International Journal of Analysis and Applications

Volume 16, Number 3 (2018), 306-316

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-16-2018-306

SOLVABILITY OF MULTI-POINT VALUE PROBLEMS WITH INTEGRAL

CONDITION AT RESONANCE

RABAH KHALDI1,∗ AND MOHAMMED KOUIDRI2

1Laboratory of Advanced Materials, Faculty of Sciences, Badji Mokhtar-Annaba University, P.O. Box 12,

23000 Annaba, Algeria

2Kasdi Merbah Ourgla University P.0.30000 Ourgla, Algeria

∗Corresponding author: rkhadi@yahoo.fr

Abstract. In this paper, we study a boundary value problem at resonance with a multi-integral boundary

conditions. By constructing suitable operators, we establish an existence theorem upon the coincidence

degree theory of Mawhin. An example is given to show the effectiveness of our results.

1. Introduction

Boundary value problem involves ordinary differential equation with non local condition appears in physi-

cal science and applied mathematics. Moreover the theory of boundary value problems with integral condition

is found in deferent areas like applied mathematics and applied physics for example plasma physics, heat

conduction, themo-elasticity, underground water flew. In recent years, the boundary value problem at reso-

nance for ordinary differential equations have been extensively studied and many results have been obtained,

we refer to [1], [2], [5]- [7], [10]- [12] and the references therein. Moreover, lots of works on multi-point

boundary value problems have appeared, for examples, see [3]- [10].

Received 2017-12-17; accepted 2018-02-07; published 2018-05-02.

2010 Mathematics Subject Classification. 34B40, 34B15.

Key words and phrases. boundary value problem at resonance; existence of solution; coincidence degree; integral condition.

c©2018 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

306

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-16-2018-306


Int. J. Anal. Appl. 16 (3) (2018) 307

The goal of this paper is to provide sufficient conditions that ensure the existence of solutions for the

following multi-point boundary value problem (BVP)

x′′(t) = f(t,x(t),x′(t)), t ∈ (0, 1) (1.1)

x(0) = 0,x(1) =

m∑
k=1

λk

∫ ηk
0

x(t)dt, (1.2)

where f : [0, 1] ×R2 → R is caratheodory function, ηk ∈ (0, 1) and
m∑
k=1

λkη
2
k = 2.

We say that the (BVP) (1.1)-(1.2) is a resonance problem if the dimension of the kernel of the linear

operator Lx = x′′ is not less than one under the boundary conditions 1.2. Otherwise, we call them a

problem at nonresonance.

In the present work, if
m∑
k=1

λkη
2
k = 2, then, (BVP) (1.1)-(1.2) is at resonance, since equation

x′′(t) = 0, t ∈ (0, 1)

with boundary condition 1.2 has nontrivial solutions x(t) = ct,c ∈ R, t ∈ [0, 1].

This paper is organized as follows, in section 2 we stated some definitions and lemmas needed in our proofs.

In section 3 we treated the existence of solution by using the coincidence degree theory of Mawhin [9], [10].

we ended by giving an example illustrating the previous results.

2. Preliminaries

For the convenience of the reader to understand the coincidence degree theory, we briefly recall some

notations, definitions and theorems which will be used later.

Definition 2.1. Let X,Y , be reaI Banach spaces, the linear operator L : domL ⊂ X → Y is said to be

a Fredholm map of index zero provided that ker L, the kernel of L, is of the same finite dimension as the

Y/ Im L, where Im L is the image of L.

Let L be a Fredholm map of index zero, and P : X → X, Q : Y → Y be continuous projectors, such

that Im P = ker L, KerQ = Im L. Then X = ker L ⊕ ker P , Y = Im L ⊕ Im Q, thus L |dom L∩ker P :

dom L∩ ker P → Im L is invertible, denote its inverse by KP .

Definition 2.2. Let L be a Fredholm map of index zero and Ω be an open bounded subset of X, such

that domL ∩ Ω 6= φ, the map N : X → Y is said to be L − compact on Ω, if QN(Ω)is bounded and

Kp(I −Q)N : Ω → X is compact.



Int. J. Anal. Appl. 16 (3) (2018) 308

Theorem 2.1. ( [13]) Let L be a Fredholm operator of index zero and let N be L−compact on Ω. Assume

that the following conditions are satisfied.

