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Several existence theorems of nonlinear m-point boundary value problem for p-Laplacian dynamic equations on time scales
J. Math. Anal. Appl. 340 (2008) 1012–1026 www.elsevier.com/locate/jmaa
Several existence theorems of nonlinear m-point boundary value problem for p-Laplacian dynamic equations on time scales ✩ Yanbin Sang a,∗ , Hua Su b a Department of Mathematics, North University of China, Taiyuan 030051, China b Department of Mathematics, Ocean University of China, Qingdao Shandong 266071, China
Received 19 December 2006 Available online 25 September 2007 Submitted by T.D. Benavides
Abstract In this paper, several existence theorems of positive solutions are established for nonlinear m-point boundary value problem for p-Laplacian dynamic equations on time scales, as an application, an example to demonstrate our results is given. The conditions we used in the paper are different from those in [H.R. Sun, W.T. Li, Positive solutions for nonlinear three-point boundary value problems on time scales, J. Math. Anal. Appl. 299 (2004) 508–524; H.R. Sun, W.T. Li, Positive solutions for nonlinear m-point boundary value problems on time scales, Acta Math. Sinica 49 (2006) 369–380 (in Chinese); Y. Wang, C. Hou, Existence of multiple positive solutions for one-dimensional p-Laplacian, J. Math. Anal. Appl. 315 (2006) 144–153; Y. Wang, W. Ge, Positive solutions for multipoint boundary value problems with one-dimensional p-Laplacian, Nonlinear Appl. 66 (6) (2007) 1246–1256]. © 2007 Elsevier Inc. All rights reserved. Keywords: Time scale; Positive solutions; Boundary value problem; Fixed point theorems
1. Introduction In this paper, we are concerned with the existence of positive solutions of the p-Laplacian dynamic equations on time scales ∇ + a(t)f t, u(t) = 0, t ∈ (0, T ), (1.1) φp u m−2 φp u (0) = ai φp u (ξi ) , i=1
u(T ) =
m−2
(1.2)
bi u(ξi ),
i=1
where φp (s) is p-Laplacian operator, i.e., φp (s) = |s|p−2 s, p > 1, φp−1 = φq , ξm−2 < ρ(T ), and ai , bi , a, f satisfy
1 p
+
1 q
= 1, 0 < ξ1 < · · · <
✩ Project supported by the National Natural Science Foundation of China (10771117), the Scientific Startup Foundation of Ocean University of China. * Corresponding author. E-mail address: [email protected] (Y. Sang).
0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.09.029
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
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m−2 m−2 m−2 (H1 ) ai , bi ∈ [0, +∞) satisfy 0 < m−2 i=1 ai < 1, and i=1 bi < 1, T i=1 bi i=1 bi ξi ; (H2 ) a(t) ∈ Cld ([0, T ], [0, +∞)) and there exists t0 ∈ (ξm−2 , T ), such that a(t0 ) > 0; (H3 ) f ∈ C([0, T ] × [0, +∞), [0, +∞)). Recently, there is much attention paid to the existence of positive solution for three point boundary value problems on time scales, see [1–5] and references therein. However, there are not many results concerning the p-Laplacian problems on time scales. A time scale T is a nonempty closed subset of R. We make the blanket assumption that 0, T are points in T. By an interval (0, T ), we always mean the intersection of the real interval (0, T ) with the given time scale; that is (0, T ) ∩ T. In [6,7], Z. He considered the existence of positive solutions of the p-Laplacian dynamic equations on time scales ∇ + a(t)f u(t) = 0, t ∈ (0, T ), (1.3) φp u satisfying the boundary conditions u(0) − B0 u (η) = 0, u (T ) = 0,
(1.4)
or u (0) = 0,
u(T ) − B1 u (η) = 0,
(1.5)
where η ∈ (0, ρ(T )). He obtained the existence of at least double and triple positive solutions of the problem (1.3)– (1.5) by using a new double fixed point theorem and triple fixed point theorem, respectively. In recent papers, D. Ma, Z. Du and W. Ge [8] have obtained the existence of monotone positive solutions for the following BVP: φp (u ) + a(t)f t, u(t) = 0, t ∈ (0, 1), (1.6) u (0) =
m−2
ai u (ξi ),
i=1
u(1) =
m−2
(1.7)
bi u(ξi ).
i=1
The main tool is the monotone iterative technique. We note that the problem (1.1) and (1.2) have been studied by Y. Wang and C. Hou [9] when T = R, (0, T ) = (0, 1) and a(t) ≡ 1. Let f (t, u) ρ : u ∈ [γρ, ρ] , = min min fγρ ξm−2 tT φp (ρ) f (t, u) ρ f0 = max max : u ∈ [0, ρ] , 0tT φp (ρ) am−2 (T − ξm−2 ) am−2 ξm−2 ξ1 γ = min , , , T − am−2 ξm−2 T T
m−2 m−2 q m−2 q −1
a ξ a ξ 1 1 i i i i i=1 i=1 1+ m= − bi ξi + , m−2 m−2 q 1 − m−2 b 1 − a 1 − i i i=1 i=1 i=1 ai i=1 m−2
m−2 m−2 q m−2 q −1
1 1 i=1 ai ξi i=1 ai ξi bi 1 + − bi ξi + . M= q 1 − m−2 1 − m−2 1 − m−2 i=1 bi i=1 i=1 ai i=1 ai i=1 They mainly obtained the following results. Theorem 1.1. Assume that one of the following conditions holds: (H4 ) There exist ρ1 , ρ2 , ρ3 ∈ (0, +∞) with ρ1 < γρ2 and ρ2 < ρ3 , such that ρ
f0 1 φp (m),
ρ2 fγρ φp (Mγ ) 2
ρ
and f0 3 φp (m);
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
(H5 ) There exist ρ1 , ρ2 , ρ3 ∈ (0, +∞) with ρ1 < ρ2 < γρ3 , such that ρ
f0 2 φp (m),
ρ1 fγρ φp (Mγ ) 1
ρ3 and fγρ φp (Mγ ). 3
Then (1.1), (1.2) has two positive solutions. Y. Wang and W. Ge [10] also considered the BVP (1.6), (1.7). They obtained the similar results. In this paper, we will establish two new theorems of twin positive solutions of (1.1) and (1.2), our work concentrates on the case when the nonlinear term does not satisfy the conditions of Theorem 1.1. Our results generalize Theorem 1.1 and [10, Theorem 3.1]. At the end of the paper, we will give an example which illustrates that our work is true. 2. Preliminaries and lemmas For convenience, we list the following definitions which can be found in [11–14,18]. Definition 2.1. A time scale T is a nonempty closed subset of real numbers R. For t < sup T and r > inf T define the forward jump operator σ and backward jump operator ρ, respectively, by σ (t) = inf{τ ∈ T | τ > t} ∈ T, ρ(r) = sup{τ ∈ T | τ < r} ∈ T, for all t, r ∈ T. If σ (t) > t, t is said to be right scattered, and if ρ(r) < r, r is said to be left scattered; if σ (t) = t, t is said to be right dense, and if ρ(r) = r, r is said to be left dense. If T has a right scattered minimum m, define Tk = T − {m}; otherwise set Tk = T. If T has a left scattered maximum M, define Tk = T − {M}; otherwise set Tk = T. Definition 2.2. For f : T → R and t ∈ Tk , the delta derivative of f at the point t is defined to be the number f (t) (provided it exists), with the property that for each > 0, there is a neighborhood U of t, such that f σ (t) − f (s) − f (t) σ (t) − s σ (t) − s , for all s ∈ U . For f : T → R and t ∈ Tk , the nabla derivative of f at t, denoted by f ∇ (t) (provided it exists) with the property that for each > 0, there is a neighborhood U of t, such that f ρ(t) − f (s) − f ∇ (t) ρ(t) − s ρ(t) − s , for all s ∈ U . Definition 2.3. A function f is left-dense continuous (i.e. ld-continuous), if f is continuous at each left-dense point in T and its right-sided limit exists at each right-dense point in T. Definition 2.4. If φ (t) = f (t), then we define the delta integral by b f (t)t = φ(b) − φ(a). a
If
F ∇ (t) = f (t),
then we define the nabla integral by
b f (t)∇t = F (b) − F (a). a
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
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To prove the main results in this paper, we will employ several lemmas. These lemmas are based on the linear BVP ∇ φp u + h(t) = 0, t ∈ (0, T ), (2.1) m−2 ai φp u (ξi ) , φp u (0) =
u(T ) =
i=1
Lemma 2.1. If
m−2
s
t u(t) = −
φq 0
bi u(ξi ).
(2.2)
i=1
ai = 1 and
i=1
m−2
m−2 i=1
bi = 1, then for h ∈ Cld [0, T ] the BVP (2.1) and (2.2) has the unique solution
h(τ )∇τ − A s + B,
(2.3)
0
where ξ ai 0 i h(τ )∇τ , 1 − m−2 i=1 ai T s s m−2 ξi i=1 bi 0 φq ( 0 h(τ )∇τ − A)s 0 φq ( 0 h(τ )∇τ − A)s − . B= 1 − m−2 i=1 bi m−2
A=−
i=1
Proof. Let u be as in (2.3). By [13, Theorem 2.10(iii)], taking the delta derivative of (2.3), we have
t u (t) = −φq
h(τ )∇τ − A , 0
moreover, we get φp u = −
t
h(τ )∇τ − A ,
0
taking the nabla derivative of this expression yields (φp (u ))∇ = −h(t). And routine calculation verify that u satisfies the boundary value conditions in (2.2), so that u given in (2.3) is a solutionof (2.1) and (2.2). m−2 It is easy to see that BVP (φp (u ))∇ = 0, φp (u (0)) = m−2 i=1 ai φp (u (ξi )), u(T ) = i=1 bi u(ξi ) has only the trivial solution. Thus u in (2.3) is the unique solution of (2.1) and (2.2). The proof is complete. 2 Lemma 2.2. Assume (H1 ) holds, for h ∈ Cld [0, T ] and h 0, then the unique solution u of (2.1) and (2.2) satisfies u(t) 0,
for t ∈ [0, T ].
