# Spectral Criteria for Solvability of Boundary Value Problems and Positivity of Solutions of Time-Fractional Differential Equations

Authors: Valentin Keyantuo; Carlos Lizama; Mahamadi Warma
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We investigate mild solutions of the fractional order nonhomogeneous Cauchy problemDtαu(t)=Au(t)+f(t),t>0, where 0<α<1. When A is the generator of a C0-semigroup (T(t))t≥0 on a Banach space X, we obtain an explicit representation of mild solutions of the above problem in terms of the semigroup. We then prove that this problem under the boundary condition u(0)=u(1) admits a unique mild solution for each f∈C([0,1];X) if and only if the operator I-Sα(1) is invertible. Here, we use the representation Sα(t)x=∫0∞‍Φα(s)T(stα)xds,t>0 in which Φα is a Wright type function. For the first order case, that is, α=1, the corresponding result was proved by Prüss in 1984. In case X is a Banach lattice and the semigroup (T(t))t≥0 is positive, we obtain existence of solutions of the semilinear problem Dtαu(t)=Au(t)+f(t,u(t)),t>0,0<α<1.


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## Body

## 1. Introduction

LetX be a complex Banach space and A the generator of a C0-semigroup (T(t))t≥0 in X. We consider the following linear differential equation:
(1)Dtαu(t)=Au(t)+f(t),t>0,0<α<1,
where Dtα is the Caputo fractional derivative.In the integer caseα=1, it is well known that there exists a strong connection between the spectrum of (T(t))t≥0 and solutions of the inhomogeneous differential equation (1) satisfying the condition u(0)=u(1), where f is a forcing term. A complete characterization of the class of generators A such that for any given f∈C([0,1];X), (1) with the condition u(0)=u(1) has a unique solution which was obtained by Prüss [1] in 1984, extending earlier results by Haraux (see [2]).Denoting the resolvent set of an operatorL by ρ(L), the result of Prüss reads as follows: 1∈ρ(T(1)) if and only if for any f∈C([0,1];X), the equation u′=Au+f admits exactly one mild solution satisfying u(0)=u(1).After Prüss theorem, many interesting consequences and related results have appeared. For example, the corresponding connection with the spectrum of strongly continuous sine functions [3], cosine functions [4], and connections with maximal Lp-regularity are discussed in [5–7].More recently, Nieto [8] studied periodic boundary valued solutions of (1) considering the scalar case. Nieto considers the Riemann-Liouville fractional derivative, and the meaning he gives to a “periodic” boundary condition is the following:
(2)limt→0+t1-αu(t)=limt→1-t1-αu(t),0<α<1.
Further results along these lines are given in [9]. One disadvantage of this condition is that continuity of u(t) for t≥0 forces the condition u(1)=0. It thus appears that Riemann-Liouville is not the most appropriate choice when one considers periodic boundary valued problems. In contrast, the Caputo derivative needs higher regularity conditions of u(t) than the Riemann-Liouville derivative.Our objective in this paper is twofold: first, we reformulate Nieto's results for the vector-valued case of (1) considering Caputo's fractional derivative and the natural periodic boundary condition:
(3)u(0)=u(1).
Even more, we are successful in extending to the range 0<α<1 the above mentionedcharacterization given by Prüss, in terms of the following strongly continuous resolvent family associated to (1):
(4)Sα(t)x=∫0∞‍Φα(s)T(stα)xds,t≥0,0<α<1,x∈X,
where Φα is a Wright type function defined by Φα(z)=∑n=0∞((-z)n/n!Γ(-αn+1-α)) for every z∈ℂ (see Section 2 for more properties of the function Φα). We observe that Φα(t) is a probability density function on [0,∞), whose Laplace transform is the Mittag-Leffler function in the whole complex plane.A remarkable consequence of our extension result given in Theorem9 is the following: if A generates a uniformly stable semigroup, then for each f∈C([0,1];X) (1) admits exactly one mild solution fulfilling the boundary conditions u(0)=u(1). In order to do this, we study mild solutions of (1) and show that any mild solution has the following representation:
(5)u(t)=Sα(t)u(0)+∫0t‍Pα(t-s)f(s)ds,t>0,
where (Sα(t)) is given by (4) and (Pα(t)) is a second operator family associated with (T(t))t≥0 (see Section 2).Secondly, we study positivity of mild solutions and obtain a simple spectral condition that ensures positivity thereof in the periodic boundary value case. More precisely, letα∈(0,1) and A be the generator of a positive C0-semigroup (T(t))t≥0. Suppose (I-Sα(1))-1x≥0 for all x∈X+ and assume that u is a mild solution of (1) and that
(6)Dtαu(t)-Au(t)≥0,u(0)=u(1).
Then, u(t)≥0 for all t∈[0,1].Finally, we study in Banach lattices existence of mild solutions for the semilinear problem:(7)Dtαu(t)=Au(t)+f(t,u(t)),t>0,0<α<1,
under the hypothesis that A generates a positive C0-semigroup. This is an extension of recent results given by Zhang [10] in the integer case α=1 (cf. Theorem 16).Typical operators to which the results apply are elliptic operators in divergence form: namely, letΩ be an open subset of ℝN. We consider on Lp(Ω) the operator formally given by
(8)Au=∑i,j=1N‍Di(ai,j(x)Dju)=div(a(x)∇u),
in which (ai,j)1≤i,j≤N are bounded real valued functions. Under various boundary conditions (including Dirichlet, Neumann, Robin, and Wentzell), the results apply (see Section 4).While in the present paper, we concentrate on periodic boundary conditions, we mention the recent papers [11–13] dealing with fractional differential equations. The first two deal with nonlocal Cauchy problems, while the third considers the fractional evolution problem governed by an almost sectional operator and proceeds to construct the corresponding evolution operators by mean of a certain functional calculus.The paper is organized as follows. In Section2, we present some preliminaries on the resolvent families needed in the sequel. In Section 3, assuming that A generates a C0-semigroup, we represent the resolvent families of Section 2 using the subordination principle. In Section 4, we study mild solutions in general and in the periodic boundary valued case in particular. Positivity of mild solutions as well as the semilinear equation are considered in Section 5.


