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We discuss the solution of Laplace’s differential equation by using operational calculus in the framework of distribution theory. We here study the solution of that differential Equation with an inhomogeneous term, and also a fractional differential equation of the type of Laplace’s differential equation.

Yosida [1,2] discussed the solution of Laplace’s differential equation (DE), which is a linear DE with coefficients which are linear functions of the variable. The DE which he takes up is

where and for are constants. His discussion is based on Mikusiński’s operational calculus [

In our preceding papers [4,5], we discuss the initial-value problem of linear fractional differential equation (fDE) with constant coefficients, in terms of distribution theory. The formulation is given in the style of primitive operational calculus, solving a Volterra integral equation with the aid of Neumann series.

Yosida [1,2] studied the homogeneous Equation (1.1), where he gave only one of the solutions by that method. One of the purposes of the present paper is to give the recipe of obtaining the solution of the inhomogeneous equation as well as the homogeneous one, in the style of operational calculus in the framework of distribution theory. With the aid of that recipe, we show how the set of two solutions of the homogeneous equation is attained.

Another purpose of this paper is to discuss the solution of an fDE of the type of Laplace’s DE, which is a linear fDE with coefficients which are linear functions of the variable. In place of (1.1), we consider

for and. Here for

is the Riemann-Liouville (R-L) fractional derivative defined in Section 2. We use to denote the set of all real numbers, and. When is equal to an integer,. When

, (1.2) is the inhomogeneous DE for (1.1). We use to denote the set of all integers, and

and for

satisfying. We use for, to denote the least integer that is not less than.

In Section 2, we prepare the definition of R-L fractional derivative and then explain how (1.2) is converted into a DE or an fDE of a distribution in distribution theory. A compact definition of distributions in the space and their fractional integral and derivative are described in Appendix A. A proof of a lemma in Section 2 is given in Appendix B. After these preparation, a recipe is given to be used in solving a DE with the aid of operational culculus 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. In Section 4, we apply the recipe to the DE where, of which special one is Kummer’s DE. This is an example which Yosida [1,2] takes up. In Section 5, we apply the recipe to the fDE with, assuming.

The discussion is done in the style of our preceding papers [4,5].

We use Heaviside’s step function, which we denote by. When is defined on, is assumed to be equal to when and to when.

Let be locally integrable on. We then define the R-L fractional integral of order by

where is the gamma function. The thus-defined is locally integrable on, and if.

We define the R-L fractional derivative of order, by

if it exists, where, and for.

We now assume that the following condition is satisfied.

Condition A is locally integrable on, and there exists for, and for are continuous and differentiable at, where. We then assume that there exists a finite value

for every.

Because of this condition, the Taylor series expansion of is given by

where is a function of as, so that as. By comparing (2.2)

and (2.4), we obtain.

We consider distributions belonging to. When a function is locally integrable on and has a support bounded on the left, it belongs to and is called a regular distribution. The distributions in are called right-sided distributions.

A compact formal definition of a distribution in and its fractional integral and derivative is given in Appendix A.

Let be a regular distribution. Then

for is also a regular distribution, and distribution is defined by

Let, and let be such a regular distribution that is continuous and differentiable on

, for every. Then is defined by

Let, for and

, be continuous and differentiable on, for every. Then

When is a regular distribution, is defined for all.

Lemma 1 For, the index law:

is valid for every.

Dirac’s delta function is the distribution defined by.

Lemma 2 Let for. Then

Proof By putting, , and in

(2.1), we obtain. By (2.5), we then have. By applying to this and using (2.6) and (2.8), we obtain (2.9).

We now adopt the following condition.

Condition B and are expressed as a linear combination of for.

Then and are expressed as

Lemma 3 Let exist for. Then the products and belong toand they are related by

Proof We obtain (2.11) from (2.4) by multiplying from the right and then applying. We first note due to (2.5).

Applying to this, we obtain the lefthand side of (2.11), and hence from the lefthand side of (2.4). We next note that

due to (2.6) and as noted after

(2.4). Thus we obtain the first term on the righthand side of (2.11) from the last term of (2.4). As to the remaining terms, we only use (2.9).

Lemma 4 Let. Then

The last derivative with respect to is taken regarding as a variable.

Proof of Lemma 4 for. Let,. Then by (2.9), we have

by using (2.9) repeatedly.

A proof of this lemma for is given in Appendix B.

The following lemma is a consequence of this lemma.

Lemma 5 Let satisfy Condition B. Then

Proof By using (2.10) and (2.13), we obtain

We now express the DE/fDE (1.2) to be solved, as follows:

where or, and. In Sections 4 and 5, we study this DE for and this fDE for, respectively.

Using Lemma 3, we express (3.1) as

where

By using (2.10) and (2.13), we express (3.2) as

where

In order to solve the Equation (3.4) for, we solve the following equation for function of real variable:

Lemma 7 The complementary solution (C-solution) of Equation (3.7) is given by, where is an arbitrary constant and

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

Lemma 8 Let be the C-solution of (3.7), and let the particular solution (P-solution) of (3.7) be when the inhomogeneous part is for. Then

where is any constant.

Since satisfies Condition B and is given by (3.6), the P-solution of (3.7) is expressed as a linear combination of for and for.

From the solution of (3.7), is obtained by substituting by. Then we confirm that (3.4) is satisfied by that applied to.

