SOLUTIONS
CEZAR AVRAMESCU Received 9 November 2001
The problem ˙x = f(t, x),x(−∞)=x(+∞), wherex(±∞) :=limt→±∞x(t)∈Rn, is considered. Some existence results for this problem are established using the fixed point method and topological degree theory.
1. Introduction
Let f :R×Rn →Rn be a continuous function; consider the boundary value problem
˙
x=f(t, x), x(−∞)=x(+∞), (1.1) where
x(±∞) := lim
t→±∞x(t)∈Rn. (1.2)
The solutions of problem (1.1) are often called, by Poincar´e,homoclinic solu- tions. They appear in certain celeste mechanics and cosmogony problems.
Problem (1.1) can be considered as a generalization of the boundary value problem
˙
x=f(t, x), x(a)=x(b), (1.3)
whena→ −∞andb→+∞.
The boundary value problems on compact intervals have been studied in nu- merous papers but the boundary value problems on noncompact intervals have been less studied. A first substantial approach of these problems, using func- tional methods are due to Kartsatos [8]. Last time, this type of results has been published in [2,3,4,5,6].
For problem (1.3), Mawhin obtained many existence results through topolog- ical degree theory; in [9,10,11] the reader can find the fundamental ideas of the
Copyright©2002 Hindawi Publishing Corporation Abstract and Applied Analysis 7:1 (2002) 1–27
2000 Mathematics Subject Classification: 34B40, 37C29, 47H11 URL:http://dx.doi.org/10.1155/S108533750200074X
method developed by Mawhin, the main results, and a rich bibliography in this field. Some approaches of Mawhin dedicated to problem (1.3) can be adjusted for problem (1.1).
The present paper is dedicated to the existence of solutions for problem (1.1);
the used method will be the reduction of problem (1.1) to a fixed point problem for a convenient operator defined in a suitable functional space. Such a space is
Cl:=
x:R−→Rn,∃x(±∞)∈Rn
. (1.4)
Section 2deals especially to praise the main properties of the spaceCl. The specified isomorphism betweenClandC([a, b],Rn) permits to obtain a com- pactness criterion inCl (see [1]). We define in Cl the notion of an associated operator to problem (1.1) and indicate the construction method of such opera- tor together with its main properties. An associated operator for problem (1.1) is an operator whose fixed points are solutions for (1.1).
InSection 3, assuming the existence and uniqueness onRof the solutions for the problem
˙
x=f(t, x), x(0)=y, (1.5)
one builds up associated operators mapping inRn; consequently, their topolog- ical degree will be a Brouwer one.
In Section 4, the continuation method is presented (see Proposition 4.1).
Through this method we obtain existence results for perturbed equations. The starting equation is chosen such that the topological degree of its associated op- erator is easy to be evaluated, and the perturbation is done through homoge- neous or “small” functions.
For further details about the construction of the associated operators, the reader can consult [12]. For the topological degree theory we recommend the delightful book [13].
2. General hypotheses and preliminary results
2.1. Introduction. Let f :R×Rn →Rn be a continuous mapping; consider problem (1.1) wherex(±∞) :=limt→±∞x(t)∈ Rn (notation used throughout this paper).
It is clear from the introduction that the aim of this paper is to find sufficient conditions to assure the existence of solutions for problem (1.1). The method will be the reduction of the existence solutions for problem (1.1) to the existence of fixed points for an adequate operator which maps in an adequate functional space.
In this section, we present the principal function spaces, their main proper- ties, the notations, and the principal theoretical results needed in what follows.
2.2. Function spaces. Denote by|·|an arbitrary norm inRnand Cc:=
x:R−→Rn, xcontinuous. (2.1) As is well known,Cc is a Fr´echet space endowed with the uniform conver- gence on compact subsets ofRwith the usual topology. LetCc1denote the linear subspace ofC1functions inCc.
The principal function spaces are Cl:=
x∈Cc,∃x(±∞)∈Rn ,
Cll:=x∈Cl, x(−∞)=x(+∞), (2.2) whereClandCllare Banach spaces with respect to the norm
x∞:=sup
t∈R
x(t), (2.3)
whereRnwill be identified naturally with the constant functions subspace. Con- siderC1l :=Cl∩Cc1,Cll1:=Cll∩C1c.
Another function space, interesting only as linear space, is the space of all Riemann integrable functions onR,
CR:= x∈Cc;
+∞
−∞x(t)dt <+∞
, (2.4)
where
+∞
−∞x(t)dt:= lim
A→−∞
0 A
x(t)dt+ lim
A→+∞
A 0
x(t)dt. (2.5)
Remark 2.1. A function x of class C1 belongs to Cl if and only if ˙x belongs toCR.
Finally, we use the spaces C(a,b):=
x: [a, b]−→RN, xcontinuous,
C[a,b]:=x∈C(a,b), x(a)=x(b), (2.6) endowed with the usual norm
x:= sup
t∈[a,b]
x(t). (2.7)
In the case of a Banach spaceX, whereX=ClorX=Cll, set B(ρ) :=
x∈X,x∞< ρ, Σ(ρ) :=
x∈Rn,|x|< ρ. (2.8)
2.3. Properties of the spaceCl. We state certain properties of the spaceCl. Proposition2.2. The spacesClandC(a,b)are isomorphic.
Proof. Indeed, considerϕ: (a, b)→Ra continuous and bijective mapping; de- fine the mappingΦ:Cl→C(a,b)by the equality
(Φx)(t) :=
xϕ(t), ift∈(a, b), x(−∞), ift=a, x(+∞), ift=b.