Proof. Let
s ϕ0 (s) = φq
h(τ )∇τ − A . 0
Since s h(τ )∇τ − A = 0
then ϕ0 (s) 0.
h(τ )∇τ + 0
ξ ai 0 i h(τ )∇τ 0, 1 − m−2 i=1 ai
m−2
s
i=1
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
According to Lemma 2.1, we get T m−2 ξi i=1 bi 0 ϕ0 (s)s 0 ϕ0 (s)s − u(0) = B = m−2 1 − i=1 bi T T m−2 T i=1 bi ( 0 ϕ0 (s)s − ξi ϕ0 (s)s) 0 ϕ0 (s)s − = 1 − m−2 i=1 bi T T m−2 i=1 bi ξi ϕ0 (s)s = ϕ0 (s)s + 0 1 − m−2 i=1 bi 0
and T u(T ) = −
ϕ0 (s)s + B 0
T
T =−
ϕ0 (s)s + 0
0
ξi i=1 bi 0 1 − m−2 i=1 bi m−2
ϕ0 (s)s −
T bi ξi ϕ0 (s)s 0. 1 − m−2 i=1 bi
ϕ0 (s)s
m−2 i=1
=
If t ∈ (0, T ), we have t ϕ0 (s)s +
u(t) = − 0
1−
T −
ϕ0 (s)s + 0
=
=
1−
1−
1 m−2 i=1
1 m−2 i=1
So u(t) 0, t ∈ [0, T ].
1−
1 m−2 i=1
1 m−2 i=1
− 1−
bi
m−2
T bi
bi
bi
i=1
ϕ0 (s)s −
m−2 i=1
0
T
bi
ϕ0 (s)s 0 ξi
bi
ϕ0 (s)s
0
T ϕ0 (s)s +
bi
ξi
i=1
0
T
i=1 m−2
ϕ0 (s)s −
m−2
0
ϕ0 (s)s − 0
m−2 i=1
ξi bi
ϕ0 (s)s 0
T ϕ0 (s)s 0.
bi ξi
2
Lemma 2.3. Assume (H1 ) holds, if h ∈ Cld [0, T ] and h 0, then the unique solution u of (2.1) and (2.2) satisfies inf u(t) γ u ,
t∈[0,T ]
where γ=
m−2
bi (T − ξi ) , m−2 T − i=1 bi ξi i=1
u = max u(t). t∈[0,T ]
Proof. It is easy to check that u (t) = −ϕ0 (t) 0, this implies that u = u(0),
min u(t) = u(T ).
t∈[0,T ]
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It is easy to see that u (t2 ) u (t1 ) for any t1 , t2 ∈ [0, T ] with t1 t2 . Hence u (t) is a decreasing function on [0, T ]. This means that the graph of u (t) is concave down on (0, T ). For each i ∈ {1, 2, . . . , m − 2}, we have u(T ) − u(0) u(T ) − u(ξi ) , T −0 T − ξi i.e., T u(ξi ) − ξi u(T ) (T − ξi )u(0), so that T
m−2
bi u(ξi ) −
i=1
m−2
bi ξi u(T )
i=1
m−2
bi (T − ξi )u(0).
i=1
With the boundary condition u(T ) = m−2 bi (T − ξi ) u(0). u(T ) i=1m−2 T − i=1 bi ξi
m−2 i=1
bi u(ξi ), we have
2
This completes the proof.
Let the norm on Cld [0, T ] be the maximum norm. Then the Cld [0, T ] is a Banach space. It is easy to see that the BVP (1.1) and (1.2) has a solution u = u(t) if and only if u is a fixed point of the operator equation t s s + B, (Au)(t) = − φq a(τ )f τ, u(τ ) ∇τ − A 0
where = − A = B
m−2
T 0
i=1
0
ai
ξi
a(τ )f (τ, u(τ ))∇τ , 1 − m−2 i=1 ai 0
ξi s s φq ( 0 a(τ )f (τ, u(τ ))∇τ − A)s − m−2 i=1 bi 0 φq ( 0 a(τ )f (τ, u(τ ))∇τ − A)s . m−2 1 − i=1 bi
Denote
K = u u ∈ Cld [0, T ], u(t) 0,
inf u(t) γ u ,
t∈[0,T ]
where γ is the same as in Lemma 2.3. It is obvious that K is a cone in Cld [0, T ]. By Lemma 2.3, A(K) ⊂ K. So by applying Arzela–Ascoli theorem on time scales [15], we can obtain that A(K) is relatively compact. In view of Lebesgue’s dominated convergence theorem on time scales [16], it is easy to prove that A is continuous. Hence, A : K → K is completely continuous. Lemma 2.4. Let s ϕ(s) = φq
a(τ )f τ, u(τ ) ∇τ − A .
0
For ξi (i = 1, . . . , m − 2), then ξi 0
ξi ϕ(s)s T
T ϕ(s)s. 0
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
Proof. Since m−2 ξi s s a a(τ )f (τ, u(τ ))∇τ = a(τ )f τ, u(τ ) ∇τ + i=1 i 0 a(τ )f τ, u(τ ) ∇τ − A 1 − m−2 i=1 ai 0
0
0, then ϕ(s) 0. For ∀t ∈ (0, T ], we have t
t tϕ(t) − 0 ϕ(s)s 0 ϕ(s)s 0. = t tσ (t) t In fact, let ψ(t) = tϕ(t) − 0 ϕ(s)s, taking the delta derivative of the above expression, we have ψ (t) = tϕ (t) 0. Hence, ψ(t) is a nondecreasing function on [0, T ], i.e. ψ(t) 0. For ∀t ∈ (0, T ], T t ϕ(s)s 0 ϕ(s)s 0 . t T By (2.4), for ξi (i = 1, . . . , m − 2), we have ξi ϕ(s)s 0
ξi T
(2.4)
T 2
ϕ(s)s. 0
Lemma 2.5. (See [17].) Let E be a Banach space, and let K ⊂ E be a cone. Assume Ω1 , Ω2 are open bounded subsets of E with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let F : K ∩ (Ω2 \ Ω1 ) → K be a completely continuous operator such that (i) F u u , u ∈ K ∩ ∂Ω1 , and F u u , u ∈ K ∩ ∂Ω2 ; or (ii) F u u , u ∈ K ∩ ∂Ω1 , and F u u , u ∈ K ∩ ∂Ω2 . Then F has a fixed point in K ∩ (Ω2 \ Ω1 ). Throughout this paper, we will assume that 0 μ ν T . Now, we introduce the following notations. Let A0 =
1−
1 m−2 i=1
s
T bi
a(τ )∇τ +
φq 0
−1 ξ ai 0 i a(τ )∇τ , s 1 − m−2 i=1 ai
m−2
0
i=1
−1 m−2 ξi m−2 ν s T m−2 i=1 ai 0 a(τ )∇τ i=1 bi − i=1 bi ξi φq a(τ )∇τ + . s B0 = T (1 − m−2 1 − m−2 i=1 bi ) i=1 ai μ
0
For l > 0, Ωl = {u ∈ K: u < l}, ∂Ωl = {u ∈ K: u = l}, α(l) = sup Au : u ∈ ∂Ωl , β(l) = inf Au : u ∈ ∂Ωl , by Lemma 2.2, α and β are well defined.
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
3. Main results Theorem 3.1. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (A1 ) pi ∈ C([0, +∞), [0, +∞)), i = 1, 2, and p1 (l) lim l→0 l p−1
p−1
< A0
,
p2 (l) lim l→∞ l p−1
p−1
< A0
;
(A2 ) ki ∈ L1 ([0, T ], [0, +∞)), i = 1, 2; (A3 ) There exist 0 < c1 c2 , 0 λ2 < p − 1 < λ1 , such that f (t, l) p1 (l) + k1 (t)l λ1 ,
(t, l) ∈ [0, T ] × [0, c1 ],
f (t, l) p2 (l) + k2 (t)l ,
(t, l) ∈ [0, T ] × [c2 , +∞);
λ2
(A4 ) There exists b > 0, such that min f (t, l): (t, l) ∈ [μ, ν] × [γ b, b] (bB0 )p−1 . Then the problem (1.1) and (1.2) has at least two positive solutions u∗1 , u∗2 satisfying 0 < u∗1 < b < u∗2 . Theorem 3.2. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (B1 ) pi ∈ C([0, +∞), [0, +∞)), i = 3, 4, and
p−1
p−1 B0 B0 p3 (l) p4 (l) lim p−1 > , lim p−1 > ; γ γ l→∞ l l→0+ l (B2 ) ki ∈ L1 ([μ, ν], [0, +∞)), i = 3, 4; (B3 ) There exist 0 < c3 c4 , 0 λ4 < p − 1 < λ3 , such that f (t, l) p3 (l) − k3 (t)l λ3 ,
(t, l) ∈ [μ, ν] × [0, c3 ],
f (t, l) p4 (l) − k4 (t)l ,
(t, l) ∈ [μ, ν] × [c4 , +∞);
λ4
(B4 ) There exists a > 0, such that max f (t, l): (t, l) ∈ [0, T ] × [0, a] (aA0 )p−1 . Then the problem (1.1) and (1.2) has at least two positive solutions u∗3 , u∗4 satisfying 0 < u∗3 < a < u∗4 . Proof of Theorem 3.1. Let p1 (l) p−1 p2 (l) 1 p−1
= min A0 − lim p−1 , A0 − lim p−1 , l→∞ l l→0 l 2 then there exist 0 < a1 c1 , c2 a2 < +∞, such that p−1 p1 (l) A0 − l p−1 , 0 l a1 , p−1 p2 (l) A0 − l p−1 , a2 l +∞. If 0 l a1 , u ∈ ∂Ωl , then 0 u(t) l, 0 t T . By condition (A3 ), we have f t, u(t) p1 u(t) + k1 (t)uλ1 (t) p−1 A0 − up−1 (t) + k1 (t)uλ1 (t) p−1 A0 − u p−1 + k1 (t) u λ1 p−1 = A0 − l p−1 + k1 (t)l λ1 .
