Multiple orthogonal polynomials associated with an exponential cubic weight

Transcription

1 Multiple orthogonal polynomials associated with an exponential cubic weight Walter Van Assche Galina Filipu Lun Zhang Abstract We consider multiple orthogonal polynomials associated with the exponential cubic weight e x over two contours in the complex plane. We study the basic properties of these polynomials, including the Rodrigues formula and nearest-neighbor recurrence relations. It turns out that the recurrence coefficients are related to a discrete Painlevé equation. The asymptotics of the recurrence coefficients, the ratio of the diagonal multiple orthogonal polynomials and the (scaled) zeros of these polynomials are also investigated. Keywords: multiple orthogonal polynomials, exponential cubic weight, Rodrigues formula, nearest-neighbor recurrence relations, string equations, discrete Painlevé equation, zeros, asymptotics Introduction and statement of the results. Orthogonal polynomials associated with an exponential cubic weight A sequence of non-constant monic polynomials {p n } with deg p n n is said to be orthogonal with respect to the exponential cubic weight e x if p n (x)x e x dx = 0, = 0,,..., n, (.) Γ where the contour Γ is chosen such that the above integral converges. These polynomials satisfy the three-term recurrence relation where β n = xp n (x) = p n+ (x) + β n p n (x) + γ np n (x), (.) Γ xp n(x)e x dx Γ p n(x)e x dx, γ n = xp Γ n(x)p n (x)e x dx, (.) Γ p n (x)e x dx Department of Mathematics, KU Leuven, Celestijnenlaan 00B box 400, BE-00 Leuven, Belgium. Walter.VanAssche@wis.uleuven.be Faculty of Mathematics, Informatics and Mechanics, University of Warsaw, Banacha, Warsaw, 0-097, Poland. filipu@mimuw.edu.pl School of Mathematical Sciences and Shanghai Key Laboratory for Contemporary Applied Mathematics, Fudan University, Shanghai 004, People s Republic of China. lunzhang@fudan.edu.cn

2 and the initial condition is taen to be γ0 p = 0. It is shown by A. Magnus [9] that the recurrence coefficients β n and γn satisfy the string equations γn+ + β n + γ n = 0, (.4) γn(β n + β n ) = n. (.5) For the convenience of the reader, we derive the string equations using ladder operators for orthogonal polynomials in the Appendix. Some variants of orthogonal polynomials associated with the exponential cubic weight have recently been studied in the context of numerical analysis [8] and random matrix theory [4]. Γ Γ 0 Γ Figure : The three rays Γ 0, Γ, Γ For our purpose, we are concerned with the polynomials for specific contours Γ. Consider the three rays (see Figure ) Γ = {z C : arg z = ω }, = 0,,, (.6) where ω = e πi/ is the primitive third root of unity and the orientations are all taen from left to right. Clearly, the integral (.) is well-defined for each Γ. We shall denote by p () n the polynomials satisfying (.) with Γ = Γ 0 Γ. The corresponding recurrence coefficients will be accordingly denoted by β n () and (γ n () ). Hence, we have p () n (x)x e x dx = 0, = 0,,..., n, (.7) Γ 0 Γ and From (.), it is readily seen that xp () n (x) = p () n+(x) + β () n p () n (x) + (γ () n ) p () n (x). (.8) β () 0 = Γ 0 Γ xe x dx Γ 0 Γ e x dx = Γ(/) Γ(/) eπi/. (.9) Thus, one can determine (β n (), (γ n () ) ) recursively from the string equations (.4) (.5) with initial condition γ () 0 = 0 and β () 0 = Γ(/) Γ(/) eπi/.

3 In a similar manner, we let p () n be the polynomials satisfying (.) with Γ = Γ 0 Γ, and denote by β n () and (γ n () ) the corresponding recurrence coefficients. To this end, one has β () Γ 0 = 0 Γ xe x dx Γ 0 Γ e x dx = β() 0 = Γ(/) Γ(/) e πi/. (.0) For the recurrence coefficients β (i) n and (γ (i) n ), i =,, the following proposition holds. Proposition.. There exist two real sequences a n and b n, n N = {0,,,,...} such that and a n, b n satisfy the coupled difference relations with initial conditions β () n = b n e πi/, (γ () n ) = a n e πi/, (.) a n + a n+ = b n, (.) a n (b n + b n ) = n, (.) a 0 = 0, b 0 = Γ(/) Γ(/). (.4) Similarly, we have with the same sequences a n and b n. β () n = b n e πi/, (γ () n ) = a n e πi/, (.5) From (.), one can easily eliminate a n in (.) and obtain n b n + b n + n + b n + b n+ = b n. (.6) This difference equation belongs to A c -type equation on the list of discrete Painlevé equations by Grammaticos and Ramani [6, 7], which has a connection with the second Painlevé equation. It is also an alternative discrete Painlevé I equation in Clarson s list [6, Appendix A.4], see also [], [0]. We give a short derivation of the string equations (.4) (.5) in the Appendix, where we also deal with the more general weight e x +tx.. Multiple orthogonal polynomials with an exponential cubic weight Multiple orthogonal polynomials are polynomials of one variable which are defined by orthogonality relations with respect to r different measures µ, µ,..., µ r, where r is a positive integer. As a generalization of orthogonal polynomials, multiple orthogonal polynomials originated from Hermite-Padé approximation in the context of irrationality and transcendence proofs in number theory. They were further developed in approximation theory, we refer to Aptearev et al. [, ], Coussement and Van Assche [8], Niishin and Soroin [, Chapter 4, ], and Ismail [8, Chapter ] for more information.

