Electronic Journal of Differential Equations, Vol. 2008(2008), No. 123, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF COUNTABLY MANY POSITIVE SOLUTIONS
FOR nTH-ORDER m-POINT BOUNDARY-VALUE PROBLEMS
ON TIME SCALES
SIHUA LIANG, JIHUI ZHANG, ZHIYONG WANG
Abstract. In this paper, we study the existence of positive solutions for the
nonlinear n-th order with m-point singular boundary-value problem. By using
the fixed point index theory and a new fixed point theorem in cones, the existence of countably many positive solutions for a nonlinear singular boundary
value problem are obtained.
1. Introduction
In this paper, by introducing a new operator, improving and generating a pLaplace operator for some p > 1, we study the existence of countably many positive
solutions for n-th order with m-point nonlinear boundary-value problems
(ϕ(u∆
n−1
)(t))∇ + a(t)f (u(t), u∆ (t), . . . , u∆
n−2
(t)) = 0,
0 < t < T,
(1.1)
subject to the boundary conditions
i
u∆ (0) = 0,
u∆
n−2
(0) =
m−2
X
i = 0, 1, . . . , n − 3,
αi u∆
n−2
(ξi ),
u∆
n−1
(1.2)
(T ) = 0,
i=1
where ϕ : R → R is the increasing homeomorphism and positive homomorphism
and ϕ(0) = 0. ξi ∈ [0, T ]T with 0 < ξ1 < ξ2 < · · · < ξm−2 < T and αi satisfy
Pm−2
αi ∈ [0, T ]T , 0 < i=1 αi < 1. a(t) : [0, T ]T → [0, +∞) and has countably many
singularities in [0, T ]T .
A projection ϕ : R → R is called an increasing homeomorphism and positive
homomorphism, if the following conditions are satisfied:
(1) if x ≤ y, then ϕ(x) ≤ ϕ(y), for all x, y ∈ R;
(2) ϕ is a continuous bijection and its inverse mapping is also continuous;
(3) ϕ(xy) = ϕ(x)ϕ(y), for all x, y ∈ [0, +∞).
In the above definition, we can replace condition (3) by the following stronger
condition:
2000 Mathematics Subject Classification. 34B18.
Key words and phrases. Time scales; positive solutions; singular boundary-value;
fixed-point index theory.
c
2008
Texas State University - San Marcos.
Submitted February 20, 2008. Published September 4, 2008.
1
2
S. LIANG, J. ZHANG
EJDE-2008/123
(4) ϕ(xy) = ϕ(x)ϕ(y), for all x, y ∈ R, where R = (−∞, +∞).
Remmark 1.1. If conditions (1), (2) and (4) hold, then ϕ is homogenous generating a p-Laplace operator; i.e., ϕ(x) = |x|p−2 x, for some p > 1.
Moreover, throughout this paper the following conditions hold:
(C1) f : [0, +∞) → [0, +∞) is continuous;
(C2) a : [0, T ]T → [0, +∞) and has countably many singularities in [0, T ]T , i.e.,
T
there exists a sequence {ti }∞
i=1 such that 0 < ti+1 < ti < 2 , limi→∞ ti =
T
t0 < 2 , and t0 ∈ [0, T ]T . limt→ti a(t) = ∞, i = 1, 2, . . . , and a(t) does not
vanish identically on any subinterval of [0, T ]T . Moreover
Z T
a(s)∇s < +∞.
0<
0
Recently, there is much attention paid to the existence of positive solutions for
