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The Power Series Expansion of Mathieu Function and Its Integral Formalism

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The Power Series Expansion of Mathieu Function and Its Integral Formalism International Journal of Differential Equations and Applications Volume 14 No. 2 2015, 81-99 ISSN: 1311-2872 AP url: http://www.ijpam.eu acadpubl.eu doi: http://dx.doi.org/10.12732/ijdea.v14i2.2092 THE POWER SERIES EXPANSION OF MATHIEU FUNCTION AND ITS INTEGRAL FORMALISM Yoon Seok Choun Baruch College The City University of New York Natural Science Department, A506 17 Lexington Avenue, New York, NY 10010, USA Abstract: By applying three term recurrence formula (Choun, 2012 [7]), power series expansions in closed forms of Mathieu equation for an infinite series and their integral forms are constructed. One interesting observation resulting from the calculations is the fact that a modified Bessel function recurs in each of sub-integral forms: the first sub- ′ integral form contains zero term of Ans, the second one contains one term of An’s, the third one contains two terms of An’s, etc. Section 5 contains two additional examples of Mathieu equation. This paper is 5th out of 10 in series “Special functions and three term recurrence formula (3TRF).” See section 6 for all the papers in the series. AMS Subject Classification: 33E10, 34B30 Key Words: Mathieu equation, integral form, three term recurrence formula, modified Bessel function 1. Introduction Mathieu ordinary differential equation is of Fuchsian types with the two regular and one irregular singularities. In contrast, Heun equation of Fuchsian types has the four regular singularities. Heun equation has the four kind of conflu- c 2015 Academic Publications, Ltd. Received: November 26, 2014 url: www.acadpubl.eu 82 Y.S. Choun ent forms: (slowromancapi@) confluent Heun (two regular and one irregular singularities), (slowromancapii@) doubly confluent Heun (two irregular singu- larities), (slowromancapiii@) biconfluent Heun (one regular and one irregular singularities) and (slowromancapiv@) triconfluent Heun equations (one irregu- lar singularity). For DLFM version [25], Mathieu equation in algebraic forms is also derived from the confluent Heun equation by changing all coefficients δ = γ 1 , ǫ 0, α q and q λ+2q . → 2 → → → 4 One example of three therm recursion relations is the Mathieu equation, introduced by Emile´ L´eonard Mathieu (1868)[22], while investigating the vi- brating elliptical drumhead. Mathieu equation, known for the elliptic cylinder equation, appears in diverse areas such as astronomy and physical problems involving Schr¨odinger equation for a periodic potentials [17], the parametric Resonance in the reheating process of universe [32], and wave equations in general relativity[4], etc. Mathieu function has been used in various areas in modern physics and mathematics. [1, 2, 3, 18, 20, 23, 26, 27, 29] Unfortunately, even though Mathieu equation has been observed in various areas mentioned above, there are no power series expansions in closed forms and their integral formalism analytically. The Mathieu functions have only been described in numerical approximations (Whittaker 1914[30], Frenkel and Portugal 2001[19]). Sips 1949[28], Frenkel and Portugal 2001[19] argued that it is not possible to represent analytically the Mathieu functions in a simple and handy way. By applying the same method I used in analyzing Heun functions [8, 9], I construct power series expansions of Mathieu equation in closed forms and their integral representations analytically using three-term recurrence formula (3TRF) [7]. Mathieu equation is written by d2y + (λ 2q cos2z) y = 0 (1) dz2 − where λ and q are parameters. This is an equation with periodic-function coefficient. Mathieu equation also can be described in algebraic forms putting x = cos2z: d2y dy 4x(1 x) + 2(1 2x) + (λ + 2q 4qx)y = 0 (2) − dx2 − dx − This equation has two regular singularities: x = 0 and x = 1; the other singu- larity x = is irregular. Assume that its solution is ∞ ∞ n+ν y(x)= cnx (3) n=0 X THE POWER SERIES EXPANSION OF MATHIEU... 83 where ν is an indicial root. Plug (3) into (2). cn = An cn + Bn cn− ; n 1 (4) +1 1 ≥ where, 4(n + ν)2 (λ + 2q) An = − (5a) 2(n +1+ ν)(2(n + ν) + 1) 4q Bn = (5b) 2(n +1+ ν)(2(n + ν) + 1) c1 = A0 c0 (5c) 1 We have two indicial roots which are ν = 0 and 2 . 