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Existence results for nonlinear functional evolution equations with delay conditions
Nonlinear Analysis 60 (2005) 1219 – 1237 www.elsevier.com/locate/na
Existence results for nonlinear functional evolution equations with delay conditions夡 Jong Soo Jung Department of Mathematics, Dong-A University, Pusan 604-714, Republic of Korea Received 24 September 2002; accepted 12 October 2004
Abstract Let X be a real Banach space, A(t) : D(A(t)) ⊂ X → 2X be an m-accretive operator, G : [0, T ] × Lp (−r, 0; X) × X → X be a mapping, Lt : Lp (−r, t; X) → X be a mapping, ut : [−r, 0] → X ¯ satisfy ut (s)=u(t +s) for every s ∈ [−r, 0], and 0 ∈ Lp (−r, 0; X) for 1 p < ∞ with 0 (0) ∈ D, ¯ where D=D(A(t)) (independent of t). The local existence of integral solutions of nonlinear functional evolution equation with delay condition du(t) + A(t)u(t) G(t, ut , Lt u), 0 t T , dt u(0) = 0 (t), −r t 0 is established in the case when the evolution operator {U (t, s)} generated by {A(t)} is equicontinuous. 䉷 2004 Elsevier Ltd. All rights reserved. Keywords: Delay condition; Equicontinuity; m-Accretive operator; Evolution operator; Compact operator; Resolvent
1. Introduction Let X be a real Banach space. We consider the abstract nonlinear functional evolution equation with delay condition du(t) + A(t)u(t) G(t, ut , Lt u), 0 t T , dt (1.1) u(0) = 0 (t), −r t 0. 夡 This
work was supported by Korea Research Foundation Grant (KRF-2001-041-D00016). E-mail addresses: [email protected], [email protected] (J.S. Jung).
0362-546X/$ - see front matter 䉷 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.10.015
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Here u : [0, T ] → X is an unknown function, A(t) : D(A(t)) ⊂ X → 2X is an m-accretive operator, G : [0, T ] × Lp (−r, 0; X) × X → X is a mapping, Lt : Lp (−r, t; X) → X is a mapping, ut : [−r, 0] → X satisfy ut (s) = u(t + s) for every s ∈ [−r, 0], and 0 ∈ ¯ where D¯ = D(A(t)) (independent of t). Lp (−r, 0; X) for 1 p < ∞ with 0 (0) ∈ D, The existence problems for this kind have been studied under various different conditions by many authors, for instance, Ha et al. [7], Ha and Shin [6], Kartsatos and Liu [9], Kartsatos and Shin [10], Mitidieri and Vrabie [11,12]. In particular, Kartsatos and Shin [10] studied nonlinear abstract functional evolution equation du(t) + A(t)u(t) G(t, ut ), dt
0t T ,
u(0) = 0 (t),
−r t 0
to prove the existence of local solutions by the compactness methods with continuity of G : [0, T ] × C([−r, 0]; X) → X and 0 ∈ C([−r, 0]; X). They utilized Schauder’s fixed point theorem in C([0, T ]; X) as in others (for instance, Mitidieri and Vrabie [11,12] in the case when A = A(t) is independent of t). Recently, using the theory of compact nonlinear semigroups and fixed point technique (Schauder’s fixed point theorem in Lp (0, T ; X) as in [17]), Ha and Shin [6] proved the existence of local solutions to problem (1.1) with Lipschitz continuity of G and Lt in the case when A = A(t) is independent of t. Their results improved those of the quasi-nonlinear t case in [7,9] when Lt u = 0 k(t, s, us ) ds for every t ∈ [0, T ] to the fully nonlinear case. In this paper, by using Schauder’s fixed point theorem in Lp (0, T ; X) for 1 p ∞, we establish the local existence of integral solutions of problem (1.1) in the case when the evolution operator U (t, s) generated by {A(t)} is equicontinuous. Further, by utilizing the relative compactness of {uf : f ∈ B} in Lp (0, T ; X) for 1 p < ∞ under the assumption that the resolvent operator J (t) = (I + A(t))−1 is compact for every > 0 and B is a bounded subset of L1 (0, T ; X) satisfying the condition of equiintegrable type, we show the existence of local solution of problem (1.1) in the case when the resolvent operator J (t) = (I + A(t))−1 is compact for every > 0. Our results improve the main results of Ha and Shin [6]. Also our work may be viewed as an attempt to develop a general existence theory for abstract delay problems of general form (1.1) under various compactness assumptions.
2. Preliminaries Let X be a real Banach space with norm and X ∗ be the dual space of X. A set-valued operator A in X with domain D(A) and range R(A) is said to be accretive if x1 −x2 x1 − x2 + (y1 − y2 ) for all > 0 and yi ∈ Ax i , i = 1, 2, . . . , A is called m-accretive if it is accretive and R(I + A) = X for all > 0. (Here I stands for the identity on X.) For each T 0, let {A(t) : 0 t T } be a family of nonlinear (possibly multivalued) operators A(t) : D(A(t)) ⊂ X → 2X with D(A(t)) = D¯ (independent of t) satisfying the following assumptions, which have been considered by Pavel [13–16]: (C1) {A(t) : 0 t T } is a family of m-accretive operators on X.
