Introduction
It has long been known that the Bose gas condenses below a critical temperature [10 , 15 ] as was verified experimentally by [1 , 14 ] in extremely low density samples of rubidium and sodium. Despite phenomenological theories that were developed by Bogolyubov, Gross, Pitaevskii, Landau and others [19 , 29 , 8 , 22 ] in the low-density regime, a proof of Bose-Einstein condensation from first principles in interacting Bose gases is lacking (very recently, in a release of machine-generated manuscripts [28 ] , a proof of Bose–Einstein condensation for the dilute hard-sphere Bose gas in the thermodynamic limit has been announced; this result is still under review, and is yet to be published in a scientific journal). Proving many other properties of interacting Bose gases including super-fluidity remains an open problem.
Over the past three decades, starting with Lieb and Yngvason’s study of its ground state energy [26 , 27 ] , mathematical physicists have worked towards filling these gaps. There have been many successes both in the Gross-Pitaevskii regime [25 , 7 ] and the mean-field regime [31 ] , and in the thermodynamic limit [32 , 16 , 17 , 6 , 5 ] . Lately there has been renewed interest in studying the hard-core Bose gas [26 , 17 , 5 , 6 ] . This is for two reasons. Firstly, it is a mathematical challenge due to the infinite gradient of the potential and secondly realistic potentials for bosonic atoms such as the Aziz potential [2 ] or the HFDHE2 potential [9 ] include a strongly repulsive singularity at the origin which is well approximated by a hard-core.
Most of the results that have been obtained in the thermodynamic limit concern the ground-state energy [32 , 16 , 17 , 5 , 6 ] and free energy [4 , 18 , 30 , 33 , 20 ] . However a description of other observables of Bose gases, such as the condensate fraction, still remains lacking [27 ] .
In 1963, Lieb introduced an effective equation for the ground state properties of the interacting Bose gas in the thermodynamic limit, known as the Simplified Approach [24 ] . In a recent sequence of papers [11 , 12 , 13 , 21 ] , the Simplified Approach has been shown to be surprisingly accurate. In particular, its prediction for the ground-state energy asymptotically matches that for the Bose gas both at low and high densities [11 ] . In addition, its predictions for the condensate fraction, the momentum distribution and the 2-point correlation function agree with Bogolyubov’s theory at low densities [12 ] . Furthermore, numerical studies showed the ground-state energy prediction of the Simplified Approach is quantitatively accurate when compared with Monte-Carlo simulations at all densities as well as the qualitative behavior of the condensate fraction [13 ] .
The existing results of the Simplified Approach have been established for integrable soft-core potentials. In this note, we develop a theory for the Simplified Approach with a hard-core potential. We prove that the ground-state energy with a hard-core potential agrees asymptotically with the many-body Bose gas at low densities (up to Lee-Huang-Yang [LHY] order; see (3) ). In addition, we show that the condensate fraction is in agreement with Bogolyubov’s theory.
The derivation of the Simplified Approach from the many-body Bose gas is non-rigorous and relies on a truncation of a hierarchy of correlation functions in terms of the 2-point correlation function g ( x ) g(x) g ( x ) [24 , 21 ] . This approximation leads to a collection of equations of varying difficulties of complexity and accuracy. On one side of the spectrum the Complete Equation was numerically verified to be accurate as mentioned previously but is exceedingly difficult to study analytically [13 ] . On the other end of the spectrum is the Simple Equation which is only accurate at high and low densities, but is simple enough to be studied analytically. In this note, we will prove properties of the Simple Equation with a hard-core potential at low densities (since there is a critical packing fraction the high-density regime is inaccessible). For a numerical analysis of the solutions see [13 ] .
Definition 1 (Simple Equation). The Simple Equation of the ground-state Bose gas in R 3 \mathbb{R}^{3} R 3 v ( x ) ∈ L 1 ( R 3 ) ∩ L p ( R 3 ) v(x) \in L^{1}(\mathbb{R}^{3}) \cap L^{p}(\mathbb{R}^{3}) v ( x ) ∈ L 1 ( R 3 ) ∩ L p ( R 3 ) p > 3 2 p > \frac{3}{2} p > 2 3
is
( − Δ + 4 e + v ( ∣ x ∣ ) ) u ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e ρ u ∗ u ( ∣ x ∣ ) (1) \left(-\Delta+ 4e + v(|x|)\right)u(|x|) = v(|x|) + 2e\rho\, u * u(|x|) \tag*{(1)} ( − Δ + 4 e + v ( ∣ x ∣ ) ) u ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e ρ u ∗ u ( ∣ x ∣ ) (1) e = ρ 2 ∫ ( 1 − u ( ∣ x ∣ ) ) v ( ∣ x ∣ ) d x (2) e = \frac{\rho}{2}\int(1-u(|x|))v(|x|)\,dx \tag*{(2)} e = 2 ρ ∫ ( 1 − u ( ∣ x ∣ )) v ( ∣ x ∣ ) d x (2) where u u u e e e
[11 ] proved the existence of a solution satisfying 0 ≤ u ≤ 1 0 \le u \le1 0 ≤ u ≤ 1
In addition, the energy-per-particle, e e e
e = 2 π ρ a [ 1 + 128 15 π ρ a 3 + o ( ρ a 3 ) ] (3) e = 2\pi\rho a\left[1+\frac{128}{15\sqrt{\pi}}\sqrt{\rho a^3}+o(\sqrt{\rho a^3})\right] \tag*{(3)} e = 2 π ρ a [ 1 + 15 π 128 ρ a 3 + o ( ρ a 3 ) ] (3) where a a a v v v [32 , 16 , 3 ] .
To introduce the hard-core Simple Equation, we note that (1) is equivalent to
( − Δ + 4 e + v ( ∣ x ∣ ) ) u ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e ρ u ∗ u ( ∣ x ∣ ) , (4) \left(-\Delta+ 4e + v(|x|)\right)u(|x|) = v(|x|) + 2e\rho\,u * u(|x|), \tag*{(4)} ( − Δ + 4 e + v ( ∣ x ∣ ) ) u ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e ρ u ∗ u ( ∣ x ∣ ) , (4) ∫ u = 1 ρ (5) \int u = \frac{1}{\rho} \tag*{(5)} ∫ u = ρ 1 (5) Indeed, when the potential is integrable, integration of the first equation of (1) yields
2 e ρ ( ∫ u ) 2 − 4 e ∫ u + ∫ [ 1 − u ] v = 0 (6) 2e\rho\left(\int u\right)^2 - 4e\int u + \int[1-u]v = 0 \tag*{(6)} 2 e ρ ( ∫ u ) 2 − 4 e ∫ u + ∫ [ 1 − u ] v = 0 (6) which implies that
e = ρ 2 ∫ ( 1 − u ) v ⟺ ∫ u = 1 ρ . (7) e = \frac{\rho}{2}\int(1-u)v \Longleftrightarrow\int u = \frac{1}{\rho}. \tag*{(7)} e = 2 ρ ∫ ( 1 − u ) v ⟺ ∫ u = ρ 1 . (7) This re-formulation naturally suggests the generalization of the definition of the Simple Equation to the hard-core potential: since the wave-function vanishes inside the hard core we impose 1 − τ ( x ) = 0 1-\tau(x)=0 1 − τ ( x ) = 0 ∣ x ∣ ≤ 1 |x|\le1 ∣ x ∣ ≤ 1
Definition 2 (Hard-Core Simple Equation).
Δ τ ( ∣ x ∣ ) = 4 e τ ( ∣ x ∣ ) − 2 e ρ h c τ ∗ τ ( ∣ x ∣ ) , ∣ x ∣ > 1 , (8) \Delta\tau(|x|) = 4e\tau(|x|) - 2e\rho_{\mathrm{hc}}\tau* \tau(|x|), \qquad|x|>1, \tag*{(8)} Δ τ ( ∣ x ∣ ) = 4 e τ ( ∣ x ∣ ) − 2 e ρ hc τ ∗ τ ( ∣ x ∣ ) , ∣ x ∣ > 1 , (8) τ ( ∣ x ∣ ) = 1 , ∣ x ∣ ≤ 1 , (9) \tau(|x|) = 1, \qquad|x|\le1, \tag*{(9)} τ ( ∣ x ∣ ) = 1 , ∣ x ∣ ≤ 1 , (9) ∫ τ = 1 ρ h c . (10) \int\tau= \frac{1}{\rho_{\mathrm{hc}}}. \tag*{(10)} ∫ τ = ρ hc 1 . (10) We can recover a formula for the ground state energy from (10) by integrating (8) over { ∣ x ∣ > 1 } \{|x|>1\} { ∣ x ∣ > 1 }
e = 8 π 3 e ρ h c − 2 π ρ h c ∂ r τ ∣ 1 + − e ρ h c 2 ∫ B 1 ( 0 ) τ ∗ τ , (11) e = \frac{8\pi}{3}e\rho_{\mathrm{hc}} - 2\pi\rho_{\mathrm{hc}}\partial_r\tau\big|_{1+} - e\rho_{\mathrm{hc}}^2\int_{B_1(0)}\tau* \tau, \tag*{(11)} e = 3 8 π e ρ hc − 2 π ρ hc ∂ r τ 1 + − e ρ hc 2 ∫ B 1 ( 0 ) τ ∗ τ , (11) where B 1 ( 0 ) B_1(0) B 1 ( 0 ) 4 π 3 = ∣ B 1 ( 0 ) ∣ \frac{4\pi}{3}=|B_1(0)| 3 4 π = ∣ B 1 ( 0 ) ∣ 4 π 4\pi 4 π (11) , we get
e = − 2 π ρ h c ∂ r τ ∣ 1 + 1 − 8 π 3 ρ h c + ρ h c 2 ∫ B 1 ( 0 ) τ ∗ τ (12) e=\frac{-2\pi\rho_{\mathrm{hc}}\partial_r\tau\big|_{1+}}{1-\frac{8\pi}{3}\rho_{\mathrm{hc}}+\rho_{\mathrm{hc}}^2\int_{B_1(0)}\tau*\tau} \tag*{(12)} e = 1 − 3 8 π ρ hc + ρ hc 2 ∫ B 1 ( 0 ) τ ∗ τ − 2 π ρ hc ∂ r τ 1 + (12) Our first main result is the following.
Theorem 1 (LHY Formula). The hard-core Simple Equation (8) has a unique integrable solution 0 ≤ τ ≤ 1 0\leq\tau\leq1 0 ≤ τ ≤ 1 and, as ρ h c → 0 \rho_{\mathrm{hc}}\to0 ρ hc → 0
e = 2 π ρ h c a 0 ( 1 + 128 15 π ρ h c a 0 3 + o ( ρ h c a 0 3 ) ) (13) e=2\pi\rho_{\mathrm{hc}}a_0\left(1+\frac{128}{15\sqrt{\pi}}\sqrt{\rho_{\mathrm{hc}}a_0^3}+o\left(\sqrt{\rho_{\mathrm{hc}}a_0^3}\right)\right) \tag*{(13)} e = 2 π ρ hc a 0 ( 1 + 15 π 128 ρ hc a 0 3 + o ( ρ hc a 0 3 ) ) (13) where a 0 = 1 a_0=1 a 0 = 1 is the scattering length of the hard-sphere potential.
In other words we recover the LHY formula (3) for the hard-core Simple Equation. The LHY formula was proved for the hard-core many-body Bose gas by [17 , 5 , 6 ] and thus we show that the energy of the hard-core Simple Equation is asymptotically accurate in the low density limit.
Condensate Fraction
The Simplified Approach also makes predictions about quantities other than the ground state energy [12 , 21 ] ; here we focus on the condensate fraction. Denote by P j P_j P j j j j
P j ψ ( x 1 , … , x N ) = ∫ d x j V ψ ( x 1 , … , x N ) , (14) P_j\psi(x_1,\ldots,x_N)=\int\frac{dx_j}{V}\psi(x_1,\ldots,x_N), \tag*{(14)} P j ψ ( x 1 , … , x N ) = ∫ V d x j ψ ( x 1 , … , x N ) , (14) so that 1 N ∑ j = 1 N P j \frac{1}{N}\sum_{j=1}^{N}P_j N 1 ∑ j = 1 N P j
H N , μ : = − 1 2 ∑ i = 1 N Δ i + ∑ i < j v ( ∣ x i − x j ∣ ) − μ N ∑ j = 1 N P j (15) \mathcal{H}_{N,\mu}:=-\frac{1}{2}\sum_{i=1}^{N}\Delta_i+\sum_{i<j}v(|x_i-x_j|)-\frac{\mu}{N}\sum_{j=1}^{N}P_j \tag*{(15)} H N , μ := − 2 1 i = 1 ∑ N Δ i + i < j ∑ v ( ∣ x i − x j ∣ ) − N μ j = 1 ∑ N P j (15) and, denoting its ground state by ψ 0 , μ \psi_{0,\mu} ψ 0 , μ E 0 , μ \mathcal{E}_{0,\mu} E 0 , μ
⟨ ψ 0 , μ , 1 N ∑ j = 1 N P j ψ 0 , μ ⟩ = − ∂ μ E 0 , μ ∣ μ = 0 . (16) \left\langle\psi_{0,\mu},\frac{1}{N}\sum_{j=1}^{N}P_j\psi_{0,\mu}\right\rangle=-\left.\partial_\mu\mathcal{E}_{0,\mu}\right|_{\mu=0}. \tag*{(16)} ⟨ ψ 0 , μ , N 1 j = 1 ∑ N P j ψ 0 , μ ⟩ = − ∂ μ E 0 , μ ∣ μ = 0 . (16) The prediction of the Simplified Approach for the condensate fraction is obtained by deriving the Simplified Approach from the modified Hamiltonian (15) (following the same prescription as described above), computing the prediction for its ground state energy e μ e_\mu e μ μ \mu μ P j P_j P j [12 (1.15)] : the perturbation enters the Simple Equation only through the constant shift − Δ ↦ − Δ + 2 μ -\Delta\mapsto-\Delta+2\mu − Δ ↦ − Δ + 2 μ
( − Δ + 2 μ + 4 e μ + v ( ∣ x ∣ ) ) u μ ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e μ ρ u μ ∗ u μ ( ∣ x ∣ ) , (17) (-\Delta+2\mu+4e_\mu+v(|x|))u_\mu(|x|)=v(|x|)+2e_\mu\rho u_\mu*u_\mu(|x|), \tag*{(17)} ( − Δ + 2 μ + 4 e μ + v ( ∣ x ∣ )) u μ ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e μ ρ u μ ∗ u μ ( ∣ x ∣ ) , (17) e μ = ρ 2 ∫ ( 1 − u μ ( ∣ x ∣ ) ) v ( ∣ x ∣ ) d x (18) e_\mu=\frac{\rho}{2}\int(1-u_\mu(|x|))v(|x|)\,dx \tag*{(18)} e μ = 2 ρ ∫ ( 1 − u μ ( ∣ x ∣ )) v ( ∣ x ∣ ) d x (18) The modified hard-core Simple Equation is obtained analogously:
Δ τ μ ( ∣ x ∣ ) = ( 4 e μ + 2 μ ) τ μ ( ∣ x ∣ ) − 2 e μ ρ h c τ μ ∗ τ μ ( ∣ x ∣ ) ∣ x ∣ > 1 (19) \Delta\tau_\mu(|x|) = (4e_\mu+ 2\mu)\tau_\mu(|x|) - 2e_\mu\rho_{\mathrm{hc}} \tau_\mu* \tau_\mu(|x|) \qquad|x| > 1 \tag*{(19)} Δ τ μ ( ∣ x ∣ ) = ( 4 e μ + 2 μ ) τ μ ( ∣ x ∣ ) − 2 e μ ρ hc τ μ ∗ τ μ ( ∣ x ∣ ) ∣ x ∣ > 1 (19) τ μ ( ∣ x ∣ ) = 1 ∣ x ∣ ≤ 1 (20) \tau_\mu(|x|) = 1 \qquad|x| \le1 \tag*{(20)} τ μ ( ∣ x ∣ ) = 1 ∣ x ∣ ≤ 1 (20) ∫ τ μ = 1 ρ h c (21) \int\tau_\mu= \frac{1}{\rho_{\mathrm{hc}}} \tag*{(21)} ∫ τ μ = ρ hc 1 (21) This yields a prediction for the condensate fraction:
η : = − ∂ μ e μ = 1 + ρ 2 ∫ v ( x ) ∂ μ u μ ( x ) ∣ μ = 0 d x , (22) \eta:= -\partial_\mu e_\mu= 1 + \frac{\rho}{2}\int v(x)\partial_\mu u_\mu(x)\big|_{\mu=0}\,dx, \tag*{(22)} η := − ∂ μ e μ = 1 + 2 ρ ∫ v ( x ) ∂ μ u μ ( x ) μ = 0 d x , (22) which [12 ] proved to scale as
η = 1 − 8 ρ a 0 3 3 π + o ( ρ ) , (23) \eta= 1 - \frac{8\sqrt{\rho a_0^3}}{3\sqrt{\pi}} + o(\sqrt{\rho}), \tag*{(23)} η = 1 − 3 π 8 ρ a 0 3 + o ( ρ ) , (23) coinciding with Lee, Huang and Yang’s predictions [23 ] . Our next main result shows that the hard-core equation also possesses the same condensate fraction.
Theorem 2 (Condensate Fraction). The hard-core Simple Equation (8) , in the dilute limit ρ h c → 0 \rho_{\mathrm{hc}} \to0 ρ hc → 0 satisfies
η = 1 − 8 ρ h c a 0 3 3 π + o ( ρ h c ) , (24) \eta= 1 - \frac{8\sqrt{\rho_{\mathrm{hc}}a_0^3}}{3\sqrt{\pi}} + o(\sqrt{\rho_{\mathrm{hc}}}), \tag*{(24)} η = 1 − 3 π 8 ρ hc a 0 3 + o ( ρ hc ) , (24) where a 0 = 1 a_0 = 1 a 0 = 1 is the scattering length.
Proof strategy: limit of strong soft cores
The main tool in the proofs is to compute the energy and the condensate fraction of the hard-core Simple Equation (8) as limits of the energy and the condensate fraction for a soft-core Simple Equation. We will use the soft-core potential
v m ( ∣ x ∣ ) : = m 1 B 1 ( 0 ) ( x ) : = { m if x ∈ B 1 ( 0 ) : = { x : ∣ x ∣ ≤ 1 } 0 otherwise . (25) v_m(|x|) := m\mathbb{1}_{B_1(0)}(x) :=
\begin{cases}
m & \text{if } x \in B_1(0) := \{x : |x| \le1\} \\
0 & \text{otherwise}.
\end{cases} \tag*{(25)} v m ( ∣ x ∣ ) := m 1 B 1 ( 0 ) ( x ) := { m 0 if x ∈ B 1 ( 0 ) := { x : ∣ x ∣ ≤ 1 } otherwise . (25) The soft-core Simple Equation with potential v m v_m v m (1) –(2) )
( − Δ + 4 e + v m ( ∣ x ∣ ) ) u ( ∣ x ∣ ) = v m ( ∣ x ∣ ) + 2 e ρ ˉ m u ∗ u ( ∣ x ∣ ) , e = ρ ˉ m 2 ∫ ( 1 − u ( ∣ x ∣ ) ) v m ( ∣ x ∣ ) d x . (26) (-\Delta+ 4e + v_m(|x|))u(|x|) = v_m(|x|) + 2e\bar{\rho}_m u * u(|x|), \qquad e = \frac{\bar{\rho}_m}{2}\int(1-u(|x|))v_m(|x|)\,dx. \tag*{(26)} ( − Δ + 4 e + v m ( ∣ x ∣ )) u ( ∣ x ∣ ) = v m ( ∣ x ∣ ) + 2 e ρ ˉ m u ∗ u ( ∣ x ∣ ) , e = 2 ρ ˉ m ∫ ( 1 − u ( ∣ x ∣ )) v m ( ∣ x ∣ ) d x . (26) We will prove that the energy and the condensate fraction for the hard-core equation are the limit m → ∞ m \to\infty m → ∞ (8) was introduced above on its own terms, not as a limit of soft-core equations: the two are independently defined objects, each with its own existence and uniqueness theory (see Lemmas 7 and 8 below for the hard-core case).
However, for technical reasons, we will have to change the point of view slightly: instead of the setting described above in which ρ h c \rho_{\mathrm{hc}} ρ hc e e e (8) –(10) , we will fix e e e ρ h c \rho_{\mathrm{hc}} ρ hc (8) –(10) . This change of point of view was also needed in [11 ] . The two points of view are conjectured to be equivalent [11 ] , but proving this is the case boils down to proving that e ↦ ρ h c ( e ) e \mapsto\rho_{\mathrm{hc}}(e) e ↦ ρ hc ( e ) e e e [12 Theorem 1.3] , but the small-e e e m → ∞ m \to\infty m → ∞ e e e e e e § 7 : for e ≤ e 0 e \le e_0 e ≤ e 0 c e ≤ ρ h c ( e ) ce \le\rho_{\mathrm{hc}}(e) ce ≤ ρ hc ( e ) ∣ e − 2 π ρ h c ( e ) ∣ ≤ C e 3 / 2 |e - 2\pi\rho_{\mathrm{hc}}(e)| \le Ce^{3/2} ∣ e − 2 π ρ hc ( e ) ∣ ≤ C e 3/2 e ↦ ρ h c ( e ) e \mapsto\rho_{\mathrm{hc}}(e) e ↦ ρ hc ( e ) e → 0 e \to0 e → 0 ρ h c → 0 \rho_{\mathrm{hc}} \to0 ρ hc → 0 e = 2 π ρ h c ( 1 + o ( 1 ) ) e = 2\pi\rho_{\mathrm{hc}}(1 + o(1)) e = 2 π ρ hc ( 1 + o ( 1 )) e e e ρ h c \rho_{\mathrm{hc}} ρ hc e ↦ ρ h c ( e ) e \mapsto\rho_{\mathrm{hc}}(e) e ↦ ρ hc ( e )
Theorem 3 (Condensate Fraction Expectation). Consider e > 0 e > 0 e > 0
(i) (Energy.) As m → ∞ m \to\infty m → ∞
ρ ‾ m ⟶ ρ h c , ∣ ρ ‾ m − ρ h c ∣ = O ( m − 1 / 2 ) . (27) \overline{\rho}_m \longrightarrow\rho_{\mathrm{hc}}, \qquad\left|\overline{\rho}_m - \rho_{\mathrm{hc}}\right| = O\left(m^{-1/2}\right). \tag*{(27)} ρ m ⟶ ρ hc , ∣ ρ m − ρ hc ∣ = O ( m − 1/2 ) . (27) Moreover, each soft-core equation satisfies the low-density expansion (3) , with a 0 a_0 a 0 a m a_m a m v m v_m v m m m m
(ii) (Condensate fraction.) Let e μ , m e_{\mu,m} e μ , m e μ , h c e_{\mu,\mathrm{hc}} e μ , hc (18) , with v = v m v = v_m v = v m ρ = ρ ‾ m \rho= \overline{\rho}_m ρ = ρ m (19) . Then
lim m → ∞ ∂ μ e μ , m ∣ μ = 0 = ∂ μ e μ , h c ∣ μ = 0 , (28) \lim_{m\to\infty} \left.\partial_\mu e_{\mu,m}\right|_{\mu=0} = \left.\partial_\mu e_{\mu,\mathrm{hc}}\right|_{\mu=0}, \tag*{(28)} m → ∞ lim ∂ μ e μ , m ∣ μ = 0 = ∂ μ e μ , hc ∣ μ = 0 , (28) that is, the soft-core condensate fractions converge to the hard-core condensate fraction. Moreover, each soft-core equation satisfies the condensate-fraction expansion (24) , with a 0 a_0 a 0 a m a_m a m m m m
Theorem 3 implies Theorems 1 and 2. For each fixed m m m [11 ] applied to v m v_m v m
ρ ‾ m = e 2 π a m ( 1 − 64 2 a m 15 π e + o ( e ) ) , (29) \overline{\rho}_m = \frac{e}{2\pi a_m}\left(1 - \frac{64\sqrt{2}a_m}{15\pi}\sqrt{e} + o(\sqrt{e})\right), \tag*{(29)} ρ m = 2 π a m e ( 1 − 15 π 64 2 a m e + o ( e ) ) , (29) where a m = 1 − tanh ( m ) m a_m = 1 - \frac{\tanh(\sqrt{m})}{\sqrt{m}} a m = 1 − m t a n h ( m ) v m v_m v m o ( e ) o(\sqrt{e}) o ( e ) uniform in m m m : it is bounded by λ ( e ) e \lambda(e)\sqrt{e} λ ( e ) e λ ( e ) → 0 \lambda(e) \to0 λ ( e ) → 0 m m m m → ∞ m \to\infty m → ∞ (27) and a m → 1 a_m \to1 a m → 1
ρ h c = e 2 π ( 1 − 64 2 15 π e + o ( e ) ) , e → 0 , \rho_{\mathrm{hc}} = \frac{e}{2\pi}\left(1 - \frac{64\sqrt{2}}{15\pi}\sqrt{e} + o(\sqrt{e})\right), \qquad e \to0, ρ hc = 2 π e ( 1 − 15 π 64 2 e + o ( e ) ) , e → 0 , which inverts, via e = 2 π ρ h c ( 1 + o ( 1 ) ) \sqrt{e} = \sqrt{2\pi\rho_{\mathrm{hc}}}(1 + o(1)) e = 2 π ρ hc ( 1 + o ( 1 )) (13) . Likewise the expansion of η m \eta_m η m § 8 is uniform in m m m m → ∞ m \to\infty m → ∞ (28) yields Theorem 2 (again via e = 2 π ρ h c ( 1 + o ( 1 ) ) e = 2\pi\rho_{\mathrm{hc}}(1 + o(1)) e = 2 π ρ hc ( 1 + o ( 1 )) m m m ∫ v m \int v_m ∫ v m m → ∞ m \to\infty m → ∞ e e e m → ∞ m \to\infty m → ∞ v m v_m v m ( 1 − ϕ m ) v m (1 - \phi_m)v_m ( 1 − ϕ m ) v m ϕ m \phi_m ϕ m v m v_m v m § 7 — whose total mass is 4 π a m ≤ 4 π 4\pi a_m \le4\pi 4 π a m ≤ 4 π ∣ S m ∣ ≤ 1 |S_m| \le1 ∣ S m ∣ ≤ 1 (68) . Finally, the rate in (27) extends uniformly in μ \mu μ (28) , carried out for the condensate fraction in Theorem 26. The convergence (27) itself is obtained from the L 1 L^1 L 1 ∫ ∣ σ m − τ ∣ → 0 \int|\sigma_m - \tau| \to0 ∫ ∣ σ m − τ ∣ → 0 ∣ ρ ‾ m − ρ h c ∣ = ρ h c ρ ‾ m ∣ ∫ ( σ m − τ ) ∣ |\overline{\rho}_m - \rho_{\mathrm{hc}}| = \rho_{\mathrm{hc}}\overline{\rho}_m\left|\int(\sigma_m - \tau)\right| ∣ ρ m − ρ hc ∣ = ρ hc ρ m ∫ ( σ m − τ )
We will now discuss the strategy to prove Theorem 3 . The Simple Equation, hard-core or not, is extremely difficult to tackle directly due to the convolutional non-linearity. For a given potential, v ( x ) v(x) v ( x ) [11 , 12 ] was to use the following iteration scheme (we will introduce other iteration schemes, and to differentiate them, we will use different letters for the iterates; the soft-core iterates from [11 , 12 ] will be denoted by u n u_n u n
( − Δ + 4 e + v ( ∣ x ∣ ) ) u n ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e ρ n − 1 u n − 1 ∗ u n − 1 ( ∣ x ∣ ) (30) (-\Delta+4e+v(|x|))u_n(|x|)=v(|x|)+2e\rho_{n-1}u_{n-1}*u_{n-1}(|x|) \tag*{(30)} ( − Δ + 4 e + v ( ∣ x ∣ )) u n ( ∣ x ∣ ) = v ( ∣ x ∣ ) + 2 e ρ n − 1 u n − 1 ∗ u n − 1 ( ∣ x ∣ ) (30) ρ n = 2 e ∫ ( 1 − u n ) v (31) \rho_n=\frac{2e}{\int(1-u_n)v} \tag*{(31)} ρ n = ∫ ( 1 − u n ) v 2 e (31) with u 0 = 0 u_0=0 u 0 = 0 u n u_n u n u u u (1) for potential v v v [11 Theorem 1.3] . This iteration, for the soft-core potential v m ≡ m 1 B 1 ( 0 ) v_m\equiv m\mathbf{1}_{B_1(0)} v m ≡ m 1 B 1 ( 0 )
( − Δ + 4 e + v m ) σ m , n = v m + 2 e ρ ˉ m , n − 1 σ m , n − 1 ∗ σ m , n − 1 , ρ ˉ m , n = 2 e m ∫ B 1 ( 0 ) ( 1 − σ m , n ) . (32) (-\Delta+4e+v_m)\sigma_{m,n}=v_m+2e\bar{\rho}_{m,n-1}\sigma_{m,n-1}*\sigma_{m,n-1},\qquad\bar{\rho}_{m,n}=\frac{2e}{m\int_{B_1(0)}(1-\sigma_{m,n})}. \tag*{(32)} ( − Δ + 4 e + v m ) σ m , n = v m + 2 e ρ ˉ m , n − 1 σ m , n − 1 ∗ σ m , n − 1 , ρ ˉ m , n = m ∫ B 1 ( 0 ) ( 1 − σ m , n ) 2 e . (32) with σ m , 0 = 0 \sigma_{m,0}=0 σ m , 0 = 0 [11 Theorem 1.3] , lim n → ∞ σ m , n ≡ σ m \lim_{n\to\infty}\sigma_{m,n}\equiv\sigma_m lim n → ∞ σ m , n ≡ σ m (26) ; the properties of these iterates that we use are collected in Lemma 13.
