Continued fractions and Bessel functions
aa r X i v : . [ m a t h . G M ] M a y [Continued fractions and Bessel functions. . . ]Adyghe State University,Pervomayskaya st., 208,385000, Maykop, Russia
12 3 ontinued fractions and Bessel functions
A.A. AllahverdyanMay 9, 2019
Abstract
Elementary transformations of equations Aψ = λψ are considered. The invert-ibility condition (Theorem 1) is established and similar transformations of Riccatiequations in the case of second order differential operator A are constructed (The-orem 2). Applications to continuous fractions for Bessel functions and Chebyshevpolynomials are established. It is shown particularly that the elementary solu-tions of Bessel equations are related to a fixed point transformations of Riccatiequations. Keywords:
Bessel functions, invertible Darboux transforms, continued frac-tions, Euler operator, Riccati equation.
Let A be a differential operator A of order nA = a ( x ) D nx + a ( x ) D n − x + ... + a n ( x ) , D x = ddx . (1)We consider transformations of this operator defined by substitutions of theform ˆ ψ = ( b D x + b ) ψ and their superpositions. In the case b = 0 thetransformation is invertible and the operator A in the considered equationtransforms in the operator ˆ A as followsˆ A = b ◦ A ◦ b − . The following theorem [1] holds true in the general case. ∗ Allahverdyan A.A. The Darboux transformations and Bessel functions † c (cid:13) ‡ Recieved ... 2019 heorem 1 (on eigenfunctions).
The equation for eigenfunctions Aψ = λψ , λ = 0 admits an invertible substitution ˆ ψ = ( D x − g ) ψ, g = (log ϕ ) x = ϕ x ϕ , Aϕ = 0 . (2)First, we prove the following lemma. Lemma 1.
The differential operator A of order n > A = D x − g iff g = (log ϕ ) x , where ϕ ∈ ker A . ◮ Let φ ∈ ker A and g = (log ϕ ) x . The formula (1) A = n X j =0 a j ( x ) D n − jx = ˜ A ( D x − f ) , ˜ A = n − X j =0 ˜ a j ( x ) D n − ( j +1) x , g = (log ϕ ) x , implies that A ( ϕ ) = 0 because ( D x − f )( ϕ ) = 0. Then by the substitution y = ϕ ˆ y we obtain an operator ˆ A with the zero coefficient a n = 0. Consequently,this polynomial is divisible by D x iff the initial operator is divisible by D x − f . ◭ Now we prove the theorem on eigenfunctions. ◮ Note that A ( ϕ ) = 0 and operator A takes the form (1): A = n X j =0 a j ( x ) D n − jx . By substituting A in Aψ = λψ we find˜ A ˆ ψ = λψ (3)From (3) we have λψ = a ˆ ψ n − + a ˆ ψ n − + ... + a n ˆ ψ. (4)Then by the substitution (2) from (4) we obtain λ ˆ ψ = ddx [ a ˆ ψ n − + a ˆ ψ n − + ... + a n ˆ ψ ] (5)If λ = 0 then the equation (5) and original equation Aψ = λψ have the sameorder, but coefficients in (5) are different.Hence, we have proved that the equation Aψ = λψ , λ = 0 admits a substi-tution ˆ ψ = ( D x − g ) ψ if g = (log ϕ ) x and this substitution is invertible. ◭ From this point on, we consider applications of Theorem 1 in the case when A is Euler operator. Note that the replacement (2) is invertible and its inverse is written by the formula(4). efinition 1. Euler operator has the form A = e mt k ( D t )where k ( D t ) is a polynomial in D t = ddt with constant coefficients. Lemma 2. If A = e mt k ( D t ), B = e nt z ( D t ) then a superposition of Euleroperators A and B takes the form: A ◦ B = e ( m + n ) t c ( D t ) , c ( D t ) = k ( D t + n ) z ( D t ) . In this case the substitution ˆ ψ = ( b D x + b ) ψ becomes an Euler operator ofthe first order ˆ ψ = e t ( D t + c ) ψ, c ∈ C . (6)Indeed, x = e − t , dx = − e − t dt , therefore D x = − e t D t . Let us consider second order equations and application of Theorem 1 in thiscase. An operator A can be described as follows A = a ( x ) D + a ( x ) D + a ( x ) (7)Using the substitution ψ = e ϕ ˆ ψ and assuming that a coefficient of D is equalto zero, we can obtain that a coefficient of D is equal to 1, i.e. A = D + q ( x ) . (8)Then (8) takes the form: A = ( D − g )( D + g ) (9)Indeed, Aψ = ( D + q ( x )) ψ = ψ ′′ + q ( x ) ψ. On the other hand, Aψ = ( D − g )( D + g ) ψ = ψ ′′ + ( − g ′ + g ) ψ = ψ ′′ + q ( x ) ψ, where q ( x ) + g ′ + g = 0. 4 efinition 2. A Riccati equation associated with the equation Aψ = λψ is the following equation for the logarithmic derivative f = ψ ′ ψ : a ( f ′ + f ) + a f + a = λ. (10)In the particular case that an operator A is of the form (9) the equation (10)can be written as follows: f ′ + f + q ( x ) = λ, f = ψ ′ ψ . Suppose that an operator A is given by A = D + 1 x D − β x . (11)Then by the substitution x = e − t : D x = ddx = − e t ddt = − e t D t , one can rewrite the equation Aψ = λψ in the following form: Aψ = e t ( D t − β ) ψ = e t ( D t − β − ◦ e t ( D t + β ) ψ = λψ. (12)Note that D x = ( − e t D t ) ◦ ( − e t D t ) = e t ( D t + D t ) . The equation for eigenfunctions of an operator A is Aψ = λψ . Here anoperator A is of the form (11). The equation considered here is called Besselequation.By applying Theorem 1 and Lemma 2 to equation (12) we obtain e t ( D t + β ) ψ = ˆ ψ, e t ( D t − β −
