Hindawi Journal of Function Spaces Volume 2018, Article ID 2193234, 10 pages https://doi.org/10.1155/2018/2193234
Research Article Solutions for Integral Boundary Value Problems of Nonlinear Hadamard Fractional Differential Equations Keyu Zhang ,1 Jianguo Wang 1 2
,1 and Wenjie Ma2
School of Mathematics, Qilu Normal University, Jinan 250013, China Department of Applied Mathematics, Shandong University of Science and Technology, Qingdao 266590, China
Correspondence should be addressed to Keyu Zhang; keyu
[email protected] Received 25 August 2018; Accepted 19 October 2018; Published 1 November 2018 Academic Editor: Xinguang Zhang Copyright Β© 2018 Keyu Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In this paper using fixed point methods we establish some existence theorems of positive (nontrivial) solutions for integral boundary value problems of nonlinear Hadamard fractional differential equations.
1. Introduction In this work we study the following integral boundary value problems of nonlinear Hadamard fractional differential equations π·π½ (ππ (π·πΌ π’ (π‘))) = π (π‘, π’ (π‘)) ,
1 < π‘ < π,
ππ (π·πΌ π’ (π)) = π β« ππ (π·πΌ π’ (π‘)) 1
dπ‘ , π‘
where πΌ, π½, and π are three positive real numbers with πΌ β (2, 3], π½ β (1, 2], and π β [0, π½), ππ (π ) = |π |πβ2 π is the πLaplacian for π > 1, π β R, and π is a continuous function on [1, π] Γ R. Moreover, let ππβ1 = ππ with 1/π + 1/π = 1. In what follows, we offer some related definitions and lemmas for Hadamard fractional calculus. Definition 1 (see [1, Page 111]). The πΌth Hadamard fractional order derivative of a function π¦ : [1, +β) σ³¨β R is defined by π·πΌ π¦ (π‘) =
π‘ 1 π‘ πΌβ1 dπ β« (log ) π¦ (π ) , Ξ (πΌ) 1 π π
(3)
Lemma 2 (see [1, Theorem 2.3]). Let πΌ > 0, π = [πΌ] + 1. Then (1)
π
πΌπΌ π¦ (π‘) =
where Ξ is the gamma function.
π’ (1) = π’σΈ (1) = π’σΈ (π) = 0, π·πΌ π’ (1) = 0,
also offer the πΌth Hadamard fractional order integral of π¦ : [1, +β) σ³¨β R which is defined by
d π π‘ 1 π‘ πβπΌβ1 dπ (π‘ ) β« (log ) π¦ (π ) , (2) Ξ (π β πΌ) dπ‘ π π 1
where πΌ > 0, π = [πΌ] + 1, and [πΌ] denotes the largest integer which is less than or equal to πΌ. Moreover, we here
πΌπΌ π·πΌ π¦ (π‘) = π¦ (π‘) + π1 (log π‘) + ππ (log π‘)
πΌβπ
,
πΌβ1
+ π2 (log π‘)
πΌβ2
+ β
β
β
(4)
where ππ β R, π = 1, 2, . . . , π. In recent years, there have been some significant developments in the study of boundary value problems for nonlinear fractional differential equations; we refer to [2β11] and the references therein. For more related works, see also [12β49]. 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, 12β15]. For differential equations with the π-Laplacian, see, for example, [6, 7, 15β20] and the references therein. In [6], Wang
2
Journal of Function Spaces sign-changing and unbounded from below. This improves and generalizes some semipositone problems [21β31].
considered the nonlinear Hadamard fractional differential equation with integral boundary condition and π-Laplacian operator π·π½ ππ (π·πΌ π’ (π‘)) = π (π‘, π’ (π‘)) ,
2. Preliminaries
π‘ β (1, π) ,
π’ (π) = ππΌπ π’ (π) ,
In this section, we first calculate Greenβs functions associated with (1) and then transform the boundary value problem into its integral form. For this, we give the following lemma.
(5)
π·πΌ π’ (1) = 0,
Lemma 3. Let πΌ, π½, π, ππ , and π·πΌ , π·π½ be as in (1). Then (1) can take the integral form
π’ (1) = 0, where π grows (π β 1)βsublinearly at +β, and by using the Schauder fixed point theorem, a solution existence result is obtained. In [7], Li and Lin used the Guo-Krasnoselβskii fixed point theorem to obtain the existence and uniqueness of positive solutions for (1) with π = 0. However, we note that these are seldom considered Hadamard fractional differential equations with the πLaplacian in the literature; in this paper we are devoted to this direction. We first utilize the Guo-Krasnoselβskii fixed point theorem to obtain two positive solutions existence theorems when π grows (π β 1)βsuperlinearly and (π β 1)βsublinearly with the π-Laplacian, and secondly by using the fixed point index, we obtain a nontrivial solution existence theorem without the π-Laplacian, but the nonlinearity can allow being
π» (π‘, π) = π»1 (π‘, π) +
π
π
1
1
π’ (π‘) = β« πΊ (π‘, π ) ππ (β« π» (π , π) π (π, π’ (π))
where πΊ (π‘, π ) =
1 Ξ (πΌ) πΌβ1
+ π2 (log π‘)
π½β2
π½β1
(9)
Note that π·πΌ π’(1) = 0 implies ππ (π·πΌ π’(1)) = 0, and then π2 = 0. Therefore, we obtain π½β1
.
