Electronic Journal of Differential Equations, Vol. 2004(2004), No. 119, pp. 1–7.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
SEMIPOSITONE m-POINT BOUNDARY-VALUE PROBLEMS
NICKOLAI KOSMATOV
Abstract. We study the m-point nonlinear boundary-value problem
−[p(t)u0 (t)]0 = λf (t, u(t)),
u0 (0) = 0,
m−2
X
0 < t < 1,
αi u(ηi ) = u(1),
i=1
where 0 < η1 < η2 < · · · < ηm−2 < 1, αi > 0 for 1 ≤ i ≤ m − 2 and
Pm−2
i=1 αi < 1, m ≥ 3. We assume that p(t) is non-increasing continuously
differentiable on (0, 1) and p(t) > 0 on [0, 1]. Using a cone-theoretic approach
we provide sufficient conditions on continuous f (t, u) under which the problem
admits a positive solution.
1. Introduction
In this note we consider the nonlinear m-point eigenvalue problem
−[p(t)u0 (t)]0 = λf (t, u(t)),
u0 (0) = 0,
m−2
X
0 < t < 1,
αi u(ηi ) = u(1),
(1.1)
(1.2)
i=1
Pm−2
where 0 < η1 < η2 < · · · < ηm−2 < 1, αi > 0 for 1 ≤ i ≤ m − 2, i=1 αi < 1. We
also assume that the function p(t) is non-increasing continuously differentiable on
(0, 1) and p(t) > 0 on [0, 1]. The inhomogeneous term in (1.1) is allowed to change
its sign. Other assumptions on f (t, u(t)) will be made later.
The study of multi-point boundary-value problems was initiated by Il’in and Moiseev in [7, 8]. Many authors since then considered nonlinear multi-point boundaryvalue problems (see, e.g., [2, 4, 5, 6, 9, 14, 15, 16, 17] and the references therein).
In particular, Ma studied in [15] positive solutions to the three-point nonlinear
boundary-value problem
−u00 (t) = a(t)f (u(t)),
u(0) = 0,
0 < t < 1,
αu(η) = u(1),
2000 Mathematics Subject Classification. 34B10, 34B18.
Key words and phrases. Green’s function; fixed point theorem; positive solutions;
multi-point boundary-value problem.
c
2004
Texas State University - San Marcos.
Submitted April 23, 2004. Published October 10, 2004.
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NICKOLAI KOSMATOV
EJDE-2004/119
where 0 < α, 0 < η < 1 and αη < 1. The results of [15] were complemented in the
works of Webb [17], Kaufmann [9], Kaufmann and Kosmatov [10], and Kaufmann
and Raffoul [11].
Among the studies dealing with semipositone multi-point boundary-value problems, we mention the papers by Cao and Ma [3] and Liu [13]. Cao and Ma considered
the boundary-value problem
−u00 (t) = λa(t)f (u(t), u0 (t)),
u(0) = 0,
m−2
X
0 < t < 1,
αi u(ηi ) = u(1).
i=1
The authors applied the Leray-Schauder fixed point theorem to obtain an interval
of eigenvalues for which at least one positive solution exists. Liu applied a fixed
point index method to obtain such an interval for
−u00 (t) = λa(t)f (u(t)),
0
u (0) = 0,
0 < t < 1,
αu(η) = u(1).
Our approach is based on Krasnosel’skiı̆’s cone-theoretic theorem [12] and enables
us to show the existence of a positive solution for the semipositone problem (1.1),
(1.2). Other applications of Krasnosel’skiı̆’s fixed point theorem to semipositone
problems can, for example, be found in [1].
2. Preliminaries
We now proceed with the auxiliaries. Consider the equation
−[p(t)u0 (t)]0 = g(t),
0 < t < 1,
(2.1)
with the boundary conditions (1.2).
Pm−2
For convenience we set α = i=1 αi . Recall that α < 1.
