Electronic Journal of Differential Equations, Vol. 2007(2007), No. 19, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
THIRD-ORDER NONLOCAL PROBLEMS WITH
SIGN-CHANGING NONLINEARITY ON TIME SCALES
DOUGLAS R. ANDERSON, CHRISTOPHER C. TISDELL
Abstract. We are concerned with the existence and form of positive solutions
to a nonlinear third-order three-point nonlocal boundary-value problem on
general time scales. Using Green’s functions, we prove the existence of at least
one positive solution using the Guo-Krasnoselskii fixed point theorem. Due
to the fact that the nonlinearity is allowed to change sign in our formulation,
and the novelty of the boundary conditions, these results are new for discrete,
continuous, quantum and arbitrary time scales.
1. Statement of the problem
We will develop an interval of λ values whereby a positive solution exists for the
following nonlinear, third-order, three-point, nonlocal boundary-value problem on
arbitrary time scales
(px∆∆ )∇ (t) = λf (t, x(t)), t ∈ [t1 , t3 ]T ,
Z ξ2
αx(ρ(t1 )) − βx∆ (ρ(t1 )) =
a(t)x(t)∇t,
∆
x (t2 ) = 0,
∆∆
(px
Z
(1.1)
ξ1
η2
)(t3 ) =
b(t)(px∆∆ )(t)∇t,
(1.2)
η1
where: p is a left-dense continuous, real-valued function on T with p > 0; λ > 0 is
a real scalar;
(H1) the real scalars α, β > 0 and the three boundary points satisfy t1 < t2 <
t3 ∈ T such that
Z σ2 (t3 ) Z t2
Z
∆r
β t2 ∆r
0<
∆u +
< ∞;
α ρ(t1 ) p(r)
ρ(t1 )
u p(r)
(H2) the points ξi , ηi ∈ T satisfy
ρ(t1 ) ≤ η1 < η2 ≤ t3 ;
ρ(t1 ) < ξ1 < ξ2 < t2 ,
2000 Mathematics Subject Classification. 34B18, 34B27, 34B10, 39A10.
Key words and phrases. Boundary value problem; time scale; three-point; Green’s function.
c
2007
Texas State University - San Marcos.
Submitted October 31, 2006. Published January 27, 2007.
Supported by grant DP0450752 from the Australian Research Council’s Discovery Projects.
1
2
D. R. ANDERSON, C. C. TISDELL
EJDE-2007/19
(H3) the left-dense continuous real-valued functions on T satisfy a, b ≥ 0 with
Z η2
Z ξ2
b(t)∇t < 1;
a(t)∇t < α and 0 <
0<
η1
ξ1
2
(H4) the continuous function f : [ρ(t1 ), σ (t3 )]T ×[0, ∞) → (−∞, ∞) is such that
lim
x→+∞
f (t, x) unif
= +∞,
x
t ∈ [ξ1 , t2 ]T ;
(H5) there exist left-dense continuous functions y, z : [ρ(t1 ), σ 2 (t3 )]T → (0, ∞)
and a continuous function h : [0, ∞) → (0, ∞) such that
Z t3
(z(s) + y(s))∇s < ∞.
−y(t) ≤ f (t, x) ≤ z(t)h(x), 0 <
ρ(t1 )
Third-order differential equations, though less common in applications than
even-order problems, nevertheless do appear, for example in the study of quantum fluids; see Gamba and Jüngel [11]. Here we approach third-order problems
on general time scales, namely on any nonempty closed subset of the real line, to
include the discrete, continuous, and quantum calculus as special cases. Boundary
value problems on time scales that utilize both delta and nabla derivatives, such
as the one here, were first introduced by Atici and Guseinov [5]. Three-point and
right-focal boundary value problems, in both the continuous and discrete cases,
have been addressed in [1, 2, 3, 4], by Eloe and McKelvey [9], and recently by
Graef and Yang [12, 13], Sun [23], and Wong [25]. For more on existence of solutions to boundary value problems, see [7, Chapters 4 and 6-9], Davis, Erbe, and
Henderson [8], Erbe and Wang [10], the text by Guo and Lakshmikantham [14],
Henderson [15], Henderson and Thompson [17], Lan [19, 20], Ma and Thompson
[22], and Zhang and Liu [26]. Problem (1.3), (1.4) is an extension of the continuous
and discrete discussions of third-order right-focal boundary value problems to time
scales, and by the addition of the nonhomogeneous nonlocal boundary conditions
and the allowance of sign changes in the nonlinearity f , problem (1.1), (1.2) is
introduced for the first time on any time scale, including R, Z, and the quantum
time scale. One could also consider a third-order problem with derivatives in the
order of nabla, nabla, delta, but the results would be similar; other permutations
of nablas and/or deltas lead to a Green function that is less easy to calculate.