three-point boundary-value problems on time scales, see [2, 4, 5, 7, 8, 13, 16, 17] and
references therein. However, there are not many results concerning the increasing
homeomorphism and positive homomorphism operator 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.
Anderson [2] discussed the dynamic equation on time scales:
u∆∇ (t) + a(t)f (u(t)) = 0,
u(0) = 0,
t ∈ (0, T ),
(1.3)
αu(η) = u(T ).
(1.4)
He obtained some results for the existence of one positive solution of the problem
f (u)
(1.3) and (1.4) based on the limits f0 = limu→0+ f (u)
u and f∞ = limu→∞ u . He
also obtained the existence of at least three positive solutions.
Kaufmann [8] studied the problem (1.3) and (1.4) and obtained existence results
of finitely many positive solutions and countably many positive solutions.
Zhou and Su [17] studied the quasi-linear equation with p-Laplacian operator:
(φp (u(n−1) ))0 + g(t)f (u(t), u0 (t), . . . , u(n−2) (t)) = 0,
u(i) (0) = 0
u
u
(n−2)
(n−2)
(1.5)
0 ≤ i ≤ n − 3,
(n−1)
(ξ)) = 0 n ≥ 3,
(n−1)
(η)) = 0 n ≥ 3.
(0) − B0 (u
(1) + B1 (u
0 < t < T,
(1.6)
They obtained the existence of one solution, and of multiple solutions by using the
fixed-point index theory.
Liu and Zhang [12] considered the existence of positive solutions of the following
quasi-linear differential equation
(ϕ(x0 ))0 + a(t)f (x(t)) = 0,
0
x(0) − βx (0) = 0,
0 < t < 1,
0
x(1) + δx (1) = 0.
(1.7)
(1.8)
Where ϕ : R → R is an increasing homeomorphism and positive homomorphism
and ϕ(0) = 0. They obtained the existence of one or two positive solutions of the
problem (1.7) and (1.8) by using a fixed-point index theorem in cones.
But whether or not we can obtain the countably many positive solutions of nthorder with m-point boundary value problem (1.1) and (1.2) still remain unknown.
EJDE-2008/123
MULTI-POINT PROBLEMS ON TIME SCALES
3
So the goal of present paper is to improve and generate p-Laplacian operator and
establish some criteria for the existence of countable many solutions.
The plan of the paper is as follows. In Section 2, for the convenience of the reader
we give some definitions. In Section 3, we present some lemmas in order to prove
our main results. Section 4 is developed in order to present and prove our main
results. In Section 5 we present the example of the increasing homeomorphism and
positive homomorphism operators.
2. Some definitions and fixed point theorems
For convenience, we list the following definitions which can be found in [1, 3, 4,
5, 7].
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 Tκ = T − {m};
otherwise set Tκ = T. If T has a left scattered maximum M , define Tκ = T − {M };
otherwise set Tκ = T.