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
So that s
a(τ )f τ, u(τ ) ∇τ − A
0
s =
m−2
a(τ )f τ, u(τ ) ∇τ +
i=1
ai
0
s
ξi
a(τ )f (τ, u(τ ))∇τ 1 − m−2 i=1 ai 0
p−1 a(τ ) A0 − l p−1 + k1 (τ )l λ1 ∇τ +
m−2 i=1
ai
ξi 0
p−1
− )l p−1 + k1 (τ )l λ1 ]∇τ . m−2 1 − i=1 ai
a(τ )[(A0
0
Therefore = Au B
1−
1−
1−
i=1
1 m−2 i=1 i=1
ϕ(s)s −
bi
i=1
1 m−2
m−2 +
T
1 m−2
m−2 i=1
0
ξi bi
ϕ(s)s 0
T bi
ϕ(s)s 0
s
T bi 0 ξi
ai
0
φq
p−1 a(τ ) A0 − l p−1 + k1 (τ )l λ1 ∇τ
0 p−1
a(τ )[(A0 1−
− )l p−1 + k1 (τ )l λ1 ]∇τ
m−2 i=1
ai
s.
It follows that T s p−1 1 α(l) φq a(τ ) A0 − + k1 (τ )l λ1 −p+1 ∇τ m−2 l 1 − i=1 bi 0 0
m−2 ξi p−1 λ1 −p+1 ]∇τ a a(τ )[A −
+ k (τ )l i 1 0 0 + i=1 s. 1 − m−2 i=1 ai Noticing λ1 − p + 1 > 0, we have 1 α(l) lim m−2 + l l→0 1 − i=1 bi =
p−1 (A0
1−
− ) m−2 i=1
s
T φq 0
1 p−1
bi
p−1 a(τ ) A0 − ∇τ +
0
s
T 0
0
i=1
ai
p−1 a(τ )(A −
)∇τ 0 0 s 1 − m−2 i=1 ai
ξi
ξ ai 0 i a(τ )∇τ s 1 − m−2 i=1 ai
m−2 a(τ )∇τ +
φq
m−2
i=1
1 p−1 = A0 − p−1 A−1 0 1 −(p−1) p−1 = 1 − A0
< 1.
Therefore, there exists 0 < a1 < a1 , such that α(a1 ) < a1 . It implies that Au < u , u ∈ ∂Ωa1 . If a2 l < +∞ and u ∈ ∂Ωl , then 0 u(t) l. Similar to the above argument, noticing that λ2 − p + 1 < 0, we can get liml→∞ α(l) l < 1. Therefore, there exists 0 < a2 < a2 , such that α(a2 ) < a2 . It implies that Au < u , u ∈ ∂Ωa2 .
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
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On the other hand, since f : [0, T ] × [0, +∞) → [0, +∞) is continuous, by condition (A4 ), there exist a1 < b1 < b < b2 < a2 , such that min f (t, l): (t, l) ∈ [μ, ν] × [γ bi , bi ] (bi B0 )p−1 ,
i = 1, 2.
If u ∈ ∂Ωb1 , then γ b1 u(t) b1 , μ t ν. Applying Lemma 2.4, it follows that T Au = max (Au)(t) −
ϕ(s)s +
0tT
0
m−2
=
m−2
=
T
T
T
m−2 ϕ(s)s −
bi − i=1 bi ξi T (1 − m−2 i=1 bi )
i=1
T (1 −
=
ϕ(s)s 0
m−2
bi − i=1 bi ξi T (1 − m−2 i=1 bi )
T
bi ξi
m−2 i=1
bi )
ϕ(s)s 0
ϕ(s)s 0
ν
i=1
ϕ(s)s μ
m−2 ξi ν s p−1 ∇τ bi − m−2 i=1 ai 0 a(τ )(b1 B0 ) i=1 bi ξi p−1 φq a(τ )(b1 B0 ) ∇τ + s m−2 T (1 − m−2 1 − i=1 ai i=1 bi )
m−2 i=1
μ
T
0
bi
T
i=1
m−2
i=1
ξi
ξ bi 0 i ϕ(s)s 1 − m−2 i=1 bi
ϕ(s)s −
m−2
m−2
ϕ(s)s −
bi
m−2
i=1
T
i=1 bi m−2 1 − i=1 bi 0
i=1
T
m−2
T
i=1 bi m−2 1 − i=1 bi 0
1−
1 m−2
bi − m−2 i=1 bi ξi b1 B0 m−2 T (1 − i=1 bi )
m−2
0
ν
i=1
= b1 B0 B0−1
s a(τ )∇τ +
φq μ
ξ ai 0 i a(τ )∇τ s 1 − m−2 i=1 ai
m−2 i=1
0
= b1 = u .
In the same way, we can prove that if u ∈ ∂Ωb2 , then Au u . Now, we consider the operator A on Ωb1 \ Ωa1 and Ωa2 \ Ωb2 , respectively. By Lemma 2.5, we assert that the operator A has two fixed points u∗1 , u∗2 ∈ K, such that a1 u∗1 b1 , b2 u∗1 a2 . Therefore, u∗i , i = 1, 2, are positive solutions of the problem (1.1) and (1.2). 2 Proof of Theorem 3.2. Let
=
p−1
p−1 1 p3 (l) p4 (l) B0 B0 , min lim p−1 − , lim p−1 − 2 γ γ l→∞ l l→0+ l
then there exist 0 < b3 c3 , c4 b4 < +∞, such that
p3 (l)
p4 (l)
B0 γ B0 γ
p−1 p−1
+ l p−1 ,
0 l b3 ,
+ l p−1 ,
b4 l +∞.
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If 0 l b3 , u ∈ ∂Ωl , then γ l u(t) l, μ t ν. By Lemma 2.4 and condition (B3 ), we have Au = max (Au)(t) 0tT
T
T
T bi − m−2 i=1 bi ξi ϕ(s)s T (1 − m−2 i=1 bi )
m−2 i=1
m−2
m−2
bi − i=1 bi ξi T (1 − m−2 i=1 bi )
0
ν
i=1
ϕ(s)s μ
s bi − i=1 bi ξi φq a(τ ) p3 u(τ ) − k3 (τ )uλ3 ∇τ m−2 T (1 − i=1 bi ) μ 0
m−2 ξi λ3 i=1 ai 0 a(τ )[p3 (u(τ )) − k3 (τ )u ]∇τ + s 1 − m−2 i=1 ai m−2
p−1 ν s T m−2 B0 i=1 bi − i=1 bi ξi p−1 λ3 ∇τ φ a(τ ) +
(γ l) − k (τ )l q 3 γ T (1 − m−2 i=1 bi ) μ 0
m−2 ξi B0 p−1 + )(γ l)p−1 − k3 (τ )l λ3 ]∇τ i=1 ai 0 a(τ )[(( γ ) + s. 1 − m−2 i=1 ai T
m−2
m−2
ν
i=1
It follows that m−2 ν s p−1 b − b ξ i i i i=1 i=1 φq a(τ ) B0 + γ p−1 − k3 (τ )l λ3 −p+1 ∇τ m−2 T (1 − i=1 bi ) μ 0
m−2 ξi p−1 p−1 +γ
− k3 (τ )l λ3 −p+1 ]∇τ i=1 ai 0 a(τ )[B0 + s. 1 − m−2 i=1 ai
β(l) T l
m−2
Noticing λ3 − p + 1 > 0, we get ν s p−1 bi − m−2 i=1 bi ξi φq a(τ ) B0 + γ p−1 ∇τ m−2 T (1 − i=1 bi ) μ 0 m−2 ξi p−1 p−1 +γ
)∇τ i=1 ai 0 a(τ )(B0 + s 1 − m−2 i=1 ai 1 p−1 = B0 + γ p−1 p−1 B0−1 1 −(p−1) p−1 = 1 + γ p−1 B0
> 1.
