Painlevé V, Painlevé XXXIV and the Degenerate Laguerre Unitary Ensemble
aa r X i v : . [ m a t h - ph ] J a n Painlev´e V, Painlev´e XXXIV and the Degenerate LaguerreUnitary Ensemble
Chao Min ∗ and Yang Chen † January 3, 2019
Abstract
In this paper, we study the Hankel determinant associated with the degenerate Laguerreunitary ensemble. This problem originates from the largest or smallest eigenvalue distributionof the degenerate Laguerre unitary ensemble. We derive the ladder operators and its compati-bility condition with respect to a general perturbed weight. By applying the ladder operatorsto our problem, we obtain two auxiliary quantities R n ( t ) and r n ( t ) and show that they satisfythe coupled Riccati equations, from which we find that R n ( t ) satisfies the Painlev´e V equation.Furthermore, we prove that σ n ( t ), a quantity related to the logarithmic derivative of the Han-kel determinant, satisfies both the continuous and discrete Jimbo-Miwa-Okamoto σ -form ofthe Painlev´e V. In the end, by using Dyson’s Coulomb fluid approach, we consider the large n asymptotic behavior of our problem at the soft edge, which gives rise to the Painlev´e XXXIVequation. Keywords : Hankel determinant; Degenerate Laguerre unitary ensemble; Ladder operators;Orthogonal polynomials; Painlev´e equations; Asymptotics.
Mathematics Subject Classification 2010 : 15B52, 42C05, 33E17. ∗ School of Mathematical Sciences, Huaqiao University, Quanzhou 362021, China; e-mail: [email protected] † Correspondence to: Yang Chen, Department of Mathematics, Faculty of Science and Technology, University ofMacau, Macau, China; e-mail: [email protected] Introduction
In random matrix theory, it is well known that the partition function of a unitary ensemble is givenby [23] ∆ N [ w ] := 1 N ! Z [ a,b ] N Y ≤ i
1, it corresponds to theprobability that all the eigenvalues are not greater than t in the dLUE. Furthermore, the γ = 0 casehas been studied by Basor and Chen [1]. They also investigated the Hankel determinant generatedby the weight ˜ w ( x, t ) = x α e − x ( x + t ) γ , x > , t >
0, which is related to the information theoryof MIMO wireless systems [2]. Note that the problem on the weight ˜ w ( x, t ) is different from ours,since our weight vanishes at a singular point t in the interior of the support. Finally, we mentionthat for γ > −
1, the weight w ( x, t ) is called the Laguerre weight with a Fisher-Hartwig singularity[12]. Rencently, Wu, Xu and Zhao [31] studied the Hankel determinant for the Gaussian weightperturbed by a Fisher-Hartwig singularity at the soft edge.We now introduce some elementary facts about the orthogonal polynomials. Let P n ( x, t ) be themonic polynomials of degree n orthogonal with respect to the weight w ( x, t ), Z ∞ P m ( x, t ) P n ( x, t ) w ( x, t ) dx = h n ( t ) δ mn , m, n = 0 , , , . . . . (1.2)We write P n ( x, t ) in the following expansion form, P n ( x, t ) = x n + p( n, t ) x n − + · · · , (1.3)and we will see that p( n, t ), the coefficient of x n − , plays a significant role in the following discussions.For the orthogonal polynomials P n ( x, t ), we have the three-term recurrence relation [29, 9] xP n ( x, t ) = P n +1 ( x, t ) + α n ( t ) P n ( x, t ) + β n ( t ) P n − ( x, t ) (1.4)with the initial conditions P ( x, t ) = 1 , β ( t ) P − ( x, t ) = 0 . An easy consequence of (1.2), (1.3) and (1.4) gives α n ( t ) = p( n, t ) − p( n + 1 , t ) (1.5)and β n ( t ) = h n ( t ) h n − ( t ) . (1.6)A telescopic sum of (1.5) yields n − X j =0 α j ( t ) = − p( n, t ) . (1.7)3inally, it is well known that [18] D n ( t ) = n − Y j =0 h j ( t ) . (1.8)The rest of this paper is organized as follows. In Sec. 2, we derive the ladder operators and thecompatibility conditions with respect to the weight w ( x ) := w ( x ) | x − t | γ ( A + Bθ ( x − t )), where w ( x ) is a general smooth weight. In Sec. 3, we apply the ladder operators for the general caseto the perturbed Laguerre weight and obtain some important identities on the auxiliary quantities R n ( t ) and r n ( t ). In Sec. 4, we show that R n ( t ) and r n ( t ) satisfy the coupled Riccati equations,which give rise to the Painlev´e V equation satisfied by R n ( t ). We also prove that a quantity σ n ( t ),allied to the logarithmic derivative of the Hankel determinant, satisfies both the continuous anddiscrete σ -form of the Painlev´e V. In Sec. 5, we consider the large n asymptotics of our problem atthe edge, from which the Painlev´e XXXIV equation appears. In the following discussions, for convenience, we shall not display the t dependence in P n ( x ), w ( x ), h n , α n and β n unless it is needed. Theorem 2.1.
