Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 29, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http:ejde.math.unt.edu
ftp ejde.math.txstate.edu
HIGHER ORDER BOUNDARY VALUE PROBLEMS AT
RESONANCE ON AN UNBOUNDED INTERVAL
ASSIA FRIOUI, ASSIA GUEZANE-LAKOUD, RABAH KHALDI
This corrected version was posted on March 4, 2016.
The original version is attached at on pages 10-18
Abstract. The aim of this paper is the solvability of a class of higher order differential equations with initial conditions and an integral boundary condition
on the half line. Using coincidence degree theory by Mawhin and constructing
suitable operators, we prove the existence of solutions for the posed resonance
boundary value problems.
1. Introduction
In this article, we are concerned with the existence of solutions of the higherorder ordinary differential equation
x(n) (t) = f (t, x(t)),
t ∈ (0, ∞),
(1.1)
with the integral boundary value conditions
x(i) (0) = 0, i = 0, 1, . . . , n − 2,
x(n−1) (∞) =
n!
ξn
Z
ξ
x(t)dt ,
(1.2)
0
where n ≥ 3 is an integer, ξ > 0 and f : [0, ∞) × R → R is a given function
satisfying certain conditions.
A boundary value problem (BVP for short) is said to be at resonance if the
corresponding homogeneous boundary value problem has a non-trivial solution.
Resonance problems can be formulated as an abstract equation Lx = N x, where L
is a noninvertible operator. When L is linear, as is known, the coincidence degree
theory of Mawhin [19] has played an important role in dealing with the existence
of solutions for these problems. For more recent results, we refer the reader to
[3, 5, 6, 6, 8, 9, 14, 20, 22, 24, 25] and the references therein.
Moreover boundary value problems on the half line arise in many applications
in physics such that in modeling the unsteady flow of a gas through semi-infinite
porous media, in plasma physics, in determining the electrical potential in an isolated neutral atom, or in combustion theory. For an extensive literature of results
2010 Mathematics Subject Classification. 34B40, 34B15.
Key words and phrases. Boundary value problem at resonance; existence of solution;
coincidence degree; integral condition.
c
2016
Texas State University.
Submitted October 20, 2015. Published January 19, 2016.
1
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A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
EJDE-2016/29
as regards boundary value problems on unbounded domains, we refer the reader to
the monograph by Agarwal and O’Regan [1].
Recently, there have been many works concerning the existence of solutions for
the boundary value problems on the half-line. For instance see [2, 4, 10, 11, 12,
13, 15, 16, 17, 18, 21, 23] and the references therein. By the way, much of work on
the existence of solutions for the boundary value problems on unbounded domains
involves second or third-order differential equations.
However, for the resonance case, there is no work done for the higher-order
boundary value problems with integral boundary conditions on the half-line, such
as BVP (1.1)-(1.2).
The remaining part of this paper is organized as follows. We present in Section
2 some notations and basic results involved in the reformulation of the problem. In
Section 3, we give the main theorem and some lemmas, then we will show that the
proof of the main theorem is an immediate consequence of these lemmas and the
coincidence degree of Mawhin.
2. Preliminaries
For the convenience of the readers, we recall some notation and two theorems
which will be used later.
Let X, Y be two real Banach spaces and let L : dom L ⊂ X → Y be a linear
operator which is Fredholm map of index zero, and let P : X → X, Q : Y → Y be
continuous projectors such that Im P = ker L, ker Q = Im L. Then X = ker L ⊕
ker P , Y = Im L ⊕ Im Q. It follows that L|dom L∩ker P : dom L ∩ ker P → Im L is
invertible, we denote the inverse of that map by KP . Let Ω be an open bounded
subset of X such that dom L ∩ Ω 6= ∅, the map N : X → Y is said to be L-compact
on Ω if the map QN (Ω) is bounded and KP (I − QN ) : Ω → X is compact.
Theorem 2.1 ([19]). Let L be a Fredholm operator of index zero and N be Lcompact on Ω. Assume that the following conditions are satisfied:
(1) Lx 6= λN x for every (x, λ) ∈ [(dom L\ ker L) ∩ ∂Ω] × (0, 1).
(2) N x ∈
/ Im L for every x ∈ ker L ∩ ∂Ω.
(3) deg(QN |ker L , Ω ∩ ker L, 0) 6= 0, where Q : Y → Y is a projection such that
Im L = ker Q.
Then the equation Lx = N x has at least one solution in dom L ∩ Ω.
Since the Arzelá-Ascoli theorem fails in the noncompact interval case, we use
the following result in order to show that KP (I − QN ) : Ω → X is compact.
Theorem 2.2 ([1]). Let F be a subset of C∞ = {y ∈ C([0, +∞)), limt→∞ y(t) exists}
that is equipped with the norm kyk∞ = sup |y(t)|. Then F is relatively compact
t∈[0,+∞)
if the following conditions hold:
(1) F is bounded in X.
(2) The functions belonging to F are equi-continuous on any compact subinterval of [0, ∞).
(3) The functions from F are equi-convergent at +∞.
Let
X = x ∈ C n−1 [0, +∞), lim e−t |x(i) (t)| exist, 0 ≤ i ≤ n − 1
t→∞
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BOUNDARY VALUE PROBLEMS AT RESONANCE
3
endowed with the norm kxk = max0≤i≤n−1 supt∈[0,+∞) e−t |x(i) (t)| . Then X is a
Banach space.
Lemma 2.3. Let M ⊂ X, then M is relatively compact in X if the following
conditions hold:
(1) M is bounded in X.
(2) The family V i = {yi : yi (t) = e−t x(i) (t), t ≥ 0, x ∈ M } is equicontinuous
on any compact subinterval of [0, +∞) for i = 0, . . . , n − 1.
(3) The family V i = {yi : yi (t) = e−t x(i) (t), t ≥ 0, x ∈ M } is equiconvergent
at ∞ for i = 0, . . . , n − 1.