(2.9)
It is clear thatΦis an isometric isomorphism and the proof ends.
Remark 2.3. The same mappingΦis an isomorphism betweenCllandC[a,b]. The property inProposition 2.2allows us to obtain a compactness criterion inCl; obviously, it will work inClltoo, sinceCllis a closed subspace ofCl. Definition 2.4. A familyA⊂Clis called equiconvergent if and only if
∀ε >0, ∃T=T(ε)>0, ∀x∈A,∀t1, t2∈R, t1t2>0, ti> T(ε), xt1
−xt2< ε. (2.10)
Proposition2.5. A familyA⊂Clis relatively compact if and only if the following three conditions are fulfilled:
(i)Ais uniformly bounded onR;
(ii)Ais equicontinuous on every compact interval ofR; (iii)Ais equiconvergent.
Proposition 2.5results immediately from the fact that the isomorphismΦ given by (2.9) transforms a setA, satisfying conditions (i), (ii), and (iii), into an equicontinuous and uniformly bounded set inC(a,b).
Definition 2.6. A familyA⊂Ccis calledCR-bounded if and only if there exists a functionα:R→R,α(t)≥0 for everyt∈R,α∈CR, such that
∀x∈A, t∈R, x(t)≤α(t). (2.11) Corollary2.7. A familyA⊂Cl∩C1c, uniformly bounded onRhaving the family of derivativesCR-bounded, is relatively compact inCl.
2.4. Operators. The first operator is the Nemitzky operator,F:Cc→Ccgener- ated by the continuous function f :R×Rn→Rnand defined by
(Fx)(t) :=ft, x(t). (2.12) Taking into accountRemark 2.1, it results that for every solutionxof (1.1) it holds
x∈Cl⇐⇒Fx∈CR. (2.13)
Similarly, for every solutionxof (1.1), x∈Cll⇐⇒
+∞
−∞(Fx)(s)ds=0. (2.14)
In what follows,X⊆Cldenotes a closed subspace ofClandD⊂X is a void set. Define onDan important category of operators calledassociated.
Definition 2.8. The operatorU :D⊂X→X isassociatedto problem (1.1) on the setDif and only if every fixed point ofUis a solution for problem (1.1).
By using the formula of a solution for (1.1), it is naturally, in the building of the operatorU, to admit
FD⊂CR. (2.15)
Remark, in addition, that ifU maps inCl, then only the fixed points satisfy condition (1.1) and ifD⊂Cllthen we haveUD⊂Cll.
ByRemark 2.1, we can easily obtain associated operators to problem (1.1).
Such an operator is, for example, (Ux)(t) :=x(b) +α(t)
+∞
−∞(Fx)(s)ds+ +∞
b
(Fx)(s)ds, (2.16) whereb∈R¯, andα:R→Ris an arbitrary continuous function withα(b)=0, andα∈Cl; this operator maps inCl. If, in addition,α(−∞)=α(+∞), thenUCl⊂ Cll.
Another possibility to construct associated operators inCll is the next: we search a linear and continuous operatorT :Cll→Cll such that the operator Lx:=x˙+Txis invertible; then
U=L−1(F+T). (2.17)
Examples of such operatorsTareTx=θ(·)x(0) orT=θ(·)x, whereθ:R→R is a continuous and strictly positive mapping with−∞+∞θ(t)dt=1.
There exist general procedures to build up the associated operators, like the one from below having a pure algebraic character.
LetX andZbe two linear spaces,L:D(L)⊂X →Za linear operator, and N:D(N)⊂X→Zan arbitrary operator.
If dimN(L)=codimR(L)<∞andP:X→X,Q:Z→Zare two projectors such thatR(P)=N(L),N(Q)=R(L), thenx∈Xis a solution for the equation
Lx=Nx (2.18)
if and only ifxis a fixed point for the operator
U=P+aQN+K(I−Q)N, (2.19)
whereIis the identity operator inZ,a∈R,a=0, andKis the right inverse ofL (more preciselyK=(L|D(L)∩N(P))−1).
This result has been successfully used by Mawhin for the building of associ- ated operators to the boundary value problem (closely related to the periodic solutions problem [9,10,11])
˙
x=f(t, x), x(0)=x(T). (2.20)
By this model in the next subsection, we briefly describe how can we construct associated operators to problem (1.1) and their properties.
2.5. Construction of associated operators. The form of associated operators depends firstly on the fundamental spaceXand next on the spaceZand on the choice of the operatorsL,Nand the choice of the projectorsP,Q; only after this Kcan be determined and also the final form of the operatorU. Having so many arbitrary elements we can find many associated operators.
In what follows, we sketch the building of associated operators in two im- portant cases:X=ClandX=Cll; further details about the construction can be found in [4,12].
In the caseX=Clwe distinguish three subcases related toLandN; this choice must be made such that the equation (L, N) does contain (1.1). The expression of projectorsPandQdepends on the considered case.
In all three cases, we haveZ=CR×Rn,D(L)=C1l. 2.5.1. The caseL1.
Lx=
˙
x, x(+∞), Nx=
Fx, x(−∞). (2.21) In this caseP=Q=0, so the operatorLis invertible and thereforeU=L−1N.
