Existence Results for Nonlinear Fractional Problems with Non-Homogeneous Integral Boundary Conditions

mathematics Article Existence Results for Nonlinear Fractional Problems with Non-Homogeneous Integral Boundary Conditions Alberto Cabada 1,*,† and Om Kalthoum Wanassi 2,† 1Departamento de Estatística, Análise Matemática e Optimización Instituto de Matemáticas, Facultade de Matemáticas, Universidade de Santiago de Compostela, 15782 Santiago de Compostela, Spain 2Department of Mathematics, University of Monastir, Monastir 5000, Tunisia; [email protected] *Correspondence: [email protected]; Tel.: +34-881-813206 † These authors contributed equally to this work. Received: 9 January 2020; Accepted: 7 February 2020; Published: 14 February 2020   Abstract: This paper deals with the study of the existence and non-existence of solutions of a three-parameter family of nonlinear fractional differential equation with mixed-integral boundary value conditions. We consider the α -Riemann-Liouville fractional derivative, with α∈( 1, 2 ] . To deduce the existence and non-existence results, we first study the linear equation, by deducing the main properties of the related Green functions. We obtain the optimal set of parameters where the Green function has constant sign. After that, by means of the index theory, the nonlinear boundary value problem is studied. Some examples, at the end of the paper, are showed to illustrate the applicability of the obtained results. Keywords: fractional equations; Green functions; integral boundary conditions; fixed-point index; existence and non-existence. 1. Introduction In recent years, fractional calculus has been applied to a huge number of fields in science, engineering, and mathematics. Some of the areas where fractional calculus has made a profound impact include biology, physics, viscoelasticity, rheology, electrical, engineering, electrochemistry and control theory, see for instance [1–7]. Integral boundary conditions have been considered, for instance, in [ 8 ], where various applications in applied fields such as chemical engineering, thermoelasticity, population dynamics are explained. The existence of solutions of nonlinear boundary value problem coupled with integral boundary conditions in ordinary and fractional cases has been widely studied by many authors, see for example [ 9 – 14 ] and the references therein. In 2009, Ahmad and Nieto [ 15 ] obtained some existence results for the following nonlinear fractional integrodifferential equations with integral boundary conditions:        CDqx(t) = f(t,x(t),(χx)(t)), 0 <t<1, 1 <q≤2, αx(0) + βx0(0) = R1 0q1(x(s))ds, αx(1) + βx0(1) = R1 0q2(x(s))ds, where f:[0, 1]×X×X−→ X, for γ:[0, 1]×[0, 1]−→ [0, ∞), (χx)(t) = Zt 0γ(t,s)x(s)ds, Mathematics 2020,8, 255; doi:10.3390/math8020255 www.mdpi.com/journal/mathematics Mathematics 2020,8, 255 2 of 13 q1 , q2:X−→ Xα≥ 0, β≥ 0 are real numbers and X is a Banach space, by employing to Guo-Krasnoselskii fixed-point theorem and contraction mapping principle. In [16], it is studied the following nonlinear problem involving nonlinear integral conditions:        CDαy(t) = f(t,y(t)),t∈[0, T], 1 <α≤2, y(0)−y0(0) = RT 0g(s,y(s)ds, y(T)−y0(T) = RT 0h(s,y(s))ds. Here, f , g and h:[ 0, T]×E−→ E are given functions that satisfy suitable assumptions and E is a Banach space. By means of the technique associated with measures of non-compactness and the fixed-point theorem of Monch type, it is proved