Hindawi Publishing Corporation
Boundary Value Problems
Volume 2011, Article ID 569191, 13 pages
doi:10.1155/2011/569191
Research Article
A Fourth-Order Boundary Value Problem with
One-Sided Nagumo Condition
Wenjing Song1, 2 and Wenjie Gao1
1
2
Institute of Mathematics, Jilin University, Changchun 130012, China
Institute of Applied Mathematics, Jilin University of Finance and Economics, Changchun 130017, China
Correspondence should be addressed to Wenjing Song, songwenjing1978@sina.com
Received 10 January 2011; Accepted 9 March 2011
Academic Editor: I. T. Kiguradze
Copyright q 2011 W. Song and W. Gao. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
The aim of this paper is to study a fourth-order separated boundary value problem with the righthand side function satisfying one-sided Nagumo-type condition. By making a series of a priori
estimates and applying lower and upper functions techniques and Leray-Schauder degree theory,
the authors obtain the existence and location result of solutions to the problem.
1. Introduction
In this paper we apply the lower and upper functions method to study the fourth-order
nonlinear equation
u4 t f t, ut, u t, u t, ut ,
0 < t < 1,
1.1
with f : 0, 1 × Ê4 → Ê being a continuous function.
This equation can be used to model the deformations of an elastic beam, and the type of
boundary conditions considered depends on how the beam is supported at the two endpoints
1, 2. We consider the separated boundary conditions
u0 u1 0,
au 0 − bu0 A,
cu 1 du 1 B
with a, b, c, d ∈ Ê 0, ∞, A, B ∈ Ê.
1.2
2
Boundary Value Problems
For the fourth-order differential equation
u4 t f t, ut, u t, u t, ut ,
0 < t < 1,
u0 u1 u 0 u 1,
1.3
the authors in 3 obtained the existence of solutions with the assumption that f satisfies the
two-sided Nagumo-type conditions. For more related works, interested readers may refer to
1–14. The one-sided Nagumo-type condition brings some difficulties in studying this kind
of problem, as it can be seen in 15–18.
Motivated by the above works, we consider the existence of solutions when f
satisfies one-sided Nagumo-type conditions. This is a generalization of the above cases.
We apply lower and upper functions technique and topological degree method to prove
the existence of solutions by making a priori estimates for the third derivative of all
solutions of problems 1.1 and 1.2. The estimates are essential for proving the existence of
solutions.
The outline of this paper is as follows. In Section 2, we give the definition of lower
and upper functions to problems 1.1 and 1.2 and obtain some a priori estimates. Section 3
will be devoted to the study of the existence of solutions. In Section 4, we give an example to
illustrate the conclusions.
2. Definitions and A Priori Estimates
Upper and lower functions will be an important tool to obtain a priori bounds on u, u , and
u . For this problem we define them as follows.
Definition 2.1. The functions α, β ∈ C4 0, 1 ∩ C3 0, 1 verifying
α t ≤ β t,
∀t ∈ 0, 1,
2.1
define a pair of lower and upper functions of problems 1.1 and 1.2 if the following
conditions are satisfied:
i α4 t ≥ ft, αt, α t, α t, α t, β4 t ≤ ft, βt, β t, β t, β t,
ii α0 ≤ 0, α1 ≤ 0, aα 0 − bα 0 ≤ A, cα 1 dα 1 ≤ B, β0 ≥ 0, β1 ≥
0, aβ 0 − bβ 0 ≥ A, cβ 1 dβ 1 ≥ B,
iii α 0 − β 0 ≤ min{β0 − β1, α1 − α0, 0}.
Remark 2.2. By integration, from iii and 2.1, we obtain
αt ≤ βt,
α t ≤ β t,
∀t ∈ 0, 1,
2.2
that is, lower and upper functions, and their first derivatives are also well ordered.
To have an a priori estimate on u , we need a one-sided Nagumo-type growth
condition, which is defined as follows.
Boundary Value Problems
3
Definition 2.3. Given a set E ⊂ 0, 1 × Ê4 , a continuous f : E → Ê is said to satisfy the
one-sided Nagumo-type condition in E if there exists a real continuous function hE : Ê0 →
k, ∞, for some k > 0, such that
ft, x0 , x1 , x2 , x3 ≤ hE |x3 |,
∀t, x0 , x1 , x2 , x3 ∈ E,
2.3
with
∞
0
s
ds ∞.
hE s
2.4
Lemma 2.4. Let Γi t, γi t ∈ C0, 1, Ê satisfy
Γi t ≥ γi t,
∀t ∈ 0, 1, i 0, 1, 2,
2.5
and consider the set
E t, x0 , x1 , x2 , x3 ∈ 0, 1 × Ê4 : γi t ≤ xi ≤ Γi t, i 0, 1, 2 .
