Received 24 January 2018, Accepted 24 May 2018, Available Online 6 January 2021.
1. Introduction
A unitary ensemble of Hermitian matrices M = (Mij)n×n has probability density
p(M)dM∝e−trv(M)vol(dM),vol(dM)=∏i=1ndMii∏i≤j≤k≤nd(ReMjk)d(ImMjk).
Here v(M) is a matrix function [21] defined via Jordan canonical form and vol(dM) is called the volume element [22]. In this paper, we consider
v(x)=v(x;t)=−αlogx+x+tx,x∈[0,∞),α>−1,t≥0.
Under an eigenvalue-eigenvector decomposition, the joint probability density function of the eigenvalues {xk}k=1n of this ensemble is given by [29]
1Dn[w]1n!∏1≤j≤n(xj−xi)2∏k=1nw(xk;t),(1.1)
where
Dn[
w] is the normalization constant and
w(
x;
t) is a positive weight function supported on [0, ∞) and defined by
w(x;t):=e−v(x;t)=xαe−x−tx,
which has finite moments
μk:=∫0∞xkw(x;t)dx, k=0,1,2,….
Denote by (s, t, n) the probability that all the eigenvalues of this ensemble are greater than s, then
(s,t,n)=1n!∫(s,∞)n∏1≤i≤j≤n(xj−xi)2∏k=1nw(xk;t)dxk1n!∫(0,∞)n∏1≤i≤j≤n(xj−xi)2∏k=1nw(xk;t)dxk=det(∫s∞xi+jw(x;t)dx)i,j=0n−1det(∫0∞xi+jw(x;t)dx)i,j=0n−1=:Dn(s,t)Dn(0,t).
Here note that the multiple integral is represented as the determinant of the Hankel (or moment) matrix. See [32].
The Hankel determinant is a fundamental object in random matrix theory [29]. It can be used to describe the eigenvalue distribution of the Gaussian, Laguerre and Jacobi unitary ensembles [4, 27, 30] and also be applied to study the outage capacity and the error probability of multiple-input multiple-output antenna wireless communication systems [9, 15].
An elementary approach to dealing with the Hankel determinant is to write it as the product of the square of the L2 norms of the non-standard orthogonal polynomials. From the supplementary conditions associated with the ladder operators for the orthogonal polynomials, we derive a series of difference and differential equations to characterize the Hankel determinant, usually involving a Painlevé equation or the associated σ form. Such a formalism has been applied to the Hankel determinant generated by the deformation of the classical Gaussian weight [8, 16], the classical Laguerre weight [1, 2], and the (shifted) classical Jacobi weight [3, 7, 17, 18].
When the order of the Hankel matrix, n, becomes large, one chooses a suitable combination of n and a parameter present in the weight function such that the combination remains finite in the limit. Under such a double scaling, from the finite n equations, one can derive a differential equation satisfied by the log-derivative of the scaled Hankel determinant in the scaled variable. From this equation and by integration, the asymptotic expansion of the scaled Hankel determinant is established. The constant term in the expansion is determined by using the linear statistics results [14] based on logarithmic potential theory with an external field, sometimes referred to in this context as Dyson’s Coulomb fluid approximation [19]. See for example [5,6,27
] for the formulation.
In this paper we shall apply the approach described above for finite n and large n to study the Hankel determinant
Dn(s,t)=det(∫s∞xi+jw(x;t)dx)i,j=0n−1,
with
s ≥ 0, and with the weight function
w(x;t)=xαe−x−tx, α>−1, t≥0.
As mentioned above, Dn(s,t) is connected with the smallest eigenvalue distribution of the singularly deformed Laguerre unitary ensemble. The singular deformation here means that the Laguerre weight or the Gamma density supported on [0, ∞) is multiplied by a factor which vanishes infinitely fast at x = 0, that is, e−tx, t > 0. The case s = 0 and the case t = 0 in Dn(s, t) have been investigated for finite n in [13] and [1] respectively through the ladder operator framework, and are discussed for large n in [5] and [27] respectively by means of the double scaling and logarithmic potential theory with an external field. This two variable problem was suggested to the third author who spoke in a Seminar at the National Taiwan University on the s = 0 problem. A person in the audience, suggested that the same weight supported on the interval [s,∞) maybe an interesting problem to study, since this would naturally lead to a PDE in the variables s and t which opens up a venue for the investigation of varying scaling limits between the ‘gap’ and the ‘time’ variables.
Let hj(s, t) be the square of the L2 norm of monic polynomial Pj(x; s, t) orthogonal with respect to w(x; t) over [s, ∞):
hj(s,t)δjk:=∫s∞Pj(x;s,t)Pk(x;s,t)w(x;t)dx, j,k=0,1,2,…,(1.2)
and define
p(
j,
s,
t) by
Pj(x;s,t)=xj+p(j,s,t)xj−1+…+Pj(0;s,t).
Then Dn(s, t) admits a representation [32]
Dn(s,t)=∏j=0n−1hj(s,t).
From the orthogonality relation, there follows the three-term recurrence relation
zPn(z;s,t)=Pn+1(z;s,t)+αn(s,t)Pn(z;s,t)+βn(s,t)Pn−1(z;s,t), n≥0,
subject to the initial conditions
P0(z;s,t):=1, β0(s,t)P−1(z;s,t):=0,
and the recurrence coefficients are given by
αn(s,t)=p(n,s,t)−p(n+1,s,t), βn(s,t)=hn(s,t)hn−1(s,t).
As an immediate consequence, we have
∑j=0n−1αj(s,t)=−p(n,s,t).
This paper is built up as follows. By using (S1) and (S′2) (given later), we express in the next section the recurrence coefficients αn(s, t) and βn(s, t) in terms of four auxiliary variables Rn⋆(s,t), Rn(s, t), rn⋆(s,t) and rn(s, t), that satisfy a system of difference equations. Combining these expressions with the equalities obtained from the differentiation of (1.2) over s and t, we establish a pair of second order partial differential equations satisfied by Rn⋆(s,t) and Rn(s, t), and a large second order sixth degree partial differential equation satisfied by Hn(s,t):=(s∂∂s+t∂∂t)logDn(s,t). Based on these results for finite n, we devote Section 3 to the large n analysis of Dn(s, t) by assuming n → ∞, s → 0 and t → 0 such that S := 4ns and T := (2n + 1 + α)t are fixed. We derive the asymptotic expansions of the scaled Hankel determinant in three cases of the new variables: S is large or small with T fixed, and T is large with S > 0 fixed. The constant term in the expansion for each case is shown to satisfy a difference equation in the last section, by using logarithmic potential theory with an external field to evaluate Pn(z; s, t) at z = 0.
