POSITIVE SOLUTIONS FOR SUPERLINEAR RIEMANN-LIOUVILLE FRACTIONAL BOUNDARY-VALUE PROBLEMS

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1 Electronic Journal of Differential Equations, Vol. 27 (27), No. 24, pp. 6. ISSN: URL: or POSITIVE SOLUTIONS FOR SUPERLINEAR RIEMANN-LIOUVILLE FRACTIONAL BOUNDARY-VALUE PROBLEMS IMED BACHAR, HABIB MÂAGLI, VICENŢIU D. RĂDULESCU Communicated by Marco Squassina Abstract. Using a perturbation argument, we establish the existence and uniqueness of a positive continuous solution for the following superlinear Riemann- Liouville fractional boundary-value problem D α u(x) u(x)ϕ(x, u(x)) =, < x <, u() = u () = lim x4 α u (x) =, u () = a >, x + where 3 < α 4 and ϕ(x, t) satisfies a suitable integrability condition.. Introduction Fractional differential equations have been of great interest recently. They can be used to model many phenomena in control theory of dynamic systems, fluid flow, electrochemistry of corrosion, rheology etc. For more applications, we refer to [5, 6, 7, 9,, 2, 3, 7, 8, 2, 23, 25, 26, 27, 28] and references therein. By means of the lower and upper solution method and fixed-point theorems, Liang and Zhang established in [4] the existence of positive solutions for the following Riemann-Liouville fractional problem D α u(x) + f(x, u(x)) =, < x <, u() = u () = u () = u () =, (.) where 3 < α 4 and f is a nonnegative continuous function satisfying some adequate conditions. Recently, Zhai et al. [29], studied problem (.) with f(x, u(x)) = g(x, u(x)) + h(x, u(x)). They proved the existence and uniqueness of positive solutions by using a fixed point theorem for a sum of operators. For further existence results related to (.), we refer to [, 2, 3, 4, 4, 2, 22, 24, 29] and the references therein. 2 Mathematics Subject Classification. 34A8, 34B5, 34B8, 34B27. Key words and phrases. Fractional differential equation; positive solution; Green s function; perturbation; Schäuder fixed point theorem. c 27 Texas State University. Submitted August 8, 27. Published October 4, 27.

2 2 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 In this paper, we are concerned with the following superlinear Riemann-Liouville fractional boundary value problem D α u(x) u(x)ϕ(x, u(x)) =, < x <, u() = u () = lim x + x 4 α u (x) =, u () = a >, (.2) where 3 < α 4 and ϕ(x, t) is a nonnegative continuous function in (, ) [, ) that is required to satisfy some appropriate integrability condition. We emphasize that the condition a > on the boundary is crucial to obtain a positive solution. Our approach is based on a perturbation argument. Notation: (i) C(X) (resp.c + (X)) is the set of continuous (resp. nonnegative continuous) functions in a metric space X; (ii) B((, )) (resp. B + ((, ))) is the set of Borel (resp., nonnegative Borel) measurable functions in (, ); (iii) L ((, )) := {q B((, )), q(r) dr < }; (iv) C+([, )) is the set of nonnegative continuously differentiable functions on [, ); (v) for 3 < α 4, K α := {q B + ((, )), (vi) for 3 < α 4 and a >, we let ω(x) = r α ( r) α 3 q(r)dr < }; (.3) a (α )(α 2) xα, x. (.4) Observe that ω(x) is the unique solution of the problem D α u(x) =, < x <, u() = u () = lim x + x4 α u (x) =, u () = a >. (.5) (vii) For 3 < α 4, we denote by G(x, t) the Green s function of the operator u D α u, with boundary conditions u() = u () = lim x + x 4 α u (x) = u () =. We have (see Lemma 2.4) G(x, t) = (viii) For each q K α, we let α q := { x α ( t) α 3 (x t) α, t x ; x α ( t) α 3, x t. sup x,t (,) It will be showed that if q K α, then α q <. (.6) G(x, r)g(r, t) q(r)dr. (.7) G(x, t) To state our main results, we need a combination of the following assumptions. (H) ϕ C + ((, ) [, )). (H2) There exists a function q K α C((, )) with α q /2 such that for each x (, ), the map t t(q(x) ϕ(x, tω(x))) is nondecreasing on [, ]. (H3) For each x (, ), the function t tϕ(x, t) is nondecreasing on [, ).

