Loading...

The Gauss-Airy functions and their properties Alireza Ansari Abstract. In this paper, in connection with the generating function of three-variable Hermite polynomials, we introduce the Gauss-Airy function Z 1 ∞ −yt2 zt3 GAi(x; y, z) = e )dt, y ≥ 0, x, z ∈ R. cos(xt + π 0 3 Some properties of this function such as behaviors of zeros, orthogonal relations, corresponding inequalities and their integral transforms are investigated. Key words and phrases. Integral transforms.

Gauss-Airy function, Airy function, Inequality, Orthogonality,

1. Introduction We consider the three-variable Hermite polynomials n

3 Hn (x, y, z)

= n!

n−3j

[3] [ 2 ] X X zj j=0 k=0

yk xn−3j−2k , j! k!(n − 2k)! 2

as the coefficient set of the generating function ext+yt e

xt+yt2 +zt3

=

∞ X

+zt3

3 Hn (x, y, z)

n=0

(1)

tn . n!

(2)

For the first time, these polynomials [8] and their generalization [9] was introduced by Dattoli et al. and later Torre [10] used them to describe the behaviors of the HermiteGaussian wavefunctions in optics. These are wavefunctions appeared as Gaussian apodized Airy polynomials [3, 7] in elegant and standard forms. Also, similar families to the generating function (2) have been stated in literature, see [11, 12]. Now in this paper, it is our motivation to introduce the Gauss-Airy function derived from operational relation 3 Hn (x, y, z)

d2

d3

= ey dx2 +z dx3 xn .

(3)

For this purpose, in view of the operational calculus of the Mellin transform [4–6], we apply the following integral representation Z ∞ 2 3 eys +zs = esξ GAi(ξ, y, z) dξ, (4) −∞

Received February 28, 2015. 119

120

A. ANSARI

Figure 1. The function GAi(x). where the function GAi(ξ, y, z) is given by Z 1 ∞ −yr2 e cos(rξ − zr3 )dr. GAi(ξ, y, z) = π 0

(5)

d If we set s = dx in (4) and incorporate it with the relation (3), we see that the threevariable Hermite polynomials is presented by the following integral representation with respect to the Gauss-Airy function as a convolution product of two distributions Z ∞ ξ n GAi(ξ − x, y, z) dξ. (6) 3 Hn (x, y, z) = −∞

The main object of this paper is devoted to the properties of Gauss-Airy function GAi, which are summarized as the follows. In Section 2, we state the Weierstrass infinite product for the entire function GAi(x) with respect to its zeros gan . We define the Gauss-Airy function zeta and state a formula for obtaining its values. In Section 3, we show that the Gauss-Airy function is logarithmically concave and state some inequality for this function using this property. In Section 4, we get a relationship between the Gauss-Airy function and Airy function, and present orthogonal relations for the Gauss-Airy function with resect to its zeros. In last section, we try to find some identities for the Gauss-Airy function in view of the Mellin, Laplace and Fourier transforms of this function. 2. Zeros of the Gauss-Airy function and the Gauss-Airy zeta function In this section, for describing the behavior of function GAi(x, y, z), we confine ourself to the function GAi(x) = GAi(x, 1, − 31 ). For the other parameters y and z, the same analysis can be applied. As we see in Figure 1, GAi(x) is an oscillatory function for negative x and tends to zero algebraically. We denote the zeros by gan , n = 1, 2, · · · , and show the first six roots of this function, see Table 1. Also, we

THE GAUSS-AIRY FUNCTIONS AND THEIR PROPERTIES

ga1 ga2 ga3 ga4 ga5 ga6 Table 1.