2. Difference Equations and Differential Equations
With the potential v(x) = v(x; t) given by
v(x)=−αlogx+x+tx,
the monic polynomials {
Pn(
x;
s,
t)} orthogonal with respect to
w(
x;
t) over [
s, ∞) satisfy a pair of ladder operators which lower or raise the index
n of
Pn(
x;
s,
t):
(∂∂z+Bn(z))Pn(z;s,t)=βn(s,t)An(z)Pn−1(z;s,t),(∂∂z−Bn(z)−v′(z))Pn−1(z;s,t)=−An−1(z)Pn(z;s,t),
where
An(
z) =
An(
z;
s,
t) and
Bn(
z) =
Bn(
z;
s,
t) are defined by
An(z):=Pn2(y;s,t)w(y;t)hn(s,t)(y−z)|y=sy=∞+1hn(s,t)∫s∞v′(z)−v′(y)z−yPn2(y;s,t)w(y;t)dy,Bn(z):=Pn(y;s,t)Pn−1(y;s,t)w(y;t)hn−1(s,t)(y−z)|y=sy=∞+1hn−1(s,t)∫s∞v′(z)−v′(y)z−yPn(y;s,t)Pn−1(y;s,t)w(y;t)dy.
From the above ladder operators and the recurrence relation for Pn(z; s, t), one can derive two compatibility conditions satisfied by An(z) and Bn(z):
Bn+1(z)+Bn(z)=(z−αn(s,t))An(z)−v′(z),(S1)
1+(z−αn(s,t))(Bn+1(z)−Bn(z))=βn+1(s,t)An+1(z)−βn(s,t)An−1(z).(S2)
(S1) and (S2) can be combined to give the ‘sum rule’:
Bn2(z)+v′(z)Bn(z)+∑j=0n−1Aj(z)=βn(s,t)An(z)An−1(z).(S′2)
Instrumental for the deductions that follow. See [11, 12, 28] for a detailed derivation. See also [23, § 3.2] for the form adapted to orthonormal polynomials.
For our problem, we have
v′(z)−v′(y)z−y=αzy+tzy2+tz2y.
Hence, from the definitions of An(z) and Bn(z), and through integration by parts, we obtain
An(z)=Rn⋆(s,t)z−s+1−Rn⋆(s,t)z+tRn(s,t)z2,Bn(z)=rn⋆(s,t)z−s+rn⋆(s,t)+nz+trn(s,t)z2.(2.1)
Here the auxiliary quantities Rn⋆(s,t), Rn(s, t), rn⋆(s,t) and rn(s, t) are defined by
Rn⋆(s,t):=w(s;t)hn(s,t)Pn2(s;s,t),Rn(s,t):=1hn(s,t)∫s∞dyyPn2(y;s,t)w(y;t),rn⋆(s,t):=w(s;t)hn−1(s,t)Pn(s;s,t)Pn−1(s;s,t),rn(s,t):=1hn−1(s,t)∫s∞dyyPn(y;s,t)Pn−1(y;s,t)w(y;t).
Note that Pn(s; s, t) is the evaluation of Pn(z; s, t) at z = s and w(s;t)=sαe−s−ts.
2.1. The recurrence coefficients in terms of the auxiliary quantities
Substituting (2.1) into (S1) and (S′2), and by equating the residues, we can establish a set of equations that will give insight into the recurrence coefficients αn(s, t), βn(s, t) and the auxiliary quantities.
From (S1), we arrive at three difference equations relating αn(s, t) to the auxiliary quantities:
rn+1⋆+rn⋆=(s−αn)Rn⋆,(2.2)
rn+1+rn=1−αnRn,(2.3)
rn+1⋆+rn⋆+2n+1=αn(1−Rn⋆)−tRn−α.(2.4)
From (S′2), we obtain the following difference equations involving βn, ∑j=0n−1Rj, ∑j=0n−1R⋆j, and the auxiliary quantities:
(rn⋆)2=βnRn⋆Rn−1⋆,(2.5)
rn2−rn=βnRnRn−1,(2.6)
n−(2α+α)rn−2rnrn⋆+rn⋆=βn(Rn(1−Rn−1⋆)+Rn−1(1−Rn⋆)),(2.7)
(rn⋆+n)2−2trnrn⋆s+trn+trn⋆s+α(rn⋆+n)+t∑j=0n−1Rj =βn(1−Rn⋆)(1−Rn−1⋆)−βntRnRn−1⋆s−βntRn⋆Rn−1s,(2.8)
−2rn⋆(rn⋆+n)+2trnrn⋆s+srn⋆−trn⋆s+αrn⋆+s∑j=0n−1Rj⋆ =βnRn⋆(1−Rn−1⋆)+βnRn−1⋆(1−Rn⋆)+βntRnRn−1⋆s+βntRn⋆Rn−1s.(2.9)
To continue, we make use of (2.2)–(2.7) to show that αn(s, t) and βn(s, t) are expressible in terms of the auxiliary quantities and in turn satisfy a system of difference equations.
Lemma 2.1
The recurrence coefficients αn(s, t) and βn(s, t) are expressed in terms of Rn⋆(s,t), Rn(s, t), rn⋆(s,t), and rn(s, t) by
αn(s,t)=2n+1+α+sRn⋆+tRn,(2.10)
βn(s,t)+1Rn(n−(2n+α)rn−2rnrn⋆+rn⋆)−rn2−rnRn2(1−Rn⋆)+(rn⋆)2Rn⋆.(2.11)
Proposition 2.1.
The auxiliary quantities Rn⋆(s,t), Rn(s, t), rn⋆(s,t) and rn(s, t) satisfy the following difference equations
rn+1⋆+rn⋆=(s−(2n+1+α+sRn⋆+tRn))Rn⋆,(2.12)
rn+1+rn=1−(2n+1+α+sRn⋆+tRn)Rn,(2.13)
(rn2−rn)Rn⋆Rn−1⋆=(rn⋆)2RnRn−1,(2.14)
(rn⋆)2(Rn(1−Rn−1⋆)+Rn−1(1−Rn⋆))=(n−(2n+α)rn−2rnrn⋆+rn⋆)Rn⋆Rn−1⋆,(2.15)
which can be solved with the initial conditions
R0⋆(s,t)=sαe−s−ts∫s∞yαe−y−tydy,R0(s,t)=∫s∞yα−1e−y−tydy∫s∞yαe−y−tydy, r0⋆(s,t)=0, r0(s,t)=0.
Proof.
Substituting (2.10) into (2.2) and (2.3) gives rise to (2.12) and (2.13) respectively. Getting rid of βn from (2.5) and (2.6), we get (2.14). Eliminating βn from (2.7) by using (2.5) yields (2.15).
Finally, we present two expressions involving ∑j=0n−1(sRj⋆(s,t)+tRj(s,t)), which we will see are crucial for the forthcoming derivations.
Lemma 1.
