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 dif- ferential 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 higher- order 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
0
x(t)dt , (1.2) 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 equationLx=N x, whereL is a noninvertible operator. WhenLis 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 iso- lated neutral atom, or in combustion theory. For an extensive literature of results
2010Mathematics 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
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 : domL ⊂X → Y be a linear operator which is Fredholm map of index zero, and letP :X →X,Q:Y →Y be continuous projectors such that ImP = kerL, kerQ= ImL. Then X = kerL⊕ kerP, Y = ImL⊕ImQ. It follows that L|domL∩kerP : domL∩kerP → ImL is invertible, we denote the inverse of that map byKP. Let Ω be an open bounded subset ofX such that domL∩Ω6=∅, the mapN :X →Y is said to beL-compact on Ω if the mapQN(Ω) is bounded andKP(I−QN) : Ω→X is compact.
Theorem 2.1 ([19]). Let L be a Fredholm operator of index zero and N be L- compact onΩ. Assume that the following conditions are satisfied:
(1) Lx6=λN xfor every(x, λ)∈[(domL\kerL)∩∂Ω]×(0,1).
(2) N x /∈ImLfor every x∈kerL∩∂Ω.
(3) deg(QN|kerL,Ω∩kerL,0)6= 0, whereQ:Y →Y is a projection such that ImL= kerQ.
Then the equationLx=N x has at least one solution indomL∩Ω.
Since the Arzel´a-Ascoli theorem fails in the noncompact interval case, we use the following result in order to show thatKP(I−QN) : Ω→X is compact.
Theorem 2.2([1]). LetF be a subset ofC∞={y∈C([0,+∞)),limt→∞y(t)exists}
that is equipped with the normkyk∞= sup
t∈[0,+∞)
|y(t)|. ThenF is relatively compact if the following conditions hold:
(1) F is bounded inX.
(2) The functions belonging toF are equi-continuous on any compact subinter- val of[0,∞).
(3) The functions from F are equi-convergent at+∞.
Let
X=
x∈Cn−1[0,+∞), lim
t→∞e−t|x(i)(t)|exist, 0≤i≤n−1
endowed with the normkxk= max0≤i≤n−1 supt∈[0,+∞)e−t|x(i)(t)|
. ThenX 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 Vi ={yi : yi(t) =e−tx(i)(t), t≥0, x ∈M} is equicontinuous on any compact subinterval of[0,+∞)fori= 0, . . . , n−1.
(3) The family Vi ={yi : yi(t) =e−tx(i)(t), t≥0, x∈ M} is equiconvergent at∞fori= 0, . . . , n−1.
Proof. Let (xk)k be a sequence inM and setyi,k(t) =e−tx(i)k (t). Since the setVi, for everyi= 0, . . . , n−1 is relatively compact inC∞(see Theorem 2.2), then from the sequence (y0,k)k ⊂V0, we can extract a subsequence denoted also by (y0,k)k, that converges toy∗0 in C∞. Setx∗0(t) =ety∗0(t), then
lim
k→∞ sup
t∈[0,∞)
e−t|xk(t)−x∗0(t)|= 0.
Now let y1,k(t) = e−tx0k(t) then the sequence (y1,k)k ⊂ V1 and we can extract from it a subsequence denoted also by (y1,k)k, that converges to y∗1 in C∞. Set x∗1(t) =ety∗1(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 thatx∗0is differentiable on [0,+∞) andx∗1= (x∗0)0. Reasoning the same way, we obtain x∗i = (x∗0)(i),i= 0, . . . , n−1, and limk→∞kxk−x∗0k= 0. ThenM is relatively compact.
LetY =L1[0,+∞) with norm kyk1=R+∞
0 |y(t)|dt. DenoteACloc[0,+∞) the space of locally absolutely continuous functions on the interval [0,+∞). Define the operatorL: domL⊂X→Y byLx=x(n), where
domL=n
x∈X, x(n−1)∈ACloc[0,+∞), x(i)(0) = 0, i= 0, n−2 x(n−1)(∞) = n!
