Electronic Journal of Differential Equations, Vol. 2004(2004), No. 18, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF SOLUTIONS TO SECOND ORDER ORDINARY
DIFFERENTIAL EQUATIONS HAVING FINITE LIMITS AT ±∞
CEZAR AVRAMESCU & CRISTIAN VLADIMIRESCU
Abstract. In this article, we study the boundary-value problem
ẍ = f (t, x, ẋ),
x(−∞) = x(+∞),
ẋ(−∞) = ẋ(+∞) .
Under adequate hypotheses and using the Bohnenblust-Karlin fixed point theorem for multivalued mappings, we establish the existence of solutions.
1. Introduction
3
Let f : R → R be a continuous mapping. Consider the infinite boundary-value
problem
ẍ = f (t, x, ẋ),
x(−∞) = x(+∞),
(1.1)
ẋ(−∞) = ẋ(+∞),
(1.2)
where x(±∞) and ẋ(±∞) denote the limits
x(±∞) = lim x(t)
t→±∞
and ẋ(±∞) = lim ẋ(t),
t→±∞
(1.3)
which are assumed to be finite. Problem (1.1)-(1.2) may be considered as a generalization of problem (1.1) with boundary condtions
x(a) = x(b),
ẋ(a) = ẋ(b),
(1.4)
as a → −∞ and b → +∞. The bilocal boundary-value problem (1.1)-(1.4) is closely
related to the problem of finding periodic solutions to (1.1). The reader is referred
to [17, 19, 20] where extensive use of topological degree theory is made to study
this problem.
Problem (1.1)-(1.2) is related to the so-called convergent solutions, i.e. the solutions defined on R+ = [0, +∞) (or R) and having finite limit to +∞ (respectively
−∞), see [4, 5, 14, 15, 16]. For studies on (1.1)-(1.2) using variational methods,
we refer the reader to [1, 2, 3, 13, 20, 21]. In [12] the existence of the solutions to
the equation (1.1) with the boundary conditions x(∞) = ẋ(∞) = 0 is studied for
f (t, u, v) = g(t)v − u + h(t, u). Through the Schauder-Tychonoff and Banach fixed
point Theorems estimates for the solutions are found.
2000 Mathematics Subject Classification. 34B15, 34B40, 34C37, 54C60.
Key words and phrases. Nonlinear boundary-value problem, set-valued mappings,
boundary-value problems on infinite intervals.
c
2004
Texas State University - San Marcos.
Submitted February 14, 2003. Published February 9, 2004.
1
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CEZAR AVRAMESCU & CRISTIAN VLADIMIRESCU
EJDE-2004/18
When f is a differentiable function, (1.1) can be written as
ẍ = a(t, x, ẋ)ẋ + b(t, x, ẋ)x + c(t),
(1.5)
R
1
∂f
where a, b : R3 → R, c : R → R, a(t, u, v) := 0 ∂u (t, su, sv) ds, b(t, u, v) :=
R 1 ∂f
(t, su, sv) ds and c(t) := f (t, 0, 0), for all t, u, v ∈ R.
0 ∂v
Sufficient conditions for the existence of solutions to the linear problem
ẍ = a(t)ẋ + b(t)x + c(t),
(1.6)
with boundary condition (1.2), were given in [11]. By using this result, in the real
Banach space
X := x ∈ C 2 (R) : (∃) x(±∞), (∃) ẋ(±∞)
endowed with the uniform convergence topology on R one defines an operator T :
X → 2X which maps u ∈ X into the set of the solutions to the problem (1.7)-(1.2),
where
ẍ = a(t, u(t), u̇(t))ẋ + b(t, u(t), u̇(t))x + c(t).
(1.7)
Next one considers the restriction of T to a bounded, convex and closed set M ,
conveniently chosen so that the Bohnenblust-Karlin Theorem can be applied. The
compactness of T (M ) is established by using a characterization developed by the
the first author in [4, 6].
