Properties of the Resolvent of a Linear Abel Integral Equation: Implications for a Complementary Fractional Equation

New and known properties of the resolvent of the kernel of linear Abel integral equations of the form x(t) = f (t) − λ t 0 (t − s) q−1 x(s) ds, (A λ) where λ > 0 and q ∈ (0, 1), are assembled and derived here. First, a priori bounds on potential solutions of the resolvent equation R(t) = λt q−1 − λ t 0 (t − s) q−1 R(s) ds (R λ) are obtained. Second, it is proven—using these bounds, Banach's contraction mapping principle, new continuation and translation results, and Schaefer's fixed point theorem—that (R λ) has a unique continuous solution on (0, ∞), which is called the resolvent in the literature and denoted here by R(t). Third, both known and new properties of R(t) are derived. Fourth, R(t) is shown to be completely monotone and the unique continuous solution of the initial value problem of fractional order q: D q x(t) = −λΓ(q)x(t), lim t→0 + t 1−q x(t) = λ, (I λ) where D q denotes the Riemann–Liouville fractional differential operator. Finally, the resolvent integral function t 0 R(s) ds is shown to be the unique continuous solution of an integral equation closely related to (R λ). Closed-form expressions for it and R(t) are derived.


Introduction
This is both an expository and research paper.It is expository in that many of the results for the resolvent of the kernel λ(t − s) q−1 have been collected from various sources and are conveniently assembled here.It is also a research paper in that many of the derivations of the properties of the resolvent have not appeared before-additionally, there are new results, such as explicit formulas for the resolvent and the resolvent integral function mentioned in the abstract and showing that the resolvent equation is equivalent to a fractional initial value problem.Our approach is to first prove that the resolvent equation has a unique continuous solution on the entire interval (0, ∞), which will then serve as a springboard for deriving important and useful properties of this solution, both known and new.
Resolvents are used to express the solutions of Volterra equations.Volterra integral and integro-differential equations aid in modeling a host of situations: ranging from the integrodifferential equations of population [23] and pharmacokinetics [35] models; to the integral equations of renewal theory [12,16]; to the partial differential equation models of heat conduction and diffusion problems that can be recast as integral equations, such as are found in [23,27]; and the list goes on.More recently, Heese and Freyberger state that the coupled Heisenberg equations of motion that appear in their paper [21, p. 4] take on the form of a Volterra integro-differential equation-after explicitly solving the dynamics of certain particles-and that its general solution can formally be expressed by means of a resolvent [2,20].Gorenflo and Vessella [18, p. 142] model the temperature u(x, t) along a one-dimensional, semi-infinite rod (x ≥ 0) with the heat equation and assume that Newtonian heating takes place at the boundary x = 0.The model of the inside boundary temperature u(0, t) simplifies to a Volterra integral equation which they then solve using the method of Laplace transforms.However, there is the alternative of expressing the solution in terms of a resolvent [4, p. 4840].
The material in this paper revolves around the resolvent equation and its fractional differential equation counterpart, where λ and q are positive constants with q ∈ (0, 1).The singularities of the functions λt q−1 and λ(t − s) q−1 present challenges, but they can be surmounted by choosing a suitable Banach space and employing a translation to circumvent the singularity at t = 0.The fact that (R λ ) plays a role in a multitude of diverse applications, especially when q = 1/2, warrants a thorough investigation of this equation.This is one in a series of papers [6][7][8][9][10] constituting a study of the scalar Volterra integral equation together with the scalar fractional differential equation subject to the initial condition lim t→0 + t 1−q x(t) = x 0 , (1.2b) where f : (0, T] × I → R denotes a continuous function and I ⊆ R an unbounded interval while x 0 , q denote constants with x 0 = 0 and q ∈ (0, 1).The symbol D q denotes the Riemann-Liouville fractional differential operator of order q, which for 0 < q < 1 is defined by where Γ is Euler's Gamma function.
One of the main results in [6] is an "equivalence theorem", which asserts that if a function x(t) is continuous on an interval (0, T] and if both x(t) and f (t, x(t)) are absolutely integrable on (0, T], then x(t) satisfies the Volterra integral equation (1.1) on (0, T] if and only if it satisfies the initial value problem (1.2) on this same interval.In [7] we proved a couple of theorems that establish the existence and uniqueness of a continuous, absolutely integrable solution of (1.1) and (1.2) on an interval (0, T] when f (t, x) satisfies either a Lipschitz condition or a generalized Lipschitz condition.We also discussed the continuation of such a solution beyond T by means of a transformation of the integral equation (1.1) followed by a translation.In [9] we obtain an existence theorem for (1.1) that is atypical of the standard existence theorems in the literature in that (i) a growth condition obviates the need for f to be bounded or even to satisfy a Lipschitz condition and (ii) the existence of solutions is dependent on the value of q.Equations with closed-form solutions were constructed to illustrate the main results in all of these papers.