β(l) T lim l→0+ l
m−2 i=1
Therefore, there exists b3 with 0 < b3 < a, such that β(b3 ) > b3 . It implies that Au > u , for u ∈ ∂Ωb3 . If b4 γ l < +∞ and u ∈ ∂Ωl , then b4 γ l u(t) l, μ t ν. Similar to the above argument, noticing that λ4 − p + 1 < 0, we can get liml→+∞ β(l) l > 1. Therefore, there exists b4 with 0 < b4 < +∞, such that β(b4 ) > b4 . It implies that Au > u , for u ∈ ∂Ωb4 . By condition (B4 ), we can see that there exist b3 < a3 < a < a4 < b4 , such that max f (t, l): (t, l) ∈ [0, T ] × [0, ai ] (ai A0 )p−1 , i = 3, 4. If u ∈ ∂Ωa3 , then 0 u(t) a3 , 0 t T , and f (t, u(t)) (a3 A0 )p−1 . It follows that
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
Au
1−
1−
1 m−2 i=1
1 m−2 i=1
1023
T bi
ϕ(s)s 0
s
T bi
a3 A0
a(τ )∇τ +
φq 0
ξ ai 0 i a(τ )∇τ s 1 − m−2 i=1 ai
m−2 i=1
0
= a3 = u . Similarly, if u ∈ ∂Ωa4 , then Au u . Now, we study the operator A on Ωa3 \ Ωb3 and Ωb4 \ Ωa4 , respectively. By Lemma 2.5, we assert that the operator A has two fixed points u∗3 , u∗4 ∈ K, such that b3 u∗1 a3 , a4 u∗1 b4 . Therefore, u∗i , i = 3, 4, are positive solutions of the problem (1.1) and (1.2). 2 4. Further discussion If the conditions of Theorems 3.1 and 3.2 are weakened, we will get the existence of single positive solution of the problem (1.1) and (1.2). Corollary 4.1. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (C1 ) p1 ∈ C([0, +∞), [0, +∞)), and liml→0+ (C2 ) k1
p1 (l) l p−1
p−1
< A0
;
∈ L1 ([0, T ], [0, +∞));
(C3 ) There exist c1 > 0, λ1 > p − 1, such that f (t, l) p1 (l) + k1 (t)l λ1 ,
(t, l) ∈ [0, T ] × [0, c1 ];
(C4 ) There exists b > 0, such that min f (t, l): (t, l) ∈ [μ, ν] × [γ b, b] (bB0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. Corollary 4.2. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (D1 ) p2 ∈ C([0, +∞), [0, +∞)), and liml→∞ (D2 ) k2
p2 (l) l p−1
p−1
< A0
;
∈ L1 ([0, T ], [0, +∞));
(D3 ) There exist c2 > 0, 0 λ2 < p − 1, such that f (t, l) p2 (l) + k2 (t)l λ2 ,
(t, l) ∈ [0, T ] × [c2 , +∞);
(D4 ) There exists b > 0, such that min f (t, l): (t, l) ∈ [μ, ν] × [γ b, b] (bB0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. Corollary 4.3. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (E1 ) p3 ∈ C([0, +∞), [0, +∞)), and liml→0+ (E2 ) k3 ∈ L1 ([μ, ν], [0, +∞));
p3 (l) l p−1
> ( Bγ0 )p−1 ;
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
(E3 ) There exist c3 > 0, λ3 > p − 1, such that f (t, l) p3 (l) − k3 (t)l λ3 ,
(t, l) ∈ [μ, ν] × [0, c3 ];
(E4 ) There exists a > 0, such that max f (t, l): (t, l) ∈ [0, T ] × [0, a] (aA0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. Corollary 4.4. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (F1 ) p4 ∈ C([0, +∞), [0, +∞)), and liml→∞
p4 (l) l p−1
> ( Bγ0 )p−1 ;
(F2 ) k4 ∈ L1 ([μ, ν], [0, +∞)); (F3 ) There exist c4 > 0, 0 λ4 < p − 1, such that f (t, l) p4 (l) − k4 (t)l λ4 ,
(t, l) ∈ [μ, ν] × [c4 , +∞);
(F4 ) There exists a > 0, such that max f (t, l): (t, l) ∈ [0, T ] × [0, a] (aA0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. The proof of the above results is similar to Theorems 3.1 and 3.2, we omit it. 5. Some examples In the section, we present a simple example to explain our results. We only study the case T = R, (0, T ) = (0, 1). Let f (t, 0) ≡ 0, Consider the following BVP:
φ3 (u ) + f t, u(t) = 0, t ∈ (0, 1),
1 1 1 1 , u(1) = u , φ3 u (0) = φ3 u 2 2 2 2
(5.1) (5.2)
where √ ⎧ 1 223u3 + min{ √t (1−t) , u2 } u5 , (t, u) ∈ [0, 1] × [0, 1], ⎪ ⎪ ⎨ (t, u) ∈ [0, 1] × [1, 3], f (t, u) = 225, √ ⎪ √ ⎪ ⎩ 74u + 3 min{ √ 1 , 2u } u3 , (t, u) ∈ [0, 1] × [3, +∞). 6 t (1−t) 3 It is easy to check that f : [0, 1] × [0, +∞) → [0, +∞) is continuous. In this case, p = 3, a(t) ≡ 1, m = 3, a1 = b1 = 12 , ξ1 = 12 , it follows from a direct calculation that A0 =
γ=
1 1−
1 1 2 0
φq s +
1 2
·
1 2 1 − 12
−1
ds
1 (1 − 1 ) 1 b1 (1 − ξ1 ) = 2 1 21 = . 1 − b1 ξ1 3 1− 2 · 2
√ √ 3( 54 + 2 ) , = 52
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
1025
Let μ = 12 , ν = 1, we have
b1 − b 1 ξ1 B0 = 1 − b1 =
1 2
−
1 2
·
1−
1 2
1 2
ν φq μ
−1 a 1 ξ1 s+ ds 1 − a1
1
s+ 1 2
1 2
·
1 2 1 − 12
−1
1 2
ds
√ 6(3 6 + 4) < 5. = 19 Choose c1 = 1, c2 = 3, b = 3, λ1 = 52 , λ2 = 32 , p1 (u) = 223u3 , p2 (u) = 74u, k1 (t) = k2 (t) = check that 5
f (t, u) p1 (u) + k1 (t)u 2 ,
√ 1 , t (1−t)
it is easy to
(t, u) ∈ [0, 1] × [0, 1],
3 2
f (t, u) p2 (u) + k2 (t)u ,
(t, u) ∈ [0, 1] × [3, +∞), √ √ p1 (u) 3( 54 + 2 ) 2 2 lim = lim = 0 < A0 = , u→0 u2 l→0 u2 52 √ √ p2 (u) 74u 3( 54 + 2 ) 2 2 lim = lim 2 = 0 < A0 = , u→∞ u2 l→∞ u 52 √ 1 3 · 6(3 6 + 4) 2 = (bB0 )2 . min f (t, u): (t, u) ∈ , 1 × [1, 3] = 225 > 2 19 223u3
It follows that f satisfies the conditions (A1 )–(A4 ) of Theorem 3.1, then problem (1.1) and (1.2) has at least two positive solutions. However, Theorem 1.1 cannot be applied to the example, hence we generalize Theorem 1.1 and [10, Theorem 3.1]. References [1] H.R. Sun, W.T. Li, Positive solutions for nonlinear three-point boundary value problems on time scales, J. Math. Anal. Appl. 299 (2004) 508–524. [2] D.R. Anderson, Solutions to second-order three-point problems on time scales, J. Difference Equ. Appl. 8 (2002) 673–688. [3] E.R. Kaufmann, Positive solutions of a three-point boundary value problem on a time scale, Electron. J. Differential Equations 82 (2003) 1–11. [4] H. Luo, Q.Z. Ma, Positive solutions to a generalized second-order three-point boundary value problem on time scales, Electron. J. Differential Equations 17 (2005) 1–14. [5] H.R. Sun, W.T. Li, Positive solutions for nonlinear m-point boundary value problems on time scales, Acta Math. Sinica 49 (2006) 369–380 (in Chinese). [6] Z.M. He, Double positive solutions of three-point boundary value problems for p-Laplacian dynamic equations on time scales, J. Comput. Appl. Math. 182 (2005) 304–315. [7] Z.M. He, Triple positive solutions of boundary value problems for p-Laplacian dynamic equations on time scales, J. Math. Anal. Appl. 321 (2006) 911–920. [8] 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. [9] Y. Wang, C. Hou, Existence of multiple positive solutions for one-dimensional p-Laplacian, J. Math. Anal. Appl. 315 (2006) 144–153. [10] Y. Wang, W. Ge, Positive solutions for multipoint boundary value problems with one-dimensional p-Laplacian, Nonlinear Appl. 66 (6) (2007) 1246–1256. [11] R.P. Agarwal, D. O’Regan, Nonlinear boundary value problems on time scales, Nonlinear Anal. 44 (2001) 527–535. [12] M. Bohner, A. Peterson, Advances in Dynamic Equations on Time Scales, Birkhäuser Boston, Cambridge, MA, 2003. [13] F.M. Atici, G.Sh. Guseinov, On Green’s functions and positive solutions for boundary value problems on time scales, J. Comput. Anal. Math. 141 (2002) 75–99. [14] M. Bohner, A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhäuser Boston, Cambridge, MA, 2001. [15] R.P. Agarwal, M. Bohner, P. Rehak, Half-linear Dynamic Equations, Nonlinear Analysis and Applications: To V. Lakshmikantham on His 80th Birthday, Kluwer Academic Publishers, Dordrecht, 2003, pp. 1–57.
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[16] B. Aulbach, L. Neidhart, Integration on measure chain, in: Proc. of the Sixth Int. Conf. on Difference Equations, CRC, Boca Raton, FL, 2004, pp. 239–252. [17] D. Guo, V. Lakshmikanthan, Nonlinear Problems in Abstract Cones, Academic Press, San Diego, 1988. [18] H. Su, B. Wang, Z. Wei, Positive solutions of four-point boundary value problems for four-order p-Laplacian dynamic equations on time scales, Electron. J. Differential Equations 78 (2006) 1–13.