We have
−∑j=0n−1(sRj⋆(s,t)+tRj(s,t))=p(n,s,t)+n(n+α)(2.16)
=srn⋆(s,t)+trn(s,t)−βn(s,t)+n(n+α),(2.17)
and a combination of (2.16) and (2.17) gives rise to
βn(s,t)=srn⋆(s,t)+trn(s,t)−p(n,s,t).(2.18)
Proof.
Replacing n by j in (2.10), we have
αj(s,t)=2j+1+α+sRj⋆(s,t)+tRj(s,t).
Summing it from
j = 0 to
n − 1, in view of
∑j=0n−1αj(s,t)=−p(n,s,t), we come to
(2.16).
Equation (2.17) results from the summation of
(2.8) and
(2.9).
2.2. s evolution and t evolution
We shall consider the differentiation of
hj(s,t)δjk:=∫s∞Pj(x;s,t)Pk(x;s,t)w(x;t)dx,
with
s and
t, with
j =
k =
n and
j =
k + 1 =
n. We see that
Rn⋆(s,t) and
Rn(
s,
t) are closely related to the log-derivative of
hn(
s,
t), while
rn⋆(s,t) and
rn(
s,
t) are connected with the derivative of
p(
n,
s,
t). With the aid of the identities in the previous subsection, we derive a system of first order partial differential equations satisfied by
Rn⋆(s,t),
Rn(
s,
t),
rn⋆(s,t) and
rn(
s,
t).
Taking the derivative of
hn(s,t)=∫s∞Pn2(x;s,t)w(x;t)dx,
with respect to
s and
t, we get
∂sloghn(s,t)=−Rn⋆(s,t),∂tloghn(s,t)=−Rn(s,t).(2.19)
From the fact βn(s, t) = hn(s, t)/hn−1(s, t) and (2.19), it follows that
∂slogβn(s,t)=Rn−1⋆(s,t)−Rn⋆(s,t),∂tlogβn(s,t)=Rn−1(s,t)−Rn(s,t).(2.20)
We proceed with the differentiation with s and t on
0=∫s∞Pn(x;s,t)Pn−1(x;s,t)w(x;t)dx,
to find,
∂sp(n,s,t)=rn⋆(s,t),∂np(n,s,t)=rn(s,t).(2.21)
Hence, in view of αn(s, t) = p(n, s, t) − p(n + 1, s, t), we find
∂sαn(s,t)=rn⋆(s,t)−rn+1⋆(s,t),∂tαn(s,t)=rn(s,t)−rn+1(s,t).(2.22)
Moreover, on combining (2.18) with (2.21), we come to a representation of βn(s, t) in terms of p(n, s, t) and its first order partial derivatives:
βn(s,t)=(s∂s+t∂t)p(n,s,t)−p(n,s,t).(2.23)
Now, we are in a position to derive analogs of coupled Riccati equations: There are four equations instead of two.
Lemma 2.3.
The auxiliary quantities Rn⋆(s,t), Rn(s, t), rn⋆(s,t) and rn(s, t) satisfy four first order partial differential equations
s∂sRn⋆+t∂sRn=2rn⋆+(2n+α+sRn⋆+tRn)Rn⋆−sRn⋆,(2.24)
s∂tRn⋆+t∂tRn=2rn+(2n+α+sRn⋆+tRn)Rn−1,(2.25)
s(∂srn⋆)+t(∂srn)=(1−Rn⋆)((rn⋆)2Rn⋆+Rn⋆Rn2(rn2−rn)) +Rn⋆Rn(2rnrn⋆−rn⋆+(2n+α)rn−n),(2.26)
s∂trn⋆+∂trn=−RnRn⋆(rn⋆)2+2−Rn⋆Rn(rn2−rn) +2rnrn⋆−rn⋆+(2n+α)rn−n.(2.27)
Proof.
We differentiate (2.10), i.e.
αn(s,t)=2n+1+α+sRn⋆(s,t)+tRn(s,t),
over
s and
t and in view of
(2.22), to find,
rn⋆−rn+1⋆=Rn⋆+s∂sRn⋆+t∂sRn,rn−rn+1=Rn+s∂tRn⋆+t∂tRn.
Getting rid of rn+1⋆ and rn+1 by using (2.12) and (2.13) respectively, we are led to (2.24) and (2.25).
Taking the derivative of (2.18), i.e.
βn(s,t)=srn⋆(s,t)+trn(s,t)−p(n,s,t),
with respect to
s and
t, and remembering that
∂sp(n,s,t)=rn⋆(s,t) and
∂t p(
n,
s,
t) =
rn(
s,
t), we find
βnRn−1⋆−βnRn⋆=s∂srn⋆+t∂srn,βnRn−1−βnRn=s∂srn⋆+t∂trn.
According to (2.5) and (2.6), we replace βnRn−1⋆ by (rn⋆)2Rn⋆ and βnRn−1 by rn2−rnRn in the above two equations, then on removing βn by using (2.11) in the resulting equations, we arrive at (2.26) and (2.27).
We now look at the Hankel determinant in question, namely,
Dn(s,t)=det(∫s∞xi+jw(x;t)dx)i,j=0n−1 =∏j=0n−1hj(s,t).
Applying s∂s + t∂t to
logDn(s,t)=∑j=0n−1loghj(s,t),
we get, because of
(2.19),
(s∂s+t∂t)logDn(s,t)=−∑j=0n−1(sRj⋆(s,t)+tRj(s,t))(2.28)
=p(n,s,t)+n(n+α),(2.29)
where the second equality comes from
(2.16). Applying
s∂s +
t∂t again to
(2.29), in view of
(2.21), we find
(s2∂ss+2st∂st+t2∂tt)logDn(s,t)+(s∂s+t∂t)logDn(s,t)=srn⋆(s,t)+trn(s,t),
which combined with
(2.29) yields
(s2∂ss+2st∂st+t2∂tt)logDn(s,t)=srn⋆(s,t)+trn(s,t)−p(n,s,t)−n(n+α) =βn(s,t)−n(n+α),(2.30)
where the second equality is due to
(2.18). Noting that
βn(s,t)=Dn+1(s,t)Dn−1(s,t)(Dn(s,t))2,
which is a consequence of
βn(
s,
t) =
hn(
s,
t)/
hn−1(
s,
t) and
hn(
s,
t) =
Dn+1(
s,
t)/
Dn(
s,
t), we finally establish the following statement from
(2.30) which can be viewed as a two variable version of the Toda-molecule equations.
Proposition 2.2.
The following second order partial differential-difference equation holds:
(s2∂ss+2st∂st+t2∂tt)logDn(s,t)=Dn+1(s,t)Dn−1(s,t)(Dn(s,t))2−n(n+α).(2.31)
By making the substitution
Dn(s,t)=(s+t)n(n+α)D^n(s,t),
this is transformed into
(s2∂ss+2st∂st+t2∂tt)logD^n(s,t)=(s+t)2D^n+1(s,t)D^n−1(s,t)(D^n(s,t))2.
