Electronic Journal of Differential Equations, Vol. 2010(2010), No. 36, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
NEUTRAL FUNCTIONAL DIFFERENTIAL EQUATIONS OF
SECOND-ORDER WITH INFINITE DELAYS
RUNPING YE, GUOWEI ZHANG
Abstract. This work shows the existence of mild solutions to neutral functional differential equations of second-order with infinite delay. The Hausdorff
measure of noncompactness and fixed point theorem are used, without assuming compactness on the associated family of operators.
1. Introduction
Differential equations with delays are often more realistic to describe natural
phenomena than those without delays, and neutral differential equations arise in
many areas of applied mathematics. These two reasons may explain, why they have
received much attention in the previous decades. Among the published works, we
have [1, 4, 5, 12, 13, 14, 16, 21] and references therein. Existence and stability have
been studied by Hale [9, 10], Travis and Webb [19], and Webb [20]. second-order
differential equations and integrodifferential equations in Banach spaces have been
studied in [2, 11] and [15], respectively.
In this article, we investigate the existence of mild solutions for the neutral
functional differential equation
d 0
(x (t) + g(t, xt )) = Ax(t) + f (t, xt ), t ∈ J = [0, b],
dt
x0 = ϕ ∈ B, x0 (0) = z ∈ X .
(1.1)
(1.2)
We also consider the second order problem
d 0
(x (t) + g(t, xt , x0 (t))) = Ax(t) + f (t, xt , x0 (t)),
dt
x0 = ϕ ∈ B, x0 (0) = z ∈ X,
t ∈ J = [0, b],
(1.3)
(1.4)
where A is the infinitesimal generator of a strongly continuous cosine family {C(t) :
t ∈ R} of bounded linear operators on a Banach space X. In both cases, the history
xt : (−∞, 0] → X, xt (θ) = x(t + θ), belongs to some abstract phase space B defined
axiomatically; g, f are appropriate functions.
2000 Mathematics Subject Classification. 34K30, 34K40, 47D09.
Key words and phrases. Neutral functional differential equations; mild solution;
Hausdorff measure of noncompactness; phase space.
c
2010
Texas State University - San Marcos.
Submitted November 24, 2009. Published March 9, 2010.
Supported by grant Z2009004 from the Science and Technology Foundation of Suqian, China.
1
2
R. YE, G. ZHANG
EJDE-2010/36
In this paper, we prove the existence of mild solution of the initial value problems
(1.1)-(1.2) and (1.3)-(1.4) under the conditions under assumptions on Hausdorff’s
measure of noncompactness.
2. Preliminaries
Now we introduce some definitions, notation and preliminary facts which are
used throughout this paper.
We say that a family {C(t) : t ∈ R} of operators in B(X) is a strongly continuous
cosine family if
(i) C(0) = I (I is the identity operator in X);
(ii) C(t + s) + C(t − s) = 2C(t)C(s) for all s, t ∈ R;
(iii) The map t → C(t)x is strongly continuous for each x ∈ X.
The strongly continuous sine family {S(t) : t ∈ R}, associated to the given
strongly continuous cosine family {C(t) : t ∈ R}, is defined by
Z t
S(t)x =
C(s)xds, x ∈ X, t ∈ R.
0
For more details on strongly continuous cosine and sine families, we refer the reader
to the books by Goldstein [7] and Fattorini [6].
The operator A is the infinitesimal generator of a strongly continuous cosine
function of bounded linear operators, (C(t))t∈R , on X and S(t) is the sine function
e certain constants such that
associated with (C(t))t∈R . We designate by N , N
e
kC(t)k ≤ N and kS(t)k ≤ N for every t ∈ J. We refer the reader to [6] for the
necessary concepts about cosine functions. Next we only mention a few results and
notations needed to establish our results. As usual we denote by D(A) the domain
of A endowed with the graph norm kxkA = kxk + kAxk, x ∈ D(A).
In this work we employ an axiomatic definition of the phase space B which is
similar to that introduced by Hale and Kato [10] and it is appropriate to treat
retarded differential equations with infinite delay.