We will use the following iteration for the hard-core equation which follows from (8) and (12) :
Δ τ n = 4 e τ n − 2 e ρ n − 1 τ n − 1 ∗ τ n − 1 , ∣ x ∣ > 1 , (33) \Delta\tau_n=4e\tau_n-2e\rho_{n-1}\tau_{n-1}*\tau_{n-1},\qquad|x|>1, \tag*{(33)} Δ τ n = 4 e τ n − 2 e ρ n − 1 τ n − 1 ∗ τ n − 1 , ∣ x ∣ > 1 , (33) τ n = 1 , ∣ x ∣ ≤ 1 , (34) \tau_n=1,\qquad|x|\le1, \tag*{(34)} τ n = 1 , ∣ x ∣ ≤ 1 , (34) with
ρ n = 2 e 2 e [ 8 π 3 − ρ n − 1 ∫ B 1 ( 0 ) τ n − 1 ∗ τ n − 1 ] − 4 π ∂ r τ n ∣ 1 + . (35) \rho_n=\frac{2e}{2e\left[\frac{8\pi}{3}-\rho_{n-1}\int_{B_1(0)}\tau_{n-1}*\tau_{n-1}\right]-4\pi\partial_r\tau_n\big|_{1+}}. \tag*{(35)} ρ n = 2 e [ 3 8 π − ρ n − 1 ∫ B 1 ( 0 ) τ n − 1 ∗ τ n − 1 ] − 4 π ∂ r τ n 1 + 2 e . (35) We initialize the iteration with Δ τ 1 = 4 e τ 1 \Delta\tau_1=4e\tau_1 Δ τ 1 = 4 e τ 1 ∣ x ∣ > 1 |x|>1 ∣ x ∣ > 1 τ 1 = 1 \tau_1=1 τ 1 = 1 B 1 ( 0 ) B_1(0) B 1 ( 0 )
τ 1 ( r ) = { 1 r ≤ 1 e − 2 e ( r − 1 ) / r r > 1 , ρ 1 = 1 2 ∫ τ 1 = 2 e 4 π ( 1 + 2 e ) + 16 π 3 e . (36) \tau_1(r)=
\begin{cases}
1 & r\le1\\
e^{-2\sqrt{e}(r-1)/r} & r>1
\end{cases},
\qquad
\rho_1=\frac{1}{2\int\tau_1}=\frac{2e}{4\pi(1+2\sqrt{e})+\frac{16\pi}{3}e}. \tag*{(36)} τ 1 ( r ) = { 1 e − 2 e ( r − 1 ) / r r ≤ 1 r > 1 , ρ 1 = 2 ∫ τ 1 1 = 4 π ( 1 + 2 e ) + 3 16 π e 2 e . (36) To show that τ n \tau_n τ n τ n \tau_n τ n ρ n \rho_n ρ n n n n Lemmas 6 and 7 ; uniqueness of the limit is Lemma 8. The monotonicity and comparison arguments throughout rest on the maximum principle in the form of Lemma 4 and Corollary 5.
The main idea is then to compare τ n \tau_n τ n σ m , n \sigma_{m,n} σ m , n n n n σ m , n → τ n \sigma_{m,n} \to\tau_n σ m , n → τ n m → ∞ m \to\infty m → ∞ e e e m m m ρ ˉ m \bar{\rho}_m ρ ˉ m B 1 ( 0 ) B_1(0) B 1 ( 0 ) σ m \sigma_m σ m m m m 1 1 1 m → ∞ m \to\infty m → ∞ ρ ˉ m , n σ m , n ∗ σ m , n \bar{\rho}_{m,n}\sigma_{m,n} * \sigma_{m,n} ρ ˉ m , n σ m , n ∗ σ m , n ρ ˉ m − 1 , n σ m − 1 , n ∗ σ m − 1 , n \bar{\rho}_{m-1,n}\sigma_{m-1,n} * \sigma_{m-1,n} ρ ˉ m − 1 , n σ m − 1 , n ∗ σ m − 1 , n σ m , n \sigma_{m,n} σ m , n ρ ˉ m , n \bar{\rho}_{m,n} ρ ˉ m , n
Figure 1.
In order to work around this impasse, we will avoid the competition between inverse monotonicities, by introducing two more iterations.
The type γ \gamma γ γ n \gamma_n γ n γ 0 = 0 \gamma_0 = 0 γ 0 = 0 n ≥ 1 n \ge1 n ≥ 1
( − Δ + 4 e + v n ) γ n = v n + 2 e ρ n − 1 γ n − 1 ∗ γ n − 1 , (37) (-\Delta+ 4e + v_n)\gamma_n = v_n + 2e\rho_{n-1}\gamma_{n-1} * \gamma_{n-1}, \tag*{(37)} ( − Δ + 4 e + v n ) γ n = v n + 2 e ρ n − 1 γ n − 1 ∗ γ n − 1 , (37) where v n = n 1 B 1 ( 0 ) v_n = n1_{B_1(0)} v n = n 1 B 1 ( 0 ) ρ n − 1 \rho_{n-1} ρ n − 1 ρ n \rho_n ρ n
The type ω \omega ω ω n \omega_n ω n ω 0 = 0 \omega_0 = 0 ω 0 = 0
( − Δ + 4 e + v n ) ω n = v n + 2 e ρ h c ω n − 1 ∗ ω n − 1 . (38) (-\Delta+ 4e + v_n)\omega_n = v_n + 2e\rho_{\mathrm{hc}}\omega_{n-1} * \omega_{n-1}. \tag*{(38)} ( − Δ + 4 e + v n ) ω n = v n + 2 e ρ hc ω n − 1 ∗ ω n − 1 . (38) where ρ h c \rho_{\mathrm{hc}} ρ hc γ \gamma γ ρ n − 1 \rho_{n-1} ρ n − 1 ρ h c \rho_{\mathrm{hc}} ρ hc
Figure 2.
Type γ \gamma γ ρ n τ n ∗ τ n \rho_n\tau_n * \tau_n ρ n τ n ∗ τ n ρ n γ n ∗ γ n \rho_n\gamma_n * \gamma_n ρ n γ n ∗ γ n τ n ≥ γ n \tau_n \ge\gamma_n τ n ≥ γ n γ n → L 1 τ \gamma_n \xrightarrow{L^1} \tau γ n L 1 τ γ n ≤ ω n ≤ τ \gamma_n \le\omega_n \le\tau γ n ≤ ω n ≤ τ ω n \omega_n ω n τ n \tau_n τ n ρ h c ≥ ρ n ∀ n \rho_{\mathrm{hc}} \ge\rho_n\ \forall n ρ hc ≥ ρ n ∀ n ρ h c ω n ∗ ω n \rho_{\mathrm{hc}}\omega_n * \omega_n ρ hc ω n ∗ ω n ρ n γ n ∗ γ n \rho_n\gamma_n * \gamma_n ρ n γ n ∗ γ n ρ h c ≥ ρ n \rho_{\mathrm{hc}} \ge\rho_n ρ hc ≥ ρ n ω n → L 1 τ \omega_n \xrightarrow{L^1} \tau ω n L 1 τ σ n , n \sigma_{n,n} σ n , n ω n \omega_n ω n σ n → L 1 τ \sigma_n \xrightarrow{L^1} \tau σ n L 1 τ σ n , n ≤ σ n \sigma_{n,n} \le\sigma_n σ n , n ≤ σ n L 1 L^1 L 1 ρ ˉ m \bar{\rho}_m ρ ˉ m ρ h c \rho_{\mathrm{hc}} ρ hc
Iterative Construction of the Hard-Core Solution
Before constructing the iterative solution scheme to the hard-core, we will introduce two lemmas will be used repeatedly in the following.
Lemma 4 (Comparison principle). Let D ⊆ R 3 D \subseteq\mathbb{R}^3 D ⊆ R 3 0 ≤ W ∈ L ∞ ( D ) 0 \le W \in L^\infty(D) 0 ≤ W ∈ L ∞ ( D ) w w w D ‾ \overline{D} D
( − Δ + W ) w ≥ 0 on D (39) (-\Delta+ W)w \ge0 \quad\text{on } D \tag*{(39)} ( − Δ + W ) w ≥ 0 on D (39) in the sense of distributions. If w ≥ 0 w \ge0 w ≥ 0 ∂ D ≡ D ‾ ∖ D \partial D \equiv\overline{D} \setminus D ∂ D ≡ D ∖ D lim inf ∣ x ∣ → ∞ w ( x ) ≥ 0 \liminf_{|x|\to\infty} w(x) \ge0 lim inf ∣ x ∣ → ∞ w ( x ) ≥ 0 D D D w ≥ 0 w \ge0 w ≥ 0 D ‾ \overline{D} D
Proof. Fix ε > 0 \varepsilon> 0 ε > 0 Σ ε = { x ∈ D : w ( x ) < − ε } \Sigma_\varepsilon= \{x \in D : w(x) < -\varepsilon\} Σ ε = { x ∈ D : w ( x ) < − ε } lim inf ∣ x ∣ → ∞ w ( x ) ≥ 0 \liminf_{|x|\to\infty} w(x) \ge0 lim inf ∣ x ∣ → ∞ w ( x ) ≥ 0 Σ ε \Sigma_\varepsilon Σ ε w ≥ 0 > − ε w \ge0 > -\varepsilon w ≥ 0 > − ε ∂ D \partial D ∂ D Σ ε ‾ ∩ ∂ D = ∅ \overline{\Sigma_\varepsilon} \cap\partial D = \varnothing Σ ε ∩ ∂ D = ∅ Σ ε ‾ \overline{\Sigma_\varepsilon} Σ ε D D D w = − ε w=-\varepsilon w = − ε ∂ Σ ε \partial\Sigma_\varepsilon ∂ Σ ε Σ ε ≠ ∅ \Sigma_\varepsilon\ne\varnothing Σ ε = ∅ Σ ε \Sigma_\varepsilon Σ ε w < 0 w<0 w < 0 W ≥ 0 W\ge0 W ≥ 0 Δ w ≤ W w ≤ 0 \Delta w\le Ww\le0 Δ w ≤ W w ≤ 0 w w w Σ ε \Sigma_\varepsilon Σ ε ∂ Σ ε \partial\Sigma_\varepsilon ∂ Σ ε w = − ε w=-\varepsilon w = − ε Σ ε \Sigma_\varepsilon Σ ε Σ ε = ∅ \Sigma_\varepsilon=\varnothing Σ ε = ∅ ε > 0 \varepsilon>0 ε > 0
Corollary 5 (Constant barriers). Let D ⊂ R 3 D\subset\mathbb{R}^{3} D ⊂ R 3 c > 0 c>0 c > 0 0 ≤ W ∈ L ∞ ( D ) 0\le W\in L^\infty(D) 0 ≤ W ∈ L ∞ ( D ) w w w D ‾ \overline{D} D
0 ≤ ( − Δ + c + W ) w ≤ F on D 0\le(-\Delta+c+W)w\le F\quad\text{on }D 0 ≤ ( − Δ + c + W ) w ≤ F on D in the sense of distributions, for a constant F ≥ 0 F\ge0 F ≥ 0 D D D
min { 0 , min ∂ D w } ≤ w ≤ max { 0 , max ∂ D w , F c } . \min\left\{0,\min_{\partial D}w\right\}\le w\le\max\left\{0,\max_{\partial D}w,\frac{F}{c}\right\}. min { 0 , ∂ D min w } ≤ w ≤ max { 0 , ∂ D max w , c F } . Proof. Set α = min { 0 , min ∂ D w } \alpha=\min\{0,\min_{\partial D}w\} α = min { 0 , min ∂ D w } M = max { 0 , max ∂ D w , F / c } M=\max\{0,\max_{\partial D}w,F/c\} M = max { 0 , max ∂ D w , F / c } w − α w-\alpha w − α α ≤ 0 \alpha\le0 α ≤ 0 ( − Δ + c + W ) ( w − α ) ≥ − ( c + W ) α ≥ 0 (-\Delta+c+W)(w-\alpha)\ge-(c+W)\alpha\ge0 ( − Δ + c + W ) ( w − α ) ≥ − ( c + W ) α ≥ 0 w − α ≥ 0 w-\alpha\ge0 w − α ≥ 0 ∂ D \partial D ∂ D w ≥ α w\ge\alpha w ≥ α M − w M-w M − w M ≥ F / c M\ge F/c M ≥ F / c W ≥ 0 W\ge0 W ≥ 0 ( − Δ + c + W ) ( M − w ) ≥ c M − F ≥ 0 (-\Delta+c+W)(M-w)\ge cM-F\ge0 ( − Δ + c + W ) ( M − w ) ≥ c M − F ≥ 0 M − w ≥ 0 M-w\ge0 M − w ≥ 0 ∂ D \partial D ∂ D
With this lemma and corollary proven, we can now prove several simple but useful properties of the iterative solutions to the hard-core Simple Equation. For n ≥ 2 n\ge2 n ≥ 2 ( − Δ + 4 e ) (-\Delta+4e) ( − Δ + 4 e ) R 3 \mathbb{R}^{3} R 3 Y 4 e ( x ) = e − 2 e ∣ x ∣ / ( 4 π ∣ x ∣ ) Y_{4e}(x)=e^{-2\sqrt{e}|x|}/(4\pi|x|) Y 4 e ( x ) = e − 2 e ∣ x ∣ / ( 4 π ∣ x ∣ ) (33) directly, we find
τ n ( r ) = a n e − 2 e ( r − 1 ) r + e ρ n − 1 2 r ∫ 1 ∞ s [ τ n − 1 ∗ τ n − 1 ] ( s ) e − 2 e ∣ s − r ∣ d s , r > 1 , (40) \tau_n(r)=a_n\frac{e^{-2\sqrt{e}(r-1)}}{r}+\frac{\sqrt{e}\rho_{n-1}}{2r}\int_1^\infty s[\tau_{n-1}*\tau_{n-1}](s)e^{-2\sqrt{e}|s-r|}\,ds,\qquad r>1, \tag*{(40)} τ n ( r ) = a n r e − 2 e ( r − 1 ) + 2 r e ρ n − 1 ∫ 1 ∞ s [ τ n − 1 ∗ τ n − 1 ] ( s ) e − 2 e ∣ s − r ∣ d s , r > 1 , (40) with
a n = 1 − e ρ n − 1 2 ∫ 1 ∞ s [ τ n − 1 ∗ τ n − 1 ] ( s ) e − 2 e ( s − 1 ) d s . (41) a_n=1-\frac{\sqrt{e}\rho_{n-1}}{2}\int_1^\infty s[\tau_{n-1}*\tau_{n-1}](s)e^{-2\sqrt{e}(s-1)}\,ds. \tag*{(41)} a n = 1 − 2 e ρ n − 1 ∫ 1 ∞ s [ τ n − 1 ∗ τ n − 1 ] ( s ) e − 2 e ( s − 1 ) d s . (41) Therefore each τ n \tau_n τ n R 3 \mathbb{R}^{3} R 3 B 1 ( 0 ) B_1(0) B 1 ( 0 ) B 1 c B_1^c B 1 c r = 1 + r=1^+ r = 1 +
∂ r τ n ∣ 1 + = − 1 − 2 e − 4 e a n . (42) \left.\partial_r\tau_n\right|_{1^+}=-1-2\sqrt{e}-4\sqrt{e}a_n. \tag*{(42)} ∂ r τ n ∣ 1 + = − 1 − 2 e − 4 e a n . (42) Lemma 6 (Properties of hard-core iterates). For all n ≥ 1 n\ge1 n ≥ 1
(i) 0 ≤ τ n ≤ 1 0\le\tau_n\le1 0 ≤ τ n ≤ 1
(ii) ∫ τ n ≤ 1 / ρ n \int\tau_n\le1/\rho_n ∫ τ n ≤ 1/ ρ n
(iii) τ n ≤ τ n + 1 \tau_n\le\tau_{n+1} τ n ≤ τ n + 1 ρ n ≤ ρ n + 1 \rho_n\le\rho_{n+1} ρ n ≤ ρ n + 1
Proof. We proceed by induction. For n = 1 n=1 n = 1 τ 1 ≤ τ 2 \tau_1\le\tau_2 τ 1 ≤ τ 2 ρ 2 ≥ ρ 1 \rho_2\ge\rho_1 ρ 2 ≥ ρ 1
Assume the statements hold for all indices up to n − 1 n-1 n − 1
Non-negativity. On { ∣ x ∣ > 1 } \{|x|>1\} { ∣ x ∣ > 1 } ( − Δ + 4 e ) τ n = 2 e ρ n − 1 τ n − 1 ∗ τ n − 1 ≥ 0 (-\Delta+4e)\tau_n=2e\rho_{n-1}\tau_{n-1}*\tau_{n-1}\ge0 ( − Δ + 4 e ) τ n = 2 e ρ n − 1 τ n − 1 ∗ τ n − 1 ≥ 0 τ n = 1 \tau_n=1 τ n = 1 ∂ B 1 \partial B_1 ∂ B 1 τ n → 0 \tau_n\to0 τ n → 0 D = { ∣ x ∣ > 1 } D=\{|x|>1\} D = { ∣ x ∣ > 1 } c = 4 e c=4e c = 4 e W = 0 W=0 W = 0 τ n ≥ 0 \tau_n\ge0 τ n ≥ 0
Upper bound τ n ≤ 1 \tau_n\le1 τ n ≤ 1 τ n − 1 ≤ 1 \tau_{n-1} \le1 τ n − 1 ≤ 1 ρ n − 1 ∫ τ n − 1 ≤ 1 \rho_{n-1}\int\tau_{n-1} \le1 ρ n − 1 ∫ τ n − 1 ≤ 1 1 − τ n = 0 1-\tau_n=0 1 − τ n = 0 ∂ B 1 \partial B_1 ∂ B 1 1 − τ n → 1 1-\tau_n\to1 1 − τ n → 1 τ n ≤ 1 \tau_n\le1 τ n ≤ 1
Monotonicity. For τ n ≤ τ n + 1 \tau_n\le\tau_{n+1} τ n ≤ τ n + 1 ∣ x ∣ > 1 |x|>1 ∣ x ∣ > 1
( − Δ + 4 e ) ( τ n + 1 − τ n ) = 2 e [ ρ n τ n ∗ τ n − ρ n − 1 τ n − 1 ∗ τ n − 1 ] ≥ 0 , (-\Delta+4e)(\tau_{n+1}-\tau_n)=2e[\rho_n\tau_n * \tau_n-\rho_{n-1}\tau_{n-1} * \tau_{n-1}]\ge0, ( − Δ + 4 e ) ( τ n + 1 − τ n ) = 2 e [ ρ n τ n ∗ τ n − ρ n − 1 τ n − 1 ∗ τ n − 1 ] ≥ 0 , by the induction hypothesis ρ n ≥ ρ n − 1 \rho_n\ge\rho_{n-1} ρ n ≥ ρ n − 1 τ n ≥ τ n − 1 \tau_n\ge\tau_{n-1} τ n ≥ τ n − 1 τ n + 1 − τ n = 0 \tau_{n+1}-\tau_n=0 τ n + 1 − τ n = 0 ∂ B 1 \partial B_1 ∂ B 1 τ n + 1 ≥ τ n \tau_{n+1}\ge\tau_n τ n + 1 ≥ τ n ∂ r τ n + 1 ∣ 1 + ≥ ∂ r τ n ∣ 1 + \partial_r\tau_{n+1}|_{1+}\ge\partial_r\tau_n|_{1+} ∂ r τ n + 1 ∣ 1 + ≥ ∂ r τ n ∣ 1 + (35) does not increase, so ρ n + 1 ≥ ρ n \rho_{n+1}\ge\rho_n ρ n + 1 ≥ ρ n
Integral bound. Integrating (33) over { ∣ x ∣ > 1 } \{|x|>1\} { ∣ x ∣ > 1 } ρ n \rho_n ρ n
2 ∫ τ n = ρ n − 1 ( ∫ τ n − 1 ) 2 + 1 ρ n ≤ ∫ τ n − 1 + 1 ρ n ≤ ∫ τ n + 1 ρ n , (43) 2\int\tau_n=\rho_{n-1}\left(\int\tau_{n-1}\right)^2+\frac{1}{\rho_n}\le\int\tau_{n-1}+\frac{1}{\rho_n}\le\int\tau_n+\frac{1}{\rho_n}, \tag*{(43)} 2 ∫ τ n = ρ n − 1 ( ∫ τ n − 1 ) 2 + ρ n 1 ≤ ∫ τ n − 1 + ρ n 1 ≤ ∫ τ n + ρ n 1 , (43) hence ∫ τ n ≤ 1 / ρ n \int\tau_n\le1/\rho_n ∫ τ n ≤ 1/ ρ n
Lemma 7 (Convergence). The limits τ = lim n → ∞ τ n \tau=\lim_{n\to\infty}\tau_n τ = lim n → ∞ τ n and ρ h c = lim n → ∞ ρ n \rho_{\mathrm{hc}}=\lim_{n\to\infty}\rho_n ρ hc = lim n → ∞ ρ n exist, satisfy 0 ≤ τ ≤ 1 0\le\tau\le1 0 ≤ τ ≤ 1 ρ h c ∫ τ = 1 \rho_{\mathrm{hc}}\int\tau=1 ρ hc ∫ τ = 1 and ∥ τ n − τ ∥ L 1 → 0 \|\tau_n-\tau\|_{L^1}\to0 ∥ τ n − τ ∥ L 1 → 0
Proof. τ n \tau_n τ n τ n → τ \tau_n\to\tau τ n → τ 0 ≤ τ n ≤ τ 0\le\tau_n\le\tau 0 ≤ τ n ≤ τ ∫ τ n ≤ 1 / ρ 1 < ∞ \int\tau_n\le1/\rho_1<\infty ∫ τ n ≤ 1/ ρ 1 < ∞ τ ∈ L 1 \tau\in L^1 τ ∈ L 1 ∥ τ − τ n ∥ L 1 = ∫ ( τ − τ n ) → 0 \|\tau-\tau_n\|_{L^1}=\int(\tau-\tau_n)\to0 ∥ τ − τ n ∥ L 1 = ∫ ( τ − τ n ) → 0 ρ n \rho_n ρ n ρ n ≤ 1 / ∫ τ 1 \rho_n\le1/\int\tau_1 ρ n ≤ 1/ ∫ τ 1 ρ n → ρ h c \rho_n\to\rho_{\mathrm{hc}} ρ n → ρ hc ρ n ∫ τ n = 1 \rho_n\int\tau_n=1 ρ n ∫ τ n = 1 (43) and the monotone limits) to the limit gives ρ h c ∫ τ = 1 \rho_{\mathrm{hc}}\int\tau=1 ρ hc ∫ τ = 1 τ \tau τ (8) follows by taking n → ∞ n\to\infty n → ∞ (33) .