(i)
Proof. Let (xk )k be a sequence in M and set yi,k (t) = e−t xk (t). Since the set V i ,
for every i = 0, . . . , n − 1 is relatively compact in C∞ (see Theorem 2.2), then from
the sequence (y0,k )k ⊂ V 0 , we can extract a subsequence denoted also by (y0,k )k ,
that converges to y0∗ in C∞ . Set x∗0 (t) = et y0∗ (t), then
lim
sup e−t |xk (t) − x∗0 (t)| = 0.
k→∞ t∈[0,∞)
Now let y1,k (t) = e−t x0k (t) then the sequence (y1,k )k ⊂ V 1 and we can extract
from it a subsequence denoted also by (y1,k )k , that converges to y1∗ in C∞ . Set
x∗1 (t) = et y1∗ (t), then limk→∞ supt∈[0,∞) e−t |x0k (t)−x∗1 (t)| = 0, from the fact that the
convergence is uniform on [0, T ], T > 0, we get that x∗0 is differentiable on [0, +∞)
and x∗1 = (x∗0 )0 . Reasoning the same way, we obtain x∗i = (x∗0 )(i) , i = 0, . . . , n − 1,
and limk→∞ kxk − x∗0 k = 0. Then M is relatively compact.
R
+∞
Let Y = L1 [0, +∞) with norm kyk1 = 0 |y(t)|dt. Denote ACloc [0, +∞) the
space of locally absolutely continuous functions on the interval [0, +∞). Define the
operator L : dom L ⊂ X → Y by Lx = x(n) , where
n
dom L = x ∈ X, x(n−1) ∈ ACloc [0, +∞), x(i) (0) = 0, i = 0, n − 2
Z
o
n! ξ
x(n−1) (∞) = n
x(t)dt, x(n) ∈ Y ⊂ X,
ξ 0
then L maps dom L into Y . Let N : X → Y be the operator N x(t) = f (t, x(t)),
t ∈ [0, +∞), then (1.1)-(1.2) can be written as Lx = N x.
3. Main results
We can now state our results on the existence of a solution for (1.1)-(1.2).
Theorem 3.1. Assume that the following conditions are satisfied:
(H1) There exists functions α, β ∈ L1 [0, ∞), such that for all x ∈ R and t ∈
[0, ∞),
|f (t, x)| ≤ e−t α(t)|x| + β(t).
(3.1)
(H2) There exists a constant M > 0, such that for x ∈ dom L, if |x(n−1) (t)| > M ,
for all t ∈ [0, ∞), then
Z ∞
Z ξ
1
f (s, x(s))ds − n
(ξ − s)n f (s, x(s))ds 6= 0.
(3.2)
ξ 0
0
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A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
EJDE-2016/29
(H3) There exists a constant M ∗ > 0, such that for any x(t) = c0 tn−1 ∈ ker L
with |c0 | > M ∗ /(n − 1)!, either
Z ξ
hZ ∞
1
c0
f (s, c0 sn−1 )ds − n
(ξ − s)n f (s, c0 sn−1 )ds < 0,
(3.3)
ξ 0
0
or
Z ξ
hZ ∞
i
1
(ξ − s)n f (s, c0 sn−1 )ds > 0.
c0
f (s, c0 sn−1 )ds − n
(3.4)
ξ 0
0
Then (1.1)-(1.2), has at least one solution in X, provided
where Mn = max0≤i≤n−1
1 − 2Mn kαk1 > 0,
supt∈[0,∞) e−t tn−1−i .
(3.5)
To prove Theorem 3.1, we need to prove some Lemmas.
Lemma 3.2. The operator L : dom L ⊂ X → Y is a Fredholm operator of index
zero. Furthermore, the linear projector operator Q : Y → Y can be defined by
Z ξ
i
hZ ∞
1
−t
(ξ − s)n y(s)ds ,
Qy(t) = ae
y(s)ds − n
ξ 0
0
where
Z ξ
n
X
1
1
n!
e−ξ
=1− n
(ξ − s)n e−s ds = 1 −
(−1)k
+ (−1)n n! n
k
a
ξ 0
(n − k)!ξ
ξ
k=0
and the linear operator KP : Im L → dom L ∩ ker P can be written as
Z t
1
Kp y(t) =
(t − s)n−1 y(s)ds, y ∈ Im L.
(n − 1)! 0
Furthermore,
kKp yk ≤ Mn kyk1 , for every y ∈ Im L.
(3.6)
Proof. It is clear that
ker L = x ∈ dom L : x = ctn−1 , c ∈ R, t ∈ [0, ∞) .
Now we show that
Im L = y ∈ Y :
Z
∞
y(s)ds −
0
1
ξn
Z
ξ
(ξ − s)n y(s)ds = 0 .
(3.7)
0
The problem
x(n) (t) = y(t)
(3.8)
(i)
has a solution x(t) that satisfies the conditions x (0) = 0, for i = 0, 1, . . . , n − 2,
Rξ
and x(n−1) (∞) = ξn!n 0 x(t)dt if and only if
Z ∞
Z ξ
1
y(s)ds − n
(ξ − s)n y(s)ds = 0.
(3.9)
ξ 0
0
In fact from (3.8) and the boundary conditions (1.2) we have
Z t
1
(t − s)n−1 y(s)ds + c0 + c1 t + c2 t2 + · · · + cn−1 tn−1
x(t) =
(n − 1)! 0
Z t
1
=
(t − s)n−1 y(s)ds + ctn−1 .
(n − 1)! 0
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BOUNDARY VALUE PROBLEMS AT RESONANCE
From x(n−1) (∞) =
n!
ξn
Rξ
0
x(t)dt, we obtain
Z
∞
y(s)ds =
0
1
ξn
Z
5
ξ
(ξ − s)n y(s)ds.
0
On the other hand, if (3.9) holds, setting
Z t
1
x(t) =
(t − s)n−1 y(s)ds + ctn−1
(n − 1)! 0
where c is an arbitrary constant, then x(t) is a solution of (3.8). Hence (3.7) holds.
