Electronic Journal of Differential Equations, Vol. 2007(2007), No. 123, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
SOLVABILITY OF A FOUR-POINT BOUNDARY-VALUE
PROBLEM FOR FOURTH-ORDER ORDINARY
DIFFERENTIAL EQUATIONS
JING GE, CHUANZHI BAI
Abstract. In this paper we investigate the existence of solutions of a class
of four-point boundary-value problems for fourth-order ordinary differential
equations. Our analysis relies on a fixed point theorem due to Krasnoselskii
and Zabreiko.
1. Introduction
In recent years, boundary-value problems for second and higher order differential
equations have been extensively studied. The monographs of Agarwal [1] and Agarwal, O’Regan, and Wong [2] contain excellent surveys of known results. Recently
an increasing interest in studying the existence of solutions and positive solutions
to boundary-value problems for higher order differential equations is observed; see
for example [3, 4, 5, 6, 7, 8].
Very recently, Zhang, Chen and Lü [10] by using the upper and lower solution
method investigated the fourth order nonlinear ordinary differential equation
u(4) (t) = f (t, u(t), u00 (t)),
0 < t < 1,
with the four-point boundary conditions
u(0) = u(1) = 0,
au00 (ξ1 ) − bu000 (ξ1 ) = 0,
cu00 (ξ2 ) + du000 (ξ2 ) = 0,
(1.1)
(1.2)
where a, b, c, d are nonnegative constants satisfying ad + bc + ac(ξ2 − ξ1 ) > 0,
0 ≤ ξ1 < ξ2 ≤ 1 and f ∈ C([0, 1] × R × R). They proved the following Lemma (a
key lemma):
Lemma 1.1 ([10, Lemma 2.2]). Suppose a, b, c, d, ξ1 , ξ2 are nonnegative constants
satisfying 0 ≤ ξ1 < ξ2 ≤ 1, b−aξ1 ≥ 0, d−c+cξ2 ≥ 0 and δ = ad+bc+ac(ξ2 −ξ1 ) 6=
0. If u(t) ∈ C 4 [0, 1] satisfies
• u(4) (t) ≥ 0, t ∈ (0, 1),
• u(0) ≥ 0, u(1) ≥ 0,
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Four-point boundary-value problem; existence of solutions;
fixed point theorem.
c
2007
Texas State University - San Marcos.
Submitted May 10, 2007. Published September 21, 2007.
Supported by grants NNSFC-10771212 and 06KJB110010.
1
2
J. GE, C. BAI
EJDE-2007/123
• au00 (ξ1 ) − bu000 (ξ1 ) ≤ 0, cu00 (ξ2 ) + du000 (ξ2 ) ≤ 0,
then u(t) ≥ 0 and u00 (t) ≤ 0 for t ∈ [0, 1].
Unfortunately this Lemma is wrong as shown below.
Counterexample to [10, Lemma 2.2]. Let u(t) = 31 t4 + 41 t3 − 43 t2 + 43 t which
1
belongs to C 4 [0, 1], ξ1 = 10
, ξ2 = 18 , a, b, c, d be nonnegative constants satisfying
15
7
1
1
ac 6= 0. Then we have
b ≥ 10 a = aξ1 , d = 16 c > 8 c = (1 − ξ2 )c and δ = ad + bc + 40
u(4) (t) = 8 > 0,
t ∈ (0, 1),
u(0) = 0, u(1) = 0,
h
h
3i
8i
3
− b 8t +
au00 (ξ1 ) − bu000 (ξ1 ) = a 4t2 + t −
2
3 t=1/10
2 t=1/10
3
143
a − 2 b ≤ 0,
= −2
300
10
and
h
h
3
8i
3i
cu00 (ξ2 ) + du000 (ξ2 ) = c 4t2 + t −
+ d 8t +
2
3 t=1/8
2 t=1/8
5
29
5 15
7
29
= − c + d = − c + · c = − c ≤ 0.
12
2
12
2 16
96
But
8
= −0.0031 < 0;
9
that is, [10, Lemma 2.2] is incorrect.