Several existence theorems of nonlinear m-point boundary value problem for p-Laplacian dynamic equations on time scales ✩ Yanbin Sang a,∗ , Hua Su b a Department of Mathematics, North University of China, Taiyuan 030051, China b Department of Mathematics, Ocean University of China, Qingdao Shandong 266071, China
Received 19 December 2006 Available online 25 September 2007 Submitted by T.D. Benavides
Abstract In this paper, several existence theorems of positive solutions are established for nonlinear m-point boundary value problem for p-Laplacian dynamic equations on time scales, as an application, an example to demonstrate our results is given. The conditions we used in the paper are different from those in [H.R. Sun, W.T. Li, Positive solutions for nonlinear three-point boundary value problems on time scales, J. Math. Anal. Appl. 299 (2004) 508–524; H.R. Sun, W.T. Li, Positive solutions for nonlinear m-point boundary value problems on time scales, Acta Math. Sinica 49 (2006) 369–380 (in Chinese); Y. Wang, C. Hou, Existence of multiple positive solutions for one-dimensional p-Laplacian, J. Math. Anal. Appl. 315 (2006) 144–153; Y. Wang, W. Ge, Positive solutions for multipoint boundary value problems with one-dimensional p-Laplacian, Nonlinear Appl. 66 (6) (2007) 1246–1256]. © 2007 Elsevier Inc. All rights reserved. Keywords: Time scale; Positive solutions; Boundary value problem; Fixed point theorems
1. Introduction In this paper, we are concerned with the existence of positive solutions of the p-Laplacian dynamic equations on time scales ∇ + a(t)f t, u(t) = 0, t ∈ (0, T ), (1.1) φp u m−2 φp u (0) = ai φp u (ξi ) , i=1
u(T ) =
m−2
(1.2)
bi u(ξi ),
i=1
where φp (s) is p-Laplacian operator, i.e., φp (s) = |s|p−2 s, p > 1, φp−1 = φq , ξm−2 < ρ(T ), and ai , bi , a, f satisfy
1 p
+
1 q
= 1, 0 < ξ1 < · · · <
✩ Project supported by the National Natural Science Foundation of China (10771117), the Scientific Startup Foundation of Ocean University of China. * Corresponding author. E-mail address: [email protected] (Y. Sang).
0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.09.029
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
1013
m−2 m−2 m−2 (H1 ) ai , bi ∈ [0, +∞) satisfy 0 < m−2 i=1 ai < 1, and i=1 bi < 1, T i=1 bi i=1 bi ξi ; (H2 ) a(t) ∈ Cld ([0, T ], [0, +∞)) and there exists t0 ∈ (ξm−2 , T ), such that a(t0 ) > 0; (H3 ) f ∈ C([0, T ] × [0, +∞), [0, +∞)). Recently, there is much attention paid to the existence of positive solution for three point boundary value problems on time scales, see [1–5] and references therein. However, there are not many results concerning the p-Laplacian problems on time scales. A time scale T is a nonempty closed subset of R. We make the blanket assumption that 0, T are points in T. By an interval (0, T ), we always mean the intersection of the real interval (0, T ) with the given time scale; that is (0, T ) ∩ T. In [6,7], Z. He considered the existence of positive solutions of the p-Laplacian dynamic equations on time scales ∇ + a(t)f u(t) = 0, t ∈ (0, T ), (1.3) φp u satisfying the boundary conditions u(0) − B0 u (η) = 0, u (T ) = 0,
(1.4)
or u (0) = 0,
u(T ) − B1 u (η) = 0,
(1.5)
where η ∈ (0, ρ(T )). He obtained the existence of at least double and triple positive solutions of the problem (1.3)– (1.5) by using a new double fixed point theorem and triple fixed point theorem, respectively. In recent papers, D. Ma, Z. Du and W. Ge [8] have obtained the existence of monotone positive solutions for the following BVP: φp (u ) + a(t)f t, u(t) = 0, t ∈ (0, 1), (1.6) u (0) =
m−2
ai u (ξi ),
i=1
u(1) =
m−2
(1.7)
bi u(ξi ).
i=1
The main tool is the monotone iterative technique. We note that the problem (1.1) and (1.2) have been studied by Y. Wang and C. Hou [9] when T = R, (0, T ) = (0, 1) and a(t) ≡ 1. Let f (t, u) ρ : u ∈ [γρ, ρ] , = min min fγρ ξm−2 tT φp (ρ) f (t, u) ρ f0 = max max : u ∈ [0, ρ] , 0tT φp (ρ) am−2 (T − ξm−2 ) am−2 ξm−2 ξ1 γ = min , , , T − am−2 ξm−2 T T
m−2 m−2 q m−2 q −1
a ξ a ξ 1 1 i i i i i=1 i=1 1+ m= − bi ξi + , m−2 m−2 q 1 − m−2 b 1 − a 1 − i i i=1 i=1 i=1 ai i=1 m−2
m−2 m−2 q m−2 q −1
1 1 i=1 ai ξi i=1 ai ξi bi 1 + − bi ξi + . M= q 1 − m−2 1 − m−2 1 − m−2 i=1 bi i=1 i=1 ai i=1 ai i=1 They mainly obtained the following results. Theorem 1.1. Assume that one of the following conditions holds: (H4 ) There exist ρ1 , ρ2 , ρ3 ∈ (0, +∞) with ρ1 < γρ2 and ρ2 < ρ3 , such that ρ
f0 1 φp (m),
ρ2 fγρ φp (Mγ ) 2
ρ
and f0 3 φp (m);
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
(H5 ) There exist ρ1 , ρ2 , ρ3 ∈ (0, +∞) with ρ1 < ρ2 < γρ3 , such that ρ
f0 2 φp (m),
ρ1 fγρ φp (Mγ ) 1
ρ3 and fγρ φp (Mγ ). 3
Then (1.1), (1.2) has two positive solutions. Y. Wang and W. Ge [10] also considered the BVP (1.6), (1.7). They obtained the similar results. In this paper, we will establish two new theorems of twin positive solutions of (1.1) and (1.2), our work concentrates on the case when the nonlinear term does not satisfy the conditions of Theorem 1.1. Our results generalize Theorem 1.1 and [10, Theorem 3.1]. At the end of the paper, we will give an example which illustrates that our work is true. 2. Preliminaries and lemmas For convenience, we list the following definitions which can be found in [11–14,18]. Definition 2.1. A time scale T is a nonempty closed subset of real numbers R. For t < sup T and r > inf T define the forward jump operator σ and backward jump operator ρ, respectively, by σ (t) = inf{τ ∈ T | τ > t} ∈ T, ρ(r) = sup{τ ∈ T | τ < r} ∈ T, for all t, r ∈ T. If σ (t) > t, t is said to be right scattered, and if ρ(r) < r, r is said to be left scattered; if σ (t) = t, t is said to be right dense, and if ρ(r) = r, r is said to be left dense. If T has a right scattered minimum m, define Tk = T − {m}; otherwise set Tk = T. If T has a left scattered maximum M, define Tk = T − {M}; otherwise set Tk = T. Definition 2.2. For f : T → R and t ∈ Tk , the delta derivative of f at the point t is defined to be the number f (t) (provided it exists), with the property that for each > 0, there is a neighborhood U of t, such that f σ (t) − f (s) − f (t) σ (t) − s σ (t) − s , for all s ∈ U . For f : T → R and t ∈ Tk , the nabla derivative of f at t, denoted by f ∇ (t) (provided it exists) with the property that for each > 0, there is a neighborhood U of t, such that f ρ(t) − f (s) − f ∇ (t) ρ(t) − s ρ(t) − s , for all s ∈ U . Definition 2.3. A function f is left-dense continuous (i.e. ld-continuous), if f is continuous at each left-dense point in T and its right-sided limit exists at each right-dense point in T. Definition 2.4. If φ (t) = f (t), then we define the delta integral by b f (t)t = φ(b) − φ(a). a
If
F ∇ (t) = f (t),
then we define the nabla integral by
b f (t)∇t = F (b) − F (a). a
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
1015
To prove the main results in this paper, we will employ several lemmas. These lemmas are based on the linear BVP ∇ φp u + h(t) = 0, t ∈ (0, T ), (2.1) m−2 ai φp u (ξi ) , φp u (0) =
u(T ) =
i=1
Lemma 2.1. If
m−2
s
t u(t) = −
φq 0
bi u(ξi ).
(2.2)
i=1
ai = 1 and
i=1
m−2
m−2 i=1
bi = 1, then for h ∈ Cld [0, T ] the BVP (2.1) and (2.2) has the unique solution
h(τ )∇τ − A s + B,
(2.3)
0
where ξ ai 0 i h(τ )∇τ , 1 − m−2 i=1 ai T s s m−2 ξi i=1 bi 0 φq ( 0 h(τ )∇τ − A)s 0 φq ( 0 h(τ )∇τ − A)s − . B= 1 − m−2 i=1 bi m−2
A=−
i=1
Proof. Let u be as in (2.3). By [13, Theorem 2.10(iii)], taking the delta derivative of (2.3), we have
t u (t) = −φq
h(τ )∇τ − A , 0
moreover, we get φp u = −
t
h(τ )∇τ − A ,
0
taking the nabla derivative of this expression yields (φp (u ))∇ = −h(t). And routine calculation verify that u satisfies the boundary value conditions in (2.2), so that u given in (2.3) is a solutionof (2.1) and (2.2). m−2 It is easy to see that BVP (φp (u ))∇ = 0, φp (u (0)) = m−2 i=1 ai φp (u (ξi )), u(T ) = i=1 bi u(ξi ) has only the trivial solution. Thus u in (2.3) is the unique solution of (2.1) and (2.2). The proof is complete. 2 Lemma 2.2. Assume (H1 ) holds, for h ∈ Cld [0, T ] and h 0, then the unique solution u of (2.1) and (2.2) satisfies u(t) 0,
for t ∈ [0, T ].