Setting
Z ξ
Z ∞
1
(ξ − s)n y(s)ds,
y(s)ds − n
Ry =
ξ 0
0
define Qy(t) = ae−t Ry, it is clear that dim Im Q = 1. We have
Z ξ
Z ∞
1
(ξ − s)n e−s ds
Q2 y = Q(Qy) = ae−t (a.Ry)
e−s ds − n
ξ 0
0
−t
= ae Ry = Qy,
which implies the operator Q is a projector. Furthermore, Im L = ker Q.
Let y = (y − Qy) + Qy, where y − Qy ∈ ker Q = Im L, Qy ∈ Im Q. It follows
from ker Q = Im L and Q2 y = Qy that Im Q ∩ Im L = {0}. Then, we have
Y = Im L ⊕ Im Q. Thus dim ker L = 1 = dim Im Q = codim Im L = 1, this means
that L is a Fredholm operator of index zero. Now we define a projector P from X
to X by setting
x(n−1) (0) n−1
t
.
P x(t) =
(n − 1)!
Then the generalized inverse KP : Im L dom L ∩ ker P of L can be written as
Z t
1
Kp y(t) =
(t − s)n−1 y(s)ds.
(n − 1)! 0
Obviously, Im P = ker L and P 2 x = P x. It follows from x = (x − P x) + P x that
X = ker P + ker L. By simple calculation, we obtain that ker L ∩ ker P = {0}.
Hence X = ker L ⊕ ker P .
From the definitions of P and KP , it is easy to see that the generalized inverse
of L is KP . In fact, for y ∈ Im L, we have
(LKp )y(t) = (Kp y(t))(n) = y(t),
and for x ∈ dom L ∩ ker P , we know that
Z t
1
(Kp L)x(t) = (Kp )x(n) (t) =
(t − s)n−1 x(n) (s)ds
(n − 1)! 0
= x(t) − [x(0) + x0 (0)t + · · · +
x(n−2) (0) n−2 x(n−1) (0) n−1
t
+
t
].
(n − 2)!
(n − 1)!
In view of x ∈ dom L ∩ ker P , x(i) (0) = 0, for i = 0, 1, . . . , n − 2, and P x = 0, thus
(Kp L)x(t) = x(t).
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A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
EJDE-2016/29
This shows that Kp = (L|dom L∩ker P )−1 . From the definition of Kp , we have for
i = 0, . . . , n − 1,
Z t
e−t
−t
(i)
e |(Kp y) (t)| ≤
(t − s)n−1−i |y(s)|ds ≤ Mn kyk1 ,
(n − 1 − i)! 0
which leads to
kKp yk =
max
0≤i≤n−1
sup e−t |(Kp y)i (t)| ≤ Mn kyk1 .
t∈[0,∞)
This completes the proof.
Lemma 3.3. Let Ω1 = {x ∈ dom L\ ker L : Lx = λN x for some λ ∈ [0, 1]}. Then
Ω1 is bounded.
Proof. Suppose that x ∈ Ω1 , and Lx = λN x. Thus λ 6= 0 and QN x = 0, so that
Z ξ
Z ∞
1
(ξ − s)n f (s, x(s))ds = 0.
f (s, x(s))ds − n
ξ
0
0
Thus, by condition (H2), there exists t0 ∈ R+ , such that |x
follows from the absolute continuity of x(n−1) that
Z t0
(n−1)
(n−1)
(t0 ) −
x(n) (s)ds,
|x
(0)| = x
(n−1)
(t0 )| ≤ M . It
0
then, we have
|x
(n−1)
Z
(0)| ≤ M +
∞
Z
|Lx(s)|ds ≤ M +
0
∞
|N x(s)|ds = M + kN xk1 .
(3.10)
0
Again for x ∈ Ω1 and x ∈ dom L\ ker L, we have (I − P )x ∈ dom L ∩ ker P and
LP x = 0; thus from Lemma 3.2,
k(I − P )xk = kKp L(I − P )xk
≤ Mn kL(I − P )xk1
(3.11)
= Mn kLxk1 ≤ Mn kN xk1 .
So
kxk ≤ kP xk + k(I − P )xk = Mn |x(n−1) (0)| + Mn kN xk1 ,
again from (3.10) and (3.11), (3.12) becomes
(3.12)
kxk ≤ Mn M + Mn kN xk1 + Mn kN xk1 ≤ Mn M + 2Mn kN xk1 .
(3.13)
On the other hand by (3.1) we have
Z ∞
kN xk1 =
|f (s, x(s))|ds ≤ kxkkαk1 + kβk1 .
(3.14)
0
Therefore, (3.13) and (3.14), it yield
kxk ≤ Mn M + 2Mn kxkkαk1 + 2Mn kβk1 ;
since 1 − 2Mn kαk1 > 0, we obtain
kxk ≤
Mn M
2Mn kβk1
+
.
1 − 2Mn kαk1
1 − 2Mn kαk1
So Ω1 is bounded.
Lemma 3.4. The set Ω2 = {x ∈ ker L : N x ∈ Im L} is bounded.
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BOUNDARY VALUE PROBLEMS AT RESONANCE
7
Proof. Let x ∈ Ω2 . Then x ∈ ker L implies x(t) = ctn−1 , c ∈ R, and QN x = 0;
therefore
Z ∞
Z ξ
1
f (s, csn−1 )ds − n
(ξ − s)n f (s, csn−1 )ds = 0.
ξ
0
0
From condition (H2), there exists t1 ∈ R+ , such as |x(n−1) (t1 )| ≤ M . We have
M
(n − 1)!|c| ≤ M so |c| ≤ (n−1)!
. On the other hand
kxk = |c| max
sup e−t (tn−1 )(i) ≤ M Mn < ∞,
0≤i≤n−1
t∈[0,∞)
so Ω2 is bounded.
Lemma 3.5. Suppose that the first part of Condition (H3) holds. Let
Ω3 = {x ∈ ker L : −λJx + (1 − λ)QN x = 0, λ ∈ [0, 1]}
where J : ker L → Im Q is the linear isomorphism given by J(ctn−1 ) = ce−t , for all
c ∈ R t ≥ 0. Then Ω3 is bounded.