The use of a multivalued operator T is motivated by the fact that one cannot determine a solution to the problem (1.7)-(1.2) through an “initial” condition
independent of u.
2. Main result
3
Let a, b : R → R, c : R → R be continuous functions, and let
α1 (t) := inf a(t, u, v) , α2 (t) := sup a(t, u, v) ,
u,v∈R
β(t) := sup
u,v∈R
b(t, u, v) ,
u,v∈R
Ai (t) := exp
Z
t
αi (s) ds ,
0
for i ∈ {1, 2} and t ∈ R. We shall assume that α1 , α2 , β are defined on R.
Consider the following hypotheses, where the integrals are considered in the
Riemann sense:
(A1) The mappings α1 and α2 are bounded on R, and limt→±∞ αi (t) = 0, for
i ∈ {1, 2}
(A2) limt→±∞ Ai (t) = 0 for i ∈ {1, 2}
(B1) 0 ≤ b(t, u, v) for every t, u, v ∈ R and limt→±∞ β(t) = 0
R +∞
Rt
(B2) −∞ Ai (t) · 0 Aβ(s)
ds dt ∈ R for i ∈ {1, 2}
i (s)
R +∞
dt < +∞, for i ∈ {1, 2}
(B3) −∞ Aβ(t)
R +∞ i (t)
(C1) −∞ |c(t)|dt < +∞
R +∞ R t |c(s)| (C2) −∞
ds dt ∈ R for i ∈ {1, 2}.
−t Ai (s)
Our main result is as follows:
Theorem 2.1. If the hypotheses (A1)–(A2), (B1)–(B3), (C1)–(C2) are satisfied,
then (1.5)-(1.2) has a solution.
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EXISTENCE OF SOLUTIONS
3
Since
lim
t→±∞
Ai (t)
1
= lim R t β(s)
R t β(s)
t→±∞
ds
Ai (t) · 0 Ai (s) ds
0 Ai (s)
is a real number by hypothesis (B3), it follows by hypothesis (B2), via a well known
convergence criterion for Riemann integrals, that for each i ∈ {1, 2},
Z +∞
Ai (t)dt < +∞ .
(2.1)
−∞
Similarly, by hypothesis (A2),
lim
β(t)
t→±∞ β(t)
Ai (t)
= 0,
lim
|c(t)|
t→±∞ |c(t)|
Ai (t)
= 0,
it follows, by hypothesis (B3), that
Z +∞
β(t)dt < +∞,
(2.2)
|c(t)|
dt < +∞,
Ai (t)
(2.3)
−∞
and, by hypothesis (C1),
Z
+∞
−∞
for each i ∈ {1, 2}.
Remark 2.2. (i) One can replace the hypothesis (B2) by
R +∞
(B2’) −∞ Ai (t)dt < +∞ .
(ii) Assumption (B2’) does not imply (C2).
R (·)
(i) Indeed, since (B3) implies the boundedness of the mapping 0 Aβ(s)
ds and
i (s)
therefore, (B2’) implies (B2).
(ii) It is sufficient to choose c(t) = Ai (t), for all t ∈ R, where i = 1 or i = 2.
For proving our main result we use the following theorem.
Theorem 2.3 (Bohnenblust-Karlin [22, p. 452]). Let X be a Banach space and
M ⊂ X be a convex closed subset of it. Suppose that T : X → 2X is a multivalued
operator on X satisfying the following hypotheses:
(a) T (M ) ⊂ 2M and T is upper semicontinuous
(b) the set T (M ) is relatively compact
(c) for every x ∈ M , T (x) is a non-empty convex closed set.
Then T admits fixed points.
Recall that T : M → 2M is upper semicontinuous if for every closed subset A of
M , the set
T −1 (A) := x ∈ M : T (x) ∩ A 6= ∅
is also a closed subset of M . Another useful result is the following Lemma.