ξn Z ξ
0
x(t)dt, x(n)∈Yo
⊂X,
then L maps domL into Y. LetN : X →Y be the operatorN x(t) =f(t, x(t)), t∈[0,+∞), then (1.1)-(1.2) can be written asLx=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−tα(t)|x|+β(t). (3.1) (H2) There exists a constantM >0, such that forx∈domL, if|x(n−1)(t)|> M,
for allt∈[0,∞), then Z ∞
0
f(s, x(s))ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, x(s))ds6= 0. (3.2)
(H3) There exists a constant M∗ >0, such that for any x(t) = c0tn−1 ∈kerL with|c0|> M∗/(n−1)!, either
c0hZ ∞ 0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)ds
<0, (3.3) or
c0
hZ ∞
0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)dsi
>0. (3.4) Then (1.1)-(1.2), has at least one solution in X, provided
1−2Mnkαk1>0, (3.5)
whereMn= max0≤i≤n−1 supt∈[0,∞)e−ttn−1−i .
To prove Theorem 3.1, we need to prove some Lemmas.
Lemma 3.2. The operator L: domL⊂X →Y is a Fredholm operator of index zero. Furthermore, the linear projector operatorQ:Y →Y can be defined by
Qy(t) =ae−thZ ∞ 0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)dsi , where
1
a = 1− 1 ξn
Z ξ
0
(ξ−s)ne−sds= 1−
n
X
k=0
(−1)k n!
(n−k)!ξk + (−1)nn!e−ξ ξn and the linear operatorKP : ImL→domL∩kerP can be written as
Kpy(t) = 1 (n−1)!
Z t
0
(t−s)n−1y(s)ds, y∈ImL.
Furthermore,
kKpyk ≤Mnkyk1, for everyy ∈ImL. (3.6) Proof. It is clear that
kerL=
x∈domL:x=ctn−1, c∈R, t∈[0,∞) . Now we show that
ImL= y∈Y :
Z ∞
0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)ds= 0 . (3.7) The problem
x(n)(t) =y(t) (3.8)
has a solution x(t) that satisfies the conditionsx(i)(0) = 0, fori= 0,1, . . . , n−2, andx(n−1)(∞) = ξn!nRξ
0 x(t)dtif and only if Z ∞
0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)ds= 0. (3.9) In fact from (3.8) and the boundary conditions (1.2) we have
x(t) = 1 (n−1)!
Z t
0
(t−s)n−1y(s)ds+c0+c1t+c2t2+· · ·+cn−1tn−1
= 1
(n−1)!
Z t
0
(t−s)n−1y(s)ds+ctn−1.
Fromx(n−1)(∞) = ξn!nRξ
0 x(t)dt, we obtain Z ∞
0
y(s)ds= 1 ξn
Z ξ
0
(ξ−s)ny(s)ds.
On the other hand, if (3.9) holds, setting x(t) = 1
(n−1)!
Z t
0
(t−s)n−1y(s)ds+ctn−1
wherecis an arbitrary constant, thenx(t) is a solution of (3.8). Hence (3.7) holds.
Setting
Ry= Z ∞
0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)ds, defineQy(t) =ae−tRy, it is clear that dim ImQ= 1. We have
Q2y=Q(Qy) =ae−t(a.Ry)Z ∞ 0
e−sds− 1 ξn
Z ξ
0
(ξ−s)ne−sds
=ae−tRy=Qy,
which implies the operatorQis a projector. Furthermore, ImL= kerQ.
Lety = (y−Qy) +Qy, where y−Qy ∈kerQ= ImL, Qy ∈ImQ. It follows from kerQ = ImL and Q2y = Qy that ImQ∩ImL = {0}. Then, we have Y = ImL⊕ImQ. Thus dim kerL= 1 = dim ImQ= codim ImL= 1, this means thatL is a Fredholm operator of index zero. Now we define a projectorP fromX toX by setting
P x(t) = x(n−1)(0) (n−1)! tn−1.
Then the generalized inverseKP : ImLdomL∩kerP ofLcan be written as Kpy(t) = 1
(n−1)!
Z t
0
(t−s)n−1y(s)ds.
Obviously, ImP = kerLand P2x=P x. It follows fromx= (x−P x) +P x that X = kerP + kerL. By simple calculation, we obtain that kerL∩kerP = {0}.
HenceX = kerL⊕kerP.
From the definitions ofP andKP, it is easy to see that the generalized inverse ofLisKP. In fact, fory∈ImL, we have
(LKp)y(t) = (Kpy(t))(n)=y(t), and forx∈domL∩kerP, we know that
(KpL)x(t) = (Kp)x(n)(t) = 1 (n−1)!