121

-3.338107410459767038489 -5.087949444130970616637 -6.52055982809555105913 -7.78670809007175899878 -8.944133587120853123138 -10.022650853340980380158 The first six real roots of GAi(x).

can write the Weierstrass infinite product for this function as ∞ Y x x GAi0 (0) GAi(x) = GAi(0)e−kx 1+ e− |gan | , k = | | = 0.1763219672. |gan | GAi(0) n=1 (7) Now, using the above infinite product and zeros gan , we define the Gauss-Airy zeta function and evaluate some values for it. Let us start with the following lemma. Lemma 2.1. The Gauss-Airy function GAi(x) satisfy the following ordinary differential equation y 00 − 2y 0 − xy = 0, x ∈ R. (8) Proof. By using the Laplace integral Z

exr v(r)dr,

y(x) =

(9)

Γ

and substituting into the ordinary differential equation (8), we get a first order differential equation for v(r) as follows v 0 (r) + (r2 − 2r)v(r) = 0.

(10)

After the deformation and normalization of integral (9), we rewrite y as follows Z i∞ 2 r3 1 exr+r − 3 dr, (11) y= 2πi −i∞ which implies that the solution of differential equation (8) is the Gauss-Airy function. Remark 2.1. By the same procedure to Lemma 2.1, we can show the Gauss-Airy function GAi(x, a, b) satisfy the following differential equation −by 00 − 2ay 0 − xy = 0,

x, b ∈ R, a > 0.

(12)

Theorem 2.2. The value of the Gauss-Airy zeta function ζga (s) =

∞ X

1 , |gan |s n=1

(13)

is given by ds (ln(GAi(x)))x=0 , dxs and in special case, we get ζga (2) = k 2 + 2k and ζga (3) = ζga (s) =

(14) 1 2

− k(k + 1)(k + 2).

122

A. ANSARI

Proof. If we take the logarithmic derivative of the Weierstrass infinite product, we obtain ∞ X 1 d 1 ln(GAi(x)) = −k + − . (15) dx x + |gan | |gan | n=1 Applying the higher order derivatives on the above relation and setting x = 0, we get the identity (14). The special values ζga (2) and ζga (3) are given by (14) and differential equation (8) simultaneously.

3. Some inequalities for the Gauss-Airy function In this section, we get some inequalities for the Gauss-Airy function. We start with the logarithmic concavity concept and state the following lemma. Lemma 3.1. The Gauss-Airy function GAi(x) is logarithmically concave over (ga1 , ∞), where ga1 is the first negative root on real axis. 2

≤ 0 over the interval (ga1 , ∞). For this Proof. It suffices to show that d ln(GAi(x)) dx2 purpose, by using the Weierstrass infinite product (7) we see that ∞ 1 d2 ln(GAi(x)) X , = − 2 dx (1 + gaxn )2 n=1

is negative except at the zeros of GAi(x).

(16)

Remark 3.1. Since the Gauss-Airy function GAi(x) is logarithmically concave over 1 (ga1 , ∞), then f (x) = GAi(x) is logarithmically convex over and satisfy the following inequality f (ux + (1 − u)y) ≤ f (x)u f (y)1−u ,

0 ≤ u ≤ 1, x, y > ga1 .

(17)

Theorem 3.2. The following inequalities hold for the Gauss-Airy function GAi(x) (1) xGAi2 (x) ≤ GAi02 (x), x ∈ (ga1 , ∞). x ∈ (ga1 , ∞). (2) GAi(x)GAi(y) ≤ GAi2 ( x+y 2 ), (3) GAiα (x)GAi1−α (0) ≤ GAi(αx), x ∈ (ga1 , ∞), 0 ≤ α ≤ 1. (4) GAi(x) ≤ |GAi(z)|, x ∈ (ga1 , ∞), z = x + iy. (5) |GAi(z)|GAi(0) ≤ GAi(x)|GAi(iy)|, x ≥ 0, z = x + iy. Proof. To prove these inequalities, we apply the approaches of [14] which generalize the results of the Airy function. For the inequality (1), we use Lemma 3.1 for the logarithmic concavity of GAi(x), which implies that d2 ln(GAi(x)) xGAi2 (x) − GAi02 (x) = ≤ 0, 2 dx GAi2 (x)

(18)

and proof is completed. The inequalities (2) and (3) are deduced from Remark 3.1 by setting u = 21 and u = α, y = 0 in (17), respectively. To prove (4), using the infinite