Definition 2.2. For f : T → R and t ∈ Tκ , 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 ∈ Tκ , the nabla derivative of f at t is the number 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. It is well-known that if f is ld-continuous, then there is a
function F (t) such that F ∇ (t) = f (t). In this case, it is defined that
Z b
f (t)∇t = F (b) − F (a).
a
If u
∆∇
(t) ≤ 0 on [0, T ], then we say u is concave on [0, T ].
Definition 2.4. Let (E, k.k) be a real Banach space. A nonempty, closed, convex
set P ⊂ E is said to be a cone provided the following are satisfied:
(a) if y ∈ P and λ ≥ 0, then λy ∈ P ;
(b) if y ∈ P and −y ∈ P , then y = 0.
4
S. LIANG, J. ZHANG
EJDE-2008/123
If P ⊂ E is a cone, we denote the order induced by P on E by ≤, that is, x ≤ y
if and only if y − x ∈ P .
Definition 2.5. Given a nonnegative continuous functional γ on a cone P of E,
for each d > 0 we define the set
P (γ, d) = {x ∈ P : γ(x) < d}.
The following fixed point theorems are fundamental and important for the proofs
of our main results.
Theorem 2.6 ([6]). Let E be a Banach space and P ⊂ T
E be a cone in E. Let
r > 0 define Ωr = {x ∈ P : kxk < r}. Assume that A : P Ωr → P is completely
continuous operator such that Ax 6= x for x ∈ ∂Ωr .
(i) If kAxk < kxk for x ∈ ∂Ωr , then i(A, Ωr , P ) = 1.
(ii) If kAxk > kxk for x ∈ ∂Ωr , then i(A, Ωr , P ) = 0.
Theorem 2.7 ([15]). Let P be a cone in a Banach space E. Let α, β and γ be
three increasing, nonnegative and continuous functionals on P , satisfying for some
c > 0 and M > 0 such that
γ(x) ≤ β(x) ≤ α(x),
kxk ≤ M γ(x)
for all x ∈ P (γ, c). Suppose there exists a completely continuous operator A :
P (γ, c) → P and 0 < a < b < c such that
(i) γ(Ax) < c, for all x ∈ ∂P (γ, c);
(ii) β(Ax) > b, for all x ∈ ∂P (β, b);
(iii) P (α, a) 6= ∅, and α(Ax) < a, for all x ∈ ∂P (α, a). Then A has at least
three fixed points x1 , x2 , x3 ∈ P (γ, c) such that
0 ≤ α(x1 ) < a < α(x2 ),
β(x2 ) < b < β(x3 ),
γ(x3 ) < c.
3. Preliminaries and Lemmas
In the rest of this article, T is closed subset of R with 0 ∈ Tκ , T ∈ Tκ . And
i
n−2
E = u ∈ Cld
[0, T ] : u∆ (0) = 0, 0 ≤ i ≤ n − 3 .
n−2
Then E is a Banach space with the norm kuk = supt∈[0,T ] |u∆ (t)|. And let
n−2
n−2
P = u ∈ E : u∆ (t) ≥ 0, u∆ (t)is concave nondecreasing on [0, T ] .
Obviously, P is a cone in E. Set Pr = {u ∈ P : kuk ≤ r}. We can easily get the
following Lemmas.
Lemma 3.1. Suppose condition (C2) holds. Then there exists a constant θ ∈
max{t ∈ T | 0 < t < T2 } that satisfies
Z T −θ
0<
a(s)∇s < +∞.
θ
Furthermore, the function
Pm−2 R t −1 R t
Z T −t1
Z T −t1
a(τ )∇τ ∆s
i=1 αi t1 ϕ
s
−1
H(t) =
ϕ
a(τ )∇τ ∆s +
Pm−2
1 − i=1 αi
t
s
EJDE-2008/123
MULTI-POINT PROBLEMS ON TIME SCALES
5
is continuous and positive on [t1 , T − t1 ]. Furthermore there exists a constant L > 0