=
π1 =
(10)
ππ (π·πΌ π’ (π)) = πΌπ½ π¦ (π) + π1
1 β€ π‘ β€ π β€ π,
π½β1
,
for π‘, π β [1, π] , 1 β€ π β€ π‘ β€ π,
π π‘ π ππ1 dπ dπ‘ π½β1 + . β« β« (log π‘ β log π) π¦ (π) π½ π π‘ Ξ (π½) 1 1
π π‘ ππ½ dπ dπ‘ π½β1 β« β« (log π‘ β log π) π¦ (π) π π‘ (π½ β π) Ξ (π½) 1 1
Substituting π1 into (10) gives (11)
ππ (π·πΌ π’ (π‘)) =
π‘ 1 dπ π½β1 β« (log π‘ β log π) π¦ (π) π Ξ (π½) 1
π½β1
dπ‘ π β« ππ (π·πΌ π’ (π‘)) π‘ 1 π
π dπ‘ π½β1 dπ‘ = π β« πΌ π¦ (π‘) + ππ1 β« (log π‘) π‘ π‘ 1 1 π½
(8)
1 β€ π‘ β€ π β€ π.
π π½ dπ π½β1 β« (1 β log π) π¦ (π) . β π (π½ β π) Ξ (π½) 1
and π
(7)
π
π
π 1 dπ π½β1 β« (1 β log π) π¦ (π) , π Ξ (π½) 1
1 β€ π β€ π‘ β€ π,
The condition ππ (π·πΌ π’(π)) = π β«1 ππ (π·πΌ π’(π‘))(dπ‘/π‘) enables us to obtain
Next, we calculate ππ (π·πΌ π’(π)) and π β«1 ππ (π·πΌ π’(π‘))(dπ‘/π‘):
= π1 +
,
(12)
,
for ππ β R, π = 1, 2.
ππ (π·πΌ π’ (π‘)) = πΌπ½ π¦ (π‘) + π1 (log π‘)
πΌβ1
and
Proof. Use π¦(π‘) to replace π(π‘, π’) in (1). Let π·π½ (ππ (π·πΌ π’(π‘))) = π¦(π‘). Then from Lemma 2 we have π½β1
πΌβ2
{(log π‘) (1 β log π ) β (log π‘ β log π ) β
{ πΌβ1 πΌβ2 {(log π‘) (1 β log π ) ,
1 {(log π‘) (1 β log π) β (log π‘ β log π) π»1 (π‘, π) = Ξ (π½) {(log π‘)π½β1 (1 β log π)π½β1 , {
ππ (π·πΌ π’ (π‘)) = πΌπ½ π¦ (π‘) + π1 (log π‘)
(6)
for π‘ β [1, π] ,
π π½β1 π½β1 (log π‘) log π (1 β log π) , (π½ β π) Ξ (π½) π½β1
dπ dπ ) , π π
β
π π½ (log π‘) dπ π½β1 β« (1 β log π) π¦ (π) π (π½ β π) Ξ (π½) 1
+
π π‘ ππ½ (log π‘) dπ dπ‘ π½β1 β« β« (log π‘ β log π) π¦ (π) π π‘ (π½ β π) Ξ (π½) 1 1
π½β1
(13)
Journal of Function Spaces
3 π½β1
Note that βππ (π·πΌ π’(π‘)) = ππ (βπ·πΌ π’(π‘)), and hence we obtain
π‘ 1 dπ (log π‘) π½β1 β β« (log π‘ β log π) π¦ (π) = π Ξ (π½) 1 Ξ (π½) π
β
β« (1 β log π)
π½β1
1
π
β π·πΌ π’ (π‘) = ππ (β« π» (π‘, π) π¦ (π) 1
π½β1
π¦ (π)
dπ (log π‘) + π Ξ (π½)
π
Then, if we let π₯(π‘) = ππ (β«1 π»(π‘, π)π¦(π)(dπ/π)), π‘ β [1, π], from Lemma 2 we obtain
π½β1
π½ (log π‘) dπ π½β1 β β
β« (1 β log π) π¦ (π) π (π½ β π) Ξ (π½) 1 β
β« (1 β log π)
π½β1
1
π
π‘
1
1
π‘
β
β« [(log π‘)
π½β1
1
π’ (π‘) = βπΌπΌ π₯ (π‘) + π1 (log π‘)
π½β1
dπ ππ½ (log π‘) + π¦ (π) π (π½ β π) Ξ (π½)
β
β« β« (log π‘ β log π)
π½β1
(1 β log π)
π½β1
+ π3 (log π‘)
β (log π‘ β log π)
π½β1
π1 =
]
π’ (π‘) = β
π½β1
π
π‘
1
1
dπ ππ½ (log π‘) + π (π½ β π) Ξ (π½)
β
β« β« (log π‘ β log π)
π½β1
, for ππ β R, π = 1, 2, 3.
π 1 dπ πΌβ2 β« (1 β log π ) π₯ (π ) . Ξ (πΌ) 1 π
π
dπ dπ‘ π π‘
for π‘ β [1, π] .
1
π
π
1
π
This completes the proof. Lemma 4. Greenβs functions πΊ, π» defined by (7) and (8) have the following properties: (i) πΊ, π» are continuous, nonnegative functions on [1, π] Γ [1, π], (ii) (log π‘)πΌβ1 [(1βlog π )πΌβ2 β(1βlog π )πΌβ1 ] β€ Ξ(πΌ)πΊ(π‘, π ) β€ (1 β log π )πΌβ2 β (1 β log π )πΌβ1 , for π‘, π β [1, π]. From [7, Lemma 7] and [8, Lemma 2.2] we easily obtain this lemma, so we omit its proof. Let π dπ πΊ1 (π‘, π ) = β« πΊ (π‘, π) π» (π, π ) , π 1
π½β1
π
β
β« β« (log π‘ β log π)
dπ ππ½ (log π‘) + π (π½ β π) Ξ (π½) π½β1
π¦ (π)
dπ‘ dπ π‘ π
π½β1
β
π π (log π‘) dπ π½β1 β« (1 β log π) π¦ (π) π (π½ β π) Ξ (π½) 1 π½β1
π
= β β« π»1 (π‘, π) π¦ (π) 1
π (log π‘) dπ + π (π½ β π) Ξ (π½)
π (π ) =
π½β1
π
π (log π‘) dπ π½ β β
β« (1 β log π) π¦ (π) π (π½ β π) Ξ (π½) 1 π
β
β« (1 β log π)
π½β1
1
π¦ (π)
β
β« [(1 β log π‘)
π
1
dπ = β β« π»1 (π‘, π) π 1
π¦ (π)
(19) πΌβ2
β (1 β log π‘)
πΌβ1
] π» (π‘, π )
dπ‘ , π‘
for π‘, π β [1, π] .