Z t
0
(t−s)n−1x(n)(s)ds
=x(t)−[x(0) +x0(0)t+· · ·+x(n−2)(0)
(n−2)! tn−2+x(n−1)(0) (n−1)! tn−1].
In view ofx∈domL∩kerP,x(i)(0) = 0, fori= 0,1, . . . , n−2, andP x= 0, thus (KpL)x(t) =x(t).
This shows that Kp = (L|domL∩kerP)−1. From the definition of Kp, we have for i= 0, . . . , n−1,
e−t|(Kpy)(i)(t)| ≤ e−t (n−1−i)!
Z t
0
(t−s)n−1−i|y(s)|ds≤Mnkyk1, which leads to
kKpyk= max
0≤i≤n−1
sup
t∈[0,∞)
e−t|(Kpy)i(t)|
≤Mnkyk1.
This completes the proof.
Lemma 3.3. Let Ω1={x∈domL\kerL:Lx=λN x for someλ∈[0,1]}. Then Ω1 is bounded.
Proof. Suppose thatx∈Ω1, andLx=λN x. Thusλ6= 0 and QN x= 0, so that Z ∞
0
f(s, x(s))ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, x(s))ds= 0.
Thus, by condition (H2), there exists t0 ∈ R+, such that |x(n−1)(t0)| ≤ M. It follows from the absolute continuity ofx(n−1)that
|x(n−1)(0)|=
x(n−1)(t0)− Z t0
0
x(n)(s)ds , then, we have
|x(n−1)(0)| ≤M+ Z ∞
0
|Lx(s)|ds≤M+ Z ∞
0
|N x(s)|ds=M+kN xk1. (3.10) Again for x ∈Ω1 and x∈ domL\kerL, we have (I−P)x∈ domL∩kerP and LP x= 0; thus from Lemma 3.2,
k(I−P)xk=kKpL(I−P)xk
≤MnkL(I−P)xk1
=MnkLxk1≤MnkN xk1.
(3.11) So
kxk ≤ kP xk+k(I−P)xk=Mn|x(n−1)(0)|+MnkN xk1, (3.12) again from (3.10) and (3.11), (3.12) becomes
kxk ≤MnM+MnkN xk1+MnkN xk1≤MnM+ 2MnkN xk1. (3.13) On the other hand by (3.1) we have
kN xk1= Z ∞
0
|f(s, x(s))|ds≤ kxkkαk1+kβk1. (3.14) Therefore, (3.13) and (3.14), it yield
kxk ≤MnM+ 2Mnkxkkαk1+ 2Mnkβk1; since 1−2Mnkαk1>0, we obtain
kxk ≤ MnM 1−2Mnkαk1
+ 2Mnkβk1 1−2Mnkαk1
.
So Ω1is bounded.
Lemma 3.4. The setΩ2={x∈kerL:N x∈ImL} is bounded.
Proof. Let x∈ Ω2. Then x∈ kerL implies x(t) = ctn−1, c ∈ R, and QN x = 0;
therefore
Z ∞
0
f(s, csn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, csn−1)ds= 0.
From condition (H2), there exists t1 ∈ R+, such as |x(n−1)(t1)| ≤ M. We have (n−1)!|c| ≤M so |c| ≤ (n−1)!M . On the other hand
kxk=|c| max
0≤i≤n−1
sup
t∈[0,∞)
e−t(tn−1)(i)
≤M Mn<∞,
so Ω2 is bounded.
Lemma 3.5. Suppose that the first part of Condition(H3) holds. Let Ω3={x∈kerL:−λJ x+ (1−λ)QN x= 0, λ∈[0,1]}
whereJ : kerL→ImQis the linear isomorphism given byJ(ctn−1) =ce−t, for all c∈Rt≥0. ThenΩ3 is bounded.
Proof. In fact x0 ∈ Ω3, means that x0 ∈ kerL i.e. x0(t) = c0tn−1 and λJ x0 = (1−λ)QN x0. Then we obtain
λc0= (1−λ)aZ ∞ 0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)ds .
Ifλ= 1, thenc0= 0. Otherwise, if|c0|> M∗/(n−1)!, in view of (3.3) one has λc20= (1−λ)ac0Z ∞
0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)ds
<0, which contradicts the fact thatλc20≥0. So|c0| ≤M∗/(n−1)!, moreover
kx0k=|c0| max
0≤i≤n−1
sup
t∈[0,∞)
e−t|(tn−1)(i)|
≤M∗Mn.
Therefore Ω3is bounded.