2.6
Let f : 0, 1 × Ê4 → Ê be a continuous function satisfying one-sided Nagumo-type condition in E.
Then, for every ρ > 0, there exists an R > 0 such that for every solution ut of problems 1.1
and 1.2 with
u 0 ≤ ρ,
u1 ≥ −ρ,
γi t ≤ ui t ≤ Γi t,
2.7
2.8
for i 0, 1, 2 and every t ∈ 0, 1, one has u ∞ < R.
Proof. Let u be a solution of problems 1.1 and 1.2 such that 2.7 and 2.8 hold. Define
η : max Γ2 1 − γ2 0, Γ2 0 − γ2 1 .
2.9
Assume that ρ ≥ η, and suppose, for contradiction, that |ut| > ρ for every t ∈ 0, 1.
If u t > ρ for every t ∈ 0, 1, then we obtain the following contradiction:
Γ2 1 − γ2 0 ≥ u 1 − u 0 1
0
u tdt >
1
ρ dt ≥ Γ2 1 − γ2 0.
2.10
0
If u t < −ρ for every t ∈ 0, 1, a similar contradiction can be derived. So there is a t ∈ 0, 1
such that |u t| ≤ ρ. By 2.4 we can take R1 > ρ such that
R1
ρ
s
ds > max Γ2 t − min γ2 t.
hE s
t∈0,1
t∈0,1
2.11
4
Boundary Value Problems
If |ut| ≤ ρ for every t ∈ 0, 1, then we have trivially |u t| < R1 . If not, then we can
take t1 ∈ 0, 1 such that u t1 < −ρ or t1 ∈ 0, 1 such that u t1 > ρ. Suppose that the first
case holds. By 2.7 we can consider t1 < t0 ≤ 1 such that
u t0 −ρ,
ut < −ρ,
∀t ∈ t1 , t0 .
2.12
Applying a convenient change of variable, we have, by 2.3 and 2.11,
−u t1 −u t0 s
ds hE s
t1
t0
t0
t1
≤
t0
−u t 4 −u t dt
hE −u t
−u t
f t, ut, u t, ut, ut dt
hE −u t
2.13
−u tdt u t1 − u t0 t1
≤ max Γ2 t − min γ2 t <
t∈0,1
R1
t∈0,1
ρ
s
ds.
hE s
Hence, u t1 > −R1 . Since t1 can be taken arbitrarily as long as ut1 < −ρ, we conclude that
u t > −R1 for every t ∈ 0, 1 provided that ut < −ρ.
In a similar way, it can be proved that u t < R1 , for every t ∈ 0, 1 if u t > ρ.
Therefore,
u t < R1 ,
∀t ∈ 0, 1.
2.14
Consider now the case η > ρ, and take R2 > η such that
R2
η
s
ds > max Γ2 t − min γ2 t.
hE s
t∈0,1
t∈0,1
2.15
In a similar way, we may show that
u t < R2 ,
∀t ∈ 0, 1.
2.16
Taking R max{R1 , R2 }, we have u ∞ < R.
Remark 2.5. Observe that the estimation R depends only on the functions hE , γ2 , Γ2 , and ρ and
it does not depend on the boundary conditions.
3. Existence and Location Result
In the presence of an ordered pair of lower and upper functions, the existence and location
results for problems 1.1 and 1.2 can be obtained.
Boundary Value Problems
5
Theorem 3.1. Suppose that there exist lower and upper functions αt and βt of problems 1.1
and 1.2, respectively. Let f : 0, 1 × Ê4 → Ê be a continuous function satisfying the one-sided
Nagumo-type conditions 2.3 and 2.4 in
E∗ t, x0 , x1 , x2 , x3 ∈ 0, 1 × Ê4 : αt ≤ x0 ≤ βt, α t ≤ x1 ≤ β t, α t ≤ x2 ≤ β t
3.1
If f verifies
f t, αt, α t, x2 , x3 ≥ ft, x0 , x1 , x2 , x3 ≥ f t, βt, β t, x2 , x3
3.2
for t, x2 , x3 ∈ 0, 1 × Ê2 and
αt, α t ≤ x0 , x1 ≤ βt, β t ,
3.3
where x0 , x1 ≤ y0 , y1 means x0 ≤ y0 and x1 ≤ y1 , then problems 1.1 and 1.2 has at least one
solution ut ∈ C4 0, 1 satisfying
αt ≤ ut ≤ βt,
α t ≤ u t ≤ β t,
α t ≤ u t ≤ β t
3.4
for t ∈ 0, 1.