(i)Lx 6= λNx, for every (x,λ) ∈ [(domL\KerL) ∩∂Ω] × (0, 1).

(ii) Nx /∈ Im L, for every x ∈ KerL∩∂Ω .

(iii) deg(QN|ker L, ker L∩ Ω, 0) 6= 0, where Q : Y → Y is a projection as above with Im L = ker Q.

Then, the equation Lx = Nx has at least one solution in domL∩ Ω .

In the following, we shall use the classical Banach spaces X = C1[0, 1] and Y = L1[0, 1] equipped

respectively with the norm ‖x‖ = max{‖x‖∞ ,‖x
′‖∞}, ‖x‖∞ = max

t∈[0,1]
|x(t)| and ‖y‖1 =

∫ 1
0
|y(t)|dt.

We will use the space AC2 [a,b] =
{
u ∈ C1 [a,b] ,u′ ∈ AC [a,b]

}
, where AC [a,b] is the space of absolutely

continuous functions on [a,b] .

3. Existence of Solutions

Define the operator L : dom L ⊂ X → Y by Lx = x′′, where

domL =

{
x ∈ W2,1(0, 1) : x(0) = 0,x(1) =

m∑
k=1

λk

∫ ηk
0

x(t)dt; ηk ∈ (0, 1),
m∑
k=1

λkη
2
k = 2

}
.

Let N : X → Y the operator

Nx(t) = f(t,x(t),x′(t)), t ∈ (0, 1).

Then, the equation 1.1, can be written as Lx = Nx.

Next, in order to apply Theorem 2.1, we need the following Lemma.

Lemma 3.1. (i) ker L = {x ∈ domL : x = ct,c ∈ R, t ∈ [0, 1]};

(ii) Im L = {y ∈ Y :
∫ 1
0

(1 −s)y(s)ds− 1
2

k=m∑
k=1

λk
∫η
0

(ηk −s)2y(s)ds = 0};

(iii) L : domL ⊂ X → Y is a Fredholm operator of index zero, and the linear continuous projector operator

Q : Y → Y can be defined as

Qy(t) = k.(Ry).t

where

Ry =

∫ 1
0

(1 −s)y(s)ds−
1

2

m∑
k=1

λk

∫ ηk
0

(ηk −s)
2y(s)ds

and

k−1 =
1

6
−

1

24

m∑
k=1

λkη
4
k.

(iv) The linear operator Kp : Im L → domL∩ ker P can be written as

Kpy =

∫ t
0

(t−s)y(s)ds,



Int. J. Anal. Appl. 16 (3) (2018) 309

Moreover, for all y ∈ Im L we have

‖Kpy‖≤‖y‖1 . (3.1)

Proof. (i) For x ∈ ker L, we have x′′ (t) = 0. Then, we obtain x(t) = a + bt, where a,b ∈ R.

From x(0) = 0, we have a = 0.

Again, since
∑m
k=1 λkη

2
k = 2, then from x(1) =

k=m∑
k=1

λk
∫ηk
0
x(t)dt, we get

ker L = {x ∈ domL : x = ct,c ∈ R, t ∈ [0, 1]}

(ii) To prove that

Im L = {y ∈ Y :
∫ 1
0

(1 −s)y(s)ds−
1

2

k=m∑
k=1

λk

∫ ηk
0

(ηk −s)
2y(s)ds = 0},

we show that, the linear equation

x′′ = y, (3.2)

has a solution x(t) satisfied, x(0) = 0, x(1) =
m∑
k=1

λk
∫ηk
0
x(t)dt,

m∑
k=1

λkη
2
k = 2, if and only if

∫ 1
0

(1 −s)y(s)ds−
1

2

m∑
k=1

λk

∫ η
0

(ηk −s)
2y(s)ds = 0.