3 EJDE-27/24 SUPERLINEAR FRACTIONAL BOUNDARY VALUE PROBLEMS 3 We first prove that if q belongs to K α C((, )) with α q /2 and f belongs to B + ((, )), then the fractional problem D α u(x) q(x)u(x) = f(x), < x <, 3 < α 4, u() = u () = lim x 4 α u (x) = u (.8) () =, x + admits a positive Green s function G(x, t). Exploiting properties of the Green s function G(x, t) and by using a perturbation argument, we establish the following result. Theorem.. Under assumptions (H), (H2), problem (.2) admits a positive solution u in C([, ]) satisfying c ω(x) u(x) ω(x), x [, ], (.9) where c is a constant in (, ). Moreover, the uniqueness of such solution is proved if, further, hypothesis (H3) is satisfied. From Theorem., we deduce the following property. Corollary.2. Let f C+([, )) be such that the map r θ(r) = rf(r) is nondecreasing on [, ). Let p C + ((, )) be such that the function x p(x) := p(x) max ξ ω(x) θ (ξ) K α. Then for λ [, 2α ep ), the problem D α u(x) λp(x)u(x)f(u(x)) =, x (, ), u() = u () = lim x4 α u (x) =, u () = a >, x + has a unique positive solution u in C([, ]) satisfying ( λα ep )ω(x) u(x) ω(x), x [, ]. (.) Note that hypotheses (H) (H3), are satisfied with ϕ(x, t) = λp(x)t σ, for σ, p C + ((, )) such that r (α )(+σ) ( r) α 3 p(r)dr <. This article is organized as follows. In section 2, some estimates on the Green s function G(x, t) are obtained. In particular, we prove that for all x, r, t (, ), G(x, r)g(r, t) G(x, t) 4(α )2 r α ( r) α 3. ( This implies that for each q K α, we have α q <. In section 3, we start by deriving the Green s function G(x, t) associated to the boundary value problem (.8). We also establish some basic estimates of this function. In particular, we show that for (x, t) [, ] [, ], we have We also prove the resolvent equation ( α q )G(x, t) G(x, t) G(x, t). V f = V q f + V q (qv f) = V q f + V (qv q f), for f B + ((, )), where the kernels V and V q are defined on B + ((, )) by V f(x) := G(x, t)f(t)dt and V q f(x) := G(x, t)f(t)dt, x [, ].

4 4 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 By using the above results and a perturbation argument, we prove Theorem.. 2. Fractional calculus and estimates on the Green s function 2.. Fractional calculus. We recall the following basic definitions and properties on fractional calculus (see [, 23, 25]). Definition 2.. Let β > and Γ(β) be the Euler Gamma function. For a measurable function f : (, ) R, the integral (provided that it exists) I β f(x) = Γ(β) (x t) β f(t)dt, x >, is called the Riemann-Liouville fractional integral of order β. Definition 2.2. Let β > and [β] be its integer part. For a measurable function f : (, ) R, the expression (provided that it exists) D β f(x) = Γ(n β) ( d dx )n (x t) n β f(t)dt = ( d dx )n I n β f(x), where n = [β] +, is called the Riemann-Liouville fractional derivative of order β. Lemma 2.3. Let β > and u C((, )) L ((, )). Then we have (i) For < γ < β, D γ I β u = I β γ u and D β I β u = u. (ii) D β u(x) = if and only if u(x) = c x β + c 2 x β c m x β m, c i R, i =,..., m, where m is the smallest integer greater than or equal to β. (iii) Assume that D β u C((, )) L ((, )). Then I β D β u(x) = u(x) + c x β + c 2 x β c m x β m, c i R, i =,..., m, where m is the smallest integer greater than or equal to β Estimates on the Green s function. In the next lemma, we give the explicit expression of the Green s function G(x, t). Lemma 2.4. If f C + ([, ]), then the fractional boundary value problem has a unique nonnegative solution where for x, t [, ], G(x, t) = D α u(x) = f(x), < x <, 3 < α 4, u() = u () = lim x + x 4 α u (x) = u () =, u(x) = (2.) G(x, t)f(t)dt, (2.2) { x α ( t) α 3 (x t) α, t x ; x α ( t) α 3, x t. Proof. Since f C([, ]), by Lemma 2.3 and Definition 2., we have u(x) = c x α + c 2 x α 2 + c 3 x α 3 + c 4 x α 4 I α f(x).

