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Multiple positive solutions for m -point boundary-value problems with a one-dimensional p -Laplacian
Nonlinear Analysis 68 (2008) 2269–2279 www.elsevier.com/locate/na
Multiple positive solutions for m-point boundary-value problems with a one-dimensional p-LaplacianI Hanying Feng a,b,∗ , Weigao Ge b , Ming Jiang a a Department of Mathematics, Shijiazhuang Mechanical Engineering College, Shijiazhuang 050003, People’s Republic of China b Department of Mathematics, Beijing Institute of Technology, Beijing 100081, People’s Republic of China
Received 13 August 2006; accepted 26 January 2007
Abstract In this paper, we consider the multipoint boundary- value problem for the one-dimensional p-Laplacian (φ p (u 0 ))0 + q(t) f (t, u) = 0,
t ∈ (0, 1),
subject to the boundary conditions: u(0) =
m−2 X
ai u(ξi ),
u(1) =
i=1
m−2 X
bi u(ξi ),
i=1
where φ p (s) = |s| p−2 s. By using a fixed point theorem in a cone, we present sufficient conditions for the existence of multiple positive solutions for the above boundary value problem. c 2007 Elsevier Ltd. All rights reserved.
MR Subject Classification: 34B10; 34B15 Keywords: Boundary value problems; Multiple positive solutions; Fixed point theorem; One-dimensional p-Laplacian
1. Introduction In this paper, we study the existence of multiple positive solutions for the multipoint boundary-value problem (BVP for short) for the one-dimensional p-Laplacian (φ p (u 0 ))0 + q(t) f (t, u) = 0, u(0) =
m−2 X i=1
ai u(ξi ),
u(1) =
t ∈ (0, 1), m−2 X
bi u(ξi ),
(1.1) (1.2)
i=1
I Supported by NNSF of China (10671012) and SRFDP of China (20050007011). ∗ Corresponding author at: Department of Mathematics, Shijiazhuang Mechanical Engineering College, Shijiazhuang 050003, People’s Republic of China. E-mail address: [email protected] (H. Feng).
c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.01.052
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where φ p (s) = |s| p−2 s, p > 1, ξi ∈ (0, 1) with 0 < ξ1 < ξ2 < · · · < ξm−2 < 1 and ai , f satisfy Pm−2 Pm−2 (H1 ) ai , bi ∈ [0, 1) satisfy 0 ≤ i=1 ai < 1, 0 ≤ i=1 bi < 1; (H2 ) Assume that f ∈ C([0, 1] × [0, +∞), (0, +∞)), q(t) ∈ L 1 [0, 1] is nonnegative on (0, 1) and q(t) 6≡ 0 on any subinterval of (0, 1). The study of multipoint boundary value problems for linear second-order ordinary differential equations was initiated by Il’in and Moiseev [10]. Since then a lot of attention has been paid to the study of nonlinear multipoint boundary value problems, see [1–3,8,9,15,16,20–22]. The equation with a p-Laplacian operator arises in the modeling of different physical and natural phenomena, non-Newtonian mechanics [4,11], nonlinear elasticity and glaciology [7,11], combustion theory [19], population biology [17,18], nonlinear flow laws [7,13,14], and system of Monge–Kantorovich partial differential equations [5]. There exist a very large number of papers devoted to the existence of solutions of the p-Laplacian operator. The problem (1.1) with different boundary conditions has been studied by many authors, see [2,3,9,12,15,21,22] and the references therein. For example, Ma, Du and Ge [15] studied the existence of positive solutions for the multipoint BVP, (φ p (u 0 ))0 + q(t) f (t, u) = 0, u 0 (0) =
m−2 X
αi u 0 (ξi ),
t ∈ (0, 1),
u(1) =
i=1
m−2 X
βi u(ξi ).
i=1
The main tool is the monotone iterative technique. In [21,22], Wang, Hou and Wang, Ge considered the multipoint BVPs for the one-dimensional p-Laplcian successively (φ p (u 0 ))0 + f (t, u) = 0, φ p (u 0 (0)) =
n−2 X
t ∈ (0, 1),
ai φ p (u 0 (ξi )),
u(1) =
i=1
n−2 X
bi u(ξi ).
i=1
and (φ p (u 0 ))0 + f (t, u) = 0, u 0 (0) =
n−2 X
αi u 0 (ξi ),
i=1
t ∈ (0, 1), u(1) =
n−2 X
βi u(ξi ).
(1.3) (1.4)
i=1
They provided sufficient conditions for the existence of multiple positive solutions for the above BVPs by applying the fixed point index theory in a cone. In the papers [2,3], the authors also investigated BVP (1.1) with the boundary value conditions u(0) = 0, u(1) = Pm−2 i=1 bi u(ξi ) (which is the special case when ai = 0 in (1.2)) and (1.3), (1.4) respectively. Unfortunately, the expressions of solutions on BVPs in [2,3] are not correct, which was pointed out by the authors of [21]. In [22], the authors have given the correct conclusions for BVP (1.3), (1.4). But the conclusions in [2] have not been reconsidered so far as we know. The main difficulties are that it is not easy to define an operator, and so it is more difficult to apply the fixed point theorem. Motivated by the works in [9,21,22] mentioned above, the purpose of this paper is to show the existence of multiple positive solutions to the BVP (1.1), (1.2). To obtain positive solutions of BVP (1.1), (1.2), the following well-known Krasnoselskii’s fixed point theorem in a cone is useful. Theorem 1.1 ([6]). Let E be a Banach space, and K ⊂ E be a cone. Assume Ω1 , Ω2 are open bounded subsets of E with 0 ∈ Ω1 , Ω 1 ⊂ Ω2 , and let T : K ∩ (Ω 2 \ Ω1 ) −→ K be a completely continuous operator such that
H. Feng et al. / Nonlinear Analysis 68 (2008) 2269–2279
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(i) kT uk ≤ kuk, u ∈ K ∩ ∂Ω1 , and kT uk ≥ kuk, u ∈ K ∩ ∂Ω2 ; or (ii) kT uk ≥ kuk, u ∈ K ∩ ∂Ω1 , and kT uk ≤ kuk, u ∈ K ∩ ∂Ω2 , Then T has at least one fixed point in K ∩ (Ω 2 \ Ω1 ). 2. Preliminaries In this paper, we assume that (H1 ) and (H2 ) hold and denote C + [1, 0] = {x ∈ C[0, 1] : x(t) ≥ 0, t ∈ [0, 1]}. is the inverse function of φ p . For any x ∈ C + [0, 1], suppose u is a solution of the following BVP
φ −1 p
(φ p (u 0 ))0 + q(t) f (t, x) = 0, u(0) =
m−2 X
ai u(ξi ),
t ∈ (0, 1), m−2 X
u(1) =
i=1
(2.1)
bi u(ξi ).
(2.2)
i=1
Then
t
Z
q(τ ) f (τ, x(τ ))dτ Ax − 0 Z t u(t) = u(0) + I (s, x)ds u (t) = 0
φ −1 p
=: I (t, x),
0
or 1
Z
I (s, t)ds.
u(t) = u(1) − t
By the boundary condition (2.2), we have Z t m−2 X Z ξi 1 u(t) = ai I (s, x)ds + I (s, x)ds m−2 P 0 0 i=1 1− ai i=1
or u(t) = − 1−
1 m−2 P
m−2 X
bi
1
Z bi
ξi
i=1
I (s, x)ds −
1
Z
I (s, x)ds.
t
i=1
Here, A x satisfies the equation 1−
m−2 P
1−
i=1 m−2 P
Hx (c) =
bi ai
m−2 X
ξi
Z ai 0
i=1
φ −1 p
s
Z c−
q(τ ) f (τ, x(τ ))dτ ds
0
i=1
+
1−
m−2 X
!Z bi 0
i=1
+
m−2 X i=1
Z bi
1 ξi
1
φ −1 p
φ −1 p
c−
c−
q(τ ) f (τ, x(τ ))dτ ds
0 s
Z
s
Z
q(τ ) f (τ, x(τ ))dτ ds = 0.