THE GAUSS-AIRY FUNCTIONS AND THEIR PROPERTIES

123

product representation for x ∈ (ga1 , ∞), we have Q∞ − |gax | x −kx n GAi(0)e 1 + n=1 |gan | e GAi(x) = x+iy Q∞ |GAi(x + iy)| 1 + x+iy e− |gan | | |GAi(0)e−k(x+iy) n=1

=

∞ Y

x gan

1+ q (1 + gaxn )2 + n=1

y2 ga2n

|gan |

,

(19)

which implies that each factor in the infinite product is less than or equal to unit, and proof is completed. Moreover, since for the each factor in the infinite product we have 1 + gaxn 1 1 q ≥q =q , (20) 2 2 y y2 y 1 + ga2 (1+ x )2 (1 + gaxn )2 + ga 1 + 2 2 ga n gan

n

and |GAi(iy)| = GAi(0)

n

∞ Y

1 q 1+ n=1

y2 ga2n

.

(21)

Therefore, we can write the following inequality GAi(0) GAi(x) ≥ , |GAi(z)| |GAi(iy)|

(22)

or equivalently |GAi(z)|GAi(0) ≤ GAi(x)|GAi(iy)|.

(23)

4. Orthogonality of the Gauss-Airy functions In this section, we intend to introduce an orthonormal basis on the interval (0, ∞) in terms of the Gauss-Airy function and its zeros. Let us start with the following lemma. Lemma 4.1. The following relation holds between the Gauss-Airy function and Airy function Ai(x) 2 (24) Ai(1 + x) = e−(x+ 3 ) GAi(x), where the Airy function is given by [1, 2] Z 1 ∞ r3 Ai(x) = cos(xr + )dr. (25) π 0 3 Proof. If we use the differential equation of Gauss-Airy function y 00 − 2y 0 − xy = 0,

x ∈ R,

(26)

−x

and set z(x) = Ce y(x), then the above differential equation changes to the following Airy type differential equation z 00 − (1 + x)z = 0,

x ∈ R,

(27)

with the solution z(x) = Ai(1 + x). After matching the Gauss-Airy function and Airy 2 function at zero point, we get C = e− 3 and the relation (24) is obtained.

124

A. ANSARI

Lemma 4.2. By the same procedure to Lemma 4.1 and applying differential equation (12), we generalize the relation (24) as follows 1

a− 3 Ai(

b2 a

4 3

1

b

+ a− 3 x) = e−( a x+γ) GAi(x; b, a),

where γ = ln

2

1

a− 3 Ai( b 4 ) a3

GAi(0; b, a)

(28)

.

(29)

Remark 4.1. The relation (24) implies that the difference between the zeros of Airy function an and the zeros of Gauss-Airy function gan is unity, i.e., an − gan = 1, n = 1, 2, · · · . n) Theorem 4.3. The functions { GAi(x+ga GAi0 (gan ) , n ∈ N} form an orthonormal basis on the interval (0, ∞) with respect to the weight function w(x) = e−2x and zeros gan .

Proof. We recall the following orthonormal basis on the interval (0, ∞) in terms of the Airy function and different zeros an and an0 [15, 16] Z ∞ Ai(x + an )Ai(x + an0 )dx = 0, n 6= n0 , n, n0 ∈ N, (30) 0 Z ∞ Ai2 (x + an )dx = Ai02 (an ). (31) 0

At this point, if we substitute the relation (24) orthonormal basis for the Gauss-Airy function weight function w(x) = e−2x Z ∞ e−2x GAi(x + gan )GAi(x + gan0 )dx 0 Z ∞ e−2x GAi2 (x + gan )dx

into the above integrals, we get an in terms of the zeros gan and the n 6= n0 , n, n0 ∈ N,

=

0,

=

GAi02 (gan ).

