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Existence Results for Nonlocal Multi-Point and Multi-Term Fractional Order Boundary Value Problems

Existence Results for Nonlocal Multi-Point and Multi-Term Fractional Order Boundary Value Problems axioms Article Existence Results for Nonlocal Multi-Point and Multi-Term Fractional Order Boundary Value Problems 1,† 2,† 1,† 1,3, ,† Bashir Ahmad , Najla Alghamdi , Ahmed Alsaedi and Sotiris K. Ntouyas * Nonlinear Analysis and Applied Mathematics (NAAM), Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia; bashirahmad_qau@yahoo.com (B.A.); aalsaedi@hotmail.com (A.A.) Department of Mathematics, Faculty of Science, University of Jeddah, P.O. Box 80327, Jeddah 21589, Saudi Arabia; nmalghamdi1@uj.edu.sa Department of Mathematics, University of Ioannina, 45110 Ioannina, Greece * Correspondence: sntouyas@uoi.gr † These authors contributed equally to this work. Received: 25 May 2020; Accepted: 21 June 2020; Published: 24 June 2020 Abstract: In this paper, we discuss the existence and uniqueness of solutions for a new class of multi-point and integral boundary value problems of multi-term fractional differential equations by using standard fixed point theorems. We also demonstrate the application of the obtained results with the aid of examples. Keywords: caputo fractional derivative; multi-term fractional differential equations; existence; fixed point 1. Introduction Fractional differential equations are found to be of great utility in improving the mathematical modeling of many engineering and scientific disciplines such as physics [1] bioengineering [2], viscoelasticy [3], ecology [4], disease models [5–7], etc. For applications of differential equations containing more than one fractional order differential operators, we refer the reader to Bagley-Torvik [8], Basset equation [9] to name a few. Fractional order boundary value problems equipped with a variety of classical and non-classical (nonlocal) boundary conditions have recently been investigated by many researchers and the literature on the topic is now much enriched, for instance, see [10–21] and the references cited therein. There has been a special focus on boundary value problems involving multi-term fractional differential equations [22–24]. The objective of the present work is to develop the existence theory for multi-term fractional differential equations equipped with nonlocal multi-point boundary conditions. Precisely, we investigate the following boundary value problem: c s+2 c s+1 c s (q D + q D + q D )x(t) = f (t, x(t)), 0 < s < 1, 0 < t < 1, (1) 2 1 0 x(0) = h(x), x(x) = j x(h ), x(1) = l x(s)ds, (2) å i i i=1 c s where D denote the Caputo fractional derivative of order s, 0 < s < 1, f : [0, 1]  R ! R, h : C([0, 1],R) ! R are given continuous functions, 0 < d < x < h < h < . . . < h < 1, l 2 R, 2 n q , q , and q are real constants with q 6= 0. One can characterize the first and second conditions 0 1 2 2 Axioms 2020, 9, 70; doi:10.3390/axioms9020070 www.mdpi.com/journal/axioms Axioms 2020, 9, 70 2 of 21 in (2) as initial-nonlocal and nonlocl multi-point ones, while the last condition in (2) can be understood in the sense that the value of the unknown function x at the right-end point of the domain (x(1)) is proportional to the average value of x on the sub-domain (0, d). Existence and uniqueness results are established by using the classical Banach and Krasnoselskii fixed point theorems and Leray–Schauder nonlinear alternative. Here, we emphasize that the results presented in this paper rely on the standard tools of the fixed point theory. However, their exposition to the given nonlocal problem for a multi-term (sequential) fractional differential equation produces new results which contributes to the related literature. The rest of the paper is organized as follows: In Section 2 we recall some preliminary concepts of fractional calculus and prove a basic lemma, helping us to transform the boundary value problem (1) and (2) into a fixed point problem. The main existence and uniqueness results for the case q 4q q > 0 2 0 are presented in details in Section 3. In Sections 4 and 5 we indicate the results for the cases 2 2 q 4q q = 0 and q 4q q < 0 respectively. Examples illustrating the obtained results are 0 2 0 2 1 1 also included. 2. Basic Results Before presenting some auxiliary results, let us recall some preliminary concepts of fractional calculus [25,26]. (m) a Definition 1. Let y, y 2 L [a, b]. Then the Riemann–Liouville fractional derivative D y of order a 2 1 a (m 1, m], m 2 N, existing almost everywhere on [a, b], is defined as m m d 1 d a ma m1a D y (t) = J y (t) = (t s) y (s)ds. a a m m dt G m a dt ( ) c a The Caputo fractional derivative D y of order a 2 (m 1, m], m 2 N is defined as m1 (t a) (t a) c a a 0 (m1) D y (t) = D y (t) y (a) y (a) . . . y (a) . a a 1! (m 1)! m c a Remark 1. If y 2 AC [a, b], then the Caputo fractional derivative D y of order a 2 (m 1, m], m 2 N, existing almost everywhere on [a, b], is defined as m1a c a ma (m) (m) D y(t) = J y (t) = (t s) y (s)ds. a a G (m a) In the sequel, the Riemann–Liouville fractional integral I and the Caputo fractional derivative c a a c a D with a = 0 are respectively denoted by I and D . Lemma 1. [25] With the given notations, the following equality holds: a c a n1 I ( D y(t)) = y(t) c c t . . . c t , t > 0, n 1 < a < n, (3) 0 1 n1 where c (i = 1, . . . , n 1) are arbitrary constants. The following lemmas associated with the linear variant of problem (1) and (2) plays an important role in the sequel. Axioms 2020, 9, 70 3 of 21 Lemma 2. For any j 2 C([0, 1],R) and q 4q q > 0, the solution of linear multi-term fractional 0 2 differential equation c s+2 c s+1 c s (q D + q D + q D )x(t) = j(t), 0 < s < 1, 0 < t < 1, (4) 2 0 supplemented with the boundary conditions (2) is given by Z Z s1 t s 1 (s u) x(t) = A(t) j(u)du ds q (m m ) G(s) 0 0 2 2 1 Z Z s1 x s (s u) +r (t) A(x) j(u)du ds G(s) 0 0 Z Z n s1 h s (s u) j A(h ) j(u)du ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +r (t) A(1) j(u)du ds (5) G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) l j(u)du ds m m G(s) 0 0 2 1 h  n n m t m x m h 2 2 2 i +h(x) e + r (t) e j + j e 1 å i å i i=1 i=1 m m d 2 2 m e le l +r (t) , where m (ks) m (ks) 2 1 A(k) = e e , k = t, 1, x and h , q q 2 2 q q 4q q q + q 4q q 1 0 2 1 0 2 1 1 m = , m = , 1 2 2q 2q 2 2 w $ (t) w $ (t) w $ (t) w $ (t) 3 2 2 2 4 1 1 1 r (t) = , r (t) = , 1 2 m m 1 1 m t m t 2 1 m (1 e ) m (1 e ) 1 2 $ (t) = , m m 1 2 m t m t 2 1 $ (t) = q (m m )(e e ), 2 2 2 1 m = w w w w 6= 0, (6) 1 1 4 2 3 h  n n m x m h 1 1 i w = m 1 j e + j e 1 å i å i m m 1 2 i=1 i=1 n n i m x m h 2 2 i m 1 j e + j e , 1 å i å i i=1 i=1 n n m x m x m h m h 1 2 1 i 2 i w = q m m )(e e j e + j e , 2 2 2 1 å i å i i=1 i=1 m m d 1 1 w = m 1 e ld + l/m (e 1) 3 2 1 m m m m d 2 2 m 1 e ld + l/m (e 1) , 1 2 m m d 1 1 w = q (m m ) (e + l/m (1 e )) 4 2 2 1 1 m m d 2 2 (e + l/m (1 e )) . 2 Axioms 2020, 9, 70 4 of 21 Proof. Applying the operator I on (4) and using (3), we get s1 (t s) (q D + q D + q )x(t) = j(s)ds + c , (7) 2 1 0 1 G(s) where c is an arbitrary constant. By the method of variation of parameters, the solution of (7) can be written as " # m t m t 1 2 m (1 e ) m (1 e ) 2 1 m t m t x(t) = c + c e + c e 1 2 3 q m m (m m ) 2 1 2 2 1 Z Z s1 t s 1 (s u) m (ts) e j(u)du ds q (m m ) G(s) 0 0 2 2 1 Z Z s1 t s 1 (s u) m (ts) + e j(u)du ds, (8) q (m m ) G(s) 0 0 2 2 1 where m and m are given by (6). Using x(0) = h(x) in (8), we get 1 2 " # m t m t 1 2 m (1 e ) m (1 e ) 2 1 m t m t m t 1 2 2 x(t) = c + c e e + h(x)e q m m (m m ) 2 1 2 2 1 " ! # Z Z t s s1 1 (s u) m (ts) m (ts) 2 1 + e e j(u)du ds , (9) q (m m ) G(s) 2 2 0 0 which together with the conditions x(x) = j x(h ) and x(1) = l x(s)ds yields the following i i i=1 system of equations in the unknown constants c and c : 1 2 c w + c w = V , (10) 1 1 2 2 1 c w + c w = V . (11) 3 2 2 1 4 where Z Z s1 x s (s u) V = A(x) j(u)du ds G(s) 0 0 Z Z n s1  n h s (s u) m h m x 2 i 2 + j A(h ) j(u)du ds + h(x) j e e , å i i å i G(s) 0 0 i=1 i=1 Z Z s1 m d m 1 s 2 2 (s u) le l m e V = A(1) j(u)du ds + h(x) G(s) m 0 0 Z Z h i d s m (ds) m (ds) s1 1 2 (e 1) (e 1) (s u) +l j(u)du ds. m m G(s) 0 0 1 2 Solving the system (10)–(11) together with the notations (6), we find that V w V w V w V w 1 4 2 2 2 1 1 3 c = , c = . 