(2.3)
0
Lemma 2.1. Assume that (H1 ), (H2 ) hold. For any x ∈ C + [0, R σ1], there exists a unique A x ∈ (−∞, +∞) satisfying (2.3). Moreover, there is a unique σx ∈ (0, 1) such that A x = 0 x q(τ ) f (τ, x(τ ))dτ .
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Proof. For any x ∈ C + [0, 1], by the definition of Hx (c), it is easy to know that Hx : R → R is continuous and strictly increasing, and ! Z 1 Hx (0) < 0, Hx q(τ ) f (τ, x(τ ))dτ > 0. 0
R 1 Therefore, there exists a unique A x ∈ 0, 0 q(τ ) f (τ, x(τ ))dτ ⊂ (−∞, +∞) satisfying (2.3). Furthermore, let Z t q(τ ) f (τ, x(τ ))dτ, F(t) = 0
then F(t) is continuous and strictly increasing on [0, 1], and F(0) = 0, F(1) = Z 1 q(τ ) f (τ, x(τ ))dτ. 0 = F(0) < A x < F(1) =
R1 0
q(τ ) f (τ, x(τ ))dτ . So
0
It follows that there exists a unique σx ∈ (0, 1) such that A x =
R σx 0
q(τ ) f (τ, x(τ ))dτ by the property of F(t).
Lemma 2.2. Assume that (H1 ), (H2 ) hold. For any x ∈ C + [0, 1], the unique solution u of (2.1), (2.2) has the following properties (i) the graph of u(t) is concave; (ii) u(t) ≥ 0; (iii) there exists a unique t0 ∈ (0, 1) such that u(t0 ) = max0≤t≤1 u(t), moreover u 0 (t0 ) = 0; (iv) σx = t0 . Proof. Suppose that u(t) is the solution of (2.1) and (2.2). (i) From the fact that (φ p (u 0 ))0 (t) = −q(t) f (t, x(t)) ≤ 0, we know that φ p (u 0 (t)) is nonincreasing. It follows that u 0 (t) is also nonincreasing. Thus, the graph of u(t) is concave down on (0, 1). (ii) Without loss of generality, we assume that u(0) = min{u(0), u(1)}. By the concavity of u, we know that u(ξi ) ≥ u(0) (i = 0, 1, 2, . . . , n). So, we get u(0) =
m−2 X
ai u(ξi ) ≥
i=1
m−2 X
ai u(0).
i=1
Pm−2 By 1 − i=1 ai > 0, it is obvious that u(0) ≥ 0. Hence, u(1) ≥ u(0) ≥ 0. So, from the concavity of u, we know that u(t) ≥ 0, t ∈ [0, 1]. (iii) Since u is strictly concave, there is a unique point t0 where the maximum is attained. From the boundary conditions (1.2), this cannot happen at 0 or at 1, i.e., t0 ∈ (0, 1) such that u(t0 ) = max0≤t≤1 u(t) and then u 0 (t0 ) = 0. R σx 0 (iv) By Lemma 2.1, it is easy to check u 0 (t) = φ −1 p t q(τ ) f (τ, x(τ ))dτ . Hence, we obtain that u (σx ) = 0, this implies σx = t0 . Lemma 2.3. Assume that (H1 ), (H2 ) hold. For x ∈ C + [0, 1] and 0 < γ < min{ξ1 , 1 − ξm−2 }, so then the unique solution u(t) of BVP (2.1), (2.2) satisfies mint∈[γ ,1−γ ] u(t) ≥ γ kuk. Proof. By Lemma 2.2, we know the unique solution u(t) of BVP (2.1), (2.2) is continuous, nonnegative and concave on [0, 1], then u(t) ≥ min{t, 1 − t}kuk. Hence mint∈[γ ,1−γ ] u(t) ≥ γ kuk. For any x(t) ∈ C + [0, 1], define an operator Z t m−2 X Z ξi 1 a I (s, x)ds + I (s, x)ds, 0 ≤ t ≤ σx , i m−2 P 0 0 i=1 ai 1 − i=1 (T x)(t) = (2.4) Z 1 m−2 X Z 1 1 − bi I (s, x)ds − I (s, x)ds, σx ≤ t ≤ 1. m−2 P ξi t i=1 bi 1− i=1
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Furthermore, we have the following result. Lemma 2.4. Let K = ω : ω ∈ C + [0, 1],
min
t∈[γ ,1−γ ]
ω(t) ≥ γ kωk ,
then T : K −→ K is completely continuous. Proof. Here we only prove the case 0 ≤ t ≤ σx . The case σx ≤ t ≤ 1 can be proved similarly. It is easy to check that K is a cone in C[0, 1]. According to Lemmas 2.2 and 2.3, we easily obtain (T x) ∈ C + [0, 1]
and
min (T x)(t) ≥ γ kT xk
t∈[γ ,1−γ ]
for x ∈ K .
This means that T K ⊂ K . Now we show that T : K −→ K is completely continuous. (1) We show that T is continuous. First, we prove that A x is continuous in x. Suppose {xn } ∈ C + [0, 1] with xn −→ x0 ∈ C + [0, 1]. Let {An } (n = 0, 1, 2, . . .) be constants decided by (2.3) corresponding to {xn } (n = 0, 1, 2, . . .). As xn −→ x0 uniformly on [0, 1] and f : [0, 1] × [0, ∞) −→ (0, ∞) is continuous, we have that for ε = 1, there exists N > 0, when n > N , for any τ ∈ [0, 1], 0 ≤ f (τ, xn (τ )) ≤ 1 + f (τ, x0 (τ )) ≤ 1 + max f (τ, x0 (τ )).
(2.5)
0≤τ ≤1
So, 1
Z An ∈ 0,
! q(τ ) f (τ, xn (τ ))dτ
⊆ 0, 1 + max f (τ, x0 (τ )) 0≤τ ≤1
0
1
Z
! q(τ )dτ ,
0
which implies An is uniformly bounded. (1) (2) (1) Assume {An } does not converge to A0 . Then, there exist two subsequences {An k } and {An k } of {An } with An k −→ (2) c1 and An k −→ c2 , since An is uniformly bounded, but c1 6= c2 . By the construction of {An } (n = 0, 1, 2, . . .), we (1) have Hx (1) (An k ) = 0. Combining (2.5) and using Lebesgue’s dominated convergence theorem, we get nk
lim H (1) (A(1) nk ) n k →∞ xn k
=H
(1)
lim xn k
n k →∞
lim A(1) n k →∞ n k
= Hx0 (c1 ) = 0.
Since {An } (n = 0, 1, 2 . . .) is unique, we have c1 = A0 . Similarly, c2 = A0 . Hence, c1 = c2 , which is a contradiction. So, for any xn −→ x0 , An −→ A0 , which means A x : C + [0, 1] −→ R is continuous. Therefore the continuity of T is obvious. (2) We prove T is compact. Let Ω ⊂ K be a bounded set. Then, there exists R, such that Ω ⊂ {x ∈ K | kxk ≤ R}. For any x ∈ Ω , we have Z 1 Z 1 0≤ q(τ ) f (τ, x(τ ))dτ ≤ max f (τ, x) q(τ )dτ =: M. τ ∈[0,1],x∈[0,R]
0
0
So we get |A x | ≤ M. Therefore, by the definition of T , one has kT xk ≤
φ −1 p (2M) 1−
m−2 P
,
k(T x)0 k ≤ φ −1 p (2M).
ai
i=1
Then T Ω is uniformly bounded and equicontinuous. By the Arzel`a–Ascoli’s theorem, T Ω is relatively compact and therefore that T is compact.
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Remark 2.1. By Lemma 2.1, (2.4) can be rewritten as the following Z σx m−2 X Z ξi 1 −1 q(τ ) f (τ, x(τ ))dτ ds a φ i p m−2 P s 0 i=1 1− ai i=1 Z t Z σx −1 + φ q(τ ) f (τ, x(τ ))dτ ds, p s 0 m−2 (T x)(t) = P Z s i=1 bi Z 1 −1 φ q(τ ) f (τ, x(τ ))dτ ds p m−2 P ξi σx 1 − bi i=1 Z s Z 1 + q(τ ) f (τ, x(τ ))dτ ds, φ −1 t
p
σx
0 ≤ t ≤ σx , (2.6)
σx ≤ t ≤ 1.