Proof. In fact x0 ∈ Ω3 , means that x0 ∈ ker L i.e. x0 (t) = c0 tn−1 and λJx0 =
(1 − λ)QN x0 . Then we obtain
Z ξ
Z ∞
1
λc0 = (1 − λ)a
f (s, c0 sn−1 )ds − n
(ξ − s)n f (s, c0 sn−1 )ds .
ξ 0
0
If λ = 1, then c0 = 0. Otherwise, if |c0 | > M ∗ /(n − 1)!, in view of (3.3) one has
Z ξ
Z ∞
1
λc20 = (1 − λ)ac0
(ξ − s)n f (s, c0 sn−1 )ds < 0,
f (s, c0 sn−1 )ds − n
ξ 0
0
which contradicts the fact that λc20 ≥ 0. So |c0 | ≤ M ∗ /(n − 1)!, moreover
kx0 k = |c0 | max
sup e−t |(tn−1 )(i) | ≤ M ∗ Mn .
0≤i≤n−1
t∈[0,∞)
Therefore Ω3 is bounded.
Lemma 3.6. Suppose that the second part of Condition (H3) holds. Let
Ω3 = {x ∈ ker L : λJx + (1 − λ)QN x = 0, λ ∈ [0, 1]}
where J : ker L → Im Q is the linear isomorphism given by J(ctn−1 ) = ce−t , for all
c ∈ R, t ≥ 0. Then Ω3 is bounded, here J is as in Lemma 3.5.
Proof. Similar to the above argument, we can verify that Ω3 is bounded.
Lemma 3.7. 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, kxk ≤ r. For x ∈ Ω, and by condition
(3.1), we obtain
Z ξ
hZ ∞
i
1
−t
|QN x| ≤ ae
|f (s, x(s))|ds + n
(ξ − s)n |f (s, x(s))|ds
ξ 0
0
hZ ∞
≤ ae−t
e−s α(s)|x(s)| + β(s)ds
1
+ n
ξ
Z
0
ξ
0
i
(ξ − s)n (e−s α(s)|x(s)| + β(s))ds
8
A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
∞
h Z
≤ ae−t r
Z
α(s)ds +
0
∞
Z
β(s)ds + r
0
EJDE-2016/29
ξ
Z
α(s)ds +
0
ξ
i
β(s)ds
0
≤ ae−t [2rkαk1 + 2kβk1 ]
≤ 2a[rkαk1 + kβk1 ];
thus,
kQN xk1 ≤ 2a[rkαk1 + kβk1 ],
(3.15)
which implies that QN (Ω) is bounded. Next, we show that KP (I − Q)N (Ω) is
compact, for this we use Lemma 2.3 . Let x ∈ Ω, by (3.1) we have
Z ∞
kN xk1 =
|f s, x(s)|ds ≤ [rkαk1 + kβk1 ];
(3.16)
0
on the other hand, from the definition of KP and together with (3.6), (3.15) and
(3.16) one obtain
kKP (I − Q)N xk ≤ Mn k(I − Q)N xk1 ≤ Mn [kN xk1 + kQN xk1 ]
≤ Mn [r(1 + 2a)kαk1 + (1 + 2a)kβk1 ].
It follows that KP (I − Q)N (Ω) is uniformly bounded.
Let us prove that T is equicontinuous. For any x ∈ Ω and any t1 , t2 ∈ [0, T ] with
t1 < t2 and T ∈ [0, ∞), we have for 0 ≤ i ≤ n − 2:
−t
e 1 (KP (I − Q)N x)(i) (t1 ) − e−t2 (KP (I − Q)N x)(i) (t2 )
Z
t2 −s
=
[e (KP (I − Q)N x)(i) (s)]0 ds
t1
t2
=
Z
[−e−s (KP (I − Q)N x)(i) (s) + e−s (KP (I − Q)N x)(i+1) (s)]ds
t1
≤ 2(t2 − t1 )kKP (I − Q)N xk
≤ 2(t2 − t1 )Mn [r(1 + 2a)kαk1 + (1 + 2a)kβk1 ] → 0,
as t1 → t2 .
For i = n − 1, we obtain
−t
e 1 (KP (I − Q)N x)(n−1) (t1 ) − e−t2 (KP (I − Q)N x)(n−1) (t2 )
Z t1
Z t2
−t
−t2
1
= e
(I − Q)N x(s)ds − e
(I − Q)N x(s)ds
0
0
Z t1
Z t2
≤
(e−t1 − e−t2 )|(I − Q)N x(s)|ds +
e−t2 |(I − Q)N x(s)|ds
0
t1
t1
Z
≤ (t2 − t1 )
Z
t2
|(I − Q)N x(s)|ds +
0
|(I − Q)N x(s)|ds → 0,
t1
as t1 → t2 . So KP (I − Q)N (Ω) is equicontinuous on every compact subinterval of
[0, ∞). In addition, we claim that KP (I − Q)N (Ω) is equiconvergent at infinity. In
fact, for x ∈ Ω, i = 0, . . . , n − 1, we have
−t
e (Kp (I − Q)N x)(i) (t)
Z t
e−t
≤
(t − s)n−1−i |(I − Q)N x(s)|ds
(n − 1 − i)! 0
Z t
−t n−1−i
|(I − Q)N x(s)|ds ≤ e−t tn−1−i k(I − Q)N xk1
≤e t
0
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BOUNDARY VALUE PROBLEMS AT RESONANCE
9
≤ e−t tn−1−i [kN xk1 + kQN xk1 ] ≤ e−t tn−1−i (1 + 2a)[rkαk1 + kβk1 ,
thus, limt→∞ e−t (Kp (I − Q)N x)(i) (t) = 0, for every i = 0, . . . , n − 1, which means
that KP (I − Q)N (Ω) is equiconvergent at infinity.