Related equations
In this paper our interest in (1.1) and (1.2) is confined to with λ > 0 and q ∈ (0, 1).For the most part, attention will be directed to the integral equation (1.1) rather than to (1.2); note that any results obtained for it will also pertain to (1.2) because of the aforementioned equivalence theorem in [6].Generally speaking, we use x(t) to denote the unknown function; however, we reserve z(t) for denoting the unknown function in (1.1) when f (t, x) has the special form (2.1).In this case, (1.1) simplifies to where x 0 = 0, λ > 0, and q ∈ (0, 1).Equation (2.2) and the resolvent equation (R λ ) are intertwined: since one can be easily converted into the other, any solutions they may have are related by a simple formula.Specifically, multiplying (R λ ) by x 0 /λ yields Then replacing (x 0 /λ)R(t) with z(t), we get is a solution of (2.2).The actual proof of the existence of a continuous solution R(t) of (R λ ) is found in the next section.Strictly speaking, the function is known as the kernel of equations (2.2) and (R λ ); nevertheless, for ease of communication, we will also refer to C(t) as the kernel.
Changing variables, we obtain the following alternate form of the resolvent equation (R λ ): 3 Existence and uniqueness of solutions has a unique continuous solution on the entire open interval (0, ∞) if the kernel A is positive, continuous, and nonincreasing on (0, ∞); A ∈ L 1 (0, 1); and for each T > 0, the quotient A(t)/A(t + T) is a nonincreasing function of t on (0, ∞).For details, consult Miller [26,Thm. 2] who gives an existence and uniqueness proof based on a result in the 1963 paper by Friedman [17, pp. 384-387]-or see Miller's monograph [27,Thm. 6.2].Since these conditions are satisfied when A(t) = λt q−1 and λ > 0, it follows that (R λ ) has a unique continuous solution on (0, ∞).More recently, a constructive proof for the equation (R λ ) that exploits λ(t − s) q−1 being weakly singular and employs iterated kernels to derive and express the unique continuous solution of (R λ ) in terms of a series can be found in [4,Thm. 4.2].
Here we present another proof of the existence and uniqueness of a continuous solution of (R λ ), which delineates a procedure that could be adapted to certain types of nonlinear Volterra integral equations to establish the local existence of a continuous solution on an interval [0, τ] and then to continue that solution beyond τ.(Recent local existence results for nonlinear equations can be found in [7,Thms. 2.7,4.1].Another result appears in [9,Thm. 3.1].)By focusing on the specific kernel λ(t − s) q−1 instead of on the kernel of (R A ), we will be able to obtain results for (R λ ) that may not necessarily pertain to the more general (R A ).
The steps that we will employ for the resolvent equation (R λ ) are the following.
(S1) Determine a priori bounds for any continuous solutions of (R λ ) that may exist.
(S2) Prove the existence and uniqueness of a continuous solution of (R λ ) on an interval (0, τ] with an appropriate fixed point mapping. (S3) Prove that if the local solution obtained in (S2) can be continued past τ, then that continuation is unique.
(S4) Insert a parameter ν ∈ (0, 1] into (R λ ) (cf. (3.12)) to set it up for the eventual use of Schaefer's fixed point theorem.Then translate this modified equation in order to bypass the singularity of the forcing function at t = 0. Determine a priori bounds for any possible solutions of the translated equation.
(S5) Apply Schaefer's theorem to prove that the translated equation with ν = 1 has a continuous solution on every finite interval [0, b].Prove each of these solutions is unique.Use this uniqueness to prove the existence of a unique continuous solution on [0, ∞).
The details of each of these steps are spelled out in the rest of this section.In the following subsection, we will establish the existence of a priori bounds for any continuous solutions of (R λ ) that may exist.

The resolvent equation and a priori bounds
Since the purpose of carrying out steps (S1)-(S6) is to prove the existence of a continuous solution of (R λ ), the first thing to consider is whether the improper Riemann integral exists when φ(s) is continuous for s > 0, and if so, whether the integral itself is continuous.This can be answered with the help of the following lemma whose proof is found in [4, (2.6)-(2. But this is not quite the result that we need here since any solution of (R λ ) is necessarily undefined at t = 0; nonetheless, it can be used to prove the following lemma for functions that are both continuous and absolutely integrable on the left-open interval (0, b].Details are provided in [6, pp. 7-8].With the help of the next lemma, we will prove that there are a priori bounds for continuous solutions of (R λ ) should any exist.Notice that if one does exist, call it R(t), then equation (R λ ) suggests that In fact, see (3.11) and the subsequent sentence.

