t ) = π 2 1 d t d ∫ 0 t t 2 − r 2 r v ( r ) d r , = ∫ 0 π /2 sin θ e − ∣ x ∣ s i n θ d θ , = π 1 ∫ 0 a [ h ( t − s ) + h ( t + s ) ] g ( s ) d s . (37) For g ∈ C 2 [ 0 , a ] g \in C^2[0,a] g ∈ C 2 [ 0 , a ] g ′ ( 0 ) = 0 g'(0)=0 g ′ ( 0 ) = 0
( A a g ) ( r ) = g ( a ) a 2 − r 2 − ∫ r a g ′ ( t ) t 2 − r 2 d t , 0 ≤ r < a . (38) (\mathcal{A}_a g)(r)=\frac{g(a)}{\sqrt{a^2-r^2}}-\int_r^a\frac{g'(t)}{\sqrt{t^2-r^2}}\,\mathrm{d}t,\qquad0\le r<a. \tag*{(38)} ( A a g ) ( r ) = a 2 − r 2 g ( a ) − ∫ r a t 2 − r 2 g ′ ( t ) d t , 0 ≤ r < a . (38) Lemma 5.4 The equation ( I − P a ) g = T v (I-\mathcal{P}_a)g=\mathcal{T}v ( I − P a ) g = T v P a \mathcal{P}_a P a
∥ P a ∥ C [ 0 , a ] → C [ 0 , a ] = ω ( a ) = 1 − 2 π ∫ 0 π / 2 e − a sin θ d θ < 1. (39) \|\mathcal{P}_a\|_{C[0,a]\to C[0,a]}=\omega(a)=1-\frac{2}{\pi}\int_0^{\pi/2}e^{-a\sin\theta}\,\mathrm{d}\theta<1. \tag*{(39)} ∥ P a ∥ C [ 0 , a ] → C [ 0 , a ] = ω ( a ) = 1 − π 2 ∫ 0 π /2 e − a s i n θ d θ < 1. (39) The solution belongs to C 2 [ 0 , a ] C^2[0,a] C 2 [ 0 , a ] g ′ ( 0 ) = 0 g'(0)=0 g ′ ( 0 ) = 0 f = A a g f=\mathcal{A}_a g f = A a g L ∗ ( f d x ) = v L^*(f\,\mathrm{d}x)=v L ∗ ( f d x ) = v f ( r ) = O ( ( a − r ) − 1 / 2 ) f(r)=O((a-r)^{-1/2}) f ( r ) = O (( a − r ) − 1/2 ) L L L ∂ B a \partial B_a ∂ B a v ( a ) v(a) v ( a ) g ( n ) = ∑ j = 0 n P a j T v g^{(n)}=\sum_{j=0}^n\mathcal{P}_a^j\mathcal{T}v g ( n ) = ∑ j = 0 n P a j T v
∥ g − g ( n ) ∥ ∞ ≤ ω ( a ) n + 1 1 − ω ( a ) ∥ T v ∥ ∞ . (40) \|g-g^{(n)}\|_\infty\le\frac{\omega(a)^{n+1}}{1-\omega(a)}\|\mathcal{T}v\|_\infty. \tag*{(40)} ∥ g − g ( n ) ∥ ∞ ≤ 1 − ω ( a ) ω ( a ) n + 1 ∥ T v ∥ ∞ . (40) Proof. Let J 0 J_0 J 0
∫ 0 t r J 0 ( ξ r ) t 2 − r 2 d r = sin ( ξ t ) ξ , \int_0^t\frac{rJ_0(\xi r)}{\sqrt{t^2-r^2}}\,\mathrm{d}r=\frac{\sin(\xi t)}{\xi}, ∫ 0 t t 2 − r 2 r J 0 ( ξ r ) d r = ξ sin ( ξ t ) , with value t t t ξ = 0 \xi=0 ξ = 0 f f f
∫ 0 a f ( r ) J 0 ( ξ r ) r d r = ∫ 0 a g ( t ) cos ( ξ t ) d t , T [ J 0 ( ξ ⋅ ) ] ( t ) = cos ( ξ t ) π 2 . \int_0^a f(r)J_0(\xi r)r\,\mathrm{d}r=\int_0^a g(t)\cos(\xi t)\,\mathrm{d}t,\qquad\mathcal{T}[J_0(\xi\cdot)](t)=\frac{\cos(\xi t)}{\pi^2}. ∫ 0 a f ( r ) J 0 ( ξ r ) r d r = ∫ 0 a g ( t ) cos ( ξ t ) d t , T [ J 0 ( ξ ⋅ )] ( t ) = π 2 cos ( ξ t ) . To invert the Coulomb multiplier insert e − ϵ ξ e^{-\epsilon\xi} e − ϵ ξ
∫ 0 ∞ e − ϵ ξ J 0 ( r ξ ) cos ( t ξ ) d ξ = ℜ ( r 2 + ( ϵ − i t ) 2 ) − 1 / 2 . \int_0^\infty e^{-\epsilon\xi}J_0(r\xi)\cos(t\xi)\,\mathrm{d}\xi=\Re\left(r^2+(\epsilon-it)^2\right)^{-1/2}. ∫ 0 ∞ e − ϵ ξ J 0 ( r ξ ) cos ( t ξ ) d ξ = ℜ ( r 2 + ( ϵ − i t ) 2 ) − 1/2 . The square root has positive real part. The angular representation of J 0 J_0 J 0 ϵ ↓ 0 \epsilon\downarrow0 ϵ ↓ 0 ( r 2 − t 2 ) − 1 / 2 (r^2-t^2)^{-1/2} ( r 2 − t 2 ) − 1/2 t < r t<r t < r t > r t>r t > r ∣ r 2 − t 2 ∣ − 1 / 2 |r^2-t^2|^{-1/2} ∣ r 2 − t 2 ∣ − 1/2 t t t r > 0 r>0 r > 0 ϵ > 0 \epsilon>0 ϵ > 0
C ∗ ( f d x ) ( r ) = 2 π ∫ 0 min ( r , a ) g ( t ) r 2 − t 2 d t , r > 0. C^*(f\,\mathrm{d}x)(r)=2\pi\int_0^{\min(r,a)}\frac{g(t)}{\sqrt{r^2-t^2}}\,\mathrm{d}t,\qquad r>0. C ∗ ( f d x ) ( r ) = 2 π ∫ 0 m i n ( r , a ) r 2 − t 2 g ( t ) d t , r > 0. One integration by parts makes the cosine integral O ( ξ − 1 ) O(\xi^{-1}) O ( ξ − 1 ) r > 0 r>0 r > 0 J 0 ( r ξ ) = O ( ξ − 1 / 2 ) J_0(r\xi)=O(\xi^{-1/2}) J 0 ( r ξ ) = O ( ξ − 1/2 ) L ∗ ( f d x ) ( r ) L^*(f\,\mathrm{d}x)(r) L ∗ ( f d x ) ( r ) ϵ ↓ 0 \epsilon\downarrow0 ϵ ↓ 0
The multiplier L ^ ( ξ ) = 2 π / 1 + ξ 2 \widehat{L}(\xi)=2\pi/\sqrt{1+\xi^2} L ( ξ ) = 2 π / 1 + ξ 2 L ∗ ( f d x ) = v L^*(f\,\mathrm{d}x)=v L ∗ ( f d x ) = v