Lemma 2.4 (Barbălat). If f : [0, +∞) → R satisfies: (a) f is uniformly continuous
R +∞
and (b) the integral 0 f (t) dt exists and is finite, then limt→+∞ f (t) = 0.
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CEZAR AVRAMESCU & CRISTIAN VLADIMIRESCU
EJDE-2004/18
The main idea of this paper is to build a multivalued operator T defined on an
adequate space which satisfies the hypotheses of the Bohnenblust-Karlin Theorem.
We define
X := x ∈ C 2 (R) : (∃) x(±∞) and ẋ(±∞) ,
which, endowed with the usual norm,
kxk := sup max |x(t)|, |ẋ(t)| ,
t∈R
becomes a real Banach space. The relative compactness of the set T (M ) be will be
proved by using the following Proposition.
Proposition 2.5 (Avramescu [4, 6]). A set A ⊂ X is relatively compact if and
only if the following conditions are fulfilled:
(a) There exist h1 , h2 ≥ 0 such that for every x ∈ A and t ∈ R, we have
|x(t)| ≤ h1 and |ẋ(t)| ≤ h2
(b) For every K = [a, b] ⊂ R and ε > 0 there exists δ = δ(K, ε) > 0 such that
for every x ∈ A and t1 , t2 ∈ K with |t1 −t2 | < δ, we have |x(t1 )−x(t2 )| < ε
and |ẋ(t1 ) − ẋ(t2 )| < ε
(c) For every ε > 0 there exists T = T (ε) > 0 such that for every t1 , t2 with
|t1 |, |t2 | > T and t1 ·t2 > 0, and for every x ∈ A, we have |x(t1 )−x(t2 )| < ε
and |ẋ(t1 ) − ẋ(t2 )| < ε.
3. Construction of the multivalued operator T
Let u ∈ C 2 (R) be arbitrary. Consider the problem
ẍ = au (t)ẋ + bu (t)x + c(t)
x(+∞) = x(−∞),
ẋ(+∞) = ẋ(−∞),
(3.1)
where au (t) := a(t, u(t), u̇(t)) and bu (t) = b(t, u(t), u̇(t)). Consider the homogeneous problem
ẍ = au (t)ẋ + bu (t)x
(3.2)
x(+∞) = x(−∞), ẋ(+∞) = ẋ(−∞).
Since
Z t
x(t) = exp
y(s)ds , t ∈ R
0
is a solution to ẍ = au (t)ẋ + bu (t)x if and only if y is a solution to
ẏ = au y + bu − y 2 ,
we have au (t)y − y 2 ≤ ẏ ≤ au (t)y + bu (t), for every t ∈ R.
Let v, w satisfy
v̇ = au (t)v − v 2
v(0) = ξ
(3.3)
(3.4)
and
ẇ = au (t)w + bu (t)
w(0) = ξ .
Hence
ẏ = au y + bu − y 2
y(0) = ξ,
(3.5)
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EXISTENCE OF SOLUTIONS
5
which implies
v(t) ≤ y(t) ≤ w(t),
if t ≥ 0,
w(t) ≤ y(t) ≤ v(t), if t ≤ 0.
Rt
Let αu (t) := exp 0 au (s)ds , for every t ∈ R. Thus
ξαu (t)
Rt
1 + ξ 0 αu (s)ds
Z t
bu (s) w(t) = αu (t) ξ +
ds .
α
u (s)
0
v(t) =
(3.6)
Therefore,
Z t
h
ξαu (t)
bu (s) i
≤ y(t) ≤ αu (t) ξ +
ds ,
Rt
1 + ξ 0 αu (s)ds
0 αu (s)
Z t
h
bu (s) i
ξαu (t)
αu (t) ξ +
ds ≤ y(t) ≤
,
Rt
α
(s)
u
1 + ξ αu (s)ds
0
if t ≥ 0,
if t ≤ 0.