1 2 m m 1 1 Substituting the value of c and c in (9), we obtain the solution (5). The converse of the lemma follows by direct computation. This completes the proof. We do not provide the proofs of the following lemmas, as they are similar to that of Lemma 2. Axioms 2020, 9, 70 5 of 21 Lemma 3. For any j 2 C([0, 1],R) and q 4q q = 0, the solution of linear multi-term fractional 0 2 differential equation c s+2 c s+1 c s (q D + q D + q D )x(t) = j(t), 0 < s < 1, 0 < t < 1, (12) 2 0 supplemented with the boundary conditions (2) is given by Z Z s1 t s 1 (s u) x(t) = B(t) j(u)du ds q G(s) 0 0 Z Z s1 x s (s u) +c (t) B(x) j(u)du ds G(s) 0 0 Z Z n s1 h s (s u) j B(h ) j(u)du ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +c (t) B(1) j(u)du ds (13) G(s) 0 0 Z Z m(ds) m(ds) s1 d s m(d s)e e + 1 (s u) l j(u)du ds m G(s) 0 0 h  n   i m md me le + l mt mx mh +h(x) e + c (t) e j e + c (t) , 1 å i 2 i=1 where m(ks) B(k) = (k s)e , k = t, 1, x and h , m = , 2q v v (t) v v (t) v v (t) v v (t) 3 2 4 1 2 1 1 2 c (t) = , c (t) = , 1 2 m m 2 2 mt mt mte e + 1 mt v (t) = , v (t) = q te , 1 2 2 mx mx n mh mh i i mx e e + 1 j (mh e e + 1) i i i=1 v = , (14) mx mh v = q x e j h e , 2 2 å i i i=1 2 m m md md m e me + m mlde + 2le 2l mld v = , 2 m md md m e lmde + le l v = q , m = v v v v 6= 0. 2 1 4 2 3 Lemma 4. For any j 2 C([0, 1],R) and q 4q q < 0, the solution of linear multi-term fractional 0 2 differential equation c s+2 c s+1 c s (q D + q D + q D )x(t) = j(t), 0 < s < 1, 0 < t < 1, (15) 2 0 supplemented with the boundary conditions (2) is given by Z Z s1 t s 1 (s u) x(t) = F(t) j(u)du ds q b G(s) 0 0 2 Axioms 2020, 9, 70 6 of 21 Z Z s1 x s (s u) +t (t) F(x) j(u)du ds G(s) 0 0 Z Z n i s1 h s (s u) j F(h ) j(u)du ds å i i G(s) 0 0 i=1 Z Z h s1 1 s (s u) +t (t) F(1) j(u)du ds (16) G(s) 0 0 Z Z d s a(ds) b be cos b(d s) 2 2 a + b 0 0 s1 (s u) a(ds) ae sin b(d s) j(u)du ds G(s) h  n at ax ah +h(x) e cos bt + t (t) e cos bx j e cos bh å i i i=1 a ad ad +t (t) e cos b (a ae cos bd + be sin bd) , 2 2 a + b where a(ks) F(k) = e sin b(k s), k = t, 1, x and h , q 4q q q 1 0 2 1 m = a bi, a = , b = , 1,2 2q 2q 2 2 p n (t) p n (t) p n (t) p n (t) 3 2 2 2 4 1 1 1 t (t) = , t (t) = , 1 2 m m 3 3 at at b be cos bt ae sin bt at n (t) = , n (t) = q be sin bt 1 2 2 2 2 a + b ax ax p = b be cos bx ae sin bx 2 2 a + b n i ah ah i i j (b be cos bh ae sin bh ) , å i i i i=1 ax ah p = q b e sin bx j e sin bh , (17) 2 2 å i i i=1 a a p = b be cos b ae sin b bld 2 2 a + b bl ad ad + (a ae cos bd + be sin bd) 2 2 a + b al ad ad (b be cos bd ae sin bd) , 2 2 a + b h i a ad ad p = q b e sin b (b be cos bd ae sin bd) , 4 2 2 2 a + b m = p p p p 6= 0. 3 1 4 2 3 3. Existence and Uniqueness Results Denote by C = C([0, 1],R) the Banach space of all continuous functions from [0, 1] to R endowed with the norm defined by kxk = supfjx(t)j : t 2 [0, 1]g. In relation to the problem (1) and (2) with q 4q q > 0, we define an operator J : C ! C by Lemma 2 as 0 2 Z Z s1 t s 1 (s u) (J x)(t) = A(t) f (u, x(u))du ds q (m m ) G(s) 0 0 2 2 1 Z Z s1 x s (s u) +r (t) A(x) f (u, x(u))du ds G(s) 0 0 Axioms 2020, 9, 70 7 of 21 Z Z n i s1 h s (s u) j A(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z h s1 1 s (s u) +r (t) A(1) f (u, x(u))du ds (18) G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) l f (u, x(u))du ds m m G(s) 0 0 2 1 h  n n m t m x m h 2 2 2 i +h(x) e + r (t) e j + j e 1 å i å i i=1 i=1 m m d 2 2 m e le l +r (t) , where A(), r (t) and r (t) are defined by (6). 1 2 Observe that the problem (1) and (2) is equivalent to the operator equation x = J x, (19) In the sequel, for the sake of computational convenience, we set rb = max jr (t)j, rb = max jr (t)j, 1 1 2 2 t2[0,1] t2[0,1] m t m t 1 2 # = max m (1 e ) m (1 e ) , 2 1 t2[0,1] n h s m x m x 1 2 a = # + rb x jm (1 e ) m (1 e )j 1 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 n i s m h m h 1 i 2 i + jj jh jm (1 e ) m (1 e )j (20) å i 1 i=1 m m 1 2 +rb jm (1 e ) m (1 e )j 2 2 1 s io d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j , 1 2 2 1 jm m j m m d 2 2 jm e j +jljje + 1j m t m x m h 2 2 2 i D = max je j + rb je j + jj jje + 1j + rb . 1 1 å i 2 jm j t2[0,1] i=1 Now the platform is set to present our main results. In the first result, we use Krasnoselskii’s fixed point theorem to prove the existence of solutions for the problem (1) and (2). Theorem 1. (Krasnoselskii’s fixed point theorem [27]). Let Y be a bounded, closed, convex, and nonempty subset of a Banach space X. Let F and F be the operators satisfying the conditions: (i) F y + F y 2 Y 1 2 1 1 2 2 whenever y , y 2 Y; (ii) F is compact and continuous; (iii) F is a contraction mapping. Then there exists 1 2 1 2 y 2 Y such that y = F y + F y. 1 2 In the forthcoming analysis, we need the following assumptions: (G ) j f (t, x) f (t, y)j  `jx yj, for all t 2 [0, 1], x, y 2 R, ` > 0. (G ) jh(x) h(y)j  Lkx yk, for all t 2 [0, 1], x, y 2 C , L > 0. (G ) j f (t, x)j  J(t), for all t 2 [0, 1], x 2 R and J 2 C([0, 1],R ). Theorem 2. Let f : [0, 1]  R ! R be a continuous function satisfying the conditions (G ) and (G ), 1 3 h : C([0, 1],R) ! R be continuous function satisfying the conditions (G ). Then the problem (1) and (2) with q 4q q > 0, has at least one solution on [0, 1] if 0 2 LD < 1, (21) 1 Axioms 2020, 9, 70 8 of 21 where D is given by (20). Proof. Setting sup jJ(t)j = kJk, we can fix t2[0,1] n h kJk s m x m x 1 2 r  # + rb x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(d+1) 2 1 2 2 1 i h n s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j + rb jm (1 e ) m (1 e )j (22) i 2 1 2 2 1 i=1 i io d jlj 2 m d 2 m d + jm (m d e + 1) m (m d e + 1)j + D khk, 1 2 1 2 1 jm m j 1 2 and consider B = fx 2 C : kxk  rg. Introduce the operators J and J on B as follows: r 1 2 r Z Z s1 t s 1 (s u) (J x)(t) = A(t) f (u, x(u))du ds q (m m ) G(s) 0 0 2 2 1 Z Z h s1 x s 1 (s u) + r (t) A(x) f (u, x(u))du ds q (m m ) G(s) 2 2 0 0 Z Z n s1 i h s (s u) j A(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +r (t) A(1) f (u, x(u))du ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) l f (u, x(u))du ds , m m G(s) 0 0 2 1 and m t m x n n m h 2 2 2 i (J x)(t) = h(x) e + r (t) e j + j e r å å 2 1 i i i=1 i=1 (23) m m d 2 2 m e le l +r (t) . Observe that J = J +J . For x, y 2 B , we have 1 2 r kJ x +J yk 1 2 = sup j(J x)(t) + (J y)(t)j 1 2 t2[0,1] Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h  n   i m m d 2 2 jm e j +jljje + 1j m t m x m h 2 2 2 i +jh(y)j je j + r (t) je j + jj jje + 1j + r (t) 1 å i 2 jm j i=1 kJk s m (ts) m (ts) 2 1 sup t e e ds jq (m m )jG(s + 1) 2 2 1 t2[0,1] Z Z h n i x h s m (xs) m (xs) s m (h s) m (h s) 2 1 2 i 1 i +jr (t)j x e e ds + jj jh e e ds 1 å i 0 0 i=1 Axioms 2020, 9, 70 9 of 21 Z Z h m (ds) m (ds) i 1 d 2 1 (e 1) (e 1) m (1s) m (1s) s 2 1 +jr (t)j e e ds +jljd ds m m 0 0 2 1 +D khk kJk s m x m x 1 2 # + r [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j i 2 1 2 2 1 å i i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + D khk  r, 1 2 1 2 1 jm m j where we used (22). Thus J x +J y 2 B . Using the assumptions (G ) (G ) together with (21), 1 2 r 1 3 we show that J is a contraction as follows: kJ xJ yk 2 2 = sup j(J x)(t) (J y)(t)j 2 2 t2[0,1] h  n   i m m d 2 2 jm e j +jljje + 1j m t m x m h 2 2 2 i jh(x) h(y)j je j + r (t) je j + jj jje + 1j + r (t) 1 å i 2 jm j i=1 LD kx yk. Note that continuity of f implies that the operator J is continuous. Also, J is uniformly bounded 1 1 on B as kJ xk = sup j(J x)(t)j 1 1 t2[0,1] kJk s m x m x 1 2 # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 2 2 1 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] . 1 2 2 1 jm m j 1 2 Now we prove the compactness of operator J . We define sup j f (t, x)j = f . Thus, (t,x)2[0,1]B for 0 < t < t < 1, we have 1 2 j(J x)(t ) (J x)(t )j 2 1 Z Z h i s1 t s 1 (s u) A(t )A(t ) f (u, x(u))du ds 2 1 jq (m m )j G(s) 0 0 2 2 1 Z Z s1 t s (s u) + A(t ) f (u, x(u))du ds G(s) t 0 Z Z x s s1 (s u) +jr (t ) r (t )j A(x) j f (u, x(u))jdu ds 1 2 1 1 G(s) 0 0 Z Z n i h s s1 (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t ) r (t )j A(1) j f (u, x(u))jdu ds 2 2 2 1 G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 Axioms 2020, 9, 70 10 of 21 s s m (t t ) m (t t ) 2 2 1 1 2 1 t t m (1 e ) m (1 e ) 1 2 1 2 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m t m t m t m t 2 2 2 1 1 2 1 1 +t m (e e ) m (e e ) 1 2 s m x m x 1 2 +jr (t ) r (t )j[x jm (1 e ) m (1 e )j 1 2 1 1 2 1 s m h m h 1 i 2 i + jj jh jm (1 e ) m (1 e )j] å i 2 1 i=1 m m +jr (t ) r (t )j[jm (1 e ) m (1 e )j 2 2 2 1 2 1 d jlj 2 m d 2 m d 2 1 + jm (m d e + 1) m (m d e + 1)j] ! 0, as t ! t , 2 1 1 2 1 2 jm m j 1 2 independent of x. Thus, J is relatively compact on B . Hence, by the Arzelá-Ascoli Theorem, J is 1 r 1 compact on B . Thus all the assumption of Theorem 1 are satisfied. So, by the conclusion of Theorem 1, the problem (1) and (2) has at least one solution on [0, 1]. The proof is completed. Remark 2. In the above theorem we can interchange the roles of the operators J and J to obtain a second result by replacing (21) by the following condition: `a < 1. Now we apply Banach’s contraction mapping principle to prove existence and uniqueness of solutions for the problem (1) and (2). Theorem 3. Assume that f : [0, 1] R ! R is a continuous function such that (G ) and (G ) are satisfied. 