Definition 2.1 ([10]). Let B be a linear space of functions mapping (−∞, 0] into
X endowed with a seminorm k · kB and that satisfies the following cinditions:
(A) If x : (−∞, σ + b] → X, b > 0, such that xσ ∈ B and x|[σ,σ+b] ∈ C([σ, σ + b] :
X), then for every t ∈ [σ, σ + b) the following conditions hold:
(i) xt is in B,
(ii) kx(t)k ≤ Hkxt kB ,
(iii) kxt kB ≤ K(t − σ) sup{kx(s)k : σ ≤ s ≤ t} + M (t + σ)kxσ kB ,
where H > 0 is a constant; K, M : [0, ∞) → [1, ∞), K is continuous, M is
locally bounded and H, K, M are independent of x(·).
(A1) For the function x(·) in (A), xt is a B-valued continuous function on [σ, σ +
b).
(B) The space B is complete.
Definition 2.2 ([3]). The Hausdorff’s measure of noncompactness is defined as
χY (B) = inf{r > 0, B can be covered by finite number of balls with radius r}. for
bounded set B in any Banach space Y .
Lemma 2.3 ([3]). Let Y be a real Banach space and B, C ⊆ Y be bounded, the
following properties are satisfied:
EJDE-2010/36
NEUTRAL FUNCTIONAL DIFFERENTIAL EQUATION
3
(1) B is pre-compact if and only if χY (B) = 0;
(2) χY (B) = χY (B) = χY (convB), where B and convB are the closure and
the convex hull of B respectively;
(3) χY (B) ≤ χY (C) when B ⊆ C;
(4) χY (B + C) ≤ χY (B) + χY (C) where B + C = {x + y : x ∈ B, y ∈ C};
(5) χY (B ∪ C) ≤ max {χY (B), χY (C)};
(6) χY (λB) = |λ|χY (B) for any λ ∈ R;
(7) If the map Q : D(Q) ⊆ Y → Z is Lipschitz continuous with constant k,
then χZ (QB) ≤ kχY (B) for any bounded subset B ⊆ D(Q), where Z is a
Banach space;
(8) If {Wn }+∞
n=1 is a decreasing sequence of bounded closed nonempty subsets of
Y and limn→∞ χY (Wn ) = 0, then ∩+∞
n=1 Wn is nonempty and compact in Y .
Definition 2.4 ([3]). The map Q : W ⊆ Y → Y is said to be a χY − contraction
if Q is bounded continuous and there exists a positive constant k < 1 such that
χY (Q(C)) ≤ kχY (C)) for any bounded closed subset C ⊆ W , where Y is a Banach
space.
Lemma 2.5 (Darbo-Sadovskii [3]). If W ⊆ Y is bounded closed and convex, the
map Q : W → W is a χY − contraction, then the map Q has at least one fixed
point in W .
In this paper we denote χ the Hausdorff’s measure of noncompactness of X, χC
the Hausdorff’s measure of noncompactness of C([0, b]; X) and χC 1 the Hausdorff’s
measure of noncompactness of C 1 ([0, b]; X). To discuss the existence results we
need the following auxiliary results.
Lemma 2.6 ([3]).
(1) If W ⊂ C([a, b]; X) is bounded, then χ(W (t)) ≤ χC (W ), for t ∈ [a, b],
where W (t) = {u(t) : u ∈ W } ⊆ X;
(2) If W is equicontinuous on [a, b], then χ(W (t)) is continuous for t ∈ [a, b],
and
χC (W ) = sup {χ(W (t)), t ∈ [a, b]};
(3) If W ⊂ C([a, b]; X) is bounded and equicontinuous, then χ(W (t)) is continuous for t ∈ [a, b], and
Z t
Z t
W (s)ds) ≤
χW (s)ds
χ(
a
for all t ∈ [a, b], where
a
Rt
a
W (s)ds = {
Rt
a
x(s)ds : x ∈ W }.
The following lemmas are easy to prove.
Lemma 2.7. If the semigroup S(t) is equicontinuous and η ∈ L([0, b]; R+ ), then the
Rt
set { 0 S(t − s)u(s)ds, ku(s)k ≤ η(s) for a.e. s ∈ [0, b]} is equicontinuous for t ∈
[0, b].