Now for convenience we use the following notations. Let f (t, u) f γ R,R = min min : u ∈ [γ R, R] , t∈[γ ,1−γ ] φ p (R) f (t, u) 0,r : u ∈ [0, r ] , f = max max t∈[0,1] φ p (r ) f (t, u) , f α = lim sup max u→α t∈[0,1] φ p (u) f (t, u) f α = lim inf min , (α := ∞, or 0+ ), u→α t∈[γ ,1−γ ] φ p (u) ξi + ξi+1 , (i = 0, 1, . . . , m − 2), ti∗ = 2 ! Z 1 1 1 −1 = φp q(τ )dτ , m−2 l P 0 1− ai i=1
1 = min Li
(Z
ti∗ ξi
φ −1 p
ti∗
Z
! q(τ )dτ ds,
s
Z
ξi+1 ti∗
φ −1 p
Z
s ti∗
!
)
q(τ )dτ ds ,
1 1 = min , 0≤i≤m−2 L i L where 0 < γ = ξ0 < ξ1 < ξ2 < · · · < ξm−2 < ξm−1 = 1 − γ < 1. Remark 2.2. Instead of ti∗ =
ξi +ξi+1 , 2
we can take any ti∗ in (ξi , ξi+1 ).
3. The existence of one positive solution We now give our results for the existence of a positive solution of BVP (1.1) and (1.2). Theorem 3.1. Assume that there exist constants r, R > 0 with r < γ R (or L R < lr ), such that the following conditions hold: (H3 ) f 0,r ≤ φ p (l), (H4 ) f γ R,R ≥ φ p (L). Then BVP (1.1), (1.2) has at least one positive solution u such that 0 < r ≤ kuk ≤ R (or 0 < R ≤ kuk ≤ r ).
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Proof. It is obvious that BVP (1.1) and (1.2) has a solution u = u(t) if and only if u solves operator equation T u = u. First, we assume r < γ R. Let K r := {u ∈ K : kuk < r }; then ∂ K r = {u ∈ K : kuk = r }. By (2.6) and (H3 ), for u ∈ ∂ K r we have kT uk = (T u)(σu ) = 1−
1 m−2 P
m−2 X
φ −1 p
ai 0
i=1
ai
ξi
Z
σx
Z
q(τ ) f (τ, x(τ ))dτ ds +
σx
Z 0
s
φ −1 p
Z
σx
q(τ ) f (τ, x(τ ))dτ ds
s
i=1 m−2 P
ai i=1 m−2 P
≤
1−
1
Z 0
ai
1
Z
φ −1 p
! q(τ ) f (τ, x(τ ))dτ ds +
1
Z 0
0
φ −1 p
1
Z
! q(τ ) f (τ, u(τ ))dτ ds
0
i=1
≤ 1−
1 m−2 P
1
Z
φ −1 p
! q(τ ) f (τ, x(τ ))dτ ds
0
ai
i=1
≤ 1−
1 m−2 P
1
Z
rlφ −1 p
! q(τ )dτ
0
ai
i=1
= r = kuk. This implies that kT uk ≤ kuk for u ∈ ∂ K r . Let K R := {u ∈ K : kuk < R}; then ∂ K R = {u ∈ K : kuk = R}. For u ∈ ∂ K R , it follows that γ R ≤ u ≤ R, γ ≤ t ≤ 1 − γ . We consider two cases. (i) u ∈ ∂ K R , ti∗ ≤ σu . By (2.6), (H4 ) and noticing (T u)(0) ≥ 0, we obtain Z σu Z t∗ i ∗ −1 (T u)(ti ) ≥ φp q(τ ) f (τ, u(τ ))dτ ds s 0 ! Z ∗ Z ∗ ti
≥ ξi
ti
φ −1 p ti∗
Z ≥ RL
ξi
q(τ ) f (τ, u(τ ))dτ ds
s ti∗
Z
φ −1 p
! q(τ )dτ ds
s
≥ R = kuk. (ii) u ∈ ∂ K R , σu < ti∗ . By (2.6), (H4 ), and noticing (T u)(1) ≥ 0 from Lemma 2.3, we arrive at Z s Z 1 ∗ −1 (T u)(ti ) ≥ φp q(τ ) f (τ, u(τ ))dτ ds σu
ti∗
Z
ξi+1
≥ ti∗
Z
φ −1 p ξi+1
≥ RL ti∗
s
Z
ti∗
φ −1 p
! q(τ ) f (τ, u(τ ))dτ ds Z
s ti∗
! q(τ )dτ ds
≥ R = kuk. This implies that kT uk ≥ kuk for u ∈ ∂ K R . Therefore, by the first part of Krasnoselskii’s fixed-point theorem, it follows that T has a fixed-point u in K ∩ (K R \ K r ). That is BVP (1.1), (1.2) have at least one positive solution such that 0 < r ≤ kuk ≤ R.
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Now we assume L R < lr . By a similar proof, one easily gets kT uk ≥ kuk for u ∈ ∂ K R and kT uk ≤ kuk for u ∈ ∂ K r . Therefore, by the second part of Krasnoselskii’s fixed-point theorem, it follows that T has a fixed-point u in K ∩ (K r \ K R ). That is, BVP (1.1), (1.2) have at least one positive solution such that 0 < R ≤ kuk ≤ r . Similarly, we can obtain the following conclusion. Theorem 3.1*. Assume that there exist constants r, R > 0 with r < γ R (or L R < lr ), such that the following conditions hold: (H3 )∗ f 0,r < φ p (l), (H4 )∗ f γ R,R > φ p (L), Then BVP (1.1) and (1.2) has at least one positive solution u such that 0 < r < kuk < R (or 0 < R < kuk < r ). Corollary 3.1. Assume the following conditions hold: (H5 ) f 0 ≤ φ p (l) and f ∞ ≥ φ p ( γL ),
(H6 ) f 0 ≥ φ p ( γL ), and f ∞ ≤ φ p (l). Then BVP (1.1), (1.2) have at least one positive solution. Proof. We show (H5 ) implies (H3 ) and (H4 ). Suppose (H5 ) holds; then there exist r and R with 0 < r < γ R, such that f (t, u) ≤ φ p (l), 0 ≤ t ≤ 1, 0 < u ≤ r φ p (u) and f (t, u) ≥ φp φ p (u)
L , γ
γ ≤ t ≤ 1 − γ , u ≥ γ R.
Hence, we obtain f (t, u) ≤ φ p (l)φ p (u) ≤ φ p (l)φ p (r ) = φ p (rl),
0 ≤ t ≤ 1, 0 < u ≤ r
and f (t, u) ≥ φ p
L L φ p (u) ≥ φ p φ p (γ R) = φ p (R L), γ γ
γ ≤ t ≤ 1 − γ , γ R ≤ u ≤ R.
Hence, (H3 ) and (H4 ) hold. Therefore, by Theorem 3.1, BVP (1.1) and (1.2) have at least one positive solution. Now suppose (H6 ) holds; then there exist 0 < r < R with Lr < l R such that L f (t, u) ≥ φp , γ ≤ t ≤ 1 − γ, 0 < u ≤ r φ p (u) γ f (t, u) ≤ φ p (l), φ p (u)
0 ≤ t ≤ 1, u ≥ R.
By (3.1), it follows that L L f (t, u) ≥ φ p φ p (u) ≥ φ p φ p (γ r ) = φ p (Lr ), γ γ
(3.1) (3.2)
for γ ≤ t ≤ 1 − γ , γ r ≤ u ≤ r.
So, the condition (H4 ) holds for r . We consider two cases for (3.2). (i) f (t, u) is bounded. In this case, there exists a constant M > 0 such that f (t, u) ≤ M, nfor 0 ≤ t ≤ o1 and 0 ≤ u < ∞. By (3.2), there exists a constant λ ≥ R with Lr < l R ≤ λl satisfying φ p (r ) ≥ max φ p (R), φ pM(l) such that f (t, u) ≤ M ≤ φ p (λl), for 0 ≤ t ≤ 1, 0 ≤ u ≤ λ, that is the condition (H3 ) holds for λ.