Lemma 8 (Uniqueness). ( τ , ρ h c ) (\tau,\rho_{\mathrm{hc}}) ( τ , ρ hc ) is the unique solution of (8) with 0 ≤ τ ≤ 1 0\le\tau\le1 0 ≤ τ ≤ 1 and τ ∈ L 1 \tau\in L^1 τ ∈ L 1
Proof. Let τ ′ \tau' τ ′ ρ ′ \rho' ρ ′ τ ′ ≥ τ n \tau'\ge\tau_n τ ′ ≥ τ n n n n τ ′ ≥ τ n − 1 \tau'\ge\tau_{n-1} τ ′ ≥ τ n − 1 Υ = { ∣ x ∣ > 1 : τ ′ ( x ) < τ n ( x ) } ≠ ∅ \Upsilon=\{|x|>1:\tau'(x)<\tau_n(x)\}\ne\varnothing Υ = { ∣ x ∣ > 1 : τ ′ ( x ) < τ n ( x )} = ∅ Υ \Upsilon Υ
Δ ( τ ′ − τ n ) = 4 e ( τ ′ − τ n ) − 2 e [ ρ ′ τ ′ ∗ τ ′ − ρ n − 1 τ n − 1 ∗ τ n − 1 ] < 0 , \Delta(\tau'-\tau_n)=4e(\tau'-\tau_n)-2e[\rho'\tau'*\tau'-\rho_{n-1}\tau_{n-1}*\tau_{n-1}]<0, Δ ( τ ′ − τ n ) = 4 e ( τ ′ − τ n ) − 2 e [ ρ ′ τ ′ ∗ τ ′ − ρ n − 1 τ n − 1 ∗ τ n − 1 ] < 0 , so τ ′ − τ n \tau'-\tau_n τ ′ − τ n Υ \Upsilon Υ ∂ Υ \partial\Upsilon ∂ Υ τ ′ = τ n = 1 \tau'=\tau_n=1 τ ′ = τ n = 1 ∂ B 1 \partial B_1 ∂ B 1 τ ′ ≥ τ n \tau'\ge\tau_n τ ′ ≥ τ n n n n τ ′ ≥ τ \tau'\ge\tau τ ′ ≥ τ ρ ′ ≥ ρ h c \rho'\ge\rho_{\mathrm{hc}} ρ ′ ≥ ρ hc ∫ τ ′ = 1 / ρ ′ \int\tau'=1/\rho' ∫ τ ′ = 1/ ρ ′ ∫ τ = 1 / ρ h c \int\tau=1/\rho_{\mathrm{hc}} ∫ τ = 1/ ρ hc ρ ′ ≤ ρ h c \rho'\le\rho_{\mathrm{hc}} ρ ′ ≤ ρ hc ρ ′ = ρ h c \rho'=\rho_{\mathrm{hc}} ρ ′ = ρ hc τ ′ = τ \tau'=\tau τ ′ = τ
Intermediate Iterative Systems
The n iteration
We recall the definition of the γ n \gamma_n γ n (37) :
γ 0 = 0 , ( − Δ + 4 e + v n ) γ n = v n + 2 e ρ n − 1 γ n − 1 ∗ γ n − 1 . (44) \gamma_0=0,\qquad(-\Delta+4e+v_n)\gamma_n=v_n+2e\rho_{n-1}\gamma_{n-1}*\gamma_{n-1}. \tag*{(44)} γ 0 = 0 , ( − Δ + 4 e + v n ) γ n = v n + 2 e ρ n − 1 γ n − 1 ∗ γ n − 1 . (44) In this section we will prove that γ n \gamma_n γ n τ \tau τ L 1 L^1 L 1
Lemma 9. For all n ≥ 0 n\ge0 n ≥ 0
(i) 0 ≤ γ n ≤ 1 0\le\gamma_n\le1 0 ≤ γ n ≤ 1
(ii) γ n ≤ τ n \gamma_n\le\tau_n γ n ≤ τ n
(iii) γ n ≤ γ n + 1 \gamma_n \le\gamma_{n+1} γ n ≤ γ n + 1
Proof. Take γ 0 = τ 0 = 0 \gamma_0=\tau_0=0 γ 0 = τ 0 = 0 n = 0 n=0 n = 0 0 ≤ γ n − 1 ≤ 1 0\le\gamma_{n-1}\le1 0 ≤ γ n − 1 ≤ 1 γ n − 1 ≤ τ n − 1 \gamma_{n-1}\le\tau_{n-1} γ n − 1 ≤ τ n − 1
Non-negativity. The right-hand side of (37) is non-negative and ( − Δ + 4 e + v n ) − 1 \left(-\Delta+4e+v_n\right)^{-1} ( − Δ + 4 e + v n ) − 1 γ n ≥ 0 \gamma_n\ge0 γ n ≥ 0
Upper bound γ n ≤ 1 \gamma_n\le1 γ n ≤ 1 On R 3 \mathbb{R}^3 R 3
( − Δ + 4 e + v n ) ( 1 − γ n ) = 4 e − 2 e ρ n − 1 γ n − 1 ∗ γ n − 1 ≥ 4 e − 2 e = 2 e > 0 , \left(-\Delta+4e+v_n\right)(1-\gamma_n)=4e-2e\rho_{n-1}\gamma_{n-1}*\gamma_{n-1}\ge4e-2e=2e>0, ( − Δ + 4 e + v n ) ( 1 − γ n ) = 4 e − 2 e ρ n − 1 γ n − 1 ∗ γ n − 1 ≥ 4 e − 2 e = 2 e > 0 , using γ n − 1 ≤ 1 \gamma_{n-1}\le1 γ n − 1 ≤ 1 ρ n − 1 ∫ γ n − 1 ≤ ρ n − 1 ∫ τ n − 1 ≤ 1 \rho_{n-1}\int\gamma_{n-1}\le\rho_{n-1}\int\tau_{n-1}\le1 ρ n − 1 ∫ γ n − 1 ≤ ρ n − 1 ∫ τ n − 1 ≤ 1 1 − γ n → 1 1-\gamma_n\to1 1 − γ n → 1 D = R 3 D=\mathbb{R}^3 D = R 3 c = 4 e c=4e c = 4 e W = v n W=v_n W = v n γ n ≤ 1 \gamma_n\le1 γ n ≤ 1
Monotonicity. Assuming γ n − 1 ≤ γ n \gamma_{n-1}\le\gamma_n γ n − 1 ≤ γ n R 3 \mathbb{R}^3 R 3
( − Δ + 4 e + v n + 1 ) ( γ n + 1 − γ n ) = ( v n + 1 − v n ) ( 1 − γ n ) + 2 e [ ρ n γ n ∗ γ n − ρ n − 1 γ n − 1 ∗ γ n − 1 ] ≥ 0 , \begin{aligned}
\left(-\Delta+4e+v_{n+1}\right)(\gamma_{n+1}-\gamma_n)
&=(v_{n+1}-v_n)(1-\gamma_n)\\
&\quad+2e[\rho_n\gamma_n*\gamma_n-\rho_{n-1}\gamma_{n-1}*\gamma_{n-1}]\ge0,
\end{aligned} ( − Δ + 4 e + v n + 1 ) ( γ n + 1 − γ n ) = ( v n + 1 − v n ) ( 1 − γ n ) + 2 e [ ρ n γ n ∗ γ n − ρ n − 1 γ n − 1 ∗ γ n − 1 ] ≥ 0 , since v n + 1 ≥ v n v_{n+1}\ge v_n v n + 1 ≥ v n γ n ≤ 1 \gamma_n\le1 γ n ≤ 1 ρ n ≥ ρ n − 1 \rho_n\ge\rho_{n-1} ρ n ≥ ρ n − 1 γ n ≥ γ n − 1 ≥ 0 \gamma_n\ge\gamma_{n-1}\ge0 γ n ≥ γ n − 1 ≥ 0 γ n + 1 − γ n → 0 \gamma_{n+1}-\gamma_n\to0 γ n + 1 − γ n → 0 D = R 3 D=\mathbb{R}^3 D = R 3 c = 4 e c=4e c = 4 e W = v n + 1 W=v_{n+1} W = v n + 1 γ n + 1 ≥ γ n \gamma_{n+1}\ge\gamma_n γ n + 1 ≥ γ n
Comparison γ n ≤ τ n \gamma_n\le\tau_n γ n ≤ τ n B 1 ( 0 ) B_1(0) B 1 ( 0 ) γ n ≤ 1 = τ n \gamma_n\le1=\tau_n γ n ≤ 1 = τ n { ∣ x ∣ > 1 } \{|x|>1\} { ∣ x ∣ > 1 }
( − Δ + 4 e ) ( τ n − γ n ) = 2 e ρ n − 1 [ τ n − 1 ∗ τ n − 1 − γ n − 1 ∗ γ n − 1 ] ≥ 0 (-\Delta+4e)(\tau_n-\gamma_n)=2e\rho_{n-1}[\tau_{n-1}*\tau_{n-1}-\gamma_{n-1}*\gamma_{n-1}]\ge0 ( − Δ + 4 e ) ( τ n − γ n ) = 2 e ρ n − 1 [ τ n − 1 ∗ τ n − 1 − γ n − 1 ∗ γ n − 1 ] ≥ 0 by the induction hypothesis, while τ n − γ n = 1 − γ n ( 1 ) ≥ 0 \tau_n-\gamma_n=1-\gamma_n(1)\ge0 τ n − γ n = 1 − γ n ( 1 ) ≥ 0 ∂ B 1 \partial B_1 ∂ B 1 τ n − γ n → 0 \tau_n-\gamma_n\to0 τ n − γ n → 0 D = { ∣ x ∣ > 1 } D=\{|x|>1\} D = { ∣ x ∣ > 1 } c = 4 e c=4e c = 4 e W = 0 W=0 W = 0 γ n ≤ τ n \gamma_n\le\tau_n γ n ≤ τ n
Lemma 10. γ n ↑ τ \gamma_n\uparrow\tau γ n ↑ τ L 1 L^1 L 1
Proof. By Lemma 9, the sequence ( γ n ) (\gamma_n) ( γ n ) τ \tau τ γ n ↑ γ ≤ τ \gamma_n\uparrow\gamma\le\tau γ n ↑ γ ≤ τ τ ∈ L 1 \tau\in L^1 τ ∈ L 1 γ ∈ L 1 \gamma\in L^1 γ ∈ L 1
∥ γ − γ n ∥ L 1 = ∫ ( γ − γ n ) → 0. \|\gamma-\gamma_n\|_{L^1}=\int(\gamma-\gamma_n)\to0. ∥ γ − γ n ∥ L 1 = ∫ ( γ − γ n ) → 0. The rest of the proof consists in showing that γ = τ \gamma=\tau γ = τ
Boundary value. Integrating (44) over R 3 \mathbb{R}^3 R 3
n ∫ B 1 ( 0 ) ( 1 − γ n ) = 4 e ∫ γ n − 2 e ρ n − 1 ∥ γ n − 1 ∥ L 1 2 ≤ 4 e ∫ τ n ≤ 4 e ρ 1 , (45) n\int_{B_1(0)}(1-\gamma_n)=4e\int\gamma_n-2e\rho_{n-1}\|\gamma_{n-1}\|_{L^1}^2\le4e\int\tau_n\le\frac{4e}{\rho_1}, \tag*{(45)} n ∫ B 1 ( 0 ) ( 1 − γ n ) = 4 e ∫ γ n − 2 e ρ n − 1 ∥ γ n − 1 ∥ L 1 2 ≤ 4 e ∫ τ n ≤ ρ 1 4 e , (45) using γ n ≤ τ n \gamma_n\le\tau_n γ n ≤ τ n ∫ τ n ≤ 1 / ρ n ≤ 1 / ρ 1 \int\tau_n\le1/\rho_n\le1/\rho_1 ∫ τ n ≤ 1/ ρ n ≤ 1/ ρ 1 ∫ B 1 ( 0 ) ( 1 − γ n ) ≤ C / n → 0 \int_{B_1(0)}(1-\gamma_n)\le C/n\to0 ∫ B 1 ( 0 ) ( 1 − γ n ) ≤ C / n → 0 γ n ( 1 ) → 1 \gamma_n(1)\to1 γ n ( 1 ) → 1 ι n = 1 − γ n ≥ 0 \iota_n=1-\gamma_n\ge0 ι n = 1 − γ n ≥ 0 B 1 ( 0 ) B_1(0) B 1 ( 0 ) x ∈ B 1 ( 0 ) x\in B_1(0) x ∈ B 1 ( 0 )
( − Δ + κ 2 ) ι n = 4 e − 2 e ρ n − 1 ( γ n − 1 ∗ γ n − 1 ) ≡ f n , κ = 4 e + n , (-\Delta+\kappa^2)\iota_n=4e-2e\rho_{n-1}(\gamma_{n-1}*\gamma_{n-1})\equiv f_n,\qquad\kappa=\sqrt{4e+n}, ( − Δ + κ 2 ) ι n = 4 e − 2 e ρ n − 1 ( γ n − 1 ∗ γ n − 1 ) ≡ f n , κ = 4 e + n , with 0 ≤ f n ≤ 4 e 0\le f_n\le4e 0 ≤ f n ≤ 4 e 2 e ρ n − 1 ( γ n − 1 ∗ γ n − 1 ) ≤ 2 e ρ n − 1 ∥ γ n − 1 ∥ L ∞ ∥ γ n − 1 ∥ L 1 ≤ 2 e ρ n − 1 ∫ τ n − 1 ≤ 2 e 2e\rho_{n-1}(\gamma_{n-1}*\gamma_{n-1})\le2e\rho_{n-1}\|\gamma_{n-1}\|_{L^\infty}\|\gamma_{n-1}\|_{L^1}\le2e\rho_{n-1}\int\tau_{n-1}\le2e 2 e ρ n − 1 ( γ n − 1 ∗ γ n − 1 ) ≤ 2 e ρ n − 1 ∥ γ n − 1 ∥ L ∞ ∥ γ n − 1 ∥ L 1 ≤ 2 e ρ n − 1 ∫ τ n − 1 ≤ 2 e ι n = ι ~ n + h n \iota_n=\tilde\iota_n+h_n ι n = ι ~ n + h n ι ~ n ∣ ∂ B 1 = 0 \tilde\iota_n|_{\partial B_1}=0 ι ~ n ∣ ∂ B 1 = 0 h n h_n h n ( − Δ + κ 2 ) h n = 0 (-\Delta+\kappa^2)h_n=0 ( − Δ + κ 2 ) h n = 0 h n ( 1 ) = ι n ( 1 ) h_n(1)=\iota_n(1) h n ( 1 ) = ι n ( 1 ) h n ( r ) = ι n ( 1 ) sinh ( κ r ) r sinh κ h_n(r)=\iota_n(1)\frac{\sinh(\kappa r)}{r\sinh\kappa} h n ( r ) = ι n ( 1 ) r s i n h κ s i n h ( κ r ) ( − Δ + κ 2 ) h n = 0 (-\Delta+\kappa^2)h_n=0 ( − Δ + κ 2 ) h n = 0 ι ~ n \widetilde{\iota}_n ι n ι n \iota_n ι n ι ~ n = 0 \widetilde{\iota}_n=0 ι n = 0 ∂ B 1 \partial B_1 ∂ B 1 D = B 1 ( 0 ) D=B_1(0) D = B 1 ( 0 ) c = κ 2 c=\kappa^2 c = κ 2 W = 0 W=0 W = 0 F = 4 e F=4e F = 4 e 0 ≤ ι ~ n ≤ 4 e / κ 2 0\leq\widetilde{\iota}_n\leq4e/\kappa^2 0 ≤ ι n ≤ 4 e / κ 2
∫ B 1 ( 0 ) h n = 4 π ι n ( 1 ) κ 2 sinh κ ( κ cosh κ − sinh κ ) ∼ 4 π ι n ( 1 ) κ ( κ → ∞ ) . \int_{B_1(0)}h_n=\frac{4\pi\iota_n(1)}{\kappa^2\sinh\kappa}(\kappa\cosh\kappa-\sinh\kappa)\sim\frac{4\pi\iota_n(1)}{\kappa}\quad(\kappa\to\infty). ∫ B 1 ( 0 ) h n = κ 2 sinh κ 4 π ι n ( 1 ) ( κ cosh κ − sinh κ ) ∼ κ 4 π ι n ( 1 ) ( κ → ∞ ) . Since ι n = ι ~ n + h n ≥ h n ≥ 0 \iota_n=\widetilde{\iota}_n+h_n\geq h_n\geq0 ι n = ι n + h n ≥ h n ≥ 0 ∫ B 1 ( 0 ) ι n = O ( 1 / n ) \int_{B_1(0)}\iota_n=O(1/n) ∫ B 1 ( 0 ) ι n = O ( 1/ n ) (45) , we obtain ι n ( 1 ) = O ( κ / n ) = O ( 1 / n ) \iota_n(1)=O(\kappa/n)=O(1/\sqrt{n}) ι n ( 1 ) = O ( κ / n ) = O ( 1/ n ) γ n ( 1 ) → 1 \gamma_n(1)\to1 γ n ( 1 ) → 1 γ \gamma γ γ ( 1 ) = lim n γ n ( 1 ) = 1 \gamma(1)=\lim_n\gamma_n(1)=1 γ ( 1 ) = lim n γ n ( 1 ) = 1 (45) , ∫ B 1 ( 0 ) ( 1 − γ ) = lim n ∫ B 1 ( 0 ) ( 1 − γ n ) = 0 \int_{B_1(0)}(1-\gamma)=\lim_n\int_{B_1(0)}(1-\gamma_n)=0 ∫ B 1 ( 0 ) ( 1 − γ ) = lim n ∫ B 1 ( 0 ) ( 1 − γ n ) = 0 γ = 1 \gamma=1 γ = 1 B 1 ( 0 ) B_1(0) B 1 ( 0 )
Continuity at the boundary. γ \gamma γ ∂ B 1 \partial B_1 ∂ B 1 n n n r > 1 r>1 r > 1 γ ( r ) ≥ γ n ( r ) \gamma(r)\geq\gamma_n(r) γ ( r ) ≥ γ n ( r ) γ n \gamma_n γ n
lim inf r → 1 + γ ( r ) ≥ γ n ( 1 ) = 1 − O ( n − 1 / 2 ) for every n , \liminf_{r\to1^+}\gamma(r)\geq\gamma_n(1)=1-O(n^{-1/2})\quad\text{for every }n, r → 1 + lim inf γ ( r ) ≥ γ n ( 1 ) = 1 − O ( n − 1/2 ) for every n , whence lim inf r → 1 + γ ( r ) ≥ 1 \liminf_{r\to1^+}\gamma(r)\geq1 lim inf r → 1 + γ ( r ) ≥ 1 γ ≤ 1 \gamma\leq1 γ ≤ 1 γ ( r ) → 1 = γ ( 1 ) \gamma(r)\to1=\gamma(1) γ ( r ) → 1 = γ ( 1 )
Limit equation. On { ∣ x ∣ > 1 } \{|x|>1\} { ∣ x ∣ > 1 } v n = 0 v_n=0 v n = 0 (44) reads ( − Δ + 4 e ) γ n = 2 e ρ n − 1 ( γ n − 1 ∗ γ n − 1 ) (-\Delta+4e)\gamma_n=2e\rho_{n-1}(\gamma_{n-1}*\gamma_{n-1}) ( − Δ + 4 e ) γ n = 2 e ρ n − 1 ( γ n − 1 ∗ γ n − 1 )
∥ γ n − 1 ∗ γ n − 1 − γ ∗ γ ∥ L ∞ ≤ ( ∥ γ n − 1 ∥ L ∞ + ∥ γ ∥ L ∞ ) ∥ γ n − 1 − γ ∥ L 1 ≤ 2 ∥ γ − γ n − 1 ∥ L 1 → 0 , \|\gamma_{n-1}*\gamma_{n-1}-\gamma*\gamma\|_{L^\infty}\leq(\|\gamma_{n-1}\|_{L^\infty}+\|\gamma\|_{L^\infty})\|\gamma_{n-1}-\gamma\|_{L^1}\leq2\|\gamma-\gamma_{n-1}\|_{L^1}\to0, ∥ γ n − 1 ∗ γ n − 1 − γ ∗ γ ∥ L ∞ ≤ ( ∥ γ n − 1 ∥ L ∞ + ∥ γ ∥ L ∞ ) ∥ γ n − 1 − γ ∥ L 1 ≤ 2∥ γ − γ n − 1 ∥ L 1 → 0 , and ρ n − 1 → ρ h c \rho_{n-1}\to\rho_{\mathrm{hc}} ρ n − 1 → ρ hc Lemma 7 ). Testing against φ ∈ C c ∞ ( { ∣ x ∣ > 1 } ) \varphi\in C^\infty_c(\{|x|>1\}) φ ∈ C c ∞ ({ ∣ x ∣ > 1 })
( − Δ + 4 e ) γ = 2 e ρ h c ( γ ∗ γ ) on { ∣ x ∣ > 1 } , γ = 1 on ∂ B 1 . (-\Delta+4e)\gamma=2e\rho_{\mathrm{hc}}(\gamma*\gamma)\quad\text{on }\{|x|>1\},\qquad\gamma=1\text{ on }\partial B_1. ( − Δ + 4 e ) γ = 2 e ρ hc ( γ ∗ γ ) on { ∣ x ∣ > 1 } , γ = 1 on ∂ B 1 . Since the right-hand side is bounded (γ ∗ γ ≤ ∥ γ ∥ L 1 \gamma*\gamma\leq\|\gamma\|_{L^1} γ ∗ γ ≤ ∥ γ ∥ L 1 γ \gamma γ { ∣ x ∣ > 1 } \{|x|>1\} { ∣ x ∣ > 1 }
Identification with τ \tau τ Via the uniqueness property of τ \tau τ Lemma 8 , it results immediately that γ = τ \gamma=\tau γ = τ
The n iteration
We recall the definition of the ω n \omega_n ω n
ω 0 = 0 , ( − Δ + 4 e + v n ) ω n = v n + 2 e ρ h c ω n − 1 ∗ ω n − 1 . (46) \omega_0=0,\qquad(-\Delta+4e+v_n)\omega_n=v_n+2e\rho_{\mathrm{hc}}\omega_{n-1}*\omega_{n-1}. \tag*{(46)} ω 0 = 0 , ( − Δ + 4 e + v n ) ω n = v n + 2 e ρ hc ω n − 1 ∗ ω n − 1 . (46) In this section we prove that ω n \omega_n ω n τ \tau τ L 1 L^1 L 1
Lemma 11. For all n ≥ 0 n\geq0 n ≥ 0 0 ≤ ω n ≤ 1 0\leq\omega_n\leq1 0 ≤ ω n ≤ 1 ω n ≤ τ \omega_n\leq\tau ω n ≤ τ ω n ≥ γ n \omega_n\geq\gamma_n ω n ≥ γ n ω n ≤ ω n + 1 \omega_n\leq\omega_{n+1} ω n ≤ ω n + 1
Proof. Take ω 0 = 0 \omega_0=0 ω 0 = 0 n = 0 n=0 n = 0 0 ≤ ω n − 1 ≤ 1 0\leq\omega_{n-1}\leq1 0 ≤ ω n − 1 ≤ 1 ω n − 1 ≤ τ \omega_{n-1}\leq\tau ω n − 1 ≤ τ ω n − 1 ≥ γ n − 1 \omega_{n-1}\geq\gamma_{n-1} ω n − 1 ≥ γ n − 1
Non-negativity. The right-hand side of (38) is non-negative and ( − Δ + 4 e + v n ) − 1 (-\Delta+4e+v_n)^{-1} ( − Δ + 4 e + v n ) − 1 ω n ≥ 0 \omega_n\geq0 ω n ≥ 0
Upper bound ω n ≤ 1 \omega_n\leq1 ω n ≤ 1 On R 3 \mathbb{R}^3 R 3
( − Δ + 4 e + v n ) ( 1 − ω n ) = 4 e − 2 e ρ h c ω n − 1 ∗ ω n − 1 ≥ 4 e − 2 e = 2 e > 0 , (-\Delta+4e+v_n)(1-\omega_n)=4e-2e\rho_{\mathrm{hc}}\omega_{n-1}*\omega_{n-1}\geq4e-2e=2e>0, ( − Δ + 4 e + v n ) ( 1 − ω n ) = 4 e − 2 e ρ hc ω n − 1 ∗ ω n − 1 ≥ 4 e − 2 e = 2 e > 0 , using ω n − 1 ≤ 1 \omega_{n-1} \le1 ω n − 1 ≤ 1 ρ h c ∫ ω n − 1 ≤ ρ h c ∫ τ = 1 \rho_{\mathrm{hc}} \int\omega_{n-1} \le\rho_{\mathrm{hc}} \int\tau= 1 ρ hc ∫ ω n − 1 ≤ ρ hc ∫ τ = 1 1 − ω n → 1 1-\omega_n \to1 1 − ω n → 1 D = R 3 D = \mathbb{R}^3 D = R 3 c = 4 e c = 4e c = 4 e W = v n W = v_n W = v n ω n ≤ 1 \omega_n \le1 ω n ≤ 1
Monotonicity. Assuming ω n − 1 ≤ ω n \omega_{n-1} \le\omega_n ω n − 1 ≤ ω n R 3 \mathbb{R}^3 R 3
( − Δ + 4 e + v n + 1 ) ( ω n + 1 − ω n ) = ( v n + 1 − v n ) ( 1 − ω n ) + 2 e ρ h c [ ω n ∗ ω n − ω n − 1 ∗ ω n − 1 ] ≥ 0 , (-\Delta+ 4e + v_{n+1})(\omega_{n+1}-\omega_n) = (v_{n+1}-v_n)(1-\omega_n) + 2e\rho_{\mathrm{hc}}[\omega_n * \omega_n-\omega_{n-1} * \omega_{n-1}] \ge0, ( − Δ + 4 e + v n + 1 ) ( ω n + 1 − ω n ) = ( v n + 1 − v n ) ( 1 − ω n ) + 2 e ρ hc [ ω n ∗ ω n − ω n − 1 ∗ ω n − 1 ] ≥ 0 , since v n + 1 ≥ v n v_{n+1} \ge v_n v n + 1 ≥ v n ω n ≤ 1 \omega_n \le1 ω n ≤ 1 ω n ≥ ω n − 1 ≥ 0 \omega_n \ge\omega_{n-1} \ge0 ω n ≥ ω n − 1 ≥ 0 ω n + 1 − ω n → 0 \omega_{n+1}-\omega_n \to0 ω n + 1 − ω n → 0 D = R 3 D = \mathbb{R}^3 D = R 3 c = 4 e c = 4e c = 4 e W = v n + 1 W = v_{n+1} W = v n + 1 ω n + 1 ≥ ω n \omega_{n+1} \ge\omega_n ω n + 1 ≥ ω n
Comparison ω n ≤ τ \omega_n \le\tau ω n ≤ τ For ∣ x ∣ ≤ 1 |x| \le1 ∣ x ∣ ≤ 1 ω n ≤ 1 = τ \omega_n \le1 = \tau ω n ≤ 1 = τ { ∣ x ∣ > 1 } \{|x| > 1\} { ∣ x ∣ > 1 } (8) gives ( − Δ + 4 e ) τ = 2 e ρ h c τ ∗ τ (-\Delta+ 4e)\tau= 2e\rho_{\mathrm{hc}} \tau* \tau ( − Δ + 4 e ) τ = 2 e ρ hc τ ∗ τ
( − Δ + 4 e ) ( τ − ω n ) = 2 e ρ h c [ τ ∗ τ − ω n − 1 ∗ ω n − 1 ] ≥ 0 (-\Delta+ 4e)(\tau-\omega_n) = 2e\rho_{\mathrm{hc}}[\tau* \tau-\omega_{n-1} * \omega_{n-1}] \ge0 ( − Δ + 4 e ) ( τ − ω n ) = 2 e ρ hc [ τ ∗ τ − ω n − 1 ∗ ω n − 1 ] ≥ 0 by the induction hypothesis, while τ − ω n = 1 − ω n ( 1 ) ≥ 0 \tau-\omega_n = 1-\omega_n(1) \ge0 τ − ω n = 1 − ω n ( 1 ) ≥ 0 ∂ B 1 \partial B_1 ∂ B 1 lim inf ∣ x ∣ → ∞ ( τ − ω n ) ≥ 0 \liminf_{|x|\to\infty}(\tau-\omega_n) \ge0 lim inf ∣ x ∣ → ∞ ( τ − ω n ) ≥ 0 τ ≥ 0 \tau\ge0 τ ≥ 0 ω n → 0 \omega_n \to0 ω n → 0 D = { ∣ x ∣ > 1 } D = \{|x| > 1\} D = { ∣ x ∣ > 1 } c = 4 e c = 4e c = 4 e W = 0 W = 0 W = 0 ω n ≤ τ \omega_n \le\tau ω n ≤ τ
Comparison ω n ≥ γ n \omega_n \ge\gamma_n ω n ≥ γ n On R 3 \mathbb{R}^3 R 3 (37) from (38) ,
( − Δ + 4 e + v n ) ( ω n − γ n ) = 2 e [ ρ h c ω n − 1 ∗ ω n − 1 − ρ n − 1 γ n − 1 ∗ γ n − 1 ] ≥ 0 , (-\Delta+ 4e + v_n)(\omega_n-\gamma_n) = 2e[\rho_{\mathrm{hc}}\omega_{n-1} * \omega_{n-1}-\rho_{n-1}\gamma_{n-1} * \gamma_{n-1}] \ge0, ( − Δ + 4 e + v n ) ( ω n − γ n ) = 2 e [ ρ hc ω n − 1 ∗ ω n − 1 − ρ n − 1 γ n − 1 ∗ γ n − 1 ] ≥ 0 , since ρ h c ≥ ρ n − 1 \rho_{\mathrm{hc}} \ge\rho_{n-1} ρ hc ≥ ρ n − 1 ω n − 1 ≥ γ n − 1 ≥ 0 \omega_{n-1} \ge\gamma_{n-1} \ge0 ω n − 1 ≥ γ n − 1 ≥ 0 ω n − γ n → 0 \omega_n-\gamma_n \to0 ω n − γ n → 0 D = R 3 D = \mathbb{R}^3 D = R 3 c = 4 e c = 4e c = 4 e W = v n W = v_n W = v n ω n ≥ γ n \omega_n \ge\gamma_n ω n ≥ γ n
Corollary 12. ∥ ω n − τ ∥ L 1 → 0 \|\omega_n-\tau\|_{L^1} \to0 ∥ ω n − τ ∥ L 1 → 0
Proof. From γ n ≤ ω n ≤ τ \gamma_n \le\omega_n \le\tau γ n ≤ ω n ≤ τ ∫ ( τ − ω n ) ≤ ∫ ( τ − γ n ) → 0 \int(\tau-\omega_n) \le\int(\tau-\gamma_n) \to0 ∫ ( τ − ω n ) ≤ ∫ ( τ − γ n ) → 0 L 1 L^1 L 1
The Soft-Core Simple Equation
We recall [11 Lemma 3.1-3.2, Equation (3.9)] :
Lemma 13. For all n ≥ 1 n \ge1 n ≥ 1 σ m , n ≥ 0 \sigma_{m,n} \ge0 σ m , n ≥ 0 σ m , n ≤ 1 \sigma_{m,n} \le1 σ m , n ≤ 1 σ m , n ≤ σ m , n + 1 \sigma_{m,n} \le\sigma_{m,n+1} σ m , n ≤ σ m , n + 1 ρ ˉ m , n ≤ ρ ˉ m , n + 1 \bar{\rho}_{m,n} \le\bar{\rho}_{m,n+1} ρ ˉ m , n ≤ ρ ˉ m , n + 1
2 ∫ σ m , n = 1 / ρ ˉ m , n + ρ ˉ m , n − 1 ∥ σ m , n − 1 ∥ L 1 2 , ∫ σ m , n ≤ 1 / ρ ˉ m , n . (47) 2\int\sigma_{m,n} = 1/\bar{\rho}_{m,n}+\bar{\rho}_{m,n-1}\|\sigma_{m,n-1}\|_{L^1}^2,\qquad\int\sigma_{m,n} \le1/\bar{\rho}_{m,n}. \tag*{(47)} 2 ∫ σ m , n = 1/ ρ ˉ m , n + ρ ˉ m , n − 1 ∥ σ m , n − 1 ∥ L 1 2 , ∫ σ m , n ≤ 1/ ρ ˉ m , n . (47) The proof uses the same tools as the proofs in section 3.