Lemma 2.8 ([8]). Let W ⊂ C 1 (J; X) be bounded and W 0 be equicontinuous, then
χC 1 (W ) = max {χC (W ), χC (W 0 )} = max {max χC (W (t)), max χC (W 0 (t))},
t∈J
0
0
where W = {u : u ∈ W }, J = [a, b].
t∈J
4
R. YE, G. ZHANG
EJDE-2010/36
3. Main results
Now we define the mild solution for the initial value problem(1.1)-(1.2).
Definition 3.1. A function x : (−∞, b] → X is a mild solution of the initial value
problem (1.1)-(1.2), if x0 = ϕ, x(·)|J ∈ C(J; X) and for t ∈ J,
Z t
Z t
x(t) = C(t)ϕ(0) + S(t)(z + g(0, ϕ)) −
C(t − s)g(s, xs )ds +
S(t − s)f (s, xs )ds.
0
0
For (1.1)-(1.2), we assume the following hypotheses:
(H1f) f : J × B → X satisfies the following two conditions:
(1) For each x : (−∞, b] → X, x0 ∈ B and x|J ∈ C(J; X), the function
t → f (t, xt ) is strongly measurable and f (t, ·) is continuous for a.e.
t ∈ J;
(2) There exist an integrable function α : J → [0, +∞) and a monotone
continuous nondecreasing function Ω : [0, +∞) → (0, +∞), such that
kf (t, v)k ≤ α(t)Ω(kvkB ), for all t ∈ J, v ∈ B;
(3) There exists an integrable function η : J → [0, +∞), such that
χ(S(s)f (t, D)) ≤ η(t)
sup
χ(D(θ))
for a.e. s, t ∈ J,
−∞≤θ≤0
where D(θ) = {v(θ) : v ∈ D}.
(H1g) The function g(·) is continuous and g(t, ·) satisfies a Lipschitz condition;
that is, there exists a positive constant Lg , such that
kg(t, v1 ) − g(t, v2 )k ≤ Lg kv1 − v2 kB , (t, vi ) ∈ J × B, i = 1, 2.
R
e b α(s)ds lim supτ →∞ Ω(τ ) ) < 1
(H1) (1) Kb (N bLg + N
τ
Rb 0
(2) Kb N Lg b + 0 η(s)ds < 1.
In this section, y : (−∞, b] → X is the function defined by y0 = ϕ and y(t) =
C(t)ϕ(0) + S(t)(z + g(0, ϕ)) on J. Clearly, kyt kB ≤ Kb kykb + Mb kϕkB , where
Kb = sup K(t),
0≤t≤b
Mb = sup M (t), kykb = sup ky(t)k.
0≤t≤b
0≤t≤b
Now we are in position to estate our main results.
Theorem 3.2. If the hypotheses (H1f), (H1g), (H1) are satisfied, then the initial
value problem (1.1)-(1.2) has at least one mild solution.
Proof. Let S(b) be the space S(b) = {x : (−∞, b] → X | x0 = 0, x|J ∈ C(J; X)}
endowed with supremum norm k · kb . Let Γ : S(b) → S(b) be the map defined by


t ∈ (−∞, 0],
0, R
t
(3.1)
(Γx)(t) = − 0 C(t − s)g(s, xs + ys )ds

 Rt
+ 0 S(t − s)f (s, xs + ys )ds, t ∈ J.
It is easy to see that kxt + yt kB ≤ Kb kykb + Mb kϕkB + Kb kxkt , where kxkt =
sup0≤s≤t kx(s)k. Thus, Γ is well defined and with values in S(b). In addition, from
the axioms of phase space, the Lebesgue dominated convergence theorem and the
conditions (H1f) (H1g), we can show that Γ is continuous.
Step 1. There exists k > 0 such that Γ(Bk ) ⊂ Bk , where Bk = {x ∈ S(b) : kxkb ≤ k}.