1 2 Then there exists a unique solution for the problem (1) and (2) on [0, 1] if `a + LD < 1, where a and D are 1 1 given by (20). a M + L D 0 1 Proof. Let us define sup j f (t, 0)j = M, sup jh(0)j = L and select r ¯  to t2[0,1] t2[0,1] 1 (`a + LD ) show that J B  B , where B = fx 2 C : kxk  r ¯g and J is defined by (18). Using the condition ¯ ¯ ¯ r r r (G ) and (G ), we have 1 2 j f (t, x)j = j f (t, x) f (t, 0) + f (t, 0)j  j f (t, x) f (t, 0)j +j f (x, 0)j (24) `kxk + M  `r ¯ + M, jh(x)j = jh(x) h(0) + h(0)j  jh(x) h(0)j +jh(0)j  Lkxk + L  Lr ¯ + L . (25) 0 0 Then, for x 2 B , we obtain r ¯ kJ (x)k = sup jJ (x)(t)j t2[0,1] Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 2 2 1 0 0 t2[0,1] Z Z h s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 i h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Axioms 2020, 9, 70 11 of 21 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h n m t m x m h 2 2 2 +jh(x)j je j + r (t)(je j + jj jje + 1j) 1 å i i=1 m m d i 2 2 jm e j +jljje + 1j +r (t) jm j (`r ¯ + M) s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n i m (h s) m (h s) 2 i 1 i + jj j e e ds å i G(s + 1) i=1 1 s m (1s) m (1s) 2 1 + jr (t)j e e ds G(s + 1) m (ds) m (ds) s 2 1 (e 1) (e 1) s +jlj ds + (Lr ¯ + L )D 0 1 m m G(s + 1) 2 1 (`r ¯ + M) s m x m x # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 2 2 1 i i 1 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + (Lr ¯ + L )D 1 2 0 1 2 1 jm m j 1 2 ¯ ¯ ¯ = (`r + M)a + (Lr + L )D  r, 0 1 which clearly shows that J x 2 B for any x 2 B . Thus J B  B . Now, for x, y 2 C and for each r ¯ r ¯ r ¯ r ¯ t 2 [0, 1], we have k(J x) (J y)k Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u)) f (u, y(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z s1 x s (s u) +jr (t)j A(x) j f (u, x(u)) f (u, y(u))jdu ds G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) j f (u, x(u)) f (u, y(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u)) f (u, y(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u)) f (u, y(u))jdu ds m m G(s) 0 0 2 1 h n m t m x m h 2 2 2 +jh(x) h(y)j je j + r (t)(je j + jj jje + 1j) 1 å i i=1 m m d i 2 2 jm e j +jljje + 1j +r (t) jm j ` s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] Axioms 2020, 9, 70 12 of 21 m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n i m (h s) m (h s) 2 i 1 i + jj j e e ds å i G(s + 1) i=1 1 s m (1s) m (1s) 2 1 +jr (t)j e e ds G(s + 1) m (ds) m (ds) s 2 1 (e 1) (e 1) s +jlj ds kx yk + LD kx yk m m G(s + 1) 2 1 s m x m x 1 2 # + r [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + r [jm (1 e ) m (1 e )j i 2 1 2 2 1 å i i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] kx yk + LD kx yk 1 2 1 2 1 jm m j 1 2 = (`a + LD )kx yk, a and D are given by (20) and depend only on the parameters involved in the problem. In view of the condition `a + LD < 1, it follows that J is a contraction. Thus, by the contraction mapping principle (Banach fixed point theorem), the problem (1) and (2) has a unique solution on [0, 1]. This completes the proof. The next existence result is based on Leray–Schauder nonlinear alternative. Theorem 4. (Nonlinear alternative for single valued maps [28]). Let E be a Banach space, C a closed, convex subset of E, U an open subset of C and 0 2 U. Suppose that F : U ! C is a continuous, compact (that is, F(U) is a relatively compact subset of C) map. Then either (i) F has a fixed point in U, or (ii) there is a u 2 ¶U (the boundary of U in C) and e 2 (0, 1) with u = eF(u). We need the following assumptions: + + ( H ) There exist a function g 2 C([0, 1],R ), and a nondecreasing function y : R ! R such that j f (t, y)j  g(t)y(kyk), 8(t, y) 2 [0, 1] R. ( H ) h : C([0, 1],R) ! R, is continuous function with h(0) = 0 and there exist constant L > 0 with 2 1 L < D , such that jh(x)j  L kxk 8 x 2 C . ( H ) There exists a constant K > 0 such that (1 L D )K 1 1 > 1. kgky(K)a Theorem 5. Let f : [0, 1] R ! R be a continuous function. Then the problem (1) and (2) has at least one solution on [0, 1], if ( H )–( H ) are satisfied. 1 3 Proof. Consider the operator J : C ! C defined by (18). We show that J maps bounded sets into bounded sets in C . For a positive number z, let E = fx 2 C : kxk  zg be a bounded set in C . Then we have kJ (x)k = sup jJ (x)(t)j t2[0,1] Axioms 2020, 9, 70 13 of 21 Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z h s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 i h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h n m t m x m h 2 2 2 i +jh(x)j je j + r (t)(je j + jj jje + 1j) 1 å i i=1 m m d i 2 2 jm e j +jljje + 1j +r (t) jm j kgky(z) s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n s m (h s) m (h s) 2 i 1 i + jj j e e ds å i G(s + 1) i=1 m (1s) m (1s) +jr (t)j e e ds G(s + 1) m (ds) m (ds) s 2 1 (e 1) (e 1) s +jlj ds + L D z 1 1 m m G(s + 1) 2 1 kgky(z) s m x m x 1 2 # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + L D z, 1 2 1 1 2 1 jm m j 1 2 which yields kgky(z) s m x m x 1 2 kJ xk  # + rb x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 2 1 2 i i + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + L D z. 1 2 1 1 2 1 jm m j 1 2 Next we show that J maps bounded sets into equicontniuous sets of C . Let t , t 2 [0, 1] with 1 2 t < t and y 2 E , where E is a bounded set of C . Then we obtain 1 2 z z j(J x)(t ) (J x)(t )j 2 1 Z Z h i s1 t s 1 (s u) A(t )A(t ) f (u, x(u))du ds 2 1 jq (m m )j G(s) 2 2 1 0 0 Axioms 2020, 9, 70 14 of 21 Z Z s1 t s (s u) + A(t ) f (u, x(u))du ds G(s) t 0 Z Z s1 x s (s u) +jr (t ) r (t )j A(x) j f (u, x(u))jdu ds 1 2 1 1 G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +jr (t ) r (t )j A(1) j f (u, x(u))jdu ds 2 2 2 1 G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h n m t m t m x m h 2 2 2 1 2 2 i +jh(x)j je e j + (r (t ) r (t ))(je j + jj jje + 1j) 1 2 1 1 å i i=1 m m d 2 2 jm e j +jljje + 1j +(r (t ) r (t )) 2 2 2 1 jm j s s m (t t ) m (t t ) 2 2 1 1 2 1 t t m (1 e ) m (1 e ) 1 2 1 2 jq m m (m m )jG(s + 1) 2 2 2 1 1 s m t m t m t m t 2 2 2 1 1 2 1 1 +t m (e e ) m (e e ) 1 2 s m x m x 1 2 +jr (t ) r (t )j[x jm (1 e ) m (1 e )j 2 2 1 1 1 1 s m h m h 1 i 2 i + jj jh jm (1 e ) m (1 e )j] å i 2 1 i=1 m m 1 2 +jr (t ) r (t )j[jm (1 e ) m (1 e )j 2 2 2 1 2 1 s o h d jlj 2 m d 2 m d m t m t 2 1 2 2 2 1 + jm (m d e + 1) m (m d e + 1)j] +jh(x)j je e j 2 2 1 jm m j 1 2 m x m h 2 2 i +(r (t ) r (t )) je j + jj jje + 1j 1 2 1 1 i i=1 m m d 2 2 jm e j +jljje + 1j +(r (t ) r (t )) , 2 2 2 1 jm j which tends to zero independently of x 2 E as t t ! 0. As J satisfies the above assumptions, z 2 1 therefore it follows by the Arzelá-Ascoli theorem that J : C ! C is completely continuous. The result will follow from the Leray–Schauder nonlinear alternative once it is shown that there exists U  C with x 6= qJ x for q 2 (0, 1) and x 2 ¶U . Let x 2 C be such that x = qJ x for q 2 [0, 1]. Then, for t 2 [0, 1], we have jx(t)j = jqJ x(t)j Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jl j f (u, x(u))jdu ds m m G(s) 0 0 2 1 Axioms 2020, 9, 70 15 of 21 h  n m t m x m h 2 2 2 i +jh(x)j je j + r (t) je j + jj jje + 1j 1 å i i=1 m m d 2 2 jm e j +jljje + 1j +r (t) jm j t s kgky(kxk) s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n i m (h s) m (h s) i 1 i + jj j e e ds å i G(s + 1) i=1 h s m (1s) m (1s) 2 1 +jr (t)j e e ds G(s + 1) m (ds) m (ds) s d 2 (e 1) (e 1) s +jlj ds +jh(x)jD m m G(s + 1) 2 1 kgky(kxk) s m x m x 1 2 # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 2 2 1 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 2 1 + jm (m d e + 1) m (m d e + 1)j] + L D kxk 2 1 1 1 1 2 jm m j 1 2 = kgky(kxk)a + L D kxk, 1 1 which implies that (1 L D )kxk 1 1 kgky(kxk)a In view of ( H ), there is no solution x such that kxk 6= K. Let us set U = fx 2 C : kxk < Kg. The operator J : U ! C is continuous and completely continuous. From the choice of U, there is no u 2 ¶U such that u = qJ (u) for some q 2 (0, 1). Consequently, by the nonlinear alternative of Leray–Schauder type [28], we deduce that J has a fixed point u 2 U which is a solution of the problem (1) and (2). Example 1. Let us consider the following boundary value problem c 12/5 c 7/5 c 2/5 1 (2 D + 3 D + D )x(t) = tan x + cos t, 0 < t < 1, (26) 4 4 + t subject the boundary condition 1/6 x(0) = sin x(t), x(1/5) = x(1/4) + 2x(1/3) + x(1/2), x(1) = 2 x(s)ds. (27) Here, q = 2, q = 3, q = 1, s = 2/5, x = 1/5, h = 1/4, h = 1/3, h = 1/2, d = 1/6, j = 1, 2 1 0 1 2 3 1 j = 2, j = 1, l = 2, t is a fixed value in [0, 1] and 2 3 f (t, x) = tan x + cos t. 4 4 + t Axioms 2020, 9, 70 16 of 21 Clearly q 4q q = 1 > 0, and 0 2 j f (t, x) f (t, y)j  jx yj, jh(x) h(y)j  kx yk. where ` = 1/8, L = 1/9. Using the given values, we found a  0.095961, D  6.9171. pe It is easy to check that j f (t, x)j  + cos t = J(t) and LD < 1. As all the condition 8 4 + t of Theorem 2 are satisfied the problem (26) and (27) has at least one solution on [0, 1]. On the other hand, `a + LD < 1 and thus there exists a unique solution for the problem (26) and (27) on [0, 1] by Theorem 3. Example 2. Consider the following fractional differential equation c 12/5 c 7/5 c 2/5 1 (2 D + 3 D + D )x(t) = p x tan x + p/2 , 0 < t < 1, (28) p 9 + t subject the boundary conditions (27). Here f (t, x) = p (x tan x + p/2). p 9 + t Clearly j f (t, x)j  p kxk + 1 , 2 9 + t with g(t) = , y(kxk) = kxk + 1. 