4 We tae r = and for (, l) N, we are interested in the monic polynomials P,l of degree + l which satisfy the orthogonality conditions x i P,l (x)e x dx = 0, i = 0,,...,, (.7) Γ 0 Γ x i P,l (x)e x dx = 0, i = 0,,..., l. (.8) Γ 0 Γ We call P,l the (type II) multiple orthogonal polynomial for the exponential cubic weight. If one of and l is equal to zero, then P,l reduce to the usual orthogonal polynomials with respect to the exponential cubic weight e x, i.e., P,0 (x) = p () (x), P 0,(x) = p () (x), (.9) where p (i), i =, are defined in Section.. It is the aim of this paper to derive some basic properties of P,l. Our main results are Theorem. (Rodrigues formula). Let n, m N = {0,,,,...}, then e x P n,n+m (x) = ( )n d n ( ) e x P n dx n 0,m (x), (.0) e x P n+m,n (x) = ( )n d n ( ) e x P n dx n m,0 (x). (.) where P 0,m (x) and P m,0 (x) are given in (.9). The polynomials P n,n (x) were already mentioned by Pólya and Szegö in their problem boo [4, Part V, Chapter, Problem 59] and Pólya investigated their zeros in [, Satz IV]. They are also a special case of polynomials introduced by Gould and Hopper [5] and were investigated, among others, by Dominici [9] and Paris []. Their multiple orthogonality (or d-orthogonality, if one only considers the diagonal polynomials) was already noted earlier, see e.g., [] and references there. In this paper we are investigating the full range of polynomials P n,m (x) and not only the diagonal polynomials, but we obtain ratio asymptotics and the distribution of the zeros for the diagonal polynomials in Section 4. For asymptotic approximations and an asymptotic expansion of P n,n (x) we refer to [9] and []. Multiple orthogonal polynomials satisfy a system of nearest-neighbor recurrence relations [8, Theorem.7]. For P,l defined in (.7) (.8) we can represent the recurrence coefficients explicitly in terms of the sequences a n and b n in Proposition., as stated in the following theorem. Theorem. (the nearest-neighbor recurrence relations). Let n, m N, then xp n,n+m (x) = P n+,n+m (x) + c n,n+m P n,n+m (x) xp n,n+m (x) = P n,n+m+ (x) + d n,n+m P n,n+m (x) + a n,n+m P n,n+m (x) + b n,n+m P n,n+m (x), (.) + a n,n+m P n,n+m (x) + b n,n+m P n,n+m (x), (.) 4

5 where c n,n+m = { Γ(/) Γ(/) eπi/, m = 0, b m e πi/, m > 0, (.4) d n,n+m = b m e πi/, (.5) { n a n,n+m = Γ(/) i, m = 0, Γ(/) (.6) nam m eπi/, m > 0, b n,n+m = { n Γ(/) Γ(/) i, m = 0, (n+m)a m m e πi/, m > 0. (.7) Similarly, where xp n+m,n (x) = P n+m+,n (x) + c n+m,n P n+m,n (x) xp n+m,n (x) = P n+m,n+ (x) + d n+m,n P n+m,n (x) + a n+m,n P n+m,n (x) + b n+m,n P n+m,n (x), (.8) + a n+m,n P n+m,n (x) + b n+m,n P n+m,n (x), (.9) c n+m,n = b m e πi/, (.0) { Γ(/) d n+m,n = Γ(/) e πi/, m = 0, (.) b m e πi/, m > 0, { n a n+m,n = Γ(/) i, m = 0, Γ(/) (n+m)a m (.) e πi/, m > 0, m { n b n+m,n = Γ(/) i, m = 0, Γ(/) (.) nam m e πi/, m > 0. Here, a n and b n are the two real sequences generated from (.) (.4). It is also easy to chec that the recurrence coefficients derived in Theorem. satisfy the partial difference equations obtained in [7, Theorem.]. The rest of this paper is organized as follows. Theorems. and. will be proved in Section. The string equation (.4) plays a particular role in the derivation of the coefficients in the nearest-neighbor recurrence relations. We then perform a numerical study of the coefficients a n, b n in Section. The study suggests that a n+ and b n, n N are all strictly positive, and the limits of a n /n / and b n /n / exist as n, and we can identify these limits explicitly. Section 4 deals with the zeros of P,l. We will give precise location and interlacing results for the zeros of the diagonal multiple orthogonal polynomials P n,n and asymptotic results for the ratio of diagonal multiple orthogonal polynomials. The latter allows us to find the asymptotic distribution of the scaled zeros for these diagonal multiple orthogonal polynomials. The zeros of P,l, with l, have a more interesting structure, which depends on the limit of the ratio /l. We investigate these zeros numerically and end this paper with some conclusions and outloo. 5

6 Proofs. Proof of Proposition. This proposition can be proved by induction on the index n. When n = 0, the relation (.) is obvious, which also gives the initial conditions (.4). Suppose we have β () = b e πi/, (γ () ) = a e πi/, (.) and (a, b ) R for n. From (.4), it follows that thus, (γ () n+ ) = ((γ () n ) + (β () n ) ) On the other hand, the equation (.5) implies that = a n e πi/ b n eπi/ = (b n a n)e πi/, (.) β () n+ = n + (γ n+) () β() n = a n+ = b n a n R. (.) ( n + a n+ b n ) e πi/, (.4) thus b n+ = n + b n R. (.5) a n+ The coupled difference equations (.) (.) are immediate from (.) and (.5). The claim for β n () and (γ n () ) can be proved similarly, we omit the details here.. Proof of Theorem. We shall only prove (.0) since the proof of (.) is similar. We first show that P n,n+m defined in (.0) is a monic polynomial of degree n + m. Observe that ( ) n d n ( ) e x P n dx n 0,m = ( )n n ( d n dx n ( e x P 0,m ) ) = (e x P n,n+m (x)), we then obtain from (.0) the following difference-differential equation for P n,n+m : P n,n+m (x) = x P n,n+m (x) P n,n+m (x). (.6) We can now use induction on n. Clearly P 0,m = p () m is a monic polynomial of degree m. Suppose that P n,m+n is a monic polynomial of degree n + m, then (.6) implies that P n,n+m is a monic polynomial of degree n + m. Next, we show that P n,n+m satisfies the orthogonality conditions (.7) (.8). With Γ 0 defined in (.6), it follows from (.0) and integration by parts times that Γ 0 x P n,n+m (x)e x dx = ( )n n Γ0 x dn dx n ( e x P 0,m (x) ) dx = ( )n+! d n ( e x P n dx n 0,m (x) ) x=0 =! +P n,n+m (0), (.7) 6