So the conclusions of [10] should be reconsidered. The aim of this paper is to
investigate the existence of solutions of the BVP (1.1)-(1.2) by using a fixed point
theorem due to Krasnoselskii and Zabreiko in [9].
u
2. Main result
First, we give some lemmas which are needed in our discussion of the main
results.
Lemma 2.1. Suppose a, b, c, d, ξ1 , ξ2 are nonnegative constants satisfying 0 ≤ ξ1 <
ξ2 ≤ 1 and δ = ad + bc + ac(ξ2 − ξ1 ) 6= 0. If h ∈ C[0, 1], then the boundary-value
problem
v 00 (t) = h(t),
0
av(ξ1 ) − bv (ξ1 ) = 0,
t ∈ [0, 1],
(2.1)
0
cv(ξ2 ) + dv (ξ2 ) = 0,
has a unique solution
Z
Z t
1 ξ2
v(t) =
(t − s)h(s)ds +
(a(ξ1 − t) − b)(c(ξ2 − s) + d)h(s)ds.
δ ξ1
ξ1
(2.2)
(2.3)
Proof. By (2.1), it is easy to know that
Z
t
(t − s)h(s)ds,
v(t) = C1 + C2 t +
0
(2.4)
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FOUR-POINT BOUNDARY-VALUE PROBLEMS
3
where C1 , C2 are any two constants. Substituting (2.4) into boundary conditions
(2.2), by a routine calculation, we get
Z ξ1
Z ξ2
1
sh(s)ds + (aξ1 − b)
C1 =
(c(ξ2 − s) + d)h(s)ds,
(2.5)
δ
0
ξ1
Z ξ1
Z
a ξ2
C2 = −
h(s)ds −
(c(ξ2 − s) + d)h(s)ds.
(2.6)
δ ξ1
0
Substituting (2.5) and (2.6) into (2.4), we obtain (2.3) which implies lemma.
Remark 2.2. Let ξ1 = 0, ξ2 = 1, then (2.3) reduces to
Z 1
v(t) = −
G(t, s)h(s)ds,
0
where
1
G(t, s) =
δ
(
(as + b)(d + c(1 − t)), 0 ≤ s ≤ t ≤ 1,
(at + b)(d + c(1 − s)), 0 ≤ t < s ≤ 1.
Remark 2.3. Let
1
((a(t − ξ1 ) + b)x3 + (c(ξ2 − t) + d)x2 ),
δ
(
s(1 − t), 0 ≤ s ≤ t ≤ 1,
G1 (t, s) =
t(1 − s), 0 ≤ t < s ≤ 1,
(
1 (a(s − ξ1 ) + b)(d + c(ξ2 − t)), ξ1 ≤ s ≤ t ≤ ξ2 ,
G2 (t, s) =
δ (a(t − ξ1 ) + b)(d + c(ξ2 − s)), ξ1 ≤ t < s ≤ ξ2 .
R(t) =
In [10, Lemma 2.2] it is claimed that
Z 1
Z
u(t) = tx1 +(1−t)x0 −
G1 (t, ξ)R(ξ)dξ+
0
1
Z
ξ2
G1 (t, ξ)
0
(2.7)
G2 (ξ, s)h(s)dsdξ, (2.8)
ξ1
is the solution of the boundary-value problem
u(4) (t) = h(t),
u(0) = x0 ,
00
000
au (ξ1 ) − bu (ξ1 ) = x2 ,
0 < t < 1,
u(1) = x1 ,
cu00 (ξ2 ) + du000 (ξ2 ) = x3 .
However (2.8) is wrong. Indeed, by Lemma 2.1, (2.8) should be replaced by
Z 1
Z 1
u(t) = tx1 + (1 − t)x0 −
G1 (t, ξ)R(ξ)dξ −
G1 (t, η)v(η)dη,
0
0
where
Z
η
(η − s)h(s)ds +
v(η) =
ξ1
1
δ
Z
ξ2
(a(ξ1 − η) − b)(c(ξ2 − s) + d)h(s)ds.