## 2. Preliminaries

The algebra of bounded linear operators on a Banach spaceX will be denoted by ℬ(X), the resolvent set of a linear operator A by ρ(A), and the spectral radius of a bounded operator S will be denoted by rσ(S). Let τ>0 be a real number. The space of continuous functions f:[0,τ]→X is denoted by C([0,τ];X) and its norm by ∥f∥∶=sup{∥f(t)∥:t∈[0,τ]}. We denote gα(t):=tα-1/Γ(α), α>0, where Γ is the usual gamma function. It will be convenient to write g0:=δ0, the Dirac measure concentrated at 0. Note the semigroup property: gα+β=gα*gβ for all α,β>0.The Riemann-Liouville fractional integral of orderα,  0<α<1, of a function u:[0,1]→X is given by
(9)Iαu(t)∶=(gα*u)(t)∶=∫0t‍gα(t-s)u(s)ds,
for example, when u is locally integrable on (0,1).The Caputo fractional derivative of order0<α<1 of a function u is defined by
(10)Dtαu(t)∶=I1-αu′(t)=∫0t‍g1-α(t-s)u′(s)ds,
where u′ is the distributional derivative of u(·), under appropriate assumptions. The definition can be extended in a natural way to α>0. Then, when α=n is a natural number, we get Dtn:=dn/dtn.The Laplace transform of a locally integrable functionf:[0,∞)→X is defined by
(11)ℒ(f)(λ)∶=f^(λ)∶=∫0∞‍e-λtf(t)dt=limR→∞∫0R‍e-λtf(t)dt,
provided that the integral converges for some λ∈ℂ. If, for example, f is exponentially bounded, that is, ∥f(t)∥≤Meωt,  t≥0 for some M>0, and ω∈ℝ, then the integral converges absolutely for Re(λ)>ω and defines an analytic function there.Regarding the fractional derivative, we have the following important property: forf∈C([0,∞);X) such that g1-α*f∈W1,1((0,∞);X),
(12)Dtαf^(λ)=λαf^(λ)-λα-1f(0).
The power function λα  is uniquely defined as λα=|λ|αeiarg(λ), with -π<arg(λ)<π.The Mittag-Leffler function (see, e.g., [14, 15]) is defined as follows:
(13)Eα,β(z):=∑n=0∞‍znΓ(αn+β)=12πi∫Ha‍eμμα-βμα-zdμ,hhhhhhhhhhhhhhhhhhhhhhhhα,β>0,z∈ℂ,
where Ha is a Hankel path, that is, a contour which starts and ends at -∞ and encircles the disc |μ|≤|z|1/α counterclockwise. It is an entire function which provides a generalization of several usual functions, for example,(i)
exponential function:E1,1(z)=ez;(ii)
cosine functions:E2,1(z2)=cosh(z) and E2,1(-z2)=cos(z);(iii)
sine functions:zE2,2(z2)=sinh(z) and zE2,2(-z2)=sin(z).Let k∈ℕ∪{0}. The Laplace transform of the kth-order derivative of the Mittag-Leffler function is given by [15]
(14)∫0∞‍e-λttαk+β-1Eα,β(k)(±ωtα)dt=k!λα-β(λα∓ω)k+1,222222222222222222222222Re(λ)>|ω|1/α.
Using this formula, we obtain for 0<α<1(15)DtαEα,1(ztα)=zEα,1(ztα),t>0,z∈ℂ,
and the identity
(16)ddtEα,1(ztα)=ztα-1Eα,α(ztα).
To see this, it is sufficient to write
(17)ℒ(tα-1Eα,α(ztα))(λ)=1λα-z=1z[λλα-1λα-z-1]
and invert the Laplace transform.The following two definitions are taken from [16, 17], respectively.Definition 1.
LetA be a closed and linear operator with domain D(A) defined on a Banach space X and α>0. We call A the generator of an (α,α)-resolvent family if there exists ω≥0 and a strongly continuous function Pα:[0,∞)→ℬ(X) (resp., Pα:(0,∞)→ℬ(X) in case 0<α<1) such that {λα:Re(λ)>ω}⊂ρ(A) and
(18)(λα-A)-1x=∫0∞‍e-λtPα(t)xdt,Re(λ)>ω,x∈X.
In this case, Pα(t) is called the (α,α)-resolvent family generated by A.Definition 2.
LetA be a closed and linear operator with domain D(A) defined on a Banach space X and α>0. We call A the generator of an (α,1)-resolvent family if there exists ω≥0 and a strongly continuous function Sα:[0,∞)→ℬ(X) such that {λα:Re(λ)>ω}⊂ρ(A) and
(19)λα-1(λα-A)-1x=∫0∞‍e-λtSα(t)xdt,Re(λ)>ω,x∈X.