Local solution of the resolvent equation
Now that we have found a priori bounds for continuous solutions of (R λ ) should any exist, we proceed to the next step (S2).We will prove that a continuous solution does in fact exist-at least on a short interval.Moreover, it is unique.In the rest of this section, R(t) designates this unique continuous solution.
Remark 3.6.The interval (0, τ] is really quite short unless λ 1.For instance, with λ = 1 and q = 1/2, Proof.For τ given by (3.3), let C(0, τ] denote the vector space of all continuous functions φ : (0, τ] → R. Define the subset X ⊂ C(0, τ] by Since one can show that M is a closed subset of the complete metric space (X, ρ), the metric space (M, ρ) is also complete.And so if we can establish that the mapping P defined by for φ ∈ M maps M into itself and is a contraction on M, then it would follow from Banach's contraction mapping principle that P has a unique fixed point in M.
First we show P : M → M.So let φ ∈ M. By virtue of being in M, it is absolutely integrable on (0, τ].As a result, Lemma 3.2 implies that Pφ ∈ C(0, τ].Furthermore, we see from (3.8) that for 0 < t ≤ τ.Let us evaluate the integral with the following formula that is derived from the Beta function (cf.[6, (4.4)]): it converges if and only if p, q > 0.Then, using τ from (3.3), we have for 0 < t ≤ τ.Thus, P : M → M.
In order to show that P is a contraction on M, select any In fact, because of the value of τ given by (3.3), α = 1/2.Thus we have In other words, |Pφ − Pψ| g ≤ α|φ − ψ| g where α ∈ (0, 1).And so P is a contraction on M.
Therefore, as P : M → M is a contraction mapping, there is a unique point R ∈ M such that PR = R.In other words, R is the only continuous function residing in the set M that satisfies the resolvent equation (R λ ) on the interval (0, τ].
Since R ∈ M, |R(t)| ≤ 2λt q−1 for 0 < t ≤ τ.Thus it is absolutely integrable on (0, τ] since Also observe from (3.10) that Hence, lim And so from we see that t 1−q R(t) → λ as t → 0 + , which proves (3.4).Now that we have established that the continuous solution R satisfies (3.4), it follows from Theorem 3.4 that R actually resides in a subset of M, namely {φ ∈ C(0, τ] : 0 ≤ φ(t) ≤ λt q−1 }.Moreover, this theorem rules out any continuous solutions existing outside this set.Thus R(t) is the only continuous solution of (R λ ) on the interval (0, τ].

Continuation of the solution
Let us move on to (S3) and show that there is only one possible way to continue the local solution of Theorem 3.5.Theorem 3.7.If, in Theorem 3.5, the local continuous solution R(t) of the resolvent equation (R λ ) can be extended beyond τ so that it remains a continuous solution of (R λ ), then that continued solution is unique.
Proof.Suppose to the contrary that R 1 and R 2 are continuous solutions of (R λ ) that separate at a point t 1 ≥ τ but which are identical on the interval (0, t 1 ].Then for δ > 0 sufficiently small, the difference R 2 (t) − R 1 (t) is nonzero and does not change sign for t ∈ (t 1 , t 1 + δ].However, this results in a contradiction since the left-and right-hand sides of Now we will move on to steps (S4)-(S6) to prove that R(t) can indeed be extended beyond (0, τ] to the entire interval (0, ∞) so that it remains a continuous solution of the resolvent equation (R λ ).Before we do so, let us summarize what has been proven so far: Moreover, it is nonnegative and bounded above by the forcing function λt q−1 .And t q−1 R(t) has the limit given by (3.4).(Cf.Theorem 3.5.)(b) If a continuous solution of (R λ ) exists on an interval (0, L] that is longer than (0, τ], then it is the only such continuous solution.Also, it is nonnegative and bounded above by the forcing function λt q−1 for 0 < t ≤ L. (Cf.Theorems 3.4 and 3.7.)

