Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 45, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
THREE SOLUTIONS FOR A FOURTH-ORDER
BOUNDARY-VALUE PROBLEM
GHASEM A. AFROUZI, SAEID SHOKOOH
Abstract. Using two three-critical points theorems, we prove the existence
of at least three weak solutions for one-dimensional fourth-order equations.
Some particular cases and two concrete examples are then presented.
1. Introduction
In this note, we consider the fourth-order boundary-value problem
u0000 h(x, u0 ) − u00 = [λf (x, u) + g(u)]h(x, u0 ),
00
in (0, 1),
00
u(0) = u(1) = 0 = u (0) = u (1),
(1.1)
where λ is a positive parameter, f : [0, 1] × R → R is an L1 -Carathéodory function,
g : R → R is a Lipschitz continuous function with the Lipschitz constant L > 0,
i.e.,
|g(t1 ) − g(t2 )| ≤ L|t1 − t2 |
for every t1 , t2 ∈ R, with g(0) = 0, and h : [0, 1] × R → [0, +∞) is a bounded and
continuous function with m := inf (x,t)∈[0,1]×R h(x, t) > 0.
Due to the importance of fourth-order two-point boundary value problems in
describing a large class of elastic deflection, many researchers have studied the
existence and multiplicity of solutions for such a problem, we refer the reader to
[1, 2, 3, 6, 11] and references therein. For example, authors in [2], using Ricceri’s
Variational Principle [10, Theorem 1], established the existence three weak solutions
for the problem
u0000 + αu00 + βu = λf (x, u) + µg(x, u),
00
in (0, 1),
00
u(0) = u(1) = 0 = u (0) = u (1),
where α, β are real constants, f, g : [0, 1] × R → R are Carathéodory functions and
λ, µ > 0.
In this article, employing two three-critical points theorems which we recall in
the next section (Theorems 2.1 and 2.2), we establish the existence three weak
solutions for (1.1). A special case of Theorem 3.1 is the following theorem.
2000 Mathematics Subject Classification. 34B15, 34B18, 58E05.
Key words and phrases. Dirichlet boundary condition; variational method; critical point.
c
2015
Texas State University - San Marcos.
Submitted November 14, 2014. Published February 17, 2015.
1
2
G. A. AFROUZI, S. SHOKOOH
EJDE-2015/45
Theorem 1.1. Let f : R → R be a non-negative continuous function such that
Z 2
Z 3√3
12
2
f (x) dx <
f (x) dx,
0
0
Rξ
lim sup
|ξ|→+∞
0
f (x) dx
≤ 0.
ξ2
Then, for each
λ∈
i 213 (π 4 + π 2 + 1) 2(π 4 + π 2 + 1) h
,
,
R2
R 3√3
π 4 0 f (x) dx π 4 0 f (x) dx
the problem
u0000 − u00 + u = f (u),
in (0, 1),
00
u(0) = u(1) = 0 = u (0) = u00 (1)
admits at least three weak solutions.
The following result is a consequence of Theorem 3.6.
Theorem 1.2. Let f : R → R be a non-negative continuous function such that
Z 2
Z 3
211
f (x) dx <
f (x) dx,
0
Z
210
f (x) dx < 27
0
0
Z
3
f (x) dx,
0
Then, for each
n
i 213 (π 4 + π 2 + 1) (π 4 + π 2 + 1)
oh
2
220
min R 2
,
, R 1024
λ∈
,
R3
4
π
π 4 0 f (x) dx
f
(x)
dx
f
(x)
dx
0
0
the problem
u0000 − u00 − u = f (u),
in (0, 1),
00
u(0) = u(1) = 0 = u (0) = u00 (1)
admits at least three weak solutions.
2. Preliminaries
We now state two critical point theorems established by Bonanno and coauthors
[4, 5] which are the main tools for the proofs of our results. The first result has been
obtained in [5] and it is a more precise version of Theorem 3.2 of [4]. The second
one has been established in [4]. In the first one the coercivity of the functional
Φ − λΨ is required, in the second one a suitable sign hypothesis is assumed.