7 for = 0,,..., n. Similarly, it is easily seen that x P n,n+m (x)e x dx = x P n,n+m (x)e x dx =! Γ Γ P n,n+m (0). (.8) + Combining (.7) and (.8) gives x P n,n+m (x)e x dx = 0, = 0,,..., n, Γ 0 Γ x P n,n+m (x)e x dx = 0, = 0,,..., n. Γ 0 Γ We still need m more orthogonality conditions to complete (.8), but these follow from ( ) x n+ P n,n+m (x)e x dx = ( )n n+ dn x e Γ 0 Γ Γ0 Γ x P n dx n 0,m (x) dx (n + )! = x e x P! n 0,m (x)dx = 0 Γ 0 Γ for = 0,,..., m, where we used the fact that P 0,m = p () m is the orthogonal polynomial for the cubic exponential weight on Γ 0 Γ.. Proof of Theorem. We will present the proof of (.) (.7), the remaining part of the theorem can be proved in a similar manner. Let us denote the coefficients of x +l and x +l in P,l by δ,l and ε,l, respectively, i.e., P,l (x) = x +l + δ,l x +l + ε,l x +l +. (.9) Substituting the above formula into (.6) and comparing the coefficients of x n+m and x n+m on both sides leads to thus, for m N. Similarly, we have which implies δ n,n+m = δ n,n+m, ε n,n+m = ε n,n+m, δ n,n+m = δ 0,m, ε n,n+m = ε 0,m, (.0) P n+m,n (x) = x P n+m,n (x) P n+m,n (x), (.) δ n+m,n = δ m,0, ε n+m,n = ε m,0. (.) If we insert (.9) into (.) (.), then the coefficients of second leading term x n+m give { δ0,0 δ c n,n+m = δ n,n+m δ n+,n+m =,0, m = 0, (.) δ 0,m δ 0,m, m >, d n,n+m = δ n,n+m δ n,n+m+ = δ 0,m δ 0,m+, (.4) 7

8 where we have also made use of the first equalities in (.0) and (.). On account of the facts that xp m,0 (x) = P m+,0 (x) + β m () P m,0 (x) + (γ m () ) P m,0 (x), (.5) xp 0,m (x) = P 0,m+ (x) + β m () 0,m(x) + (γ m () P 0,m (x), (.6) (see (.9) and (.8)), it is immediate that δ m,0 = δ m+,0 + β m () = δ m+,0 + b m e πi/, (.7) δ 0,m = δ 0,m+ + β m () 0,m+ + b m e πi/, (.8) in view of (.) and (.5). The values for c n,n+m, d n,n+m in (.4) (.5) then follow from combining (.), (.4) and (.7) (.8). We now establish the equalities (.6) (.7) for a n,n+m and b n,n+m. Multiplying both sides of (.) by x n+m e x and integrating the equality over Γ 0 Γ, the orthogonality condition (.8) implies x n+m P n,n+m (x)e x dx Γ b n,n+m = 0 Γ. (.9) x n+m P n,n+m (x)e x dx Γ 0 Γ By (.0), (.) and integrating by parts, we find that x n+m P n,n+m (x)e x dx = ( )n Γ 0 Γ n and Γ 0 Γ x n P n,n (x)e x = = (n + m)! n m! (n + m)! n m! dx = ( )n n = (n )! n ( ) e x P dx n 0,m (x) dx x m P 0,m (x)e x dx Γ 0 Γ P0,m dx, (.0) (x)e x Γ 0 Γ n+m dn x Γ0 Γ ( ) n dn x e dx Γ0 Γ x P n,0 (x) dx P,0 (x)e x dx. (.) Γ 0 Γ Hence, we can simplify (.9) as n Γ 0 Γ e x dx Γ 0 Γ P,0 (x)e x dx, m = 0, b n,n+m = n + m Γ 0 Γ P0,m(x)e x dx m Γ 0 Γ P0,m (x)e x dx, m > 0. (.) Note that Γ 0 Γ P 0,m (x)e x dx = (γ () γ() γ m () ) e x dx, m > 0, (.) Γ 0 Γ 8

9 and straightforward calculations give us dx = Γ(/) ( ω ), (.4) Γ 0 Γ e x P,0 (x)e x dx = (x β () 0 )e x dx Γ 0 Γ Γ 0 Γ = Γ(/) ( ω) Γ(/)β() 0 ( ω ) = Γ(/) ( ω)( ( + ω)e πi/ ). (.5) See (.9) for the value of β () 0. Inserting (.) (.5) into (.), we arrive at n Γ(/) i, m = 0, Γ(/) b n,n+m = n+m m (γ() m ), m > 0, (.6) which is (.7) by (.5). We can also represent a n,n+m as a ratio of two integrals. Indeed, by performing similar strategies above, it is easily seen that x n P n,n+m (x)e x dx Γ a n,n+m = 0 Γ = n P 0,m (x)e x dx Γ 0 Γ. (.7) x n P n,n+m (x)e x dx P 0,m+ (x)e x dx Γ 0 Γ Γ 0 Γ Unfortunately, this representation is not suitable for direct calculation except for m = 0, which gives a n,n = n P 0,0 (x)e x dx Γ 0 Γ P 0, (x)e x dx Γ 0 Γ = n e x dx Γ 0 Γ (x β () 0 )e x dx Γ 0 Γ = b n,n = n Γ(/) i. (.8) Γ(/) For m > 0 the integrals of P 0,m over Γ 0 Γ are involving polynomials orthogonal on the contour Γ 0 Γ, hence it is then difficult to deal with them. So we proceed in another way and we calculate the sum a n,n+m + b n,n+m. Recall the notation δ,l and ε,l in (.9). By comparing the coefficient of x n+m on both sides of (.), it follows from (.0) that a n,n+m + b n,n+m = ε n,n+m ε n+,n+m c n,n+m δ n,n+m = ε 0,m ε 0,m c n,n+m δ 0,m (.9) 9