The translated equation and a priori bounds
In order to prove that the solution R(t) can be continued beyond the point τ (cf.(3.3)), we first insert a parameter ν ∈ (0, 1] into the resolvent equation (R λ ), thereby obtaining the equation This is to get ready for the eventual application of Schaefer's fixed point theorem (cf.Theorem 3.10).Actually we will work with a translation of (3.12) (cf.(S4)) so as to bypass the singularity of the forcing function νλt q−1 at t = 0 (more on this later).Note that (R λ ) becomes (3.12) when λ is replaced with νλ.Consequently, for a given ν ∈ (0, 1], items (a) and (b) in the summary also pertain to (3.12) if in those statements λ and R(t) are replaced with νλ and r(t), respectively, where the corresponding value of τ is Of course, with ν = 1, then r(t) is precisely R(t) and (3.12) is the resolvent equation To obtain the translation of (3.12) to which we just alluded, first let Then replace t in (3.12) with t + T: With the change of variable u = s − T, we get Now use (3.14) to express this in terms of x: And so we have where Recall from the discussion before (3.13) with regard to items (a) and (b) that for each value of ν ∈ (0, 1] there is a unique continuous solution of (3.12) on the interval (0, τ] ≡ (0, 2T], where τ is given by (3.13).Letting r(t) denote this solution, it trivially follows from the preceding work that (3.15) always has a solution on the interval [0, T], namely (3.14), which is identical to the piece of r(t) on the interval [T, 2T].We show next that if (3.12) has a solution x(t) extending beyond T, then it and r(t) can be spliced together to obtain a continuation of the solution r(t).Theorem 3.8.For a given ν ∈ (0, 1], let r(t) be the unique continuous solution of (3.12) on the interval (0, τ] where τ is given by (3.13).Let T = τ/2.If for some b > T there exists a continuous solution x(t) of the translated equation (3.15) on the interval [0, b], then the function is a continuation of r(t) to the interval (0, b + T].Furthermore, it is the only continuous solution of (3.12) on this interval and 0 ≤ r c (t) ≤ νλt q−1 (3.18) for all 0 < t ≤ b + T.
Proof.Since r c ≡ r on (0, T), it follows from (3.12) that Thus, r c (t) is a continuous solution of (3.12) on (0, T).Now consider r c to the right of T. By hypothesis, x(t) is a continuous solution of (3.15) on [0, b]; so With the change of variable u = s + T, this becomes

L. C. Becker
Expressed in terms of the function r c , this simplifies to This and the continuity of r(t) and x(t − T) on their respective intervals in (3.17In the context of Theorem 3.8, the uniqueness of the solutions r(t) and r c (t) on their respective domains implies that r c (t) is identical to r(t) on the interval (0, 2T].A corollary of this theorem is that there are a priori bounds for solutions of (3.15)-for those existing on [0, T] and for any possibly existing on a longer interval-and that the bounds are independent of the interval.Corollary 3.9.Suppose, for a given ν ∈ (0, 1], a solution x of the translated equation (3.15) exists on an interval [0, b], where the function r in the integrand of (3.16) is the unique continuous solution of (3.12) on the interval (0, τ] with τ given by (3.13).Then, irrespective of the values of b and ν, for 0 ≤ t ≤ b, where T = τ/2.

The translated equation and solutions
At this juncture, we move on to (S5).Now our objective is to prove that there actually is a continuous solution of (3.15) when ν = 1 on every finite interval [0, b].This will be achieved with Corollary 3.9 and the version of Schaefer's fixed point theorem [33] for normed spaces [34, p. 29] that is stated below in Theorem 3.10.Then we show that this equation has a unique continuous solution on [0, ∞).
So we will address existence as a fixed point problem and work with the following Banach space and mapping.For a given b > 0, let B denote the Banach space of continuous functions Define the mapping P on B by Expressed in terms of this operator, (3.15) is Referring back to Corollary 3.9, we see that the set of all continuous solutions of (3.24) is bounded, irrespective of the length of the interval [0, b] or the value of ν ∈ (0, 1].Consequently, for the mapping P defined by (3.23), this rules out the alternative labeled (ii) in the following statement of Schaefer's fixed point theorem.Theorem 3.10 (Schaefer).Let (B, • ) be a normed space, P a continuous mapping of B into B which is compact on each bounded subset Ω of B. Then either (i) the equation x = νP x has a solution for ν = 1, or (ii) the set of all solutions of x = νP x for 0 < ν < 1 is unbounded.
This leaves (i), namely, the existence of a continuous solution of the equation x = Px on the interval [0, b], provided P satisfies the hypotheses of Schaefer's theorem.Showing this then becomes our present task.Let us begin with the next two lemmas to prove that the function F in (3.23) is uniformly continuous on [0, ∞).Recall that the function r appearing in its definition (cf.(3.16)) is the unique continuous solution of (3.12) on (0, τ] ≡ (0, 2T] and that it is nonnegative and bounded above by νλt q−1 .

Lemma 3.11. The function
r(s) ds is uniformly continuous on [a, ∞) for each a > 0.