In this case, Sα(t) is called the (α,1)-resolvent family generated by A.In the above definitions, the integrals involved are taken in the sense of Riemann, more precisely as improper Riemann integrals.By the uniqueness theorem for the Laplace transform, a(1,1)-resolvent family is the same as a C0-semigroup; a (2,2)-resolvent family corresponds to the concept of sine family, while a (2,1)-resolvent family corresponds to a cosine family. See, for example, [18] and the references therein for an overview on these concepts. A systematic study in the fractional case is carried out in [17].Some properties of(Pα(t)) and (Sα(t)) are included in the following lemma. Their proof uses techniques from the general theory of (a,k)-regularized resolvent families [19] (see also [16, 17]). It will be of crucial use in the investigation of mild solutions in Section 4.Lemma 3.
The following properties hold.(i)
Sα(0)=I.(ii)
Sα(t)D(A)⊂D(A) and ASα(t)x=Sα(t)Ax for all x∈D(A),  t≥0.(iii)
For allx∈D(A):Sα(t)x=x+∫0t‍gα(t-s)ASα(s)xds,  t≥0.(iv)
For allx∈X:(gα*Sα)(t)x∈D(A) and
(20)Sα(t)x=x+A∫0t‍gα(t-s)Sα(s)xds,t≥0.(v)
Pα(t)D(A)⊂D(A) and APα(t)x=Pα(t)Ax for all x∈D(A),  t>0.(vi)
For allx∈D(A):Pα(t)x=gα(t)x+∫0t‍gα(t-s)APα(s)xds,  t>0.(vii)
For allx∈X:(gα*Pα)(t)x∈D(A) and
(21)Pα(t)x=gα(t)x+A∫0t‍gα(t-s)Pα(s)xds,t>0.(viii)
For0<α<1,  limt→0t1-αPα(t)=(1/Γ(α))I, and P1(0)=I and for α>1,  Pα(0)=0.Proof.
Letω be as in Definitions 1 or 2. Let μ,  λ>ω and x∈D(A). Then, x=(I-μ-αA)-1y for some y∈X. Since (I-λ-αA)-1 and (I-μ-αA)-1 are bounded and commute and since the operator A is closed, we obtain from the definition of Pα(t)(22)P^α(λ)x=∫0∞‍e-λtPα(t)xdt=P^α(λ)(I-μ-αA)-1y=λ-α(I-λ-αA)-1(I-μ-αA)-1y=(I-μ-αA)-1λ-α(I-λ-αA)-1y=(I-μ-αA)-1P^α(λ)y=∫0∞‍e-λt(I-μ-αA)-1Pα(t)ydt
and, analogously, from the definition of Sα(t)(23)S^α(λ)x=∫0∞‍e-λtSα(t)xdt=S^α(λ)(I-μ-αA)-1y=λ-1(I-λ-αA)-1(I-μ-αA)-1y=(I-μ-αA)-1λ-1(I-λ-αA)-1y=(I-μ-αA)-1S^α(λ)y=∫0∞‍e-λt(I-μ-αA)-1Sα(t)ydt.
Hence, by uniqueness of the Laplace transform,
(24)Pα(t)x=(I-μ-αA)-1Pα(t)(I-μ-αA)x,Sα(t)x=(I-μ-αA)-1Sα(t)(I-μ-αA)x,
for all t>0. From these two equalities and the continuity of Sα on [0,∞), we immediately get (ii) and (v).
On the other hand, from the convolution theorems we obtain, forx∈D(A),
(25)∫0∞‍e-λtgα(t)xdt=λ-αx=P^α(λ)(I-λ-αA)x=P^α(λ)x-λ-αP^α(λ)Ax=∫0∞‍e-λt[Pα(t)x-∫0t‍gα(t-s)Pα(s)Axds]dt,∫0∞‍e-λtxdt=λ-1x=S^α(λ)(I-λ-αA)x=S^α(λ)x-λ-αS^α(λ)Ax=∫0∞‍e-λt[Sα(t)x-∫0t‍gα(t-s)Sα(s)Axds]dt.
The uniqueness theorem for the Laplace transform yields (iii) and (vi).
We now prove (iv) and (vii). Letx∈X and define y=(λ-A)-1x∈D(A), where λ∈ρ(A) is fixed. Let z=(gα*Sα)(t)x,  t≥0. We have to show that z∈D(A) and Az=Sα(t)x-x. Indeed, using (ii) and (iii), we obtain that
(26)z=(λ-A)(gα*Sα)(t)y=λ(gα*Sα)(t)y-A(gα*Sα)(t)y=λ(gα*Sα)(t)y-(Sα(t)y-y)∈D(A),Az=λA(gα*Sα)(t)y-ASα(t)y+Ay=λ(gα*ASα)(t)y-Sα(t)Ay+(λy-x)=λ(gα*ASα)(t)y-Sα(t)(λy-x)+λy-x=λ[(gα*ASα)(t)y-Sα(t)y+y]+Sα(t)x-x=Sα(t)x-x,
proving the claim. Analogously, we prove that (gα*Pα)(t)x∈D(A) and
(27)Pα(t)x=gα(t)x+A∫0t‍gα(t-s)Pα(s)xds.
From the continuity of Sα on [0,∞) and from (iv), we obtain (i). That P1(0)=I and Pα(0)=0 for α>1 follow from (vi) by using the fact that g1(0)=1, gα(0)=0 for α>1, and that the operator A is closed. We notice that (iv) implies that the domain D(A) of the operator A is necessarily dense in X. Now, if x∈D(A), the first assertion in (viii), that is, limt→0t1-αPα(t)x=(1/Γ(α))x, for 0<α<1, follows from (vi), and we obtain that limt→0t1-αPα(t)=(1/Γ(α))I, for 0<α<1, by using this and the fact that D(A) is dense in X. The proof of the lemma is finished.Note that it follows from (vii) and (viii) that(Pα(t)) exhibits a singular behavior at the origin if 0<α<1. However, t↦∥Pα(t)∥ is in Lloc1[0,∞) since by (viii), we have that if t is near zero, then
(28)∥Pα(t)∥≤1Γ(α)t1-α.
