Electronic Journal of Differential Equations, Vol. 2008(2008), No. 52, 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)
EXISTENCE OF SOLUTIONS FOR A FOURTH-ORDER
BOUNDARY-VALUE PROBLEM
YANG LIU
Abstract. In this paper, we use the lower and upper solution method to
obtain an existence theorem for the fourth-order boundary-value problem
u(4) (t) = f (t, u(t), u0 (t), u00 (t), u000 (t)), 0 < t < 1,
Z
` 1 00
´
u(0) = u0 (1) = u00 (0) = 0, u000 (1) = g
u (t)dθ(t) ,
0
where f : [0, 1] × R4 → R, g : R → R are continuous and may be nonlinear,
R
and 01 u00 (t)dθ(t) denotes the Riemann-Stieltjes integral.
1. Introduction
It is well known that the bending of an elastic beam can be described with
fourth-order boundary-value problems. Recently, many authors have investigated
the existence of solutions for fourth-order boundary-value problems subject to a
variety of boundary-value conditions, see for example [1, 3, 4, 6, 7].
Very recently, Bai [2] used the lower and upper solution method to obtain the
existence of solutions for the problem
u(4) (t) = f (t, u(t), u0 (t), u00 (t), u000 (t)),
0
00
0 < t < 1,
000
u(0) = u (1) = u (0) = u (1) = 0,
where f : [0, 1] × R4 → R is increasing.
Motivated by the above-mentioned papers and the main ideas in [5], in this
paper, we use the lower and upper solution method to establish the existence of
solutions for the fourth-order boundary-value problem
u(4) (t) = f (t, u(t), u0 (t), u00 (t), u000 (t)), 0 < t < 1,
Z 1
0
00
000
u(0) = u (1) = u (0) = 0, u (1) = g
u00 (t)dθ(t) ,
(1.1)
0
4
where f : [0, 1] × R → R, g : R → R are continuous and may be nonlinear.
θ : [0, 1] → R is increasing nonconstant function defined on [0,1] and θ(0) = 0.
2000 Mathematics Subject Classification. 34B15.
Key words and phrases. Fourth-order boundary-value problem; upper and lower solution;
Riemann-Stieltjies integral; Nagumo-type condition.
c
2008
Texas State University - San Marcos.
Submitted October 19, 2007. Published April 10, 2008.
1
2
Y. LIU
EJDE-2008/52
If g(x) = x, x ∈ R, then the problem (1.1) educes a three-point boundary-value
problem by applying the following property of the Riemann-Stieltjes integral.
Lemma 1.1. Assume that
(1) u(t) is a bounded function value on C 2 [a, b], i.e., there exist c, C ∈ R such
that c ≤ u00 (t) ≤ C, ∀t ∈ [a, b];
(2) θ(t) is increasing on [a, b];
Rb
(3) the Riemann-Stieltjes integral a u00 (t)dθ(t) exists.
Then there is a number v ∈ R with c ≤ v ≤ C such that
Z b
u00 (t)dθ(t) = v(θ(b) − θ(a)).
a
For any continuous solution u(t) of (1.1), by Lemma 1.1, there exists η ∈ (0, 1)
such that
Z b
u00 (t)dθ(t) = u00 (η)(θ(1) − θ(0)) = u00 (η)θ(1).
a
Let σ = θ(1) and g(x) = x, x ∈ R. Then problem (1.1) can be rewritten as the
three-point boundary-value problem
u(4) (t) = f (t, u(t), u0 (t), u00 (t), u000 (t)),
0
00
u(0) = u (1) = u (0) = 0,
0 < t < 1,
000
u (1) = σu00 (η),
(1.2)
The paper is organized as follows. In the next section, we present some preliminaries and lemmas. Section 3 is devote to our main results.
2. Preliminaries
Definition 2.1. Let α ∈ C 3 [0, 1] ∩ C 4 (0, 1). We say α is a lower solution of (1.1)
if
α(4) (t) ≤ f (t, α(t), α0 (t), α00 (t), α000 (t)),
0 < t < 1,
0
α(0) ≤ 0,
α (1) ≤ 0,
Z 1
α000 (1) ≥ g(
α00 (t)dθ(t)).
α00 (0) ≥ 0,
0
