Electronic Journal of Differential Equations, Vol. 2004(2004), No. 47, 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)
NONLINEAR TRIPLE-POINT PROBLEMS ON TIME SCALES
DOUGLAS R. ANDERSON
Abstract. We establish the existence of multiple positive solutions to the
nonlinear second-order triple-point boundary-value problem on time scales,
u∆∇ (t) + h(t)f (t, u(t)) = 0,
u(a) = αu(b) + δu∆ (a),
βu(c) + γu∆ (c) = 0
for t ∈ [a, c] ⊂ T, where T is a time scale, β, γ, δ ≥ 0 with β + γ > 0,
and b ∈ (a, c) ⊂ T.
0 < α < c−a
c−b
1. Introduction to the boundary-value problem
We are concerned with proving the existence of multiple positive solutions to
the second-order triple-point nonlinear boundary-value problem on a time scale T
given by the time-scale dynamic equation
u∆∇ (t) + h(t)f (t, u(t)) = 0,
t ∈ (a, c) ⊂ T
(1.1)
with boundary conditions
u(a) = αu(b) + δu∆ (a),
βu(c) + γu∆ (c) = 0,
(1.2)
c−a
c−b
where β, γ, δ ≥ 0 with β +γ > 0, 0 < α <
and b ∈ (a, c) ⊂ T for a ∈ Tκ , c ∈ Tκ .
The function h ∈ Cld [a, c] is nonnegative with h(t0 ) > 0 for at least one t0 ∈ (a, b],
and the nonlinearity f : [a, c] × [0, ∞) → [0, ∞) is continuous such that f (t, ·) > 0
on any subset of T containing t0 . This problem is related to that first studied in
the case T = R on the unit interval by He and Ge [14] and Ma [19, 20, 21],
u00 + f (t, u) = 0,
t ∈ (0, 1) u(0) = 0,
αu(η) = u(1)
(1.3)
where 0 < η < 1 and 0 < α < 1/η. The boundary-value problem (1.3) has since
been extended to general time scales in Anderson [2] and Kaufmann [17] as
u∆∇ (t) + f (t, u(t)) = 0,
u(0) = 0, αu(η) = u(T ),
and in a slightly different way by Sun and Li [22]
u∆∇ (t) + a(t)f (t, u(t)) = 0,
u∆ (0) = 0, αu(η) = u(T ).
In this paper there is a nexus between the boundary conditions at a and b instead of
at b = η and c = T , and we add the δ ≥ 0 term as well. Consequently these results
2000 Mathematics Subject Classification. 34B10, 34B15, 39A10.
Key words and phrases. Fixed-point theorems, time scales, dynamic equations, cone.
c
2004
Texas State University - San Marcos.
Submitted March 7, 2004. Published April 6, 2004.
1
2
DOUGLAS R. ANDERSON
EJDE-2004/47
are new for differential and difference equations as well as for dynamic equations
on general time scales.
For more on time scales and positive solutions, see the books by Bohner and
Peterson [8, 9] and the following articles: [1, 3, 4, 5, 6, 10, 11, 12, 15, 16].
2. preliminary results
To prove the main existence results we will employ several straightforward lemmas. These lemmas are based on the linear boundary-value problem
u∆∇ (t) + y(t) = 0
∆
u(a) = αu(b) + δu (a),
t ∈ (a, c) ⊂ T
∆
βu(c) + γu (c) = 0.
(2.1)
(2.2)
Lemma 2.1. Let
d := γ(1 − α) + β [(c − a) − α(c − b) + δ] .
