Electronic Journal of Differential Equations, Vol. 2007(2007), No. 95, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
THREE POSITIVE SOLUTIONS FOR P-LAPLACIAN
FUNCTIONAL DYNAMIC EQUATIONS ON TIME SCALES
DA-BIN WANG
Abstract. In this paper, we establish the existence of three positive solutions
to the following p-Laplacian functional dynamic equation on time scales,
[Φp (u∆ (t))]∇ + a(t)f (u(t), u(µ(t))) = 0,
t ∈ (0, T )T ,
t ∈ [−r, 0]T ,
u0 (t) = ϕ(t),
u(0) − B0 (u∆ (η)) = 0,
u∆ (T ) = 0, .
using the fixed-point theorem due to Avery and Peterson [8]. An example is
given to illustrate the main result.
1. Introduction
Let T be a time scale; i.e., T is a nonempty closed subset of R. Let 0, T be points
in T, an interval (0, T )T denotes time scales interval, that is, (0, T )T := (0, T ) ∩ T.
Other types of intervals are defined similarly.
The theory of dynamic equations on time scales has been a new important mathematical branch (see, for example, [1, 2, 9, 10, 17]) since it was initiated by Hilger
[16]. At the same time, boundary value problems (BVPs) for dynamic equation
on time scales have received considerable attention [3, 4, 5, 6, 11, 12, 13, 14, 15,
18, 19, 20, 21, 22]. However, to the best of our knowledge, few papers can be
found in the literature on bvps of p-Laplacian dynamic equations on time scales
[5, 14, 15, 19, 20, 22], especially for p-Laplacian functional dynamic equations on
time scales [19].
This paper concerns the existence of positive solutions for the p-Laplacian functional dynamic equation on time scale,
[Φp (u∆ (t))]∇ + a(t)f (u(t), u(µ(t))) = 0,
t ∈ (0, T )T ,
u0 (t) = ϕ(t),
t ∈ [−r, 0]T ,
∆
u(0) − B0 (u (η)) = 0,
u (T ) = 0,
p−2
where Φp (s) is p-Laplacian operator, i.e., Φp (s) = |s|
1
1
p + q = 1, η ∈ (0, ρ(T ))T and
s, p > 1, (Φp )−1 = Φq ,
(C1) f : (R+ )2 → R+ is continuous;
2000 Mathematics Subject Classification. 39A10, 34B15.
Key words and phrases. Time scale; p-Laplacian functional dynamic equation;
boundary value problem; positive solution; fixed point.
c
2007
Texas State University - San Marcos.
Submitted May 17, 2007. Published June 29, 2007.
1
(1.1)
∆
2
D.-B. WANG
EJDE-2007/95
(C2) a : T → R+ is left dense continuous (i.e., a ∈ Cld (T, R+ )) and does not
vanish identically on any closed subinterval of [0, T ], where Cld (T, R+ )
denotes the set of all left dense continuous functions from T to R+ ;
(C3) ϕ : [−r, 0]T → R+ is continuous and r > 0;
(C4) µ : [0, T ]T → [−r, T ]T is continuous, µ(t) ≤ t for all t;
(C5) B0 : R → R is continuous and there exist constant A ≥ 1, B > 0 such that
Bv ≤ B0 (v) ≤ Av, for all v ≥ 0.
In [19], by using a double-fixed-point theorem due to Avery et al. [7] in a cone,
Song and Xiao considered the problem (1.1) and obtained the existence of two
positive solutions.
In paper [15], Hong studied the problem (1.1) when ϕ(t) = 0, t ∈ [−r, 0]T and
the nonlinear term is not involved u(µ(t)). He imposed conditions on f to yield
at least three positive solutions to the problem (1.1), by applying the fixed-point
theorem due to Avery and Peterson [8].
Motivated by [15, 19], we shall show that the problem (1.1), has at least three
positive solutions by means of the fixed point theorem due to Avery and Peterson.