2. Power series expansion for infinite series ∞ n+ν By putting a power series y(x)= n=0 cnx into linear ODEs, the recurrence relation between successive coefficients starts to appear. In general, the recur- P rence relation for a 3-term is given by (4) where c = A c and c = 0. There 1 0 0 0 6 are an infinite series and three types of polynomials in three-term recurrence relation of a linear ordinary differential equation: (slowromancapi@) polyno- mial which makes Bn term terminated: An term is not terminated, designated as ‘a polynomial of type 1,’ (slowromancapii@) polynomial which makes An term terminated: Bn term is not terminated, denominated as ‘a polynomial of type 2’ and (slowromancapiii@) polynomial which makes An and Bn terms terminated at the same time, referred as ‘a complete polynomial.’ For n = 0, 1, 2, 3, in (4), the sequence cn is expanded to combinations of · · · An and Bn terms. I suggest that a sub-power series yl(x) where l = 0, 1, 2, 3, · · · is constructed by observing the term of sequence cn which includes l terms of ′ Ans [7]. The power series solution is described by sums of each yl(x) such as ∞ y(x) = n=0 yn(x). By allowing An in the sequence cn is the leading term of each sub-power series yl(x), the general summation formulas of the 3-term P recurrence relation in a linear ODE are constructed for an infinite series and a polynomial of type 1, designated as ‘three term recurrence formula (3TRF).’ As we see (5b), there is no way to make Bn term terminated at certain value of an index summation n. Because the numerator of (5b) is just consist of a constant q parameter. So there are only two kind of power series expansions such as an infinite series and a polynomial of type 2. In this paper I construct an analytic solution of Mathieu equation for an infinite series. And its local solutions for a polynomial of type 2 are available in chapter 7 of Ref.[16]. 84 Y.S. Choun In Ref.[7] the general expression of a power series y(x) for an infinite series is defined by ∞ y(x) = yn(x)= y (x)+ y (x)+ y (x)+ y (x)+ 0 1 2 3 · · · n=0 X ∞ i0−1 2i0+λ = c0 B2i1+1 x ( i i ! X0=0 Y1=0 ∞ i0−1 ∞ i2−1 2i2+1+λ + A2i0 B2i1+1 B2i3+2 x i ( i i i i i !) X0=0 Y1=0 X2= 0 3Y= 0 ∞ ∞ i0−1 + A2i0 B2i1+1 N ( i ( i X=2 X0=0 Y1=0 N−1 ∞ i2k−1 A i k B × 2 2k+ 2i2k+1+(k+1) i i i i ! kY=1 2k=X2(k−1) 2k+1Y= 2(k−1) ∞ i2N −1 B x2i2N +N+λ (6) × 2i2N+1+(N+1) i i i i !)) ) 2N =X2(N−1) 2N+1Y= 2(N−1) Substitute (5a)-(5c) into (6). In this article Pochhammer symbol (x)n is used Γ(x+n) to represent the rising factorial: (x)n = Γ(x) . The general expression of a power series of Mathieu equation for an infinite series is given by ∞ y(x) = yn(x)= y (x)+ y (x)+ y (x)+ y (x)+ 0 1 2 3 · · · n=0 X ∞ ν 1 i0 = c0x ν 3 ν η (i (1 + 2 )i0 ( 4 + 2 )i0 X0=0 ∞ ν 2 1 (i0 + 2 ) 42 (λ + 2q) 1 + 1 ν− 1 ν ν 3 ν (i (i0 + 2 + 2 )(i0 + 4 + 2 ) (1 + 2 )i0 ( 4 + 2 )i0 X0=0 ∞ 3 ν 5 ν ( + )i ( + )i 2 2 0 4 2 0 ηi1 x × 3 ν 5 ν i i ( 2 + 2 )i1 ( 4 + 2 )i1 ) X1= 0 ∞ ∞ ν 2 1 (i0 + 2 ) 42 (λ + 2q) 1 + 1 ν− 1 ν ν 3 ν n=2 (i (i0 + 2 + 2 )(i0 + 4 + 2 ) (1 + 2 )i0 ( 4 + 2 )i0 X X0=0 THE POWER SERIES EXPANSION OF MATHIEU... 85 n−1 ∞ k ν 2 1 (ik + + ) 2 (λ + 2q) 2 2 − 4 × (i + k + 1 + ν )(i + k + 1 + ν ) i i − k 2 2 2 k 2 4 2 kY=1 k=Xk 1 (1 + k + ν ) ( 3 + k + ν ) 2 2 ik−1 4 2 2 ik−1 × k ν 3 k ν (1 + 2 + 2 )ik ( 4 + 2 + 2 )ik ) ∞ n ν 3 n ν (1 + + )i − ( + + )i − 2 2 n 1 4 2 2 n 1 ηin xn (7) × (1 + n + ν ) ( 3 + n + ν ) i i − 2 2 in 4 2 2 in n=Xn 1 1 2 where η = 4 qx . Put c0= 1 as ν = 0 for the first independent solution of 1 Mathieu equation and ν = 2 for the second one in (7). Remark 1. The power series expansion of Mathieu equation of the first kind for an infinite series about x = 0 using 3TRF is given by 1 y(x) = MF q, λ; η = qx2,x = cos2z 4 ∞ 1 i0 = 3 η i (1)i0 ( 4 )i0 X0=0 ∞ 2 1 ∞ 3 5 i 2 (λ + 2q) ( ) ( ) 0 4 1 2 i0 4 i0 i1 + − 1 1 3 3 5 η x (i (i0 + 2 )(i0 + 4 ) (1)i0 ( 4 )i0 i i ( 2 )i1 ( 4 )i1 ) X0=0 X1= 0 ∞ ∞ 2 1 i0 42 (λ + 2q) 1 + − 1 1 3 n=2 (i (i0 + 2 )(i0 + 4 ) (1)i0 ( 4 )i0 X X0=0 n−1 ∞ k 2 1 k 3 k (ik + ) 2 (λ + 2q) (1 + )i − ( + )i − 2 − 4 2 k 1 4 2 k 1 × (i + k + 1 )(i + k + 1 ) (1 + k ) ( 3 + k ) i i − k 2 2 k 2 4 2 ik 4 2 ik kY=1 k=Xk 1 ∞ n 3 n (1 + )in− ( + )in− 2 1 4 2 1 ηin xn (8) × (1 + n ) ( 3 + n ) i i − 2 in 4 2 in n=Xn 1 Remark 2.
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