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Multiple positive solutions of a singular fractional differential equation with negatively perturbed term
Mathematical and Computer Modelling 55 (2012) 1263–1274
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Multiple positive solutions of a singular fractional differential equation with negatively perturbed term Xinguang Zhang a,∗ , Lishan Liu b,c , Yonghong Wu c a
School of Mathematical and Informational Sciences, Yantai University, Yantai 264005, Shandong, People’s Republic of China
b
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China
c
Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia
article
info
Article history: Received 5 April 2011 Accepted 5 October 2011 Keywords: Fractional differential equation Multiple positive solutions Green function Perturbed term
abstract Let Dα0+ be the standard Riemann–Liouville derivative. We discuss the existence of multiple positive solutions for the following fractional differential equation with a negatively perturbed term
−Dα0+ u(t ) = p(t )f (t , u(t )) − q(t ), u(0) = u′ (0) = u(1) = 0,
0 < t < 1,
where 2 < α ≤ 3 is a real number, the perturbed term q : (0, 1) → [0, +∞) is Lebesgue integrable and may be singular at some zero measures set of [0,1], which implies the nonlinear term may change sign. © 2011 Elsevier Ltd. All rights reserved.
1. Introduction In this paper, we discuss the existence of multiple positive solutions for the following fractional differential equation with a negatively perturbed term
−Dα0+ u(t ) = p(t )f (t , u(t )) − q(t ), u(0) = u′ (0) = u(1) = 0,
0 < t < 1,
(1.1)
where Dα0+ is the standard Riemann–Liouville derivative, 2 < α ≤ 3 is a real number, q : (0, 1) → [0, +∞) is Lebesgue integrable and does not vanish identically on any subinterval of (0, 1). Fractional calculus has played a significant role in engineering, science, economics, and other fields. During the last few decades, many papers and books on fractional calculus and fractional differential equations have appeared (see [1–8]). It should be noted that most of the papers and books on fractional calculus are devoted to the solvability of linear initial value fractional differential equations in terms of special functions, for detail, see [2,4,9,10]. Recently there have been some papers, for example [11–14], dealing with the existence and multiplicity of solutions (or positive solutions) of nonlinear initial value fractional differential equations by using techniques of nonlinear analysis. In particular, there are a few papers [6,15–17] that consider the Dirichlet-type problem for ordinary differential equations of fractional order. Jiang and Yuan [17] investigated the existence and multiplicity of positive solutions of nonlinear fractional differential equation Dirichlet-type boundary value problem
∗
Dα0+ u(t ) + f (t , u(t )) = 0, u(0) = u(1) = 0.
0 < t < 1,
(1.2)
Corresponding author. Tel.: +86 535 6902406. E-mail addresses: [email protected], [email protected] (X. Zhang), [email protected] (L. Liu), [email protected] (Y. Wu).
0895-7177/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2011.10.006
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He proved that if f (t , u) is continuous on [0, 1] × [0, +∞), and there exist g C ((0, 1), [0, +∞)) such that q1 (t )g (y) ≤ f (t , t α−2 y) ≤ q2 (t )g (y),
∈ C ([0, +∞), [0, +∞)), q1 , q2 ∈
t ∈ (0, 1), y ∈ [0, +∞)
1
and 0 qi (s)ds < +∞, i = 1, 2, g satisfies some local conditions and superlinear or sublinear condition, then the problem (1.2) has at least one positive solution or two solutions. However, all of the above mentioned works are based on a key assumption, i.e., the nonlinear term is required to be non-negative. When nonlinear fractional differential equations include a sign-changing coefficient a(t ), by using the nonlinear alternative of Leray–Schauder type, Bai [18] investigated the existence of a positive solution for the nonlinear fractional equation Ds u(t ) = λa(t )f (u),
0
with initial value u(0) = 0 where 0 < s < 1, Ds is the standard Riemann–Liouville fractional derivative, λ > 0 is a parameter, and f : [0, +∞) → [0, +∞) is continuous and f (0) > 0, a : [0, 1] → R is continuous, a(0) ̸= 0. And then under the similar conditions, Zhang [19] considered the existence of nonnegative solutions of an initial value problem for singular fractional differential equation
Ds u(t ) = λa(t )f (u, Dα u), u(0) = 0,
0 < t < 1,
where 0 < s, α < 1. To the best of our knowledge, the multiplicity of positive solutions of a fractional differential with a negatively perturbed term, however, has not been studied previously. In the present work, motivated by [17–19], we study the existence of multiple positive solutions for the fractional differential equation with a negatively perturbed term q. It should be emphasized that q is the only needed Lebesgue integrable, thus the nonnegative of the nonlinearity is no longer required. 2. Preliminaries and lemmas For the convenience of the reader, we also present here the necessary definitions from fractional calculus theory. Definition 2.1. The fractional integral of a function u : (0, +∞) → R with order α > 0 is given by D−α 0+ u(t ) =
t
∫
1
Γ (α)
(t − s)α−1 u(s)ds
0
provided that the right-hand side is pointwise defined on (0, +∞). Definition 2.2. The fractional derivative of a continuous function u : (0, +∞) → R with order α > 0 is given by Dα0+ u(t ) =
1
Γ (n − α)
d dt
n ∫
t
(t − s)n−α−1 u(s)ds, 0
where n = [α] + 1, [α] denotes the integer part of number α , provided that the right-hand side is pointwise defined on (0, +∞). Remark 2.1. If u, v : (0, +∞) → R with order α > 0, then Dα0+ (u(t ) + v(t )) = Dα0+ u(t ) + Dα0+ v(t ). Lemma 2.1. Let α > 0, u(t ) be integrable, then α α−1 D−α + c2 t α−2 + · · · + cn t α−n , 0+ D0+ u(t ) = u(t ) + c1 t
where ci ∈ R (i = 1, 2, . . . , n), n is the smallest integer greater than or equal to α . Set G(t , s) =
[t (1 − s)]α−1 , 0 ≤ t ≤ s ≤ 1, α−1 − (t − s)α−1 , 0 ≤ s ≤ t ≤ 1. Γ (α) [t (1 − s)] 1
(2.1)
X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
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Lemma 2.2. Given y(t ) ∈ C (0, 1) ∩ L(0, 1), then the unique solution of the problem
Dα0+ u(t ) + y(t ) = 0, 0 < t < 1, u(0) = u′ (0) = u(1) = 0
(2.2)
is u( t ) =
1
∫
G(t , s)y(s)ds,
(2.3)
0
where G(t , s) is the Green function of the boundary value problem (2.2). Proof. Deduced from Lemma 2.1, we have α−1 u(t ) = −D−α + c2 t α−2 + c3 t α−3 . 0+ y(t ) + c1 t
Consequently, the solution of (2.2) is u( t ) = −
t
∫
1
Γ (α)
(t − s)α−1 y(s)ds + c1 t α−1 + c2 t α−2 + c3 t α−3 .
0
By u(0) = u′ (0) = 0, there is c2 = c3 = 0. On the other hand, u(1) = −
1
∫
1
Γ (α)
(1 − s)α−1 y(s)ds + c1
0
yields c1 =
1
∫
1
Γ (α)
(1 − s)α−1 y(s)ds.
0
Therefore, the solution of (2.2) is 1
u( t ) =
∫
t
[t (1 − s)]α−1 y(s)ds −
Γ (α) 0 ∫ 1 = G(t , s)y(s)ds.
1
Γ (α)
t
∫
(t − s)α−1 y(s)ds
0
0
Conversely, let u(t ) be given by formula (2.3). It is easy to get that u(t ) is the solution of (2.2). This completes the proof.
Set
ϱ(t ) = t (1 − t )α−1 .
(2.4)
Lemma 2.3. The function G(t , s) has the following properties: (1) (2) (3) (4)
G(t , s) = G(1 − s, 1 − t ). G(t , s) > 0, for t , s ∈ (0, 1). ϱ(1 − t )ϱ(s) ≤ Γ (α)G(t , s) ≤ (α − 1)ϱ(s), for t , s ∈ [0, 1]. ϱ(1 − t )ϱ(s) ≤ Γ (α)G(t , s) ≤ (α − 1)ϱ(1 − t ),for t , s ∈ [0, 1].