Per-n n n
Lemma 14. For each fixed n ≥ 1 n \ge1 n ≥ 1 ∥ σ m , n − τ n ∥ L 1 → 0 \|\sigma_{m,n}-\tau_n\|_{L^1} \to0 ∥ σ m , n − τ n ∥ L 1 → 0 ρ ˉ m , n → ρ n \bar{\rho}_{m,n} \to\rho_n ρ ˉ m , n → ρ n m → ∞ m \to\infty m → ∞
Proof. By induction on n n n n = 1 n = 1 n = 1 (32) with σ m , 0 = 0 \sigma_{m,0}=0 σ m , 0 = 0 ∂ r σ m , 1 ( r ) ∣ r = 0 = 0 \partial_r\sigma_{m,1}(r)|_{r=0}=0 ∂ r σ m , 1 ( r ) ∣ r = 0 = 0 σ m , 1 \sigma_{m,1} σ m , 1
and radially symmetric), lim r → ∞ σ m , 1 ( r ) = 0 \lim_{r\to\infty}\sigma_{m,1}(r)=0 lim r → ∞ σ m , 1 ( r ) = 0 σ m , 1 ( r ) \sigma_{m,1}(r) σ m , 1 ( r ) ∂ r σ m , 1 ( r ) \partial_r\sigma_{m,1}(r) ∂ r σ m , 1 ( r ) r = 1 r=1 r = 1
σ m , 1 ( r ) = { m κ 2 ( 1 − 1 + 2 e κ cosh κ + 2 e sinh κ sinh ( κ r ) r ) , r < 1 , σ m , 1 ( 1 ) e − 2 e ( r − 1 ) r , r > 1 , κ = 4 e + m . \sigma_{m,1}(r)=
\begin{cases}
\frac{m}{\kappa^2}\left(1-\frac{1+2\sqrt{e}}{\kappa\cosh\kappa+2\sqrt{e}\sinh\kappa}\frac{\sinh(\kappa r)}{r}\right), & r<1,\\[6pt]
\sigma_{m,1}(1)\frac{e^{-2\sqrt{e}(r-1)}}{r}, & r>1,
\end{cases}
\qquad\kappa=\sqrt{4e+m}. σ m , 1 ( r ) = ⎩ ⎨ ⎧ κ 2 m ( 1 − κ c o s h κ + 2 e s i n h κ 1 + 2 e r s i n h ( κ r ) ) , σ m , 1 ( 1 ) r e − 2 e ( r − 1 ) , r < 1 , r > 1 , κ = 4 e + m . with σ m , 1 ( 1 ) = m κ 2 κ cosh κ − sinh κ κ cosh κ + 2 e sinh κ = 1 − O ( 1 / m ) \sigma_{m,1}(1)=\frac{m}{\kappa^2}\frac{\kappa\cosh\kappa-\sinh\kappa}{\kappa\cosh\kappa+2\sqrt{e}\sinh\kappa}=1-O(1/\sqrt{m}) σ m , 1 ( 1 ) = κ 2 m κ c o s h κ + 2 e s i n h κ κ c o s h κ − s i n h κ = 1 − O ( 1/ m )
m ∫ B 1 ( 0 ) ( 1 − σ m , 1 ( ∣ x ∣ ) ) d x = 4 e ∫ R 3 σ m , 1 ( ∣ x ∣ ) d x (48) m\int_{B_1(0)}(1-\sigma_{m,1}(|x|))\,dx=4e\int_{\mathbb{R}^3}\sigma_{m,1}(|x|)\,dx \tag*{(48)} m ∫ B 1 ( 0 ) ( 1 − σ m , 1 ( ∣ x ∣ )) d x = 4 e ∫ R 3 σ m , 1 ( ∣ x ∣ ) d x (48) so in particular ∫ B 1 ( 0 ) ∣ σ m , 1 − τ 1 ∣ ≡ ∫ B 1 ( 0 ) ( 1 − σ m , 1 ) = O ( 1 / m ) \int_{B_1(0)}|\sigma_{m,1}-\tau_1|\equiv\int_{B_1(0)}(1-\sigma_{m,1})=O(1/m) ∫ B 1 ( 0 ) ∣ σ m , 1 − τ 1 ∣ ≡ ∫ B 1 ( 0 ) ( 1 − σ m , 1 ) = O ( 1/ m ) σ m , 1 ( 1 ) = 1 + O ( 1 / m ) \sigma_{m,1}(1)=1+O(1/\sqrt{m}) σ m , 1 ( 1 ) = 1 + O ( 1/ m ) ∥ σ m , 1 − τ 1 ∥ L 1 = O ( 1 / m ) \|\sigma_{m,1}-\tau_1\|_{L^1}=O(1/\sqrt{m}) ∥ σ m , 1 − τ 1 ∥ L 1 = O ( 1/ m ) m ∫ B 1 ( 0 ) ( 1 − σ m , 1 ) = 4 e ∫ σ m , 1 → 4 e ∫ τ 1 = 2 e / ρ 1 m\int_{B_1(0)}(1-\sigma_{m,1})=4e\int\sigma_{m,1}\to4e\int\tau_1=2e/\rho_1 m ∫ B 1 ( 0 ) ( 1 − σ m , 1 ) = 4 e ∫ σ m , 1 → 4 e ∫ τ 1 = 2 e / ρ 1 (36) ), and therefore by the second of (32), ρ ‾ m , 1 → ρ 1 \overline{\rho}_{m,1}\to\rho_1 ρ m , 1 → ρ 1
Estimates inside the core. We mimic the Boundary value step in the proof of Lemma 10, with the roles of n n n m m m ι m , n = 1 − σ m , n ≥ 0 \iota_{m,n}=1-\sigma_{m,n}\geq0 ι m , n = 1 − σ m , n ≥ 0 B 1 ( 0 ) B_1(0) B 1 ( 0 ) x ∈ B 1 ( 0 ) x\in B_1(0) x ∈ B 1 ( 0 )
( − Δ + κ 2 ) ι m , n = 4 e − 2 e ρ ‾ m , n − 1 ( σ m , n − 1 ∗ σ m , n − 1 ) ≡ f , κ = 4 e + m , (-\Delta+\kappa^2)\iota_{m,n}=4e-2e\overline{\rho}_{m,n-1}(\sigma_{m,n-1}*\sigma_{m,n-1})\equiv f,\qquad\kappa=\sqrt{4e+m}, ( − Δ + κ 2 ) ι m , n = 4 e − 2 e ρ m , n − 1 ( σ m , n − 1 ∗ σ m , n − 1 ) ≡ f , κ = 4 e + m , with 0 ≤ f ≤ 4 e 0\leq f\leq4e 0 ≤ f ≤ 4 e 2 e ρ ‾ m , n − 1 ( σ m , n − 1 ∗ σ m , n − 1 ) ≤ 2 e ρ ‾ m , n − 1 ∥ σ m , n − 1 ∥ L ∞ ∥ σ m , n − 1 ∥ L 1 ≤ 2 e 2e\overline{\rho}_{m,n-1}(\sigma_{m,n-1}*\sigma_{m,n-1})\leq2e\overline{\rho}_{m,n-1}\|\sigma_{m,n-1}\|_{L^\infty}\|\sigma_{m,n-1}\|_{L^1}\leq2e 2 e ρ m , n − 1 ( σ m , n − 1 ∗ σ m , n − 1 ) ≤ 2 e ρ m , n − 1 ∥ σ m , n − 1 ∥ L ∞ ∥ σ m , n − 1 ∥ L 1 ≤ 2 e ι m , n = ι ~ m , n + h m , n \iota_{m,n}=\widetilde{\iota}_{m,n}+h_{m,n} ι m , n = ι m , n + h m , n ι ~ m , n ∣ ∂ B 1 = 0 \widetilde{\iota}_{m,n}|_{\partial B_1}=0 ι m , n ∣ ∂ B 1 = 0 h m , n h_{m,n} h m , n ( − Δ + κ 2 ) h m , n = 0 (-\Delta+\kappa^2)h_{m,n}=0 ( − Δ + κ 2 ) h m , n = 0 h m , n ( 1 ) = ι m , n ( 1 ) h_{m,n}(1)=\iota_{m,n}(1) h m , n ( 1 ) = ι m , n ( 1 ) h m , n ( r ) = ι m , n ( 1 ) sinh ( κ r ) r sinh κ h_{m,n}(r)=\iota_{m,n}(1)\frac{\sinh(\kappa r)}{r\sinh\kappa} h m , n ( r ) = ι m , n ( 1 ) r s i n h κ s i n h ( κ r ) ( − Δ + κ 2 ) h m , n = 0 (-\Delta+\kappa^2)h_{m,n}=0 ( − Δ + κ 2 ) h m , n = 0 ι ~ m , n \widetilde{\iota}_{m,n} ι m , n ι m , n \iota_{m,n} ι m , n ι ~ m , n = 0 \widetilde{\iota}_{m,n}=0 ι m , n = 0 ∂ B 1 \partial B_1 ∂ B 1 D = B 1 ( 0 ) D=B_1(0) D = B 1 ( 0 ) c = κ 2 c=\kappa^2 c = κ 2 W = 0 W=0 W = 0 F = 4 e F=4e F = 4 e 0 ≤ ι ~ m , n ≤ 4 e / κ 2 0\leq\widetilde{\iota}_{m,n}\leq4e/\kappa^2 0 ≤ ι m , n ≤ 4 e / κ 2
∫ B 1 ( 0 ) h m , n = 4 π ι m , n ( 1 ) κ 2 sinh κ ( κ cosh κ − sinh κ ) ∼ 4 π ι m , n ( 1 ) κ ( κ → ∞ ) . (49) \int_{B_1(0)}h_{m,n}=\frac{4\pi\iota_{m,n}(1)}{\kappa^2\sinh\kappa}(\kappa\cosh\kappa-\sinh\kappa)\sim\frac{4\pi\iota_{m,n}(1)}{\kappa}\qquad(\kappa\to\infty). \tag*{(49)} ∫ B 1 ( 0 ) h m , n = κ 2 sinh κ 4 π ι m , n ( 1 ) ( κ cosh κ − sinh κ ) ∼ κ 4 π ι m , n ( 1 ) ( κ → ∞ ) . (49) Integrating (32) over R 3 \mathbb{R}^3 R 3
m ∫ B 1 ( 0 ) ι m , n = 4 e ∫ σ m , n − 2 e ρ ‾ m , n − 1 ∥ σ m , n − 1 ∥ L 1 2 ≤ 4 e ρ ‾ m , 1 , m\int_{B_1(0)}\iota_{m,n}=4e\int\sigma_{m,n}-2e\overline{\rho}_{m,n-1}\|\sigma_{m,n-1}\|_{L^1}^2\leq\frac{4e}{\overline{\rho}_{m,1}}, m ∫ B 1 ( 0 ) ι m , n = 4 e ∫ σ m , n − 2 e ρ m , n − 1 ∥ σ m , n − 1 ∥ L 1 2 ≤ ρ m , 1 4 e , and since ρ ‾ m , 1 → ρ 1 > 0 \overline{\rho}_{m,1}\to\rho_1>0 ρ m , 1 → ρ 1 > 0 m m m ∫ B 1 ( 0 ) ι m , n = O ( 1 / m ) \int_{B_1(0)}\iota_{m,n}=O(1/m) ∫ B 1 ( 0 ) ι m , n = O ( 1/ m ) ι m , n = ι ~ m , n + h m , n ≥ h m , n \iota_{m,n}=\widetilde{\iota}_{m,n}+h_{m,n}\geq h_{m,n} ι m , n = ι m , n + h m , n ≥ h m , n ∫ B 1 ( 0 ) h m , n = O ( 1 / m ) \int_{B_1(0)}h_{m,n}=O(1/m) ∫ B 1 ( 0 ) h m , n = O ( 1/ m ) ι m , n ( 1 ) = O ( κ / m ) = O ( 1 / m ) \iota_{m,n}(1)=O(\kappa/m)=O(1/\sqrt{m}) ι m , n ( 1 ) = O ( κ / m ) = O ( 1/ m ) 1 − σ m , n ( 1 ) = O ( m − 1 / 2 ) 1-\sigma_{m,n}(1)=O(m^{-1/2}) 1 − σ m , n ( 1 ) = O ( m − 1/2 ) ∫ B 1 ( 0 ) ( 1 − σ m , n ) = O ( m − 1 ) \int_{B_1(0)}(1-\sigma_{m,n})=O(m^{-1}) ∫ B 1 ( 0 ) ( 1 − σ m , n ) = O ( m − 1 )
Convergence outside B 1 ( 0 ) B_1(0) B 1 ( 0 ) Let w = τ n − σ m , n w=\tau_n-\sigma_{m,n} w = τ n − σ m , n { ∣ x ∣ > 1 } \{|x|>1\} { ∣ x ∣ > 1 } w w w w ( 1 ) = 1 − σ m , n ( 1 ) w(1)=1-\sigma_{m,n}(1) w ( 1 ) = 1 − σ m , n ( 1 )
( − Δ + 4 e ) w = 2 e ( ρ n − 1 τ n − 1 ∗ τ n − 1 − ρ ‾ m , n − 1 σ m , n − 1 ∗ σ m , n − 1 ) ≡ β n on { ∣ x ∣ > 1 } . (-\Delta+4e)w=2e\left(\rho_{n-1}\tau_{n-1}*\tau_{n-1}-\overline{\rho}_{m,n-1}\sigma_{m,n-1}*\sigma_{m,n-1}\right)\equiv\beta_n\quad\text{on }\{|x|>1\}. ( − Δ + 4 e ) w = 2 e ( ρ n − 1 τ n − 1 ∗ τ n − 1 − ρ m , n − 1 σ m , n − 1 ∗ σ m , n − 1 ) ≡ β n on { ∣ x ∣ > 1 } . Since w w w Y 4 e ( x ) = e − 2 e ∣ x ∣ / ( 4 π ∣ x ∣ ) Y_{4e}(x)=e^{-2\sqrt{e}|x|}/(4\pi|x|) Y 4 e ( x ) = e − 2 e ∣ x ∣ / ( 4 π ∣ x ∣ ) − Δ + 4 e -\Delta+4e − Δ + 4 e R 3 \mathbb{R}^3 R 3 β n \beta_n β n R n \mathbb{R}^n R n B 1 ( 0 ) B_1(0) B 1 ( 0 ) W : = Y 4 e ∗ ( β n 1 B 1 c ) W := Y_{4e} * (\beta_n 1_{B_1^c}) W := Y 4 e ∗ ( β n 1 B 1 c ) ( − Δ + 4 e ) W = β n 1 B 1 c (-\Delta+ 4e)W = \beta_n 1_{B_1^c} ( − Δ + 4 e ) W = β n 1 B 1 c R 3 \mathbb{R}^3 R 3 { ∣ x ∣ > 1 } \{|x| > 1\} { ∣ x ∣ > 1 } β n \beta_n β n Y 4 e Y_{4e} Y 4 e
W ( r ) = 1 4 e r ∫ 1 ∞ s β n ( s ) ( e − 2 e ∣ r − s ∣ − e − 2 e ( r + s ) ) d s W(r) = \frac{1}{4\sqrt{e}r} \int_1^\infty s\beta_n(s)\left(e^{-2\sqrt{e}|r-s|} - e^{-2\sqrt{e}(r+s)}\right)\,ds W ( r ) = 4 e r 1 ∫ 1 ∞ s β n ( s ) ( e − 2 e ∣ r − s ∣ − e − 2 e ( r + s ) ) d s so it is C 2 C^2 C 2 w w w W W W ( − Δ + 4 e ) u = β n (-\Delta+ 4e)u = \beta_n ( − Δ + 4 e ) u = β n W ( 1 ) = w ( 1 ) W(1) = w(1) W ( 1 ) = w ( 1 ) w ( r ) = W ( r ) w(r) = W(r) w ( r ) = W ( r ) r ≥ 1 r \ge1 r ≥ 1 r ≥ 1 r \ge1 r ≥ 1
w ( r ) = Y 4 e ∗ ( β n 1 B 1 ( 0 ) c ) . (50) w(r) = Y_{4e} * (\beta_n 1_{B_1(0)^c}). \tag*{(50)} w ( r ) = Y 4 e ∗ ( β n 1 B 1 ( 0 ) c ) . (50) We then estimate ∥ w ∥ L 1 ( B 1 c ) ≤ ∥ w ∥ L 1 \|w\|_{L^1(B_1^c)} \le\|w\|_{L^1} ∥ w ∥ L 1 ( B 1 c ) ≤ ∥ w ∥ L 1 ∥ Y 4 e ∥ L 1 = 1 / ( 4 e ) \|Y_{4e}\|_{L^1} = 1/(4e) ∥ Y 4 e ∥ L 1 = 1/ ( 4 e )
∥ w ∥ L 1 ( B 1 c ) ≤ ∥ β n ∥ L 1 4 e \|w\|_{L^1(B_1^c)} \le\frac{\|\beta_n\|_{L^1}}{4e} ∥ w ∥ L 1 ( B 1 c ) ≤ 4 e ∥ β n ∥ L 1 By the induction hypothesis and Young’s inequality,
∥ β n ∥ L 1 ≤ 2 e ∣ ρ n − 1 − ρ ˉ m , n − 1 ∣ ∥ τ n − 1 ∥ L 1 2 + 2 e ρ ˉ m , n − 1 ( ∥ τ n − 1 ∥ L 1 + ∥ σ m , n − 1 ∥ L 1 ) ∥ τ n − 1 − σ m , n − 1 ∥ L 1 → 0. \|\beta_n\|_{L^1} \le2e|\rho_{n-1}-\bar{\rho}_{m,n-1}|\|\tau_{n-1}\|_{L^1}^2+2e\bar{\rho}_{m,n-1}\left(\|\tau_{n-1}\|_{L^1}+\|\sigma_{m,n-1}\|_{L^1}\right)\|\tau_{n-1}-\sigma_{m,n-1}\|_{L^1}\to0. ∥ β n ∥ L 1 ≤ 2 e ∣ ρ n − 1 − ρ ˉ m , n − 1 ∣∥ τ n − 1 ∥ L 1 2 + 2 e ρ ˉ m , n − 1 ( ∥ τ n − 1 ∥ L 1 + ∥ σ m , n − 1 ∥ L 1 ) ∥ τ n − 1 − σ m , n − 1 ∥ L 1 → 0. Combining with ∫ B 1 ( 0 ) ∣ τ n − σ m , n ∣ ≡ ∫ B 1 ( 0 ) ( 1 − σ m , n ) = O ( 1 / m ) \int_{B_1(0)}|\tau_n-\sigma_{m,n}| \equiv\int_{B_1(0)}(1-\sigma_{m,n}) = O(1/m) ∫ B 1 ( 0 ) ∣ τ n − σ m , n ∣ ≡ ∫ B 1 ( 0 ) ( 1 − σ m , n ) = O ( 1/ m ) ∥ σ m , n − τ n ∥ L 1 → 0 \|\sigma_{m,n}-\tau_n\|_{L^1}\to0 ∥ σ m , n − τ n ∥ L 1 → 0
Density convergence. By the integral relation (47) and the L 1 L^1 L 1
1 ρ ˉ m , n = 2 ∫ σ m , n − ρ ˉ m , n − 1 ( ∫ σ m , n − 1 ) 2 ⟶ 2 ∫ τ n − ρ n − 1 ( ∫ τ n − 1 ) 2 = 1 ρ n , \frac{1}{\bar{\rho}_{m,n}}=2\int\sigma_{m,n}-\bar{\rho}_{m,n-1}\left(\int\sigma_{m,n-1}\right)^2 \longrightarrow2\int\tau_n-\rho_{n-1}\left(\int\tau_{n-1}\right)^2=\frac{1}{\rho_n}, ρ ˉ m , n 1 = 2 ∫ σ m , n − ρ ˉ m , n − 1 ( ∫ σ m , n − 1 ) 2 ⟶ 2 ∫ τ n − ρ n − 1 ( ∫ τ n − 1 ) 2 = ρ n 1 , the last equality being (43); hence ρ ˉ m , n → ρ n \bar{\rho}_{m,n}\to\rho_n ρ ˉ m , n → ρ n
Convergence of the Converged Soft-Core Solution
Let σ m = lim n → ∞ σ m , n \sigma_m=\lim_{n\to\infty}\sigma_{m,n} σ m = lim n → ∞ σ m , n ρ ˉ m = lim n → ∞ ρ ˉ m , n \bar{\rho}_m=\lim_{n\to\infty}\bar{\rho}_{m,n} ρ ˉ m = lim n → ∞ ρ ˉ m , n [11 Lemma 3.3] , ( σ m , ρ ˉ m ) (\sigma_m,\bar{\rho}_m) ( σ m , ρ ˉ m )
( − Δ + 4 e + v m ) σ m = v m + 2 e ρ ˉ m σ m ∗ σ m (51) (-\Delta+4e+v_m)\sigma_m=v_m+2e\bar{\rho}_m\sigma_m*\sigma_m \tag*{(51)} ( − Δ + 4 e + v m ) σ m = v m + 2 e ρ ˉ m σ m ∗ σ m (51) with
ρ ˉ m = 2 e m ∫ B 1 ( 0 ) ( 1 − σ m ) . (52) \bar{\rho}_m=\frac{2e}{m\int_{B_1(0)}(1-\sigma_m)}. \tag*{(52)} ρ ˉ m = m ∫ B 1 ( 0 ) ( 1 − σ m ) 2 e . (52) In particular, we have ρ ˉ m ∫ σ m = 1 \bar{\rho}_m\int\sigma_m=1 ρ ˉ m ∫ σ m = 1
Lemma 15. The map m ↦ ρ ˉ m m\mapsto\bar{\rho}_m m ↦ ρ ˉ m ρ ˉ m ≥ ρ h c \bar{\rho}_m\ge\rho_{\mathrm{hc}} ρ ˉ m ≥ ρ hc m m m
Proof. The derivative formula. For f ∈ L 1 f\in L^1 f ∈ L 1 C f C_f C f C f g : = f ∗ g C_f g:=f*g C f g := f ∗ g
K m = ( − Δ + 4 e + v m − 4 e ρ ˉ m C σ m ) − 1 . (53) K_m=\left(-\Delta+4e+v_m-4e\bar{\rho}_m C_{\sigma_m}\right)^{-1}. \tag*{(53)} K m = ( − Δ + 4 e + v m − 4 e ρ ˉ m C σ m ) − 1 . (53) The operator K m − 1 K_m^{-1} K m − 1 L 2 L^2 L 2 v m v_m v m σ m \sigma_m σ m C σ m C_{\sigma_m} C σ m ∥ σ m ∥ L 1 = 1 / ρ ˉ m \|\sigma_m\|_{L^1}=1/\bar{\rho}_m ∥ σ m ∥ L 1 = 1/ ρ ˉ m K m [ v m ] : = ∫ 0 ∞ e − t K m − 1 v m d t K_m[v_m]:=\int_0^\infty e^{-tK_m^{-1}}v_m\,dt K m [ v m ] := ∫ 0 ∞ e − t K m − 1 v m d t Φ m : = K m [ v m ] \Phi_m:=K_m[v_m] Φ m := K m [ v m ] 0 ≤ Φ m ≤ 1 0\leq\Phi_m\leq1 0 ≤ Φ m ≤ 1 Φ m < 1 \Phi_m<1 Φ m < 1 [12 Lemma 7.5] .