Lemma 3.6. Suppose that the second part of Condition(H3) holds. Let Ω3={x∈kerL:λJ x+ (1−λ)QN x= 0, λ∈[0,1]}
whereJ : kerL→ImQis the linear isomorphism given byJ(ctn−1) =ce−t, for all c∈R,t≥0. Then Ω3 is bounded, hereJ is as in Lemma 3.5.
Proof. Similar to the above argument, we can verify that Ω3 is bounded.
Lemma 3.7. Suppose that Ωis an open bounded subset of X such that dom(L)∩ Ω6=∅. Then N isL-compact onΩ.
Proof. Suppose that Ω⊂X is a bounded set. Without loss of generality, we may assume that Ω =B(0, r), then for anyx∈Ω,kxk ≤r. Forx∈Ω, and by condition (3.1), we obtain
|QN x| ≤ae−thZ ∞ 0
|f(s, x(s))|ds+ 1 ξn
Z ξ
0
(ξ−s)n|f(s, x(s))|dsi
≤ae−thZ ∞ 0
e−sα(s)|x(s)|+β(s)ds + 1
ξn Z ξ
0
(ξ−s)n(e−sα(s)|x(s)|+β(s))dsi
≤ae−th r
Z ∞
0
α(s)ds+ Z ∞
0
β(s)ds+r Z ξ
0
α(s)ds+ Z ξ
0
β(s)dsi
≤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 . Letx∈Ω, by (3.1) we have
kN xk1= Z ∞
0
|f s, x(s)|ds≤[rkαk1+kβk1]; (3.16) 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 ≤Mnk(I−Q)N xk1≤Mn[kN xk1+kQN xk1]
≤Mn[r(1 + 2a)kαk1+ (1 + 2a)kβk1].
It follows thatKP(I−Q)N(Ω) is uniformly bounded.
Let us prove thatT is equicontinuous. For anyx∈Ω and anyt1, t2∈[0, T] with t1< t2 andT ∈[0,∞), we have for 0≤i≤n−2:
e−t1(KP(I−Q)N x)(i)(t1)−e−t2(KP(I−Q)N x)(i)(t2)
=
Z t2
t1
[e−s(KP(I−Q)N x)(i)(s)]0ds
=
Z t2
t1
[−e−s(KP(I−Q)N x)(i)(s) +e−s(KP(I−Q)N x)(i+1)(s)]ds
≤2(t2−t1)kKP(I−Q)N xk
≤2(t2−t1)Mn[r(1 + 2a)kαk1+ (1 + 2a)kβk1]→0, ast1→t2. Fori=n−1, we obtain
e−t1(KP(I−Q)N x)(n−1)(t1)−e−t2(KP(I−Q)N x)(n−1)(t2)
= e−t1
Z t1
0
(I−Q)N x(s)ds−e−t2 Z t2
0
(I−Q)N x(s)ds
≤ Z t1
0
(e−t1−e−t2)|(I−Q)N x(s)|ds+ Z t2
t1
e−t2|(I−Q)N x(s)|ds
≤(t2−t1) Z t1
0
|(I−Q)N x(s)|ds+ Z t2
t1
|(I−Q)N x(s)|ds→0,
ast1→t2. SoKP(I−Q)N(Ω) is equicontinuous on every compact subinterval of [0,∞). In addition, we claim thatKP(I−Q)N(Ω) is equiconvergent at infinity. In fact, forx∈Ω,i= 0, . . . , n−1, we have
e−t(Kp(I−Q)N x)(i)(t)
≤ e−t (n−1−i)!
Z t
0
(t−s)n−1−i|(I−Q)N x(s)|ds
≤e−ttn−1−i Z t
0
|(I−Q)N x(s)|ds≤e−ttn−1−ik(I−Q)N xk1
≤e−ttn−1−i[kN xk1+kQN xk1]≤e−ttn−1−i(1 + 2a)[rkαk1+kβk1
, thus, limt→∞e−t(Kp(I−Q)N x)(i)(t) = 0, for everyi= 0, . . . , n−1, which means thatKP(I−Q)N(Ω) is equiconvergent at infinity.
Now we are able to give the proof of Theorem 3.1, which is an immediate con- sequence of Theorem 2.1 and the above lemmas.