(32) (33)

0

Also, in connection with the orthogonality of Airy functions on interval (−∞, ∞) [16] Z ∞ δ(b − a), α = β, ξ+b ξ+a (34) )Ai( )dξ = Ai( b−a α β −∞ |αβ| 1 Ai , β 6= α, 1 3 3 3 3 |β −α | 3

(β −α ) 3

we can extend the orthogonality of the Gauss-Airy functions as follows. Theorem 4.4. The functions GAi(x) are orthogonal on the interval (−∞, ∞) with respect to the weight function of exponential form as follows Z ∞ ξ(α+β) ξ+a−α ξ+b−β δ(b − a), α = β, − αβ e GAi( )GAi( )dξ = (35) α β I, β 6= α, −∞ where I=

|αβ| 1

|β 3 − α3 | 3

e

a b 1 α+β−3−

b−a 1 (β 3 −α3 ) 3

1

GAi

b − a − (β 3 − α3 ) 3 1

(β 3 − α3 ) 3

! .

(36)

THE GAUSS-AIRY FUNCTIONS AND THEIR PROPERTIES

125

5. Integral Transforms of the Gauss-Airy function In this section, we want to obtain the integral transforms type of the Gauss-Airy function. We focus on the Mellin, Laplace and Fourier transforms of this function in view of the Meijer G functions and Airy functions. Theorem 5.1. The Mellin transform of the Gauss-Airy function is given by the following relation ! Γ(s) cos(s π2 ) − s 1,3 3 ∆(3, s+1 ) 2 √ 3 2 G3,2 | M{GAi(x); s} = 4 ∆(1, 2), ∆(1, − 21 ) 2 π Γ(s) sin(s π2 ) − s 1,3 3 ∆(3, s+1 ) 2 2 √ − | 3 G3,2 , (37) 4 ∆(1, 1), ∆(1, 0) 2 π a+k−1 where ∆ is denoted by ∆(k, a) = ka , a+1 . k ,··· , k

Proof. If we use the definition of Mellin transform Z ∞ M{f (x); s} = xs−1 f (x)dx,

(38)

0

for the Gauss-Airy function GAi(x) Z 1 ∞ −r2 1 GAi(x) = e cos(rx + r3 )dr, π 0 3 and apply the following facts Z ∞ π Γ(s) cos(s ), xs−1 cos(rx)dx = s r 2 Z0 ∞ Γ(s) π xs−1 sin(rx)dx = sin(s ), s r 2 0 then, we get the Mellin transform as follows

(39) (40)

Z Γ(s) cos(s π2 ) ∞ −s −r2 1 r e cos( r3 )dr x GAi(x)dx = π 3 0 0 π Z ∞ Γ(s) sin(s 2 ) 2 1 − r−s e−r sin( r3 )dr π 3 0 Z Γ(s) cos(s π2 ) ∞ − s+1 −r 1 3 = r 2 e cos( r 2 )dr 2π 3 0 π Z ∞ Γ(s) sin(s 2 ) s+ 1 3 − r− 2 1 e−r sin( r 2 )dr. 2π 3 0 At this point, we evaluate the improper integrals in above equation in terms of the Meijer G functions for <(s) < 2 as follows [13] (page 82) ! Z ∞ Γ(s) cos(s π2 ) − s 1,3 3 ∆(3, s+1 ) s−1 2 √ x GAi(x)dx = 3 2 G3,2 | 4 ∆(1, 2), ∆(1, − 21 ) 2 π 0 Γ(s) sin(s π2 ) − s 1,3 3 ∆(3, s+1 ) 2 √ − 3 2 G3,2 | , 4 ∆(1, 1), ∆(1, 0) 2 π Z

∞

s−1

which confirms the relation (37).

126

A. ANSARI

In order to obtain the Laplace and Fourier transforms of the Gauss-Airy function, we recall a well-known integral representation for the product of Airy functions as follows [16] Z ∞ 2 1 u + v i(u−v)ξ Ai(2 3 (ξ 2 + Ai(u)Ai(v) = 1 ))e dξ. (41) 2 2 3 π −∞ Theorem 5.2. The following Laplace transform type of the Gauss-Airy function holds Z ∞ √ 1 u 1 (42) e−ux √ GAi(x; , )dx = 4 2πe−γ Ai2 (u), 2 4 x 0 where the constant γ is given by (29). Proof. Setting u = v in the relation (41), we obtain Z ∞ 2 1 Ai2 (u) = 1 Ai(2 3 (ξ 2 + u))dξ. 2 3 π −∞ In this case, after substituting a = 41 , b = of variable we easily arrive at (42).