(2.3)
If d 6= 0, then for y ∈ Cld [a, c] the boundary-value problem (2.1), (2.2) has the
unique solution
hZ c
1
u(t) = (γ + β(c − t))
(s − a + δ)y(s)∇s
d
a
(2.4)
Z c
i Z c
−α
(s − b)y(s)∇s −
(s − t)y(s)∇s.
b
t
Proof. Let u be as in (2.4). Then the delta derivative of u is given by
Z c
Z c
Z c
β
u∆ (t) = −
(s − a + δ)y(s)∇s − α
(s − b)y(s)∇s +
y(s)∇s
d a
b
t
and
u∆∇ (t) = −y(t),
so that u given in (2.4) is a solution of (2.1). It is routine to check that the boundary
conditions (2.2) are met by u in (2.4) as well.
Lemma 2.2. If d > 0 and y ∈ Cld [a, c] with y ≥ 0, the unique solution u of (2.1),
(2.2) given in (2.4) satisfies
u(t) ≥ 0,
t ∈ [a, c] ⊂ T.
Proof. From the fact that u∆∇ (t) = −y(t) ≤ 0, we know that if u(a) ≥ 0 and
u(c) ≥ 0, then u(t) ≥ 0 for t ∈ [a, c]. For 0 < α ≤ 1,
Z c
Z c
γ
u(c) =
(s − a + δ)y(s)∇s − α
(s − b)y(s)∇s
d a
b
Z c
Z c
γ
≥
(1 − α)
(s − b)y(s)∇s + δ
y(s)∇s
d
b
a
≥ 0.
EJDE-2004/47
TRIPLE-POINT PROBLEMS ON TIME SCALES
For 1 < α <
3
c−a
c−b ,
γ
u(c) =
d
"Z
γ
≥
d
"Z
b
Z
(s − a + δ)y(s)∇s +
a
#
c
(s − a − α(s − b) + δ)y(s)∇s
b
b
Z
(s − a + δ)y(s)∇s +
a
#
c
(c − a − α(c − b) + δ)y(s)∇s
b
≥0
since α < (c − a)/(c − b). Finally,
Z c
hZ c
i dZ c
1
u(a) = (γ + β(c − a))
(s − a + δ)y(s)∇s − α
(s − b)y(s)∇s −
(s − a)y(s)∇s
d
d a
a
b
Z b
Z c
α
α
= (γ + β(c − b))
(s − a)y(s)∇s + (b − a)
(γ + β(c − s))y(s)∇s
d
d
a
b
Z c
δ
+
(γ + β(c − s))y(s)∇s
d a
≥ 0.
Lemma 2.3. Let d > 0. If y ∈ Cld [a, c] with y nonnegative but nontrivial, then the
unique solution u as in (2.4) of (2.1), (2.2) satisfies
inf u(t) ≥ kkuk,
t∈[a,b]
where
k := min
kuk := sup |u(t)|,
(2.5)
t∈[a,c]
o
n α(c − b) c − b
α(b − a)
,
,
∈ (0, 1).
c − a c − a c − a − α(c − b)
(2.6)
Proof. Note that u∆∇ (t) = −y(t) ≤ 0 for all t ∈ (a, c), so that
min u(t) = min{u(a), u(b)}.
t∈[a,b]
Then for any τ ∈ [a, c),
c−t
u(τ )
c−τ
satisfies η(τ ) = 0, η(c) = u(c) ≥ 0, and η ∆∇ (t) = u∆∇ (t) ≤ 0 on [τ, c). In
particular,
u(t)
u(τ )
≥
c−t
c−τ
for all t ∈ [τ, c). Now fix τ ∈ [a, c) such that u(τ ) = kuk. If a ≤ τ ≤ b and
u(a) ≤ u(b), then
αu(b)
αu(τ )
αu(τ )
≥
≥
,
c−b
c−τ
c−a
rewritten with the boundary condition at b as
c − b
u(a) − δu∆ (a) ≥
αkuk.
c−a
Therefore,
c − b
min u(t) = u(a) ≥
αkuk,
c−a
t∈[a,b]
η(t) := u(t) −
4
DOUGLAS R. ANDERSON
EJDE-2004/47
since u∆ (a) ≥ 0 in this case. If a ≤ τ ≤ b and u(b) ≤ u(a), then
c−b
c − b
u(b) ≥
u(τ ) ≥
u(τ ),
c−τ
c−a
so that
c − b
min u(t) = u(b) ≥
kuk.