Proof of Theorem 3.1. We shall prove that all conditions of Theorem 2.1 are satis- fied. Set Ω to be an open bounded subset of X such that∪3i=1Ωi ⊂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) Lx6=λN xpour tout (x, λ)∈[(domL\kerL)∩∂Ω]×(0,1);
(ii) N x /∈ImLpour toutx∈kerL∩∂Ω.
At last we prove that condition (iii) of Theorem 2.1 is satisfied. To this end, let H(x, λ) =±λJ x+ (1−λ)QN x
By the definition of Ω we know that Ω3 ⊂ Ω, thus H(x, λ) 6= 0 for every x ∈ kerL∩∂Ω. Then, by the homotopy property of degree, we obtain
deg(QN|kerL,Ω∩kerL,0) = deg(H(·,0),Ω∩kerL,0)
= deg(H(·,1),Ω∩kerL,0)
= deg(±J,Ω∩kerL,0)6= 0.
So, the third assumption of Theorem 2.1 is fulfilled and Lx=N x has at least one solution in domL∩Ω; i.e. (1.1)-(1.2) has at least one solution inX. 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
0
x(t)dt , (11.2) wheren≥3 is an integer,ξ >0 andf ∈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 equationLx=N x, whereL is a noninvertible operator. WhenLis 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 iso- lated 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 : domL ⊂X → Y be a linear operator which is Fredholm map of index zero, and letP :X →X,Q:Y →Y be continuous projectors such that ImP = kerL, kerQ= ImL. Then X = kerL⊕ kerP, Y = ImL⊕ImQ. It follows that L|domL∩kerP : domL∩kerP → ImL is invertible, we denote the inverse of that map byKP. Let Ω be an open bounded subset ofX such that domL∩Ω6=∅, the mapN :X →Y is said to beL-compact on Ω if the mapQN(Ω) is bounded andKP(I−QN) : Ω→X is compact.
Theorem 12.1 ([19]). Let L be a Fredholm operator of index zero and N be L- compact onΩ. Assume that the following conditions are satisfied:
(1) Lx6=λN xfor every(x, λ)∈[(domL\kerL)∩∂Ω]×(0,1).
(2) N x /∈ImLfor every x∈kerL∩∂Ω.
(3) deg(QN|kerL,Ω∩ ∩kerL,0) 6= 0, where Q : Y → Y is a projection such that ImL= kerQ.
Then the equationLx=N x has at least one solution indomL∩Ω.
Since the Arzel´a-Ascoli theorem fails in the noncompact interval case, we use the following result in order to show thatKP(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 inX.
(2) The functions belonging toF are equi-continuous on any compact interval of [0,∞).
(3) The functions from F are equi-convergent at+∞.
Let AC[0,+∞) denote the space of locally absolutely continuous functions on the interval [0,+∞). Let
X =
x∈Cn−1[0,+∞) :x(n−1)∈ACloc[0,+∞), lim
t→∞e−t|x(t)|exists endowed with the norm kxk = supt∈[0,+∞) e−t|x(t)|. Let Y = L1[0,+∞) with normkyk1=R+∞
0 |y(t)|dt.
Define the operatorL: domL⊂X→Y byLx=x(n), where domL=
x∈X :x(i)(0) = 0, i= 0, n−2, x(n−1)(∞) = n!
ξn Z ξ
0
x(t)dt . 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 asLx=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−tα(t)|x|+β(t). (13.1) (H2) There exists a constantM >0, such that forx∈domL, if|x(n−1)(t)|> M,
for allt∈[0,∞), then Z ∞
0
f(s, x(s))ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, x(s))ds6= 0. (13.2) (H3) There exists a constant M∗ >0, such that for any x(t) = c0tn−1 ∈kerL
with|c0|> M∗/(n−1)!, either c0
hZ ∞
0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)ds
<0, (13.3) or
c0hZ ∞ 0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)dsi
>0. (13.4) Then (11.1)-(11.2), has at least one solution in C[0,∞), provided
1−2Mnkαk1>0, (13.5)
whereMn= supt∈[0,∞)e−ttn−1= (n−1e )n−1.
To prove Theorem 13.1, we need to prove some Lemmas.
Lemma 13.2. The operator L: domL⊂X →Y is a Fredholm operator of index zero. Furthermore, the linear projector operatorQ:Y →Y can be defined by
Qy(t) =ae−thZ ∞ 0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)dsi ,
where
1/a= 1−
n
X
k=0
(−1)k n!
(n−k)!ξk
and the linear operatorKP : ImL→domL∩kerP can be written as Kpy(t) = 1
(n−1)!
