# Solutions for Integral Boundary Value Problems of Nonlinear Hadamard Fractional Differential Equations.

1. IntroductionIn this work we study the following integral boundary value problems of nonlinear Hadamard fractional differential equations

[mathematical expression not reproducible] (1)

where [alpha], [beta], and [mu] are three positive real numbers with [alpha] [member of] (2,3], [beta] [member of] (1,2], and [mu] [member of] [0, [beta]), ([[phi].sub.p](s) = [[absolute value of s].sup.p-2]s is the p-Laplacian for p > 1, s [member of] R, and f is a continuous function on [1, e] x R. Moreover, let [[phi].sup.-1.sub.p] = [[phi].sub.q] with 1/p + 1/q = 1. In what follows, we offer some related definitions and lemmas for Hadamard fractional calculus.

Definition 1 (see [1, Page 111]). The [alpha]th Hadamard fractional order derivative of a function y :[ 1, +[infinity]) [right arrow] R is defined by

[mathematical expression not reproducible], (2)

where [alpha] > 0, n = [[alpha]] + 1, and [[alpha]] denotes the largest integer which is less than or equal to a. Moreover, we here also offer the ath Hadamard fractional order integral of y : [1, +[infinity]) [right arrow] R which is defined by

[mathematical expression not reproducible], (3)

where [GAMMA] is the gamma function.

Lemma 2 (see [1, Theorem 2.3]). Let [alpha] > 0, n = [[alpha]] + 1. Then

[mathematical expression not reproducible]. (4)

In recent years, there have been some significant developments in the study of boundary value problems for nonlinear fractional differential equations; we refer to [2-11] and the references therein. For more related works, see also [12-49]. For example, by using monotone iterative methods, Wang et al. [3] investigated a class of boundary value problems of Hadamard fractional differential equations involving nonlocal multipoint discrete and Hadamard integral boundary conditions and established monotone iterative sequences, which can converge to the unique positive solution of their problems. Similar methods are also applied in [4, 5, 12-15].

For differential equations with the p-Laplacian, see, for example, [6,7,15-20] and the references therein. In [6], Wang considered the nonlinear Hadamard fractional differential equation with integral boundary condition and p-Laplacian operator

[mathematical expression not reproducible], (5)

where f grows (p - 1)-sublinearly at +[infinity], and by using the Schauder fixed point theorem, a solution existence result is obtained. In [7], Li and Lin used the Guo-Krasnosel'skii fixed point theorem to obtain the existence and uniqueness of positive solutions for (1) with [mu] = 0.

However, we note that these are seldom considered Hadamard fractional differential equations with the p-Laplacian in the literature; in this paper we are devoted to this direction. We first utilize the Guo-Krasnosel'skii fixed point theorem to obtain two positive solutions existence theorems when f grows (p - 1)-superlinearly and (p - 1)-sublinearly with the p-Laplacian, and secondly by using the fixed point index, we obtain a nontrivial solution existence theorem without the p-Laplacian, but the nonlinearity can allow being sign-changing and unbounded from below. This improves and generalizes some semipositone problems [21-31].

2. Preliminaries

In this section, we first calculate Green's functions associated with (1) and then transform the boundary value problem into its integral form. For this, we give the following lemma.

Lemma 3. Let [alpha], [beta], [mu], [[phi].sub.p], and [D.sup.[alpha]], [D.sup.[beta]] be as in (1). Then (1) can take the integral form

[mathematical expression not reproducible], (6)

where

[mathematical expression not reproducible], (7)

and

[mathematical expression not reproducible]. (8)

Proof. Use y(t) to replace f(t, u) in (1). Let [D.sup.[beta]]([[phi].sub.p]([D.sup.[alpha]]u(t))) = y(t). Then from Lemma 2 we have

[mathematical expression not reproducible]. (9)

Note that [D.sup.[alpha]]u(1) = 0 implies [[phi].sub.p]([D.sup.[alpha]]u(1)) = 0, and then [c.sub.2] = 0. Therefore, we obtain

[[phi].sub.p] ([D.sup.[alpha]]u(t)) = [I.sup.[beta]] y(t) + [c.sub.1] [(log t).sup.[beta]-1]. (10)

Next, we calculate [[phi].sub.p] ([D.sup.[alpha]]u(e)) and [mu] [[integral].sup.e.sub.1] [[phi].sub.p] ([D.sup.[alpha]]u(t))(dt/t):