Clearly there are other approaches to the existence of positive solutions for dynamic equations on time scales than those featured in this work; for alternative approaches to the existence of solutions and multiple solutions to dynamic equations
on time scales, consult, for example, Bohner and Luo [6], Henderson [16], Ma, Du,
and Ge [21], and Tisdell, Drábek, and Henderson [24]. Underlying our technique,
however, will be Green’s function for the homogeneous, third-order, three-point
boundary-value problem
(px∆∆ )∇ (t) = 0,
∆
t ∈ [t1 , t3 ]T ,
∆
∆∆
αx(ρ(t1 )) − βx (ρ(t1 )) = x (t2 ) = (px
(1.3)
)(t3 ) = 0.
(1.4)
Green’s function for (1.3), (1.4) will be defined on [ρ(t1 ), σ 2 (t3 )]T , nonnegative on
[ρ(t1 ), σ 2 (t3 )]T contingent on the distance between boundary points, nondecreasing
on [ρ(t1 ), t2 ]T , and nonincreasing on [t2 , σ 2 (t3 )]T , as will be shown in the following
lemmas.
EJDE-2007/19
THIRD-ORDER THREE-POINT NONLOCAL PROBLEMS
3
Lemma 1.1. Green’s function corresponding to the problem (1.3), (1.4) is given
by
G(t, s)





s ∈ [ρ(t1 ), t2 ]T



=





s ∈ [t2 , σ 2 (t3 )]T


R
R
R s ∆r
β s
∆r
 t
∆u + α
:t<s
ρ(t1 ) p(r)
ρ(t1 ) u p(r)
: Rs Rs
R
s
β
∆r
∆r

∆u + α ρ(t1 ) p(r) : t ≥ s
ρ(t ) u p(r)
R 1 R
R
t2 ∆r
β t2
∆r
 t
∆u + α
ρ(t1 ) u p(r)
ρ(t1 ) p(r)
: R t R t2
R t R u ∆r
R
β t2
∆r

∆u + s s p(r)
∆u + α
ρ(t1 )
ρ(t1 ) u p(r)
:t<s
∆r
p(r)
: t ≥ s.
(1.5)
Proof. We follow the approach given, for example, in Kelley and Peterson [18,
Chapter 5]. As the Cauchy function y(·, s) satisfies the homogeneous time-scale
initial-value problem
(py ∆∆ (·, s))∇ (t) = 0,
y(s, s) = 0, y ∆ (s, s) = 0, y ∆∆ (ρ(s), s) = 1/p(ρ(s))
R t R u ∆r
it is easy to verify that y(t, s) = s s p(r)
∆u. Thus the Green function takes the
form
(

u1 (t, s) : t < s



s ∈ [ρ(t1 ), t2 ]T
:


v1 (t, s) : t ≥ s
(
G(t, s) =


u2 (t, s) : t < s

s ∈ [t2 , σ 2 (t3 )]T :

v2 (t, s) : t ≥ s,
where ui (t, s)+y(t, s) = vi (t, s) and vi (·, s) satisfy (px∆∆ )∇ (t) = 0, for i = 1, 2. Let
s ∈ [ρ(t1 ), t2 ]T . Then the boundary conditions are αu1 (ρ(t1 ), s) − βu∆
1 (ρ(t1 ), s) = 0
for t < s and v1∆ (t2 , s) = (pv1∆∆ (·, s))(t3 ) = 0 for s ≤ t. Solving for v1 , we see that
v1 (t, s) = k(s) for some function k. Since u1 = v1 − y, αu1 (ρ(t1 ), s) = βu∆
1 (ρ(t1 ), s)
for these s implies that
Z s Z s
Z
∆r
β s ∆r
k(s) =
∆u +
.
α ρ(t1 ) p(r)
ρ(t1 ) u p(r)
Thus for s ∈ [ρ(t1 ), t2 ]T ,
(R t
G(t, s) =
R s ∆r
∆u
ρ(t1 ) u p(r)
R s R s ∆r
∆u
ρ(t1 ) u p(r)
+
+
β
α
β
α
Rs
∆r
p(r)
∆r
ρ(t1 ) p(r)
1)
Rρ(t
s
:t<s
: t ≥ s.
