Existence Results for A Perturbed Fourth-Order Equation | Heidari Tavani | Journal of the Indonesian Mathematical Society 498 1423 1 PB

J. Indones. Math. Soc.
Vol. 23, No. 2 (2017), pp. 55–65.

EXISTENCE RESULTS FOR A PERTURBED
FOURTH-ORDER EQUATION
MOHAMMAD REZA HEIDARI TAVANI1
1
Department of Mathematics,
Ramhormoz branch, Islamic Azad University, Ramhormoz, Iran

m.reza.h56@gmail.com

Abstract. The existence of at least three weak solutions for a class of perturbed
fourth-order problems with a perturbed nonlinear term is investigated. Our approach is based on variational methods and critical point theory.
Key words and Phrases: Fourth-order equation, weak solution, critical point theory,
variational methods.

Abstrak. Paper ini memeriksa keujudan sekurang-kurangnya tiga solusi lemah dari
kelas masalah orde ke-empat dipertubasi dengan sebuah suku non linier di pertubasi.
Pendekatan yang kami lakukan berdasarkan metode variasional dan teori titik kritis.
Kata kunci: Persamaan orde ke-empat, solusi lemah, teori titik kritis, metode variasional.

1. INTRODUCTION
Consider the following fourth-order problem

 u(iυ) (x) = λf (x, u(x)) + h(u(x)),
u(0) = u′ (0) = 0,
 ′′
u (1) = 0, u′′′ (1) = µ g(u(1)),

x ∈ [0, 1],

(1)

where λ and µ are a positive parameters, f : [0, 1] × R → R is L1 -Carath´eodory
function, g : R → R is a continuous function and h : R → R is a Lipschitz continuous
function with the Lipschitz constant 0 < L < 1, i.e.,
|h(t1 ) − h(t2 )| ≤ L|t1 − t2 |

for every t1 , t2 ∈ R, and h(0) = 0. According to definition of function h we see that
|h(t)| ≤ L|t|

2000 Mathematics Subject Classification: 34B15, 58E05.
Received: 7 May 2017, revised: 21 June 2017, accepted: 21 June 2017.
55

56

Tavani, M.H.

for each t ∈ R.
The problem (1) is related to the deflections of elastic beams based on nonlinear
elastic . In relation with the problem (1), there is an interesting physical description.
An elastic beam of length d = 1, which is clamped at its left side x = 0, and resting
on a kind of elastic bearing at its right side x = 1 which is given by µg . Along
its length, a load λf + h, is added to cause deformations. if u = u(x) denotes the
configuration of the deformed beam, then since u′′′ (1) represents the shear force at
x = 1, the condition u′′′ (1) = µg(u(1)) means that the vertical force is equal to
µg(u(1)), which denotes a relation, possibly nonlinear, between the vertical force
and the displacement u(1).
Classical bending theory of elastic beams are very important in engineering sciences

and so many studies have been done on a variety of problems like this.
For example in [12], authors considered iterative solutions for problem (1) in the
case of λ = µ = 1 with nonlinear boundary conditions. In [9, 11] Some authors by
using the critical point theory studied Existence and multiplicity results for this
kind of problem . In particular, by using a variational methods the existence of
non-zero solutions for problem (1) in the case of h(t) ≡ 0 has been established in
[4].
In the present paper, using one kind of three critical points theorem obtained in
[5] which we recall in the next section , we establish the existence of at least three
weak solutions for the problem (1). For example this theorem used to ensure the
existence of at least three solutions for perturbed boundary value problems in the
papers [2, 3, 7, 10].

2. Preliminaries
Our main tools is the following three critical points Theorem. In this
following Theorem the coercivity of the functional Φ − λΨ is required.

