Electronic Journal of Differential Equations, Vol. 2010(2010), No. 139, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLVABILITY OF A THREE-POINT NONLINEAR
BOUNDARY-VALUE PROBLEM
ASSIA GUEZANE-LAKOUD, SMAIL KELAIAIA
Abstract. Using the Leray Schauder nonlinear alternative, we prove the existence of a nontrivial solution for the three-point boundary-value problem
u00 + f (t, u) = 0,
0
0<t<1
u(1) = βu0 (η),
u(0) = αu (0),
where η ∈ (0, 1), α, β ∈ R, f ∈ C([0, 1] × R, R). Some examples are given to
illustrate the results obtained.
1. Introduction and preliminaries
During the previous years, many authors have studied three-point boundaryvalue problems (BVP) for second order differential equations. Such problems have
potential applications in physics, biology, chemistry, etc. For example, a secondorder three-point (BVP) is used as a model for the membrane response of a spherical cap [10] in nonlinear diffusion generated by nonlinear sources and in chemical
reactor theory.
In this article, we investigate the existence of a nontrivial solution for the secondorder three-point boundary-value problem
u00 + f (t, u) = 0,
0
0 < t < 1,
0
u(0) = αu (0),
u(1) = βu (η)
(1.1)
(1.2)
where η ∈ (0, 1), α, β ∈ R, f ∈ C([0, 1] × R, R). We do not assume any monotonicity condition on the nonlinearity f , and the parameters α and β belong to R, so
our conditions are more general than the conditions found in the literature. Such
problem arises in the study of the equilibrium states of a heated bar. In this situation two controllers at t = 0 and t = η alter the heat according to the temperatures
detected by a sensor at t = 1.
This study is motivated by Il’in and Moiseev’s results [5], on similar boundary
value problems for certain linear ordinary differential equations. Many of the results
involving nonlocal boundary value problems are studied in [2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12]. In [12] the author used the Leray-Shauder nonlinear alternative to
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Fixed point theorem; three-point boundary-value problem;
non trivial solution.
c 2010 Texas State University - San Marcos.
Submitted March 19, 2010. Published September 27, 2010.
1
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A. GUEZANE-LAKOUD, S. KELAIAIA
EJDE-2010/139
establish some results on the existence of solutions for the equation (1.1) subject
to the conditions
u(0) = 0, u(1) = αu(η)
Gupta [3] studied certain three-point boundary-value problem with the above nonlocal conditions. Similar boundary value problem with the conditions
u(0) = αu0 (0),
u(1) = βu(η)
is considered in [11]. Under some conditions on the nonlinearity of f and by using Leray Schauder nonlinear alternative, we establish the existence of nontrivial
solution of the BVP (1.1)-(1.2). Some examples illustrate our results.
This article is organized as follows. First, we list some preliminary material to be
used later. Then in Section 2, we present and prove our main results which consist
in existence theorems and corollaries. We end our work with some illustrating
examples.
Let E = C[0, 1], with the supremum norm ky(t)k = supt∈[0,1] |y(t)|, for all y ∈ E.
Now we state two preliminary results.
Lemma 1.1. Let y ∈ E. If β(η + α) 6= α + 1, then the three-point BVP
u00 (t) + y(t) = 0,
0
u(1) = βu0 (η)
u(0) = αu (0),
has a unique solution
Z t
u(t) = −
(t − s)y(s)ds +
0
t+α
1+α−β
0<t<1
Z
1
(1 − s)y(s)ds − β
0
t+α
1+α−β
Z
η
y(s)ds.
0
Proof. Rewriting the differential equation as u00 (t) = −y(t), and integrating twice,
we obtain
Z t
u(t) = −
(t − s)y(s)ds + C1 t + C.
(1.3)
0
Then u0 (0) = C1 ,
Z
0
u (η) = −
η
y(s)ds + C1 ,
0
1
Z
u(1) = −
(1 − s)y(s)ds + C1 + C2 ,
0
u(0) = C2 = αC1 .