ξ1
Remark 2.4. In [10, Theorem 3.1], the operator T : C[0, 1] → C[0, 1] is defined as
Z 1
Z ξ2
T u(t) =
G1 (t, η)
G2 (η, s)f (s, u(s), u00 (s))dsdη,
0
ξ1
4
J. GE, C. BAI
EJDE-2007/123
where G1 (t, s) and G2 (t, s) are as in Remark 2.2. By 2.1 and Remark 2.2, the
definition of T is incorrect. In fact, the operator T should be defined as
Z 1
Z η
T u(t) =
G1 (t, η)
(s − η)f (s, u(s), u00 (s))dsdη
0
+
ξ1
1
δ
1
Z
Z
ξ2
G1 (t, η)
(b − a(ξ1 − η))(c(ξ2 − s) + d)f (s, u(s), u00 (s))dsdη.
ξ1
0
The following well-known fixed point theorem [9] will play an important role in
the proof of our theorem.
Lemma 2.5. Let X be a Banach space, and F : X → X be completely continuous. Assume that A : X → X is a bounded linear operator such that 1 is not an
eigenvalue of A and
kF (x) − A(x)k
lim
= 0.
kxk
kxk→∞
Then F has a fixed point in X.
Let X = C 2 [0, 1] be endowed with the norm by
kuk0 = max{kuk, ku00 k},
where kuk = max0≤t≤1 |u(t)|.
We are now in a position to present and prove our main result. Let
(H1) a, b, c, d, ξ1 , ξ2 are nonnegative constants satisfying 0 ≤ ξ1 < ξ2 ≤ 1, b −
aξ1 ≥ 0 and δ = ad + bc + ac(ξ2 − ξ1 ) 6= 0,
(H2) f (t, u, v) = p(t)g(u) + q(t)h(v), where g, h : R → R are continuous with
g(u)
h(v)
= λ, lim
= µ,
u→∞ u
v→∞ v
where p, q ∈ C[0, 1]. Moreover, there exists some t0 ∈ [0, 1] such that
p(t0 )g(0)+q(t0 )h(0) 6= 0, and there exists a continuous nonnegative function
w : [0, 1] → R+ such that |p(s)| + |q(s)| ≤ w(s) for each s ∈ [0, 1].
Theorem 2.6. Assume (H1)–(H2). If max{|λ|, |µ|} < min L11 , L12 , where
Z
Z 1
1 h ξ1 3
L1 =
τ (2 − τ )w(τ )dτ +
(1 − τ )3 (1 + τ )w(τ )dτ
12 0
ξ1
Z
i
2(b − aξ1 ) + a ξ2
+
(c(ξ2 − τ ) + d)w(τ )dτ ,
δ
ξ1
lim
and
Z
1
(1 − τ )w(τ )dτ +
L2 =
ξ1
1
δ
Z
ξ2
(b + a(1 − ξ1 ))(c(ξ2 − τ ) + d)w(τ )dτ,
ξ1
then BVP (1.1) and (1.2) has at least one nontrivial solution u ∈ C 2 [0, 1].
Proof. Define an operator F : C 2 [0, 1] → C 2 [0, 1] by
Z 1
Z s
F u(t) :=
G1 (t, s)
(τ − s)[p(τ )g(u(τ )) + q(τ )h(u00 (τ )]dτ ds
0
ξ1
Z ξ2
1
G1 (t, s)
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)
δ 0
ξ1
× p(τ )g(u(τ )) + q(τ )h(u00 (τ ) dτ ds,
Z
+
1
(2.9)
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FOUR-POINT BOUNDARY-VALUE PROBLEMS
5
where G1 (t, s) is as in (2.7). Then by Lemma 2.1 and Remark 2.4, we easily know
that the fixed points of F are the solutions to the boundary-value problem (1.1)
and (1.2). It is well known that the operator F is a completely continuous operator.
Now, we consider the following boundary-value problem
u(4) (t) = λp(t)u(t) + µq(t)u00 (t),
0<t<1
u(0) = u(1) = 0,
00
000
au (ξ1 ) − bu (ξ1 ) = 0,
00
(2.10)
000
cu (ξ2 ) + du (ξ2 ) = 0.