Z t
0
(t−s)n−1y(s)ds, y∈ImL.
Furthermore
kKpyk ≤ Mn
(n−1)!kyk1, for everyy∈ImL. (13.6) Proof. It is clear that
kerL={x∈domL:x=ctn−1, c∈R, t∈[0,∞) . Now we show that
ImL= y∈Y :
Z ∞
0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)ds= 0 . (13.7) The problem
x(n)(t) =y(t) (13.8)
has a solution x(t) that satisfies the conditionsx(i)(0) = 0, fori= 0,1, . . . , n−2, andx(n−1)(∞) = ξn!n
Rξ
0 x(t)dtif and only if Z ∞
0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)ds= 0. (13.9) In fact from (13.8) and the boundary conditions (11.2) we have
x(t) = 1 (n−1)!
Z t
0
(t−s)n−1y(s)ds+c0+c1t+c2t2+· · ·+cn−1tn−1
= 1
(n−1)!
Z t
0
(t−s)n−1y(s)ds+ctn−1. Fromx(n−1)(∞) = ξn!n
Rξ
0 x(t)dt, we obtain Z ∞
0
y(s)ds= 1 ξn
Z ξ
0
(ξ−s)ny(s)ds.
On the other hand, if (13.9) holds, setting x(t) = 1
(n−1)!
Z t
0
(t−s)n−1y(s)ds+ctn−1
where c is an arbitrary constant, then x(t) is a solution of (13.8). Hence (13.7) holds. Setting
Ry= Z ∞
0
y(s)ds− 1 ξn
Z ξ
0
(ξ−s)ny(s)ds, defineQy(t) =ae−tRy, it is clear that dim ImQ= 1. We have
Q2y=Q(Qy) =ae−t(a.Ry)(
Z ∞
0
e−sds− 1 ξn
Z ξ
0
(ξ−s)ne−sds)
=ae−tRy=Qy,
that implies the operatorQis a projector. Furthermore, ImL= kerQ.
Lety = (y−Qy) +Qy, where y−Qy ∈kerQ= ImL, Qy ∈ImQ. It follows from kerQ = ImL and Q2y = Qy that ImQ∩ImL = {0}. Then, we have Y = ImL⊕ImQ. Thus dim kerL= 1 = dim ImQ= codim ImL= 1, this means
thatL is a Fredholm operator of index zero. Now we define a projectorP fromX toX by setting
P x(t) = x(n−1)(0) (n−1)! tn−1.
Then the generalized inverseKP : ImL→domL∩kerP ofLcan be written as Kpy= 1
(n−1)!
Z t
0
(t−s)n−1y(s)ds.
Obviously, ImP = kerLand P2x=P x. It follows fromx= (x−P x) +P x that X = kerP + kerL. By simple calculation, we obtain that kerL∩kerP = {0}.
HenceX = kerL⊕kerP.
From the definitions ofP andKP, it is easy to see that the generalized inverse ofLisKP. In fact, fory∈ImL, we have
(LKp)y(t) = (Kpy(t))(n)=y(t), and forx∈domL∩kerP, we know that
(KpL)x(t) = (Kp)x(n)(t) = 1 (n−1)!
Z t
0
(t−s)n−1x(n)(s)ds
=x(t)−[x(0) +x0(0)t+. . . .x(n−2)(0)
(n−2)! tn−2+x(n−1)(0) (n−1)! tn−1].
In view ofx∈domL∩kerP,x(i)(0) = 0, fori= 0,1, . . . , n−2, andP x= 0, thus (KpL)x(t) =x(t).
This shows thatKp= (L|domL∩kerP)−1. From the definition ofKp, we have kKpyk= sup
t∈[0,∞)
e−t|Kpy| ≤ sup
t∈[0,∞)
e−t (n−1)!
Z t
0
(t−s)n−1|y(s)|ds
< Mn
(n−1)!
Z ∞
0
|y(s)|ds= Mn
(n−1)!kyk1.
This completes the proof.
Lemma 13.3. Let Ω1 = {x ∈ domL\kerL : Lx = λN xfor someλ ∈ [0,1]}.
ThenΩ1 is bounded.
Proof. Suppose thatx∈Ω1, andLx=λN x. Thusλ6= 0 and QN x= 0, so that Z ∞
0
f(s, x(s))ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, x(s))ds= 0.