Now let s ∈ [t2 , σ 2 (t3 )]T , so that the boundary conditions are αu2 (ρ(t1 ), s) −
∆
∆∆
βu∆
2 (ρ(t1 ), s) = u2 (t2 , s) = 0 for t < s and (pv2 (·, s))(t3 ) = 0 for s ≤ t. Clearly
Z
Z
Z
t2
t
∆r
β t2 ∆r u2 (t, s) = −q(s)
∆u +
α ρ(t1 ) p(r)
ρ(t1 ) u p(r)
for some function q. Using the fact that v2 = u2 + y and the remaining boundary
condition yields q(s) ≡ −1 and
R
R t2 ∆r
R
β t2
∆r
 t
∆+ α
:t<s
ρ(t1 ) u p(r)
ρ(t1 ) p(r)
G(t, s) = R t R t2
R
R
R
t u ∆r
β t2
∆r
∆r

∆ + s s p(r) ∆u + α ρ(t1 ) p(r) : t ≥ s
ρ(t1 ) u p(r)
for s ∈ [t2 , σ 2 (t3 )]T .
4
D. R. ANDERSON, C. C. TISDELL
EJDE-2007/19
Remark 1.2. As in (1.5) and the proof above, throughout the rest of the paper
we take
Z t Z t2
Z
∆r
β t2 ∆r
u2 (t) :=
∆u +
.
(1.6)
α ρ(t1 ) p(r)
ρ(t1 ) u p(r)
Lemma 1.3. Green’s function (1.5) corresponding to the problem (1.3), (1.4) satisfies
0 < G(t, s) ≤ G(t2 , s) ≤ u2 (t2 )
for (t, s) ∈ [ρ(t1 ), σ 2 (t3 )]T × (ρ(t1 ), σ 2 (t3 )]T if and only if (H1) holds; that is,
u2 (σ 2 (t3 )) > 0.
R
β s
∆r
> 0 and u1 (·, s) is
Proof. Fix s ∈ (ρ(t1 ), t2 ]T . Then u1 (ρ(t1 ), s) = α
ρ(t1 ) p(r)
increasing, so that 0 < u1 (t, s) ≤ u1 (s, s) for t ∈ [ρ(t1 ), s)T . But u1 (s, s) ≡ v1 (t, s)
for t ∈ [t2 , σ 2 (t3 )]T . It follows that G(t2 , s) ≥ G(t, s) > 0 for s ∈ (ρ(t1 ), t2 ]T
and t ∈ [ρ(t1 ), σ 2 (t3 )]T . Now fix s ∈ [t2 , σ 2 (t3 )]T . The branch u2 is positive at
ρ(t1 ), increases until t2 , and then decreases until s. We then switch to branch v2 ,
which continues to decrease, so that v2 (t, s) ≥ v2 (σ 2 (t3 ), s). As a function of s,
v2 (σ 2 (t3 ), s) is also decreasing, whence v2 (t, s) ≥ v2 (σ 2 (t3 ), σ 2 (t3 )) = u2 (σ 2 (t3 )) for
u2 given in (1.6). Thus G(t2 , s) ≥ G(t, s) for s ∈ [t2 , σ 2 (t3 )]T and t ∈ [ρ(t1 ), σ 2 (t3 )]T
as well, and G(t, s) > 0 for s ∈ [t2 , σ 2 (t3 )]T and t ∈ [ρ(t1 ), σ 2 (t3 )]T if and only if
(H1) holds.
Remark 1.4. If T = Z, α = 1, β = 0, and p(t) ≡ 1, then the necessary and
sufficient condition for the Green function to be positive is t2 − t1 − 1 ≥ t3 − t2 ; see
[2].
Lemma 1.5. Assume (H1). For any (t, s) ∈ [ρ(t1 ), σ 2 (t3 )]T × (ρ(t1 ), σ 2 (t3 )]T , the
Green function (1.5) corresponding to the problem (1.3), (1.4) satisfies, using (1.6),
G(t, s)
u2 (t)
≤
≤ 1.
u2 (t2 )
G(t2 , s)
Proof. The right-hand inequality follows from the previous lemma. For the lefthand inequality, we proceed by analyzing branches of the Green function (1.5). For
fixed t ∈ [ρ(t1 ), s)T and s ∈ (t, t2 ]T ,
R t R s ∆r
R
β s
∆r
∆u + α
G(t, s)
ρ(t1 ) u p(r)
ρ(t1 ) p(r)
= R s R s ∆r
=: φ(s).