g ( t ) + 2 π ∫ 0 a g ( s ) ∫ 0 ∞ ( ξ 1 + ξ 2 − 1 ) cos ( ξ t ) cos ( ξ s ) d ξ d s = T v ( t ) . g(t)+\frac{2}{\pi}\int_0^a g(s)\int_0^\infty\left(\frac{\xi}{\sqrt{1+\xi^2}}-1\right)\cos(\xi t)\cos(\xi s)\,\mathrm{d}\xi\,\mathrm{d}s=\mathcal{T}v(t). g ( t ) + π 2 ∫ 0 a g ( s ) ∫ 0 ∞ ( 1 + ξ 2 ξ − 1 ) cos ( ξ t ) cos ( ξ s ) d ξ d s = T v ( t ) . The inner integral is absolutely convergent. The substitution τ = sin 2 θ \tau=\sin^2\theta τ = sin 2 θ
1 − z 1 + z = 1 π ∫ 0 1 τ / ( 1 − τ ) z + τ d τ , z > 0 , (41) 1-\sqrt{\frac{z}{1+z}}=\frac{1}{\pi}\int_0^1\frac{\sqrt{\tau/(1-\tau)}}{z+\tau}\,\mathrm{d}\tau,\qquad z>0, \tag*{(41)} 1 − 1 + z z = π 1 ∫ 0 1 z + τ τ / ( 1 − τ ) d τ , z > 0 , (41) and with the cosine transform of ( ξ 2 + τ ) − 1 (\xi^2+\tau)^{-1} ( ξ 2 + τ ) − 1
∫ 0 ∞ ( 1 − ξ 1 + ξ 2 ) cos ( ξ x ) d ξ = h ( x ) . \int_0^\infty\left(1-\frac{\xi}{\sqrt{1+\xi^2}}\right)\cos(\xi x)\,\mathrm{d}\xi=h(x). ∫ 0 ∞ ( 1 − 1 + ξ 2 ξ ) cos ( ξ x ) d ξ = h ( x ) . With the product-to-sum identity, the integral equation becomes ( I − P a ) g = T v (I-P_a)g=\mathcal{T}v ( I − P a ) g = T v
The function h h h [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) ∫ R h = π \int_{\mathbb{R}}h=\pi ∫ R h = π
∫ 0 a [ h ( t − s ) + h ( t + s ) ] d s = ∫ t − a t + a h ( u ) d u ≤ ∫ − a a h ( u ) d u . \int_0^a [h(t-s)+h(t+s)]\,\mathrm{d}s=\int_{t-a}^{t+a}h(u)\,\mathrm{d}u\leq\int_{-a}^a h(u)\,\mathrm{d}u. ∫ 0 a [ h ( t − s ) + h ( t + s )] d s = ∫ t − a t + a h ( u ) d u ≤ ∫ − a a h ( u ) d u . Equality holds at t = 0 t=0 t = 0 h h h (39) and summing the geometric series bounds the remainder as in (40) .
The kernel h h h − π / 2 -\pi/2 − π /2 ( π / 2 ) δ 0 (\pi/2)\delta_0 ( π /2 ) δ 0 g g g [ − a , a ] [-a,a] [ − a , a ] ( 0 , a ) (0,a) ( 0 , a ) g ∈ C 2 [ 0 , a ] g\in C^2[0,a] g ∈ C 2 [ 0 , a ] g ′ ( 0 ) = 0 g'(0)=0 g ′ ( 0 ) = 0 ∣ g ′ ( t ) ∣ ≤ C t |g'(t)|\leq Ct ∣ g ′ ( t ) ∣ ≤ C t (38) bounds ∣ f ( r ) ∣ |f(r)| ∣ f ( r ) ∣ C ( a 2 − r 2 ) − 1 / 2 C(a^2-r^2)^{-1/2} C ( a 2 − r 2 ) − 1/2 [ 0 , a ) [0,a) [ 0 , a ) f f f L \mathcal{L} L f f f R 2 \mathbb{R}^2 R 2 u = L ∗ ( f d x ) − v u=\mathcal{L}*(f\,\mathrm{d}x)-v u = L ∗ ( f d x ) − v [ 0 , a ] [0,a] [ 0 , a ] L ∗ ( f d x ) \mathcal{L}*(f\,\mathrm{d}x) L ∗ ( f d x ) C ∗ ( f d x ) \mathcal{C}*(f\,\mathrm{d}x) C ∗ ( f d x ) ξ ( 1 + ξ 2 ) − 1 / 2 − 1 = O ( ξ − 2 ) \xi(1+\xi^2)^{-1/2}-1=O(\xi^{-2}) ξ ( 1 + ξ 2 ) − 1/2 − 1 = O ( ξ − 2 ) T \mathcal{T} T T ( L ∗ ( f d x ) ) = ( I − P a ) g = T v \mathcal{T}(\mathcal{L}*(f\,\mathrm{d}x))=(I-P_a)g=\mathcal{T}v T ( L ∗ ( f d x )) = ( I − P a ) g = T v
∫ 0 t r u ( r ) ( t 2 − r 2 ) − 1 / 2 d r \int_0^t ru(r)(t^2-r^2)^{-1/2}\,\mathrm{d}r ∫ 0 t r u ( r ) ( t 2 − r 2 ) − 1/2 d r has zero derivative on ( 0 , a ) (0,a) ( 0 , a ) t ∥ u ∥ ∞ t\|u\|_\infty t ∥ u ∥ ∞ t ( s 2 − t 2 ) − 1 / 2 t(s^2-t^2)^{-1/2} t ( s 2 − t 2 ) − 1/2 0 < t < s 0<t<s 0 < t < s π 2 ∫ 0 s r u ( r ) d r = 0 \frac{\pi}{2}\int_0^s ru(r)\,\mathrm{d}r=0 2 π ∫ 0 s r u ( r ) d r = 0 s ≤ a s\leq a s ≤ a u = 0 u=0 u = 0 [ 0 , a ] [0,a] [ 0 , a ] L \mathcal{L} L v v v v ( a ) v(a) v ( a )
Positivity of densities
We expand T j R \mathcal{T}j_R T j R t t t r r r ∣ r ∣ < R |r|<R ∣ r ∣ < R ( R − r e i θ ) ( R − r e − i θ ) (R-re^{i\theta})(R-re^{-i\theta}) ( R − r e i θ ) ( R − r e − i θ ) r e 1 re_1 r e 1 ( − ∞ , 0 ] (-\infty,0] ( − ∞ , 0 ] j R j_R j R [ 0 , R ) [0,R) [ 0 , R ) L L L 0 ≤ t < R 0\leq t<R 0 ≤ t < R
π 2 ( T j R ) ( t ) = ∑ n = 0 ∞ ( Δ n L ) ( R ) ( 2 n ) ! t 2 n . (42) \pi^2(\mathcal{T}j_R)(t)=\sum_{n=0}^{\infty}\frac{(\Delta^nL)(R)}{(2n)!}t^{2n}. \tag*{(42)} π 2 ( T j R ) ( t ) = n = 0 ∑ ∞ ( 2 n )! ( Δ n L ) ( R ) t 2 n . (42) The coefficient of r 2 n r^{2n} r 2 n j R ( r ) j_R(r) j R ( r ) ( Δ n L ) ( R ) / ( 4 n ( n ! ) 2 ) (\Delta^nL)(R)/(4^n(n!)^2) ( Δ n L ) ( R ) / ( 4 n ( n ! ) 2 ) π 2 T \pi^2\mathcal{T} π 2 T 4 n ( n ! ) 2 / ( 2 n ) ! 4^n(n!)^2/(2n)! 4 n ( n ! ) 2 / ( 2 n )!