[mathematical expression not reproducible], (11)

and

[mathematical expression not reproducible]. (12)

The condition [[phi].sub.p] ([D.sup.[alpha]]u(e)) = [mu] [[integral].sup.e.sub.1] [[phi].sub.p] ([D.sup.[alpha]]u(t))(dt/t) enables us to obtain

[mathematical expression not reproducible]. (13)

Substituting [c.sub.1] into (10) gives

[mathematical expression not reproducible]. (14)

Note that -[[phi].sub.p] ([D.sup.[alpha]]u(t)) = [[phi].sub.p] (-[D.sup.[alpha]]u(t)), and hence we obtain

[mathematical expression not reproducible]. (15)

Then, if we let x(t) = [[phi].sub.q] ([[integral].sup.e.sub.1] H(t, [tau])y([tau])(d[tau]/ [tau])), t [member of] [1, e], from Lemma 2 we obtain

[mathematical expression not reproducible]. (16)

The condition u(1) = u(1) = 0 implies that [c.sub.2] = [c.sub.3] = 0. Then we substitute e into the first derivative of u, and we calculate c1 as follows:

[c.sub.1] = 1/[GAMMA]([alpha]) [[integral].sup.e.sub.1] [(1 - log s).sup.[alpha]-2] x(s) ds/s. (17)

As a result, from (16) we have

[mathematical expression not reproducible]. (18)

This completes the proof.

Lemma 4. Green's functions G, H defined by (7) and (8) have the following properties:

(i) G,H are continuous, nonnegative functions on [1, e] x [1, e],

(ii) [(log t).sup.[alpha]-1] [(1 - log s).sup.[alpha]-2] - [(1 - log s).sup.[alpha]-1]] [less than or equal to] [GAMMA]([alpha])G(t, s) [less than or equal to] [(1 - log s).sup.[alpha]-2] - [(1 -log s).sup.[alpha]-1], for t, s [member of] [1, e].

From [7, Lemma 7] and [8, Lemma 2.2] we easily obtain this lemma, so we omit its proof.

Let

[mathematical expression not reproducible]. (19)

Then we obtain the following lemma.

Lemma 5. There exist [[kappa].sub.1] = [[integral].sup.e.sub.1] [(1 - log s).sup.[alpha]-1] [phi](t)(dt/t), [[kappa].sub.1] = [[integral].sup.e.sub.1][phi](t)(dt/t) such that

[mathematical expression not reproducible].

Proof. We only prove the left inequality above. From Lemma 4(ii) we have

[mathematical expression not reproducible]. (21)

This completes the proof.

Let E = C[1, e] be the Banach space equipped with the norm [absolute value of u] = [max.sub.t[member of][1,e]] [absolute value of M(f)]. Then we define two sets on E as follows:

[mathematical expression not reproducible]. (22)

Consequently, P, [P.sub.0] are cones on E. From Lemma 3 we can define an operator A on E as follows:

[mathematical expression not reproducible]. (23)

The continuity of G, H, f implies that A : E [right arrow] E is a completely continuous operator and the existence of solutions for (1) if and only if the existence of fixed points for A.

Lemma 6 (see [50]). Let E be a Banach space and [OMEGA] a bounded open set in E. Suppose that A : [OMEGA] [right arrow] E is a continuous compact operator. If there exists [u.sub.0] [member of] E \ [0} such that

u - Au [not equal to] [mu][u.sub.0], [for all]u [member of] [partial derivative][OMEGA], [mu] [greater than or equal to] 0, (24)

then the topological degree deg(I - A, [OMEGA], 0) = 0.

Lemma 7 (see [50]). Let E be a Banach space and [OMEGA] a bounded open set in E with 0 [member of] [OMEGA]. Suppose that A : [OMEGA] [right arrow] E is a continuous compact operator. If

Au [not equal to] [mu]u, [for all]u [member of] [partial derivative][OMEGA], [mu] [greater than or equal to] 1, (25)

then the topological degree deg(I - A, [OMEGA], 0) = 1.