R
β s
∆r
G(t2 , s)
∆u + α
ρ(t1 ) u p(r)
ρ(t1 ) p(r)
Then
φ∇ (s)
R
R R
R s ∆r
R
t
s s ∆r
β s
β
∆r
(t − s) ρ(t1 ) u p(r)
∆u + α
+
t
−
ρ(t
)
+
∆u
1
α
ρ(t1 ) p(r)
t u p(r)
.
=
R s R s ∆r
R
R
R
R
ρ(s) ρ(s) ∆r
β s
β ρ(s) ∆r
∆r
pρ (s) ρ(t1 ) u p(r) ∆u + α
∆u
+
p(r)
α ρ(t1 ) p(r)
ρ(t1 ) p(r)
ρ(t1 ) u
The denominator of φ∇ is clearly positive, so consider the numerator,
ψ(s) := ψ1 (s) + ψ2 (s)
Z t Z
:= (t − s)
s
Z sZ s
∆r
∆r
∆u + (t − ρ(t1 ))
∆u
ρ(t1 ) u p(r)
t
u p(r)
Z s
Z sZ s
β
∆r
∆r
+
(t − s)
+
∆u .
α
ρ(t1 ) p(r)
t
u p(r)
EJDE-2007/19
THIRD-ORDER THREE-POINT NONLOCAL PROBLEMS
5
Note that ψ(t) = 0; the first part satisfies
Z t Z s
Z t Z ρ(s)
(t − ρ(t1 ))ν(s)
∆r
∆r
ψ1∇ (s) =
−
∆u
=
−
∆u ≤ 0
pρ (s)
p(r)
ρ(t1 ) u p(r)
ρ(t1 ) u
for s ∈ (t, t2 ]T , while the second part satisfies
Z s
Z sZ s
Z t
β
∆r
∆r
β
∆r
(t − s)
ψ2 (s) ≤
+
∆u = (t − s)
≤ 0.
α
p(r)
p(r)
α
p(r)
ρ(t1 )
t
t
ρ(t1 )
Therefore, the whole numerator satisfies ψ(s) ≤ 0, so that φ∇ ≤ 0 and φ is nonincreasing as a function of s. Thus φ(s) ≥ φ(t2 ) for s ∈ (t, t2 ]; in other words,
G(t, t2 )
u2 (t)
G(t, s)
≥
=
.
G(t2 , s)
G(t2 , t2 )
u2 (t2 )
For s ∈ [ρ(t1 ), t2 ]T and t ∈ [s, σ 2 (t3 )]T ,
u2 (t)
G(t, s)
≡1≥
.
G(t2 , s)
u2 (t2 )
If s ∈ [t2 , σ 2 (t3 )]T and t ∈ [ρ(t1 ), s)T , then
u2 (t)
G(t, s)
=
.
G(t2 , s)
u2 (t2 )
Finally, if s ∈ [t2 , σ 2 (t3 )]T and t ∈ [s, σ 2 (t3 )]T , then
R t R u ∆r
u2 (t) + s s p(r)
∆u
u2 (t)
G(t, s)
=
≥
.
G(t2 , s)
u2 (t2 )
u2 (t2 )
2. Exploring the nonlocal problem
In this section we turn our attention to the problem
(px∆∆ )∇ (t) = λy(t),
t ∈ [t1 , t3 ]T ,
(2.1)
with nonlocal boundary conditions (1.2), where y is as in (H5), and λ > 0. Assume
(H2) and (H3), and and use (1.6) to define
Z η2
Z ξ2
D := u2 (t2 ) 1 −
b(t)∇t
a(t)∇t − α < 0.