1) ˆ ψ = λψ. (13)As a result, we have that the equation Aψ = λψ takes the form e t ( D t − ˆ β ) ˆ ψ = λ ˆ ψ, ˆ β def = β + 1 . (14)Rewriting now equations (13) in terms of f β = (log ψ ) t and f ˆ β = (log ˆ ψ ) t oneobtains (see Definition 2):ˆ ψ = e t ( f β + β ) ψ, λψ = e t ( f ˆ β − ˆ β ) , ( f β + β )( f ˆ β − ˆ β ) = λ · e − t = λ · x (15)Without loss of generality we put λ = 1 in the last equation and prove themain theorem. 5 heorem 2. Let f = f β be a solution of the Riccati equation f t + f = β + x and the function ˆ f = f ˆ β be defined by the following equation( f + β )( ˆ f − ˆ β ) = x , ˆ β = β + 1 , (16)then this equation states the equivalence of two Riccati equations f t + f = β + x ⇔ ˆ f t + ˆ f = ˆ β + x . ◭ Let the function µ = µ ( t ) satisfies the differential equation µ t = − µ and f β be a solution of the Riccati equation f t + f = β + µ ( t ). Then the function f ˆ β is defined by the formula for ˆ f as followsˆ f = µf + β + ˆ β. (17)By differentiating (17) with respect to t :ˆ f t = − µ ( f + β ) − f t µ ( f + β ) = − µf + β − f t µ ( f + β ) . (18)Note that f t = β − f + µ. (19)By substituting (19) in (18),ˆ f t = − µf + β + µ f − βf + β − µ ( f + β ) . (20) µ ( f + β ) = − ˆ f t − µf + β + µ f − βf + β . (21)By squaring both sides of (17) we haveˆ f = µ ( f + β ) + 2 ˆ β µf + β + ˆ β . Indeed, µ ( f + β ) = ˆ f − β µf + β − ˆ β . (22)Henceforth, from equations (21) and (22) we can obtain: − ˆ f t − µf + β + µ f − βf + β = ˆ f − β µf + β − ˆ β , ˆ f t + ˆ f = ˆ β + µ − f − β + 2( β + 1) f + β . So, ˆ f t + ˆ f = ˆ β + µ ( t ) . ◮ orollary of Theorem 2. The mapping A → ˆ A defined in Theorem 2has a fixed point: ˆ f = f, ( ˆ β ) = β . (23) ◭ In the case (23), (16) we find by solving quadratic equations that f = f ± = 12 ± x, β = − , ˆ β = 12 ,f t + f = 14 + x , ( x = e − t ) . (24)It easy as well to see that ˆ f ± = f ± ◮ . In the case of Chebyshev polynomials T n ( x ) T n ( x ) = cos nθ, x = cos θ, D θ = −√ − x D x (25)we have T n +1 + T n − = 2 xT n , T = 1 , T = x. (26)This recurrent relation totally defines polynomials T n ( x ) in x -variable andrewriting (26) one obtains (Cf. [4]):( f n − x )( f n +1 + x ) = − , f n = T n − T n + x, f ( x ) = x + 1 x (27)whicn looks very similar to (16). Generally speaking, recurrent relations (27)and (26) are equivalent. Moreover, by reversing in a certain sense Theorem 2one may obtain from its proof and (27) the second order differential equation D x T n = x · D x − x T n − n − x T n for Chebyshev’s polynomials T n ( x ). In the case of Bessel functions we canchoose, as a basic one, an analog of the linear recurrent relation (26) (see (15)and [3]), but the recurrent relation in the “Riccati” form (16) provides someadvantages (Cf [2]) and yields the formulae (24) used below in order to obtainrational in x -variable solutions of the eq. (16). Introducing a numbering wedenote (Cf. (16)) β = 12 , β j +1 = β j + 1 , j = 1 , , . . . . roposition. Let β j = j − . Then the formula as follows f j +1 = β j +1 + x f j + β j , f = − x + 12 , (28)provides rational solutions of the Riccati equation of Theorem 2 f = 32 + x − x . Similarly determined f , f ... : f = 52 + x x − x + 3 ; f = 72 + x x − x + 15... Conclusion
Theorem 1 and equation (6) reduce the spectral problem Aψ = λψ with Euleroperator A to an algebraic one. This allows us to investigate a generalizationof the results of § References [1] A. Shabat, “Symmetries of Spectral Problems”,
Lecture notes inPhysics , (1): 139-173, 2009.[2] P. Flajolet, R. Schott, “Non-overlapping Partitions, Continued Frac-tions, Bessel Functions and a Divergent Series”, Europ. J. Combinatorics , : 421–432, 1990.[3] G.N. Watson, “A treatise on the theory of Bessel functions”, 1945.[4] P. Chebyshev, “Decomposition using continuous fractions”, Sbornik:Mathematics ,1