Electronic Journal of Differential Equations, Vol. 2007(2007), No. 107, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF SOLUTIONS TO A THIRD-ORDER
MULTI-POINT PROBLEM ON TIME SCALES
DOUGLAS R. ANDERSON, JOAN HOFFACKER
Abstract. We are concerned with the existence and form of solutions to
nonlinear third-order three-point and multi-point boundary-value problems
on general time scales. Using the corresponding Green function, we prove
the existence of at least one positive solution using the Guo-Krasnosel’skii
fixed point theorem. Moreover, a third-order multi-point eigenvalue problem
is formulated, and eigenvalue intervals for the existence of a positive solution
are found.
1. introduction
We will establish the corresponding Green function whereby conditions can be
given such that a positive solution exists for the following nonlinear third-order
three-point boundary value problem on arbitrary time scales
(px∆∆ )∇ (t) + a(t)f (x(t)) = 0,
x(ρ(t1 )) = x∆ (ρ(t1 )) = 0,
t ∈ [t1 , t3 ]T ,
(1.1)
x∆ (σ(t3 )) − αx∆ (t2 ) = 0,
(1.2)
where: p is a right-dense continuous, real-valued function with 0 < p(t) ≤ 1 on T;
the boundary points from T satisfy t1 < t2 < t3 , with t2 /α ∈ T such that
(H1) the constants d and α satisfy
Z
σ(t3 )
d :=
ρ(t1 )
∆τ
−α
p(τ )
Z
t2
ρ(t1 )
∆τ
>0
p(τ )
R σ(t3 )
and
∆τ
ρ(t ) p(τ )
1 < α < R t2 1
∆τ
ρ(t1 ) p(τ )
;
(H2) the continuous function f : [0, ∞) → [0, ∞) is such that the following exist:
f0 := lim+
x→0
f (x)
x
f∞ := lim
x→∞
f (x)
.
x
(H3) the left-dense continuous function a : [ρ(t1 ), σ(t3 )]T → [0, ∞) is such that a
is not identically zero on [t2 /α, t2 ]T .
2000 Mathematics Subject Classification. 34B18, 34B27, 34B10, 39A10.
Key words and phrases. Boundary value problem; time scale; third order; Green function.
c
2007
Texas State University - San Marcos.
Submitted July 8, 2007. Published August 7, 2007.
1
2
D. R. ANDERSON, J. HOFFACKER
EJDE-2007/107
If T = R, then (1.1), (1.2) is the ordinary third-order three-point boundary value
problem
(px00 )0 (t) + a(t)f (x(t)) = 0,
0
x(t1 ) = x (t1 ) = 0,
t ∈ [t1 , t3 ]R ,
0
x (t3 ) − αx0 (t2 ) = 0.
If T = Z, then (1.1), (1.2) is the discrete third-order three-point boundary value
problem
∇ p∆2 x)(t) + a(t)f (x(t) = 0, t ∈ [t1 , t3 ]Z ,
x(t1 − 1) = ∆x(t1 − 1) = 0,
∆x(t3 + 1) − α∆x(t2 ) = 0,
where ∆y(t) = y(t + 1) − y(t) and ∇y(t) = y(t) − y(t − 1). As a final illustration, if
T is a quantum time scale for some real q > 1, then (1.1), (1.2) is the third-order
three-point quantum boundary value problem
Dq (pDq (Dq x)) (t) + a(t)f (x(t)) = 0,
x(t1 /q) = Dq x(t1 /q) = 0,
t ∈ [t1 , t3 ]T ,
Dq x(qt3 ) − αDq x(t2 ) = 0,
where the quantum derivatives are given by the difference quotients
Dq y(t) =
y(qt) − y(t)
(q − 1)t
and Dq y(t) =
y(t) − y(t/q)
.
(1 − 1/q)t
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 [5]. Here we approach a third-order three-point
problem 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.
Of late there have been several papers on third-order boundary value problems.
Hopkins and Kosmatov [9]; Li [10]; Liu, Ume, and Kang [11, 12]; and Minghe and
Chang [13] have all recently considered third-order problems. All of these papers,
however, were two-point problems with T = R. Graef and Yang [6], Sun [16],
and Wong [17] consider three-point focal problems, while Palamides and Smyrlis
consider the three-point boundary conditions
x(0) = x00 (η) = x(1) = 0,
T = [0, 1]R .
On general time scales there are also a few papers on third-order problems. Sun
[15] considers a third-order two-point boundary value problem; a couple of papers
on third-order three-point boundary value problems considered on general time
scales are [1, 2] in the right-focal case. Note that 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 [3]. For more on existence of solutions
to boundary value problems, see [4, Chapters 4 and 6-9], the text by Guo and
Lakshmikantham [7], and Zhang and Liu [18].
Problem (1.1), (1.2) is an extension of the unit interval boundary value problem
x000 (t) + a(t)f (x(t)) = 0,
0
x(0) = x (0) = 0,
t ∈ (0, 1)R ,
0
αx (η) = x0 (1),
to arbitrary time scales [8]; in other words, take T = R, p ≡ 1, t1 = 0, t2 = η and
t3 = 1 in (1.1), (1.2) to get the results in [8]. One could also consider a third-order
problem with derivatives in the order of nabla, nabla, delta, but the results would
EJDE-2007/107
THIRD-ORDER THREE-POINT BVP
3
be similar; other permutations of nablas and/or deltas lead to a Green function
that is less easy to calculate.
2. preliminary lemmas
Underlying our technique will be the Green function for the homogeneous, thirdorder, three-point boundary-value problem
−(px∆∆ )∇ (t) = 0,
∆
x(ρ(t1 )) = x (ρ(t1 )) = 0,
t ∈ [t1 , t3 ]T ,
∆
∆
αx (t2 ) = x (σ(t3 )).
(2.1)
(2.2)
The Green function for (2.1), (2.2) will be defined, nonnegative, and bounded above
on [ρ(t1 ), σ 2 (t3 )]T × [t1 , σ(t3 )]T , as will be shown in the following lemmas.
Lemma 2.1. For y ∈ Cld [ρ(t1 ), σ(t3 )]T , the boundary value problem
(px∆∆ )∇ (t) + y(t) = 0,
∆
t ∈ [t1 , t3 ]T ,
∆
(2.3)
∆
x(ρ(t1 )) = x (ρ(t1 )) = 0, αx (t2 ) = x (σ(t3 ))
(2.4)
R σ(t3 )
has a unique solution x(t) = ρ(t1 ) G(t, s)y(s)∇s, where the Green function corresponding to the problem (2.1), (2.2) is given by
 R
R t2 ∆τ R t R u
σ(t3 ) ∆τ
1
∆τ