Theorem 2.1 ([5, Theorem 2.6]). Let X be a reflexive real Banach space; Φ :
X → R be a sequentially weakly lower semicontinuous, coercive and continuously
Gâteaux differentiable functional whose Gâteaux derivative admits a continuous inverse on X ∗ , Ψ : X → R be a sequentially weakly upper semicontinuous, continuously Gâteaux differentiable functional whose Gâteaux derivative is compact, such
that
Φ(0) = Ψ(0) = 0 .
Assume that there exist r > 0 and x̄ ∈ X, with r < Φ(x̄) such that
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THREE SOLUTIONS FOR A BOUNDARY-VALUE PROBLEM
3
(i) supΦ(x)≤r Ψ(x) < rΨ(x̄)/Φ(x̄),
(ii) for each λ in
h
i Φ(x̄)
r
,
Λr :=
,
Ψ(x̄) supΦ(x)≤r Ψ(x)
the functional Φ − λΨ is coercive.
Then, for each λ ∈ Λr the functional Φ − λΨ has at least three distinct critical
points in X.
Theorem 2.2 ([4, Theorem 3.2]). Let X be a reflexive real Banach space; Φ :
X → R be a convex, coercive and continuously Gâteaux differentiable functional
whose Gâteaux derivative admits a continuous inverse on X ∗ , Ψ : X → R be a
continuously Gâteaux differentiable functional whose Gâteaux derivative is compact,
such that
inf Φ = Φ(0) = Ψ(0) = 0 .
X
Assume that there exist two positive constants r1 , r2 > 0 and x̄ ∈ X, with 2r1 <
Φ(x̄) < r2 /2, such that
(j) supΦ(x)≤r1 Ψ(x)/r1 < (2/3)Ψ(x̄)/Φ(x̄),
(jj) supΦ(x)≤r2 Ψ(x)/r2 < (1/3)Ψ(x̄)/Φ(x̄),
(jjj) for each λ in
i 3 Φ(x̄)
h
r2
r1
,
, min
Λ∗r1 ,r2 :=
2 Ψ(x̄)
supΦ(x)≤r1 Ψ(x) 2 supΦ(x)≤r2 Ψ(x)
and for every x1 , x2 ∈ X, which are local minima for the functional Φ−λΨ,
and such that Ψ(x1 ) ≥ 0 and Ψ(x2 ) ≥ 0, one has inf t∈[0,1] Ψ(tx1 + (1 −
t)x2 ) ≥ 0.
Then, for each λ ∈ Λ∗r1 ,r2 the functional Φ − λΨ has at least three distinct critical
points which lie in Φ−1 (] − ∞, r2 [).
Let us introduce some notation which will be used later. Define
H01 ([0, 1]) := u ∈ L2 ([0, 1]) : u0 ∈ L2 ([0, 1]), u(0) = u(1) = 0 ,
H 2 ([0, 1]) := u ∈ L2 ([0, 1]) : u0 , u00 ∈ L2 ([0, 1]) .
Take X = H 2 ([0, 1]) ∩ H01 ([0, 1]) endowed with the usual norm
Z 1
1/2
kuk :=
|u00 (x)|2 dx
.
0
We recall the following Poincaré type inequalities (see, for instance, [8, Lemma
2.3]):
1
kuk2 ,
π2
1
≤ 4 kuk2
π
ku0 k2L2 ([0,1]) ≤
(2.1)
kuk2L2 ([0,1])
(2.2)
for all u ∈ X. For the norm in C 1 ([0, 1]),
n
o
kuk∞ := max max |u(x)|, max |u0 (x)| ,
x∈[0,1]
we have the following relation.
x∈[0,1]
4
G. A. AFROUZI, S. SHOKOOH
EJDE-2015/45
Proposition 2.3. Let u ∈ X. Then
1
kuk.
