Asymptotics of recurrence coefficients for a modified

Asymptotics of recurrence coefficients for a modified Laguerre weight via nonlinear difference equations Yang Chen∗, Galina Filipuk†and Maria das Neves Rebocho

‡

Abstract In this paper we derive second and third order nonlinear difference equations for one of the recurrence coefficients in the three term recurrence relation of polynomials orthogonal with respect to the modified Laguerre weight. We use these equations to derive asymptotic behavior of the recurrence coefficients for large n.

2010 MSC: 33C47, 39A99, 34E05, 42C05 Key words: Orthogonal polynomials; difference equations; asymptotic expansions.

1

Introduction

In recent years there has been a considerable interest to derive nonlinear difference equations for the recurrence coefficients of orthogonal polynomials (e.g., [1, 2, 3] and the references therein). In this paper we derive second and third order nonlinear difference equations satisfied by one of the recurrence coefficients of a one parameter family of polynomials orthogonal with respect to the weight w(x) = w(x, s) := xα e−x e−s/x ,

(1)

where 0 ≤ x < ∞, α > 0, s > 0. The weight (1) can be represented as w(x, s) = w0 (x)e−s/x , where w0 (x) = xα e−x is the Laguerre weight. The deformed (by a factor e−s/x ) Laguerre weight was extensively studied in [1, 5, 10] (see other references therein). It is well-known [6, 9] that monic orthogonal polynomials, which also depend on the parameter s in case of the weight (1), Pn (x, s) = xn + p1 (n, s)xn−1 + . . . + Pn (0, s), where P0 (x, s) = 1, p1 (0, s) := 0 satisfy a three-term recurrence relation xPn (x) = Pn+1 (x) + αn (s)Pn (x) + βn (s)Pn−1 (x) ∗

(2)

Faculty of Science and Technology, Department of Mathematics, University of Macau, Av. Padre Tom´ as Pereiro, Taipa, Macau, China. Email: [email protected] † Corresponding author. Faculty of Mathematics, Informatics and Mechanics University of Warsaw, Banacha 2 Warsaw, 02-097, Poland. Email: [email protected] ‡ Department of Mathematics, University of Beira Interior, 6201-001 Covilh˜a, Portugal; CMUC, Department of Mathematics, University of Coimbra, Apartado 3008, EC Santa Cruz, 3001-501 Coimbra, Portugal. Email: [email protected]

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with initial conditions P0 (x) = 1, β0 P−1 (x) = 0. Recurrence coefficients αn (s) and βn (s) satisfy αn (s) ∈ R, n = 0, 1, . . . , and βn (s) > 0, n = 1, 2, . . .. The coefficient p1 (n, s) P satisfies p1 (n, s) − p1 (n + 1, s) = αn (s), or, equivalently, p1 (n, s) = − n−1 α (s). It is j=0 j shown in [1] that for fixed n the recurrence coefficients are related to the solutions of the third Painlev´e equation (as functions of s). The relation of the recurrence coefficients for certain weights to the solutions of the Painlev´e equations are known in other cases as well (e.g., [8]). In this paper we derive nonlinear difference equations for αn = αn (s). In particular, we obtain a second order equation F (αn+1 , αn , αn−1 , s) = 0 and a third order equation G(αn+2 , αn+1 , αn , αn−1 , s) = 0. We use these equations to study the asymptotic behavior of the recurrence coefficients αn and βn and of the coefficient p1 (n, s) as n → ∞ in cases s 6= 0. In case s = 0 the recurrence coefficients can be found explicitly. We also show that the coefficient βn is expressed in terms of αn+1 , αn and αn−1 .

2

Nonlinear difference equations

First, we shall briefly recall necessary formulas. We shall use the notation 0 = d/dx. The weight (1) is a member of the so-called semi-classical class, as it satisfies a homogeneous differential equation with polynomial coefficients [8]. Indeed, for the weight (1) we have Aw0 = Cw with A(x) = x2 , C(x) = −x2 + αx + s .