0
We write
gu (t) ≤ y(t) ≤ Gu (t),
for t ∈ R,
(3.7)
where
gu (t) :=


ξαu (t)
Rt
,
α (s)ds
0h u
R t bu (s) i
αu (t) ξ +
ds ,
0 αu (s)
1+ξ
if t ≥ 0
(3.8)
if t ≤ 0
and
Gu (t) :=

h
R
αu (t) ξ + t
0
bu (s)
αu (s) ds
i
, if t ≥ 0
ξαu (t)

R
,
1+ξ 0t αu (s)ds
(3.9)
if t ≤ 0 .
Let yu denote the solution to the equation (3.3) with the initial condition
yu (0) = ξ .
Hence, gu (t) ≤ yu (t) ≤ Gu (t), for every t ∈ R. From (3.6) we see that yu is defined
for all t ∈ R if and only if
1
1
, R0
ξ ∈ − R +∞
:= (λu , µu ).
α
(s)ds
α
(s)ds
u
u
0
−∞
We let λ := supu∈C 2 (R) {λu } and µ := inf u∈C 2 (R) {µu }. Since
A1 (t) ≤ αu (t) ≤ A2 (t),
for every t ≥ 0 and u ∈ C 2 (R)
A2 (t) ≤ αu (t) ≤ A1 (t),
for every t ≤ 0 and u ∈ C 2 (R)
it follows that
− R +∞
0
1
A1 (t)dt
≤ − R +∞
0
1
αu (s)ds
≤ − R +∞
0
1
A2 (t)dt
:= λ
and
µ := R 0
−∞
1
A1 (t)dt
≤ R0
−∞
1
αu (s)ds
≤ R0
−∞
1
A2 (t)dt
.
(3.10)
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CEZAR AVRAMESCU & CRISTIAN VLADIMIRESCU
EJDE-2004/18
Therefore,
− R +∞
0
1
1
:= λ < 0 < µ := R 0
A2 (t)dt
−∞
.
(3.11)
A1 (t)dt
Let
g(t) :=
inf
u∈C 2 (R)
gu (t) and G(t) :=
sup Gu (t), for t ∈ R.
u∈C 2 (R)
For t ≤ 0, we have
Z
h
gu (t) ≥ αu (t) λ +
0
t
Z t
h
bu (s) i
β(s) i
ds ≥ A1 (t) λ +
ds
αu (s)
0 A2 (s)
and for t ≥ 0,
gu (t) ≥
λαu (t)
λA2 (t)
≥
.
Rt
Rt
1 + λ 0 αu (s)ds
1 + λ 0 A2 (s)ds
Thus


λA2 (t)
Rt
,
A (s)ds
0h 2
R t β(s) i
A1 (t) λ +
ds ,
0 A2 (s)
g(t) :=
1+λ
if t ≥ 0
if t ≤ 0 .
(3.12)
Similarly

h
R
A2 (t) µ + t
0
G(t) :=
β(s)
A1 (s) ds
i
,
µA1 (t)

R
,
1+µ 0t A1 (s)ds
if t ≥ 0
if t ≤ 0 .
(3.13)
By hypothesis (A2), one has g(±∞) = G(±∞) = 0. Thus for every ξ ∈ (λ, µ) and
for every y solution to the equation (3.3) with the initial condition y(0) = ξ, we
have
g(t) ≤ y(t) ≤ G(t), for every t ∈ R.
(3.14)
Let ξ1 , ξ2 ∈ (λ, µ), ξ1 6= ξ2 be arbitrary, and yiu be the solution to the problem
ẏ = au (t)y + bu (t) − y 2
y(0) = ξi
Rt
where i ∈ {1, 2} and u ∈ C 2 (R). Let xui (t) := exp( 0 yiu (s)ds), for t ∈ R, i ∈ {1, 2}
and u ∈ C 2 (R). Then xui (0) = 1, ẋui (0) = ξi , ẋui (t) = yiu (t) · xui (t), for t ∈ R,
i ∈ {1, 2} and u ∈ C 2 (R).