(2.3)
2π
Proof. Taking (2.1) into account, the conclusion follows from the well-known in
equality kuk∞ ≤ 21 ku0 kL2 ([0,1]) .
kuk∞ ≤
For an excellent overview of the most significant mathematical methods employed
in this paper we refer to [7, 9].
Let g : R → R is a Lipschitz continuous function with the Lipschitz constant
L > 0, i.e.,
|g(t1 ) − g(t2 )| ≤ L|t1 − t2 |
for every t1 , t2 ∈ R, and g(0) = 0, h : [0, 1] × R → [0, +∞) is a bounded and
continuous function with m := inf (x,t)∈[0,1]×R h(x, t) > 0, and f : [0, 1] × R → R be
an L1 -Carathéodory function.
We recall that f : [0, 1] × R → R is an L1 -Carathéodory function if
(a) the mapping x 7→ f (x, ξ) is measurable for every ξ ∈ R;
(b) the mapping ξ 7→ f (x, ξ) is continuous for almost every x ∈ [0, 1];
(c) for every ρ > 0 there exists a function lρ ∈ L1 ([0, 1]) such that
sup |f (x, ξ)| ≤ lρ (x)
|ξ|≤ρ
for almost every x ∈ [0, 1].
Corresponding to f, g and h we introduce the functions F : [0, 1] × R → R, G : R →
R and H : [0, 1] × R → [0, +∞), respectively, as follows
Z t
Z t
F (x, t) :=
f (x, ξ) dξ, G(t) := −
g(ξ) dξ,
0
0
Z tZ τ
1
dδ dτ
H(x, t) :=
0
0 h(x, δ)
for all x ∈ [0, 1] and t ∈ R.
In the following, we let M := sup(x,t)∈[0,1]×R h(x, t) and suppose that the Lipschitz constant L of the function g satisfies 0 < L < π 4 .
We say that a function u ∈ X is a weak solution of (1.1) if
Z 1
Z 1 Z u0 (x)
1
00
00
u (x)v (x) dx +
dτ v 0 (x) dx
h(x, τ )
0
0
0
Z 1
Z 1
−λ
f (x, u(x))v(x) dx −
g(u(x))v(x) dx = 0
0
0
holds for all v ∈ X.
3. Main results
Put
π4 − L
π 2 + m(π 4 + L)
,
B
:=
,
2π 4
2mπ 4
and suppose that B ≤ 4Aπ 2 . We formulate our main results as follows.
A :=
Theorem
√ 3.1. Assume that there exist two positive constants c, d, satisfying c <
32d/(3 3π), such that
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5
(A1) F (x, t) ≥ 0 for all (x, t) ∈ ([0, 3/8] ∪ [5/8, 1]) × [0, d];
(A2)
R 5/8
R1
max|t|≤c F (x, t) dx
27 3/8 F (x, d) dx
0
<
;
c2
4096
d2
(A3)
R1
supx∈[0,1] F (x, ξ)
π 4 A 0 max|t|≤c F (x, t) dx
lim sup
≤
.
ξ2
B
c2
|ξ|→+∞
Then, for every λ in
i
h
4096Bd2
Bc2
Λ :=
, R1
,
R 5/8
27 3/8 F (x, d) dx 0 max|t|≤c F (x, t) dx
problem (1.1) has at least three distinct weak solutions.
Proof. Fix λ as in the conclusion. Our aim is to apply Theorem 2.1 to our problem.
To this end, for every u ∈ X, we introduce the functionals Φ, Ψ : X → R by setting
Z 1
Z 1
1
Φ(u) := kuk2 +
H(x, u0 (x)) dx +
G(u(x)) dx,
2
0
0
Z 1
Ψ(u) :=
F (x, u(x)) dx,
0
and putting
Iλ (u) := Φ(u) − λΨ(u) ∀ u ∈ X.