−
α
∆u

s p(τ )
ρ(t1 ) ρ(t1 ) p(τ )

 d R tsR u p(τ )

∆τ


: s ≤ min{t2 , t}
− s s p(τ ) ∆u


R
 R σ(t3 )
R
R
t
t2 ∆τ
u
1
∆τ
∆τ
∆u
: t ≤ s ≤ t2
G(t, s) = d s
p(τ ) − α s p(τ )
ρ(t1 ) ρ(t1 ) p(τ )


R
R
R
R
R

t u ∆τ
σ(t3 ) ∆τ
t
u
1
∆τ


∆u − s s p(τ

d
p(τ )
) ∆u : t2 ≤ s ≤ t
s
ρ(t1 ) ρ(t1 ) p(τ )

 R
R

Ru

σ(t3 ) ∆τ
t
∆τ
1
∆u
: max{t2 , t} ≤ s.
d
p(τ )
s
ρ(t1 ) ρ(t1 ) p(τ )
(2.5)
Proof. We follow the approach given in the case T = R in [8]. If ρ(t1 ) ≤ t ≤ t2 ,
then
Z σ(t3 )
x(t) =
G(t, s)y(s)∇s
ρ(t1 )
Z
Z t2
Z u
∆τ
∆τ t
∆τ
−α
∆u
p(τ )
p(τ ) ρ(t1 ) ρ(t1 ) p(τ )
ρ(t1 ) d
s
s
Z tZ u
i
∆τ
−
∆u y(s)∇s
s
s p(τ )
Z t2 h Z σ(t3 )
Z t2
Z
Z u
i
1
∆τ
∆τ t
∆τ
+
−α
∆u y(s)∇s
d s
p(τ )
p(τ ) ρ(t1 ) ρ(t1 ) p(τ )
t
s
Z σ(t3 ) h Z σ(t3 )
Z t Z u
i
1
∆τ
∆τ
+
∆u y(s)∇s
d s
p(τ ) ρ(t1 ) ρ(t1 ) p(τ )
t2
Z t2 Z σ(t3 )
Z t2
Z σ(t3 ) Z σ(t3 )
h
i
1
∆τ
∆τ ∆τ
=
−α
y(s)∇s +
y(s)∇s
d ρ(t1 )
p(τ )
p(τ )
p(τ )
s
s
s
t2
h Z t Z u ∆τ
i Z t Z t Z u ∆τ
×
∆u −
∆u y(s)∇s.
ρ(t1 )
s
s p(τ )
ρ(t1 ) ρ(t1 ) p(τ )
Z
=
t
h1Z
σ(t3 )
4
D. R. ANDERSON, J. HOFFACKER
EJDE-2007/107
If σ 2 (t3 ) ≥ t ≥ t2 , then
Z σ(t3 )
G(t, s)y(s)∇s
x(t) =
ρ(t1 )
Z u
Z
Z t2
∆τ
∆τ
∆τ t
=
−α
∆u
p(τ )
p(τ ) ρ(t1 ) ρ(t1 ) p(τ )
ρ(t1 ) d
s
s
Z tZ u
i
∆τ
−
∆u y(s)∇s
s p(τ )
s
Z u
Z
Z tZ u
Z t h Z σ(t3 )
i
∆τ
∆τ t
∆τ
1
∆u −
∆u y(s)∇s
+
p(τ ) ρ(t1 ) ρ(t1 ) p(τ )
s
s p(τ )
s
t2 d
Z σ(t3 ) h Z σ(t3 )
Z
Z
u
i
1
∆τ t
∆τ
+
∆u y(s)∇s
d s
p(τ ) ρ(t1 ) ρ(t1 ) p(τ )
t
Z
Z
Z t2
Z σ(t3 ) Z σ(t3 )
i
1 h t2 σ(t3 ) ∆τ
∆τ ∆τ
=
−α
y(s)∇s +
y(s)∇s
d ρ(t1 )
p(τ )
p(τ )
p(τ )
s
s
t2
s
i Z t Z t Z u ∆τ
h Z t Z u ∆τ
∆u −
∆u y(s)∇s.
×
ρ(t1 ) ρ(t1 ) p(τ )
ρ(t1 )
s
s p(τ )
Z
t2
h1Z
σ(t3 )
For the remainder of the proof let
Z
Z
Z t2
Z σ(t3 ) Z σ(t3 )
i
1 h t2 σ(t3 ) ∆τ
∆τ ∆τ
k(s) :=
−α
y(s)∇s +
y(s)∇s .
d ρ(t1 )
p(τ )
p(τ )
p(τ )
s
s
t2
s
Thus for all t ∈ [ρ(t1 ), σ 2 (t3 )]T ,
Z t Z u
Z t Z t Z u
∆τ
∆τ
x(t) = k(s)
∆u −
∆u y(s)∇s.
ρ(t1 ) ρ(t1 ) p(τ )
ρ(t1 )
s
s p(τ )
Note that x(ρ(t1 )) = 0. Taking a delta derivative,
Z t Z t
Z t
∆τ
∆τ
−
y(s)∇s.
x∆ (t) = k(s)
ρ(t1 )
ρ(t1 ) p(τ )
s p(τ )
Again it is easy to see that x∆ (ρ(t1 )) = 0. To verify the third boundary condition,
check that
x∆ (σ(t3 )) − αx∆ (t2 )
Z σ(t3 ) Z
= dk(s) −
ρ(t1 )
s
σ(t3 )
∆τ y(s)∇s + α
p(τ )
Z
t2
ρ(t1 )
Z
s
t2
∆τ y(s)∇s
p(τ )
= 0.
It follows that the boundary conditions (2.4) are satisfied. To finish the proof,
another delta derivative yields