(2.2)
η1
ξ1
Lemma 2.1. Assume (H1) through (H5). Then the nonhomogeneous dynamic
equation (2.1) with boundary conditions (1.2) has a unique solution x∗ , where for
t ∈ [ρ(t1 ), σ 2 (t3 )]T ,
Z t3
∗
x (t) = λ
G(t, s)y(s)∇s + A(y)u2 (t) + B(y) (u2 (t2 ) − u2 (t))
(2.3)
ρ(t1 )
holds, where: G(t, s) is the Green function (1.5) of the boundary-value problem
(1.3), (1.4); and the functionals A and B are defined using (1.6) by
R η2
R ξ2
R ξ2
b(t)∇t
−
1
a(t)u
(t)∇t
+
u
(t
)
α
−
a(t)∇t
2
2
2
1 η1
ξ1
ξ1 ,
A(y) :=
R ξ2
R t3
R η2
R t3
D
b(t)
y(s)∇s
∇t
a(t)
G(t,
s)y(s)∇s
∇t
η1
t
ξ1
ρ(t1 )
(2.4)
6
D. R. ANDERSON, C. C. TISDELL
1
B(y) :=
D
R η2
b(t)∇t − 1
η1
R η
2 b(t) R t3 y(s)∇s ∇t
η1
t
EJDE-2007/19
R ξ2
a(t)u
(t)∇t
2
ξ
1
. (2.5)
R ξ2
R t3
a(t) ρ(t1 ) G(t, s)y(s)∇s ∇t
ξ1
Proof. For y as in (H5), we show that the function x∗ given in (2.3) is a solution
of (2.1) with conditions (1.2) only if A(y) and B(y) are given by (2.4) and (2.5),
respectively. If x∗ is a solution of (2.1), (1.2), then
Z t
Z t3
x∗ (t) = λ
G(t, s)y(s)∇s + λ
G(t, s)y(s)∇s + Au2 (t) + B(u2 (t2 ) − u2 (t))
ρ(t1 )
t
for some constants A and B. Taking the delta derivative with respect to t yields
Z t2
Z t3
Z t
Z t2
∆r
∆r
−B
;
G∆ (t, s)y(s)∇s+A
G∆ (t, s)y(s)∇s+λ
x∗∆ (t) = λ
p(r)
p(r)
t
t
ρ(t1 )
t
since p times the delta derivative of this expression is
Z t3
(px∗∆∆ )(t) = −λ
y(s)∇s − A + B,
t
we see that (2.1) holds. It is also clear that x∗∆ (t2 ) = 0 is satisfied. To meet the
other two boundary conditions in (1.2), we must have at ρ(t1 ) that
Z ξ2
Z t3
αBu2 (t2 ) =
a(t) λ
G(t, s)y(s)∇s+Au2 (t)+B(u2 (t2 )−u2 (t)) ∇t, (2.6)
ξ1
ρ(t1 )
while at t3 we have
Z
η2
−A + B =
Z
b(t) − λ
η1
t3
y(s)∇s − A + B ∇t.
(2.7)
t
Combining (2.6) and (2.7), we arrive at the system of equations
Z ξ2
h Z ξ2
i
A
a(t)u2 (t)∇t + B
a(t) (u2 (t2 ) − u2 (t)) ∇t − αu2 (t2 )
ξ1
ξ1
Z
ξ2
= −λ
ξ1
Z
a(t)
t3
G(t, s)y(s)∇s ∇t
ρ(t1 )
and
hZ
A
η2
η1
Z
i
h
b(t)∇t − 1 + B 1 −
η2
η1
Z
i
b(t)∇t = −λ
η2
η1
Z
b(t)
t3
y(s)∇s ∇t.
t
The determinant of the coefficients of A and B is D, given by (2.2), which is
negative, and by elementary linear algebra we verify (2.4) and (2.5) with λ factored
out. Also note that A(y) > B(y) > 0 since D < 0 and
R
R η2
t
b(t) t 3 y(s)∇s ∇t
η1
Rη
.
A(y) − B(y) =
1 − η12 b(t)∇t
Corollary 2.2. Assume (H1) through (H5). Then the unique solution x∗ as in
(2.3) of the problem (2.1), (1.2) satisfies x∗ (t) ≥ 0 for t ∈ [ρ(t1 ), σ 2 (t3 )].
EJDE-2007/19
THIRD-ORDER THREE-POINT NONLOCAL PROBLEMS
7
Proof. From Lemma 1.3 we know that on the Green function (1.5) satisfies G(t, s) ≥
0. Assumption (H3) applied to (2.4) and (2.5) imply that A(y) > B(y) > 0.
Lemma 2.3. Assume (H1) through (H5). Then the unique solution x∗ as in (2.3)
of the problem (2.1), (1.2) satisfies
θkx∗ k ≤ x∗ (t) ≤ λθΘ
[ρ(t1 ), σ 2 (t3 )]T ,
on
where using (1.6) we take
u2 (ρ(t1 )) u2 (σ 2 (t3 ))
θ := min
,
∈ (0, 1),
u2 (t2 )
u2 (t2 )
kx∗ k :=
max
t∈[ρ(t1 ),σ 2 (t3 )]T
x∗ (t) = x∗ (t2 ),
(2.8)
and
Θ :=
1
u2 (t2 ) 1 + Ā
θ
Z
t3
y(s)∇s
(2.9)
ρ(t1 )
for
1
Ā :=
D
R
R
R
η2
η b(t)∇t − 1 ξξ2 a(t)u2 (t)∇t + u2 (t2 ) α − ξξ2 a(t)∇t 1
1
R1
.