Lemma 8 (see [50]). Let E be a Banach space and P [subset] E a cone in E. Assume that [[OMEGA].sub.1], [[OMEGA].sub.2] are open subsets of E with 0 [member of] [[OMEGA].sub.1] [subset] [[bar.[OMEGA]].sub.1] [subset] [[OMEGA].sub.2], and let A : P [integral] ([[bar.[OMEGA]].sub.1] \ [[OMEGA].sub.1]) [right arrow] P be a completely continuous operator such that either

(G1) [mathematical expression not reproducible],

or

(G2) [mathematical expression not reproducible].

Then A has a fixed point in P [intersection] ([[bar.[OMEGA]].sub.1] \ [[OMEGA].sub.1]).

3. Positive Solutions for (1)

Let [mathematical expression not reproducible]. Now, we first list our assumptions on /:

(H1) f [member of] C([0, 1] x [R.sup.+], [R.sup.+]),

(H2) there exist [mathematical expression not reproducible],

(H3) there exists [mathematical expression not reproducible],

(H4) [mathematical expression not reproducible],

(H5) there exist [mathematical expression not reproducible],

where

Lemma 9. Suppose that (H1) holds. Then A(P) [subset] [P.sub.0].

Proof. If u [member of] P, from Lemma 4 we have

[mathematical expression not reproducible]. (27)

On the other hand,

[mathematical expression not reproducible]. (28)

This completes the proof.

Remark 10. Our aim is to find operator equation u = Au has fixed points in P, and from Lemma 9, these fixed points must belong to the cone [P.sub.0]. Therefore, our work space can be chosen [P.sub.0] rather than P.

In what follows, we discuss the existence of positive solutions for (1) in [P.sub.0].

Theorem 11. Suppose that (H1)-(H3) hold. Then (1) has at least two positive solutions.

Proof. From (H3), when [mathematical expression not reproducible], we have (Am) (f)

[mathematical expression not reproducible]. (29)

Hence, we obtain

[mathematical expression not reproducible]. (30)

On the other hand, by the second limit inequality in (H2), there exists [r.sub.1] [member of] (0, [[rho].sub.1]) such that

f(t, u) [greater than or equal to] [[phi].sub.p] ([N.sub.2]u), [for all]m e [0, [r.sub.1]], t [member of] [[[delta].sub.1], e], (31)

Note that if [mathematical expression not reproducible], from the definition of [P.sub.0] we have

u(t) [greater than or equal to] [(1 - log [[delta].sub.1]).sup.[alpha]-1] [parallel] u [parallel]. (32)

This, together with (31), implies that

[mathematical expression not reproducible]. (33)

By the first limit inequality in (H2), there exist [R.sub.1] > [[rho].sub.1] and [C.sub.1] > 0 such that

f(t, u) [greater than or equal to] [[phi].sub.p] ([N.sub.1] u) - [C.sub.1], [for all]u [member of]e [R.sup.+], t [member of] [[[delta].sub.1], e]. (34)

Note that [R.sub.1] can be chosen large enough, and if [mathematical expression not reproducible], together with (32), there exists [C.sub.2] >0 such that

f(t, u) [greater than or equal to] [[phi].sub.p] ([N.sub.1] [(log [[delta].sub.1]).sup.[alpha]-1] [R.sub.1] - [C.sub.2]), (35)

Combining this and (33), we find

[mathematical expression not reproducible], (36)

where [mathematical expression not reproducible]. Consequently, we have

[mathematical expression not reproducible]. (37)

In summary, from (30), (33), and (37) with [R.sub.1] > [[rho].sub.1] > [r.sub.1], Lemma 8 enables us to obtain that (1) has at least two positive solutions in [mathematical expression not reproducible]. This completes the proof.

Theorem 12. Suppose that (H1), (H4)-(H5) hold. Then (1) has at least two positive solutions.

Proof. If [mathematical expression not reproducible]. Hence, from (H5) we obtain

[mathematical expression not reproducible]. (38)

This indicates that

[mathematical expression not reproducible]. (39)

On the other hand, by the second limit inequality in (H4), there exists [r.sub.2] [member of] (0, [[rho].sub.2]) such that

f(t, u) [less than or equal to] [[phi].sub.p] ([M.sub.2] u), [for all]u e [0, [r.sub.2]], t [member of] [1, e]. (40)

This, if [mathematical expression not reproducible], implies that

[mathematical expression not reproducible]. (41)

This gives

[mathematical expression not reproducible]. (42)

By the first limit inequality in (H4), there exist [R.sub.2] > [[rho].sub.2] and [C.sub.4] > 0 such that