Theorem 2.1 ([5], Theorem 2.6). Let X be a reflexive real Banach space, Φ : X −→
R be a coercive continuously Gˆ
ateaux differentiable and sequentially weakly lower

semicontinuous functional whose Gˆ
ateaux derivative admits a continuous inverse
on X ∗ , Ψ : X −→ R be a continuously Gˆ
ateaux differentiable functional whose
Gˆ
ateaux derivative is compact such that Φ(0) = Ψ(0) = 0.
Assume that there exist r > 0 and w ∈ X, with r < Φ(w) such that
"
#
supΦ(u)≤r Ψ(u)
Ψ(w)
r
Φ(w)
<
, (a2 ) for each λ ∈ Λr :=
,
(a1 )
r
Φ(w)
Ψ(w) supΦ(u)≤r Ψ(u)

the functional Φ − λΨ is coercive.
Then, for each λ ∈ Λr the functional Φ − λΨ has at least three distinct critical
points in X.
Let us recall some basic concepts. Denote
X := {u ∈ H 2 ([0, 1])|u(0) = u′ (0) = 0},

Existence Results

57

where H 2 ([0, 1]) is the sobolev space of all functions u : [0, 1] → R such that u
and its distributional derivative u′ are absolutely continuous and u′′ belongs to
L2 ([0, 1]).It is clearly that X is a Hilbert space with the usual norm
Z 1
1/2
′′
2
′
2
2

||u||X =
(|u (x)| + |u (x)| + |u(x)| )dx
0

which is equivalent to the norm

||u|| =

Z

1
′′

2

(|u (x)| dx
0

1/2

.

Also according to [13] the embedding X ֒→ C 1 ([0, 1]) is compact and we have
||u||C 1 ([0,1]) = max{||u||∞ , ||u′ ||∞ } ≤ ||u||

(2)

for each u ∈ X. We suppose that the Lipschitz constant L of the function h satisfies
L < 1. A function u : [0, 1] → R is a weak solution to the problem (1) if u ∈ X and
Z 1
Z 1
Z 1
h(u(x))v(x)dx = 0
f (x, u(x))v(x)dx − µg(u(1))v(1) −
u′′ (x)v ′′ (x)dx − λ
0

0

0

for every v ∈ X. Standard methods (see [13, Lemma 2.1]) show that if f is continuous, then each weak solution u of the problem (1) is a classical solution.
Put
Z
t

f (x, ξ)dξ for all (x, t) ∈ [0, 1] × R,
Z t
G(t) =
g(ξ)dξ for all t ∈ R,

F (x, t) =

0

0

Gβ = max G(t)
|t|≤β

and
H(t) =

Z

for all β > 0,

t

h(ξ)dξ
0

for all t ∈ R.

Now we express the following proposition that, in the proof of the main theorem
of this paper is required.
Proposition 2.2. Let T : X → X ∗ be the operator defined by
Z 1
Z 1
T (u)(v) =

h(u(x))v(x)dx
u′′ (x)v ′′ (x)dx −
0

0

for every u, v ∈ X. Then T admits a continuous inverse on X ∗ .

Proof. For every u, v ∈ X we have,
Z
Z 1
|u′′ (x)−v ′′ (x)|2 dx−L
hT (u)−T (v), u−vi ≥
0

1
0

|u(x)−v(x)|2 dx ≥ (1−L)||u−v||2 .

So T is uniformly monotone. By [[14], Theorem 26.A(d)], we have that T −1 exists
and is continuous.


Tavani, M.H.

58

3. Main Results
In order to introduce our first result, fixing two positive constants θ and δ
such that
Z
Z 1
2 3 2 1
2
4
F (x, δ)dx,
(3)
sup F (x, t)dx < (1 − L)θ
8π (1 + L)( ) δ
3
3
0 |t|≤θ
4
and taking




 4(1 + L)( 2 )3 π 4 δ 2

(1 − L)θ2


3
λ ∈ Λ :=  Z 1
, Z 1
,


F (x, δ)dx
sup F (x, t)dx
2
3
4

(4)

0 |t|≤θ

and set ηλ,g given by

ηλ,g :=











(1−L)θ 2 −2λ

Z

1

sup F (x, t)dx
0 |t|≤θ
2Gθ

if G(δ) ≥ 0

Z 1
Z 1



2
4 2
3 3

sup F (x, t)dx 4(1+L)π δ −( 2 ) λ
F (x, δ)dx o

n (1−L)θ −2λ

3

0 |t|≤θ
 min
4
,
2Gθ
( 3 )3 G(δ)
2

if G(δ) < 0.
(5)