Proof. Define the auxiliary functions
⎧
⎪
αi t,
⎪
⎪
⎪
⎨
δi t, xi xi ,
⎪
⎪
⎪
⎪
⎩
βi t,
xi < αi t,
αi t ≤ xi ≤ βi t,
i 0, 1, 2,
3.5
xi > βi t.
For λ ∈ 0, 1, consider the homotopic equation
u4 t λf t, δ0 t, ut, δ1 t, u t , δ2 t, ut , ut u t − λδ2 t, ut ,
3.6
with the boundary conditions
u0 u1 0,
u 0 λ au 0 − A ,
b
u 1 λ
B − cu 1 .
d
3.7
6
Boundary Value Problems
Take r1 > 0 large enough such that, for every t ∈ 0, 1,
−r1 < α t ≤ β t < r1 ,
3.8
f t, αt, α t, α t, 0 − r1 − α t < 0,
3.9
f t, βt, β t, β t, 0 r1 − β t > 0,
3.10
|A|
< r1 ,
a
|B|
< r1 .
c
3.11
Step 1. Every solution ut of problems 3.6 and 3.7 satisfies
i u t < r1 ,
∀t ∈ 0, 1
3.12
for i 0, 1, 2, for some r1 independent of λ ∈ 0, 1.
Assume, for contradiction, that the above estimate does not hold for i 2. So there
exist λ ∈ 0, 1, t ∈ 0, 1, and a solution u of 3.6 and 3.7 such that |u t| ≥ r1 . In the case
u t ≥ r1 define
max u t : u t0 ≥ r1 .
t∈0,1
3.13
If t0 ∈ 0, 1, then ut0 0 and u4 t0 ≤ 0. Then, by 3.2 and 3.10, for λ ∈ 0, 1,
the following contradiction is obtained:
0 ≥ u4 t0 λf t0 , δ0 t0 , ut0 , δ1 t0 , u t0 , δ2 t0 , u t0 , ut0 u t0 − λδ2 t0 , u t0 λf t0 , δ0 t0 , ut0 , δ1 t0 , u t0 , β t0 , 0 u t0 − λβ t0 ≥ λf t0 , βt0 , β t0 , β t0 , 0 u t0 − λβ t0 λ f t0 , βt0 , β t0 , β t0 , 0 r1 − β t0 u t0 − λr1 > 0.
3.14
For λ 0,
0 ≥ u4 t0 u t0 ≥ r1 > 0.
3.15
max u t : u 0 ≥ r1 > 0
3.16
If t0 0, then
t∈0,1
and u 0 u0 ≤ 0. If λ 0, then u 0 0 and so u4 0 ≤ 0. Therefore, the above
computations with t0 replaced by 0 yield a contradiction. For λ ∈ 0, 1, by 3.11, we get the
Boundary Value Problems
7
following contradiction:
0 ≥ u 0 λ
λ au 0 − A ≥ ar1 − A > 0.
b
b
3.17
The case t0 1 is analogous. Thus, u t < r1 for every t ∈ 0, 1. In a similar way, we
may prove that u t > −r1 for every t ∈ 0, 1.
By the boundary condition 3.7 there exists a ξ ∈ 0, 1, such that u ξ 0. Then by
integration we obtain
t u t u sds < r1 |t − ξ| ≤ r1 ,
ξ
3.18
t
|ut| u sds < r1 t ≤ r1 .
0
Step 2. There is an R > 0 such that for every solution ut of problems 3.6 and 3.7
u t < R,
∀t ∈ 0, 1,
3.19
with R independent of λ ∈ 0, 1.
Consider the set
Er1 t, x0 , x1 , x2 , x3 ∈ 0, 1 × Ê4 : −r1 ≤ xi ≤ r1 , i 0, 1, 2
and for λ ∈ 0, 1 the function Fλ : Er1 →
3.20
Ê given by
Fλ t, x0 , x1 , x2 , x3 λft, δ0 t, x0 , δ1 t, x1 , δ2 t, x2 , x3 x2 − λδ2 t, x2 .
