Solution of Inhomogeneous Fractional Differential Equations with Polynomial Coefficients in Terms of the Green’s Function, in Nonstandard Analysis

Discussions are presented by Morita and Sato in Mathematics 2017; 5, 62: 1–24, on the problem of obtaining the particular solution of an inhomogeneous ordinary differential equation with polynomial coefficients in terms of the Green’s function, in the framework of distribution theory. In the present paper, a compact recipe in nonstandard analysis is presented, which is applicable to an inhomogeneous ordinary and also fractional differential equation with polynomial coefficients. The recipe consists of three theorems, each of which provides the particular solution of a differential equation for an inhomogeneous term, satisfying one of three conditions. The detailed derivation of the applications of these theorems is given for a simple fractional differential equation and an ordinary differential equation.


Introduction
We consider a fractional differential equation, which takes the following form: where n ∈ Z >−1 , a l (t) for l ∈ Z [0,n] are polynomials of t, ρ l ∈ C for l ∈ Z [0,n] satisfy Re ρ 0 > Re ρ 1 ≥ · · · ≥ Re ρ n and Re ρ 0 > 0. We use Heaviside's step function H(t), which is equal to 1 if t > 0 and, to 0 if t ≤ 0. Here, R D Remark 1. In solving Equation (1), we assume that f (t) satisfies f (t) = f (t)H(t), and we use the solution u(t) = 0 for t ≤ 0. As a result, the solution u(t) satisfies u(t) = u(t)H(t). In this notation, we use the following definition for τ = 0.
Here, Z, R and C are the sets of all integers, all real numbers and all complex numbers, respectively, and 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 also use R >a = {x ∈ R|x > a} for a ∈ R, and C + = {z ∈ C|Re z > 0}.
In accordance with Definition 1, we adopt the following: for ν ∈ C\Z <1 . Here R D t is used in place of usually used notation 0 D R , in order to show that the variable is t.
In [3,4], discussions are made of an ordinary differential equation, which is expressed by (1) for ρ l = n − l, in terms of distribution theory, and with the aid of the analytic continuation of Laplace transform, respectively. In those papers, solutions are given of differential equations with an inhomogeneous term f (t), which satisfies one of the following three conditions.

Recipe of Solution of Differential Equation, in Distribution Theory
In a recent paper [5], the solution of Euler's differential equation in distribution theory is compared with the solution in nonstandard analysis. In distribution theory [6][7][8], we use distributionH(t), which corresponds to function H(t), differential operator D and distribution δ(t) = DH(t), which is called Dirac's delta function.

Lemma 2.
In [3], it was shown that if f (t) satisfies Condition 1 (i) and G 0 (t, τ) is the one given in Lemma 1, u f (t), given by the following, is a particular solution of Equation (1).
A proof of this lemma is given to show that if G 0 (t, τ) satisfies (7), u f (t) given by (8) is a particular solution of (1), in Section 1.4.

Preliminaries on Nonstandard Analysis
In the present paper, we use nonstandard analysis [9], where infinitesimal numbers are used. We denote the set of all infinitesimal real numbers by R 0 . We also use R 0 >0 and N ∈ Z >0 , then < 1 N . We use R ns , which has subsets R and R 0 . If x ∈ R ns and x / ∈ R, x is expressed as x 1 + by x 1 ∈ R and ∈ R 0 , where x 1 may be 0 ∈ R. Equation x y for x ∈ R ns and y ∈ R ns , is used, when x − y ∈ R 0 . We denote the set of all infinitesimal complex numbers by C 0 , which is the set of complex numbers z, which satisfy |Re z| + |Im z| ∈ R 0 . We use C ns , which has subsets C and C 0 . If z ∈ C ns and z / ∈ C, z is expressed as z 1 + by z 1 ∈ C and ∈ C 0 , where z 1 may be 0 ∈ C. Remark 3. In nonstandard analysis [9], in addition to infinitesimal numbers, we use unlimited numbers, which are often called infinite numbers. In the present paper, we do not use them, but if we use them, we have to consider sets R ∞ and C ∞ such that if ω ∈ R ∞ , there exists ∈ R 0 satisfying ω = 1 , and if ω ∈ C ∞ , there exists ∈ C 0 satisfying ω = 1 , and to include these sets as subsets of R ns and C ns , respectively.
In place of (4), we now use the following: for all ρ ∈ C and ν ∈ C.
Remark 4. When ∈ R 0 or ∈ C 0 , we often ignore terms of O( ) compared with a term of O( 0 ). For instance, when ν ∈ R >0 and ν − ρ ∈ R >0 , we adopt 1 in place of (9). In the following, we often use "=" in place of " ".
which is