5 EJDE-27/24 SUPERLINEAR FRACTIONAL BOUNDARY VALUE PROBLEMS 5 Using the fact that u() = u () = lim x + x 4 α u (x) = u () =, we obtain c 2 = c 3 = c 4 = and c = ( t)α 3 f(t)dt. Then the unique solution of problem (2.) is u(x) = = This completes the proof. x α ( t) α 3 f(t)dt G(x, t)f(t)dt. (x t) α f(t)dt Next, we establish sharp estimates on the Green s function G(x, t). Proposition 2.5. The following properties hold: (i) For x, t [, ], H (x, t) G(x, t) where H (x, t) = x α 2 ( t) α 3 min(x, t). (ii) For x, t [, ], xα t( t) α 3 G(x, t) (iii) For x (, ] and t [, ), 2(α ) H (x, t), 2(α ) x α 2 t( t) α 3. (α ) (α )(α 2) H(x, t) G(x, t) H(x, t), x where H(x, t) = x α 3 ( t) α 3 min(x, t). (iv) For x (, ] and t [, ), (α )(α 2)(α 3) H(x, t) 2 (α )(α 2) G(x, t) H(x, t), x2 where H(x, t) = x α 4 ( t) α 4 min(x, t)( max(x, t)). Proof. We first observe that for λ, µ (, ) and c, t [, ], we have min(, µ λ )( ctλ ) ct µ max(, µ λ )( ctλ ). (2.3) (i) By Lemma 2.4, for x, t [, ], we have G(x, t) = Since max(x t,) x( t) { x α ( t) α 3 (x t) α, t x ; x α ( t) α 3, x t, = xα ( t) α 3 (max(x t, )) α, = xα ( t) α 3[ ( t) 2( max(x t, ) x( t) ) α ]. [, ], for x (, ] and t [, ), the required result follows from (2.3) with µ = α, λ = and c = ( t) 2. (ii) The assertion follows from (i) and the elementary inequalities xt min(x, t) t, for x, t [, ].

6 6 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 (iii) For all x, t [, ] we have α G(x, t) = x { x α 2 ( t) α 3 (x t) α 2, t x ; x α 2 ( t) α 3, x t, = α xα 2 ( t) α 3[ ( max(x t, ) ) α 2 ] ( t). x( t) Then the required result follows from (2.3) with µ = α 2, λ = and c = ( t). (iv) For all x (, ] and t [, ) we have { 2 (α )(α 2) x α 3 ( t) α 3 (x t) α 3, t x ; G(x, t) = x2 x α 3 ( t) α 3, x t, = (α )(α 2) x α 3 ( t) α 3[ ( max(x t, ) x( t) ) α 3 ]. Then the required result follows again from (2.3) with µ = α 3, λ = and c =. This completes the proof. From Proposition 2.5 (ii), we deduce the following characterization property. Corollary 2.6. Let f B + ((, )). Then x V f(x) C([, ]) t( t) α 3 f(t)dt <. Proposition 2.7. Let 3 < α < 4 and assume that the map t t( t) α 3 f(t) C((, )) L ((, )). Then V f is the unique solution in C([, ]) of the problem D α u(x) = f(x), < x <, u() = u () = lim x + x 4 α u (x) = u () =. Proof. Using Corollary 2.6 we deduce that V f C([, ]). I 4 α (V f ) is finite on [, ]. So, by Fubini s theorem, I 4 α (V f)(x) = Γ(4 α) = Γ(4 α) = (x t) 3 α V f(t)dt ( K(x, s)f(s)ds, (x t) 3 α G(t, s)dt)f(s)ds (2.4) This implies that where K(x, s) := x Γ(4 α) (x t)3 α G(t, s)dt. Next, we aim to give an explicit expression of the kernel K(x, s). To this end, observe that by making the substitution t = s + (x s)θ, we obtain for γ, ν >, s (x t) γ (t s) ν dt = Using this fact and (.6), we deduce that Γ(γ + )Γ(ν + ) (x s) γ+ν+. (2.5) Γ(γ + ν + 2) ( s)α 3 K(x, s) = (x t) 3 α t α dt Γ(4 α) x (x t) 3 α (max(t s, )) α dt Γ(4 α)

7 EJDE-27/24 SUPERLINEAR FRACTIONAL BOUNDARY VALUE PROBLEMS 7 = 6 x3 ( s) α 3 Γ(4 α) Now, assume that s x. Then by (2.5) we have (x t) 3 α (max(t s, )) α dt = s (x t) 3 α (max(t s, )) α dt (x t) 3 α (t s) α dt Γ(4 α) = (x s) 3. 6 On the other hand, if x s and t (, x), we have So, combining (2.6) and (2.7), we obtain Hence for x (, ), we have 6I 4 α (V f)(x) = 6 We claim that (2.6) (x t) 3 α (max(t s, )) α dt =. (2.7) K(x, s) = 6 x3 ( s) α 3 6 (max(x s, ))3. K(x, s)f(s)ds = x 3 [( s) α 3 ]f(s)ds + 3x 2 sf(s)ds 3x s 2 f(s)ds + =: J (x) + J 2 (x) + J 3 (x) + J 4 (x) + J 5 (x). s 3 f(s)ds + x 3 ( s) α 3 f(s)ds D α (V f)(x) := d4 dx 4 (I4 α (V f))(x) = f(x), for x (, ). Since the function s sf(s) is continuous and integrable near and the function s ( s) α 3 f(s) is continuous and integrable near, then the functions J 2 (x), J 3 (x), J 4 (x) and J 5 (x) are differentiable. On the other hand, since ( s) α 3 = O(s) near, it follows that J (x) is differentiable. By simple computation, we obtain d dx (6I4 α (V f))(x) = 3x 2 3x Using similar arguments as above, we obtain [( s) α 3 ]f(s)ds + 6x x sf(s)ds s 2 f(s)ds + 3x 2 ( s) α 3 f(s)ds. d 4 dx 4 (I4 α (V f))(x) = f(x), for x (, ). Next, we need to verify the boundary conditions. Since G(, t) = and V f(x) C([, ]), then it follows that V f() =. On the other hand, by Proposition 2.5 (iii), there exists a constant c > such that for all x, t [, ], we have x G(x, t) c min(x, t)( t)α 3 ct( t) α 3. This implies, by Lebesgue s theorem, that (V f) () =. x