Since ρ ˉ m ∫ σ m = 1 \bar{\rho}_m\int\sigma_m=1 ρ ˉ m ∫ σ m = 1
K m − 1 1 = 4 e + v m − 4 e ρ ˉ m ∫ σ m = v m = K m − 1 Φ m , K_m^{-1}1=4e+v_m-4e\bar{\rho}_m\int\sigma_m=v_m=K_m^{-1}\Phi_m, K m − 1 1 = 4 e + v m − 4 e ρ ˉ m ∫ σ m = v m = K m − 1 Φ m , so 1 − Φ m 1-\Phi_m 1 − Φ m
( − Δ + 4 e + v m ) ( 1 − Φ m ) = 4 e ρ ˉ m σ m ∗ ( 1 − Φ m ) . (54) (-\Delta+4e+v_m)(1-\Phi_m)=4e\bar{\rho}_m\sigma_m*(1-\Phi_m). \tag*{(54)} ( − Δ + 4 e + v m ) ( 1 − Φ m ) = 4 e ρ ˉ m σ m ∗ ( 1 − Φ m ) . (54) Multiplying (52) by 1 − Φ m 1-\Phi_m 1 − Φ m σ m \sigma_m σ m R 3 \mathbb{R}^3 R 3
∫ ( 1 − Φ m ) ( − Δ + 4 e + v m ) σ m = ∫ ( 1 − Φ m ) v m + 2 e ρ ˉ m ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) , (55) \int(1-\Phi_m)(-\Delta+4e+v_m)\sigma_m=\int(1-\Phi_m)v_m+2e\bar{\rho}_m\int(1-\Phi_m)(\sigma_m*\sigma_m), \tag*{(55)} ∫ ( 1 − Φ m ) ( − Δ + 4 e + v m ) σ m = ∫ ( 1 − Φ m ) v m + 2 e ρ ˉ m ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) , (55) ∫ σ m ( − Δ + 4 e + v m ) ( 1 − Φ m ) = 4 e ρ ˉ m ∫ σ m ( σ m ∗ ( 1 − Φ m ) ) . (56) \int\sigma_m(-\Delta+4e+v_m)(1-\Phi_m)=4e\bar{\rho}_m\int\sigma_m(\sigma_m*(1-\Phi_m)). \tag*{(56)} ∫ σ m ( − Δ + 4 e + v m ) ( 1 − Φ m ) = 4 e ρ ˉ m ∫ σ m ( σ m ∗ ( 1 − Φ m )) . (56) All integrals converge: the right-hand side of (52) lies in L 1 L^1 L 1 0 ≤ 1 − Φ m ≤ 1 0\leq1-\Phi_m\leq1 0 ≤ 1 − Φ m ≤ 1 σ m ∈ L 1 \sigma_m\in L^1 σ m ∈ L 1 σ m ∗ ( 1 − Φ m ) \sigma_m*(1-\Phi_m) σ m ∗ ( 1 − Φ m ) 4 e + v m 4e+v_m 4 e + v m σ m ( x ) = O ( ∣ x ∣ − 4 ) \sigma_m(x)=O(|x|^{-4}) σ m ( x ) = O ( ∣ x ∣ − 4 ) [11 ] along with
∇ σ m = ∇ Y 4 e ∗ ( v m ( 1 − σ m ) + 2 e ρ ˉ m σ m ∗ σ m ) = O ( ∣ x ∣ − 4 ) (57) \nabla\sigma_m=\nabla Y_{4e}*(v_m(1-\sigma_m)+2e\bar{\rho}_m\sigma_m*\sigma_m)=O(|x|^{-4}) \tag*{(57)} ∇ σ m = ∇ Y 4 e ∗ ( v m ( 1 − σ m ) + 2 e ρ ˉ m σ m ∗ σ m ) = O ( ∣ x ∣ − 4 ) (57) while ∣ Φ m ∣ ≤ 1 |\Phi_m|\leq1 ∣ Φ m ∣ ≤ 1
∇ Φ m = ∇ Y 4 e ∗ ( v m ( 1 − Φ m ) + 4 e ρ ˉ m σ m ∗ Φ m ) = O ( 1 ) . (58) \nabla\Phi_m=\nabla Y_{4e}*(v_m(1-\Phi_m)+4e\bar{\rho}_m\sigma_m*\Phi_m)=O(1). \tag*{(58)} ∇ Φ m = ∇ Y 4 e ∗ ( v m ( 1 − Φ m ) + 4 e ρ ˉ m σ m ∗ Φ m ) = O ( 1 ) . (58) Thus,
∫ ( 1 − Φ m ) v m + 2 e ρ ˉ m ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) = 4 e ρ ˉ m ∫ σ m ( σ m ∗ ( 1 − Φ m ) ) . (59) \int(1-\Phi_m)v_m+2e\bar{\rho}_m\int(1-\Phi_m)(\sigma_m*\sigma_m)=4e\bar{\rho}_m\int\sigma_m(\sigma_m*(1-\Phi_m)). \tag*{(59)} ∫ ( 1 − Φ m ) v m + 2 e ρ ˉ m ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) = 4 e ρ ˉ m ∫ σ m ( σ m ∗ ( 1 − Φ m )) . (59) Furthermore, Fubini’s theorem and the evenness of the radial function σ m \sigma_m σ m
∫ σ m ( x ) ( σ m ∗ ( 1 − Φ m ) ) ( x ) d x = ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) \int\sigma_m(x)(\sigma_m*(1-\Phi_m))(x)\,dx=\int(1-\Phi_m)(\sigma_m*\sigma_m) ∫ σ m ( x ) ( σ m ∗ ( 1 − Φ m )) ( x ) d x = ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) and so
∫ ( 1 − Φ m ) v m = 2 e ρ ˉ m ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) . (60) \int(1-\Phi_m)v_m=2e\bar{\rho}_m\int(1-\Phi_m)(\sigma_m*\sigma_m). \tag*{(60)} ∫ ( 1 − Φ m ) v m = 2 e ρ ˉ m ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) . (60) Now differentiate in m m m [12 Section 3] . Fix m ′ ≠ m m'\ne m m ′ = m δ m : = m ′ − m \delta m:=m'-m δ m := m ′ − m δ σ : = σ m ′ − σ m \delta\sigma:=\sigma_{m'}-\sigma_m δ σ := σ m ′ − σ m δ ρ : = ρ ˉ m ′ − ρ ˉ m \delta\rho:=\bar{\rho}_{m'}-\bar{\rho}_m δ ρ := ρ ˉ m ′ − ρ ˉ m σ ~ : = 1 2 ( σ m + σ m ′ ) \tilde{\sigma}:=\frac{1}{2}(\sigma_m+\sigma_{m'}) σ ~ := 2 1 ( σ m + σ m ′ )
K ~ − 1 : = − Δ + 4 e + v m ′ − 4 e ρ ˉ m ′ C σ ~ ; \widetilde{K}^{-1}:=-\Delta+4e+v_{m'}-4e\bar{\rho}_{m'}C_{\tilde{\sigma}}; K − 1 := − Δ + 4 e + v m ′ − 4 e ρ ˉ m ′ C σ ~ ; by [12 Section 3] there is a bounded radial function Φ ~ \widetilde{\Phi} Φ K ~ − 1 Φ ~ = v m ′ \widetilde{K}^{-1}\widetilde{\Phi}=v_{m'} K − 1 Φ = v m ′ 0 ≤ Φ ~ ≤ 1 0\leq\widetilde{\Phi}\leq1 0 ≤ Φ ≤ 1 Φ m \Phi_m Φ m ∇ Φ ~ \nabla\widetilde{\Phi} ∇ Φ m m m m ′ m' m ′ σ m ′ ∗ σ m ′ − σ m ∗ σ m = 2 δ σ ∗ σ ~ \sigma_{m'}\ast\sigma_{m'}-\sigma_m\ast\sigma_m=2\delta\sigma\ast\widetilde{\sigma} σ m ′ ∗ σ m ′ − σ m ∗ σ m = 2 δ σ ∗ σ v m ′ − v m = δ m 1 B 1 ( 0 ) v_{m'}-v_m=\delta m\,\mathbb{1}_{B_1(0)} v m ′ − v m = δ m 1 B 1 ( 0 )
K ~ − 1 δ σ = δ m 1 B 1 ( 0 ) ( 1 − σ m ) + 2 e δ ρ ( σ m ∗ σ m ) , (61) \widetilde{K}^{-1}\delta\sigma=\delta m\,\mathbb{1}_{B_1(0)}(1-\sigma_m)+2e\,\delta\rho\,(\sigma_m\ast\sigma_m), \tag*{(61)} K − 1 δ σ = δ m 1 B 1 ( 0 ) ( 1 − σ m ) + 2 e δ ρ ( σ m ∗ σ m ) , (61) while subtracting 2 e / ρ ‾ m = m ∫ B 1 ( 0 ) ( 1 − σ m ) 2e/\overline{\rho}_m=m\int_{B_1(0)}(1-\sigma_m) 2 e / ρ m = m ∫ B 1 ( 0 ) ( 1 − σ m ) 2 e / ρ ‾ m ′ = m ′ ∫ B 1 ( 0 ) ( 1 − σ m ′ ) 2e/\overline{\rho}_{m'}=m'\int_{B_1(0)}(1-\sigma_{m'}) 2 e / ρ m ′ = m ′ ∫ B 1 ( 0 ) ( 1 − σ m ′ )
− 2 e δ ρ ρ ‾ m ρ ‾ m ′ = δ m ∫ B 1 ( 0 ) ( 1 − σ m ) − ∫ v m ′ δ σ . (62) -\frac{2e\,\delta\rho}{\overline{\rho}_m\overline{\rho}_{m'}}=\delta m\int_{B_1(0)}(1-\sigma_m)-\int v_{m'}\,\delta\sigma. \tag*{(62)} − ρ m ρ m ′ 2 e δ ρ = δ m ∫ B 1 ( 0 ) ( 1 − σ m ) − ∫ v m ′ δ σ . (62) Multiplying K ~ − 1 Φ ~ = v m ′ \widetilde{K}^{-1}\widetilde{\Phi}=v_{m'} K − 1 Φ = v m ′ δ σ \delta\sigma δ σ δ σ \delta\sigma δ σ δ σ , ∇ δ σ = O ( ∣ x ∣ − 4 ) \delta\sigma,\nabla\delta\sigma=O(|x|^{-4}) δ σ , ∇ δ σ = O ( ∣ x ∣ − 4 ) Φ ~ , ∇ Φ ~ \widetilde{\Phi},\nabla\widetilde{\Phi} Φ , ∇ Φ σ ~ \widetilde{\sigma} σ ∫ v m ′ δ σ = ∫ Φ ~ K ~ − 1 δ σ \int v_{m'}\,\delta\sigma=\int\widetilde{\Phi}\,\widetilde{K}^{-1}\delta\sigma ∫ v m ′ δ σ = ∫ Φ K − 1 δ σ
∫ v m ′ δ σ = ∫ Φ ~ [ δ m 1 B 1 ( 0 ) ( 1 − σ m ) + 2 e δ ρ ( σ m ∗ σ m ) ] . (63) \int v_{m'}\,\delta\sigma=\int\widetilde{\Phi}\left[\delta m\,\mathbb{1}_{B_1(0)}(1-\sigma_m)+2e\,\delta\rho\,(\sigma_m\ast\sigma_m)\right]. \tag*{(63)} ∫ v m ′ δ σ = ∫ Φ [ δ m 1 B 1 ( 0 ) ( 1 − σ m ) + 2 e δ ρ ( σ m ∗ σ m ) ] . (63) Using ∫ ( σ m ∗ σ m ) = ( ∫ σ m ) 2 = 1 / ρ ‾ m 2 \int(\sigma_m\ast\sigma_m)=(\int\sigma_m)^2=1/\overline{\rho}_m^2 ∫ ( σ m ∗ σ m ) = ( ∫ σ m ) 2 = 1/ ρ m 2 δ ρ = ρ ‾ m ′ − ρ ‾ m \delta\rho=\overline{\rho}_{m'}-\overline{\rho}_m δ ρ = ρ m ′ − ρ m
− 2 e δ ρ ∫ σ m ∗ σ m + 2 e ( δ ρ ) 2 ρ ‾ m ρ ‾ m ′ = − 2 e δ ρ ρ ‾ m 2 + 2 e δ ρ ( ρ ‾ m ′ − ρ ‾ m ) ρ ‾ m 2 ρ ‾ m ′ = − 2 e δ ρ ρ ‾ m ρ ‾ m ′ . (64) -2e\,\delta\rho\int\sigma_m\ast\sigma_m+\frac{2e(\delta\rho)^2}{\overline{\rho}_m\overline{\rho}_{m'}}=-\frac{2e\,\delta\rho}{\overline{\rho}_m^2}+\frac{2e\,\delta\rho(\overline{\rho}_{m'}-\overline{\rho}_m)}{\overline{\rho}_m^2\overline{\rho}_{m'}}=-\frac{2e\,\delta\rho}{\overline{\rho}_m\overline{\rho}_{m'}}. \tag*{(64)} − 2 e δ ρ ∫ σ m ∗ σ m + ρ m ρ m ′ 2 e ( δ ρ ) 2 = − ρ m 2 2 e δ ρ + ρ m 2 ρ m ′ 2 e δ ρ ( ρ m ′ − ρ m ) = − ρ m ρ m ′ 2 e δ ρ . (64) so, inserting (62) and (63) into (61), we find
0 = δ m ∫ B 1 ( 0 ) ( 1 − Φ ~ ) ( 1 − σ m ) + 2 e δ ρ [ ∫ ( 1 − Φ ~ ) ( σ m ∗ σ m ) − δ ρ ρ ‾ m 2 ρ ‾ m ′ ] . 0=\delta m\int_{B_1(0)}(1-\widetilde{\Phi})(1-\sigma_m)+2e\,\delta\rho\left[\int(1-\widetilde{\Phi})(\sigma_m\ast\sigma_m)-\frac{\delta\rho}{\overline{\rho}_m^2\overline{\rho}_{m'}}\right]. 0 = δ m ∫ B 1 ( 0 ) ( 1 − Φ ) ( 1 − σ m ) + 2 e δ ρ [ ∫ ( 1 − Φ ) ( σ m ∗ σ m ) − ρ m 2 ρ m ′ δ ρ ] . Solving for the difference quotient, we arrive at the exact identity
δ ρ δ m = ρ ‾ m ′ − ρ ‾ m m ′ − m = − ∫ B 1 ( 0 ) ( 1 − Φ ~ ) ( 1 − σ m ) 2 e [ ∫ ( 1 − Φ ~ ) ( σ m ∗ σ m ) − δ ρ ρ ‾ m 2 ρ ‾ m ′ ] . (65) \frac{\delta\rho}{\delta m}=\frac{\overline{\rho}_{m'}-\overline{\rho}_m}{m'-m}=-\frac{\int_{B_1(0)}(1-\widetilde{\Phi})(1-\sigma_m)}{2e\left[\int(1-\widetilde{\Phi})(\sigma_m\ast\sigma_m)-\frac{\delta\rho}{\overline{\rho}_m^2\overline{\rho}_{m'}}\right]}. \tag*{(65)} δ m δ ρ = m ′ − m ρ m ′ − ρ m = − 2 e [ ∫ ( 1 − Φ ) ( σ m ∗ σ m ) − ρ m 2 ρ m ′ δ ρ ] ∫ B 1 ( 0 ) ( 1 − Φ ) ( 1 − σ m ) . (65) Now, following [12 Section 3 and Lemma 7.7] , we can show that as m ′ → m m'\to m m ′ → m δ ρ → 0 \delta\rho\to0 δ ρ → 0 Φ ~ → Φ m \widetilde{\Phi}\to\Phi_m Φ → Φ m [12 Section 3 and Lemma 7.7] apply almost exactly using the fact that the dependence of K − 1 K^{-1} K − 1 m m m 0 ≤ Φ ~ ≤ 1 0\leq\widetilde{\Phi}\leq1 0 ≤ Φ ≤ 1 2 e ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) = m ρ ‾ m ∫ B 1 ( 0 ) ( 1 − Φ m ) 2e\int(1-\Phi_m)(\sigma_m\ast\sigma_m)=\frac{m}{\overline{\rho}_m}\int_{B_1(0)}(1-\Phi_m) 2 e ∫ ( 1 − Φ m ) ( σ m ∗ σ m ) = ρ m m ∫ B 1 ( 0 ) ( 1 − Φ m )
∂ m ρ ‾ m = − ρ ‾ m m ∫ B 1 ( 0 ) ( 1 − Φ m ) ( 1 − σ m ) ∫ B 1 ( 0 ) ( 1 − Φ m ) . (66) \partial_m\overline{\rho}_m=-\frac{\overline{\rho}_m}{m}\frac{\int_{B_1(0)}(1-\Phi_m)(1-\sigma_m)}{\int_{B_1(0)}(1-\Phi_m)}. \tag*{(66)} ∂ m ρ m = − m ρ m ∫ B 1 ( 0 ) ( 1 − Φ m ) ∫ B 1 ( 0 ) ( 1 − Φ m ) ( 1 − σ m ) . (66) We can now prove that ρ ‾ \overline{\rho} ρ 1 − Φ m ≥ 0 1-\Phi_m\geq0 1 − Φ m ≥ 0 ( − Δ + 4 e + v m ) − 1 (-\Delta+4e+v_m)^{-1} ( − Δ + 4 e + v m ) − 1 1 − Φ m > 0 1-\Phi_m>0 1 − Φ m > 0 ∫ B 1 ( 0 ) ( 1 − σ m ) = 2 e / ( m ρ ‾ m ) > 0 \int_{B_1(0)}(1-\sigma_m)=2e/(m\overline{\rho}_m)>0 ∫ B 1 ( 0 ) ( 1 − σ m ) = 2 e / ( m ρ m ) > 0 ∂ m ρ ‾ m < 0 \partial_m\overline{\rho}_m<0 ∂ m ρ m < 0
Comparison with the hard-core density. By Lemma 13 the iterates σ m , n \sigma_{m,n} σ m , n n n n ρ ‾ m , n = 2 e / ( m ∫ B 1 ( 0 ) ( 1 − σ m , n ) ) \overline{\rho}_{m,n}=2e/(m\int_{B_1(0)}(1-\sigma_{m,n})) ρ m , n = 2 e / ( m ∫ B 1 ( 0 ) ( 1 − σ m , n )) n n n ρ ‾ m \overline{\rho}_m ρ m ρ ‾ m ≥ ρ ‾ m , n \overline{\rho}_m\geq\overline{\rho}_{m,n} ρ m ≥ ρ m , n n n n ρ ‾ m , n → ρ n \overline{\rho}_{m,n}\to\rho_n ρ m , n → ρ n m → ∞ m\to\infty m → ∞ lim inf m → ∞ ρ ‾ m ≥ ρ n \liminf_{m\to\infty}\overline{\rho}_m\geq\rho_n lim inf m → ∞ ρ m ≥ ρ n n n n n → ∞ n\to\infty n → ∞ lim inf m → ∞ ρ ‾ m ≥ ρ h c \liminf_{m\to\infty}\overline{\rho}_m\geq\rho_{\mathrm{hc}} lim inf m → ∞ ρ m ≥ ρ hc ρ ‾ m \overline{\rho}_m ρ m m m m lim m → ∞ ρ ‾ m \lim_{m\to\infty}\overline{\rho}_m lim m → ∞ ρ m ≥ ρ h c \geq\rho_{\mathrm{hc}} ≥ ρ hc ρ ‾ m ≥ ρ h c \overline{\rho}_m\geq\rho_{\mathrm{hc}} ρ m ≥ ρ hc m m m
Lemma 16. For all m ≥ 0 m \ge0 m ≥ 0 σ m ≥ ω m \sigma_m \ge\omega_m σ m ≥ ω m
Proof. Induction on m m m m = 0 m=0 m = 0 σ m − 1 ≥ ω m − 1 \sigma_{m-1} \ge\omega_{m-1} σ m − 1 ≥ ω m − 1 v m = m 1 B 1 ( 0 ) v_m=m\mathbf{1}_{B_1(0)} v m = m 1 B 1 ( 0 )
( − Δ + 4 e + v m ) ( σ m − ω m ) = 2 e [ ρ ˉ m − 1 σ m − 1 ∗ σ m − 1 − ρ h c ω m − 1 ∗ ω m − 1 ] ≥ 0 , (-\Delta+4e+v_m)(\sigma_m-\omega_m)=2e\left[\bar{\rho}_{m-1}\sigma_{m-1}*\sigma_{m-1}-\rho_{\mathrm{hc}}\omega_{m-1}*\omega_{m-1}\right]\ge0, ( − Δ + 4 e + v m ) ( σ m − ω m ) = 2 e [ ρ ˉ m − 1 σ m − 1 ∗ σ m − 1 − ρ hc ω m − 1 ∗ ω m − 1 ] ≥ 0 , because ρ ˉ m − 1 ≥ ρ h c \bar{\rho}_{m-1}\ge\rho_{\mathrm{hc}} ρ ˉ m − 1 ≥ ρ hc σ m − 1 ≥ ω m − 1 \sigma_{m-1}\ge\omega_{m-1} σ m − 1 ≥ ω m − 1 ( − Δ + 4 e + v m ) − 1 (-\Delta+4e+v_m)^{-1} ( − Δ + 4 e + v m ) − 1
Theorem 17. ∥ σ m − τ ∥ L 1 → 0 \|\sigma_m-\tau\|_{L^1}\to0 ∥ σ m − τ ∥ L 1 → 0 ρ ˉ m → ρ h c \bar{\rho}_m\to\rho_{\mathrm{hc}} ρ ˉ m → ρ hc m → ∞ m\to\infty m → ∞
Proof. From Lemma 16 we have σ m ≥ ω m \sigma_m\ge\omega_m σ m ≥ ω m
1 ρ ˉ m = ∫ σ m ≥ ∫ ω m ⟶ 1 ρ h c , \frac{1}{\bar{\rho}_m}=\int\sigma_m\ge\int\omega_m\longrightarrow\frac{1}{\rho_{\mathrm{hc}}}, ρ ˉ m 1 = ∫ σ m ≥ ∫ ω m ⟶ ρ hc 1 , by Corollary 12. Hence lim sup m → ∞ ρ ˉ m ≤ ρ h c \limsup_{m\to\infty}\bar{\rho}_m\le\rho_{\mathrm{hc}} lim sup m → ∞ ρ ˉ m ≤ ρ hc ρ ˉ m → ρ h c \bar{\rho}_m\to\rho_{\mathrm{hc}} ρ ˉ m → ρ hc L 1 L^1 L 1
∥ σ m − τ ∥ L 1 ≤ ∥ σ m − ω m ∥ L 1 + ∥ ω m − τ ∥ L 1 . \|\sigma_m-\tau\|_{L^1}\le\|\sigma_m-\omega_m\|_{L^1}+\|\omega_m-\tau\|_{L^1}. ∥ σ m − τ ∥ L 1 ≤ ∥ σ m − ω m ∥ L 1 + ∥ ω m − τ ∥ L 1 . The second term tends to zero by Corollary 12. Since σ m ≥ ω m \sigma_m\ge\omega_m σ m ≥ ω m
∥ σ m − ω m ∥ L 1 = ∫ σ m − ∫ ω m ⟶ 0. \|\sigma_m-\omega_m\|_{L^1}=\int\sigma_m-\int\omega_m\longrightarrow0. ∥ σ m − ω m ∥ L 1 = ∫ σ m − ∫ ω m ⟶ 0. Hence ∥ σ m − τ ∥ L 1 → 0 \|\sigma_m-\tau\|_{L^1}\to0 ∥ σ m − τ ∥ L 1 → 0
Rate of Convergence
Theorem 18 (Proof of Theorem 3 (i)). ∣ ρ ˉ m − ρ h c ∣ = O ( m − 1 / 2 ) |\bar{\rho}_m-\rho_{\mathrm{hc}}|=O(m^{-1/2}) ∣ ρ ˉ m − ρ hc ∣ = O ( m − 1/2 )
Proof. From (66) , since the ratio appearing there is an average of 1 − σ m 1-\sigma_m 1 − σ m ( 1 − Φ m ) 1 B 1 ( 0 ) (1-\Phi_m)\mathbf{1}_{B_1(0)} ( 1 − Φ m ) 1 B 1 ( 0 )
∣ ∂ m ρ ˉ m ∣ ≤ ρ ˉ m m sup B 1 ( 0 ) ( 1 − σ m ) . |\partial_m\bar{\rho}_m|\le\frac{\bar{\rho}_m}{m}\sup_{B_1(0)}(1-\sigma_m). ∣ ∂ m ρ ˉ m ∣ ≤ m ρ ˉ m B 1 ( 0 ) sup ( 1 − σ m ) . We estimate the supremum as in the core estimate of Lemma 14, now at the level of the converged solution. Set ι m = 1 − σ m ≥ 0 \iota_m=1-\sigma_m\ge0 ι m = 1 − σ m ≥ 0 B 1 ( 0 ) B_1(0) B 1 ( 0 ) (32) at convergence reads