Let us prove that, for every i ∈ {1, 2} and u ∈ C 2 (R), xui (±∞), ẋui (±∞), exist
and are finite. Indeed, by relation (2.1),
Z +∞
xui (+∞) = exp
yiu (t)dt
0
Z t
Z +∞
β(s) ≤ exp
A2 (t) µ +
ds dt
0
0 A1 (s)
Z
Z
+∞
n +∞
h
β(s) io
≤ exp (
A2 (t)dt · µ +
ds < +∞,
A1 (s)
0
0
and
xui (−∞)
= exp
Z
−∞
yiu (t)dt
0
≤ exp
n Z
0
−∞
Z
h
A1 (t)dt · λ +
0
−∞
β(s) io
ds < +∞,
A1 (s)
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EXISTENCE OF SOLUTIONS
7
for every i ∈ {1, 2} and u ∈ C 2 (R). For i ∈ {1, 2} and u ∈ C 2 (R),
Z t
|xui (t)| = exp
yiu (s)ds .
0
Hence, for t ≥ 0,
Z t
nZ
exp
yiu (s)ds ≤ exp (
0
+∞
0
and for t ≤ 0,
Z t
nZ
u
exp
yi (s)ds ≤ exp (
0
+∞
β(s) io
ds =: δ1
A1 (s)
−∞
β(s) io
ds =: δ2 .
A1 (s)
Z
h
A2 (t)dt) · µ +
0
−∞
h
Z
A1 (t)dt) · λ +
0
0
u
|xi (t)|
Therefore, taking M1 := max {δ1 , δ2 } > 0, we have
≤ M1 , for every t ∈ R,
i ∈ {1, 2}, and u ∈ C 2 (R).
Since g and G are continuous with g(±∞) = G(±∞) = 0 it follows that they
are bounded on R. But
g(t) ≤ yiu (t) ≤ G(t),
for every t ∈ R, i ∈ {1, 2} and u ∈ C 2 (R).
Hence, there exists a constant δ3 > 0 such that
|yiu (t)| ≤ δ3 ,
for t ∈ R, i ∈ {1, 2} and u ∈ C 2 (R)
and so
|ẋui (t)| ≤ M1 · δ3 =: M2 ,
for t ∈ R, i ∈ {1, 2} and u ∈ C 2 (R).
For u ∈ C 2 (R) the general solution to the nonhomogeneous equation
ẍ = au (t)ẋ + bu (t)x + c(t)
is
x(t) = γ1u xu1 (t) + γ2u xu2 (t) + xu2 (t) ·
Z
0
Z
t
xu1 (s) ·
c(s)
ds
(ξ2 − ξ1 )αu (s)
t
c(s)
− xu1 (t) ·
xu2 (s) ·
ds,
(ξ2 − ξ1 )αu (s)
0
with γ1u , γ2u ∈ R. From the condition x(+∞) = x(−∞), we have
γ1u · [xu1 (+∞) − xu1 (−∞)] + γ2u · [xu2 (+∞) − xu2 (−∞)]
Z +∞
c(s)
= xu1 (+∞) ·
xu2 (s) ·
ds
(ξ2 − ξ1 )αu (s)
0
Z −∞
c(s)
− xu1 (−∞) ·
xu2 (s) ·
ds
(ξ
−
ξ1 )αu (s)
2
0
Z −∞
c(s)
u
+ x2 (−∞) ·
xu1 (s) ·
ds
(ξ
−
ξ1 )αu (s)
2
0
Z +∞
c(s)
− xu2 (+∞) ·
xu1 (s) ·
ds.