In fact, by integrating equation 3.2 and tacking and account that x(0) = 0, we get

x(t) = x′(0)t +

∫ t
0

(t−s)y(s)ds

again from x(1) =
k=m∑
k=1

λk
∫ηk
0
x(t)dt, we obtain

x′(0) +

∫ 1
0

(1 −s)y(s)ds =
m∑
k=1

λk

∫ ηk
0

[
x′(0)t +

∫ t
0

(t−s)y(s)ds
]
dt

= x′(0) +

m∑
k=1

λk

[∫ ηk
0

∫ t
0

(t−s)y(s)dsdt
]

= x′(0) +
1

2

m∑
k=1

λk

[∫ ηk
0

(ηk −s)
2y(s)ds

]
which implies ∫ 1

0

(1 −s)y(s)ds−
1

2

m∑
k=1

λk

[∫ ηk
0

(ηk −s)
2y(s)ds

]
= 0

(iii) For y ∈ Y , we take the projector Q as

Qy = k

(∫ 1
0

(1 −s)y(s)ds−
1

2

m∑
k=1

λk

[∫ ηk
0

(ηk −s)
2y(s)ds

])
t,



Int. J. Anal. Appl. 16 (3) (2018) 310

where

k−1 =

∫ 1
0

(1 −s)sds−
1

2

m∑
k=1

λk

[∫ ηk
0

(ηk −s)
2sds

]
=

1

6
−

1

24

m∑
k=1

λkη
4
k.

It follows from ker Q = Im L that Y = Im L⊕ Im Q, thus, co dim L = dim ker L = 1.

Hence, L is a Fredholm operator of index zero.

(iv) Taking P : X → X as follows,

Px = x′(0)t

then, the generalized inverse Kp : Im L → domL∩ ker P of L can be written as

Kpy =

∫ t
0

(t−s)y(s)ds

In fact, for y ∈ Im L, we have

(LKp)y(t) = y(t)

and, for x ∈ domL∩ ker P , we know

(KpL)x(t) =

∫ t
0

(t−s)x′′(s)ds = x(t)

This shows that Kp = (L|domL∩ker P )
−1.

Finally from the definition of Kp, we get

‖Kpy‖∞ ≤
∫ 1
0

(1 −s) |y(s)|ds ≤
∫ 1
0

|y(s)|ds = ‖y‖1 .

�

Now, we give the result on the existence of a solution for the problem (1.1)-(1.2).

Theorem 3.1. Assume that the following conditions are satisfied :

(H1) There exists nonnegative functions α,β,γ ∈ L1[0, 1], such that, for all (x,y) ∈ R2, t ∈ [0, 1],

satisfying the following inequalities:

|f(t,x,y)| ≤ α(t) |x| + β(t) |y| + γ(t) (3.3)

(H2) There exists a constant M > 0, such that, for x ∈ domL, if |x′(t)| > M, for all t ∈ [0, 1], then,

∫ 1
0

(1 −s)f(s,x(s),x′(s))ds−
1

2

m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),x′(s))ds 6= 0

(H3) There exists a constant M
∗ > 0, such that, for any x(t) = ct ∈ ker L with |c| > M ∗either



Int. J. Anal. Appl. 16 (3) (2018) 311

c

[∫ 1
0

(1 −s)f(s,x(s),c)ds−
1

2

m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),c)ds

]
< 0

or

c

[∫ 1
0

(1 −s)f(s,x(s),c)ds−
1

2

k=m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),c)ds

]
> 0

then BVP (1.1)-(1.2) has at least one solution in C1 [0, 1] , provided ‖α(t)‖ + ‖β(t)‖≤ 1
2
.

Next, in order to prove Theorem 3.1, we need the following Lemma.

Lemma 3.2. Suppose that Ω is an open bounded subset of X such that dom L∩Ω 6= ∅. Then N is L−compact

on Ω.

Proof. Suppose that Ω ⊂ X is a bounded set. Without loss of generality, we may assume that Ω = B (0,r) ,

then for any x ∈ Ω, ‖x‖≤ r. For x ∈ Ω, and by condition 3.3, we obtain

|QNx| ≤ k
∫ 1
0

|f(s,x(s),x′(s))|ds +
k

2

m∑
k=1

λk

∫ ηk
0

|f(s,x(s),x′(s))|ds

≤ k
∫ 1
0

|α(s)| |x(s)| + |β(s)| |x′(s)| + |γ(t)|ds +
k

2

m∑
k=1

λk

∫ ηk
0

|α(s)| |x(s)| + |β(s)| |x′(s)| + |γ(t)|ds

≤

(
k +

k

2

m∑
k=1

λk

)
[r (‖α(t)‖1 + ‖β(t)‖1) + ‖γ(t)‖1]