Δ ( e − r r − m ) = e − r ( r − m + ( 2 m − 1 ) r − m − 1 + m 2 r − m − 2 ) , m ≥ 1. \Delta(e^{-r}r^{-m})=e^{-r}\left(r^{-m}+(2m-1)r^{-m-1}+m^2r^{-m-2}\right),\qquad m\geq1. Δ ( e − r r − m ) = e − r ( r − m + ( 2 m − 1 ) r − m − 1 + m 2 r − m − 2 ) , m ≥ 1. Consequently π 2 ( T j R ) ( 0 ) = e − R / R \pi^2(\mathcal{T}j_R)(0)=e^{-R}/R π 2 ( T j R ) ( 0 ) = e − R / R ( T j R ) ′ ( t ) > 0 (\mathcal{T}j_R)'(t)>0 ( T j R ) ′ ( t ) > 0 0 < t < R 0<t<R 0 < t < R
Lemma 5.5 Let 0 < a < R 0<a<R 0 < a < R k > 0 k>0 k > 0 ( I − P a ) g = T ( 1 − k j R ) (I-P_a)g=\mathcal{T}(1-kj_R) ( I − P a ) g = T ( 1 − k j R ) f = A a g f=A_ag f = A a g f ≥ 0 f\geq0 f ≥ 0 ( 0 , a ) ⟺ g ( a ) ≥ 0 (0,a)\Longleftrightarrow g(a)\geq0 ( 0 , a ) ⟺ g ( a ) ≥ 0
Under either condition, − g ′ ( t ) > 0 -g'(t)>0 − g ′ ( t ) > 0 0 < t ≤ a 0<t\leq a 0 < t ≤ a f ( r ) > 0 f(r)>0 f ( r ) > 0 0 ≤ r < a 0\leq r<a 0 ≤ r < a
Proof. Put v = 1 − k j R v=1-kj_R v = 1 − k j R
d ( t , s ) = h ( t − s ) − h ( t + s ) , ( Q a u ) ( t ) = 1 π ∫ 0 a d ( t , s ) u ( s ) d s . d(t,s)=h(t-s)-h(t+s),\qquad(\mathcal{Q}_a u)(t)=\frac{1}{\pi}\int_0^a d(t,s)u(s)\,ds. d ( t , s ) = h ( t − s ) − h ( t + s ) , ( Q a u ) ( t ) = π 1 ∫ 0 a d ( t , s ) u ( s ) d s . The kernel is nonnegative and strictly positive for t , s > 0 t,s>0 t , s > 0 ∥ Q a ∥ ≤ ω ( a ) < 1 \lVert\mathcal{Q}_a\rVert\leq\omega(a)<1 ∥ Q a ∥ ≤ ω ( a ) < 1 g = T v + P a g g=\mathcal{T}v+\mathcal{P}_a g g = T v + P a g s s s
( I − Q a ) ( − g ′ ) ( t ) = − ( T v ) ′ ( t ) + d ( t , a ) π g ( a ) . (43) (I-\mathcal{Q}_a)(-g')(t)=-(\mathcal{T}v)'(t)+\frac{d(t,a)}{\pi}g(a). \tag*{(43)} ( I − Q a ) ( − g ′ ) ( t ) = − ( T v ) ′ ( t ) + π d ( t , a ) g ( a ) . (43) Here d ( t , 0 ) = 0 d(t,0)=0 d ( t , 0 ) = 0 ∂ t [ h ( t − s ) + h ( t + s ) ] = − ∂ s d ( t , s ) \partial_t[h(t-s)+h(t+s)]=-\partial_s d(t,s) ∂ t [ h ( t − s ) + h ( t + s )] = − ∂ s d ( t , s ) T 1 = 1 / π 2 \mathcal{T}1=1/\pi^2 T 1 = 1/ π 2 − ( T v ) ′ = k ( T j R ) ′ > 0 -(\mathcal{T}v)'=k(\mathcal{T}j_R)'>0 − ( T v ) ′ = k ( T j R ) ′ > 0 ( 0 , a ] (0,a] ( 0 , a ] g ( a ) ≥ 0 g(a)\geq0 g ( a ) ≥ 0 t > 0 t>0 t > 0 I − Q a I-\mathcal{Q}_a I − Q a − g ′ ( t ) > 0 -g'(t)>0 − g ′ ( t ) > 0 (38) are then nonnegative, the integral positive for every r < a r<a r < a a 2 − r 2 f ( r ) → g ( a ) \sqrt{a^2-r^2}\,f(r)\to g(a) a 2 − r 2 f ( r ) → g ( a ) r ↑ a r\uparrow a r ↑ a f ≥ 0 f\geq0 f ≥ 0 g ( a ) ≥ 0 g(a)\geq0 g ( a ) ≥ 0
For data 1 − k j R 1-kj_R 1 − k j R
g 1 = ( I − P a ) − 1 T 1 , ϕ a = A a g 1 , g_1=(I-\mathcal{P}_a)^{-1}\mathcal{T}1,\qquad\phi_a=\mathcal{A}_a g_1, g 1 = ( I − P a ) − 1 T 1 , ϕ a = A a g 1 , g 2 = ( I − P a ) − 1 T j R , ψ a , R = A a g 2 . (44) g_2=(I-\mathcal{P}_a)^{-1}\mathcal{T}j_R,\qquad\psi_{a,R}=\mathcal{A}_a g_2. \tag*{(44)} g 2 = ( I − P a ) − 1 T j R , ψ a , R = A a g 2 . (44) and set
F a = K ∗ ( ϕ a d x ) , H a , R = K ∗ σ R − K ∗ ( ψ a , R d x ) . (45) F_a=K*(\phi_a\,dx),\qquad H_{a,R}=K*\sigma_R-K*(\psi_{a,R}\,dx). \tag*{(45)} F a = K ∗ ( ϕ a d x ) , H a , R = K ∗ σ R − K ∗ ( ψ a , R d x ) . (45) On the disk, the discrepancy F a + k H a , R − 1 F_a+kH_{a,R}-1 F a + k H a , R − 1 ( 1 − Δ ) u = 0 (1-\Delta)u=0 ( 1 − Δ ) u = 0 I 0 I_0 I 0 k k k a a a
Lemma 5.6 For every 0 < a < R 0<a<R 0 < a < R ϕ a > 0 \phi_a>0 ϕ a > 0 B a B_a B a F a ( x ) < 1 F_a(x)<1 F a ( x ) < 1 H a , R ( x ) > 0 H_{a,R}(x)>0 H a , R ( x ) > 0 x ∈ R 2 x\in\mathbb{R}^2 x ∈ R 2
Proof. The convolution of the indicators of two centered intervals is even and nonincreasing on [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) g 1 g_1 g 1 1 / π 2 1/\pi^2 1/ π 2 [ − a , a ] [-a,a] [ − a , a ] g 1 g_1 g 1 g 1 ( a ) > 0 g_1(a)>0 g 1 ( a ) > 0 ϕ a \phi_a ϕ a L L L J ϕ J_{\phi} J ϕ J ϕ < 1 J_{\phi}<1 J ϕ < 1 ( 1 − Δ ) − 1 (1-\Delta)^{-1} ( 1 − Δ ) − 1 F a = ( 1 − Δ ) − 1 J ϕ < 1 F_a=(1-\Delta)^{-1}J_{\phi}<1 F a = ( 1 − Δ ) − 1 J ϕ < 1
For the second sign apply Lemma 5.2 with ψ = ψ a , R \psi=\psi_{a,R} ψ = ψ a , R j R − L ∗ ( ψ a , R d x ) j_R-L*(\psi_{a,R}\,dx) j R − L ∗ ( ψ a , R d x ) H a , R H_{a,R} H a , R