(ξ
−
ξ1 )αu (s)
2
0
(3.15)
(3.16)
(3.17)
Now we prove that the relation (3.17) is satisfied by infinitely many pairs (γ1u , γ2u ),
u ∈ C 2 (R). Indeed, if we denote by
d1 := xu1 (+∞) − xu1 (−∞),
d2 := xu2 (+∞) − xu2 (−∞),
and d3 the right hand side of (3.17), then we have to consider only three cases.
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CEZAR AVRAMESCU & CRISTIAN VLADIMIRESCU
EJDE-2004/18
Case 1. If d1 6= 0 and d2 = 0, it follows that γ1u = dd13 and γ2u ∈ R; similarly, if
d1 = 0 and d2 6= 0, it follows that γ1u ∈ R and γ2u = dd32 .
Case 2. If d1 6= 0 and d2 6= 0, it follows that
γ1u =
d3 − d2 γ2u
d1
and γ2u ∈ R.
Case 3. If d1 = 0 and d2 = 0, we show that d3 = 0 (and so the solutions are γ1u ,
γ2u ∈ R).
Indeed, in this case, xu1 (+∞) = xu1 (−∞) and xu2 (+∞) = xu2 (−∞), and we have
to prove that
Z +∞
Z +∞
c(s)
c(s)
ds = xu2 (+∞) ·
xu1 (s) ·
ds.
(3.18)
xu1 (+∞) ·
xu2 (s) ·
α
(s)
α
u
u (s)
−∞
−∞
To prove (3.18) we shall apply Lemma 2.4 to the mapping f : [0, +∞) → R, defined
by
Z +t
Z +t
c(s)
c(s)
ds − xu2 (t) ·
xu1 (s) ·
ds.
f (t) := xu1 (t) ·
xu2 (s) ·
αu (s)
αu (s)
−t
−t
Thus
Z +t
Z +t
df
c(s)
c(s)
u
u
u
(t) = ẋ1 (t) ·
ds − ẋ2 (t) ·
ds
x2 (s) ·
xu1 (s) ·
dt
α
(s)
α
u
u (s)
−t
−t
c(−t)
[xu (t) · xu2 (−t) − xu2 (t) · xu1 (−t)] .
+
αu (−t) 1
Since ẋui (±∞) = xui (±∞) · yiu (±∞) = 0, i ∈ {1, 2}, the mapping
R (see hypothesis (C2)), and
c
αu
is bounded on
lim [xu1 (t) · xu2 (−t) − xu2 (t) · xu1 (−t)] = 0,
t→+∞
it follows that limt→+∞ df
dt (t) = 0. Therefore f is uniformly continuous on [0, +∞),
being Lipschitz on [0, +∞). Since xui , i ∈ {1, 2} are bounded, from (C2) it follows
R +∞
that 0 f (t) dt exists and is finite. Hence, by Lemma 2.4 we obtain
lim f (t) = 0.
t→+∞
Now we define the multivalued operator T : X → 2X , by
Z (·)
n
c(s)
xu1 (s) ·
T u := γ1u xu1 (·) + γ2u xu2 (·) + xu2 (·) ·
ds
(ξ2 − ξ1 )αu (s)
0
Z (·)
c(s)
− xu1 (·) ·
xu2 (s) ·
ds,
(ξ
−
ξ1 )αu (s)
2
0
o
with |γ1u | + |γ2u | ≤ 1, γ1u , γ2u satisfying (3.17) ,
for every u ∈ X. By (3.15)-(3.16) we have
Z
Z
M1 t u
c(s) t u
c(s) |x(t)| ≤ 2M1 +
x1 (s)
ds +
x2 (s)
ds .
|ξ2 − ξ1 |
αu (s)
αu (s)
0
0
Hence |x(t)| ≤ k1 , for every t ∈ R, where
Z +∞
Z 0
n
2M12
|c(s)|
2M12
|c(s)| o
k1 := max 2M1 +
ds, 2M1 +
ds .