In the remainder of this section we list the following well known definitions which
can be found in [2, 6, 9, 10].
Definition 1.1. For t < sup T and r > inf T, define the forward jump operator σ
and the backward jump operator ρ,
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 sad
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 1.2. For x : T →R and t ∈ Tk , we define the delta derivative of x(t),
x∆ (t), to be the number (when it exists), with the property that, for any ε > 0,
there is a neighborhood U of t such that
|[x(σ(t)) − x(s)] − x∆ (t)[σ(t) − s]| < ε|σ(t) − s|,
for all s ∈ U . For x : T →R and t ∈ Tk , we define the nabla derivative of x(t),
x∇ (t), to be the number (when it exists), with the property that, for any ε > 0,
there is a neighborhood V of t such that
|[x(ρ(t)) − x(s)] − x∇ (t)[ρ(t) − s]| < ε|ρ(t) − s|,
for all s ∈ V . If T = R, then x∆ (t) = x∇ (t) = x0 (t). If T = Z, then x∆ (t) =
x(t + 1) − x(t) is the forward difference operator while x∇ (t) = x(t) − x(t − 1) is
the backward difference operator.
Definition 1.3. If F ∆ (t) = f (t), then we define the delta integral by
Z t
f (s)∆s = F (t) − F (a).
a
If Φ∇ (t) = f (t), then we define the nabla integral by
Z t
f (s)∇s = Φ(t) − Φ(a).
a
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THREE POSITIVE SOLUTIONS
3
Throughout this papers, we assume T is closed subset of R with 0 ∈ Tk and
T ∈ Tk .
Lemma 1.4 ([6]). The following formulas hold:
R
∆
t
(i)
f
(s)∆s
= f (t),
a
∇
R
t
(ii)
f (s)∆s
= f (ρ(t)),
a
R
∆
t
(iii)
f (s)∇s
= f (σ(t)),
a
R
∇
t
(iv)
f (s)∇s
= f (t).
a
2. Preliminaries
In this section, we provide some background materials from the theory of cones
in Banach spaces and we then state the fixed-point theorem due to Avery and
Peterson.
Definition 2.1. Let E be a real Banach space. A nonempty, closed, convex set
P ⊂ E is said to be a cone provided the following conditions are satisfied:
(i) if x ∈ P and λ ≥ 0, then λx ∈ P ;
(ii) if x ∈ P and −x ∈ P , then x = 0.
Every cone P ⊂ E induces an ordering in E given by
x≤y
if y − x ∈ P.
Definition 2.2. Given a cone P in a real Banach space E, the map ς : P → [0, ∞)
is called a nonnegative continuous concave function on cone P provided that ς is
continuous and
ς(tx + (1 − t)y) ≥ tς(x) + (1 − t)ς(y),
for x, y ∈ P and 0 ≤ t ≤ 1.
Dual to this, we call the map τ : P → [0, ∞) is called a nonnegative continuous
convex function on cone P provided that τ is continuous and
τ (tx + (1 − t)y) ≤ tτ (x) + (1 − t)τ (y),
for x, y ∈ P and 0 ≤ t ≤ 1.
Let γ and θ be nonnegative continuous convex functions on P , α be a nonnegative
continuous concave function on P and ψ be a nonnegative continuous function on
P . Then, for positive real numbers a, b, c and d, we define the following convex
sets
P (γ, d) = {x ∈ P : γ(x) < d},
P (γ, α, b, d) = {x ∈ P : b ≤ α(x), γ(x) ≤ d} ,
P (γ, θ, α, b, c, d) = {x ∈ P : b ≤ α(x), θ(x) ≤ c, γ(x) ≤ d},
and a closed set
R(γ, ψ, a, d) = {x ∈ P : a ≤ ψ(x), γ(x) ≤ d}.
To prove our main results, we need the following fixed-point theorem due to
Avery and Peterson in [8].