This case gives us the easiest associated operators, Ux=x(+∞) +
(·)
−∞(Fx)(s)ds (2.22)
and the symmetric form
Ux=x(−∞) + (·)
+∞(Fx)(s)ds. (2.23)
2.5.2. The caseL2.
Lx=( ˙x,0), Nx=Fx, x(+∞)−x(−∞). (2.24) In this case, sinceR(L)=CR×{0}the projectorQmay be
Q(y, c)=(0, c). (2.25)
ForP, we take
Px=x(b), x∈R¯, (2.26)
or
Px= +∞
−∞e(t)x(t)dt, (2.27)
where
e:R−→R, econtinuous, +∞
−∞e(s)ds=1. (2.28) ForU, we can construct
Ux=x(b) +ax(+∞)−x(−∞)+ (·)
b
(Fx)(s)ds (2.29) or
Ux=ax(+∞)−x(−∞) +
+∞
−∞
x(s)−1
2(Fx)(s)
e−2|s|ds+ (·)
0
(Fx)(s)ds. (2.30) 2.5.3. The caseL3.
Lx=x, x(+˙ ∞)−x(−∞), Nx=(Fx,0). (2.31) In this case,
R(L)=
(y, c)∈CR×Rn|c= +∞
−∞y(s)ds
(2.32) and hence the projectorQmust be changed; we can take for example
Q(y, c)= 0, c−
+∞
−∞y(s)ds
(2.33) and therefore,
Ux=x(b) +a +∞
−∞(Fx)(s)ds+ (·)
b
(Fx)(s)ds (2.34)
and other more complicated forms.
In the caseX=Cll, we haveD(L)=C1ll,Z=CR,Lx=x˙and consequently, R(L)=
y∈CR,
+∞
−∞y(s)ds=0
. (2.35)
We can take, for example,
(Qy)(t)=e(t) +∞
−∞y(s)ds. (2.36)
In general, in this case the expression ofU is more complicated since all its values must be inCll. For example, for (2.26) we get
(Ux)(t)=x(b) +
ae−2|t|− t
b
e−2|s|ds
· +∞
−∞(Fx)(s)ds+ t
b
(Fx)(s)ds (2.37) and with (2.27), wheree(t)=e−2|t|,
(Ux)(t)= t
0
e−2|s|ds+ae−2|t|− t
0
e−2|s|
1−1
2e−2|s|
ds
−1 2
+∞
−∞e−2|s|(Fx)(s)ds +∞
−∞(Fx)(s)ds+ t
0
(Fx)(s)ds.
(2.38)
2.6. Admissible operators. It is obvious that this construction of the associated operators has an algebraic character; the condition
FD⊂CR (2.39)
is sufficient for the existence of these operators, but it is not sufficient to confer their important topological properties.
Definition 2.9. An associated operator on the setD⊂X to problem (1.1), con- structed as inSection 2.5, is calledadmissibleif and only ifU :D⊂X→Xis compact.
Proposition2.10. LetXbe a subspace ofClandD⊂Clbe a bounded subset. If FDisCR-bounded, then every associated operator constructed as inSection 2.5is compact.
The proof of this proposition is complicated in calculus, but it is basically an easy application of the elementary known properties of uniform convergence, which allows to establish immediately the continuity of the operatorU which contains finite rank projectors and application of type
x−→
+∞
−∞(Fx)(s)ds, x−→
(·)
b
(Fx)(s)ds, b∈R¯. (2.40)
The compactness of the operatorUis an immediate consequence ofCorollary 2.7. At least for the operators U given by (2.22), (2.23), (2.29), (2.30), (2.34), (2.37), and (2.38) the verification of compactity is immediate.
Remark that if f satisfies the condition
f(t, x)≤θ(t)·β|x|, (2.41) whereθ∈CR,θ≥0,β≥0,β:R→Ris continuous, thenFDisCR-bounded for every bounded setD⊂X; indeed, we have
(Fx)(t)≤ρθ(t), (2.42)
where
ρ:=supβ(u),|u| ≤r, r:=supx∞, x∈D. (2.43) The situation is more complicated in the case when (1.1) proceeds from a second-order equation
¨
y=ht, y,y˙, (2.44)
whereh:R×Rk×Rk→Rkis a continuous function. Substituting (2.44) in (1.1), where
x=x1, x2
, x1=y, x2=y,˙ f(t, x)=x2, ht, x1, x2
, (2.45)
then
x∈Cl⇐⇒y, y˙∈Cl. (2.46)
Since
y∈Cl⇐⇒y˙∈CR, (2.47)
it results that
y˙∈Cl∩CR (2.48)
and hence
t→±∞lim y(t)˙ =0. (2.49)
Therefore, the boundary value problem defining the homoclinic solutions for (2.34) has the form
¨
y=ht, y,y˙, y(−∞)=y(+∞), y(˙ −∞)=y(+˙ ∞)=0. (2.50) We give an example to obtain theCR-boundedness ofF(D) in this case.
Let α1, α2 ∈ CR, α1, α2 be positive; in addition, suppose that α2(±∞)= 0.
Letγ, β:R→Rbe two continuous and positive functions. We take as funda- mental spaceX=Cl×CRl, whereCRl:=CR∩Cl={x∈CR, x(±∞) =0}.
LetD1be a bounded set inCRand letD=D1×D2, where D2=
x∈CR,x(t)≤α2(t), t∈R
. (2.51)
It is easy to check that if
ht, x1, x2≤k1α1(t)γx1+k2x2βx2, (2.52) thenF(D) isCR-bounded (more precisely,CR×CR-bounded).