Z t
y(s)
k(s)
∆∆
−
∇s,
x (t) =
p(t)
ρ(t1 ) p(t)
which results in (px∆∆ )∇ (t) = −y(t), so that (2.3) is satisfied as well.
EJDE-2007/107
THIRD-ORDER THREE-POINT BVP
5
We now seek bounds on the Green function given in (2.5). For later reference,
set
Z
Z
s ∆τ σ(t3 ) ∆τ 1
g(s) := (α + 1) σ 2 (t3 ) − ρ(t1 )
.
(2.6)
d
p(τ )
s
ρ(t1 ) p(τ )
Lemma 2.2. Assume (H1). The Green function (2.5) corresponding to the problem
(2.1), (2.2) satisfies
0 ≤ G(t, s) ≤ g(s)
2
for (t, s) ∈ [ρ(t1 ), σ (t3 )]T × [t1 , σ(t3 )]T .
Proof. First we show that G(t, s) is nonnegative. For t ≤ s ≤ t2 , consider the
coefficient
Z
Z t2
1 σ(t3 ) ∆τ
∆τ c1 (s) :=
.
−α
d s
p(τ )
p(τ )
s
α−1
Since c1 (ρ(t1 )) = 1 and c∆
1 (s) = dp(s) > 0, c1 (s) ≥ 1 for all s ≥ t1 . It follows that
branches 1,2, and 4 of G(t, s) in (2.5) are nonnegative, so we consider the third
branch of G(t, s):
Z
Z
Z u
Z tZ u
i
1 h σ(t3 ) ∆τ t
∆τ
∆τ
∆u − d
∆u .
d
p(τ ) ρ(t1 ) ρ(t1 ) p(τ )
s
s
s p(τ )
For t2 ≤ s ≤ t let
v(t, s) :=
σ(t3 )
Z
s
∆τ p(τ )
t
Z
ρ(t1 )
Z
u
ρ(t1 )
∆τ
∆u − d
p(τ )
Z tZ
s
s
u
∆τ
∆u.
p(τ )
Then
Z tZ s
i
∆τ
∆τ
v(t, s) =
∆u +
∆u
s
ρ(t1 ) p(τ )
s
ρ(t1 ) ρ(t1 ) p(τ )
Z
Z
Z
σ(t3 ) ∆τ
t u ∆τ
∆u
−d
+
p(τ )
s
s
s p(τ )
Z tZ s
Z σ(t3 ) ∆τ h Z s Z u ∆τ
i
∆τ
=
∆u +
∆u
p(τ )
s
ρ(t1 ) ρ(t1 ) p(τ )
s
ρ(t1 ) p(τ )
Z t2 ∆τ Z t Z u ∆τ
Z s ∆τ Z t Z u ∆τ
+α
∆u −
∆u
ρ(t1 ) p(τ )
s
ρ(t1 ) p(τ )
s
s p(τ )
s p(τ )
Z σ(t3 ) ∆τ Z s Z u ∆τ
Z t2 ∆τ Z t Z u ∆τ
=
∆u + α
∆u
p(τ )
s
ρ(t1 ) ρ(t1 ) p(τ )
ρ(t1 ) p(τ )
s
s p(τ )
Z s ∆τ h Z σ(t3 ) ∆τ Z t ∆τ Z t Z u ∆τ
i
+
−
∆u .
p(τ )
ρ(t1 ) p(τ )
s
s p(τ )
s
s p(τ )
Z
σ(t3 )
∆τ h
p(τ )
Z
s
Z
u
Clearly this is nonnegative if the last term in brackets is nonnegative. For the
remainder of the proof set
Z σ(t3 ) ∆τ Z t ∆τ Z t Z u ∆τ
j(t) :=
−
∆u
p(τ )
s
s p(τ )
s
s p(τ )
for t2 ≤ s ≤ t. Then j(s) = 0, and
1
j (t) =
p(t)
∆
Z
s
σ(t3 )
∆τ
−
p(τ )
Z
s
t
∆τ
≥0
p(τ )
6
D. R. ANDERSON, J. HOFFACKER
EJDE-2007/107
since 0 < p(t) ≤ 1 on T by assumption. Thus j(t) is nonnegative, guaranteeing overall that v(t, s) is nonnegative, so that ultimately G(t, s) is nonnegative as
claimed.
Now we show that G(t, s) ≤ g(s) on [ρ(t1 ), σ 2 (t3 )]T ×[t1 , σ(t3 )]T , where g is given
in (2.6). For any fixed s ∈ [t1 , σ(t3 )]T , a delta derivative of G(t, s) with respect to
t yields
 R σ(t )
R t2 ∆τ R t
R t ∆τ
3
∆τ
∆τ
1

−
α
− s p(τ
: s ≤ min{t2 , t}

d
p(τ )
)
s
s p(τ )
ρ(t1 ) p(τ )