Annular inequality
Lemma 5.7 Let 0 < a < R 0 < a < R 0 < a < R ν \nu ν B a ‾ ∪ ∂ B R \overline{B_a} \cup\partial B_R B a ∪ ∂ B R U = K ∗ ν U = K * \nu U = K ∗ ν J = L ∗ ν J = L * \nu J = L ∗ ν U ∈ C 1 U \in C^1 U ∈ C 1 U = γ U = \gamma U = γ B a ‾ \overline{B_a} B a U ( R ) = γ U(R) = \gamma U ( R ) = γ J J J a a a J ( a ) = γ J(a) = \gamma J ( a ) = γ U ( r ) > γ U(r) > \gamma U ( r ) > γ a < r < R a < r < R a < r < R ν \nu ν B R ‾ \overline{B_R} B R
Proof. Set h = J − γ h = J - \gamma h = J − γ u = U − γ u = U - \gamma u = U − γ Δ J > 0 \Delta J > 0 Δ J > 0 (8) , so t ↦ h ( e t ) t \mapsto h(e^t) t ↦ h ( e t ) h ( a ) = 0 h(a) = 0 h ( a ) = 0 h ( R − ) = + ∞ h(R-) = +\infty h ( R − ) = + ∞
( 1 − Δ ) u = h , u ( a ) = u ′ ( a ) = u ( R ) = 0. (1-\Delta)u=h,\qquad u(a)=u'(a)=u(R)=0. ( 1 − Δ ) u = h , u ( a ) = u ′ ( a ) = u ( R ) = 0. Let v v v v ′ ′ + v ′ / r − v = 0 v'' + v'/r - v = 0 v ′′ + v ′ / r − v = 0 v ( R ) = 0 v(R) = 0 v ( R ) = 0 v ′ ( R ) = − 1 v'(R) = -1 v ′ ( R ) = − 1 v > 0 v > 0 v > 0 [ a , R ) [a,R) [ a , R )
[ r ( v u ′ − v ′ u ) ] ′ = − r v h \left[r(vu'-v'u)\right]'=-rvh [ r ( v u ′ − v ′ u ) ] ′ = − r v h has integrable right-hand side, since h h h R R R U ′ U' U ′
∫ a R r v h d r = 0. \int_a^R rvh\,dr=0. ∫ a R r v h d r = 0. By strict convexity in log r \log r log r h h h ∫ a r s v ( s ) h ( s ) d s \int_a^r sv(s)h(s)\,ds ∫ a r s v ( s ) h ( s ) d s r < R r < R r < R ( u / v ) ′ > 0 (u/v)' > 0 ( u / v ) ′ > 0 u ( a ) = 0 u(a) = 0 u ( a ) = 0 u > 0 u > 0 u > 0
Equilibrium construction
Theorem 5.8 Fix R > R 2 R > R_2 R > R 2 0 < a < R 0 < a < R 0 < a < R (44) –(45) and put
k ( a , R ) = 1 − F a ( 0 ) H a , R ( 0 ) > 0. k(a,R)=\frac{1-F_a(0)}{H_{a,R}(0)}>0. k ( a , R ) = H a , R ( 0 ) 1 − F a ( 0 ) > 0. There is a unique a ∈ ( 0 , R ) a \in(0,R) a ∈ ( 0 , R )
F a ( R ) + k ( a , R ) H a , R ( R ) = 1 , g 1 ( a ) − k ( a , R ) g 2 ( a ) ≥ 0. (46) F_a(R)+k(a,R)H_{a,R}(R)=1,\qquad g_1(a)-k(a,R)g_2(a)\geq0. \tag*{(46)} F a ( R ) + k ( a , R ) H a , R ( R ) = 1 , g 1 ( a ) − k ( a , R ) g 2 ( a ) ≥ 0. (46) With k = k ( a , R ) k=k(a,R) k = k ( a , R ) g = g 1 − k g 2 g=g_1-kg_2 g = g 1 − k g 2
M = k + 2 π ∫ 0 a g ( t ) d t . (47) M=k+2\pi\int_0^a g(t)\,dt. \tag*{(47)} M = k + 2 π ∫ 0 a g ( t ) d t . (47) The maximizing probability measure and maximum diversity are
μ R = A a g ( ∣ x ∣ ) M 1 B a ( x ) d x + k M σ R , D max ( B R ‾ ) = M . (48) \mu_R=\frac{A_ag(|x|)}{M}\mathbf{1}_{B_a}(x)\,dx+\frac{k}{M}\sigma_R,\qquad D_{\max}(\overline{B_R})=M. \tag*{(48)} μ R = M A a g ( ∣ x ∣ ) 1 B a ( x ) d x + M k σ R , D m a x ( B R ) = M . (48) Its equilibrium level is M − 1 M^{-1} M − 1 0 < k / M < 1 0 < k/M < 1 0 < k / M < 1 B R B_R B R B a B_a B a (40) .
Proof. By Lemma 5.6 the denominator and numerator defining k k k ( 1 − Δ ) F a = 1 (1-\Delta)F_a=1 ( 1 − Δ ) F a = 1 ( 1 − Δ ) H a , R = 0 (1-\Delta)H_{a,R}=0 ( 1 − Δ ) H a , R = 0
F a ( r ) = 1 + ( F a ( 0 ) − 1 ) I 0 ( r ) , H a , R ( r ) = H a , R ( 0 ) I 0 ( r ) . (49) F_a(r)=1+(F_a(0)-1)I_0(r),\qquad H_{a,R}(r)=H_{a,R}(0)I_0(r). \tag*{(49)} F a ( r ) = 1 + ( F a ( 0 ) − 1 ) I 0 ( r ) , H a , R ( r ) = H a , R ( 0 ) I 0 ( r ) . (49) The definition of k k k
For existence take the unique equilibrium, and with Proposition 3.11 and (23) denote its inner radius, nonnegative density, outer mass and level by a R a_R a R f R f_R f R q R q_R q R γ R \gamma_R γ R a R > 0 a_R>0 a R > 0
L ∗ ( f R d x ) = γ R − q R j R L^*(f_R\,dx)=\gamma_R-q_Rj_R L ∗ ( f R d x ) = γ R − q R j R almost everywhere. The interior part of the measure has finite L L L γ R \gamma_R γ R L L L ( a R 2 − r 2 ) − 1 / 2 (a_R^2-r^2)^{-1/2} ( a R 2 − r 2 ) − 1/2 π 2 \pi^2 π 2 L L L 2 π / 1 + ∣ ξ ∣ 2 2\pi/\sqrt{1+|\xi|^2} 2 π / 1 + ∣ ξ ∣ 2 L L L
f R γ R = ϕ a R − q R γ R ψ a R , R . \frac{f_R}{\gamma_R}=\phi_{a_R}-\frac{q_R}{\gamma_R}\psi_{a_R,R}. γ R f R = ϕ a R − γ R q R ψ a R , R . Contact at the origin sets q R / γ R q_R/\gamma_R q R / γ R k ( a R , R ) k(a_R,R) k ( a R , R ) (46) , and nonnegativity with the Abel edge limit proves its inequality.