The case of second-order equation is different from the first-order equation;
this is why it will not be treated here, but it will make the object of a future note.
2.7. Remarks on the topological degree of the admissible operators. LetΩ⊂X be an open and bounded set, whereXisClorCll.
Suppose thatF(Ω) isCR-bounded, for an admissible operatorU, if
x=Ux, x∈∂Ω, (2.53)
where∂Ωis the boundary ofΩ, we can consider its topological degree
deg(I−U,Ω,0). (2.54)
If this degree is nonzero, thenUadmits fixed points and so problem (1.1) has solutions.
As we said, the results contained in this section are based on the ones by Mawhin related to the boundary value problem
˙
x=f(t, x), x(0)=x(T). (2.55)
This author proves that the associated operators to problem (2.55) in the spaceC(0,T)orC[0,T]are compact on the bounded sets without supplementary conditions on the mapping f as it was to be expected. Moreover, these operators have the same topological degree which does not depend on the choice ofL,N, P,Q. In addition, if in particularf(t, x)=g(x), then for each associated operator Uto problem (2.55) on the bounded and open setΩfromC(0,T)(orC[0,T]), we have
deg(I−U,Ω,0)=(−1)ndegBg,Ω∩Rn,0, (2.56) where degBdenotes the Brouwer degree.
The associated operators to problem (1.1) onΩfromCl orCllhave the de- grees invariant with respect toL,N,P,Q; the proof, based on the invariance of topological degree to homeomorphisms, is essentially simple but complicated to achieve. As we do not use this property in the present paper, we renounce to its proof.
Finally we make only a remark on the isomorphismΦgiven by (2.9).
LetΩ⊂Xbe an open and bounded set inX(X=ClorCll) and letUbe an admissible operator onΩfor problem (1.1) fulfilling (2.56).
Set
ΩΦ:=Ω(Φ), UΦ:=ΦUΦ−1, (2.57) whereΦis given by (2.9) witha=0,b=T.
ThenΩΦis open and bounded,Φ(∂Ω)=∂ΩΦandΩΦ⊂C(0,T)(resp.,ΩΦ⊂ C[0,T]).
Furthermore,UΦis compact and since∂ΩΦ=Φ(∂Ω), we have
x=Ux, x∈∂Ω⇐⇒x=UΦ, x∈∂ΩΦ. (2.58) Hence
deg(I−U,Ω,0)=degI−UΦ,ΩΦ,0. (2.59) 3. Existence results in the hypothesis of uniqueness of solutions
3.1. Introduction. Letf :R×Rn→Rnbe a continuous function; consider again problem (1.1).
We research the existence of solutions for problem (1.1) in the hypothesis that the Cauchy problem
˙
x=f(t, x), x(0)=y (3.1)
has a unique solution defined on the whole real axisR, for everyGa bounded set inRnand for everyy∈G; denote the solution of (3.1) by
x(t;y), y∈G. (3.2)
The uniqueness condition is fulfilled in particular if f(t, x) is locally Lipschitz with respect tox. Condition (2.41) is sufficient to assure the existence onRof the solution (3.2), it is in particular fulfilled in conditions of type (2.41) and even more general.
It is known that the uniqueness condition assures the continuous dependence of the functionx(t;·); this property would be stated as: for every [a, b]⊂Rand for everyyn∈G,yn→y∈G, the sequencex(t;yn) converges uniformly on [a, b]
tox(t;y).
In this section, we present certain existence results for problem (1.1), exploit- ing this continuous dependence with respect to initial data.
3.2. Generalized Poincar´e operator. LetΩ⊂Clbe a bounded and open set; let G:=
y∈Rn, x(·;y)∈Ω. (3.3) Obviously,Gis a bounded and open set.
Theorem3.1. Suppose that (i)FΩ¯ isCR-bounded;
(ii)for everyy∈∂Gand for everyt >0
x(t;y)=x(−t;y); (3.4)
(iii)for everyy∈∂G
f(0, y)=0; (3.5)
(iv)for everyy∈∂G
degBf(0, y), G,0=0. (3.6) Then problem (1.1) has solutions inΩ.¯
Proof. By hypothesis (i) it results thatx(·;y)∈Cl, for everyy∈G; set¯ P y=1
2
x(+∞;y)−x(−∞;y) (3.7) (we callPthegeneralized Poincar´e operator). It is easy to check that the solution x(·;y)∈Cllif and only ifP y=0.
We want to show thatP: ¯G→G¯is continuous; for this aim we remark that P y=1
2 +∞
−∞ fs, x(s;y)ds. (3.8)
By hypothesis (i) it results that the integral in (3.8) is uniformly convergent with respect toy∈G; on the other hand, since the mapping¯ y→x(·;y) is con- tinuous (as mentioned in the previous paragraph) we conclude the continuity of the mapping (t, y)→ f(t, x(t;y)) on every set of type [−A, A]×G. Hence the¯ mappingy→P yis continuous on ¯G.
Define the applicationh: ¯G×[0,1]→Rnby
h(y, λ) :=
1 2λ
x λ
1−λ;y−x λ
λ−1;y, λ∈(0,1), y∈G,¯
P y, λ=1,
f(0, y), λ=0.
(3.9)
By L’Hospital rule,
limλ↓0h(y, λ)=f(0, λ). (3.10) Since
limλ↑1h(y, λ)=P y, (3.11)
it follows thathis continuous.