R
R

R

t
t2 ∆τ

∆τ
∆τ
 d1 sσ(t3 ) p(τ
: t ≤ s ≤ t2
) − α s p(τ )
ρ(t1 ) p(τ )
∆t
R
R
G (t, s) =
R
σ(t3 ) ∆τ
t
t ∆τ
∆τ
1


− s p(τ
: t2 ≤ s ≤ t

p(τ )
)
ρ(t1 ) p(τ )
d s


R
R

σ(t3 ) ∆τ
t
∆τ
1
: max{t , t} ≤ s.
d
s
p(τ )
2
ρ(t1 ) p(τ )
Then rewriting we see that
0 ≤ G∆t (t, s)
h
i
R s
Rt
∆τ
∆τ

d
+
(α
−
1)

p(τ
)
p(τ
)
ρ(t1 )

 ρ(t1 )
h
i

Rt
Rs


∆τ
∆τ

d
+
(α
−
1)
ρ(t1 ) p(τ )
ρ(t1 ) p(τ )
1
R
R
R
R
=
s
t ∆τ
t2
σ(t3 ) ∆τ
∆τ
d

+
α

p(τ )
ρ(t1 )
ρ(t1 ) p(τ )
s p(τ )
t




∆τ
 R σ(t3 ) ∆τ R t
s
p(τ )
ρ(t1 ) p(τ )
and rewriting again we obtain
R
 R σ(t )
s
3
α
∆τ
∆τ


d
p(τ )
s
ρ(t1 ) p(τ )



R
Rs


∆τ
∆τ
 αd sσ(t3 ) p(τ
)
ρ(t1 ) p(τ )
∆t
R
R
0 ≤ G (t, s) ≤
σ(t3 ) ∆τ
s
α+1
∆τ



d
p(τ
)
s
ρ(t1 ) p(τ )
 