( − Δ + κ 2 ) ι m = 4 e − 2 e ρ ˉ m ( σ m ∗ σ m ) ≡ f , κ = 4 e + m , (-\Delta+\kappa^2)\iota_m=4e-2e\bar{\rho}_m(\sigma_m*\sigma_m)\equiv f,\qquad\kappa=\sqrt{4e+m}, ( − Δ + κ 2 ) ι m = 4 e − 2 e ρ ˉ m ( σ m ∗ σ m ) ≡ f , κ = 4 e + m , with 0 ≤ f ≤ 4 e 0\le f\le4e 0 ≤ f ≤ 4 e σ m ≤ 1 \sigma_m\le1 σ m ≤ 1 ρ ˉ m ∫ σ m = 1 \bar{\rho}_m\int\sigma_m=1 ρ ˉ m ∫ σ m = 1 ρ ˉ m ( σ m ∗ σ m ) ≤ 1 \bar{\rho}_m(\sigma_m*\sigma_m)\le1 ρ ˉ m ( σ m ∗ σ m ) ≤ 1 ι m = ι ~ m + h m \iota_m=\tilde{\iota}_m+h_m ι m = ι ~ m + h m ι ~ m ∣ ∂ B 1 = 0 \tilde{\iota}_m|_{\partial B_1}=0 ι ~ m ∣ ∂ B 1 = 0 h m ( r ) = ι m ( 1 ) sinh ( κ r ) r sinh κ h_m(r)=\iota_m(1)\frac{\sinh(\kappa r)}{r\sinh\kappa} h m ( r ) = ι m ( 1 ) r s i n h κ s i n h ( κ r ) h m ( 1 ) = ι m ( 1 ) h_m(1)=\iota_m(1) h m ( 1 ) = ι m ( 1 ) Corollary 5 with D = B 1 ( 0 ) D=B_1(0) D = B 1 ( 0 ) c = κ 2 c=\kappa^2 c = κ 2 W = 0 W=0 W = 0 F = 4 e F=4e F = 4 e 0 ≤ ι ~ m ≤ 4 e / κ 2 0\le\tilde{\iota}_m\le4e/\kappa^2 0 ≤ ι ~ m ≤ 4 e / κ 2 h m ≤ ι m ( 1 ) h_m\le\iota_m(1) h m ≤ ι m ( 1 ) B 1 ( 0 ) B_1(0) B 1 ( 0 ) r ↦ sinh ( κ r ) / r r\mapsto\sinh(\kappa r)/r r ↦ sinh ( κ r ) / r
sup B 1 ( 0 ) ( 1 − σ m ) ≤ ( 1 − σ m ( 1 ) ) + 4 e κ 2 . \sup_{B_1(0)}(1-\sigma_m)\le(1-\sigma_m(1))+\frac{4e}{\kappa^2}. B 1 ( 0 ) sup ( 1 − σ m ) ≤ ( 1 − σ m ( 1 )) + κ 2 4 e . For the boundary value, the definition of ρ ˉ m \bar{\rho}_m ρ ˉ m ∫ B 1 ( 0 ) ι m = 2 e / ( m ρ ˉ m ) ≤ 2 e / ( m ρ h c ) \int_{B_1(0)} \iota_m = 2e/(m\bar{\rho}_m) \le2e/(m\rho_{\mathrm{hc}}) ∫ B 1 ( 0 ) ι m = 2 e / ( m ρ ˉ m ) ≤ 2 e / ( m ρ hc ) ρ ˉ m ≥ ρ h c \bar{\rho}_m \ge\rho_{\mathrm{hc}} ρ ˉ m ≥ ρ hc ∫ B 1 ( 0 ) h m ∼ 4 π ι m ( 1 ) / κ \int_{B_1(0)} h_m \sim4\pi\iota_m(1)/\kappa ∫ B 1 ( 0 ) h m ∼ 4 π ι m ( 1 ) / κ κ → ∞ \kappa\to\infty κ → ∞ ι m = ι ~ m + h m ≥ h m ≥ 0 \iota_m = \tilde{\iota}_m+h_m \ge h_m \ge0 ι m = ι ~ m + h m ≥ h m ≥ 0 ι m ( 1 ) = O ( κ / m ) = O ( m − 1 / 2 ) \iota_m(1) = O(\kappa/m) = O(m^{-1/2}) ι m ( 1 ) = O ( κ / m ) = O ( m − 1/2 )
sup B 1 ( 0 ) ( 1 − σ m ) = O ( m − 1 / 2 ) . \sup_{B_1(0)}(1-\sigma_m)=O(m^{-1/2}). B 1 ( 0 ) sup ( 1 − σ m ) = O ( m − 1/2 ) . Since ρ ˉ m \bar{\rho}_m ρ ˉ m ρ ˉ m ≤ ρ ˉ 1 \bar{\rho}_m \le\bar{\rho}_1 ρ ˉ m ≤ ρ ˉ 1 m ≥ 1 m \ge1 m ≥ 1 ∣ ∂ m ρ ˉ m ∣ ≤ C m − 3 / 2 |\partial_m\bar{\rho}_m| \le Cm^{-3/2} ∣ ∂ m ρ ˉ m ∣ ≤ C m − 3/2 M > m M>m M > m ρ ˉ M − ρ ˉ m = ∫ m M ∂ m ′ ρ ˉ m ′ d m ′ \bar{\rho}_M-\bar{\rho}_m=\int_m^M \partial_{m'}\bar{\rho}_{m'}\,\mathrm{d}m' ρ ˉ M − ρ ˉ m = ∫ m M ∂ m ′ ρ ˉ m ′ d m ′ M → ∞ M\to\infty M → ∞ ρ h c − ρ ˉ m \rho_{\mathrm{hc}}-\bar{\rho}_m ρ hc − ρ ˉ m
∣ ρ ˉ m − ρ h c ∣ ≤ ∫ m ∞ C ( m ′ ) − 3 / 2 d m ′ = 2 C m . |\bar{\rho}_m-\rho_{\mathrm{hc}}|\le\int_m^\infty C(m')^{-3/2}\,\mathrm{d}m'=\frac{2C}{\sqrt{m}}. ∣ ρ ˉ m − ρ hc ∣ ≤ ∫ m ∞ C ( m ′ ) − 3/2 d m ′ = m 2 C . The LHY Formula
The scattering solution ϕ m \phi_m ϕ m v m v_m v m
( − Δ + v m ) ϕ m = v m , ϕ m ( x ) → 0 as ∣ x ∣ → ∞ ; (-\Delta+v_m)\phi_m=v_m,\qquad\phi_m(x)\to0\text{ as }|x|\to\infty; ( − Δ + v m ) ϕ m = v m , ϕ m ( x ) → 0 as ∣ x ∣ → ∞ ; equivalently, ψ m : = 1 − ϕ m \psi_m:=1-\phi_m ψ m := 1 − ϕ m − Δ ψ m + v m ψ m = 0 -\Delta\psi_m+v_m\psi_m=0 − Δ ψ m + v m ψ m = 0 ψ m → 1 \psi_m\to1 ψ m → 1 v m = m 1 B 1 ( 0 ) v_m=m\mathbb{1}_{B_1(0)} v m = m 1 B 1 ( 0 ) ∂ ψ m ( 0 ) = 0 \partial\psi_m(0)=0 ∂ ψ m ( 0 ) = 0 ψ m \psi_m ψ m ∂ m ψ \partial_m\psi ∂ m ψ ψ m ( r ) = sinh ( m r ) r m cosh m \psi_m(r)=\frac{\sinh(\sqrt{m}r)}{r\sqrt{m}\cosh\sqrt{m}} ψ m ( r ) = r m c o s h m s i n h ( m r ) r ≤ 1 r\le1 r ≤ 1 ψ m ( r ) = 1 − a 0 ( m ) / r \psi_m(r)=1-a_0(m)/r ψ m ( r ) = 1 − a 0 ( m ) / r r ≥ 1 r\ge1 r ≥ 1 r = 1 r=1 r = 1
a 0 ( m ) = 1 − tanh m m . (67) a_0(m)=1-\frac{\tanh\sqrt{m}}{\sqrt{m}}. \tag*{(67)} a 0 ( m ) = 1 − m tanh m . (67) which increases to 1 1 1 m → ∞ m\to\infty m → ∞ 0 ≤ ϕ m ≤ 1 0\le\phi_m\le1 0 ≤ ϕ m ≤ 1
We adopt the Fourier transform convention
f ^ ( k ) = ∫ e i k ⋅ x f ( x ) d x , f ( x ) = 1 ( 2 π ) 3 ∫ e − i k ⋅ x f ^ ( k ) d k . \hat{f}(k)=\int e^{ik\cdot x}f(x)\,\mathrm{d}x,\qquad f(x)=\frac{1}{(2\pi)^3}\int e^{-ik\cdot x}\hat{f}(k)\,\mathrm{d}k. f ^ ( k ) = ∫ e ik ⋅ x f ( x ) d x , f ( x ) = ( 2 π ) 3 1 ∫ e − ik ⋅ x f ^ ( k ) d k . For each m > 0 m>0 m > 0
S m ( k ) = ρ ˉ m 2 e ∫ e i k ⋅ x ( 1 − σ m ( x ) ) v m ( x ) d x . (68) S_m(k)=\frac{\bar{\rho}_m}{2e}\int e^{ik\cdot x}(1-\sigma_m(x))v_m(x)\,\mathrm{d}x. \tag*{(68)} S m ( k ) = 2 e ρ ˉ m ∫ e ik ⋅ x ( 1 − σ m ( x )) v m ( x ) d x . (68) and let w m w_m w m − Δ w m = ( 1 − σ m ) v m -\Delta w_m=(1-\sigma_m)v_m − Δ w m = ( 1 − σ m ) v m w ^ m ( k ) = 2 e S m ( k ) / ( ρ ˉ m k 2 ) \hat{w}_m(k)=2eS_m(k)/(\bar{\rho}_m k^2) w ^ m ( k ) = 2 e S m ( k ) / ( ρ ˉ m k 2 ) 2 e ρ ˉ m σ ^ m 2 − ( k 2 + 4 e ) σ ^ m + 2 e ρ ˉ m S m = 0 2e\bar{\rho}_m\hat{\sigma}_m^2-(k^2+4e)\hat{\sigma}_m+\frac{2e}{\bar{\rho}_m}S_m=0 2 e ρ ˉ m σ ^ m 2 − ( k 2 + 4 e ) σ ^ m + ρ ˉ m 2 e S m = 0 ∣ k ∣ → ∞ |k|\to\infty ∣ k ∣ → ∞ [12 Eq. (1.43)]
σ ^ m ( k ) = 1 ρ ˉ m ( k 2 4 e + 1 − ( k 2 4 e + 1 ) 2 − S m ( k ) ) . (69) \hat{\sigma}_m(k)=\frac{1}{\bar{\rho}_m}\left(\frac{k^2}{4e}+1-\sqrt{\left(\frac{k^2}{4e}+1\right)^2-S_m(k)}\right). \tag*{(69)} σ ^ m ( k ) = ρ ˉ m 1 4 e k 2 + 1 − ( 4 e k 2 + 1 ) 2 − S m ( k ) . (69) (As a consistency check, note that σ ^ m ( 0 ) = 1 / ρ ˉ m \hat{\sigma}_m(0)=1/\bar{\rho}_m σ ^ m ( 0 ) = 1/ ρ ˉ m ρ ˉ m ∫ σ m = 1 \bar{\rho}_m\int\sigma_m=1 ρ ˉ m ∫ σ m = 1
Lemma 19. S m ( k ) S_m(k) S m ( k ) ∣ S m ( k ) ∣ ≤ 1 |S_m(k)| \le1 ∣ S m ( k ) ∣ ≤ 1 k k k m m m S m ( 0 ) = 1 S_m(0)=1 S m ( 0 ) = 1
Proof. ( 1 − σ m ) v m (1-\sigma_m)v_m ( 1 − σ m ) v m S m S_m S m ∣ S m ( k ) ∣ ≤ S m ( 0 ) = ρ ˉ m 2 e ∫ ( 1 − σ m ) v m = 1 |S_m(k)| \le S_m(0) = \frac{\bar{\rho}_m}{2e}\int(1-\sigma_m)v_m = 1 ∣ S m ( k ) ∣ ≤ S m ( 0 ) = 2 e ρ ˉ m ∫ ( 1 − σ m ) v m = 1 ρ ˉ m \bar{\rho}_m ρ ˉ m [12 Sec. 5] ). □ \square □
Lemma 20. Let g ( k ~ , s ) : = k ~ 2 + 1 − ( k ~ 2 + 1 ) 2 − s − s 2 k ~ 2 g(\tilde{k},s):=\tilde{k}^{2}+1-\sqrt{(\tilde{k}^{2}+1)^{2}-s}-\frac{s}{2\tilde{k}^{2}} g ( k ~ , s ) := k ~ 2 + 1 − ( k ~ 2 + 1 ) 2 − s − 2 k ~ 2 s h ( k ~ ) : = g ( k ~ , − 1 ) − g ( k ~ , 1 ) h(\tilde{k}):=g(\tilde{k},-1)-g(\tilde{k},1) h ( k ~ ) := g ( k ~ , − 1 ) − g ( k ~ , 1 ) h ∈ L 1 ( R 3 ) h\in L^{1}(\mathbb{R}^{3}) h ∈ L 1 ( R 3 ) x ∈ R 3 x\in\mathbb{R}^{3} x ∈ R 3 m > 0 m>0 m > 0
∣ σ m ( x ) − w m ( x ) ∣ ≤ C 0 e 3 / 2 ρ ˉ m , C 0 : = 1 π 3 ∫ h ( k ~ ) d 3 k ~ . |\sigma_m(x)-w_m(x)|\le\frac{C_0e^{3/2}}{\bar{\rho}_m},\qquad C_0:=\frac{1}{\pi^{3}}\int h(\tilde{k})\,d^{3}\tilde{k}. ∣ σ m ( x ) − w m ( x ) ∣ ≤ ρ ˉ m C 0 e 3/2 , C 0 := π 3 1 ∫ h ( k ~ ) d 3 k ~ . Proof. Set k ~ = k / ( 2 e ) \tilde{k}=k/(2\sqrt{e}) k ~ = k / ( 2 e ) w ^ m \widehat{w}_m w m
σ m ( x ) − w m ( x ) = e 3 / 2 ρ ˉ m π 3 ∫ e − 2 i e k ~ ⋅ x g ( k ~ , S m ( 2 e k ~ ) ) d 3 k ~ . (70) \sigma_m(x)-w_m(x)=\frac{e^{3/2}}{\bar{\rho}_m\pi^{3}}\int e^{-2i\sqrt{e}\tilde{k}\cdot x}g(\tilde{k},S_m(2\sqrt{e}\tilde{k}))\,d^{3}\tilde{k}. \tag*{(70)} σ m ( x ) − w m ( x ) = ρ ˉ m π 3 e 3/2 ∫ e − 2 i e k ~ ⋅ x g ( k ~ , S m ( 2 e k ~ )) d 3 k ~ . (70) (cf. [12 Eq. (5.13)] for the analogous representation). The map s ↦ g ( k ~ , s ) s\mapsto g(\tilde{k},s) s ↦ g ( k ~ , s ) [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] ∂ s g = 1 2 ( k ~ 2 + 1 ) 2 − s − 1 2 k ~ 2 ≤ 0 \partial_sg=\frac{1}{2\sqrt{(\tilde{k}^{2}+1)^{2}-s}}-\frac{1}{2\tilde{k}^{2}}\le0 ∂ s g = 2 ( k ~ 2 + 1 ) 2 − s 1 − 2 k ~ 2 1 ≤ 0 g ( k ~ , 0 ) = 0 g(\tilde{k},0)=0 g ( k ~ , 0 ) = 0 ∣ g ( k ~ , S m ) ∣ ≤ max { − g ( k ~ , 1 ) , g ( k ~ , − 1 ) } ≤ h ( k ~ ) |g(\tilde{k},S_m)|\le\max\{-g(\tilde{k},1),g(\tilde{k},-1)\}\le h(\tilde{k}) ∣ g ( k ~ , S m ) ∣ ≤ max { − g ( k ~ , 1 ) , g ( k ~ , − 1 )} ≤ h ( k ~ ) g ( k ~ , − 1 ) ≥ 0 ≥ g ( k ~ , 1 ) g(\tilde{k},-1)\ge0\ge g(\tilde{k},1) g ( k ~ , − 1 ) ≥ 0 ≥ g ( k ~ , 1 ) [12 Eq. (5.14)] ). As k ~ → 0 \tilde{k}\to0 k ~ → 0 h ( k ~ ) = O ( k ~ − 2 ) h(\tilde{k})=O(\tilde{k}^{-2}) h ( k ~ ) = O ( k ~ − 2 ) k ~ → ∞ \tilde{k}\to\infty k ~ → ∞ h ( k ~ ) = O ( k ~ − 4 ) h(\tilde{k})=O(\tilde{k}^{-4}) h ( k ~ ) = O ( k ~ − 4 ) h ∈ L 1 ( R 3 ) h\in L^{1}(\mathbb{R}^{3}) h ∈ L 1 ( R 3 ) □ \square □
Lemma 21. (See also [11] Eq. (4.21 and 4.40))
e − 2 π ρ ˉ m a 0 ( m ) = ρ ˉ m 2 ∫ ( w m − σ m ) ( 1 − ϕ m ) v m d x . (71) e-2\pi\bar{\rho}_ma_0(m)=\frac{\bar{\rho}_m}{2}\int(w_m-\sigma_m)(1-\phi_m)v_m\,dx. \tag*{(71)} e − 2 π ρ ˉ m a 0 ( m ) = 2 ρ ˉ m ∫ ( w m − σ m ) ( 1 − ϕ m ) v m d x . (71) Proof. From e = ρ ˉ m 2 ∫ ( 1 − σ m ) v m e=\frac{\bar{\rho}_m}{2}\int(1-\sigma_m)v_m e = 2 ρ ˉ m ∫ ( 1 − σ m ) v m 2 π ρ ˉ m a 0 ( m ) = ρ ˉ m 2 ∫ ( 1 − ϕ m ) v m 2\pi\bar{\rho}_ma_0(m)=\frac{\bar{\rho}_m}{2}\int(1-\phi_m)v_m 2 π ρ ˉ m a 0 ( m ) = 2 ρ ˉ m ∫ ( 1 − ϕ m ) v m
e − 2 π ρ ˉ m a 0 ( m ) = ρ ˉ m 2 ∫ ( ϕ m − σ m ) v m . e-2\pi\bar{\rho}_ma_0(m)=\frac{\bar{\rho}_m}{2}\int(\phi_m-\sigma_m)v_m. e − 2 π ρ ˉ m a 0 ( m ) = 2 ρ ˉ m ∫ ( ϕ m − σ m ) v m . Set ξ m = w m − ϕ m \xi_m=w_m-\phi_m ξ m = w m − ϕ m ( − Δ + v m ) ξ m = − ( σ m − w m ) v m (-\Delta+v_m)\xi_m=-(\sigma_m-w_m)v_m ( − Δ + v m ) ξ m = − ( σ m − w m ) v m ( − Δ + v m ) − 1 (-\Delta+v_m)^{-1} ( − Δ + v m ) − 1 ( − Δ + v m ) − 1 v m = ϕ m (-\Delta+v_m)^{-1}v_m=\phi_m ( − Δ + v m ) − 1 v m = ϕ m ∫ ξ m v m = ∫ ( w m − σ m ) ϕ m v m \int\xi_mv_m=\int(w_m-\sigma_m)\phi_mv_m ∫ ξ m v m = ∫ ( w m − σ m ) ϕ m v m ϕ m − σ m = ( w m − σ m ) − ξ m \phi_m-\sigma_m=(w_m-\sigma_m)-\xi_m ϕ m − σ m = ( w m − σ m ) − ξ m [11 Eq. (4.21 and 4.40)] ). □ \square □
Proof of Theorem 1. Let A : = 32 2 15 π 2 = − π − 3 ∫ R 3 g ( k ~ , 1 ) d 3 k ~ A:=\frac{32\sqrt{2}}{15\pi^{2}}=-\pi^{-3}\int_{\mathbb{R}^{3}}g(\tilde{k},1)\,d^{3}\tilde{k} A := 15 π 2 32 2 = − π − 3 ∫ R 3 g ( k ~ , 1 ) d 3 k ~ ( k ~ 2 + 1 ) 2 − 1 = k ~ k ~ 2 + 2 \sqrt{(\tilde{k}^{2}+1)^{2}-1}=\tilde{k}\sqrt{\tilde{k}^{2}+2} ( k ~ 2 + 1 ) 2 − 1 = k ~ k ~ 2 + 2 [11 ] ), and define
ρ ˉ m f e , m ( x ) : = σ m ( x ) − w m ( x ) + A e 3 / 2 ρ ˉ m . (72) \sqrt{\bar{\rho}_m}f_{e,m}(x):=\sigma_m(x)-w_m(x)+\frac{Ae^{3/2}}{\bar{\rho}_m}. \tag*{(72)} ρ ˉ m f e , m ( x ) := σ m ( x ) − w m ( x ) + ρ ˉ m A e 3/2 . (72) By the representation Eq. (70) and the definition of A A A
f e , m ( x ) = e 3 / 2 ρ ˉ m 3 / 2 π 3 ∫ R 3 ( e − 2 i e k ~ ⋅ x g ( k ~ , S m ( 2 e k ~ ) ) − g ( k ~ , 1 ) ) d 3 k ~ . (73) f_{e,m}(x)=\frac{e^{3/2}}{\bar{\rho}_m^{3/2}\pi^{3}}\int_{\mathbb{R}^{3}}\left(e^{-2i\sqrt{e}\tilde{k}\cdot x}g(\tilde{k},S_m(2\sqrt{e}\tilde{k}))-g(\tilde{k},1)\right)\,d^{3}\tilde{k}. \tag*{(73)} f e , m ( x ) = ρ ˉ m 3/2 π 3 e 3/2 ∫ R 3 ( e − 2 i e k ~ ⋅ x g ( k ~ , S m ( 2 e k ~ )) − g ( k ~ , 1 ) ) d 3 k ~ . (73) We claim that
λ ( e ) : = π − 3 ∫ R 3 sup m > 0 , ∣ x ∣ ≤ 1 ∣ e − 2 i e k ~ ⋅ x g ( k ~ , S m ( 2 e k ~ ) ) − g ( k ~ , 1 ) ∣ d 3 k ~ ⟶ 0 ( e → 0 ) . \lambda(e):=\pi^{-3}\int_{\mathbb{R}^{3}}\sup_{m>0,\ |x|\le1}\left|e^{-2i\sqrt{e}\tilde{k}\cdot x}g(\tilde{k},S_m(2\sqrt{e}\tilde{k}))-g(\tilde{k},1)\right|\,d^{3}\tilde{k}\longrightarrow0\qquad(e\to0). λ ( e ) := π − 3 ∫ R 3 m > 0 , ∣ x ∣ ≤ 1 sup e − 2 i e k ~ ⋅ x g ( k ~ , S m ( 2 e k ~ )) − g ( k ~ , 1 ) d 3 k ~ ⟶ 0 ( e → 0 ) . Indeed, d ν m ( y ) : = ρ ˉ m 2 ( 1 − σ m ( y ) ) v m ( y ) d y d\nu_m(y) := \frac{\bar{\rho}_m}{2}(1-\sigma_m(y))v_m(y)\,dy d ν m ( y ) := 2 ρ ˉ m ( 1 − σ m ( y )) v m ( y ) d y B 1 ( 0 ) B_1(0) B 1 ( 0 ) Lemma 19 , and S m ( k ) = ∫ B 1 ( 0 ) e i k ⋅ y d ν m ( y ) S_m(k) = \int_{B_1(0)} e^{ik\cdot y}\,d\nu_m(y) S m ( k ) = ∫ B 1 ( 0 ) e ik ⋅ y d ν m ( y ) ∣ e − 2 i e k ~ ⋅ x − 1 ∣ ≤ 2 e ∣ k ~ ∣ \lvert e^{-2i\sqrt{e}\,\tilde{k}\cdot x}-1\rvert\le2\sqrt{e}\lvert\tilde{k}\rvert ∣ e − 2 i e k ~ ⋅ x − 1 ∣ ≤ 2 e ∣ k ~ ∣ ∣ S m ( 2 e k ~ ) − 1 ∣ ≤ 2 e ∣ k ~ ∣ \lvert S_m(2\sqrt{e}\,\tilde{k})-1\rvert\le2\sqrt{e}\lvert\tilde{k}\rvert ∣ S m ( 2 e k ~ ) − 1 ∣ ≤ 2 e ∣ k ~ ∣ m m m ∣ x ∣ ≤ 1 \lvert x\rvert\le1 ∣ x ∣ ≤ 1 s ↦ g ( k ~ , s ) s\mapsto g(\tilde{k},s) s ↦ g ( k ~ , s ) [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] k ~ ≠ 0 \tilde{k}\ne0 k ~ = 0 λ ( e ) \lambda(e) λ ( e ) k ~ \tilde{k} k ~ e → 0 e\to0 e → 0 h ( k ~ ) h(\tilde{k}) h ( k ~ ) s ↦ g ( k ~ , s ) s\mapsto g(\tilde{k},s) s ↦ g ( k ~ , s ) g ( k ~ , − 1 ) ≥ 0 ≥ g ( k ~ , 1 ) g(\tilde{k},-1)\ge0\ge g(\tilde{k},1) g ( k ~ , − 1 ) ≥ 0 ≥ g ( k ~ , 1 ) h ∈ L 1 ( R 3 ) h\in L^1(\mathbb{R}^3) h ∈ L 1 ( R 3 ) Lemma 20 ; dominated convergence gives λ ( e ) → 0 \lambda(e)\to0 λ ( e ) → 0
Next we bound e / ρ ˉ m e/\bar{\rho}_m e / ρ ˉ m ρ ˉ m ≥ ρ ˉ m , 1 \bar{\rho}_m\ge\bar{\rho}_{m,1} ρ ˉ m ≥ ρ ˉ m , 1 σ m , 1 \sigma_{m,1} σ m , 1 α = 2 e \alpha=2\sqrt{e} α = 2 e κ = 4 e + m \kappa=\sqrt{4e+m} κ = 4 e + m