|ξ2 − ξ1 | 0
A1 (s)
|ξ2 − ξ1 | −∞ A2 (s)
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EXISTENCE OF SOLUTIONS
9
Similarly
Z t
Z t
c(s)
c(s) u u
u u
u
u
u
|ẋ(t)| = γ1 ẋ1 (t) + γ2 ẋ2 (t) + ẋ2 (t)
x1 (s)
ds − ẋ1 (t)
xu2 (s)
ds,
αu (s)
αu (s)
0
0
and there exists another constant k2 ≥ 0,
Z
Z
n
2M1 M2 0 |c(s)| o
2M1 M2 +∞ |c(s)|
k2 := max 2M2 +
ds, 2M2 +
ds ,
|ξ2 − ξ1 | 0
A1 (s)
|ξ2 − ξ1 | −∞ A2 (s)
such that |ẋ(t)| ≤ k2 , for every t ∈ R. Remark that, by relation (2.3), k1 , k2 are
finite. We let k := max{k1 , k2 }, and
M := x ∈ C 2 (R), |x(t)| ≤ k, |ẋ(t)| ≤ k, for every t ∈ R .
4. Proof of main result
To prove Theorem 2.1 it is sufficient to prove that the operator T has a fixed
point. We do this in three steps.
Step 1: For every u ∈ M , T (u) is a non-empty convex closed set. Let u ∈ M be
arbitrary.
From the definition of T we see that T (u) is non-empty and convex.
Let (xn )n∈N ⊂ T (u) be such that xn → x and ẋn → ẋ uniformly on R as n → ∞.
We have
u
u
xn (t) := γ1,n
xu1 (t) + γ2,n
xu2 (t) + H u (t),
u
u
u
u
, γ2,n
satisfying (3.17), and
| ≤ 1, γ1,n
for every n ∈ N, with |γ1,n
| + |γ2,n
Z t
Z t
c(s)
c(s)
H u (t) := xu2 (t) ·
xu1 (s) ·
ds − xu1 (t) ·
xu2 (s) ·
ds.
(ξ2 − ξ1 )αu (s)
(ξ2 − ξ1 )αu (s)
0
0
u
u
Then there exist subsequences such that γ1,k
→ γ1u and γ2,k
→ γ2u , as n → ∞.
n
n
kn
u u
u u
Since (x )n∈N converges uniformly to y := γ1 x1 + γ2 x2 + H u , it follows that
x = y. Also
ẋkn → ẏ = ẋ,
as n → ∞.
So x ∈ T (u), that is T (u) is a closed set.
Step 2: T (M ) is relatively compact. The relative compactness of T (M ) will be
proved by using Proposition 2.5.
From the definitions of T and M we see that |x(t)| ≤ k, |ẋ(t)| ≤ k, for all t ∈ R.
Thus the first condition of Proposition 2.5 is fulfilled with h1 = h2 = k.
Conditions (b) and (c) of Proposition 2.5 are implied by the following assumption:
(d) There exist f1 , f2 : R → R+ integrable on R such that for every x ∈ A
|ẋ(t)| ≤ f1 (t) and
|ẍ(t)| ≤ f2 (t),
for t ∈ R.
This last assertion follows from the fact that, for every t1 , t2 ∈ R,
Z t2
Z t2
x(t1 ) − x(t2 ) =
ẋ(t)dt and ẋ(t1 ) − ẋ(t2 ) =
ẍ(t) dt .
t1
t1
For i ∈ {1, 2} let
o
n

R
|ξi |A2 (t)
max A2 (t) µ + t β(s) ds ,
Rt
,
t≥0
A
(s)
0 1
|1+ξi 0 A2 (s)ds|
n
g1i (t) :=
R 0 β(s) o
|ξ
|A
(t)
i
1
max
R
, A1 (t) − λ + t A2 (s) ds , t ≤ 0 .