In fact, if we assume that the assertion is false, then for k > 0 there exist xk ∈ Bk
EJDE-2010/36
NEUTRAL FUNCTIONAL DIFFERENTIAL EQUATION
5
and tk ∈ I such that k < kΓxk (tk )k. This yields
k < kΓxk (tk )k
Z tk
Z
e
≤N
(Lg kxks + ys kB + kg(s, 0)k)ds + N
0
tk
α(s)Ω(kxks + ys kB )ds
0
b
Z
Lg (Kb kykb + Mb kϕkB + Kb k + kg(s, 0)k)ds
≤N
0
Z
b
α(s)dsΩ(Kb kykb + Mb kϕkB + Kb k)
e
+N
0
which implies
b
Z
e
1 < Kb N bLg + N
α(s)ds lim sup
k→∞
0
Ω(Kb kykb + Mb kϕkB + Kb k)
k
b
Z
e
≤ Kb (N bLg + N
α(s)ds lim sup
τ →∞
0
Ω(τ )
) < 1,
τ
which is a contradiction.
Step 2. Next, we show that Γ is χ − contraction. To clarify this, we decompose Γ
in the form Γ = Γ1 + Γ2 , for t ≥ 0, where
Z t
Γ1 x(t) = −
C(t − s)g(s, xs + ys )ds,
0
Z t
Γ2 x(t) =
S(t − s)f (s, xs + ys )ds.
0
First, we show the Γ1 is Lipschitz continuous. For arbitrary x1 , x2 ∈ Bk , from
Definition 2.1 and hypotheses, we obtain
Z t
kΓ1 x1 (t) − Γ1 x2 (t)k ≤ k
C(t − s)(g(s, x1s + ys ) − g(s, x2s + ys ))dsk
0
≤ N Lg bkx1t − x2t kB ≤ Kb N Lg bkx1t − x2t kb ;
that is, kΓ1 x1 (t) − Γ1 x2 (t)kb ≤ Kb N Lg bkx1t − x2t kb ; hence, Γ1 is Lipschitz continuous, with Lipschitz constant L0 = Kb N Lg b.
Next, taking W ⊂ Γ(Bk ). Obviously, S(t) is equicontinuous. From Lemma 2.7,
W is equicontinuous. As χC (W ) = sup {χ(W (t)), t ∈ J}, we have
Z t
χ(Γ2 W (t)) = χ(
S(t − s)f (s, Ws + ys )ds)
0
Z t
≤
η(s) sup χ(W (s + θ) + y(s + η))ds
−∞<θ≤0
0
Z
≤
t
η(s) sup χW (τ )ds
0
0≤τ ≤s
Z
≤ χC (W )
t
η(s)ds,
0
for each bounded set W ∈ C(J; X). Since
χC (ΓW ) = χC (Γ1 W + Γ2 W )
≤ χC (Γ1 W ) + χC (Γ2 W )
6
R. YE, G. ZHANG
≤ (L0 +
Z
EJDE-2010/36
t
η(s)ds))χC (W ) ≤ χC (W ),
0
the function Γ is χ-contraction. In view of Lemma 2.5, Darbo-Sadovskii fixed point
theorem, we conclude that Γ has at least one fixed point in W . Let x be a fixed
of Γ on S(b), then z = x + y is a mild solution of (1.1)-(1.2). So we deduce the
existence of a mild solution of (1.1)-(1.2).
For (1.3)-(1.4), it is possible to establish similar results as those given in the first
part of this section. Furthermore, we denote by C 1 the space of smooth functions
in the sense above described endowed with the norm kuk1 = kuk + ku0 k.
Now we define the mild solution for the initial value problem (1.3)-(1.4).
Definition 3.3. A function x : (−∞, b] → X is a mild solution of the initial value
problem (1.3)-(1.4) if x0 = ϕ, x(·)|J ∈ C 1 (J; X) and for t ∈ J,
Z t
x(t) = C(t)ϕ(0) + S(t)(z + g(0, ϕ, z)) −
C(t − s)g(s, xs , x0 (t))ds
0
Z t
+
S(t − s)f (s, xs , x0 (t))ds.