4
D.-B. WANG
EJDE-2007/95
Theorem 2.3. Let P be a cone in a real Banach space E. Let γ and θ be nonnegative continuous convex functionals on P , α be a nonnegative continuous concave
functionals on P and ψ be a nonnegative continuous functional on P satisfying
ψ(λx) ≤ λψ(x) for 0 ≤ λ ≤ 1, such that for some positive numbers h and d,
α(x) ≤ ψ(x)
and
kxk ≤ hγ(x),
for all x ∈ P (γ, d). Suppose that
F : P (γ, d) → P (γ, d)
is completely continuous and there exist positive numbers a, b and c with a < b such
that:
(i) {x ∈ P (γ, θ, α, b, c, d) : α(x) > b} =
6 ∅ and α(F x) > b for x in the set
P (γ, θ, α, b, c, d);
(ii) α(F x) > b for x ∈ P (γ, α, b, d) with θ(F x) > c;
(iv) 0 ∈
/ R(γ, ψ, a, d) and ψ(F x) < a for x ∈ R(γ, ψ, a, d) with ψ(x) = a.
Then F has at least three fixed points x1 , x2 , x3 ∈ P (γ, d) such that γ(xi ) ≤ d for
i = 1, 2, 3, and b < α(x1 ), a < ψ(x2 ) with α(x2 ) < b and ψ(x3 ) < a.
3. Existence of Three Positive Solutions
We note that u(t) is a solution of (1.1) if and only if

RT

B (Φ ( a(r)f (u(r), u(µ(r)))∇r))

 0R t q ηR T
u(t) = + 0 Φq ( s a(r)f (u(r), u(µ(r)))∇r)∆s, t ∈ [0, T ]T ,



ϕ(t),
t ∈ [−r, 0]T .
∆
Let E = Cld
([0, T ]T , R) be endowed with the norm
kuk = max
max |u(t)|, max |u∆ (t)| ,
t∈[0,T ]T
t∈[0,T ]Tk
so E is a Banach space. Define cone P ⊂ E by
P = u ∈ E : u is concave and nonnegative valued on [0, T ]T , and u∆ (T ) = 0 .
For each u ∈ E, extend u(t) to [−r, T ]T with u(t) = ϕ(t) for t ∈ [−r, 0]T . Define a
completely continuous operator F : P → E by
Z T
(F u)(t) =B0 (Φq (
a(r)f (u(r), u(µ(r)))∇r))
η
Z
+
0
t
Z
Φq (
T
a(r)f (u(r), u(µ(r)))∇r)∆s,
s
We seek for a point, u1 , of F in the cone P . Define
(
u1 (t), t ∈ [0, T ]T ,
u(t) =
ϕ(t), t ∈ [−r, 0]T .
Then u(t) denotes a positive solution of (1.1).
Lemma 3.1. If u ∈ P , then
(i) u(t) ≥ Tt maxt∈[0,T ]T |u(t)|, t ∈ [0, T ]T .
(ii) u(t) is increasing on t ∈ [0, T ]T .
(iii) u∆ (t) is decreasing on t ∈ [0, T ]Tk .
t ∈ [0, T ]T .
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THREE POSITIVE SOLUTIONS
5
Proof. Part (i) is of [14, Lemma 3.1]. Parts (ii) and (iii) are easy, so we omit them
here.
Lemma 3.2. The operator F maps P into P .
Proof. For each u ∈ P , F u ∈ E and (F u)(t) ≥ 0, for all t ∈ [0, T ]T . It follows from
Lemma 1.4 that
Z T
(F u)∆ (t) = Φq (
a(r)f (u(r), ϕ(r))∇r).
t
∆
Obviously (F u) (t) is a continuous function and (F u)∆ (t) ≥ 0, that is (F u)(t)
is increasing on [0, T ]T . Note that Φq is increasing, we have that (F u)∆ (t) is
decreasing.
