# Existence of Positive Solutions for an Elastic Beam Equation with Nonlinear Boundary Conditions

Authors: Ruikuan Liu; Ruyun Ma
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We study the existence and nonexistence of positive solutions for the following fourth-order two-point boundary value problem subject to nonlinear boundary conditionsu
′′′′
(
t
)
=
λ
f
(
t
,
u
(
t
)
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
0
,
u
′
(
0
)
=
μ
h
(
u
(
0
)
)
,
u
′′
(
1
)
=
0
,
u
′′′
(
1
)
=
μ
g
(
u
(
1
)
), where λ
>
0
,
μ
≥
0 are parameters, and f
:
0
,
1
×
0
,
+
∞
→
0
,
+
∞
,
h
:
0
,
+
∞
→
0
,
+
∞, and g
:
0
,
+
∞
→
-
∞
,
0 are continuous. By using the fixed-point index theory, we prove that the problem has at least one positive solution for λ
,
μ sufficiently small and has no positive solution for λ large enough.


---

## Body

## 1. Introduction

In this paper, we study the existence and nonexistence of positive solutions for the following fourth-order boundary value problem with nonlinear boundary conditions(1)
u
′′′′
(
t
)
=
λ
f
(
t
,
u
(
t
)
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
0
,
u
′
(
0
)
=
μ
h
(
u
(
0
)
)
,
u
′′
(
1
)
=
0
,
u
′′′
(
1
)
=
μ
g
(
u
(
1
)
)
,
where λ
>
0, μ
≥
0 are parameters and f
∈
C
(
[
0,1
]
×
ℝ
,
ℝ
) and h
,
g
∈
C
(
ℝ
,
ℝ
) are real functions. If λ
=
1, μ
=
0, in mechanics, problem (1) is called cantilever beam equation [1, 2]. The equation describes the deflection of an elastic beam fixed at the left and freed at the right. There are some papers discussing the existence of solutions of the equation by using various methods, such as the lower and upper solution method, the Leray-Schauder continuation method, fixed-point theory, and the monotone iterative method; see [3–11]. If μ
≠
0, problem (1) has also been studied; see [12–17].In the caseμ
=
1, h
(
u
(
0
)
)
=
0, Yang et al. [12] obtained sufficient conditions of the existence of two solutions of problem (1) by using variational technique and a three-critical-point theorem. Recently, Li and Zhang [13] are concerned with the existence and uniqueness of monotone positive solution of problem (1) with λ
=
μ
=
1, h
(
u
(
0
)
)
=
0 by using a new fixed-point theorem of generalized concave operators, but some monotone assumptions on f and g are needed.In 2013, by using a three-critical-point theorem, Cabada and Tersian [14] studied the existence and multiplicity of solutions of problem (1) with λ
=
μ, h
(
u
(
0
)
)
=
0.A natural question is what would happen ifλ
≠
μ and h
(
u
(
0
)
)
≠
0?Motivated by the above papers, we will prove the existence and nonexistence of positive solution for problem (1) by using the fixed-point index theory with the nonlinearity f satisfying superlinear growth condition at infinity.We make the following assumptions.(A1)
f
:
[
0,1
]
×
[
0
,
+
∞
)
→
(
0
,
+
∞
) is continuous.
(A2)
h
:
[
0
,
+
∞
)
→
[
0
,
+
∞
) is continuous and g
:
[
0
,
+
∞
)
→
(
-
∞
,
0
] is continuous.
(A3)
For anyt
∈
[
0,1
], f
∞
=
lim
⁡
u
→
∞
(
f
(
t
,
u
)
/
u
)
=
+
∞.The main results of the present paper are summarized as follows.Theorem 1.
Assume that (A1)–(A3) hold. Then problem (1) has at least one positive solution for λ, μ sufficiently small.Theorem 2.
Assume that (A1)–(A3) hold. Then problem (1) has no positive solution for λ large enough.Remark 3.
The results obtained in this paper are not a consequence of the previous theorem in the previous literature. Clearly, the boundary condition of (1) is more general than the above pieces of literature and problem (1) considers two different parameters which is more extensive.Remark 4.
It is pointed out that we do not need any monotone assumption onf, g, and h, which is weaker than the corresponding assumptions on f, g in [13]. References [12, 14] are only considered with the existence of solution for problem (1) with nonlinear boundary condition; however, we study the existence and nonexistence of positive solution of (1).The remainder of this paper is arranged as follows. Section2 presents some preliminaries. The proofs of Theorems 1 and 2 are given in Section 3. Finally, we give a simple example to illustrate our main results.