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(ii) f (t, u) is unbounded. In this case, there exist t0 ∈ [0, 1] and λ1 ≥ R, with Lr < l R ≤ λ1l such that f (t, u) ≤ f (t0 , λ1 ) for 0 ≤ t ≤ 1, 0 ≤ u ≤ λ1 . This yields f (t, u) ≤ f (t0 , λ1 ) ≤ φ p (λ1l) for 0 ≤ t ≤ 1, 0 ≤ u ≤ λ1 , and then the condition (H3 ) holds for λ1 . Therefore, the conclusion follows from Theorem 3.1. Remark 3.1. It is easy to see that Corollary 3.1 holds if f satisfies condition f 0 = 0, f ∞ = ∞ or f 0 = ∞, f ∞ = 0. 4. The existence of multiple positive solutions In this section, we give some conclusions about the existence of multiple positive solutions. Theorem 4.1. Assume the following conditions hold: (H3 )∗ f 0,r < φ p (l), (H7 ) f 0 ≥ φ p ( γL ) and f ∞ ≥ φ p ( γL ). Then BVP (1.1), (1.2) have at least two positive solutions such that 0 < ku 1 k < r < ku 2 k. Proof. By the proof of Corollary 3.1, we can take 0 < r1 < r < γ r2 such that f (t, u) ≥ φ p (r1 L), for γ ≤ t ≤ 1 − γ , γ r1 ≤ u ≤ r1 and f (t, u) ≥ φ p (r2 L), for γ ≤ t ≤ 1 − γ , γ r2 ≤ u ≤ r2 . So, by Theorem 3.1, it follows that BVP (1.1), (1.2) have at least two positive solutions such that 0 < ku 1 k < r < ku 2 k. Theorem 4.2. Assume the following conditions hold: (H4 )∗ f γ R,R > φ p (L), (H8 ) f 0 ≤ φ p (M) and f ∞ ≤ φ p (M). Then BVP (1.1), (1.2) have at least two positive solutions such that 0 < ku 1 k < R < ku 2 k. Theorem 4.3. Assume (H5 ) (or (H6 )) holds, and there exist constants r1 , r2 > 0 with r1 L < r2l (or r1 < γ r2 ) such that (H3 )∗ holds for r = r2 (or r = r1 ) and (H4 )∗ holds for R = r1 (or R = r2 ). Then BVP (1.1), (1.2) have at least three positive solutions such that 0 < ku 1 k < r1 < ku 2 k < r2 < ku 3 k. The proof of Theorems 4.2 and 4.3 is similar to that provided for Theorem 4.1. We omit it here. From Theorems 4.1–4.3, it is easy to see that if conditions (H3 )∗ , (H4 )∗ , (H5 )–(H8 ) are combined properly, then BVP (1.1) and (1.2) has arbitrarily many positive solutions. Theorem 4.4. Let n = 2k + 1, k ∈ N . Assume (H5 ) (or (H6 )) holds. If there exist constants r1 , r2 , . . . , rn−1 > 0 with r2i < γ r2i+1 , for 1 ≤ i ≤ k − 1 and r2i−1 L < r2i l, for 1 ≤ i ≤ k (or with r2i−1 < γ r2i , for 1 ≤ i ≤ k and r2i L < r2i+1l, for 1 ≤ i ≤ k − 1) such that (H4 )∗ (or (H3 )∗ ) holds for r2i−1 , 1 ≤ i ≤ k and (H3 )∗ (or (H4 )∗ ) holds for r2i , 1 ≤ i ≤ k. Then BVP (1.1), (1.2) have at least n positive solutions u 1 , u 2 , . . . , u n such that 0 < ku 1 k < r1 < ku 2 k < r2 < · · · < ku n−1 k < rn−1 < ku n k. Theorem 4.5. Let n = 2k, k ∈ N . Assume (H7 ) (or (H8 )) holds. If there exist constants r1 , r2 , . . . , rn−1 > 0 with r2i−1 < γ r2i and r2i L < r2i+1l, for 1 ≤ i ≤ k − 1 (or with r2i < γ r2i+1 and r2i−1 L < r2i l, for 1 ≤ i ≤ k − 1) such that (H3 )∗ (or (H4 )∗ ) holds for r2i−1 , 1 ≤ i ≤ k and (H4 )∗ (or (H3 )∗ ) holds for r2i , 1 ≤ i ≤ k − 1. Then BVP (1.1) and (1.2) have at least n positive solutions u 1 , u 2 , . . . , u n such that 0 < ku 1 k < r1 < ku 2 k < r2 < · · · < ku n−1 k < rn−1 < ku n k. 5. Examples Example 5.1. Consider the following BVP (|u 0 (t)|u 0 (t))0 + e−t g(u) = 0, 1 1 1 2 u(0) = u + u , 4 3 4 3
t ∈ (0, 1),
(5.1)
1 1 1 2 u(1) = u + u , 3 3 3 3
(5.2)
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where 2 u , 0 ≤ u ≤ 1, 1 679 615u − 1 679 614, 1 ≤ u ≤ 2, g(u) = 1 679 616, 2 ≤ u ≤ 12, u 2 10u + 1 679 496 , u ≥ 12. u + 132 √ Choose r = 1/2, R = 12, take γ = 1/6. Then it is easily checked that l = 2; L = 36 3. One has r < γ R and f (t, u) = e−t u 2 ≤ 1 = φ3 (rl), 0 ≤ t ≤ 1, 0 ≤ u ≤ 1; √ 1 679 616 1 5 f (t, u) = 1 679 616e−t > = (36 3 × 12)2 = φ3 (L R), ≤ t ≤ , 2 ≤ u ≤ 12. 3 6 6 Then all conditions of Theorem 3.1 hold. Thus BVP (5.1), (5.2) have at least one positive solution u such that 1 ≤ kuk ≤ 12. Example 5.2. In Example 5.1 replace g there by 2 u + 3, 0 ≤ u ≤ 1, 140u − 136, 1 ≤ u ≤ 2, g(u) = 144, 2 ≤ u ≤ 6, u 2 10u + 84 , u ≥ 6. u + 30 Choosing r = 6, R =
1√ , 36 3
one obtains R L < rl and
f (t, u) = e−t u 2 ≤ 144 = 122 = φ3 (rl), 0 ≤ t ≤ 1, 0 ≤ u ≤ 6; ! ! 2 2 1 1 1 −t f (t, u) = e +3 > + 3 > 1 = φ3 (R L), √ √ 3 216 3 216 3 5 1 1 1 ≤t ≤ , √ ≤u≤ √ . 6 6 216 3 36 3 Then Theorem 3.1 applies. Acknowledgment The authors wish to thank the referee for his (or her) valuable suggestions on the original manuscript. References [1] R.I. Avery, J. Henderson, Existence of three positive pseudo-symmetric solutions for a one-dimensional p-Laplacian, J. Math. Anal. Appl. 277 (2003) 395–404. [2] C. Bai, J. Fang, Existence of multiple positive solutions for nonlinear m-point boundary value problems, J. Math. Anal. Appl. 281 (2003) 76–85. [3] C. Bai, J. Fang, Existence of multiple positive solutions for nonlinear m-point boundary value problems, Appl. Math. Comput. 140 (2003) 297–305. [4] J. Diaz, F. de Thelin, On a nonlinear parabolic problem arising in some models related to turbulent flows, SIAM J. Math. Anal. 25 (4) (1994) 1085–1111. [5] L. Evans, W. Gangbo, Differential equations methods for the Monge–Kantorovich mass transfer problem, Mem. Amer. Math. Soc. 137 (653) (1999). [6] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Boston, 1988. [7] R. Glowinski, J. Rappaz, Approximation of a nonlinear elliptic problem arising in a non-Newtonian fluid flow model in glaciology, Math. Model. Numer. Anal. 37 (1) (2003) 175–186. [8] J. Henderson, H. Wang, Positive solutions for nonlinear eigenvalue problems, J. Math. Anal. Appl. 208 (1997) 252–259. [9] X. He, W. Ge, Existence of positive solutions for the one-dimension p-Laplacian equation, Acta Math. Sinica 46 (4) (2003) 805–810 (in Chinese). [10] V.A. Il’in, E.I. Moiseev, Nonlocal boundary value problem of the second kind for a Sturm–Liouville operator, Differ. Equ. 23 (1987) 979–987.
