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

Abstract

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.

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Morita, T. and Sato, K. (2014) Solution of Differential Equations with the Aid of an Analytic Continuation of Laplace Transform. Applied Mathematics, 5, 1229-1239. doi: 10.4236/am.2014.58115.

1. 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

(1.1)

for and. Here for are constants, and are the Riemann-Liouville fractional derivatives to be defined in Section 2.

We use, and, to denote the sets of all real numbers, of all integers and of all complex numbers. We also use, , and

for and satisfying. If, , and

. We use for, 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.

Yosida [1] [2] studied the Equation (1.1) for with, 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 exists only when is locally integrable on, and the integral converges.

Practically, we adopt Condition B in [3] , which is Condition 1. and in (1) are expressed as a linear combination of for.

Here is defined by

(1.2)

for, where is the gamma function.

We then express as follows:

(1.3)

where are constants, and is a set of.

In a recent review [8] , we discussed the analytic continuations of fractional derivative, where an analytic continuation of Riemann-Liouville fractional derivative of function is such that the fractional derivative exists when is locally integrable on, even when the integral diverges.

In [4] , we adopted this analytic continuation of Riemann-Liouville fractional derivative, and the following condition, in place of Condition 1.

Condition 2. and in (1.1) are expressed as a linear combination of for, where S is a set of for some.

We then express as follows;

(1.4)

In [3] [4] , we take up Kummer’s differential equation as an example, which is

(1.5)

where are constants. If, one of the solutions given in [9] [10] is

(1.6)

where for and, and. The other solution is

(1.7)

In [3] , if, we obtain both of the solutions. But when, (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 to be determined.

When we consider the Laplace transform of a function which is locally integrable on, we assume the following condition.

Condition 3. There exists some such that as.

Let be locally integrable on and satisfy Condition 3, and the integral converge. We then denote its Laplace transform by, so that

(1.8)

The Laplace transform, , of for is then given by

(1.9)

Let expressed by (1.3) satisfy Condition 3, and let its Laplace transform be given by

(1.10)

Then we can show that we are able to solve the problems solved in [3] , with the aid of Laplace transform.

When 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, and that if expressed by (1.4) satisfies Condition 3, and its analytic continuation of Laplace transform of, which we denote by, is given by

(1.11)

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 and its fractional derivative in (1.1) are converted into the corresponding equation for the analytic continuation of Laplace transform, , of, and also how is converted back into. 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 and. When, 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 and, of which special one is Kummer’s differential equation. In Section 5, we apply the recipe to the fractional differential equation with, assuming.

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.

2. Formulas

Lemma 1. Let be defined by (1.2). Then for,

(2.1)

Proof. By (1.2), for,. ,

2.1. Analytic Continuation of Riemann-Liouville Fractional Derivative

Let a function be locally integrable on for, and let exist. We then define the Riemann-Liouville fractional integral, , of order by

(2.2)

We then define the Riemann-Liouville fractional derivative, , of order, by

(2.3)

if it exists, where, and for.

For and, we have

(2.4)

If we assume that takes a complex value, by definition (2.2) is analytic function of in the domain, and defined by (2.3) is its analytic continuation to the whole complex plane. If we assume that also takes a complex value, defined by (2.3) is an analytic function of in the domain. The analytic continuation as a function of was also studied. The argument is concluded that (2.4) should apply for the analytic continuation via Pochhammer’s contour, even in except at the points where; see [8] .

We now adopt this analytic continuation of to represent, and hence we accept the following lemma.

Lemma 2. Formula (2.4) holds for every,.

By (1.4) and (2.4), we have

(2.5)

For defined by (1.4), we note that is locally integrable on.

2.2. Analytic Continuation of Laplace Transform

The gamma function for satisfying, is defined by Euler’s second integral:

(2.6)

The analytic continuation of for is given by Hankel’s formula:

(2.7)

where is Hankel’s contour shown in Figure 1(a).

We now define an integral transform of a function which satisfies the following condition.