8 8 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 By Proposition 2.5 (iv), there exists c > such that each x, t [, ], we have x 4 α 2 x 2 G(x, t) c min(x, t)( t) α 3 ct( t) α 3. Hence, by Lebesgue s theorem, we deduce that lim x + x 4 α (V f) (x) =. Let η (, ). Again by Proposition 2.5 (iv), there exists a constant c >, such that for x [η, ] and t (, ), we have 2 x 2 G(x, t) cη α 4 t( t) α 4 ( max(x, t)) cη α 4 t( t) α 3. So by the Lebesgue s theorem, we deduce that (V f) () =. Finally, we need to prove the uniqueness. Let u, v C([, ]) be two solutions of (2.4) and put z = u v. Then z C([, ]) and D α z =. Hence, by Lemma 2.3 (iii), we deduce that z(x) = c x α + c 2 x α 2 + c 3 x α 3 + c 4 x α 4. Using the fact that z() = z () = lim x + x 4 α z (x) = z () =, we deduce that z = and therefore u = v. This completes the proof. Remark 2.8. Note that the conclusion of Proposition 2.7 is also valid for α = 4. Proposition 2.9. For each x, r, t (, ), we have G(x, r)g(r, t) G(x, t) 4(α )2 r α ( r) α 3. (2.8) Proof. Using Proposition 2.5 (i), for each x, r, t (, ), we have G(x, r)g(r, t) G(x, t) 4(α )2 r α 2 α 3 min(x, r) min(r, t) ( r). min(x, t) So the result follows from this fact and that min(x, r) min(r, t) r. min(x, t) This completes the proof. Proposition 2.. Let q K α. Then (i) α q 4(α )2 (ii) For x [, ], we have where ω(x) = r α ( r) α 3 q(r)dr <. (2.9) G(x, t)ω(t)q(t)dt α q ω(x), (2.) a (α )(α 2) xα. Proof. Let q K α. (i) Using (.7) and (2.8), we obtain inequality (2.9). G(t,r) (ii) For all x, t (, ], we have lim r G(x,r) = tα x. Thus, by Fatou s lemma α and (.7), we deduce that G(x, t) ω(t) q(t)dt lim inf ω(x) r G(x, t) G(t, r) G(x, r) q(t)dt α q.

9 EJDE-27/24 SUPERLINEAR FRACTIONAL BOUNDARY VALUE PROBLEMS 9 This implies that for x [, ], which completes the proof. G(x, t)ω(t)q(t)dt α q ω(x), 3. Proof of main results Let q K α with α q <. Define the function G(x, t) on [, ] [, ] by G(x, t) = ( ) n G n (x, t), (3.) where G (x, t) = G(x, t) and G n (x, t) = n= G(x, r)g n (r, t)q(r)dr, n. (3.2) Lemma 3.. Let q K α with α q <. For all n and (x, t) [, ] [, ], we have (i) G n (x, t) α n q G(x, t). In particular, G(x, t) is well defined in [, ] [, ]. (ii) The following inequalities hold: (iii) (iv) where L n x α t( t) α 3 G n (x, t) R n x α 2 t( t) α 3, (3.3) L n = () n+ ( R n = ( 2α 2 )n+ ( G n+ (x, t) = G(x, r)g(r, t)q(r)dr = r α ( r) α 3 q(r)dr) n, r α ( r) α 3 q(r)dr) n. G n (x, r)g(r, t)q(r)dr. G(x, r)g(r, t)q(r)dr. Proof. (i) Clearly, inequality in (i) holds for n =. Assume that inequality in (i) is valid for some n, then by using (3.2) and (.7), we obtain G n+ (x, t) α n q G(x, r)g(r, t)q(r)dr αq n+ G(x, t). Since G n (x, t) αq n G(x, t), it follows that G(x, t) is well defined in [, ] [, ]. (ii) We can prove (3.3) by using Proposition 2.5 (ii), (3.2) and using a standard induction argument. (iii) We proceed by induction. The equality is true for n =. Assume that for a given integer n and (x, t) [, ] [, ], we have G n (x, t) = G n (x, r)g(r, t)q(r)dr. (3.4)