## 2. Preliminaries

In this section we collect some preliminary results that will be used in subsequent sections. LetX
=
C
[
0,1
]; then X is a Banach space under the norm ∥
u
∥
=
max
⁡
t
∈
[
0,1
]
|
u
(
t
)
|. Define P
=
{
u
∈
X
∣
u
(
t
)
≥
0
,
t
∈
[
0,1
]
}; then P
⊂
X is a nonnegative cone.Lemma 5 (see [11, Lemma 2.2]).
Leth
∈
C
[
0,1
]; then the solution u
(
t
) of the problem
(2)
u
′′′′
(
t
)
=
h
(
t
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
u
′
(
0
)
=
u
′′
(
1
)
=
u
′′′
(
1
)
=
0
is
(3)
u
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
h
(
s
)
d
s
,
where
(4)
G
(
t
,
s
)
=
{
1
6
t
2
(
3
s
-
t
)
,
0
≤
t
≤
s
≤
1
,
1
6
s
2
(
3
t
-
s
)
,
0
≤
s
≤
t
≤
1
.Now, let us set(5)
ϕ
(
t
)
=
t
2
2
-
t
3
6
,
Φ
(
t
)
=
h
(
u
(
0
)
)
t
-
g
(
u
(
1
)
)
ϕ
(
t
)
,
0000000000000000000000000000000000
∀
t
∈
[
0,1
]
.
It follows from (3) and (5) with simple computation that problem (1) is equivalent to integral equation
(6)
u
(
t
)
=
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
+
μ
Φ
(
t
)
:
=
T
u
(
t
)
.Note that for anyu
∈
X the function (
T
u
)
(
t
) satisfies the boundary conditions in (1) by the definition of Green’s function G
(
t
,
s
). In view of Lemma 5, it is easy to see that u
∈
X is a fixed point of the operator T if and only if u
∈
X is a solution of problem (1).Lemma 6 (see [6, Lemma 2.2], [13, Lemma 2.1]).
G
(
t
,
s
), ϕ
(
t
) have the following properties:(i)
G
(
t
,
s
)
>
0, ∀
0
<
t, s
<
1;
(ii)
max
⁡
t
∈
[
0,1
]
G
(
t
,
s
)
≤
G
(
1
,
s
)
=
(
1
/
6
)
s
2
(
3
-
s
), ∀
0
≤
s
≤
1;
(iii)
G
(
t
,
s
)
≥
t
2
G
(
1
,
s
)
≥
(
t
2
/
3
)
G
(
1
,
s
), ∀
0
≤
t
,
s
≤
1;
(iv)
t
2
/
3
≤
ϕ
(
t
)
≤
t
2
/
2, ∀
0
≤
t
≤
1.Lemma 7.
Φ
(
t
) has the following properties:(i)
Φ
(
t
)
≤
Φ
(
1
)
≤
h
(
u
(
0
)
)
-
g
(
u
(
1
)
), ∀
0
<
t
<
1;
(ii)
Φ
(
t
)
≥
Φ
(
t
2
/
3
)
≥
0, ∀
0
<
t
<
1.Proof.
First, we will prove (i). For anyt
∈
[
0,1
], it follows that
(7)
Φ
′
(
t
)
=
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
(
t
-
1
2
t
2
)
.
From (A2), Φ
′
(
t
)
≥
0. So Φ
(
t
) is nondecreasing for t
∈
[
0,1
]. Moreover,
(8)
Φ
(
t
)
≤
Φ
(
1
)
=
h
(
u
(
0
)
)
-
1
3
g
(
u
(
1
)
)
≤
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
,
so (i) holds.
Now, we prove (ii). For anyt
∈
[
0,1
], obviously, t
≥
t
2
/
3
≥
0.
From (i),Φ
(
t
) is nondecreasing for t
∈
[
0,1
]; then Φ
(
t
)
≥
Φ
(
t
2
/
3
)
≥
Φ
(
0
)
=
0.For any fixedτ
∈
(
0,1
), define the cone
(9)
K
=
{
u
∈
P
∣
min
⁡
t
∈
[
τ
,
1
]
u
(
t
)
≥
τ
2
3
∥
u
∥
}
.
Then K
⊂
X is a positive cone; letting K
r
=
{
u
∈
K
∣
∥
u
∥
<
r
}, ∂
K
r
=
{
u
∈
K
∣
∥
u
∥
=
r
} with r
>
0 is a constant.Lemma 8.
Assume that (A1) and (A2) hold; thenT
:
K
→
K is completely continuous.Proof.
Obviously, for anyt
∈
[
0,1
], u
∈
K, it follows that T
u
(
t
)
≥
0.
For anyu
∈
K, from (6) and Lemma 7(i), we have
(10)
max
⁡
t
∈
[
0,1
]
T
(
u
(
t
)
)
=
max
⁡
t
∈
[
0,1
]
(
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
+
μ
Φ
(
t
)
)
max
⁡
t
∈
[
0,1
]
T
(
u
(
t
)
)
≤
λ
∫
0
1
‍
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
max
⁡
t
∈
[
0,1
]
T
(
u
(
t
)
)
=
+
μ
(
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
)
.
Hence
(11)
∥
T
u
∥
≤
λ
∫
0
1
‍
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
+
μ
(
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
)
.
For any fixedτ
∈
(
0,1
), (11) together with Lemma 7(ii), we get
(12)
min
⁡
τ
≤
t
≤
1
T
(
u
(
t
)
)
=
min
⁡
τ
≤
t
≤
1
(
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
000000
+
μ
(
h
(
u
(
0
)
)
t
-
g
(
u
(
1
)
)
ϕ
(
t
)
)
λ
∫
0
1
‍
)
≥
τ
2
3
(
λ
∫
0
1
‍
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
+
μ
(
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
)
λ
∫
0
1
‍
)
≥
τ
2
3
∥
T
u
∥
.
So, T
(
K
)
⊂
K.
According to (A1), (A2), and Arzela-Ascoli theorem, it is not difficult to verify thatT
:
K
→
K is completely continuous.The following well-known fixed-point index theorem in cones is crucial to our arguments.Lemma 9 (see [18]).
LetE be a Banach space and let K
⊂
E be a cone. For r
>
0, define K
r
=
{
u
∈
K
∣
∥
u
∥
<
r
}. Assume that T
:
K
¯
r
→
K is compact map such that T
u
≠
u for u
∈
∂
K
r
=
{
u
∈
K
∣
∥
u
∥
=
r
}.(i)
If∥
T
u
∥
≥
∥
u
∥ for u
∈
∂
K
r, then i
(
T
,
K
r
,
K
)
=
0.
(ii)
If∥
T
u
∥
≤
∥
u
∥ for u
∈
∂
K
r, then i
(
T
,
K
r
,
K
)
=
1.