c−a
t∈[a,b]
If b < τ < c, then u(a) = mint∈[a,b] u(t), and by the concavity of u and a secant
line we have
u(b) − u(a) u(τ ) ≤ u(a) +
(c − a)
b−a
[c − a − α(c − b)] u(a) − δ(c − a)u∆ (a)
=
α(b − a)
[c − a − α(c − b)] u(a)
≤
,
α(b − a)
since again u∆ (a) ≥ 0. Consequently,
min u(t) = u(a) ≥
t∈[a,b]
α(b − a)
kuk.
c − a − α(c − b)
3. Existence of at Least Two Positive Solutions
We apply the Avery-Henderson Fixed Point Theorem [7] to prove the existence of
at least two positive solutions to the nonlinear boundary-value problem (1.1), (1.2),
where h ∈ Cld [a, c] is nonnegative with h(t0 ) > 0 for at least one t0 ∈ (a, b],and
the nonlinearity f : [a, c] × [0, ∞) → [0, ∞) is continuous such that f (t, ·) > 0 on
any subset of T containing t0 . The solutions are the fixed points of the operator A
defined by
hZ c
1
Au(t) = (γ + β(c − t))
(s − a + δ)h(s)f (s, u(s))∇s
d
a
Z c
i Z c
−α
(s − b)h(s)f (s, u(s))∇s −
(s − t)h(s)f (s, u(s))∇s
b
t
by Lemma 2.1. Notationally, the cone P has subsets of the form P (φ, r) := {u ∈
P : φ(u) < r} for a given functional φ.
Theorem 3.1. [7] Let P be a cone in a real Banach space S. If η and φ are
increasing, nonnegative continuous functionals on P , let θ be a nonnegative continuous functional on P with θ(0) = 0 such that, for some positive constants r and
M,
φ(u) ≤ θ(u) ≤ η(u) and ||u|| ≤ M φ(u)
for all u ∈ P (φ, r). Suppose that there exist positive numbers p < q < r such that
θ(λu) ≤ λθ(u),
for all 0 ≤ λ ≤ 1 and u ∈ ∂P (θ, q).
If A : P (φ, r) → P is a completely continuous operator satisfying
(i) φ(Au) > r for all u ∈ ∂P (φ, r),
(ii) θ(Au) < q for all u ∈ ∂P (θ, q),
(iii) P (η, p) 6= ∅ and η(Au) > p for all u ∈ ∂P (η, p),
EJDE-2004/47
TRIPLE-POINT PROBLEMS ON TIME SCALES
5
then A has at least two fixed points u1 and u2 such that
p < η(u1 ) with θ(u1 ) < q
and
q < θ(u2 ) with φ(u2 ) < r.
Let S denote the Banach space C[ρ(a), σ(c)] with the supremum norm. Define
the cone P ⊂ S by
P = {u ∈ S : u(t) ≥ 0, u is concave, and min u(t) ≥ kkuk},
(3.1)
t∈[a,b]
where k is given in (2.6). Finally, let the nonnegative, increasing, continuous functionals φ, θ, and η be defined on the cone P by
φ(u) := min u(t),
θ(u) := max u(t),
t∈[a,b]
η(u) := max u(t).
t∈[b,c]
t∈[a,c]
Observe that, for each u ∈ P , the concavity of u implies
φ(u) ≤ θ(u) ≤ η(u),
1
1
1
1
kuk =≤
min u(t) = φ(u) ≤ θ(u) ≤ η(u).