Combining these with the second equality in condition (1.2), we obtain
Z 1
Z η
−
(1 − s)y(s)ds + C1 (1 + α) = −β
y(s)ds + C1 β,
0
0
which is equivalent to
Z
C1 (1 + α − β) =
1
Z
(1 − s)y(s)ds − β
0
η
y(s)ds.
0
Since 1 + α − β 6= 0,
C1 =
Z 1
Z η
1
( (1 − s)y(s)ds − β
y(s)ds).
1+α−β 0
0
C2 = αC1 .
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SOLVABILITY
3
Substituting C1 and C2 by their values in (1.3), we obtain the solution in the
statement of the lemma. This completes the proof.
We define the integral operator T : E → E, by
Z 1
Z t
t+α
(1 − s)f (s, u(s))ds
T u(t) = −
(t − s)f (s, u(s))ds +
1+α−β 0
0
Z η
t+α
−β
f (s, u(s))ds
1+α−β 0
(1.4)
By Lemma 1.1, BVP (1.1)-(1.2) has a solution if and only if the operator T has
a fixed point in E. By Ascoli Arzela theorem we prove that T is a completely
continuous operator. Now we cite the Leray Schauder nonlinear alternative.
Lemma 1.2 ([1]). Let F be a Banach space and Ω a bounded open subset of F ,
0 ∈ Ω. T : Ω → F be a completely continuous operator. Then, either there exists
x ∈ ∂Ω, λ > 1 such that T (x) = λx, or there exists a fixed point x∗ ∈ Ω.
2. Main Results
In this section, we present and prove our main results.
Theorem 2.1. We assume that f (t, 0) 6= 0, 1 + α 6= β and there exist nonnegative
functions k, h ∈ L1 [0, 1] such that
|f (t, x)| ≤ k(t)|x| + h(t), a.e. (t, x) ∈ [0, 1] × R,
Z 1
Z η
|1 + α|
1+α
(1 +
)
(1 − s)k(s)ds + |β
|
k(s)ds < 1
|1 + α − β| 0
1+α−β 0
(2.1)
(2.2)
Then BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C[0, 1].
Proof. Setting
Z 1
Z η
|1 + α|
1+α
)
|
(1 − s)k(s)ds + |β
k(s)ds,
|1 + α − β| 0
1+α−β 0
Z 1
Z η
|1 + α|
1+α
N = (1 +
)
|
(1 − s)h(s)ds + |β
(η − s)h(s)ds.
|1 + α − β| 0
1+α−β 0
M = (1 +
By (2.2), we have M < 1. Since f (t, 0) 6= 0, then there exists an interval [σ, τ ] ⊂
[0, 1] such that min |f (t, 0)| > 0 and as h(t) ≥ |f (t, 0)|, for all t ∈ [0, 1] then N > 0.
σ≤t≤r
N
Let m = (1−M
) , Ω = {u ∈ C[0, 1] : kuk < m}. Assume that u ∈ ∂Ω, λ > 1 such
T u = λu, then
λm = λkuk = kT uk = max |(T u)(t)|
0≤t≤1
Z 1
Z η
h
|1 + α|
1+α
≤ kuk (1 +
)
(1 − s)k(s)ds + |β
|
k(s)ds
|1 + α − β| 0
1+α−β 0
Z 1
Z η
i
|1 + α|
1+α
+ (1 +
)
(1 − s)h(s)ds + |β
|
h(s)ds
|1 + α − β| 0
1+α−β 0
= M kuk + N
From this we obtain
λ≤M+
N
N
=M+
= M + (1 − M ) = 1,
m
N (1 − M )−1
4
A. GUEZANE-LAKOUD, S. KELAIAIA
EJDE-2010/139
this contradicts λ > 1. By Lemma 1.2 we conclude that operator T has a fixed
point u∗ ∈ Ω and then BVP (1.1)-(1.2) has a nontrivial solution u∗ ∈ C[0, 1].