Conversely, for a radius satisfying (46) , the density f = A a ( g 1 − k g 2 ) f=A_a(g_1-kg_2) f = A a ( g 1 − k g 2 ) ν = f d x + k σ R \nu=f\,dx+k\sigma_R ν = f d x + k σ R (49) and on the outer circle by (46) , and by Lemma 5.4 its L L L ν \nu ν K ∗ ν ∈ C 1 K*\nu\in C^1 K ∗ ν ∈ C 1 K K K Lemma 5.7 then proves the strict annular inequality and optimality after normalization. Two selected radii would determine the same unique probability measure, and the strictly positive interior densities force the radii to coincide.
Integrating the Abel formula, we obtain ∫ B a f d x = 2 π ∫ 0 a g ( t ) d t \int_{B_a}f\,dx=2\pi\int_0^a g(t)\,dt ∫ B a f d x = 2 π ∫ 0 a g ( t ) d t (47) . On its support the normalized potential is M − 1 M^{-1} M − 1 M − 1 M^{-1} M − 1 M M M
Inner edge asymptotics
Theorem 5.9 Fix R > R 2 R>R_2 R > R 2
μ R = f R ( ∣ x ∣ ) 1 B a ( x ) d x + q R σ R , a = a R , \mu_R=f_R(|x|)\mathbf{1}_{B_a}(x)\,dx+q_R\sigma_R,\qquad a=a_R, μ R = f R ( ∣ x ∣ ) 1 B a ( x ) d x + q R σ R , a = a R , be the equilibrium probability measure from Theorem 5.8 , and set γ R = I ( μ R ) \gamma_R=I(\mu_R) γ R = I ( μ R ) g R = ( g 1 − k g 2 ) / M g_R=(g_1-kg_2)/M g R = ( g 1 − k g 2 ) / M k k k M M M Theorem 5.8 . The density has the Abel representation
f R ( r ) = g R ( a ) a 2 − r 2 − ∫ r a g R ′ ( t ) t 2 − r 2 d t . f_R(r)=\frac{g_R(a)}{\sqrt{a^2-r^2}}-\int_r^a\frac{g_R'(t)}{\sqrt{t^2-r^2}}\,dt. f R ( r ) = a 2 − r 2 g R ( a ) − ∫ r a t 2 − r 2 g R ′ ( t ) d t . Then g R ( a ) > 0 g_R(a)>0 g R ( a ) > 0 C R = g R ( a ) / 2 a C_R=g_R(a)/\sqrt{2a} C R = g R ( a ) / 2 a U R = K ∗ μ R U_R=K*\mu_R U R = K ∗ μ R J R = L ∗ μ R J_R=L*\mu_R J R = L ∗ μ R
f R ( a − t ) = C R t − 1 / 2 + O ( t 1 / 2 ) , (50) f_R(a-t)=C_Rt^{-1/2}+O(t^{1/2}), \tag*{(50)} f R ( a − t ) = C R t − 1/2 + O ( t 1/2 ) , (50) J R ( a + s ) = γ R − 4 π C R s 1 / 2 + O ( s 3 / 2 ) , (51) J_R(a+s)=\gamma_R-4\pi C_Rs^{1/2}+O(s^{3/2}), \tag*{(51)} J R ( a + s ) = γ R − 4 π C R s 1/2 + O ( s 3/2 ) , (51) U R ( a + s ) − γ R = 16 π C R 15 s 5 / 2 + O ( s 7 / 2 ) (52) U_R(a+s)-\gamma_R=\frac{16\pi C_R}{15}s^{5/2}+O(s^{7/2}) \tag*{(52)} U R ( a + s ) − γ R = 15 16 π C R s 5/2 + O ( s 7/2 ) (52) as s , t ↓ 0 s,t\downarrow0 s , t ↓ 0
Proof. By Lemma 5.4 , g R ∈ C 2 [ 0 , a ] g_R \in C^2[0,a] g R ∈ C 2 [ 0 , a ] ∣ f R ( r ) ∣ ≤ C ( a 2 − r 2 ) − 1 / 2 |f_R(r)| \le C(a^2-r^2)^{-1/2} ∣ f R ( r ) ∣ ≤ C ( a 2 − r 2 ) − 1/2 g R ( a ) ≥ 0 g_R(a) \ge0 g R ( a ) ≥ 0 g R ′ ( a ) < 0 g_R'(a) < 0 g R ′ ( a ) < 0 g R ( a ) = 0 g_R(a)=0 g R ( a ) = 0
Put
G = ( L − C ) ∗ ( f R 1 B a ) + q R j R . G=(L-C)*(f_R\mathbf{1}_{B_a})+q_Rj_R. G = ( L − C ) ∗ ( f R 1 B a ) + q R j R . Near ∂ B a \partial B_a ∂ B a G G G C 2 C^2 C 2 L − C = − 1 + ∣ x ∣ / 2 + O ( ∣ x ∣ 2 ) L-C=-1+|x|/2+O(|x|^2) L − C = − 1 + ∣ x ∣/2 + O ( ∣ x ∣ 2 ) C / ∣ x ∣ C/|x| C /∣ x ∣ (36) , applied with f = f R f=f_R f = f R δ \delta δ
sup x ∫ ∣ x − y ∣ < δ ∣ f R ( y ) ∣ 1 B a ( y ) ∣ x − y ∣ d y ≤ C δ 1 / 2 . \sup_x\int_{|x-y|<\delta}\frac{|f_R(y)|\mathbf{1}_{B_a}(y)}{|x-y|}\,dy\le C\delta^{1/2}. x sup ∫ ∣ x − y ∣ < δ ∣ x − y ∣ ∣ f R ( y ) ∣ 1 B a ( y ) d y ≤ C δ 1/2 . The second-derivative integrals with a continuous cutoff near the origin are continuous. The displayed estimate bounds the omitted part uniformly by C δ 1 / 2 C\delta^{1/2} C δ 1/2 ∂ B a \partial B_a ∂ B a a < R a<R a < R
Extend g R g_R g R C 2 C^2 C 2 g ~ \widetilde g g [ 0 , a + δ ] [0,a+\delta] [ 0 , a + δ ] Lemma 5.4 ,
C ∗ ( f R 1 B a ) ( r ) = 2 π ∫ 0 min ( r , a ) g R ( t ) r 2 − t 2 d t . C*(f_R\mathbf{1}_{B_a})(r)=2\pi\int_0^{\min(r,a)}\frac{g_R(t)}{\sqrt{r^2-t^2}}\,dt. C ∗ ( f R 1 B a ) ( r ) = 2 π ∫ 0 m i n ( r , a ) r 2 − t 2 g R ( t ) d t . Hence the C 2 C^2 C 2