10 for m > 0. From (.6), we have This, together with (.5), (.4) and (.9), implies ε 0,m = ε 0,m+ + β () m δ 0,m + (γ () m ). (.0) a n,n+m + b n,n+m = ε 0,m ε 0,m c n,n+m δ 0,m = β () m δ 0,m (γ () m ) + β () m δ 0,m = β () m (δ 0,m δ 0,m ) (γ () m ) = (β () m ) (γ () m ) = (γ () m ), m > 0, (.) where we have made use of (.8) in the fourth equality and the string equation (.4) in the last step. A combination of (.6), (.) and (.8) finally gives n Γ(/) i, m = 0, Γ(/) a n,n+m = (.) n m (γ() m ), m > 0, which is (.6), on account of (.5). Asymptotics of a n and b n From Theorem., it is clear that the coefficients in the nearest-neighbor recurrence relations are determined by a n and b n generated from (.) (.4). It is then interesting to study their large n behavior. In Figure we have plotted the values of a n /n / and b n /n / for n from 0 to 70, from which we see that a n+ and b n are all strictly positive for n N. We actually have the following conjecture concerning this observation Figure : The values of a n /n / (left) and b n /n / (right) for n from 0 to 70. Conjecture.. There is a unique positive solution of the recurrence relations (.) (.) with a 0 = 0 and a n+ > 0, b n > 0 for n N. This solution corresponds to the initial condition b 0 = Γ(/)/Γ(/). 0

11 The numerical study further suggests that the limits of a n /n / and b n /n / exist as n, which we can identify in the proposition below. Proposition.. Every positive solution of (.) (.) has the property that lim a n/n / = n /, lim b n/n / = n /. Proof. This can be proved by an argument which was already used by Freud in [, ]. First we show that (a n /n / ) n N is a bounded sequence. From (.) and the positivity of a n+ we find that a n b n. From (.) and the positivity of b n we find a n b n n and thus also 9a n b n n. Together this gives 9a n n, so that 0 a n /n / /9 /. Let a = lim inf n a n /n / and A = lim sup n a n /n /, then 0 a A <. From (.) and the positivity of b n we find b n = a n + a n+. Insert this in (.) to find a n ( an + a n+ + a n + a n ) = n. (.) Let n in (.) through a subsequence for which a n /n / a, then one finds 6a a + A. If n through a subsequence for which a n /n / A, then 6A a + A. Together this gives 6A a + A 6a a + A. If a + A = 0 then one automatically has a = A = 0 so that the limit exists (further on we will see that a 0 so that this case does not happen). If a + A > 0 then one finds A a, and together with a A we see that also in this case a = A and the limit exists. From (.) we then find that lim n b n /n / = a. If we use that information in (.), then 6a a =, so that a = (/6) / = /( / ). The limit for b n /n / follows immediately from this Figure : Zeros of P 45,0 (after scaling) 4 Zeros The formulas in Theorems. and. can be used to generate the multiple orthogonal polynomials P,l defined in (.7) and (.8). We investigate the distribution of their zeros numerically. If one of and l is zero, the polynomials are orthogonal for the exponential

12 Figure 4: Zeros of P 5,5 (after scaling) cubic weight on the curve (Γ 0 Γ or Γ 0 Γ ) in the complex plane. The zeros of P 45,0 (45 / x) are plotted in Figure. It is nown that, in this case, the zeros of the polynomials, after proper scaling, will accumulate on an analytic contour in the complex plane that possesses the so-called S property; cf. [4, 5]. The zero distribution was investigated earlier by Deaño, Huybrechs and Kuijlaars [8], who in fact used the weight e ix. However, a simple rotation x = ye πi/6 is enough to transform their results to the exponential cubic e y which we are using. Suppose that = l = n. It follows from Theorem. that ( P n,n (x) = ( )n dn x e e x). (4.) n dx n We can describe the asymptotic distribution of the zeros of the diagonal polynomials P n,n in more detail. The main reason is that the zeros of P n,n are all located on the three rays Γ 0 Γ Γ, which simplifies matters considerably (see Figure 4). We have the following result for the diagonal polynomials. Observe that this result is the solution of Problem 59 [4, Part V, Chapter ] for the polynomial R n and q =. Proposition 4.. The polynomials P n,n (x) satisfy the symmetry property P n,n (ωx) = ω n P n,n (x), where ω = e πi/ is the primitive third root of unity. In particular n/ j=0 a jx j, n 0 mod, P n,n (x) = x (n )/ j=0 b j x j, n mod, x (4.) (n )/+ j=0 c j x j, n mod, where (a j ) j, (b j ) j, (c j ) j are real sequences depending on n. Furthermore the number of strictly positive real zeros of P n,n is n, if n 0 mod, (n ), if n mod, (n ) +, if n mod, and P n,n (x) has a zero of multiplicity one at x = 0 when n mod and a zero of multiplicity two at x = 0 when n mod.