Proof. Let
s ϕ0 (s) = φq
h(τ )∇τ − A . 0
Since s h(τ )∇τ − A = 0
then ϕ0 (s) 0.
h(τ )∇τ + 0
ξ ai 0 i h(τ )∇τ 0, 1 − m−2 i=1 ai
m−2
s
i=1
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
According to Lemma 2.1, we get T m−2 ξi i=1 bi 0 ϕ0 (s)s 0 ϕ0 (s)s − u(0) = B = m−2 1 − i=1 bi T T m−2 T i=1 bi ( 0 ϕ0 (s)s − ξi ϕ0 (s)s) 0 ϕ0 (s)s − = 1 − m−2 i=1 bi T T m−2 i=1 bi ξi ϕ0 (s)s = ϕ0 (s)s + 0 1 − m−2 i=1 bi 0
and T u(T ) = −
ϕ0 (s)s + B 0
T
T =−
ϕ0 (s)s + 0
0
ξi i=1 bi 0 1 − m−2 i=1 bi m−2
ϕ0 (s)s −
T bi ξi ϕ0 (s)s 0. 1 − m−2 i=1 bi
ϕ0 (s)s
m−2 i=1
=
If t ∈ (0, T ), we have t ϕ0 (s)s +
u(t) = − 0
1−
T −
ϕ0 (s)s + 0
=
=
1−
1−
1 m−2 i=1
1 m−2 i=1
So u(t) 0, t ∈ [0, T ].
1−
1 m−2 i=1
1 m−2 i=1
− 1−
bi
m−2
T bi
bi
bi
i=1
ϕ0 (s)s −
m−2 i=1
0
T
bi
ϕ0 (s)s 0 ξi
bi
ϕ0 (s)s
0
T ϕ0 (s)s +
bi
ξi
i=1
0
T
i=1 m−2
ϕ0 (s)s −
m−2
0
ϕ0 (s)s − 0
m−2 i=1
ξi bi
ϕ0 (s)s 0
T ϕ0 (s)s 0.
bi ξi
2
Lemma 2.3. Assume (H1 ) holds, if h ∈ Cld [0, T ] and h 0, then the unique solution u of (2.1) and (2.2) satisfies inf u(t) γ u ,
t∈[0,T ]
where γ=
m−2
bi (T − ξi ) , m−2 T − i=1 bi ξi i=1
u = max u(t). t∈[0,T ]
Proof. It is easy to check that u (t) = −ϕ0 (t) 0, this implies that u = u(0),
min u(t) = u(T ).
t∈[0,T ]
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
1017
It is easy to see that u (t2 ) u (t1 ) for any t1 , t2 ∈ [0, T ] with t1 t2 . Hence u (t) is a decreasing function on [0, T ]. This means that the graph of u (t) is concave down on (0, T ). For each i ∈ {1, 2, . . . , m − 2}, we have u(T ) − u(0) u(T ) − u(ξi ) , T −0 T − ξi i.e., T u(ξi ) − ξi u(T ) (T − ξi )u(0), so that T
m−2
bi u(ξi ) −
i=1
m−2
bi ξi u(T )
i=1
m−2
bi (T − ξi )u(0).
i=1
With the boundary condition u(T ) = m−2 bi (T − ξi ) u(0). u(T ) i=1m−2 T − i=1 bi ξi
m−2 i=1
bi u(ξi ), we have
2
This completes the proof.
Let the norm on Cld [0, T ] be the maximum norm. Then the Cld [0, T ] is a Banach space. It is easy to see that the BVP (1.1) and (1.2) has a solution u = u(t) if and only if u is a fixed point of the operator equation t s s + B, (Au)(t) = − φq a(τ )f τ, u(τ ) ∇τ − A 0
where = − A = B
m−2
T 0
i=1
0
ai
ξi
a(τ )f (τ, u(τ ))∇τ , 1 − m−2 i=1 ai 0
ξi s s φq ( 0 a(τ )f (τ, u(τ ))∇τ − A)s − m−2 i=1 bi 0 φq ( 0 a(τ )f (τ, u(τ ))∇τ − A)s . m−2 1 − i=1 bi
Denote
K = u u ∈ Cld [0, T ], u(t) 0,
inf u(t) γ u ,
t∈[0,T ]
where γ is the same as in Lemma 2.3. It is obvious that K is a cone in Cld [0, T ]. By Lemma 2.3, A(K) ⊂ K. So by applying Arzela–Ascoli theorem on time scales [15], we can obtain that A(K) is relatively compact. In view of Lebesgue’s dominated convergence theorem on time scales [16], it is easy to prove that A is continuous. Hence, A : K → K is completely continuous. Lemma 2.4. Let s ϕ(s) = φq
a(τ )f τ, u(τ ) ∇τ − A .
0
For ξi (i = 1, . . . , m − 2), then ξi 0
ξi ϕ(s)s T
T ϕ(s)s. 0
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
Proof. Since m−2 ξi s s a a(τ )f (τ, u(τ ))∇τ = a(τ )f τ, u(τ ) ∇τ + i=1 i 0 a(τ )f τ, u(τ ) ∇τ − A 1 − m−2 i=1 ai 0
0
0, then ϕ(s) 0. For ∀t ∈ (0, T ], we have t
t tϕ(t) − 0 ϕ(s)s 0 ϕ(s)s 0. = t tσ (t) t In fact, let ψ(t) = tϕ(t) − 0 ϕ(s)s, taking the delta derivative of the above expression, we have ψ (t) = tϕ (t) 0. Hence, ψ(t) is a nondecreasing function on [0, T ], i.e. ψ(t) 0. For ∀t ∈ (0, T ], T t ϕ(s)s 0 ϕ(s)s 0 . t T By (2.4), for ξi (i = 1, . . . , m − 2), we have ξi ϕ(s)s 0
ξi T
(2.4)
T 2
ϕ(s)s. 0
Lemma 2.5. (See [17].) Let E be a Banach space, and let K ⊂ E be a cone. Assume Ω1 , Ω2 are open bounded subsets of E with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let F : K ∩ (Ω2 \ Ω1 ) → K be a completely continuous operator such that (i) F u u , u ∈ K ∩ ∂Ω1 , and F u u , u ∈ K ∩ ∂Ω2 ; or (ii) F u u , u ∈ K ∩ ∂Ω1 , and F u u , u ∈ K ∩ ∂Ω2 . Then F has a fixed point in K ∩ (Ω2 \ Ω1 ). Throughout this paper, we will assume that 0 μ ν T . Now, we introduce the following notations. Let A0 =
1−
1 m−2 i=1
s
T bi
a(τ )∇τ +
φq 0
−1 ξ ai 0 i a(τ )∇τ , s 1 − m−2 i=1 ai
m−2
0
i=1
−1 m−2 ξi m−2 ν s T m−2 i=1 ai 0 a(τ )∇τ i=1 bi − i=1 bi ξi φq a(τ )∇τ + . s B0 = T (1 − m−2 1 − m−2 i=1 bi ) i=1 ai μ
0
For l > 0, Ωl = {u ∈ K: u < l}, ∂Ωl = {u ∈ K: u = l}, α(l) = sup Au : u ∈ ∂Ωl , β(l) = inf Au : u ∈ ∂Ωl , by Lemma 2.2, α and β are well defined.
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
3. Main results Theorem 3.1. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (A1 ) pi ∈ C([0, +∞), [0, +∞)), i = 1, 2, and p1 (l) lim l→0 l p−1
p−1
< A0
,
p2 (l) lim l→∞ l p−1
p−1
< A0
;
(A2 ) ki ∈ L1 ([0, T ], [0, +∞)), i = 1, 2; (A3 ) There exist 0 < c1 c2 , 0 λ2 < p − 1 < λ1 , such that f (t, l) p1 (l) + k1 (t)l λ1 ,
(t, l) ∈ [0, T ] × [0, c1 ],
f (t, l) p2 (l) + k2 (t)l ,
(t, l) ∈ [0, T ] × [c2 , +∞);
λ2
(A4 ) There exists b > 0, such that min f (t, l): (t, l) ∈ [μ, ν] × [γ b, b] (bB0 )p−1 . Then the problem (1.1) and (1.2) has at least two positive solutions u∗1 , u∗2 satisfying 0 < u∗1 < b < u∗2 . Theorem 3.2. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (B1 ) pi ∈ C([0, +∞), [0, +∞)), i = 3, 4, and
p−1
p−1 B0 B0 p3 (l) p4 (l) lim p−1 > , lim p−1 > ; γ γ l→∞ l l→0+ l (B2 ) ki ∈ L1 ([μ, ν], [0, +∞)), i = 3, 4; (B3 ) There exist 0 < c3 c4 , 0 λ4 < p − 1 < λ3 , such that f (t, l) p3 (l) − k3 (t)l λ3 ,
(t, l) ∈ [μ, ν] × [0, c3 ],
f (t, l) p4 (l) − k4 (t)l ,
(t, l) ∈ [μ, ν] × [c4 , +∞);
λ4
(B4 ) There exists a > 0, such that max f (t, l): (t, l) ∈ [0, T ] × [0, a] (aA0 )p−1 . Then the problem (1.1) and (1.2) has at least two positive solutions u∗3 , u∗4 satisfying 0 < u∗3 < a < u∗4 . Proof of Theorem 3.1. Let p1 (l) p−1 p2 (l) 1 p−1
= min A0 − lim p−1 , A0 − lim p−1 , l→∞ l l→0 l 2 then there exist 0 < a1 c1 , c2 a2 < +∞, such that p−1 p1 (l) A0 − l p−1 , 0 l a1 , p−1 p2 (l) A0 − l p−1 , a2 l +∞. If 0 l a1 , u ∈ ∂Ωl , then 0 u(t) l, 0 t T . By condition (A3 ), we have f t, u(t) p1 u(t) + k1 (t)uλ1 (t) p−1 A0 − up−1 (t) + k1 (t)uλ1 (t) p−1 A0 − u p−1 + k1 (t) u λ1 p−1 = A0 − l p−1 + k1 (t)l λ1 .
