Solution of Differential Equations with Polynomial Coefficients with the Aid of an Analytic Continuation of Laplace Transform

In a series of papers, we discussed the solution of Laplace’s differential equation (DE) by using fractional calculus, operational calculus in the framework of distribution theory, and Laplace transform. The solutions of Kummer’s DE, which are expressed by the confluent hypergeometric functions, are obtained with the aid of the analytic continuation (AC) of Riemann–Liouville fractional derivative (fD) and the distribution theory in the spaceD′ R or the AC of Laplace transform. We now obtain the solutions of the hypergeometric DE, which are expressed by the hypergeometric functions, with the aid of the AC of Riemann–Liouville fD, and the distribution theory in the space D′ r,R, which is introduced in this paper, or by the term-by-term inverse Laplace transform of AC of Laplace transform of the solution expressed by a series.


Introduction
Stimulated by Yosida's work [1,2], in which the solution of Laplace's differential equation (DE) is obtained with the aid of the operational calculus of Mikusi ński [3], we are concerned in [4][5][6], with the DE or fractional DE of the form: where σ = 1 2 or σ = 1, m = 2, and a l ∈ C and b l ∈ C are constants.In solving the DE, we assume that the solution u(t) and the inhomogeneous part f (t) for t > 0 are expressed as a linear combination of for ν ∈ C\Z <1 , where Γ(ν) is the gamma function.
In [5,6], 0 D β R u(t) is the analytic continuation (AC) of Riemann-Liouville fractional derivative (fD), which was introduced in [7,8] and is reviewed in [9].It is defined for u(t) and f (t) satisfying the following conditions.We now adopt Condition A. We then express u(t) as follows: where When Throughout the present paper, the equations involving β are valid for β ∈ C, but in the applications given in Sections 4-7, we use them only for β = n ∈ Z >−1 , when 0 D n R g ν (t) = d n dt n g ν (t).We use R, C and Z to denote the sets of all real numbers, of all complex numbers and of all integers, respectively.We also use R >r := {x ∈ R|x > r} for r ∈ R, + C := {z ∈ C|Re z > 0}, Z >a := {n ∈ Z|n > a}, Z <b := {n ∈ Z|n < b} and Z [a,b] := {n ∈ Z|a ≤ n ≤ b} for a, b ∈ Z satisfying a < b.We use Heaviside's step function H(t), which is defined such that (i) H(t) = 1 for t > 0 and = 0 for t ≤ 0, and (ii) when f (t) is defined on R >r , f (t)H(t − r) is equal to f (t) when t > r and to 0 when t ≤ r.
In [4][5][6], we take up a modified Kummer's DE as an example, which is where a ∈ C, b ∈ C and c ∈ C are constants.Kummer's DE is this DE with b = 1 [10,11].If c / ∈ Z, the basic solutions of Equation ( 5) are given by u 1 (t) := 1 F 1 (a; c; bt), Here 1 F 1 (a; c; z) = ∑ ∞ k=0 (a) k k!(c) k z k is the confluent hypergeometric series, where z ∈ C, (a) n = ∏ n−1 k=0 (a + k) for a ∈ C and n ∈ Z >0 , and (a) 0 = 1.These solutions are expressed as linear combinations of g ν (t).
In [4,5], we consider the theory of distributions in the space D R , which is presented in [12,13] and is explained briefly in Section 3.3.The solution of Equation ( 1) with the aid of distributions in D R is presented in [5], assuming that the solution satisfies Condition A. Both of the solutions given by Equations ( 6) and ( 7), of Equation ( 5), satisfy Condition A, and we can obtain both of them, by solving Equation ( 5) by this method.
In [4], we adopt the following condition.4).In this case, the DE given by Equation (1) in terms of the distribution theory in space D R is presented in [4].The solutions of fractional DE with constant coefficients are presented in [12,13].In [4], the solution given by Equation (6) of Equation ( 5), satisfies Condition B, and hence we can obtain it by solving Equation ( 5) by this method.However, the solution given by Equation (7) satisfies Condition B only when 1 − c > −1, and hence we can obtain it only when 1 − c > −1, by this method.
In [6], it was mentioned that, when Conditions B and C are satisfied, the Laplace transform of u(t) exists and the DE is solved with the aid of Laplace transform, and that the solutions of Equation ( 5), satisfying Condition B, satisfy Condition C and hence are obtained by using the Laplace transform.
In [6], the AC of Laplace (AC-Laplace) transform is introduced as in Section 1.1 given below, and it is shown that, when Conditions A and C are satisfied, the AC-Laplace transform of u(t), which is denoted by û(s) = L H [u(t)] = L H [u(t)](s), exists and the DE given by Equation ( 1) is solved with the aid of the AC-Laplace transform.In fact, the AC-Laplace transform of g ν (t) and of u(t) given by Equation ( 3) are expressed as We review the solution in terms of the AC-Laplace transform in Section 2, and the solution with the aid of the distribution theory in Section 3. In Section 4 we confirm the following lemma.Lemma 1.Both of the solutions given by Equations ( 6) and ( 7), of Equation ( 5), satisfy Conditions A and C, and hence we can obtain both of them, by solving Equation (5) by using distribution theory in the space D R and also by using the AC-Laplace transform.
In Section 5, we consider the hypergeometric DE, which is given by where a ∈ C, b ∈ C and c ∈ C are constants.If c / ∈ Z, the basic solutions of Equation (10) in [10,11] are given by Remark 2. These solutions of Equation ( 10) converge only at t satisfying |t| < 1, and do not satisfy Condition A, and they are not obtained by the methods stated above.
We introduce the theory of distributions in the space D r,R , in Section 3.4.We now use the step function H r (t), which is defined for r ∈ R >0 such that (i) H r (t) = 1 for 0 < t < r and = 0 for t ≤ 0 or t ≥ r, and (ii) when f (t) is defined on R >0 ∩ R <r , f (t)H r (t) is equal to f (t) for 0 < t < r and to 0 for t ≤ 0 or t ≥ r.