3.21
In the following we will prove that the function Fλ satisfies the one-sided Nagumo-type
conditions 2.3 and 2.4 in Er1 independently of λ ∈ 0, 1. Indeed, as f verifies 2.3 in
E∗ , then
Fλ t, x0 , x1 , x2 , x3 λft, δ0 t, x0 , δ1 t, x1 , δ2 t, x2 , x3 x2 − λδ2 t, x2 ≤ hE∗ |x3 | r1 − λα t ≤ hE∗ |x3 | 2r1 .
3.22
So, defining hEr1 t hE∗ |x3 | 2r1 in Ê0 , we see that Fλ verifies 2.3 with E and hE replaced
by Er1 and hEr1 , respectively. The condition 2.4 is also verified since
∞
0
s
ds hEr1 s
∞
0
s
hE∗ s2r1
ds ≥
1
1 2r1 /k
∞
0
s
hE∗ s
ds ∞.
3.23
8
Boundary Value Problems
Therefore, Fλ satisfies the one-sided Nagumo-type condition in Er1 with hE replaced by hEr1 ,
with r1 independent of λ ∈ 0, 1.
Moreover, for
ρ : max
ar1 |A| |B| cr1
,
,
b
d
3.24
every solution u of 3.6 and 3.7 satisfies
u0 λ
λ au 0 − A ≤ ar1 |A| ≤ ρ,
b
b
λ
λ
u 1 B − cu 1 ≥ − |B| cr1 ≥ −ρ.
d
d
3.25
Define
γi t : −r1 ,
Γi t : r1 ,
for i 0, 1, 2.
3.26
The hypotheses of Lemma 2.4 are satisfied with E replaced by Er1 . So there exists an R > 0,
depending on r1 and hEr1 , such that |u t| < R for every t ∈ 0, 1. As r1 and hEr1 do not
depend on λ, we see that R is maybe independent of λ.
Step 3. For λ 1, the problems 3.6 and 3.7 has at least one solution u1 t.
Define the operators
L : C4 0, 1 ⊂ C3 0, 1 −→ C0, 1 × Ê4
3.27
Lu u4 − u , u0, u1, u0, u1
3.28
by
and for λ ∈ 0, 1, Nλ : C3 0, 1 → C0, 1 × Ê4 by
Nλ u λf t, δ0 t, ut, δ1 t, u t , δ2 t, u t , ut − λδ2 t, ut , 0, 0, Aλ, Bλ ,
3.29
with
Aλ :
λ au 0 − A ,
b
λ
Bλ :
B − cu 1 .
d
3.30
Boundary Value Problems
9
Observe that L has a compact inverse. Therefore, we can consider the completely continuous
operator
Tλ : C3 0, 1, Ê −→ C3 0, 1, Ê
3.31
Tλ u L−1 Nλ u.
3.32
Ω x ∈ C3 0, 1 : xi < r1 , i 0, 1, 2, x ∞ < R .
3.33
given by
For R given by Step 2, take the set
∞
By Steps 1 and 2, degree dI −Tλ , Ω, 0 is well defined for every λ ∈ 0, 1 and by the invariance
with respect to a homotopy
dI − T0 , Ω, 0 dI − T1 , Ω, 0.
3.34
The equation x T0 x is equivalent to the problem
u4 t u t,
3.35
u0 u1 u 0 u 1 0
and has only the trivial solution. Then, by the degree theory,
dI − T0 , Ω, 0 ±1.
3.36
So the equation T1 x x has at least one solution, and therefore the equivalent problem
u4 t f t, δ0 t, ut, δ1 t, u t , δ2 t, u t , u t u t − δ2 t, u t ,
u0 u1 0,
au 0 − bu0 A,
cu 1 du 1 B
has at least one solution u1 t in Ω.
Step 4. The function u1 t is a solution of the problems 1.1 and 1.2.
3.37
10
Boundary Value Problems
The proof will be finished if the above function u1 t satisfies the inequalities
α t ≤ u1 t ≤ β t,
αt ≤ u1 t ≤ βt,
α t ≤ u1 t ≤ β t.
3.38
Assume, for contradiction, that there is a t ∈ 0, 1 such that u1 t > β t, and define
max u1 t − β t : u1 t2 − β t2 > 0.
3.39
t∈0,1
4
4
If t2 ∈ 0, 1, then u
1 t2 β t2 and u1 t2 ≤ β t2 . Therefore, by 3.2 and
Definition 2.1, we obtain the contradiction
4
u1 t2 f t2 , δ0 t2 , u1 t2 , δ1 t2 , u1 t2 , δ2 t2 , u1 t2 , u
1 t2 u1 t2 − δ2 t2 , u1 t2 f t2 , δ0 t2 , u1 t2 , δ1 t2 , u1 t2 , β t2 , β t2 u1 t2 − β t2 3.40
≥ f t2 , βt2 , β t2 , β t2 , β t2 ≥ β4 t2 .