π dπ π½β1 ββ« (log π‘) β
π¦ (π) π 1 (π½ β π) Ξ (π½) π½β1
1 Ξ (πΌ)
π
π
β
log π (1 β log π)
(18)
π
dπ dπ = β« πΊ (π‘, π ) ππ (β« π» (π , π) π¦ (π) ) , π π 1 1
π π (log π‘) dπ π½β1 β« (1 β log π) π¦ (π) π (π½ β π) Ξ (π½) 1
= β β« π»1 (π‘, π) π¦ (π)
(16)
(17)
π 1 dπ πΌβ1 πΌβ2 β« (log π‘) (1 β log π ) π₯ (π ) Ξ (πΌ) 1 π
π½β1
β
πΌβ2
π‘ 1 dπ πΌβ1 β« (log π‘ β log π ) π₯ (π ) Ξ (πΌ) 1 π
+
π¦ (π)
+ π2 (log π‘)
As a result, from (16) we have
π 1 dπ π½β1 π½β1 β β« (log π‘) (1 β log π) β
π¦ (π) π Ξ (π½) π‘
β
π¦ (π)
πΌβ3
πΌβ1
The condition π’(1) = π’σΈ (1) = 0 implies that π2 = π3 = 0. Then we substitute π into the first derivative of π’, and we calculate π1 as follows:
1 dπ dπ‘ =β π π‘ Ξ (π½)
π¦ (π)
(15)
for πΌ β (2, 3] , π‘ β [1, π] .
π
π
dπ ), π
Then we obtain the following lemma. π
Lemma 5. There exist π
1 = β«1 (log π‘)πΌβ1 π(π‘)(dπ‘/π‘), π
2 = π β«1 π(π‘)(dπ‘/π‘) such that
π
dπ = β β« π» (π‘, π) π 1
π
π
1 π (π ) β€ β« πΊ1 (π‘, π ) π (π‘)
dπ β
π¦ (π) . π
1
(14)
dπ‘ β€ π
2 π (π ) , π‘
(20) for π β [1, π] .
4
Journal of Function Spaces
Proof. We only prove the left inequality above. From Lemma 4(ii) we have π
β« πΊ1 (π‘, π ) π (π‘) 1
β
π» (π, π ) π
π
1
1
π π dt = β« β« πΊ (π‘, π) π‘ 1 1
dπ dπ‘ 1 π (π‘) β₯ π π‘ Ξ (πΌ)
β
β« β« (log π‘)
πΌβ1
(21)
[(1 β log π)
πΌβ2
β (1 β log π)
πΌβ1
]
dπ‘ dπ π (π‘) = π
1 π (π ) . π π‘ This completes the proof. β
π» (π, π )
Let E = πΆ[1, π] be the Banach space equipped with the norm βπ’β = maxπ‘β[1,π] |π’(π‘)|. Then we define two sets on E as follows: π = {π’ β E : π’ (π‘) β₯ 0, βπ‘ β [1, π]} , π0 = {π’ β E : π’ (π‘) β₯ (log π‘)
πΌβ1
βπ’β , βπ‘ β [1, π]} .
(22)
Consequently, π, π0 are cones on E. From Lemma 3 we can define an operator π΄ on E as follows: π
= β« πΊ (π‘, π ) ππ (β« π» (π , π) π (π, π’ (π)) 1
Let π΅σ° fl {π’ β E : βπ’β < σ°} for σ° > 0. Now, we first list our assumptions on π: (H1) π β πΆ([0, 1] Γ R+ , R+ ), (H2) there exist πΏ1 β (1, π), π‘0 β (1, π) such that lim inf π’σ³¨β+β (π(π‘, π’)/ππ (π’)) β₯ ππ (π1 ), lim inf π’σ³¨β0+ (π(π‘, π’)/ ππ (π’)) β₯ ππ (π2 ), uniformly on π‘ β [πΏ1 , π], where 2π1β1 , π π π2β1 β (0, (log πΏ1 )πΌβ1 β«1 πΊ(π‘0 , π )ππ (β«πΏ π»(π , π)(dπ/π))(dπ /π )), 1 (H3) there exists π1 > 0 such that π(π‘, π’) β€ ππ (π3 π1 ), π π βπ’ β [0, π1 ], π‘ β [1, π], where π3β1 > β«1 πΊ(π, π )ππ (β«1 π»(π , π)(dπ/π))(dπ /π ), (H4) lim supπ’σ³¨β+β (π(π‘, π’)/ππ (π’)) β€ ππ (π1 ), lim supπ’σ³¨β0+ (π(π‘, π’)/ππ (π’)) β€ ππ (π2 ), uniformly on π π π‘ β [1, π], where (2π1 )β1 , π2β1 > β«1 πΊ(π, π )ππ (β«1 π»(π , π)(dπ/ π))(dπ /π ), (H5) there exist π2 > 0, πΏ1 β (1, π), π‘0 β (1, π) such that π(π‘, π’) β₯ ππ (π3 π2 ), βπ’ β [(log πΏ1 )πΌβ1 π2 , π2 ], π‘ β [πΏ1 , π], where π
π
1
πΏ1
π3β1 β (0, β« πΊ (π‘0 , π ) ππ (β« π» (π , π)
dπ dπ ) ). π π
(26)
Lemma 9. Suppose that (H1) holds. Then π΄(π) β π0 .