Note that the weak solutions of (1.1) are exactly the critical points of Iλ . The
functionals Φ, Ψ satisfy the regularity assumptions of Theorem 2.1. Indeed, by
standard arguments, we have that Φ is Gâteaux differentiable and sequentially
weakly lower semicontinuous and its Gâteaux derivative is the functional Φ0 (u) ∈
X ∗ , given by
Z 1
Z 1
Z 1 Z u0 (x)
1
dτ v 0 (x) dx−
Φ0 (u)(v) =
g(u(x))v(x) dx
u00 (x)v 00 (x) dx+
h(x, τ )
0
0
0
0
for any v ∈ X. Furthermore, the differential Φ0 : X → X ∗ is a Lipschitzian
operator. Indeed, taking (2.1) and (2.2) into account, for any u, v ∈ X, there holds
kΦ0 (u) − Φ0 (v)kX ∗ = sup |(Φ0 (u) − Φ0 (v), w)|
kwk≤1
1
Z
|u00 (x) − v 00 (x)||w00 (x)| dx
≤ supkwk≤1
0
Z
1
+ sup
kwk≤1
0
Z
v 0 (x)
u0 (x)
1
dτ |w0 (x)| dx
h(x, τ )
1
|g(u(x)) − g(v(x))||w(x)| dx
+ sup
kwk≤1
Z
0
≤ ku − vk +
1
sup ku0 − v 0 kL2 (0,1) kw0 kL2 (0,1)
m kwk≤1
+ L sup ku − vkL2 (0,1) kwkL2 (0,1)
kwk≤1
≤ 1+
1
L
+ 4 ku − vk = 2Bku − vk.
2
mπ
π
6
G. A. AFROUZI, S. SHOKOOH
EJDE-2015/45
Recalling that g is Lipschitz continuous and h is bounded away from zero, the claim
is true. In particular, we derive that Φ is continuously differentiable. Also, for any
u, v ∈ X, we have
Z 1 Z v0 (x)
1
(Φ0 (u) − Φ0 (v), u − v) = ku − vk2 +
dτ (u0 (x) − v 0 (x)) dx
0
u0 (x) h(x, τ )
Z 1
(g(u(x)) − g(v(x)))(u(x) − v(x)) dx
−
0
1 0
ku − v 0 k2L2 (0,1) − Lku − vk2L2 (0,1)
M
L
≥ ku − vk2 − 4 ku − vk2 = 2Aku − vk2 .
π
By the assumption L < π 4 , it turns out that Φ0 is a strongly monotone operator.
So, by applying Minty-Browder theorem [12, Theorem 26.A], Φ0 : X → X ∗ admits
a Lipschitz continuous inverse. On the other hand, the fact that X is compactly
embedded into C 0 ([0, 1]) implies that the functional Ψ is well defined, continuously
Gâteaux differentiable and with compact derivative, whose Gâteaux derivative is
given by
Z 1
0
Ψ (u)(v) =
f (x, u(x))v(x) dx
≥ ku − vk2 +
0
for any v ∈ X.
Since g is Lipschitz continuous and satisfies g(0) = 0, while h is bounded away
from zero, the inequalities (2.1) and (2.2) yield for any u ∈ X the estimate
Akuk2 ≤ Φ(u) ≤ Bkuk2 .
(3.1)
2
We will verify (i) and (ii) of Theorem 2.1. Put r = Bc . Taking (2.3) into
account, for every u ∈ X such that Φ(u) ≤ r, one has maxx∈[0,1] |u(x)| ≤ c.
Consequently,
Z 1
sup Ψ(u) ≤
max F (x, t) dx;
Φ(u)≤r
0
|t|≤c
that is,
supΦ(u)≤r Ψ(u)
≤
r
R1
0
max|t|≤c F (x, t) dx
.
Bc2
Hence,
supΦ(u)≤r Ψ(u)
1
< .
r
λ
(3.2)
Put

64d
3
2

− 9 (x − 4 x),
w(x) = d,

 64d 2 5
− 9 (x − 4 x + 41 ),
x ∈ [0, 83 [,
x ∈ [ 38 , 58 ],
x ∈] 58 , 1].