Thus, by condition (H2), there exists t0 ∈ R+, such that |x(n−1)(t0)| ≤ M. It follows from the absolute continuity ofx(n−1)that
|x(n−1)(0)|=
x(n−1)(t0)− Z t0
0
x(n)(s)ds ,
then, we have
|x(n−1)(0)| ≤M+ Z ∞
0
|Lx(s)|ds≤M+ Z ∞
0
|N x(s)|ds=M +kN xk1. (13.10)
Again for x ∈Ω1 and x∈ domL\kerL, we have (I−P)x∈ domL∩kerP and LP x= 0; thus from Lemma 13.2,
k(I−P)xk=kKpL(I−P)xk
≤ Mn
(n−1)!kL(I−P)xk1
= Mn
(n−1)!kLxk1≤ Mn
(n−1)!kN xk1.
(13.11)
So
kxk ≤ kP xk+k(I−P)xk=Mn|x(n−1)(0)|+ Mn
(n−1)!kN xk1, (13.12) again from (13.10) and (13.11), (13.12) becomes
kxk ≤MnM +MnkN xk1+ Mn
(n−1)!kN xk1≤MnM+ 2MnkN xk1. (13.13) On the other hand by (13.1) we have
kN xk1= Z ∞
0
|f(s, x(s))|ds≤ kxkkαk1+kβk1. (13.14) Therefore, (13.13) and (13.14), it yield
kxk ≤MnM+ 2Mnkxkkαk1+ 2Mnkβk1; since 1−2Mnkαk1>0, we obtain
kxk ≤ MnM 1−2Mnkαk1
+ 2Mnkβk1
1−2Mnkαk1
.
So Ω1is bounded.
Lemma 13.4. The setΩ2={x∈kerL:N x∈ImL} is bounded.
Proof. Let x ∈ Ω2, then x∈ kerL implies x(t) = ctn−1, c ∈ R, and QN x = 0;
therefore
Z ∞
0
f(s, csn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, csn−1)ds= 0.
From condition (H2), there existst1∈R+, such as |x(n−1)(t1)| ≤M. We have (n−1)!|c| ≤M
so|c| ≤ (n−1)!M . On the other hand kxk= sup
t∈[0∞)
e−t|x(t)|=|c| sup
t∈[0∞)
e−ttn−1=|c|Mn,
i.e. kxk ≤ (n−1)!MnM <∞, so Ω2 is bounded.
Lemma 13.5. Suppose that the first part of Condition(H3)holds. Let Ω3={x∈kerL:−λJ x+ (1−λ)QN x= 0, λ∈[0,1]}
where J : kerL→ ImQ is the linear isomorphism given byJ(ctn−1) =ctn−1, for allc∈Rt≥0. ThenΩ3 is bounded.
Proof. In fact x0 ∈ Ω3, means that x0 ∈ kerL i.e. x0(t) = c0tn−1 and λJ x0 = (1−λ)QN x0. Then we obtain
λc0tn−1= (1−λ)ae−tZ ∞ 0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)ds .
Ifλ= 1, thenc0= 0. Otherwise, if|c0|> M∗, in view of (13.3) one has λc20tn−1= (1−λ)ae−tc0(
Z ∞
0
f(s, c0sn−1)ds− 1 ξn
Z ξ
0
(ξ−s)nf(s, c0sn−1)ds)<0, which contradicts the fact thatλc20≥0. So|c0| ≤M∗, moreover
kx0k= supe−t|c0|tn−1=|c0|Mn≤M∗Mn.
Therefore Ω3is bounded.
Lemma 13.6. Suppose that the second part of Condition(H3) holds. Let Ω3={x∈kerL:λJ x+ (1−λ)QN x= 0, λ∈[0,1]}
where J : kerL→ ImQ is the linear isomorphism given byJ(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 thatΩ3 is bounded.
Lemma 13.7. Suppose thatΩis an open bounded subset ofX such thatdom(L)∩ Ω6=∅. Then N isL-compact onΩ.