Recall the definition of the Wright type function [20, page 10, Formula (28)]:
(29)Φα(z)∶=∑n=0∞‍(-z)nn!Γ(-αn+1-α)=12πi∫γ‍μα-1eμ-zμαdμ,0<α<1,
where γ is a contour which starts and ends at -∞ and encircles the origin once counterclockwise. By [17, page 14], Φα(t) is a probability density function, that is,
(30)Φα(t)≥0,t>0,∫0∞‍Φα(t)dt=1.
We also have Φα(0)=1/Γ(1-α), and as t→+∞, Φα has the following asymptotic expansion:
(31)Φα(t)=Yα-1/2e-Y(∑m=0M-1‍AmY-m+O(Y-M)),
for any M∈ℕ, with Y=(1-α)(ααt)1/(1-α), where Am are real numbers.The following identity holds: for everyα∈(0,1) and s>0,
(32)e-λαs=ℒ(αstα+1Φα(st-α))(λ).
See [20, Formulas (40) and (41)]. We note that the above Laplace transform was formerly first given by Pollard and Mikusinski (see [20] and references therein). For more details on the Wright type functions, we refer to the monographs [17, 20–22] and the references therein.LetX be a Banach lattice with positive cone X+. We recall that a semigroup (T(t))t≥0 on X is positive if for any x∈X+ and t≥0,  T(t)x≥0. Similarly, an operator (A,D(A)) is resolvent positive if there is ω∈ℝ such that (ω,∞)⊂ρ(A) and (λ-A)-1x≥0 for all λ>ω and any x∈X+.It is a well-known fact that a strongly continuous semigroup is positive if and only if its generator is resolvent positive. We finally will need the following result due to Zhang [10].Theorem 4.
LetX be a Banach lattice and A:X→X a nonlinear operator. Suppose that there exists a positive linear bounded operator B:X→X with rσ(B)<1 and
(33)-B(x-y)≤A(x)-A(y)≤B(x-y),
for all x,y∈X,  x≥y. Then, the equation x=A(x) has a unique solution in X.


## 3. Subordination

LetA be a linear closed densely defined operator in a complex Banach space X. If A generates a C0-semigroup (T(t))t≥0 then, A generates an (α,1)-resolvent family (Sα(t)) for all 0<α<1 and they are related by the following formula [17]:
(34)Sα(t)x=∫0∞‍t-αΦα(st-α)T(s)xds,t≥0,x∈X.
A change of variables shows that the above is equivalent to
(35)Sα(t)x=∫0∞‍Φα(τ)T(τtα)xdτ,t≥0,x∈X.
In particular, it follows from the above representation formulas that (Sα(t)) is analytic and Sα(0)=I.Concerning(α,α)-resolvent families, we prove the following important theorem, which is the main result of this section.Theorem 5.
Let0<α<1. If A generates a C0-semigroup (T(t))t≥0, then A generates an (α,α)-resolvent family (Pα(t)) given for every x∈X and t>0 by
(36)Pα(t)x=α∫0∞‍stα+1Φα(st-α)T(s)xds=α∫0∞‍τt1-αΦα(τ)T(τtα)xdτ.
Moreover, for all x∈D(A),  Pα(t)x∈D(A), and
(37)Sα′(t)x=APα(t)x,t>0,x∈X,(38)(g1-α*Pα)(t)x=x+∫0tAPα(s)xds=Sα(t)x,t≥0.Proof.
SinceA generates a C0-semigroup (T(t))t≥0, there exists ω>0 such that {μ:Re(μ)>ω}⊂ρ(A) and
(39)(μ-A)-1x=∫0∞‍e-μtT(t)xdt,x∈X,Re(μ)>ω.
In particular, {λα:Re(λ)>ω1/α}⊂ρ(A). It is clear that (Pα(t)) is strongly continuous (and in fact analytic) for t>0.
We next show thatP^α(λ)=(λα-A)-1 for Re(λ) large enough. In fact, by (32) and Fubini's theorem, we obtain for every x∈X,
(40)P^α(λ)x=∫0∞‍e-λtα∫0∞‍stα+1Φα(st-α)T(s)xdsdt=∫0∞‍(∫0∞‍αe-λtstα+1Φα(st-α)dt)T(s)xds=∫0∞‍e-λαsT(s)xds=(λα-A)-1x,
for all Re(λ) sufficiently large, proving the claim. We conclude that Pα(t) is an (α,α) resolvent family with generator A.
On the other hand, from (35) and the fact that A is closed, we obtain for all x∈D(A) that Pα(t)x∈D(A) and the identity
(41)Sα′(t)x=∫0∞‍ατtα-1Φα(τ)T′(τtα)xdτ=∫0∞‍ατtα-1Φα(τ)AT(τtα)xdτ=APα(t)x,t>0,
proving (37). Integrating the above identity, we obtain
(42)x+∫0t‍AP(s)xds=Sα(t)x,t≥0.
Finally, from (vi) in Lemma 3, we get
(43)(g1-α*Pα)(t)x=(g1-α*gα)(t)x+(g1-α*gα*APα)(t)x=x+(1*APα)(t)x,
proving the theorem.Remark 6.
We observe that the paper [13] uses a different approach for the evolution operators Sα and Pα. More precisely, the authors consider an almost sectorial operator A in a Banach space and give a direct construction using the Mittag-Leffler functions.