∆τ
 α R σ(t3 ) ∆τ R s
d
≤
σ 2 (t
s
p(τ )
ρ(t1 ) p(τ )
: s ≤ min{t2 , t}
: t ≤ s ≤ t2
∆τ
p(τ )
: t2 ≤ s ≤ t
: max{t2 , t} ≤ s,
: s ≤ min{t2 , t}
: t ≤ s ≤ t2
: t2 ≤ s ≤ t
(2.7)
: max{t2 , t} ≤ s
g(s)
.
3 ) − ρ(t1 )
Delta integration from ρ(t1 ) to t yields G(t, s) ≤ g(s) for g(s) given in (2.6).
Lemma 2.3. Assume (H1). The Green function (2.5) corresponding to the problem
(2.1), (2.2) satisfies
G(t, s) ≥ γg(s),
min{α − 1, α}
γ :=
R t2 /α R u
∆τ
∆u
ρ(t1 ) p(τ )
R σ(t3 ) ∆τ
ρ(t1 ) ρ(t1 ) p(τ )
ρ(t1 )
(α + 1) σ 2 (t3 ) −
∈ (0, 1)
(2.8)
for (t, s) ∈ [t2 /α, t2 ]T × [t1 , σ(t3 )]T , where g(s) is given in (2.6).
Proof. If s = σ(t3 ), or if t1 is a left-dense point and s = t1 , then the result follows
from (2.6). Thus consider the cases where (t, s) ∈ [t2 /α, t2 ]T × (ρ(t1 ), σ(t3 ))T . For
EJDE-2007/107
THIRD-ORDER THREE-POINT BVP
7
s ≤ t ≤ t2 ,
G(t, s)
=
g(s)
R
d
σ(t3 ) ∆τ
p(τ )
s
R
=
−α
(α − 1)
≥
∆τ
p(τ )
s
Ru
t
∆τ
∆u
ρ(t1 ) ρ(t1 ) p(τ )
R
s
(α − 1) ρ(t1 )
+
R t2
(α + 1)
−
dg(s)
R
∆τ
p(τ )
R
Ru
t
∆τ
∆u
ρ(t1 ) ρ(t1 ) p(τ )
dg(s)
R t R u ∆τ
∆u
s p(τ )
s
−d
RtRu
s
∆τ
∆u
s p(τ )
Ru
t
∆τ
∆u
ρ(t1 ) ρ(t1 ) p(τ )
dg(s)
Ru
∆τ
Rt
∆u
ρ(t1 ) ρ(t1 ) p(τ )
R σ(t3 ) ∆τ
σ 2 (t3 ) − ρ(t1 ) s
p(τ )
≥ γ.
For t ≤ s ≤ t2 ,
G(t, s)
=
g(s)
R
σ(t3 ) ∆τ
p(τ )
s
−α
R t2
∆τ
p(τ )
s
R
Ru
t
∆τ
∆u
ρ(t1 ) ρ(t1 ) p(τ )
dg(s)
Rt
Ru
∆τ
∆u
ρ(t1 ) ρ(t1 ) p(τ )
d
=
+ (α − 1)
R
s
∆τ
ρ(t1 ) p(τ )
R
Ru
t
∆τ
∆u
ρ(t1 ) ρ(t1 ) p(τ )
dg(s)
Rt
≥
(α
Ru
∆τ
(α − 1) ρ(t1 ) ρ(t1 ) p(τ
) ∆u
R σ(t3 ) ∆τ
+ 1) σ 2 (t3 ) − ρ(t1 ) s
p(τ )
≥ γ.
For t ≤ t2 ≤ s,
Rt Ru
∆τ
∆u
G(t, s)
ρ(t1 ) ρ(t1 ) p(τ )
Rs
=
g(s)
(α + 1) σ 2 (t3 ) − ρ(t1 ) ρ(t1 )
∆τ
p(τ )
≥ γ.
In all cases the statement holds.
3. an existence result on cones
Let B denote the Banach space C[ρ(t1 ), σ 2 (t3 )]T with the norm
kxk =
sup
|x(t)|.
t∈[ρ(t1 ),σ 2 (t3 )]T
Define the cone P ⊂ B by
P = {x ∈ B : x(t) ≥ 0 for t ∈ [ρ(t1 ), σ 2 (t3 )]T , x(t) ≥ γkxk on [t2 /α, t2 ]T }.
For x ∈ P, define
Z
σ(t3 )
Ax(t) :=
t ∈ [ρ(t1 ), σ 2 (t3 )]T .
G(t, s)a(s)f (x(s))∇s,
ρ(t1 )
From Lemma 2.2, for t ∈ [ρ(t1 ), σ 2 (t3 )]T ,
Z σ(t3 )
Z
0 ≤ Ax(t) =
G(t, s)a(s)f (x(s))∇s ≤
ρ(t1 )
σ(t3 )
g(s)a(s)f (x(s))∇s,
ρ(t1 )
so that
Z
σ(t3 )
kAxk ≤
g(s)a(s)f (x(s))∇s.
ρ(t1 )
(3.1)
8
D. R. ANDERSON, J. HOFFACKER
By Lemma 2.3 and (3.1), for t ∈ [t2 /α, t2 ]T ,
Z
Z σ(t3 )
G(t, s)a(s)f (x(s))∇s ≥ γ
Ax(t) =
EJDE-2007/107
σ(t3 )
g(s)a(s)f (x(s))∇s ≥ γkAxk,
ρ(t1 )
ρ(t1 )
giving us
Ax(t) ≥ γkAxk,
[t2 /α, t2 ]T ,
and AP ⊂ P. Furthermore, it is straightforward to verify that A : P → P is a
completely continuous operator whose fixed points are solutions of (1.1), (1.2).
To establish an existence result we will employ the following fixed point theorem
due to Guo and Krasnosel’skii [7], and seek a fixed point of T in P.
Theorem 3.1. 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 3.2. Assume (H1), (H2), and (H3) hold. Then the boundary value problem (1.1), (1.2) has at least one positive solution if
(i) f0 = 0 and f∞ = ∞ (f is superlinear), or
(ii) f0 = ∞ and f∞ = 0 (f is sublinear).
Proof. The proof in the general time-scale setting is similar to that given in the
case T = R in [8] and is omitted.
4. nonlinear multi-point problem
In the next two sections we consider the related multi-point boundary value
problem
(px∆∆ )∇ (t) + λa(t)f (x(t)) = 0,
x(ρ(t1 )) = x∆ (ρ(t1 )) = 0,
t ∈ [t1 , t3 ]T ,
x∆ (σ(t3 )) − αx∆ (t2 ) =
n
X
(4.1)
αi x∆ (ξi ),
(4.2)
i=1
where: p is a right-dense continuous, real-valued function with 0 < p(t) ≤ 1 on
T; λ > 0 is a real scalar; the boundary points from T satisfy t1 < t2 < t3 , with
t2 /α ∈ T such that
(K1) the constants d and α satisfy
Z
σ(t3 )
d :=
ρ(t1 )
∆τ
−α
p(τ )
Z
t2
ρ(t1 )
∆τ
>0
p(τ )
R σ(t3 )
and
∆τ
ρ(t ) p(τ )
1 < α < R t2 1
∆τ
ρ(t1 ) p(τ )
;
(K2) the coefficients satisfy αi ≥ 0 for i = 1, 2, · · · , n and the points ξi ∈