L. C. Becker
Proof.Let a > 0 be given.Choose > 0. Since the function t q−1 is continuous on [a, ∞) and t q−1 → 0 as t → ∞, it is uniformly continuous on [a, ∞).Hence, for a given η > 0, there is a γ > 0 such that distinct t 1 , t 2 ∈ [a, ∞) and Hence for these t i , Since r(t) is nonnegative and bounded above by Choose any η < q /(νλT q ).Therefore, for the given > 0, there is a γ > 0 such that Lemma 3.12.The function F defined by (3.16), namely is uniformly continuous on [0, ∞).
Proof.First observe that the term (t + T) q−1 is not only continuous on [0, ∞) but it is also uniformly continuous on this interval because (t + T) q−1 → 0 as t → ∞.Solving for F(t) in (3.15), we get where Since r(t) denotes the unique continuous solution of (3.12) on (0, 2T], the function x(t), defined by Thus G(t) is uniformly continuous on [0, T].By Lemma 3.11 it is also uniformly continuous on [T/2, ∞).Consequently, G(t) is uniformly continuous on [0, ∞).Therefore, because both (t + T) q−1 and G(t) are uniformly continuous on [0, ∞), so is F(t).
We will also use the following lemma to help establish the uniqueness of the solution in Theorem 3.14 below.Lemma 3.13.The only continuous solution of on [0, ∞) is the trivial solution x(t) ≡ 0.
Proof.First we prove with the contraction mapping principle that x(t) ≡ 0 is the only continuous solution on an interval [0, b] if b is sufficiently small.To this end, take b = (q/(2λ)) 1/q .Then let B be the Banach space of continuous functions on [0, b] with the sup norm (3.22).For φ ∈ B, define the mapping Z by It follows from Lemma 3.1 that Z : B → B.Moreover, Z is a contraction on B since for φ, ψ ∈ B we have And so Thus Z has a unique fixed point in B. So it must be the trivial fixed point.Now suppose there is another continuous solution y(t) on the entire interval [0, ∞) besides x(t) ≡ 0.Then, in view of the uniqueness of the latter on [0, b], there are t 1 ≥ b and δ > 0 such that y(t) ≡ 0 on [0, t 1 ] but which is strictly positive or strictly negative on (t 1 , t 1 + δ].As a result, a contradiction since the sides are opposite in sign. Finally we are poised to prove that the translated equation (3.15) with ν = 1 has a unique continuous solution on the entire interval [0, ∞) and that it is bounded and nonnegative.

L. C. Becker
To prove P is continuous on B, choose any φ 1 ∈ B. Choose > 0. Let δ = (q )/(λb q ).Then for φ 2 ∈ B with φ 1 − φ 2 < δ, we have Thus, Thus, P : B → B is continuous.Next we will prove that P is compact on each bounded subset of B, to wit: the image PΩ of each bounded set Ω ⊂ B is contained in a compact subset of B. We will show that for each bounded set Ω it follows that PΩ is bounded, equicontinuous, and contained in a closed set.Application of Arzelà-Ascoli's theorem will then complete the proof of compactness of the mapping P. First let us establish that the set PΩ of functions is uniformly bounded and equicontinuous on [0, b].
Choose a > 0 large enough so that the closed ball Hence, as Ω ⊂ B a , for all φ ∈ Ω, which concludes the proof of equicontinuity.
Since the set PΩ of functions is equicontinuous on [0, b], so is its closure PΩ.This is easily seen with the following /3 argument.Let > 0. Suppose ψ is a limit point of PΩ.Then there is a sequence {Pφ n } ∞ n=1 ⊂ PΩ with Pφ n − ψ → 0 as n → ∞.Thus the sequence converges uniformly on [0, b] to ψ. Choose n 1 sufficiently large so that As a result, ] is not only an upper bound for all of the functions in PΩ but also for all of the limit points of PΩ.
It follows from the Arzelà-Ascoli theorem that the set PΩ is compact since it is uniformly bounded and equicontinuous on [0, b] and of course closed.Thus we have shown that P maps every bounded set Ω in B into a compact subset of B, namely into PΩ.
Therefore, we have established that the mapping P fulfills all of the conditions of Schaefer's fixed point theorem.Consequently we have alternative (i) in Theorem 3.10 since (ii) has already been ruled out.In other words, we conclude that (3.26) has a continuous solution on the interval [0, b], no matter the value of b > 0.Moreover, Corollary 3.9 shows that this solution is nonnegative and bounded above by λT q−1 , which is the assertion of (3.27).

L. C. Becker
As for uniqueness, suppose that x(t) and y(t) are continuous solutions of (3.26) We have just proved equation (3.26) has a unique continuous solution on every finite interval [0, b].In fact, because of this uniqueness, it has a unique continuous solution on the entire interval [0, ∞).We justify this assertion with the following argument.Consider the sequence {x n } ∞ n=1 , where x n denotes the unique continuous solution of (3.26) on the interval [0, n].Define the function x : [0, ∞) → R as follows: For a given t ≥ 0, choose n ∈ N (set of natural numbers) so that n > t; then let x(t) := x n (t).This function is well-defined because x n+p ≡ x n on the interval [0, n] for each p ∈ N.
Also, for every n ∈ N, define the function Clearly the sequence { x n } ∞ n=1 of functions converges uniformly to x on compact subsets of [0, ∞).Thus x is continuous on [0, n] for each n ∈ N; and consequently on [0, ∞).