Proof. It is obvious that (1) and (2) hold. In the following, we will only prove (3) as (4) can be deduced directly from (1) and (3). When 0 ≤ s ≤ t ≤ 1, we have
Γ (α)G(t , s) = [t (1 − s)]α−1 − (t − s)α−1 = (α − 1)
t (1−s)
∫
t −s
≤ (α − 1)[t (1 − s)]α−2 [t (1 − s) − (t − s)] ≤ (α − 1)(1 − s)α−2 s(1 − t ) ≤ (α − 1)(1 − s)α−1 s = (α − 1)ϱ(s). On the other hand, since 0 < α − 2 ≤ 1, we have
Γ (α)G(t , s) = ≥ = ≥
[t (1 − s)]α−1 − (t − s)α−1 [t (1 − s)]α−2 [t (1 − s) − (t − s)] t α−2 (1 − s)α−2 s(1 − t ) t α−1 (1 − s)α−1 s(1 − t ) = ϱ(1 − t )ϱ(s).
xα−2 dx
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When 0 ≤ t ≤ s ≤ 1, we have
Γ (α)G(t , s) = t α−1 (1 − s)α−1 ≤ sα−1 (1 − s)α−1 ≤ (α − 1)s(1 − s)α−1 = (α − 1)ϱ(s). On the other hand
Γ (α)G(t , s) = t α−1 (1 − s)α−1 ≥ t α−1 (1 − s)α−1 s(1 − t ) = ϱ(1 − t )ϱ(s). In the rest of the paper, we always suppose that the following assumptions hold: (B1) p : (0, 1) → [0, +∞) is continuous such that 0<
1
∫
ϱ(s)(p(s) + q(s))ds < +∞. 0
(B2) f : [0, 1] × [0, +∞) → [0, +∞) is continuous. By Lemmas 2.2 and 2.3, we have Lemma 2.4. The boundary value problem
−Dα0+ u(t ) = 2q(t ), 0 < t < 1, u(0) = u′ (0) = u(1) = 0, 1 has a unique solution ω(t ) = 2 0 G(t , s)q(s)ds with ∫ 1 2(α − 1) ω(t ) ≤ ϱ(1 − t ) q(s)ds. Γ (α) 0
(2.5)
(2.6)
Let E = C [0, 1] be endowed with the maximum norm ‖u‖ = max0≤t ≤1 |u(t )|. Define the cone P by
P =
u ∈ E : u(t ) ≥
ϱ(1 − t ) ‖u‖, t ∈ [0, 1] . α−1
Next we consider the following singular nonlinear boundary value problem:
−Dα0+ u(t ) = p(t )f (t , [u(t ) − ω(t )]∗ ) + q(t ), u(0) = u′ (0) = u(1) = 0,
0 < t < 1,
(2.7)
where [u(t ) − ω(t )]∗ = max{u(t ) − ω(t ), 0}. By Remark 2.1, it is easy to know if u(t ) ≥ w(t ) for any t ∈ [0, 1] is a positive solution of the BVP (2.7), then u − w is a positive solution of the differential equation (1.1). Let
(Tu)(t ) =
1
∫
G(t , s)[p(s)f (s, [u(s) − ω(s)]∗ ) + q(s)]ds. 0
Lemma 2.5. T : P → P is a completely continuous operator. Proof. For any u ∈ P, Lemma 2.3 implies that
(Tu)(t ) ≥
ϱ(1 − t ) Γ (α)
1
∫
ϱ(s)[f (s, [u(s) − ω(s)]∗ ) + q(s)]ds. 0
On the other hand
α−1 ‖Tu‖ = max (Tu)(t ) ≤ 0≤t ≤1 Γ (α)
1
∫
ϱ(s)[f (s, [u(s) − ω(s)]∗ ) + q(s)]ds. 0
ϱ(1−t )
Then (Tu)(t ) ≥ α−1 ‖Tu‖, which implies T : P → P. According to the Ascoli–Arzela theorem, we can easily get that T : P → P is a completely continuous operator.
Lemma 2.6 ([20]). Let E be a real Banach space and P ⊂ E be a cone. Assume Ω1 , Ω2 are two bounded open subsets of E with θ ∈ Ω1 , Ω 1 ⊂ Ω2 , and let T : P ∩ (Ω 2 \ Ω1 ) → P be a completely continuous operator such that either (1) ‖Tu‖ ≤ ‖u‖, u ∈ P ∩ ∂ Ω1 and ‖Tu‖ ≥ ‖u‖, u ∈ P ∩ ∂ Ω2 , or (2) ‖Tu‖ ≥ ‖u‖, u ∈ P ∩ ∂ Ω1 and ‖Tu‖ ≤ ‖u‖, u ∈ P ∩ ∂ Ω2 . Then T has a fixed point in P ∩ (Ω 2 \ Ω1 ).
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3. Main results Fixed [a, b] ⊂ (0, 1), define several constants: K =
(α − 1)
1 0
ϱ(s)[p(s) + q(s)]ds , Γ (α)
µ=
a(1 − b)α−1 2(α − 1)
,
l=
a(1 − b)α−1
Γ (α)
∫
b
ϱ(s)p(s)ds.
a
Theorem 3.1. Suppose (B1) and (B2) hold. In addition, assume that the following conditions are satisfied: (S1) There exists a constant
r > max 2K ,
2(α − 1)2
1 0
q(s)ds
Γ (α)
such that for any (t , u) ∈ [0, 1] × [0, r ], r − 1. f (t , u) ≤ K (S2) There exists a constant R > 2r such that, for any (t , u) ∈ [0, 1] × [µR, R], f (t , u) ≥
R l
.
(S3) lim
max
f (t , u)
u→+∞ t ∈[0,1]
u
= 0.
Then the differential equation (1.1) has at least two positive solutions x1 , x2 , and there exist two positive constants m, n such that x1 (t ) ≥ mϱ(1 − t ), x2 (t ) ≥ nϱ(1 − t ) for any t ∈ [0, 1]. Proof. Let Ω1 = {u ∈ P : ‖u‖ < r }. Then, for any u ∈ ∂ Ω1 , s ∈ [0, 1], we have
[u(s) − w(s)]∗ ≤ u(s) ≤ ‖u‖ = r . It follows from (S1) that
‖Tu‖ = max (Tu)(t ) t ∈[0,1] ∫ 1 = max G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds 0 ≤t ≤1
0
(α − 1)ϱ(s) r ≤ − 1 p(s) + q(s) ds Γ (α) K 0 1 (α − 1)r 0 ϱ(s)[p(s) + q(s)]ds ≤ K Γ (α) = r = ‖u‖. ∫
1
Therefore, u ∈ P ∩ ∂ Ω1 .
‖Tu‖ ≤ ‖u‖,
On the other hand, let Ω2 = {u ∈ P : ‖u‖ < R} and ∂ Ω2 = {u ∈ P : ‖u‖ = R}. Then for any u ∈ ∂ Ω2 , t ∈ [0, 1], noticing R > 2r and (2.6), we have u(t ) − w(t ) = u(t ) − 2
1
∫
G(t , s)q(s)ds 0
≥ u(t ) −
2(α − 1)
Γ (α)
ϱ(1 − t )
q(s)ds 0
r
1
R
2
≥ u(t ) − u(t ) ≥
1
∫
u( t )
ϱ(1 − t ) ϱ(1 − t ) ‖ u‖ = R. 2(α − 1) 2(α − 1) So by (3.1), for any u ∈ ∂ Ω2 , t ∈ [a, b], we have ≥
µR =
a(1 − b)α−1 2(α − 1)
R ≤ u(t ) − w(t ) ≤ R.
(3.1)
(3.2)
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
It follows from (S2), (3.2) and Lemma 2.2 that, for any u ∈ ∂ Ω2 , t ∈ [a, b], 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ ≥ 0 1
∫
G(t , s)p(s)f (s, [u(s) − w(s)]∗ )ds
≥ 0
∫ ϱ(1 − t ) b ϱ(s)p(s)f (s, u(s) − w(s))ds Γ (α) a ∫ b R a(1 − b)α−1 ϱ(s)p(s)ds ≥ Γ (α) l a = R = ‖u‖. ≥
So, we have u ∈ P ∩ ∂ Ω2 .