It is easy to verify that w ∈ X and, in particular,
4096 2
kwk2 =
d .
27
So, taking (3.1) into account, we deduce
4096 2
4096 2
Ad ≤ Φ(w) ≤
Bd .
27
27
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THREE SOLUTIONS FOR A BOUNDARY-VALUE PROBLEM
7
Hence, from c < 3√323π d and B ≤ 4Aπ 2 , we obtain r < Φ(w).
Since 0 ≤ w(x) ≤ d for each x ∈ [0, 1], assumption (A1) ensures that
Z 1
Z 3/8
F (x, w(x)) dx ≥ 0,
F (x, w(x)) dx +
5/8
0
and so
5/8
Z
Ψ(w) ≥
F (x, d) dx.
3/8
Therefore, we obtain
Ψ(w)
27
≥
Φ(w)
4096
R 5/8
3/8
F (x, d) dx
Bd2
>
1
.
λ
(3.3)
Therefore, from (3.2) and (3.3), condition (i) of Theorem 2.1 is fulfilled.
Now, to prove the coercivity of the functional Iλ . By (A3), we have
supx∈[0,1] F (x, ξ)
π4 A
<
.
ξ2
λ
|ξ|→+∞
lim sup
So, we can fix ε > 0 satisfying
supx∈[0,1] F (x, ξ)
π4 A
<
ε
<
.
ξ2
λ
|ξ|→+∞
lim sup
Then, there exists a positive constant θ such that
F (x, t) ≤ ε|t|2 + θ
∀x ∈ [0, 1], ∀t ∈ R .
Taking into account (2.2) and (3.1), we have
λε kuk2 − λθ
π4
for all u ∈ X. So, the functional Iλ is coercive. Now, the conclusion of Theorem
2.1 can be used. It follows that for every
i Φ(w)
h
r
λ∈Λ⊆
,
,
Ψ(w) supΦ(u)≤r Ψ(u)
Iλ (u) = Φ(u) − λΨ(u) ≥ Akuk2 − λεkuk2L2 [0,1] − λθ ≥ A −
the functional Iλ has at least three distinct critical points in X, which are the weak
solutions of the problem (1.1). This completes the proof.
Now, we present a consequence of Theorem 3.1.
Corollary 3.2. Let α ∈ L1 ([0, 1]) be a non-negative and non-zero function and let
R1
R 5/8
γ : R → R be a continuous function. Put α0 := 3/8 α(x)dx, kαk1 := 0 α(x) dx
Rt
and Γ(t) = 0 γ(ξ)dξ for all t ∈ R, and assume that there exist two positive constants c, d, with c < 3√323π d, such that
(A1’) Γ(t) ≥ 0 for all t ∈ [0, d];
(A2’)
max|t|≤c Γ(t)
27 α0 Γ(d)
<
;
c2
4096 kαk1 d2
(A3’) lim sup|ξ|→+∞ Γ(ξ)/ξ 2 ≤ 0.
8
G. A. AFROUZI, S. SHOKOOH
EJDE-2015/45
Then, for every
i 4096 Bd2
h
Bc2
,
,
27 α0 Γ(d) kαk1 max|t|≤c Γ(t)
λ∈
the problem
u0000 h(x, u0 ) − u00 = [λα(x)γ(u) + g(u)]h(x, u0 ),
00
in (0, 1),
00
u(0) = u(1) = 0 = u (0) = u (1)
(3.4)
has at least three weak solutions.
The proof of the above corollary follows from Theorem 3.1 by choosing f (x, t) :=
α(x)γ(t) for all (x, t) ∈ [0, 1] × R.
Remark 3.3. Clearly, if γ is non-negative then assumption (A1’) is satisfied and
(A2’) becomes
Γ(c)
27 α0 Γ(d)
<
.
c2
4096 kαk1 d2
Remark 3.4. Theorem 1.1 in the introduction is an immediate
consequence of
√
Corollary 3.2, on choosing g(u) = −u, h ≡ 1, c = 2 and d = 3 3.