R ξ2
η2
a(t)∇t
b(t)∇t
u
(t
)
η1
2 2
ξ1
Proof. ¿From previous work, it is clear that for all t ∈ [ρ(t1 ), σ 2 (t3 )]T ,
Z t3
∗
∗
G(t2 , s)y(s)∇s + A(y)u2 (t2 ) .
x (t) ≤ x (t2 ) = λ
ρ(t1 )
2
For t ∈ [ρ(t1 ), σ (t3 )]T , from Lemma 1.3 and Lemma 1.5, the Green function (1.5)
satisfies
n u (ρ(t )) u (σ 2 (t )) o
G(t, s)
u2 (t)
2
1
2
3
≥
≥ min
,
= θ ∈ (0, 1)
G(t2 , s)
u2 (t2 )
u2 (t2 )
u2 (t2 )
by (H1) and (1.6), and
x∗ (t)
Z
=λ
t3
G(t, s)
u2 (t)
G(t2 , s)y(s)∇s + A(y)
u2 (t2 ) + B(y)(u2 (t2 ) − u2 (t))
u2 (t2 )
ρ(t1 ) G(t2 , s)
Z t3
≥λ
θG(t2 , s)y(s)∇s + A(y)θu2 (t2 )
ρ(t1 )
= θkx∗ k.
Consequently, θkx∗ k ≤ x∗ (t) for all t ∈ [ρ(t1 ), σ 2 (t3 )]T . For D in (2.2) and A(y) in
(2.4),
A(y)
R η2
R ξ2
R ξ2
b(t)∇t
−
1
a(t)u
(t)∇t
+
u
(t
)
α
−
a(t)∇t
2
2 2
1 η1
ξ1
ξ1 =
R η2
R t3
R ξ2
R t3
D
b(t)
y(s)∇s
∇t
a(t)
G(t,
s)y(s)∇s
∇t
η1
t
ξ1
ρ(t1 )
R
R ξ2
R ξ2
η2
Z
1 η1 b(t)∇t − 1 ξ1 a(t)u2 (t)∇t + u2 (t2 ) α − ξ1 a(t)∇t t3
≤
y(s)∇s
R η2
Rξ
D b(t)∇t
u2 (t2 ) ξ12 a(t)∇t
ρ(t1 )
η1
Z t3
y(s)∇s < ∞.
= Ā
ρ(t1 )
8
D. R. ANDERSON, C. C. TISDELL
EJDE-2007/19
As a result, for t ∈ [ρ(t1 ), σ 2 (t3 )]T ,
x∗ (t) ≤ λ
Z
t3
G(t2 , s)y(s)∇s + A(y)u2 (t2 )
ρ(t1 )
t3
Z
≤ λu2 (t2 )(1 + Ā)
y(s)∇s
ρ(t1 )
≤ λθΘ
using (2.8), (2.9), and (2.10).
3. An existence result on cones
Let B denote the Banach space C[ρ(t1 ), σ 2 (t3 )]T with the norm
kxk =
|x(t)|.
sup
t∈[ρ(t1 ),σ 2 (t3 )]T
Define the cone P ⊂ B by
P = {x ∈ B : x(t) ≥ θkxk on [ρ(t1 ), σ 2 (t3 )]T },
where θ is given in (2.8). Consider the related boundary-value problem
(px∆∆ )∇ (t) = λf ∗ (t, x(t)),
Z
∆
αx(ρ(t1 )) − βx (ρ(t1 )) =
t ∈ [t1 , t3 ]T ,
ξ2
a(t)x(t)∇t,
ξ1
x∆ (t2 ) = 0,
(px∆∆ )(t3 ) =
Z
η2
b(t)(px∆∆ )(t)∇t,
η1
where
f ∗ (t, x(t)) := f (t, x† (t)) + y(t),
x† (t) := max{x(t) − x∗ (t), 0},
such that x∗ given in (2.3) is the solution of (2.1), (1.2), and y is from (H5).
For any fixed x ∈ P, x† ≤ x ≤ kxk and by (H5),
Z
t3
G(t, s)f ∗ (s, x(s))∇s
ρ(t1 )
Z t3
≤
G(t2 , s) z(s)h(x† (s)) + y(s) ∇s
ρ(t1 )
≤
max h(τ ) + 1
0≤τ ≤kxk
Z
t3
G(t2 , s) (z(s) + y(s)) ∇s < ∞.