u 2

(43)

into (28) and using the suitable change

Theorem 5.3. The following Fourier transform types hold for the Gauss-Airy function Z ∞ −yp2 +izp3 e = e−ipξ GAi(ξ, y, z) dξ, (44) −∞ Z ∞ 2 e−γ u 1 (45) Ai(u) = e−2uξ −iuξ GAi(ξ 2 , , )dξ, 2πAi(0) −∞ 2 4 where γ is given by (29). Proof. To prove (44), it suffice to substitute s = −ip into (4). For the relation (45), if we set v = 0 in the relation (41), we obtain Z ∞ 2 1 Ai(u) = 1 Ai(2 3 (ξ 2 + u))eiuξ dξ. (46) 3 2 πAi(0) −∞ Now, by substituting the relation (28) into the above integral with a = 41 , b = get the relation (45).

u 2,

we

References [1] A. Ansari, M.R. Masomi, Some results involving integral transforms of Airy functions, Integral Transforms and Special Functions 26(7) (2015), 539–549. [2] A. Ansari, The fractional Airy transform, Hacettepe Journal of Mathematics and Statistics 44 (4) (2015), 761–766. [3] A. Ansari, H. Askari, On fractional calculus of A2n+1 (x) function, Applied Mathematics and Computation 232 (2014), 487–497. [4] A. Ansari, Some inverse fractional Legendre transforms of gamma function form, Kodai Mathematical Journal 38(3) (2015), 658–671. [5] A. Ansari, M. Ahmadi Darani, M. Moradi, On fractional Mittag-Leffler operators, Reports on Mathematical Physics 70(1) (2012), 119–131. [6] A. Ansari, A. Refahi, S. Kordrostami, On the generating function ext+yφ(t) and its fractional calculus, Central European Journal of Physics 11(10) (2013), 1457–1462. [7] D. Babusci, G. Dattoli, D. Sacchetti, The Airy transform and associated polynomials, Central European Journal of Physics 9(6) (2011), 1381–1386.

THE GAUSS-AIRY FUNCTIONS AND THEIR PROPERTIES

127

[8] G. Dattoli, S. Lorenzutta, H.M. Svrivastava, A. Torre, Some families of mixed generating functions and generalized polynomials, Le Matematiche 54(1) (1999), 147–157. [9] G. Dattoli, B. Germano, P.E. Ricci, Hermite polynomials with more than two variables and associated bi-orthogonal functions, Integral Transforms and Special Functions 20(1) (2009), 17–22. [10] A. Torre, Airy polynomials, three-variable Hermite polynomials and the paraxial wave equation, Journal of Optics A: Pure and Applied Optics 14 (2012) 045704 (24pp). [11] M.A. Khan, A.H. Khan, N. Ahmad, A note on a new three variable analogue of Hermite polynomials of I kind, Thai Journal of Mathematics 9(2) (2011), 391–404. [12] M.A. Khan, G.S. Abukhammash, On Hermite polynomials of two variables suggested by S.F. Ragabs Laguerre polynomials of two variables, Bulletin Calcutta Mathematical Society 90 (1998), 61–76. [13] A.P. Prudnikov, Y.A. Brychkov, O.I. Marichev, Integrals and series, direct Laplace transforms vol. 4, Amsterdam, Gordon and Breach, 1992. [14] M. Salmassi, Inequalities satisfied by the Airy functions, Journal of Mathematical Analysis and Applications 240 (1999), 574–582. [15] E.C. Titchmarsh, Eigenfunction expansions, Clarendon Press, Oxford, 1962. [16] O. Vallee, M. Soares, Airy functions and applications to physics, Imperial College Press, London, 2004. (Alireza Ansari) 1- Department of Applied Mathematics, Faculty of Mathematical Sciences, Shahrekord University, P.O. Box 115, Shahrekord, Iran 2- Photonics Research Group, Shahrekord University, P.O. Box 115, Shahrekord, Iran E-mail address: alireza [email protected]

Loading...