such that
L=
min H(t) > 0.
t∈[t1 ,T −t1 ]
Proof. At first, it is easily seen that H(t) is continuous on [t1 , T − t1 ]. Let
Z T −t1
Z T −t1
−1
H1 (t) =
ϕ
a(τ )∇τ ∆s,
t
s
Pm−2 R t −1 R t
a(τ
)∇τ
∆s
i=1 αi t1 ϕ
s
H2 (t) =
.
Pm−2
1 − i=1 αi
Then from condition (C2), we know that H1 (t) strictly monotone decreasing on
[t1 , T − t1 ] and H1 (T − t1 ) = 0. Similarly function H2 (t) is strictly monotone
increasing on [t1 , T − t1 ] and H2 (t1 ) = 0. Since H1 (t) and H2 (t) are not equal
to zero at the same time. So the function H(t) = H1 (t) + H2 (t) is positive on
[t1 , T − t1 ], which implies L = mint∈[t1 ,T −t1 ] H(t) > 0.
Lemma 3.2. If u ∈ P . Then
θ
kuk, t ∈ [θ, T − θ],
T
The proof of the above lemma is similar to the proof of in [7], so we omit it.
n−1
Now, we define a mapping F : P → Cld
[0, T ] by
Z t Z ζ1
Z ζn−3
(F u)(t) =
...
w(ζn−2 )∆ζn−2 ∆ζn−3 . . . ∆ζ1 ,
(3.1)
u∆
n−2
0
0
−1
Z
(t) ≥
0
where
Z
ζn−2
w(ζn−2 ) =
ϕ
0
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆
n−2
(τ ))∇τ ∆s
s
Pm−2
+
T
i=1
αi
R ξi
0
Then it is easy to see that
Z t
Z
n−2
(F u)∆ (t) =
ϕ−1
ϕ−1
RT
s
a(τ )f (u(τ ), u0 (τ ), . . . , u(n−2) (τ ))∇τ ∆s
.
Pm−2
1 − i=1 αi
T
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
0
s
Pm−2 R ξi −1 R T
0
(n−2)
α
ϕ
a(τ
)f
(u(τ
),
u
(τ
),
.
.
.
,
u
(τ
))∇τ
∆s
i
i=1
0
s
+
Pm−2
1 − i=1 αi
≥ 0, 0 ≤ t ≤ T.
∆n−1
(F u)
(t) = ϕ
−1
Z
T
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆
n−2
(τ ))∇τ
t
≥ 0,
0 ≤ t ≤ T.
We also have
[ϕ((F u)∆
n−1
)(t)]∇ = −a(t)f (u(t), u∆ (t), . . . , u∆
n−2
∆n−2
(t)) ≤ 0.
Together with ϕ is a increasing operator, we know (F u)
is a concave function.
This shows F (P ) ⊂ P .
Using the Arzela-Ascoli Theorem, we obtain the following lemma.
6
S. LIANG, J. ZHANG
EJDE-2008/123
Lemma 3.3. The operator F : P → P is completely continuous.
Lemma 3.4. Suppose that conditions (C1), (C2) hold. Then the solution u(t) ∈ P
of (1.1), (1.2) satisfies
u(t) ≤ u∆ (t) ≤ · · · ≤ u∆
n−3
0 ≤ t ≤ T,
(t),
and for θ ∈ (0, T2 ) in Lemma 3.1, we have
u∆
n−3
(t) ≤
T ∆n−2
u
(t),
θ
θ ≤ t ≤ T − θ.
The proof of the above lemma is similar to the proof of in [17, lemma 2.4].
4. Main results
For notational convenience, we define
Pm−2
(1 − i=1 αi )
.
λ2 = R T
RT
−1
ϕ
a(τ
)∇τ
∆s
0
s
1
λ1 = ,
L
The main results of this paper are the following.
Theorem 4.1. Suppose that conditions (C1)-(C2) hold. Let {θk }∞
k=1 be such that
∞
θk ∈ (tk+1 , tk ) (k = 1, 2, . . . ). Let {rk }∞
k=1 and {Rk }k=1 be such that
θk
rk < rk < mrk < Rk , mrk ≤ M Rk , k = 1, 2, . . . .
T
Furthermore for each natural number k we assume that f satisfy:
(C3) f (v1 , v2 , . . . , vn−1 ) ≥ ϕ(mrk ) for all 0 ≤ v1 , v2 , . . . , vn−2 ≤ θTk rk , θTk rk ≤
vn−1 ≤ rk ;
(C4) f (v1 , v2 , . . . , vn−1 ) ≤ ϕ(M Rk ) for all 0 ≤ v1 , v2 , . . . , vn−1 ≤ Rk .