thus,

‖QNx‖1 ≤

(
k +

k

2

m∑
k=1

λk

)
[r (‖α(t)‖1 + ‖β(t)‖1) + ‖γ(t)‖1] , (3.4)

which implies that QN
(
Ω
)

is bounded. Next, we show that KP (I −Q) N
(
Ω
)

is compact. For x ∈ Ω, by

condition 3.3 we have

‖Nx‖1 =
∫ 1
0

|f(t,x(s),x′(s))|ds ≤ r (‖α(t)‖1 + ‖β(t)‖1) + ‖γ(t)‖1 . (3.5)

On the other hand, from the definition of KP and together with (3.1), (3.4) and (3.5) one gets

‖KP (I −Q) Nx‖ ≤ ‖(I −Q) Nx‖1 ≤‖Nx‖1 + ‖QNx‖1

≤

(
1 + k +

k

2

m∑
k=1

λk

)
(r (‖α(t)‖1 + ‖β(t)‖1) + ‖γ(t)‖1) .

It follows that KP (I −Q) N
(
Ω
)

is uniformly bounded.

Let us prove that KP (I −Q) N
(
Ω
)

is equicontinuous. For any x ∈ Ω, and any t1,t2 ∈ [0, 1] , t1 < t2, we

have

(KP (I −Q) Nx) (t1) − (KP (I −Q) Nx) (t2) =

=

∣∣∣∣
∫ t1
0

(t1 −s) (I −Q) Nx (s) ds



Int. J. Anal. Appl. 16 (3) (2018) 312

−
∫ t2
0

(t2 −s) (I −Q) Nx (s) ds
∣∣∣∣

≤
[∫ t1

0

(t2 − t1) |(I −Q) Nx (s)|ds

+

∫ t2
t1

(t2 −s) |(I −Q) Nx (s)|ds
]
→ 0,

as t1 → t2. On the other hand we have

∣∣(KP (I −Q) Nx)′ (t1) − (KP (I −Q) Nx)′ (t2)∣∣
=

∣∣∣∣
∫ t1
0

(I −Q) Nx (s) ds−
∫ t2
0

(I −Q) Nx (s) ds
∣∣∣∣ → 0,

as t1 → t2. So KP (I −Q) N
(
Ω
)

is equicontinuous. so, the Ascoli-Arzela theorem ensure that Kp(I−Q)N :

Ω → X is compact . The proof is complete �

Now we give the proof of Theorem 3.1

Proof. Firstly, we need to construct the set Ω satisfying all the conditions in Theorem 2.1, which is separated

into the following three steps.

Step 1. First we show that the following set

Ω1 = {x ∈ domL\ker L : Lx = λNx,forsomeλ ∈ [0, 1]},

is bounded.

In fact, let x ∈ Ω1, we have Lx = λNx and Lx 6= 0, so λ 6= 0, thus QNx = 0, from which it yields

∫ 1
0

(1 −s)f(s,x(s),x′(s))ds−
1

2

k=m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),x′(s))ds = 0,

thus, from condition (H2) of Theorem 3.1, there exists t0 ∈ [0, 1], such that |x′(t0)| ≤ M.

In view of

x′(0) = x′(t0) −
t0∫
0

x′′(t)dt

then,

‖Px‖ = |x′(0)| ≤ M + ‖x′′‖1 ≤ M + ‖Nx‖1 (3.6)

Again for x ∈ Ω1, x ∈ domL\ker L, then (I −P)x ∈ domL∩ ker P and LPx = 0,

thus from Lemma 3.1 , we know

‖(I −P)x‖ = ‖KpL(I −P)x‖≤‖L(I −P)x‖1 = ‖Lx‖1 ≤‖Nx‖1 (3.7)



Int. J. Anal. Appl. 16 (3) (2018) 313

From (3.6) and (3.7), we have

‖x‖≤‖Px‖ + ‖(I −P)x‖≤ M + 2‖Nx‖1 (3.8)

If condition 3.3 holds, then from (3.8) , we obtain

‖x‖≤ 2
[
‖α‖1 ‖x‖∞ + ‖β‖1 ‖x

′‖∞ + ‖γ‖1 +
M

2

]
. (3.9)