Condition D.
Condition A with H(t) replaced by H r (t) is valid.
In Section 5, we show that when u(t) satisfies Condition D, we can solve Equation (10) with the aid of distributions in D r,R .Definition 1.Let u(t) be given by Equation (3) and satisfy Condition D. Then, we define its AC of Riemann-Liouville fD of order β ∈ C, by which satisfies Condition D.
Definition 2. Let u(t) be given by Equation (3) and satisfy Condition D. Then, we define û(s We call this the AC-Laplace transform series of u(t).
For the solutions of Equation ( 10), the existence of the AC-Laplace transform is not guaranteed, but we can define û(s) by Equation (14).We can then set up a DE satisfied by the thus-defined û(s).In Section 5, we write the DE for the û(s), and its solution in the form of Equation ( 14) is obtained.We find that the obtained series converges for no value of s, and yet we obtain the solution u(t) by the term-by-term inverse Laplace transform of the series û(s).
The solutions given by Equations ( 11) and ( 12), of Equation ( 10), satisfy Condition D for r = 1, and we show that they are obtained by using the distribution theory in the space D r,R and also by using the AC of Laplace transform series, in Section 5.
In Section 6, we show that the Bessel functions J ±ν (t) are the solusions of Bessel's DE with the aid of the AC-Laplace transform.In Section 7, some discussions are given on Hermite's DE.Concluding remarks are given in Section 8.