Z ( r ) = 2 π ∫ 0 π / 2 g ~ ( r sin θ ) d θ + G ( r ) Z(r)=2\pi\int_0^{\pi/2}\widetilde g(r\sin\theta)\,d\theta+G(r) Z ( r ) = 2 π ∫ 0 π /2 g ( r sin θ ) d θ + G ( r ) equals γ R \gamma_R γ R r < a r<a r < a s > 0 s>0 s > 0
J R ( a + s ) = Z ( a + s ) − 2 π ∫ a a + s g ~ ( t ) ( a + s ) 2 − t 2 d t , Z ( a + s ) = γ R + o ( s 2 ) . (53) J_R(a+s)=Z(a+s)-2\pi\int_a^{a+s}\frac{\widetilde g(t)}{\sqrt{(a+s)^2-t^2}}\,dt,\qquad Z(a+s)=\gamma_R+o(s^2). \tag*{(53)} J R ( a + s ) = Z ( a + s ) − 2 π ∫ a a + s ( a + s ) 2 − t 2 g ( t ) d t , Z ( a + s ) = γ R + o ( s 2 ) . (53) If g R ( a ) = 0 g_R(a)=0 g R ( a ) = 0 g ~ ( t ) = g R ′ ( a ) ( t − a ) + O ( ( t − a ) 2 ) \widetilde g(t)=g_R'(a)(t-a)+O((t-a)^2) g ( t ) = g R ′ ( a ) ( t − a ) + O (( t − a ) 2 )
J R ( a + s ) − γ R = − 8 π g R ′ ( a ) 3 2 a s 3 / 2 + o ( s 3 / 2 ) . J_R(a+s)-\gamma_R=-\frac{8\pi g_R'(a)}{3\sqrt{2a}}s^{3/2}+o(s^{3/2}). J R ( a + s ) − γ R = − 3 2 a 8 π g R ′ ( a ) s 3/2 + o ( s 3/2 ) . The leading coefficient is positive. On the annulus u = U R − γ R u=U_R-\gamma_R u = U R − γ R
u ′ ′ + 1 r u ′ − u = γ R − J R , u ( a ) = u ′ ( a ) = 0. u''+\frac{1}{r}u'-u=\gamma_R-J_R,\qquad u(a)=u'(a)=0. u ′′ + r 1 u ′ − u = γ R − J R , u ( a ) = u ′ ( a ) = 0. Integrating twice would make u ( a + s ) u(a+s) u ( a + s ) s > 0 s>0 s > 0 Proposition 3.11 . So g R ( a ) > 0 g_R(a)>0 g R ( a ) > 0
Now (50) follows from the Abel formula. In (53) , the constant term in the expansion of g ~ \widetilde g g
2 π ∫ a a + s g ~ ( t ) ( a + s ) 2 − t 2 d t = 4 π g R ( a ) 2 a s 1 / 2 + O ( s 3 / 2 ) , 2\pi\int_a^{a+s}\frac{\widetilde g(t)}{\sqrt{(a+s)^2-t^2}}\,dt=\frac{4\pi g_R(a)}{\sqrt{2a}}s^{1/2}+O(s^{3/2}), 2 π ∫ a a + s ( a + s ) 2 − t 2 g ( t ) d t = 2 a 4 π g R ( a ) s 1/2 + O ( s 3/2 ) , and (51) follows. Integrating the radial equation twice from a a a u ( a ) = u ′ ( a ) = 0 u(a)=u'(a)=0 u ( a ) = u ′ ( a ) = 0
u ( a + s ) = 4 π C R ∫ 0 s ( s − t ) t 1 / 2 d t + O ( s 7 / 2 ) = 16 π C R 15 s 5 / 2 + O ( s 7 / 2 ) . u(a+s)=4\pi C_R\int_0^s(s-t)t^{1/2}\,dt+O(s^{7/2})=\frac{16\pi C_R}{15}s^{5/2}+O(s^{7/2}). u ( a + s ) = 4 π C R ∫ 0 s ( s − t ) t 1/2 d t + O ( s 7/2 ) = 15 16 π C R s 5/2 + O ( s 7/2 ) . Threshold asymptotics
Let U ∂ ( R ) = 2 π ∫ 0 π / 2 e − 2 R sin t d t U_{\partial}(R)=\frac{2}{\pi}\int_{0}^{\pi/2}e^{-2R\sin t}\,dt U ∂ ( R ) = π 2 ∫ 0 π /2 e − 2 R s i n t d t R R R F ( R ) = U ∂ ( R ) − e − R \mathcal{F}(R)=U_{\partial}(R)-e^{-R} F ( R ) = U ∂ ( R ) − e − R R 2 R_2 R 2
d ∗ = 1 − e − R 2 , c ∗ = e − R 2 ( 1 − 1 R 2 ) . d_* = 1-e^{-R_2}, \qquad c_* = e^{-R_2}\left(1-\frac{1}{R_2}\right). d ∗ = 1 − e − R 2 , c ∗ = e − R 2 ( 1 − R 2 1 ) . Theorem 5.10 Let m R = μ R ( B a R ) m_R=\mu_R(B_{a_R}) m R = μ R ( B a R ) ν R \nu_R ν R
μ R = m R ( T a R ) # ν R + ( 1 − m R ) σ R , T a ( z ) = a z . \mu_R=m_R(T_{a_R})_{\#}\nu_R+(1-m_R)\sigma_R,\qquad T_a(z)=az. μ R = m R ( T a R ) # ν R + ( 1 − m R ) σ R , T a ( z ) = a z . Then R 2 > 1 R_2>1 R 2 > 1 F ′ ( R 2 ) > 0 \mathcal{F}'(R_2)>0 F ′ ( R 2 ) > 0 ε = R − R 2 ↓ 0 \varepsilon=R-R_2\downarrow0 ε = R − R 2 ↓ 0
m R = F ′ ( R 2 ) d ∗ ε + o ( ε ) , (54) m_R=\frac{\mathcal{F}'(R_2)}{d_*}\varepsilon+o(\varepsilon), \tag*{(54)} m R = d ∗ F ′ ( R 2 ) ε + o ( ε ) , (54) a R = π F ′ ( R 2 ) 2 c ∗ d ∗ ε + o ( ε ) , (55) a_R=\frac{\pi\mathcal{F}'(R_2)}{2c_*d_*}\varepsilon+o(\varepsilon), \tag*{(55)} a R = 2 c ∗ d ∗ π F ′ ( R 2 ) ε + o ( ε ) , (55) $\nu_R\rightharpoonup\mu_C$.
The equilibrium level and maximum diversity satisfy
γ R = U ∂ ( R ) − F ( R ) 2 d ∗ + o ( ε 2 ) , (56) \gamma_R=U_{\partial}(R)-\frac{\mathcal{F}(R)^2}{d_*}+o(\varepsilon^2), \tag*{(56)} γ R = U ∂ ( R ) − d ∗ F ( R ) 2 + o ( ε 2 ) , (56) D_{\max}(\overline{B_R})-\frac{1}{U_{\partial}(R)} =\frac{e^{2R_2}\mathcal{F}'(R_2)^2}{d_*}\varepsilon^2+o(\varepsilon^2).