## 4. Boundary Conditions

In this section, we give a spectral characterization of existence of mild solutions of (1) with the boundary condition u(0)=u(1). The approach is based on the representation of solutions using the solution families (Sα(t)) and (Pα(t)) of the previous section. Assume that A generates a C0 semigroup (T(t))t≥0. Let Sα(t) and Pα(t) be given by (34) and (36), respectively. The linear fractional equation
(44)Dtαu(t)=Au(t)+f(t),t>0,
with initial condition
(45)u(0)=u0,
has the unique (classical) solution u given by
(46)u(t)=Sα(t)u0+∫0t‍Pα(t-s)f(s)ds,
whenever f∈W1,1(ℝ+;X) and u0∈D(A). Indeed, note that by (37),
(47)u′(t)=Sα′(t)u0+Pα(t)f(0)+(Pα*f′)(t),t>0.
Hence, using (37), (38), and Lemma 3, we obtain
(48)Dtαu(t)=(g1-α*u′)(t)=(g1-α*APα)(t)u0+(g1-α*Pα)(t)f(0)+(g1-α*Pα*f′)(t)=ASα(t)u0+Sα(t)f(0)+(Sα*f′)(t)=ASα(t)u0+f(t)+(Sα′*f)(t)=ASα(t)u0+A(Pα*f)(t)+f(t)=Au(t)+f(t).Note that forα=1, the representation (46) is nothing else but the well-known variation of constant formula for the abstract Cauchy problem of first order, S1(t)≡P1(t) and corresponds exactly to T(t), the C0-semigroup generated by A.Definition 7.
LetA with domain D(A) be a closed linear operator on a Banach space X. Let f∈Lloc1((0,∞);X) and 0<α<1. Let Iα be the operator defined in (9). A function u∈C([0,∞);X) is called a mild solution of the equation
(49)Dtαu(t)=Au(t)+f(t),t>0,u(0)=x,
if Iαu(t)∈D(A),t≥0 and
(50)u(t)=x+A∫0t‍gα(t-s)u(s)ds+∫0t‍gα(t-s)f(s)ds.
Equivalently,
(51)u(t)=x+AIαu(t)+Iαf(t),t≥0.We have the following representation of mild solutions.Lemma 8.
Suppose that the operatorA generates a C0-semigroup (T(t))t≥0, and let f∈Lloc1((0,∞);X) such that the mapping t↦∫0t‍f(s)ds is exponentially bounded. Let u∈C([0,∞);X), 0<α<1 and u(0)∶=x∈X. Then, the following assertions are equivalent:(i)
Iαu(t)∈D(A),  t≥0 and u(t)=x+AIαu(t)+Iαf(t),  t≥0, that is, u is a mild solution of (49),(ii)
u(t)=Sα(t)x+∫0t‍Pα(t-s)f(s)ds for all t>0.Note that the mappingt↦∫0t‍f(s)ds is exponentially bounded if, for example, the function f∈∪p≥1Lp((0,∞);X) or f itself is exponentially bounded.Proof of Lemma8.
(i)⇒ (ii): Assume that assertion (i) holds. Then, u(t)-u(0)=AIαu(t)+Iαf(t). Taking the Laplace transform of this equality, we get that, u^(λ)-1/λu(0)=Aλ-αu^(λ)+λ-αf^(λ), that is, u^(λ)-Aλ-αu^(λ)=1/λu(0)+λ-αf^(λ). Therefore, (I-Aλ-α)u^(λ)=1/λu(0)+λ-αf^(λ), and λ-α(λα-A)u^(λ)=1/λu(0)+λ-αf^(λ). Hence,
(52)u^(λ)=1λλα(λα-A)-1u(0)+λαλ-α(λα-A)-1f^(λ)=λα-1(λα-A)-1u(0)+(λα-A)-1f^(λ).
Taking the inverse Laplace transform of this equality, we get the assertion (ii).
(ii)⇒ (i): As a consequence of  (iv) and (vii) in Lemma 3, we have
(53)Iαu(t)=(gα*u)(t)=(gα*Sα)(t)x+(gα*Pα*f)(t)∈D(A),A(gα*u)(t)=A(gα*Sα)(t)x+A(gα*Pα*f)(t)=Sα(t)x-x+[A(gα*Pα)*f](t)=Sα(t)x-x+(Pα-gα)*f(t)=Sα(t)x-x+(Pα*f)(t)-(gα*f)(t)=u(t)-x-(gα*f)(t),
proving the lemma.Uniqueness of the classical solution follows from the lemma upon observing that any classical solution is necessarily a mild solution.The following problem was considered by Prüss [1] when α=1 and A generates a strongly continuous semigroup. If one starts with f∈C([0,1];X) and solves the problem u′(t)=Au(t)+f(t) with the boundary condition u(0)=u(1), then the resulting solution can be extended to a periodic continuous function on ℝ. We observe that Haraux [2] had considered similar problems earlier.For the fractional differential Equation (44), we obtain a mild solution on [0,∞). In the next result (Theorem 9), we obtain a necessary and sufficient condition that the mild solution will satisfy the boundary condition u(0)=u(1).We remark that the concept of periodic boundary valued solutions for fractional differential equations has been introduced in the literature by  Belmekki et al.  in the paper [9] as described in the introduction. In this line of research, we note that the paper [23] by Kaslik and Sivasundaram considers existence and nonexistence of periodic solutions of fractional differential equations for various definitions of the fractional derivative.We consider the following problem(54)Dtαu(t)=Au(t)+f(t),t∈(0,1),0<α<1,hhhhhhhhhhhhhhhhhhhhhhhhhhu(0)=u(1).Theorem 9.