If t ∈ [0, T ]Tk ∩Tk , then from [6, Theorem 2.3] it follows that (F u)∆∇ (t) ≤ 0;
i.e., F u is concave on [0, T ]T . This implies that F u ∈ P and F : P → P .
Let l ∈ T be fixed such that 0 < η < l < T , and set
Y1 = t ∈ [0, T ]T : µ(t) ≤ 0 ; Y2 = t ∈ [0, T ]T : µ(t) > 0 ; Y3 = Y1 ∩ [l, T ]T .
R
Throughout this paper, we assume Y3 6= φ and Y3 a(r)∇r > 0.
Define the nonnegative continuous concave functionals α, the nonnegative continuous convex functionals θ, γ, and the nonnegative continuous functionals ψ on
the cone P respectively as
γ(u) = kuk,
max u∆ (t),
θ(u) =
t∈[l,T ]Tk
α(u) = min u(t),
ψ(u) =
t∈[η,l]T
min u(t).
t∈[η,T ]T
In addition, by Lemma 3.1, we have α(u) = ψ(u) = u(η), θ(u) = u∆ (l) for each
u ∈ P . For convenience, we define
Z T
Z
ρ = (A + T )Φq (
a(r)∇r),
a(r)∇r), δ = (B + η)Φq (
0
Y3
Z
λ = (A + η)Φq (
T
a(r)∇r).
0
We now state growth conditions on f so that (1.1) has at least three positive
solutions.
Theorem 3.3. Let 0 < Tη a < b < d, ρb < δd, and suppose that f satisfies the
following conditions:
(H1) f (u, ϕ(s)) ≤ Φp ( dρ ), if 0 ≤ u ≤ d, uniformly in s ∈ [−r, 0]T ; f (u1 , u2 ) ≤
Φp ( dρ ), if 0 ≤ ui ≤ d, i = 1, 2,
(H2) f (u, ϕ(s)) > Φp ( δb ), if b ≤ u ≤ d, uniformly in s ∈ [−r, 0]T ,
(H3) f (u, ϕ(s)) < Φp ( λa ), if 0 ≤ u ≤ Tη a, uniformly in s ∈ [−r, 0]T ; f (u1 , u2 ) <
Φp ( λa ), if 0 ≤ ui ≤ Tη a, i = 1, 2.
Then (1.1) has at least three positive solutions of the form
(
ui (t), t ∈ [0, T ]T , i = 1, 2, 3,
u(t) =
ϕ(t), t ∈ [−r, 0]T ,
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D.-B. WANG
EJDE-2007/95
where γ(ui ) ≤ d for i = 1, 2, 3, b < α(u1 ), a < ψ(u2 ) with α(u2 ) < b and
ψ(u3 ) < a.
Proof. We first assert that F : P (γ, d) → P (γ, d). Let u ∈ P (γ, d), then γ(u) =
kuk ≤ d, consequently, 0 ≤ u(t) ≤ d for t ∈ [0, T ]T . From (H1), we have
|(F u)(t)|
Z
= B 0 Φq
T
a(r)f (u(r), u(µ(r)))∇r
η
t
Z
+
Φq
Z
0
T
a(r)f (u(r), u(µ(r)))∇r ∆s
s
Z T
a(r)f (u(r), u(µ(r)))∇r + T Φq
a(r)f (u(r), u(µ(r)))∇r
0
Z 0
i
hZ
= (A + T )Φq
a(r)f (u(r), ϕ(µ(r)))∇r +
a(r)f (u(r), u(µ(r)))∇r
≤ AΦq
T
Z
Y1
T
≤ (A + T )Φq
Y2
Z
a(r)∇r
d
0
ρ
= d,
and
|(F u)∆ (t)| = Φq
Z
T
a(r)f (u(r), u(µ(r)))∇r
a(r)f (u(r), u(µ(r)))∇r
t
≤ Φq
Z
T
0
= Φq
hZ
Z
a(r)f (u(r), ϕ(µ(r)))∇r +
Y1
T
≤ Φq
Z
a(r)f (u(r), u(µ(r)))∇r
i
Y2
a(r)∇r
0
d
ρ
d
≤ d.