Lemma 2.1. If g ∈ C[0, 1] and g(t) ≥ 0 on [0, 1], then
Z 1 Z 1
Z tZ t
dτ 1
dτ u(t) = −
g(s) ds +
g(s) ds
1−α 0
0
s p(τ )
s p(τ )
Z ηi Z ηi
m−2
1 X
dτ
−
αi
g(s) ds
1 − α i=1
p(τ )
0
s
(2.2)
is the unique nonnegative solution on [0, 1] of the problem (2.1), (1.2).
Proof. Integration of (2.1) from 0 to t with the use of the boundary condition (1.2)
at 0 yields
Z t
1
u0 (t) = −
g(s) ds ≤ 0.
p(t) 0
Integrating again we get
Z t
Z
Z tZ t
1 s
dτ u(t) = −
g(τ ) dτ ds + A = −
g(s) ds + A.
0 p(s)
0
0
s p(τ )
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SEMIPOSITONE m-POINT BOUNDARY-VALUE PROBLEMS
3
Using the multi-point condition in (1.2) we determine A and obtain (2.2). Since
u0 (t) ≤ 0,
u(t) ≥ u(1)
Z ηi Z ηi
m−2
1 X
dτ dτ g(s) ds −
αi
g(s) ds
1 − α i=1
p(τ )
0
s p(τ )
0
s
Z
Z
Z ηi Z ηi
m−2
i
1 X h 1 1 dτ dτ =
αi
g(s) ds −
g(s) ds ≥ 0
1 − α i=1
p(τ )
0
s p(τ )
0
s
α
=
1−α
1
Z
1
Z
on [0, 1] and the proof is complete.
For g(t) = 1 on [0, 1], we denote by u0 (t) the unique solution (2.2). Then we
have
C = max u0 (t) = u0 (0)
t∈[0,1]
1
=
1−α
Z
0
1
Z
s
1
Z ηi Z ηi
m−2
dτ 1 X
dτ g(s)ds −
αi
g(s)ds.
p(τ )
1 − α i=1
p(τ )
0
s
The Green’s function for −[p(t)u0 (t)]0 = 0 with (1.2) is given by
Z 1
dτ
1
G(t, s) =
1 − α s p(τ )
(
(R t
R ηi dτ
Pm−2
dτ
1
,
s
≤
t
i=1 αi χi (s) s p(τ ) , s ≤ ηm−2
1−α
s p(τ )
−
−
0,
s>t
0,
s > ηm−2 ,
where
(
1, s ≤ ηi
χi (s) =
0, s > ηi .
Note that
Z
1
max
t∈[0,1]
G(t, s) ds = C.
(2.3)
0
The integral operator T : B → B associated with (1.1), (1.2) is defined by
Z 1
T u(t) =
G(t, s)f (s, u(s)) ds
0
A routine argument shows that T is completely continuous.
Definition 2.2. Let B be a Banach space and let C ⊂ B be closed and nonempty.
Then C is said to be a cone if
(1) αu + βv ∈ C for all u, v ∈ C and for all α, β ≥ 0, and
(2) u, −u ∈ C implies u ≡ 0.
Our Banach space, B, is the space C[0, 1] with the norm kuk = maxt∈[0,1] |u(t)|.
We will show now that the unique solution (2.2) satisfies
min u(t) ≥ γkuk,
t∈[0,1]
where
γ=
max
1≤i≤m−2
αi (1 − ηi )
.
1 − αi ηi
(2.4)
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NICKOLAI KOSMATOV
EJDE-2004/119
To this end, note that the solution (2.2) is concave, since g(t) ≥ 0 and u0 (t), p0 (t) ≤ 0
on [0, 1]. By concavity and since u(1) > αi u(ηi ) for each 1 ≤ i ≤ m − 2,
kuk = u(0)
u(1) − u(ηi )
(0 − 1)
1 − ηi
1 − αi ηi
< u(1)
αi (1 − ηi )
1 − αi ηi
=
min u(t)
αi (1 − ηi ) t∈[0,1]
≤ u(1) +
and hence (2.4) holds.
The estimate (2.4) is used for defining our cone C ⊂ B by
C = {u(t) ∈ B : u(t) ≥ 0 on [0, 1], min u(t) ≥ γkuk}.