0
For (1.3)-(1.4), we assume the following hypotheses:
(H2f) f : J × B × X → X satisfies the following conditions:
(1) For each x : (−∞, b] → X, x0 = ϕ ∈ B and x|J ∈ C 1 , the function
t → f (t, xt , x0 (t)) is strongly measurable and f (t, ·, ·) is continuous for
a.e. t ∈ J;
(2) There exist an integrable function α : J → [0, +∞) and a monotone
continuous nondecreasing function Ω : [0, +∞) → (0, +∞), such that
kf (t, v, w)k ≤ α(t)Ω(kvkB + kwk),
t ∈ J, (v, w) ∈ B × X;
(3) There exist integrable functions ηi : J → [0, +∞), i = 1, 2, such that
χ(S(s)f (t, D1 , D2 )) ≤ η1 (t)
sup
χ(D1 (θ)),
−∞≤θ≤0
χ(C(s)f (t, D1 , D2 )) ≤ η2 (t)
sup
χ(D2 (θ))
for a.e. s, t ∈ J,
−∞≤θ≤0
where Di (θ) = {Di (θ) : v ∈ D}, i = 1, 2.
(H2g) There exists a positive constant Lg such that
kg(t, v1 , w1 ) − g(t, v2 , w2 )k ≤ Lg (kv1 − v2 kB + kw1 − w2 k),
(t, vi , wi ) ∈ J × B × X, i = 1, 2.
R
e b) + (N + N
e ) b α(s)ds lim supτ →∞ Ω(τ ) ) <
(H2) (1) (Kb + 1)(Lg (N b + 1 + kAkN
τ
0
1;
R
R
e b)(Kb + 1) + max { b η1 (s)ds, b η2 (s)ds} < 1.
(2) Lg (N b + 1 + kAkN
0
0
In this section, y : (−∞, b] → X is the function defined by y0 = ϕ and y(t) =
C(t)ϕ(0) + S(t)(z + g(0, ϕ, z)) on J. Clearly, kyt kB ≤ Kb kykb + Mb kϕkB , where
Kb = sup0≤t≤b K(t), Mb = sup0≤t≤b M (t), kykb = sup0≤t≤b ky(t)k.
Theorem 3.4. If the hypotheses (H2f) (H2g), (H2) are satisfied, then the initial
value problem (1.3)-(1.4) has at least one mild solution.
EJDE-2010/36
NEUTRAL FUNCTIONAL DIFFERENTIAL EQUATION
7
Proof. Let S 1 (b) be the space
S 1 (b) = {x : (−∞, b] → X : x0 = 0, x|J ∈ C 1 (J; X), x0 (0) = −g(0, ϕ, z)}
endowed with supremum norm k · k1b . Let Γ : S 1 (b) → S 1 (b) be the map defined
by


t ∈ (−∞, 0],
0, R
t
(Γx)(t) = − 0 C(t − s)g(s, xs + ys , x0 (s) + y 0 (s))ds
(3.2)

 Rt
0
0
+ 0 S(t − s)f (s, xs + ys , x (s) + y (s))ds, t ∈ J,
where y0 = ϕ and y(t) = C(t)ϕ(0) + S(t)(z + g(0, ϕ, z)) on J. It is easy to see that
kxt + yt kB ≤ Kb kykb + Mb kϕkB + Kb kxkt ,
where kxkt = sup0≤s≤t kx(s)k. Thus, Γ is well defined and with values in S 1 (b),
and
Z t
0
0
0
(Γx) (t) = −g(t, xt + yt , x (t) + y (t)) −
AS(t − s)g(s, xs + ys , x0 (s) + y 0 (s))ds
0
Z t
+
C(t − s)f (s, xs + ys , x0 (s) + y 0 (s))ds, t ∈ J.
0
In addition, from the axioms of phase space, the Lebesgue dominated convergence
theorem and the conditions (H2f), (H2g), we can show that Γ and Γ0 are continuous.