2 9+t Then by using the condition ( H ), we find that K > 0.241877 (we have used a = 0.27045). Thus, the conclusion of Theorem 5 applies to problem (28) and (27). 4. Existence Results for Problem (1) and (2) with q 4q q = 0 0 2 In view of Lemma 3, we can transform problem (1) and (2) into equivalent fixed point problem as follows: x = Hx, (29) where the operator H : C ! C is defined by Z Z s1 t s 1 (s u) (Hx)(t) = B(t) f (u, x(u))du ds q G(s) 0 0 R R s1 x s (su) +c (t) B(x) f (u, x(u))du ds 0 0 G(s) R R s1 h s (su) å j B(h ) f (u, x(u))du ds i i i=1 0 0 G(s) (30) R R s1 1 s (su) +c (t) B(1) f (u, x(u))du ds 0 0 G(s) R R m(ds) m(ds) s1 d s m(ds)e e +1 (su) l f (u, x(u))du ds 0 0 G(s) h    i m md n me le +l mt mx mh +h(x) e + c (t) e j e + c (t) , 1 i 2 i=1 where B(), c (t) and c (t) are defined by (14). We set 1 2 cb = max jc (t)j, cb = max jc (t)j, 1 1 2 2 t2[0,1] t2[0,1] Axioms 2020, 9, 70 17 of 21 n h m m s mx mx b = (1 + cb )jme e + 1j + cb x jmx e e + 1j 2 1 jq jm G(s + 1) i o n s jljd c s mh mh md md i i + jj jh jmh e e + 1j + md(e + 1) + 2(1 e ) , (31) å i i jmj i=1 m md jme j +jljje + 1j mt mx mh D = max je j + cb (je j + jj jje j) + cb . 2 1 å i 2 jmj t2[0,1] i=1 Now we present our main results for problem (1) and (2) with q 4q q = 0. Since the methods 0 2 for proof of these results are similar to the ones obtained in Section 3, so we omit the proofs. Theorem 6. Let f : [0, 1] R ! R be a continuous function satisfying the conditions (G )-(G ). Then the 1 3 problem (1) and (2) with q 4q q = 0, has at least one solution on [0, 1] if 0 2 LD < 1, (32) where D is given by (31). Theorem 7. Assume that f : [0, 1] R ! R is a continuous function such that (G ) is satisfied. Then there exists a unique solution for problem (1) and (2) with q 4q q = 0, on [0, 1] if `b + LD < 1, where b and 0 2 2 D are given by (31). Theorem 8. Let f : [0, 1]R ! R be a continuous function. Then the problem (1) and (2) with q 4q q = 0 2 0, has at least one solution on [0, 1], if ( H ), ( H ) and the following condition hold: 1 2 ( H ) There exists a constant K > 0 such that (1 L D )K 1 2 1 > 1, kgky(K )b where b is defined by (31). Example 3. Consider the sequential fractional differential equation jxj c 12/5 c 7/5 c 2/5 t (2 D + 4 D + 2 D )x(t) = + e , 0 < t < 1, (33) (t + 6)(jxj + 1) subject the boundary conditions (27). Here jxj f (t, x) = + e . (t + 6)(jxj + 1) Clearly q 4q q = 0, and 0 2 j f (t, x) f (t, y)j  jx yj, jh(x) h(y)j  kx yk. where ` = 1/6, L = 1/9. Using the given values, we find that b  0.29913, b  0.15022 and D  5.135. It is easy to check that j f (t, x)j  + e = J(t) and LD < 1. As all the conditions of t + 6 Theorem 6 are satisfied, the problem (27)–(33) has at least one solution on [0, 1]. On the other hand, `b + LD < 1 and thus there exists a unique solution for the problem (27)–(33) on [0, 1] by Theorem 7. 2 Axioms 2020, 9, 70 18 of 21 5. Existence Results for Problem (1) and (2) with q 4q q < 0 0 2 In view of Lemma 4, we can transform problem (1) and (2) into equivalent fixed point problem as follows: x = Kx, (34) where the operator K : C ! C is defined by Z Z s1 t s 1 (s u) (Kx)(t) = F(t) f (u, x(u))du ds q b G(s) 0 0 Z Z s1 x s (s u) +t (t) F(x) y(u)du ds G(s) 0 0 Z Z n i s1 h s (s u) j F(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z h s1 1 s (s u) +t (t) F(1) f (u, x(u))du ds G(s) 0 0 Z Z d s a(ds) b be cos b(d s) 2 2 a + b 0 0 s1 (s u) a(ds) ae sin b(d s) f (u, x(u))du ds G(s) h n at ax ah +h(x) e cos bt + t (t)(e cos bx j e cos bh ) å i i i=1 a ad ad +t (t)(e cos b (a ae cos bd + be sin bd)) , 2 2 a + b where F(), t (t) and t (t) are defined by (17). We set 1 2 tb = max jt (t)j, tb = max jt (t)j 1 1 2 2 t2[0,1] t2[0,1] n h i a a g = (1 + tb ) jb be cos b ae sin bj 2 2 jq b(a + b )jG(s + 1) s ax ax s ah +tb x jb be cos bx ae sin bxj + jj jh jb be cos bh 1 å i i i=1 i h io ah s ad ae sin bh j +jljd tb jbd e sin bdj , (35) i 2 at ax ah D = max je cos btj + tb (je cos bxj + jj jje cos bh j) 3 1 å i i t2[0,1] i=1 jlj a ad ad +tb je cos bj + (ja ae cos bd + be sin bdj) . 2 2 a + b Here are the existence and uniqueness results for problem (1) and (2) with q 4q q < 0. As argued 0 2 in the last section, we do not provide the proofs for these results. Theorem 9. Let f : [0, 1] R ! R be a continuous function satisfying the conditions (G )–(G ). Then the problem (1) and (2) with p 4 p p < 0, has at least one solution on [0, 1] if 0 2 LD < 1, (36) where g and D are given by (35). 1 Axioms 2020, 9, 70 19 of 21 Theorem 10. Assume that f : [0, 1] R ! R is a continuous function such that (G ) and (G ) are satisfied. 1 2 Then there exists a unique solution for the problem (1) and (2) with q 4q q < 0, on [0, 1] if `g + LD < 1, 0 2 3 where g and D are given by (35). Theorem 11. Let f : [0, 1]R ! R be a continuous function. Then the problem (1) and (2) with q 4q q < 0 2 0, has at least one solution on [0, 1], if ( H ), ( H ) and the following condition are satisfied: 1 2 ( H ) There exists a constant K > 0 such that (1 L D )K 1 3 2 > 1, kgky(K )g where g and D are defined by (35). Example 4. Consider the following boundary value problem 2t 1 e c 12/5 c 7/5 c 2/5 (2 D + 3 D + 2 D )x(t) = cos x + , 0 < t < 1, (37) (t + 4) 13 subject the boundary condition 1/6 x(0) = x(t), x(1/5) = x(1/4) + 2x(1/3) + x(1/2), x(1) = 2 x(s)ds. (38) Here, s = 2/5, x = 1/5, h = 1/4, h = 1/3, h = 1/2, d = 1/6, j = 1, j = 2, j = 1, l = 2, t is 1 2 3 1 2 3 a fixed value in [0, 1] and 2t 1 e f (t, x) = cos x + . (t + 4) 13 Clearly q 4q q = 7 < 0, and 0 2 j f (t, x) f (t, y)j  jx yj, jh(x) h(y)j  kx yk, where ` = 1/16, L = 1/8. Using the given values, it is found that g  0.34744, g  0.17937 and D  1.8499. 2t 1 e Obviously j f (t, x)j  + = J(t) and LD < 1. As the hypothesis of Theorem 9 holds (t + 4) 13 true, the problem (37) and (38) has at least one solution on [0, 1]. Furthermore, we have `g + LD < 1, which implies that there exists a unique solution for the problem (37) and (38) on [0, 1] by Theorem 10. 6. Conclusions We have presented a detailed analysis for a multi-term fractional differential equation supplemented with nonlocal multi-point integral boundary conditions. The existence and uniqueness results are given for all three cases depending on the coefficients of the multi-term fractional differential 2 2 2 equation: (i) q 4q q > 0, (ii) q 4q q = 0 and (iii) q 4q q < 0. Existence results are 0 2 0 2 0 2 1 1 1 proved by means of Krasnoselskii fixed point theorem and Leray–Schauder nonlinear alternative, while Banach contraction mapping principle is applied to establish the uniqueness of solutions for the given problem. The obtained results are well-illustrated with examples. Our results are new and enrich the literature on nonlocal integro-multipoint boundary problems for multi-term Caputo type fractional differential equations. Axioms 2020, 9, 70 20 of 21 Author Contributions: Conceptualization, B.A. and S.K.N.; Formal analysis, B.A., N.A., A.A. and S.K.N.; Funding acquisition, A.A.; Methodology, B.A., N.A., A.A. and S.K.N. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Conflicts of Interest: The authors declare no conflict of interest. References 1. Herrmann, R. Fractional Calculus: An Introduction for Physicists; World Scientific: Singapore, 2011. 2. Magin, R.L. 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A system of coupled multi-term fractional differential equations with three-point coupled boundary conditions. Fract. Calc. Appl. Anal. 2019, 22, 601–618. [CrossRef] 24. Ruzhansky, M.; Tokmagambetov, N.; Torebek, B.T. On a non-local problem for a multi-term fractional diffusion-wave equation. Fract. Calc. Appl. Anal. 2020, 23, 324–355. [CrossRef] 25. Kilbas, A.A.; Srivastava, H.M.; Trujillo, J.J. Theory and Applications of Fractional Differential Equations; In North-Holland Mathematics Studies; Elsevier Science B.V.: Amsterdam, The Netherlands, 2006; Volume 2014. 26. Zhou, Y. Basic theory of Fractional Differential Equations; World Scientific Publishing Co. Pte. Ltd.: Hackensack, NJ, USA, 2014. 27. Krasnoselskii, M.A. Two remarks on the method of successive approximations. Uspekhi Mat. Nauk 1955, 10, 123–127. 28. Granas, A.; Dugundji, J. Fixed Point Theory; Springer: New York, NY, USA, 2005. c 2020 by the authors. Licensee MDPI, Basel, Switzerland. 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Existence Results for Nonlocal Multi-Point and Multi-Term Fractional Order Boundary Value Problems