‖Tu‖ ≥ ‖u‖,
Next, let us choose ε > 0 such that
ε
1
∫
ϱ(s)p(s)ds < 1. 0
Then for the above ε , by (S3), there exists N > R > 0 such that, for any t ∈ [0, 1] and for any u ≥ N, f (t , u) ≤ ε u. Take
max
R∗ =
(t ,u)∈[0,1]×[0,N ]
f ( t , u) + 1
Γ (α) α−1
1
1
0
0
ϱ(s)[p(s) + q(s)]ds + 1 − ε 0 ϱ(s)p(s)ds
ϱ(s)q(s)ds + N,
then R∗ > N > R. Now let Ω3 = {u ∈ P : ‖u‖ < R∗ } and ∂ Ω3 = {u ∈ P : ‖u‖ = R∗ }. Then, for any u ∈ P ∩ ∂ Ω3 , we have 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ = max 0≤t ≤1
0
∫ α−1 1 ϱ(s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds Γ (α) 0 ∫ 1 α−1 ≤ max f ( t , u) + 1 ϱ(s)[p(s) + q(s)]ds (t ,u)∈[0,1]×[0,N ] Γ (α) 0 ∫ 1 + ϱ(s)[p(s)ε[u(s) − w(s)]∗ + q(s)]ds ≤
0
∫ 1 α−1 ≤ max f ( t , u) + 1 ϱ(s)[p(s) + q(s)]ds (t ,u)∈[0,1]×[0,N ] Γ (α) 0 ∫ 1 + ϱ(s)[p(s)ε‖u‖ + q(s)]ds 0
∫ 1 α−1 = max f ( t , u) + 1 ϱ(s)[p(s) + q(s)]ds (t ,u)∈[0,1]×[0,N ] Γ (α) 0 ∫ 1 ∫ 1 + ϱ(s)q(s)ds + ε ϱ(s)p(s)dsR∗ 0
0
< R∗ = ‖u‖, which implies that
‖Tu‖ ≤ ‖u‖,
u ∈ P ∩ ∂ Ω3 .
By Lemma 2.6, T has two fixed points u1 , u2 such that r ≤ ‖u1 ‖ ≤ R ≤ ‖u2 ‖.
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It follows from r >
2(α − 1)2
1 0
q(s)ds
Γ (α)
that u1 (t ) − w(t ) ≥ u1 (t ) −
≥ u1 (t ) − =
≥
1−
2(α − 1)
Γ (α)
1
∫
q(s)ds 0
2(α − 1)2 u1 (t )
1
∫
Γ (α)r 0 1 2(α − 1)2 0 q(s)ds Γ (α)r
r
ϱ(1 − t )
1−
α−1
2(α − 1)2
1 0
q(s)ds u1 ( t )
q(s)ds
Γ (α)r
= mϱ(1 − t ) > 0,
ϱ(t )
t ∈ (0, 1).
(3.3)
As for (3.3), we also find a positive constant n such that u2 (t ) − w(t ) ≥ nϱ(1 − t ) > 0,
t ∈ (0, 1).
Let xi (t ) = ui (t ) − w(t ) (i = 1, 2), then xi (t ) > 0,
t ∈ (0, 1), i = 1, 2.
Thus the differential equation (1.1) has at least two positive solutions x1 , x2 satisfying x1 (t ) ≥ mϱ(1 − t ),
u2 (t ) ≥ nϱ(1 − t ),
t ∈ [0, 1]
for some positive constants m, n. The proof of Theorem 3.1 is completed.
Theorem 3.2. Suppose (B1) and (B2) hold. In addition, assume that the following conditions are satisfied: (S4) There exists a constant r >
4(α − 1)2
1
∫
Γ (α)
q(s)ds 0
such that, for any (t , u) ∈ [0, 1] × [µr , r ], r f (t , u) ≥ . l (S5) There exists a constant R > max r ,
r l
+ 1 K such that, for any (t , u) ∈ [0, 1] × [0, R],
R − 1, K where r is defined by (S4). f (t , u) ≤
(S6) lim
min
u→+∞ t ∈[a,b]
f ( t , u) u
= +∞.
Then the differential equation (1.1) has at least two positive solutions x1 , x2 , and there exist two positive constants m, n such that x1 (t ) ≥ mϱ(1 − t ), x2 (t ) ≥ nϱ(1 − t ) for any t ∈ [0, 1]. Proof. Firstly, let Ω1 = {u ∈ P : ‖u‖ < r }. Then, for any u ∈ ∂ Ω1 , t ∈ [a, b], by (2.6), we have u(t ) − w(t ) = u(t ) − 2
1
∫
G(t , s)q(s)ds 0
≥ u(t ) − ≥ u(t ) −
2(α − 1)
Γ (α)
ϱ(1 − t )
2(α − 1)2 u(t )
Γ (α)r
1
∫
q(s)ds 0
1
∫
q(s)ds 0
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≥
1− 1
≥
2
2(α − 1)2
Γ (α)r
u(t ) ≥
1
∫
q(s)ds u(t ) 0
ϱ(1 − t ) 2(α − 1)
‖u‖ =
ϱ(1 − t ) 2(α − 1)
R.
(3.4)
So for any u ∈ Ω1 , t ∈ [a, b], by (3.4), we have
µr ≤ u(t ) − w(t ) ≤ r .
(3.5)
It follows from (S4), (3.5) and Lemma 2.2 that, for any u ∈ ∂ Ω2 , t ∈ [a, b], 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ ≥ 0 1
∫
G(t , s)p(s)f (s, [u(s) − w(s)]∗ )ds
≥ 0
∫ ϱ(1 − t ) b ≥ ϱ(s)p(s)f (s, x(s) − w(s))ds Γ (α) a ∫ b a(1 − b)α−1 r ≥ ϱ(s)p(s)ds Γ (α) l a = r = ‖u‖. Therefore, we have
‖Tu‖ ≥ ‖u‖,
u ∈ P ∩ ∂ Ω1 .
Next, by (S5), we have R > r and R K
−2>
r l
> 0.
Let Ω2 = {u ∈ P : ‖u‖ < R}. Then, for any u ∈ ∂ Ω2 , s ∈ [0, 1], we have
[u(s) − w(s)]∗ ≤ u(s) ≤ ‖u‖ = R. It follows from (S5) that
‖Tu‖ = max (Tu)(t ) t ∈[0,1] ∫ 1 = max G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds 0≤t ≤1 0 [ ] ∫ 1 R (α − 1)ϱ(s) − 1 p(s) + q(s) ds ≤ Γ (α) K 0 1 (α − 1)R 0 ϱ(s)[p(s) + q(s)]ds ≤ K Γ (α) = r = ‖u‖. Therefore,
‖Tu‖ ≤ ‖u‖,
u ∈ P ∩ ∂ Ω2 .
On the other hand, choose a real number M > 0 such that
µMa(1 − b)α−1 Γ (α)
∫
b
ϱ(s)p(s)ds ≥ 1.
a
From (S6), there exists N > R such that, for any t ∈ [a, b], f (t , u) ≥ Mu,
u ≥ N.
Take R∗ > max
N
µ
,R ,
(3.6)
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then R∗ > R > r. Let Ω3 = {u ∈ P : ‖u‖ < R∗ }, for any u ∈ P ∩ ∂ Ω3 and for any t ∈ [a, b], we have u(t ) − w(t ) ≥ u(t ) − 2
∫
1
G(t , s)q(s)ds
0
≥ u(t ) −
2(α − 1)
Γ (α)
ϱ(1 − t )
1
∫
q(s)ds 0
∫ 2(α − 1)2 u(t ) 1 ≥ u(t ) − q(s)ds Γ (α)R∗ 0 ∫ 1 2(α − 1)2 ≥ 1− q(s)ds u(t ) Γ (α)R∗ 0 1 ϱ(1 − t ) ϱ(1 − t ) ∗ ≥ u( t ) ≥ ‖ u‖ = R 2 2(α − 1) 2(α − 1) ≥ µR∗ ≥ N > 0.
(3.7)
So for any u ∈ P ∩ ∂ Ω3 , t ∈ [a, b], by (3.6), (3.7), we have 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ ≥ 0
≥ ≥ = = ≥
∫ ϱ(1 − t ) b ϱ(s)p(s)f (s, u(s) − w(s))ds Γ (α) a ∫ b a(1 − b)α−1 ϱ(s)p(s)M u(s) − w(s) ds Γ (α) a ∫ b Ma(1 − b)α−1 ϱ(s)p(s)ds × µR∗ Γ (α) a ∫ µMa(1 − b)α−1 b ϱ(s)p(s)ds × R∗ Γ (α) a R∗ = ‖u‖.