Let w ( x ) be a smooth weight function defined on [ a, b ] , and w ( a ) = w ( b ) = 0 . Themonic orthogonal polynomials with respect to w ( x ) := w ( x ) | x − t | γ ( A + Bθ ( x − t )) , t ∈ [ a, b ] , γ > satisfy the lowering operator equation (cid:18) ddz + B n ( z ) (cid:19) P n ( z ) = β n A n ( z ) P n − ( z ) , (2.1) where A n ( z ) := 1 h n Z ba v ′ ( z ) − v ′ ( y ) z − y P n ( y ) w ( y ) dy + a n ( z, t ) ,a n ( z, t ) := γh n Z ba P n ( y )( z − y )( y − t ) w ( y ) dy ; B n ( z ) := 1 h n − Z ba v ′ ( z ) − v ′ ( y ) z − y P n ( y ) P n − ( y ) w ( y ) dy + b n ( z, t ) ,b n ( z, t ) := γh n − Z ba P n ( y ) P n − ( y )( z − y )( y − t ) w ( y ) dy and v ( z ) = − ln w ( z ) . roof. Since P n ( z ) is a polynomial of degree n , we have P ′ n ( z ) = n − X k =0 C n,k P k ( z )and the coefficient C n,k = 1 h k Z ba P ′ n ( y ) P k ( y ) w ( y ) dy. (2.2)After integration by parts and noting that w ( a ) = w ( b ) = 0, we find C n,k = − h k Z ba P n ( y ) P k ( y ) w ′ ( y ) dy = − h k Z ba P n ( y ) P k ( y ) (cid:26) w ′ ( y ) | y − t | γ ( A + Bθ ( y − t )) dy + w ( y ) | y − t | γ Bδ ( y − t )+ w ( y ) (cid:20) δ ( y − t )(( y − t ) γ − ( t − y ) γ ) + γ | y − t | γ y − t (cid:21) ( A + Bθ ( y − t )) (cid:27) dy = − h k Z ba P n ( y ) P k ( y )( − v ′ ( y )) w ( y ) dy − γh k Z ba P n ( y ) P k ( y ) w ( y ) y − t dy (2.3)= − h k Z ba P n ( y ) P k ( y )(v ′ ( z ) − v ′ ( y )) w ( y ) dy − γh k Z ba P n ( y ) P k ( y ) w ( y ) y − t dy, where we have used the formula [3] ∂ y | y − t | γ = δ ( y − t )(( y − t ) γ − ( t − y ) γ ) + γ | y − t | γ y − t , which is obtained by writing | y − t | γ = ( y − t ) γ θ ( y − t ) + ( t − y ) γ θ ( t − y ).It follows that P ′ n ( z ) = − Z ba P n ( y ) n − X k =0 P k ( z ) P k ( y ) h k (v ′ ( z ) − v ′ ( y )) w ( y ) dy − γ Z ba P n ( y ) n − X k =0 P k ( z ) P k ( y ) h k w ( y ) y − t dy. By using the Christoffel-Darboux formula, n − X k =0 P k ( z ) P k ( y ) h k = P n ( z ) P n − ( y ) − P n ( y ) P n − ( z ) h n − ( z − y ) , we arrive at equation (2.1). Proposition 2.2.
We have the following two important identities: Z ba P n ( y )v ′ ( y ) w ( y ) dy = γ Z ba P n ( y ) w ( y ) y − t dy, (2.4)1 h n − Z ba P n ( y ) P n − ( y )v ′ ( y ) w ( y ) dy = n + γh n − Z ba P n ( y ) P n − ( y ) w ( y ) y − t dy. (2.5)5 roof. From (2.2), we see that C n,n = 0 , C n,n − = n. The equations (2.4) and (2.5) follow from (2.3) if we replace k by n and n −
1, respectively.
Theorem 2.3.
The functions A n ( z ) and B n ( z ) satisfy the equations: B n +1 ( z ) + B n ( z ) = ( z − α n ) A n ( z ) − v ′ ( z ) , ( S )1 + ( z − α n )( B n +1 ( z ) − B n ( z )) = β n +1 A n +1 ( z ) − β n A n − ( z ) . ( S ) Proof.
From the definition of B n ( z ), we have B n +1 ( z ) + B n ( z ) = Z ba P n ( y ) (cid:18) P n +1 ( y ) h n + P n − ( y ) h n − (cid:19) v ′ ( z ) − v ′ ( y ) z − y w ( y ) dy + γ Z ba P n ( y ) (cid:18) P n +1 ( y ) h n + P n − ( y ) h n − (cid:19) w ( y )( z − y )( y − t ) dy. (2.6)It follows from the three-term recurrence relation (1.4) and (1.6) that P n +1 ( y ) h n + P n − ( y ) h n − = ( y − α n ) P n ( y ) h n . Substituting it into (2.6) gives B n +1 ( z )+ B n ( z ) = 1 h n Z ba ( y − α n ) P n ( y ) v ′ ( z ) − v ′ ( y ) z − y w ( y ) dy + γh n Z ba ( y − α n ) P n ( y ) w ( y )( z − y )( y − t ) dy. Using the definition of A n ( z ), it follows that B n +1 ( z ) + B n ( z ) − ( z − α n ) A n ( z ) = − h n Z ba (v ′ ( z ) − v ′ ( y )) P n ( y ) w ( y ) dy − γh n Z ba P n ( y ) w ( y ) y − t dy. From the orthogonality (1.2) and (2.4), we find B n +1 ( z ) + B n ( z ) − ( z − α n ) A n ( z ) = − v ′ ( z ) , which is just ( S ).We now turn to prove ( S ). Similarly, by using the definition of B n ( z ), we have( z − α n )( B n +1 ( z ) − B n ( z )) = " Z ba (v ′ ( z ) − v ′ ( y )) P n ( y ) (cid:18) P n +1 ( y ) h n − P n − ( y ) h n − (cid:19) w ( y ) dy + γ Z ba P n ( y ) (cid:18) P n +1 ( y ) h n − P n − ( y ) h n − (cid:19) w ( y ) y − t dy + " Z ba v ′ ( z ) − v ′ ( y ) z − y ( y − α n ) P n ( y ) (cid:18) P n +1 ( y ) h n − P n − ( y ) h n − (cid:19) w ( y ) dy + γ Z ba ( y − α n ) P n ( y ) (cid:18) P n +1 ( y ) h n − P n − ( y ) h n − (cid:19) w ( y )( z − y )( y − t ) dy , (2.7)6here we write z − α n = ( z − y ) + ( y − α n ) to get two parts in (2.7).From (1.4), we have ( y − α n ) P n ( y ) = P n +1 ( y ) + β n P n − ( y ) . Substituting it into (2.7), we find( z − α n )( B n +1 ( z ) − B n ( z )) = Z ba (v ′ ( z ) − v ′ ( y )) P n ( y ) (cid:18) P n +1 ( y ) h n − P n − ( y ) h n − (cid:19) w ( y ) dy + γ Z ba P n ( y ) (cid:18) P n +1 ( y ) h n − P n − ( y ) h n − (cid:19) w ( y ) y − t dy + Z ba v ′ ( z ) − v ′ ( y ) z − y (cid:18) P n +1 ( y ) h n − β n P n − ( y ) h n − (cid:19) w ( y ) dy + γ Z ba (cid:18) P n +1 ( y ) h n − β n P n − ( y ) h n − (cid:19) w ( y )( z − y )( y − t ) dy. Using the definition of A n ( z ), it follows that β n +1 A n +1 ( z ) − β n A n − ( z ) − ( z − α n )( B n +1 ( z ) − B n ( z ))= Z ba (cid:18) P n ( y ) P n − ( y ) h n − − P n +1 ( y ) P n ( y ) h n (cid:19) (v ′ ( z ) − v ′ ( y )) w ( y ) dy + γ Z ba (cid:18) P n ( y ) P n − ( y ) h n − − P n +1 ( y ) P n ( y ) h n (cid:19) w ( y ) y − t dy = − n − [ − ( n + 1)]= 1 , where we have used the orthogonality (1.2) and (2.5). The proof is complete.The combination of ( S ) and ( S ) produces a sum rule. Theorem 2.4. A n ( z ) , B n ( z ) and P n − j =0 A j ( z ) satisfy the equation B n ( z ) + v ′ ( z ) B n ( z ) + n − X j =0 A j ( z ) = β n A n ( z ) A n − ( z ) . ( S ′ ) Proof.