Step 1. There exists k > 0 such that Γ(Bk ) ⊂ Bk := {x ∈ S 1 (b) : kxk1b ≤ k}. In
fact, if we assume that the assertion are false, then for k > 0 there exist xk ∈ Bk
and tk ∈ J such that k < kΓxk (tk )k1 . This yields
k < kΓxk (tk )k1
= kΓxk (tk )k + k(Γxk )0 (tk )k
Z tk
≤N
(Lg (kxks + ys kB + kx0k (s) + y 0 (t)k) + kg(s, 0, 0)k)ds
0
Z tk
e α(s)Ω(kxks + ys kB + kx0k (s) + y 0 (t)k)ds
+
N
0
+ Lg (kxktk + ytk kB + kx0k (tk ) + y 0 (tk )k) + kg(tk , 0, 0)k
Z tk
e (Lg (kxks + ys kB + kx0 (s) + y 0 (s)k) + kg(s, 0, 0)k)ds
+
kAkN
k
0
Z tk
+
N α(s)Ω(kxks + ys kB + kx0k (s) + y 0 (s)k)ds
0
≤ bN Lg (Kb k + Kb kykb + Mb kϕkB + k + ky 0 kb ) + N
Z
b
kg(s, 0, 0)kds
0
Z
b
α(s)dsΩ(Kb k + Kb kykb + Mb kϕkB + k + ky 0 kb )
e
+N
0
+ Lg (Kb k + Kb kykb + Mb kϕkB + k + ky 0 kb ) + kg(tk , 0, 0)k
e Lg (Kb k + Kb kykb + Mb kϕkB + k + ky 0 kb ) + kAkN
e
+ bkAkN
Z
kg(s, 0, 0)kds
0
Z
+N
0
b
α(s)dsΩ(Kb k + Kb kykb + Mb kϕkB + k + ky 0 kb ),
b
8
R. YE, G. ZHANG
EJDE-2010/36
which implies
fb)
1 < Lg (Kb + 1)(N b + 1 + kAkN
Z b
Ω(Kb k + Kb kykb + Mb kϕkB + k + ky 0 kb )
e)
+ (N + N
α(s)ds lim sup
)
k
k→∞
0
Z b
Ω(τ )
e b) + (N + N
e)
≤ (Kb + 1)(Lg (N b + 1 + kAkN
α(s)ds lim sup
) < 1,
τ
τ →∞
0
which is a contradiction.
Step 2. Next we show that Γ is χ − contraction. To clarify this, we decompose Γ
in the form Γ = Γ1 + Γ2 , for t ≥ 0, where
Z t
C(t − s)g(s, xs + ys , x0 (s) + y 0 (s))ds,
Γ1 x(t) = −
0
Z t
Γ2 x(t) =
S(t − s)f (s, xs + ys , x0 (s) + y 0 (s))ds.
0
First, we show the Γ1 is Lipschitz continuous. For arbitrary x1 , x2 ∈ Bk , from
Definition 2.1 and hypotheses conditions, we obtain
kΓ1 x1 (t) − Γ1 x2 (t)k1
≤ kΓ1 x1 (t) − Γ1 x2 (t)k + k(Γ1 x1 )0 (t) − (Γ1 x2 )0 (t)k
Z t
≤k
C(t − s)(g(s, x1s + ys , x01 (s) + y 0 (s)) − g(s, x2s + ys , x02 (s) + y 0 (s)))dsk
0
+ kg(t, x1t + yt , x01 (t) + y 0 (t)) − g(t, x2t + yt , x02 (t) + y 0 (t))k
Z t
+k
AS(t − s)(g(s, x1s + ys , x01 (s) + y 0 (s)) − g(s, x2s + ys , x02 (s) + y 0 (s)))dsk
0
Z t
≤N
Lg (kx1s − x2s kB + kx01 (s) − x02 (s)k)ds
0
+ Lg (kx1t − x2t kB + kx01 (t) − x02 (t)k)
Z t
e
+ kAkN
Lg (kx1s − x2s kB + kx01 (s) − x02 (s)k)ds
0
Z t
≤N
Lg (K(s) sup kx1 (τ ) − x2 (τ )k + kx01 (s) − x02 (s)k)ds
0≤τ ≤s
0
+ Lg (K(t) sup kx1 (τ ) − x2 (τ )k + kx01 (t) − x02 (t)k)
0≤τ ≤t
Z
e
+ kAkN
0
t
Lg (K(s) sup kx1 (τ ) − x2 (τ )k + kx01 (s) − x02 (s)k)ds
0≤τ ≤s
e b) sup (K(t)kx1 (τ ) − x2 (τ )k + kx01 (t) − x02 (t)k)
≤ Lg (N b + 1 + kAkN
0≤τ ≤t
e b)(Kb + 1)kx1 − x2 k1 ;
≤ Lg (N b + 1 + kAkN
that is,
e b)(Kb + 1)kx1 − x2 k1b .