10 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 Using (3.2) and the Fubini-Tonelli theorem, we obtain G n+ (x, t) = = = G(x, r)( ( G n (r, ξ)g(ξ, t)q(ξ)dξ)q(r)dr G(x, r)g n (r, ξ)q(r)dr)g(ξ, t)q(ξ)dξ G n (x, ξ)g(ξ, t)q(ξ)dξ. (iv) Let n and x, r, t [, ]. By Lemma 3. (i) we have G n (x, r)g(r, t)q(r) α n q G(x, r)g(r, t)q(r). Hence the series n G n(x, r)g(r, t)q(r)dr converges. By the dominated convergence theorem and Lemma 3. (iii), we deduce that G(x, r)g(r, t)q(r)dr = ( ) n G n (x, r)g(r, t)q(r)dr This completes the proof. = = n= n= ( ) n G(x, r)g n (r, t)q(r)dr G(x, r)g(r, t)q(r)dr. Proposition 3.2. Let q K α with α q <. Then the function (x, t) G(x, t) is continuous on [, ] [, ]. Proof. We claim that for n, the function (x, t) G n (x, t) is continuous on [, ] [, ]. Clearly, G (x, t) is continuous on [, ] [, ]. Assume that the function (x, t) G n (x, t) is continuous on [, ] [, ]. So, for each r [, ], the function (x, t) G(x, r)g n (r, t) is continuous on [, ] [, ]. By using Lemma 3. (i) and Proposition 2.5 (ii), we have for each (x, t, r) [, ] [, ] [, ], G(x, r)g n (r, t)q(r) αq n G(x, r)g(r, t)q(r) ( 2(α ) ) 2r α ( r) α 3 q(r). We deduce by (3.2) and the dominated convergence theorem, that the function (x, t) G n (x, t) is continuous on [, ] [, ]. This proves our claim. Using again Lemma 3. (i) and Proposition 2.5 (ii), we have for all x, t [, ], G n (x, t) α n q G(x, t) 2(α ) αq n. Therefore the series n ( ) n G n (x, t) is uniformly convergent on [, ] [, ] and hence the function (x, t) G(x, t) is continuous on [, ] [, ]. This completes the proof. Lemma 3.3. Let q K α with α q /2. Then for all (x, t) [, ] [, ], we have ( α q )G(x, t) G(x, t) G(x, t). (3.5)

11 EJDE-27/24 SUPERLINEAR FRACTIONAL BOUNDARY VALUE PROBLEMS Proof. Since α q /2, we deduce by Lemma 3. (i) that Note that G(x, t) (α q ) n G(x, t) = G(x, t). (3.6) α q n= G(x, t) = G(x, t) ( ) n G n+ (x, t). (3.7) n= Since the series n G(x, r)g n(r, t)q(r)dr is convergent, we deduce by (3.7) and (3.2) that Therefore, G(x, t) = G(x, t) = G(x, t) ( ) n G(x, r)g n (r, t)q(r)dr n= G(x, r)( ( ) n G n (r, t))q(r)dr. n= G(x, t) = G(x, t) V (qg(, t))(x). (3.8) Using (3.6) and Lemma 3. (i) (with n = ), we deduce that V (qg(., t))(x) α q V (qg(., t))(x) = α q G (x, t) α q α q G(x, t). This implies by (3.8) that G(x, t) G(x, t) α q α q G(x, t) = 2α q α q G(x, t). So G(x, t) G(x, t) and by (3.8) and Lemma 3. (i) (with n = ), we have The proof is now complete. G(x, t) G(x, t) V (qg(, t))(x) ( α q )G(x, t). Using Proposition 3.2, (3.5) and Proposition 2.5 (ii), we obtain the following property. Corollary 3.4. Let q K α with α q 2 and let f B+ ((, )). Then the following characterization property holds: x V q f(x) C([, ]) t( t) α 3 f(t)dt <. Lemma 3.5. Let q K α with α q 2 and f B+ ((, )). Then we have In particular, if V (qf) <, we have Here, V (q.)(f) := V (qf). V f = V q f + V q (qv f) = V q f + V (qv q f). (3.9) (I V q (q.))(i + V (q.))f = (I + V (q.))(i V q (q.))f = f. (3.)