Where m ∈ (λ1 , ∞), M ∈ (0, λ2 ). Then the boundary-value problem (1.1), (1.2)
has infinitely many solutions {uk }∞
k=1 such that
Rk+1 <
rk ≤ kuk k ≤ Rk ,
Proof. Since 0 < t0 < tk+1 < θk < tk <
by the Lemma 3.2 we have
u∆
n−2
(t) ≥
Consider the sequences {Ω1,k }∞
k=1
T
2
k = 1, 2, . . . .
, k = 1, 2, . . . , for any k ∈ N and u ∈ P ,
θk
kuk, t ∈ [θk , T − θk ].
(4.1)
T
and {Ω2,k }∞
k=1 of open subsets of E defined by
Ω1,k = {u ∈ P : kuk < rk },
k = 1, 2, . . . ,
Ω2,k = {u ∈ P : kuk < Rk },
k = 1, 2, . . . .
For a fixed k and u ∈ ∂Ω1,k . From (4.1) we have
θk
θk
kuk = rk , t ∈ [θk , T − θk ].
T
T
Since (t1 , T − t1 ) ⊂ [θk , T − θk ], in the following we consider three cases:
(i) If ξ1 ∈ [t1 , T − t1 ]. In this case, from (3.1), condition (C3) and Lemma 3.1, we
have
rk = kuk ≥ u∆
kF uk = (F u)∆
n−2
(T )
n−2
(t) ≥
EJDE-2008/123
MULTI-POINT PROBLEMS ON TIME SCALES
Z
T
7
T
Z
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
s
0
Pm−2 R ξi −1 R T
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
i=1 αi 0 ϕ
s
+
Pm−2
1 − i=1 αi
Z T −t1
Z T −t1
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
≥
ϕ−1
s
ξ1
Pm−2 R ξ1 −1 R ξ1
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
i=1 αi t1 ϕ
s
+
Pm−2
1 − i=1 αi
Z T −t1
h Z T −t1
−1
a(τ )∇τ ∆s
≥ (mrk )
ϕ
ϕ−1
=
s
ξ1
Pm−2
i=1 αi
P
m−2
i=1 αi
+
Z
ξ1
ϕ
−1
Z
ξ1
i
a(τ )∇τ ∆s
1−
t1
s
= mrk H(ξ1 ) > mrk L > rk = kuk.
(ii) If ξ1 ∈ [0, t1 ]. In this case, from (3.1), condition (C3 ) and Lemma3.1, we have
Z T
Z T
n−2
−1
kF uk ≥
ϕ
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
0
Z
s
T −t1
≥
ϕ−1
T −t1
Z
t1
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆
n−2
(τ ))∇τ ∆s
s
hZ
≥ (mrk )
T −t1
t1
ϕ−1
Z
T −t1
i
a(τ )∇τ ∆s
s
= mrk H(t1 ) > mrk L > rk = kuk.
(iii) If ξ1 ∈ [T − t1 , T ]. In this case, from (3.1), condition (C3 ) and Lemma3.1, we
have
Pm−2 R T −t1 −1 R T −t1
n−2
ϕ
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
i=1 αi t1
s
kF uk ≥
Pm−2
1 − i=1 αi
h Pm−2 α Z t1
Z T −t1
i
i
−1
i=1
≥ (mrk )
ϕ
a(τ )∇τ ∆s
Pm−2
1 − i=1 αi T −t1
s
= mrk H(T − t1 ) > mrk L > rk = kuk.
Thus in all cases, an application of Theorem 2.6 implies
i(F, Ω1,k , P ) = 0.
On the another hand, let u(t) ∈ ∂Ω2,k , we have u
have
∆n−2
(4.2)
(t) ≤ kuk = Rk , by (C4 ) we
kF uk = (F u)(n−2) (T )
Z T
Z T
n−2
=
ϕ−1
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
0
s
Pm−2 R ξi −1 R T
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
i=1 αi 0 ϕ
s
+
Pm−2
1 − i=1 αi
8
S. LIANG, J. ZHANG
Z
≤
T
ϕ−1
T
Z
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆
n−2
(τ ))∇τ ∆s
s
0
Pm−2
+
EJDE-2008/123
αi
i=1
R ξm−2
0
1
Pm−2
≤ M Rk
1−
= Rk = kuk.
i=1
RT
ϕ−1
hZ
αi
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆
Pm−2