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axioms Article Existence Results for Nonlocal Multi-Point and Multi-Term Fractional Order Boundary Value Problems 1,† 2,† 1,† 1,3, ,† Bashir Ahmad , Najla Alghamdi , Ahmed Alsaedi and Sotiris K. Ntouyas * Nonlinear Analysis and Applied Mathematics (NAAM), Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia; bashirahmad_qau@yahoo.com (B.A.); aalsaedi@hotmail.com (A.A.) Department of Mathematics, Faculty of Science, University of Jeddah, P.O. Box 80327, Jeddah 21589, Saudi Arabia; nmalghamdi1@uj.edu.sa Department of Mathematics, University of Ioannina, 45110 Ioannina, Greece * Correspondence: sntouyas@uoi.gr † These authors contributed equally to this work. Received: 25 May 2020; Accepted: 21 June 2020; Published: 24 June 2020 Abstract: In this paper, we discuss the existence and uniqueness of solutions for a new class of multi-point and integral boundary value problems of multi-term fractional differential equations by using standard fixed point theorems. We also demonstrate the application of the obtained results with the aid of examples. Keywords: caputo fractional derivative; multi-term fractional differential equations; existence; fixed point 1. Introduction Fractional differential equations are found to be of great utility in improving the mathematical modeling of many engineering and scientific disciplines such as physics [1] bioengineering [2], viscoelasticy [3], ecology [4], disease models [5–7], etc. For applications of differential equations containing more than one fractional order differential operators, we refer the reader to Bagley-Torvik [8], Basset equation [9] to name a few. Fractional order boundary value problems equipped with a variety of classical and non-classical (nonlocal) boundary conditions have recently been investigated by many researchers and the literature on the topic is now much enriched, for instance, see [10–21] and the references cited therein. There has been a special focus on boundary value problems involving multi-term fractional differential equations [22–24]. The objective of the present work is to develop the existence theory for multi-term fractional differential equations equipped with nonlocal multi-point boundary conditions. Precisely, we investigate the following boundary value problem: c s+2 c s+1 c s (q D + q D + q D )x(t) = f (t, x(t)), 0 < s < 1, 0 < t < 1, (1) 2 1 0 x(0) = h(x), x(x) = j x(h ), x(1) = l x(s)ds, (2) å i i i=1 c s where D denote the Caputo fractional derivative of order s, 0 < s < 1, f : [0, 1]  R ! R, h : C([0, 1],R) ! R are given continuous functions, 0 < d < x < h < h < . . . < h < 1, l 2 R, 2 n q , q , and q are real constants with q 6= 0. One can characterize the first and second conditions 0 1 2 2 Axioms 2020, 9, 70; doi:10.3390/axioms9020070 www.mdpi.com/journal/axioms Axioms 2020, 9, 70 2 of 21 in (2) as initial-nonlocal and nonlocl multi-point ones, while the last condition in (2) can be understood in the sense that the value of the unknown function x at the right-end point of the domain (x(1)) is proportional to the average value of x on the sub-domain (0, d). Existence and uniqueness results are established by using the classical Banach and Krasnoselskii fixed point theorems and Leray–Schauder nonlinear alternative. Here, we emphasize that the results presented in this paper rely on the standard tools of the fixed point theory. However, their exposition to the given nonlocal problem for a multi-term (sequential) fractional differential equation produces new results which contributes to the related literature. The rest of the paper is organized as follows: In Section 2 we recall some preliminary concepts of fractional calculus and prove a basic lemma, helping us to transform the boundary value problem (1) and (2) into a fixed point problem. The main existence and uniqueness results for the case q 4q q > 0 2 0 are presented in details in Section 3. In Sections 4 and 5 we indicate the results for the cases 2 2 q 4q q = 0 and q 4q q < 0 respectively. Examples illustrating the obtained results are 0 2 0 2 1 1 also included. 2. Basic Results Before presenting some auxiliary results, let us recall some preliminary concepts of fractional calculus [25,26]. (m) a Definition 1. Let y, y 2 L [a, b]. Then the Riemann–Liouville fractional derivative D y of order a 2 1 a (m 1, m], m 2 N, existing almost everywhere on [a, b], is defined as m m d 1 d a ma m1a D y (t) = J y (t) = (t s) y (s)ds. a a m m dt G m a dt ( ) c a The Caputo fractional derivative D y of order a 2 (m 1, m], m 2 N is defined as m1 (t a) (t a) c a a 0 (m1) D y (t) = D y (t) y (a) y (a) . . . y (a) . a a 1! (m 1)! m c a Remark 1. If y 2 AC [a, b], then the Caputo fractional derivative D y of order a 2 (m 1, m], m 2 N, existing almost everywhere on [a, b], is defined as m1a c a ma (m) (m) D y(t) = J y (t) = (t s) y (s)ds. a a G (m a) In the sequel, the Riemann–Liouville fractional integral I and the Caputo fractional derivative c a a c a D with a = 0 are respectively denoted by I and D . Lemma 1. [25] With the given notations, the following equality holds: a c a n1 I ( D y(t)) = y(t) c c t . . . c t , t > 0, n 1 < a < n, (3) 0 1 n1 where c (i = 1, . . . , n 1) are arbitrary constants. The following lemmas associated with the linear variant of problem (1) and (2) plays an important role in the sequel. Axioms 2020, 9, 70 3 of 21 Lemma 2. For any j 2 C([0, 1],R) and q 4q q > 0, the solution of linear multi-term fractional 0 2 differential equation c s+2 c s+1 c s (q D + q D + q D )x(t) = j(t), 0 < s < 1, 0 < t < 1, (4) 2 0 supplemented with the boundary conditions (2) is given by Z Z s1 t s 1 (s u) x(t) = A(t) j(u)du ds q (m m ) G(s) 0 0 2 2 1 Z Z s1 x s (s u) +r (t) A(x) j(u)du ds G(s) 0 0 Z Z n s1 h s (s u) j A(h ) j(u)du ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +r (t) A(1) j(u)du ds (5) G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) l j(u)du ds m m G(s) 0 0 2 1 h  n n m t m x m h 2 2 2 i +h(x) e + r (t) e j + j e 1 å i å i i=1 i=1 m m d 2 2 m e le l +r (t) , where m (ks) m (ks) 2 1 A(k) = e e , k = t, 1, x and h , q q 2 2 q q 4q q q + q 4q q 1 0 2 1 0 2 1 1 m = , m = , 1 2 2q 2q 2 2 w $ (t) w $ (t) w $ (t) w $ (t) 3 2 2 2 4 1 1 1 r (t) = , r (t) = , 1 2 m m 1 1 m t m t 2 1 m (1 e ) m (1 e ) 1 2 $ (t) = , m m 1 2 m t m t 2 1 $ (t) = q (m m )(e e ), 2 2 2 1 m = w w w w 6= 0, (6) 1 1 4 2 3 h  n n m x m h 1 1 i w = m 1 j e + j e 1 å i å i m m 1 2 i=1 i=1 n n i m x m h 2 2 i m 1 j e + j e , 1 å i å i i=1 i=1 n n m x m x m h m h 1 2 1 i 2 i w = q m m )(e e j e + j e , 2 2 2 1 å i å i i=1 i=1 m m d 1 1 w = m 1 e ld + l/m (e 1) 3 2 1 m m m m d 2 2 m 1 e ld + l/m (e 1) , 1 2 m m d 1 1 w = q (m m ) (e + l/m (1 e )) 4 2 2 1 1 m m d 2 2 (e + l/m (1 e )) . 2 Axioms 2020, 9, 70 4 of 21 Proof. Applying the operator I on (4) and using (3), we get s1 (t s) (q D + q D + q )x(t) = j(s)ds + c , (7) 2 1 0 1 G(s) where c is an arbitrary constant. By the method of variation of parameters, the solution of (7) can be written as " # m t m t 1 2 m (1 e ) m (1 e ) 2 1 m t m t x(t) = c + c e + c e 1 2 3 q m m (m m ) 2 1 2 2 1 Z Z s1 t s 1 (s u) m (ts) e j(u)du ds q (m m ) G(s) 0 0 2 2 1 Z Z s1 t s 1 (s u) m (ts) + e j(u)du ds, (8) q (m m ) G(s) 0 0 2 2 1 where m and m are given by (6). Using x(0) = h(x) in (8), we get 1 2 " # m t m t 1 2 m (1 e ) m (1 e ) 2 1 m t m t m t 1 2 2 x(t) = c + c e e + h(x)e q m m (m m ) 2 1 2 2 1 " ! # Z Z t s s1 1 (s u) m (ts) m (ts) 2 1 + e e j(u)du ds , (9) q (m m ) G(s) 2 2 0 0 which together with the conditions x(x) = j x(h ) and x(1) = l x(s)ds yields the following i i i=1 system of equations in the unknown constants c and c : 1 2 c w + c w = V , (10) 1 1 2 2 1 c w + c w = V . (11) 3 2 2 1 4 where Z Z s1 x s (s u) V = A(x) j(u)du ds G(s) 0 0 Z Z n s1  n h s (s u) m h m x 2 i 2 + j A(h ) j(u)du ds + h(x) j e e , å i i å i G(s) 0 0 i=1 i=1 Z Z s1 m d m 1 s 2 2 (s u) le l m e V = A(1) j(u)du ds + h(x) G(s) m 0 0 Z Z h i d s m (ds) m (ds) s1 1 2 (e 1) (e 1) (s u) +l j(u)du ds. m m G(s) 0 0 1 2 Solving the system (10)–(11) together with the notations (6), we find that V w V w V w V w 1 4 2 2 2 1 1 3 c = , c = . 