(ρ(t1 ), σ 2 (t3 ))T are such that
Z ξi
n
X
∆τ
ξ1 < ξ2 < · · · < ξn and d −
αi
> 0;
p(τ
)
ρ(t1 )
i=1
EJDE-2007/107
THIRD-ORDER THREE-POINT BVP
9
(K3) the continuous function f : [0, ∞) → [0, ∞) is such that the following exist:
f0 := lim+
x→0
f (x)
,
x
f (x)
;
x→∞ x
f∞ := lim
(K4) the left-dense continuous function a : [ρ(t1 ), σ(t3 )]T → [0, ∞) is such that
∃t∗ ∈ [t2 /α, t2 ]T 3 a(t∗ ) > 0.
(4.3)
By the novelty of the multi-point boundary conditions, problem (4.1), (4.2) is introduced for the first time on any time scale, including R, Z, and the quantum time
scale.
We now turn our attention to the problem
(px∆∆ )∇ (t) + λy(t) = 0,
t ∈ [t1 , t3 ]T ,
(4.4)
with multi-point boundary conditions (4.2), where y : [ρ(t1 ), σ 2 (t3 )]T → (0, ∞) is a
left-dense continuous function, and λ > 0.
Lemma 4.1. Assume (K1) and (K2). If y ∈ Cld [ρ(t1 ), σ 2 (t3 )] with y ≥ 0, then
the nonhomogeneous dynamic equation (4.4) with boundary conditions (4.2) has a
unique solution x∗ on t ∈ [ρ(t1 ), σ 2 (t3 )]T given by
Z t Z u
Z σ(t3 )
∆τ
∆u ,
(4.5)
x∗ (t) = λ
G(t, s)y(s)∇s + B(y)
ρ(t1 )
ρ(t1 ) ρ(t1 ) p(τ )
where: G(t, s) is the Green function (2.5) of the boundary value problem (2.1), (2.2)
and the functional B is defined by
Z ξi
Z σ(t3 )
n
n
X
∆τ −1 X
αi
αi
B(y) := d −
G∆t (ξi , s)y(s)∇s.
(4.6)
p(τ
)
ρ(t
)
ρ(t
)
1
1
i=1
i=1
Proof. Let y ∈ Cld [ρ(t1 ), σ 2 (t3 )] with y ≥ 0; we show that the function x∗ given in
(4.5) is a solution of (4.4) with conditions (4.2) only if B(y) is given by (4.6). If x∗
is a solution of (4.4), (4.2), then
Z t
Z σ(t3 )
Z t Z u
∆τ
∆u
x∗ (t) = λ
G(t, s)y(s)∇s + λ
G(t, s)y(s)∇s + B
ρ(t1 )
t
ρ(t1 ) ρ(t1 ) p(τ )
for some constant B. Taking the delta derivative with respect to t yields
Z t
Z σ(t3 )
Z t
∆τ
x∗∆ (t) = λ
G∆ (t, s)y(s)∇s + λ
G∆ (t, s)y(s)∇s + B
;
ρ(t1 )
t
ρ(t1 ) p(τ )
clearly x∗ (ρ(t1 )) = x∗∆ (ρ(t1 )) = 0 as G(t, s) satisfies (1.2). Since p times the delta
derivative of this expression is
Z t h Z σ(t3 )
Z t2
i
1
∆τ
∆τ (px∗∆∆ )(t) = λ
−α
− 1 y(s)∇s
p(τ )
p(τ )
ρ(t1 ) d
s
s
Z t2 h Z σ(t3 )
Z t2
1
∆τ
∆τ i
+λ
−α
y(s)∇s + B
d s
p(τ )
p(τ )
t
s
Z
Z
λ σ(t3 ) σ(t3 ) ∆τ +
y(s)∇s,
d t2
p(τ )
s
10
D. R. ANDERSON, J. HOFFACKER
EJDE-2007/107
we see that
∗∆∆ ∇
(px
) (t) = λ
h1Z
σ(t3 )
∆τ
−α
p(τ )
d t
h 1 Z σ(t3 )
−λ
d t
Z
t2
i
∆τ − 1 y(t)
p(τ )
t
Z t2
∆τ
∆τ i
y(t) = −λy(t)
−α
p(τ )
p(τ )
t
and (4.4) holds; this works regardless of the placement of t in [t1 , t3 ]T . To meet the
other boundary condition in (4.2), we must have that
Bd =
n
X
αi
hZ
σ(t3 )
Z
∆
ρ(t1 )
ρ(t1 )
i=1
ξi
G (ξi , s)y(s)∇s + B
∆τ i
,
p(τ )
from which (4.6) follows.
Corollary 4.2. Assume (K1) and (K2). If y ∈ Cld [ρ(t1 ), σ 2 (t3 )] with y ≥ 0, then
the unique solution x∗ as in (4.5) of the problem (4.4), (4.2) satisfies x∗ (t) ≥ 0 for
t ∈ [ρ(t1 ), σ 2 (t3 )].
Proof. From Lemma 2.2 we know that on [ρ(t1 ), σ 2 (t3 )]T × [t1 , σ(t3 )]T the Green
function (2.5) satisfies G(t, s) ≥ 0. ¿From equation (2.7) and (K2) we have that
B(y) ≥ 0 for y ≥ 0.
Lemma 4.3. Assume (K1) and (K2). If y ∈ Cld [ρ(t1 ), σ 2 (t3 )] with y ≥ 0, then the
unique solution x∗ as in (4.5) of the problem (4.4), (4.2) satisfies
γkx∗ k ≤ x∗ (t),
t ∈ [t2 /α, t2 ]T
for γ given in (2.8).
Proof. Let y ∈ Cld [ρ(t1 ), σ 2 (t3 )] with y ≥ 0. From previous work in Lemma 2.2, it
is clear that for all t ∈ [ρ(t1 ), σ 2 (t3 )]T and B(y) given in (4.6),
∗
x (t) ≤ λ
Z
σ(t3 )
Z
σ 2 (t3 )
Z
u
g(s)y(s)∇s + B(y)
ρ(t1 )
ρ(t1 )
ρ(t1 )
∆τ
∆u .
p(τ )
(4.7)
For t ∈ [t2 /α, t2 ]T , from Lemma 2.3 and the definition of γ in (2.8) we have
∗
Z
σ(t3 )
Z
t
Z
u
∆τ
∆u
ρ(t1 )
ρ(t1 ) ρ(t1 ) p(τ )
Z
Z
t2 /α Z u
σ(t3 )
∆τ
≥λ
γg(s)y(s)∇s + B(y)
∆u
ρ(t1 )
ρ(t1 )
ρ(t1 ) p(τ )
R
σ 2 (t ) R σ(t ) ∆τ
Z σ(t3 )
(α + 1) ρ(t1 )3 ρ(t13) p(τ
) ∆u
= γλ
g(s)y(s)∇s + B(y)
min{α − 1, α}
ρ(t1 )
x (t) = λ
G(t, s)y(s)∇s + B(y)
(4.8)
≥ γkx∗ k.