|1+ξ t A (s)ds|
i
0
1
10
CEZAR AVRAMESCU & CRISTIAN VLADIMIRESCU
EJDE-2004/18
Hence |ẋui | is bounded by the integrable function M1 · g1i , i ∈ {1, 2}. Furthermore,
since
Z
t u
c(s)
ds
x2 (s) ·
(ξ
−
ξ
)
·
α
(s)
2
1
u
0
R +∞ |c(s)|
M1
is bounded (on the positive semiaxis by |ξ2 −ξ1 | · 0
A1 (s) ds and on the negative
R 0 |c(s)|
M1
u
semiaxis by |ξ2 −ξ1 | · −∞ A1 (s) ds), and |ẋ1 | is bounded by an integrable function,
we see that
Z (·)
u
c(s)
ẋ1 ·
xu2 (s) ·
ds
(ξ2 − ξ1 ) · αu (s)
0
is bounded by an integrable function. Similarly,
Z (·)
u
c(s)
ẋ2 ·
ds
xu1 (s) ·
(ξ2 − ξ1 ) · αu (s)
0
is bounded by an integrable function. Therefore, the existence of f1 in assertion
(d) follows. Now, since
ẍ(t) = au (t)ẋ(t) + bu (t)x + c(t),
au is bounded (hypothesis (A1)), |ẋ| is bounded by an integrable function, |x| is
bounded (by k), bu is integrable on R (by relation (2.2), hypothesis (B1), and |c| is
integrable on R (by hypothesis (C1)), we see that |ẍ| is bounded by an integrable
function. This proves the existence of f2 , and hence assertion (d) is verified.
Step 3: T is upper semicontinuous. Let A be a closed subset of M . Hence if
(un )n ⊂ A such that un → u and u̇n → u̇ uniformly on R, as n → ∞, it follows
that u ∈ A.
Let zn ∈ T −1 (A) be such that zn → z and żn → ż uniformly on R, as n → ∞.
We have to prove that z ∈ T −1 (A). Since zn ∈ T −1 (A) there exists xn ∈ A,
xn ∈ T zn . Thus
ẍn = a(t, zn , żn )ẋn + x(t, zn , żn )xn + c(t),
n∈N
(4.1)
and
xn (+∞) = xn (−∞),
ẋn (+∞) = ẋn (−∞),
n ∈ N.
(4.2)
Since xn ∈ T (M ) and T (M ) is relatively compact, the sequence xn contains subsequence converging in C 2 to some x. One can assume that xn → x, ẋn → ẋ
uniformly on R, as n → ∞.
Since a(t, zn (t), żn (t)) → a(t, z(t), ż(t)) and b(t, zn (t), żn (t)) → b(t, z(t), ż(t)),
uniformly on compact subsets of R, it follows that x is solution to the equation
ẍ = a(t, z(t), ż(t))ż + b(t, z(t), ż(t))z + c(t),
with
x(0) = lim xn (0) and ẋ(0) = lim ẋn (0).
n→∞
n→∞
Furthermore, by (4.2) we find, by passing to the limit as n → ∞,
x(+∞) = x(−∞) and ẋ(+∞) = ẋ(−∞).
Since the set A is closed, x ∈ A. Therefore, z ∈ T −1 (A), which completes the proof
of Theorem 2.1.
EJDE-2004/18
EXISTENCE OF SOLUTIONS
11
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Cezar Avramescu
Centre for Nonlinear Analysis and its Applications, University of Craiova, Al.I. Cuza
Street, No. 13, Craiova RO-200585, Romania
E-mail address: zarce@central.ucv.ro, cezaravramescu@hotmail.com
12
CEZAR AVRAMESCU & CRISTIAN VLADIMIRESCU
EJDE-2004/18
Cristian Vladimirescu
Department of Mathematics, University of Craiova, Al.I. Cuza Street, No. 13, Craiova
RO-200585, Romania
E-mail address: cvladi@central.ucv.ro, vladimirescucris@hotmail.com