LetX be a Banach space, and assume that A generates a C0 semigroup (T(t))t≥0. Let (Sα(t)) be the subordinated (α,1)-resolvent family. Then, 1∈ρ(Sα(1)) if and only if for any f∈C([0,1];X); (54) admits precisely one mild solution.Proof.
Suppose that1∈ρ(Sα(1)). Note that if the solution u of the differential equation in (54) satisfies the condition u(0)=u(1), then Lemma 8(ii) implies
(55)(I-Sα(1))u(0)=∫01‍Pα(1-s)f(s)ds.
Hence,
(56)u(0)=(I-Sα(1))-1∫01‍Pα(1-s)f(s)ds.
We notice that the existence of solutions follows, if one defines u(0) by (56).
Conversely, defineKα:C([0,1];X)→C([0,1];X) by means of (Kαf)(t)=u(t), where u(t) denotes the unique mild solution of (54). It is clear that Kα is linear and everywhere defined. Moreover, it is not difficult to show, using the closed graph theorem, that Kα is bounded. Now, for x∈X, consider
(57)fα(t)x∶=(α-1)g2-α(t)x-α(g1-α*Sα)(t)x
and define Qαx∶=(Kαfα)(0)x. Clearly, Qα:X→X is linear and bounded. We claim that
(58)∫0t‍Pα(t-s)fα(s)xds=tSα(t)x,t>0.
Indeed, the Laplace transform of fα is given by
(59)f^α(λ)=(α-1)λα-2-αλα-1λα-1(λα-A)-1.
Let If(t) be the left-hand side of (58). Then, taking the Laplace transform and using (59), we get that
(60)I^f(λ)=(λα-A)-1f^α(λ)=(α-1)λα-2(λα-A)-1-αλα-1λα-1(λα-A)-2=ddλ(λα-1(λα-A)-1)=ddλ(Sα(λ)).
From the uniqueness of the Laplace transform, we obtain (58) and the claim is proved. Now, using (55) and (58), we get that
(61)(I-Sα(1))(Qαx+x)=∫01‍Pα(1-s)fα(s)xds+x-Sα(1)x=Sα(1)x+x-Sα(1)x=x.
This shows that I-Sα(1) is surjective. Now, let x0∈X  be  such that (I-Sα(1))x0=0. Then, using (58), we get that
(62)x0=Sα(1)x0=∫01‍Pα(1-s)fα(s)x0ds.
Using (62), (55) (which follows from Lemma 8(ii)), and (58) again, we get that
(63)0=(I-Sα(1))x0=∫01‍Pα(1-s)fα(s)x0ds=Sα(1)x0.
We have shown that x0=0 proving that I-Sα(1) is injective. Hence, I-Sα(1) is invertible and the proof is finished.Remark 10.
An alternative proof of the injectivity of(I-Sα(1)) in the preceding proof runs as follows: let x0∈X  be  such that (I-Sα(1))x0=0 and set u(t):=Sα(t)x0. Then, u is a mild solution of (54) with the forcing term f=0. Since the function u=0 is also of mild solution of (54) (with the forcing term f=0), the uniqueness of the solution yields x0=u(0)=0, proving that I-Sα(1) is injective.We remark that the condition1∈ρ(Sα(1)) is trivially satisfied if ∥Sα(1)∥<1.Corollary 11.
Suppose that the operatorA generates a C0-semigroup (T(t))t≥0 satisfying limt→∞∥T(t)∥=0. Then, for f∈C([0,1];X), (54) admits exactly one mild solution.Proof.
First, observe thatSα(1)x=∫0∞‍Φα(s)T(s)xds,  x∈X and recall that Φα(s)≥0,  s≥0, and ∫0∞‍Φα(s)ds=1. Since Φα(z) is a nonzero analytic function, it follows that for each τ>0, we have ∫τ∞‍Φα(s)ds>0. We first assume that (T(t))t≥0 is contractive, that is, ∥T(t)∥≤1,  t≥0. Then, for x∈X,
(64)∥Sα(1)x∥=∥∫0∞‍Φα(s)T(s)xds∥≤∫0∞‍Φα(s)∥T(s)x∥ds≤∫0τ‍Φα(s)ds∥x∥+∫τ∞‍Φα(s)∥T(s)x∥ds.
Let 0<ɛ<1. We can choose τ>0 such that ∥T(s)∥<ɛ for all s≥τ. It follows that
(65)∥Sα(1)x∥≤∫0τ‍Φα(s)ds∥x∥+∫τ∞‍Φα(s)∥T(s)x∥ds≤(∫0τ‍Φα(s)ds+ɛ∫τ∞‍Φα(s)ds)∥x∥≤(1+(ɛ-1)∫τ∞‍Φα(s)ds)∥x∥.
Therefore, ∥Sα(1)∥<1 and hence, I-Sα(1) is invertible. If (T(t))t≥0 is not contractive, we renorm the space X with ∥|x|∥=supt≥0∥T(t)x∥. This norm is equivalent to the original one and ∥|T(t)|∥≤1,t≥0. The proof is complete.Note that by [24, Prop. V.1.2 & V.1.7], the assumption on (T(t))t≥0 is equivalent to the fact that (T(t))t≥0 is exponentially stable.In the following examples, the semigroups are exponentially stable.Example 12.
(1) LetΩ⊂ℝN be a bounded open set. Define the operator AD on L2(Ω) by
(66)D(AD)∶={u∈H01(Ω),Δu∈L2(Ω)},ADu∶=Δu.