the existence of solutions of the problem. In [ 17 ], it is considered the following nonlinear fractional differential equations with boundary value conditions (Dαu(t) + g(t)f(t,u(t)) = 0, 0 <t<1, u(0) = 0, u0(1) = R1 0h(t)u(t)dt, where 1 <α≤ 2, g∈C(( 0, 1 ) , [ 0, ∞)) and g may be singular at t= 0 or/and at t= 1, h∈L1[ 0, 1 ] and f∈C[ 0, 1 ]×[ 0, 1 ] , [ 0, ∞)) . The authors derive the Green function associated with the above problem and sharp estimates on it are established. Thus, by using fixed-point theorem in cones, they proved some results on the existence of positive solutions. In this paper, the following nonlinear fractional differential equation with non-homogeneous integral boundary conditions is considered:    Dαu(t)−λu(t) + f(t,t2−αu(t)) = 0, t∈I:= [0, 1], lim t→0+t2−αu(t) = µR1 0u(s)ds,u0(1) = ηR1 0u(s)ds.(1) Here λ∈R , µ , η≥ 0, Dα , 1 <α≤ 2, is the Riemann-Liouville fractional derivative and f∈C([0, 1]×[0, ∞),[0, ∞)). We look for solutions u:I→R such that function t2−αu(t)∈C1(I) . Thus, as a direct consequence, we deduce that, in particular, u∈C1(( 0, 1 ]) . Moreover, it may be discontinuous at t=0. We are interested in to prove the existence and non-existence of solutions of the treated problem. To this end, we will use the classical index theory [ 18 – 20 ], in the line of the papers [ 21 – 25 ], where it is used for ordinary differential equations. The main tool is the construction of the Green function related to the corresponding linear problem    Dαu(t)−λu(t) + y(t) = 0, t∈I, lim t→0+t2−αu(t) = µR1 0u(s)ds,u0(1) = ηR1 0u(s)ds.(2) It is important to mention that the results obtained in [ 26 ] are fundamental in the development of our results. In such reference we consider the homogeneous case µ=η= 0, i.e., with no integral boundary conditions. There, we obtain the explicit expression of its related Green function (denoted by G1 on this paper). Moreover, we deduce some sharp properties concerning its constant sign, compiled on Lemmas 1and 2. The main properties of the Green function G related to the considered problem (2) are derived from the ones proved in [ 26 ] for G1 . Therefore, we use the qualitative properties obtained on those reference and study the parameter relationship between α , λ , µ and η that ensure the constant sign of the Green function related to the linear problem (2) . We follow similar arguments to the ones used on [12,27–30]. Mathematics 2020,8, 255 3 of 13 The paper is scheduled as follows: after some introductory results, we study, in Section 3, the related linear equation and deduce suitable properties on the qualitative behavior and constant sign of the related Green function. Next section is devoted to ensuring the existence and non-existence of solutions of the considered nonlinear boundary value problem. The results follow from index theory. Finally, in last section, some examples are given to point out the applicability of the obtained results. 