1 2 m m 1 1 Substituting the value of c and c in (9), we obtain the solution (5). The converse of the lemma follows by direct computation. This completes the proof. We do not provide the proofs of the following lemmas, as they are similar to that of Lemma 2. Axioms 2020, 9, 70 5 of 21 Lemma 3. For any j 2 C([0, 1],R) and q 4q q = 0, the solution of linear multi-term fractional 0 2 differential equation c s+2 c s+1 c s (q D + q D + q D )x(t) = j(t), 0 < s < 1, 0 < t < 1, (12) 2 0 supplemented with the boundary conditions (2) is given by Z Z s1 t s 1 (s u) x(t) = B(t) j(u)du ds q G(s) 0 0 Z Z s1 x s (s u) +c (t) B(x) j(u)du ds G(s) 0 0 Z Z n s1 h s (s u) j B(h ) j(u)du ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +c (t) B(1) j(u)du ds (13) G(s) 0 0 Z Z m(ds) m(ds) s1 d s m(d s)e e + 1 (s u) l j(u)du ds m G(s) 0 0 h  n   i m md me le + l mt mx mh +h(x) e + c (t) e j e + c (t) , 1 å i 2 i=1 where m(ks) B(k) = (k s)e , k = t, 1, x and h , m = , 2q v v (t) v v (t) v v (t) v v (t) 3 2 4 1 2 1 1 2 c (t) = , c (t) = , 1 2 m m 2 2 mt mt mte e + 1 mt v (t) = , v (t) = q te , 1 2 2 mx mx n mh mh i i mx e e + 1 j (mh e e + 1) i i i=1 v = , (14) mx mh v = q x e j h e , 2 2 å i i i=1 2 m m md md m e me + m mlde + 2le 2l mld v = , 2 m md md m e lmde + le l v = q , m = v v v v 6= 0. 2 1 4 2 3 Lemma 4. For any j 2 C([0, 1],R) and q 4q q < 0, the solution of linear multi-term fractional 0 2 differential equation c s+2 c s+1 c s (q D + q D + q D )x(t) = j(t), 0 < s < 1, 0 < t < 1, (15) 2 0 supplemented with the boundary conditions (2) is given by Z Z s1 t s 1 (s u) x(t) = F(t) j(u)du ds q b G(s) 0 0 2 Axioms 2020, 9, 70 6 of 21 Z Z s1 x s (s u) +t (t) F(x) j(u)du ds G(s) 0 0 Z Z n i s1 h s (s u) j F(h ) j(u)du ds å i i G(s) 0 0 i=1 Z Z h s1 1 s (s u) +t (t) F(1) j(u)du ds (16) G(s) 0 0 Z Z d s a(ds) b be cos b(d s) 2 2 a + b 0 0 s1 (s u) a(ds) ae sin b(d s) j(u)du ds G(s) h  n at ax ah +h(x) e cos bt + t (t) e cos bx j e cos bh å i i i=1 a ad ad +t (t) e cos b (a ae cos bd + be sin bd) , 2 2 a + b where a(ks) F(k) = e sin b(k s), k = t, 1, x and h , q 4q q q 1 0 2 1 m = a bi, a = , b = , 1,2 2q 2q 2 2 p n (t) p n (t) p n (t) p n (t) 3 2 2 2 4 1 1 1 t (t) = , t (t) = , 1 2 m m 3 3 at at b be cos bt ae sin bt at n (t) = , n (t) = q be sin bt 1 2 2 2 2 a + b ax ax p = b be cos bx ae sin bx 2 2 a + b n i ah ah i i j (b be cos bh ae sin bh ) , å i i i i=1 ax ah p = q b e sin bx j e sin bh , (17) 2 2 å i i i=1 a a p = b be cos b ae sin b bld 2 2 a + b bl ad ad + (a ae cos bd + be sin bd) 2 2 a + b al ad ad (b be cos bd ae sin bd) , 2 2 a + b h i a ad ad p = q b e sin b (b be cos bd ae sin bd) , 4 2 2 2 a + b m = p p p p 6= 0. 3 1 4 2 3 3. Existence and Uniqueness Results Denote by C = C([0, 1],R) the Banach space of all continuous functions from [0, 1] to R endowed with the norm defined by kxk = supfjx(t)j : t 2 [0, 1]g. In relation to the problem (1) and (2) with q 4q q > 0, we define an operator J : C ! C by Lemma 2 as 0 2 Z Z s1 t s 1 (s u) (J x)(t) = A(t) f (u, x(u))du ds q (m m ) G(s) 0 0 2 2 1 Z Z s1 x s (s u) +r (t) A(x) f (u, x(u))du ds G(s) 0 0 Axioms 2020, 9, 70 7 of 21 Z Z n i s1 h s (s u) j A(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z h s1 1 s (s u) +r (t) A(1) f (u, x(u))du ds (18) G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) l f (u, x(u))du ds m m G(s) 0 0 2 1 h  n n m t m x m h 2 2 2 i +h(x) e + r (t) e j + j e 1 å i å i i=1 i=1 m m d 2 2 m e le l +r (t) , where A(), r (t) and r (t) are defined by (6). 1 2 Observe that the problem (1) and (2) is equivalent to the operator equation x = J x, (19) In the sequel, for the sake of computational convenience, we set rb = max jr (t)j, rb = max jr (t)j, 1 1 2 2 t2[0,1] t2[0,1] m t m t 1 2 # = max m (1 e ) m (1 e ) , 2 1 t2[0,1] n h s m x m x 1 2 a = # + rb x jm (1 e ) m (1 e )j 1 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 n i s m h m h 1 i 2 i + jj jh jm (1 e ) m (1 e )j (20) å i 1 i=1 m m 1 2 +rb jm (1 e ) m (1 e )j 2 2 1 s io d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j , 1 2 2 1 jm m j m m d 2 2 jm e j +jljje + 1j m t m x m h 2 2 2 i D = max je j + rb je j + jj jje + 1j + rb . 1 1 å i 2 jm j t2[0,1] i=1 Now the platform is set to present our main results. In the first result, we use Krasnoselskii’s fixed point theorem to prove the existence of solutions for the problem (1) and (2). Theorem 1. (Krasnoselskii’s fixed point theorem [27]). Let Y be a bounded, closed, convex, and nonempty subset of a Banach space X. Let F and F be the operators satisfying the conditions: (i) F y + F y 2 Y 1 2 1 1 2 2 whenever y , y 2 Y; (ii) F is compact and continuous; (iii) F is a contraction mapping. Then there exists 1 2 1 2 y 2 Y such that y = F y + F y. 1 2 In the forthcoming analysis, we need the following assumptions: (G ) j f (t, x) f (t, y)j  `jx yj, for all t 2 [0, 1], x, y 2 R, ` > 0. (G ) jh(x) h(y)j  Lkx yk, for all t 2 [0, 1], x, y 2 C , L > 0. (G ) j f (t, x)j  J(t), for all t 2 [0, 1], x 2 R and J 2 C([0, 1],R ). Theorem 2. Let f : [0, 1]  R ! R be a continuous function satisfying the conditions (G ) and (G ), 1 3 h : C([0, 1],R) ! R be continuous function satisfying the conditions (G ). Then the problem (1) and (2) with q 4q q > 0, has at least one solution on [0, 1] if 0 2 LD < 1, (21) 1 Axioms 2020, 9, 70 8 of 21 where D is given by (20). Proof. Setting sup jJ(t)j = kJk, we can fix t2[0,1] n h kJk s m x m x 1 2 r  # + rb x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(d+1) 2 1 2 2 1 i h n s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j + rb jm (1 e ) m (1 e )j (22) i 2 1 2 2 1 i=1 i io d jlj 2 m d 2 m d + jm (m d e + 1) m (m d e + 1)j + D khk, 1 2 1 2 1 jm m j 1 2 and consider B = fx 2 C : kxk  rg. Introduce the operators J and J on B as follows: r 1 2 r Z Z s1 t s 1 (s u) (J x)(t) = A(t) f (u, x(u))du ds q (m m ) G(s) 0 0 2 2 1 Z Z h s1 x s 1 (s u) + r (t) A(x) f (u, x(u))du ds q (m m ) G(s) 2 2 0 0 Z Z n s1 i h s (s u) j A(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +r (t) A(1) f (u, x(u))du ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) l f (u, x(u))du ds , m m G(s) 0 0 2 1 and m t m x n n m h 2 2 2 i (J x)(t) = h(x) e + r (t) e j + j e r å å 2 1 i i i=1 i=1 (23) m m d 2 2 m e le l +r (t) . Observe that J = J +J . For x, y 2 B , we have 1 2 r kJ x +J yk 1 2 = sup j(J x)(t) + (J y)(t)j 1 2 t2[0,1] Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h  n   i m m d 2 2 jm e j +jljje + 1j m t m x m h 2 2 2 i +jh(y)j je j + r (t) je j + jj jje + 1j + r (t) 1 å i 2 jm j i=1 kJk s m (ts) m (ts) 2 1 sup t e e ds jq (m m )jG(s + 1) 2 2 1 t2[0,1] Z Z h n i x h s m (xs) m (xs) s m (h s) m (h s) 2 1 2 i 1 i +jr (t)j x e e ds + jj jh e e ds 1 å i 0 0 i=1 Axioms 2020, 9, 70 9 of 21 Z Z h m (ds) m (ds) i 1 d 2 1 (e 1) (e 1) m (1s) m (1s) s 2 1 +jr (t)j e e ds +jljd ds m m 0 0 2 1 +D khk kJk s m x m x 1 2 # + r [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j i 2 1 2 2 1 å i i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + D khk  r, 1 2 1 2 1 jm m j where we used (22). Thus J x +J y 2 B . Using the assumptions (G ) (G ) together with (21), 1 2 r 1 3 we show that J is a contraction as follows: kJ xJ yk 2 2 = sup j(J x)(t) (J y)(t)j 2 2 t2[0,1] h  n   i m m d 2 2 jm e j +jljje + 1j m t m x m h 2 2 2 i jh(x) h(y)j je j + r (t) je j + jj jje + 1j + r (t) 1 å i 2 jm j i=1 LD kx yk. Note that continuity of f implies that the operator J is continuous. Also, J is uniformly bounded 1 1 on B as kJ xk = sup j(J x)(t)j 1 1 t2[0,1] kJk s m x m x 1 2 # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 2 2 1 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] . 1 2 2 1 jm m j 1 2 Now we prove the compactness of operator J . We define sup j f (t, x)j = f . Thus, (t,x)2[0,1]B for 0 < t < t < 1, we have 1 2 j(J x)(t ) (J x)(t )j 2 1 Z Z h i s1 t s 1 (s u) A(t )A(t ) f (u, x(u))du ds 2 1 jq (m m )j G(s) 0 0 2 2 1 Z Z s1 t s (s u) + A(t ) f (u, x(u))du ds G(s) t 0 Z Z x s s1 (s u) +jr (t ) r (t )j A(x) j f (u, x(u))jdu ds 1 2 1 1 G(s) 0 0 Z Z n i h s s1 (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t ) r (t )j A(1) j f (u, x(u))jdu ds 2 2 2 1 G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 Axioms 2020, 9, 70 10 of 21 s s m (t t ) m (t t ) 2 2 1 1 2 1 t t m (1 e ) m (1 e ) 1 2 1 2 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m t m t m t m t 2 2 2 1 1 2 1 1 +t m (e e ) m (e e ) 1 2 s m x m x 1 2 +jr (t ) r (t )j[x jm (1 e ) m (1 e )j 1 2 1 1 2 1 s m h m h 1 i 2 i + jj jh jm (1 e ) m (1 e )j] å i 2 1 i=1 m m +jr (t ) r (t )j[jm (1 e ) m (1 e )j 2 2 2 1 2 1 d jlj 2 m d 2 m d 2 1 + jm (m d e + 1) m (m d e + 1)j] ! 