13 Proof. We use induction on n. The symmetry property follows easily from the Rodrigues formula, so we only need to prove the result about the positive real zeros. Observe that P 0,0 (x) =, P, (x) = x, P, (x) = x(x /), so that the result is true for n = 0,,. Suppose that the result is true for n and let x > x > > x > 0 be the positive real zeros of P n,n. Clearly the sign of P n,n (x j ) is ( ) j+ for j, hence from P n,n (x) = x P n,n (x) P n,n (x) (4.) we find that the sign of P n,n (x j ) is ( ) j, hence P n,n changes sign times and Rolle s theorem guarantees that there are at least zeros y > y > > y with x j < y j < x j, where x 0 = +. If n 0 mod then n mod and the induction hypothesis says that = (n )/ + is odd and P n,n (x) has a zero of multiplicity one at x = 0. The sign of P n,n (0) is ( ) = so that the sign of P n,n (0) is ( ) + =, hence there is also a zero y + of P n,n between 0 and x, giving a total of + = n/ positive real zeros. The ω-symmetry gives another n/ zeros on Γ and n/ zeros on Γ, which is a total of n zeros. Hence there are no other zeros of P n,n. If n mod then n 0 mod and the induction hypothesis gives = (n )/ and P n,n (x) has no zero at x = 0. Hence there will not be an additional zero between 0 and x so that there are = (n )/ positive real zeros for P n,n. There is double zero of P n,n (x) at x = 0. The ω-symmetry gives another (n )/ zeros on Γ and (n )/ zeros on Γ, hence the total number of zeros is (n ) + = n so that there are no other zeros. If n mod then n mod and the induction hypothesis gives = (n )/ is even and a double zero for P n,n (x) at x = 0. Then (4.) implies that P n,n (x) has a single zero at 0. The polynomial P n,n (x)/x of degree n 4 has positive zeros and the sign of this polynomial as x 0 is ( ) =, so that P n,n (x)/x has sign ( ) + = as x 0. Hence P n,n (x)/x has a zero y + between 0 and x, giving a total of + = (n )/+ positive real zeros. The ω-symmetry gives another (n )/ + zeros on Γ and another (n )/ + zeros on Γ, hence together with the single zero at x = 0 this gives a total of (n ) + 4 = n zeros for P n,n so that there are no other zeros. Observe that the proof also shows that the zeros of P n,n and P n,n interlace in the sense that x < y <, x j < y j < x j for j =,...,, x < y < and 0 < y + < x (the latter only when n 0 mod and n mod ). We can now prove the following results Theorem 4.. Let K be a compact set in C \ (Γ 0 Γ Γ ), then lim n P n,n (n / x) n / P n,n (n / x) = Φ(x),

14 holds uniformly for x K, where Furthermore Φ(x) = e πi/ ( holds uniformly for x K. Proof. Consider the ratio N lim n + 9 4x ) / ( + e πi/ P n,n (n/ x) n / P n,n (n / x) = x Φ(x), d P dx n,n(n / x) = P n,n(n / x) P n,n (N / x) N / P n,n (N / x) = N ) /. 9 4x + x n j= x x j,n /N /, where {x j,n, j n} are the zeros of P n,n which are all on the set Γ 0 Γ Γ, then if x K we have P n,n (N/ x) N / P n,n (N / x) n N x x j,n /N / n Nδ, j= where δ = inf{ x y : x K, y Γ 0 Γ Γ } > 0 is the minimal distance between K and Γ 0 Γ Γ. If n/n we then see that the family of analytic functions P n,n (N/ x) N / P n,n (N / x), is uniformly bounded on K. By Montel s theorem there exists a subsequence (n ) N such that P n lim,n (n / x) n n / = F(x), (4.4) P n,n (n / x) uniformly for x K, where F is an analytic function on K for which F(x) = /x+o(/x ) as x. This function F may depend on the selected subsequence, so our aim is to prove that it is independent of the subsequence. Now consider (4.) for P n,n, then N / P n,n (N / x) P n,n (N / x) = x N / P n,n (N/ x) P n,n (N / x), hence (4.4) implies (with N = n ) lim n n / P n,n (n / x) = Φ(x), (4.5) P n,n (n / x) uniformly on K, where Φ(x) = x F(x)/. This uniform convergence of analytic functions implies also the uniform convergence of the derivatives, hence ( ) Φ P n,n (x) = lim (n / x) n n / P n,n (n / ( x) P n,n = lim (n / x) / n P n,n (n / x) n n / P n,n (n / x) P n,n (n / x) n / P n,n (n / x) ) P n,n (n / x), 4

15 but this means that lim n n / P n,n (n / x) = F(x), (4.6) P n,n (n / x) uniformly on K, with the same limit as in (4.4). But then (4.) implies that lim n n / P n,n (n / x) = Φ(x), (4.7) P n,n (n / x) uniformly on K, with the same limit as in (4.5). We can repeat this reasoning once more and find that P n lim,n (n / x) = F(x), (4.8) n and lim n n / n / P n,n (n / x) P n +,n +(n / x) P n,n (n / x) = Φ(x), (4.9) uniformly on K. We will show that the function Φ satisfies a cubic equation, from which we can determine Φ and hence also F uniquely, so that Φ and F do not depend on the selected subsequence (n ) N. Consider the nearest neighbor recurrence relations for the diagonal n = m xp n,n (x) = P n+,n (x) + c n,n P n,n (x) + a n,n P n,n (x) + b n,n P n,n (x) (4.0) xp n,n (x) = P n,n+ (x) + d n,n P n,n (x) + a n,n P n,n (x) + b n,n P n,n (x). (4.) Subtracting (4.0) and (4.) gives P n+,n (x) P n,n+ (x) = (d n,n c n,n )P n,n (x). Use this for n n to eliminate P n,n (x) in (4.0) to find xp n,n (x) = P n+,n (x) + c n,n P n,n (x) + (a n,n + b n,n )P n,n (x) From Theorem. we have + a n,n (c n,n d n,n )P n,n (x). c n,n = Γ(/) Γ(/) eπi/, d n,n = Γ(/) Γ(/) e πi/, so that c n,n d n,n = i Γ(/)/Γ(/). Furthermore a n,n = ni Γ(/) Γ(/), b n,n = ni Γ(/) Γ(/), so that the recurrence relation becomes xp n,n (x) = P n+,n (x) + Γ(/) Γ(/) eπi/ P n,n (x) + n P n,n (x). (4.) 5