If fory∈∂Gwe haveP y=0, thenx(·;y) is a solution for (1.1).
Suppose thenP y=0, for everyy∈∂G; by hypotheses (ii) and (iii) it results that
h(y, λ)=0, ∀λ∈[0,1],∀y∈∂G. (3.12) By homotopic invariance property of the topological degree it results that degB(h(·, λ), G,0) is constant forλ∈[0,1]; in particular,
degBh(·,0), G,0=degBh(·,1), G,0, (3.13) that is,
degB(P, G,0)=degBf(·, y), G,0 (3.14) and hence, by (3.6)
degB(P, G,0)=0, (3.15)
which assures the existence ofy∈GwithP y=0. The theorem is proved.
3.3. The caseΩconnected. The advantage of the previous result is that the topological degrees appearing are Brouwer degrees; the drawback is that con- dition (3.4) is not easy to be checked. We state now another existence result.
As usual, suppose thatΩ⊂Clis a bounded and open set; define on ¯Ωthe operators
Hx= (·)
0
(Fx)(s)ds, S=I−H. (3.16)
Lemma3.2. If
(i) f :R×Rn→Rnis locally Lipschitz with respect to the second variable;
(ii)F( ¯Ω)isCR-bounded, thenS: ¯Ω→Clis injective.
Proof. Letx,z∈Ω¯ such that
S(x)=S(z). (3.17)
Ifx=z, then there existst0∈Rsuch thatx(t0)=z(t0); we can assume that t0>0. LetA >0 be such thatt0∈[0, A] andr=max{x∞,y∞}.
Since
∃Lr>0,∀u, v∈Σ(r), f(t, u)−f(t, v)≤Lr|u−v|, x(t)−z(t)≤Lr
t
0
u(s)−v(s)ds, t∈[0, A], (3.18)
we obtain by using Gronwall’s lemma
x(t)=z(t), ∀t∈[0, A]. (3.19)
Remark 3.3. The mappingS: ¯Ω→S( ¯Ω) is a homeomorphism. In addition, since His a compact operator,S−1is a compact perturbation of identity, too.
Remark 3.4. Ify∈Rn∩S( ¯Ω), then
S−1y=x(·;y). (3.20)
Set
Πx:=x(0)−Px. (3.21)
Observe that the operator
U:=Π+H (3.22)
is just the admissible operator (2.22), whereb=0 anda=1/2.
Remark 3.5. The following identity holds:
I−U=
I−ΠS−1S. (3.23)
Theorem3.6. Assume that the following hypotheses are fulfilled:
(i) f :R×Rn→Rnis locally Lipschitz;
(ii)Ω⊂Clis a connected, open, and bounded set;
(iii)F( ¯Ω)isCR-bounded;
(iv)the following relations hold:
x=Ux, x∈∂Ω,
y=P y, y∈∂SΩ¯∩Rn
. (3.24)
Then
deg(I−U,Ω,0)=±degBP, S(Ω)∩Rn,0. (3.25) In addition, if
degBP, S(Ω)∩Rn,0=0, (3.26) then problem (1.1) admits solutions.
Proof. By identity (3.22) and applying the Leray-Schauder result for topological degree of product operators, we get
deg(I−U,Ω,0)=deg(S−y,Ω,0)·degI−ΠS−1, S(Ω),0, (3.27)
where y ∈ S(Ω) is arbitrary. SinceS(Ω) is connected andS : ¯Ω→S( ¯Ω) is a homeomorphism, then
deg(S−y,Ω,0)=1, ∀y∈S(Ω). (3.28) SinceΠS−1takes values inRn, we have
degI−ΠS−1, S(Ω),0=degBI−ΠS−1, S(Ω)∩Rn,0 (3.29) and since
ΠS−1=I−P, (3.30)
it results (3.25).
Finally, if (3.26) is satisfied, thenUadmits fixed points and sinceUis associ- ated to (1.1), every fixed point is a solution for problem (1.1).
Remark 3.7. If condition (3.4) is fulfilled for everyt∈R¯ andy∈∂G, then deg(I−U,Ω,0)=±degBf(·, y), G,0. (3.31) Remark 3.8. Formula (3.31) is available for every operator U : ¯Ω ⊂Cl →Cl
admissible for problem (1.1).
3.4. Existence results using Miranda’s theorem. LetK =Πni=1[−l, l]⊂Rnand Φ:K→Rnbe a continuous function; denote byΦitheith component ofΦand byyitheith component ofy∈Rn. DefineL+i, L−i ⊂Rnby
L+i :=
y1, . . . , yi−1, l, yi+1, . . . , yn , L−i :=
y1, . . . , yi−1,−l, yi+1, . . . , yn, i∈1, n. (3.32) Remark that if we take inRnthe norm
|y|=max
1≤i≤nyi, (3.33)
then, if|yj| ≤l,j=i, it results thatK=Σ(l) andL+i,L−i are on two contrary faces of a hypercubeK(soL+i, L−i ∈∂K).
Miranda’s theorem states that, if Φi
L+i≤0, Φi
L−i≥0, i∈1, n,
yj≤l, j=i, (3.34)
then the equation
Φ(y)=0 (3.35)
admits solutions inK.