Now we are able to give the proof of Theorem 3.1, which is an immediate consequence of Theorem 2.1 and the above lemmas.
Proof of Theorem 3.1. We shall prove that all conditions of Theorem 2.1 are satisfied. Set Ω to be an open bounded subset of X such that ∪3i=1 Ωi ⊂ Ω. We know
that L is a Fredholm operator of index zero and N is L-compact on Ω. By the
definition of Ω we have
(i) Lx 6= λN x pour tout (x, λ) ∈ [(dom L\ ker L) ∩ ∂Ω] × (0, 1);
(ii) N x ∈
/ Im L pour tout x ∈ ker L ∩ ∂Ω.
At last we prove that condition (iii) of Theorem 2.1 is satisfied. To this end, let
H(x, λ) = ±λJx + (1 − λ)QN x
By the definition of Ω we know that Ω3 ⊂ Ω, thus H(x, λ) 6= 0 for every x ∈
ker L ∩ ∂Ω. Then, by the homotopy property of degree, we obtain
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.
So, the third assumption of Theorem 2.1 is fulfilled and Lx = N x has at least one
solution in dom L ∩ Ω; i.e. (1.1)-(1.2) has at least one solution in X. The prove is
complete.
Acknowledgements. The authors would like to thank the anonymous reader for
pointing out several mistakes in the original article, and for checking the corrections.
We want to also to thank the managing editor Prof. Julio G. Dix for allowing us
to correct our article.
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Next, for record keeping we include the original article that has several mistakes.
11. Introduction
In this paper, we are concerned with the existence of solutions of the higher-order
ordinary differential equation
x(n) (t) = f (t, x(t)),
t ∈ (0, ∞),
(11.1)
with the integral boundary value conditions
x(i) (0) = 0, i = 0, 1, . . . , n − 2,
x(n−1) (∞) =
n!
ξn
Z
ξ
x(t)dt ,
(11.2)
0
where n ≥ 3 is an integer, ξ > 0 and f ∈ C([0, ∞) × R, R).
A boundary value problem (BVP for short) is said to be at resonance if the
corresponding homogeneous boundary value problem has a non-trivial solution.
Resonance problems can be formulated as an abstract equation Lx = N x, where L
is a noninvertible operator. When L is linear, as is known, the coincidence degree
theory of Mawhin [19] has played an important role in dealing with the existence
of solutions for these problems. For more recent results, we refer the reader to
[3, 5, 6, 6, 8, 9, 14, 20, 22, 24, 25] and the references therein.
EJDE-2016/29
BOUNDARY VALUE PROBLEMS AT RESONANCE
11
Moreover boundary value problems on the half line arise in many applications
in physics such that in modeling the unsteady flow of a gas through semi-infinite
porous media, in plasma physics, in determining the electrical potential in an isolated neutral atom, or in combustion theory. For an extensive literature of results
as regards boundary value problems on unbounded domains, we refer the reader to
the monograph by Agarwal and O’Regan [1].
Recently, there have been many works concerning the existence of solutions for
the boundary value problems on the half-line. For instance see [2, 4, 10, 11, 12,
13, 15, 16, 17, 18, 21, 23] and the references therein. By the way, much of work on
the existence of solutions for the boundary value problems on unbounded domains
involves second or third-order differential equations.
However, for the resonance case, there is no work done for the higher-order
boundary value problems with integral boundary conditions on the half-line, such
as BVP (11.1)-(11.2).
The remaining part of this paper is organized as follows. We present in Section
2 some notations and basic results involved in the reformulation of the problem. In
Section 3, we give the main theorem and some lemmas, then we will show that the
proof of the main theorem is an immediate consequence of these lemmas and the
coincidence degree of Mawhin.
12. Preliminaries
For the convenience of the readers, we recall some notation and two theorems
which will be used later.
Let X, Y be two real Banach spaces and let L : dom L ⊂ X → Y be a linear
operator which is Fredholm map of index zero, and let P : X → X, Q : Y → Y be
continuous projectors such that Im P = ker L, ker Q = Im L. Then X = ker L ⊕
ker P , Y = Im L ⊕ ImQ. It follows that L|dom L∩ker P : dom L ∩ ker P → Im L is
invertible, we denote the inverse of that map by KP . Let Ω be an open bounded
subset of X such that dom L ∩ Ω 6= ∅, the map N : X → Y is said to be L-compact
on Ω if the map QN (Ω) is bounded and KP (I − QN ) : Ω → X is compact.
Theorem 12.1 ([19]). Let L be a Fredholm operator of index zero and N be Lcompact on Ω. Assume that the following conditions are satisfied:
(1) Lx 6= λN x for every (x, λ) ∈ [(dom L\ ker L) ∩ ∂Ω] × (0, 1).
(2) N x ∈
/ Im L for every x ∈ ker L ∩ ∂Ω.
(3) deg(QN |ker L , Ω ∩ ∩ ker L, 0) 6= 0, where Q : Y → Y is a projection such
that Im L = ker Q.
Then the equation Lx = N x has at least one solution in dom L ∩ Ω.
Since the Arzelá-Ascoli theorem fails in the noncompact interval case, we use
the following result in order to show that KP (I − QN ) : Ω → X is compact.
Theorem 12.2 ([1]). Let F ⊂ X. Then F is relatively compact if the following
conditions hold:
(1) F is bounded in X.
(2) The functions belonging to F are equi-continuous on any compact interval
of [0, ∞).
(3) The functions from F are equi-convergent at +∞.
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A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
EJDE-2016/29
Let AC[0, +∞) denote the space of locally absolutely continuous functions on
the interval [0, +∞). Let
X = x ∈ C n−1 [0, +∞) : x(n−1) ∈ ACloc [0, +∞), lim e−t |x(t)| exists
t→∞
−t
endowed with the norm kxk = supt∈[0,+∞) e |x(t)|. Let Y = L1 [0, +∞) with
R +∞
norm kyk1 = 0 |y(t)|dt.