Since ‖x‖∞ ≤‖x‖, then from (3.9) it yields

‖x‖∞ ≤
2

1 − 2‖α‖1

[
‖β‖1 ‖x

′‖∞ + ‖γ‖1 +
M

2

]
(3.10)

By using ‖x′‖∞ ≤‖x‖, (3.9) and (3.10) one has

‖x′‖∞

[
1 −

2‖β‖1
1 − 2‖α‖1

]
≤

2

1 − 2‖α‖1

[
‖γ‖1 +

M

2

]
therefore,

‖x′‖∞

[
1 − 2‖α‖1 − 2‖β‖1

1 − 2‖α‖1

]
≤

1

1 − 2‖α‖1
[2‖γ‖1 + M]

i.e.

‖x′‖∞ ≤
2
[
‖γ‖1 +

M
2

]
1 − 2‖α‖1 − 2‖β‖1

, (3.11)

thus, from (3.10) and (3.11), there exist M1 > 0, such that

‖x‖≤ M1

which proves that Ω1 is bounded.

Step 2. We will show that the set

Ω2 = {x ∈ ker L : Nx ∈ Im L}

is bounded.

Let x ∈ Ω2, then x(t) = bt,b ∈ R, t ∈ [0, 1], and QNx = 0, therefor

∫ 1
0

(1 −s)f(s,bs,b)ds−
1

2

k=m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,bs,b)ds = 0

In view of condition (H2) of Theorem 3.1, there exists t0 ∈ [0, 1], such that |x′(t0)| ≤ M that is |b| ≤ M,

so

‖x′‖∞ = |b| ≤ M,

from which, we get

‖x‖ = max{‖x‖∞ ,‖x
′‖∞} = |b| ≤ M,



Int. J. Anal. Appl. 16 (3) (2018) 314

which implies that Ω2 is bounded.

Step 3. In the next, we show the boundedness of the following set

Ω3 = {x ∈ ker L : −λJx + (1 −λ)QNx = 0,λ ∈ [0, 1]},

where, J : ker L → Im Q is the linear isomorphism given by J(ct) = ct,∀c ∈ R, t ∈ [0, 1].

If the first part of condition (H3) of Theorem 3.1 holds, that is, there exists M
∗ > 0, such that, for any

c ∈ R, if |c| > M∗, then

c

[∫ 1
0

(1 −s)f(s,x(s),c)ds−
1

2

m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),c)ds

]
< 0 ((3.12))

Let x ∈ Ω3, then x(t) = ct and λJx = (1 −λ)QNx, that is equivalently written as

λc = (1 −λ)k

[∫ 1
0

(1 −s)f(s,x(s),c)ds−
1

2

m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),c)ds

]
,

if λ = 1, then c = 0. Otherwise, if |c| > M∗, then in view of (3.12) and together with the fact that k > 0,

we get

λc2 = (1 −λ)kc

[∫ 1
0

(1 −s)f(s,x(s),c)ds−
1

2

m∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),c)ds

]
< 0

which contradicts λc2 ≥ 0. Thus, Ω3 ⊂{x ∈ KerL : ‖x‖≤ M∗} is bounded.

On the other hand, If the second part of condition (H3) of Theorem 3.1 holds, by applying similar reasoning

as above, we can prove that Ω3 is bounded.

Finally, we shall prove that all conditions of Theorem 2.1 are satisfied.

Let Ω be a bounded open subset of X, such that ∪3i=1Ωi ⊂ Ω. It follows from Lemma 3.1 that L is a

Fredholm operator of index zero. By Lemma 3.2, we have N is L-compact on Ω. By virtue of the definition

of Ω, the assumptions (i) and (ii) are satisfied.

Now we prove that condition (iii) of Theorem 2.1 is satisfied. Let H (x,λ) = ±λJx + (1 −λ) QNx. Since

Ω3 ⊂ Ω, then H (x,λ) 6= 0 for every x ∈ ker L∩∂Ω. By the homotopy property of degree, we get

deg (QN |ker L, Ω ∩ ker L, 0) = deg ( H (·, 0) , Ω ∩ ker L, 0)

= deg ( H (·, 1) , Ω ∩ ker L, 0)

= deg ( ±J, Ω ∩ ker L, 0) 6= 0.