Theorem 2.2. Assume (2.1) and one of the following four conditions:
(1) There exists a constant p > 1 such that
Z 1
ip
h
(1 + q)1/q
1 1
k p (s)ds <
( + = 1);
1+α
1+α
1/q
p
q
1
+
|
|
+
(|β
|)(η(1
+
q))
0
1+α−β
1+α−β
(2.3)
(2) There exists constant µ > −1 such that
k(s) <
(µ + 1)(µ + 2)
sµ
1+α
1+α
1 + | 1+α−β
| + (|β 1+α−β
|)(µ + 2)η µ+1
(2.4)
and
meas s ∈ [0, 1] : k(s) <
(µ + 1)(µ + 2)
sµ > 0;
1+α
1+α
1 + | 1+α−β
| + |β 1+α−β
|(µ + 2)η µ+1
(3) The function k(s) satisfies
k(s) <
|1 + α − β|
|1 + α| + |1 + α − β| + |β(1 + α)|η
(2.5)
and
meas s ∈ [0, 1] : k(s) <
|1 + α − β|
|1 + α| + |1 + α − β| + |β(1 + α)|η
> 0;
(4) The function f (t, x) satisfies
ω = lim sup max |
|x|→∞ t∈[0,1]
f (t, x)
1
|1 + α − β|
|< (
).
x
2 |1 + α| + |1 + α − β| + |β(1 + α)|η
(2.6)
Then BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C[0, 1].
Proof. Let M be defined as in the proof of Theorem 2.1. To prove Theorem 2.2,
we only need to prove that M < 1.
(1) By using Hölder inequality, we obtain
Z
1/p Z 1
1/q
1 + α 1 p
M ≤ 1+|
|
k (s)ds
(1 − s)q ds
1+α−β
0
Z 0
1/p Z η 1/q
1+α η p
|
+ |β
k (s)ds
ds
.
1+α−β
0
0
Then
M≤
Z
1
k p (s)ds
1/p h
0
1+α
+ |β
1+α−β
1 + α |
1+α−β
Z
η 1/q i
|
ds
.
Z
1+|
1
(1 − s)q ds
1/q
0
0
Integrating, it yields
M<
(1 + q)1/q
1+α
1+α
(1 + | 1+α−β
|) + |β 1+α−β
|(η(1 + q))1/q
h
i
1 1/q
1+α
|1 + α|
× (1 +
)(
) + |β
|η 1/q = 1.
|1 + α − β| 1 + q
1+α−β
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SOLVABILITY
5
(2) Using the same reasoning as in the proof of the first statement we obtain
(µ + 1)(µ + 2)
1+α
1+α
1 + | 1+α−β
| + (|β 1+α−β
|)(µ + 2)η µ+1
Z
Z η
h
i
1+α
1+α 1
(1 − s)sµ ds + |β
|
|
sµ ds
× 1+|
1+α−β
1+α−β 0
0
1+α (µ + 1)(µ + 2)
1
1+|
=
|
1+α
1+α
µ+1
1 + α − β (µ + 1)(µ + 2)
1 + | 1+α−β | + |β 1+α−β |(µ + 2)η
M<
+ β
1+α
1+α−β
η µ+1 =1
µ+1
(3) we have
M = (1 +
|1 + α|
)
|1 + α − β|
Z
1
(1 − s)k(s)ds + |β
0
1+α
|
1+α−β
Z
η
k(s)ds.
0
Then
|1 + α − β|
|(1 + α)| + |1 + α − β| + |β(1 + α)|η
Z
Z η
1+α 1
1+α
× 1+|
|
(1 − s)ds + |β
|
ds
1+α−β
1+α−β 0
0
1
|1+α−β|
1+α
1+α
= ( |(1+α)|+|1+α−β|+|β(1+α)|η
)(1 + | 1+α−β
|) + |β 1+α−β
|η = 1
2
M<
(4) From ω = lim sup|x|→∞ maxt∈[0,1] | f (t,x)
x | we deduce that there exists c > 0
such that for |x| > c we have
|f (t, x)| ≤ (ω + ε)|x|
∀ε > 0.
Set
h(t) = max{|f (t, x)| : (t, x) ∈ [0, 1] × (−c, c)}.