Then, AD is a realization of the Laplace operator on L2(Ω) with Dirichlet boundary conditions and it generates a C0-semigroup on L2(Ω) which is exponentially stable. Moreover, the semigroup interpolates on all Lp(Ω) and each semigroup on Lp(Ω)  (1≤p<∞) is also exponentially stable (for a complete description we refer, e.g., to [25]).
(2) Assume thatΩ⊂ℝN is a bounded domain with a Lipschitz continuous boundary ∂Ω and let γ∈L∞(∂Ω) satisfy γ(x)≥γ0>0 for some constant γ0. Define the bilinear form aγ on L2(Ω) by
(67)aγ(u,v)=∫Ω‍∇u∇vdx+∫∂Ω‍γuvdσ,u,v∈H1(Ω).
Then, the operator Aγ on L2(Ω) associated with the form aγ in the sense that
(68)D(Aγ)={u∈H1(Ω),∃f∈L2(Ω),aγ(u,v)=(f,v)L2(Ω)∀v∈H1(Ω)},Aγu=-f
is a realization of the Laplace operator with Robin boundary conditions. As in part (1), this operator generates an exponentially stable C0-semigroup in L2(Ω) which interpolates on Lp(Ω) and each semigroup is exponentially stable in Lp(Ω)  (1≤p<∞).
(3) LetΩ and γ be as in part (2). Let A be an elliptic operator in divergence form of the type
(69)Au=∑i,j=1N‍Di(aij(x)Dju)=div(a(x)∇u),
where aij (i,j=1,…,N) are real valued bounded measurable functions such that aij(x)=aji(x) and there exists a constant α0>0 such that ∑i,j=1Naij(x)ξiξj≥α0|ξ|2 holds for all ξ∈ℝN and almost every x in Ω. Let ΔΓu=divΓ(∇Γu) be the Laplace-Beltrami operator on the boundary, where ∇Γu denotes the tangential gradient at the boundary ∂Ω. Define the bilinear symmetric form 𝒜 with domain ℋ1(Ω)∶={U=(u,u∣∂Ω):u∈H1(Ω),u∣∂Ω∈H1(∂Ω)} on the product space L2(Ω)×L2(∂Ω) by
(70)𝒜(u,v)=∑i,j=1N‍∫Ω‍aij(x)DiuDjvdx+∫∂Ω‍∇Γu∇Γvdσ+∫∂Ω‍γ(x)uvdσ.
It is straightforward to show that 𝒜 is closed and the operator on L2(Ω)×L2(∂Ω) associated with it generates a contraction C0-semigroup with generator A2 given by
(71)D(A2)={U∈ℋ1(Ω),Au∈L2(Ω),ΔΓu-∂νau-γuexistsinL2(∂Ω)},A2u=(Au,ΔΓu-∂νau-γu),
while ν→ is the unit outer normal and ∂νau∶=(a(x)∇u)·ν→ is the conormal derivative of u with respect to the matrix a(x)=(aij(x))1≤i,j≤N. Moreover, the semigroup interpolates on all Lp(Ω)×Lp(∂Ω), and each semigroup is contractive and exponentially stable for every p∈[1,∞).


## 5. Positivity of Solutions and the Semilinear Equation

Throughout this section,X will be a real Banach lattice.It was shown by Nieto [8, 26] that if α∈(0,1), λ∈ℝ,  Eα,α(λ)<1/Γ(α), and u  is  such that
(72)𝔻tαu(t)-λu(t)≥0,
where 𝔻tα is the Riemann-Liouville fractional derivative; then u(t)≥0 for t∈(0,1]. Motivated by Nieto's result, we show in this section that if we consider the Caputo fractional derivative in (72) then the same type of result holds assuming that Eα,1(λ)<1 instead of Eα,α(λ)<1/Γ(α).We begin with the following maximum principle based only on initial conditions.Proposition 13.
Letα∈(0,1) and A be the generator of a positive C0-semigroup (T(t))t≥0. Assume that u is a mild solution of (49) such that
(73)Dtαu(t)-Au(t)≥0,u(0)≥0.
Then, u(t)≥0 for all t∈[0,1].Proof.
Letx∈X+. By the subordination formulae (34) and (36) and the fact that Φα is a probability density on [0,∞), we obtain Sα(t)x≥0 and Pα(t)x≥0, respectively. Using the representation of the solution given in Lemma 8(ii), we see that u(t)≥0 for all t∈[0,1].Proposition 14.
Letα∈(0,1) and A be the generator of a positive C0-semigroup (T(t))t≥0. Suppose that (I-Sα(1))-1x≥0 for all x∈X+. Assume that u is a mild solution of (49) and that
(74)Dtαu(t)-Au(t)≥0,u(0)=u(1).
Then, u(t)≥0 for all t∈[0,1].Proof.
Under the given hypothesis, we haveu(0)≥0 by (56). Hence, the result follows from Proposition 13.We mention that the condition in (73) or in (74) is equivalent to f(t)≥0, since by hypothesis, u is a mild solution of (49).Note that if the spectral radius ofSα(1), that is, rσ(Sα(1)), is less than 1, then the inverse (I-Sα(1))-1 is given by the Neumann series ∑n=0∞(Sα(1))n, from which it follows that ∑n=0∞(Sα(1))n≥0, since Sα(1)≥0.The following corollary shows that the result obtained by Nieto (mentioned at the beginning of this section) using the Riemann-Liouville fractional derivative also holds for the Caputo fractional derivative.Corollary 15.
Letα∈(0,1) and λ∈ℝ. Suppose that Eα,1(λ)<1. Assume that u is a mild solution of (54) with A=λI and satisfies
(75)Dtαu(t)-λu(t)≥0,∀t∈(0,1),u(0)=u(1).