The following lemma will be crucial in our arguments.
Lemma 3.5. Assume that f (x, t) ≥ 0 for all (x, t) ∈ [0, 1] × R. If u is a weak
solution of (1.1), then u(x) ≥ 0 for all x ∈ [0, 1].
Proof. Arguing by contradiction, if we assume that u is negative at a point of [0, 1],
the set
Ω := {x ∈ [0, 1] : u(x) < 0},
is non-empty and open. Let us consider v̄ := min{u, 0}, one has, v̄ ∈ X. So, taking
into account that u is a weak solution and by choosing v = v̄, from our assumptions,
one has
Z
0≥λ
f (x, u(x))u(x) dx
Ω
Z
00
2
|u (x)| dx +
=
Ω
4
Z Z
Ω
0
u0 (x)
1
dτ u0 (x) dx −
h(x, τ )
Z
g(u(x))u(x) dx
Ω
π −L
kuk2H 2 (Ω)∩H 1 (Ω) .
0
π4
Therefore, kukH 2 (Ω)∩H01 (Ω) = 0 which is absurd. Hence, the conclusion is achieved.
≥
Our other main result is as follows.
Theorem
3.6. √Assume that there exist three positive constants c1 , c2 , d, satisfying
√
3π
3 √3
3 √
c
<
d
< 64
c , such that
1
16 2
2 2
(B1) f (x, t) ≥ 0 for all (x, t) ∈ [0, 1] × [0, c2 ];
(B2)
R 5/8
R1
F (x, d) dx
F
(x,
c
)
dx
9
1
3/8
0
<
;
2
c1
2048
d2
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THREE SOLUTIONS FOR A BOUNDARY-VALUE PROBLEM
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(B3)
R1
0
R 5/8
F (x, c2 ) dx
9
<
2
c2
4096
3/8
F (x, d) dx
d2
.
Let
Λ0 :=
Bd2
i 2048
9
R 5/8
3/8
F (x, d) dx
n
, B min R 1
0
c21
oh
c22
, R1
.
F (x, c1 ) dx 2 0 F (x, c2 ) dx
0
Then, for every λ ∈ Λ the problem (1.1) has at least three weak solutions ui ,
i = 1, 2, 3, such that 0 < kui k∞ ≤ c2 .
Proof. Without loss of generality, we can assume f (x, t) ≥ 0 for all (x, t) ∈ [0, 1]×R.
Fix λ as in the conclusion and take X, Φ and Ψ as in the proof of Theorem 3.1. Put
w as in Theorem 3.1, r1 = Bc21 and r2 = Bc22 . Therefore, one has 2r1 < Φ(w) < r22
and we have
Z 1
1
1
1
sup Ψ(u) ≤
F (x, c1 ) dx <
2
r1 Φ(u)<r1
Bc1 0
λ
R 5/8
9 3/8 F (x, d) dx
<
2048
Bd2
2 Ψ(w)
,
≤
3 Φ(w)
and
2
r2
sup Ψ(u) ≤
Φ(u)<r2
2
Bc22
Z
1
F (x, c2 ) dx <
0
1
λ
R 5/8
9 3/8 F (x, d) dx
2048
Bd2
2 Ψ(w)
≤
.
3 Φ(w)
<
So, conditions (j) and (jj) of Theorem 2.2 are satisfied. Finally, let u1 and u2 be
two local minima for Φ − λΨ. Then, u1 and u2 are critical points for Φ − λΨ, and
so, they are weak solutions for the problem (1.1). Hence, owing to Lemma 3.5, we
obtain u1 (x) ≥ 0 and u2 (x) ≥ 0 for all x ∈ [0, 1]. So, one has Ψ(su1 + (1 − s)u2 ) ≥ 0
for all s ∈ [0, 1]. From Theorem 2.2 the functional Φ−λΨ has at least three distinct
critical points which are weak solutions of (1.1). This complete the proof.