Multiplying ( S ) by A n ( z ) on both sides, we have A n ( z ) + ( z − α n ) A n ( z )( B n +1 ( z ) − B n ( z )) = β n +1 A n +1 ( z ) A n ( z ) − β n A n ( z ) A n − ( z ) . Using ( S ), the above becomes A n ( z ) + ( B n +1 ( z ) + B n ( z ) + v ′ ( z )) ( B n +1 ( z ) − B n ( z )) = β n +1 A n +1 ( z ) A n ( z ) − β n A n ( z ) A n − ( z ) , A n ( z ) + B n +1 ( z ) − B n ( z ) + v ′ ( z )( B n +1 ( z ) − B n ( z )) = β n +1 A n +1 ( z ) A n ( z ) − β n A n ( z ) A n − ( z ) . A telescopic sum gives the desired result.
Theorem 2.5.
The monic orthogonal polynomials P n ( z ) satisfy the raising operator equation (cid:18) ddz − B n ( z ) − v ′ ( z ) (cid:19) P n − ( z ) = − A n − ( z ) P n ( z ) . (2.8) Proof.
From (2.1) we replace n by n −
1, it reads P ′ n − ( z ) = β n − A n − ( z ) P n − ( z ) − B n − ( z ) P n − ( z ) . (2.9)The recurrence relation (1.4) gives β n − P n − ( z ) = ( z − α n − ) P n − ( z ) − P n ( z ) . Substituting it into (2.9), we obtain P ′ n − ( z ) = [( z − α n − ) A n − ( z ) − B n − ( z )] P n − ( z ) − A n − ( z ) P n ( z )= ( B n ( z ) + v ′ ( z )) P n − ( z ) − A n − ( z ) P n ( z ) . where we have made use of ( S ). This completes the proof. Theorem 2.6.
The monic orthogonal polynomials P n ( z ) satisfy the second order differential equa-tion P ′′ n ( z ) − (cid:18) v ′ ( z ) + A ′ n ( z ) A n ( z ) (cid:19) P ′ n ( z ) + B ′ n ( z ) − B n ( z ) A ′ n ( z ) A n ( z ) + n − X j =0 A j ( z ) ! P n ( z ) = 0 . Proof.
Solving for P n − ( z ) from (2.1) gives P n − ( z ) = P ′ n ( z ) + B n ( z ) P n ( z ) β n A n ( z ) . Substituting it into (2.8), we obtain P ′′ n ( z ) − (cid:18) v ′ ( z ) + A ′ n ( z ) A n ( z ) (cid:19) P ′ n ( z ) + (cid:18) B ′ n ( z ) − B n ( z ) A ′ n ( z ) A n ( z )+ β n A n ( z ) A n − ( z ) − B n ( z ) − v ′ ( z ) B n ( z ) (cid:19) P n ( z ) = 0 . Using ( S ′ ), we obtain the desired result. 8 emark 1 . In Theorem 2.1, a, b could be −∞ or ∞ . Remark 2 . The three identities ( S ), ( S ) and ( S ′ ) are valid for z ∈ C ∪ {∞} . Remark 3 . The ladder operator approach has been widely applied to the study of orthogonalpolynomials, Hankel determinants and random matrix theory; see [5, 8, 10, 11, 24, 26, 30] forreference.
In this section, we apply the ladder operators and its supplementary conditions to our problem.For the problem at hand, w ( x ) = w ( x ) | x − t | γ ( A + Bθ ( x − t )) , x ∈ [0 , ∞ ) , t ∈ [0 , ∞ ) , where w ( x ) = x α e − x , v ( x ) = x − α ln x. It is easy to see that w (0) = w ( ∞ ) = 0 andv ′ ( z ) − v ′ ( y ) z − y = αzy . From Theorem 2.1 we have A n ( z ) = αzh n Z ∞ P n ( y ) w ( y ) y dy + a n ( z, t ) , (3.1) a n ( z, t ) = γh n Z ∞ P n ( y )( z − y )( y − t ) w ( y ) dy ; B n ( z ) = αzh n − Z ∞ P n ( y ) P n − ( y ) w ( y ) y dy + b n ( z, t ) , (3.2) b n ( z, t ) = γh n − Z ∞ P n ( y ) P n − ( y )( z − y )( y − t ) w ( y ) dy. Theorem 3.1. As z → ∞ , we have A n ( z ) = 1 z + γ + tR n ( t ) z + γα n + γt + t R n ( t ) z + O (cid:18) z (cid:19) , (3.3) B n ( z ) = − nz + tr n ( t ) z + γβ n + t r n ( t ) z + O (cid:18) z (cid:19) , (3.4)9 here R n ( t ) := γh n Z ∞ P n ( y ) y − t w ( y ) dy,r n ( t ) := γh n − Z ∞ P n ( y ) P n − ( y ) y − t w ( y ) dy. Proof.