Global solution of the resolvent equation
Now that we have established the existence of a unique continuous solution of the translated equation (3.26) on [0, ∞), we splice it to the unique continuous solution of (R λ ) on the short interval (0, τ].As a result, we obtain the following global existence and uniqueness result. Theorem 3.15.Let R(t) be the unique continuous solution of the resolvent equation (R λ ) on (0, τ], where τ is defined by (3.3).Let x(t) be the unique continuous solution of the translated equation is the unique continuous solution of the resolvent equation (R λ ) on (0, ∞).Furthermore, for t > 0.
Proof.This follows from Theorem 3.14, Theorem 3.8 with ν = 1, and a "sequence of functions" argument like the one at the end of the proof of Theorem 3.14.

Note.
Henceforth, the function R(t) defined by (3.30) will be renamed R(t).That is, for the remainder of this paper, R(t) will denote the unique continuous solution of the equation (R λ ), equivalently of (R a λ ), on the entire interval (0, ∞).In the literature of integral equations, R(t) is known either as the resolvent [27, p. 21] or the resolvent kernel [23, p. 38] of C(t) = λt q−1 .We prefer the term resolvent and will usually omit the mention of C(t) since it is the only kernel considered in this paper.
The next result follows from the discussion about the function (2.3) and Theorem 3.15.It is also valid when x 0 = 0 (cf.Lemma 3.13).
Corollary 3.16.Let x 0 ∈ R and λ > 0. The function is the unique continuous solution of (2.2) on (0, ∞).In particular, is the unique continuous solution of

Properties of the resolvent
We now prove, based on our work in Section 3, some known properties of the resolvent R that have been used in some of the references at the end of this paper.We also obtain some new results.For example, we improve property (i) below significantly with Corollary 4.6; and with Theorem 4.5 we improve upon the bounds of t 0 R(s) ds in (viii) by sandwiching the integral between functions that approach 1 as t → ∞.In Theorem 4.10 we shall see that the kernel C of (R λ ) is the unique continuous solution of a complementary integral equation.Additional new results for R will be derived in the remaining sections of this paper.
Define the function u by Multiplying this by the integrating factor exp λt q−1 dt = e λ q t q , we obtain e λ q t q du dt + λt q−1 e λ q t q u ≥ 0 or d dt e λ q t q u(t) ≥ 0 for all t > 0. Hence, for a given t > 0 and an arbitrary δ ∈ (0, t), e λ q t q u(t) ≥ e λ q δ q u(δ) ≥ u(δ). Consequently, Since this is true for every δ ∈ (0, t), e λ q t q u(t) ≥ 1. Therefore, for every t > 0. Note this holds for t = 0 as well.
Later on we appeal to either the following theorem or to its corollary to justify interchanging the order of integration in some of the proofs.
Proof.For a given t > 0, let Ω t denote the interior of the triangular region Define the function f t on the rectangle [0, t] × [0, t] by Observe that f t is discontinuous on the diagonal {(u, u) : 0 ≤ u ≤ t}.It is also discontinuous on the right side of the rectangle if ϕ(t) = 0 and on the bottom side unless ϕ is defined at t = 0 with ϕ(0) = 0.At all other points it is continuous.So f t is continuous almost everywhere since these line segments have measure 0 in R 2 .Consequently, where the inequalities are consequences of t q being an increasing function and the integrability of ϕ on every interval (0, t).Then the finiteness of this iterated integral and the measurability of f t imply both ).The first iterated integral can be written as (4.3) and the second one as According to Theorems 3.15 and 4.1, R is continuous and nonnegative on (0, ∞) and integrable on (0, 1).So it satisfies the conditions of Theorem 4.3.
Proof.We only need to prove the left-hand inequality since the right-hand inequality has already been established with Lemma 4.2.Integrating (R λ ) and using Corollary 4.4 to interchange the order of integration, we obtain Thus, as t q is increasing, we have for all t > 0. Theorem 4.5 enables us to greatly improve upon the function bounding R(t) from above in Theorem 4.1 (i).Later on, with Theorem 7.3, we show that R(t) is strictly positive.Thus, the integrals in (iii) and (viii) of Theorem 4.1 never assume the value zero on the interval (0, ∞).Corollary 4.6.For given q ∈ (0, 1) and λ > 0, 0 < R(t) ≤ q q + λt q λt q−1 (4.5) for all t > 0.
Since the functions bounding the integral in Theorem 4.5 approach 1 as t → ∞, we have the following important result.