If t2 0, then we have
max u1 t − β t : u1 0 − β 0 > 0,
t∈0,1
3.41
u
1 0
− β 0 u
1 0 − β 0 ≤ 0.
By Definition 2.1 this yields a contradiction
u
1 0 1
1 au1 0 − A > aβ 0 − A ≥ β 0.
b
b
3.42
0 and, by similar arguments, we prove that t2 /
1. Thus,
Then t2 /
u1 t ≤ β t,
∀t ∈ 0, 1.
3.43
Using an analogous technique, it can be deduced that α t ≤ u1 t for every t ∈ 0, 1. So we
have
α t ≤ u1 t ≤ β t.
3.44
On the other hand, by 1.2,
0 u1 1 − u1 0 1
0
u1 tdt
1
t
1 t
u1 0 u1 sds dt u1 0 u1 sds dt, 3.45
0
0
0
0
Boundary Value Problems
11
that is,
u1 0
−
1 t
0
0
u1 sds dt.
3.46
Applying the same technique, we have
−
1 t
0
β sds dt −
0
1
β tdt β 0 β0 − β1 β 0,
3.47
0
and then by Definition 2.1 iii, 3.44 and 3.46, we obtain
α 0 ≤ β 0 − β1 β0
−
1 t
0
β sds dt ≤ −
0
1 t
0
0
u1 sds dt u1 0,
3.48
β 0 ≥ α 0 − α1 α0
−
1 t
0
0
α sds dt ≥ −
1 t
0
0
u1 sds dt u1 0,
that is,
α 0 ≤ u1 0 ≤ β 0.
3.49
Since, by 3.44, β t − u1 t is nondecreasing, we have by 3.49
β t − u1 t ≥ β 0 − u1 0 ≥ 0,
3.50
and, therefore, β t ≥ u 1 t for every t ∈ 0, 1. By the monotonicity of βt − u1 t,
βt − u1 t ≥ β0 − u1 0 β0 ≥ 0,
3.51
and so βt ≥ u1 t for every t ∈ 0, 1.
The inequalities u1 t ≥ α t and u1 t ≥ αt for every t ∈ 0, 1 can be proved in the
same way. Then u1 t is a solution of problems 1.1 and 1.2.
4. An Example
The following example shows the applicability of Theorem 3.1 when f satisfies only the onesided Nagumo-type condition.
12
Boundary Value Problems
Example 4.1. Consider now the problem
2 4
u4 t −3 ut eu t u t − 2 − u t ,
4.1
u0 u1 0,
u 0 − u 0 A,
4.2
u 1 u 1 B
with A, B ∈ Ê. The nonlinear function
4.3
ft, x0 , x1 , x2 , x3 −3 x0 ex1 x2 − 22 − x3 4
is continuous in 0, 1 × Ê4 . If A, B ∈ −2, 2, then the functions α, β : 0, 1 →
αt −t2 − t,
Ê defined by
βt t2 t
4.4
are, respectively, lower and upper functions of 4.1 and 4.2. Moreover, define
E t, x0 , x1 , x2 , x3 ∈ 0, 1 × Ê4 : −t2 − t ≤ x0 ≤ t2 t, −2t − 1 ≤ x1 ≤ 2t 1, −2 ≤ x2 ≤ 2 .
4.5
Then f satisfies condition 3.2 and the one-sided Nagumo-type condition with hE |x3 | 1,
in E.
Therefore, by Theorem 3.1, there is at least one solution ut of Problem 4.1 and 4.2
such that, for every t ∈ 0, 1,
−t2 − t ≤ ut ≤ t2 t,
−2t − 1 ≤ u t ≤ 2t 1,
−2 ≤ u t ≤ 2.
4.6
Notice that the function
ft, x0 , x1 , x2 , x3 −3 x0 ex1 x2 − 22 − x3 4
4.7
does not satisfy the two-sided Nagumo condition.
Acknowledgments
The authors would like to thank the referees for their valuable comments on and suggestions
regarding the original manuscript. This work was supported by NSFC 10771085, by Key
Lab of Symbolic Computation and Knowledge Engineering of Ministry of Education, and by
the 985 Program of Jilin University.
Boundary Value Problems
13
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