[mathematical expression not reproducible]. (43)

Consequently, if [mathematical expression not reproducible] large enough, we obtain

[mathematical expression not reproducible], (44)

where [C.sub.5] = [C.sub.4] [[integral].sup.e.sub.1] G(e, s) [[phi].sub.p] ([[integral].sup.e.sub.1] H(s, [tau])(d[tau]/[tau]))(ds/s). Hence, we have

[mathematical expression not reproducible]. (45)

In a word, from (39), (42), and (45) with [R.sub.2] > [[rho].sub.2] > [r.sub.2], Lemma 8 enables us to obtain that (1) has at least two positive solutions in [mathematical expression not reproducible]. This completes the proof.

Example 13. Let

[mathematical expression not reproducible], (46)

where [[gamma].sub.1] [member of] (p - 1, +[infinity]), [[gamma].sub.2] [member of] (0, p - 1), and [N.sub.3], [[rho].sub.1] are defined by (H3). Then

[mathematical expression not reproducible]. (47)

Moreover, for u [member of] [0, [[rho].sub.1], t [member of] [1, e] we have

[mathematical expression not reproducible]. (49)

Therefore, (H1)-(H3) hold.

Example 14. Let

where [[gamma].sub.3] [member of] (0, p - 1), [[gamma].sub.4] [member of] (p - 1, +[infinity]), and [M.sub.3], [[rho].sub.1] are defined by (H5). Then

[mathematical expression not reproducible]. (50)

Moreover, [mathematical expression not reproducible] we have

[mathematical expression not reproducible]. (51)

Therefore, (H1), (H4)-(H5) hold.

4. Nontrivial Solutions for (1)

In this section we consider the boundary value problem (1) without the p-Laplacian, i.e., p = 2. In this case, (1) can be transformed into its integral form as follows:

[mathematical expression not reproducible]. (52)

As said in Section 3, we define an operator, still denoted by A, as follows:

(Au)(t) = [[integral].sup.e.sub.1] [G.sub.1](t, s) f (s, u(s)) ds/s,

for u [member of] E, t [member of] [1, e]. (53)

In what follows, we aim to find the existence of fixed points of A. For this, we list our assumptions on f:

(H6) f [member of] C([1, e] x R, R),

(H7) There exist nonnegative functions a(t), b(t) [member of] E with b [??] 0 and K(u) [member of] C(R, [R.sup.+]) such that

f (t, u) [greater than or equal to] - a(t) - b(t)K(u), [for all]u [member of] R, t [member of] [1, e]. (54)

Moreover,

[mathematical expression not reproducible], (55)

(H8) [mathematical expression not reproducible],

(H9) [mathematical expression not reproducible].

Theorem 15. Suppose that (H6)-(H9) hold. Then (1) has at least one nontrivial solution.

Proof. From (H9) there exist [[epsilon].sub.3] [member of] (0, [[kappa].sup.-1.sub.2]) and [r.sub.3] > 0 such that

[mathematical expression not reproducible]. (56)

For this [r.sub.3], we show that

[mathematical expression not reproducible]. (57)

If otherwise, there exist [mathematical expression not reproducible] such that

[Au.sub.1] = [[mu].sub.1][u.sub.1], (58)

and hence, we obtain

[mathematical expression not reproducible]. (59)

Multiply both sides of the above inequality by [phi](t) and integrate from 1 to e and together with Lemma 5 we obtain

[mathematical expression not reproducible]. (60)

This implies that [[integral].sup.e.sub.1] [absolute value of [u.sub.1](t)][phi](t)(dt/t) = 0, and [u.sub.1] [equivalent to] 0 for the fact that [phi](t) [??] 0,for t e [1, e], which contradicts [mathematical expression not reproducible]. Therefore, (57) is true, and from Lemma 7 we obtain

[mathematical expression not reproducible]. (61)

On the other hand, by (H8), there exist [e.sub.4] > 0 and [X.sub.0] > 0 such that

[mathematical expression not reproducible]. (62)

For every fixed e with [[parallel]b[parallel].sub.[epsilon]] [member of] (0, [[epsilon].sub.4]), [[parallel]b[parallel] = [max.sub.t[member of][1,e]] [absolute value of b(t)], and from (H7), there exists [X.sub.1] > [X.sub.0] such that

K(u) [less than or equal to] [epsilon] [absolute value of u], [for all][absolute value of u] > [X.sub.1]. (63)