(π΄π’) (π‘) π
3. Positive Solutions for (1)
1
dπ dπ ) , π π
(23)
(π΄π’) (π‘) β€
for π’ β E, π‘ β [1, π] . The continuity of πΊ, π», π implies that π΄ : E σ³¨β E is a completely continuous operator and the existence of solutions for (1) if and only if the existence of fixed points for π΄. Lemma 6 (see [50]). Let E be a Banach space and Ξ© a bounded open set in E. Suppose that π΄ : Ξ© σ³¨β E is a continuous compact operator. If there exists π’0 β πΈ \ {0} such that π’ β π΄π’ =ΜΈ ππ’0 ,
βπ’ β πΞ©, π β₯ 0,
π
dπ dπ ) , π π
β
ππ (β« π» (π , π) π (π, π’ (π)) 1
On the other hand, πΌβ1
(π΄π’) (π‘) β₯ (log π‘) π
(25)
then the topological degree deg(πΌ β π΄, Ξ©, 0) = 1. Lemma 8 (see [50]). Let E be a Banach space and π β E a cone in E. Assume that Ξ©1 , Ξ©2 are open subsets of E with 0 β Ξ©1 β Ξ©1 β Ξ©2 , and let π΄ : π β© (Ξ©2 \ Ξ©1 ) σ³¨β π be a completely continuous operator such that either (G1) βπ΄π’β β€ βπ’β, π’ β πΞ©1 β© π, and βπ΄π’β β₯ βπ’β, π’ β πΞ©2 β© π, or (G2) βπ΄π’β β₯ βπ’β, π’ β πΞ©1 β© π, and βπ΄π’β β€ βπ’β, π’ β πΞ©2 β© π. Then π΄ has a fixed point in π β© (Ξ©2 \ Ξ©1 ).
(27) βπ‘ β [1, π] .
β
β« [(1 β log π )
Lemma 7 (see [50]). Let E be a Banach space and Ξ© a bounded open set in E with 0 β Ξ©. Suppose that π΄ : Ξ© σ³¨β E is a continuous compact operator. If βπ’ β πΞ©, π β₯ 1,
π 1 πΌβ2 πΌβ1 β« [(1 β log π ) β (1 β log π ) ] Ξ (πΌ) 1
β
(24)
then the topological degree deg(πΌ β π΄, Ξ©, 0) = 0.
π΄π’ =ΜΈ ππ’,
Proof. If π’ β π, from Lemma 4 we have
πΌβ2
1
1 Ξ (πΌ) β (1 β log π )
πΌβ1
] (28)
π
dπ dπ β
ππ (β« π» (π , π) π (π, π’ (π)) ) π π 1 β₯ (log π‘)
πΌβ1
βπ΄π’β ,
βπ‘ β [1, π] .
This completes the proof. Remark 10. Our aim is to find operator equation π’ = π΄π’ has fixed points in π, and from Lemma 9, these fixed points must belong to the cone π0 . Therefore, our work space can be chosen π0 rather than π. In what follows, we discuss the existence of positive solutions for (1) in π0 . Theorem 11. Suppose that (H1)-(H3) hold. Then (1) has at least two positive solutions.
Journal of Function Spaces
5
Proof. From (H3), when π’ β ππ΅π1 β© π0 , we have
Combining this and (33), we find
(π΄π’) (π‘)
π
π
1
πΏ1
βπ΄π’β β₯ β« πΊ (π‘0 , π ) ππ (β« π» (π , π) π
π
dπ dπ β€ max β« πΊ (π‘, π ) ππ (β« π» (π , π) π (π, π’ (π)) ) π‘β[1,π] 1 π π 1 π
π
β€ β« πΊ (π, π ) ππ (β« π» (π , π) ππ (π3 π1 ) 1
1
π
dπ dπ ) π π
β
ππ (π1 (log πΏ1 ) (29)
= (π1 (log πΏ1 )
π
dπ dπ < π1 , = π3 π1 β« πΊ (π, π ) ππ (β« π» (π , π) ) π π 1 1
π
πΏ1
dπ dπ ) π π
π
1 β πΆ2 )
π
(36)
π
1 β πΆ2 ) β« πΊ (π‘0 , π ) 1
dπ dπ ) β₯ 2π
1 β πΆ3 , π π
π
π
where πΆ3 = πΆ2 β«1 πΊ(π‘0 , π )ππ (β«πΏ π»(π , π)(dπ/π))(dπ /π ). Conse1 quently, we have
Hence, we obtain for π’ β ππ΅π1 β© π0 .
πΌβ1
β
ππ (β« π» (π , π)
βπ‘ β [1, π] .
βπ΄π’β < βπ’β ,
πΌβ1
βπ΄π’β > βπ’β , for ππ΅π
1 β© π0 , if βπ’β σ³¨β β.
(30)
(37)
(31)
In summary, from (30), (33), and (37) with π
1 > π1 > π1 , Lemma 8 enables us to obtain that (1) has at least two positive solutions in (π΅π
1 \π΅π1 )β©π0 and (π΅π1 \π΅π1 )β©π0 . This completes the proof.
Note that if π’ β ππ΅π1 β© π0 , π‘ β [πΏ1 , π], from the definition of π0 we have
Theorem 12. Suppose that (H1), (H4)-(H5) hold. Then (1) has at least two positive solutions.