The proof is complete.
EJDE-2007/107
THIRD-ORDER THREE-POINT BVP
11
5. eigenvalue intervals
To establish eigenvalue intervals for the eigenvalue problem (4.1), (4.2) we will
employ Theorem 3.1. To that end, let B denote the Banach space C[ρ(t1 ), σ 2 (t3 )]T
with the norm kxk = supt∈[ρ(t1 ),σ2 (t3 )]T |x(t)|. Define the cone P ⊂ B by
P = {x ∈ B : x(t) ≥ 0 for t ∈ [ρ(t1 ), σ 2 (t3 )]T , x(t) ≥ γkxk on [t2 /α, t2 ]T },
where γ is given in (2.8). When x ∈ P define the operator T : P → B for t ∈
[ρ(t1 ), σ 2 (t3 )]T by
Z t Z u
Z σ(t3 )
∆τ
(T x)(t) := λ
G(t, s)a(s)f (x(s))∇s + B(af (x))
∆u , (5.1)
ρ(t1 )
ρ(t1 ) ρ(t1 ) p(τ )
using (4.6). We seek a fixed point of T in P by establishing the hypotheses of
Theorem 3.1.
Lemma 5.1. Assume (K1) through (K4). Then T : P → P is completely continuous.
Proof. Consider the integral operator T in (5.1). If x ∈ P, then by Lemma 2.2 we
have, as in (4.7) and (4.8),
Z σ2 (t3 ) Z u
Z σ(t3 )
∆τ
∆u ,
(T x)(t) ≤ λ
g(s)a(s)f (x(s))∇s + B(af (x))
ρ(t1 )
ρ(t1 )
ρ(t1 ) p(τ )
so that for t ∈ [t2 /α, t2 ]T ,
Z
Z σ(t3 )
(T x)(t) ≥ λ
γg(s)a(s)f (x(s))∇s + B(af (x))
ρ(t1 )
= γλ
Z
t2 /α
ρ(t1 )
(α + 1)
σ(t3 )
g(s)a(s)f (x(s))∇s + B(af (x))
u
Z
∆τ
∆u
ρ(t1 ) p(τ )
R σ2 (t3 ) R σ(t3 ) ∆τ
ρ(t1 )
∆u
ρ(t1 ) p(τ )
min{α − 1, α}
ρ(t1 )
≥ γkT xk.
Therefore, T : P → P. Moreover, T is completely continuous by a typical application of the Ascoli-Arzela Theorem.
For G(t, s) in (2.5) and B (4.6), define the constant
Z σ(t3 )
Z σ2 (t3 ) Z
J :=
g(s)a(s)∇s + B(a)
ρ(t1 )
ρ(t1 )
u
ρ(t1 )
∆τ
∆u.
p(τ )
(5.2)
Theorem 5.2. Assume (K1) through (K4). Then for each λ satisfying
1
f∞ γ
R t2
t2 /α
G(σ 2 (t
3 ), s)a(s)∇s
<λ<
1
f0 J
(5.3)
there exists at least one positive solution of (4.1), (4.2) in P.
Proof. Let J be as in (5.2), λ as in (5.3), and let > 0 be such that
1
(f∞ − )γ
R t2
G(τ, s)a(s)∇s
t2 /α
≤λ≤
1
.
(f0 + )J
(5.4)
12
D. R. ANDERSON, J. HOFFACKER
EJDE-2007/107
First consider f0 . There exists an R1 > 0 such that f (x) ≤ (f0 + )x for 0 <
x ≤ R1 by the definition of f0 . Pick x ∈ P with kxk = R1 . ¿From (4.6) we have
|B(af (x))| ≤ B(a)kf (x)k. Using Lemma 2.2 we have
Z t Z u
Z σ(t3 )
∆τ
(T x)(t) = λ
G(t, s)a(s)f (x(s))∇s + B(af (x))
∆u
ρ(t1 )
ρ(t1 ) ρ(t1 ) p(τ )
2
Z
Z
Z
u
σ (t3 )
σ(t3 )
∆τ
∆u
≤ λkf (x)k
g(s)a(s)∇s + B(a)
ρ(t1 ) p(τ )
ρ(t1 )
ρ(t1 )
≤ λ(f0 + )kxkJ ≤ kxk
from the right side of (5.4). As a result, kT xk ≤ kxk. Thus, take
Ω1 := {x ∈ B : kxk < R1 }
so that kT xk ≤ kxk for x ∈ P ∩ ∂Ω1 .
Next consider f∞ . Again by definition there exists an R20 > R1 such that f (x) ≥
(f∞ − )x for x ≥ R20 ; take R2 = max{2R1 , R20 /Γ}. If x ∈ P with kxk = R2 , then
for s ∈ [t2 /α, t2 ]T we have
x(s) ≥ γkxk = γR2 .
(5.5)
Define Ω2 := {x ∈ B : kxk < R2 }; using (5.5) for s ∈ [t2 /α, t2 ] we get
(T x)(σ 2 (t3 ))
Z
Z σ(t3 )
2
=λ
G(σ (t3 ), s)a(s)f (x(s))∇s + B(af (x))
Z
ρ(t1 )
t2
≥λ
G(σ 2 (t3 ), s)a(s)f (x(s))∇s ≥ λ(f∞ − )
t2 /α
Z
σ 2 (t3 )
ρ(t1 )
t2
Z
u
ρ(t1 )
∆τ
∆u
p(τ )
G(σ 2 (t3 ), s)a(s)x(s)∇s
t2 /α
Z
t2
G(σ 2 (t3 ), s)a(s)∇s ≥ R2 = kxk,
≥ λ(f∞ − )γR2
t2 /α
where we have used the left side of (5.4). Hence we have shown that
kT xk ≥ kxk,
x ∈ P ∩ ∂Ω2 .
An application of Theorem 3.1 yields the conclusion of the theorem; in other words,
T has a fixed point x in P ∩ (Ω2 \ Ω1 ) with R1 ≤ kxk ≤ R2 .
Theorem 5.3. Assume (K1) through (K4). Then for each λ satisfying
1
f0 γ
R t2
t2 /α
G(σ 2 (t
3 ), s)a(s)∇s
<λ<
1
f∞ J
(5.6)
there exists at least one positive solution of (4.1), (4.2) in P.
Proof. Let J be as in (5.2), λ as in (5.6), and let > 0 be such that
1
(f0 − )γ
R t2
t2 /α
G(τ, s)a(s)∇s
≤λ≤
1
.
(f∞ + )J
(5.7)
Again let T be the operator defined in (5.1). We once more seek a fixed point of T
in P by establishing the hypotheses of Theorem 3.1.
EJDE-2007/107
THIRD-ORDER THREE-POINT BVP
13
First consider f0 . There exists an R1 > 0 such that f (x) ≥ (f0 − )x for