Then for (t, x) ∈ [0, 1] × R, with ε = ω, we obtain
|f (t, x)| ≤ 2ω|x| + h(t)
≤
|1 + α − β|
|x| + h(t).
|(1 + α)| + |1 + α − β| + |β(1 + α)|η
Setting
k(t) <
|1 + α − β|
,
|(1 + α)| + |1 + α − β| + |β(1 + α)|η
by applying the above statement we complete the proof.
Corollary 2.3. Assume the conditions of Theorem 2.2, and one of the following
four conditions:
(1) There exists a constant p > 1 such that
Z 1
h
ip
(1 + q)1/q
k p (s)ds <
,
1+α
1+α
1 + | 1+α−β
| + |β 1+α−β
|(1 + q)1/q
0
1 1
( + = 1).
p q
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A. GUEZANE-LAKOUD, S. KELAIAIA
EJDE-2010/139
(2) There exists a constant µ > −1 such that
k(s) <
1+
(µ + 1)(µ + 2)
sµ ,
1+α
+ (|β 1+α−β
|)(µ + 2)
1+α
| 1+α−β
|
meas s ∈ [0, 1] : k(s) <
1+
(µ + 1)(µ + 2)
sµ > 0
1+α
+ |β 1+α−β
|(µ + 2)
1+α
| 1+α−β
|
(3) The function k(s) satisfies
|1 + α − β|
,
|(1 + α)| + |1 + α − β| + |β(1 + α)|η
|1 + α − β|
meas s ∈ [0, 1], k(s) <
>0
|(1 + α)| + |1 + α − β| + |β(1 + α)|η
k(s) <
(4) The function f (t, x) satisfies
ω = lim sup max |
|x|→∞ t∈[0,1]
< 1/2(
f (t, x)
|
x
|1 + α − β|
)
|(1 + α)| + |1 + α − β| + |β(1 + α)|
Then the BVP (1.1)-(1.2) has at least one nontrivial solution u∗ ∈ C[0, 1].
Proof. Taking into account
Z
Z η
1+α
|1 + α| 1
|
(1 − s)k(s)ds + |β
k(s)ds
|1 + α − β| 0
1+α−β 0
Z
Z 1
|1 + α| 1
1+α
≤ 1+
|
(1 − s)k(s)ds + |β
k(s)ds.
|1 + α − β| 0
1+α−β 0
M = 1+
the proof follows.
3. Examples
To illustrate our results, we give the following examples.
Example 3.1. Consider the three-point BVP
√
u
u00 + (t sin u − e−t cos u) + tet = 0, 0 < t < 1
2
1
1
u(0) = u0 (0), u(1) = u0 ( 12 ).
2
2
Where α = β = η = 1/2, +α − β = 1/2 6= 0 and
√
x
f (t, x) = (t sin x − e−t cos x) + tet ,
2
It is easy to see that
|f (t, x)| ≤
1
(t + e−t )|x| + tet ,
2
(t, x) ∈ [0, 1] × R,
set
k(t) =
1
(t + e−t ) ≥ 0,
2
h(t) = tet ≥ 0,
(3.1)
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SOLVABILITY
7
we have k, h ∈ L1 [0, 1], f (t, 0) = tet 6= 0 and
Z
Z 1
5 1
3 2
M=
(1 − s)k(s)ds +
k(s)ds
2 0
4 0
Z
Z
3 1/2
5 1
(1 − s)(s + e−s )ds +
(s + e−s )ds = 0.86261.
=
4 0
8 0
Then, by Theorem 2.1, the BVP (3.1) has at least one nontrivial solution u∗ in
C[0, 1].