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[11] U. Janfalk, On Certain Problem Concerning the p-Laplace Operator, in: Likping Studies in Sciences and Technology, Dissertations, vol. 326, 1993. [12] B. liu, Positive solutions of singular three-point boundary value problems for the one-dimensional p-Laplacian, Comput. Math. Appl. 48 (2004) 913–925. [13] C. Liu, Weak solutions for a viscous p-Laplacian equation, Electron. J. Differential Equations (63) (2003) 1–11. [14] I. Ly, D. Seck, Isoperimetric inequality for an interior free boundary problem with p-Laplacian operator, Electron. J. Differential Equations (109) (2004) 1–12. [15] D. Ma, Z. Du, W. Ge, Existence and iteration of monotone positive solutions for multipoint boundary value problem with p-Laplacian operator, Comput. Math. Appl. 50 (2005) 729–739. [16] R. Ma, Positive solutions for multipoint boundary value problem with a one-dimensional p-Laplacian, Comput. Math. Appl. 42 (2001) 755–765. [17] S. Oruganti, J. Shi, R. Shivaji, Diffusive logistic equation with constant yield harvesting, I: Steady-states, Trans. Amer. Math. Soc. 354 (9) (2002) 3601–3619. [18] S. Oruganti, J. Shi, R. Shivaji, Logistic equation with p-Laplacian and constant yield harvesting, Abstr. Appl. Anal. 9 (2004) 723–727. [19] M. Ramaswamy, R. Shivaji, Multiple positive solutions for classes of p-Laplacian equations, Differential Integral Equations 17 (11–12) (2004) 1255–1261. [20] F. Wong, Existence of positive solutions for m-Laplacian boundary value problems, Appl. Math. Lett. 12 (1999) 11–17. [21] Y. Wang, C. Hou, Existence of multiple positive solutions for one-dimensional p-Laplacian, J. Math. Anal. Appl. 315 (2006) 144–153. [22] Y. Wang, W. Ge, Positive solutions for multipoint boundary value problems with a one-dimensional p-Laplacian, Nonlinear Anal. 66 (2007) 1246–1256.
Multiple positive solutions for m-point boundary-value problems with a one-dimensional p-LaplacianI Hanying Feng a,b,∗ , Weigao Ge b , Ming Jiang a a Department of Mathematics, Shijiazhuang Mechanical Engineering College, Shijiazhuang 050003, People’s Republic of China b Department of Mathematics, Beijing Institute of Technology, Beijing 100081, People’s Republic of China
Received 13 August 2006; accepted 26 January 2007
Abstract In this paper, we consider the multipoint boundary- value problem for the one-dimensional p-Laplacian (φ p (u 0 ))0 + q(t) f (t, u) = 0,
t ∈ (0, 1),
subject to the boundary conditions: u(0) =
m−2 X
ai u(ξi ),
u(1) =
i=1
m−2 X
bi u(ξi ),
i=1
where φ p (s) = |s| p−2 s. By using a fixed point theorem in a cone, we present sufficient conditions for the existence of multiple positive solutions for the above boundary value problem. c 2007 Elsevier Ltd. All rights reserved.
MR Subject Classification: 34B10; 34B15 Keywords: Boundary value problems; Multiple positive solutions; Fixed point theorem; One-dimensional p-Laplacian
1. Introduction In this paper, we study the existence of multiple positive solutions for the multipoint boundary-value problem (BVP for short) for the one-dimensional p-Laplacian (φ p (u 0 ))0 + q(t) f (t, u) = 0, u(0) =
m−2 X i=1
ai u(ξi ),
u(1) =
t ∈ (0, 1), m−2 X
bi u(ξi ),
(1.1) (1.2)
i=1
I Supported by NNSF of China (10671012) and SRFDP of China (20050007011). ∗ Corresponding author at: Department of Mathematics, Shijiazhuang Mechanical Engineering College, Shijiazhuang 050003, People’s Republic of China. E-mail address: [email protected] (H. Feng).
c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.01.052
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where φ p (s) = |s| p−2 s, p > 1, ξi ∈ (0, 1) with 0 < ξ1 < ξ2 < · · · < ξm−2 < 1 and ai , f satisfy Pm−2 Pm−2 (H1 ) ai , bi ∈ [0, 1) satisfy 0 ≤ i=1 ai < 1, 0 ≤ i=1 bi < 1; (H2 ) Assume that f ∈ C([0, 1] × [0, +∞), (0, +∞)), q(t) ∈ L 1 [0, 1] is nonnegative on (0, 1) and q(t) 6≡ 0 on any subinterval of (0, 1). The study of multipoint boundary value problems for linear second-order ordinary differential equations was initiated by Il’in and Moiseev [10]. Since then a lot of attention has been paid to the study of nonlinear multipoint boundary value problems, see [1–3,8,9,15,16,20–22]. The equation with a p-Laplacian operator arises in the modeling of different physical and natural phenomena, non-Newtonian mechanics [4,11], nonlinear elasticity and glaciology [7,11], combustion theory [19], population biology [17,18], nonlinear flow laws [7,13,14], and system of Monge–Kantorovich partial differential equations [5]. There exist a very large number of papers devoted to the existence of solutions of the p-Laplacian operator. The problem (1.1) with different boundary conditions has been studied by many authors, see [2,3,9,12,15,21,22] and the references therein. For example, Ma, Du and Ge [15] studied the existence of positive solutions for the multipoint BVP, (φ p (u 0 ))0 + q(t) f (t, u) = 0, u 0 (0) =
m−2 X
αi u 0 (ξi ),
t ∈ (0, 1),
u(1) =
i=1
m−2 X
βi u(ξi ).
i=1
The main tool is the monotone iterative technique. In [21,22], Wang, Hou and Wang, Ge considered the multipoint BVPs for the one-dimensional p-Laplcian successively (φ p (u 0 ))0 + f (t, u) = 0, φ p (u 0 (0)) =
n−2 X
t ∈ (0, 1),
ai φ p (u 0 (ξi )),
u(1) =
i=1
n−2 X
bi u(ξi ).
i=1
and (φ p (u 0 ))0 + f (t, u) = 0, u 0 (0) =
n−2 X
αi u 0 (ξi ),
i=1
t ∈ (0, 1), u(1) =
n−2 X
βi u(ξi ).
(1.3) (1.4)
i=1
They provided sufficient conditions for the existence of multiple positive solutions for the above BVPs by applying the fixed point index theory in a cone. In the papers [2,3], the authors also investigated BVP (1.1) with the boundary value conditions u(0) = 0, u(1) = Pm−2 i=1 bi u(ξi ) (which is the special case when ai = 0 in (1.2)) and (1.3), (1.4) respectively. Unfortunately, the expressions of solutions on BVPs in [2,3] are not correct, which was pointed out by the authors of [21]. In [22], the authors have given the correct conclusions for BVP (1.3), (1.4). But the conclusions in [2] have not been reconsidered so far as we know. The main difficulties are that it is not easy to define an operator, and so it is more difficult to apply the fixed point theorem. Motivated by the works in [9,21,22] mentioned above, the purpose of this paper is to show the existence of multiple positive solutions to the BVP (1.1), (1.2). To obtain positive solutions of BVP (1.1), (1.2), the following well-known Krasnoselskii’s fixed point theorem in a cone is useful. Theorem 1.1 ([6]). Let E be a Banach space, and K ⊂ E be a cone. Assume Ω1 , Ω2 are open bounded subsets of E with 0 ∈ Ω1 , Ω 1 ⊂ Ω2 , and let T : K ∩ (Ω 2 \ Ω1 ) −→ K be a completely continuous operator such that
H. Feng et al. / Nonlinear Analysis 68 (2008) 2269–2279
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(i) kT uk ≤ kuk, u ∈ K ∩ ∂Ω1 , and kT uk ≥ kuk, u ∈ K ∩ ∂Ω2 ; or (ii) kT uk ≥ kuk, u ∈ K ∩ ∂Ω1 , and kT uk ≤ kuk, u ∈ K ∩ ∂Ω2 , Then T has at least one fixed point in K ∩ (Ω 2 \ Ω1 ). 2. Preliminaries In this paper, we assume that (H1 ) and (H2 ) hold and denote C + [1, 0] = {x ∈ C[0, 1] : x(t) ≥ 0, t ∈ [0, 1]}. is the inverse function of φ p . For any x ∈ C + [0, 1], suppose u is a solution of the following BVP
φ −1 p
(φ p (u 0 ))0 + q(t) f (t, x) = 0, u(0) =
m−2 X
ai u(ξi ),
t ∈ (0, 1), m−2 X
u(1) =
i=1
(2.1)
bi u(ξi ).