Hence, the assumption (iii) of Theorem 2.1 holds.

Since all hypothesis of Theorem 2.1 are satisfied. Therefore, equation Lx = Nx has at least one solution in

dom L∩Ω; i.e. boundary value problem (1.1)-(1.2) has at least one solution in X. The proof is completed. �



Int. J. Anal. Appl. 16 (3) (2018) 315

4. An Illustrative Example

In this section we give an example to illustrate the usefulness of our main results.

Consider the multi-point boundary value problem

(P)




x′′ (t) = f (t,x (t) ,x′(t)) , t ∈ (0, 1) ,

x (0) = 0, x(1) = 4

1
2∫
0

x(t)dt + 16
9

2
3∫
0

x(t)dt

with

f(t,x,y) =
t

4
x +

(
1 − t2

4

)
y + t.

Since
2∑
k=1

λkη
2
k = 4

(
1
2

)2
+ 16

9

(
3
4

)2
= 2, the problem (P) is at resonance.

We have

|f (t,x,y)| ≤ α (t) |x| + β (t) |y| + γ(t),

where α (t) = t
4
,β (t) = 1−t

2

4
and γ(t) = t, then α, β are nonnegative and belong to L1 [0, 1], so, hypothesis

(H1) of Theorem 3.1 is satisfied.

We claim that condition (H2) of Theorem 3.1 is satisfied, indeed, for M = 1. 821 4 > 0 and x ∈ domL,

x(t) = ct, if |x′(t)| > M, for all t ∈ [0, 1], then,

∫ 1
0

(1 −s)f(s,x(s),x′(s))ds−
1

2

2∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),x′(s))ds

=

∫ 1
0

(1 −s)(
c

4
+ s)ds−

1

2

(
4

∫ 1
2

0

(
1

2
−s)2(

c

4
+ s)ds +

16

9

∫ 3
4

0

(
3

4
−s)2(

c

4
+ s)ds

)

=
7

96
c +

17

128
6= 0.

Now, for M∗ = 2 > 0 and any x (t) = ct ∈ ker L with |c| > M∗, we have

c

[∫ 1
0

(1 −s)f(s,x(s),x′(s))ds−
1

2

2∑
k=1

λk

∫ ηk
0

(ηk −s)
2f(s,x(s),x′(s))ds

]

=
7

96
c2 +

17

128
c > 0,

consequently, condition (H3) of Theorem 3.1 is satisfied.

Finally, a simple calculus gives ‖α‖1 +‖β‖1 =
1
8

+ 1
6
≤ 1

2
. We conclude from Theorem 3.1 that the problem

(P) has at least one solution in C1[0, 1].

References

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Math. Sci. 2012 (2012), Art. ID 513189.

[4] A. Guezane-Lakoud, R. Khaldi, Study of a third-order three-point boundary value problem, AIP Conf. Proc., 1309(2010),

329-335.

[5] C. P. Gupta, Solvability of multi-point boundary value problems at resonance, Results Math. 28(1995), 270-276.

[6] C. P. Gupta, A second order m-point boundary value problem at resonance, Nonlinear Anal. 24 (1995), 1483-1489.

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Math. Sci. 18 (1995), no. 4, 705-710.

[8] N. Kosmatov, A multi-point boundary value problem with two critical conditions. Nonlinear Anal. 65 (2006), no. 3, 622–633.

[9] S. K. Ntouyas and P. Ch. Tsamatos, Multi-point boundary value problems for ordinary differential equations. An. Ştiinţ.

Univ. Al. I. Cuza Iaşi. Mat. (N.S.) 45 (1999), no. 1, 57–64 (2000).

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(2011), 1690 – 1700.

[11] R. Ma, Multiplicity results for a third order value problem at resonance, Nonlinear Anal. 32 (1998), no. 4, 493–499.

[12] R. K. Nagle and K. L. Pothoven, On a third-order nonlinear boundary value problems at resonance, J. Math. Anal. Appl.

195 (1995), no 1, 148-159.

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Mathematics, 40. American Mathematical Society, Providence, R.I., 1979.

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	1. Introduction
	2. Preliminaries
	3. Existence of Solutions
	4. An Illustrative Example
	References