To close this subsection, we apply (2.10), (2.20) and (2.22), namely the expressions involving the recurrence coefficients and their first order partial derivatives, to derive the results below.
Proposition 2.3.
The following partial differential relations hold for the recurrence coefficients αn(s, t) and βn(s, t):
(s∂s+t∂t−1)αn(s,t)=βn(s,t)−βn+1(s,t),(2.32)
(s∂s+t∂t)logβn(s,t)=αn−1(s,t)−αn(s,t)+2,(2.33)
and, as a consequence, we have
(s2∂ss+2st∂st+t2∂tt)logβn(s,t)=βn−1(s,t)−2βn(s,t)+βn+1(s,t)−2.(2.34)
Proof.
To prove (2.32), we apply s∂s + t∂t to
αn(s,t)=p(n,s,t)−p(n+1,s,t).
According to (2.21), we get
(s∂s+t∂t)αn(s,t)=srn⋆(s,t)+trn(s,t)−(srn+1⋆(s,t)+trn+1(s,t))=βn(s,t)+p(n,s,t)−(βn+1(s,t)+p(n+1,s,t))=βn(s,t)−βn+1(s,t)+αn(s,t),
where the second equality results from
(2.18).
From (2.20), it follows that
(s∂s+t∂t)logβn(s,t)=sRn−1⋆(s,t)+tRn−1(s,t)−(sRn⋆(s,t)+tRn(s,t)),
which, in view of
(2.10), leads to
(2.33).
We differentiate (2.33) over s and t to obtain
(∂s+s∂ss+t∂st)logβn(s,t)=∂sαn−1(s,t)−∂sαn(s,t),(s∂ts+∂t+t∂tt)logβn(s,t)=∂tαn−1(s,t)−∂tαn(s,t).
Multiplying the first equality by s and the second by t, and adding them, we find
(s2∂ss+2st∂st+t2∂tt)logβn(s,t)+(s∂s+t∂t)logβn(s,t) =(s∂s+t∂t)αn−1(s,t)−(s∂s+t∂t)αn(s,t),
which, combined with
equations (2.32) and
(2.33), establishes
(2.34).
2.3. Partial differential equations satisfied by Rn⋆(s,t), Rn(s, t) and (s∂s + t∂t)logDn(s, t)
Bear in mind that the auxiliary quantities Rn⋆(s,t), Rn(s, t), rn⋆(s,t) and rn(s, t) satisfy four analogs of Riccati equations given by (2.24)–(2.27). We solve for rn⋆(s,t) from (2.24) and rn(s, t) from (2.25), and substitute them into (2.26) and (2.27), to establish two second order partial differential equations for Rn⋆(s,t) and Rn(s, t).
Theorem 2.1.
The auxiliary quantities Rn⋆(s,t) and Rn(s, t) satisfy a pair of second order nonlinear partial differential equations:
0=s(s∂ss+t∂st)Rn⋆+t(s∂ss+t∂st)Rn+Rn⋆−12Rn⋆(Rn(s))2+(Rn⋆−1)Rn⋆2Rn2(Rn(t))2 −Rn⋆RnRn(s)Rn(t)−tRnRn(s)+tRn⋆Rn(t)+(s∂s+t∂t)Rn⋆−st(∂sRn) −Rn⋆(sRn⋆+tRn)2+sRn⋆(sRn⋆+tRn)−(2n+1+α)Rn⋆(sRn⋆+tRn−s) +s22Rn⋆(Rn⋆−1)+Rn⋆(1−Rn⋆)2Rn2−αRn⋆Rn,(2.35a)
and
0=s(t∂tt+s∂st)Rn⋆+t(t∂tt+s∂st)Rn+Rn2Rn⋆(Rn(s))2+(Rn⋆−2)2Rn(Rn(t))2 −Rn(s)Rn(t)−sRnRn(s)+tRn⋆Rn(t)+(s∂s+t∂t)Rn−s2(∂tRn⋆) −Rn(sRn⋆+tRn)2−(2n+1+α)Rn(sRn⋆+tRn)+s22Rn⋆Rn+2−Rn⋆2Rn−α.(2.35b)
Here,
Rn(s)(s,t):=s∂sRn⋆(s,t)+t∂sRn(s,t),Rn(t)(s,t):=s∂tRn⋆(s,t)+t∂tRn(s,t).(2.36)
Define
Hn(s,t):=(s∂s+t∂t)logDn(s,t),
then it follows from
(2.28) and
(2.29) that
Hn(s,t)=−∑j=0n−1(sRj⋆(s,t)+tRj(s,t)) =p(n,s,t)+n(n+α).(2.37)
Since ∂sp(n,s,t)=rn⋆(s,t) and ∂t p(n, s, t) = rn(s, t) due to (2.21), we find from (2.37)
∂sHn(s,t)=rn⋆(s,t),∂tHn(s,t)=rn(s,t).(2.38)
In order to characterize the Hankel determinant, we first establish the connection between Hn(s, t) and the quantities Rn⋆(s,t) and Rn(s, t), and then make use of (2.35) to derive the partial differential equation satisfied by Hn(s,t).
Theorem 2.2.
The quantity Hn(s, t) is expressible in terms of Rn⋆(s,t) and Rn(s, t) as
Hn(s,t)=14Rn⋆(Rn(s))2+1−Rn⋆4Rn2(Rn(t))2+Rn(s)Rn(t)2Rn −14(sRn⋆+tRn)2−(n+α2)(sRn⋆+tRn)+s24Rn⋆ −α24+t2+α2Rn+Rn⋆−14Rn2,(2.39)
where Rn(s)(s,t) and Rn(t)(s,t) have the same meaning as in (2.36). Moreover, Hn(
s,
t)
satisfies the following second order sixth degree partial differential equation:
[((t∂tt+s∂st)Hn)2+4((∂tHn)−1)(∂tHn)((s∂s+t∂t)Hn−Hn+n(n+α))]2 ⋅[((s∂ss+t∂st)Hn)2+4(∂sHn)2((s∂s+t∂t)Hn−Hn+n(n+α))] =[((t∂tt+s∂st)Hn)2((∂sHn)2+(s∂s+t∂t)Hn−Hn+n(n+α)) +((s∂ss+t∂st)Hn)2((∂tHn)−1)(∂tHn) −((t∂tt+s∂st)Hn)((s∂ss+t∂st)Hn)((2n+α)(∂tHn)+(∂sHn)(2(∂tHn)−1)−n) +((s∂s+t∂t)Hn−Hn+n(n+α)) ⋅(−2(2(∂tHn)−1)((2n+α)(∂tHn)−n)(∂sHn)−(n−(2n+α)(∂tHn))2 −(∂sHn)2+4((∂tHn)−1)(∂tHn)((s∂s+t∂t)Hn−Hn+n(n+α)))]2.(2.40)
Proof.