On the other hand, by the second limit inequality in (H2), there exists π1 β (0, π1 ) such that π (π‘, π’) β₯ ππ (π2 π’) ,
βπ’ β [0, π1 ] , π‘ β [πΏ1 , π] .
π’ (π‘) β₯ (log πΏ1 )
πΌβ1
βπ’β .
(32)
This, together with (31), implies that π
βπ΄π’β
βπ΄π’β = max (π΄π’) (π‘) β₯ (π΄π’) (π‘0 ) = β« πΊ (π‘0 , π ) π‘β[1,π]
1
dπ dπ β
ππ (β« π» (π , π) π (π, π’ (π)) ) π π 1 π
π
1
πΏ1
dπ dπ πΌβ1 β
ππ (π2 (log πΏ1 ) βπ’β) ) π π
π
β
ππ (β« π» (π , π) πΏ1
π
1
1
π
π
1
πΏ1
β₯ β« πΊ (π‘0 , π ) ππ (β« π» (π , π) ππ (π3 π2 )
β₯ β« πΊ (π‘0 , π ) ππ (β« π» (π , π)
πΌβ1
π
β₯ β« πΊ (π‘0 , π ) ππ (β« π» (π , π) π (π, π’ (π))
π
= π2 (log πΏ1 )
Proof. If π’ β ππ΅π2 β© π0 , we have βπ’β = π2 , and π’ β [(log πΏ1 )πΌβ1 π2 , π2 ], for π’ β π0 , π‘ β [πΏ1 , π]. Hence, from (H5) we obtain
(33)
βπ’β β« πΊ (π‘0 , π )
πΏ1
(38)
dπ dπ ) > π2 . π π
for π’ β ππ΅π2 β© π0 .
(39)
On the other hand, by the second limit inequality in (H4), there exists π2 β (0, π2 ) such that
dπ dπ ) > βπ’β , π π
π (π‘, π’) β€ ππ (π2 π’) ,
for π’ β ππ΅π1 β© π0 . By the first limit inequality in (H2), there exist π
1 > π1 and πΆ1 > 0 such that +
βπ’ β R , π‘ β [πΏ1 , π] .
(34)
Note that π
1 can be chosen large enough, and if π’ β ππ΅π
1 β©π0 , together with (32), there exists πΆ2 > 0 such that πΌβ1
1
βπ΄π’β > βπ’β ,
1
π (π‘, π’) β₯ ππ (π1 (log πΏ1 )
π
dπ dπ ) π π
This indicates that
π
π (π‘, π’) β₯ ππ (π1 π’) β πΆ1 ,
π
β₯ π3 π2 β« πΊ (π‘0 , π ) ππ (β« π» (π , π)
dπ dπ ) π π
βπ‘ β [πΏ1 , π] .
(35)
(40)
This, if π’ β ππ΅π2 β© π0 , implies that βπ΄π’β π
π
1
1
π
π
β€ β« πΊ (π, π ) ππ (β« π» (π , π) ππ (π2 π’ (π)) β€ β« πΊ (π, π ) ππ (β« π» (π , π) ππ (π2 π2 ) 1
π
1 β πΆ2 ) ,
βπ’ β [0, π2 ] , π‘ β [1, π] .
1
π
π
1
1
= π2 π2 β« πΊ (π, π ) ππ (β« π» (π , π)
dπ dπ ) π π
dπ dπ ) π π
dπ dπ ) < π2 . π π
(41)
6
Journal of Function Spaces solutions in (π΅π
2 \π΅π2 )β©π0 and (π΅π2 \π΅π2 )β©π0 . This completes the proof.
This gives βπ΄π’β < βπ’β ,
for π’ β ππ΅π2 β© π0 .
(42)
By the first limit inequality in (H4), there exist π
2 > π2 and πΆ4 > 0 such that βπ’ β R+ , π‘ β [1, π] .
π (π‘, π’) β€ ππ (π1 π’ + πΆ4 ) ,
Example 13. Let π (π‘, π’) πβ1βπΎ
(43)
Consequently, if π’ β ππ΅π
2 β©π0 with π
2 large enough, we obtain π
βπ΄π’β β€ β« πΊ (π, π ) β
ππ (β« π» (π , π) ππ (π1 π
2 + πΆ4 ) 1
π
lim inf
dπ dπ ) π π
π’σ³¨β+β
dπ dπ = (π1 π
2 + πΆ4 ) β« πΊ (π, π ) ππ (β« π» (π , π) ) π π 1 1
for π’ β ππ΅π
2 β© π0 , if βπ’β σ³¨β β.
(47)
πβ1
π3 π’πΎ2
π’πβ1
= +β
β(πΌβ1)πΎ
πβ1βπΎ
Moreover, for π’ β [0, π1 ], π‘ β [1, π] we have πβ1βπΎ2
(45)
In a word, from (39), (42), and (45) with π
2 > π2 > π2 , Lemma 8 enables us to obtain that (1) has at least two positive πβ1
3 3 π2 π3 π’πΎ3 , {(log πΏ1 ) π (π‘, π’) = { β(πΌβ1)πΎ4 πβ1βπΎ4 πβ1 π2 π3 π’πΎ4 , {(log πΏ1 )
π (π‘, π’) β€ π1
π (π‘, π’) lim sup π’σ³¨β+β ππ (π’)
πβ1 πΎ
π3 π1 2 = (π3 π1 )
πβ1
.