It is easy to show that ηλ,g > 0 (also see [4]) .Now we express the main results.
Theorem 3.1. Assume that there exist two positive constants θ and δ with θ <
8π 2
√ δ, such that
3 3
(A1 ) F (x, t) ≥ 0, for each (x, t) ∈ [0, 34 [×[0, δ];
Z
Z 1
2 3 2
2
4
sup F (x, t)dx < (1 − L)θ
(A2 ) 8π (1 + L)( 3 ) δ
(A3 )

0 |t|≤θ
supx∈[0,1] F (x,t)
lim sup|t|→+∞
t2

1

F (x, δ)dx;
3
4

≤ 0.

Then, for each λ ∈ Λ and for every continuous function g : R → R there exists ηλ,g > 0 given by (5) such that, for each µ ∈]0, ηλ,g [, the problem (1) admits at
least three weak solutions in X.
Proof. Fix λ ∈ Λ and µ ∈]0, ηλ,g [. Let Φ, Ψ : X → R be defined by
Z 1
Z
1 1 ′′
2
Φ(u) =
H(u(x))dx
|u (x)| dx −
2 0
0

(6)

Existence Results

and
Z

59

1

µ
G(u(1))
(7)
λ
0
for every u ∈ X. It is well known that Ψ is a differentiable functional whose
differential at the point u ∈ X is
Z 1
µ
f (x, u(x))v(x)dx + g(u(1))v(1),
Ψ′ (u)(v) =
λ
0
Ψ(u) =

F (x, u(x))dx +

as well as is sequentially weakly upper semicontinuous. Furthermore, Ψ′ : X → X ∗
is a compact operator[see [13]-page 1602]. Moreover, Φ is continuously differentiable
whose differential at the point u ∈ X is
Z 1
Z 1
h(u(x))v(x)dx
u′′ (x)v ′′ (x)dx −
Φ′ (u)(v) =
0

0

for every v ∈ X, while Proposition 2.2 gives that Φ′ admits a continuous inverse
on X ∗ . Furthermore, Φ is sequentially weakly lower semicontinuous. Put
r :=

(1 − L) 2
θ
2

and

 0
δ cos2 ( 4π3 x )
w(x) :=

δ

We clearly observe that w ∈ X and,

if x ∈ [0, 83 ]
if x ∈] 83 , 34 [
if x ∈ [ 43 , 1].

(8)

(9)

2
||w||2 = 8π 4 δ 2 ( )3 .
3
Now according to (2) and definition of functional Φ we have,
(1 + L)
(1 − L)
||u||2 ≤ Φ(u) ≤
||u||2
2
2
and in particular,
2
2
4(1 − L)π 4 δ 2 ( )3 ≤ Φ(w) ≤ 4(1 + L)π 4 δ 2 ( )3 .
3
3
8π 2
Now using θ < √ δ, and (10) we observe that
3 3
0 < r < Φ(w).
Since

(1−L)
2
2 ||u||

≤ Φ(u), for each u ∈ X , we see that

Φ−1 (] − ∞, r])

=
=
⊆

{u ∈ X; Φ(u) ≤ r}


(1 − L)
2
||u|| ≤ r
u ∈ X;
2
{u ∈ X; |u(x)| ≤ θ for each x ∈ [0, 1]} ,

(10)

Tavani, M.H.

