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

We discuss the solution of Laplace’s differential equation and a fractional differential equation of that type, by using analytic continuations of Riemann-Liouville fractional derivative and of Laplace transform. We show that the solutions, which are obtained by using operational calculus in the framework of distribution theory in our preceding papers, are obtained also by the present method.


Introduction
Yosida [1] [2] discussed the solution of Laplace's differential equation, which is a linear differential equation, with coefficients which are linear functions of the variable.In recent papers [3] [4], we discussed the solution of that equation, and fractional differential equation of that type.The differential equations are expressed as We use  ,  and  , to denote the sets of all real numbers, of all integers and of all complex numbers.We also use { } ( ) ( ) We use x     for x ∈  , to denote the least integer that is not less than x.In the present paper, the variable t is always assumed to take values on 0 >  .Yosida [1] [2] studied the Equation (1.1) for 1 σ = with ( ) 0 f t = , by using Mikusiński's operational calculus [5].In [3] [4], operational calculus in terms of distribution theory is used, which was developed for the initial-value problem of fractional differential equation with constant coefficients in our preceding papers [6] [7].In [3], the derivative is the ordinary Riemann-Liouville fractional derivative, so that the fractional derivative of a function ( ) u t exists only when ( ) u t is locally integrable on 0 >  , and the integral ( ) Practically, we adopt Condition B in [3], which is ( ) (1) are expressed as a linear combination of ( ) , where ( ) ν Γ is the gamma function.
We then express ( ) u t as follows: ( ) ( ) and 1 S is a set of 0 ν > ∈  .In a recent review [8], we discussed the analytic continuations of fractional derivative, where an analytic continuation of Riemann-Liouville fractional derivative of function ( ) u t is such that the fractional derivative exists when ( ) u t is locally integrable on 0 >  , even when the integral ( ) In [4], we adopted this analytic continuation of Riemann-Liouville fractional derivative, and the following condition, in place of Condition 1.

Condition 2. ( )
( ) f t in (1.1) are expressed as a linear combination of ( ) for some We then express ( ) u t as follows; ( ) ( ) In [3] [4], we take up Kummer's differential equation as an example, which is where , a c ∈  are constants.If c ∉  , one of the solutions given in [9] [10] is where ( ) ( ) ( ) In [3], if 2 c < , we obtain both of the solutions.But when 2 c ≥ , (1.7) does not satisfy Condition 1 and we could not get it in [3].In [4], we always obtain both of the solutions.In [1] [2], Yosida obtained only the solution (1.7).
We now study the solution of a differential equation with the aid of Laplace transform.Then it is required that there exists the Laplace transform of the function ( ) The Laplace transform, ( ) , of ( ) Let ( ) u t expressed by (1.3) satisfy Condition 3, and let its Laplace transform ( ) u s  be given by ( ) Then we can show that we are able to solve the problems solved in [3], with the aid of Laplace transform.
When ( ) u t satisfies Conditions 2 and 3, Laplace transform is not applicable.In [4], we adopted an analytic continuation of Riemann-Liouville fractional derivative, by which we could solve the differential equation assuming Condition 2. The analytic continuation is achieved with the aid of Pochhammer's contour, which is used in the analytic continuation of the beta function.
We now introduce the analytic continuation of Laplace transform with the aid of Hankel's contour, which is used in the analytic continuation of the gamma function.We then show that (1.9) is valid for that if ( ) u t expressed by (1.4) satisfies Condition 3, and its analytic continuation of Laplace transform of ( ) u t , which we denote by ( ) then we can solve the problems solved in [4], with the aid of the analytic continuation of Laplace transform.
In Section 2, we prepare the definition of analytic continuations of Riemann-Liouville fractional derivative and of Laplace transform, and then explain how the equation for the function ( ) u t and its fractional derivative in (1.1) are converted into the corresponding equation for the analytic continuation of Laplace transform, ( ) of ( ) u t , and also how ( ) û s is converted back into ( ) u t .After these preparations, a recipe is given to be used in solving the fractional differential Equation (1.1) with the aid of the analytic continuation of Laplace transform in Section 3. In this recipe, the solution is obtained only when 2 0 a ≠ and 2 0 b = .When 1 2 σ = , 1 0 b = is also required.An explanation of this fact is given in Appendices C and D of [3].In Section 4, we apply the recipe to (1.1) where 1 σ = and 0 0 a = , of which special one is Kummer's differential equation.In Section 5, we apply the recipe to the fractional differential equation with 1 2 σ = , assuming 0 0 a = .
In Section 6, comments are given on the relation of the present method with the preceding one developed in [4], and on the application of the analytic continuation of Laplace transform to the differential equations with constant coefficients.