(48)
Therefore, (H1)-(H3) hold. Example 14. Let π’ β [(log πΏ1 )
πΌβ1
π’ β [0, (log πΏ1 )
where πΎ3 β (0, πβ1), πΎ4 β (πβ1, +β), and π3 , π2 are defined by (H5). Then
π2 , +β) , π‘ β [1, π] ,
πΌβ1
(49)
π2 ) , π‘ β [1, π] ,
= (π3 π2 )
πβ1
. (51)
Therefore, (H1), (H4)-(H5) hold. β(πΌβ1)πΎ3
πβ1βπΎ3
π2
π’πβ1
π’σ³¨β+β
4. Nontrivial Solutions for (1)
πβ1
π3 π’πΎ3
=0
β€ ππ (π1 ) ,
(50)
π (π‘, π’) lim sup π’σ³¨β0+ ππ (π’) (log πΏ1 )
= +β
β₯ ππ (π2 ) .
where πΆ5 = πΆ4 β«1 πΊ(π, π )ππ (β«1 π»(π , π)(dπ/π))(dπ /π ). Hence, we have
= lim sup
π’πβ1
πβ1βπΎ2
π
(log πΏ1 )
πβ1
π3 π’πΎ1
β₯ ππ (π1 ) , π1 π (π‘, π’) = lim inf lim inf π’σ³¨β0+ ππ (π’) π’σ³¨β0+
1 β€ π
2 + πΆ5 , 2 π
π π (π‘, π’) = lim inf 1 π’σ³¨ β+β ππ (π’)
(44)
π
(46)
π’ β [0, π1 ] , π‘ β [1, π] ,
πβ1βπΎ1
π
= lim sup
π’ β (π1 , +β) , π‘ β [1, π] ,
where πΎ1 β (π β 1, +β), πΎ2 β (0, π β 1), and π3 , π1 are defined by (H3). Then
1
βπ΄π’β < βπ’β ,
πβ1
1 π3 π’πΎ1 , {π1 = { πβ1βπΎ πβ1 2 π3 π’πΎ2 , {π1
In this section we consider the boundary value problem (1) without the π-Laplacian, i.e., π = 2. In this case, (1) can be transformed into its integral form as follows: π
π
1 π
1
π’ (π‘) = β« πΊ (π‘, π ) β« π» (π , π) π (π, π’ (π)) β(πΌβ1)πΎ4
+
π’σ³¨β0
πβ1βπΎ4 πβ1 π2 π3 π’πΎ4 π’πβ1
=0
β€ ππ (π2 ) . Moreover, for π’ β [(log πΏ1 )πΌβ1 π2 , π2 ], π‘ β [πΏ1 , π] we have π (π‘, π’) β₯ (log πΏ1 )
β(πΌβ1)πΎ3
πβ1βπΎ3
π2
πβ1
π3 π’πΎ3
dπ dπ π π
(52) dπ , for π‘ β [1, π] . π 1 As said in Section 3, we define an operator, still denoted by π΄, as follows: π dπ (π΄π’) (π‘) = β« πΊ1 (π‘, π ) π (π , π’ (π )) , π 1 (53) = β« πΊ1 (π‘, π ) π (π , π’ (π ))
for π’ β E, π‘ β [1, π] .
Journal of Function Spaces
7
In what follows, we aim to find the existence of fixed points of π΄. For this, we list our assumptions on π: (H6) π β πΆ([1, π] Γ R, R), (H7) There exist nonnegative functions π(π‘), π(π‘) β E with π β‘ΜΈ 0 and πΎ(π’) β πΆ(R, R+ ) such that π (π‘, π’) β₯ βπ (π‘) β π (π‘) πΎ (π’) ,
βπ’ β R, π‘ β [1, π] .
Multiply both sides of the above inequality by π(π‘) and integrate from 1 to π and together with Lemma 5 we obtain π
dπ‘ σ΅¨ σ΅¨ β« σ΅¨σ΅¨σ΅¨π’1 (π‘)σ΅¨σ΅¨σ΅¨ π (π‘) π‘ 1 π
(54)
π
dπ‘ σ΅¨ σ΅¨ dπ β€ (π
2β1 β π3 ) β« β« πΊ1 (π‘, π ) σ΅¨σ΅¨σ΅¨π’1 (π )σ΅¨σ΅¨σ΅¨ π (π‘) π π‘ 1 1
(60)
π
Moreover, lim
|π’|σ³¨ββ
dπ‘ σ΅¨ σ΅¨ β€ (π
2β1 β π3 ) π
2 β« σ΅¨σ΅¨σ΅¨π’1 (π‘)σ΅¨σ΅¨σ΅¨ π (π‘) . π‘ 1
πΎ (π’) = 0, |π’|
(55)
(H8) lim inf |π’|σ³¨ββ (π(π‘, π’)/|π’|) > π
1β1 , uniformly in π‘ β [1, π], (H9) lim inf |π’|σ³¨β0 (|π(π‘, π’)|/|π’|) < π
2β1 , uniformly in π‘ β [1, π]. Theorem 15. Suppose that (H6)-(H9) hold. Then (1) has at least one nontrivial solution. Proof. From (H9) there exist π3 β that σ΅¨σ΅¨ σ΅¨ β1 σ΅¨σ΅¨π (π‘, π’)σ΅¨σ΅¨σ΅¨ β€ (π
2 β π3 ) |π’| ,
(0, π
2β1 )
and π3 > 0 such
π
This implies that β«1 |π’1 (π‘)|π(π‘)(dπ‘/π‘) = 0, and π’1 β‘ 0 for the fact that π(π‘) β‘ΜΈ 0, for π‘ β [1, π], which contradicts π’1 β ππ΅π3 . Therefore, (57) is true, and from Lemma 7 we obtain deg (πΌ β π΄, π΅π3 , 0) = 1.
On the other hand, by (H8), there exist π4 > 0 and π0 > 0 such that π (π‘, π’) β₯ (π
1β1 + π4 ) |π’| , βπ‘ β [1, π] , |π’| > π0 .