ρ(t1 )
For A in (2.4) and using (2.10), we have
Z
t3
A(z + y) ≤ Ā
(z(s) + y(s))∇s;
ρ(t1 )
(3.1)
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THIRD-ORDER THREE-POINT NONLOCAL PROBLEMS
9
likewise for B in (2.5) and using (H5),
B(z + y)
R η2
R ξ2
b(t)∇t
−
1
a(t)u
(t)∇t
1 2
Rηt13
R ξ2
R t3 ξ1
=
R η2
D η b(t) t (z(s) + y(s))∇s ∇t ξ a(t) ρ(t ) G(t, s)(z(s) + y(s))∇s ∇t
1
1
1
Z
R η
R ξ2
2
t3
a(t)u2 (t)∇t 1 η1 b(t)∇t − 1
ξ1
Rξ
(z(s) + y(s))∇s
≤
R
D ηη2 b(t)∇t
u2 (t2 ) ξ 2 a(t)∇t ρ(t1 )
1
1
< ∞.
This allows us to define for y ∈ P the operator T : P → B for t ∈ [ρ(t1 ), σ 2 (t3 )]T
by
Z t3
(T x)(t) := λ
G(t, s)f ∗ (s, x(s))∇s + A(f ∗ )u2 (t) + B(f ∗ )(u2 (t2 ) − u2 (t)) ,
ρ(t1 )
(3.2)
using (2.4), (2.5), and (3.1).
Lemma 3.1. Assume (H1) through (H5). Then T : P → P is completely continuous.
Proof. For any x ∈ P, Lemmas 1.3, 1.5 and Lemma 2.3 imply that (T x)(t) ≥ θkT xk
on [ρ(t1 ), σ 2 (t3 )]T , so that T (P) ⊆ P. By a standard application of the ArzelaAscoli Theorem, T is completely continuous.
To establish an existence result we will employ the following fixed point theorem
due to Guo and Krasnoselskii [14], and seek a fixed point of T in P.
Theorem 3.2. Let E be a Banach space, P ⊆ E be a cone, and suppose that S1 ,
S2 are bounded open balls of E centered at the origin with S 1 ⊂ S2 . Suppose further
that L : P ∩ (S 2 \ S1 ) → P is a completely continuous operator such that either
(i) kLyk ≤ kyk, y ∈ P ∩ ∂S1 and kLyk ≥ kyk, y ∈ P ∩ ∂S2 , or
(ii) kLyk ≥ kyk, y ∈ P ∩ ∂S1 and kLyk ≤ kyk, y ∈ P ∩ ∂S2
holds. Then L has a fixed point in P ∩ (S 2 \ S1 ).
Theorem 3.3. Assume (H1) through (H5). Then there exists λ∗ > 0 such that the
third-order nonlocal time scale boundary value problem (1.1), (1.2) has at least one
positive solution in P for any λ ∈ (0, λ∗ ).
Proof. By Lemma 3.1, T : P → P given by (3.2) is completely continuous. Take
S1 := {x ∈ B : kxk < Θ} for Θ given in (2.9), and let
R t3
)
(
y(s)∇s
ρ(t
)
1
R
λ∗ := min 1, .
t3
θ max h(τ ) + 1 ρ(t
(z(s) + y(s))∇s
1)
0≤τ ≤Θ
Then for any x ∈ P ∩ ∂S1 ,
0 ≤ x† (s) ≤ x(s) ≤ kxk = Θ,
s ∈ [ρ(t1 ), σ 2 (t3 )]T ,
10
D. R. ANDERSON, C. C. TISDELL
EJDE-2007/19
and, for Ā as in the statement of Lemma 2.3,
Z t3
(T x)(t) ≤ λ
G(t2 , s)f ∗ (s, x(s))∇s + A(f ∗ )u2 (t2 )
ρ(t1 )
≤λ
max h(τ ) + 1
Z
0≤τ ≤kxk
t3
G(t2 , s) (z(s) + y(s)) ∇s
ρ(t1 )
Z
+ λĀ
max h(τ ) + 1
0≤τ ≤kxk
t3
z(s) + y(s) ∇s u2 (t2 )
ρ(t1 )
max h(τ ) + 1
≤ λ∗ u2 (t2 ) 1 + Ā
0≤τ ≤kxk
Z
t3
(z(s) + y(s)) ∇s
ρ(t1 )
≤ Θ = kxk.
Hence kT xk ≤ kxk for x ∈ P ∩ ∂S1 . Pick Υ ∈ R such that Υ > 0 and
Z t2
λΥθ
1≤
G(ξ1 , s)∇s.