(2.2)
i=1
Then
t
Z
q(τ ) f (τ, x(τ ))dτ Ax − 0 Z t u(t) = u(0) + I (s, x)ds u (t) = 0
φ −1 p
=: I (t, x),
0
or 1
Z
I (s, t)ds.
u(t) = u(1) − t
By the boundary condition (2.2), we have Z t m−2 X Z ξi 1 u(t) = ai I (s, x)ds + I (s, x)ds m−2 P 0 0 i=1 1− ai i=1
or u(t) = − 1−
1 m−2 P
m−2 X
bi
1
Z bi
ξi
i=1
I (s, x)ds −
1
Z
I (s, x)ds.
t
i=1
Here, A x satisfies the equation 1−
m−2 P
1−
i=1 m−2 P
Hx (c) =
bi ai
m−2 X
ξi
Z ai 0
i=1
φ −1 p
s
Z c−
q(τ ) f (τ, x(τ ))dτ ds
0
i=1
+
1−
m−2 X
!Z bi 0
i=1
+
m−2 X i=1
Z bi
1 ξi
1
φ −1 p
φ −1 p
c−
c−
q(τ ) f (τ, x(τ ))dτ ds
0 s
Z
s
Z
q(τ ) f (τ, x(τ ))dτ ds = 0.
(2.3)
0
Lemma 2.1. Assume that (H1 ), (H2 ) hold. For any x ∈ C + [0, R σ1], there exists a unique A x ∈ (−∞, +∞) satisfying (2.3). Moreover, there is a unique σx ∈ (0, 1) such that A x = 0 x q(τ ) f (τ, x(τ ))dτ .
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Proof. For any x ∈ C + [0, 1], by the definition of Hx (c), it is easy to know that Hx : R → R is continuous and strictly increasing, and ! Z 1 Hx (0) < 0, Hx q(τ ) f (τ, x(τ ))dτ > 0. 0
R 1 Therefore, there exists a unique A x ∈ 0, 0 q(τ ) f (τ, x(τ ))dτ ⊂ (−∞, +∞) satisfying (2.3). Furthermore, let Z t q(τ ) f (τ, x(τ ))dτ, F(t) = 0
then F(t) is continuous and strictly increasing on [0, 1], and F(0) = 0, F(1) = Z 1 q(τ ) f (τ, x(τ ))dτ. 0 = F(0) < A x < F(1) =
R1 0
q(τ ) f (τ, x(τ ))dτ . So
0
It follows that there exists a unique σx ∈ (0, 1) such that A x =
R σx 0
q(τ ) f (τ, x(τ ))dτ by the property of F(t).
Lemma 2.2. Assume that (H1 ), (H2 ) hold. For any x ∈ C + [0, 1], the unique solution u of (2.1), (2.2) has the following properties (i) the graph of u(t) is concave; (ii) u(t) ≥ 0; (iii) there exists a unique t0 ∈ (0, 1) such that u(t0 ) = max0≤t≤1 u(t), moreover u 0 (t0 ) = 0; (iv) σx = t0 . Proof. Suppose that u(t) is the solution of (2.1) and (2.2). (i) From the fact that (φ p (u 0 ))0 (t) = −q(t) f (t, x(t)) ≤ 0, we know that φ p (u 0 (t)) is nonincreasing. It follows that u 0 (t) is also nonincreasing. Thus, the graph of u(t) is concave down on (0, 1). (ii) Without loss of generality, we assume that u(0) = min{u(0), u(1)}. By the concavity of u, we know that u(ξi ) ≥ u(0) (i = 0, 1, 2, . . . , n). So, we get u(0) =
m−2 X
ai u(ξi ) ≥
i=1
m−2 X
ai u(0).
i=1
Pm−2 By 1 − i=1 ai > 0, it is obvious that u(0) ≥ 0. Hence, u(1) ≥ u(0) ≥ 0. So, from the concavity of u, we know that u(t) ≥ 0, t ∈ [0, 1]. (iii) Since u is strictly concave, there is a unique point t0 where the maximum is attained. From the boundary conditions (1.2), this cannot happen at 0 or at 1, i.e., t0 ∈ (0, 1) such that u(t0 ) = max0≤t≤1 u(t) and then u 0 (t0 ) = 0. R σx 0 (iv) By Lemma 2.1, it is easy to check u 0 (t) = φ −1 p t q(τ ) f (τ, x(τ ))dτ . Hence, we obtain that u (σx ) = 0, this implies σx = t0 . Lemma 2.3. Assume that (H1 ), (H2 ) hold. For x ∈ C + [0, 1] and 0 < γ < min{ξ1 , 1 − ξm−2 }, so then the unique solution u(t) of BVP (2.1), (2.2) satisfies mint∈[γ ,1−γ ] u(t) ≥ γ kuk. Proof. By Lemma 2.2, we know the unique solution u(t) of BVP (2.1), (2.2) is continuous, nonnegative and concave on [0, 1], then u(t) ≥ min{t, 1 − t}kuk. Hence mint∈[γ ,1−γ ] u(t) ≥ γ kuk. For any x(t) ∈ C + [0, 1], define an operator Z t m−2 X Z ξi 1 a I (s, x)ds + I (s, x)ds, 0 ≤ t ≤ σx , i m−2 P 0 0 i=1 ai 1 − i=1 (T x)(t) = (2.4) Z 1 m−2 X Z 1 1 − bi I (s, x)ds − I (s, x)ds, σx ≤ t ≤ 1. m−2 P ξi t i=1 bi 1− i=1
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Furthermore, we have the following result. Lemma 2.4. Let K = ω : ω ∈ C + [0, 1],
min
t∈[γ ,1−γ ]
ω(t) ≥ γ kωk ,
then T : K −→ K is completely continuous. Proof. Here we only prove the case 0 ≤ t ≤ σx . The case σx ≤ t ≤ 1 can be proved similarly. It is easy to check that K is a cone in C[0, 1]. According to Lemmas 2.2 and 2.3, we easily obtain (T x) ∈ C + [0, 1]
and
min (T x)(t) ≥ γ kT xk
t∈[γ ,1−γ ]
for x ∈ K .
This means that T K ⊂ K . Now we show that T : K −→ K is completely continuous. (1) We show that T is continuous. First, we prove that A x is continuous in x. Suppose {xn } ∈ C + [0, 1] with xn −→ x0 ∈ C + [0, 1]. Let {An } (n = 0, 1, 2, . . .) be constants decided by (2.3) corresponding to {xn } (n = 0, 1, 2, . . .). As xn −→ x0 uniformly on [0, 1] and f : [0, 1] × [0, ∞) −→ (0, ∞) is continuous, we have that for ε = 1, there exists N > 0, when n > N , for any τ ∈ [0, 1], 0 ≤ f (τ, xn (τ )) ≤ 1 + f (τ, x0 (τ )) ≤ 1 + max f (τ, x0 (τ )).
(2.5)
0≤τ ≤1
So, 1
Z An ∈ 0,
! q(τ ) f (τ, xn (τ ))dτ
⊆ 0, 1 + max f (τ, x0 (τ )) 0≤τ ≤1
0
1
Z
! q(τ )dτ ,
0
which implies An is uniformly bounded. (1) (2) (1) Assume {An } does not converge to A0 . Then, there exist two subsequences {An k } and {An k } of {An } with An k −→ (2) c1 and An k −→ c2 , since An is uniformly bounded, but c1 6= c2 . By the construction of {An } (n = 0, 1, 2, . . .), we (1) have Hx (1) (An k ) = 0. Combining (2.5) and using Lebesgue’s dominated convergence theorem, we get nk
lim H (1) (A(1) nk ) n k →∞ xn k
=H
(1)
lim xn k
n k →∞
lim A(1) n k →∞ n k
= Hx0 (c1 ) = 0.