Define the operator L : dom L ⊂ X → Y by Lx = x(n) , where
Z
n! ξ
x(t)dt .
dom L = x ∈ X : x(i) (0) = 0, i = 0, n − 2, x(n−1) (∞) = n
ξ 0
Let N : X → Y be the operator N x = f (t, x(t)), t ∈ [0, +∞), then the BVP
(11.1)–(11.2) can be written as Lx = N x.
13. Main results
We can now state our results on the existence of a solution for (11.1)-(11.2).
Theorem 13.1. Assume that the following conditions are satisfied:
(H1) There exists functions α, β ∈ L1 [0, ∞), such that for all x ∈ R and t ∈
[0, ∞),
|f (t, x)| ≤ e−t α(t)|x| + β(t).
(13.1)
(H2) There exists a constant M > 0, such that for x ∈ dom L, if |x(n−1) (t)| > M ,
for all t ∈ [0, ∞), then
Z ∞
Z ξ
1
f (s, x(s))ds − n
(ξ − s)n f (s, x(s))ds 6= 0.
(13.2)
ξ
0
0
(H3) There exists a constant M ∗ > 0, such that for any x(t) = c0 tn−1 ∈ ker L
with |c0 | > M ∗ /(n − 1)!, either
Z ξ
hZ ∞
1
n−1
c0
f (s, c0 s
)ds − n
(ξ − s)n f (s, c0 sn−1 )ds < 0,
(13.3)
ξ 0
0
or
Z ξ
i
hZ ∞
1
c0
(ξ − s)n f (s, c0 sn−1 )ds > 0.
(13.4)
f (s, c0 sn−1 )ds − n
ξ 0
0
Then (11.1)-(11.2), has at least one solution in C[0, ∞), provided
1 − 2Mn kαk1 > 0,
−t n−1
where Mn = supt∈[0,∞) e t
=
(13.5)
n−1
( n−1
.
e )
To prove Theorem 13.1, we need to prove some Lemmas.
Lemma 13.2. The operator L : dom L ⊂ X → Y is a Fredholm operator of index
zero. Furthermore, the linear projector operator Q : Y → Y can be defined by
Z ξ
hZ ∞
i
1
−t
Qy(t) = ae
y(s)ds − n
(ξ − s)n y(s)ds ,
ξ 0
0
where
1/a = 1 −
n
X
(−1)k
k=0
n!
(n − k)!ξ k
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BOUNDARY VALUE PROBLEMS AT RESONANCE
13
and the linear operator KP : Im L → dom L ∩ ker P can be written as
Z t
1
(t − s)n−1 y(s)ds, y ∈ Im L.
Kp y(t) =
(n − 1)! 0
Furthermore
kKp yk ≤
Mn
kyk1 ,
(n − 1)!
for every y ∈ Im L.
(13.6)
Proof. It is clear that
ker L = {x ∈ dom L : x = ctn−1 , c ∈ R, t ∈ [0, ∞) .
Now we show that
Z
Im L = y ∈ Y :
0
∞
1
y(s)ds − n
ξ
Z
ξ
(ξ − s)n y(s)ds = 0 .
(13.7)
0
The problem
x(n) (t) = y(t)
(13.8)
(i)
has a solution x(t) that satisfies the conditions x (0) = 0, for i = 0, 1, . . . , n − 2,
Rξ
and x(n−1) (∞) = ξn!n 0 x(t)dt if and only if
Z ξ
Z ∞
1
(ξ − s)n y(s)ds = 0.
(13.9)
y(s)ds − n
ξ 0
0
In fact from (13.8) and the boundary conditions (11.2) we have
Z t
1
x(t) =
(t − s)n−1 y(s)ds + c0 + c1 t + c2 t2 + · · · + cn−1 tn−1
(n − 1)! 0
Z t
1
(t − s)n−1 y(s)ds + ctn−1 .
=
(n − 1)! 0
Rξ
From x(n−1) (∞) = ξn!n 0 x(t)dt, we obtain
Z ∞
Z ξ
1
y(s)ds = n
(ξ − s)n y(s)ds.
ξ
0
0
On the other hand, if (13.9) holds, setting
Z t
1
x(t) =
(t − s)n−1 y(s)ds + ctn−1
(n − 1)! 0
where c is an arbitrary constant, then x(t) is a solution of (13.8). Hence (13.7)
holds. Setting
Z ξ
Z ∞
1
Ry =
y(s)ds − n
(ξ − s)n y(s)ds,
ξ
0
0
define Qy(t) = ae−t Ry, it is clear that dim Im Q = 1. We have
Z ∞
Z ξ
1
Q2 y = Q(Qy) = ae−t (a.Ry)(
e−s ds − n
(ξ − s)n e−s ds)
ξ
0
0
= ae−t Ry = Qy,
that implies the operator Q is a projector. Furthermore, Im L = ker Q.
Let y = (y − Qy) + Qy, where y − Qy ∈ ker Q = Im L, Qy ∈ Im Q. It follows
from ker Q = Im L and Q2 y = Qy that Im Q ∩ Im L = {0}. Then, we have
Y = Im L ⊕ Im Q. Thus dim ker L = 1 = dim Im Q = codim Im L = 1, this means
14
A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
EJDE-2016/29
that L is a Fredholm operator of index zero. Now we define a projector P from X
to X by setting
x(n−1) (0) n−1
P x(t) =
t
.
(n − 1)!
Then the generalized inverse KP : Im L → dom L ∩ ker P of L can be written as
Z t
1
(t − s)n−1 y(s)ds.
Kp y =
(n − 1)! 0
Obviously, Im P = ker L and P 2 x = P x. It follows from x = (x − P x) + P x that
X = ker P + ker L. By simple calculation, we obtain that ker L ∩ ker P = {0}.
Hence X = ker L ⊕ ker P .