Using integration by parts, we find αh n Z ∞ P n ( y ) w ( y ) y dy = 1 h n Z ∞ P n ( y )e − y | y − t | γ ( A + Bθ ( y − t )) dy α = 1 − γh n Z ∞ P n ( y ) y − t w ( y ) dy = 1 − R n ( t )and αh n − Z ∞ P n ( y ) P n − ( y ) w ( y ) y dy = − n − r n ( t ) . As z → ∞ , 1 z − y = 1 z · − yz = 1 z (cid:18) yz + (cid:16) yz (cid:17) + O (cid:18) z (cid:19)(cid:19) = 1 z + yz + y z + O (cid:18) z (cid:19) . Then we have a n ( z, t ) = γzh n Z ∞ P n ( y ) y − t w ( y ) dy + γz h n Z ∞ yP n ( y ) y − t w ( y ) dy + γz h n Z ∞ y P n ( y ) y − t w ( y ) dy + O (cid:18) z (cid:19) ,b n ( z, t ) = γzh n − Z ∞ P n ( y ) P n − ( y ) y − t w ( y ) dy + γz h n − Z ∞ yP n ( y ) P n − ( y ) y − t w ( y ) dy + γz h n − Z ∞ y P n ( y ) P n − ( y ) y − t w ( y ) dy + O (cid:18) z (cid:19) . (3.5)By the definitions of R n ( t ) and r n ( t ), and using the orthogonality (1.2), also the recurrence relation(1.4), we obtain a n ( z, t ) = R n ( t ) z + γ + tR n ( t ) z + γα n + γt + t R n ( t ) z + O (cid:18) z (cid:19) ,b n ( z, t ) = r n ( t ) z + tr n ( t ) z + γβ n + t r n ( t ) z + O (cid:18) z (cid:19) . According to (3.1) and (3.2), the theorem is established.10ubstituting (3.3) and (3.4) into ( S ), and comparing the coefficients of z and z on both sidesrespectively, we obtain the following two equations: α n = 2 n + 1 + α + γ + tR n ( t ) , (3.6) r n +1 ( t ) + r n ( t ) = γ + ( t − α n ) R n ( t ) . (3.7)Similarly, substituting (3.3) and (3.4) into ( S ) gives rise to another two equations: β n +1 − β n = tr n +1 ( t ) − tr n ( t ) + α n , (3.8)( t − α n )( r n +1 ( t ) − r n ( t )) = β n +1 R n +1 ( t ) − β n R n − ( t ) . (3.9)Using (1.7), a telescopic sum of (3.8) gives β n = tr n ( t ) − p( n, t ) . (3.10)Multiplying both sides of (3.9) by R n ( t ) and using (3.7), we have( r n +1 ( t ) + r n ( t ) − γ )( r n +1 ( t ) − r n ( t )) = β n +1 R n +1 ( t ) R n ( t ) − β n R n ( t ) R n − ( t )or r n +1 ( t ) − r n ( t ) − γ ( r n +1 ( t ) − r n ( t )) = β n +1 R n +1 ( t ) R n ( t ) − β n R n ( t ) R n − ( t ) . A telescopic sum produces r n ( t ) − γr n ( t ) = β n R n ( t ) R n − ( t ) . (3.11)Substituting (3.3) and (3.4) into ( S ′ ), noting that the coefficient of z is 0 and comparing thecoefficients of z and z on both sides respectively, we find the following two equations: n ( n + α + γ ) + tr n ( t ) + t n − X j =0 R j ( t ) = β n , (3.12) nγt + ( t − nt − αt ) r n ( t ) + γ n − X j =0 α j + t n − X j =0 R j ( t ) = β n ( γ + tR n − ( t ) + tR n ( t )) . (3.13)It follows from (1.7) and (3.10) that n − X j =0 α j = β n − tr n ( t ) . nγ + ( t − n − α − γ ) r n ( t ) + t n − X j =0 R j ( t ) = β n R n − ( t ) + β n R n ( t ) . (3.14)Eliminating t P n − j =0 R j ( t ) from (3.12) and (3.14), we obtain β n R n − ( t ) + β n R n ( t ) = β n − (2 n + α + γ ) r n ( t ) − n ( n + α ) . (3.15)In the end, we mention that the above identities obtained from ( S ), ( S ) and ( S ′ ) are very importantfor the derivation of the fifth Painlev´e equation in next section. σ -Form We start from taking a derivative with respect to t in the following equation Z ∞ P n ( x, t ) x α e − x | x − t | γ ( A + Bθ ( x − t )) dx = h n ( t ) , n = 0 , , , . . . , which gives h ′ n ( t ) = − γ Z ∞ P n ( x ) w ( x ) x − t dx. It follows that ddt ln h n ( t ) = − R n ( t ) . (4.1)Using (1.6) we have ddt ln β n ( t ) = R n − ( t ) − R n ( t ) . That is, β ′ n ( t ) = β n R n − ( t ) − β n R n ( t ) . (4.2)We define a quantity allied to the Hankel determinant, H n ( t ) := t ddt ln D n ( t ) . (4.3)It is easy to see from (1.8) and (4.1) that H n ( t ) = − t n − X j =0 R j ( t ) . (4.4)12n the other hand, taking a derivative with respect to t in the equation Z ∞ P n ( x, t ) P n − ( x, t ) x α e − x | x − t | γ ( A + Bθ ( x − t )) dx = 0 , n = 0 , , , . . . , we obtain ddt p( n, t ) = r n ( t ) . (4.5) Theorem 4.1.
The auxiliary quantities R n ( t ) and r n ( t ) satisfy the following coupled Riccati equa-tions: tR ′ n ( t ) = tR n ( t ) + (2 n + α + γ − t ) R n ( t ) + 2 r n ( t ) − γ, (4.6) tr ′ n ( t ) = r n ( t ) − γr n ( t ) R n ( t ) − r n ( t ) − γr n ( t ) + [(2 n + α + γ ) r n ( t ) + n ( n + α )] R n ( t )1 − R n ( t ) . (4.7) Proof.