16 Use this for n / x and divide by P n,n (n / x), then x = P n+,n (n / x) + Γ(/) n / P n,n (n / x) n / Γ(/) eπi/ + and by using (4.7) we find lim n n / n P n,n (n / x), n / P n,n (n / x) P n +,n (n / x) P n,n (n / x) = x Φ(x), (4.) uniformly on K. We can repeat the reasoning for n n and use (4.5) to find lim n n / P n,n (n / x) P n,n (n / x) = x Φ(x), (4.4) uniformly on K. But then the uniform convergence also holds for the derivative, and as before (4.6) then implies that lim n n / P n,n (n / x) = F(x). P n,n (n / x) Use (4.) for P n+,n (n / x) and divide by P n,n (n / x), then the latter asymptotic result gives P n +,n lim (n / x) n n / P n,n (n / x) = x F(x) = Φ(x), (4.5) uniformly on K. In a similar way as before, the nearest neighbor recurrence relations for (n +, n) can be transformed to xp n+,n (x) = P n+,n+ (x) + d n+,n P n+,n (x) + (a n+,n + b n+,n )P n,n (x) From Theorem. we now use d n+,n = b 0 e πi/, c n+,n = b e πi/, + b n+,n (d n,n c n,n )P n,n (x). so that d n,n c n,n = (b 0 +b )e πi/ = e πi/ /(a ), where we used (.) with n =. We also have a n+,n = (n + )a e πi/, b n+,n = na e πi/, so that the recurrence relation becomes xp n+,n (x) = P n+,n+ (x) Γ(/) Γ(/) eπi/ P n+,n (x) + a e πi/ P n,n (x) + n P n,n (x). (4.6) Consider this for n / x and divide by P n+,n (n / x) then x = P n+,n+ (n / x) Γ(/) n / P n+,n (n / x) n / Γ(/) eπi/ + a P n,n(n / x) n /e πi/ P n+,n (n / x) 6 + n n / P n,n (n / x) P n+,n (n / x)

17 and by using (4.) and (4.5) we find uniformly on K. Now use the relation lim n n / P n +,n +(n / x) P n +,n (n / x) = x Φ(x), (4.7) P n+,n+ (n / x) = P n+,n+ (n / x) n / P n,n (n / x) n / P n+,n (n / x) P n+,n (n / x) n / P n,n (n / x) and let n through the subsequence (n ) N, then (4.9), (4.) and (4.7) show that Φ(x) = ( x ). (4.8) Φ(x) The cubic equation (4.8) has one solution Φ which behaves for x as Φ (x) = x + O(/x), x. There are two other solutions Φ, which behave as /(x) as x Φ (x) = x + 7x 5/ + O(/x4 ), Φ (x) = x 7x 5/ + O(/x4 ), x. Recall that our Φ satisfies Φ(x) = x F(x)/, where F(x) = O(/x), so that we need the solution Φ. The discriminant of (4.8) is (4x 9)/7 so that Φ has branch points at (9/4) /, (9/4) / e ±πi/, which are three points on Γ 0, Γ, Γ respectively, and since all the zeros of P n,n are on Γ 0 Γ Γ, we conclude that the scaled zeros x j,n /n / are dense on the three segments [0, (9/4) / ] [0, (9/4) / e πi/ ] [0, (9/4) / e πi/ ]. The cubic equation can be solved explicitly by using Cardano s formula: let y = x /(Φ) and z = /y, then the cubic equation (4.8) becomes z xz + = 0, and the solutions are z = ω j u / +ω j v / (j = 0,, ), where ω = e πi/, u +v = and uv = x, i.e., u = + 9 4x, v = 9 4x = x + 9 4x. The solution Φ corresponds to the solution with z(x) = /x + O(/x 4 ), and this is z(x) = ω u / + ω v /, and since Φ = y, we find Φ (x) = (ω u / + ω v / ) = ωu / + ω v / + x. 7

18 Corollary 4.. Let {x j,n, j =,,..., n} be the zeros of P n,n and µ n be the normalized counting measure of the scaled zeros x j,n /n /, µ n = n n δ xj,n /n /. Then the sequence (µ n ) n converges wealy to the probability measure µ for which f(x)dµ(x) = where ω = e πi/ and with (9/4) / 0 v(x) = v(x)f(x) dx + j= ω(9/4) / 0 v(x)f(x) dx + ω (9/4) / 0 v(x)f(x) dx, ( ) + x[a(x) + b(x)] [b(x) a(x)], (4.9) 4π ( )/ 9 4x a(x) =, b(x) = Proof. The Stieltjes transform of the measure µ n is ( + 9 4x x t dµ n(t) = P n,n(n / x) n / P n,n (n / x), )/. (4.0) hence Theorem 4. gives lim n x t dµ n(t) = F(x) = ( ) x Φ(x), uniformly on compact sets of C \ (Γ 0 Γ Γ ). The Grommer-Hamburger theorem [] then implies that µ n converges wealy to a measure µ for which x t dµ(t) = ( ) x Φ(x). The function Φ is analytic in C \ ([0, (9/4) / ] [0, ω(9/4) / ] [0, ω (9/4) / ]), hence the measure µ is supported on [0, (9/4) / ] [0, ω(9/4) / ] [0, ω (9/4) / ]. Furthermore it is absolutely continuous and we can find the density by using the Stieltjes inversion formula v(x) = ( ) π lim I (x + iɛ) Φ(x + iɛ). ɛ 0+ Due to the ω-symmetry, it is sufficient to determine v(x) for x [0, (9/4) / ]. Clearly v(x) = π lim IΦ(x + iɛ) = ɛ 0+ π I (ω a + ω b), with a and b given in (4.0). Then by some elementary (complex) calculus, using (a ab +b )(a + b) = a +b and ab = x, one finds the expression (4.9) for the density v. 8