Definition of the AC-Laplace Transform
The AC-Laplace transform f (s) = L H [ f (t)] of a function f (t) is defined in [6] as follows.
where γ ∈ C\Z <1 , and f 1 (z) is analytic on the neighborhood of R >0 .Definition 3. Let f γ (z) and u(t) = f γ (t) satisfy Conditions E and C, respectively.Then, we define the AC-Laplace transform fγ (s Here, C H is the contour which appears in Hankel's formula giving the AC of the gamma function Γ(z), so that C H is the contour which starts from ∞ + i , goes to δ + i , encircles the origin counterclockwise, goes to δ − i , and then to ∞ − i , where δ ∈ R >0 and ∈ R >0 satisfy 1 and δ < 1, see [14] (Section 12.22).
Remark 3. fγ (s) defined by Definition 3 is an analytic continuation of the Laplace transform defined by fγ (s) = ∞ 0 f γ (t)e −st dt for Re γ > 0, as a function of γ.

Remarks on Recent Developments
Here, we call attention to recent developments on the solutions of differential equations related with fractional calculus and perturbation method, which are based on He's variational iteration method (VIM) [15].By using the VIM, He gave the fD and fI which involve the terms determined by the initial or boundary condition.Liu et al. [16] discussed the solution of heat conduction in a fractal medium with the aid of He's fD.Kumar et al. discussed the solution of partial differential equations involving time-fD by using Laplace transform and perturbation method based on the VIM; see [17,18] and references in them.In [19,20], discussions are given on the fractional complex transform, which reduces an equation involving fD to an equation involving only integer-order derivatives.

AC-Laplace Transform
In the present section and Section 2.1, we assume that u(t) satisfies Conditions C and A, and we put û(s where (ν) n is Pochhammer's symbol, so that (ν) n = Γ(ν+n) Γ(ν) .
Applying Lemma 3 to Equation ( 4), with the aid of Equation ( 8), we obtain where Proof Here, s β ĝν (s) z is so defined that Equation (20) with Equation ( 21) represents Equation (19). where Proof We confirm these with the aid of Equation ( 9) and Lemmas 3 and 4.
Then, Equation ( 24) gives û(s) z = 0, and Theorem 1.Let u(t) satisfy Conditions C and A, and be expressed by Equation (3).Then, the AC-Laplace transform û(s) is expressed as Equation ( 9) and there exists M ∈ R >0 such that the series given by Equation ( 9) converges for |s| > M.

Recipe of Solving Differential Equation with Polynomial Coefficients
We now give a recipe of solving the DE with polynomial coefficients, which is given by where a l ∈ C, b l ∈ C and c l ∈ C for l ∈ Z [0,2] are constants.We then introduce the function p(t, s) by and express Equation (27) as In Sections 4 and 5, we discuss modified Kummer's DE given by Equation ( 5) and the hypergeometric DE given by Equation (10).
We obtain the following theorems, with the aid of Lemma 5.
Theorem 2. Let u(t) in the form of Equation (3) be the solution of Equation (29).Then û(s) given by Equation ( 9) is a solution of the DE: Theorem 3. Let û(s) in the form of Equation ( 9) be a solution of Equation (30).Then, the corresponding u(t) given by Equation ( 3) is a solution of Equation (29).
Corollary 4. Let û(s) be a solution of If the obtained û(s) satisfies then u(t) given by Equation ( 3) is a solution of Equation ( 29), that is of Equation (27).

Term-by-Term Operators for u(t) and û(s)
The AC of Riemann-Liouville fD: 0 D β R u(t) and the AC-Laplace transform series: L S [u(t)], of u(t) in the form of Equation ( 3), are defined by Equation ( 13) and Equation (14).We now define the operators which appear in Lemma 5.

Definition 4.
Let n ∈ Z >0 , β ∈ C and γ ∈ C, and let u(t) and û(s) be expressed as Equation (3) and Equation ( 14), respectively.Then, we adopt We now consider the theory of distributions in the spaces D R and D r,R , which are explained briefly in Sections 3.3 and 3.4, respectively.