From (1.5) and (4.5) we have α ′ n ( t ) = r n ( t ) − r n +1 ( t ) . (4.8)Eliminating α n from (3.6) and (3.7) gives r n +1 ( t ) = γ + ( t − n − − α − γ − tR n ( t )) R n ( t ) − r n ( t ) . (4.9)Substituting (3.6) and (4.9) into (4.8), we obtain (4.6).From (3.10) and (4.5), we have β ′ n ( t ) = tr ′ n ( t ) . Then (4.2) becomes tr ′ n ( t ) = β n R n − ( t ) − β n R n ( t ) , (4.10)or tr ′ n ( t ) = r n ( t ) − γr n ( t ) R n ( t ) − β n R n ( t ) , (4.11)where we have made use of (3.11).From (3.15), we find the expression of β n in terms of R n ( t ) and r n ( t ) with the aid of (3.11), β n = r n ( t ) − γr n ( t ) R n ( t )(1 − R n ( t )) + (2 n + α + γ ) r n ( t ) + n ( n + α )1 − R n ( t ) . (4.12)Substituting it into (4.11), we arrive at (4.7). 13 heorem 4.2. The quantity R n ( t ) satisfies a non-linear second order differential equation, t R n (1 − R n ) R ′′ n − t (1 − R n )( R ′ n ) + 2 tR n (1 − R n ) R ′ n + 2 t R n + t (4 n + 2 + 2 α + 2 γ − t ) R n − t (2 n + 1 + α + γ − t ) R n − [ t − n + 1 + α + γ ) t + α − γ ] R n − γ R n + γ = 0 . (4.13) Let S n ( t ) := R n ( t ) R n ( t ) − , then S n ( t ) satisfies the second order differential equation, S ′′ n = (3 S n − S ′ n ) S n ( S n − − S ′ n t + ( S n − t (cid:18) α S n − γ S n (cid:19) − (2 n + 1 + α + γ ) S n t − S n ( S n + 1)2( S n − . (4.14) which is a particular Painlev´e V, P V (cid:16) α , − γ , − (2 n + 1 + α + γ ) , − (cid:17) , following the convention of[16].Proof. Solving for r n ( t ) from (4.6) and substituting it into (4.7), we obtain (4.13). After the linearfractional transformation R n ( t ) = S n ( t ) S n ( t ) − or S n ( t ) = R n ( t ) R n ( t ) − , we arrive at (4.14). Remark 4 . Solving for R n ( t ) from (4.7) and substituting it into (4.6), we can obtain the secondorder differential equation satisfied by r n ( t ). Since this equation is too complicated, we decide notto write it down. Usually the differential equation for r n ( t ) is related to the Chazy type equation;see [22, 25, 24] for reference. Theorem 4.3.
The Hankel determinant D n ( t ) admits the following two alternative integral repre-sentations in terms of R n ( t ) and S n ( t ) , ln D n ( t ) D n (0) = Z t sR n ( s ) ( R n ( s ) − n s ( R ′ n ( s )) − s R n ( s ) − s (2 n + α + γ − s ) R n ( s )+ (cid:2) s (2 n + γ ) − ( α + γ − s ) (cid:3) R n ( s ) + 2 γ ( α + γ − s ) R n ( s ) − γ o ds = Z t s S n ( s ) ( S n ( s ) − n s ( S ′ n ( s )) − α S n ( s ) − (cid:2) (2 n + α ) s − α + αγ ) (cid:3) S n ( s ) − (cid:2) s − s (2 n + α − γ ) + α − αγ + γ (cid:3) S n ( s ) − γ ( α − γ − s ) S n ( s ) − γ o ds. Proof.
Combining (3.12) and (4.4), we have H n ( t ) = n ( n + α + γ ) + tr n ( t ) − β n . (4.15)Inserting (4.12) into the above and using (4.6) to eliminate r n ( t ), we obtain the expression of H n ( t )in terms of R n ( t ) and R ′ n ( t ), H n ( t ) = 14 R n ( t ) ( R n ( t ) − n t ( R ′ n ( t )) − t R n ( t ) − t (2 n + α + γ − t ) R n ( t )+ (cid:2) t (2 n + γ ) − ( α + γ − t ) (cid:3) R n ( t ) + 2 γ ( α + γ − t ) R n ( t ) − γ o . (4.16)14ince S n ( t ) = R n ( t ) R n ( t ) − , we also have H n ( t ) = 14 S n ( t ) ( S n ( t ) − n t ( S ′ n ( t )) − α S n ( t ) − (cid:2) (2 n + α ) t − α + αγ ) (cid:3) S n ( t ) − (cid:2) t − t (2 n + α − γ ) + α − αγ + γ (cid:3) S n ( t ) − γ ( α − γ − t ) S n ( t ) − γ o . In view of H n ( t ) = t ddt ln D n ( t ), the theorem is established. Theorem 4.4.
The quantity H n ( t ) satisfies a non-linear second order differential equation ( tH ′′ n ) = [ nγ − H n − (2 n + α + γ − t ) H ′ n ] − n ( n + α + γ ) − H n + tH ′ n ] (cid:2) ( H ′ n ) − γH ′ n (cid:3) (4.17) with the initial conditions H n (0) = 0 , H ′ n (0) = − nγα + γ , and also satisfies a non-linear second orderdifference equation n nγt − tH n − [ n ( n + α + γ ) − H n ] ( H n − − H n +1 ) on ( t − n − α − γ ) γt − tH n − [ n ( n + α + γ ) + γt − H n ] ( H n − − H n +1 ) o = h nγt + n ( n + α + γ )(2 n + α + γ − t ) − (2 n + α + γ ) H n i · (2 n + α + γ − t + H n − − H n +1 )( H n − H n +1 )( H n − − H n ) . (4.18) Proof.