Keeping (2.11) in mind, we get from (2.18) and (2.37)
Hn=srn⋆+trn−βn+n(n+α) =srn⋆+trn−1Rn(n−(2n+α)rn−2rnrn⋆+rn⋆) +r2−rnRn2(1−Rn⋆)−(rn⋆)2Rn⋆+n(n+α).
Eliminating rn⋆ and rn from the second equality above by using (2.24) and (2.25) respectively, after simplification, we come to (2.39).
Now that we have the expression for Hn in terms of Rn⋆ and Rn, in order to establish the differential equation for Hn, it suffices to derive the representations of Rn⋆ and Rn in terms of Hn. To achieve the desired results, we first remove Rn−1⋆ and Rn−1 from (2.20) by using (2.5) and (2.6), and get
∂s(logβn)=(rn⋆)2βnRn⋆−Rn⋆,∂t(logβn)=rn2−rnβnRn−Rn.
Solving for Rn⋆ and Rn from these two quadratic equations in Rn⋆ and Rn respectively, we obtain
Rn⋆=−12∂slogβn+12(∂slogβn)2+4(rn⋆)2βn,Rn=−12∂tlogβn+12(∂tlogβn)2+4(rn2−rn)βn.(2.41)
Here take note that there should be two solutions for both Rn⋆ and Rn, however, the solutions with minus sign before the square roots in (2.41) are rejected, since from (2.20) we see that
Rn⋆+12∂slogβn+12(Rn−1⋆+Rn⋆)≥0,Rn+12∂tlogβn+12(Rn−1+Rn)≥0,
where we have used the fact that
Rn⋆≥0 and
Rn ≥ 0 which is a direct consequence of their definitions.
From (2.23) and (2.37), we see that
βn=(s∂s+t∂t)Hn−Hn+n(n+α).
Substituting this equation and (2.38) into (2.41), we obtain two equations expressing Rn⋆ and Rn in terms of Hn and its derivatives. Plugging these expressions into (2.39), we establish (2.40).
3. Double Scaling for Large n and Asymptotic Expansions for the Scaled Hankel Determinant
Based on the large partial differential equations satisfied by Rn⋆(s,t), Rn(s, t) and Hn(s, t), we proceed with the study of the Hankel determinant as n gets large, under the assumption that n → ∞, s → 0 and t → 0 such that S and T defined by
S:=4ns, T:=(2n+1+α)t,
are fixed. This scaling is motivated as follows: for
t = 0 and under the scaling
S = 4
ns, the limiting behavior of the smallest eigenvalue distribution of Laguerre unitary ensemble is characterized by the Bessel kernel [
33]; for
s = 0 and under the scaling
T = (2
n + 1 +
α)
t, the eigenvalue correlation kernel of the singularly perturbed Laguerre unitary ensemble is given by the Ψ-kernel [
34, Corollary 1].
To proceed further, we make the Ansatz that Rn⋆(s,t), Rn(s, t), and their first and second order partial derivatives over s or t, have limits as n → ∞. Define
2R⋆(S,T):=limn→∞Rn⋆(s,t),R(S,T):=limn→∞Rn(s,t).(3.1)
Then, on replacing s by S4n and t by T2n+1+α in (2.24) and (2.25), and by sending n to ∞, we find
limn→∞rn⋆(s,t)n=−2R⋆(S,T),limn→∞rn(s,t)n=−R(S,T).(3.2)
Moreover, we derive the following limiting equations from (2.35).
Theorem 3.1.
The quantities R⋆(S, T) and R(S, T) satisfy a couple of second order first degree partial differential equations
0=S(S∂SS+T∂ST)R⋆+T(S∂SS+T∂ST)R+2R⋆−12R⋆(S(∂SR⋆)+(∂SR))2+(2R⋆−1)R⋆2R2(S(∂TR⋆)+T(∂TR))2−2R⋆R(S(∂SR⋆)+T(∂SR))(S(∂TR⋆)+T(∂TR))+(S∂S+T∂T)R⋆−R⋆(SR⋆+TR)+S2R⋆+R⋆(1−2R⋆)2R2−αR⋆R,(3.3a)
and
0=S(R∂TT+S∂TS)R⋆+T(T∂TT+S∂TS)R+RR⋆(S(∂SR⋆)+T(∂SR))2+R⋆−1R(S(∂TR⋆)+T(∂TR))2−2(S(∂SR⋆)+T(∂SR))(S(∂TR⋆)+T(∂TR))+(S∂S+T∂T)R−R(SR⋆+TR)+1−R⋆R−α.(3.3b)
Proof.
Adopting ideas in [30], we proceed as follows.
From (3.1), i.e. Rn⋆(s,t)∼2R⋆(s,t), Rn(s, t) ∼ R(S, T) as n → ∞, it follows that
∂sRn⋆∼2∂SR⋆⋅4n,∂sRn∼∂SR⋅4n,∂tRn⋆∼2∂TR⋆⋅(2n+1+α),∂tRn∼∂SR⋅(2n+1+α),∂stRn⋆∼2∂STR⋆⋅4n⋅(2n+1+α),∂stRn∼∂STR⋅4n⋅(2n+1+α),∂ssRn⋆∼2∂SSR⋆⋅(4n)2,∂ssRn∼∂SSR⋅(4n)2,∂ttRn⋆∼2∂TTR⋆⋅(2n+1+α)2,∂ttRn∼∂TTR⋅(2n+1+α)2.
Replacing the first and second order partial derivatives of Rn⋆(s,t) and Rn(s, t) in (2.35) by using the above relations, and substituting 2R⋆, R, S4n, T2n+1+α for Rn⋆, Rn, s, t respectively, by taking the series expansion in n of the resulting equation and with the aid of the software Mathematica, we obtain (3.3) by retaining leading order terms when n → ∞.
The next proposition gives us the lower and upper bounds for R⋆(S, T) and R(S, T), which result from the definitions of Rn⋆(s,t) and Rn(s, t).
Proposition 3.1.
For α > 0, the auxiliary quantities Rn⋆(s,t) and Rn(s, t) are bounded by
0≤Rn⋆(s,t)<1,0<Rn(s,t)≤1α,
with the lower bound for Rn⋆(s,t) and the upper bound for Rn(
s,
t)
attained only for s = 0
and s =
t = 0
respectively. As a consequence, we have for α > 0,
0≤R⋆(S,T)≤12,0<R(S,T)≤1α.
Proof.
Recall that
Rn⋆(s,t)=w(s;t)hn(s,t)Pn2(s;s,t)≥0,Rn(s,t)=1hn(s,t)∫s∞dyyPn2(y;s,t)w(y;t)>0.
Through integration by parts, we find
0<αRn(s,t)=1hn(s,t)∫s∞Pn2(x;s,t)e−x−txdxα=1−Rn⋆(s,t)−1hn(s,t)∫s∞dyy2Pn2(y;s,t)w(y;t)≤1−Rn⋆(s,t)≤1.