L. C. Becker
Theorem 4.10.Let R(t) be the resolvent of C(t) = λt q−1 .Then C(t) is the unique continuous solution of on the interval (0, ∞).
Proof.With Theorem 3.15 we established that the resolvent equation (R λ ) has a unique continuous solution on (0, ∞), which we now denote by R(t).Thus, from (R λ ) we have for all t > 0. With the change of variable u = t − s, this becomes verifying that C(t) is a solution of (4.6) on (0, ∞).
As for uniqueness, suppose D(t) is a continuous solution of (4.6) on an interval (0, b].Then where • denotes the sup norm (3.22).Thus, This implies D ≡ C on (0, b] since b 0 R(s) ds < 1 (cf.Theorem 4.5).In other words, the only continuous solution of (4.6) on an interval (0, b], regardless of the value of b > 0, is C(t) = λt q−1 .It then follows from a "sequence of solutions" argument such as the one concluding the proof of Theorem 3.14 that C(t) is the unique continuous solution of (4.6) on (0, ∞).

A complementary fractional differential equation
In this section, we show that for a given λ > 0 and q ∈ (0, 1) that the unique continuous solution of the resolvent equation (R λ ) is also the unique continuous solution of a related linear fractional differential equation.This result is a consequence of the following equivalence theorem, which is taken from [6, Thm.6.2].Theorem 5.1.Let q ∈ (0, 1) and x 0 = 0. Let f (t, x) be a function that is continuous on the set where I ⊆ R is an unbounded interval.Suppose a function x : (0, T] → I is continuous and both x(t) and f (t, x(t)) are absolutely integrable on (0, T].Then x(t) satisfies the Volterra integral equation on (0, T] if and only if it satisfies the initial value problem As we stated earlier in Section 1, the focus of this paper is on (1.1) and (1.2) when f (t, x) is the linear function (2.1), where q ∈ (0, 1) and λ > 0 throughout this paper.In this case, Theorem 5.1 implies the following.
Theorem 5.2.The resolvent R(t) is the unique continuous solution of the initial value problem D q x(t) = −λΓ(q)x(t), lim Proof.We see from Corollary 4.8 that the resolvent R(t) is absolutely integrable on (0, ∞).

The resolvent formula
There is a simple closed-form formula for the resolvent of a Volterra integral equation with a separable kernel (cf.[5,23]).Is there a formula for the resolvent of the kernel of equation (R λ )?The answer is yes, but it is not simple.We will now derive that formula using Laplace transforms, where L{ f (t)}(s) denotes the Laplace transform of a function f .Theorem 6.1.The Laplace transform of the resolvent R(t) exists and is L{R(t)}(s) = λΓ(q) s q + λΓ(q) (6.1) for s ≥ 0.
Let us now derive the resolvent formula for values of q ∈ (0, 1).Theorem 6.2.For a given q ∈ (0, 1), the resolvent is R(t) = λΓ(q)t q−1 E q,q (−λΓ(q)t q ) (6.4) where E m,n (t) denotes the two-parameter Mittag-Leffler function: where m, n are positive constants.That is, (6.4) is the unique continuous solution of on (0, ∞).It is also the unique continuous solution of the initial value problem D q x(t) = −λΓ(q)x(t), lim Proof.In terms of the constant a := λΓ(q), the right-hand side of (6.1) is if s q > a (a > 0 as λ > 0).Since we have With the intent of justifying interchanging the order of summation and integration, consider the series of absolute values: Rewriting this as we see that it is a geometric series that converges if s q > a.Thus, by Levi's theorem ([1, p. 269], [32, p. 36]) for series, the interchange is allowed.As a result, by (6.5) we have

Remark 6.4.
There is yet another way to derive (6.4) and that is to use the series representation for the resolvent that is found in [4,Thm. 4.2], which was derived by means of fixed point theorems.It is shown there that where β = 0 and q ∈ (0, 1), defines a unique continuous function that satisfies the equation on (0, ∞).Let us convert this equation to the form of (R λ ) by first rewriting it as for any λ = 0. Expressed in terms of λ and R, the series (6.9) is Changing the index of summation to k = i − 1, (6.11) becomes = λΓ(q)t q−1 E q,q (−λΓ(q)t q ) , which is (6.4).Note that even though we have assumed throughout this paper that λ > 0, we see here that (6.4) is the unique continuous solution of (R λ ) for any λ = 0.
Next we use Theorem 6.2 with q = 1/2 to express the resolvent in terms of the error function (cf.(6.12) below).An equivalent formula was derived in 1965 in a paper by Brakhage et al. [11, p. 297] using methods different from those in this paper.Instead of "resolvent" the authors used the term "lösender Kern" (German for "solvent kernel").Other derivations can be found in [3,Ex. 3.2] and [4,Ex. 5.1].This suggests that is a solution of (A λ ).In fact, assuming that f is continuous on [0, ∞), we will prove that it is the only continuous solution of (A λ ) on [0, ∞).Our proof relies on the following lemma.
Proof.For a given t ∈ (0, b], Since this integral converges, then so does the integral Ψ(t).Consequently, Ψ has a finite value at each t ∈ [0, b], where Ψ(0) = 0.In fact, it follows from Theorem 4.1 (viii) that Since R(t) is positive and decreasing for t > 0, Then, because of Theorem 4.5,