Since {An } (n = 0, 1, 2 . . .) is unique, we have c1 = A0 . Similarly, c2 = A0 . Hence, c1 = c2 , which is a contradiction. So, for any xn −→ x0 , An −→ A0 , which means A x : C + [0, 1] −→ R is continuous. Therefore the continuity of T is obvious. (2) We prove T is compact. Let Ω ⊂ K be a bounded set. Then, there exists R, such that Ω ⊂ {x ∈ K | kxk ≤ R}. For any x ∈ Ω , we have Z 1 Z 1 0≤ q(τ ) f (τ, x(τ ))dτ ≤ max f (τ, x) q(τ )dτ =: M. τ ∈[0,1],x∈[0,R]
0
0
So we get |A x | ≤ M. Therefore, by the definition of T , one has kT xk ≤
φ −1 p (2M) 1−
m−2 P
,
k(T x)0 k ≤ φ −1 p (2M).
ai
i=1
Then T Ω is uniformly bounded and equicontinuous. By the Arzel`a–Ascoli’s theorem, T Ω is relatively compact and therefore that T is compact.
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Remark 2.1. By Lemma 2.1, (2.4) can be rewritten as the following Z σx m−2 X Z ξi 1 −1 q(τ ) f (τ, x(τ ))dτ ds a φ i p m−2 P s 0 i=1 1− ai i=1 Z t Z σx −1 + φ q(τ ) f (τ, x(τ ))dτ ds, p s 0 m−2 (T x)(t) = P Z s i=1 bi Z 1 −1 φ q(τ ) f (τ, x(τ ))dτ ds p m−2 P ξi σx 1 − bi i=1 Z s Z 1 + q(τ ) f (τ, x(τ ))dτ ds, φ −1 t
p
σx
0 ≤ t ≤ σx , (2.6)
σx ≤ t ≤ 1.
Now for convenience we use the following notations. Let f (t, u) f γ R,R = min min : u ∈ [γ R, R] , t∈[γ ,1−γ ] φ p (R) f (t, u) 0,r : u ∈ [0, r ] , f = max max t∈[0,1] φ p (r ) f (t, u) , f α = lim sup max u→α t∈[0,1] φ p (u) f (t, u) f α = lim inf min , (α := ∞, or 0+ ), u→α t∈[γ ,1−γ ] φ p (u) ξi + ξi+1 , (i = 0, 1, . . . , m − 2), ti∗ = 2 ! Z 1 1 1 −1 = φp q(τ )dτ , m−2 l P 0 1− ai i=1
1 = min Li
(Z
ti∗ ξi
φ −1 p
ti∗
Z
! q(τ )dτ ds,
s
Z
ξi+1 ti∗
φ −1 p
Z
s ti∗
!
)
q(τ )dτ ds ,
1 1 = min , 0≤i≤m−2 L i L where 0 < γ = ξ0 < ξ1 < ξ2 < · · · < ξm−2 < ξm−1 = 1 − γ < 1. Remark 2.2. Instead of ti∗ =
ξi +ξi+1 , 2
we can take any ti∗ in (ξi , ξi+1 ).
3. The existence of one positive solution We now give our results for the existence of a positive solution of BVP (1.1) and (1.2). Theorem 3.1. Assume that there exist constants r, R > 0 with r < γ R (or L R < lr ), such that the following conditions hold: (H3 ) f 0,r ≤ φ p (l), (H4 ) f γ R,R ≥ φ p (L). Then BVP (1.1), (1.2) has at least one positive solution u such that 0 < r ≤ kuk ≤ R (or 0 < R ≤ kuk ≤ r ).
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Proof. It is obvious that BVP (1.1) and (1.2) has a solution u = u(t) if and only if u solves operator equation T u = u. First, we assume r < γ R. Let K r := {u ∈ K : kuk < r }; then ∂ K r = {u ∈ K : kuk = r }. By (2.6) and (H3 ), for u ∈ ∂ K r we have kT uk = (T u)(σu ) = 1−
1 m−2 P
m−2 X
φ −1 p
ai 0
i=1
ai
ξi
Z
σx
Z
q(τ ) f (τ, x(τ ))dτ ds +
σx
Z 0
s
φ −1 p
Z
σx
q(τ ) f (τ, x(τ ))dτ ds
s
i=1 m−2 P
ai i=1 m−2 P
≤
1−
1
Z 0
ai
1
Z
φ −1 p
! q(τ ) f (τ, x(τ ))dτ ds +
1
Z 0
0
φ −1 p
1
Z
! q(τ ) f (τ, u(τ ))dτ ds
0
i=1
≤ 1−
1 m−2 P
1
Z
φ −1 p
! q(τ ) f (τ, x(τ ))dτ ds
0
ai
i=1
≤ 1−
1 m−2 P
1
Z
rlφ −1 p
! q(τ )dτ
0
ai
i=1
= r = kuk. This implies that kT uk ≤ kuk for u ∈ ∂ K r . Let K R := {u ∈ K : kuk < R}; then ∂ K R = {u ∈ K : kuk = R}. For u ∈ ∂ K R , it follows that γ R ≤ u ≤ R, γ ≤ t ≤ 1 − γ . We consider two cases. (i) u ∈ ∂ K R , ti∗ ≤ σu . By (2.6), (H4 ) and noticing (T u)(0) ≥ 0, we obtain Z σu Z t∗ i ∗ −1 (T u)(ti ) ≥ φp q(τ ) f (τ, u(τ ))dτ ds s 0 ! Z ∗ Z ∗ ti
≥ ξi
ti
φ −1 p ti∗
Z ≥ RL
ξi
q(τ ) f (τ, u(τ ))dτ ds
s ti∗
Z
φ −1 p
! q(τ )dτ ds
s
≥ R = kuk. (ii) u ∈ ∂ K R , σu < ti∗ . By (2.6), (H4 ), and noticing (T u)(1) ≥ 0 from Lemma 2.3, we arrive at Z s Z 1 ∗ −1 (T u)(ti ) ≥ φp q(τ ) f (τ, u(τ ))dτ ds σu
ti∗
Z
ξi+1
≥ ti∗
Z
φ −1 p ξi+1
≥ RL ti∗
s
Z
ti∗
φ −1 p
! q(τ ) f (τ, u(τ ))dτ ds Z
s ti∗
! q(τ )dτ ds
≥ R = kuk. This implies that kT uk ≥ kuk for u ∈ ∂ K R . Therefore, by the first part of Krasnoselskii’s fixed-point theorem, it follows that T has a fixed-point u in K ∩ (K R \ K r ). That is BVP (1.1), (1.2) have at least one positive solution such that 0 < r ≤ kuk ≤ R.
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Now we assume L R < lr . By a similar proof, one easily gets kT uk ≥ kuk for u ∈ ∂ K R and kT uk ≤ kuk for u ∈ ∂ K r . Therefore, by the second part of Krasnoselskii’s fixed-point theorem, it follows that T has a fixed-point u in K ∩ (K r \ K R ). That is, BVP (1.1), (1.2) have at least one positive solution such that 0 < R ≤ kuk ≤ r . Similarly, we can obtain the following conclusion. Theorem 3.1*. Assume that there exist constants r, R > 0 with r < γ R (or L R < lr ), such that the following conditions hold: (H3 )∗ f 0,r < φ p (l), (H4 )∗ f γ R,R > φ p (L), Then BVP (1.1) and (1.2) has at least one positive solution u such that 0 < r < kuk < R (or 0 < R < kuk < r ). Corollary 3.1. Assume the following conditions hold: (H5 ) f 0 ≤ φ p (l) and f ∞ ≥ φ p ( γL ),
(H6 ) f 0 ≥ φ p ( γL ), and f ∞ ≤ φ p (l). Then BVP (1.1), (1.2) have at least one positive solution. Proof. We show (H5 ) implies (H3 ) and (H4 ). Suppose (H5 ) holds; then there exist r and R with 0 < r < γ R, such that f (t, u) ≤ φ p (l), 0 ≤ t ≤ 1, 0 < u ≤ r φ p (u) and f (t, u) ≥ φp φ p (u)
L , γ
γ ≤ t ≤ 1 − γ , u ≥ γ R.