Now, we present a consequence of Theorem 3.6.
Corollary 3.7. Let α ∈ L1 ([0, 1]) be such that α(x) ≥ 0 a.e. x ∈ [0, 1], α 6≡ 0, and
R1
R 5/8
let γ : R → R be a continuous function. Put α0 := 3/8 α(x)dx, kαk1 := 0 α(x) dx
Rt
and Γ(t) = 0 γ(ξ)dξ for all t ∈ R, and assume that there exist three positive
√
√
3 √3
constants c1 , c2 , d, with 316√3π2 c1 < d < 64
c , such that
2 2
(B1’) γ(t) ≥ 0 for all t ∈ [0, c2 ];
(B2’)
Γ(c1 )
9 α0 Γ(d)
<
;
c21
2048 kαk1 d2
10
G. A. AFROUZI, S. SHOKOOH
EJDE-2015/45
(B3’)
Γ(c2 )
9 α0 Γ(d)
.
<
c22
4096 kαk1 d2
Then, for every
λ∈
oh
i 2048 Bd2
n
c21
c22
,
, B min
,
9 α0 Γ(d)
kαk1 Γ(c1 ) 2kαk1 Γ(c2 )
the problem (3.4) has at least three weak solutions ui , i = 1, 2, 3, such that 0 <
kui k∞ ≤ c2 .
The proof of the above corollary follows from Theorem 3.6 by choosing f (x, t) :=
α(x)γ(t) for all (x, t) ∈ [0, 1] × R.
Remark 3.8. Theorem 1.2 in the introduction is an immediate consequence of
Corollary 3.7, on choosing g(u) = u, h ≡ 1, c1 = 2, c2 = 210 , and d = 3.
Finally, we present the following examples to illustrate our results.
Example 3.9. Consider the following problem
u0000 − u00 (2 + x + cos u0 ) + u = λf (u),
00
in (0, 1),
00
u(0) = u(1) = 0 = u (0) = u (1),
(3.5)
where f : R → R is defined by

−10

2
f (t) = 2−10 t4

 20 −2
2 t
if |t| ≤ 1,
if 1 < |t| ≤ 32,
if |t| > 32.
Here, g(t) = −t and h(x, t) = (2 + x + cos t)−1 for all x ∈ [0, 1] and t ∈ R. It is easy
to verify that (A2’) and (A3’) are satisfied with c = 1 and d = 32. From Corollary
3.2, for each parameter
i 48(π 4 + 4π 2 + 1) 512(π 4 + 4π 2 + 1) h
λ∈
,
,
π4
π4
problem (3.5) admits at least three weak solutions.
Example 3.10. Consider the problem
u0000 − u00 (3 + sin u0 ) − 2u = λf (u),
00
in (0, 1),
00
u(0) = u(1) = 0 = u (0) = u (1),
(3.6)
where f : R → R is defined by

−20

2
4
f (t) = t

 −2
t
if |t| ≤ 2−5 ,
if 2−5 < |t| ≤ 1,
if |t| > 1.
Here, g(t) = 2t and h(x, t) = (3 + sin t)−1 for all x ∈ [0, 1] and t ∈ R. It is easy to
verify that (B2’) and (B3’) are satisfied with c1 = 2−5 , d = 1 and c2 = 210 . From
Corollary 3.7, for each parameter
i 2276(π 4 + 4π 2 + 2) 214 (π 4 + 4π 2 + 2) h
,
,
λ∈
π4
π4
problem (3.6) admits at least three weak solutions ui , i = 1, 2, 3, such that 0 <
kui k∞ ≤ 1024.
EJDE-2015/45
THREE SOLUTIONS FOR A BOUNDARY-VALUE PROBLEM
11
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Ghasem A. Afrouzi
Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran
E-mail address: afrouzi@umz.ac.ir
Saeid Shokooh
Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran
E-mail address: saeid.shokooh@stu.umz.ac.ir