Suppose that f satisfies the following hypotheses:
(H1) for everyl >0, problem (3.1) has a unique solution defined on the whole R, for everyy∈K;
(H2) the functionsαi(t) :=infx∈Rn{fi(t, x)},βi(t) :=supx∈Rn{fi(t, x)}(where f =(fi)i∈1,n) are defined onRand
αi, βi∈L1(R), i∈1, n; (3.36) (H3) there exists a constantc > 0 such that for everyi∈1, nand for every
(t, y)∈R×Rnwith|yi|> c, we have
yi·fi(t, y)≥0. (3.37)
Theorem3.9. Assume that the hypotheses (H1), (H2), and (H3) are fulfilled. Then problem (1.1) admits solutions.
Proof. Consider the operatorPonKgiven by (3.8), that is, P y=1
2 +∞
−∞ fs;x(s, y)ds. (3.38)
The operatorPis well defined since hypotheses (H1), (H2), and (H3) are as- sumed; in addition, as remarked, it is continuous onRn.
Set
ai:=inf
t∈R
t
0
αi(s)ds, bi:=sup
t∈R
t
0
βi(s)ds, a:=max
1≤i≤n
ai
, b:=min
1≤i≤n
bi .
(3.39)
Then we have for every solutionx(t;y)=(xi(t;y))i∈1,n,
yi+a≤xi(t;y)≤yi+b, i∈1, n. (3.40) Consideringl≥0 such that
l≥max{c−a, c+b}, (3.41) we obtain
xi
t;L+i≥c, xi
t;L−i≤ −c, ∀i∈1, n,∀L+i, L−i ∈∂K. (3.42) If relation (3.41) is fulfilled, it follows from (H3),
PiL+i≤0, PiL−i≥0, ∀i∈1, n,∀L+i, L−i ∈∂K, (3.43) whereP=(Pi)i∈1,n.
Applying Miranda’s theorem, it results thatP has a zero inK. The proof is
now complete.
4. Continuation method
4.1. Introduction. In this section f :R×Rn→Ris a continuous function,Xis the spaceClorCllandΩ⊂Xis an open and bounded set. IfFΩ¯ isCR-bounded, as remarked inSection 2, one can associate to problem (1.1) operatorsU: ¯Ω→ Xwhich are compact and whose fixed points coincide with the solutions of (1.1).
In particular, if
x=Ux, x∈∂Ω, (4.1)
then we can define the topological degree ofUand if
deg(I−U,Ω,0)=0, (4.2)
thenUadmits fixed points inΩ.
However, when we face to check condition (4.2), then we can use the so-called continuation method, which is based on the well-known homotopic invariance property of the topological degree (used inSection 3).
One of the most used forms of this method is the following. Leth:R×Rn× [0,1]→Rnbe a continuous and CR-bounded on ¯Ωfunction in the sense that there existsθ∈CR,θ >0, such that for everyx∈Ω¯ and for everyλ∈[0,1] we have|h(t, x(t), λ)| ≤θ(t),t∈R.
Consider the problem
˙
x=h(t, x, λ), x(+∞)=x(−∞). (4.3) We can associate to problem (4.3) an operatorUλwhich in addition is com- pact for everyλ.
If the condition
x=Uλx, x∈∂Ω, λ∈[0,1] (4.4)
is fulfilled, then we can define the degree deg(I−Uλ,Ω,0); but a homotopic in- variance property tells us that this degree is constant with respect toλ. In partic- ular,
degI−U0,Ω,0=degI−U1,Ω,0. (4.5) Equality (4.5) is useful if U0 is an associated operator to problem (1.1) (h(t, x,0)= f(t, x)) and the degree ofI−U1 is easier to be computed, for ex- ample, when it is a Brouwer degree.
Condition (4.4) can be formulated under the following form:for everyλ ∈ [0,1] problem (4.3)has no solutionsx(·;λ)withx∈∂Ω. If this condition is ful- filled,everyassociated operatorUλsatisfies (4.4) because the fixed points of an
associated operator coincide with the set of solutions for the problem whose it is associated.
We get therefore the following proposition.
Proposition4.1. Assume that
(i)there existsθ∈CR,θ(t)≥0, such that|h(t, x(t), λ)| ≤θ(t), for everyx∈Ω,¯ for everyλ∈[0,1];
(ii)for everyλ∈[0,1], problem (4.3) does not admit solutionsx(·)withx∈∂Ω;
(iii)h(t, x,1)=f(t, x);
(iv) deg(I−U0,Ω,0)=0.
Then problem (1.1) admits solutions.
The question that problem (4.3) has no solutions in∂Ωcan be formulated under the following form.
“A priori estimates”: for every possible solutionx(·) of problem (4.3) with x∈Ω¯ we havex∈Ω.
Another form of the same condition is the next.
“A priori bound”: there exists a numberr >0 such that problem (4.3) does not admit solutionsx(·) withx∞=r.
In this case we setΩ:={x∈X,x< r}.
Another variant of the same condition is the following.
“Bounded set condition”: for everyλ ∈ [0,1] for which problem (4.3) has solutionsx(·) withx(t)∈D,¯ t∈R¯, we havex(t)∈D, for everyt∈R¯.
In this case whenD⊂Rnis an open and bounded set we takeΩ:={x∈X, x(t)∈D}.
In this section, we indicate certain simple functions candidates to be homo- topic linked throughhwith f, functions for which the computation of their topological degree is more advantageously.
The most difficult problem remains to establish the fact that problem (4.3) has no solutions in∂Ω; in what follows we consider certain cases when this thing is easy to be checked.
4.2. Homotopy with a linear equation. In this paragraph considerX=Cll. LetA:R→Mn(R) be a continuous quadratic matrix; denote by| · |an arbi- trary norm for the constant matrices.