Θ + 1 ξ1
By (H4), for any t ∈ [ξ1 , t2 ]T , there exists a constant K > 0 such that f (t, y) > Υy
. If S2 := {y ∈ B : kyk < Q},
for y > K. Pick Q := max λ(Θ + 1), Θ + 1, K(Θ+1)
θ
then for any x ∈ P ∩ ∂S2 and t ∈ [ρ(t1 ), σ 2 (t3 )]T ,
λΘ
x(t)
Q
λΘ λΘ ≥ 1−
x(t) ≥ 1 −
x(t)
Q
λ(Θ + 1)
x(t)
≥ 0.
=
Θ+1
x(t) − x∗ (t) ≥ x(t) − λθΘ ≥ x(t) −
Thus
min (x(t) − x∗ (t)) ≥
t∈[ξ1 ,t2 ]T
min
t∈[ξ1 ,t2 ]T
x(t)
θQ
≥
≥ K,
Θ+1
Θ+1
so that
min (T x)(t)
t∈[ξ1 ,t2 ]T
=
min
λ
Z
t∈[ξ1 ,t2 ]T
Z
t2
≥λ
t3
G(t, s)f ∗ (s, x(s))∇s + A(f ∗ )u2 (t) + B(f ∗ )(u2 (t2 ) − u2 (t))
ρ(t1 )
G(ξ1 , s)f ∗ (s, x(s))∇s
ξ1
Z
t2
≥ λΥ
G(ξ1 , s)(x(s) − x∗ (s))∇s ≥
ξ1
λΥθkxk
=
Θ+1
Z
λΥθQ
Θ+1
Z
t2
G(ξ1 , s)∇s
ξ1
t2
G(ξ1 , s)∇s ≥ kxk.
ξ1
Hence for x ∈ P ∩ ∂S2 we have kT xk ≥ kxk. By Theorem 3.2, T has a fixed point
x such that Θ ≤ kxk ≤ Q. But then
x(t) − x∗ (t) ≥ θΘ − λθΘ ≥ (1 − λ)θΘ ≥ 0.
EJDE-2007/19
THIRD-ORDER THREE-POINT NONLOCAL PROBLEMS
11
As a consequence, this x solves the boundary-value problem
∇
px∆∆ (t) = λ(f (t, x(t) − x∗ (t)) + y(t)), t ∈ [t1 , t3 ]T ,
Z ξ2
∆
αx(ρ(t1 )) − βx (ρ(t1 )) =
a(t)x(t)∇t,
x∆ (t2 ) = 0,
(px∆∆ )(t3 ) =
ξ1
η2
Z
b(t)(px∆∆ )(t)∇t.
η1
Now set X(t) := x(t) − x∗ (t) for x∗ given in (2.3). Then px∆∆
∇
px∗∆∆ . As x∗ is the solution of (2.1), (1.2), we see that
∇
pX ∆∆ (t) = λf (t, X(t)), t ∈ [t1 , t3 ]T ,
Z ξ2
∆
a(t)X(t)∇t,
αX(ρ(t1 )) − βX (ρ(t1 )) =
X ∆ (t2 ) = 0,
pX
∆∆
Z
∇
= pX ∆∆
∇
+
ξ1
η2
(t3 ) =
b(t) pX ∆∆ (t)∇t,
η1
in other words, X is a positive solution of the third-order nonlocal time scale boundary value problem (1.1), (1.2).
As remarked in the Introduction, the results in this paper are new for ordinary
differential equations (when T = R) and for difference equations (when T = Z).
We now provide an example to illustrate that conditions (H1)–(H5) are naturally
satisfied.
Example 3.4. Consider for T = R and the following choices: t1 = 0, t2 = 1/2,
t3 = 1; p = 1; α = 1 = β; f (t, x) = t + x2 ; ξ1 = 1/8, ξ2 = 1/4; η1 = 5/6, η2 = 7/8;
a(t) = t = b(t). Then, for h(x) := 1 + x2 with y = 1 and z = 1, the boundary
value problem (1.1), (1.2) has at least one positive solution in P for any λ ∈ (0, λ∗ ),
where λ∗ ≈ 0.232513.
With these choices, (1.1), (1.2) reduces to a third-order BVP involving an ordinary differential equation. It is not difficult to verify that conditions (H1)–(H5) are
satisfied. In particular, note that
o
n
5
5
=
≈ 0.232513.
λ∗ = min 1,
2
8(2 + Θ )
8(2 + Θ2 )
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Douglas R. Anderson
Department of Mathematics and Computer Science, Concordia College, Moorhead, MN
56562, USA
E-mail address: andersod@cord.edu
Christopher C. Tisdell
School of Mathematics and Statistics, University of New South Wales, Sydney, UNSW
2052, Australia
E-mail address: cct@unsw.edu.au