60

and it follows that
sup

Ψ(u)

=

sup

u∈Φ−1 (]−∞,r])

u∈Φ−1 (]−∞,r])

Z

≤

1

Z

1

F (x, u(x))dx +
0

sup F (x, t)dx +
0 |t|≤θ

µ θ
G .
λ


µ
G(u(1))
λ

On the other hand, in view of (A1 ), since w(x) ∈ [0, δ] for each x ∈ [0, 1], we have
Z 1
µ
Ψ(w) =
F (x, w(x))dx + G(w(1))
λ
0
Z 1
µ
≥
F (x, δ)dx + G(δ).
3
λ
4
Hence, we have
sup

Ψ(u)

u∈Φ−1 (]−∞,r])

r

≤

Ψ(w)
Φ(w)

R1

Z

1

sup F (x, t)dx +
0 |t|≤θ

µ θ
G
λ

(1−L) 2
2 θ

,

(11)

and
≥

3
4

F (x, δ)dx + µλ G(δ)

4(1 + L)π 4 δ 2 ( 32 )3

.

Now if G(δ) ≥ 0 then from (11), since µ < ηλ,g and λ ∈ Λ one has
Z 1
Z 1
µ
ηλ,g θ
G
sup F (x, t)dx +
sup F (x, t)dx + Gθ
λ
λ
1
0 |t|≤θ
0 |t|≤θ
<
= <
(1−L) 2
(1−L) 2
λ
θ
θ
2
2
R1
R1
µ
3 F (x, δ)dx +
3 F (x, δ)dx
λ G(δ)
4
4
≤
.
4(1 + L)π 4 δ 2 ( 23 )3
4(1 + L)π 4 δ 2 ( 32 )3
In the other hand if G(δ) < 0 and Gθ > 0 then
Z 1
Z 1
ηλ,g θ
µ
sup F (x, t)dx +
G
sup F (x, t)dx + Gθ
λ
λ
1
|t|≤θ
|t|≤θ
0
0
<
≤
(1−L) 2
(1−L) 2
λ
2 θ
2 θ
and also if G(δ) < 0 and Gθ = 0 then we have ,
Z 1
Z 1
ηλ,g θ
µ θ
sup F (x, t)dx +
G
sup F (x, t)dx + G
λ
λ
1
0 |t|≤θ
0 |t|≤θ
<
< .
(1−L) 2
(1−L) 2
λ
2 θ
2 θ
However, the following inequality can be used again.
R1
R1
µ
3 F (x, δ)dx
3 F (x, δ)dx +
1
λ G(δ)
4
4
<
≤
.
2
λ
4(1 + L)π 4 δ 2 ( 3 )3
4(1 + L)π 4 δ 2 ( 32 )3

(12)

Existence Results

So we have shown that the following inequality is established.
Z 1
µ
R1
sup F (x, t)dx + Gθ
µ
3 F (x, δ)dx +
λ
1
λ G(δ)
0 |t|≤θ
4
<
.
<
(1−L) 2
4
2
λ
4(1 + L)π δ ( 23 )3
2 θ

61

(13)

Hence from (11) ,(12) and (13) we observe that
supΦ(u)≤r Ψ(u)
1
Ψ(w)
< <
.
r
λ
Φ(w)
Hence the condition (a1 ) of Theorem 2.1 is satisfied .
Finally, we will show that for every λ ∈ Λ, functional Φ − λΨ, is coercive. For this
1−L
purpose, fix 0 < ǫ <
. From (A3 ) there is a function Kǫ ∈ L1 ([0, 1]) such that
2λ
F (x, t) ≤ ǫt2 + Kǫ (x),

(14)

for every x ∈ [0, 1] and t ∈ R. Now, for each u ∈ X, we have
Z 1
1
F (x, u(x))dx − µG(u(1)) ≥
(1 − L)kuk2 − λ
Φ(u) − λΨ(u) ≥
2
0
Z 1
Z u(1)
1−L
(
− λǫ)kuk2 − λ
Kǫ (x)dx − µ
g(ξ)dξ.
2
0
0

Now if we suppose that M := maxξ∈[0, |u(1)| ] |g(ξ)|, then by using (2) we have ,
Φ(u) − λΨ(u) ≥ (

1−L
− λǫ)kuk2 − λ||Kǫ ||L1 [0,1] − µM ||u||
2

and thus
lim

kuk→+∞

(Φ(u) − λΨ(u)) = +∞,

which means the functional Φ − λΨ is coercive, and the condition (a2 ) of Theorem
2.1 is verified. Finally, since the weak solutions of the problem (1) are exactly the
solutions of the equation Φ′ (u) − λΨ′ (u) = 0, Theorem 2.1 ensures the conclusion.