k t∈[a,b]
k
k
k
(3.2)
(3.3)
Theorem 3.2. Let d > 0 and k be as in (2.6). Suppose there exist positive numbers
0 < p < q < r such that the function f satisfies the following conditions:
(i) f (s, u) > pM for s ∈ [a, b] and u ∈ [kp, p],
(ii) f (s, u) < qm for s ∈ [a, c] and u ∈ [0, q/k],
(iii) f (s, u) > rM for s ∈ [a, b] and u ∈ [r, r/k]
for some positive constants m and M . Then the second-order boundary-value problem (1.1), (1.2), has at least two positive solutions u1 and u2 such that
p < max u1 (t)
with max u1 (t) < q,
q < max u2 (t)
with min u2 (t) < r.
t∈[a,c]
t∈[b,c]
t∈[b,c]
t∈[a,b]
Proof. For u ∈ P , u(t) ≥ kkuk for all t ∈ [a, b]. By Lemma 2.3, A(P ) ⊂ P .
Standard arguments show that A : P → P is completely continuous. For any
u ∈ P , (3.2) and (3.3) imply
φ(u) ≤ θ(u) ≤ η(u),
kuk ≤
1
φ(u).
k
It is clear that θ(0) = 0, and for all u ∈ P , λ ∈ [0, 1] we have
θ(λu) = max (λu)(t) = λ max u(t) = λθ(u).
t∈[b,c]
t∈[b,c]
Since 0 ∈ P and p > 0, P (η, p) 6= ∅.
In the following claims, we verify the remaining conditions of Theorem 3.1.
Claim 1: If u ∈ ∂P (η, p), then η(Au) > p: Since u ∈ ∂P (η, p), kp ≤ u(t) ≤ kuk = p
for t ∈ [a, b]. Define
Z b
−1
M := d min{α, 1}(γ + β(c − b))
(s − a + δ)h(s)∇s
.
(3.4)
a
6
DOUGLAS R. ANDERSON
EJDE-2004/47
Thus
η(Au) = max Au(t)
t∈[a,c]
≥ Au(b)
=
hZ c
1
(γ + β(c − b))
(s − a + δ)h(s)f (s, u(s))∇s
d
a
Z c
i Z c
−α
(s − b)h(s)f (s, u(s))∇s −
(s − b)h(s)f (s, u(s))∇s
b
b
Z
γ + β(c − b) b
(s − a + δ)h(s)f (s, u(s))∇s
=
d
Z ca
b−a+δ
+
(γ + β(c − s))h(s)f (s, u(s))∇s
d
b
Z
γ + β(c − b) b
≥
(s − a + δ)h(s)f (s, u(s))∇s
d
a
Z
γ + β(c − b) b
> pM
(s − a + δ)h(s)∇s
d
a
≥p
by hypothesis (i) and (3.4).
Claim 2: If u ∈ ∂P (θ, q), then θ(Au) < q: Note that u ∈ ∂P (θ, q) and (3.3) imply
that 0 ≤ u(t) ≤ kuk ≤ q/k for t ∈ [a, c]. Define
Z c
−1
(s − a + δ)h(s)∇s
m := d (γ + β(c − b))
.
(3.5)
a
Then
θ(Au) = max Au(t)
t∈[b,c]
hZ c
1
≤ max (γ + β(c − t))
(s − a + δ)h(s)f (s, u(s))∇s
t∈[b,c] d
a
Z c
i
−α
(s − b)h(s)f (s, u(s))∇s
b
Z c
1
≤ (γ + β(c − b))
(s − a + δ)h(s)f (s, u(s))∇s
d
a
Z c
qm
<
(γ + β(c − b))
(s − a + δ)h(s)∇s
d
a
=q
using (2.4), hypothesis (ii), and (3.5).