From (3.10) and (4.15), we havep( n, t ) = − n ( n + α + γ ) + H n ( t ) . Taking a derivative with respect to t on both sides and noting (4.5), we find r n ( t ) = H ′ n ( t ) . (4.19)It follows from (4.15) that β n = n ( n + α + γ ) − H n ( t ) + tH ′ n ( t ) . (4.20)From (3.15) and (4.10), it is easy to get2 β n R n ( t ) = β n − (2 n + α + γ ) r n ( t ) − n ( n + α ) − tr ′ n ( t ) , β n R n − ( t ) = β n − (2 n + α + γ ) r n ( t ) − n ( n + α ) + tr ′ n ( t ) . The product of the above two equations gives4 β n ( r n ( t ) − γr n ( t )) = [ β n − (2 n + α + γ ) r n ( t ) − n ( n + α )] − ( tr ′ n ( t )) , (4.21)15here we have made use of (3.11).Substituting (4.19) and (4.20) into (4.21), we obtain (4.17). The initial conditions come from (4.4)and the fact that R n (0) = γα + γ .We now turn to prove the difference equation satisfied by H n ( t ). From (4.4) we have tR n ( t ) = H n ( t ) − H n +1 ( t ) , (4.22) tR n − ( t ) = H n − ( t ) − H n ( t ) . (4.23)Multiplying both sides of (3.15) by t and substituting (4.15), (4.22) and (4.23) into it, we obtainthe expression of r n ( t ) in terms of H n ( t ), tr n ( t ) = nγt − tH n ( t ) − [ n ( n + α + γ ) − H n ( t )] ( H n − ( t ) − H n +1 ( t ))2 n + α + γ − t + H n − ( t ) − H n +1 ( t ) . (4.24)It follows from (4.15) that β n = nγt + n ( n + α + γ )(2 n + α + γ − t ) − (2 n + α + γ ) H n ( t )2 n + α + γ − t + H n − ( t ) − H n +1 ( t ) . (4.25)Finally, multiplying (3.11) by t on both sides and substituting (4.22), (4.23), (4.24) and (4.25) intoit, we obtain (4.18). The proof is complete.From the above theorem, we readily have the following results, which connect our problem withthe Painlev´e equations. Theorem 4.5.
Let σ n ( t ) := H n ( t ) − nγ , then σ n ( t ) satisfies the Jimbo-Miwa-Okamoto σ -form ofthe Painlev´e V [21], ( tσ ′′ n ) = (cid:2) σ n − tσ ′ n + 2( σ ′ n ) + ( ν + ν + ν + ν ) σ ′ n (cid:3) − ν + σ ′ n )( ν + σ ′ n )( ν + σ ′ n )( ν + σ ′ n ) , where ν = 0 , ν = n, ν = n + α, ν = − γ , with the initial conditions σ n (0) = − nγ, σ ′ n (0) = − nγα + γ .The quantity σ n ( t ) also satisfies a non-linear second order difference equation (cid:2) tσ n − ( n + nα − σ n )(2 γ − σ n − + σ n +1 ) (cid:3)(cid:2) (2 n + α + γ − t ) γt + tσ n − ( n + nα + γt − σ n )(2 γ − σ n − + σ n +1 ) (cid:3) = (cid:2) n ( n + α )(2 n + α + γ − t ) − (2 n + α + γ ) σ n (cid:3) · (2 n + α − γ − t + σ n − − σ n +1 )( γ − σ n + σ n +1 )( γ − σ n − + σ n ) , which is the discrete σ -form of the Painlev´e V. emark 5 . If γ = 0, then the results in Theorem 4.4 or Theorem 4.5 are coincident with Theorem8 in Basor and Chen [1].In the end of this section, we show the relation of our Hankel determinant with the Toda moleculeequation in the following theorem. Theorem 4.6.
The Hankel determinant D n ( t ) satisfies the following differential-difference equation, t d dt ln D n ( t ) = − n ( n + α + γ ) + D n +1 ( t ) D n − ( t ) D n ( t ) . (4.26) Furthermore, let ˜ D n ( t ) := t − n ( n + α + γ ) D n ( t ) , then ˜ D n ( t ) satisfies the Toda molecule equation [28] d dt ln ˜ D n ( t ) = ˜ D n +1 ( t ) ˜ D n − ( t )˜ D n ( t ) . (4.27) Proof.
From (1.6) and (1.8), we have β n = D n +1 ( t ) D n − ( t ) D n ( t ) . (4.28)On the other hand, from (4.20) and (4.3) we find β n = n ( n + α + γ ) + t d dt ln D n ( t ) . (4.29)The combination of (4.28) and (4.29) gives (4.26). The equation (4.27) follows from the transfor-mation ˜ D n ( t ) = t − n ( n + α + γ ) D n ( t ). This completes the proof. Remark 6 . The Hankel determinant D n ( t ) is related to the τ -function of the Painlev´e V [27]. Seealso [13, 14] on the discussion of the τ -functions and the Painlev´e equations. In the limit of large n , the eigenvalues (particles) of the Hermitian matrices from a unitary ensemblecan be approximated as a continuous fluid with a density σ ( x ) supported in J (a subset of R ).When the potential v( x ) := − ln w ( x ) is convex and v ′′ ( x ) > σ ( x ) issupported in a single interval ( a, b ). See [4, 6] for detail.The equilibrium density σ ( x ) is found to satisfy the following singular integral equation,v ′ ( x ) − P Z ba σ ( y ) x − y dy = 0 , P denotes the principal value integral.The solution subject to the boundary condition σ ( a ) = σ ( b ) = 0 reads, σ ( x ) = p ( b − x )( x − a )2 π P Z ba v ′ ( y )( y − x ) p ( b − y )( y − a ) dy with two supplementary conditions Z ba v ′ ( x ) p ( b − x )( x − a ) dx = 0 , (5.1) Z ba x v ′ ( x ) p ( b − x )( x − a ) dx = 2 πn. (5.2)For our problem, w ( x, t ) = x α e − x | x − t | γ ( A + Bθ ( x − t )) =: e − v( x ) , where v( x ) = x − α ln x − γ ln | x − t | − θ ( x − t ) ln( A + B ) − θ ( t − x ) ln A. (5.3)Substituting (5.3) into (5.1) and (5.2) and noting that ddx θ ( x − t ) = δ ( x − t ), we obtain two equationsfor the endpoints a and b (0 < a < b ):1 − α √ ab + c p ( b − t )( t − a ) = 0 , (5.4) a + b − α − γ + c t p ( b − t )( t − a ) = 2 n, (5.5)where c := π ln AA + B and we have used the following formulas [7, 15], Z ba p ( b − x )( x − a ) dx = π, Z ba x p ( b − x )( x − a ) dx = a + b π, Z ba x p ( b − x )( x − a ) dx = π √ ab , (0 < a < b ) ,P Z ba x − t ) p ( b − x )( x − a ) dx = 0 . Since α n ∼ a + b as n → ∞ [4], we denote ˜ α n := a + b . From (5.4) and (5.5) we obtain a quinticequation satisfied by ˜ α n ,( ˜ α n − n − α − γ ) (cid:2) (2 ˜ α n − t )( ˜ α n − t − n − α − γ ) − α t (cid:3) − c t ( ˜ α n − t − n − α − γ ) = 0 . (5.6)18n view of the relation (3.6), letting ˜ α n = 2 n + α + γ + t ˜ R n ( t ), we have ˜ R n ( t ) ∼ R n ( t ) as n → ∞ .It follows from (5.6) that2 t ˜ R n ( t ) − t (5 t − n − α − γ ) ˜ R n ( t ) + 4 t ( t − n − α − γ ) ˜ R n ( t ) − ( t − nt − αt − γt + α + c ) ˜ R n ( t ) + 2 c ˜ R n ( t ) − c = 0 . (5.7)In the end, we consider the case when t approaches the soft edge, i.e., n → ∞ , t = 4 n + 2 n s and s is fixed. The asymptotic behavior of R n ( t ), r n ( t ) and H n ( t ) is obtained in the followingtheorem. Theorem 5.1.