kΓ1 x1 (t) − Γ1 x2 (t)k1b ≤ Lg (N b + 1 + kAkN
Hence, Γ1 is Lipschitz continuous with Lipschitz constant L0 = Lg (N b + 1 +
e b)(Kb + 1).
kAkN
EJDE-2010/36
NEUTRAL FUNCTIONAL DIFFERENTIAL EQUATION
9
Next, taking W ⊂ Γ(Bk ). Obviously, S(t) is equicontinuous. From Lemma 2.7,
W is equicontinuous and χC (W ) = sup {χ(W (t)), t ∈ [a, b]}, we have
χC 1 (Γ2 W (t))
Z t
= χC 1 (
S(t − s)f (s, Ws + ys , W 0 (s) + y 0 (s))ds)
0
Z t
= max max χC (
S(t − s)f (s, Ws + ys , W 0 (s) + y 0 (s))) ds),
t∈J
0
Z t
max χC (
C(t − s)f (s, Ws + ys , W 0 (s) + y 0 (s))ds)}
t∈J
0
Z t
≤ max {max
η1 (s) sup χC (W (s + θ) + y(s + θ))ds),
t∈J
Z
max
t∈J
−∞<θ≤0
0
t
η2 (s)
χC (W 0 (s + θ) + y 0 (s + θ))ds)}
sup
−∞<θ≤0
0
Z
t
≤ max {
t
Z
η2 (s) sup χC (W 0 (τ ))ds}
η1 (s) sup χC (W (τ ))ds,
0≤τ ≤s
0
Z
b
≤ max {
Z
η2 (s)ds} max { sup χC (W (τ )), sup χC (W 0 (τ ))}
η1 (s)ds,
0
Z
0≤τ ≤b
0
b
≤ max {
Z
Z
≤ max {
η2 (s)ds} max {χC (W ), χC (W 0 )}
0
b
Z
η1 (s)ds,
0
0≤τ ≤b
b
η1 (s)ds,
0
0≤τ ≤s
0
b
b
η2 (s)ds}χC 1 (W ),
0
for each bounded set W ∈ C 1 (J; X). Since
χC 1 (ΓW ) = χC 1 (Γ1 W + Γ2 W ) ≤ χC 1 (Γ1 W ) + χC 1 (Γ2 W )
Z b
Z b
e b) + max {
≤ (Lg (N b + 1 + kAkN
η1 (s)ds,
η2 (s)ds})χC 1 (W ).
0
0
The function Γ is χ-contraction. In view of Lemma 2.5, we conclude that Γ has
at least one fixed point in W . Let x be a fixed of Γ on S 1 (b), then z = x + y
is a mild solution of (1.3)-(1.4). So we deduce the existence of a mild solution of
(1.3)-(1.4).
4. Examples
4.1. The phase space Cr × L2 (h, X). Let h(·) : (−∞, −r] → R be a positive
Lebesgue integrable function and B := Cr × L2 (h; X), r ≥ 0, be the space formed
of all classes of functions ϕ : (−∞, 0] → X such that ϕ|[−r,0] ∈ C([−r, 0], X), φ(·) is
Lebesgue-measurable on (−∞, −r] and h|ϕ|2 is Lebesgue integrable on (−∞, −r].
The seminorm in k · kB is defined by
Z −r
kϕkB := sup |ϕ(θ)| + (
h(θ)|ϕ(θ)|2 dθ)1/2 .
θ∈[−r,0]
−∞
Assume that h(·) satisfies [17, conditions (g-6) and (g-7)], function G is locally
bounded on (−∞, 0]. Proceeding as in the proof of [17, Theorem 1.3.8] it follows
that B is a phase space which satisfies the axioms (A) and (B). Moreover, when
10
R. YE, G. ZHANG
EJDE-2010/36
r = 0 this space coincides with X × L2 (h, X) and the parameter H = 1, as in [17,
R0
Theorem 1.3.8]; M (t) = G(−t)1/2 and K(t) = 1 + ( −t h(τ )dτ )1/2 , for t ≥ 0 (see
[17]).
Let X = L2 ([0, π]) and let A be the operator Af = f 00 with domain
D(A) := {f ∈ L2 ([0, π]) : f 00 ∈ L2 ([0, π]), f (0) = f (π) = 0}.