0, as t ! t , 2 1 1 2 1 2 jm m j 1 2 independent of x. Thus, J is relatively compact on B . Hence, by the Arzelá-Ascoli Theorem, J is 1 r 1 compact on B . Thus all the assumption of Theorem 1 are satisfied. So, by the conclusion of Theorem 1, the problem (1) and (2) has at least one solution on [0, 1]. The proof is completed. Remark 2. In the above theorem we can interchange the roles of the operators J and J to obtain a second result by replacing (21) by the following condition: `a < 1. Now we apply Banach’s contraction mapping principle to prove existence and uniqueness of solutions for the problem (1) and (2). Theorem 3. Assume that f : [0, 1] R ! R is a continuous function such that (G ) and (G ) are satisfied. 1 2 Then there exists a unique solution for the problem (1) and (2) on [0, 1] if `a + LD < 1, where a and D are 1 1 given by (20). a M + L D 0 1 Proof. Let us define sup j f (t, 0)j = M, sup jh(0)j = L and select r ¯  to t2[0,1] t2[0,1] 1 (`a + LD ) show that J B  B , where B = fx 2 C : kxk  r ¯g and J is defined by (18). Using the condition ¯ ¯ ¯ r r r (G ) and (G ), we have 1 2 j f (t, x)j = j f (t, x) f (t, 0) + f (t, 0)j  j f (t, x) f (t, 0)j +j f (x, 0)j (24) `kxk + M  `r ¯ + M, jh(x)j = jh(x) h(0) + h(0)j  jh(x) h(0)j +jh(0)j  Lkxk + L  Lr ¯ + L . (25) 0 0 Then, for x 2 B , we obtain r ¯ kJ (x)k = sup jJ (x)(t)j t2[0,1] Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 2 2 1 0 0 t2[0,1] Z Z h s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 i h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Axioms 2020, 9, 70 11 of 21 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h n m t m x m h 2 2 2 +jh(x)j je j + r (t)(je j + jj jje + 1j) 1 å i i=1 m m d i 2 2 jm e j +jljje + 1j +r (t) jm j (`r ¯ + M) s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n i m (h s) m (h s) 2 i 1 i + jj j e e ds å i G(s + 1) i=1 1 s m (1s) m (1s) 2 1 + jr (t)j e e ds G(s + 1) m (ds) m (ds) s 2 1 (e 1) (e 1) s +jlj ds + (Lr ¯ + L )D 0 1 m m G(s + 1) 2 1 (`r ¯ + M) s m x m x # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 2 2 1 i i 1 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + (Lr ¯ + L )D 1 2 0 1 2 1 jm m j 1 2 ¯ ¯ ¯ = (`r + M)a + (Lr + L )D  r, 0 1 which clearly shows that J x 2 B for any x 2 B . Thus J B  B . Now, for x, y 2 C and for each r ¯ r ¯ r ¯ r ¯ t 2 [0, 1], we have k(J x) (J y)k Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u)) f (u, y(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z s1 x s (s u) +jr (t)j A(x) j f (u, x(u)) f (u, y(u))jdu ds G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) j f (u, x(u)) f (u, y(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u)) f (u, y(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u)) f (u, y(u))jdu ds m m G(s) 0 0 2 1 h n m t m x m h 2 2 2 +jh(x) h(y)j je j + r (t)(je j + jj jje + 1j) 1 å i i=1 m m d i 2 2 jm e j +jljje + 1j +r (t) jm j ` s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] Axioms 2020, 9, 70 12 of 21 m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n i m (h s) m (h s) 2 i 1 i + jj j e e ds å i G(s + 1) i=1 1 s m (1s) m (1s) 2 1 +jr (t)j e e ds G(s + 1) m (ds) m (ds) s 2 1 (e 1) (e 1) s +jlj ds kx yk + LD kx yk m m G(s + 1) 2 1 s m x m x 1 2 # + r [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + r [jm (1 e ) m (1 e )j i 2 1 2 2 1 å i i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] kx yk + LD kx yk 1 2 1 2 1 jm m j 1 2 = (`a + LD )kx yk, a and D are given by (20) and depend only on the parameters involved in the problem. In view of the condition `a + LD < 1, it follows that J is a contraction. Thus, by the contraction mapping principle (Banach fixed point theorem), the problem (1) and (2) has a unique solution on [0, 1]. This completes the proof. The next existence result is based on Leray–Schauder nonlinear alternative. Theorem 4. (Nonlinear alternative for single valued maps [28]). Let E be a Banach space, C a closed, convex subset of E, U an open subset of C and 0 2 U. Suppose that F : U ! C is a continuous, compact (that is, F(U) is a relatively compact subset of C) map. Then either (i) F has a fixed point in U, or (ii) there is a u 2 ¶U (the boundary of U in C) and e 2 (0, 1) with u = eF(u). We need the following assumptions: + + ( H ) There exist a function g 2 C([0, 1],R ), and a nondecreasing function y : R ! R such that j f (t, y)j  g(t)y(kyk), 8(t, y) 2 [0, 1] R. ( H ) h : C([0, 1],R) ! R, is continuous function with h(0) = 0 and there exist constant L > 0 with 2 1 L < D , such that jh(x)j  L kxk 8 x 2 C . ( H ) There exists a constant K > 0 such that (1 L D )K 1 1 > 1. kgky(K)a Theorem 5. Let f : [0, 1] R ! R be a continuous function. Then the problem (1) and (2) has at least one solution on [0, 1], if ( H )–( H ) are satisfied. 1 3 Proof. Consider the operator J : C ! C defined by (18). We show that J maps bounded sets into bounded sets in C . For a positive number z, let E = fx 2 C : kxk  zg be a bounded set in C . Then we have kJ (x)k = sup jJ (x)(t)j t2[0,1] Axioms 2020, 9, 70 13 of 21 Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z h s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 i h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h n m t m x m h 2 2 2 i +jh(x)j je j + r (t)(je j + jj jje + 1j) 1 å i i=1 m m d i 2 2 jm e j +jljje + 1j +r (t) jm j kgky(z) s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n s m (h s) m (h s) 2 i 1 i + jj j e e ds å i G(s + 1) i=1 m (1s) m (1s) +jr (t)j e e ds G(s + 1) m (ds) m (ds) s 2 1 (e 1) (e 1) s +jlj ds + L D z 1 1 m m G(s + 1) 2 1 kgky(z) s m x m x 1 2 # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + L D z, 1 2 1 1 2 1 jm m j 1 2 which yields kgky(z) s m x m x 1 2 kJ xk  # + rb x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 1 2 2 1 s m h m h m m 1 2 1 2 i i + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 1 2 + jm (m d e + 1) m (m d e + 1)j] + L D z. 1 2 1 1 2 1 jm m j 1 2 Next we show that J maps bounded sets into equicontniuous sets of C . Let t , t 2 [0, 1] with 1 2 t < t and y 2 E , where E is a bounded set of C . Then we obtain 1 2 z z j(J x)(t ) (J x)(t )j 2 1 Z Z h i s1 t s 1 (s u) A(t )A(t ) f (u, x(u))du ds 2 1 jq (m m )j G(s) 2 2 1 0 0 Axioms 2020, 9, 70 14 of 21 Z Z s1 t s (s u) + A(t ) f (u, x(u))du ds G(s) t 0 Z Z s1 x s (s u) +jr (t ) r (t )j A(x) j f (u, x(u))jdu ds 1 2 1 1 G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z 1 s s1 (s u) +jr (t ) r (t )j A(1) j f (u, x(u))jdu ds 2 2 2 1 G(s) 0 0 Z Z m (ds) m (ds) s1 d s 2 1 (e 1) (e 1) (s u) +jlj j f (u, x(u))jdu ds m m G(s) 0 0 2 1 h n m t m t m x m h 2 2 2 1 2 2 i +jh(x)j je e j + (r (t ) r (t ))(je j + jj jje + 1j) 1 2 1 1 å i i=1 m m d 2 2 jm e j +jljje + 1j +(r (t ) r (t )) 2 2 2 1 jm j s s m (t t ) m (t t ) 2 2 1 1 2 1 t t m (1 e ) m (1 e ) 1 2 1 2 jq m m (m m )jG(s + 1) 2 2 2 1 1 s m t m t m t m t 2 2 2 1 1 2 1 1 +t m (e e ) m (e e ) 1 2 s m x m x 1 2 +jr (t ) r (t )j[x jm (1 e ) m (1 e )j 2 2 1 1 1 1 s m h m h 1 i 2 i + jj jh jm (1 e ) m (1 e )j] å i 2 1 i=1 m m 1 2 +jr (t ) r (t )j[jm (1 e ) m (1 e )j 2 2 2 1 2 1 s o h d jlj 2 m d 2 m d m t m t 2 1 2 2 2 1 + jm (m d e + 1) m (m d e + 1)j] +jh(x)j je e j 2 2 1 jm m j 1 2 m x m h 2 2 i +(r (t ) r (t )) je j + jj jje + 1j 1 2 1 1 i i=1 m m d 2 2 jm e j +jljje + 1j +(r (t ) r (t )) , 2 2 2 1 jm j which tends to zero independently of x 2 E as t t ! 0. As J satisfies the above assumptions, z 2 1 therefore it follows by the Arzelá-Ascoli theorem that J : C ! C is completely continuous. The result will follow from the Leray–Schauder nonlinear alternative once it is shown that there exists U  C with x 6= qJ x for q 2 (0, 1) and x 2 ¶U . Let x 2 C be such that x = qJ x for q 2 [0, 1]. Then, for t 2 [0, 1], we have jx(t)j = jqJ x(t)j Z Z s1 t s 1 (s u) sup A(t) j f (u, x(u))jdu ds jq (m m )j G(s) 0 0 2 2 1 t2[0,1] Z Z s1 x s (s u) +jr (t)j A(x) j f (u, x(u))jdu ds G(s) 0 0 Z Z n s1 h s (s u) + jj j A(h ) j f (u, x(u))jdu ds å i i G(s) 0 0 i=1 Z Z s1 1 s (s u) +jr (t)j A(1) j f (u, x(u))jdu ds G(s) 0 0 Z Z m (ds) m (ds)  s1 i d s 2 1 (e 1) (e 1) (s u) +jl j f (u, x(u))jdu ds m m G(s) 0 0 2 1 Axioms 2020, 9, 70 15 of 21 h  n m t m x m h 2 2 2 i +jh(x)j je j + r (t) je j + jj jje + 1j 1 å i i=1 m m d 2 2 jm e j +jljje + 1j +r (t) jm j t s kgky(kxk) s m (ts) m (ts) 2 1 sup e e ds jq (m m )j G(s + 1) 2 2 1 t2[0,1] m (xs) m (xs) 2 1 +jr (t)j e e ds G(s + 1) n i m (h s) m (h s) i 1 i + jj j e e ds å i G(s + 1) i=1 h s m (1s) m (1s) 2 1 +jr (t)j e e ds G(s + 1) m (ds) m (ds) s d 2 (e 1) (e 1) s +jlj ds +jh(x)jD m m G(s + 1) 2 1 kgky(kxk) s m x m x 1 2 # + rb [x jm (1 e ) m (1 e )j 1 2 1 jq m m (m m )jG(s + 1) 2 2 2 1 1 s m h m h m m 1 i 2 i 1 2 + jj jh jm (1 e ) m (1 e )j] + rb [jm (1 e ) m (1 e )j å i 2 1 2 2 1 i=1 d jlj 2 m d 2 m d 2 1 + jm (m d e + 1) m (m d e + 1)j] + L D kxk 2 1 1 1 1 2 jm m j 1 2 = kgky(kxk)a + L D kxk, 1 1 which implies that (1 L D )kxk 1 1 kgky(kxk)a In view of ( H ), there is no solution x such that kxk 6= K. Let us set U = fx 2 C : kxk < Kg. The operator J : U ! C is continuous and completely continuous. From the choice of U, there is no u 2 ¶U such that u = qJ (u) for