0 < x ≤ R1 by the definition of f0 . Pick x ∈ P with kxk = R1 . For s ∈ [t2 /α, t2 ]T
we have
x(s) ≥ γkxk = γR1 .
(5.8)
Using the left side of (5.7) and (5.8) we get, for s ∈ [t2 /α, t2 ]T ,
(T x)(σ 2 (t3 ))
Z
Z σ(t3 )
=λ
G(σ 2 (t3 ), s)a(s)f (x(s))∇s + B(af (x))
ρ(t1 )
σ 2 (t3 )
ρ(t1 )
Z
Z
u
ρ(t1 )
∆τ
∆u
p(τ )
t2
G(σ 2 (t3 ), s)a(s)x(s)∇s
≥ λ(f0 − )
t2 /α
Z
t2
≥ λ(f0 − )R1 γ
G(σ 2 (t3 ), s)a(s)∇s ≥ R1 = kxk,
t2 /α
Therefore kT xk ≥ kxk. This motivates us to define
Ω1 := {x ∈ B : kxk < R1 },
whereby our work above confirms
kT xk ≥ kxk,
x ∈ P ∩ ∂Ω1 .
Next consider f∞ . Again by definition there exists an R20 > R1 such that f (x) ≤
(f∞ + )x for x ≥ R20 ; take R2 = max{2R1 , R20 /Γ}. If f is bounded, there exists
M > 0 with f (x) ≤ M for all x ∈ (0, ∞). Let
Z σ2 (t3 ) Z u
o
n
Z σ(t3 )
∆τ
∆u .
R2 := max 2R20 , λM
g(s)a(s)∇s + B(a)
ρ(t1 )
ρ(t1 )
ρ(t1 ) p(τ )
If x ∈ P with kxk = R2 , then we have
Z
Z σ(t3 )
(T x)(t) ≤ λM
g(s)a(s)∇s + B(a)
ρ(t1 )
σ 2 (t3 )
ρ(t1 )
Z
u
ρ(t1 )
∆τ
∆u ≤ R2 = kxk.
p(τ )
As a result, kT xk ≤ kxk. Thus, take
Ω2 := {x ∈ B : kxk < R2 }
so that kT xk ≤ kxk for x ∈ P ∩ ∂Ω2 . If f is unbounded, take R2 := max{2R1 , R20 }
such that f (x) ≤ f (R2 ) for 0 < x ≤ R2 . If x ∈ P with kxk = R2 , then we have
Z t Z u
Z σ(t3 )
∆τ
(T x)(t) = λ
G(t, s)a(s)f (x(s))∇s + B(af (x))
∆u
ρ(t1 )
ρ(t1 ) ρ(t1 ) p(τ )
2
Z
Z
Z
σ (t3 )
u
σ(t3 )
∆τ
∆u
≤ λf (R2 )
g(s)a(s)∇s + B(a)
ρ(t1 )
ρ(t1 )
ρ(t1 ) p(τ )
≤ λ(f∞ + )R2 J ≤ R2 = kxk,
where we have used the left side of (5.7). Hence we have shown that
kT xk ≤ kxk,
x ∈ P ∩ ∂Ω2
if we take
Ω2 := {x ∈ B : kxk < R2 }.
As before an application of Theorem 3.1 yields the conclusion that T has a fixed
point x in P ∩ (Ω2 \ Ω1 ) with R1 ≤ kxk ≤ R2 .
14
D. R. ANDERSON, J. HOFFACKER
EJDE-2007/107
Corollary 5.4. Assume (K1) through (K4). If f is sublinear (i.e., f0 = ∞ and
f∞ = 0), or if f is superlinear (i.e., f0 = 0 and f∞ = ∞), then for any λ > 0 the
boundary value problem (4.1), (4.2) has at least one positive solution in P.
Proof. For the superlinear claim, use (5.3) of Theorem 5.2; for the sublinear claim,
use (5.6) of Theorem 5.3.
As remarked earlier, the results in this section 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 (K1) through (K4) are
naturally satisfied.
Example 5.5. Consider for T = R and the following choices: t1 = 0, t2 = 1/2,
t3 = 1; p ≡ 1; continuous f (x) such that f0 and f∞ exist; α = 3/2; α1 = 1/2;
ξ1 = 1/4; a(t) = t. Then the boundary value problem (4.1), (4.2) has at least one
positive solution in P for any
0.824
1718.45
<λ<
.
f∞
f0
With these choices, (4.1), (4.2) reduces to a third-order four-point boundary value
problem
x000 (t) + λtf (x(t)) = 0,
t ∈ [0, 1]R ,
0
x(0) = x (0) = 0,
0
x (1) − 3x0 (1/2)/2 = x0 (1/4)/2.
It is not difficult to verify that conditions (K1) through (K4) are satisfied. Some
calculations lead to γ = 1/90, g(s) = 10s(1 − s), B(s) = 73/96, J = 233/192, and
R 1/2
sG(1, s)ds = 181/3456, so that we can find the interval in (5.3) to be
1/3
192
311040
<λ<
,
181f∞
233f0
which is approximately
1718.45
0.824
<λ<
.
f∞
f0
References
[1] D. R. Anderson and A. Cabada, Third-order right-focal multi-point problems on time scales,
J. Difference Equations Appl., 12:9 (2006) 919–935.
[2] D. R. Anderson and C. C. Tisdell, Third-order nonlocal problems with sign-changing nonlinearity on time scales, Electronic J. Differential Equations, 2007 (2007), No. 19, 1–12.
[3] F. M. Atici and G. Sh. Guseinov, On Green’s functions and positive solutions for boundary
value problems on time scales, J. Comput. Appl. Math., 141 (2002) 75–99.
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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
Joan Hoffacker
Department of Mathematical Sciences, Clemson University, Clemson, SC 29634 USA
E-mail address: johoff@clemson.edu