2 e ρ ˉ m , 1 = m ∫ B 1 ( 0 ) ( 1 − σ m , 1 ) = 4 π m ( α 2 3 κ 2 + ( 1 + α ) m ( κ cosh κ − sinh κ ) κ 4 ( κ cosh κ + α sinh κ ) ) ≤ 4 π ( α 2 3 + 1 + α ) . \frac{2e}{\bar{\rho}_{m,1}}=m\int_{B_1(0)}(1-\sigma_{m,1})=4\pi m\left(\frac{\alpha^2}{3\kappa^2}+\frac{(1+\alpha)m(\kappa\cosh\kappa-\sinh\kappa)}{\kappa^4(\kappa\cosh\kappa+\alpha\sinh\kappa)}\right)\le4\pi\left(\frac{\alpha^2}{3}+1+\alpha\right). ρ ˉ m , 1 2 e = m ∫ B 1 ( 0 ) ( 1 − σ m , 1 ) = 4 π m ( 3 κ 2 α 2 + κ 4 ( κ cosh κ + α sinh κ ) ( 1 + α ) m ( κ cosh κ − sinh κ ) ) ≤ 4 π ( 3 α 2 + 1 + α ) . using m / κ 2 < 1 m/\kappa^2<1 m / κ 2 < 1 0 ≤ ( κ cosh κ − sinh κ ) / ( κ cosh κ + α sinh κ ) ≤ 1 0\le(\kappa\cosh\kappa-\sinh\kappa)/(\kappa\cosh\kappa+\alpha\sinh\kappa)\le1 0 ≤ ( κ cosh κ − sinh κ ) / ( κ cosh κ + α sinh κ ) ≤ 1 e 0 > 0 e_0>0 e 0 > 0 c > 0 c>0 c > 0 m m m ρ ˉ m ≥ c e \bar{\rho}_m\ge ce ρ ˉ m ≥ ce 0 < e ≤ e 0 0<e\le e_0 0 < e ≤ e 0 m > 0 m>0 m > 0 (73) ,
sup m > 0 sup ∣ x ∣ ≤ 1 ∣ f e , m ( x ) ∣ ≤ c − 3 / 2 λ ( e ) ⟶ 0 ( e → 0 ) . \sup_{m>0}\sup_{\lvert x\rvert\le1}\lvert f_{e,m}(x)\rvert\le c^{-3/2}\lambda(e)\longrightarrow0\qquad(e\to0). m > 0 sup ∣ x ∣ ≤ 1 sup ∣ f e , m ( x )∣ ≤ c − 3/2 λ ( e ) ⟶ 0 ( e → 0 ) . Substituting (72) into the identity of Lemma 21 gives
e − 2 π ρ ˉ m a 0 ( m ) = A e 3 / 2 2 ∫ ( 1 − ϕ m ) v m − ρ ˉ m 3 / 2 2 ∫ f e , m ( 1 − ϕ m ) v m . e-2\pi\bar{\rho}_m a_0(m)=\frac{Ae^{3/2}}{2}\int(1-\phi_m)v_m-\frac{\bar{\rho}_m^{3/2}}{2}\int f_{e,m}(1-\phi_m)v_m. e − 2 π ρ ˉ m a 0 ( m ) = 2 A e 3/2 ∫ ( 1 − ϕ m ) v m − 2 ρ ˉ m 3/2 ∫ f e , m ( 1 − ϕ m ) v m . Since ∫ ( 1 − ϕ m ) v m = 4 π a 0 ( m ) \int(1-\phi_m)v_m=4\pi a_0(m) ∫ ( 1 − ϕ m ) v m = 4 π a 0 ( m ) [11 Lemma 4.2] ), the first term equals 64 2 a 0 ( m ) 15 π e 3 / 2 \frac{64\sqrt{2}a_0(m)}{15\pi}e^{3/2} 15 π 64 2 a 0 ( m ) e 3/2
∣ ρ ˉ m 3 / 2 2 ∫ f e , m ( 1 − ϕ m ) v m ∣ ≤ 2 π a 0 ( m ) ρ ˉ m 3 / 2 sup ∣ x ∣ ≤ 1 ∣ f e , m ( x ) ∣ , \left\lvert\frac{\bar{\rho}_m^{3/2}}{2}\int f_{e,m}(1-\phi_m)v_m\right\rvert\le2\pi a_0(m)\bar{\rho}_m^{3/2}\sup_{\lvert x\rvert\le1}\lvert f_{e,m}(x)\rvert, 2 ρ ˉ m 3/2 ∫ f e , m ( 1 − ϕ m ) v m ≤ 2 π a 0 ( m ) ρ ˉ m 3/2 ∣ x ∣ ≤ 1 sup ∣ f e , m ( x )∣ , Define R m ( e ) : = − ρ ˉ m 3 / 2 2 ∫ f e , m ( 1 − ϕ m ) v m d x R_m(e):=-\frac{\bar{\rho}_m^{3/2}}{2}\int f_{e,m}(1-\phi_m)v_m\,dx R m ( e ) := − 2 ρ ˉ m 3/2 ∫ f e , m ( 1 − ϕ m ) v m d x a 0 ( m ) ≤ 1 a_0(m)\le1 a 0 ( m ) ≤ 1 (67) , and sup ∣ x ∣ ≤ 1 ∣ f e , m ∣ ≤ e 3 / 2 ρ ˉ m − 3 / 2 λ ( e ) \sup_{\lvert x\rvert\le1}\lvert f_{e,m}\rvert\le e^{3/2}\bar{\rho}_m^{-3/2}\lambda(e) sup ∣ x ∣ ≤ 1 ∣ f e , m ∣ ≤ e 3/2 ρ ˉ m − 3/2 λ ( e ) (73) , the bound just displayed gives ∣ R m ( e ) ∣ ≤ 2 π ρ ˉ m 3 / 2 ⋅ e 3 / 2 ρ ˉ m − 3 / 2 λ ( e ) = 2 π λ ( e ) e 3 / 2 \lvert R_m(e)\rvert\le2\pi\bar{\rho}_m^{3/2}\cdot e^{3/2}\bar{\rho}_m^{-3/2}\lambda(e)=2\pi\lambda(e)e^{3/2} ∣ R m ( e )∣ ≤ 2 π ρ ˉ m 3/2 ⋅ e 3/2 ρ ˉ m − 3/2 λ ( e ) = 2 π λ ( e ) e 3/2 m > 0 m>0 m > 0 e ≤ e 0 e\le e_0 e ≤ e 0
e = 2 π ρ ˉ m a 0 ( m ) + 64 2 a 0 ( m ) 15 π e 3 / 2 + R m ( e ) , ∣ R m ( e ) ∣ ≤ 2 π λ ( e ) e 3 / 2 , (74) e=2\pi\bar{\rho}_m a_0(m)+\frac{64\sqrt{2}a_0(m)}{15\pi}e^{3/2}+R_m(e),\qquad\lvert R_m(e)\rvert\le2\pi\lambda(e)e^{3/2}, \tag*{(74)} e = 2 π ρ ˉ m a 0 ( m ) + 15 π 64 2 a 0 ( m ) e 3/2 + R m ( e ) , ∣ R m ( e )∣ ≤ 2 π λ ( e ) e 3/2 , (74) where λ ( e ) → 0 \lambda(e)\to0 λ ( e ) → 0 e → 0 e\to0 e → 0 λ \lambda λ m m m λ ( e ) ≤ C 0 \lambda(e)\le C_0 λ ( e ) ≤ C 0 e e e C 0 C_0 C 0 Lemma 20 , since the integrand defining λ \lambda λ h h h m 0 > 0 m_0>0 m 0 > 0 m ≥ m 0 m\ge m_0 m ≥ m 0 a 0 ( m ) ≥ a 0 ( m 0 ) > 0 a_0(m)\ge a_0(m_0)>0 a 0 ( m ) ≥ a 0 ( m 0 ) > 0 a 0 a_0 a 0 (67) ; this lower bound on a 0 ( m ) a_0(m) a 0 ( m ) m 0 m_0 m 0
The uniform expansion. Define ϵ m ( e ) \epsilon_m(e) ϵ m ( e ) e 2 π ρ ˉ m a 0 ( m ) = 1 + ϵ m ( e ) \frac{e}{2\pi\bar{\rho}_m a_0(m)}=1+\epsilon_m(e) 2 π ρ ˉ m a 0 ( m ) e = 1 + ϵ m ( e ) a 0 ( m ) ≤ 1 a_0(m)\le1 a 0 ( m ) ≤ 1 λ ( e ) ≤ C 0 \lambda(e)\le C_0 λ ( e ) ≤ C 0 (74) gives ∣ e − 2 π ρ ˉ m a 0 ( m ) ∣ ≤ C 1 e 3 / 2 \lvert e-2\pi\bar{\rho}_m a_0(m)\rvert\le C_1e^{3/2} ∣ e − 2 π ρ ˉ m a 0 ( m )∣ ≤ C 1 e 3/2 C 1 : = 64 2 15 π + 2 π C 0 C_1:=\frac{64\sqrt{2}}{15\pi}+2\pi C_0 C 1 := 15 π 64 2 + 2 π C 0 2 π ρ ˉ m a 0 ( m ) ≥ 2 π c e a 0 ( m 0 ) 2\pi\bar{\rho}_m a_0(m)\ge2\pi c e a_0(m_0) 2 π ρ ˉ m a 0 ( m ) ≥ 2 π ce a 0 ( m 0 ) ρ ˉ m ≥ c e \bar{\rho}_m\ge ce ρ ˉ m ≥ ce a 0 ( m ) ≥ a 0 ( m 0 ) a_0(m)\ge a_0(m_0) a 0 ( m ) ≥ a 0 ( m 0 )
∣ ϵ m ( e ) ∣ ≤ C 2 e ( m ≥ m 0 , e ≤ e 0 ) , C 2 : = C 1 2 π c a 0 ( m 0 ) . \lvert\epsilon_m(e)\rvert\le C_2\sqrt{e}\qquad(m\ge m_0,\ e\le e_0),\qquad C_2:=\frac{C_1}{2\pi c a_0(m_0)}. ∣ ϵ m ( e )∣ ≤ C 2 e ( m ≥ m 0 , e ≤ e 0 ) , C 2 := 2 π c a 0 ( m 0 ) C 1 . Dividing (74) by 2 π ρ ˉ m a 0 ( m ) 2\pi\bar{\rho}_m a_0(m) 2 π ρ ˉ m a 0 ( m ) e 3 / 2 = ( 2 π ρ ˉ m a 0 ( m ) ) 3 / 2 ( 1 + ϵ m ( e ) ) 3 / 2 e^{3/2}=(2\pi\bar{\rho}_m a_0(m))^{3/2}(1+\epsilon_m(e))^{3/2} e 3/2 = ( 2 π ρ ˉ m a 0 ( m ) ) 3/2 ( 1 + ϵ m ( e ) ) 3/2
e 2 π ρ ˉ m a 0 ( m ) = 1 + 64 2 a 0 ( m ) 15 π 2 π ρ ˉ m a 0 ( m ) ( 1 + ϵ m ( e ) ) 3 / 2 + R m ( e ) 2 π ρ ˉ m a 0 ( m ) . \frac{e}{2\pi\bar{\rho}_m a_0(m)}=1+\frac{64\sqrt{2}a_0(m)}{15\pi}\sqrt{2\pi\bar{\rho}_m a_0(m)}(1+\epsilon_m(e))^{3/2}+\frac{R_m(e)}{2\pi\bar{\rho}_m a_0(m)}. 2 π ρ ˉ m a 0 ( m ) e = 1 + 15 π 64 2 a 0 ( m ) 2 π ρ ˉ m a 0 ( m ) ( 1 + ϵ m ( e ) ) 3/2 + 2 π ρ ˉ m a 0 ( m ) R m ( e ) . Since 64 2 15 π 2 π = 128 15 π \frac{64\sqrt{2}}{15\pi}\sqrt{2\pi}=\frac{128}{15\sqrt{\pi}} 15 π 64 2 2 π = 15 π 128 ∣ ( 1 + ϵ ) 3 / 2 − 1 ∣ ≤ 2 ∣ ϵ ∣ \lvert(1+\epsilon)^{3/2}-1\rvert\leq2\lvert\epsilon\rvert ∣( 1 + ϵ ) 3/2 − 1 ∣ ≤ 2 ∣ ϵ ∣ ∣ ϵ ∣ ≤ 1 2 \lvert\epsilon\rvert\leq\frac{1}{2} ∣ ϵ ∣ ≤ 2 1 e ≤ e 0 e\leq e_0 e ≤ e 0 e 0 e_0 e 0 a 0 ( m 0 ) ≤ a 0 ( m ) ≤ 1 a_0(m_0)\leq a_0(m)\leq1 a 0 ( m 0 ) ≤ a 0 ( m ) ≤ 1 ∣ R m ( e ) ∣ ≤ 2 π λ ( e ) e 3 / 2 ≤ 2 π λ ( e ) c − 3 / 2 ρ ˉ m 3 / 2 \lvert R_m(e)\rvert\leq2\pi\lambda(e)e^{3/2}\leq2\pi\lambda(e)c^{-3/2}\bar{\rho}_m^{3/2} ∣ R m ( e )∣ ≤ 2 π λ ( e ) e 3/2 ≤ 2 π λ ( e ) c − 3/2 ρ ˉ m 3/2 e ≤ ρ ˉ m / c e\leq\bar{\rho}_m/c e ≤ ρ ˉ m / c m ≥ m 0 m\geq m_0 m ≥ m 0 e ≤ e 0 e\leq e_0 e ≤ e 0
∣ e 2 π ρ ˉ m a 0 ( m ) − 1 − 128 15 π ρ ˉ m a 0 ( m ) 3 / 2 ∣ ≤ Λ ( e ) ρ ˉ m , Λ ( e ) : = 256 C 2 15 π e + c − 3 / 2 a 0 ( m 0 ) λ ( e ) , (75) \left\lvert\frac{e}{2\pi\bar{\rho}_m a_0(m)}-1-\frac{128}{15\sqrt{\pi}}\sqrt{\bar{\rho}_m}a_0(m)^{3/2}\right\rvert\leq\Lambda(e)\sqrt{\bar{\rho}_m},\qquad\Lambda(e):=\frac{256C_2}{15\sqrt{\pi}}\sqrt{e}+\frac{c^{-3/2}}{a_0(m_0)}\lambda(e), \tag*{(75)} 2 π ρ ˉ m a 0 ( m ) e − 1 − 15 π 128 ρ ˉ m a 0 ( m ) 3/2 ≤ Λ ( e ) ρ ˉ m , Λ ( e ) := 15 π 256 C 2 e + a 0 ( m 0 ) c − 3/2 λ ( e ) , (75) where Λ ( e ) → 0 \Lambda(e)\to0 Λ ( e ) → 0 e → 0 e\to0 e → 0 Λ \Lambda Λ m m m Theorem 3 .
Passage to the hard core. Fix e ≤ e 0 e\leq e_0 e ≤ e 0 m → ∞ m\to\infty m → ∞ (75) : ρ ˉ m → ρ h c \bar{\rho}_m\to\rho_{\mathrm{hc}} ρ ˉ m → ρ hc a 0 ( m ) → 1 a_0(m)\to1 a 0 ( m ) → 1 m m m
∣ e 2 π ρ h c − 1 − 128 15 π ρ h c ∣ ≤ Λ ( e ) ρ h c for all e ≤ e 0 , \left\lvert\frac{e}{2\pi\rho_{\mathrm{hc}}}-1-\frac{128}{15\sqrt{\pi}}\sqrt{\rho_{\mathrm{hc}}}\right\rvert\leq\Lambda(e)\sqrt{\rho_{\mathrm{hc}}}\qquad\text{for all }e\leq e_0, 2 π ρ hc e − 1 − 15 π 128 ρ hc ≤ Λ ( e ) ρ hc for all e ≤ e 0 , and passing to the limit in ρ ˉ m ≥ c e \bar{\rho}_m\geq ce ρ ˉ m ≥ ce ρ h c ≥ c e \rho_{\mathrm{hc}}\geq ce ρ hc ≥ ce ρ h c → 0 \rho_{\mathrm{hc}}\to0 ρ hc → 0 e → 0 e\to0 e → 0 Λ ( e ) → 0 \Lambda(e)\to0 Λ ( e ) → 0 ρ h c → 0 \rho_{\mathrm{hc}}\to0 ρ hc → 0 2 π ρ h c 2\pi\rho_{\mathrm{hc}} 2 π ρ hc
e = 2 π ρ h c ( 1 + 128 15 π ρ h c + o ( ρ h c ) ) ( ρ h c → 0 ) , e=2\pi\rho_{\mathrm{hc}}\left(1+\frac{128}{15\sqrt{\pi}}\sqrt{\rho_{\mathrm{hc}}}+o\left(\sqrt{\rho_{\mathrm{hc}}}\right)\right)\qquad(\rho_{\mathrm{hc}}\to0), e = 2 π ρ hc ( 1 + 15 π 128 ρ hc + o ( ρ hc ) ) ( ρ hc → 0 ) , as claimed.
Condensate Fraction
Soft-core condensate fraction
Until now we have fixed the energy e e e μ ≥ 0 \mu\geq0 μ ≥ 0 ( σ μ , m , e μ , m ) (\sigma_{\mu,m},e_{\mu,m}) ( σ μ , m , e μ , m )
( − Δ + 2 μ + 4 e μ , m + v m ) σ μ , m = v m + 2 e μ , m ρ ˉ m ( σ μ , m ∗ σ μ , m ) , e μ , m = ρ ˉ m 2 ∫ v m ( 1 − σ μ , m ) , (76) (-\Delta+2\mu+4e_{\mu,m}+v_m)\sigma_{\mu,m}=v_m+2e_{\mu,m}\bar{\rho}_m(\sigma_{\mu,m}\ast\sigma_{\mu,m}),\qquad e_{\mu,m}=\frac{\bar{\rho}_m}{2}\int v_m(1-\sigma_{\mu,m}), \tag*{(76)} ( − Δ + 2 μ + 4 e μ , m + v m ) σ μ , m = v m + 2 e μ , m ρ ˉ m ( σ μ , m ∗ σ μ , m ) , e μ , m = 2 ρ ˉ m ∫ v m ( 1 − σ μ , m ) , (76) where the density ρ ˉ m \bar{\rho}_m ρ ˉ m e μ , m e_{\mu,m} e μ , m m m m μ \mu μ [12 Section 5] . At μ = 0 \mu=0 μ = 0 e μ , m e_{\mu,m} e μ , m (22) , its μ \mu μ depletion ; we set
ϑ μ , m : = ∂ μ e μ , m , ϑ m : = ϑ 0 , m , η m : = 1 − ϑ m , (77) \vartheta_{\mu,m}:=\partial_\mu e_{\mu,m},\qquad\vartheta_m:=\vartheta_{0,m},\qquad\eta_m:=1-\vartheta_m, \tag*{(77)} ϑ μ , m := ∂ μ e μ , m , ϑ m := ϑ 0 , m , η m := 1 − ϑ m , (77) the soft-core depletion and condensate fraction.
Hard-core condensate fraction
For μ ≥ 0 \mu\ge0 μ ≥ 0 ( τ μ , e μ , h c ) (\tau_\mu,e_{\mu,\mathrm{hc}}) ( τ μ , e μ , hc )
Δ τ μ = ( 4 e μ , h c + 2 μ ) τ μ − 2 e μ , h c ρ h c ( τ μ ∗ τ μ ) on { ∣ x ∣ > 1 } , τ μ = 1 on B 1 , ρ h c ∫ τ μ = 1 , (78) \Delta\tau_\mu=(4e_{\mu,\mathrm{hc}}+2\mu)\tau_\mu-2e_{\mu,\mathrm{hc}}\rho_{\mathrm{hc}}(\tau_\mu\ast\tau_\mu)\quad\text{on }\{|x|>1\},\quad\tau_\mu=1\text{ on }B_1,\quad\rho_{\mathrm{hc}}\int\tau_\mu=1, \tag*{(78)} Δ τ μ = ( 4 e μ , hc + 2 μ ) τ μ − 2 e μ , hc ρ hc ( τ μ ∗ τ μ ) on { ∣ x ∣ > 1 } , τ μ = 1 on B 1 , ρ hc ∫ τ μ = 1 , (78) with e μ , h c e_{\mu,\mathrm{hc}} e μ , hc (12) . Analogously, the hard-core depletion ϑ h c \vartheta_{\mathrm{hc}} ϑ hc η h c \eta_{\mathrm{hc}} η hc
ϑ h c : = ∂ μ e μ , h c ∣ μ = 0 , η h c : = 1 − ϑ h c . (79) \vartheta_{\mathrm{hc}}:=\left.\partial_\mu e_{\mu,\mathrm{hc}}\right|_{\mu=0},\qquad\eta_{\mathrm{hc}}:=1-\vartheta_{\mathrm{hc}}. \tag*{(79)} ϑ hc := ∂ μ e μ , hc ∣ μ = 0 , η hc := 1 − ϑ hc . (79) Uniform estimates and limit exchange
We first compute the depletion at μ = 0 \mu=0 μ = 0 [12 Section 5] , whose results apply to v = v m v=v_m v = v m m m m ( 1 + ∣ x ∣ 4 ) v m ∈ L 1 ( R 3 ) ∩ L 2 ( R 3 ) (1+|x|^4)v_m\in L^1(\mathbb{R}^3)\cap L^2(\mathbb{R}^3) ( 1 + ∣ x ∣ 4 ) v m ∈ L 1 ( R 3 ) ∩ L 2 ( R 3 ) m m m
K m : = ( − Δ + 4 e + v m − 4 e ρ ˉ m C σ m ) − 1 , Y m : = ( − Δ + 4 e − 4 e ρ ˉ m C σ m ) − 1 K_m:=\left(-\Delta+4e+v_m-4e\bar{\rho}_m C_{\sigma_m}\right)^{-1},\qquad Y_m:=\left(-\Delta+4e-4e\bar{\rho}_m C_{\sigma_m}\right)^{-1} K m := ( − Δ + 4 e + v m − 4 e ρ ˉ m C σ m ) − 1 , Y m := ( − Δ + 4 e − 4 e ρ ˉ m C σ m ) − 1 Φ m : = K m [ v m ] , ξ m : = Y m [ ρ ˉ m σ m ] , \Phi_m:=K_m[v_m],\qquad\xi_m:=Y_m[\bar{\rho}_m\sigma_m], Φ m := K m [ v m ] , ξ m := Y m [ ρ ˉ m σ m ] , the operators K e K_e K e Y e Y_e Y e K e v K_ev K e v ξ \xi ξ [12 (1.8), (1.40), (5.11)] . By [12 Theorem 1.6, (1.30)] ,
ϑ m = N m 1 − 2 N m + D m , N m : = ρ ˉ m ∫ Φ m σ m d x , D m : = ρ ˉ m 2 ∫ Φ m ( σ m ∗ σ m ) d x (80) \vartheta_m=\frac{N_m}{1-2N_m+D_m},\qquad N_m:=\bar{\rho}_m\int\Phi_m\sigma_m\,dx,\qquad D_m:=\bar{\rho}_m^2\int\Phi_m(\sigma_m\ast\sigma_m)\,dx \tag*{(80)} ϑ m = 1 − 2 N m + D m N m , N m := ρ ˉ m ∫ Φ m σ m d x , D m := ρ ˉ m 2 ∫ Φ m ( σ m ∗ σ m ) d x (80) and by [12 (5.10)] , N m = ∫ v m ( 1 − Φ m ) ξ m d x N_m=\int v_m(1-\Phi_m)\xi_m\,dx N m = ∫ v m ( 1 − Φ m ) ξ m d x
Fourier representation of ξ m \xi_m ξ m With S m S_m S m (68) , k = 2 e κ k=2\sqrt{e}\,\kappa k = 2 e κ F ( κ , s ) : = κ 2 + 1 ( κ 2 + 1 ) 2 − s F(\kappa,s):=\dfrac{\kappa^2+1}{\sqrt{(\kappa^2+1)^2-s}} F ( κ , s ) := ( κ 2 + 1 ) 2 − s κ 2 + 1 [12 (5.13), (5.16)] read
ξ ^ m ( 2 e κ ) = F ( κ , S m ( 2 e κ ) ) − 1 4 e , ξ m ( x ) = e 4 π 3 ∫ R 3 e − 2 i e κ ⋅ x ( F ( κ , S m ( 2 e κ ) ) − 1 ) d 3 κ . (81) \widehat{\xi}_m(2\sqrt{e}\,\kappa)=\frac{F(\kappa,S_m(2\sqrt{e}\,\kappa))-1}{4e},\qquad\xi_m(x)=\frac{\sqrt{e}}{4\pi^3}\int_{\mathbb{R}^3}e^{-2i\sqrt{e}\,\kappa\cdot x}\bigl(F(\kappa,S_m(2\sqrt{e}\,\kappa))-1\bigr)\,d^3\kappa. \tag*{(81)} ξ m ( 2 e κ ) = 4 e F ( κ , S m ( 2 e κ )) − 1 , ξ m ( x ) = 4 π 3 e ∫ R 3 e − 2 i e κ ⋅ x ( F ( κ , S m ( 2 e κ )) − 1 ) d 3 κ . (81) By [12 (5.14)] , ∣ F ( κ , s ) − 1 ∣ ≤ h 1 ( κ ) : = F ( κ , 1 ) − F ( κ , − 1 ) |F(\kappa,s)-1|\le h_1(\kappa):=F(\kappa,1)-F(\kappa,-1) ∣ F ( κ , s ) − 1∣ ≤ h 1 ( κ ) := F ( κ , 1 ) − F ( κ , − 1 ) ∣ s ∣ ≤ 1 |s|\le1 ∣ s ∣ ≤ 1 h 1 ∈ L 1 ( R 3 ) h_1\in L^1(\mathbb{R}^3) h 1 ∈ L 1 ( R 3 ) [12 (5.15)] , 1 4 π 3 ∫ ( F ( κ , 1 ) − 1 ) d 3 κ = 2 3 π 2 \dfrac{1}{4\pi^3}\int(F(\kappa,1)-1)\,d^3\kappa=\dfrac{\sqrt{2}}{3\pi^2} 4 π 3 1 ∫ ( F ( κ , 1 ) − 1 ) d 3 κ = 3 π 2 2 sup x ∣ ξ m ( x ) ∣ ≤ c 2 e \sup_x|\xi_m(x)|\le c_2\sqrt{e} sup x ∣ ξ m ( x ) ∣ ≤ c 2 e m m m c 2 : = 1 4 π 3 ∫ h 1 d 3 κ c_2:=\dfrac{1}{4\pi^3}\int h_1\,d^3\kappa c 2 := 4 π 3 1 ∫ h 1 d 3 κ