Claim 3: If u ∈ ∂P (φ, r), then φ(Au) > r: Since u ∈ ∂P (φ, r), from (3.3) we have
that min u(t) = r and r ≤ kuk ≤ kr . Since δ ≥ 0, as in the proof of Claim 1 we
t∈[a,b]
have
Au(b) =
Z
γ + β(c − b) b
(s − a + δ)h(s)f (s, u(s))∇s
d
Z ca
b−a+δ
+
(γ + β(c − s))h(s)f (s, u(s))∇s.
d
b
EJDE-2004/47
TRIPLE-POINT PROBLEMS ON TIME SCALES
7
In a similar fashion from the proof of Lemma 2.2 we have
h γ + β(c − b) Z b
Au(a) = α
(s − a)h(s)f (s, u(s))∇s
d
a
Z
i
b−a c
+
(γ + β(c − s))h(s)f (s, u(s))∇s
d
b
Z
δ c
+
(γ + β(c − s))h(s)f (s, u(s))∇s.
d a
By concavity,
min Au(t) = min{Au(a), Au(b)}.
t∈[a,b]
If α ≥ 1, then Au(a) ≥ Au(b). If 0 < α < 1, then Au(a) ≥ αAu(b). It follows that
φ(Au) = min Au(t)
t∈[a,b]
≥ min{α, 1}Au(b)
≥ min{α, 1}
γ + β(c − b) d
b
Z
(s − a + δ)h(s)f (s, u(s))∇s
a
γ + β(c − b) > min{α, 1}rM
d
≥r
Z
b
(s − a + δ)h(s)∇s
a
by hypothesis (iii) and (3.4). Therefore the hypotheses of Theorem 3.1 are satisfied
and there exist at least two positive fixed points u1 and u2 of A in P (φ, r). Thus,
the second-order triple-point boundary-value problem (1.1), (1.2), has at least two
positive solutions u1 and u2 such that
p < η(u1 )
with θ(u1 ) < q,
q < θ(u2 ) with φ(u2 ) < r.
Corollary 3.3. Let d > 0. If there exists q > 0 such that the function f satisfies
the following conditions:
f (s, u)
(i) lim inf
> M/k for s ∈ [a, b],
u
u→0+
(ii) f (s, u) < qm for s ∈ [a, c] and u ∈ [0, q/k],
f (s, u)
(iii) lim inf
> M for s ∈ [a, b],
u→∞
u
then the second-order boundary-value problem (1.1), (1.2), has at least two positive
solutions u1 and u2 such that
p < max u1 (t)
with max u1 (t) < q,
q < max u2 (t)
with min u2 (t) < r.
t∈[a,c]
t∈[b,c]
t∈[b,c]
t∈[a,b]
Proof. From assumption (i) of the corollary we know there exists p ∈ (0, q) such
that
f (s, u)
M
>
, u ∈ (0, p], s ∈ [a, b].
u
k
8
DOUGLAS R. ANDERSON
EJDE-2004/47
In particular,
f (s, u) > uM/k ≥ pk(M/k) = pM,
u ∈ [kp, p], s ∈ [a, b],
and (i) of Theorem 3.2 holds. Now let
f∞ := lim inf
u→∞
f (s, u) ,
u
s∈[a,b]
min
and η ∈ (M, f∞ ). Then there exists r0 > q such that mins∈[a,b] f (s, u) ≥ ηu,
u ∈ [r0 , ∞). Set
$ := min min f (s, u) : u ∈ [0, r0 ]
s∈[a,b]
and take
r > max r0 ,
$ .
η−M
Then
min f (s, u) ≥ ηu − $ ≥ ηr − $ > rM,
u ∈ [r, ∞),
s∈[a,b]
so that (iii) of Theorem 3.2 holds. The conclusion follows.
Similar to Corollary 3.3, one can prove the following statement.
Corollary 3.4. Let d > 0. If there exists q > 0 such that the function f satisfies
the following conditions:
f (s, u)
< m for s ∈ [a, c],
(i) lim sup
u
u→0+
(ii) f (s, u) > qM for s ∈ [a, b] and u ∈ [kq, q],
f (s, u)
(iii) lim sup
< mk for s ∈ [a, c],
u
u→∞
then the second-order boundary-value problem (1.1), (1.2), has at least two positive
solutions.