Similarly, β ∈ C 3 [0, 1] ∩ C 4 (0, 1) is an upper solution of (1.1), if β satisfies similar
inequalities in the reverse order.
If we denote by k(t, s) the Green’s function of
−u00 (t) = 0,
0 < t < 1,
0
u(0) = u (1) = 0,
then
(
t, 0 ≤ t ≤ s ≤ 1,
k(t, s) =
s, 0 ≤ s ≤ t ≤ 1.
Setting −u00 = v, by standard calculation, we get
Z 1
u(t) =
k(t, s)v(s)ds =: (Av)(t),
0
0
Z
u (t) =
1
v(s)ds =: (Bv)(t).
t
(2.1)
EJDE-2008/52
FOURTH-ORDER BOUNDARY-VALUE PROBLEM
3
Obviously, A, B are monotone increasing operators.
We assume that g is an odd function on R. Then problem (1.1) is equivalent to
the following integral-differential boundary-value problem
−v 00 (t) = f (t, (Av)(t), (Bv)(t), −v(t), −v 0 (t)), 0 < t < 1,
Z 1
0
v(t)dθ(t)).
v(0) = 0, v (1) = g(
(2.2)
0
For v ∈ C[0, 1], we define the operator fˆ by
fˆ(v(t), v 0 (t)) = f (t, (Av)(t), (Bv)(t), −v(t), −v 0 (t)).
Then (2.2) is equivalent to
−v 00 (t) = fˆ(v(t), v 0 (t)),
Z
0
v(0) = 0, v (1) = g(
0 < t < 1,
(2.3)
1
v(t)dθ(t)).
0
Suppose α, β are the lower and upper solutions of BVP (1.1) such that α00 ≥ β 00
and let ψ = −β 00 , φ = −α00 . Then we have
Z 1
00
0
0
ˆ
−φ (t) ≤ f (φ(t), φ (t)), φ(0) ≤ 0, φ (1) ≤ g(
φ(t)dθ(t)),
0
−ψ 00 (t) ≥ fˆ(ψ(t), ψ 0 (t)),
ψ(0) ≥ 0,
Z
ψ 0 (1) ≥ g(
1
ψ(t)dθ(t)).
0
Since A, B are monotone continuous operators, there exists M such that
M = sup {kAvk∞ , kBvk∞ } > 0.
φ≤v≤ψ
Definition 2.2 ([2]). Let f ∈ C([0, 1] × R4 , R), φ, ψ ∈ C([0, 1], R) and φ(t) ≤
ψ(t), t ∈ [0, 1]. We say that f (t, x1 , x2 , x3 , x4 ) satisfies a Nagumo-type condition
with respect to φ, ψ if there exists a positive continuous function h(s) on [0, ∞)
satisfying
|f (t, x1 , x2 , x3 , x4 )| ≤ h(|x4 |),
(2.4)
for all (t, x1 , x2 , x3 , x4 ) ∈ [0, 1] × [−M, M ]2 × [φ(t), ψ(t)] × R, and
Z ∞
s
ds > max ψ(t) − min φ(t),
(2.5)
0≤t≤1
0≤t≤1
h(s)
λ
where λ = max{|ψ(1) − φ(0)|, |ψ(0) − φ(1)|}.
Lemma 2.3. Suppose f satisfies the Nagumo-type condition with respect to φ, ψ ∈
C 2 [0, 1] and φ ≤ ψ. If BVP (2.3) has a solution v(t) such that φ(t) ≤ v(t) ≤ ψ(t),
then there exists N > 0 such that |v 0 (t)| ≤ N , for t ∈ [0, 1].
The proof of the above lemma is similar to that in [2], therefore, we omit it.
3. Main results
Theorem 3.1. Suppose α, β are lower and upper solutions to BVP (1.1) such that
α00 (t) ≥ β 00 (t) and f satisfies a Nagumo-type condition with respect to α00 , β 00 . In
addition, we assume that g is odd, continuous and increasing on R, θ is increasing
on [0, 1] and θ(0) = 0. Then BVP (1.1) has a solution u(t) such that
α(t) ≤ u(t) ≤ β(t),
α00 (t) ≥ u00 (t) ≥ β 00 (t).
4
Y. LIU
EJDE-2008/52
Proof. Since f satisfies the Nagumo-type condition with respect to φ = −α00 , ψ =
−β 00 , there exists a constant N > 0 depending on φ, ψ, h such that
Z N
s
ds > max ψ(t) − min φ(t).
(3.1)
0≤t≤1
0≤t≤1
h(s)
λ
Take C > max{N, kφ0 k, kψ 0 k} and p(v 0 ) = max{−C, min{v 0 , C}}. By modifying
fˆ and g with respect to φ, ψ, we aim at obtaining a second-order boundary-value
problem and reformulating the new problem as an integral equation. We show
that solutions of the modified problem lie in the region where fˆ, g are unmodified and hence are solutions of problem (2.3). Let ε > 0 be a fixed small numR1
ber and F (v(t), v 0 (t)), G( 0 v(t)dθ(t)) are the modifications of fˆ(v(t), v 0 (t)) and
R1
g( 0 v(t)dθ(t)) as follows
F (v(t), v 0 (t))