Define
1
Z
Au(t) :=
s
Z
(τ − s)[λp(τ )u(τ ) + µq(τ )u00 (τ )dηds
G1 (t, s)
0
1
+
δ
ξ1
Z
1
Z
ξ2
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)
G1 (t, s)
(2.11)
ξ1
0
[λp(τ )u(τ ) + µq(τ )u00 (τ )]dηds.
Obviously, A is a bounded linear operator. Furthermore, the fixed point of A is a
solution of the BVP (2.10) and conversely.
We now assert that 1 is not an eigenvalue of A. In fact, if λ = 0 and µ = 0, then
the BVP (2.10) has no nontrivial solution. If λ 6= 0 or µ 6= 0, suppose the BVP
(2.10) has a nontrivial solution u and kuk0 > 0, then
Z 1
Z s
|Au(t)| ≤
G1 (t, s)
|(τ − s)[λp(τ )u(τ ) + µq(τ )u00 (τ )]|dτ ds
0
ξ1
1
+
δ
Z
1
ξ2
Z
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)
G1 (t, s)
0
ξ1
× [λp(τ )u(τ ) + µq(τ )u00 (τ )]dτ ds
Z 1
Z s
≤
s(1 − s)
(s − τ )[|λ||p(τ )||u(τ )| + |µ||q(τ )||u00 (τ )|]dτ ds
0
+
ξ1
1
δ
Z
1
Z
ξ2
s(1 − s)
0
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)
ξ1
× [|λ||p(τ )||u(τ )| + |µ||q(τ )||u00 (τ )|]dτ ds
Z
Z s
hZ 1
1 1
≤
s(1 − s)
(s − τ )[|λ||p(τ )| + |µ||q(τ )|]dτ ds +
s(1 − s)
δ 0
0
ξ1
Z ξ2
i
×
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)[|λ||p(τ )| + |µ||q(τ )|]dτ ds kuk0
ξ1
Z
1 h ξ1 3
τ (2 − τ )(|λ||p(τ )| + |µ||q(τ )|)dτ
=
12 0
Z 1
+
(1 − τ )3 (1 + τ )(|λ||p(τ )| + |µ||q(τ )|)dτ
ξ1
Z
i
2(b − aξ1 ) + a ξ2
+
(c(ξ2 − τ ) + d)(|λ||p(τ )| + |µ||q(τ )|)dτ kuk0
δ
ξ1
Z 1
Z ξ1
h
1
τ 3 (2 − τ )w(τ )dτ +
(1 − τ )3 (1 + τ )w(τ )dτ
≤ max{|λ|, |µ|}
12 0
ξ1
6
J. GE, C. BAI
+
2(b − aξ1 ) + a
δ
Z
ξ2
EJDE-2007/123
i
(c(ξ2 − τ ) + d)w(τ )dτ kuk0 ,
t ∈ [0, 1],
ξ1
which implies that
|Au(t)| ≤ max{|λ|, |µ|}L1 kuk0 <
1
L1 kuk0 = kuk0 .
L1
On the other hand, we have
Z
t
|(Au)00 (t)| = (s − t)[λp(s)u(s) + µq(s)u00 (s)]ds
ξ1
Z
1 ξ2
(b − a(ξ1 − t))(c(ξ2 − s) + d)[λp(s)u(s) + µq(s)u00 (s)]ds
+
δ ξ1
hZ 1
≤
(1 − s)(|λ||p(s)| + |µ||q(s)|)ds
ξ1
1
+
δ
Z
ξ2
i
(b + a(1 − ξ1 ))(c(ξ2 − s) + d)(|λ||p(s)| + |µ||q(s)|)ds kuk0
ξ1
≤ max{|λ|, |µ|}L2 kuk0 <
1
L2 kuk0 = kuk0 ,
L2
t ∈ [0, 1].
Then kAuk0 < kuk0 . This contradiction means that BVP (2.10) has no nontrivial
solution. Hence, 1 is not an eigenvalue of A.