Thus u ∈ P ∩ ∂ Ω3 .
‖Tu‖ ≥ ‖u‖,
(3.8)
By Lemma 2.6, T has two fixed points u1 , u2 such that r ≤ ‖u1 ‖ ≤ R ≤ ‖u2 ‖. Noticing that r >
4(α − 1)2
1
∫
Γ (α)
q(s)ds,
0
we have u1 (t ) − w(t ) ≥ u1 (t ) −
≥ u1 (t ) − = ≥
1− 1 2
2(α − 1)
Γ (α)
ϱ(1 − t ) ∫
1
Γ (α)r 0 1 2(α − 1)2 0 q(s)ds Γ (α)r r 2(α − 1)
= mϱ(1 − t ) > 0,
q(s)ds 0
2(α − 1)2 u1 (t )
u1 (t ) ≥
1
∫
q(s)ds u1 ( t )
ϱ(1 − t )
t ∈ (0, 1).
As for (3.9), we also can find a positive constant n such that u2 (t ) − w(t ) ≥ nϱ(1 − t ) > 0, Let xi (t ) = ui (t ) − w(t ) (i = 1, 2), then xi (t ) > 0,
t ∈ (0, 1), i = 1, 2.
t ∈ (0, 1).
(3.9)
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
So the differential equation (1.1) has at least two positive solutions x1 , x2 satisfying x1 (t ) ≥ mϱ(1 − t ),
x2 (t ) ≥ nϱ(1 − t ),
t ∈ [0, 1]
for some positive constants m, n. The proof of Theorem 3.2 is completed.
4. Applications Example 4.1. Consider the following problem
5 2
−D0+ u(t ) = Γ
| ln t | , −√ 3 t (1 − t ) t (1 − t ) 2 u(1) = 0,
5 2
u(0) = u′ (0) = 0,
f (u)
0 < t < 1,
(4.1)
where
1 u, 3
0 ≤ u ≤ 90,
1230u − 110670, 90 ≤ u ≤ 100, f (u) = u + 12230, 100 ≤ u ≤ 6400, 1863 u 12 , u ≥ 6400. 8 Then the BVP (4.1) has at least two positive solutions. Proof. Let p(t ) = Take
Γ
Γ 25 | ln t | q(t ) = √ . t (1 − t )
5 2 3
t (1 − t ) 2
,
, 34 ⊂ [0, 1], by direct calculation, we have ∫ 3 1 | ln s| 361 1 3 K = s(1 − s) 2 , +√ ds = 3 2 0 150 s ( 1 − s ) s(1 − s) 2 ∫ 3 4 1 1 1 µ= ds = , l= ,
1
4
96
32
64
1 4
and 2(α − 1)2
1 0
q(s)ds
1
∫
| ln s| ds √ s( 1 − s) √ ∫ 1 ∫ 2 | ln s| 9 2 ≤ √ ds + =
Γ (α)
9
2
0
2
s
0
1 1 2
√
| ln s| ds √ 1−s
√ = 18 ln 2 + 18 2 ln( 2 + 1) + 36 < 90. Choose r = 90, then
90 = r > max 2K ,
2(α − 1)2
1 0
Γ (α)
q(s)ds
,
and for any (t , u) ∈ [0, 1] × [0, 90], f (t , u) ≤ 30 ≤
r K
− 1 ≈ 37.
So the condition (S1) is satisfied. On the other hand, we take R = 6400, then R > 2r = 180, and for any (t , u) ∈ [0, 1] × [µR, R] = [0, 1] × [100, 6400], we have f (t , u) ≥ 12330 ≥
R l
= 12288,
X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
1273
this implies that the condition (S2) holds. Next, we have lim
f ( t , u)
max
u→+∞ t ∈[0,1]
u
1863
= lim
1
u 2 = 0.
8
u→+∞
Thus (S3) also holds. Moreover it is clear that (B1)–(B2) hold. Thus, by Theorem 3.1, the BVP (4.1) has at least two positive solutions. Example 4.2. Consider the following problem
7 4
D0+ u(t ) + Γ
f ( u) =
−√ 4
3
4
u(0) = u (0) = 0,
where
f (u)
7
′
t (1 − t ) 4 u(1) = 0,
1
= 0,
1−t
√ √ 2 , 288 2u, 0 ≤ u ≤ 8 √ √ 2688 + 112 2 u + 20636 − 336 2 , 287
0 < t < 1,
√ 2
287
8
≤ u ≤ 3,
20 18760 − u+ , 3 ≤ u ≤ 190, 187 187 2 2 u , u ≥ 190. 1805
The BVP (4.2) has at least two positive solutions u1 (t ) and u2 (t ) such that u1 (t ) ≥
8 3
3
t 4 (1 − t ),
u2 (t ) ≥
380 3
3
t 4 (1 − t ),
t ∈ [0, 1].
Proof. In fact, let
Γ
Γ
7
4 q(t ) = √ , 4 1−t
p(t ) =
7 4 3
t (1 − t ) 4
.
, 34 ⊂ [0, 1], then 1 ∫ (α − 1) 0 ϱ(s)[p(s) + q(s)]ds 3 1 1 1 19 3 = s(1 − s) 4 √ + , K = ds = 3 4 Γ (α) 4 0 20 1−s 4 s(1 − s) √ √ ∫ b a(1 − b)α−1 2 a(1 − b)α−1 2 µ= = , l= ϱ(s)p(s)ds = . 2(α − 1) 24 Γ (α) 32 a Set r = 4, then ∫ 1 ∫ 4(α − 1)2 9 1 ds r =4> q(s)ds = = 3, √ 4 Γ (α) 4 1−s 0 0 √ and for any (t , u) ∈ [0, 1] × 82 , 3 , Take
1 4
f (t , u) ≥ 72 ≥
r
√ = 48 2.
l On the other hand, take a constant R = 190, then r R = 190 > max r , + 1 K = {3, 69.35}, l and for any (t , u) ∈ [0, 1] × [0, 190], f (t , u) ≤ 100 ≤
R K
−1=
3800 19
− 1 = 199.
Finally, we have lim
min
f ( t , u)
u→+∞ t ∈[a,b]
Thus (S6) also holds.
u
= lim
u→+∞
min
t∈
1 3 , 4 4
f (t , u) u
= lim
u→+∞
2 1805
u = +∞.
(4.2)
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
(B1)–(B2) are obvious, it follows from Theorem 3.2 that the BVP (4.2) has at least two positive solutions, u1 (t ) and u2 (t ) such that u1 ( t ) ≥
8 3
3
t 4 (1 − t ),
u2 ( t ) ≥
380 3
3
t 4 (1 − t ),
t ∈ [0, 1].
Acknowledgments The first and second authors were supported financially by the National Natural Science Foundation of China (11071141) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017). The third author was supported financially by the Australia Research Council through an ARC Discovery Project Grant. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
O. Agrawal, Formulation of Euler–Lagrange equations for fractional variational problems, J. Math. Anal. Appl. 272 (2002) 368–379. D. Delbosco, L. Rodino, Existence and uniqueness for a nonlinear fractional differential equation, J. Math. Appl. 204 (1996) 609–625. R. Leggett, L. Williams, Multiple positive solutions of nonlinear operators on ordered Banach spaces, Indiana Univ. Math. J. 28 (1979) 673–688. I. Podlubny, Fractional Differential Equations, in: Mathematics in Science and Engineering, vol. 198, Academic Press, New Tork, London, Toronto, 1999. S. Samko, A. Kilbas, O. Marichev, Fractional Integral and Derivatives, in: Theory and Applications, Gordon and Breach, Switzerland, 1993. Z. Bai, H. Lv, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl. 311 (2005) 495–505. S. Zhang, The existence of a positive solution for a nonlinear fractional differential equation, J. Math. Anal. Appl. 252 (2000) 804–812. S. Zhang, Existence of positive solution for some class of nonlinear fractional differential equations, J. Math. Anal. Appl. 278 (1) (2003) 136–148. K. Miller, B. Ross, An Introduction to the Fractional Calculus and Fractional Differential Equations, Wiley, New York, 1993. A. Kilbas, H. Srivastava, J.J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier B.V., Netherlands, 2006. A. Kilbas, J. Trujillo, Differential equations of fractional order: methods, results and problems-I, Appl. Anal. 78 (2001) 153–192. A. Kilbas, J. Trujillo, Differential equations of fractional order: methods, results and problems-II, Appl. Anal. 81 (2002) 435–493. A. Babakhani, V. Gejji, Existence of positive solutions of nonlinear fractional differential equations, J. Math. Anal. Appl. 278 (2003) 434–442. C. Yu, G. Gao, On the solution of nonlinear fractional order differential equation, Nonlinear Anal. TMA 63 (1998) 971–976. H. Jafari, V. Gejji, Positive solutions of nonlinear fractional boundary value problems using Adomian decomposition method, Appl. Math. Comput. 180 (2006) 700–706. S. Zhang, Positive solutions for boundary value problems of nonlinear fractional differential equations, Electron. J. Differential Equations 2006 (2006) 1–12. D. Jiang, C. Yuan, The positive properties of the green function for Dirichlet-type boundary value problems of nonlinear fractional differential equations and its application, Nonlinear Anal. TMA 72 (2010) 710–719. C. Bai, Positive solutions for nonlinear fractional differential equations with coefficient that changes sign, Nonlinear Anal. TMA 64 (2006) 677–685. S. Zhang, Nonnegative solution for singular nonlinear fractional differential equation with coefficient that changes sign, Positivity 12 (2006) 711–724. K. Deimling, Nonlinear Functional Analysis, Springer, Berlin, 1985.