Assume that n → ∞ , t = 4 n + 2 n s and s is fixed. Then the large n asymptoticsof R n ( t ) , r n ( t ) and H n ( t ) are given by R n ( t ) = n − u ( s ) + n − v ( s ) + O ( n − ) ,r n ( t ) = n u ( s ) + 2 − u ′ ( s ) + v ( s ) + γ O ( n − ) , and H n ( t ) = 2 γn − ( u ′ ( s )) + 16 u ( s ) − s u ( s ) − γ u ( s ) n + O ( n ) , respectively. Here u ( s ) and v ( s ) satisfy the second order differential equations (5.11) and (5.12),and the large s behavior is given by (5.13) and (5.14). In addition, ˜ u ( s ) := − u ( s ) satisfies thePainlev´e XXXIV equation [17] ˜ u ′′ ( s ) = (˜ u ′ ( s )) u ( s ) + 4˜ u ( s ) + 2 s ˜ u ( s ) − γ u ( s ) . (5.8) Proof.
Let ˆ R n ( s ) := R n (4 n + 2 n s ) . After change of variable, equation (4.13) becomes2 (cid:16) n + s (cid:17) (cid:16) − ˆ R n ( s ) (cid:17) ˆ R n ( s ) ˆ R ′′ n ( s ) − (cid:16) n + s (cid:17) (cid:16) − R n ( s ) (cid:17) (cid:16) ˆ R ′ n ( s ) (cid:17) + (cid:16) n + 2 s (cid:17) (1 − ˆ R n ( s )) ˆ R n ( s ) ˆ R ′ n ( s ) + 2 (cid:16) n + 2 n s (cid:17) ˆ R n ( s )+ (cid:16) n + 2 n s (cid:17) (cid:16) α + 2 γ + 2 − n − × n s (cid:17) ˆ R n ( s ) − (cid:16) n + 2 n s (cid:17) (cid:16) α + γ − n + 1 − n s (cid:17) ˆ R n ( s )+ (cid:20) α − γ − α + γ + 2 n + 1) (cid:16) n + 2 n s (cid:17) + (cid:16) n + 2 n s (cid:17) (cid:21) ˆ R n ( s ) − γ ˆ R n ( s ) + γ = 0 . (5.9)19e suppose ˆ R n ( s ) = n − u ( s ) + n − v ( s ) + O ( n − ) , (5.10)which is obtained by observing from the real solution of the algebraic equation (5.7) after changingvariable t to s .Substituting (5.10) into (5.9), we obtain u ( s ) u ′′ ( s ) − − ( u ′ ( s )) + 2 u ( s ) − s u ( s ) + 2 − γ + n − h u ( s ) v ′′ ( s ) − u ′ ( s ) v ′ ( s ) + u ′′ ( s ) v ( s )+ 12 × u ( s ) v ( s ) − s u ( s ) v ( s ) + 2 ( α + γ + 1) u ( s ) i + O ( n − ) = 0 . It follows that u ( s ) and v ( s ) satisfy the following second order differential equations u ( s ) u ′′ ( s ) − − ( u ′ ( s )) + 2 u ( s ) − s u ( s ) + 2 − γ = 0 , (5.11) u ( s ) v ′′ ( s ) − u ′ ( s ) v ′ ( s ) + u ′′ ( s ) v ( s ) + 12 × u ( s ) v ( s ) − s u ( s ) v ( s ) + 2 ( α + γ + 1) u ( s ) = 0 . (5.12)From (5.11), we obtain the large s asymptotic of u ( s ). As s → ∞ , u ( s ) = s / + 1 − γ × / s − γ − γ + 916 × / s − γ − γ + 876 γ − × / s + O (cid:18) s (cid:19) . (5.13)Substituting (5.13) into (5.12), we find the large s asymptotic of v ( s ). As s → ∞ , v ( s ) = −
14 ( α + γ + 1) − (4 γ −
1) ( α + γ + 1)8 s − γ − γ + 9) ( α + γ + 1)32 s − γ − γ + 876 γ − α + γ + 1)32 s + O (cid:18) s (cid:19) . (5.14)Let ˆ r n ( s ) := r n (4 n + 2 n s ) . From (4.6), we obtain the expression of ˆ r n ( s ) in terms of ˆ R n ( s ),2ˆ r n ( s ) = (2 n + s ) ˆ R ′ n ( s ) − (4 n + 2 n s ) ˆ R n ( s ) − ( α + γ − n − n s ) ˆ R n ( s ) + γ. (5.15)Substituting (5.10) into (5.15) givesˆ r n ( s ) = n u ( s ) + 2 − u ′ ( s ) + v ( s ) + γ O ( n − ) . Similarly, we have the expression of ˆ H n ( s ) := H n (4 n + 2 n s ) in terms of ˆ R n ( s ) from (4.16). Using(5.10), we obtainˆ H n ( s ) = 2 γn − ( u ′ ( s )) + 16 u ( s ) − s u ( s ) − γ u ( s ) n + O ( n ) .