1 − i=1 αi
Z T
i
a(τ )∇τ ∆s
ϕ−1
n−2
s
T
(τ ))∇τ ∆s
s
0
Thus Theorem 2.6 implies
i(T, Ω2,k , P ) = 1.
(4.3)
Hence since rk < Rk for k ∈ N, (4.2) and (4.3), it follows from additivity of the
fixed-point index that
i(T, Ω2,k \Ω1,k , P ) = 1
for k ∈ N.
Thus F has a fixed point in Ω2,k \Ω1,k such that rk ≤ kuk k ≤ Rk . Since k ∈ N was
arbitrary, the proof is complete.
To use Theorem 2.7, let θk < rk < 1 − θk and θk of Theorem 4.1, we define the
nonnegative, increasing, continuous functionals
γk (u) = max u∆
n−2
θk ≤t≤rk
βk (u) =
αk (u) =
min
rk ≤t≤T −θk
max
θk ≤t≤T −θk
u∆
u∆
(t) = u∆
n−2
n−2
n−2
(t) = u∆
(t) = u∆
(rk ),
n−2
n−2
(rk ),
(T − θk ).
It is obvious that for each u ∈ P ,
γk (u) ≤ βk (u) ≤ αk (u).
In addition, by Lemma 3.2, for each u ∈ P ,
γk (u) = u∆
Thus
kuk ≤
n−2
θk
kuk.
T
(rk ) ≥
T
γk (u)
θk
for all u ∈ P.
For convenience, we denote
λ=
1−
hZ
1
Pm−2
T
ϕ
Z
0
i=1 αi
rk
Z T −θk
−1
Z
ηk =
−1
ϕ
θk
T
i
a(τ )∇τ ∆s ,
s
a(τ )∇τ ∆s.
s
Theorem 4.2. Suppose (C1)-(C2) hold. Let {θk }∞
k=1 be such that θk ∈ (tk+1 , tk )
∞
∞
(k = 1, 2, . . . ). Let {ak }∞
k=1 , {bk }k=1 and {ck }k=1 be such that
θk
bk < bk < ck , and ρk bk < ηk ck , for k = 1, 2, . . . .
T
Furthermore for each natural number k we assume that f satisfies:
(C5) f (v1 , v2 , . . . , vn−1 ) < ϕ( cλk ), for all 0 ≤ v1 , v2 , . . . , vn−1 ≤ θTk ck ;
ck+1 < ak <
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MULTI-POINT PROBLEMS ON TIME SCALES
9
(C6) f (v1 , v2 , . . . , vn−1 ) > ϕ( ηbkk ), for all 0 ≤ v1 , v2 , . . . , vn−2 ≤ θTk bk , bk ≤
vn−1 (t) ≤ θTk bk ;
(C7) f (v1 , v2 , . . . , vn−1 ) < ϕ( aλk ), for all 0 ≤ v1 , v2 , . . . , vn−1 ≤ θTk ak .
Then the boundary-value problem (1.1), (1.2) has three infinite families of solutions
∞
∞
{u1k }∞
k=1 {u2k }k=1 and {u3k }k=1 satisfying
0 ≤ αk (u1k ) < ak < αk (u2k ),
βk (u2k ) < bk < βk (u3k ),
γ(u3k ) < ck ,
for n ∈ N.
Proof. We define the completely continuous operator F by 3.1. So it is easy to
check that F : P (γk , ck ) → P , for k ∈ N.
We now show that all the conditions of Theorem 2.7 are satisfied. To make
use of property (i) of Theorem 2.7, we choose u ∈ ∂P (γk , ck ). Then γk (u) =
n−2
n−2
n−2
maxθk ≤t≤rk u∆ (t) = u∆ (rk ) = ck , this implies that 0 ≤ u∆ (t) ≤ ck for
[0, rk ]. If we recall that kuk ≤ θTk γk (u) = θTk ck . So we have
T
ck ,
θk
i
0 ≤ u∆ (t) ≤
0 ≤ t ≤ T, i = 0, 1, . . . , n − 1.
Then assumption (C5) implies
f (u(t), u∆ (t), . . . , u∆
n−2
(t)) < ϕ
ck ,
λ
0 ≤ t ≤ T.
Therefore
γk (F u) = max (F u)∆
θk ≤t≤rk
Z
T
Z
n−2
(t) = (F u)∆
n−2
(rk )
T
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
0
s
Pm−2 R ξm−2 −1 R T
∆
∆n−2
α
ϕ
a(τ
)f
(u(τ
),
u
(τ
),
.
.
.
,
u
(τ
))∇τ
∆s
i 0
i=1
s
+
Pm−2
1 − i=1 αi
Z T
Z T
h
i
1
ck
≤
ϕ−1
a(τ )∇τ ∆s
Pm−2
λ 1 − i=1 αi 0