2. Preliminary Results In this section, we introduce some notations and definitions that which we need in later. Definition 1 ([ 31 ]) . The Riemann-Liouville fractional integral of order α> 0for a measurable function f:(0, +∞)→Ris defined as Iαf(t) = 1 Γ(α)Zt 0(t−s)α−1f(s)ds,t>0, where Γis the Euler Gamma function, assuming that the right-hand side is point-wise defined on (0, +∞). Definition 2 ([ 31 ]) . The Riemann-Liouville fractional derivative of order α> 0for a measurable function f:(0, +∞)→Ris defined as Dαf(t) = 1 Γ(n−α)(d dt )nZt 0(t−s)n−α−1f(s)ds = ( d dt )nIn−αf(t), provided that the right-hand side is point-wise defined on the interval ( 0, +∞) . Here n= [α] + 1, where [α] denotes the integer part of the real number α. Let C(I) be the Banach space of all continuous functions defined on I endowed with the maximum norm kfk=: max{|f(t)|:t∈I} . Define for t∈I , fγ(t) = tγf(t) , if t> 0 and fγ( 0 ) = limt→0+tγf(t) , provided it exists. Let Cγ(I) , γ≥ 0 be the space of all functions f such that fγ∈C(I) . It is well known that Cγ(I)is a Banach space endowed with the norm kfkγ=: max{tγ|f(t)|:t∈I}. 3. Linear Problem This section is devoted to the study of the linear problem (2) . More concisely, we deduce the exact values of the parameter λ for which the Green function G satisfies a strong positiveness condition. In a first moment we introduce the concept of Mittag-Leffler function. Definition 3 ([ 31 ], p. 42) . A two-parameter function of the Mittag − Leffler Eα,β(x) is defined by the series expansion Eα,β(x) = ∞ ∑ k=0 xk Γ(αk+β),α,β>0, x∈R. For β=1, Eα,1 coincides with the usual Mittag−Leffler function Eα. It is not difficult to verify that, provided Eα,α−1(λ)6=0, v1(t) = Γ(α−1)tα−2Eα,α−1(λtα)−Eα,α−2(λ) Eα,α−1(λ)tα−1Eα,α(λtα)(3) Mathematics 2020,8, 255 4 of 13 is the unique solution of the problem    Dαv1(t)−λv1(t) = 0, t∈I, lim t→0+t2−αv1(t) = 1, v0 1(1) = 0, (4) and v2(t) = tα−1Eα,α(λtα) Eα,α−1(λ). (5) the unique one of    Dαv2(t)−λv2(t) = 0, t∈I, lim t→0+t2−αv2(t) = 0, v0 2(1) = 1. (6) Moreover, as it is showed in Theorem 6 in [26], if Eα,α−1(λ)6=0, the unique solution of problem    Dαv(t)−λv(t) + y(t) = 0, t∈I, lim t→0+t2−αv(t) = v0(1) = 0, follows the expression v(t) = Z1 0G1(t,s)y(s)ds, with G1(t,s) =    tα−1Eα,α(λtα)Eα,α−1(λ(1−s)α) (1−s)2−αEα,α−1(λ)−(t−s)α−1Eα,α(λ(t−s)α), 0 ≤s≤t≤1, tα−1Eα,α(λtα)Eα,α−1(λ(1−s)α) (1−s)2−αEα,α−1(λ), 0 ≤t<s<1. (7) In order to characterize the uniqueness of solutions of Problem (2) , we denote (See Figures 1 and 2) θ≡θ(α,λ):=Z1 0v1(t)dt and σ≡σ(α,λ):=Z1 0v2(t)dt. Theorem 1. Let y∈C( 0, 1 ]∩L∞( 0, 1 ) ,1 <α≤ 2, µ , η≥ 0and λ∈R be such that Eα,α−1(λ)6= 0and 1−µθ −ησ 6=0. Then problem (2) has a unique solution u ∈C1 2−α(I), given by u(t) = Z1 0G(t,s)y(s)ds, where G(t,s) = G1(t,s) + (µv1(t) + ηv2(t)) (1−µθ −ησ)Z1 0G1(r,s)dr, (8) with v1, v2and G1given in (3),(5)and (7)respectively. Proof. Arguing in a similar way as in Theorem 6 in [26], we deduce that u(t) = Z1 0G1(t,s)y(s)ds +µZ1 0u(s)dsv1(t) + ηZ1 0u(s)dsv2(t). (9) Let us denote R1 0u(s)ds =A. Then, from the previous equality, we deduce that A=Z1 0u(t)dt =Z1 0Z1 0G1(t,s)y(s)dsdt +A(µθ +ησ), Mathematics 2020,8, 255 5 of 13 that is, since 1 −µθ −ησ 6=0, A=R1 0R1 0G1(t,s)y(s)dsdt (1−µθ −ησ). Replacing Ain (9), we obtain the following expression of the function u u(t) = Z1 0G1(t,s)y(s)ds +R1 0R1 0G1(t,s)y(s)dsdt (1−µθ −ησ)(µv1(t) + ηv2(t)). (10) According to Fubini’s Theorem, we have u(t) = Z1 0G1(t,s) + µv1(t) + ηv2(t) (1−µθ −ησ)Z1 0G1(t,s)dty(s)ds =Z1 0G(t,s)y(s)ds, and the result is concluded. In our approach, we need the following properties of G1(t,s)proved in Lemma 8 in [26]. Lemma 1. Let G1 be the Green function given in (7) and λ∗ 1 be the first negative zero of Eα,α−1(λ) = 0. Then for 1<α≤2, it is satisfied that G1(t,s)>0f or all t,s∈(0, 1)i f and only i f λ>λ∗ 1. Lemma 2. Let G1 be the Green function given in (7), 1 <α≤ 2and λ>λ∗ 1 . Then there exists a positive constant M and a continuous function m such