=
(A + T )
Therefore F (u) ∈ P (γ, d), i.e., F : P (γ, d) → P (γ, d).
Secondly, we assert that {u ∈ P (γ, θ, α, b, c, d) : α(u) > b} =
6 φ and α(F u) > b
for u ∈ P (γ, θ, α, b, c, d).
Let u(t) = kb with k = ρδ > 1, then u(t) = kb > b and θ(u) = 0 < b. Furthermore, by ρb < δd we have γ(u) ≤ d. Let c = kb, then
{u ∈ P (γ, θ, α, b, c, d) :
α(u) > b} =
6 ∅.
Moreover, for all u ∈ P (γ, θ, α, b, kb, d), we have b ≤ u(t) ≤ d, t ∈ [η, T ]T . From
(H2), we see that
α(F u)
= (F u)(η)
Z
= B0 (Φq (
Z
≥ BΦq (
η
η
T
T
Z
η
Z
Φq (
a(r)f (u(r), u(µ(r)))∇r)) +
0
T
a(r)f (u(r), u(µ(r)))∇r)∆s
s
Z
a(r)f (u(r), u(µ(r)))∇r) + ηΦq (
η
T
a(r)f (u(r), u(µ(r)))∇r)
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THREE POSITIVE SOLUTIONS
7
Z T
a(r)f (u(r), u(µ(r)))∇r)
≥ (B + η)Φq (
l
Z
≥ (B + η)Φq (
a(r)f (u(r), ϕ(µ(r)))∇r)
Y3
Z
b
> (B + η)Φq (
a(r)∇r) = b,
δ
Y3
as required.
Thirdly, we assert that α(F u) > b for u ∈ P (γ, α, b, d) with θ(F u) > c. For all
u ∈ P (γ, α, b, d) with θ(F u) > kb, from Lemma 3.1 we have
Z T
θ(F u) = (F u)∆ (l) = Φq
a(r)f (u(r), u(µ(r)))∇r > kb.
l
So,
α(F u)
= (F u)(η)
Z
= B 0 Φq
T
a(r)f (u(r), u(µ(r)))∇r
+
η
≥ BΦq
Z
T
η
Z
0
a(r)f (u(r), u(µ(r)))∇r)∆s
s
Z
a(r)f (u(r), u(µ(r)))∇r + ηΦq (
η
T
Z
Φq (
T
a(r)f (u(r), u(µ(r)))∇r)
η
≥ (B + η)Φq
Z
T
a(r)f (u(r), u(µ(r)))∇r
l
ρ
> (B + η)kb = (B + η) b
δ
≥ (A + T )b > b.
This implies that α(F u) > b for u ∈ P (γ, α, b, d) with θ(F u) > c.
Finally, we assert that 0 ∈
/ R(γ, ψ, a, d) and ψ(F u) < a for u ∈ R(γ, ψ, a, d) with
ψ(u) = a .
Since ψ(0) = 0 < a, we have 0 ∈
/ R(γ, ψ, a, d). For all u ∈ R(γ, ψ, a, d) with
ψ(u) = mint∈[η,T ]T u(t) = u(η) = a, by Lemma 3.1 we have 0 ≤ u(t) ≤ Tη a, for
t ∈ [0, T ]T . From (H3), we have
ψ(F u) = (F u)(η)
Z
= B 0 Φq
T
a(r)f (u(r), u(µ(r)))∇r
η
Z
+
η
Φq
Z
0
T
a(r)f (u(r), u(µ(r)))∇r ∆s
s
Z T
a(r)f (u(r), u(µ(r)))∇r + ηΦq
a(r)f (u(r), u(µ(r)))∇r
0
Z 0
hZ
i
= (A + η)Φq
a(r)f (u(r), ϕ(µ(r)))∇r +
a(r)f (u(r), u(µ(r)))∇r
≤ AΦq
Z
T
Y1
T
< (A + η)Φq
Z
0
Y2
a(r)∇r
a
λ
= a,
which shows that condition (iv) of Theorem 2.3 is fulfilled.