Example 3.2. Consider the three-point BVP
u4
u00 + √
+ cos et (1 − sin t) = 0, 0 < t < 1,
2 t3 + 1(1 + u3 )
u(0) = −2u0 (0), u(1) = 2u0 (1/3)
(3.2)
then
x4
f (t, x) = √
+ cos et (1 − sin t),
2 t3 + 1(1 + x3 )
1
|x| + cos et (1 − sin t)
|f (t, x)| ≤ √
4 t3 + 1
= k(t)|x| + h(t),
where
1
k(t) = √
≥ 0,
4 t3 + 1
M=
=
4
3
Z
1
3
Z
1
(1 − s)k(s)ds +
0
1
(1 − s) √
0
2
3
h(t) = cos et (1 − sin t) ≥ 0
Z
η
k(s)ds
0
1
2
ds +
16
s3 + 1
Z
1/3
0
√
1
ds = 0.201 41 < 1
s3 + 1
Applying the third statement of Theorem 2.2 we obtain
max k(s) =
s
1
|1 + α − β|
9
<
= 0.642 86 ,
4
|(1 + α)| + |1 + α − β| + |β(1 + α)|η 14
|1 + α − β|
}
|(1 + α)| + |1 + α − β| + |β(1 + α)|η
9
= meas{s ∈ [0, 1] : k(s) <
= 0.642 86}
14
= meas[0, 1] > 0 .
meas{s ∈ [0, 1] : k(s) <
Hence, by Theorem 2.2, BVP (3.2) has at least one nontrivial solution u∗ in C[0, 1].
Example 3.3. Consider the three-point BVP
√
u3
−
t sin t = 0
2
2(1 + e−t )4 (1 + u2 )
1
−1 0 1
u(0) = u0 (0), u(1) =
u ( ).
3
6
4
u00 +
(3.3)
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A. GUEZANE-LAKOUD, S. KELAIAIA
EJDE-2010/139
We have
√
x3
− t sin t,
2 4
−t
2
2(1 + e ) (1 + x )
2
(1 + e−t )−4 √
|f (t, x)| ≤ |x|
+ t sin t = k(t)|x| + h(t)
2
√
2
t sin t
f (t, x)
(1 + e−t )−4
1
|
|≤
+
≤ 0.142 +
.
x
2
|x|
|x|
Applying the fourth statement in Theorem 2.2 we obtain
f (t, x)
1
ω = lim sup max |
| = lim sup(0.142 +
)
x
|x|
|x|→∞ t∈[0,1]
|x|→∞
f (t, x) =
= 0.142 <
1
(
2 43 +
13
9
13
1
9 + 27
) = 0.25658
Hence, BVP (3.3) has at least one nontrivial solution u∗ in C[0, 1].
Example 3.4. Consider the three-point BVP
sin u
u00 + √
− t2 cos t + t cos(t2 ) = 0, 0 < t < 1
2 4+t
−1 0 1
−1 0
u (0), u(1) =
u ( ),
u(0) =
4
6
5
where
sin x
f (t, x) = √
− t2 cos t + t cos(t2 ),
2 4+t
|x|
|f (t, x)| ≤ √
− t2 cos t + t cos(t2 ) = k(t)|x| + h(t),
2 4+t
1
k(t) = √
≥ 0, h(t) = −t2 cos t + t cos(t2 ) ≥ 0
2 4+t
(3.4)
We see that k, h ∈ L1 [0, 1], f (t, 0) = −t2 cos t + t cos(t2 ) 6= 0. The first statement
of Corollary 2.3 for p = 2 yields
Z
Z 1
Z 1
2
1 1 1
1
√
dt = 0.055786
k 2 (s)ds =
dt =
4 0 4+t
2( 4 + t)
0
0
h
ip
(1 + q)1/q
<
= 0.71083.
1+α
1+α
1 + | 1+α−β | + |β 1+α−β |(1 + q)1/q
Hence, BVP (3.4) has at least one nontrivial solution u∗ in C[0, 1].
Acknowledgements. The authors are grateful to the anonymous referees for their
helpful comments and fruitful remarks. The second author would like to express his
thanks to the Laboratory of advanced material at Annaba university for its support
and hospitality.
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SOLVABILITY
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Assia Guezane-Lakoud
Department of Mathematics, Faculty of Sciences, University Badji Mokhtar, B.P. 12,
23000, Annaba, Algeria
E-mail address: a guezane@yahoo.fr
Smail Kelaiaia
Department of Mathematics, Faculty of Sciences, University Badji Mokhtar, B.P. 12,
23000, Annaba, Algeria
E-mail address: kelaiaiasmail@yahoo.fr