#### A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition

Boundary Value Problems
Hindawi Publishing Corporation
A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition
Wenjing Song 0 1
Wenjie Gao 1
I. T. Kiguradze
0 Institute of Applied Mathematics, Jilin University of Finance and Economics , Changchun 130017 , China
1 Institute of Mathematics, Jilin University , Changchun 130012 , China
Correspondence should be addressed to Wenjing Song, 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
nonlinear equation
f t, u t , u t , u t , u t ,
with f : 0, 1 × 4 → 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
au 0 − bu
with a, b, c, d ∈
0, ∞ , A, B ∈ .
For the fourth-order differential equation
f t, u t , u t , u t , u t , 0 < t < 1,
define a pair of lower and upper functions of problems 1.1 and 1.2 if the following
conditions are satisfied:
iii α 0 − β 0 ≤ min{β 0 − β 1 , α 1 − α 0 , 0}.
Remark 2.2. By integration, from iii and 2.1 , we obtain
dα 1 ≤ B, β 0 ≥ 0, β 1 ≥
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.
α t ≤ β t , ∀t ∈ 0, 1 ,
α t ≤ β t , α t ≤ β t , ∀t ∈ 0, 1 ,
that is, lower and upper functions, and their first derivatives are also well ordered.
and consider the set
Lemma 2.4. Let Γi t , γi t ∈ C 0, 1 ,
f t, x0, x1, x2, x3 ≤ hE |x3| ,
∀ t, x0, x1, x2, x3 ∈ E,
∞.
∀t ∈ 0, 1 , i
Definition 2.3. Given a set E ⊂ 0, 1 × 4 , a continuous f : E → is said to satisfy the
one-sided Nagumo-type condition in E if there exists a real continuous function hE : 0 →
k, ∞ , for some k > 0, such that
ds > max Γ2 t − min γ2 t .
t∈ 0,1 t∈ 0,1
t, x0, x1, x2, x3 ∈ 0, 1 ×
4 : γi t ≤ xi ≤ Γi t , i
γi t ≤ u i t ≤ Γi t ,
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 .
Assume that ρ ≥ η, and suppose, for contradiction, that |u t | > 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
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
−u t1
−u 4 t dt
f t, u t , u t , u t , u t dt
−u t
−u t
t0 hE −u t
t1 hE −u t
−u t dt u t1 − u t0
≤ tm∈0a,1x Γ2 t − tm∈0i,n1 γ2 t < ρ hE s
If |u t | ≤ 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
−ρ, u t < −ρ, ∀t ∈ t1, t0 .
Applying a convenient change of variable, we have, by 2.3 and 2.11 ,
In a similar way, we may show that
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.
u t < R1, ∀t ∈ 0, 1 .
η hE s ds > tm∈0a,1x Γ2 t − tm∈0i,n1 γ2 t .
u t < R2, ∀t ∈ 0, 1 .
Theorem 3.1. Suppose that there exist lower and upper functions α t and β t of problems 1.1
and 1.2 , respectively. Let f : 0, 1 × 4 → 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 ×
≤ x0, x1 ≤ β t , β t ,
for t ∈ 0, 1 .
Proof. Define the auxiliary functions
α i t ≤ xi ≤ β i t , i
u t − λδ2 t, u t ,
with the boundary conditions
Step 1. Every solution u t of problems 3.6 and 3.7 satisfies
−r1 < α t ≤ β t < r1,
f t, α t , α t , α t , 0 − r1 − α t < 0,
∀t ∈ 0, 1
0 ≥ u 4 t0
≥ λf t0, β t0 , β t0 , β t0 , 0
0 ≥ u 4 t0
u t0 ≥ r1 > 0.
and u 0 u 0 ≤ 0. If λ 0, then u 0 0 and so u 4 0 ≤ 0. Therefore, the above
computations with t0 replaced by 0 yield a contradiction. For λ ∈ 0, 1 , by 3.11 , we get the
following contradiction:
0 ≥ u 0
u s ds < r1|t − ξ| ≤ r1,
u s ds < r1t ≤ r1.
u t < R, ∀t ∈ 0, 1 ,
Step 2. There is an R > 0 such that for every solution u t of problems 3.6 and 3.7
with R independent of λ ∈ 0, 1 .
Consider the set
and for λ ∈ 0, 1 the function Fλ : Er1 →
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
≤ hE∗ |x3|
So, defining hEr1 t hE∗ |x3| 2r1 in 0 , 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
∞.
ρ : max ar1 |A| , |B| cr1 ,
b d
every solution u of 3.6 and 3.7 satisfies
γi t : −r1, Γi t : r1, for i 0, 1, 2.
B − cu 1 .
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.
Define the operators
and for λ ∈ 0, 1 , Nλ : C3 0, 1
Observe that L has a compact inverse. Therefore, we can consider the completely continuous
operator
For R given by Step 2, take the set
∞ < R .
By Steps 1 and 2, degree d I −Tλ, Ω, 0 is well defined for every λ ∈ 0, 1 and by the invariance
with respect to a homotopy
The equation x
T0 x is equivalent to the problem
and has only the trivial solution. Then, by the degree theory,
So the equation T1 x
x has at least one solution, and therefore the equivalent problem
au 0 − bu 0
Step 4. The function u1 t is a solution of the problems 1.1 and 1.2 .
The proof will be finished if the above function u1 t satisfies the inequalities
α t ≤ u1 t ≤ β t .
tm∈0a,1x u1 t − β t : u1 t2 − β t2 > 0.
≥ f t2, β t2 , β t2 , β t2 , β t2 ≥ β 4 t2 .
If t2 0, then we have
By Definition 2.1 this yields a contradiction
tm∈0a,1x u1 t − β t : u1 0 − β 0 > 0,
u1 0 − β 0 ≤ 0.
1 1
b au1 0 − A > b aβ 0 − A ≥ β 0 .
Then t2 / 0 and, by similar arguments, we prove that t2 / 1. Thus,
u1 t ≤ β t , ∀t ∈ 0, 1 .
α t ≤ u1 t ≤ β t .
On the other hand, by 1.2 ,
0 u1 1 − u1 0
Applying the same technique, we have
and then by Definition 2.1 iii , 3.44 and 3.46 , we obtain
1 t
β s ds dt ≤ − 0 0 u1 s ds dt u1 0 ,
1 t
α s ds dt ≥ − 0 0 u1 s ds dt u1 0 ,
α 0 ≤ u1 0 ≤ β 0 .
β t − u1 t ≥ β 0 − u1 0 ≥ 0,
β t − u1 t ≥ β 0 − u1 0
4. An Example
The following example shows the applicability of Theorem 3.1 when f satisfies only the
onesided Nagumo-type condition.
with A, B ∈ . The nonlinear function
are, respectively, lower and upper functions of 4.1 and 4.2 . Moreover, define
t, x0, x1, x2, x3 ∈ 0, 1 ×
Then f satisfies condition 3.2 and the one-sided Nagumo-type condition with hE |x3|
in E.
−t2 − t ≤ u t ≤ t2 t,
−2t − 1 ≤ u t ≤ 2t 1,
−2 ≤ u t ≤ 2.
Example 4.1. Consider now the problem
u 0 − u 0
f t, x0, x1, x2, x3
−t2 − t,
Notice that the function
f t, x0, x1, x2, x3
− 3 x0 ex1 x2 − 2 2 − x3 4
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.
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