Hence, we have
0<αRn(s,t)≤1−Rn⋆(s,t)≤1,
where the first and second equality sign hold if and only if
t = 0 and
s = 0 respectively. Therefore, we see that
0≤Rn⋆(s,t)<1 and 0<Rn(s,t)≤1α,
where the equality signs are valid only for
s = 0 and
s =
t = 0 respectively. Finally, by using the property of the limit of a function, we obtain
0≤R⋆(S,T)≤12limn→∞Rn⋆(s,t)≤12,0<R(S,T)=limn→∞Rn(s,t)≤1α.
To continue, for s = S/(4n) and t = T/(2n + 1 + α), we let
H(S,T):=limn→∞Hn(s,t)=(S∂S+T∂T)logΔ(S,T).
Here recall that Hn(s, t) := (s∂s + t∂t)Dn(s, t) and Δ(S, T) is defined by
Δ(S,T):=limn→∞Dn(S4n,T2n+1+α)Dn(0,0).
It follows from (2.38) and (3.2) that
∂SH(S,T)=limn→∞rn⋆(s,t)4n=−12R⋆(S,T),∂TH(S,T)=limn→∞rn(s,t)2n+1+α=−12R(S,T).(3.4)
Furthermore, by an argument similar to the one used in the proof of Theorem 3.1. we establish the following statement from (2.39) and (2.40).
Theorem 3.2.
The quantity H(S, T) is expressed in terms of R⋆(S, T) and R(S, T) by
H(S,T)=−12R⋆(S(∂SR⋆)+T(∂SR))2+1−2R⋆4R2(S(∂TR⋆)+T(∂TR))2+1R(S(∂SR⋆)+T(∂SR))(S(∂TR⋆)+T(∂TR))−12(SR⋆+TR)−α24+α2R+2R⋆−14R2,(3.5)
and the following second order second degree partial differential equation holds:
0=4(∂TH)2∂SH((S∂SS+T∂ST)H)2+(4(∂SH)+1)((T∂TT+S∂TS)H)2−8(∂TH)((S∂SS+T∂ST)H)((T∂TT+S∂TS)H)−(α(∂TH)+12)2+4((S∂S+T∂T−1)H)(∂TH)2−∂SH.(3.6)
3.1. Asymptotic expansions for large S and fixed T
We assume H(S, T) with T fixed has an expansion for large S of the following form
H(S,T)=∑j=0∞aj(T)S1−j2, S→∞.(3.10)
As is pointed out in , the special case H(S, 0) is investigated in [27] where the symbol σ(s) was used instead. According to (3.13) in [27], H(S, 0) has the following expansion
H(S,0)=−S4+α2S−α24−α16S−α216S−(α316+9α256)S−32−(α416+9α264)S−2−(α516+45α3128+225α2048)S−52−(α616+45α464+27α232)S−3+O(S−72)=∑j=0∞aj(0)S1−j2, S→∞,
which provides initial conditions satisfied by
aj(
T) at
T = 0. For example, we have
a0(0)=−14,
a1(0)=α2 and
a2(0)=−α24.
Substituting (3.10) into (3.6), on comparing the coefficients of the highest order term in S on both sides, we get a0(T)a′0(T) = 0 which indicates that a0(T) must be a constant. Hence we have a0(T)=a0(0)=−14. Successive coefficients aj(T), j = 1, 2,···, are computed similarly, and we finally arrive at
H(S,T)=−S4+α2S−α24−(T2+α16)1S−α216S−(T16+α316+9α256)S−32−(α8T+α416+9α264)S−2−((3α216+27256)T+α516+45α3128+225α2048)S−52−(T216+(α34+9α16)T+α616+45α464+27α232)S−3+O(S−72), S→∞.(3.11)
Hence, we find
R⋆(S,T)=−2∂SH(S,T)=12−α2S−(T2+α16)S−32−α28S2−(3T16+3α316+27α256)S−52−(α2T+α44+9α216)S−3−((15α216+135256)T+5α516+225α3128+1125α2048)S−72−(3T28+(3α32+27α8)T+3α616+135α432+81α216)S−4+O(S−92), S→∞,
and
R(S,T)=−2∂TH(S,T)=1S+18S32+α4S2+(3α28+27128)S−52+(T4+α32+9α8)S−3+O(S−72), S→∞.
Since R⋆(S, 0) corresponds to 12R(s) in [27], we remark that our expansion formula for R⋆(S, 0) agrees with the one for R(s) (equation (3.12), [27]). a
Noting that H(S, T) = (S∂S + T∂T)logΔ(S, T, α), by integration of (3.11), we obtain the following asymptotic expansion of Δ(S, T, α).
Theorem 3.3.
We have, for large S and fixed T,
logΔ(S,T,α)=C1(TS,α)−S4+αS−α24logS+(α8−T)1S+α216S+(α324+3α128+T8)S−32+(α432+9α2128+α8T)S−2+O(S−52),(3.12)
where C1(TS,α) is an arbitrary function of TS and α.
3.2. Asymptotic expansions for small S and fixed T > 0
Motivated by (3.9), we write for fixed T > 0,
H(S,T)=−T2S+∑j=0∞bj(T)Sj2, S→0.
Substituting it into (3.6), by comparing the coefficients on both sides, we come to
H(S,T)=−T2S+116−α24+α2S−S4+S3216T−αS5216T2+S316T2+(α216−27256)S72T3−αS48T3+(−α316+99α256+T16)S92T4+(α216−964)3S5T4+O(S112), S→0.(3.13)
It follows that
R⋆(S,T)=−2∂SH(S,T)=−T2S32−α2S+12−3S16T+5αS3216T2−3S28T2−(α216−27256)7S52T3+αS3T3−(−α316+99α256+T16)9S72T4−(α28−932)15S4T4+O(S92), S→0,
and
R(S,T)=−2∂SH(S,T)=1S+S328T2−αS524T3+S34T3+(α28−27128)3S72T4−3S44T4+(−α32+99α32+3T8)S92T5+(α22−98)3S4T5+O(S112), S→0.
Since H(S, T) = (S∂S + T∂T)(logΔ(S, T, α)), we establish the asymptotic expansion of Δ(S, T, α) by integrating of (3.13).
Theorem 3.4.
For small S and fixed T > 0, we have
logΔ(S,T,α)=C2(TS,α)−TS+(116−α24)−logS+αS−S4+S328T−αS528T2+S316T2+(α28−27128)S72T3−αS48T3+O(S92),(3.14)
where C2(TS,α) is an arbitrary function of TS and α.
3.3. Asymptotic expansions for fixed S > 0 and large T
According to (3.9), we suppose H(S, T) with S > 0 fixed has the expansion for large T of the following form
H(S,T)=−T2S+∑j=0∞cj(S)T−j, T→∞.