s
= ck .
≤
ϕ−1
Hence condition (i) is satisfied.
Secondly, we show that (ii) of Theorem 2.7 is fulfilled. For this we select u ∈
n−2
n−2
∂P (βk , bk ). Then βk (u) = minrk ≤t≤T −θk u∆ (t) = u∆ (rk ) = bk , this fact
n−2
implies that u∆ (t) ≥ bk , for rk ≤ t ≤ T . Noticing that kuk ≤ θTk γk (u) ≤
T
T
θk βk (u) = θk bk , we have
bk ≤ u ∆
n−2
(t) ≤
T
bk ,
θk
for rk ≤ t ≤ T.
By (C6), we have
f (u(t), u∆ (t), . . . , u∆
n−2
(t)) > ϕ
Therefore,
βk (F u) =
min
rk ≤t≤T −θk
(F u)∆
n−2
(t) = (F u)∆
n−2
(rk )
bk .
ηk
10
S. LIANG, J. ZHANG
Z
rk
EJDE-2008/123
T
Z
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
s
0
Pm−2 R ξi −1 R T
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
i=1 αi 0 ϕ
s
+
Pm−2
1 − i=1 αi
Z rk
Z T −θk
n−2
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆ (τ ))∇τ ∆s
ϕ−1
≥
=
ϕ−1
s
θk
Z
Z
i
bk h rk −1 T −θk
=
ϕ
a(τ )∇τ ∆s
ηk θk
s
= bk .
Hence condition (ii) is satisfied.
n−2
Finally, we verify that (iii) of Theorem 2.7 is satisfied. Noting that u∆ (t) ≡
ak
ak
4 , 0 ≤ t ≤ T is a member of P (αk , ak ) and αk (u) = 4 < ak . So P (αk , ak ) 6= ∅.
n−2
n−2
Now let u ∈ ∂P (αk , ak ). Then αk (u) = maxθk ≤t≤T −θk u∆ (t) = u∆ (T −
n−2
θk ) = ak . This implies that 0 ≤ u∆ (t) ≤ ak , 0 ≤ t ≤ T − θk . Noticing that
kuk ≤ θTk γk (u) ≤ θTk αk (u) = θTk ak . Then we get
i
ak
, 0 ≤ t ≤ T, i = 0, 1, . . . , n − 1.
0 ≤ u∆ (t) ≤
rk
Then assumption (C7) implies
n−2
ak f (u(t), u∆ (t), . . . , u∆ (t)) < ϕ
, 0 ≤ t ≤ T.
λ
As before, we get
αk (F u) =
max
θk ≤t≤T −θk
Z
≤
T
ϕ−1
n−2
(T − θk )
T
Z
0
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆
n−2
(τ ))∇τ ∆s
s
Pm−2
+
(F u)(t) = (F u)∆
i=1
αi
R ξm−2
0
h
ak
1
Pm−2
λ 1 − i=1 αi
= ak .
ϕ−1
Z
T
≤
0
RT
a(τ )f (u(τ ), u∆ (τ ), . . . , u∆
Pm−2
1 − i=1 αi
Z T
i
ϕ−1
a(τ )∇τ ∆s
s
n−2
(τ ))∇τ ∆s
s
Thus (iii) of Theorem 2.7 is satisfied. Since all hypotheses of Theorem 2.7 are
satisfied, the assertion follows.
n−2
Remmark 4.3. If we add the condition of a(t)f (u(t), u∆ (t), . . . , u∆ (t)) 6≡ 0,
t ∈ [0, T ], to Theorem 4.2 we can get three infinite families of positive solutions
∞
∞
{u1k }∞
k=1 , {u2k }k=1 , and {u3k }k=1 satisfying
0 < αk (u1k ) < ak < αk (u2k ),
βk (u2k ) < bk < βk (u3k ),
γ(u3k ) < ck ,
for n ∈ N.
Remmark 4.4. The same conclusions of Theorem 4.1 and Theorem 4.2 hold when
conditions (1), (2) and (4) are satisfied. Especially, for p-Laplacian operator ϕ(x) =
|x|p−2 x, for some p > 1, our conclusions are also true and new.
EJDE-2008/123
MULTI-POINT PROBLEMS ON TIME SCALES
11
5. Applications
There exists a function a(t) satisfying condition (C2).
Example 5.1. Let T ≡ 1 and
π2
9
− ,
3
4
15
,
32
i
X
1
, i = 1, 2, . . . .
2(k + 1)4
k=1
P∞
Consider the function a(t) : [0, 1] → (0, ∞), a(t) = i=1 ai (t), t ∈ [0, 1], where