(3)

According to [8] (see also [7]), the polynomials Pn satisfy the structure relations, for all n ≥ 1,

0 APn−1

APn0 = (ln−1 − C/2)Pn + Θn−1 Pn−1 ,   Θn−2 C Θn−2 Pn + ln−2 + (x − αn−1 ) − Pn−1 , =− βn−1 βn−1 2

(4) (5)

where ln , Θn are polynomials of degree two and one, respectively, satisfying the initial conditions Θ−1 = D , Θ0 = A + (x − α0 )(C/2 − l0 ) , l−1 = C/2 , l0 = −C/2 − (x − α0 )D , where D = x + α0 − 1 − α. We shall also use later on the following notation: ln (x) = ln,2 x2 + ln,1 x + ln,0 and Θn (x) = Θn,1 x + Θn,0 . Similar to (4)–(5) expressions are given in [5]. They are as follows: Pn0 = −Bn Pn + βn An Pn−1 , 0 Pn−1 = −An−1 Pn + (Bn + v 0 )Pn−1 , 2

(6) (7)

where v 0 = −w0 /w = −C/A. Here, An and Bn are functions that satisfy the following fundamental compatibility conditions (see also [4]): Bn+1 (x) + Bn (x) = (x − αn )An (x) − v 0 (x), 1 + (x − αn )(Bn+1 (x) − Bn (x)) = βn+1 An+1 (x) − βn An−1 (x), Bn2 (x) + v 0 (x)Bn (x) +

n−1 X

Aj (x) = βn An (x)An−1 (x).

(S1 ) (S2 ) (S20 )

j=0

Using (4), (6) and (5), (7), we have βn AAn = Θn−1 ,

ABn = (C/2 − ln−1 ) .

We also have A(Bn + v 0 ) = ln−2 + (x − αn−1 )Θn−2 /βn−1 − C/2, which holds identically by using the following formulas from [7]: Θn−2 , n ≥ 1, (8) βn−1 Θn−1 Θn−2 ln + (x − αn ) = ln−2 + (x − αn−1 ) , n ≥ 1, (9) βn βn−1 Θn−1 = 0, n ≥ 0. (10) ln + ln−1 + (x − αn ) βn   Pn+1 and write the recurrence relation (2) in the Remark 1. Let us define Pn = Pn following matrix form:   x − αn −βn Pn = An Pn−1 , An = , n ≥ 1, (11) 1 0   x − α0 with initial conditions P0 = . Also, let us write (4)–(5) in the matrix form 1 A = (x − αn )(ln − ln−1 ) + Θn − βn

APn0 = Bn Pn , where



l − C/2 Bn = n −Θn−1 /βn

n ≥ 0,

(12)

 Θn . ln−1 + (x − αn )Θn−1 /βn − C/2

From the compatibility relation for the Lax pair (11)–(12) we get AA0n = Bn An − An Bn−1 ,

n ≥ 1.

(13)

Now, (13) yields two non-trivial equations from positions (1, 1) and (1, 2). They are given by Eqs. (8) and (9), respectively. Note, that after basic computations, (9) implies (10). In turn, (10) gives tr(Bn ) = −C . Additionally, (8) implies, after some basic computations, det Bn = det B0 + A

n X Θk−1 k=1

with det B0 = AD − (C/2)2 . 3

βk

,

n ≥ 1,

(14)

It can also be shown that relations (S1 ) and (S2 ) hold by using (8), (9) and (10). Furthermore, (S20 ) agrees with (14). Since in [5] 1 an An (x) = + 2 x x and n bn Bn = − + 2 , x x we have Θn = βn+1 (x + an+1 ), ln = C/2 + (n + 1)x − bn+1 . Using [7, Lemma 3.1], we obtain an = −2n − 1 + αn − α ,

(15)

bn = s/2 − ln−1,0 = βn − α0 − ηn−1 , where p1 (n, s) = −(ηn−1 + α0 ). It is shown in [5] that an and bn satisfy

(b2n

bn+1 + bn = s − (2n + 1 + α + an )an , − sbn )(an + an−1 ) = (ns − (2n + α)bn )an an−1 .