Finally, we prove that
lim
kuk0 →∞
According to limu→∞
R > 0 such that
g(u)
u
kF u − Auk0
= 0.
kuk0
= λ and limv→∞
|g(u) − λu| < ε|u|,
h(v)
v
= µ, for any ε > 0, there must be
|h(v) − µv| < ε|v|,
|u|, |v| > R.
Set R∗ = max{max|u|≤R |g(u)|, max|v|≤R |h(v)|} and select M > 0 such that R∗ +
max{|λ|, |µ|} < εM . Denote
E1 = {t ∈ [0, 1] : |u(t)| ≤ R, |v(t)| > R},
E2 = {t ∈ [0, 1] : |u(t)| > R, |v(t)| ≤ R},
E3 = {t ∈ [0, 1] : max{|u(t)|, |v(t)|} ≤ R},
E4 = {t ∈ [0, 1] : min{|u(t)|, |v(t)|} > R}.
Thus for any u ∈ C 2 [0, 1] with kuk0 > M , when t ∈ E1 , we have
|g(u(t)) − λu(t)| ≤ |g((u(t))| + |λ||u(t)| ≤ R∗ + |λ|R < εM < εkuk0 ,
and
|h(v(t)) − µv(t)| < ε|v(t)| ≤ εkvk0 .
Similarly, we conclude that for any u ∈ C 2 [0, 1] with kuk0 > M , when t ∈ Ei
(i = 2, 3, 4), we also have that
|g(u(t)) − λu(t)| < εkuk0 ,
|h(v(t)) − µv(t)| < εkvk0 .
EJDE-2007/123
FOUR-POINT BOUNDARY-VALUE PROBLEMS
7
Hence, we get
|F u(t) − Au(t)|
Z s
Z
1
(τ − s)(p(τ )[g(u(τ )) − λu(τ )] + q(τ )[h(u00 (τ )) − µu00 (τ )])dτ ds
G1 (t, s)
=
ξ1
0
+
1
δ
1
Z
Z
ξ2
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)
G1 (t, s)
0
ξ1
× (p(τ )[g(u(τ )) − λu(τ )] + q(τ )[h(u00 (τ )) − µu00 (τ )])dτ ds
Z s
hZ 1
≤
G1 (s, s)
(s − τ )(|p(τ )| + |q(τ )|)dτ ds
0
ξ1
Z ξ2
i
1
G1 (s, s)
(b − a(ξ1 − s))(c(ξ2 − τ ) + d)(|p(τ )| + |q(τ )|)dτ ds εkuk0
+
δ 0
ξ1
Z ξ1
Z 1
h
1
≤
τ 3 (2 − τ )w(τ )dτ +
(1 − τ )3 (1 + τ )w(τ )dτ
12 0
ξ1
Z
i
2(b − aξ1 ) + a ξ2
(c(ξ2 − τ ) + d)w(τ )dτ εkuk0 .
+
δ
ξ1
Z
1
= εL1 kuk0 .
(2.12)
On the other hand, we have
|(F u − Au)00 (t)|
Z
t
(s − t)(p(s)[g(u(s)) − λu(s)] + q(s)[h(u00 (s)) − µu00 (s)])ds
=
ξ1
+
1
δ
Z
ξ2
(b − a(ξ1 − t))(c(ξ2 − s) + d)
ξ1
× (p(s)[g(u(s)) − λu(s)] + q(s)[h(u00 (s)) − µu00 (s)])ds
Z
hZ 1
i
1 ξ2
≤
(1 − s)w(s)ds +
(b + a(1 − ξ1 ))(c(ξ2 − s) + d)w(s)ds εkuk0
δ ξ1
ξ1
= εL2 kuk0 .
Combining the above inequality with (2.12), we have
lim
kuk0 →∞
kF u − Auk0
= 0.
kuk0
Lemma 2.5 now guarantees that BVP (1.1) and (1.2) has a solution u∗ ∈ C 2 [0, 1].
Obviously, u∗ 6= 0 when p(t0 )g(0) + q(t0 )h(0) 6= 0 for some t0 ∈ [0, 1]. In fact, if
u∗ = 0, then (0)(4) = p(t0 )g(0) + q(t0 )h(0) 6= 0 will lead to a contradiction. This
completes the proof.