Hence, we obtain f (t, u) ≤ φ p (l)φ p (u) ≤ φ p (l)φ p (r ) = φ p (rl),
0 ≤ t ≤ 1, 0 < u ≤ r
and f (t, u) ≥ φ p
L L φ p (u) ≥ φ p φ p (γ R) = φ p (R L), γ γ
γ ≤ t ≤ 1 − γ , γ R ≤ u ≤ R.
Hence, (H3 ) and (H4 ) hold. Therefore, by Theorem 3.1, BVP (1.1) and (1.2) have at least one positive solution. Now suppose (H6 ) holds; then there exist 0 < r < R with Lr < l R such that L f (t, u) ≥ φp , γ ≤ t ≤ 1 − γ, 0 < u ≤ r φ p (u) γ f (t, u) ≤ φ p (l), φ p (u)
0 ≤ t ≤ 1, u ≥ R.
By (3.1), it follows that L L f (t, u) ≥ φ p φ p (u) ≥ φ p φ p (γ r ) = φ p (Lr ), γ γ
(3.1) (3.2)
for γ ≤ t ≤ 1 − γ , γ r ≤ u ≤ r.
So, the condition (H4 ) holds for r . We consider two cases for (3.2). (i) f (t, u) is bounded. In this case, there exists a constant M > 0 such that f (t, u) ≤ M, nfor 0 ≤ t ≤ o1 and 0 ≤ u < ∞. By (3.2), there exists a constant λ ≥ R with Lr < l R ≤ λl satisfying φ p (r ) ≥ max φ p (R), φ pM(l) such that f (t, u) ≤ M ≤ φ p (λl), for 0 ≤ t ≤ 1, 0 ≤ u ≤ λ, that is the condition (H3 ) holds for λ.
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(ii) f (t, u) is unbounded. In this case, there exist t0 ∈ [0, 1] and λ1 ≥ R, with Lr < l R ≤ λ1l such that f (t, u) ≤ f (t0 , λ1 ) for 0 ≤ t ≤ 1, 0 ≤ u ≤ λ1 . This yields f (t, u) ≤ f (t0 , λ1 ) ≤ φ p (λ1l) for 0 ≤ t ≤ 1, 0 ≤ u ≤ λ1 , and then the condition (H3 ) holds for λ1 . Therefore, the conclusion follows from Theorem 3.1. Remark 3.1. It is easy to see that Corollary 3.1 holds if f satisfies condition f 0 = 0, f ∞ = ∞ or f 0 = ∞, f ∞ = 0. 4. The existence of multiple positive solutions In this section, we give some conclusions about the existence of multiple positive solutions. Theorem 4.1. Assume the following conditions hold: (H3 )∗ f 0,r < φ p (l), (H7 ) f 0 ≥ φ p ( γL ) and f ∞ ≥ φ p ( γL ). Then BVP (1.1), (1.2) have at least two positive solutions such that 0 < ku 1 k < r < ku 2 k. Proof. By the proof of Corollary 3.1, we can take 0 < r1 < r < γ r2 such that f (t, u) ≥ φ p (r1 L), for γ ≤ t ≤ 1 − γ , γ r1 ≤ u ≤ r1 and f (t, u) ≥ φ p (r2 L), for γ ≤ t ≤ 1 − γ , γ r2 ≤ u ≤ r2 . So, by Theorem 3.1, it follows that BVP (1.1), (1.2) have at least two positive solutions such that 0 < ku 1 k < r < ku 2 k. Theorem 4.2. Assume the following conditions hold: (H4 )∗ f γ R,R > φ p (L), (H8 ) f 0 ≤ φ p (M) and f ∞ ≤ φ p (M). Then BVP (1.1), (1.2) have at least two positive solutions such that 0 < ku 1 k < R < ku 2 k. Theorem 4.3. Assume (H5 ) (or (H6 )) holds, and there exist constants r1 , r2 > 0 with r1 L < r2l (or r1 < γ r2 ) such that (H3 )∗ holds for r = r2 (or r = r1 ) and (H4 )∗ holds for R = r1 (or R = r2 ). Then BVP (1.1), (1.2) have at least three positive solutions such that 0 < ku 1 k < r1 < ku 2 k < r2 < ku 3 k. The proof of Theorems 4.2 and 4.3 is similar to that provided for Theorem 4.1. We omit it here. From Theorems 4.1–4.3, it is easy to see that if conditions (H3 )∗ , (H4 )∗ , (H5 )–(H8 ) are combined properly, then BVP (1.1) and (1.2) has arbitrarily many positive solutions. Theorem 4.4. Let n = 2k + 1, k ∈ N . Assume (H5 ) (or (H6 )) holds. If there exist constants r1 , r2 , . . . , rn−1 > 0 with r2i < γ r2i+1 , for 1 ≤ i ≤ k − 1 and r2i−1 L < r2i l, for 1 ≤ i ≤ k (or with r2i−1 < γ r2i , for 1 ≤ i ≤ k and r2i L < r2i+1l, for 1 ≤ i ≤ k − 1) such that (H4 )∗ (or (H3 )∗ ) holds for r2i−1 , 1 ≤ i ≤ k and (H3 )∗ (or (H4 )∗ ) holds for r2i , 1 ≤ i ≤ k. Then BVP (1.1), (1.2) have at least n positive solutions u 1 , u 2 , . . . , u n such that 0 < ku 1 k < r1 < ku 2 k < r2 < · · · < ku n−1 k < rn−1 < ku n k. Theorem 4.5. Let n = 2k, k ∈ N . Assume (H7 ) (or (H8 )) holds. If there exist constants r1 , r2 , . . . , rn−1 > 0 with r2i−1 < γ r2i and r2i L < r2i+1l, for 1 ≤ i ≤ k − 1 (or with r2i < γ r2i+1 and r2i−1 L < r2i l, for 1 ≤ i ≤ k − 1) such that (H3 )∗ (or (H4 )∗ ) holds for r2i−1 , 1 ≤ i ≤ k and (H4 )∗ (or (H3 )∗ ) holds for r2i , 1 ≤ i ≤ k − 1. Then BVP (1.1) and (1.2) have at least n positive solutions u 1 , u 2 , . . . , u n such that 0 < ku 1 k < r1 < ku 2 k < r2 < · · · < ku n−1 k < rn−1 < ku n k. 5. Examples Example 5.1. Consider the following BVP (|u 0 (t)|u 0 (t))0 + e−t g(u) = 0, 1 1 1 2 u(0) = u + u , 4 3 4 3
t ∈ (0, 1),
(5.1)
1 1 1 2 u(1) = u + u , 3 3 3 3
(5.2)
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where 2 u , 0 ≤ u ≤ 1, 1 679 615u − 1 679 614, 1 ≤ u ≤ 2, g(u) = 1 679 616, 2 ≤ u ≤ 12, u 2 10u + 1 679 496 , u ≥ 12. u + 132 √ Choose r = 1/2, R = 12, take γ = 1/6. Then it is easily checked that l = 2; L = 36 3. One has r < γ R and f (t, u) = e−t u 2 ≤ 1 = φ3 (rl), 0 ≤ t ≤ 1, 0 ≤ u ≤ 1; √ 1 679 616 1 5 f (t, u) = 1 679 616e−t > = (36 3 × 12)2 = φ3 (L R), ≤ t ≤ , 2 ≤ u ≤ 12. 3 6 6 Then all conditions of Theorem 3.1 hold. Thus BVP (5.1), (5.2) have at least one positive solution u such that 1 ≤ kuk ≤ 12. Example 5.2. In Example 5.1 replace g there by 2 u + 3, 0 ≤ u ≤ 1, 140u − 136, 1 ≤ u ≤ 2, g(u) = 144, 2 ≤ u ≤ 6, u 2 10u + 84 , u ≥ 6. u + 30 Choosing r = 6, R =
1√ , 36 3
one obtains R L < rl and
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