v(t)−ψ(t)

fˆ(ψ(t), ψ 0 (t)) + 1+|v(t)−ψ(t)|
,




0
0


fˆ(ψ(t), p(v )) + [fˆ(ψ(t), ψ (t)) − fˆ(ψ(t), p(v 0 (t)))



v(t)−ψ(t)
v(t)−ψ(t)


,
+ 1+|v(t)−ψ(t)| ] ×
ε

0
= fˆ(v(t), p(v (t))),




fˆ(φ(t), p(v 0 (t))) + [fˆ(φ(t), φ0 (t)) − fˆ(φ(t), p(v 0 (t)))



φ(t)−v(t)


+ 1+|φ(t)−v(t)|
] × φ(t)−v(t)
,

ε



φ(t)−v(t)
fˆ(φ(t), φ0 (t)) +
1+|φ(t)−v(t)| ,
if v(t) ≥ ψ(t) + ε,
if ψ(t) ≤ v(t) < ψ(t) + ε,
if φ(t) ≤ v(t) ≤ ψ(t),
if φ(t) − ε < v(t) ≤ φ(t),
if v(t) ≤ φ(t) − ε,
and
Z
G
1
v(t)dθ(t)
0
 R1

g( ψ(t)dθ(t)) +

 R0
1
= g( 0 v(t)dθ(t)),


g(R 1 φ(t)dθ(t)) +
0
R1
R
v(t)dθ(t)− 01 ψ(t)dθ(t)
0R
R
,
1+| 01 v(t)dθ(t)− 01 ψ(t)dθ(t)|
if v(t) > ψ(t),
if φ(t) ≤ v(t) ≤ ψ(t),
R1
R
φ(t)dθ(t)− 01 v(t)dθ(t)
0R
R
,
1
1+| 0 φ(t)dθ(t)− 01 v(t)dθ(t)|
if v(t) < φ(t).
Obviously, F : R × R → R and G : R → R are continuous and bounded. Consider
the modified problem
−v 00 (t) = F (v(t), v 0 (t)),
Z
v(0) = 0, v 0 (1) = G(
0 < t < 1,
1
v(t)dθ(t)).
(3.2)
0
Then, the BVP (3.2) is equivalent to the integral equation
Z 1
Z 1
v(t) = G(
v(t)dθ(t))t +
k(t, s)F (v(s), v 0 (s))ds.
0
(3.3)
0
Since F and G are continuous and bounded, there exist M > C, m > 0 such that
|F (v(t), v 0 (t))| < M on R × R,
Z 1
|G(
v(t)dθ(t))| < m on R.
0
EJDE-2008/52
FOURTH-ORDER BOUNDARY-VALUE PROBLEM
5
Choose M̃ ≥ δ + m + M and consider the open bounded and convex set
Ω = {v ∈ C 1 [0, 1] : kvk < M̃ , kv 0 k < M̃ }.
Define F̃ : C 1 [0, 1] × R → C[0, 1] and G̃ : C 1 [0, 1] → C[0, 1] by
Z 1
k(t, s)F (v(s), v 0 (s))ds,
F̃ (v)(t) =
0
Z
G̃(v)(t) = G(
1
v(t)dθ(t))t.
0
It is obvious that F̃ , G̃ are compact. Let T̃ = G̃ + F̃ , it is easy to see that (3.2) is
equivalent to the fixed point equation
T̃ v = v.
(3.4)
Then, it follows from Schauder fixed point theorem that the integral equation (3.4)
has a fixed point v∗ . In other words, the BVP (3.2) has a solution v∗ . Also, from
the definitions of φ, ψ, F and G and the choice of C, we have
−φ00 (t) ≤ fˆ(φ(t), φ0 (t)) = F (φ(t), φ0 (t)), 0 ≤ t ≤ 1,
Z 1
Z 1
φ(0) ≤ 0, φ0 (1) ≤ g(
φ(t)dθ(t)) = G(
φ(t)dθ(t))
0
0
and