(16) (17)

In addition, b2n − sbn = βn an an−1 .

(18)

From (17) and (18) we have bn =

ns − (an + an−1 )βn . 2n + α

(19)

Hence, equations (16) and (17) give equations for the recurrence coefficients αn and βn by using (15) and (19). Explicitly, ns + (4n + 2α − αn−1 − αn )βn 2n + α (n + 1)s + (4n + 4 + 2α − αn − αn+1 )βn+1 , (20) = 2n + 2 + α (2n − 1 + α − αn−1 )(2n + 1 + α − αn )βn (21) (ns + (4n + 2α − αn−1 − αn )βn )(s(n + α) + (αn−1 + αn − 2α − 4n)βn ) =− . (2n + α)2

s + αn (2n + 1 + α − αn ) −

To derive nonlinear difference equations we shall also use equations (8), (9) and (10). Using [7, Lemma 3.1] and from (10), evaluated at x = αn , we get 2 + βn+1 + βn+2 − (2n + 4 + α)αn+1 − s − 2α0 )/2 , ηn = (αn+1

hence, p1 (n, s) = −(αn2 + βn + βn+1 − (2n + 2 + α)αn − s)/2 .

4

(22)

Evaluating (8) and (9) at x = αn we get, respectively, (αn−1 − 2n + 1 − α)βn + αn (αn + βn ) + (2n + 3 + α)βn+1 = (αn + αn+1 )βn+1 , 2 αn−1 − (2n − 2 + α)αn−1 + (2n + 2 + α)αn − αn2 = βn+1 − βn−1 .

(23) (24)

Next, we shall use equations (23), (24), (20) and (22) to derive the third order difference equation for αn . From (23) we express βn+1 and from (20) we express βn . The compatibility between these two equations give an expression of αn in terms of αn+1 , αn and αn−1 . Hence, the following statements hold. Theorem 1. The recurrence coefficient αn satisfies the following third order difference equation: G = G(αn+2 , αn+1 , αn , αn−1 , s) = 0, where G=

3 X 4 X 1 X

! p r αnq αn−1 dp,q,r αn+1

αn+2 −

4 X 4 X 1 X

p r αnq αn−1 . cp,q,r αn+1

p=0 q=0 r=0

p=0 q=0 r=0

The list of the non-zero coefficients cp,q,r and dp,q,r is given in the Appendix. Theorem 2. The recurrence coefficient βn can be expressed in terms of αn+1 , αn and αn−1 as follows: f1 (αn+1 , αn , ) βn = , (25) f2 (αn+1 , αn , αn−1 ) where

f1 = αn+1 (α + 2n)αn2 − α(α + 2) + 4n2 + 4(α + 1)n αn − αs


+αn 8n3 + 4(3α + 5)n2 + 2 3α2 + 10α + 6 n + α α2 + 5α − s + 6 +(α + 2n)αn3 − (2α(α + 2n) + 4n(α + 2n) − 4(α + 2n))αn2 +2αns + α2 s + 3αs and f2 = 2αn2 + (2αn−1 − 12n − 6α − 6) αn + 8n(α + 2n) +4α(α + 2n) + 8(α + 2n) − 3(α + 2n + 2)αn−1 + (2αn−1 − 3(α + 2n) + 2αn ) αn+1 . Substituting βn+1 and βn found before into (24) and (22), the first one gives the third order equation for αn and the second one, with n replaced by n + 1, gives a second order equation for αn . Theorem 3. The recurrence coefficient αn satisfies a second order difference equation F (αn+1 , αn , αn−1 , s) =

2 X 8 X 2 X p=0 q=0 r=0

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p r = 0. hp,q,r αn+1 αnq αn−1

Due to very cumbersome expressions, the coefficients of F are not presented in this paper explicitly. Remark 2. Let us take (14) as −ln2 + Θn

n X Θk−1 Θn−1 = AD − C 2 /4 + A . βn β k k=1

(26)