From the definitions of P and KP , it is easy to see that the generalized inverse
of L is KP . In fact, for y ∈ Im L, we have
(LKp )y(t) = (Kp y(t))(n) = y(t),
and for x ∈ dom L∩ ker P , we know that
Z t
1
(n)
(t − s)n−1 x(n) (s)ds
(Kp L)x(t) = (Kp )x (t) =
(n − 1)! 0
= x(t) − [x(0) + x0 (0)t + . . . . . . .
x(n−2) (0) n−2 x(n−1) (0) n−1
t
+
t
].
(n − 2)!
(n − 1)!
In view of x ∈ dom L ∩ ker P , x(i) (0) = 0, for i = 0, 1, . . . , n − 2, and P x = 0, thus
(Kp L)x(t) = x(t).
This shows that Kp = (L|dom L∩ker P )−1 . From the definition of Kp , we have
Z t
e−t
−t
(t − s)n−1 |y(s)|ds
kKp yk = sup e |Kp y| ≤ sup
t∈[0,∞)
t∈[0,∞) (n − 1)! 0
Z ∞
Mn
Mn
<
kyk1 .
|y(s)|ds =
(n − 1)! 0
(n − 1)!
This completes the proof.
Lemma 13.3. Let Ω1 = {x ∈ dom L\ ker L : Lx = λN x for some λ ∈ [0, 1]}.
Then Ω1 is bounded.
Proof. Suppose that x ∈ Ω1 , and Lx = λN x. Thus λ 6= 0 and QN x = 0, so that
Z ξ
Z ∞
1
(ξ − s)n f (s, x(s))ds = 0.
f (s, x(s))ds − n
ξ
0
0
Thus, by condition (H2), there exists t0 ∈ R+ , such that |x
follows from the absolute continuity of x(n−1) that
Z t0
(n−1)
(n−1)
|x
(0)| = x
(t0 ) −
x(n) (s)ds,
(n−1)
(t0 )| ≤ M . It
0
then, we have
|x
(n−1)
Z
(0)| ≤ M +
∞
Z
|Lx(s)|ds ≤ M +
0
∞
|N x(s)|ds = M + kN xk1 .
0
(13.10)
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BOUNDARY VALUE PROBLEMS AT RESONANCE
15
Again for x ∈ Ω1 and x ∈ dom L\ ker L, we have (I − P )x ∈ dom L ∩ ker P and
LP x = 0; thus from Lemma 13.2,
k(I − P )xk = kKp L(I − P )xk
Mn
≤
kL(I − P )xk1
(n − 1)!
Mn
Mn
=
kLxk1 ≤
kN xk1 .
(n − 1)!
(n − 1)!
(13.11)
So
kxk ≤ kP xk + k(I − P )xk = Mn |x(n−1) (0)| +
Mn
kN xk1 ,
(n − 1)!
(13.12)
again from (13.10) and (13.11), (13.12) becomes
kxk ≤ Mn M + Mn kN xk1 +
Mn
kN xk1 ≤ Mn M + 2Mn kN xk1 .
(n − 1)!
On the other hand by (13.1) we have
Z ∞
kN xk1 =
|f (s, x(s))|ds ≤ kxkkαk1 + kβk1 .
(13.13)
(13.14)
0
Therefore, (13.13) and (13.14), it yield
kxk ≤ Mn M + 2Mn kxkkαk1 + 2Mn kβk1 ;
since 1 − 2Mn kαk1 > 0, we obtain
kxk ≤
2Mn kβk1
Mn M
+
.
1 − 2Mn kαk1
1 − 2Mn kαk1
So Ω1 is bounded.
Lemma 13.4. The set Ω2 = {x ∈ ker L : N x ∈ Im L} is bounded.
Proof. Let x ∈ Ω2 , then x ∈ ker L implies x(t) = ctn−1 , c ∈ R, and QN x = 0;
therefore
Z ξ
Z ∞
1
(ξ − s)n f (s, csn−1 )ds = 0.
f (s, csn−1 )ds − n
ξ 0
0
From condition (H2 ), there exists t1 ∈ R+ , such as |x(n−1) (t1 )| ≤ M . We have
(n − 1)!|c| ≤ M
so |c| ≤
M
(n−1)! .
On the other hand
kxk = sup e−t |x(t)| = |c| sup e−t tn−1 = |c|Mn ,
t∈[0∞)
i.e. kxk ≤
Mn M
(n−1)!
t∈[0∞)
< ∞, so Ω2 is bounded.
Lemma 13.5. Suppose that the first part of Condition (H3) holds. Let
Ω3 = {x ∈ ker L : −λJx + (1 − λ)QN x = 0, λ ∈ [0, 1]}
where J : ker L → Im Q is the linear isomorphism given by J(ctn−1 ) = ctn−1 , for
all c ∈ R t ≥ 0. Then Ω3 is bounded.
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A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
EJDE-2016/29
Proof. In fact x0 ∈ Ω3 , means that x0 ∈ ker L i.e. x0 (t) = c0 tn−1 and λJx0 =
(1 − λ)QN x0 . Then we obtain
Z ξ
Z ∞
1
f (s, c0 sn−1 )ds − n
(ξ − s)n f (s, c0 sn−1 )ds .
λc0 tn−1 = (1 − λ)ae−t
ξ 0
0
If λ = 1, then c0 = 0. Otherwise, if |c0 | > M ∗ , in view of (13.3) one has
Z ∞
Z ξ
1
λc20 tn−1 = (1 − λ)ae−t c0 (
f (s, c0 sn−1 )ds − n
(ξ − s)n f (s, c0 sn−1 )ds) < 0,
ξ
0
0
which contradicts the fact that λc20 ≥ 0. So |c0 | ≤ M ∗ , moreover
kx0 k = sup e−t |c0 |tn−1 = |c0 |Mn ≤ M ∗ Mn .
Therefore Ω3 is bounded.
Lemma 13.6. Suppose that the second part of Condition (H3) holds. Let
Ω3 = {x ∈ ker L : λJx + (1 − λ)QN x = 0, λ ∈ [0, 1]}
where J : ker L → Im Q is the linear isomorphism given by J(ctn−1 ) = ctn−1 , for
all c ∈ R, t ≥ 0. Then Ω3 is bounded here J as in Lemma 13.5. Similar to the
above argument, we can verify that Ω3 is bounded.