Z 1
(1−L)θ 2 −2λ
sup F (x, t)dx
0 |t|≤θ
Remark 3.1. In Theorem 3.1 we read
= +∞ when
2Gθ
Gθ = 0.
In the following, we will express another form of Theorem 3.1, which will be
achieved by reversing the role of λ and µ.
8π 2
Theorem 3.2. Assume that there exist two positive constants θ ,δ with θ < √ δ,
3 3
such that
(B1 ) 8π 4 (1 + L)( 32 )3 δ 2 Gθ < (1 − L)θ2 G(δ);

Tavani, M.H.

62

Then, for each

4(1 + L)( 32 )3 π 4 δ 2 (1 − L)θ2
µ ∈ Λ :=
,
G(δ)
2Gθ
1
and for every L -Carath´eodory function f : [0, 1] × R → R satisfied in conditions
of (A1) and (A3) of Theorem 3.1 there exists η ′ λ,g > 0 where


′

η ′ λ,g =

(1 − L)θ2 − 2µ Gθ
Z 1
2
sup F (x, t)dx
0 |t|≤θ

such that, for each λ ∈]0, η
in X.

′

λ,g [,

the problem (1) admits at least three weak solutions

ˆ : X → R be defined by
Proof. Fix µ ∈ Λ′ and λ ∈]0, η ′ λ,g [ . Let Ψ
Z 1
λ
ˆ
F (x, u(x))dx + G(u(1))
Ψ(u)
=
µ 0
for every u ∈ X. It is clearly that
ˆ
Φ(u) − λΨ(u) = Φ(u) − µΨ(u)

(15)

for every u ∈ X. Choose r and w as given in (8) and (9).Now, as proof of the
Theorem 3.1 ,we have
R
λ 1
ˆ
Ψ(w)
G(δ)
1
µ 43 F (x, δ)dx + G(δ)
≥
>
≥
(16)
2
2
4
2
4
2
3
3
Φ(w)
µ
4(1 + L)π δ ( 3 )
4(1 + L)π δ ( 3 )

and
ˆ
Ψ(u)

sup

λ
µ

u∈Φ−1 (]−∞,r])

≤

r
η ′ λ,g
µ

Z

Z

1

sup F (x, t)dx + Gθ
0 |t|≤θ
(1−L) 2
2 θ

≤

(17)

1

sup F (x, t)dx + Gθ
0 |t|≤θ

<

(1−L) 2
2 θ

1
.
µ

Therefore from (16) ,(17) we observe that
ˆ
Ψ(u)
sup

ˆ
1
Ψ(w)
<
.
r
µ
Φ(w)
On the other hand as proof of the Theorem 3.1 it is easy to show that for ev′
ˆ is coercive. Hence, since for each µ ∈ Λ′ ⊆
ery
− µΨ,
 µ ∈ Λ , functional Φ 
u∈Φ−1 (]−∞,r])

 Φ(w)
,
ˆ
Ψ(w)

r

sup

u∈Φ−1 (]−∞,r])

<


 the assumptions of Theorem 2.1 are fulfilled, the conˆ
Ψ(u)

clusion follows from Theorem 2.1.



Existence Results

63

Remark 3.2. In Theorem 3.1 and Theorem 3.2 we can replace the condition(A3 )
with the following subquadratic condition:
There exists α ∈ (0, 2) and M > 0 such that 0 < tf (x, t) ≤ α F (x, t) for all |t| ≥ M
and x ∈ [0, 1] which is stronger than (A3 ) and known as Ambrosetti-Rabinowitz
condition. Under this condition by standard computions there is a positive constant
C such that F (x, t) ≤ C|t|α for all |t| ≥ M and x ∈ [0, 1].
Now, we point out the following consequence of Theorem 3.1.
Corollary 3.3. Let f : R → R be a continuous function. Put F (t) :=