{#eq:56}
Proof. The two chord bounds
sin t ≥ 3 t π ( 0 ≤ t ≤ π / 6 ) , sin t ≥ 1 4 + 3 t 2 π ( π / 6 ≤ t ≤ π / 2 ) \sin t\ge\frac{3t}{\pi}\quad(0\le t\le\pi/6),\qquad
\sin t\ge\frac14+\frac{3t}{2\pi}\quad(\pi/6\le t\le\pi/2) sin t ≥ π 3 t ( 0 ≤ t ≤ π /6 ) , sin t ≥ 4 1 + 2 π 3 t ( π /6 ≤ t ≤ π /2 ) give
U ∂ ( 1 ) < 1 + e − 1 − 2 e − 2 3 < e − 1 . U_{\partial}(1)<\frac{1+e^{-1}-2e^{-2}}{3}<e^{-1}. U ∂ ( 1 ) < 3 1 + e − 1 − 2 e − 2 < e − 1 . The last inequality uses e < 1 + 3 e<1+\sqrt{3} e < 1 + 3 R 2 > 1 R_2>1 R 2 > 1 χ ( R ) = e R U ∂ ( R ) \chi(R)=e^R U_{\partial}(R) χ ( R ) = e R U ∂ ( R ) χ ( 0 ) = χ ( R 2 ) = 1 \chi(0)=\chi(R_2)=1 χ ( 0 ) = χ ( R 2 ) = 1 χ ′ ( 0 ) < 0 \chi'(0)<0 χ ′ ( 0 ) < 0 χ ′ ( R 2 ) > 0 \chi'(R_2)>0 χ ′ ( R 2 ) > 0 F ′ ( R 2 ) = e − R 2 χ ′ ( R 2 ) > 0 \mathcal{F}'(R_2)=e^{-R_2}\chi'(R_2)>0 F ′ ( R 2 ) = e − R 2 χ ′ ( R 2 ) > 0 c ∗ , d ∗ > 0 c_*,d_*>0 c ∗ , d ∗ > 0
Let R j ↓ R 2 R_j\downarrow R_2 R j ↓ R 2 μ R j \mu_{R_j} μ R j η \eta η B R 2 ‾ \overline{B_{R_2}} B R 2 ν \nu ν B R 2 ‾ \overline{B_{R_2}} B R 2 T j x = ( R j / R 2 ) x T_jx=(R_j/R_2)x T j x = ( R j / R 2 ) x
I ( η ) = lim j I ( μ R j ) ≤ lim j I ( ( T j ) # ν ) = I ( ν ) . I(\eta)=\lim_j I(\mu_{R_j})\le\lim_j I((T_j)_{\#}\nu)=I(\nu). I ( η ) = j lim I ( μ R j ) ≤ j lim I (( T j ) # ν ) = I ( ν ) . Uniqueness at R 2 R_2 R 2 η = σ R 2 \eta=\sigma_{R_2} η = σ R 2 μ R ⇀ σ R 2 \mu_R\rightharpoonup\sigma_{R_2} μ R ⇀ σ R 2 R ↓ R 2 R\downarrow R_2 R ↓ R 2 γ R → e − R 2 \gamma_R\to e^{-R_2} γ R → e − R 2
The limiting circle potential has the expansion
K ∗ σ R 2 ( r ) = e − R 2 + c ∗ 4 r 2 + O ( r 4 ) . K*\sigma_{R_2}(r)=e^{-R_2}+\frac{c_*}{4}r^2+O(r^4). K ∗ σ R 2 ( r ) = e − R 2 + 4 c ∗ r 2 + O ( r 4 ) . A positive subsequential limit of a R a_R a R c ∗ > 0 c_*>0 c ∗ > 0 a R → 0 a_R\to0 a R → 0 σ R 2 \sigma_{R_2} σ R 2 m R → 0 m_R\to0 m R → 0 J R = γ R J_R=\gamma_R J R = γ R ∣ e − u − 1 ∣ ≤ u |e^{-u}-1|\le u ∣ e − u − 1∣ ≤ u L L L ∣ z ∣ < 1 |z|<1 ∣ z ∣ < 1
m R a R ( C ∗ ν R ) ( z ) = γ R − ( 1 − m R ) e − R R + O ( m R + a R 2 ) . (57) \frac{m_R}{a_R}(C\ast\nu_R)(z)=\gamma_R-(1-m_R)\frac{e^{-R}}{R}+O(m_R+a_R^2). \tag*{(57)} a R m R ( C ∗ ν R ) ( z ) = γ R − ( 1 − m R ) R e − R + O ( m R + a R 2 ) . (57) Integrate against μ C \mu_C μ C
π 2 m R a R = γ R − ( 1 − m R ) e − R R + O ( m R + a R 2 ) ⟶ c ∗ . \frac{\pi}{2}\frac{m_R}{a_R}=\gamma_R-(1-m_R)\frac{e^{-R}}{R}+O(m_R+a_R^2)\longrightarrow c_*. 2 π a R m R = γ R − ( 1 − m R ) R e − R + O ( m R + a R 2 ) ⟶ c ∗ . Therefore m R / a R → 2 c ∗ / π m_R/a_R\to2c_*/\pi m R / a R → 2 c ∗ / π (57) C ∗ ν R → π / 2 C\ast\nu_R\to\pi/2 C ∗ ν R → π /2 ν R \nu_R ν R E C ( ν R ) → π / 2 E_C(\nu_R)\to\pi/2 E C ( ν R ) → π /2 [15 Lemma 2.2.1(e)] and uniqueness in Lemma 5.3 imply that every weak limit equals μ C \mu_C μ C
For the rates let
A R = ∫ e − a R ∣ z ∣ d ν R ( z ) , X R = ∫ ( K ∗ σ R ) ( a R ∣ z ∣ ) d ν R ( z ) . A_R=\int e^{-a_R|z|}\,d\nu_R(z),\qquad X_R=\int(K\ast\sigma_R)(a_R|z|)\,d\nu_R(z). A R = ∫ e − a R ∣ z ∣ d ν R ( z ) , X R = ∫ ( K ∗ σ R ) ( a R ∣ z ∣ ) d ν R ( z ) . The contact equations at the center and on the outer circle yield
( 1 − m R ) F ( R ) = m R ( A R − X R ) . (1-m_R)F(R)=m_R(A_R-X_R). ( 1 − m R ) F ( R ) = m R ( A R − X R ) . Here A R = 1 + O ( a R ) A_R=1+O(a_R) A R = 1 + O ( a R ) X R = e − R + O ( a R 2 ) X_R=e^{-R}+O(a_R^2) X R = e − R + O ( a R 2 ) m R / F ( R ) → 1 / d ∗ m_R/F(R)\to1/d_* m R / F ( R ) → 1/ d ∗ F F F m R / a R m_R/a_R m R / a R
Finally put
Y R = ∬ e − a R ∣ z − w ∣ d ν R ( z ) d ν R ( w ) = 1 + O ( a R ) . Y_R=\iint e^{-a_R|z-w|}\,d\nu_R(z)\,d\nu_R(w)=1+O(a_R). Y R = ∬ e − a R ∣ z − w ∣ d ν R ( z ) d ν R ( w ) = 1 + O ( a R ) . The energy decomposes as
γ R = ( 1 − m R ) 2 U ∂ ( R ) + 2 m R ( 1 − m R ) X R + m R 2 Y R . \gamma_R=(1-m_R)^2U_{\partial}(R)+2m_R(1-m_R)X_R+m_R^2Y_R. γ R = ( 1 − m R ) 2 U ∂ ( R ) + 2 m R ( 1 − m R ) X R + m R 2 Y R . Since a R , m R = O ( ε ) a_R,m_R=O(\varepsilon) a R , m R = O ( ε )
γ R = U ∂ ( R ) − 2 m R F ( R ) + m R 2 [ 1 − 2 e − R + U ∂ ( R ) ] + O ( ε 3 ) . \gamma_R=U_{\partial}(R)-2m_RF(R)+m_R^2[1-2e^{-R}+U_{\partial}(R)]+O(\varepsilon^3). γ R = U ∂ ( R ) − 2 m R F ( R ) + m R 2 [ 1 − 2 e − R + U ∂ ( R )] + O ( ε 3 ) . Substituting m R = F ( R ) / d ∗ + o ( ε ) m_R=F(R)/d_*+o(\varepsilon) m R = F ( R ) / d ∗ + o ( ε ) 1 − 2 e − R 2 + U ∂ ( R 2 ) = d ∗ 1-2e^{-R_2}+U_{\partial}(R_2)=d_* 1 − 2 e − R 2 + U ∂ ( R 2 ) = d ∗ U ∂ ( R 2 ) = e − R 2 U_{\partial}(R_2)=e^{-R_2} U ∂ ( R 2 ) = e − R 2