Contents lists available at SciVerse ScienceDirect
Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Multiple positive solutions of a singular fractional differential equation with negatively perturbed term Xinguang Zhang a,∗ , Lishan Liu b,c , Yonghong Wu c a
School of Mathematical and Informational Sciences, Yantai University, Yantai 264005, Shandong, People’s Republic of China
b
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, People’s Republic of China
c
Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia
article
info
Article history: Received 5 April 2011 Accepted 5 October 2011 Keywords: Fractional differential equation Multiple positive solutions Green function Perturbed term
abstract Let Dα0+ be the standard Riemann–Liouville derivative. We discuss the existence of multiple positive solutions for the following fractional differential equation with a negatively perturbed term
−Dα0+ u(t ) = p(t )f (t , u(t )) − q(t ), u(0) = u′ (0) = u(1) = 0,
0 < t < 1,
where 2 < α ≤ 3 is a real number, the perturbed term q : (0, 1) → [0, +∞) is Lebesgue integrable and may be singular at some zero measures set of [0,1], which implies the nonlinear term may change sign. © 2011 Elsevier Ltd. All rights reserved.
1. Introduction In this paper, we discuss the existence of multiple positive solutions for the following fractional differential equation with a negatively perturbed term
−Dα0+ u(t ) = p(t )f (t , u(t )) − q(t ), u(0) = u′ (0) = u(1) = 0,
0 < t < 1,
(1.1)
where Dα0+ is the standard Riemann–Liouville derivative, 2 < α ≤ 3 is a real number, q : (0, 1) → [0, +∞) is Lebesgue integrable and does not vanish identically on any subinterval of (0, 1). Fractional calculus has played a significant role in engineering, science, economics, and other fields. During the last few decades, many papers and books on fractional calculus and fractional differential equations have appeared (see [1–8]). It should be noted that most of the papers and books on fractional calculus are devoted to the solvability of linear initial value fractional differential equations in terms of special functions, for detail, see [2,4,9,10]. Recently there have been some papers, for example [11–14], dealing with the existence and multiplicity of solutions (or positive solutions) of nonlinear initial value fractional differential equations by using techniques of nonlinear analysis. In particular, there are a few papers [6,15–17] that consider the Dirichlet-type problem for ordinary differential equations of fractional order. Jiang and Yuan [17] investigated the existence and multiplicity of positive solutions of nonlinear fractional differential equation Dirichlet-type boundary value problem
∗
Dα0+ u(t ) + f (t , u(t )) = 0, u(0) = u(1) = 0.
0 < t < 1,
(1.2)
Corresponding author. Tel.: +86 535 6902406. E-mail addresses: [email protected], [email protected] (X. Zhang), [email protected] (L. Liu), [email protected] (Y. Wu).
0895-7177/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2011.10.006
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
He proved that if f (t , u) is continuous on [0, 1] × [0, +∞), and there exist g C ((0, 1), [0, +∞)) such that q1 (t )g (y) ≤ f (t , t α−2 y) ≤ q2 (t )g (y),
∈ C ([0, +∞), [0, +∞)), q1 , q2 ∈
t ∈ (0, 1), y ∈ [0, +∞)
1
and 0 qi (s)ds < +∞, i = 1, 2, g satisfies some local conditions and superlinear or sublinear condition, then the problem (1.2) has at least one positive solution or two solutions. However, all of the above mentioned works are based on a key assumption, i.e., the nonlinear term is required to be non-negative. When nonlinear fractional differential equations include a sign-changing coefficient a(t ), by using the nonlinear alternative of Leray–Schauder type, Bai [18] investigated the existence of a positive solution for the nonlinear fractional equation Ds u(t ) = λa(t )f (u),
0
with initial value u(0) = 0 where 0 < s < 1, Ds is the standard Riemann–Liouville fractional derivative, λ > 0 is a parameter, and f : [0, +∞) → [0, +∞) is continuous and f (0) > 0, a : [0, 1] → R is continuous, a(0) ̸= 0. And then under the similar conditions, Zhang [19] considered the existence of nonnegative solutions of an initial value problem for singular fractional differential equation
Ds u(t ) = λa(t )f (u, Dα u), u(0) = 0,
0 < t < 1,
where 0 < s, α < 1. To the best of our knowledge, the multiplicity of positive solutions of a fractional differential with a negatively perturbed term, however, has not been studied previously. In the present work, motivated by [17–19], we study the existence of multiple positive solutions for the fractional differential equation with a negatively perturbed term q. It should be emphasized that q is the only needed Lebesgue integrable, thus the nonnegative of the nonlinearity is no longer required. 2. Preliminaries and lemmas For the convenience of the reader, we also present here the necessary definitions from fractional calculus theory. Definition 2.1. The fractional integral of a function u : (0, +∞) → R with order α > 0 is given by D−α 0+ u(t ) =
t
∫
1
Γ (α)
(t − s)α−1 u(s)ds
0
provided that the right-hand side is pointwise defined on (0, +∞). Definition 2.2. The fractional derivative of a continuous function u : (0, +∞) → R with order α > 0 is given by Dα0+ u(t ) =
1
Γ (n − α)
d dt
n ∫
t
(t − s)n−α−1 u(s)ds, 0
where n = [α] + 1, [α] denotes the integer part of number α , provided that the right-hand side is pointwise defined on (0, +∞). Remark 2.1. If u, v : (0, +∞) → R with order α > 0, then Dα0+ (u(t ) + v(t )) = Dα0+ u(t ) + Dα0+ v(t ). Lemma 2.1. Let α > 0, u(t ) be integrable, then α α−1 D−α + c2 t α−2 + · · · + cn t α−n , 0+ D0+ u(t ) = u(t ) + c1 t
where ci ∈ R (i = 1, 2, . . . , n), n is the smallest integer greater than or equal to α . Set G(t , s) =
[t (1 − s)]α−1 , 0 ≤ t ≤ s ≤ 1, α−1 − (t − s)α−1 , 0 ≤ s ≤ t ≤ 1. Γ (α) [t (1 − s)] 1
(2.1)
X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
1265
Lemma 2.2. Given y(t ) ∈ C (0, 1) ∩ L(0, 1), then the unique solution of the problem
Dα0+ u(t ) + y(t ) = 0, 0 < t < 1, u(0) = u′ (0) = u(1) = 0
(2.2)
is u( t ) =
1
∫
G(t , s)y(s)ds,
(2.3)
0
where G(t , s) is the Green function of the boundary value problem (2.2). Proof. Deduced from Lemma 2.1, we have α−1 u(t ) = −D−α + c2 t α−2 + c3 t α−3 . 0+ y(t ) + c1 t
Consequently, the solution of (2.2) is u( t ) = −
t
∫
1
Γ (α)
(t − s)α−1 y(s)ds + c1 t α−1 + c2 t α−2 + c3 t α−3 .
0
By u(0) = u′ (0) = 0, there is c2 = c3 = 0. On the other hand, u(1) = −
1
∫
1
Γ (α)
(1 − s)α−1 y(s)ds + c1
0
yields c1 =
1
∫
1
Γ (α)
(1 − s)α−1 y(s)ds.