(62)
For every fixed π with βπβπ β (0, π4 ), βπβ = maxπ‘β[1,π] |π(π‘)|, and from (H7), there exists π1 > π0 such that
βπ‘ β [1, π] , |π’| β [0, π3 ) . (56)
For this π3 , we show that
(61)
πΎ (π’) β€ π |π’| ,
β |π’| > π1 .
(63)
Combining the two inequalities above, (H7) enables us to find
π΄π’ =ΜΈ ππ’,
π’ β ππ΅π3 , π β₯ 1.
(57)
β₯ (π
1β1 + π4 ) |π’| β π (π‘) β ππ (π‘) |π’|
If otherwise, there exist π’1 β ππ΅π3 , π1 β₯ 1 such that π΄π’1 = π1 π’1 ,
π (π‘, π’) β₯ (π
1β1 + π4 ) |π’| β π (π‘) β π (π‘) πΎ (π’)
(58)
(64)
β₯ (π
1β1 + π4 β βπβ π) |π’| β π (π‘) , β |π’| > π1 , π‘ β [1, π] .
and hence, we obtain
If we take πΆ6 = (π
1β1 + π4 β βπβπ)π1 + maxπ‘β[1,π],|π’|β€π1 |π(π‘, π’)|, πΎβ = max|π’|β€π1 πΎ(π’). Then we easily have
σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨ 1 σ΅¨σ΅¨ σ΅¨σ΅¨π’1 (π‘)σ΅¨σ΅¨σ΅¨ = σ΅¨(π΄π’1 ) (π‘)σ΅¨σ΅¨σ΅¨ β€ σ΅¨σ΅¨σ΅¨(π΄π’1 ) (π‘)σ΅¨σ΅¨σ΅¨ π1 σ΅¨ π
σ΅¨ σ΅¨ dπ β€ β« πΊ1 (π‘, π ) σ΅¨σ΅¨σ΅¨π (π , π’1 (π ))σ΅¨σ΅¨σ΅¨ π 1
(59)
π (π‘, π’) β₯ (π
1β1 + π4 β βπβ π) |π’| β π (π‘) β πΆ6 , βπ’ β R, π‘ β [1, π] .
π
σ΅¨ σ΅¨ dπ β€ (π
2β1 β π3 ) β« πΊ1 (π‘, π ) σ΅¨σ΅¨σ΅¨π’1 (π )σ΅¨σ΅¨σ΅¨ . π 1
Note that π can be chosen arbitrarily small, and we let
π
π
β1 β { (π
1 + 2 (π4 β βπβ π)) β«1 π (π ) (π (π ) + βπβ πΎ + πΆ6 ) (dπ /π ) β«1 π (π ) (π (π ) + βπβ πΎβ + πΆ6 ) (dπ /π ) } π
3 β₯ max { , π π }, (π4 β βπβ π) Ξ (πΌ) β βπβ π (π
1β1 + 2 (π4 β βπβ π)) β«1 π (π ) (dπ /π ) Ξ (πΌ) β βπβ π β«1 π (π ) (dπ /π ) { }
π
where π(π ) = β«1 (1 β log π)πΌβ2 π»(π, π )(dπ/π), for π β [1, π]. Now we prove that π’ β π΄π’ =ΜΈ ππ,
βπ’ β ππ΅π
3 , π β₯ 0,
(67)
where π is defined by (19). Indeed, if (67) is not true, then there exists π’2 β ππ΅π
3 and π0 > 0 such that π’2 β π΄π’2 = π0 π.
(65)
(68)
π
(66)
Let π’Μ(π‘) = β«1 πΊ1 (π‘, π )[π(π ) + π(π )πΎ(π’2 (π )) + πΆ6 ](dπ /π ). Then π’Μ β π0 and π
π’Μ (π‘) = β« πΊ1 (π‘, π ) [π (π ) + π (π ) πΎ (π’2 (π )) + πΆ6 ] 1
π
π
1
1
β€β« β«
dπ π
1 dπ πΌβ1 πΌβ2 (log π‘) (1 β log π) π» (π, π ) Ξ (πΌ) π
8
Journal of Function Spaces β
[π (π ) + π (π ) πΎ (π’2 (π )) + πΆ6 ]
dπ π
On the other hand, we have π
π
1β1 β« πΊ1 (π‘, π ) [π’2 (π ) + π’Μ (π )]
π π dπ 1 πΌβ1 πΌβ2 (log π‘) β« β« (1 β log π) π» (π, π ) β€ Ξ (πΌ) π 1 1
1
π
+ (π4 β βπβ π) β« πΊ1 (π‘, π ) π’2 (π )
dπ β
[π (π ) + π (π ) πΎ (π’2 (π )) + πΆ6 ] π
1
β
π 1 πΌβ1 (log π‘) β« π (π ) Ξ (πΌ) 1
=
1
+
π
(π4 β βπβ π) β« πΊ1 (π‘, π ) π’2 (π ) 1
π
dπ 1 β« π (π ) [π (π ) + π (π ) πΎ (π’2 (π )) + πΆ6 ] Ξ (πΌ) 1 π
β
|π’2 |>π1
π (π ) πΎ (π’2 (π ))
π
1
(70)
π
dπ π
β (π
1β1 + π4 β βπβ π) β« πΊ1 (π‘, π ) π’Μ (π )
dπ π
π
π’β) β« πΊ1 (π‘, π ) (log π ) β₯ (π4 β βπβ π) (π
3 β βΜ
π
1
π
β1 + π4 β βπβ π β 1 Ξ (πΌ)
dπ . π
π
β
β« π (π ) (π (π ) + βπβ πΎβ + βπβ ππ
3 + πΆ6 )
Plus π’Μ into (68) gives
1
π
π’2 (π‘) + π’Μ (π‘) = (π΄π’2 ) (π‘) + π’Μ (π‘) + π0 π (π‘)
β
β« πΊ1 (π‘, π ) (log π )
πΌβ1
1
π