|Ψ(t
When Equation (8.5) in the next theorem is known as the linear variation of parameters formula for (8.4).In this regard, see Miller's monograph [27]; in particular, note that equations (1.3) and (1.4) in Chapter IV of this work simplify to (1.1) and (1.2), respectively, when the nonlinear term G(t, x) ≡ 0. Also, see Burton [13, pp. 17, 24].Theorem 8.3 (Linear variation of parameters).Let f : [0, ∞) → R be continuous.Let C(t) = λt q−1 where λ > 0 and q ∈ (0, 1).The Abel integral equation has a unique continuous solution on [0, ∞), namely, This solution expressed in terms of the Mittag-Leffler function E q,q is A slight alteration of the proof of Theorem 4.3 justifies the following change in the order of integration: From the resolvent equation (R λ ), we have Therefore, , (25.5)].This can be seen from (9.11), which is derived in the next section.

The resolvent integral function
In this concluding section, we will use the resolvent and variation of parameters formulas to derive some useful results.As has been the case throughout the paper, R(t) denotes the resolvent of the kernel C(t) = λt q−1 where q ∈ (0, 1) and λ > 0.
Proof.Using the variation of parameters formula (8.5) with f (t) = λt q /q, we find that the unique continuous solution of (9.5) is x(t) = λ q t q − λ q t 0 (t − s) q R(s) ds.(s − u) q−1 R(u) du ds = λ q t q − λ t 0 t u (s − u) q−1 ds R(u) du = λ q t q − λ q t 0 (t − u) q R(u) du.
This and (9.6) imply that (9.4) is the unique continuous solution.
This concludes the proof of (9.7).

Lemma 3 . 1 .
8)].(It is also very similar to the proof of Lemma 8.1 that appears later in this paper.)If a function φ is continuous on an interval [0, b], then the improper Riemann integral Φ(t) defined by (3.1) defines a function that is also continuous on [0, b].Furthermore,

Lemma 3 . 2 .
If a function φ is continuous and absolutely integrable on (0, b], then the integral Φ(t) defined by (3.1) is continuous on (0, b]. ) imply r c (t) is continuous on the entire interval (0, b + T].Therefore, it follows from (3.19) and (3.20) that r c (t) is a continuous solution of (3.12) on (0, b + T].As we discussed earlier, item (b) applies not only to (R λ ) but to (3.12) as well.Hence, r c (t) is the only continuous solution of this equation on (0, b + T].Moreover, from (b) we also have(3.18).

( 3 .
14) as the translation of r(t) to the left by T units, is continuous on (−T, T].It then follows from Lemma 3.1 that X(t) is continuous on the subinterval [0, T].Hence F(t) is continuous on [0, T].This and the continuity of the first term in (3.16) imply that G(t) is also continuous on [0, T].

.
Choose any b > 0. As before, let (B, • ) denote the Banach space of continuous functions on [0, b] with the sup norm • .And let P be the mapping on B defined by (3.23).It follows from Lemmas 3.1 and 3.12 that the function Pφ is continuous on [0, b] for each φ ∈ B. Thus, P : B → B.

3 )
for s > 0. Thus T 0 e −st R(t) dt is convergent by the comparison test.Similarly, as b T e −st R(t) dt ≤ λ b T e −st t q−1 dt L. C. Becker for b > T, the integral ∞ T e −st R(t) dt is also convergent.Therefore, for each s > 0, the integral ∞ 0 e −st R(t) dt is convergent.As for s = 0, it follows from Corollary 4.8 that ∞ 0 e −st R(t) dt = ∞ 0 R(t) dt = 1.Since we have established the existence of the Laplace transform L{R(t)}(s) := ∞ 0 e −st R(t) dt for s ≥ 0, we can now take the Laplace transform of equation (R λ ):

Lemma 8 . 1 .
If a function ψ is continuous on an interval [0, b], then the integral function Ψ defined by
.6) Proof.Since the function f is continuous on [0, ∞), it follows from Lemma 8.1 that t 0 C(t − s)y(s) ds