Summary of Sections 2-6
In Section 2, we give recipe of solution of a differential equation, in nonstandard analysis, where the differential equation is the following: and the inhomogeneous termf (t) satisfies one of the following three conditions.
A complementary solution of Equation (1) is also that of Equation (15).
In [10], an ordinary differential equation is expressed in terms of blocks of classified terms. When the equation is expressed by two blocks of classified terms, the complementary solutions are obtained by using Frobenius' method. In the present case, a block of classified terms D ρ u(t) for ρ ∈ C and takes the following form: where µ ∈ C + and k x ∈ Z >−1 . An equation which consists of two blocks of classified terms, is expressed by the following: where ρ ∈ C + and l ∈ Z >0 . In [11,12], discussions are made of fractional differential equations with constant coefficients, where the differential equation for the Green's function is solved either by an operational calculus giving the Neumann series, or by changing it to an integral equation which is solved by iterations, following the discussion for the ordinary differential equation given in the book [13]. Kim and O [14] present the corresponding argument for a fractional differential equation of the form, where an example of a simple equation for which the Green's function was given.
It is the primary purpose of this paper to present the recipe in Section 2 and to give a derivation of full expressions of the Green's functions of a fractional differential equation: and of an associated ordinary differential equation: where a ∈ C and b ∈ C are constants. These studies are given in Sections 2.2 and 3-3.4, respectively. In Section 3.4, the operational calculus or the method of iterations is used. Equations (18) and (19) for b = 0 are expressed by the following: which take the form (17) with µ = 1, ρ = 3 2 and l = 2, and with µ = 1, ρ = 1 and l = 2, respectively. The equation, which Kim and O treated in [14], is Equation (20) for a = −1. The particular solutions of Equations (20) and (21) with the aid of Theorems 2 and 3, are presented in Sections 5-5.5 and 5.7.
In Section 4.1, we consider the following differential equation satisfying Condition 2 (iii): where ρ ∈ C + , λ ∈ R >0 , α ∈ R >0 and β ∈ C. This equation takes the form (17) and its solution is derived by the operational calculus or the method of iterations.

Proof of Lemma 2
We write (7) as , and then by using (8), we have the following: By taking the derivatives of the third and the last member in this equation with respect to t, we confirm that (1) is satisfied by u(t) = u f (t).

Solution of Equation (15) When Condition 2 (i) Is Satisfied
In obtaining a particular solution of Equation (15) forf (t) satisfying Condition 2 (i), in place of the Green's function defined in Remark 2, we use it as defined in the following definition.
Proof. We conclude this since in the limit (24) and (25) tend to 0 and 1, respectively, at t > τ.
In Sections 3.2-3.5, we use the following theorem.
Theorem 1. Let Condition 2 (i) be satisfied and G 1 (t, τ) and G 0 (t, τ) be given as in Lemma 5. Then u f (t), which is given by the following: in the limit 1 → 0, is a particular solution of Equation (15) for the termf (t).

Solution of Equation (15) When Condition 2 (ii) or (iii) Is Satisfied
Definition 3. When Condition 2 (ii) or (iii) is satisfied, we introduce a transformed differential equation of Equation (15), by the following: where Lemma 6. Let Condition 2 (ii) or (iii) be satisfied. Then, by Definition 3, when (28) holds, Definition 4. For ∈ R 0 >0 , the Green's function G β, (t, τ) of Equation (28) satisfies the following: Lemma 7. Let G β, (t, τ) be defined by Definition 4. Then, a particular solution of Equation (28) for the termf β (t) is given by the following: if Condition 2 (ii) is satisfied, and by w Proof. When Condition 2 (ii) is satisfied, by using Equations (32), (30) and (13), we obtain the following: is a particular solution of (28).
Remark 5. Lemma 9 shows that we have the following relations: By using these relations, we can obtain any one of G β, (t, τ), G β, , 1 (t, τ) and G β, ,0 (t, τ), from any other. Lemma 10. Let Condition 2 (ii) be satisfied, and G β, , 1 (t, τ) be defined in Lemma 9. Then a particular solution of Equation (34) for the term R D is given by the following: Lemma 8 shows that in this case, w f (t) given by w f (t) = R D − + 1 t w , 1 (t) is a particular solution of Equation (28) for the termf β (t).