It is well known that A is the infinitesimal generator of a C0 -semigroup and of
a strongly continuous cosine function on X, which will be denoted by (C(t))t∈R .
Moreover, A has discrete spectrum, the eigenvalues are −n2 , n ∈ N , with corresponding normalized eigenvectors zn (ξ) := ( π2 )1/2 sin(nξ) and the following properties hold:
(a) {zn : n ∈ N} is an orthonormal
P∞ basis of X.
(b) For f ∈ X, (−A)−1/2 fP= n=1 n1 hf, zn izn and k(−A)−1/2 k = 1.
∞
(c) For f ∈ X, C(t)f = n=1 cos(nt)hf, zn izn . Moreover, it follow from this
P∞ sin(nt)
expression that S(t)ϕ =
hϕ, zn izn , that S(t) is compact for
n=1
n
t > 0 and that kC(t)k = 1 and kS(t)k = 1 for every t ∈ R.
(d) If Φ denotes the group of translations on X defined by Φ(t)x(ξ) = x
e(ξ + t),
where x
e is the extension of x with period 2π, then C(t) = 21 (Φ(t) + Φ(−t));
A = B 2 where B is the infinitesimal generator of the group Φ and E =
{x ∈ H 1 (0, π) : x(0) = x(π) = 0, see [6] for details.
In the next applications, B will be the phase space X × L2 (h, X).
4.2. A second order neutral equation. Now we discuss the existence of solutions for the second order neutral differential equation
Z t Z π
∂ ∂u(t, ξ)
(
+
b(t − s, η, ξ)u(s, η)dηds)
∂t
∂t
−∞ 0
(4.1)
Z t
∂ 2 u(t, ξ)
=
+
F
(t,
t
−
s,
ξ,
u(s,
ξ))ds,
t
∈
[0,
a],
ξ
∈
[0,
π],
∂ξ 2
−∞
u(t, 0) = u(t, π) = 0, t ∈ [0, a],
(4.2)
u(τ, ξ) = ϕ(τ, ξ),
τ ≤ 0, 0 ≤ ξ ≤ π,
(4.3)
2
where ϕ ∈ X × L (h; X), and
(a) The functions b(s, η, ξ), ∂b(s,η,ξ)
are measurable, b(s, η, π) = b(s, η, 0) = 0
∂ξ
and
Z πZ 0 Z π
1 ∂ i b(s, η, ξ) 2
(
Lg := max {(
) dηdsdξ)1/2 : i = 0, 1} < ∞;
i
h(s)
∂ξ
0
−∞ 0
(b) The function F : R4 → R is continuous and there is continuous function
µ : R2 → R such that
Z 0
µ(t, s)2
ds < ∞
−∞ h(s)
and |F (t, s, ξ, x)| ≤ µ(t, s)|x|, (t, s, ξ, x) ∈ R4 ;
Assuming that conditions (a),(b) are satisfied, problem (4.1)-(4.3) can be modelled as the abstract Cauchy problem (1.1)-(1.2) by defining
Z 0 Z π
g(t, ψ)(ξ) :=
b(s, ν, ξ)ψ(s, ν)dνds,
(4.4)
−∞
0
EJDE-2010/36
NEUTRAL FUNCTIONAL DIFFERENTIAL EQUATION
Z
11
0
f (t, ψ)(ξ) :=
F (t, s, ξ, ψ(s, ξ))ds.
(4.5)
−∞
R0
Moreover, kf (t, ψ)k ≤ d(t)kψkB for every t ∈ [0, a], where d(t) := ( −∞
is a Lebesgue integrable function.
The next result is a consequence of Theorem 3.2.
µ(t,s)2
1/2
h(s) ds)
Proposition 4.1. Let the previous conditions be satisfied. If
Z 0
Z a
(1 + (
h(τ )dτ )1/2 )(aLg +
d(t)dt) < 1,
−a
0
then there exists a mild solution of (4.1)-(4.3).
Acknowledgements. The authors express their gratitude to the anonymous referee for his or her valuable comments and suggestions.
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Runping Ye
Education Department of Suqian College, Jiangsu, 223800, China
E-mail address: yeziping168@sina.com
Guowei Zhang
Department of Mathematics, Northeastern University, Shenyang, 110004, China
E-mail address: gwzhangneum@sina.com