Substituting it into (3.6) and comparing the coefficients on both sides, we obtain
H(S,T)=−T2S−14(S−α)2+116+S3216T+S5216T2(S−α)+S7216T3((S−α)2−2716)+S9216T4((S−α)3−274S+99α16)+S11216T5((S−α)4−98(S−α)(15S−13α)+2277128)+O(T−6), T→∞,(3.15)
Therefore, we have
R*(S,T)=−2∂SH(S,T)+T2S32+12−α2S−3S16T−S3216T2(6S−5α)−S5216T3((S−α)(9S−7α)−18916)−S7216T4(3(S−α)2(4S−3α)−1352S+89116α)+O(T−5), T→∞,
and
R(S,T)=−2∂SH(S,T)=1S+S328T2+S524T3(S−α)+3S728T4((S−α)2−2716)+S922T5((S−α)3−274S+9916α)+5S1128T6((S−α)4−98(S−α)(15S−13α)+2277128)+O(T−7), T→∞.
Using the fact that H(S, T) = (S∂S + T∂T)logΔ(S, T, α) again, after integration of (3.15), we obtain the asymptotic expansion of Δ(S, T, α).
Theorem 3.5.
We have, for large T and fixed S > 0,
logΔ(S,T,α)=C3(TS,α)−TS−S4+αS(116−α24)logS+S328T+S3−2αS5216T2+S7224T3(S−3αS+3α2−8116)+S9232T4(S32−4αS+(6α2−272)S−4α3+994α)+O(T−5).(3.16)
Here C3(TS,α) is an arbitrary function of TS and α.
To summarize, we have established the asymptotic expansions of the scaled Hankel determinant Dn(S4n,T2n+1+α,α), n → ∞, for large S with T ≥ 0 fixed, for small S with T > 0 fixed, and for large T with S > 0 fixed. In each case, there is a function term Cj(TS,α) to be determined. In order to compute them, we turn our attention back to the monic polynomials Pn(x; s, t, α) orthogonal with respect to w(x;t,α)=xαe−x−tx over [s, ∞) with α > −1, t ≥ 0, s ≥ 0.
We have the following determinant representation for Pn(z; s, t, α) in terms of the moments (see Szegö [32], Chapter II)
Pn(z;s,t,α)=1Dn(s,t,α)det(∫s∞xi+j(z−x)xαe−x−txdx)i,j=0n−1.(3.17)
Keeping in mind that Dn(s, t, α) is the Hankel determinant generated by the moments of w(x; t, α), i.e.
Dn(s,t,α)=det(∫s∞xi+jxαe−x−txdx)i,j=0n−1,
we readily see from
(3.17) that
Pn(0;s,t,α)=(−1)nDn(s,t,α+1)Dn(s,t,α).
Hence, by recalling that
Δ(S,T,α)=limn→∞Dn(S4n,T2n+1+α,α)Dn(0,0,α),
we find, as
n → ∞,
Pn(0;S4n,T2n+1+α,α)Pn(0;0,0,α)∼Δ(S,T,α+1)Δ(S,T,α)∼exp[Cj(TS,α+1)−Cj(TS,α)]+⋯.(3.18)
Note that (3.18) results from (3.12), (3.14) and (3.16) for j = 1, 2, 3, and obviously the subsequent terms of it can be written out explicitly.
We shall apply logarithmic potential theory with an external field in the next section to deduce an approximation of Pn(0;S4n,T2n+1+α,α) for large n and hence get the asymptotic formula for Pn(0;S4n,T2n+1+α,α)/Pn(0;0,0,α). By comparing the result with (3.18), we find
Cj(TS,α+1)−Cj(TS,α)=log(Γ(1+α)2π)+T2S, j=1,2,3.
4. The Evaluation of Pn(z; s, t, α) at z = 0 via Logarithmic Potential Theory with an External Field
Denote by {xk}k=1n the eigenvalues of the unitary ensembles of n × n Hermitian matrices with the weight function
w(x;t,α)=xαe−x−tx, x∈[0,∞), α>−1, t≥0,
and with the potential
v(x)=−logw(x;t,α)=−αlogx+x+tx,
and we suppose
xk ≥
s for
k = 1, 2,...,
n. If we interpret
{xk}k=1n as the positions of
n identically charged particles, then, for sufficiently large
n, the particles can be approximated as a continuous fluid with a density
σ(
x). Since
v(
x) is convex for
x ∈ [
s, ∞), we know that
σ(
x) is supported on a single interval [
s,
b]. See [
10] for a detailed analysis.
In Dyson’s works [19], such σ(x) is determined via the constrained minimization
minσF[σ] subject to ∫sbσ(x)dx=n,
with
F[σ]:=∫sbσ(x)v(x)dx−∫sb∫sbσ(x)log|x−y|σ(y)dxdy.
According to Theorem 1.3 in Chapter I.1 of [31], the equilibrium density σ(x) satisfies the condition
v(x)−2∫sblog|x−y|σ(y)dy=A, x∈[s,b],
where
A is the Lagrange multiplier that fixes
∫sbσ(x)dx=n. Differentiation of this equation with respect to
x gives rise to a singular integral equation
2P∫sbσ(y)dyx−y=v′(x), x∈[s,b],
where
P denotes the Cauchy principal value. Based on the theory of singular integral equations (see
Theorem 3.1 in Chapter IV.3, [
31]), we find
σ(x)=12π2b−xx−sP∫sbv′(y)y−xy−sb−ydy, x∈(s,b).(4.1)
From which the normalization condition ∫sbσ(x)dx=n reads
12π∫sby−sb−yv′(y)dy=n.(4.2)
It should be pointed out that (4.1) and (4.2) are not applicable in the case s = 0. Hence in our subsequent considerations, we assume s > 0 and and therefore S > 0.
Noting that
v′(x)=−αx+1−tx2,
with the aid of integral identities listed in the
Appendix, we obtain from
(4.1)
σ(x)=12πb−xx−s[1+(t2sb−(α+t2b)sb)1x−sb⋅tx2].(4.3)
Here we require sb−t2(1s+1b)>α to guarantee that σ(x) > 0 for s < x < b. Substituting (4.3) into the normalization condition ∫sbσ(x)dx=n, or by using (4.2) directly, we find for s > 0,
b−s4−α2+12sb(αs+t2(sb−1))=n.
Clearing the square root, we come to the next lemma.
Lemma 4.1.
For s > 0 and t ≥ 0, the upper edge b of the support of the equilibrium density satisfies an algebraic equation of degree five
b3(b−2(2n+α)−s)2=s((2α−ts)b+t)2.(4.4)
Substituting b=2(2n+α)+s+b˜ into (4.4), one sees that, by sending n → ∞, b˜=O(n−1/2). Hence, for large n, we suppose b has the expansion of the form
b=2(2n+α)+s+∑j=1∞dj(s,t)n−j2.(4.5)
Plugging this into (4.4) and by comparing the coefficient of the highest order term in n, we get d1(s,t)=±(αs−t2s). According to (3.19) of [27], we know that d1(s,0)=−as. Hence we choose d1(s,t)=t2s−αs. Successive coefficients dj(s, t) are uniquely determined, which ultimately gives the expansion of b for large n.