(2.5)
t∈[0,1]
It turns out that our operator T is cone-preserving. Fixed points of T are solutions
of (1.1), (1.2). The existence of a fixed point of T follows from a fixed point theorem
due to Krasnosel’skiı̆ [12], which we now state.
Theorem 2.3. Let B be a Banach space and let C ⊂ B be a cone in B. Assume
that Ω1 , Ω2 are open with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let
T : C ∩ (Ω2 \ Ω1 ) → C
be a completely continuous operator such that either
(i) kT uk ≤ kuk, u ∈ C ∩ ∂Ω1 , and kT uk ≥ kuk, u ∈ C ∩ ∂Ω2 , or
(ii) kT uk ≥ kuk, u ∈ C ∩ ∂Ω1 , and kT uk ≤ kuk, u ∈ C ∩ ∂Ω2 .
Then T has a fixed point in C ∩ (Ω2 \ Ω1 ).
The following assumptions will stand throughout the remainder of this note:
(A1) f (t, z) is a continuous function on [0, 1] × [0, ∞)
(A2) There exists M > 0 such that f (t, z) + M ≥ 0 on [0, 1] × [0, ∞)
(A3) There exist continuous nonnegative nondecreasing on [0, ∞) functions ψa (z)
and ψb (z) with ψb (z) ≤ f (t, z) + M ≤ ψa (z) on [0, 1] × [0, ∞).
3. Positive solutions
We now state our main results.
Theorem 3.1. Let the assumptions (A1)-(A3) be satisfied. Assume, in addition,
that
ψb (z)
ψa (z)
= 0 and lim
= ∞.
lim
z→∞
z
z
z→0+
Then, for a sufficiently small λ > 0, the problem (1.1), (1.2) has a positive solution.
Proof. Consider the equation
−[p(t)u0 (t)]0 = λfp (t, u(t) − uλ (t)),
0 < t < 1,
with the boundary conditions (1.2), where
(
f (t, z) + M, z ≥ 0
fp (t, z) =
f (t, 0) + M, z ≤ 0
(3.1)
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SEMIPOSITONE m-POINT BOUNDARY-VALUE PROBLEMS
5
and uλ (t) = λM u0 (t) (u0 (t) is given by (2.2) for g ≡ 1). Our objective is to show
that the problem (3.1), (1.2) has a positive solution.
Our completely continuous and cone-preserving operator associated with (3.1),
(1.2) is defined by
Z 1
Tλ u(t) = λ
G(t, s)fp (s, u(s) − uλ (s)) ds
0
Since limz→0+
ψa (z)
z
= 0, there exists R1 > 0 such that
ψa (z) ≤
1
z
λC
for all z ≤ R1 .
Define Ω1 = {u ∈ B : kuk < R1 }, then for u ∈ C ∩ ∂Ω1 we have
ψa (u(s)) ≤ ψa (kuk) ≤
1
R1
λC
(3.2)
for all s ∈ [0, 1], since ψa (z) is nondecreasing. Now, if u(s) ≥ uλ (s) for s ∈ [0, 1],
then
fp (s, u(s) − uλ (s)) = f (s, u(s) − uλ (s)) + M ≤ ψa (u(s) − uλ (s)) ≤ ψa (u(s)).
If u(s) ≤ uλ (s), then
fp (s, u(s) − uλ (s)) = f (s, 0) + M ≤ ψa (0) ≤ ψa (u(s))
(we know that u(s) ≥ 0 as an element of C). Combining both cases and using (3.2)
and (2.3), we get
Z 1
kTλ uk = max λ
G(t, s)fp (s, u(s) − uλ (s)) ds
t∈[0,1]
0
Z
1
≤ max λ
t∈[0,1]
G(t, s)ψa (u(s)) ds
0
Z 1
≤ λ max
t∈[0,1]
G(t, s) ds
0
1
R 1 = R1 ,
λC
that is, kTλ uk ≤ kuk on C ∩ ∂Ω1 .
C)
Since limz→∞ ψbz(z) = ∞, then also limz→∞ ψb (γz−λM
= ∞. Thus, there exists
z
λM C
R2 > 0 large enough (so that R2 > γ and R2 > R1 ) such that
ψb (γz − λM C) ≥
1
z
λC
for all z ≥ R2 . In fact,
ψb (γR2 − λM C) ≥
1
R2 .