Proof. Suppose that Ω⊂X is a bounded set. Without loss of generality, we may assume that Ω =B(0, r), then for anyx∈Ω,kxk ≤r. Forx∈Ω, and by condition (13.1), we obtain
e−t|QN x| ≤ae−2thZ ∞ 0
|f(s, x(s))|ds+ 1 ξn
Z ξ
0
(ξ−s)n|f(s, x(s))|dsi
≤ae−2thZ ∞ 0
e−sα(s)|x(s)|+β(s)ds + 1
ξn Z ξ
0
(ξ−s)n(e−sα(s)|x(s)|+β(s))dsi
≤ae−2th r
Z ∞
0
α(s)ds+ Z ∞
0
β(s)ds+r Z ξ
0
α(s)ds+ Z ξ
0
β(s)dsi
≤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. Forx∈Ω, by (13.1) we have
kN xk1= Z ∞
0
|f s, x(s)|ds≤[rkαk1+kβk1]; (13.16) on the other hand, from the definition ofKP and together with (13.6), (13.15) and (13.16) one gets
kKP(I−Q)Nk ≤Mnk(I−Q)Nk1≤Mn[kN xk1+kQN xk1]
≤Mn[r(1 + 2a)kαk1+ (1 + 2a)kβk1].
It follows thatKP(I−Q)N(Ω) is uniformly bounded.
Let us prove thatT is equicontinuous. For anyx∈Ω and anyt1, t2∈[0, T] with t1< t2 andT ∈[0,∞), we have
|e−t1KP(I−Q)N x(t1)−e−t2KP(I−Q)N x(t2)|
= 1
(n−1)!
Z t1
0
e−t1(t1−s)n−1(I−Q)N x(s)ds
− Z t2
0
e−t2(t2−s)n−1(I−Q)N x(s)ds
≤ 1
(n−1)![ Z t1
0
e−t2(t2−s)n−1−e−t1(t1−s)n−1|(I−Q)N x(s)|ds +
Z t2
t1
e−t2(t2−s)n−1|(I−Q)N x(s)|ds]
≤ 1
(n−1)!
hZ t1
0
(e−(t2−s)(t2−s)n−1−e−(t1−s)(t1−s)n−1)
× e−s|(I−Q)N x(s)|ds +
Z t2
t1
e−(t2−s)(t2−s)n−1e−s|(I−Q)N x(s)|dsi
≤ 1
(n−1)![Mn0(t2−t1) Z t1
0
e−s|(I−Q)N x(s)|ds
+e−t2(t2−t1)n−1 Z t2
t1
|(I−Q)N x(s)|dsi
→0, ast1→t2.
SoKP(I−Q)N(Ω) is equicontinuous on every compact subset of [0,∞). In addition, we claim thatKP(I−Q)N(Ω) is equiconvergent at infinity. In fact,
|e−tKp(I−Q)N x(t)|
≤ 1
(n−1)!
Z t
0
e−(t−s)(t−s)n−1e−s|(I−Q)N x(s)|ds
≤ Mn
(n−1)!
Z t
0
|(I−Q)N x(s)|ds≤ Mn
(n−1)!k(I−Q)N xk1
≤ Mn
(n−1)![kN xk1+kQN xk1]<∞;
thus, limt→∞|e−tKp(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=1Ωi ⊂ Ω. We know thatL is a Fredholm operator of index zero and N is L-compact on Ω. By the definition of Ω we have
(i) Lx6=λN xpour tout (x, λ)∈[(domL\kerL)∩∂Ω]×(0,1);
(ii) N x /∈ImLpour toutx∈kerL∩∂Ω.
At last we prove that condition (iii) of Theorem 12.1 is satisfied. To this end, let H(x, λ) =±λJ x+ (1−λ)QN x
By the definition of Ω we know that Ω3 ⊂ Ω, thus H(x, λ) 6= 0 for every x ∈ kerL∩∂Ω. Then, by the homotopy property of degree, we obtain
deg(QN|kerL,Ω∩ ∩kerL,0) = deg(H(·,0),Ω∩ ∩kerL,0)
= deg(H(·,1),Ω∩ ∩kerL,0)
= deg(±J,Ω∩ ∩kerL,0)6= 0.
So, the third assumption of Theorem 12.1 is fulfilled andLx=N xhas at least one solution in domL∩Ω; i.e. (11.1)-(11.2) has at least one solution inX. 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:[email protected]
Assia Guezane-Lakoud
Laboratory of Advanced Materials, Faculty of Sciences, University Badji Mokhtar- Annaba, P.O. Box 12, 23000, Annaba, Algeria
E-mail address:a [email protected]
Rabah Khaldi
Laboratory of Advanced Materials. Faculty of Sciences, University Badji Mokhtar- Annaba, P.O. Box 12, 23000, Annaba, Algeria
E-mail address:[email protected]