0
Therefore, the solution of (2.2) is 1
u( t ) =
∫
t
[t (1 − s)]α−1 y(s)ds −
Γ (α) 0 ∫ 1 = G(t , s)y(s)ds.
1
Γ (α)
t
∫
(t − s)α−1 y(s)ds
0
0
Conversely, let u(t ) be given by formula (2.3). It is easy to get that u(t ) is the solution of (2.2). This completes the proof.
Set
ϱ(t ) = t (1 − t )α−1 .
(2.4)
Lemma 2.3. The function G(t , s) has the following properties: (1) (2) (3) (4)
G(t , s) = G(1 − s, 1 − t ). G(t , s) > 0, for t , s ∈ (0, 1). ϱ(1 − t )ϱ(s) ≤ Γ (α)G(t , s) ≤ (α − 1)ϱ(s), for t , s ∈ [0, 1]. ϱ(1 − t )ϱ(s) ≤ Γ (α)G(t , s) ≤ (α − 1)ϱ(1 − t ),for t , s ∈ [0, 1].
Proof. It is obvious that (1) and (2) hold. In the following, we will only prove (3) as (4) can be deduced directly from (1) and (3). When 0 ≤ s ≤ t ≤ 1, we have
Γ (α)G(t , s) = [t (1 − s)]α−1 − (t − s)α−1 = (α − 1)
t (1−s)
∫
t −s
≤ (α − 1)[t (1 − s)]α−2 [t (1 − s) − (t − s)] ≤ (α − 1)(1 − s)α−2 s(1 − t ) ≤ (α − 1)(1 − s)α−1 s = (α − 1)ϱ(s). On the other hand, since 0 < α − 2 ≤ 1, we have
Γ (α)G(t , s) = ≥ = ≥
[t (1 − s)]α−1 − (t − s)α−1 [t (1 − s)]α−2 [t (1 − s) − (t − s)] t α−2 (1 − s)α−2 s(1 − t ) t α−1 (1 − s)α−1 s(1 − t ) = ϱ(1 − t )ϱ(s).
xα−2 dx
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When 0 ≤ t ≤ s ≤ 1, we have
Γ (α)G(t , s) = t α−1 (1 − s)α−1 ≤ sα−1 (1 − s)α−1 ≤ (α − 1)s(1 − s)α−1 = (α − 1)ϱ(s). On the other hand
Γ (α)G(t , s) = t α−1 (1 − s)α−1 ≥ t α−1 (1 − s)α−1 s(1 − t ) = ϱ(1 − t )ϱ(s). In the rest of the paper, we always suppose that the following assumptions hold: (B1) p : (0, 1) → [0, +∞) is continuous such that 0<
1
∫
ϱ(s)(p(s) + q(s))ds < +∞. 0
(B2) f : [0, 1] × [0, +∞) → [0, +∞) is continuous. By Lemmas 2.2 and 2.3, we have Lemma 2.4. The boundary value problem
−Dα0+ u(t ) = 2q(t ), 0 < t < 1, u(0) = u′ (0) = u(1) = 0, 1 has a unique solution ω(t ) = 2 0 G(t , s)q(s)ds with ∫ 1 2(α − 1) ω(t ) ≤ ϱ(1 − t ) q(s)ds. Γ (α) 0
(2.5)
(2.6)
Let E = C [0, 1] be endowed with the maximum norm ‖u‖ = max0≤t ≤1 |u(t )|. Define the cone P by
P =
u ∈ E : u(t ) ≥
ϱ(1 − t ) ‖u‖, t ∈ [0, 1] . α−1
Next we consider the following singular nonlinear boundary value problem:
−Dα0+ u(t ) = p(t )f (t , [u(t ) − ω(t )]∗ ) + q(t ), u(0) = u′ (0) = u(1) = 0,
0 < t < 1,
(2.7)
where [u(t ) − ω(t )]∗ = max{u(t ) − ω(t ), 0}. By Remark 2.1, it is easy to know if u(t ) ≥ w(t ) for any t ∈ [0, 1] is a positive solution of the BVP (2.7), then u − w is a positive solution of the differential equation (1.1). Let
(Tu)(t ) =
1
∫
G(t , s)[p(s)f (s, [u(s) − ω(s)]∗ ) + q(s)]ds. 0
Lemma 2.5. T : P → P is a completely continuous operator. Proof. For any u ∈ P, Lemma 2.3 implies that
(Tu)(t ) ≥
ϱ(1 − t ) Γ (α)
1
∫
ϱ(s)[f (s, [u(s) − ω(s)]∗ ) + q(s)]ds. 0
On the other hand
α−1 ‖Tu‖ = max (Tu)(t ) ≤ 0≤t ≤1 Γ (α)
1
∫
ϱ(s)[f (s, [u(s) − ω(s)]∗ ) + q(s)]ds. 0
ϱ(1−t )
Then (Tu)(t ) ≥ α−1 ‖Tu‖, which implies T : P → P. According to the Ascoli–Arzela theorem, we can easily get that T : P → P is a completely continuous operator.
Lemma 2.6 ([20]). Let E be a real Banach space and P ⊂ E be a cone. Assume Ω1 , Ω2 are two bounded open subsets of E with θ ∈ Ω1 , Ω 1 ⊂ Ω2 , and let T : P ∩ (Ω 2 \ Ω1 ) → P be a completely continuous operator such that either (1) ‖Tu‖ ≤ ‖u‖, u ∈ P ∩ ∂ Ω1 and ‖Tu‖ ≥ ‖u‖, u ∈ P ∩ ∂ Ω2 , or (2) ‖Tu‖ ≥ ‖u‖, u ∈ P ∩ ∂ Ω1 and ‖Tu‖ ≤ ‖u‖, u ∈ P ∩ ∂ Ω2 . Then T has a fixed point in P ∩ (Ω 2 \ Ω1 ).
X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
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3. Main results Fixed [a, b] ⊂ (0, 1), define several constants: K =
(α − 1)
1 0
ϱ(s)[p(s) + q(s)]ds , Γ (α)
µ=
a(1 − b)α−1 2(α − 1)
,
l=
a(1 − b)α−1
Γ (α)
∫
b
ϱ(s)p(s)ds.
a
Theorem 3.1. Suppose (B1) and (B2) hold. In addition, assume that the following conditions are satisfied: (S1) There exists a constant
r > max 2K ,
2(α − 1)2
1 0
q(s)ds
Γ (α)
such that for any (t , u) ∈ [0, 1] × [0, r ], r − 1. f (t , u) ≤ K (S2) There exists a constant R > 2r such that, for any (t , u) ∈ [0, 1] × [µR, R], f (t , u) ≥
R l
.
(S3) lim
max
f (t , u)
u→+∞ t ∈[0,1]
u
= 0.
Then the differential equation (1.1) has at least two positive solutions x1 , x2 , and there exist two positive constants m, n such that x1 (t ) ≥ mϱ(1 − t ), x2 (t ) ≥ nϱ(1 − t ) for any t ∈ [0, 1]. Proof. Let Ω1 = {u ∈ P : ‖u‖ < r }. Then, for any u ∈ ∂ Ω1 , s ∈ [0, 1], we have
[u(s) − w(s)]∗ ≤ u(s) ≤ ‖u‖ = r . It follows from (S1) that
‖Tu‖ = max (Tu)(t ) t ∈[0,1] ∫ 1 = max G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds 0 ≤t ≤1
0
(α − 1)ϱ(s) r ≤ − 1 p(s) + q(s) ds Γ (α) K 0 1 (α − 1)r 0 ϱ(s)[p(s) + q(s)]ds ≤ K Γ (α) = r = ‖u‖. ∫
1
Therefore, u ∈ P ∩ ∂ Ω1 .
‖Tu‖ ≤ ‖u‖,
On the other hand, let Ω2 = {u ∈ P : ‖u‖ < R} and ∂ Ω2 = {u ∈ P : ‖u‖ = R}. Then for any u ∈ ∂ Ω2 , t ∈ [0, 1], noticing R > 2r and (2.6), we have u(t ) − w(t ) = u(t ) − 2
1
∫
G(t , s)q(s)ds 0
≥ u(t ) −
2(α − 1)
Γ (α)
ϱ(1 − t )
q(s)ds 0
r
1
R
2
≥ u(t ) − u(t ) ≥
1
∫
u( t )
ϱ(1 − t ) ϱ(1 − t ) ‖ u‖ = R. 2(α − 1) 2(α − 1) So by (3.1), for any u ∈ ∂ Ω2 , t ∈ [a, b], we have ≥
µR =
a(1 − b)α−1 2(α − 1)
R ≤ u(t ) − w(t ) ≤ R.