12 2 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 Proof. Let (x, t) [, ] [, ]. Then by (3.8), we have G(x, t) = G(x, t) + V (qg(, t))(x), which implies by the Fubini-Tonelli theorem that for all f B + ((, )), V f(x) = (G(x, t) + V (qg(, t))(x))f(t)dt = V q f(x) + V (qv q f)(x). Using Lemma 3. (iii) and the Fubini-Tonelli theorem, we obtain that for all f B + ((, )) and x [, ], It follows that We deduce that This completes the proof. G(x, r)g(r, t)q(r)f(t)drdt = V q (qv f)(x) = V (qv q f)(x). V f = V q f + V (qv q f) = V q f + V q (qv f)(x). G(x, r)g(r, t)q(r)f(t)drdt. Proposition 3.6. Let q K α C((, )) such that α q /2 and f B + ((, )) such that t t( t) α 3 f(t) C((, )) L ((, )). Then V q f C + ([, ]) and it is the unique solution of problem (.8) satisfying ( α q )V f V q f V f. (3.) Proof. By Corollary 3.4, we deduce that x V q f(x) C + ([, ]). Therefore, the function x q(x)v q f(x) C((, )). Using (3.9) and Proposition 2.5 (ii), there exists a constant c such that V q f(x) V f(x) So we deduce that 2(α ) t( t) α 3 q(t)v q f(t)dt c x α 2 t( t) α 3 f(t)dt = cx α 2. (3.2) t α ( t) α 3 q(t)dt <. Hence by Proposition 2.7, the function u = V q f = V f V (qv q f) satisfies the equation D α u(x) = f(x) + q(x)u(x), x (, ), u() = u () = lim x + x4 α u (x) = u () =. By integrating inequalities (3.5), we obtain (3.). For the uniqueness, assume that v is another nonnegative solution in C([, ]) of problem (.8) satisfying (3.). Since the function t q(t)v(t) is of class C((, )) and by (3.), (3.2), the function t t( t) α 3 q(t)v(t) is in L ((, )), it follows by Proposition 2.7 that the function ṽ := v + V (qv) satisfies D α ṽ(x) + f(x) =, x (, ), ṽ() = ṽ () = lim x + x4 α ṽ (x) = ṽ () =.

13 EJDE-27/24 SUPERLINEAR FRACTIONAL BOUNDARY VALUE PROBLEMS 3 Using Proposition 2.7, we deduce that hence ṽ := v + V (qv) = V f, (I + V (q.))(v u) =. Using (3.), (3.2) and Proposition 2.5 (i) we have V (q v u )(x) 4c(α ) 4c(α ) t α 2 ( t) α 3 min(x, t)q(t)dt t α ( t) α 3 q(t)dt <. So by (3.), we deduce that u = v. This completes the proof. 3.. Proof of Theorem.. Let a > and recall that a ω(x) = (α )(α 2) xα, for x. Since ϕ satisfies (H 2 ), there exists a function q in K α C((, )) such that α q /2 and for each x (, ), the map t t(q(x) ϕ(x, tω(x))) is nondecreasing on [, ]. Let Λ := {u B + ((, )) : ( α q )ω u ω}. Define the operator T on Λ by By (3.9) and (2.), we have By (H2), we obtain T u = ω V q (qω) + V q ((q ϕ(, u))u). Using (3.4) and (3.3), we have that for all u Λ, V q (qω) V (qω) α q ω ω. (3.3) ϕ(., u) q, for all u Λ. (3.4) T u ω V q (qω) + V q (qu) ω, T u ω V q (qω) ( α q )ω. Therefore. T (Λ) Λ. Next, we prove that the operator T is nondecreasing on Λ. Indeed, let u, v Λ be such that u v. Since the map t t(q(x) ϕ(x, tω(x))) is nondecreasing on [, ], we obtain that for all x (, ), T v T u = V q ([v(q ϕ(, v)) u(q ϕ(., u))]). Now, we consider the sequence {u n } defined by u = ( α q )ω and u n+ = T u n, for n N. Since Λ is invariant under T, we have u = T u u and by the monotonicity of T, we deduce that ( α q )ω = u u u n u n+ ω. Hence by dominated convergence theorem, (H), and (H2), we conclude that the sequence {u n } converges to a function u Λ satisfying It follows that u = (I V q (q.))ω + V q ((q ϕ(, u))u). (I V q (q.))u = (I V q (q.))ω V q (uϕ(., u)).