Then, u(t)≥0 for all t∈[0,1].Proof.
By Definition2 and formula (14), we have Sα(t)=Eα,1(λtα). Moreover, the semigroup generated by A is given by T(t)=eλtI which is positive. The condition Eα,1(λ)<1 implies that (I-Sα(1))-1x=(1-Eα,1(λ))-1x≥0 for all x∈X+. Hence, the result follows from Proposition 14.Now, we consider on a Banach latticeX the following semilinear problem:
(76)Dtαu(t)=Au(t)+f(t,u(t)),0≤t≤τ,0<α<1,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhu(0)=x∈X,
where A is the generator of a positive semigroup (T(t))t≥0, f:[0,τ]×X→X is a locally integrable given function, and τ>0 is a fixed real number.Ifu∈C([0,τ];X) satisfies the integral equation
(77)u(t)=Sα(t)u(0)+∫0t‍Pα(t-s)f(s,u(s))ds,t>0,
then u is called the mild solution of the semilinear problem (76) on [0,τ]. The following is the main result in this section.Theorem 16.
LetA generate a positive C0-semigroup (T(t))t≥0 on a Banach lattice X. Let τ>0 be a fixed real number. Suppose that there exist constants M≥0, N≥0 such that for all s∈[0,τ] and x1,  x2∈X with x1≥x2,
(78)-M(x1-x2)≤f(s,x1)-f(s,x2)≤N(x1-x2).
Then, (76) has a unique mild solution in C([0,τ];X).Proof.
Define the operatorQα:C([0,τ];X)→C([0,τ];X) by
(79)(Qαu)(t)=Sα(t)u(0)+∫0t‍Pα(t-s)f(s,u(s))ds,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhht∈[0,τ].
Then, u∈C([0,τ];X) is the mild solution of (76) if and only if u=Qαu. Thus, existence of mild solutions is achieved by proving that Qα has a fixed point.
For anyu,  v∈C([0,τ];X),  u≤v, and t∈[0,τ], we have
(80)(Qαv)(t)-(Qαu)(t)≥-∫0t‍Pα(t-s)M(v(s)-u(s))ds,(Qαv)(t)-(Qαu)(t)≤∫0t‍Pα(t-s)N(v(s)-u(s))ds.
Let C∶=max{M,N}; then,
(81)-Bα(v(t)-u(t))≤(Qαv)(t)-(Qαu)(t)≤Bα(v(t)-u(t)),t∈[0,τ],
where
(82)(Bαx)(t)∶=∫0t‍Pα(t-s)x(s)ds.
Note that since the semigroup (T(t))t≥0 is positive, then (Pα(t)) is positive, and hence, Bα is a positive operator. We will show that rσ(Bα)=0. Indeed, using (28), we have that for any t∈[0,τ],
(83)∥(Bαx)(t)∥≤∫0t‍∥Pα(t-s)∥∥x(s)∥ds≤1Γ(α)sups∈[0,τ]∥x(s)∥∫0t‍(t-s)α-1ds=tαΓ(α)α∥x∥.
Let 𝔹 denote the beta function defined by
(84)𝔹(γ,ρ)=∫01‍sγ-1(1-s)ρ-1ds.
It is well known that 𝔹(γ,ρ)=Γ(γ)Γ(ρ)/Γ(γ+ρ). Therefore,
(85)∥(Bα2x)(t)∥≤∫0t‍∥Pα(t-s)∥∥(Bαx)(s)∥ds≤1Γ(α)α∥x∥∫0t‍(t-s)α-1sαds=t2αΓ(α)α∥x∥∫01‍(1-s)α-1sαds=t2αΓ(α)α∥x∥𝔹(α+1,α)=t2αΓ(α)αΓ(α+1)Γ(α)Γ(2α+1)∥x∥=t2αΓ(α)Γ(2α+1)∥x∥.
By induction, it is easy to prove that for any n∈ℕ, we have
(86)∥(Bαnx)(t)∥≤(tα)n(Γ(α))n-1Γ(nα+1)∥x∥,t∈[0,τ].
Hence, ∥Bαnx∥≤((τα)n(Γ(α))n-1/Γ(nα+1))∥x∥, and consequently ∥Bαn∥≤(τα)n(Γ(α))n-1/Γ(nα+1). Therefore,
(87)rσ(Bα)∶=limn→∞∥Bαn∥1/n≤limn→∞τα(Γ(α))1-1/n(Γ(nα+1))1/n=0.
Hence, by Theorem 4 the result follows.Remark 17.
The above result remains valid if instead off(t,u(t)) we consider the more general nonlinearity of [10], namely,
(88)f(t,u(t),∫0t‍k1(t,s)g1(s,u(s))ds,∫0τ‍k2(t,s)g2(s,u(s))ds),
where k1 and k2 are nonnegative and satisfy k1∈C(D), k2∈C([0,τ]×[0,τ]), with D∶={(t,s)∈[0,τ]×[0,τ]:s≤t, the functions gj(s,·)(s∈[0,τ],j=1,2) are nondecreasing, and there are constants Mj,Nj,L≥0 such that for every s∈[0,τ] and xj,yj,zj∈X(j=1,2),  x1≥x2,  y1≥y2,  and z1≥z2, we have
(89)-M1(x2-x1)-M2(y2-y1)≤f(s,x2,y2,z2)-f(s,x1,y1,z1),f(s,x2,y2,z2)-f(s,x1,y1,z1)≤N1(x2-x1)+N2(y2-y1),g1(s,x2)-g1(s,x1)≤L(x2-x1).



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