1019
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
So that s
a(τ )f τ, u(τ ) ∇τ − A
0
s =
m−2
a(τ )f τ, u(τ ) ∇τ +
i=1
ai
0
s
ξi
a(τ )f (τ, u(τ ))∇τ 1 − m−2 i=1 ai 0
p−1 a(τ ) A0 − l p−1 + k1 (τ )l λ1 ∇τ +
m−2 i=1
ai
ξi 0
p−1
− )l p−1 + k1 (τ )l λ1 ]∇τ . m−2 1 − i=1 ai
a(τ )[(A0
0
Therefore = Au B
1−
1−
1−
i=1
1 m−2 i=1 i=1
ϕ(s)s −
bi
i=1
1 m−2
m−2 +
T
1 m−2
m−2 i=1
0
ξi bi
ϕ(s)s 0
T bi
ϕ(s)s 0
s
T bi 0 ξi
ai
0
φq
p−1 a(τ ) A0 − l p−1 + k1 (τ )l λ1 ∇τ
0 p−1
a(τ )[(A0 1−
− )l p−1 + k1 (τ )l λ1 ]∇τ
m−2 i=1
ai
s.
It follows that T s p−1 1 α(l) φq a(τ ) A0 − + k1 (τ )l λ1 −p+1 ∇τ m−2 l 1 − i=1 bi 0 0
m−2 ξi p−1 λ1 −p+1 ]∇τ a a(τ )[A −
+ k (τ )l i 1 0 0 + i=1 s. 1 − m−2 i=1 ai Noticing λ1 − p + 1 > 0, we have 1 α(l) lim m−2 + l l→0 1 − i=1 bi =
p−1 (A0
1−
− ) m−2 i=1
s
T φq 0
1 p−1
bi
p−1 a(τ ) A0 − ∇τ +
0
s
T 0
0
i=1
ai
p−1 a(τ )(A −
)∇τ 0 0 s 1 − m−2 i=1 ai
ξi
ξ ai 0 i a(τ )∇τ s 1 − m−2 i=1 ai
m−2 a(τ )∇τ +
φq
m−2
i=1
1 p−1 = A0 − p−1 A−1 0 1 −(p−1) p−1 = 1 − A0
< 1.
Therefore, there exists 0 < a1 < a1 , such that α(a1 ) < a1 . It implies that Au < u , u ∈ ∂Ωa1 . If a2 l < +∞ and u ∈ ∂Ωl , then 0 u(t) l. Similar to the above argument, noticing that λ2 − p + 1 < 0, we can get liml→∞ α(l) l < 1. Therefore, there exists 0 < a2 < a2 , such that α(a2 ) < a2 . It implies that Au < u , u ∈ ∂Ωa2 .
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
1021
On the other hand, since f : [0, T ] × [0, +∞) → [0, +∞) is continuous, by condition (A4 ), there exist a1 < b1 < b < b2 < a2 , such that min f (t, l): (t, l) ∈ [μ, ν] × [γ bi , bi ] (bi B0 )p−1 ,
i = 1, 2.
If u ∈ ∂Ωb1 , then γ b1 u(t) b1 , μ t ν. Applying Lemma 2.4, it follows that T Au = max (Au)(t) −
ϕ(s)s +
0tT
0
m−2
=
m−2
=
T
T
T
m−2 ϕ(s)s −
bi − i=1 bi ξi T (1 − m−2 i=1 bi )
i=1
T (1 −
=
ϕ(s)s 0
m−2
bi − i=1 bi ξi T (1 − m−2 i=1 bi )
T
bi ξi
m−2 i=1
bi )
ϕ(s)s 0
ϕ(s)s 0
ν
i=1
ϕ(s)s μ
m−2 ξi ν s p−1 ∇τ bi − m−2 i=1 ai 0 a(τ )(b1 B0 ) i=1 bi ξi p−1 φq a(τ )(b1 B0 ) ∇τ + s m−2 T (1 − m−2 1 − i=1 ai i=1 bi )
m−2 i=1
μ
T
0
bi
T
i=1
m−2
i=1
ξi
ξ bi 0 i ϕ(s)s 1 − m−2 i=1 bi
ϕ(s)s −
m−2
m−2
ϕ(s)s −
bi
m−2
i=1
T
i=1 bi m−2 1 − i=1 bi 0
i=1
T
m−2
T
i=1 bi m−2 1 − i=1 bi 0
1−
1 m−2
bi − m−2 i=1 bi ξi b1 B0 m−2 T (1 − i=1 bi )
m−2
0
ν
i=1
= b1 B0 B0−1
s a(τ )∇τ +
φq μ
ξ ai 0 i a(τ )∇τ s 1 − m−2 i=1 ai
m−2 i=1
0
= b1 = u .
In the same way, we can prove that if u ∈ ∂Ωb2 , then Au u . Now, we consider the operator A on Ωb1 \ Ωa1 and Ωa2 \ Ωb2 , respectively. By Lemma 2.5, we assert that the operator A has two fixed points u∗1 , u∗2 ∈ K, such that a1 u∗1 b1 , b2 u∗1 a2 . Therefore, u∗i , i = 1, 2, are positive solutions of the problem (1.1) and (1.2). 2 Proof of Theorem 3.2. Let
=
p−1
p−1 1 p3 (l) p4 (l) B0 B0 , min lim p−1 − , lim p−1 − 2 γ γ l→∞ l l→0+ l
then there exist 0 < b3 c3 , c4 b4 < +∞, such that
p3 (l)
p4 (l)
B0 γ B0 γ
p−1 p−1
+ l p−1 ,
0 l b3 ,
+ l p−1 ,
b4 l +∞.
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
If 0 l b3 , u ∈ ∂Ωl , then γ l u(t) l, μ t ν. By Lemma 2.4 and condition (B3 ), we have Au = max (Au)(t) 0tT
T
T
T bi − m−2 i=1 bi ξi ϕ(s)s T (1 − m−2 i=1 bi )
m−2 i=1
m−2
m−2
bi − i=1 bi ξi T (1 − m−2 i=1 bi )
0
ν
i=1
ϕ(s)s μ
s bi − i=1 bi ξi φq a(τ ) p3 u(τ ) − k3 (τ )uλ3 ∇τ m−2 T (1 − i=1 bi ) μ 0
m−2 ξi λ3 i=1 ai 0 a(τ )[p3 (u(τ )) − k3 (τ )u ]∇τ + s 1 − m−2 i=1 ai m−2
p−1 ν s T m−2 B0 i=1 bi − i=1 bi ξi p−1 λ3 ∇τ φ a(τ ) +
(γ l) − k (τ )l q 3 γ T (1 − m−2 i=1 bi ) μ 0
m−2 ξi B0 p−1 + )(γ l)p−1 − k3 (τ )l λ3 ]∇τ i=1 ai 0 a(τ )[(( γ ) + s. 1 − m−2 i=1 ai T
m−2
m−2
ν
i=1
It follows that m−2 ν s p−1 b − b ξ i i i i=1 i=1 φq a(τ ) B0 + γ p−1 − k3 (τ )l λ3 −p+1 ∇τ m−2 T (1 − i=1 bi ) μ 0
m−2 ξi p−1 p−1 +γ
− k3 (τ )l λ3 −p+1 ]∇τ i=1 ai 0 a(τ )[B0 + s. 1 − m−2 i=1 ai
β(l) T l
m−2
Noticing λ3 − p + 1 > 0, we get ν s p−1 bi − m−2 i=1 bi ξi φq a(τ ) B0 + γ p−1 ∇τ m−2 T (1 − i=1 bi ) μ 0 m−2 ξi p−1 p−1 +γ
)∇τ i=1 ai 0 a(τ )(B0 + s 1 − m−2 i=1 ai 1 p−1 = B0 + γ p−1 p−1 B0−1 1 −(p−1) p−1 = 1 + γ p−1 B0
> 1.