ξ m ( x ) = 2 e 3 π 2 + r m ( x ) , sup m > 0 ∣ x ∣ ≤ 1 ∣ r m ( x ) ∣ ≤ e Ω ( e ) , (82) \xi_m(x)=\frac{\sqrt{2e}}{3\pi^2}+r_m(x),\qquad\sup_{\substack{m>0\\ |x|\le1}}|r_m(x)|\le\sqrt{e}\,\Omega(e), \tag*{(82)} ξ m ( x ) = 3 π 2 2 e + r m ( x ) , m > 0 ∣ x ∣ ≤ 1 sup ∣ r m ( x ) ∣ ≤ e Ω ( e ) , (82) where Ω ( e ) : = 1 4 π 3 sup m > 0 ∣ x ∣ ≤ 1 ∫ ∣ e − 2 i e κ ⋅ x ( F ( κ , S m ( 2 e κ ) ) − 1 ) − ( F ( κ , 1 ) − 1 ) ∣ d 3 κ \Omega(e):=\dfrac{1}{4\pi^3}\sup_{\substack{m>0\\ |x|\le1}}\int\left|e^{-2i\sqrt{e}\,\kappa\cdot x}\bigl(F(\kappa,S_m(2\sqrt{e}\,\kappa))-1\bigr)-(F(\kappa,1)-1)\right|\,d^3\kappa Ω ( e ) := 4 π 3 1 sup m > 0 ∣ x ∣ ≤ 1 ∫ e − 2 i e κ ⋅ x ( F ( κ , S m ( 2 e κ )) − 1 ) − ( F ( κ , 1 ) − 1 ) d 3 κ m m m m m m e → 0 e\to0 e → 0 [12 (5.12), (5.16)–(5.19)] . The same proof gives Ω ( e ) → 0 \Omega(e)\to0 Ω ( e ) → 0 m m m 2 h 1 2h_1 2 h 1 ∣ e − 2 i e κ ⋅ x − 1 ∣ ≤ 2 e ∣ κ ∣ |e^{-2i\sqrt{e}\,\kappa\cdot x}-1|\le2\sqrt{e}|\kappa| ∣ e − 2 i e κ ⋅ x − 1∣ ≤ 2 e ∣ κ ∣ ∣ x ∣ ≤ 1 |x|\le1 ∣ x ∣ ≤ 1 [12 (5.18)] , 0 ≤ F ( κ , 1 ) − F ( κ , S m ) ≤ ( κ 2 + 1 ) ( 1 − S m ( 2 e κ ) ) 2 ( ( κ 2 + 1 ) 2 − 1 ) 3 / 2 0\le F(\kappa,1)-F(\kappa,S_m)\le\dfrac{(\kappa^2+1)(1-S_m(2\sqrt{e}\,\kappa))}{2((\kappa^2+1)^2-1)^{3/2}} 0 ≤ F ( κ , 1 ) − F ( κ , S m ) ≤ 2 (( κ 2 + 1 ) 2 − 1 ) 3/2 ( κ 2 + 1 ) ( 1 − S m ( 2 e κ )) 1 − S m ( k ) ≤ ∣ k ∣ 2 / 2 1-S_m(k)\le|k|^2/2 1 − S m ( k ) ≤ ∣ k ∣ 2 /2 ρ ˉ m 2 e ( 1 − σ m ) v m d x \dfrac{\bar{\rho}_m}{2e}(1-\sigma_m)v_m\,dx 2 e ρ ˉ m ( 1 − σ m ) v m d x B 1 ( 0 ) B_1(0) B 1 ( 0 )
Lemma 22. Uniformly in m > 0 m > 0 m > 0 e ≤ e 0 e \le e_0 e ≤ e 0 0 ≤ D m ≤ C e 1 / 4 0 \le D_m \le C e^{1/4} 0 ≤ D m ≤ C e 1/4 ∣ N m ∣ ≤ C e 1 / 2 |N_m| \le C e^{1/2} ∣ N m ∣ ≤ C e 1/2
Consequently
ϑ m = ∫ v m ( 1 − Φ m ) ξ m d x + O ( e 3 / 4 ) . (83) \vartheta_m = \int v_m(1-\Phi_m)\,\xi_m\,dx + O(e^{3/4}). \tag*{(83)} ϑ m = ∫ v m ( 1 − Φ m ) ξ m d x + O ( e 3/4 ) . (83) Proof. This is similar to [12 (5.9)–(5.10)] except it is uniform in ∥ v m ∥ 1 \|v_m\|_1 ∥ v m ∥ 1 (60) and ∫ σ m ∗ σ m = ρ ˉ m − 2 \int\sigma_m * \sigma_m = \bar{\rho}_m^{-2} ∫ σ m ∗ σ m = ρ ˉ m − 2 ∫ v m ( 1 − Φ m ) = 2 e ρ ˉ m ( 1 − D m ) ≤ 2 e ρ ˉ m \int v_m(1-\Phi_m) = \frac{2e}{\bar{\rho}_m}(1-D_m) \le\frac{2e}{\bar{\rho}_m} ∫ v m ( 1 − Φ m ) = ρ ˉ m 2 e ( 1 − D m ) ≤ ρ ˉ m 2 e [12 (1.41)] , Φ m = Y m [ v m ( 1 − Φ m ) ] \Phi_m = Y_m[v_m(1-\Phi_m)] Φ m = Y m [ v m ( 1 − Φ m )] [12 Lemma 7.3] , which uses only (69) and ∣ S m ∣ ≤ 1 |S_m| \le1 ∣ S m ∣ ≤ 1 ∥ Y m f ∥ 2 ≤ C e − 1 / 4 ∥ f ∥ 1 \|Y_m f\|_2 \le C e^{-1/4}\|f\|_1 ∥ Y m f ∥ 2 ≤ C e − 1/4 ∥ f ∥ 1 C C C m m m ∥ Φ m ∥ 2 ≤ C e − 1 / 4 2 e ρ ˉ m \|\Phi_m\|_2 \le C e^{-1/4}\frac{2e}{\bar{\rho}_m} ∥ Φ m ∥ 2 ≤ C e − 1/4 ρ ˉ m 2 e
Using ∥ σ m ∗ σ m ∥ 2 ≤ ∥ σ m ∥ 1 ∥ σ m ∥ 2 \|\sigma_m * \sigma_m\|_2 \le\|\sigma_m\|_1\|\sigma_m\|_2 ∥ σ m ∗ σ m ∥ 2 ≤ ∥ σ m ∥ 1 ∥ σ m ∥ 2 ρ ˉ m ∥ σ m ∥ 1 = 1 \bar{\rho}_m\|\sigma_m\|_1 = 1 ρ ˉ m ∥ σ m ∥ 1 = 1 ∥ σ m ∥ 2 ≤ ρ ˉ m − 1 / 2 \|\sigma_m\|_2 \le\bar{\rho}_m^{-1/2} ∥ σ m ∥ 2 ≤ ρ ˉ m − 1/2 D m ≤ ρ ˉ m ∥ Φ m ∥ 2 ≤ C e 3 / 4 ρ ˉ m − 1 / 2 ≤ C e 1 / 4 D_m \le\sqrt{\bar{\rho}_m}\|\Phi_m\|_2 \le C e^{3/4}\bar{\rho}_m^{-1/2} \le C e^{1/4} D m ≤ ρ ˉ m ∥ Φ m ∥ 2 ≤ C e 3/4 ρ ˉ m − 1/2 ≤ C e 1/4 ρ ˉ m ≥ c e \bar{\rho}_m \ge ce ρ ˉ m ≥ ce ∣ N m ∣ ≤ ∫ v m ( 1 − Φ m ) sup x ∣ ξ m ∣ ≤ 2 e ρ ˉ m c 2 e ≤ C e 1 / 2 |N_m| \le\int v_m(1-\Phi_m)\sup_x|\xi_m| \le\frac{2e}{\bar{\rho}_m} c_2\sqrt{e} \le C e^{1/2} ∣ N m ∣ ≤ ∫ v m ( 1 − Φ m ) sup x ∣ ξ m ∣ ≤ ρ ˉ m 2 e c 2 e ≤ C e 1/2 ϑ m − N m = N m ( 2 N m − D m ) / ( 1 − 2 N m + D m ) = O ( e 3 / 4 ) \vartheta_m-N_m = N_m(2N_m-D_m)/(1-2N_m+D_m)=O(e^{3/4}) ϑ m − N m = N m ( 2 N m − D m ) / ( 1 − 2 N m + D m ) = O ( e 3/4 )
Combining (83) , (82) and ∫ v m ( 1 − Φ m ) = 2 e ρ ˉ m ( 1 − D m ) \int v_m(1-\Phi_m) = \frac{2e}{\bar{\rho}_m}(1-D_m) ∫ v m ( 1 − Φ m ) = ρ ˉ m 2 e ( 1 − D m ) 2 e ρ ˉ m = 4 π a 0 ( m ) ( 1 + ϵ m ( e ) ) \frac{2e}{\bar{\rho}_m}=4\pi a_0(m)(1+\epsilon_m(e)) ρ ˉ m 2 e = 4 π a 0 ( m ) ( 1 + ϵ m ( e )) § 7 , where 4 π a 0 ( m ) = ∫ v m ( 1 − ϕ m ) 4\pi a_0(m)=\int v_m(1-\phi_m) 4 π a 0 ( m ) = ∫ v m ( 1 − ϕ m ) [11 Lemma 4.2] , we obtain, uniformly in m m m
ϑ m = 2 e 3 π 2 ⋅ 4 π a 0 ( m ) ( 1 + O ( e 1 / 4 ) ) + O ( e Ω ( e ) ) + O ( e 3 / 4 ) = 4 2 3 π e a 0 ( m ) + o ( e ) , (84) \vartheta_m = \frac{\sqrt{2e}}{3\pi^2}\cdot4\pi a_0(m)(1+O(e^{1/4}))+O(\sqrt{e}\,\Omega(e))+O(e^{3/4}) = \frac{4\sqrt{2}}{3\pi}\sqrt{e}\,a_0(m)+o(\sqrt{e}), \tag*{(84)} ϑ m = 3 π 2 2 e ⋅ 4 π a 0 ( m ) ( 1 + O ( e 1/4 )) + O ( e Ω ( e )) + O ( e 3/4 ) = 3 π 4 2 e a 0 ( m ) + o ( e ) , (84) which is [12 (5.25)] with the error uniform in m m m
Lemma 23 (Uniform convergence of the energy). There exists K > 0 K > 0 K > 0 μ \mu μ m m m μ ∈ [ 0 , μ 0 ] \mu\in[0,\mu_0] μ ∈ [ 0 , μ 0 ] m ≥ m 0 m \ge m_0 m ≥ m 0
∣ e μ , m − e μ , h c ∣ ≤ K m . (85) |e_{\mu,m}-e_{\mu,\mathrm{hc}}| \le\frac{K}{\sqrt{m}}. \tag*{(85)} ∣ e μ , m − e μ , hc ∣ ≤ m K . (85) Proof. For each fixed μ \mu μ 18 apply verbatim to (76) and (78) , with 4 e 4e 4 e 4 e μ , m + 2 μ 4e_{\mu,m}+2\mu 4 e μ , m + 2 μ 2 e 2e 2 e 2 e μ , m 2e_{\mu,m} 2 e μ , m ∣ e μ , m − e μ , h c ∣ ≤ K ( μ ) m − 1 / 2 |e_{\mu,m}-e_{\mu,\mathrm{hc}}| \le K(\mu)m^{-1/2} ∣ e μ , m − e μ , hc ∣ ≤ K ( μ ) m − 1/2 ρ ˉ m \bar{\rho}_m ρ ˉ m μ \mu μ K ( μ ) K(\mu) K ( μ ) μ \mu μ e μ , m e_{\mu,m} e μ , m e μ , h c e_{\mu,\mathrm{hc}} e μ , hc 2 μ 2\mu 2 μ ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) e μ , m = e + μ ϑ μ ∗ , m e_{\mu,m}=e+\mu\vartheta_{\mu^*,m} e μ , m = e + μ ϑ μ ∗ , m μ ∗ ∈ ( 0 , μ ) \mu^* \in(0,\mu) μ ∗ ∈ ( 0 , μ ) ∣ ϑ μ ∗ , m ∣ ≤ ∣ ϑ m ∣ + ω ( μ 0 ) |\vartheta_{\mu^*,m}| \le|\vartheta_m|+\omega(\mu_0) ∣ ϑ μ ∗ , m ∣ ≤ ∣ ϑ m ∣ + ω ( μ 0 ) ∣ ϑ m ∣ ≤ C e |\vartheta_m| \le C\sqrt{e} ∣ ϑ m ∣ ≤ C e (84) , so that ∣ e μ , m − e ∣ ≤ μ 0 ( C e + ω ( μ 0 ) ) |e_{\mu,m}-e| \le\mu_0(C\sqrt{e}+\omega(\mu_0)) ∣ e μ , m − e ∣ ≤ μ 0 ( C e + ω ( μ 0 )) m ≥ m 0 m \ge m_0 m ≥ m 0 e μ , h c e_{\mu,\mathrm{hc}} e μ , hc μ ∈ [ 0 , μ 0 ] \mu\in[0,\mu_0] μ ∈ [ 0 , μ 0 ] μ \mu μ K : = sup μ ∈ [ 0 , μ 0 ] K ( μ ) < ∞ K:=\sup_{\mu\in[0,\mu_0]}K(\mu)<\infty K := sup μ ∈ [ 0 , μ 0 ] K ( μ ) < ∞
Lemma 24 (Equicontinuity in μ \mu μ There is a function ω \omega ω ω ( δ ) → 0 \omega(\delta)\to0 ω ( δ ) → 0 δ → 0 \delta\to0 δ → 0 m m m ∣ ϑ μ , m − ϑ ν , m ∣ ≤ ω ( ∣ μ − ν ∣ ) |\vartheta_{\mu,m}-\vartheta_{\nu,m}| \le\omega(|\mu-\nu|) ∣ ϑ μ , m − ϑ ν , m ∣ ≤ ω ( ∣ μ − ν ∣ ) μ , ν ∈ [ 0 , μ 0 ] \mu,\nu\in[0,\mu_0] μ , ν ∈ [ 0 , μ 0 ] m ≥ m 0 m\ge m_0 m ≥ m 0
Proof. For μ ∈ [ 0 , μ 0 ] \mu\in[0,\mu_0] μ ∈ [ 0 , μ 0 ] K μ , m K_{\mu,m} K μ , m Y μ , m Y_{\mu,m} Y μ , m Φ μ , m \Phi_{\mu,m} Φ μ , m ξ μ , m \xi_{\mu,m} ξ μ , m N μ , m N_{\mu,m} N μ , m D μ , m D_{\mu,m} D μ , m K m K_m K m Y m Y_m Y m Φ m \Phi_m Φ m ξ m \xi_m ξ m N m N_m N m D m D_m D m (80) , with 4 e 4e 4 e 4 e μ , m + 2 μ 4e_{\mu,m}+2\mu 4 e μ , m + 2 μ e e e e μ , m e_{\mu,m} e μ , m (76) in μ \mu μ ρ ˉ m \bar{\rho}_m ρ ˉ m Φ μ , m \Phi_{\mu,m} Φ μ , m (80) gives the exact formula
ϑ μ , m = N μ , m 1 − 2 N μ , m + D μ , m , N μ , m = ∫ v m ( 1 − Φ μ , m ) ξ μ , m d x , \vartheta_{\mu,m}=\frac{N_{\mu,m}}{1-2N_{\mu,m}+D_{\mu,m}},\qquad N_{\mu,m}=\int v_m(1-\Phi_{\mu,m})\,\xi_{\mu,m}\,dx, ϑ μ , m = 1 − 2 N μ , m + D μ , m N μ , m , N μ , m = ∫ v m ( 1 − Φ μ , m ) ξ μ , m d x , so it suffices to show that N μ , m N_{\mu,m} N μ , m D μ , m D_{\mu,m} D μ , m μ \mu μ m m m ∣ N μ , m ∣ + D μ , m ≤ 1 4 |N_{\mu,m}|+D_{\mu,m}\le\frac14 ∣ N μ , m ∣ + D μ , m ≤ 4 1 μ 0 \mu_0 μ 0 (81) holds at parameter $\mu with the rescaling k = 2 e ~ μ , m k = 2\sqrt{\widetilde{e}_{\mu,m}} k = 2 e μ , m e ~ μ , m : = e μ , m + μ / 2 \widetilde{e}_{\mu,m} := e_{\mu,m} + \mu/2 e μ , m := e μ , m + μ /2 e ~ μ , m 3 / 2 / ( 4 π 3 e μ , m ) \widetilde{e}_{\mu,m}^{3/2}/(4\pi^{3}e_{\mu,m}) e μ , m 3/2 / ( 4 π 3 e μ , m ) S m S_m S m θ μ , m S μ , m \theta_{\mu,m}S_{\mu,m} θ μ , m S μ , m θ μ , m : = ( e μ , m / e ~ μ , m ) 2 \theta_{\mu,m} := (e_{\mu,m}/\widetilde{e}_{\mu,m})^{2} θ μ , m := ( e μ , m / e μ , m ) 2 S μ , m S_{\mu,m} S μ , m [12 (5.14)] and the bound 1 − S μ , m ( k ) ≤ ∣ k ∣ 2 / 2 1-S_{\mu,m}(k) \leq\lvert k\rvert^{2}/2 1 − S μ , m ( k ) ≤ ∣ k ∣ 2 /2 μ ↦ ξ μ , m \mu\mapsto\xi_{\mu,m} μ ↦ ξ μ , m L ∞ ( B 1 ( 0 ) ) L^{\infty}(B_{1}(0)) L ∞ ( B 1 ( 0 )) m m m (82) . The same holds for μ ↦ Φ μ , m \mu\mapsto\Phi_{\mu,m} μ ↦ Φ μ , m L 2 L^{2} L 2 D μ , m D_{\mu,m} D μ , m L 1 → L 2 L^{1} \to L^{2} L 1 → L 2 [12 Lemma 7.3] , whose constant depends on μ \mu μ e μ , m e_{\mu,m} e μ , m
∫ v m ( 1 − Φ μ , m ) ≤ 2 ( e μ , m + μ 2 + 4 μ e μ , m ) ρ ‾ m . \int v_m(1-\Phi_{\mu,m}) \leq\frac{2(e_{\mu,m}+\sqrt{\mu^{2}+4\mu e_{\mu,m}})}{\overline{\rho}_m}. ∫ v m ( 1 − Φ μ , m ) ≤ ρ m 2 ( e μ , m + μ 2 + 4 μ e μ , m ) . obtained as (59) was, from K μ , m − 1 1 = v m + 2 μ 2 + 4 μ e μ , m K_{\mu,m}^{-1}1=v_m+2\sqrt{\mu^{2}+4\mu e_{\mu,m}} K μ , m − 1 1 = v m + 2 μ 2 + 4 μ e μ , m ∣ N μ , m ∣ + D μ , m ≤ 1 4 \lvert N_{\mu,m}\rvert+D_{\mu,m}\leq\frac14 ∣ N μ , m ∣ + D μ , m ≤ 4 1 μ = 0 \mu=0 μ = 0 μ 0 \mu_0 μ 0 e e e
Lemma 25 (Differentiability of the hard-core energy). e μ , h c e_{\mu,\mathrm{hc}} e μ , hc μ = 0 \mu=0 μ = 0 ϑ h c \vartheta_{\mathrm{hc}} ϑ hc
Proof. Set s = ∂ μ τ μ ∣ μ = 0 s=\left.\partial_\mu\tau_\mu\right|_{\mu=0} s = ∂ μ τ μ ∣ μ = 0 μ = 0 \mu=0 μ = 0 s s s s s s μ = 0 \mu=0 μ = 0 ϑ h c \vartheta_{\mathrm{hc}} ϑ hc ∂ r s ∣ 1 + \left.\partial_rs\right|_{1+} ∂ r s ∣ 1 + ∫ B 1 ( 0 ) ( τ ∗ s ) \int_{B_1(0)}(\tau*s) ∫ B 1 ( 0 ) ( τ ∗ s )
Theorem 26 (Limit exchange; proof of part (ii) of Theorem 3). lim m → ∞ ϑ m = ϑ h c \lim_{m\to\infty}\vartheta_m=\vartheta_{\mathrm{hc}} lim m → ∞ ϑ m = ϑ hc lim m → ∞ η m = η h c \lim_{m\to\infty}\eta_m=\eta_{\mathrm{hc}} lim m → ∞ η m = η hc
Proof. Fix ε > 0 \varepsilon>0 ε > 0 δ 0 > 0 \delta_0>0 δ 0 > 0 0 < δ < δ 0 0<\delta<\delta_0 0 < δ < δ 0
∣ e δ , h c − e 0 , h c δ − ϑ h c ∣ < ε . \left\lvert\frac{e_{\delta,\mathrm{hc}}-e_{0,\mathrm{hc}}}{\delta}-\vartheta_{\mathrm{hc}}\right\rvert<\varepsilon. δ e δ , hc − e 0 , hc − ϑ hc < ε . Fix such a δ \delta δ m ≥ m 0 m\geq m_0 m ≥ m 0 δ ∗ ∈ ( 0 , δ ) \delta^*\in(0,\delta) δ ∗ ∈ ( 0 , δ )
e δ , m − e 0 , m δ = ϑ δ ∗ , m , \frac{e_{\delta,m}-e_{0,m}}{\delta}=\vartheta_{\delta^*,m}, δ e δ , m − e 0 , m = ϑ δ ∗ , m , so by Lemma 24,
∣ ϑ m − e δ , m − e 0 , m δ ∣ = ∣ ϑ 0 , m − ϑ δ ∗ , m ∣ ≤ ω ( δ ) , \left\lvert\vartheta_m-\frac{e_{\delta,m}-e_{0,m}}{\delta}\right\rvert=\lvert\vartheta_{0,m}-\vartheta_{\delta^*,m}\rvert\leq\omega(\delta), ϑ m − δ e δ , m − e 0 , m = ∣ ϑ 0 , m − ϑ δ ∗ , m ∣ ≤ ω ( δ ) , while Lemma 23 gives
∣ e δ , m − e 0 , m δ − e δ , h c − e 0 , h c δ ∣ ≤ 2 K δ m . \left\lvert\frac{e_{\delta,m}-e_{0,m}}{\delta}-\frac{e_{\delta,\mathrm{hc}}-e_{0,\mathrm{hc}}}{\delta}\right\rvert\leq\frac{2K}{\delta\sqrt{m}}. δ e δ , m − e 0 , m − δ e δ , hc − e 0 , hc ≤ δ m 2 K . Combining the three estimates,
∣ ϑ m − ϑ h c ∣ ≤ ω ( δ ) + 2 K δ m + ε . \lvert\vartheta_m-\vartheta_{\mathrm{hc}}\rvert\leq\omega(\delta)+\frac{2K}{\delta\sqrt{m}}+\varepsilon. ∣ ϑ m − ϑ hc ∣ ≤ ω ( δ ) + δ m 2 K + ε . Letting m → ∞ m\to\infty m → ∞ δ \delta δ lim sup m → ∞ ∣ ϑ m − ϑ h c ∣ ≤ ω ( δ ) + ε \limsup_{m\to\infty}\lvert\vartheta_m-\vartheta_{\mathrm{hc}}\rvert\leq\omega(\delta)+\varepsilon lim sup m → ∞ ∣ ϑ m − ϑ hc ∣ ≤ ω ( δ ) + ε δ < δ 0 \delta<\delta_0 δ < δ 0 ε > 0 \varepsilon>0 ε > 0 ϑ m → ϑ h c \vartheta_m\to\vartheta_{\mathrm{hc}} ϑ m → ϑ hc η m = 1 − ϑ m → 1 − ϑ h c = η h c \eta_m=1-\vartheta_m\to1-\vartheta_{\mathrm{hc}}=\eta_{\mathrm{hc}} η m = 1 − ϑ m → 1 − ϑ hc = η hc
Corollary 27 (Proof of Theorem 2). Each soft-core equation satisfies (24) with a 0 a_0 a 0 a 0 ( m ) a_0(m) a 0 ( m ) m m m
η h c = 1 − 8 3 π ρ h c + o ( ρ h c ) , (86) \eta_{\mathrm{hc}}=1-\frac{8}{3\sqrt{\pi}}\sqrt{\rho_{\mathrm{hc}}}+o(\sqrt{\rho_{\mathrm{hc}}}), \tag*{(86)} η hc = 1 − 3 π 8 ρ hc + o ( ρ hc ) , (86) which is Theorem 2.
Proof. By (84) , ϑ m = 4 2 3 π e a 0 ( m ) + o ( e ) \vartheta_m = \frac{4\sqrt{2}}{3\pi}\sqrt{e}\,a_0(m) + o(\sqrt{e}) ϑ m = 3 π 4 2 e a 0 ( m ) + o ( e ) m ≥ m 0 m \ge m_0 m ≥ m 0 (75) , e = 2 π ρ ˉ m a 0 ( m ) ( 1 + o ( 1 ) ) e = 2\pi\bar{\rho}_m a_0(m)(1+o(1)) e = 2 π ρ ˉ m a 0 ( m ) ( 1 + o ( 1 )) m m m e = 2 π ρ ˉ m a 0 ( m ) ( 1 + o ( 1 ) ) \sqrt{e} = \sqrt{2\pi\bar{\rho}_m a_0(m)}(1+o(1)) e = 2 π ρ ˉ m a 0 ( m ) ( 1 + o ( 1 ))
η m = 1 − 8 3 π ρ ˉ m a 0 ( m ) 3 + o ( ρ ˉ m ) , \eta_m = 1 - \frac{8}{3\sqrt{\pi}}\sqrt{\bar{\rho}_m a_0(m)^3} + o(\sqrt{\bar{\rho}_m}), η m = 1 − 3 π 8 ρ ˉ m a 0 ( m ) 3 + o ( ρ ˉ m ) , uniformly in m ≥ m 0 m \ge m_0 m ≥ m 0 (24) with a 0 a_0 a 0 a 0 ( m ) a_0(m) a 0 ( m ) m → ∞ m \to\infty m → ∞ e e e Theorem 26 , a 0 ( m ) → 1 a_0(m) \to1 a 0 ( m ) → 1 ρ ˉ m → ρ h c \bar{\rho}_m \to\rho_{\mathrm{hc}} ρ ˉ m → ρ hc ϑ h c = 4 2 3 π e + o ( e ) \vartheta_{\mathrm{hc}} = \frac{4\sqrt{2}}{3\pi}\sqrt{e} + o(\sqrt{e}) ϑ hc = 3 π 4 2 e + o ( e ) e = 2 π ρ h c ( 1 + o ( 1 ) ) \sqrt{e} = \sqrt{2\pi\rho_{\mathrm{hc}}}(1+o(1)) e = 2 π ρ hc ( 1 + o ( 1 )) (86) .
Acknowledgements. We thank Elliott Lieb for many useful advice and discussions. IJ gratefully acknowledges support through NSF Grant DMS-2349077.
AI disclosure. AI was used for formatting and grammatical corrections to the paper.