20n the end, letting ˜ u ( s ) := − u ( s ), then it follows from (5.11) that ˜ u ( s ) satisfies the Painlev´eXXXIV equation (5.8). Remark 7 . The Painlev´e XXXIV equation (5.8) also appeared in the study of the critical edgebehavior in quite general unitary random matrix ensembles by Its, Kuijlaars and ¨Ostensson [19, 20],and the study of perturbed Gaussian unitary ensemble with a Fisher-Hartwig singularity by Wu,Xu and Zhao [31].
Acknowledgments
Chao Min was supported by the Scientific Research Funds of Huaqiao University under grant number600005-Z17Y0054. Yang Chen was supported by the Macau Science and Technology DevelopmentFund under grant numbers FDCT 130/2014/A3, FDCT 023/2017/A1 and by the University ofMacau under grant numbers MYRG 2014-00011-FST, MYRG 2014-00004-FST.
References [1] E. L. Basor and Y. Chen,
Painlev´e V and the distribution function of a discontinuous linearstatistic in the Laguerre unitary ensembles , J. Phys. A: Math. Theor. (2009) 035203.[2] E. L. Basor and Y. Chen, Perturbed Laguerre unitary ensembles, Hankel determinants, andinformation theory , Math. Meth. App. Sci. (2015) 4840-4851.[3] Y. Chen and M. V. Feigin, Painlev´e IV and degenerate Gaussian unitary ensembles , J. Phys.A: Math. Gen. (2006) 12381–12393.[4] Y. Chen and M. E. H. Ismail, Thermodynamic relations of the Hermitian matrix ensembles , J.Phys. A: Math. Gen. (1997) 6633–6654.[5] Y. Chen and A. Its, Painlev´e III and a singular linear statistics in Hermitian random matrixensembles, I , J. Approx. Theory (2010) 270–297.[6] Y. Chen, N. Lawrence,
On the linear statistics of Hermitian random matrices , J. Phys. A: Math.Gen. (1998) 1141–1152. 217] Y. Chen and M. R. McKay, Coulomb fluid, Painlev´e transcendents, and the information theoryof MIMO systems , IEEE Trans. Inf. Theory (2012) 4594–4634.[8] Y. Chen and L. Zhang, Painlev´e VI and the unitary Jacobi ensembles , Stud. Appl. Math. (2010) 91–112.[9] T. S. Chihara,
An introduction to orthogonal polynomials , Dover, New York, 1978.[10] D. Dai and L. Zhang,
Painlev´e VI and Hankel determinants for the generalized Jacobi weight ,J. Phys. A: Math. Theor. (2010) 055207.[11] G. Filipuk, W. Van Assche and L. Zhang, The recurrence coefficients of semi-classical Laguerrepolynomials and the fourth Painlev´e equation , J. Phys. A: Math. Theor. (2012) 205201.[12] M. E. Fisher and R. E. Hartwig, Toeplitz determinants: some applications, theorems, andconjectures , Advan. Chem. Phys. (1968) 333-353.[13] P. J. Forrester, Log-Gases and Random Matrices , Princeton University Press, Princeton, 2010.[14] P. J. Forrester and N. S. Witte,
Application of the τ -function theory of Painlev´e equations torandom matrices: P V , P III , the LUE, JUE and CUE , Commun. Pure Appl. Math. (2002)679–727.[15] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products: Seventh Edition ,Academic Press, New York, 2007.[16] V. I. Gromak, I. Laine and S. Shimomura,
Painlev´e Differential Equations in the ComplexPlane , Walter de Gruyter, Berlin, 2002.[17] E. L. Ince,
Ordinary Differential Equations , Dover, New York, 1956.[18] M. E. H. Ismail,
Classical and Quantum Orthogonal Polynomials in One Variable , Encyclopediaof Mathematics and its Applications 98, Cambridge University Press, Cambridge, 2005.[19] A. R. Its, A. B. J. Kuijlaars and J. ¨Ostensson,
Critical edge behavior in unitary random matrixensembles and the thirty-fourth Painlev´e transcendent , Int. Math. Res. Notices (2008) 1–67.2220] A. R. Its, A. B. J. Kuijlaars and J. ¨Ostensson,
Asymptotics for a special solution of the thirtyfourth Painlev´e equation , Nonlinearity (2009) 1523–1558.[21] M. Jimbo and T. Miwa, Monodromy preserving deformation of linear ordinary differentialequations with rational coefficients. II , Physica D (1981) 407–448.[22] S. Lyu and Y. Chen, The largest eigenvalue distribution of the Laguerre unitary ensemble , ActaMath. Sci. (2017) 439–462.[23] M. L. Mehta,
Random Matrices , 3rd edn., Elsevier, New York, 2004.[24] C. Min and Y. Chen,
Gap probability distribution of the Jacobi unitary ensemble: an elementarytreatment, from finite n to double scaling , Stud. Appl. Math. (2018) 202–220.[25] C. Min and Y. Chen, Painlev´e transcendents and the Hankel determinants generated by adiscontinuous Gaussian weight , Math. Meth. Appl. Sci. (2019) 301–321.[26] C. Min, S. Lyu and Y. Chen, Painlev´e III ′ and the Hankel determinant generated by a singularlyperturbed Gaussian weight , Nucl. Phys. B (2018) 169–188.[27] K. Okamoto, Studies on the Painlev´e equations II. Fifth Painlev´e equation P V , Japan J. Math.(N. S.) (1987) 47–76.[28] K. Sogo, Time-dependent orthogonal polynomials and theory of soliton-Applications to matrixmodel, vertex model and level statistics , J. Phys. Soc. Japan (1993) 1887–1894.[29] G. Szeg˝o, Orthogonal Polynomials , 4th edn., AMS Colloquium Publications, Vol. 23, Provi-dence, RI, 1975.[30] W. Van Assche,
Orthogonal Polynomials and Painlev´e Equations , Australian MathematicalSociety Lecture Series 27, Cambridge University Press, Cambridge, 2018.[31] X.-B. Wu, S.-X. Xu and Y.-Q. Zhao,
Gaussian unitary ensemble with boundary spectrum sin-gularity and σ -form of the Painlev´e II equation , Stud. Appl. Math.140