Combining the two inequalities above, (H7) enables us to find

[mathematical expression not reproducible]. (64)

If we take [mathematical expression not reproducible]. Then we easily have

f(t, u) [greater than or equal to] ([kappa].sup.-1.sub.1] + [[epsilon].sub.4] - [parallel]b[parallel] [epsilon] [absolute value of u] - a(t) - [C.sub.6]

[for all]u [member of] R, t [member of] [1, e]. (65)

Note that e can be chosen arbitrarily small, and we let

[mathematical expression not reproducible], (66)

where W(s) = [[integral].sup.e.sub.1] [(1 - log [tau]).sup.[alpha]-2] H([tau], s)(d[tau]/[tau]), for s [member of] [1, e]. Now we prove that

[mathematical expression not reproducible], (67)

where f is defined by (19). Indeed, if (67) is not true, then there exists [mathematical expression not reproducible] such that

[u.sub.2] - [Au.sub.2] = [[mu].sub.0][phi]. (68)

Let [??](t) = [[integral].sup.e.sub.1] [G.sub.1](t, s)[a(s) + b(s)K([u.sub.2](s)) + [C.sub.6]](ds/s). Then [??] [member of] [P.sub.0] and

[mathematical expression not reproducible]. (69)

Consequently, we have

[mathematical expression not reproducible]. (70)

Plus [??] into (68) gives

[mathematical expression not reproducible]. (71)

Note that f(s, [u.sub.2](s)) + a(s) + b(s)K([u.sub.2](s)) + [C.sub.6] [member of] P, s [member of] [1, e] and [phi] [member of] [P.sub.0]. Lemma 9 enables us to know that [u.sub.2] + [??] [member of] [P.sub.0] From (65) we have

[mathematical expression not reproducible]. (72)

On the other hand, we have

[mathematical expression not reproducible]. (73)

This inequality holds if

[mathematical expression not reproducible]. (74)

Indeed, [mathematical expression not reproducible]. Consequently,

[mathematical expression not reproducible]. (75)

As a result, we have

[mathematical expression not reproducible], (76)

where (Tu)(t) = [[integral].sup.e.sub.1] [G.sub.1](t, s)u(s)(ds/s), for u [member of] E, t [member of] [1, e]. Using (68) we obtain

[mathematical expression not reproducible]. (77)

Define

[[mu].sup.*] = sup {[mu] > 0 : [u.sub.2] + [??] [less than or equal to] [mu][phi]}. (78)

Note that [[mu].sub.0] [member of] > 0 : [u.sub.2] + [??] [greater than or equal to] [mu][phi]}, and then [[mu].sup.*] [greater than or equal to] [[mu].sub.0], [u.sub.2] + [??] [greater than or equal to] [[mu].sup.*][phi]. From Lemma 5 we have

[mathematical expression not reproducible], (79)

and hence

[mathematical expression not reproducible], (80)

which contradicts the definition of [[mu].sup.*]. Therefore, (67) holds, and from Lemma 6 we obtain

[mathematical expression not reproducible]. (81)

This, together with (61), implies that

[mathematical expression not reproducible]. (82)

Therefore the operator A has at least one fixed point in [mathematical expression not reproducible], and (1) has at least one nontrivial solution. This completes the proof.

Example 16. Let [mathematical expression not reproducible]. Therefore, (H6)-(H9) hold.

https://doi.org/10.1155/2018/2193234

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no competing interests.

Acknowledgments

This work is supported by Natural Science Foundation of Shandong Province (ZR2018MA011, ZR2018MA009, and ZR2015AM014).

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Keyu Zhang (iD), (1) Jianguo Wang (iD), (1) and Wenjie Ma (2)

(1) School of Mathematics, Qilu Normal University, Jinan 250013, China

(2) Department of Applied Mathematics, Shandong University of Science and Technology, Qingdao 266590, China

Correspondence should be addressed to Keyu Zhang; keyu_292@163.com

Received 25 August 2018; Accepted 19 October 2018; Published 1 November 2018

Academic Editor: Xinguang Zhang

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Title Annotation: | Research Article |
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Author: | Zhang, Keyu; Wang, Jianguo; Ma, Wenjie |

Publication: | Journal of Function Spaces |

Date: | Jan 1, 2018 |

Words: | 4810 |

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