β(l) T lim l→0+ l
m−2 i=1
Therefore, there exists b3 with 0 < b3 < a, such that β(b3 ) > b3 . It implies that Au > u , for u ∈ ∂Ωb3 . If b4 γ l < +∞ and u ∈ ∂Ωl , then b4 γ l u(t) l, μ t ν. Similar to the above argument, noticing that λ4 − p + 1 < 0, we can get liml→+∞ β(l) l > 1. Therefore, there exists b4 with 0 < b4 < +∞, such that β(b4 ) > b4 . It implies that Au > u , for u ∈ ∂Ωb4 . By condition (B4 ), we can see that there exist b3 < a3 < a < a4 < b4 , such that max f (t, l): (t, l) ∈ [0, T ] × [0, ai ] (ai A0 )p−1 , i = 3, 4. If u ∈ ∂Ωa3 , then 0 u(t) a3 , 0 t T , and f (t, u(t)) (a3 A0 )p−1 . It follows that
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
Au
1−
1−
1 m−2 i=1
1 m−2 i=1
1023
T bi
ϕ(s)s 0
s
T bi
a3 A0
a(τ )∇τ +
φq 0
ξ ai 0 i a(τ )∇τ s 1 − m−2 i=1 ai
m−2 i=1
0
= a3 = u . Similarly, if u ∈ ∂Ωa4 , then Au u . Now, we study the operator A on Ωa3 \ Ωb3 and Ωb4 \ Ωa4 , respectively. By Lemma 2.5, we assert that the operator A has two fixed points u∗3 , u∗4 ∈ K, such that b3 u∗1 a3 , a4 u∗1 b4 . Therefore, u∗i , i = 3, 4, are positive solutions of the problem (1.1) and (1.2). 2 4. Further discussion If the conditions of Theorems 3.1 and 3.2 are weakened, we will get the existence of single positive solution of the problem (1.1) and (1.2). Corollary 4.1. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (C1 ) p1 ∈ C([0, +∞), [0, +∞)), and liml→0+ (C2 ) k1
p1 (l) l p−1
p−1
< A0
;
∈ L1 ([0, T ], [0, +∞));
(C3 ) There exist c1 > 0, λ1 > p − 1, such that f (t, l) p1 (l) + k1 (t)l λ1 ,
(t, l) ∈ [0, T ] × [0, c1 ];
(C4 ) There exists b > 0, such that min f (t, l): (t, l) ∈ [μ, ν] × [γ b, b] (bB0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. Corollary 4.2. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (D1 ) p2 ∈ C([0, +∞), [0, +∞)), and liml→∞ (D2 ) k2
p2 (l) l p−1
p−1
< A0
;
∈ L1 ([0, T ], [0, +∞));
(D3 ) There exist c2 > 0, 0 λ2 < p − 1, such that f (t, l) p2 (l) + k2 (t)l λ2 ,
(t, l) ∈ [0, T ] × [c2 , +∞);
(D4 ) There exists b > 0, such that min f (t, l): (t, l) ∈ [μ, ν] × [γ b, b] (bB0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. Corollary 4.3. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (E1 ) p3 ∈ C([0, +∞), [0, +∞)), and liml→0+ (E2 ) k3 ∈ L1 ([μ, ν], [0, +∞));
p3 (l) l p−1
> ( Bγ0 )p−1 ;
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Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
(E3 ) There exist c3 > 0, λ3 > p − 1, such that f (t, l) p3 (l) − k3 (t)l λ3 ,
(t, l) ∈ [μ, ν] × [0, c3 ];
(E4 ) There exists a > 0, such that max f (t, l): (t, l) ∈ [0, T ] × [0, a] (aA0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. Corollary 4.4. Assume (H1 )–(H3 ) hold, and assume that the following conditions hold: (F1 ) p4 ∈ C([0, +∞), [0, +∞)), and liml→∞
p4 (l) l p−1
> ( Bγ0 )p−1 ;
(F2 ) k4 ∈ L1 ([μ, ν], [0, +∞)); (F3 ) There exist c4 > 0, 0 λ4 < p − 1, such that f (t, l) p4 (l) − k4 (t)l λ4 ,
(t, l) ∈ [μ, ν] × [c4 , +∞);
(F4 ) There exists a > 0, such that max f (t, l): (t, l) ∈ [0, T ] × [0, a] (aA0 )p−1 . Then the problem (1.1) and (1.2) has at least one positive solution. The proof of the above results is similar to Theorems 3.1 and 3.2, we omit it. 5. Some examples In the section, we present a simple example to explain our results. We only study the case T = R, (0, T ) = (0, 1). Let f (t, 0) ≡ 0, Consider the following BVP:
φ3 (u ) + f t, u(t) = 0, t ∈ (0, 1),
1 1 1 1 , u(1) = u , φ3 u (0) = φ3 u 2 2 2 2
(5.1) (5.2)
where √ ⎧ 1 223u3 + min{ √t (1−t) , u2 } u5 , (t, u) ∈ [0, 1] × [0, 1], ⎪ ⎪ ⎨ (t, u) ∈ [0, 1] × [1, 3], f (t, u) = 225, √ ⎪ √ ⎪ ⎩ 74u + 3 min{ √ 1 , 2u } u3 , (t, u) ∈ [0, 1] × [3, +∞). 6 t (1−t) 3 It is easy to check that f : [0, 1] × [0, +∞) → [0, +∞) is continuous. In this case, p = 3, a(t) ≡ 1, m = 3, a1 = b1 = 12 , ξ1 = 12 , it follows from a direct calculation that A0 =
γ=
1 1−
1 1 2 0
φq s +
1 2
·
1 2 1 − 12
−1
ds
1 (1 − 1 ) 1 b1 (1 − ξ1 ) = 2 1 21 = . 1 − b1 ξ1 3 1− 2 · 2
√ √ 3( 54 + 2 ) , = 52
Y. Sang, H. Su / J. Math. Anal. Appl. 340 (2008) 1012–1026
1025
Let μ = 12 , ν = 1, we have
b1 − b 1 ξ1 B0 = 1 − b1 =
1 2
−
1 2
·
1−
1 2
1 2
ν φq μ
−1 a 1 ξ1 s+ ds 1 − a1
1
s+ 1 2
1 2
·
1 2 1 − 12
−1
1 2
ds
√ 6(3 6 + 4) < 5. = 19 Choose c1 = 1, c2 = 3, b = 3, λ1 = 52 , λ2 = 32 , p1 (u) = 223u3 , p2 (u) = 74u, k1 (t) = k2 (t) = check that 5
f (t, u) p1 (u) + k1 (t)u 2 ,
√ 1 , t (1−t)
it is easy to
(t, u) ∈ [0, 1] × [0, 1],
3 2
f (t, u) p2 (u) + k2 (t)u ,
(t, u) ∈ [0, 1] × [3, +∞), √ √ p1 (u) 3( 54 + 2 ) 2 2 lim = lim = 0 < A0 = , u→0 u2 l→0 u2 52 √ √ p2 (u) 74u 3( 54 + 2 ) 2 2 lim = lim 2 = 0 < A0 = , u→∞ u2 l→∞ u 52 √ 1 3 · 6(3 6 + 4) 2 = (bB0 )2 . min f (t, u): (t, u) ∈ , 1 × [1, 3] = 225 > 2 19 223u3
It follows that f satisfies the conditions (A1 )–(A4 ) of Theorem 3.1, then problem (1.1) and (1.2) has at least two positive solutions. However, Theorem 1.1 cannot be applied to the example, hence we generalize Theorem 1.1 and [10, Theorem 3.1]. References [1] H.R. Sun, W.T. Li, Positive solutions for nonlinear three-point boundary value problems on time scales, J. Math. Anal. Appl. 299 (2004) 508–524. [2] D.R. Anderson, Solutions to second-order three-point problems on time scales, J. Difference Equ. Appl. 8 (2002) 673–688. [3] E.R. Kaufmann, Positive solutions of a three-point boundary value problem on a time scale, Electron. J. Differential Equations 82 (2003) 1–11. [4] H. Luo, Q.Z. Ma, Positive solutions to a generalized second-order three-point boundary value problem on time scales, Electron. J. Differential Equations 17 (2005) 1–14. [5] H.R. Sun, W.T. Li, Positive solutions for nonlinear m-point boundary value problems on time scales, Acta Math. Sinica 49 (2006) 369–380 (in Chinese). [6] Z.M. He, Double positive solutions of three-point boundary value problems for p-Laplacian dynamic equations on time scales, J. Comput. Appl. Math. 182 (2005) 304–315. [7] Z.M. He, Triple positive solutions of boundary value problems for p-Laplacian dynamic equations on time scales, J. Math. Anal. Appl. 321 (2006) 911–920. [8] 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. [9] Y. Wang, C. Hou, Existence of multiple positive solutions for one-dimensional p-Laplacian, J. Math. Anal. Appl. 315 (2006) 144–153. [10] Y. Wang, W. Ge, Positive solutions for multipoint boundary value problems with one-dimensional p-Laplacian, Nonlinear Appl. 66 (6) (2007) 1246–1256. [11] R.P. Agarwal, D. O’Regan, Nonlinear boundary value problems on time scales, Nonlinear Anal. 44 (2001) 527–535. [12] M. Bohner, A. Peterson, Advances in Dynamic Equations on Time Scales, Birkhäuser Boston, Cambridge, MA, 2003. [13] F.M. Atici, G.Sh. Guseinov, On Green’s functions and positive solutions for boundary value problems on time scales, J. Comput. Anal. Math. 141 (2002) 75–99. [14] M. Bohner, A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhäuser Boston, Cambridge, MA, 2001. [15] R.P. Agarwal, M. Bohner, P. Rehak, Half-linear Dynamic Equations, Nonlinear Analysis and Applications: To V. Lakshmikantham on His 80th Birthday, Kluwer Academic Publishers, Dordrecht, 2003, pp. 1–57.
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