Consider the system
˙
x=A(t)x (4.6)
and denote byX=X(t) its fundamental matrix withX(0)=I. In [5], the follow- ing result is proved.
Proposition4.2. Assume that +∞
−∞
A(t)dt <∞, (4.7)
then there existsX(±∞)=limt→±∞X(t). If in addition
rankX(+∞)−X(−∞)=n, (4.8) then the problem
˙
x=A(t)x, x(+∞)=x(−∞) (4.9)
admits only the zero solution.
Theorem4.3. Assume that
(i)A:R→Mn(R)is a continuous matrix such that conditions (4.7), (4.8) are fulfilled;
(ii) f :R×Rn→Rnis a continuous function fulfilling the condition f(t, x)≤θ(t)·ω|x|
, (4.10)
whereθ:R→R,ω:R→Rare continuous and positive functions,θ∈CR; (iii)there existsr >0such that for everyλ∈[0,1]the problem
˙
x=(1−λ)A(t)x+λ f(t, x), x(+∞)=x(−∞) (4.11) has no solutionx(·)such thatx∞=r.
Then problem (1.1) admits solutions.
Proof. Set
h(t, x, λ)=(1−λ)A(t)x+λ f(t, x), Ω=B(r) :=
x∈Cll,x∞< r. (4.12) For everyx∈Ω¯, we have
ht, x(t), λ≤ρA(t)+ω(ρ)θ(t), (4.13) where
ρ=sup
|u|≤rω(u) (4.14)
and so hypothesis (i) ofProposition 4.1is satisfied; obviously (iii) is satisfied, too.
Forλ=0, problem (4.3) becomes
˙
x=A(t)x, x(+∞)=x(−∞) (4.15) which, byProposition 4.2, admits only the zero solution; that means every oper- atorU0attached to problem (4.15) is injective. SinceU0is linear and compact, then after a known property,
degI−U0,Ω,0=±1. (4.16)
This ends the proof.
Hypothesis (ii) ofProposition 4.1is difficult to be checked in practice. In the next theorems it will be fulfilled.
Consider the problem
˙
x=A(t)x+g(t, x) +p(t), x(−∞)=x(+∞), (4.17) whereg:R×Rn→Rnandp:Rn→Rnare continuous functions.
Theorem4.4. Assume that
(i)conditions (4.7), (4.8) are fulfilled;
(ii) p∈L1(R)∩Cc;
(iii)there existsα∈(0,1)such that
g(t, x)=kαg(t, x), ∀k >0,∀t∈R,∀x∈Rn; (4.18) (iv)the following inequality holds:
g(t, x)≤θ(t), ∀t∈R,∀x∈Rn,|x| ≤1, (4.19) whereθ∈CR,θ(t)≥0,(∀)t∈R.
Then problem (4.17) admits solutions.
Proof. Set inProposition 4.1
h(t, x, λ) :=(1−λ)A(t)x+λA(t)x+p(t) +h(t, x), Ω=B(ρ) :=
x∈Cll,x∞< ρ. (4.20) By (4.18) it results that
|x| ≤ρ=⇒g(t, x)≤ραθ(t), (4.21) which shows that forx∈Ω,¯
h(t, x, λ)≤2ρA(t)+ραθ(t) +p(t)∈CR. (4.22) Obviously, to applyProposition 4.1, it remains to check only hypothesis (ii).
For this aim, we will show that there existsρ0>0 such that for everyλ∈[0,1]
and for everyρ > ρ0problem (4.3) has no solutionx(·) withx∞=ρ.
Indeed, if not, then we could find a sequenceλk∈[0,1], a sequenceρk→ ∞, such that the problem
˙
x=ht, x, λk
, x(−∞)=x(+∞) (4.23)
admits solutionsxk(·) with
xk∞=ρk. (4.24)
Setting
uk:= 1
xk∞xk= 1
ρkxk, (4.25)
we have
uk∞=1,
˙
uk=1−λkA(t)uk+λkA(t)uk+ρα−1k gt, uk+ρ−1k p(t). (4.26) ByCorollary 2.7, we get the compactness of the sequence (uk)kinCll. Letu∈(uk)k,λ∈(λk)k; by using the classical properties of uniform conver- gence we obtain, after computations,
˙
u=A(t)u, u(−∞)=u(+∞),u∞=1, (4.27)
which contradictsProposition 4.2.
4.3. Auxiliary results. InSection 4.2, the homotopy has been achieved through a linear mapping for which it was easy to evaluate its topological degree. We give rise to another case when the topological degree computation is not too difficult in the sense that it becomes a Brouwer degree. This result will be a consequence of a more general result which links the existence of solutions for problem (1.1) to the existence of solutions for the problems of the type
˙
y=g(t, y), y(0)=y(T),0< T <∞. (4.28) Letθ:R→R,θ∈CR,θ(t)>0, for everyt∈R; set
ψ(t) := t
−∞θ(s)ds, ϕ:=ψ−1, T:= +∞
−∞θ(s)ds. (4.29) Obviously, through (2.9),ϕ: (0, T)→Rdetermines by (2.9) an isomorphism betweenClandC(0,T)(or betweenCllandC[0,T]).
Proposition4.5. Suppose that f :R×Rn→Rnis a continuous function such that
t→±∞lim 1
θ(t)f(t, y)=γ±(y), y∈Rn, (4.30) the convergence being uniform with respect toyon every compact subset ofRn.