4. Existence of at Least Three Positive Solutions
To prove the existence of at least three positive solutions to (1.1), (1.2) we will
use the Leggett-Williams fixed point theorem [13, 18]:
Theorem 4.1. Let P be a cone in the real Banach space S, A : Pr → Pr be
completely continuous and φ be a nonnegative continuous concave functional on P
with φ(u) ≤ kuk for all u ∈ Pr . Suppose there exists 0 < p < q < ` ≤ r such that
the following conditions hold:
(i) {u ∈ P (φ, q, `) : φ(u) > q} =
6 ∅ and φ(Au) > q for all u ∈ P (φ, q, `);
(ii) kAuk < p for kuk ≤ p;
(iii) φ(Au) > q for u ∈ P (φ, q, r) with kAuk > `.
Then A has at least three fixed points u1 , u2 , and u3 in Pr satisfying:
ku1 k < p,
φ(u2 ) > q,
p < ku3 k with φ(u3 ) < q.
Again define the continuous concave functional φ : P → [0, ∞) to be φ(u) :=
mint∈[a,b] u(t), the cone P as in (3.1), M as in (3.4), and
Z c
−1
m := d (γ + β(c − a))
(s − a + δ)h(s)∇s
.
a
EJDE-2004/47
TRIPLE-POINT PROBLEMS ON TIME SCALES
9
Moreover, we take
Pr := {u ∈ P : kuk < r},
P (φ, p, q) := {u ∈ P : p ≤ φ(u), kuk ≤ q}.
Theorem 4.2. Let d > 0. Suppose that there exist constants 0 < p < q < q/k < r
such that
(D1) f (s, u) ≤ rm for s ∈ [a, c], u ∈ [0, r];
(D2) f (s, u) ≥ qM for s ∈ [a, b], u ∈ [q, q/k];
(D3) f (s, u) < pm for s ∈ [a, c], u ∈ [0, p],
where k is given in (2.6). Then the boundary-value problem (1.1), (1.2) has at least
three positive solutions u1 , u2 , u3 satisfying
ku1 k < p,
q < min u2 (t),
t∈[a,b]
ku3 k > p with min u3 (t) < q.
t∈[a,b]
Proof. Again the solutions are the fixed points of the operator A defined by
hZ c
1
(s − a + δ)h(s)f (s, u(s))∇s
Au(t) = (γ + β(c − t))
d
a
Z c
i Z c
−α
(s − b)h(s)f (s, u(s))∇s −
(s − t)h(s)f (s, u(s))∇s.
b
t
The conditions of Theorem 4.1 will now be shown to be satisfied. For all u ∈ P
we have φ(u) ≤ kuk. If u ∈ Pr , then kuk ≤ r and assumption (D1) implies
f (s, u(s)) ≤ rm for s ∈ [a, c]. Consequently,
kAuk = max Au(t)
t∈[a,c]
hZ c
1
≤ max (γ + β(c − t))
(s − a + δ)h(s)f (s, u(s))∇s
t∈[a,c] d
a
Z c
i
−α
(s − b)h(s)f (s, u(s))∇s
b
Z c
1
≤ (γ + β(c − a))
(s − a + δ)h(s)f (s, u(s))∇s
d
a
Z c
rm
<
(γ + β(c − a))
(s − a + δ)h(s)∇s
d
a
= r.
This proves that A : Pr → Pr . Similarly, if u ∈ Pp , then assumption (D3) yields
f (s, u(s)) < pm for s ∈ [a, c]. Just as above, we have A : Pp → Pp . It follows that
condition (ii) of Theorem 4.1 is satisfied.