14 4 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 On the other hand, by (3.3), we have V (qu) V (qω) ω <. Then by applying the operator (I + V (q.)) on both sides of the above equality and using (3.9) and (3.), we conclude that u satisfies u = ω V (uϕ(, u)). (3.5) We claim that u is the required solution. From (3.4), we have a u(t)ϕ(t, u(t)) q(t)ω(t) = (α )(α 2) tα q(t). (3.6) So t( t)α 3 u(t)ϕ(t, u(t))dt <. Therefore by Corollary 2.6, the function x V (uϕ(., u))(x) C([, ]) and from (3.5), we conclude that u C([, ]). Since by (H) and (3.6), the function t t( t) α 3 u(t)ϕ(t, u(t)) belongs to C((, )) L ((, )), we deduce by Proposition 2.7 that u is the required solution. To prove uniqueness, assume (H3) and let v be another nonnegative solution in C([, ]) of problem (.2) satisfying (.9). Since v satisfies (.9), we deduce by (3.4)) and (3.6) that a v(t)ϕ(t, v(t)) q(t)ω(t) = (α )(α 2) tα q(t). So the function t t( t) α 3 v(t)ϕ(t, v(t)) belongs to C((, )) L ((, )), and by Proposition 2.7, we deduce that the function ṽ := v + V (vϕ(, v)) satisfies Hence We deduce that D α ṽ(x) =, < x <, ṽ() = ṽ () = lim x + x4 α ṽ (x) =, ṽ () = a >. ṽ := v + V (vϕ(, v)) = ω. v = ω V (vϕ(, v)). (3.7) Let h be the function defined on (, ) by { v(x)ϕ(x,v(x)) u(x)ϕ(x,u(x)) v(x) u(x), if v(x) u(x), h(x) =, if v(x) = u(x). Then by (H3), h B + ((, )) and by (3.5) and (3.7), we have (I + V (h.))(v u) =. On the other hand, by (H2), we remark that h q and by (2.) we deduce that V (h v u ) 2V (qω) 2α q ω <. Hence by (3.), we conclude that u = v. This completes the proof. Example 3.7. Let 3 < α 4 and a >. Let σ, and p C + ((, )) such that r (α )(+σ) ( r) α 3 p(r)dr <. Let p(x) := (σ + )p(x)(ω(x)) σ. Since p K α, then for λ [, 2α ep ), the problem D α u(x) λp(x)u σ+ (x) =, < x <, u() = u () = lim x + x4 α u (x) =, u () = a,

15 EJDE-27/24 SUPERLINEAR FRACTIONAL BOUNDARY VALUE PROBLEMS 5 has a unique positive solution u in C([, ]) satisfying ( λα ep )ω(x) u(x) ω(x), x [, ]. Example 3.8. Let 3 < α 4 and a >. Let σ, γ > and p C + ((, )) such that r (α )+(α )(σ+γ) ( r) α 3 p(r)dr <. Let θ(s) = s σ+ log( + s γ ) and p(t) := p(t) max ξ ω(t) θ (ξ). Since p K α, then for λ [, 2α ep ), the problem D α u(x) λp(x)u σ+ (x) log( + u γ (x)) =, < x <, u() = u () = lim x + x4 α u (x) =, u () = a, has a unique positive solution u in C([, ]) satisfying ( λα ep )ω(x) u(x) ω(x), x [, ]. Acknowledgements. The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding this Research group NO (RG ). References [] I. Bachar, H. Mâagli; Superlinear Singular Fractional Boundary Value Problems, Electron. J. Differential Equations, Vol. 26 (26), No. 8, pp. -5. [2] I. Bachar, H. Mâagli, V. D. Rădulescu; Fractional Navier boundary value problems, Bound. Value Problems 26:79 (26) DOI.86/s [3] I. Bachar, H. Mâagli, F. Toumi, Z. Zine el Abidine; Existence and global asymptotic behavior of positive solutions for sublinear and superlinear fractional boundary value problems, Chinese Annals of Mathematics, Series B, 37 (26), no., -28. [4] R. Demarque, O. H. Miyagaki; Radial solutions of inhomogeneous fourth order elliptic equations and weighted Sobolev embeddings, Adv. Nonlinear Anal. 4 (2) (25), [5] K. Diethelm, A. D. Freed; On the solution of nonlinear fractional order differential equations used in the modeling of viscoplasticity, In Scientific computing in Chemical Engineering. II- Computational Fluid Dynamics, Reaction Engineering and Molecular Properties (F. Keil, W. Mackens, H. Voss, J. Werther, eds.), pp , Springer, Heidelberg, 999. [6] L. Gaul, P. Klein, S. Kempfle; Damping description involving fractional operators, Mech. Syst. Signal Process., 5 (99), [7] W. G. Glockle, T. F. Nonnenmacher; A fractional calculus approach of self-similar protein dynamics, Biophys. J., 68 (995), [8] J. R. Graef, L. Kong, Q. Kong, M. Wang; Existence and uniqueness of solutions for a fractional boundary value problem with Dirichlet boundary condition, Electron. J. Qual. Theory Differ. Equ. (23), No. 55, -. [9] R. Hilfer; Applications of Fractional Calculus in Physics, World Scientific, Singapore, 2. [] E. R. Kaufmann, E. Mboumi; Positive solutions of a boundary value problem for a nonlinear fractional differential equation, Electron. J. Qual. Theory Differ. Equ., 3 (28),. [] A. Kilbas, H. Srivastava, J. Trujillo; Theory and Applications of Fractional Differential Equations, North-Holland Mathematics Studies, Vol. 24, Elsevier, Amsterdam, 26. [2] A. A. Kilbas, J. J. Trujillo; Differential equations of fractional order: methods, results and problems, I, Appl. Anal., 78 (2), [3] A. A. Kilbas, J. J. Trujillo; Differential equations of fractional order: methods, results and problems, II, Appl. Anal., 8 (22), [4] S. Liang, J. Zhang; Positive solutions for boundary value problems of nonlinear fractional differential equation, Nonlinear Anal., 7 (29),