Analytic Continuation of Riemann-Liouville Fractional Derivative
Let a function ( ) We then define the Riemann-Liouville fractional derivative, defined by (2.3) is its analytic continuation to the whole complex plane.If we assume that ν also takes a complex value, defined by (2.3) is an analytic function of ν in the domain Re 0 ν > .The analytic continuation as a function of ν was also studied.The argument is concluded that (2.4) should apply for the analytic continuation via Pochhammer's contour, even in Re 0 ν ≤ except at the points where 1 ν < ∈  ; see [8].We now adopt this analytic continuation of , and hence we accept the following lemma.
For ( ) u t defined by (1.4), we note that

Analytic Continuation of Laplace Transform
The gamma function ( ) for z ∈  satisfying Re 0 z > , is defined by Euler's second integral: The analytic continuation of ( ) for z ∈  is given by Hankel's formula: where H C is Hankel's contour shown in Figure 1(a).
We now define an integral transform ( ) ( ) f t which satisfies the following condition.

Condition 4. ( )
f z is expressed as ( ) ( ) ( ) Let ( ) f t satisfy Conditions 3 and 4. Then we define ( ) Lemma 3. Let ( ) f t satisfy Condition 4. Then ( ) f s defined by (2.8) is an analytic continuation of ( ) f s  , which is defined by (1.8) for Re 0 γ > , as a function of γ .
Proof.The equality ( ) ( ) when Re 0 γ > is proved in the same way as the equality of ( ) z Γ given by (2.6) and by (2.7) for 0 > R z e ; see e.g. ( [11], Section 12.22).The analyticity of ( ) f s and of ( ) Proof.Formula (2.8) for ( ) ( )  .The last equality in (2.13) is due to (1.9).By using (2.1) and (2.10), the lefthand side of (2.11) is expressed as . By replacing s , γ and ζ in (2.13), by t , , respectively, we obtain the first equality of (2.12) for ν ∉  , with the aid of the formula ( ) ( ) ( ) ( ) ( ) where C I is a contour shown in Figure 1(b).Here it is assumed that ( ) f s is analytic to the right of the vertical line on C I , and is so above and below the upper and lower horizontal lines, respectively, on C I .
Proof.For ( ) f t , the usual Laplace inversion formula applies, so that ( ) ( ) ( ) where C B is a contour shown in Figure 1(b).Here it is assumed that ( ) f s is analytic to the right of the contour C B .By using this with (2.10) and (2.12), we confirm (2.15). Lemma 5. Let ( ) f t satisfy Conditions 3 and 4 with an entire function ( ) û s be expressed in the form of (1.11).Then the Laplace inversion ( ) provided that the obtained ( ) u t satisfies the conditions for ( ) f t in Lemma 5, or it is a linear combination of such functions.

Lemma 7. Let ( )
u t satisfy the conditions for ( ) Proof.By using (2.5) and Lemma 4, we obtain these results.

Recipe of Solving Laplace's Differential Equation and Fractional Differential Equation of That Type
We now express the differential Equation (1.1) to be solved, as follows: We now apply the integral transform H  to (3.1).By using (2.18) and (2.19), we then obtain , Lemma 8.The complementary solution (C-solution) of Equation (3.2) is given by ( ) ( ) , where C 1 is an arbitrary constant and where the integral is the indefinite integral and C 2 is any constant.where C 3 is any constant.
Since ( ) f t in (3.1) satisfies Condition 2 and ( ) v s is given by (3.4), the P-solution ( ) 2) is expressed as a linear combination of ( ) ∈  , and of ( ) The solution ( ) û s of (3.2) is converted to a solution ( ) u t of (3.1) for 0 t > , with the aid of Lemma 6.

Particular Solution of (3.2)
We now obtain the P-solution of (3.2), when the term is equal to s ν − for ν ∈  .

σ
are the Riemann-Liouville fractional derivatives to be defined in Section 2.

a
= .The other solution is t be locally integrable on b >  for b ∈  , and let then define the Riemann-Liouville fractional integral,
. Then the Laplace inversion formula is given by the contour integral: In Sections 4 and 5, we study this differential equation for 1 σ = and this fractional differential equation for 1 2 σ = , respectively.

1 )
In order to obtain the C-solution ( ) ˆs φ of (3.2) by using (3.5), we express ( ) ( ) B s A s as follows:

9 )>∉
We then confirm that the expression (4.8) for 1 c  agrees with (1.7), which is one of the C-solutions of Kummer's differential equation given in[9] [10].
proved in[3].In fact, (4.11) is the partial fraction expansion of applying Lemma 6 to this, we obtain the C-solution of (5.1):