We now consider condition (i) of Theorem 4.1; pick uP (t) ≡ q/k for t ∈ [a, c],
for k given in (2.6). Then uP ∈ P (φ, q, q/k) and φ(uP ) = φ(q/k) > q, so that {u ∈
P (φ, q, q/k) : φ(u) > q} =
6 ∅. Consequently, if u ∈ P (φ, q, q/k), then q ≤ u(s) ≤ q/k
when s ∈ [a, b]. From assumption (D2) we have that
f (s, u(s)) ≥ qM
for s ∈ [a, b]. As in Claim 3 of the proof of Theorem 3.2,
min Au(t) = min{Au(a), Au(b)}.
t∈[a,b]
10
DOUGLAS R. ANDERSON
EJDE-2004/47
If α ≥ 1, then Au(a) ≥ Au(b); if 0 < α < 1, then Au(a) ≥ αAu(b). It follows that
φ(Au) = min Au(t)
t∈[a,b]
≥ min{α, 1}Au(b)
γ + β(c − b) ≥ min{α, 1}
d
Z
b
(s − a + δ)h(s)f (s, u(s))∇s
a
γ + β(c − b) > min{α, 1}qM
d
≥q
b
Z
(s − a + δ)h(s)∇s
a
by hypothesis (D2) and (3.4). Therefore,
φ(Au) > q,
u ∈ P (φ, q, q/k),
so that condition (i) of Theorem 4.1 holds. To check on Theorem 4.1 (iii), we
suppose that u ∈ P (φ, q, r) with kAuk > q/k. Then, Lemma 2.3 and the definition
of φ yield
φ(Au) = min Au(t) ≥ kkAuk > kq/k = q.
t∈[a,b]
Using the ideas in the proof of the above theorem, we can establish the existence
of an arbitrary odd number of positive solutions of (1.1), (1.2), assuming the right
conditions on the nonlinearity f .
Theorem 4.3. Let d > 0. Suppose that there exist constants
0 < p1 < q1 < q1 /k < p2 < q2 < q2 /k < p3 < · · · < pn ,
n ∈ {2, 3, 4, · · · },
such that
(D1) f (s, u) ≥ qi M for s ∈ [a, b], u ∈ [qi , qi /k];
(D2) f (s, u) < pi m for s ∈ [a, c], u ∈ [0, pi ],
where k is given in (2.6). Then the boundary-value problem (1.1), (1.2) has at least
2n − 1 positive solutions.
5. Example
Let T = {1 − (1/2) } ∪ {1}. Taking a = 0, b = 31/32, c = β = 1, α = 20,
γ = 1/19, and δ = 3/4, we have d = 1/8 and k = 1/32. If we let h(s) ≡ 1, then
m = 57/680 and M = 77824/71145. Suppose
N0
2000eu
, t ∈ [0, 1], u ≥ 0.
99e700 + eu
Clearly f is always increasing. If we take p = 701, q = 705, and r = 24000, then
f (t, u) = f (u) :=
0 < p < q < q/k < r.
We check that (D1), (D2), and (D3) of Theorem 4.2 are satisfied. Since f (715) ≈
1999.94 and lim f (u) = 2000,
f (u) ≤ 24000m ≈ 2011.76,
u ∈ [0, r],
so that (D1) is met. To verify (D2), note that f (705) ≈ 1199.72, so that
f (u) ≥ 705M ≈ 771.18,
u ∈ [q, 32q].
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TRIPLE-POINT PROBLEMS ON TIME SCALES
11
Lastly, as f (701) ≈ 53.45,
f (u) < 701m ≈ 58.76,
u ∈ [0, p]
and (D3) holds. Therefore by Theorem 4.2, the boundary-value problem
3
u(0) = 20u(31/32) + u∆ (0),
4
has at least three positive solutions u1 , u2 , u3 satisfying
u∆∇ (t) + f (u(t)) = 0,
ku1 k < 701,
705 <
min
t∈[0,31/32]
u2 (t),
ku3 k > 701
with
u(1) +
1 ∆
u (1) = 0
19
min
t∈[0,31/32]
u3 (t) < 705.
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Douglas R. Anderson
Department of Mathematics and Computer Science, Concordia College, Moorhead,
Minnesota 56562 USA
E-mail address: andersod@cord.edu