Lemma 13.7. 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, kxk ≤ r. For x ∈ Ω, and by condition
(13.1), we obtain
Z ξ
hZ ∞
i
1
−t
−2t
e |QN x| ≤ ae
|f (s, x(s))|ds + n
(ξ − s)n |f (s, x(s))|ds
ξ 0
0
hZ ∞
≤ ae−2t
e−s α(s)|x(s)| + β(s)ds
1
+ n
ξ
Z
0
ξ
i
(ξ − s)n (e−s α(s)|x(s)| + β(s))ds
0
∞
h Z
≤ ae−2t r
Z
α(s)ds +
0
∞
Z
β(s)ds + r
0
ξ
Z
α(s)ds +
0
ξ
i
β(s)ds
0
≤ ae−2t [2rkαk1 + 2kβk1 ]
≤ 2a[rkαk1 + kβk1 ];
thus,
kQN xk1 ≤ 2a[rkαk1 + kβk1 ],
(13.15)
which implies that QN (Ω) is bounded. Next, we show that KP (I − Q)N (Ω) is
compact. For x ∈ Ω, by (13.1) we have
Z ∞
kN xk1 =
|f s, x(s)|ds ≤ [rkαk1 + kβk1 ];
(13.16)
0
on the other hand, from the definition of KP and together with (13.6), (13.15) and
(13.16) one gets
kKP (I − Q)N k ≤ Mn k(I − Q)N k1 ≤ Mn [kN xk1 + kQN xk1 ]
≤ Mn [r(1 + 2a)kαk1 + (1 + 2a)kβk1 ].
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It follows that KP (I − Q)N (Ω) is uniformly bounded.
Let us prove that T is equicontinuous. For any x ∈ Ω and any t1 , t2 ∈ [0, T ] with
t1 < t2 and T ∈ [0, ∞), we have
|e−t1 KP (I − Q)N x(t1 ) − e−t2 KP (I − Q)N x(t2 )|
Z t1
1
e−t1 (t1 − s)n−1 (I − Q)N x(s)ds
=
(n − 1)! 0
Z t2
e−t2 (t2 − s)n−1 (I − Q)N x(s)ds
−
0
Z t1
1
≤
e−t2 (t2 − s)n−1 − e−t1 (t1 − s)n−1 |(I − Q)N x(s)|ds
[
(n − 1)! 0
Z t2
+
e−t2 (t2 − s)n−1 |(I − Q)N x(s)|ds]
t1
≤
h Z t1
1
(e−(t2 −s) (t2 − s)n−1 − e−(t1 −s) (t1 − s)n−1 )
(n − 1)! 0
× e−s |(I − Q)N x(s)|ds
Z t2
i
e−(t2 −s) (t2 − s)n−1 e−s |(I − Q)N x(s)|ds
+
t1
Z t1
1
0
[M (t2 − t1 )
≤
e−s |(I − Q)N x(s)|ds
(n − 1)! n
0
Z t2
i
+ e−t2 (t2 − t1 )n−1
|(I − Q)N x(s)|ds → 0,
as t1 → t2 .
t1
So KP (I−Q)N (Ω) is equicontinuous on every compact subset of [0, ∞). In addition,
we claim that KP (I − Q)N (Ω) is equiconvergent at infinity. In fact,
|e−t Kp (I − Q)N x(t)|
Z t
1
e−(t−s) (t − s)n−1 e−s |(I − Q)N x(s)|ds
≤
(n − 1)! 0
Z t
Mn
Mn
≤
k(I − Q)N xk1
|(I − Q)N x(s)|ds ≤
(n − 1)! 0
(n − 1)!
Mn
[kN xk1 + kQN xk1 ] < ∞;
≤
(n − 1)!
thus, limt→∞ |e−t Kp (I − Q)N x(t)| < ∞. Which means that KP (I − Q)N (Ω) is
equiconvergent
Now we are able to give the proof of Theorem 13.1, which is an immediate
consequence of Theorem 12.1 and the above lemmas.
Proof of Theorem 13.1. We shall prove that all conditions of Theorem 12.1 are
satisfied. Set Ω to be an open bounded subset of X such that ∪3i=1 Ωi ⊂ Ω. We
know that L is a Fredholm operator of index zero and N is L-compact on Ω. By
the definition of Ω we have
(i) Lx 6= λN x pour tout (x, λ) ∈ [(dom L\ ker L) ∩ ∂Ω] × (0, 1);
(ii) N x ∈
/ Im L pour tout x ∈ ker L ∩ ∂Ω.
18
A. FRIOUI, A. GUEZANE-LAKOUD, R. KHALDI
EJDE-2016/29
At last we prove that condition (iii) of Theorem 12.1 is satisfied. To this end, let
H(x, λ) = ±λJx + (1 − λ)QN x
By the definition of Ω we know that Ω3 ⊂ Ω, thus H(x, λ) 6= 0 for every x ∈
ker L ∩ ∂Ω. Then, by the homotopy property of degree, we obtain
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.
So, the third assumption of Theorem 12.1 is fulfilled and Lx = N x has at least one
solution in dom L ∩ Ω; i.e. (11.1)-(11.2) has at least one solution in X. The prove
is complete.
Assia Frioui
Laboratory of applied mathematics and modeling, University 08 Mai 45-Guelma, P.O.
Box 401, Guelma 24000, Algeria
E-mail address: frioui.assia@yahoo.fr
Assia Guezane-Lakoud
Laboratory of Advanced Materials, Faculty of Sciences, University Badji MokhtarAnnaba, P.O. Box 12, 23000, Annaba, Algeria
E-mail address: a guezane@yahoo.fr
Rabah Khaldi
Laboratory of Advanced Materials. Faculty of Sciences, University Badji MokhtarAnnaba, P.O. Box 12, 23000, Annaba, Algeria
E-mail address: rkhadi@yahoo.fr