19 x~ Figure 5: Histogram of the real zeros of P 600,600 and the density v(x) on [0, (9/4) / ] In Figure 5 we have given a histogram of the 400 real zeros of P 600,600 together with the density v, scaled so as to have total mass one for all the real zeros. There are 400 zeros on the interval [0, ω(9/4) / ] and 400 zeros on the interval [0, ω (9/4) / ] and these zeros are obtained by rotating the real zeros over an angle ±π/. The density v has a finite non-zero value at the origin v(0) = / /4π = and tends to zero as (9/4) / x when x (9/4) /. If l, the rotational symmetry of the zeros is broen. Suppose that l, then we see numerically that zeros of P,l lie on the line containing Γ (some zeros are in fact on Γ ), while the other zeros are distributed on a complex contour in the lower half plane; see Figure 6. Similarly, if l, then l zeros of P,l lie on the line containing Γ (again some zeros are on Γ ), and the other zeros are distributed on a complex contour in the upper half plane, as illustrated in Figure 7. Indeed, from Theorem., it is easily seen that the zeros of P,l are complex conjugates of the zeros of P l,. We expect that the asymptotic zero distribution of P,l will depend on the limit of the ratio /l. Figure 6: Zeros of P 7,0 (left) and P 4,5 (right) after scaling 9

20 Figure 7: Zeros of P 0,7 (left) and P 6,4 (right) after scaling 5 Conclusions and outloo In this paper, we have introduced the multiple orthogonal polynomials associated with an exponential cubic weight e x over two contours in the complex plane. The basic properties of these polynomials are studied, which include the Rodrigues formula and nearest-neighbor recurrence relations. These results then allow us to perform numerical studies of the recurrence coefficients and zero distributions of the multiple orthogonal polynomials. Moreover, the recurrence coefficients are related to a discrete Painlevé equation. One can also consider the more general exponential cubic weight e x +tx, where t R, and the associated multiple orthogonal polynomials have similar Rodrigues formulas and nearest-neighbor recurrence relations. Indeed, with e x replaced by e x +tx in Theorem., the difference-differential equations (.6) and (.) now read This then implies that P n,n+m (x) = (x t )P n,n+m (x) P n,n+m (x), P n+m,n (x) = (x t )P n+m,n (x) P n+m,n (x). δ n,n+m = δ 0,m, ε n,n+m = ε 0,m t, δ n+m,n = δ m,0, ε n+m,n = ε m,0 t, where δ,l and ε,l are defined in (.9) and now depend on t. Following the same strategy as in the proof of Theorem. and using the string equation (A.7) at the final stage, we obtain (.) and (.) with { β () 0 (t), m = 0, c n,n+m = β () m d n,n+m = β m () (t), a n,n+m = b n,n+m = (t), m > 0, { n β () 0 (t) β() 0 n m { (γ() n β () 0 (t) β() 0 (n+m) m (γ() (t), m = 0, m (t)) t, m > 0, 0 (t), m = 0, m (t)), m > 0,

21 where β n () (t), (γ n () ) (t) are the recurrence coefficients of the monic orthogonal polynomials with respect to the weight e x +tx on Γ 0 Γ, and β n () (t), (γ n () ) (t) are the recurrence coefficients of the monic orthogonal polynomials with respect to the weight e x +tx on Γ 0 Γ. The recurrence relations (.8) and (.9) for this general case can be obtained similarly, but we omit the results here. Clearly, in the general case, we lose the nice structure of the recurrence coefficients as stated in Theorem., and more importantly we lose the symmetry given in Proposition 4., which is why we focus on the weight e x in this paper. The challenging problem is to establish the asymptotic zero distribution of P,l (x) for the non-symmetric case. At present we are unable to find an analogue of Theorem 4. and Corollary 4. because of two reasons: first one needs the asymptotic behavior of the recurrence coefficients and at present we can only conjecture the behavior (see Proposition.). If we assume this to be correct, then the proof of Theorem 4. can be used to find the asymptotic behavior, away from the set where the zeros of the multiple orthogonal polynomials accumulate, in terms of an algebraic function Φ satisfying a cubic equation. But the second reason is that we don t now where the zeros of the multiple orthogonal polynomials accumulate. The discriminant of the cubic equation is a quartic polynomial in x and the four roots are branch points of the algebraic function Φ. The zeros will accumulate on two curves, each connecting two points in the complex plane, see Figures 6 and 7. One of the curves is a straight line, the other is a curved line connecting two of the four branch points. The straight line, however, does not connect the other two branch points but starts from one branch point and stops before the second branch point is reached. This suggests that a vector equilibrium problem is involved, for two measures living on curves connecting four branch points, with an external field x induced by the weight e x. In order to characterize the limiting zero distribution, one may need to extend the concept of S-property (cf. [4, 5]) for orthogonal polynomials and equilibrium measures to this setting for multiple orthogonal polynomials and vector equilibrium problems. Once that is obtained, one may be able to use the Riemann-Hilbert method to find the asymptotic behavior of the multiple orthogonal polynomials. Appendix A Derivation of the string equations In this appendix, we give an alternative proof of the string equations (.4) (.5) using ladder operators for orthogonal polynomials. Note that the ladder operators for multiple orthogonal polynomials and their compatibility conditions can be found in [0]. Following the general set-up (cf. [5]), if the weight function w vanishes at the endpoints of the orthogonality interval, the lowering and raising ladder operators for the associated monic polynomials p n are given by ( ) d dx + B n(x) p n (x) = γn A n(x)p n (x), (A.) ( ) d dx B n(x) v (x) p n (x) = A n (x)p n (x), (A.)