some q 2 (0, 1). Consequently, by the nonlinear alternative of Leray–Schauder type [28], we deduce that J has a fixed point u 2 U which is a solution of the problem (1) and (2). Example 1. Let us consider the following boundary value problem c 12/5 c 7/5 c 2/5 1 (2 D + 3 D + D )x(t) = tan x + cos t, 0 < t < 1, (26) 4 4 + t subject the boundary condition 1/6 x(0) = sin x(t), x(1/5) = x(1/4) + 2x(1/3) + x(1/2), x(1) = 2 x(s)ds. (27) Here, q = 2, q = 3, q = 1, s = 2/5, x = 1/5, h = 1/4, h = 1/3, h = 1/2, d = 1/6, j = 1, 2 1 0 1 2 3 1 j = 2, j = 1, l = 2, t is a fixed value in [0, 1] and 2 3 f (t, x) = tan x + cos t. 4 4 + t Axioms 2020, 9, 70 16 of 21 Clearly q 4q q = 1 > 0, and 0 2 j f (t, x) f (t, y)j  jx yj, jh(x) h(y)j  kx yk. where ` = 1/8, L = 1/9. Using the given values, we found a  0.095961, D  6.9171. pe It is easy to check that j f (t, x)j  + cos t = J(t) and LD < 1. As all the condition 8 4 + t of Theorem 2 are satisfied the problem (26) and (27) has at least one solution on [0, 1]. On the other hand, `a + LD < 1 and thus there exists a unique solution for the problem (26) and (27) on [0, 1] by Theorem 3. Example 2. Consider the following fractional differential equation c 12/5 c 7/5 c 2/5 1 (2 D + 3 D + D )x(t) = p x tan x + p/2 , 0 < t < 1, (28) p 9 + t subject the boundary conditions (27). Here f (t, x) = p (x tan x + p/2). p 9 + t Clearly j f (t, x)j  p kxk + 1 , 2 9 + t with g(t) = , y(kxk) = kxk + 1. 2 9+t Then by using the condition ( H ), we find that K > 0.241877 (we have used a = 0.27045). Thus, the conclusion of Theorem 5 applies to problem (28) and (27). 4. Existence Results for Problem (1) and (2) with q 4q q = 0 0 2 In view of Lemma 3, we can transform problem (1) and (2) into equivalent fixed point problem as follows: x = Hx, (29) where the operator H : C ! C is defined by Z Z s1 t s 1 (s u) (Hx)(t) = B(t) f (u, x(u))du ds q G(s) 0 0 R R s1 x s (su) +c (t) B(x) f (u, x(u))du ds 0 0 G(s) R R s1 h s (su) å j B(h ) f (u, x(u))du ds i i i=1 0 0 G(s) (30) R R s1 1 s (su) +c (t) B(1) f (u, x(u))du ds 0 0 G(s) R R m(ds) m(ds) s1 d s m(ds)e e +1 (su) l f (u, x(u))du ds 0 0 G(s) h    i m md n me le +l mt mx mh +h(x) e + c (t) e j e + c (t) , 1 i 2 i=1 where B(), c (t) and c (t) are defined by (14). We set 1 2 cb = max jc (t)j, cb = max jc (t)j, 1 1 2 2 t2[0,1] t2[0,1] Axioms 2020, 9, 70 17 of 21 n h m m s mx mx b = (1 + cb )jme e + 1j + cb x jmx e e + 1j 2 1 jq jm G(s + 1) i o n s jljd c s mh mh md md i i + jj jh jmh e e + 1j + md(e + 1) + 2(1 e ) , (31) å i i jmj i=1 m md jme j +jljje + 1j mt mx mh D = max je j + cb (je j + jj jje j) + cb . 2 1 å i 2 jmj t2[0,1] i=1 Now we present our main results for problem (1) and (2) with q 4q q = 0. Since the methods 0 2 for proof of these results are similar to the ones obtained in Section 3, so we omit the proofs. Theorem 6. Let f : [0, 1] R ! R be a continuous function satisfying the conditions (G )-(G ). Then the 1 3 problem (1) and (2) with q 4q q = 0, has at least one solution on [0, 1] if 0 2 LD < 1, (32) where D is given by (31). Theorem 7. Assume that f : [0, 1] R ! R is a continuous function such that (G ) is satisfied. Then there exists a unique solution for problem (1) and (2) with q 4q q = 0, on [0, 1] if `b + LD < 1, where b and 0 2 2 D are given by (31). Theorem 8. Let f : [0, 1]R ! R be a continuous function. Then the problem (1) and (2) with q 4q q = 0 2 0, has at least one solution on [0, 1], if ( H ), ( H ) and the following condition hold: 1 2 ( H ) There exists a constant K > 0 such that (1 L D )K 1 2 1 > 1, kgky(K )b where b is defined by (31). Example 3. Consider the sequential fractional differential equation jxj c 12/5 c 7/5 c 2/5 t (2 D + 4 D + 2 D )x(t) = + e , 0 < t < 1, (33) (t + 6)(jxj + 1) subject the boundary conditions (27). Here jxj f (t, x) = + e . (t + 6)(jxj + 1) Clearly q 4q q = 0, and 0 2 j f (t, x) f (t, y)j  jx yj, jh(x) h(y)j  kx yk. where ` = 1/6, L = 1/9. Using the given values, we find that b  0.29913, b  0.15022 and D  5.135. It is easy to check that j f (t, x)j  + e = J(t) and LD < 1. As all the conditions of t + 6 Theorem 6 are satisfied, the problem (27)–(33) has at least one solution on [0, 1]. On the other hand, `b + LD < 1 and thus there exists a unique solution for the problem (27)–(33) on [0, 1] by Theorem 7. 2 Axioms 2020, 9, 70 18 of 21 5. Existence Results for Problem (1) and (2) with q 4q q < 0 0 2 In view of Lemma 4, we can transform problem (1) and (2) into equivalent fixed point problem as follows: x = Kx, (34) where the operator K : C ! C is defined by Z Z s1 t s 1 (s u) (Kx)(t) = F(t) f (u, x(u))du ds q b G(s) 0 0 Z Z s1 x s (s u) +t (t) F(x) y(u)du ds G(s) 0 0 Z Z n i s1 h s (s u) j F(h ) f (u, x(u))du ds å i i G(s) 0 0 i=1 Z Z h s1 1 s (s u) +t (t) F(1) f (u, x(u))du ds G(s) 0 0 Z Z d s a(ds) b be cos b(d s) 2 2 a + b 0 0 s1 (s u) a(ds) ae sin b(d s) f (u, x(u))du ds G(s) h n at ax ah +h(x) e cos bt + t (t)(e cos bx j e cos bh ) å i i i=1 a ad ad +t (t)(e cos b (a ae cos bd + be sin bd)) , 2 2 a + b where F(), t (t) and t (t) are defined by (17). We set 1 2 tb = max jt (t)j, tb = max jt (t)j 1 1 2 2 t2[0,1] t2[0,1] n h i a a g = (1 + tb ) jb be cos b ae sin bj 2 2 jq b(a + b )jG(s + 1) s ax ax s ah +tb x jb be cos bx ae sin bxj + jj jh jb be cos bh 1 å i i i=1 i h io ah s ad ae sin bh j +jljd tb jbd e sin bdj , (35) i 2 at ax ah D = max je cos btj + tb (je cos bxj + jj jje cos bh j) 3 1 å i i t2[0,1] i=1 jlj a ad ad +tb je cos bj + (ja ae cos bd + be sin bdj) . 2 2 a + b Here are the existence and uniqueness results for problem (1) and (2) with q 4q q < 0. As argued 0 2 in the last section, we do not provide the proofs for these results. Theorem 9. Let f : [0, 1] R ! R be a continuous function satisfying the conditions (G )–(G ). Then the problem (1) and (2) with p 4 p p < 0, has at least one solution on [0, 1] if 0 2 LD < 1, (36) where g and D are given by (35). 1 Axioms 2020, 9, 70 19 of 21 Theorem 10. Assume that f : [0, 1] R ! R is a continuous function such that (G ) and (G ) are satisfied. 1 2 Then there exists a unique solution for the problem (1) and (2) with q 4q q < 0, on [0, 1] if `g + LD < 1, 0 2 3 where g and D are given by (35). Theorem 11. Let f : [0, 1]R ! R be a continuous function. Then the problem (1) and (2) with q 4q q < 0 2 0, has at least one solution on [0, 1], if ( H ), ( H ) and the following condition are satisfied: 1 2 ( H ) There exists a constant K > 0 such that (1 L D )K 1 3 2 > 1, kgky(K )g where g and D are defined by (35). Example 4. Consider the following boundary value problem 2t 1 e c 12/5 c 7/5 c 2/5 (2 D + 3 D + 2 D )x(t) = cos x + , 0 < t < 1, (37) (t + 4) 13 subject the boundary condition 1/6 x(0) = x(t), x(1/5) = x(1/4) + 2x(1/3) + x(1/2), x(1) = 2 x(s)ds. (38) Here, s = 2/5, x = 1/5, h = 1/4, h = 1/3, h = 1/2, d = 1/6, j = 1, j = 2, j = 1, l = 2, t is 1 2 3 1 2 3 a fixed value in [0, 1] and 2t 1 e f (t, x) = cos x + . (t + 4) 13 Clearly q 4q q = 7 < 0, and 0 2 j f (t, x) f (t, y)j  jx yj, jh(x) h(y)j  kx yk, where ` = 1/16, L = 1/8. Using the given values, it is found that g  0.34744, g  0.17937 and D  1.8499. 2t 1 e Obviously j f (t, x)j  + = J(t) and LD < 1. As the hypothesis of Theorem 9 holds (t + 4) 13 true, the problem (37) and (38) has at least one solution on [0, 1]. Furthermore, we have `g + LD < 1, which implies that there exists a unique solution for the problem (37) and (38) on [0, 1] by Theorem 10. 6. Conclusions We have presented a detailed analysis for a multi-term fractional differential equation supplemented with nonlocal multi-point integral boundary conditions. The existence and uniqueness results are given for all three cases depending on the coefficients of the multi-term fractional differential 2 2 2 equation: (i) q 4q q > 0, (ii) q 4q q = 0 and (iii) q 4q q < 0. Existence results are 0 2 0 2 0 2 1 1 1 proved by means of Krasnoselskii fixed point theorem and Leray–Schauder nonlinear alternative, while Banach contraction mapping principle is applied to establish the uniqueness of solutions for the given problem. The obtained results are well-illustrated with examples. Our results are new and enrich the literature on nonlocal integro-multipoint boundary problems for multi-term Caputo type fractional differential equations. Axioms 2020, 9, 70 20 of 21 Author Contributions: Conceptualization, B.A. and S.K.N.; Formal analysis, B.A., N.A., A.A. and S.K.N.; Funding acquisition, A.A.; Methodology, B.A., N.A., A.A. and S.K.N. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Conflicts of Interest: The authors declare no conflict of interest. References 1. Herrmann, R. Fractional Calculus: An Introduction for Physicists; World Scientific: Singapore, 2011. 2. Magin, R.L. 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