Together with the boundary formula in Proposition 3.4, (56) shows that D max ( B R ‾ ) D_{\max}(\overline{B_R}) D m a x ( B R ) R 2 R_2 R 2
Discussion
How many interior support components can occur in dimensions N ≥ 4 N\ge4 N ≥ 4 J ≤ γ J\le\gamma J ≤ γ Δ L \Delta L Δ L (3) changes sign for N ≥ 4 N\ge4 N ≥ 4 N = 4 , 5 N=4,5 N = 4 , 5
Maximum diversity is continuous in Gromov–Hausdorff distance on compact positive definite metric spaces [7 ] . For a nonempty compact set E ⊂ R N E \subset\mathbb{R}^{N} E ⊂ R N h h h E h ⊂ E E_h \subset E E h ⊂ E
D max ( E h ) ≤ D max ( E ) ≤ e 2 h D max ( E h ) . D_{\max}(E_h) \leq D_{\max}(E) \leq e^{2h}D_{\max}(E_h). D m a x ( E h ) ≤ D m a x ( E ) ≤ e 2 h D m a x ( E h ) . A measurable nearest point map T : E → E h T:E \to E_h T : E → E h ∣ T x − T y ∣ ≥ ∣ x − y ∣ − 2 h \lvert Tx-Ty\rvert\geq\lvert x-y\rvert- 2h ∣ T x − T y ∣ ≥ ∣ x − y ∣ − 2 h e 2 h e^{2h} e 2 h
Leinster and Roff also ask about cubes. On a box with the ℓ 1 \ell_1 ℓ 1 [9 ] . For the Euclidean metric, the proof of Lemma 3.6 uses only that U ≥ γ U \geq\gamma U ≥ γ J ≤ γ J \leq\gamma J ≤ γ N = 2 , 3 N=2,3 N = 2 , 3 J J J J J J Lemma 3.8 ) and the radial monotonicity that puts the whole boundary in the support (Lemma 3.7), and its support is yet to be determined.
Does the interior support component remain separated from the outer boundary under a small smooth perturbation of a three-dimensional ball with R > R 3 R>R_3 R > R 3
In the planar construction the contraction factor tends to one as the inner radius grows. Direct Neumann iteration gets slow on large disks, and the inverse square root edge law suggests putting the singular factor into the numerical basis. Quantitative error bounds for the quadrature and scalar root are needed as the inner radius grows. The planar large-radius asymptotics also remain to be determined.
Declarations
Funding
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The author declares no competing interests.
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Adam Dajani is the sole author and is responsible for the research and manuscript.
No generative AI was used in this work.
Proof of analyticity
Proof of Lemma 3.9 Set g = H − c g = H - c g = H − c B a B_a B a B ( c ) = c B(c) = c B ( c ) = c B g Bg B g B a B_a B a g g g R 2 ∖ B a \mathbb{R}^{2} \setminus B_a R 2 ∖ B a
B g ( x ) = − ∫ R 2 ∖ B a g ( y ) n ( x − y ) d y . Bg(x) = -\int_{\mathbb{R}^{2}\setminus B_a} g(y)n(x-y)\,\mathrm{d}y. B g ( x ) = − ∫ R 2 ∖ B a g ( y ) n ( x − y ) d y . The interchange is absolutely convergent because n n n ∣ z ∣ ≥ δ > 0 |z| \ge\delta> 0 ∣ z ∣ ≥ δ > 0 ∣ g ( y ) ∣ ≤ ∣ H ( y ) ∣ + ∣ c ∣ |g(y)| \le|H(y)| + |c| ∣ g ( y ) ∣ ≤ ∣ H ( y ) ∣ + ∣ c ∣
For analyticity we extend this integral to a complex neighborhood. Fix x 0 ∈ B a x_0 \in B_a x 0 ∈ B a δ > 0 \delta> 0 δ > 0 dist ( x 0 , R 2 ∖ B a ) ≥ δ \operatorname{dist}(x_0,\mathbb{R}^{2}\setminus B_a) \ge\delta dist ( x 0 , R 2 ∖ B a ) ≥ δ z ∈ C 2 z \in\mathbb{C}^{2} z ∈ C 2 ∣ z − x 0 ∣ < δ / 4 |z-x_0| < \delta/4 ∣ z − x 0 ∣ < δ /4 y ∉ B a y \notin B_a y ∈ / B a d ( z , y ) = ( z − y ) ⋅ ( z − y ) d(z,y) = \sqrt{(z-y)\cdot(z-y)} d ( z , y ) = ( z − y ) ⋅ ( z − y ) u = x 0 − y u = x_0-y u = x 0 − y ∣ u ∣ 2 |u|^{2} ∣ u ∣ 2 ( 9 / 16 ) ∣ u ∣ 2 (9/16)|u|^{2} ( 9/16 ) ∣ u ∣ 2 R e d ( z , y ) ≥ ( 7 / 4 ) ∣ u ∣ \mathop{\mathrm{Re}}d(z,y) \ge(\sqrt{7}/4)|u| Re d ( z , y ) ≥ ( 7 /4 ) ∣ u ∣ ∣ d ( z , y ) ∣ ≥ ( 7 / 4 ) ∣ u ∣ |d(z,y)| \ge(\sqrt{7}/4)|u| ∣ d ( z , y ) ∣ ≥ ( 7 /4 ) ∣ u ∣ ∣ H ( y ) ∣ + ∣ c ∣ |H(y)| + |c| ∣ H ( y ) ∣ + ∣ c ∣ ( 1 + d ) e − d / ( 2 π d 3 ) (1+d)e^{-d}/(2\pi d^{3}) ( 1 + d ) e − d / ( 2 π d 3 )
Numerical values and experiments
The left table is for R 3 \mathbb{R}^{3} R 3 γ \gamma γ
Table 1.
Table 2.
Table 3.
For N = 4 , 5 N = 4, 5 N = 4 , 5 R R R − 10 − 15 -10^{-15} − 1 0 − 15 σ R \sigma_R σ R e − R e^{-R} e − R U ∂ ( R ) U_{\partial}(R) U ∂ ( R ) ∂ B R \partial B_R ∂ B R
Further spheres appear at the center, and a minimization on a grid of 800 radii, without continuation, finds the same number of spheres between successive birth radii. The intervals between successive birth radii shrink by the factors in the table, so the births appear to accumulate near the listed accumulation radii. Up to those radii the computed support is a finite union of concentric spheres whose number grows with the radius. At larger radii the discrete minimizers spread mass over a central region. We were unable to distinguish concentric spheres from a central ball.
References 