## 3. Proof of Main Results

In this section, we will prove our main results.Proof of Theorem1.
Let(13)
M
=
max
⁡
(
t
,
s
)
∈
[
0,1
]
×
[
0,1
]
G
(
t
,
s
)
>
0
.
Let q
>
0; set
(14)
I
(
q
)
=
M
max
⁡
u
∈
K
,
∥
u
∥
=
q
(
∫
0
1
‍
f
(
s
,
u
(
s
)
)
d
s
)
>
0
.
For any number r
1
>
0, let K
r
1
=
{
u
∈
K
∣
∥
u
∥
≤
r
1
}. Choose σ
>
0 so that
(15)
σ
≤
r
1
2
I
(
r
1
)
,
∀
u
∈
∂
K
r
1
,
σ
(
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
)
≤
r
1
2
,
∀
u
∈
∂
K
r
1
.
Then, for λ
≤
σ, μ
≤
σ and u
∈
∂
K
r
1, combining (6) with Lemma 7, we have
(16)
T
u
(
t
)
≤
λ
M
∫
0
1
‍
f
(
s
,
u
(
s
)
)
d
s
+
μ
(
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
)
≤
σ
M
∫
0
1
‍
f
(
s
,
u
(
s
)
)
d
s
+
σ
(
h
(
u
(
0
)
)
-
g
(
u
(
1
)
)
)
≤
σ
I
(
r
1
)
+
r
1
2
≤
r
1
,
which implies that
(17)
∥
T
u
∥
≤
r
1
=
∥
u
∥
,
u
∈
∂
K
r
1
.
Thus, Lemma 9 implies that
(18)
i
(
T
,
K
r
1
,
K
)
=
1
.
According to (A3), for anyt
∈
[
0,1], there exists a p
>
0 such that
(19)
f
(
t
,
u
)
≥
η
u
,
∀
u
≥
p
,
where η is chosen so that
(20)
λ
(
τ
2
3
)
2
η
∫
τ
1
‍
G
(
1
,
s
)
d
s
≥
1
,
∀
τ
∈
(
0,1
)
.
For any τ
∈
(
0,1
), choose r
2
>
max
⁡
{
(
3
/
τ
2
)
p
,
2
r
1
}, and set K
r
2
=
{
u
∈
K
∣
∥
u
∥
≤
r
2
}. If u
∈
∂
K
r
2, then
(21)
min
⁡
τ
≤
t
≤
1
u
(
t
)
≥
τ
2
3
r
2
>
p
.
It follows from (A2) and Lemma 6 that
(22)
min
⁡
t
∈
[
τ
,
1
]
T
(
u
(
t
)
)
≥
min
⁡
t
∈
[
τ
,
1
]
(
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
)
≥
λ
τ
2
3
∫
0
1
‍
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
λ
τ
2
3
∫
0
1
‍
G
(
1
,
s
)
η
u
(
s
)
d
s
≥
λ
τ
2
3
η
∫
τ
1
‍
G
(
1
,
s
)
u
(
s
)
d
s
≥
λ
(
τ
2
3
)
2
η
∫
τ
1
‍
G
(
1
,
s
)
d
s
∥
u
∥
≥
∥
u
∥
,
which implies that
(23)
∥
T
(
u
)
∥
≥
∥
u
∥
,
u
∈
∂
K
r
2
.
Hence, from Lemma 9, we get
(24)
i
(
T
,
K
r
2
,
K
)
=
0
.
From (18) and (24), it follows that
(25)
i
(
T
,
K
r
2