Let
g(t, y) :=
˙
ϕ(t)fϕ(t), y, ift∈(0, T), y∈Rn, γ−(y), ift=0, y∈Rn, γ+(y), ift=1, y∈Rn.
(4.31)
Then problem (1.1) admits solutions if and only if (4.28) admits solutions.
Proof. Remark that ifD⊂Clis a bounded set, thenFD⊂CR; indeed, for|t| ≥A, we have
f(t, y)≤(a+m)θ(t), (4.32)
where
m:=maxsupγ+(u), γ−(u), u∈D∩Rn
. (4.33)
Set
r:=supx∞, x∈D, α(t) :=sup
|y|≤r
f(t, y), t∈[−A, A], β(t) :=
α(t), |t|< A, (a+m)θ(t), |t| ≥A.
(4.34)
We obtainβ∈CRand
f(t, x)≤β(t), ∀t∈R,∀x∈D. (4.35) Letx(t) be a solution for (1.1); theny=Φ(x) is a solution for the differential equation appearing in (4.28) on the interval (0, T). Sincey(t) has limits in 0 and T, it can be prolonged as solution on [0, T]; but by definition ofy(t) it follows that
y(0)=ϕx(−∞)=ϕx(+∞)=y(T). (4.36) The converse is proved by using the isomorphismΦ−1.
LetΩ⊂Clbe an open and bounded set. Hypothesis (4.30) allows us, as re- marked, to associate to problem (1.1) the operator
U: ¯Ω⊂Cl−→Cl, Ux=x(+∞) + (·)
−∞(Fx)(s)ds, (4.37) which, from (4.35), is compact.
The operatorUΦdefined in (2.15) is
UΦ:ΩΦ⊂C(0,T)−→C(0,T), UΦy:=y(T) + (·)
0
gτ, y(τ)dτ. (4.38) But the operatorUΦ is associated to problem (4.28). By using the remarks from 2.4 we obtain the following result.
Corollary4.6. Ifx=Ux, for everyx∈∂Ω, then
deg(I−U,Ω,0)=degI−UΦ,ΩΦ,0. (4.39) (Obviously, the first degree is computed inCl, the second inC(0,T).)
An important particular case is
f(t, x)=θ(t)·g(x). (4.40)
In this case, when (4.30) is fulfilled, (4.24) becomes
˙
y=g(y), y(0)=y(T). (4.41)
As it is proved in [7], for every associated operator to problem (4.41) inC(0,T) orC[0,T](so forUΦ, too) we have
degI−UΦ,ΩΦ,0=±degBg,Ω∩Rn,0. (4.42) We obtain therefore the following proposition.
Proposition4.7. Suppose that
(i)θ:R→R,θ∈CR,θ(t)≥0, for everyt∈R; (ii)g:Rn→Rn,gcontinuous.
Consider the problem
˙
x=θ(t)g(x), x(−∞)=x(+∞). (4.43) Let Ω⊂Cl be an open and bounded set. If for the operator U associated to problem (4.43), we have
x=Ux, x∈∂Ω, (4.44)
then
deg(I−U,Ω,0)=±degBg,Ω∩Rn,0. (4.45) Furthermore, if
degBg,Ω∩Rn,0=0, (4.46)
it results that (4.43) admits solutions.
4.4. Homotopies with nonlinear equations. We consider the problem
˙
x=f(t, x) +p(t), x(−∞)=x(+∞). (4.47) Suppose that the following hypotheses are fulfilled:
(a1) f :R×Rn→Rnis a continuous function;
(a2)|f(t, x)| ≤β(t),x∈Rn,|x| ≤1,t∈R,β∈CR∩Cc;
(a3) there existsα∈(0,1), f(t, kx)=kα·f(t, x),k >0,t∈R,x∈Rn; (a4)|p| ∈CR.
In addition, letθ∈CR,θ(t)>0, with−∞+∞θ(t)dt=1.
Set
g(x) := +∞
−∞ f(s, x)ds. (4.48)
Theorem4.8. Assume that hypotheses (a1), (a2), (a3), and (a4) are fulfilled. Con- sider the problem
˙
x=(1−λ)θ(t)g(x) +λf(t, x) +p(t), x(−∞)=x(+∞). (4.49) Then
(1)if
g(y)=0, y∈Rn,y=1, (4.50) it results that there existsρ >0such that for everyλ∈[0,1], problem (4.49) has no solutionx(·)withx∞=ρ0;
(2)if for thisρ0
degBg,Σρ0
,0=0, (4.51)
then (4.47) admits solutions.
Proof. If conclusion (1) is not true, then there would exist the sequences (ρk)k⊂ (0,∞), (xk)k⊂X, withxk∞=ρk,λk∈[0,1],ρk> kand
˙
xk=1−λkθ(t)gxk+λkft, xk+p(t), x(−∞)=x(+∞). (4.52) Setting
uk=xk
ρk, (4.53)
we get
˙
uk=ρ−1k 1−λkθ(t)guk+λkft, uk+ρ−1k λkp(t),
uk=1, uk(−∞)=uk(+∞). (4.54) Based onCorollary 2.7, it results that (uk)kis relatively compact inX. We can assume, up to subsequences, thatuk→u,λk→λ; we have
u∞=1. (4.55)
By (4.50) it results that ˙uk→0, inX. Thereforeu∈Rn. On the other hand, since
+∞
−∞u˙k(s)ds=0 (4.56)