Condition 4. is expressed as on a neighborhood of, for, where, and is analytic on the neighborhood of.

Let satisfy Conditions 3 and 4. Then we define for, by, where

(2.8)

for, and

(2.9)

for.

Lemma 3. Let satisfy Condition 4. Then defined by (2.8) is an analytic continuation of, which is defined by (1.8) for, as a function of.

Proof. The equality when is proved in the same way as the equality of given by (2.6) and by (2.7) for; see e.g. ([11] , Section 12.22). The analyticity of and of is proved as in ([11] , Sections 5.31, 5.32). Lemma 4. For,

(a)(b)

Figure 1. (a) Hankel’s contour CH, and (b) contours CI, −CH and CB which appear in (2.12), (2.15) and (2.16).

(2.10)

(2.11)

(2.12)

where and are two of the contours shown in Figure 1(b).

Proof. Formula (2.8) for gives

(2.13)

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, and in (2.13), by, and, respectively, we obtain the first equality of (2.12) for, with the aid of the formula. The equality for is obtained by continuity. Theorem 1. Let satisfy Conditions 3 and 4 for, and be expressed as

(2.14)

where is a finite set of. Then the Laplace inversion formula is given by the contour integral:

(2.15)

where CI is a contour shown in Figure 1(b). Here it is assumed that is analytic to the right of the vertical line on CI, and is so above and below the upper and lower horizontal lines, respectively, on CI.

Proof. For, the usual Laplace inversion formula applies, so that

(2.16)

where CB is a contour shown in Figure 1(b). Here it is assumed that is analytic to the right of the contour CB. By using this with (2.10) and (2.12), we confirm (2.15). Lemma 5. Let satisfy Conditions 3 and 4 with an entire function. Then

(2.17)

Lemma 6. Let be expressed in the form of (1.11). Then the Laplace inversion is given by (1.4), provided that the obtained satisfies the conditions for in Lemma 5, or it is a linear combination of such functions.

Lemma 7. Let satisfy the conditions for in Lemma 5. Then

(2.18)

(2.19)

Proof. By using (2.5) and Lemma 4, we obtain these results. ,

3. 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:

(3.1)

where or, and. In Sections 4 and 5, we study this differential equation for and this fractional differential equation for, respectively.

We now apply the integral transform to (3.1). By using (2.18) and (2.19), we then obtain

(3.2)

where

(3.3)

(3.4)

Here.

Lemma 8. The complementary solution (C-solution) of Equation (3.2) is given by, where C1 is an arbitrary constant and

(3.5)

where the integral is the indefinite integral and C2 is any constant.

Lemma 9. Let be the C-solution of (3.2), and be the particular solution (P-solution) of (3.2), when the inhomogeneous term is for. Then

(3.6)

where C3 is any constant.

Since in (3.1) satisfies Condition 2 and is given by (3.4), the P-solution of (3.2) is expressed as a linear combination of for for, and of for, respectively.

The solution of (3.2) is converted to a solution of (3.1) for, with the aid of Lemma 6.

4. Laplace’s and Kummer’s Differential Equations

We now consider the case of, , , , and. Then (3.1) reduces to

(4.1)

By (3.3) and (3.4), , and are

(4.2)

(4.3)

4.1. Complementary Solution of (3.2) and (4.1)

In order to obtain the C-solution of (3.2) by using (3.5), we express as follows:

(4.4)

where

(4.5)

By using (3.5), we obtain

(4.6)

in the form of (2.17) or (1.11), where for and are the binomial coefficients.

If, we obtain a C-solution of (4.1), by using Lemma 6:

(4.7)

(4.8)

Remark 1. In Introduction, Kummer’s differential equation is given by (1.5). It is equal to (4.1) for, , and. In this case,

(4.9)

We then confirm that the expression (4.8) for agrees with (1.7), which is one of the C-solutions of Kummer’s differential equation given in [9] [10] .