−ψ 00 (t) ≥ fˆ(ψ(t), ψ 0 (t)) = F (ψ(t), ψ 0 (t)), 0 ≤ t ≤ 1,
Z 1
Z 1
0
ψ(0) ≥ 0, ψ (1) ≥ g(
ψ(t)dθ(t)) = G(
ψ(t)dθ(t)).
0
0
That is, φ and ψ are the lower and upper solutions of (3.2).
We claim that the solution v∗ of (3.2) satisfies φ(t) ≤ v∗ (t) ≤ ψ(t) for t ∈ [0, 1].
We only prove φ(t) ≤ v∗ (t), t ∈ [0, 1], the other part is proved in a similar way. Let
w(t) = φ(t) − v∗ (t) for t ∈ [0, 1]. Assume that w(t0 ) = max w(t) > 0. We divide
0≤t≤1
the proof into three cases.
Case 1. t0 = 0. Then we have w(0) = φ(0) − v∗ (0) = φ(0) > 0. It contradict the
definition of φ.
Case 2. t0 = 1. Then w(1) > 0 and w0 (1) ≥ 0. The boundary value conditions of
(3.2) imply
Z 1
Z 1
w0 (1) = φ0 (1) − v∗ 0 (1) ≤ g(
φ(t)dθ(t)) − G(
v∗ (t)dθ(t)).
0
0
If v∗ (t) < φ(t), then
Z
G(
0
1
Z
v∗ (t)dθ(t)) = g(
1
0
0
Z
> g(
0
R1
φ(t)dθ(t) − 0 v∗ (t)dθ(t)
φ(t)dθ(t)) +
R1
R1
1 + 0 φ(t)dθ(t) − 0 v∗ (t)dθ(t)
R1
1
φ(t)dθ(t)),
6
Y. LIU
EJDE-2008/52
which implies w0 (1) < 0. It is a contradiction. If v∗ (t) > ψ(t), then
R1
R1
Z 1
Z 1
v (t)dθ(t) − 0 ψ(t)dθ(t)
0 ∗
ψ(t)dθ(t)) +
v∗ (t)dθ(t)) = g(
G(
R1
R1
1 + 0 v∗ (t)dθ(t) − 0 ψ(t)dθ(t)
0
0
Z 1
> g(
ψ(t)dθ(t))
0
1
Z
≥ g(
φ(t)dθ(t)),
0
we can also get w0 (1) < 0, which is a contradiction. Hence, φ(t) ≤ v∗ (t) ≤ ψ(t). So
Z 1
Z 1
Z 1
φ(t)dθ(t)),
v∗ (t)dθ(t)) ≥ g(
v∗ (t)dθ(t)) = g(
G(
0
0
0
0
0
which implies w (1) ≤ 0. If w (1) < 0, it is a contradiction. So we have w0 (1) = 0.
Since t0 6= 0, there exists t1 ∈ [0, 1) such that w(t1 ) = 0 and w(t) > 0 on (t1 , 1].
Then for each t ∈ [t1 , 1], we have
h
w(t) i
w00 (t) = φ00 (t) − v∗ 00 (t) ≥ −fˆ(φ(t), φ0 (t)) + fˆ(φ(t), φ0 (t)) +
> 0.
1 + w(t)
Thus, by w0 (1) = 0, we get w0 (t) ≤ 0 on [t1 , 1], which implies that w is decreasing
on [t1 , 1] and hence w(1) ≤ 0, it is a contradiction.
Case 3. t0 ∈ (0, 1). Then, we have w0 (t0 ) = 0 and w00 (t0 ) ≤ 0. However, for
0 < w(t0 ) < ε, we have
w00 (t0 ) = φ00 (t0 ) − v∗ 00 (t0 )
≥ −fˆ(φ(t0 ), φ0 (t0 )) + F (v∗ (t0 ), v∗ 0 (t0 ))
=
w2 (t0 )
> 0,
(1 + w(t0 ))ε
a contradiction. For w(t0 ) ≥ ε, we obtain
w00 (t0 ) = φ00 (t0 ) − v∗ 00 (t0 ) ≥
w(t0 )
> 0,
1 + w(t0 )