(3.1)
(3.2)
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
It follows from (S2), (3.2) and Lemma 2.2 that, for any u ∈ ∂ Ω2 , t ∈ [a, b], 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ ≥ 0 1
∫
G(t , s)p(s)f (s, [u(s) − w(s)]∗ )ds
≥ 0
∫ ϱ(1 − t ) b ϱ(s)p(s)f (s, u(s) − w(s))ds Γ (α) a ∫ b R a(1 − b)α−1 ϱ(s)p(s)ds ≥ Γ (α) l a = R = ‖u‖. ≥
So, we have u ∈ P ∩ ∂ Ω2 .
‖Tu‖ ≥ ‖u‖,
Next, let us choose ε > 0 such that
ε
1
∫
ϱ(s)p(s)ds < 1. 0
Then for the above ε , by (S3), there exists N > R > 0 such that, for any t ∈ [0, 1] and for any u ≥ N, f (t , u) ≤ ε u. Take
max
R∗ =
(t ,u)∈[0,1]×[0,N ]
f ( t , u) + 1
Γ (α) α−1
1
1
0
0
ϱ(s)[p(s) + q(s)]ds + 1 − ε 0 ϱ(s)p(s)ds
ϱ(s)q(s)ds + N,
then R∗ > N > R. Now let Ω3 = {u ∈ P : ‖u‖ < R∗ } and ∂ Ω3 = {u ∈ P : ‖u‖ = R∗ }. Then, for any u ∈ P ∩ ∂ Ω3 , we have 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ = max 0≤t ≤1
0
∫ α−1 1 ϱ(s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds Γ (α) 0 ∫ 1 α−1 ≤ max f ( t , u) + 1 ϱ(s)[p(s) + q(s)]ds (t ,u)∈[0,1]×[0,N ] Γ (α) 0 ∫ 1 + ϱ(s)[p(s)ε[u(s) − w(s)]∗ + q(s)]ds ≤
0
∫ 1 α−1 ≤ max f ( t , u) + 1 ϱ(s)[p(s) + q(s)]ds (t ,u)∈[0,1]×[0,N ] Γ (α) 0 ∫ 1 + ϱ(s)[p(s)ε‖u‖ + q(s)]ds 0
∫ 1 α−1 = max f ( t , u) + 1 ϱ(s)[p(s) + q(s)]ds (t ,u)∈[0,1]×[0,N ] Γ (α) 0 ∫ 1 ∫ 1 + ϱ(s)q(s)ds + ε ϱ(s)p(s)dsR∗ 0
0
< R∗ = ‖u‖, which implies that
‖Tu‖ ≤ ‖u‖,
u ∈ P ∩ ∂ Ω3 .
By Lemma 2.6, T has two fixed points u1 , u2 such that r ≤ ‖u1 ‖ ≤ R ≤ ‖u2 ‖.
X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
1269
It follows from r >
2(α − 1)2
1 0
q(s)ds
Γ (α)
that u1 (t ) − w(t ) ≥ u1 (t ) −
≥ u1 (t ) − =
≥
1−
2(α − 1)
Γ (α)
1
∫
q(s)ds 0
2(α − 1)2 u1 (t )
1
∫
Γ (α)r 0 1 2(α − 1)2 0 q(s)ds Γ (α)r
r
ϱ(1 − t )
1−
α−1
2(α − 1)2
1 0
q(s)ds u1 ( t )
q(s)ds
Γ (α)r
= mϱ(1 − t ) > 0,
ϱ(t )
t ∈ (0, 1).
(3.3)
As for (3.3), we also find a positive constant n such that u2 (t ) − w(t ) ≥ nϱ(1 − t ) > 0,
t ∈ (0, 1).
Let xi (t ) = ui (t ) − w(t ) (i = 1, 2), then xi (t ) > 0,
t ∈ (0, 1), i = 1, 2.
Thus the differential equation (1.1) has at least two positive solutions x1 , x2 satisfying x1 (t ) ≥ mϱ(1 − t ),
u2 (t ) ≥ nϱ(1 − t ),
t ∈ [0, 1]
for some positive constants m, n. The proof of Theorem 3.1 is completed.
Theorem 3.2. Suppose (B1) and (B2) hold. In addition, assume that the following conditions are satisfied: (S4) There exists a constant r >
4(α − 1)2
1
∫
Γ (α)
q(s)ds 0
such that, for any (t , u) ∈ [0, 1] × [µr , r ], r f (t , u) ≥ . l (S5) There exists a constant R > max r ,
r l
+ 1 K such that, for any (t , u) ∈ [0, 1] × [0, R],
R − 1, K where r is defined by (S4). f (t , u) ≤
(S6) lim
min
u→+∞ t ∈[a,b]
f ( t , u) u
= +∞.
Then the differential equation (1.1) has at least two positive solutions x1 , x2 , and there exist two positive constants m, n such that x1 (t ) ≥ mϱ(1 − t ), x2 (t ) ≥ nϱ(1 − t ) for any t ∈ [0, 1]. Proof. Firstly, let Ω1 = {u ∈ P : ‖u‖ < r }. Then, for any u ∈ ∂ Ω1 , t ∈ [a, b], by (2.6), we have u(t ) − w(t ) = u(t ) − 2
1
∫
G(t , s)q(s)ds 0
≥ u(t ) − ≥ u(t ) −
2(α − 1)
Γ (α)
ϱ(1 − t )
2(α − 1)2 u(t )
Γ (α)r
1
∫
q(s)ds 0
1
∫
q(s)ds 0
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
≥
1− 1
≥
2
2(α − 1)2
Γ (α)r
u(t ) ≥
1
∫
q(s)ds u(t ) 0
ϱ(1 − t ) 2(α − 1)
‖u‖ =
ϱ(1 − t ) 2(α − 1)
R.
(3.4)
So for any u ∈ Ω1 , t ∈ [a, b], by (3.4), we have
µr ≤ u(t ) − w(t ) ≤ r .
(3.5)
It follows from (S4), (3.5) and Lemma 2.2 that, for any u ∈ ∂ Ω2 , t ∈ [a, b], 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ ≥ 0 1
∫
G(t , s)p(s)f (s, [u(s) − w(s)]∗ )ds
≥ 0
∫ ϱ(1 − t ) b ≥ ϱ(s)p(s)f (s, x(s) − w(s))ds Γ (α) a ∫ b a(1 − b)α−1 r ≥ ϱ(s)p(s)ds Γ (α) l a = r = ‖u‖. Therefore, we have
‖Tu‖ ≥ ‖u‖,
u ∈ P ∩ ∂ Ω1 .
Next, by (S5), we have R > r and R K
−2>
r l
> 0.
Let Ω2 = {u ∈ P : ‖u‖ < R}. Then, for any u ∈ ∂ Ω2 , s ∈ [0, 1], we have
[u(s) − w(s)]∗ ≤ u(s) ≤ ‖u‖ = R. It follows from (S5) that
‖Tu‖ = max (Tu)(t ) t ∈[0,1] ∫ 1 = max G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds 0≤t ≤1 0 [ ] ∫ 1 R (α − 1)ϱ(s) − 1 p(s) + q(s) ds ≤ Γ (α) K 0 1 (α − 1)R 0 ϱ(s)[p(s) + q(s)]ds ≤ K Γ (α) = r = ‖u‖. Therefore,
‖Tu‖ ≤ ‖u‖,
u ∈ P ∩ ∂ Ω2 .
On the other hand, choose a real number M > 0 such that
µMa(1 − b)α−1 Γ (α)
∫
b
ϱ(s)p(s)ds ≥ 1.
a
From (S6), there exists N > R such that, for any t ∈ [a, b], f (t , u) ≥ Mu,
u ≥ N.