Operational Calculus in the Spaces D R
In the theory of distributions in D R , a regular distribution is such a distribution that can be regarded as a function which is locally integrable on R. In the present paper, when the product u(t)H(t) ∈ L loc (R), we consider the regular distribution ũ(t) which is regarded to be equal to u(t)H(t).We then denote this correspondence between a distribution and a function by ũ(t) •−• u(t)H(t).Here, u(t)H(t) ∈ L loc (R) denotes that u(t)H(t) is locally integrable on R. From Equation (2), we have g 1 (t) = 1.We put H(t) = g1 (t), and then we obtain H(t) •−• g 1 (t)H(t) = H(t).Definition 5.In the space D R , the differential operator D λ of order λ ∈ C is defined as follows: The index law stated in the following lemma is valid.Lemma 6.Let h(t) ∈ D R .Then, the index law: is valid for every pair of λ ∈ C and µ ∈ C.
Dirac's delta function δ(t) ∈ D R is the distribution defined by δ(t) = D H(t).
We now introduce the correspondence even between a distribution, which is not a regular one, and a function, as follows.
We now define gν (t) for ν ∈ C by From Equations ( 2) and ( 4), we have g Remark 5. Note here that when Re ν ≤ −1, gν (t) is not a regular distribution.
where the derivative with respect to D is taken regarding D as a variable.
By Equations (37) and (36), we have Comparing this with Equations ( 4) and (38), we have where Definition 7. Let u(t) be expressed by Equation (3).Then, we define the distribution ũ(t) by in accordance with Equation (37). where then u(t) given by Equation ( 3) is a solution of Equation (29).

Operational Calculus in the Space D r,R
For the function H r (t) defined below Equation (12) in the Introduction section, we define the regular distribution Hr (t) in the space D r,R , so that Hr (t) •−• g 1 (t)H r (t) = H r (t), and then define δ r (t) ∈ D r,R by δ r (t) = D Hr (t) = δ(t) − δ(t − r).We can use the formulas presented in Section 3.1 for the space D R , also in the space D r,R , if we replace H(t), H(t) and δ(t), by H r (t), Hr (t) and δ r (t), respectively.
In Section 4, Theorem 1 applies.In Section 5, the rhs of Equation ( 44) for n = 0 is u(t)H r (t) = ∑ ν∈S u ν−1 g ν (t)H r (t), which converges at |t| < 1.Although û(s) converges for no value of s, we regard that the rhs of Equation (44) converges at |t| < 1, and the lhs of Equation ( 44) represents the corresponding regular distribution.The result obtained for u(t) is then the one obtained by the term-by-term inverse Laplace transform of û(s), which is expressed as Equation ( 14).

Distributions in the Space D R
Distributions in the space D are first introduced [21][22][23][24].The distributions are either regular ones or their derivatives.A regular distribution in D corresponds to a function which is locally integrable on R. The space D, that is dual to D , is the space of testing functions, which are infinitely differentiable on R and have a compact support.A distribution h ∈ D is a functional, to which h, φ ∈ C is associated with every φ ∈ D.
Definition 9. Let f (t) be a regular distribution in D and f (t) be the function corresponding to it.
for every φ(t) ∈ D , where Lemma 11.D n for n ∈ Z >−1 are operators in the space D .
In the present study, we consider functions g ν (t) and 0 D −λ R g ν (t) for ν ∈ R >0 and λ ∈ R >0 , and the distributions gν (t) and D −λ gν (t) corresponding to them.We then desire to introduce the operator D −λ W such that D −λ h, φ = h, D −λ W φ for h ∈ D and φ ∈ D. However, we find that D −λ W φ does not belong to D.
In this situation, we consider the problem in the space D R [12,13].A regular distribution in D R is such a distribution that it corresponds to a function which is locally integrable on R and has a support bounded on the left.The space D R , that is dual to D R , is the space of testing functions, which are infinitely differentiable on R and have a support bounded on the right.