Equating coefficients in (26) we get, for the coefficient of x1 and x0 , respectively, −2ln,1 ln,0 + Θn,0

Θn−1,1 Θn−1,0 + Θn,1 = −αs/2 , βn βn

2 −ln,0 + Θn,0

Θn−1,0 = −s2 /4 . βn

Thus, after basic computations, βn+1 (αn+1 + αn − 2 − 2n − α) = (α + 2n + 2)

n X

! αj

+ (n + 1)s ,

(27)

j=0

−

n X

s αj − βn+1 + 2 j=0

!2 + βn+1 (2n + 3 − αn+1 + α) (2n + 1 − αn + α) = −

s2 . (28) 4

P Eliminating nj=0 αj between these equations and using the expression for βn from Theorem 2 and then the expression for αn+2 from Theorem 1, we get the second order difference equation for αn from Theorem 3, which gives an alternative way to derive this equation. Remark 3. When s = 0 we get the classical Laguerre polynomials. In this case A(x) = x and C(x) = −x + α. Equations (8), (9) and (10) give αn + βn − βn−1 = 0, (2 + αn−1 − αn )βn + n = 0, αn = 2n + 1 + α, which give βn = n(n + α) and, hence, p1 (n, 0) = −n(n + α).

3

Asymptotics of recurrence coefficients

In this section we shall obtain different asymptotic expansions of αn , βn and p1 (n, s) via nonlinear difference equations. The formulas are found by substituting an appropriate Ansatz for the expansion and by determining unknown coefficients step by step. Theorem 4. For fixed s and n → ∞ the following expansions hold:   s2/3 1 αs1/3 1 αn (s) = 2n + 1 + α + 1/3 1/3 − +O , 2 n 3 · 22/3 n2/3 n4/3   s2/3 n2/3 1 1/3 1/3 1 1 2 2 βn = n + αn + − 2 αs + (6α − 1) + O , 2 · 21/3 3 36 n1/3
3s2/3 n2/3 p1 (n, s) = −n2 − αn − + O n1/3 . 1/3 2·2 6

Remark 4. Expansions in Theorem 4 for αn and βn also appear in [10] (see also [1]). However, [10] they are obtained by using Riemann-Hilbert problem and not directly via the nonlinear difference equations. The expression for p1 (n, s) is new to authors’ knowledge. We also note that we can search for epxansions of the form αn = 2n + 1 + α +

s2/3 αs1/3 − + .... (2n + 1 + α)1/3 3(2n + 1 + α)2/3

Using the second order difference equation F (αn+1 , αn , αn−1 , 0) = 0 (with s = 0) or the third order equation G(αn+2 , αn+1 , αn , αn−1 , 0) = 0 we can also obtain expansion   c(2c − α − 1) 1 c , +O αn = 2n + 1 + α + 2 + 3 n n n4 where c is arbitrary. For s = 0 there exist also algebraic expansions of the form   c 18c2 243c3 1 αn = 2n + 1 + α + 4/3 + 5/3 + , +O n n n2 n7/3 where c is arbitrary.