Take R∗ > max
N
µ
,R ,
(3.6)
X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
1271
then R∗ > R > r. Let Ω3 = {u ∈ P : ‖u‖ < R∗ }, for any u ∈ P ∩ ∂ Ω3 and for any t ∈ [a, b], we have u(t ) − w(t ) ≥ u(t ) − 2
∫
1
G(t , s)q(s)ds
0
≥ u(t ) −
2(α − 1)
Γ (α)
ϱ(1 − t )
1
∫
q(s)ds 0
∫ 2(α − 1)2 u(t ) 1 ≥ u(t ) − q(s)ds Γ (α)R∗ 0 ∫ 1 2(α − 1)2 ≥ 1− q(s)ds u(t ) Γ (α)R∗ 0 1 ϱ(1 − t ) ϱ(1 − t ) ∗ ≥ u( t ) ≥ ‖ u‖ = R 2 2(α − 1) 2(α − 1) ≥ µR∗ ≥ N > 0.
(3.7)
So for any u ∈ P ∩ ∂ Ω3 , t ∈ [a, b], by (3.6), (3.7), we have 1
∫
G(t , s)[p(s)f (s, [u(s) − w(s)]∗ ) + q(s)]ds
‖Tu‖ ≥ 0
≥ ≥ = = ≥
∫ ϱ(1 − t ) b ϱ(s)p(s)f (s, u(s) − w(s))ds Γ (α) a ∫ b a(1 − b)α−1 ϱ(s)p(s)M u(s) − w(s) ds Γ (α) a ∫ b Ma(1 − b)α−1 ϱ(s)p(s)ds × µR∗ Γ (α) a ∫ µMa(1 − b)α−1 b ϱ(s)p(s)ds × R∗ Γ (α) a R∗ = ‖u‖.
Thus u ∈ P ∩ ∂ Ω3 .
‖Tu‖ ≥ ‖u‖,
(3.8)
By Lemma 2.6, T has two fixed points u1 , u2 such that r ≤ ‖u1 ‖ ≤ R ≤ ‖u2 ‖. Noticing that r >
4(α − 1)2
1
∫
Γ (α)
q(s)ds,
0
we have u1 (t ) − w(t ) ≥ u1 (t ) −
≥ u1 (t ) − = ≥
1− 1 2
2(α − 1)
Γ (α)
ϱ(1 − t ) ∫
1
Γ (α)r 0 1 2(α − 1)2 0 q(s)ds Γ (α)r r 2(α − 1)
= mϱ(1 − t ) > 0,
q(s)ds 0
2(α − 1)2 u1 (t )
u1 (t ) ≥
1
∫
q(s)ds u1 ( t )
ϱ(1 − t )
t ∈ (0, 1).
As for (3.9), we also can find a positive constant n such that u2 (t ) − w(t ) ≥ nϱ(1 − t ) > 0, Let xi (t ) = ui (t ) − w(t ) (i = 1, 2), then xi (t ) > 0,
t ∈ (0, 1), i = 1, 2.
t ∈ (0, 1).
(3.9)
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
So the differential equation (1.1) has at least two positive solutions x1 , x2 satisfying x1 (t ) ≥ mϱ(1 − t ),
x2 (t ) ≥ nϱ(1 − t ),
t ∈ [0, 1]
for some positive constants m, n. The proof of Theorem 3.2 is completed.
4. Applications Example 4.1. Consider the following problem
5 2
−D0+ u(t ) = Γ
| ln t | , −√ 3 t (1 − t ) t (1 − t ) 2 u(1) = 0,
5 2
u(0) = u′ (0) = 0,
f (u)
0 < t < 1,
(4.1)
where
1 u, 3
0 ≤ u ≤ 90,
1230u − 110670, 90 ≤ u ≤ 100, f (u) = u + 12230, 100 ≤ u ≤ 6400, 1863 u 12 , u ≥ 6400. 8 Then the BVP (4.1) has at least two positive solutions. Proof. Let p(t ) = Take
Γ
Γ 25 | ln t | q(t ) = √ . t (1 − t )
5 2 3
t (1 − t ) 2
,
, 34 ⊂ [0, 1], by direct calculation, we have ∫ 3 1 | ln s| 361 1 3 K = s(1 − s) 2 , +√ ds = 3 2 0 150 s ( 1 − s ) s(1 − s) 2 ∫ 3 4 1 1 1 µ= ds = , l= ,
1
4
96
32
64
1 4
and 2(α − 1)2
1 0
q(s)ds
1
∫
| ln s| ds √ s( 1 − s) √ ∫ 1 ∫ 2 | ln s| 9 2 ≤ √ ds + =
Γ (α)
9
2
0
2
s
0
1 1 2
√
| ln s| ds √ 1−s
√ = 18 ln 2 + 18 2 ln( 2 + 1) + 36 < 90. Choose r = 90, then
90 = r > max 2K ,
2(α − 1)2
1 0
Γ (α)
q(s)ds
,
and for any (t , u) ∈ [0, 1] × [0, 90], f (t , u) ≤ 30 ≤
r K
− 1 ≈ 37.
So the condition (S1) is satisfied. On the other hand, we take R = 6400, then R > 2r = 180, and for any (t , u) ∈ [0, 1] × [µR, R] = [0, 1] × [100, 6400], we have f (t , u) ≥ 12330 ≥
R l
= 12288,
X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
1273
this implies that the condition (S2) holds. Next, we have lim
f ( t , u)
max
u→+∞ t ∈[0,1]
u
1863
= lim
1
u 2 = 0.
8
u→+∞
Thus (S3) also holds. Moreover it is clear that (B1)–(B2) hold. Thus, by Theorem 3.1, the BVP (4.1) has at least two positive solutions. Example 4.2. Consider the following problem
7 4
D0+ u(t ) + Γ
f ( u) =
−√ 4
3
4
u(0) = u (0) = 0,
where
f (u)
7
′
t (1 − t ) 4 u(1) = 0,
1
= 0,
1−t
√ √ 2 , 288 2u, 0 ≤ u ≤ 8 √ √ 2688 + 112 2 u + 20636 − 336 2 , 287
0 < t < 1,
√ 2
287
8
≤ u ≤ 3,
20 18760 − u+ , 3 ≤ u ≤ 190, 187 187 2 2 u , u ≥ 190. 1805
The BVP (4.2) has at least two positive solutions u1 (t ) and u2 (t ) such that u1 (t ) ≥
8 3
3
t 4 (1 − t ),
u2 (t ) ≥
380 3
3
t 4 (1 − t ),
t ∈ [0, 1].
Proof. In fact, let
Γ
Γ
7
4 q(t ) = √ , 4 1−t
p(t ) =
7 4 3
t (1 − t ) 4
.
, 34 ⊂ [0, 1], then 1 ∫ (α − 1) 0 ϱ(s)[p(s) + q(s)]ds 3 1 1 1 19 3 = s(1 − s) 4 √ + , K = ds = 3 4 Γ (α) 4 0 20 1−s 4 s(1 − s) √ √ ∫ b a(1 − b)α−1 2 a(1 − b)α−1 2 µ= = , l= ϱ(s)p(s)ds = . 2(α − 1) 24 Γ (α) 32 a Set r = 4, then ∫ 1 ∫ 4(α − 1)2 9 1 ds r =4> q(s)ds = = 3, √ 4 Γ (α) 4 1−s 0 0 √ and for any (t , u) ∈ [0, 1] × 82 , 3 , Take
1 4
f (t , u) ≥ 72 ≥
r
√ = 48 2.
l On the other hand, take a constant R = 190, then r R = 190 > max r , + 1 K = {3, 69.35}, l and for any (t , u) ∈ [0, 1] × [0, 190], f (t , u) ≤ 100 ≤
R K
−1=
3800 19
− 1 = 199.
Finally, we have lim
min
f ( t , u)
u→+∞ t ∈[a,b]
Thus (S6) also holds.
u
= lim
u→+∞
min
t∈
1 3 , 4 4
f (t , u) u
= lim
u→+∞
2 1805
u = +∞.
(4.2)
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X. Zhang et al. / Mathematical and Computer Modelling 55 (2012) 1263–1274
(B1)–(B2) are obvious, it follows from Theorem 3.2 that the BVP (4.2) has at least two positive solutions, u1 (t ) and u2 (t ) such that u1 ( t ) ≥
8 3
3
t 4 (1 − t ),
u2 ( t ) ≥
380 3
3
t 4 (1 − t ),
t ∈ [0, 1].
Acknowledgments The first and second authors were supported financially by the National Natural Science Foundation of China (11071141) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017). The third author was supported financially by the Australia Research Council through an ARC Discovery Project Grant. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
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