ti+1 +ti
1
,

(2i−1)(2i+1)(ti+1 +ti ) , 0 ≤ t <
2



t
+t
1
i+1
i

≤ t < ti ,

2
 δ(ti −t)1/2 ,
δ=2
t∗ =
ai (t) =
t i = t∗ −
1
δ(t−ti )1/2




0,



ti < t ≤
,
ti +ti−1
2
1
2(2i−1)(2i+1)(1−t1 ) ,
It is easy to check that t1 =
P∞ 1
π4
i=1 i4 = 90 ), and
7
16
ti +ti−1
,
2
<
1
2,
< t ≤ t1 .
t1 ≤ t ≤ 1.
ti − ti+1 =
1
2(i+2)4 ,
i = 1, 2, . . . (denote
∞
t0 = lim ti =
i→∞
P∞
15 X
1
31
π4
1
−
=
−
>
32
2(k + 1)4
32 180
5
k=1
2
2
and because i=1 1/i = π /6, we have
∞ Z 1
X
ai (t)∇t
i=1
0
Z
Z ti +t2i−1
∞
i
1 X h ti
1
1
1
+
∇t
+
∇t
=
1
ti+1 +ti
(2i − 1)(2i + 1) δ i=1
(ti − t)1/2
(t − ti ) 2
ti
2
i=1
√ ∞ h
i
1
2X
= +
(ti − ti+1 )1/2 + (ti−1 − ti )1/2
2
δ i=1
∞
X
∞
=
i
1 1 Xh
1
1
+
+
2 δ i=1 (i + 2)2
(i + 1)2
9
1 1 π2
+
−
= 1.
2 δ 3
4
Therefore,
Z 1
=
a(t)∇t =
0
∞ Z
X
i=1
1
ai (t)∇t = 1 < ∞.
0
Which implies that Condition (C2).
Example 5.2. As an example we mention the boundary-value problem
2
[ϕ(u∆ )]∇ + a(t)f (u(t)) = 0,
t ∈ [0, 1]T ,
(5.1)
∆
u∆
2
u(0) = u (0) = 0,
3
1 2 1
1 2 1
(0) = u∆ ( ) + u∆ ( ), u∆ (1) = 0,
4
4
2
2
(5.2)
12
S. LIANG, J. ZHANG
EJDE-2008/123
where
(
ϕ(u) =
u7
1+u2 ,
2
u ≤ 0,
u > 0,
u ,
and


M 2 R12 ,
u ∈ (R1 , +∞),


2
2

M 2 Rk
−m2 rk

2
2

(u − rk ),
u ∈ [rk , Rk ],

Rk −rk
m rk +
2 2
m
r
,
u ∈ ( θTk rk , rk ),
f (u(t)) =
k
2 2
2 2

m r −M Rk+1

2

+ θkk
(u − Rk+1 ), u ∈ (Rk+1 , θTk rk ],
M 2 Rk+1


T rk −Rk+1


0,
u = 0.
Since H 0 (t) ≤ 0, So it is easy to see by calculating that
Pm−2
3
8
1 2
i=1 αi
L = min H(t) = H(1 − t1 ) =
(1 − 2t1 ) 2 = .
P
m−2
1
−
t
3
9
[t1 ,1−t1 ]
1
1 − i=1 αi
Then
1
9
2
= , λ2 = .
L
8
5
Therefore, we take m = 10 ∈ ( 98 , +∞), M = 51 ∈ (0, 52 ) and let
λ1 =
k+1
For Rk =
1
800k
k
θ k = t∗ −
X
1 X
1
1
15
+
∈ (0, )
2 i=1 2(i + 1)4 i=1 2(i + 1)4
32
and rk =
1
,
300×800k
k = 1, 2, . . . we have
θk
1
m
1
1
<
<
<
<
.
800k+1
300 × 800k
300 × 800k
300 × 800k
800k
After some simple calculation we have
f (u) ≥ ϕ(mrk ) = m2 rk2
f (u) ≤ ϕ(M Rk ) = M
2
Rk2
for u ∈ [µk rk , rk ];
for u ∈ [0, Rk ].
Then by Theorem 4.1, the boundary-value problem (5.1) and (5.2) has infinitely
many solutions {uk }∞
k=1 such that
1
1
≤ kuk k ≤
,
300 × 800k
800k
k = 1, 2, . . . .
Remmark 5.3. From the Example 5.2, we can see that ϕ is not odd, then the
boundary value problem with p-Laplacian operator [7, 17] do not apply to Example
5.2. So, we generalize a p-Laplace operator for some p > 1 and the function ϕ which
we defined above is more comprehensive and general than p-Laplace operator.
Acknowledgment. The authors are indebted to the anonymous referee whose
suggestions have improved greatly this article.
This project was supported by supported by NSFC, Foundation of Major Project
of Science and Technology of Chinese Education Ministry, SRFDP of Higher Education, NSF of Education Committee of Jiangsu Province and Project of Graduate
Education Innovation of Jiangsu Province.
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MULTI-POINT PROBLEMS ON TIME SCALES
13
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Sihua Liang
Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, 210097, Jiangsu, China.
College of Mathematics, Changchun Normal University, Changchun 130032, Jilin, China
E-mail address: liangsihua@163.com
Jihui Zhang
Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, 210097, Jiangsu, China
E-mail address: jihuiz@jlonline.com
Zhiyong Wang
Department of Mathematics, Nanjing University of Information Science and Technology, Nanjing 210044, Jiangsu, China
E-mail address: mathswzhy@126.com