it is also a contradiction. Thus, φ(t) ≤ v∗ (t), t ∈ [0, 1]. By the similar discussion,
we can get v∗ (t) ≤ ψ(t).
According to the Lemma 2.3 and the choice of C, for the solution v∗ of (3.2)
with φ(t) ≤ v∗ (t) ≤ ψ(t), t ∈ [0, 1], we have
|v∗ 0 (t)| ≤ N < C.
Thus,
F (v∗ (t), v∗ 0 (t)) = fˆ(v∗ (t), v∗ 0 (t)),
Z 1
Z 1
G(
v∗ (t)dθ(t)) = g(
v∗ (t)dθ(t)).
0
0
Hence, the solution v∗ of (3.2) with φ(t) ≤ v∗ (t) ≤ ψ(t), t ∈ [0, 1], is a solution of
(2.3). The proof is complete.
Using the Theorem 3.1, we can prove the following result.
EJDE-2008/52
FOURTH-ORDER BOUNDARY-VALUE PROBLEM
7
Corollory 3.2. Suppose α, β are lower and upper solutions to (1.2) such that
α00 (t) ≥ β 00 (t) and f satisfies a Nagumo-type condition with respect to α00 , β 00 . Then
(1.2) has a solution u(t) such that
α(t) ≤ u(t) ≤ β(t),
α00 (t) ≥ u00 (t) ≥ β 00 (t).
References
[1] C. Bai, D. Yang, H. Zhu; Existence of solutions for fourth order differential equation with
four-point boundary conditions, Appl. Math. Lett., 20 (2007) 1131-1136.
[2] Z. Bai; The upper and lower solution method for some fourth-order boundary-value problem,
Nonlinear Anal., 67 (2007) 1704-1709.
[3] Z. Bai, H. Wang; On positive solutions of some nonlinear fourth-order beam equations, J.
Math. Appl., 270 (2002) 357-368.
[4] A. Cabada, J.Á. Cid, L. Sanchez; Positivity and lower and upper solutions for fourth order
boundary-value problems, Nonlinear Anal., 67 (2007) 1599-1612.
[5] R. A. Khan, M. Rafique; Existence and multiplicity results for some three-point boundaryvalue problems, Nonlinear Anal., 66 (2007) 1686-1697.
[6] B. Liu; Positive solutions of fourth-order two point boundary-value problems, Appl. Math.
Comput., 148 (2004) 407-420.
[7] F. Li, Q. Zhang, Z. Liang; Existence and multiplicity of solutions of a kind of fourth-order
boundary-value problem, Nonlinear Anal., 16 (2005) 803-816.
Yang Liu
Department of Mathematics, Yanbian University, Yanji, Jilin 133000, China.
Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 223300, China
E-mail address: liuyang19830206@yahoo.com.cn