that m(t)> 0on ( 0, 1 ] and m( 0 ) = 0, for which the following inequalities are fulfilled: m(t)≤t2−αG1(t,s) s(1−s)α−2≤M,f or all t,s∈(0, 1). (11) Next, we prove the following properties for the Green function G(t , s) . To this end, in Table 1, by means of numerical approach, we give some values of θ and σ for 1 <α≤ 2 for which we ensure that the Green function G(t,s)has a constant sign. Table 1. Somevalues of θand σfor 1 <α≤2. α1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 λ−5−4.5 −4−3.5 −3−2.5 −2−1.5 −1−0.5 θ(α,λ)12.647 5.62054 2.83501 1.47577 0.657725 −0.0336574 −1.21724 54.6033 2.13745 1.20846 σ(α,λ)−4.52049 −1.94567 −0.99204 −0.688973 −0.638201 −0.777128 −1.40827 38.7227 1.22996 0.630732 Lemma 3. Let G be the Green function related to problem (2) and λ∗ 1 be the first negative zero of Eα,α−1(λ) = 0. Then for (1−µθ −ησ)>0and 1<α≤2, the following properties hold: 1. G is a continuous function on (0, 1]×[0, 1). 2. If λ>λ∗ 1then G(t,s)>0for all t,s∈(0, 1) 3. Consider the function m(t) and the positive content M , introduced in Lemma 2. Then the following inequality holds: m(t)≤t2−αG(t,s) s(1−s)α−2≤M0,for all t,s∈(0, 1), (12) with M0=M1+L (1−µθ −ησ)(α−1) Mathematics 2020,8, 255 6 of 13 and L=µkv1k2−α+ηkv2k2−α. -205 -10 20 Figure 1. Graph of θfor α=1.1 (blue) and α=1.5 (orange). -205 -4 2 Figure 2. Graph of σfor α=1.1 (blue) and α=1.5 (orange). Proof. 1. It is obvious from the continuity of G1,v1and v2. 2. From Lemma 1, G1(t , s)> 0 for all t , s∈( 0, 1 ) . Moreover, from Lemma 18 in [ 26 ], we have that v1and v2are positive on (0, 1]. And so, Property 2 holds immediately from expression (8). Mathematics 2020,8, 255 7 of 13 3. From Lemma 2and for t∈(0, 1]and s∈(0, 1), since (1−µθ −ησ)>0, we have t2−αG(t,s) = t2−αG1(t,s) + t2−α(µv1(t) + ηv2(t)) (1−µθ −ησ)Z1 0G1(t,s)dt ≥t2−αG1(t,s) ≥s(1−s)α−2m(t). Now, using again Lemma 2, from equation (8) we obtain t2−αG(t,s)≤s(1−s)α−2M+L (1−µθ −ησ)s(1−s)α−2MZ1 0rα−2dr =s(1−s)α−2M1+L (1−µθ −ησ)(α−1). Which completes the proof. 4. Nonlinear Problem 4.1. Existence of Solutions This section is devoted to proving the existence of at least one solution of the nonlinear fractional differential equation with non-homogeneous integral boundary conditions (1). To this end, we will apply the classical index theory. Let K be a cone in a Banach space X . If Ω is a bounded open subset of K , we denote by Ω and ∂Ω the closure and the boundary relative to K . When D is an open bounded subset of X we write DK=D∩K, an open subset of K. The following result is well-known in fixed index theory for completely continuous operators T (i.e., continuous and T(S) compact for each bounded subset S⊂K ). See for example [ 18 , 20 , 32 ] for further information. Lemma 4. Let D be an open bounded set with DK6=∅ and DK6=K . Assume that T:DK→K is a completely continuous operator such that x6=Tx for x∈∂DK . Then the fixed-point index iK(T , DK) has the following properties. (1) If there exists e ∈K\{0}such that x 6=Tx +µe for all x ∈∂DKand all µ>0, then iK(T,DK) = 0. (2) If γx6=Tx for all x ∈∂DKand all γ≥1, then iK(T,DK) = 1. (3) Let D1be open in X such that D1⊂DK. Then iK(T,DK) = iK(T,D1) + iK(T,DK\D1). (4) If iK(T,DK)6=0then there exists u ∈DKsuch that u =Tu. We assume the following regularity for the nonlinear part of the equation: (H1)f:[0, 1]×[0, ∞)−→ [0, ∞)is a continuous function. Next, we define the operator T:C2−α[0, 1]→C2−α[0, 1]by Tu(t) = Z1 0G(t,s)f(s,s2−αu(s))ds, 0 <t≤1, (13) where Gis given by expression (8). Mathematics 2020,8, 255 8 of 13 Now, fix c1∈(0, 1), denote m0=min t∈[c1,1]m(t)>0 and c=m0/M0, and define the following cone K={u∈C2−α(I):u≥0 on (0, 1], min t∈[c1,1]{t2−αu(t)} ≥ ckuk2−α}. (14) To use the properties showed in Theorem 4, it is not difficult to verify that T is a completely continuous operator on Ksuch that T(K)⊂K(see Lemma 12 in [26] for details). To prove the existence of solutions of problem (1), we need to prove that iK(T , DK) = 0 for an open set DK⊂K . Therefore, we construct a relatively open set DK=Ωρ for which Ωρ6=Ks for each s> 0 and show that iK(T , Ωρ) = 0. This allows f to satisfy weaker conditions than those used in [ 26 ]. Definition 4. Let us define the following sets for every ρ>0: Kρ={u∈K:kuk2−α<ρ}, and Ωρ={u∈K:t2−αu(t)<ρf or all t ∈[c1, 1]}. It is clear that Kρ⊂Ωρ⊂Kρ c, and, in particular, both Kρand Ωρare open and bounded sets of C2−α(I)for all ρ>0. In the two following lemmas some sufficient conditions are given to ensure that for a suitable ρ>0, the index is either 1 or 0. Lemma 5. Let 0<H=max 0≤t≤1{R1 0t2−αG(t,s)ds}−1 ∈R and fρ:=max{f(t,u) ρ; 0 ≤t≤1, 0 ≤u≤ρ}. If there exists ρ>0such that f ρ<H,then iK(T,Kρ) = 1. Proof. To apply Lemma 4(2), we will show that Tu 6=γufor all u∈∂Kρand every γ≥1. Suppose, on the contrary, that there exists u∈∂Kρand γ≥1 such that γt2−αu(t) = t2−αZ1 0G(t,s)f(s,s2−αu(s))ds. Taking the maximum for t∈I, we obtain γρ =γkuk2−α=max t∈I{t2−αZ1 0G(t,s)f(s,s2−αu(s))ds} ≤ρfρmax t∈I{t2−αZ1 0G(t,s)ds} =ρfρ H<ρ, which contradicts the fact that γ≥1. Therefore, the result is proved. Lemma 6. Let m1=min c1≤t≤1{R1 c1t2−αG(t,s)ds}−1 Mathematics 2020,8, 255 9 of 13 and f1 ρ:=min{f(t,u) ρ;c1≤t≤1, 0 ≤u≤ρ c}. If there exists ρ>0such that f1 ρ>m1,then iK(T,Ωρ) = 0. Proof. Let us see that there exists e∈K\{ 0 } such that u6=Tu +γe for all x∈∂Ωρ and all γ> 0. Indeed, take e(t) = tα−2+rin I, with r∈(0, 1)such that cr 1>m0/M0. It is clear that e∈K\{0}. Assume, on the contrary, that there is u∈∂Ωρ and γ> 0 such that u=Tu +γe . Then, for all t∈[c1, 1], the following inequalities hold: t2−αu(t) = t2−αZ1 0G(t,s)f(s,s2−αu(s))ds +γtα−2+r ≥Z1 0t2−αG(t,s)f(s,s2−αu(s))ds ≥Z1 c1 t2−αG(t,s)f(s,s2−αu(s))ds Now, since ρ≥s2−αu(s)≥m0 M0kuk2−α=ckuk2−α, for all s∈[c1, 1], we have that previous expression is bigger than or equals to ρf1 ρZ1 c1 t2−αG(t,s)ds >ρf1 ρ m1 >ρ and we arrive at a contradiction. Thus, from Lemma 4(1), we deduce that iK(T,Ωρ) = 0. The previous results allow us to deduce the following new result on existence of solutions for problem (1). Theorem 2. Let H and m1 be as in Lemmas 5and 6and 0 <ρ1<cρ2 . Suppose that fρ2<H and f1 ρ1>m1 . Then Problem (1) has at least one solution u such that kuk2−α≤ρ2 and there is t0∈[c1 , 1 ] for which t2−α 0u(t0)≥ρ1. Proof. As we have proved along this section, the solutions of Problem (1) coincide with the fixed points of operator T. Of course, if T has a fixed-point u∈K , such that kuk2−α=ρ2 , we have that Problem (1) has a solution satisfying such property. Moreover, we have, for any t∈I, t2−αu(t)≥Z1 c1 t2−αG(t,s)f(s,s2−αu(s))ds. Therefore, if t2−αu(t)<ρ1 for all t∈[c1 , 1 ] , we arrive at a contradiction as in the proof of Lemma 6. Therefore, suppose that u6=Tu for all u∈∂Kρ2 . By Lemmas 5and 6, it is fulfilled that iK(T , Kρ2) = 1 and iK(T , Ωρ1) = 0. In addition, since ρ1<cρ2 , we have that Ωρ1⊂Kρ1 c⊂Kρ2 . Therefore, from Lemma 4(3), we have that iK(T,Kρ2\Ωρ1) = iK(T,Kρ2)−iK(T,Ωρ1) = 1,