Lemma 4.2.
For s > 0 and t ≥ 0, the following asymptotic expansion of b holds
b=4n+2α+s+(t2s−αs)n−12 +18(αs32+(2α2−3t2)s−αts)n−32 −18(α2s−αt+t24s)n−2+O(n−52), n→∞,(4.6)
from which, on replacing s by S4n and t by T2n+1+α, and by sending n to ∞
, we find
b=4n+2α+(S4−α2S+T2s)n−1 +(α28S−(3α+2)T8S)n−2+O(n−3), n→∞.
Based on logarithmic potential theory with an external field and the linear statistics formula, it is proved in [14] that the monic polynomial Pn(x; s, t, α) orthogonal with respect to w(x; t, α) over [s, ∞) can be approximated as follows:
Pn(z;s,t,α)∼e−S1(z;s,t,α)−S2(z;s,t,α), z∈ℂ\[s,b],
where
S1(z;s,t,α)=14log[16(z−s)(z−b)(b−s)2(z−s−z+bz−s+z−b)2],
S2(z;s,t,α)=−nlog(z−s+z+b2)2+12π∫sbv(x)dx(b−x)(x−s)[z−s−z+bx−z+1],
and
v(x)=−logw(x;t,α)=−αlogx+x+tx. We point out here a simpler representation for e
−S1(z; s, t, α):
e−S1(z;s,t,α)=12[(z−bz−s)14+(z−sz−b)14].
The above result also appeared in the context of one dimensional free probability [26]. We now give an evaluation of Pn(0; s, t, α).
Theorem 4.1.
The monic polynomials Pn(x; s, t, α) orthogonal with respect to xαe−x−tx over [s, ∞) with α > −1, t ≥ 0, s > 0, is evaluated at x = 0 by
(−1)nPn(0;s,t,α)∼nn+α2+14(4s)−α2−14(1+(1+2α)s2n+O(n−1))⋅exp[−n+4ns−s2+t4s+O(n−12)], n→∞,(4.7)
so that, we have for s=S4n and t=T2n+1+α,
(−1)nPn(0;S4n,T2n+1+α,α)∼nn+α+12S−α2−14(1+(1+2α)(S4+α8)n−1+O(n−2))⋅exp[−n+S+T2SO(n−1)], n→∞.(4.8)
As a consequence, we obtain as n → ∞,
Pn(0;S4n,T2n+1+α,α)Pn(0;0,0,α)∼Γ(1+α)2πS−α2−14eS+T2S.(4.9)
Proof.
In what follows, the symbol ∼ refers to ’asymptotic to’ for large n, in the sense that the ratio tends to 1. We have
Pn(0;s,t,α)∼e−S1(0;s,t,α)−S2(0;s,t,α),
where
e−S1(0;s,t,α)=12[(bs)14+(sb)14],
and
S2(0;s,t,α)=−nlog(s+−b2)2+12π∫sb−αlogx+x+tx(b−x)(x−s)[(−s)(−b)x+1]dx.
By using the integral formulas given in the Appendix and taking the branch −s = seπi and −b = beπi, we arrive at
e−S2(0;s,t,α)=(−1)n4−n−α(b+s)2n((bs)14+(sb)14)2α⋅exp[(tsb−1)(b−s2)2].
By plugging the expansion of b given by (4.6) into e−S1(0; s, t, α) and e−S2(0; s, t, α), we obtain (4.7). Upon substituting S4n for s and T2n+1+α for t in (4.7), and by sending n to ∞, expression (4.8) follows.
Since Dn(0, 0, α) has a closed-form expression and reads (see [29], p.321)
Dn(0,0,α)=∏j=0n−1Γ(j+1)Γ(j+α+1),
we have
(−1)nPn(0;0,0,α)=Dn(0,0,α+1)Dn(0,0,α)=Γ(n+1+α)Γ(1+α).
Hence, it follows from (4.8) that
Pn(0;S4n,T2n+1+α,α)Pn(0;0,0,α)∼Γ(1+α)nn+α+12e−nΓ(n+1+α)S−α2−14eS+T2S∼Γ(1+α)2πS−α2−14eS+T2S.
Here in the last step, we make use of the approximation
(ne)n∼Γ(n+1+α)2πn−α−12,
which results from Stirling’s formula [
25]
n!∼2πn(ne)n,
and the standard asymptotic approximation for Gamma function
Γ(n+α)∼Γ(n)nα, α∈ℂ.
The proof is completed.
Bearing in mind that
Pn(0;S4n,T2n+1+α,α)Pn(0;0,0,α)∼Δ(S,T,α+1)Δ(S,T,α), n→∞,
we establish the following statement.
Corollary 4.1.
The undetermined terms that appear in the asymptotic expansions of logΔ(S, T, α) in the previous section satisfy a difference equation
Cj(TS,α+1)−Cj(TS,α)=log(Γ(1+α)2π)+T2S, j=1,2,3,(4.10)
and therefore, up to a constant that is independent of α,S and T the term Cj(TS,α) has the form
logG(α+1)(2π)α/2+α2⋅TS,
where G(·)
denotes the Barnes-G function defined by
G(z+1)=Γ(z)G(z), G(1)=1.
Proof.
Since the proofs of (4.10) for j = 1, 2, 3 are similar, we only prove the case j = 1 with S large and T ≥ 0 fixed. We find from (3.12)
limn→∞Pn(0;S4n,T2n+1+α,α)Pn(0;0,0,α)=Δ(S,T,α+1)Δ(S,T,α) ∼exp[C1(TS,α+1)−C1(TS,α)+S−(α2+14)logS].
Comparing this with (4.9) leads to
C1(TS,α+1)−C1(TS,α)=log(Γ(1+α)2π)T2S.
Appendix A.
Some Relevant Integral Identities
We state the following integral identities for 0 < a < b, which are relevant to our derivation and could be found in [9], [15] and [20].
∫abdx(b−x)(x−a)=π,∫ab xdx(b−x)(x−a)=(a+b)π2,∫abdxx(b−x)(x−a)=πab,∫abdxx2(b−x)(x−a)=(a+b)π2(ab)3/2,∫ablogxdx(b−x)(x−a)=2πlog(a+b2),∫ablogxdxx(b−x)(x−a)=2πablog(2aba+b).
Acknowledgement
The financial support of the Macau Science and Technology Development Fund under grant number FDCT 130/2014/A3 and FDCT 023/2017/A1 is gratefully acknowledged. We would also like to thank the University of Macau for generous support: MYRG 2014–00011 FST, MYRG 2014–00004 FST.
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