∖

K
¯
r
1
,
K
)
=
i
(
T
,
K
r
2
,
K
)
-
i
(
T
,
K
r
1
,
K
)
=
-
1
.
Therefore, T has a fixed point in K
r
2
∖
K
¯
r
1; that is, problem (1) has at least one positive solution.Proof of Theorem2.
From (A1) and (A3), for anyt
∈
[
0,1
], there is a constant c such that
(26)
f
(
t
,
u
)
≥
c
u
,
∀
u
≥
0
.
Let u
∈
X be a positive solution of (1); then u
∈
K satisfies (6). Choose λ large enough such that
(27)
λ
(
τ
2
3
)
2
c
∫
τ
1
‍
G
(
1
,
s
)
d
s
>
1
,
∀
τ
∈
(
0,1
)
.
For any τ
∈
(
0,1
), we have
(28)
min
⁡
t
∈
[
τ
,
1
]
u
(
t
)
≥
min
⁡
t
∈
[
τ
,
1
]
(
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
)
≥
λ
τ
2
3
∫
τ
1
‍
G
(
1
,
s
)
c
u
(
s
)
d
s
≥
λ
(
τ
2
3
)
2
c
∫
τ
1
‍
G
(
1
,
s
)
d
s
∥
u
∥
>
∥
u
∥
,
which is a contradiction. Therefore, problem (1) has no positive solution for λ large enough.Finally, we give an example to illustrate our main result.Example 10.
Consider the following fourth-order two-point problem with nonlinear boundary conditions:(29)
u
′′′′
(
t
)
=
λ
(
u
3
(
t
)
(
3
+
sin
(
u
(
t
)
)
)
+
t
2
+
1
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
0
,
u
′
(
0
)
=
μ
(
2
[
u
(
0
)
]
2
+
1
)
,
u
′′
(
1
)
=
0
,
u
′′′
(
1
)
=
μ
(
[
-
(
u
(
1
)
)
]
1
/
2
-
1
)
.
Clearly, the nonlinearity
(30)
f
(
t
,
u
)
=
u
3
(
t
)
(
3
+
sin
(
u
(
t
)
)
)
+
t
2
+
1
,
0000000000000000
(
t
,
u
)
∈
[
0,1
]
×
[
0
,
∞
)
,
h
(
s
)
=
2
s
2
+
1
,
g
(
s
)
=
-
s
1
/
2
-
1
,
∀
s
∈
[
0
,
+
∞
)
.It is easy to check that (A1)–(A3) are satisfied. By simple computation, we haveM
=
1
/
2. Set q
=
2; then I
(
q
)
=
17.Ifr
1
=
4 (where r
1 can be any real number greater than 0), u
∈
∂
K
r
1, then r
1
/
2
I
(
q
)
=
2
/
17, Φ
(
t
)
≤
36, and take σ
=
1
/
20; then ∥
T
(
u
)
∥
≤
∥
u
∥ obviously holds for λ
,
μ
<
1
/
20.Choosingr
2
=
9, it is easy to verify that
(31)
∥
T
(
u
)
∥
≥
∥
u
∥
,
∀
u
∈
∂
K
r
2
.From Theorem1, T has a fixed point in K
r
2

∖

K
¯
r
1; that is, problem (29) has at least one positive solution.For anyt
∈
[
0,1
], there is a constant c
=
3, so that
(32)
f
(
t
,
u
)
≥
3
u
,
∀
u
≥
0
.
If τ
=
1
/
2, then problem (29) has no positive solution for λ
>
450.



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*Source: 972135-2014-03-20.xml*