8
D.-B. WANG
EJDE-2007/95
Thus, all the conditions of Theorem 2.3 are satisfied. Hence, F has at least three
fixed points u1 , u2 , u3 satisfying γ(ui ) ≤ d for i = 1, 2, 3, b < α(u1 ), a < ψ(u2 )
with α(u2 ) < b and ψ(u3 ) < a. Let
(
ui (t), t ∈ [0, T ]T , i = 1, 2, 3,
u(t) =
ϕ(t), t ∈ [−r, 0]T ,
which are three positive solutions of (1.1).
4. Example
Let T = [− 34 , − 14 ] ∪ {0, 34 } ∪ {( 12 )N0 }, where N0 denotes the set of all nonnegative
integers. Consider the following p-Laplacian functional dynamic equation on time
scale T,
8u3 (t)
= 0, t ∈ (0, 1)T ,
u3 (t) + u3 (t − 43 ) + 1
3
u0 (t) = ϕ(t) ≡ 0, t ∈ [− , 0]T ,
4
1
u(0) − B0 (u∆ ( )) = 0, u∆ (1) = 0,
4
[Φp (u∆ (t))]∇ +
(4.1)
where T = 1, p = 23 , a(t) ≡ 1, B = 1, A = 1, µ : [0, 1]T → [− 43 , 1]T and µ(t) = t− 43 ,
r = 34 , η = 14 , l =
3
4 ]T ,
1
2
and f (u, ϕ(s)) =
( 34 , 1]T ,
8u3
u3 +1 ,
f (u1 , u2 ) =
[ 12 , 34 ]T .
8u31
.
u31 +u32 +1
We deduce that
Y2 =
Y3 =
Y1 = [0,
5
Thus it is easy to see by calculating that ρ = 2, δ = 64
, λ = 45 . Choose a =
b = 1, d = 140, then we have 0 < Tη a < b < d, ρb < δd, then
r
d
140
f (u, ϕ(s)) < 8 < Φp ( ) =
≈ 8.3666, 0 ≤ u ≤ 140;
ρ
2
r
140
d
f (u1 , u2 ) < 8 < Φp ( ) =
≈ 8.3666, 0 ≤ u ≤ 140,
ρ
2
r
b
64
f (u, ϕ(s)) ≥ 4 > Φp ( ) =
≈ 3.5777, 1 ≤ u ≤ 140,
δ
5
r
8
a
1
1
f (u, ϕ(s)) ≤
≈ 0.008 < Φp ( ) =
≈ 0.1414, 0 ≤ u ≤
;
1001
λ
50
10
r
8
a
1
1
f (u1 , u2 ) ≤
≈ 0.008 < Φp ( ) =
≈ 0.1414, 0 ≤ u ≤
,
1002
λ
50
10
1
40 ,
Thus by Theorem 3.3, the (4.1) has at least three positive solutions of the form
(
ui (t), t ∈ [0, 1]T , i = 1, 2, 3,
u(t) =
ϕ(t), t ∈ [− 34 , 0]T ,
where γ(ui ) ≤ 140 for i = 1, 2, 3, 1 < α(u1 ),
1
ψ(u3 ) < 40
.
1
40
< ψ(u2 ) with α(u2 ) < 1 and
Acknowledgment. The author would like to thank the anonymous referees for
their valuable suggestions which led to an improvement of this paper.
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THREE POSITIVE SOLUTIONS
9
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Da-Bin Wang
Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou,
Gansu, 730050, China
E-mail address: wangdb@lut.cn