Example 2.7. Consider the fourth-order four-point boundary-value problem
u(4) (t) =
t sin 2πt
1
u(t) − tecos t cos u00 (t),
2
t +1
2
u(0) = u(1) = 0,
u00 (1/3) − u000 (1/3) = 0,
0 < t < 1,
u00 (2/3) + u000 (2/3) = 0.
(2.13)
8
J. GE, C. BAI
EJDE-2007/123
To show (2.13) has at least one nontrivial solution we apply Theorem 2.6 with
2πt
1 cos t
p(t) = t sin
, g(u) = u, h(u) = cos u, a = b = c = d = 1, ξ1 = 1/3
t2 +1 , q(t) = 2 te
and ξ2 = 2/3. Clearly (H1) is satisfied. Obviously,
p(t0 )g(0) + q(t0 )h(0) =
1 cos t0
t0 e
6= 0,
2
t0 ∈ (0, 1].
Since |p(s)| + |q(s)| ≤ ( 2e + 1)s := w(s) for each s ∈ [0, 1], we have
Z 1
e
h Z 1/3
4
2 +1
L1 =
τ (2 − τ )dτ + (e + 1)
(1 − τ )3 (1 + τ )τ dτ
12
0
1/3
Z 2/3
i
5
+
( − τ )τ dτ ,
1/3 3
L2 =
h
e
+1
2
Z
1
τ (1 − τ )dτ +
1/3
5
7
Z
2/3
1/3
i
5
− τ τ dτ .
3
By simple calculation we easily know that
L1 < L2 <
1 e
+ 1 < 1.
3 2
Notice
g(u)
= 1,
u→∞ u
λ = lim
h(u)
= 0,
u→∞ u
µ = lim
we have
max{λ, µ} < 1 < min{
1 1
,
}.
L1 L2
So (H2) is satisfied. Thus, Theorem 2.6 now guarantees that BVP (2.13) has at
least one nontrivial solution u ∈ C 2 [0, 1].
References
[1] R. P. Agarwal, Focal Boundary Value Problems for Differential and Difference Equations,
Kluwer Academic, Dordrecht, 1998.
[2] R. P. Agarwal, D. O’Regan, P. J.Y . Wong; Positive Solutions of Differential, Difference,
and Integral Equations, Kluwer Academic, Dordrecht, 1998.
[3] D. R. Anderson, J. M. Davis; Multiple solutions and eigenvalues for third-order right focal
boundary-value problem, J. Math. Anal. Appl. 267 (2002), 135C157.
[4] Z. Bai, H. Wang; On positive solutions of some nonlinear fourth-ordr beam equations, J.
Math. Anal. Appl. 270 (2002) 357-368.
[5] J. R. Graef, B. Yang; On a nonlinear boundary-value problem for fourth order equations,
Appl. Anal. 72 (1999) 439-448.
[6] J. R. Graef, B. Yang; Positive solutions to a multi-point higher order boundary-value problem,
J. Math. Anal. Appl. 316 (2006), 409-421.
[7] J. R. Graef, J. Henderson, B. Yang; Positive solutions of a nonlinear higher order boundaryvalue problem, Electron. J. Differential Equations, Vol. 2007(2007), No. 45, pp. 1-10.
[8] Z. Hao, L. Liu, L. Debnath; A necessary and sufficiently condition for the existence of positive
solution of fourth-order singular boundary-value problems, Appl. Math. Lett. 16 (2003) 279285.
[9] M. A. Krasnoselskii, P. P. Zabreiko; Geometrical Methods of Nonlinear Analysis, SpringerVerlag, New York, 1984.
[10] Q. Zhang, S. Chen, J. Lü; Upper and lower solution method for fourth-order four-point
boundary-value problems, J. Comput. Appl. Math. 196 (2006) 387-393.
EJDE-2007/123
FOUR-POINT BOUNDARY-VALUE PROBLEMS
9
Jing Ge
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsi 223300, China
E-mail address: gejing0512@163.com
Chuanzhi Bai
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsi 223300, China
E-mail address: czbai8@sohu.com