Z

t

f (ξ)dξ
0

for each t ∈ R. Fix δ > 0 and assume that F (δ) > 0 and F (ξ) ≥ 0 for each ξ ∈ [0, δ]
and also suppose that
lim inf
ξ→0

F (ξ)
F (ξ)
= lim sup 2 = 0.
ξ2
ξ
ξ→+∞

Then, there is λ∗ > 0 such that for each λ > λ∗ and for every nonnegative contin′
′
uous function g : R → R there exists δλ,g > 0 such that, for each µ ∈ [0, δλ,g [, the
problem

 u(iυ) (x) = λf (u(x)) + h(u(x)), x ∈ [0, 1],
(18)
u(0) = u′ (0) = 0,
 ′′
u (1) = 0, u′′′ (1) = µ g(u(1)),
admits at least three weak solutions.
Proof. Fix λ > λ∗ :=

16(1+L)( 32 )3 π 4 δ 2
F (δ)

. From the condition

lim inf
ξ→0

F (ξ)
= 0,
ξ2

there is a sequence {θn } ⊂]0, +∞[ such that lim θn = 0 and
n→∞

sup F (ξ)
lim

|ξ|≤θn

= 0.

θn2

n→∞

Therefore, there exists θ > 0 such that
sup F (ξ)
|ξ|≤θ

θ
and θ <
each

< min

2



1−L
F (δ)(1 − L)
,
32(1 + L)( 32 )3 π 4 δ 2 2λ



(19)

8π 2
√ δ . Since G(δ) is positive hence according to the Theorem 3.1 for
3 3
2

#

′

µ ∈ 0, δλ,g =

(1 − L)θ − 2λ sup F (t) "
|t|≤θ

2Gθ

problem (18) admits at least three weak solutions.

,


Tavani, M.H.

64

Remark 3.3. In the Corollary 3.3, in calculating the upper bound for the µ , if λ
value increases, then according to (19) , the value of sup F (t), will be reduced. In
|t|≤θ

fact, the choice of θ, is dependent on the value of λ. Actually for each fix λ > λ∗
8π 2
2
we can consider θ such that θ < √ δ and (1 − L)θ − 2λ sup F (t) > 0 .
3 3
|t|≤θ
Now, we present the following example to illustrate corollary 3.3.
Example 3.4. Consider the problem (1) where

0
2
f (t) =
e−t (10t9 − 2t11 )

if t ≤ 0
if t > 0

and √
g(t) = et and h(t) = 0.001tanht for all t ∈ R. We can consider L = 0.001. Let
δ = 3 . According to definition of f and g we have,

0
if t ≤ 0
F (t) =
10 −t2
t e
if t > 0

and G(t) = et − 1 .It is clear that
lim inf
ξ→0

F (ξ)
F (ξ)
= lim sup 2 = 0.
2
ξ
ξ
ξ→+∞

So by applying Corollary 3.3, for every fix
λ > λ∗ =

128.128 π 4
≃ 114.6247
2187 e−3

8π 2
2
according to θ < √ δ and (1 − L)θ − 2λ sup F (t) > 0 we can consider θ such
3 3
|t|≤θ
2
8π 2
that θ < √ δ ≃ 26.32 and 0.999(θ)2 − 2λ sup (t10 e−t ) > 0 and hence for every
3 3
|t|≤θ
2

#

µ ∈ 0,

0.999(θ)2 − 2λ sup (t10 e−t ) "
|t|≤θ

2(eθ

− 1)

,

the problem (1) has at least three weak solutions.

References
[1] Amster, P., and Mariani, M.C., A fixed point operator for a nonlinear boundary value problem, J. Math. Anal. Appl., 266(2001), 160–168.
[2] Bonanno, G., and A. Chinn`ı, A., Existence of three solutions for a perturbed two-point boundary value problem, Appl. Math. Lett., 23 (2010), 807–811.
[3] Bonanno, G., and D’Agu`ı, G., Multiplicity results for a perturbed elliptic Neumann problem,
Abstract and Applied Analysis 2010 (2010), doi:10.1155/2010/564363, 10 pages.

Existence Results

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