λC
(3.3)
Define Ω2 = {u ∈ B : kuk < R2 }, then for u ∈ C ∩ ∂Ω2 we have
u(s) − uλ (s) ≥ γkuk − λM u0 (s) ≥ γR2 − λM C > 0.
Now, for all s ∈ [0, 1],
fp (s, u(s) − uλ (s)) = f (s, u(s) − uλ (s)) + M ≥ ψb (u(s) − uλ (s)) ≥ ψb (γR2 − λM C),
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NICKOLAI KOSMATOV
EJDE-2004/119
since ψb (z) is nondecreasing. Therefore, by (3.3) and (2.3),
Z 1
kTλ uk = max λ
G(t, s)fp (s, u(s) − uλ (s)) ds
t∈[0,1]
0
Z
1
≥ max λ
t∈[0,1]
G(t, s)ψb (γR2 − λM C) ds
0
Z 1
≥ λ max
t∈[0,1]
G(t, s) ds
0
1
R 2 = R2 ,
λC
that is, kTλ uk ≤ kuk on C ∩ ∂Ω2 .
Since the assumptions of Theorem 2.3 are satisfied, we conclude that the problem
(3.1), (1.2) has a positive solution in C ∩ (Ω2 \ Ω1 ), which we denote by up .
C
Let λ be small enough so that R1 > λM
γ . Now we have up (t) ≥ γkup k ≥ γR1 >
λM C ≥ uλ (t) for all t ∈ [0, 1]. Set u(t) = up (t) − uλ (t), then
−[p(t)u0 (t)]0 = −[p(t)u0p (t)]0 − λM
= λfp (t, up (t) − uλ (t)) − λM
= λ(f (t, up (t) − uλ (t)) + M ) − λM
= λf (t, u(t)),
which shows that u(t) is a positive solution of (1.1), (1.2). The proof is complete.
Example. To illustrate our main result, we consider the inhomogeneous term in
the form of the function
f (t, z) = −1 + z 2 (2 + sin (4πz(1 + t3 ))).
The function f (t, z) is continuous and, setting M = 1, we get f (t, z) + M ≥ 0
on [0, 1] × [0, ∞). In addition, for ψb (z) = z 2 and ψa (z) = 3z 2 , we have ψb (z) ≤
f (t, z) + M ≤ ψa (z) and
ψa (z)
ψb (z)
= 0 and lim
= ∞.
z→∞
z
z
Thus, Theorem 3.1 applies.
With only minor adjustments to the argument above one can prove our next
theorem.
lim
z→0+
Theorem 3.2. Let the assumptions (A1)-(A3) be satisfied. Assume, in addition,
that
ψa (z)
ψb (z)
= ∞ and lim
= 0.
lim+
z→∞
z
z
z→0
Then, for a sufficiently small λ > 0, the problem (1.1), (1.2) has a positive solution.
Remark. If problem (1.1), (1.2) has a positive solution for some λ1 > 0, there is
also a positive solution for each λ ∈ (0, λ1 ].
We say that a function ψ(z) is sublinear if
lim
ψ(z)
= ∞ and
z
lim+
ψ(z)
= 0 and
z
z→0+
lim
z→∞
ψ(z)
= 0.
z
On the other hand, if
z→0
lim
z→∞
ψ(z)
= ∞,
z
EJDE-2004/119
SEMIPOSITONE m-POINT BOUNDARY-VALUE PROBLEMS
7
then the function ψ is called superlinear.
If in the assumption (A3) we take ψa (z) = ψb (z), then the following corollary to
Theorems 3.1 and 3.2 becomes immediate.
Corollary 3.3. Let the assumptions (A1)-(A3) be satisfied. Assume, in addition,
that ψa (z) = ψb (z) is either sublinear or superlinear. Then, for a sufficiently small
λ > 0, the problem (1.1), (1.2) has a positive solution.
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Department of Mathematics and Statistics, University of Arkansas at Little Rock,
Little Rock, AR 72204-1099, USA
E-mail address: nxkosmatov@ualr.edu