Complementary Solutions of Equations (18) and (19)
Lemma 11. The complementary solution of Equation (19) is y(t) = Cy 1 (t), where the following holds: and C is a constant. When b = 0, this becomes the corresponding solution of Equation (21) given by the following: Proof. We see the following holds: which gives (44).
With the aid of (44) and (45), we obtain the following lemma.

Lemma 12.
The complementary solution of (18) is given by u(t) = Cu 1 (t), where the following holds: When b = 0, this becomes the corresponding solution of Equation (20) given by the following: Proof. When b = 0, by using (45), we obtain the following: from which we obtain (48) by using two of the formulas given in the following lemma.
Proof. Following Lemma 5, G y,0 (t, τ) is chosen to be the complementary solution of G y,0 (t, τ) = 1 at t > τ. The third member in Equation (55) is expressed as the following: Lemma 15. Let Condition 2 (i) be satisfied, and G y,0 (t, τ) be given by (55). Then Theorem 1 shows that a particular solution of Equation (19) is given by y f (t) = t 0 G y,0 (t, τ)f (τ)dτ.

Green's Function G u (t, τ) for Equation (18)
Corresponding to Equations (24) and (25), we define the Green's function G u (t, τ) for Equation (18), so that it satisfies the following: We note that these equations are obtained from Equations (53) and (54), by replacing G y (t, τ) by R D 1/2 t G u (t, τ). As a consequence, we obtain the following lemma by using Lemma 14.

Remark 6.
The Green's function given by (59) for b = 0 is not in agreement with the first several terms obtained in [14], except the leading term.

Solution of Integral Equations
We note that these equations are obtained from Equations (60) and (61), by replacing y(x) by R D 1/2 xũ (x). In [14], the solution of (62) is obtained by transforming it to an integral equation and then solving it by iterations. By (63), the integral equation is the following: In the case of Equation (60), the integral equation is obtained with the aid of (61), as follows:ỹ This integral equation is an example of the type of equation which was discussed in the book [13]. Remark 7. By the operational calculus described in [11][12][13], applied to (65), the particular solution of (60) is given by the following: We can write this as the following: whereỹ We can regard the solution given by (67) with (68), as the solution of Equation (65) obtained by iterations.
In the case of Equation (62), in place of Remark 7, we have the following remark.

Remark 11.
Here we give another expression of y f (t), by using (55) for b = 0, in the second member of (77): Since this y f (t) must be equal to y f (t) given by (76), when ν = −β + , we have the following: Lemma 21. When −β ∈ C + , the solution of Equation (74), which is obtained with the aid of Theorem 1, is given by the following: Proof. By using G u,0 (t, τ) given in Lemma 16 for b = 0, andf (t) = 1 Γ(ν) t ν−1 H(t), in u f (t) given in Lemma 17, we have the following: By using (80) in this equation, when ν = −β + , we have (81). The last equality in (81) is due to Formula (51) for ν = 1 − β + .

Solution of Equation (22) Satisfying Condition 2 (iii) by Iterations
Lemma 22. Let u(t) be a solution of (22) and y(t) = R D λ−α+ρ t u(t). Then, y(t) satisfies the following: In order to obtain a solution, we use the method adopted in Remarks 7 and 9.
Lemma 24. Let C λ,k (γ) be given as in Lemma 23. Then the solution of (22) is given by the following:
Lemma 25. The solution of Equation (75) is given by the following: This agrees with (76), which is derived for the case of −β ∈ C + .

Complementary Solutions of Equations
Lemma 29. The complementary solutions of Equation (100) are given by the following: Lemma 30. The complementary solutions of Equation (99) are given by the following: Proof. By using (104) and (105), with the aid of formulas given in Lemma 13, we obtain (106) and (107), respectively.

Solutions of Equations (75) and (74) with the Aid of Theorem 3
When Condition 2 (iii) is satisfied, Equations (21) and (20) are expressed by (75) and (74), respectively. It is stated in Lemmas 32 and 34 that particular solutions of them are obtained in terms of the Green's functions G v, 1 (t, 0) and G w, 1 (t, 0), which are given in Lemmas 31 and 33.