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Appendix The list of non-zero coefficients of G(αn+2 , αn+1 , αn , αn−1 , s) in Theorem 1: c0,0,0 = 8αs(α + 2n + 2), c0,0,1 = αs(α + 2n + 1),
c0,1,0 = −4α4 − 20α3 − 64n4 − 32(4α + 5)n3 − 16 6α2 + 15α + 2 n2
+2n −16α3 − 60α2 + α(s − 16) + 32 + α2 (s − 8) + α(32 − 5s),
c0,1,1 = 4α3 + 24α2 + 32n3 + 48(α + 2)n2 + 8 3α2 + 12α + 8 n − α(s − 32),
c0,2,0 = 11α3 + 57α2 + 88n3 + 12(11α + 19)n2 + 2 33α2 + 114α + 52 n − α(s − 52), c0,2,1 = −7α2 − 36α − 28n2 − 4(7α + 18)n − 32, c0,3,0 = −10α2 − 48α − 40n2 − 8(5α + 12)n − 32, c0,3,1 = 3α + 6n + 12, c0,4,0 = 3α + 6n + 12,
4 3 c1,0,0 = 4α + 44α + 64n4 + 32(4α + 11)n3 + 16 6α2 + 33α + 38 n2
+n 32α3 + 264α2 − 2α(s − 304) + 320 − α2 (s − 152) + α(160 − 9s),
c1,0,1 = −3α3 − 33α2 − 24n3 − 12(3α + 11)n2 − 6 3α2 + 22α + 38 n − α(s + 114) − 120, c1,1,0 = −48α2 − 192α − 192n2 − 192(α + 2)n − 120, c1,1,1 = − − 4α2 − 16n2 − 16αn + 40, c1,2,0 = −12α2 − 16α − 48n2 − 16(3α + 2)n + 40, c1,2,1 = 8α + 16n + 18, c1,3,0 = 10α + 20n + 18, c1,3,1 = −2, c1,4,0 = −2,
c2,0,0 = −11α3 − 75α2 − 88n3 − 12(11α + 25)n2 − 2 33α2 + 150α + 124 n + α(s − 124), c2,0,1 = 8α2 + 54α + 32n2 + 4(8α + 27)n + 88, c2,1,0 = 12α2 + 80α + 48n2 + 16(3α + 10)n + 88, c2,1,1 = −2α − 4n − 16, c2,2,0 = −16, c2,2,1 = −2, c2,3,0 = −2, 2 c3,0,0 = 10α + 32α + 40n2 + 8(5α + 8)n, c3,0,1 = −7α − 14n − 22, c3,1,0 = −10α − 20n − 22, c3,1,1 = 2, c3,2,0 = 2, c4,0,0 = −3α − 6n, c4,0,1 = 2, c4,1,0 = 2,

8

d0,0,0 = αs(α + 2n + 3), 3

2

3


d0,1,0 = −3α − 3α − 24n − 12(3α + 1)n2 − 6 3α2 + 2α − 2 n − α(s − 6), d0,1,1 = 3α2 + 6α + 12n2 + 12(α + 1)n, d0,2,0 = 8α2 + 10α + 32n2 + 4(8α + 5)n, d0,2,1 = −5α − 10n − 6, d0,3,0 = −7α − 14n − 6, d0,3,1 = 2, d0,4,0 = 2,
d1,0,0 = 4α3 + 24α2 + 32n3 + 48(α + 2)n2 + 8 3α2 + 12α + 8 n − α(s − 32), d1,0,1 − 3α2 − 18α − 12n2 − 12(α + 3)n − 24, d1,1,0 = −4α2 − 32α − 16n2 − 16(α + 4)n − 24, d1,1,1 = 8, d1,2,0 = −2α − 4n + 8, d1,2,1 = 2, d1,3,0 = 2, 2 d2,0,0 = −7α − 20α − 28n2 − 4(7α + 10)n, d2,0,1 = 5α + 10n + 14, d2,1,0 = 8α + 16n + 14, d2,1,1 = −2, d2,2,0 = −2, d3,0,0 = 3α + 6n, d3,0,1 = −2, d3,1,0 = −2.

Acknowledgments The authors thank D. Dai for fruitful discussions. Y. Chen would like to thank the Science and Technology Fund of the Macau SAR for generous support: FDCT 077/2012/A3 and FDCT 130/2014/A3. G. Filipuk acknowledges the support of Alexander von Humboldt Foundation. The support of the Warsaw Center of Mathematics and Computer Science is also gratefully acknowledged. The work of M.N. Rebocho is supported by the Centre for Mathematics of the University of Coimbra – UID/MAT/00324/2013, funded by the Portuguese Government through FCT/MCTES and co-funded by the European Regional Development Fund through the Partnership Agreement PT2020.

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