4.2. Particular Solution of (3.2)

We now obtain the P-solution of (3.2), when the inhomogeneous term is equal to for.

When the C-solution of (3.2) is, the P-solution of (3.2) is given by (3.6). By using (4.2) and (4.6), the following result is obtained in [3] :

(4.10)

where

(4.11)

Lemma 10. When, defined by (4.11) is expressed as

(4.12)

where

(4.13)

This lemma is proved in [3] . In fact, (4.11) is the partial fraction expansion of given by (4.12) as a function of.

Applying Lemma 6 to (4.10), and using (4.12), we obtain Theorem 2. Let, , and let for. Then we have a P-solution of (4.1), given by

(4.14)

where

(4.15)

(4.16)

Here.

4.3. Complementary Solution of (4.1)

By (4.3) and (4.5),. When and, the P-solution of (3.2) is given by

(4.17)

By using (4.14) for, if, we obtain a C-solution of (4.1):

(4.18)

In Section 4.1, we have (4.7), that is another C-solution of (4.1). If we compare (4.7) with (4.15), when, it can be expressed as

(4.19)

Proposition 1. When, the complementary solution of (4.1) is given by the sum of the righthand sides of (4.8) and of (4.18), which are equal to and, respectively.

Remark 2. As stated in Remark 1, for Kummer’s differential equation, and are given in (4.9), and

(4.20)

We then confirm that if, the set of (4.8) and (4.18) agrees with the set of (1.6) and (1.7).

5. Solution of Fractional Differential Equation (3.1) for

In this section, we consider the case of, , , , and. Then the Equation (3.1) to be solved is

(5.1)

Now (3.3) and (3.4) are expressed as

(5.2)

(5.3)

5.1. Complementary Solution of (3.2)

By using (5.2), is expressed as

(5.4)

where

(5.5)

By (3.5), the C-solution of (3.2) is given by

(5.6)

If, by applying Lemma 6 to this, we obtain the C-solution of (5.1):

(5.7)

5.2. Particular Solution of (3.2) and (5.1)

By using the expressions of and given by (5.2) and (5.6) in (3.6), we obtain the P-solution of (3.2), when the inhomogeneous term is for satisfying:

(5.8)

where is defined by (4.11) and is given by (4.12).

By applying Lemma 6 to this, we obtain the following theorem.

Theorem 3. Let, , and let. Then we have a P-solution of (5.1), given by

(5.9)

where

(5.10)

5.3. Complementary Solution of (5.1)

We obtain the solution only for. When is given by (5.3) with nonzero values of, Theorem 2 does not give a solution of (5.1). Hence given by (5.7) is the only C-solution of (5.1).

If we compare (5.7) with (5.10), we obtain the following proposition.

Proposition 2. Let. Then the C-solution of (5.1) is given by

(5.11)

6. Concluding Remarks

6.1. Solution with the Aid of Distribution Theory

In [4] , the solution of (3.1) is assumed to be expressed as (1.4). In distribution theory, the differential equation for the distribution is set up, where for and for. Then it is expressed as, where is the derivative of order in the space of

of distributions. In [4] , after obtaining, is obtained by using Neumann series expansion. In the present paper, is the analytic continuation of Laplace transform of. After obtaining, we obtain for by Laplace inversion.

The steps of solution in [4] and the present paper are closely related with each other, and one may use a favorite one. One difference is that Condition 3 is assumed in the present paper but is not required in [4] .

6.2. Solutions of Differential Equations with Constant Coefficients

We now consider the differential equation given by (3.1), where. Then assuming that the solution and the inhomogeneous term satisfy Conditions 2 and 3, we show that the analytic continuation of Laplace transform of that equation is given by (3.2) with. We then obtain the analytic continuation of Laplace transform of,. If it can be expressed as (1.11), then is given by its Laplace inverse (1.4). If we take account of Section 6.1, we confirm that the results obtained in [7] are obtained by Laplace inversion.

Conflicts of Interest

The authors declare no conflicts of interest.

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