= β« πΊ1 (π‘, π ) [π (π , π’2 (π )) + π (π ) + π (π ) πΎ (π’2 (π )) (71) 1 dπ + π0 π (π‘) . π Note that π(π , π’2 (π )) + π(π ) + π(π )πΎ(π’2 (π )) + πΆ6 β π, π β [1, π] and π β π0 . Lemma 9 enables us to know that π’2 + π’Μ β π0 . From (65) we have + πΆ6 ]
π
(π΄π’2 ) (π‘) + π’Μ (π‘) = β« πΊ1 (π‘, π ) [π (π , π’2 (π )) + π (π ) 1
π dπ β₯ β« πΊ1 (π‘, π ) π 1 π
σ΅¨ σ΅¨ dπ β₯ β« πΊ1 (π‘, π ) (π
1β1 + π4 β βπβ π) σ΅¨σ΅¨σ΅¨π’2 (π )σ΅¨σ΅¨σ΅¨ π 1
+ π4 β βπβ π) π’2 (π )
dπ . π
dπ β₯ 0. π
As a result, we have π
(π΄π’2 ) (π‘) + π’Μ (π‘) β₯ π
1β1 β« πΊ1 (π‘, π ) [π’2 (π ) + π’Μ (π )] 1
fl
π
1β1 π (π’2
+ π’Μ) (π‘) ,
dπ π
(76)
βπ‘ β [1, π] ,
π
where (ππ’)(π‘) = β«1 πΊ1 (π‘, π )π’(π )(dπ /π ), for π’ β E, π‘ β [1, π]. Using (68) we obtain (77)
β₯ π0 π. (72)
(75)
dπ π
π’2 + π’Μ = π΄π’2 + π’Μ + π0 π β₯ π
1β1 π (π’2 + π’Μ) + π0 π
π
dπ β₯ β« πΊ1 (π‘, π ) π 1
β
[π (π , π’2 (π )) + π (π ) + πΆ6 ]
dπ π
πΌβ1
1
β
β« π (π ) (π (π ) + βπβ πΎβ + βπβ ππ
3 + πΆ6 )
β
1
1
dπ βπβ π + π Ξ (πΌ)
1 σ΅¨ σ΅¨ dπ β€ π (π ) σ΅¨σ΅¨σ΅¨π’2 (π )σ΅¨σ΅¨σ΅¨ π Ξ (πΌ) |π’2 |>π1
(π
1β1
dπ β₯ 0. β« πΊ1 (π‘, π ) π’Μ (π ) π 1
π
β
β«
+ π (π ) πΎ (π’2 (π )) + πΆ6 ]
(74)
π
(π4 β βπβ π) β« πΊ1 (π‘, π ) [π’2 (π ) + π’Μ (π )]
1 dπ )β€ π Ξ (πΌ)
β
β« π (π ) (π (π ) + βπβ πΎβ + πΆ6 )
π
1β1
dπ π
Indeed, π’2 + π’Μ β π0 implies that π’2 (π‘) + π’Μ(π‘) β₯ (log π‘)πΌβ1 βπ’2 + π’Μβ β₯ (log π‘)πΌβ1 (βπ’2 β β βΜ π’β), for π‘ β [1, π]. Consequently,
dπ βπβ (β« π (π ) πΎ (π’2 (π )) Ξ (πΌ) |π’2 |β€π1 π
+β«
dπ . π
This inequality holds if
Consequently, we have
π 1 dπ β« π (π ) (π (π ) + πΆ6 ) β€ Ξ (πΌ) 1 π
(73)
dπ β« πΊ1 (π‘, π ) π’Μ (π ) π 1
π
(69)
dπ π
π
π
1β1
β₯ π
1β1 β« πΊ1 (π‘, π ) [π’2 (π ) + π’Μ (π )]
dπ β
[π (π ) + π (π ) πΎ (π’2 (π )) + πΆ6 ] . π
π’β β€ βΜ
dπ π
Define πβ = sup {π > 0 : π’2 + π’Μ β₯ ππ} .
(78) β
Note that π0 β {π > 0 : π’2 + π’Μ β₯ ππ}, and then π β₯ π0 , π’2 + π’Μ β₯ πβ π. From Lemma 5 we have π
1β1 π (π’2 + π’Μ) β₯ πβ π
1β1 ππ β₯ πβ π,
(79)
Journal of Function Spaces
9
and hence π’2 + π’Μ β₯
π
1β1 π (π’2
β
+ π’Μ) + π0 π β₯ (π0 + π ) π,
(80)
which contradicts the definition of πβ . Therefore, (67) holds, and from Lemma 6 we obtain deg (πΌ β π΄, π΅π
3 , 0) = 0.
(81)
This, together with (61), implies that deg (πΌ β π΄, π΅π
3 \ π΅π3 , 0) = deg (πΌ β π΄, π΅π
3 , 0) β deg (πΌ β π΄, π΅π3 , 0) = β1.
(82)
Therefore the operator π΄ has at least one fixed point in π΅π
3 \ π΅π3 , and (1) has at least one nontrivial solution. This completes the proof. Example 16. Let π(π‘, π’) = π|π’| β ππ(π’), π(π’) = ln(|π’| + 1), π’ β R, π‘ β [1, π], where π β (π
1β1 , +β) and π β (π, π + π
2β1 ). Then lim|π’|σ³¨β+β (π(π’)/|π’|) = 0, and lim|π’|σ³¨β+β ((π|π’| β π ln(|π’| + 1))/|π’|) = π > π
1β1 , lim|π’|σ³¨β0 (|π|π’| β π ln(|π’| + 1)|/|π’|) = |π β π| < π
2β1 . Therefore, (H6)-(H9) hold.
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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