Solution of Modified Kummer's DE
We now study the modified Kummer's DE given by Equation ( 5).We define p K (t, s) by Then Equation ( 5) is expressed as p K (t, d dt )u(t) = 0, t > 0, and the DE which corresponds to Equation (30) is where the rhs is evaluated by using the first equality and Equations ( 24) and ( 26) as 50) is 0, and the solution of Equation ( 50) is given by where 1 F 0 (a; s) is of the form of Equation ( 9), and we can take its inverse Laplace transform.Then, choosing C = Γ(2 − c), we obtain by using the formula . This is the solution given by Equation (7).By Corollary 4, we confirm the following lemma.
The derivation of this lemma based on the distribution theory is as follows.When 1 − c / ∈ Z <0 , ũ(t) = û(D)δ(t) takes the form of Equation (43), and satisfies p and then the corresponding solution u(t) is given by Equation (53), by Corollary 8.

Solution Satisfying u 0 = 1
In [4][5][6], the other solution given by Equation (6) of Equation ( 5) is obtained by solving the inhomogeneous Equation (50) for b = 0, 1 − c = 0 and u 0 = 1.We note here that the method using the following lemma with the aid of the solution given by Equation (53), which is already obtained, is easier.
is also a solution of the same equation.

Proof
We put u(t) = t λ w(t) in Equation ( 5), and then we obtain Equation ( 5) with a, c and u replaced by 1 + a − c, 2 − c and w, respectively, when λ = 1 − c.

Solution of the Hypergeometric DE
In solving the hypergeometric DE given by Equation (10), we introduce p H (t, s), by Now , we put û(s) = L S [u(t)] and then the DE which corresponds to Equation (30) is where the rhs is evaluated by using the first equality and Equations ( 24) and ( 26) as Lemma 16.Let û(s) be a solution of Equation (60), û(s) = s a−1 ŵ(s) and Then , ŵ(s) is a solution of the following DE: Proof We put û(s) = s λ v(s) in Equation (60).By using d ds [s λ v(s)] = s λ [λs −1 v(s) + d ds v(s)] and we obtain When λ = a and ŵ(s) = s • v(s), this gives Equation (63).
where K is any constant, and 2 F 0 (a, b; Proof For p W (t, s) given by Equation (62), we choose the DE for w(t) as follows: The solution of this DE is where C 1 is an arbitrary constant.ŵ(s) given by Equation ( 65) is the term-by-term Laplace transform of this w(t).Theorem 2 and Proposition 5 show that Equation ( 63) for (1 − c)u 0 = 0 is satisfied by this ŵ(s), since the first equality of Equation (63) shows p W (− d ds , s) ŵ(s) z = 0, If a − c / ∈ Z <0 , this w(t) takes the form of Equation ( 3), and the corresponding w(t), expressed as Equation ( 43), is given by w(t) = ŵ(D)δ 1 (t), when ŵ(s) is given by Equation ( 65) and Remark 7. We note that ŵ(s) given by Equation (65) converges for no value of s, but w(t) = ŵ(D)δ 1 (t) obtained by using this ŵ(s) corresponds to Equation (67).Moreover, we note that Equation ( 67) is obtained from Equation (65) by term-by-term inverse Laplace transform.

Solution Satisfying u 0 = 1
The other solution given by Equation ( 11) is obtained by solving the inhomogeneous Equation (63) for 1 − c = 0 and u 0 = 1.We note here that the method using the following lemma with the solution given by Equation (12), which is already obtained, is easier.
is also a solution of the same equation.