Finally the obtained expression of is expanded into the sum of terms of negative powers of, and then we obtain the solution of (3.4). If the obtained is a linear combination of for, is converted to the solution of (3.2) by using (2.10) and (2.9). It becomes a solution of (3.1) for.

1) We prepare the data: by (2.10), and, and by (3.5) and (3.6).

2) We obtain by (3.8). If, the Csolution of (3.1) is given by

3) If or, we obtain given by (3.9).

4) If, the C-solution of (3.1) is given by

where are constants.

5) If, the P-solution of (3.1)

is given by

where and are constants.

In the above recipe, we first obtain the C-solution of (3.7), that is. It gives the C-solution of (3.4) and hence the C-solutions of (3.2) and of (3.1).

We next obtain the P-solution of (3.7) when the inhomogeneous part is for. As noted above, the P-solutions of (3.7) for and for, are expressed as a linear combination of for and of for, respectively. The sum of the P-solutions of (3.7) for and for gives the P-solution of (3.4) and hence the P-solution of (3.2). The C-solution of (3.1) comes from the C-solution of (3.7) and the P-solution of (3.7) for.

When we obtain at the end of Section 3.2, we must examine whether it is compatible with Condition B. We will find that if for, the obtained is not acceptable. Hence we have to solve the problem, assuming that for all.

When and, we put. When

and, we put. Discussion of this problem is given in Appendices C and D.

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

By (3.5) and (3.6), , and are

where.

In order to obtain the C-solution of (3.7) by using (3.8), we express as follows:

where

is now expressed as

.

By using (3.8), we obtain

where for and

are the binomial coefficients. Here

for and, and.

The C-solution of (3.4) is given by

If, Condition B is satisfied. Then by using (2.9), we obtain the C-solution of (3.2):

Remark 1 In [6,7], Kummer’s DE is given, which is equal to the DE (4.1) for, , and. In this case,

We then confirm that the expression (4.8) agrees with one of the C-solutions of Kummer’s DE given in those books.

We now obtain the P-solution of (3.7) when the inhomogeneous part is equal to for.

When the C-solution of (3.7) is given by (4.6), the P-solution of (3.7) is given by (3.9). By using (4.2) and (4.6), we obtain

where

Lemma 9 defined by (4.11) is expressed as

Proof Equation (4.10) shows that the P-solution of (3.7) is now expressed as

where. Substituting this into (3.7), we obtain an equation which states that a power series of is equal to 0. By the condition that the coefficient of every power must be 0, we obtain a recurrence equation for the coefficients:

By using this repeatedly, we have

By comparing (4.10), (4.13) and (4.16), we obtain (4.12).

Equation (4.10) shows that if the inhomogeneous part is for, the P-solution of (3.2) is given by

By using (4.12) in (4.17), we obtain

By (4.3),. When the inhomogeneous part is, the P-solution of (3.7) is given by

By using (4.18) for, we obtain

Proposition 1 Let and.

Then the complementary solution of (4.1), multiplied by, is given by the sum of the righthand sides of (4.8) and of (4.21).

Remark 2 As stated in Remark 1, in [6,7], the result for, , and, is given. In this case, and are given in (4.9), and

We then confirm that the set of (4.8) and (4.21) agrees with the set of two C-solutions of Kummer’s DE given in those books.

In this section, we consider the case of, ,

, , , and.

Then the Equation (3.1) to be solved is

Then (3.5) and (3.6) are expressed as

where.

By using (5.2), is expressed as

where

By (3.8), the C-solution of (3.7) is given by

The C-solution of (3.2) is given by

By Condition B, we have to require

.

Then by using (2.9) in (5.7), we obtain

The C-solution of (5.1) is equal to this for.

By using the expressions of and given by (5.2) and (5.6) in (3.9), we obtain the P-solution of (3.7) when the inhomogeneous part is:

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

By using (4.12) in (5.9), we can show that if the inhomogeneous part is for, the P-solution of (3.2) is given by

This for gives the P-solution of (5.1)when the inhomogeneous part is for.

A right-sided distribution is a functional for which a number is associated with all, where is the space of infinitely differentiable functions which is defined on and has a support bounded on the right.

A regular right-sided distribution is a locally integrable function on, which has a support bounded on the left, and is given by

Let. If, the fractional integral is

and if, the fractional derivative is given by

where. We set, and

for.

In this place, we can confirm that the index law

is valid for every.

For a distribution, for is defined by

The index law (2.8) follows from (A.4) by (A.5).

Dirac’s delta function is defined by, as stated just below Lemma 1, and hence

It is customary to use the notation:

Let and. Then is defined by

Here we give a proof of Lemma 4 for, with the aid of notations explained in Appendix A.

Let, and. Then

Using Lemma 4 for in the last member, we obtain

Formula (2.12) for follows from this.

We now consider the DE (3.1) for and. Then (3.5) and (3.6) are expressed as

where

In solving (3.7), we express as

where and are constants. In Section 4.1, we assume that and obtain the C-solution given by (4.8) which satisfies Condition B. In the presence of the first term on the righthand side of (C.5), we will see that we cannot obtain a solution satisfying Condition B. Hence we have to assume.

In this section, we consider the fDE (3.1) for

and. Then (3.5) and (3.6) are expressed as

where are given by (C.4).

In solving (3.7), we express as

where and are constants. In Section 5.2, we assume that and obtain the C-solution given by (5.8) which satisfies Condition B. In the presence of the first two terms on the righthand side of (D.4), we will see that we cannot obtain a solution satisfying Condition B. Hence we have to assume.