22 with and where v(x) := lnw(x), A n (x) := v (x) v (y) [p n (y)] w(y)dy, (A.) h n x y B n (x) := v (x) v (y) p n (y)p n (y)w(y)dy, (A.4) h n x y p m (x)p n (x)ω(x)dx = h n δ m,n, m, n = 0,,,.... (A.5) Note that A n and B n are not independent, but satisfy the following compatibility conditions [8, Lemma.. and Theorem..4]. Proposition A.. The functions A n and B n defined in (A.) and (A.4) satisfy B n+ (x) + B n (x) = (x β n )A n (x) v (x), (S ) + (x β n )[B n+ (x) B n (x)] = γ n+a n+ (x) γ na n (x). (S ) Now we consider a more general exponential cubic weight e x +tx, with parameter t R. Then v(x) = lnw(x) = x tx, and It then follows from (A.) (A.5) that v (x) v (y) x y = (x + y). A n (x) = (x + β n ), B n (x) = γ n. (A.6) Substituting (A.6) into (S ) and comparing the coefficients of the constant term, we have γ n + γ n+ + β n t = 0. (A.7) From (S ) we similarly get γ n(β n + β n ) = n. (A.8) Note that in this case, the recurrence coefficients β n and γ n all depend on t. By setting t = 0 in (A.7) and (A.8), we recover the string equations (.4) and (.5). The weight e x +tx is a modification of the weight e x with an exponential factor e tx, and as a consequence the recurrence coefficients satisfy the Toda equations [8,.8] d dt γ n = γ n (β n β n ), d dt β n = γn+ γn. If we differentiate (A.0) and then use (A.9), we find β n(t) = γ n+(β n+ β n ) γ n(β n β n ). (A.9) (A.0)

23 Then use (A.7) and (A.8) to find β n (t) = β n t β n + n +, which is the Painlevé II equation. The equations (A.7) and (A.8) give n + n + + βn β n + β n β n+ + β = t n which also follows from the Bäclund transformation of the second Painlevé equation (see [] and [7]), hence it is not surprising to find that β n satisfies the Painlevé II equation. If we write x n (t) = aβ n ( at), where a = (/) /, then x n(t) = x n + tx n + n +, which is the Painlevé II equation in standard form and with parameter α = n +. The second Painlevé equation is closely related to the Airy equation and has special solutions in terms of Airy functions for the parameter values α = n +, with n Z [6, 7.]. This relation with the Airy function was to be expected since one has the integral representation Ai(t) = e πi/ e z / tz dz, πi e πi/ (see Eq of the NIST Digital Library of Mathematical Functions ) which contains (a slight variation of) the weight e z +tz. The special solution of Painlevé II in terms of Airy functions is x n (t) = τ n(t) τ n (t) τ n+ (t) τ n+ (t), where τ n is the Hanel matrix ϕ(t) ϕ (t) ϕ (n ) (t) ϕ (t) ϕ (t) ϕ (n) (t) τ n (t) =.. ϕ (n ) (t) ϕ (n) (t)., ϕ (n ) (t) and ϕ is a solution of the Airy equation ϕ + tϕ = 0. This solution of P II coincides with the well nown solution β n = δ n δ n+, where δ n is the coefficient of x n for the monic orthogonal polynomial P n (x) = x n + δ n x n +. One has δ n = n / n, where n is the Hanel matrix containing the moments m 0 m m n m m m n n =... m n m n m n

24 and n is a similar determinant but with the last column replaced by m n, m n+,..., m n respectively. Even though this is an explicit solution, it is not convenient for finding the recurrence coefficients when n is large because of the high number of computations involved, whereas the relations (A.7) (A.8) have a low computational complexity. The explicit solution is also not convenient for obtaining the asymptotic behavior of β n and a n as n, which is easier to obtain from the string equations (see Proposition.). Acnowledgements The authors are grateful to the referees for their careful reading and constructive suggestions. WVA is supported by KU Leuven research grant OT//07, FWO research grant G.094. and the Belgian Interuniversity Attraction Poles Programme P7/8. GF is supported by the MNiSzW Iuventus Plus grant Nr 04/IP/0/7. LZ was a Postdoctoral Fellow of FWO, and is also partially supported by The Program for Professor of Special Appointment (Eastern Scholar) at Shanghai Institutions of Higher Learning (No. SHH4007) and by Grant SGST DZ 7800 from Fudan University. References [] A.I. Aptearev, Multiple orthogonal polynomials, J. Comput. Appl. Math. 99 (998), [] A.I. Aptearev, A. Branquinho and W. Van Assche, Multiple orthogonal polynomials for classical weights, Trans. Amer. Math. Soc. 55 (00), [] Y. Ben Cheih and N. Ben Romdhane, On d-symmetric classical d-orthogonal polynomials, J. Comput. Appl. Math. 6 (0), [4] P. Bleher and A. Deaño, Topological expansion in the cubic random matrix model, Int. Math. Res. Not. 0, no., [5] Y. Chen and M. Ismail, Jacobi polynomials from compatibility conditions, Proc. Amer. Math. Soc. (005), [6] P.A. Clarson, Painlevé equations Nonlinear special functions, in Orthogonal Polynomials and Special Functions: Computation and Applications (F. Marcellán, W. Van Assche, Eds.), Lecture Notes in Mathematics 88, Springer-Verlag, Berlin, 006, pp. 4. [7] P. Clarson, E.L. Mansfield, H.N. Webster, On the relation between discrete and continuous Painlevé equations, Theoret. and Math. Phys. (000), no., 6. [8] A. Deaño, D. Huybrechs and A.B.J. Kuijlaars, Asymptotic zero distribution of complex orthogonal polynomials associated with Gaussian quadrature, J. Approx. Theory 6 (00), 0 4. [9] D. Dominici, Asymptotic analysis of generalized Hermite polynomials, Analysis 8 (008),

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