16 6 I. BACHAR, H. MÂAGLI, V.D. RĂDULESCU EJDE-27/24 [5] H. Mâagli, N. Mhadhebi, N. Zeddini; Existence and estimates of positive solutions for some singular fractional boundary value problems, Abstract and Applied Analysis Vol. 24 (24), Article ID 278, 7 pages. [6] H. Mâagli, N. Mhadhebi, N. Zeddini; Existence and exact asymptotic behavior of positive solutions for a fractional boundary value problem, Abstract and Applied Analysis Vol. 23 (23), Article ID 4254, 6 pages. [7] F. Mainardi; Fractional calculus: Some basic problems in continuum and statical mechanics, in: A. Carpinteri, F. Mainardi (eds.) Fractals and fractional calculus in continuum calculus Mechanics, pp Springer, Vienna, 997. [8] F. Mainardi; Fractional diffusive waves in viscoelastic solids, in: Wegner, JL, Norwood, FR (eds.) Nonlinear Waves in Solids, ASME/AMR, Fairfield, 995. [9] R. Metzler, J. Klafter; Boundary value problems for fractional diffusion equations, Physica A, 278 (2), [2] K. Miller, B. Ross; An Introduction to the Fractional Calculus and Fractional Differential Equations, Wiley, New York, 993. [2] G. Molica Bisci, V. D. Rădulescu, R. Servadei; Variational Methods for Nonlocal Fractional Problems, Encyclopedia of Mathematics and its Applications, Vol. 62, Cambridge University Press, Cambridge, 26. [22] G. Molica Bisci, D. Repovš; Existence and localization of solutions for nonlocal fractional equations, Asymptot. Anal. 9 (24), no. 3-4, [23] I. Podlubny; Fractional Differential Equations, in : Mathematics in Sciences and Engineering, vol. 98, Academic Press, San Diego, 999. [24] V. D. Rădulescu; Qualitative Analysis of Nonlinear Elliptic Partial Differential Equations, Hindawi Publishing Corporation, 28. [25] S. Samko, A. Kilbas, O. Marichev; Fractional Integrals and Derivatives, Theory and Applications, Gordon and Breach, Yverdon, 993. [26] H. Scher, E. Montroll; Anomalous transit-time dispersion in amorphous solids, Phys. Rev. B, 2 (975), [27] V. Tarasov; Fractional Dynamics: Applications of Fractional Calculus to Dynamics of Particles, Fields and Media, Springer-Verlag, New York, 2. [28] S. Timoshenko, J. M. Gere; Theory of Elastic Stability, McGraw-Hill, New York, 96. [29] C. Zhai, W. Yan, C. Yang; A sum operator method for the existence and uniqueness of positive solutions to Riemann Liouville fractional differential equation boundary value problems, Commun. Nonlinear Sci. Numer. Simul., 8 (23), Imed Bachar (corresponding author) King Saud University, Department of Mathematics, College of Science, P.O.Box 2455, Riyadh 45, Saudi Arabia address: abachar@ksu.edu.sa Habib Mâagli Department of Mathematics, College of Sciences and Arts, King Abdulaziz University, Rabigh Campus P.O. Box 344, Rabigh 29, Saudi Arabia. Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire, 292 Tunis, Tunisia address: abobaker@kau.edu.sa, habib.maagli@fst.rnu.tn Vicenţiu D. Rădulescu Department of Mathematics, Faculty of Sciences, King Abdulaziz University, P.O. Box 823, Jeddah 2589, Saudi Arabia. Department of Mathematics, University of Craiova, 2585 Craiova, Romania address: vicentiu.radulescu@math.cnrs.fr