Proof
We put u(t) = t λ w(t) in Equation (10), and then we obtain Equation ( 10) with a, b, c and u replaced by 1 + a − c, 1 + b − c, 2 − c and w, respectively, when λ = 1 − c.
The solution satisfying νv 0 = 0 is given by v where C is an arbitrary constant.By the inverse Laplace transform, we obtain This satisfies

Solution of Hermite's DE
We now take up Hermite's DE: where n ∈ Z >−1 is a constant.The Hermite polynomial He n (t) and the Hermite function of the second kind he n (t) are solutions of this DE.In [11] (p. 82), they are expressed as for l ∈ Z >−1 .
In order to obtain these solutions by the present method, we put x = t 2 and u(t) = v(x) in Equation (78).Then, we obtain the DE for v(x): This DE is Equation (5) with t, u(t), c, b and a replaced by x, v(x), 1 2 , 1 2 and − 1 2 n, respectively.Corresponding to the solutions given by Equations ( 6) and (7) of Equation ( 5), we have the following solutions of Equation (81): ).
The solutions given by Equations ( 79) and (80) of Equation ( 78) are given by for l ∈ Z >−1 , where C k for k ∈ Z [1,4] are constants.

Conclusions
In the present paper, we are concerned with the problem of obtaining a solution u(t) of a DE with polynomial coefficients.
We know that by the basic method of solution [14] (Section 10.4), we usually obtain solutions in the form of Equation (54) which are a power of t multiplied by a power series in t.In the present paper, we are interested in obtaining the solutions with the aid of AC of Riemann-Liouville fD along with distribution theory or the Laplace transform or its AC.
We then set up the DE satisfied by the AC-Laplace transform, û(s), of u(t).The obtained DE for û(s) is found to be a DE with polynomial coefficients.
We now obtain the solution in the form of Equation (55) which is a power of s multiplied by a power series in s −1 , by solving the DE for û(s).When it converges at large |s|, it is the Laplace transform of a solution of the DE for u(t) or its AC.
In Section 4, we obtain such a solution û(s) that the series in it converges at |s| > |b| for b = 0.Then, we obtain the solution u(t) by term-by-term inverse Laplace transform by writing the distribution associated with the solution u(t), by using the obtained û(s).In Section 6, another example is given.
In Section 5, we obtain such a solution û(s) that the series in it converges for no value of s.Then, we obtain the solution u(t) by writing the distribution ũ(t) = û(D)δ 1 (t) associated with the solution u(t), by using the obtained non-convergent series of s −1 for û(s).The result is seen to be obtained by term-by-term inverse Laplace transform of û(s), as mentioned in Section 3.2.
We may conclude the study in this paper as follows.When we desire to obtain the solution u(t) of a linear DE with polynomial coefficients, in the form of Equation (3), we can obtain it, by setting up the DE for the AC-Laplace transform û(s), obtaining its solution in the form of Equation ( 9), and then taking its term-by-term inverse Laplace transform.
We can obtain the solution of the DE for û(s) by the basic method, obtaining it in the form of Equation (55).Comparing the solutions of the DE for u(t) and û(s), obtained by the basic method of solution, we find that the solution û(s) is the term-by-term Laplace transform of u(t).Thus, we obtain u(t) by the term-by-term inverse Laplace transform of û(s), when the latter is obtained.

(− 1 )Proposition 5 . 3 .
n d n ds n [s β û(s)] = ∑ ν∈S [(−1) n d n ds n (s β u ν−1 ĝν (s))].(35) Let the operators on the lhs of the equations in Definition 2, 3 and 4 be defined by the respective rhs.Then, Theorems 2 and 3 and Corollary 4 are valid.Operational Calculus in the Spaces D R and D r,R

Definition 10 .
Definition 9 with D and D replaced by D R and D R , respectively, is valid.Lemma 12. D ν for ν ∈ C are operators in the space D R .

Definition 11 .
to a function which is locally integrable on (−∞, r), and has a support bounded on the left.Definition 9 with D and D replaced by D r,R and D r,R , respectively, is valid.Lemma 13.D ν for ν ∈ C are operators in the space D r,R .Now the rhs of Equation (48) may be expressed as r
Lemma 9. Let u(t) be expressed by Equation (3), n ∈ Z >−1 and β ∈ C .Then ] z are defined by Equation (24) with − d ds and s replaced by − ∂ ∂D and D, respectively.