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(C2) There exist a continuous function q : [0, T ] → X, and a nondecreasing bounded (on bounded subsets) function L : [0, ∞) → [0, ∞) such that y1 − y2 , x1 − x2 s − q(t) − q(s)x1 − x2 L(max{x1 , x2 }) for all 0 s t T , y1 ∈ A(t)x1 and y2 ∈ A(s)x2 . Here the function ·, ·s : X×X → R is defined by y, xs =sup{x ∗ (y) : x ∗ ∈ J (x)}, where J : X → X∗ is the duality mapping of X. Recall that {A(t)} which satisfies conditions (C1) and (C2) generates an evolution operator U (t, s) via the following formula: U (t, s)x ≡ lim
n→∞
n
I+
j =1
−1 t −s t −s A s+j x n n
(2.1)
for all x ∈ D¯ and all 0 s t T and U (t, s) is an evolution operator in the following sense: (i) (ii) (iii) (iv)
¯ U (s, s) = I, 0 s t T , U (t, s) : D¯ → D, U (t, s)U (s, r) = U (t, r), 0 r s t T , the function (t, s, x) → U (t, s)x is continuous, ¯ U (t, s)x − U (t, s)y x − y, 0 s t, x, y ∈ D.
We say that the evolution operator U is compact if U (t, s) maps bounded subsets of D¯ into relatively compact subsets of D¯ for all 0 s t T . We refer to [8] for an example. Next, we consider the initial value problem du(t) + A(t)u(t) f (t), dt
0t T ,
u(0) = u0 ,
(2.2)
¯ where A(t) is m-accretive in X, f ∈ L1 (0, T ; X) and u0 ∈ D. ¯ is said to be an integral solution Definition 2.1 (Pavel [13]). A function u ∈ C([0, T ]; D) of problem (2.2) if u(0) = u0 and the following inequalities are satisfied: t (f () − y, u() − xs u(t) − x2 − u(s) − x2 2 s
+ Cu() − xq() − q(r)) d
(2.3)
for all 0 s t T , r ∈ [0, T ], x ∈ D(A(r)) and y ∈ A(r)x, q and L as in (C2), and C = L(max{u, x}), where u = sup0 t T u(t). It can be shown that if (C1) and (C2) are fulfilled, then for each u0 ∈ D¯ and f ∈ problem (2.2) has a unique integral solution. Moreover if v is the integral solution of the problem L1 (0, T ; X),
dv(t) + A(t)v(t) g(t), dt
v(0) = v0
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¯ then we refer to for g ∈ L1 (0, T ; X) and v0 ∈ D, t f () − g() d u(t) − v(t) u(s) − v(s) + s
(2.4)
for all 0 s t T as “Bénilan’s inequality” [2]. And the function t → U (t, s)u0 is an integral solution to the problem u (t) + A(t)u(t) 0, u(0) = u0 , 0 s t T . For these facts, we refer the reader to [13] or [14, Chapter 1]. Now we mention the relation between the evolution operator U (t, s) and the resolvent operator J (t). To this end, let U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1) and J (t) = (I + A(t))−1 be the operator from X into D¯ for each > 0. It is known that the following inequalities: t 2 U (t, s)x − x U (, s)x − x d + J (s)x − x t −s t −s s t C + q() − q(s) d (2.5) t −s s and
t −s J (s)x − x + (t − s)L(J (s)x) U (t, s)x − x 2 + × sup{q(t) − q(s) : t, s ∈ [0, T ], |t − s| T }
(2.6)
¯ 0 s t T and > 0, where C = C(x, s) is a positive constant hold for every x ∈ D, with property that C = C(x, s) is bounded on x-bounded set, and L and q are as in (C2). For proof of this result, known as the extension of Brézis’s inequalities [3], we refer to [14, Chapter 1; 16]. We prepare the following lemma, which improves that in [18] (cf. [19]). Lemma 2.2. Assume that {A(t)} satisfies conditions (C1) and (C2). Let U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1), f ∈ L1 (0, T ; X), and u0 ∈ ¯ Let u be the unique integral solution of (2.2) on [0, T ] corresponding to f and u0 . Then D. for all t ∈ (0, T ], l ∈ [0, T ) and h > 0 with t − h ∈ [0, T ], l + h ∈ [0, T ], t f () d (2.7) U (t, t − h)u(t − h) − u(t) t−h
and
U (l + h, l)u(l) − u(l + h)
l+h l
f () d
hold. Furthermore, the following inequalities hold: h f () d u(l) − u(l + h) U (h, 0)u0 − u0 + 0 l f ( + h) − f () d + 0
(2.8)
(2.9)
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
and
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h
f () d u(t) − u(t − h) U (h, 0)u0 − u0 + 0 t + f ( − h) − f () d. h
(2.10)
Proof. We prove only (2.7) since (2.8) follows in a similar way. Let v h : [t − h, t] → D¯ be a function defined by v h = U (, t − h)u(t − h), Then
vh
t − h t.
is the unique integral solution of the problem
dv h () + A()v h () 0, t − h t, v h (t − h) = u(t − h). d Applying (2.4) to v h and to u on [t − h, t], we obtain t f () d, U (t, t − h)u(t − h) − u(t) t−h
which proves (2.7). Next we prove (2.9). Eq. (2.10) follows in the same method. Let w : [0, T − h] → X be the function defined by w(l) = u(l + h),
0 l T − h.
Then w is the integral solution of problem dw(l) + A(l)w(l) f (l + h), dl From (2.4) we deduce
0 l T − h,
l
u(l) − w(l) u(0) − w(0) +
w(0) = u(h).
f ( + h) − f () d.
0
Since u(0) − w(0) = u0 − u(h), we obtain u(l) − u(l + h) U (h, 0)u0 − u0 + u(h) − U (h, 0)u0 l + f ( + h) − f () d. 0
Applying (2.8) to dominate u(h) − U (h, 0)u0 , we have (2.9), which completes the proof. Finally, let U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1). We recall that U (t, s) is said to be equicontinuous if for every bounded subset B ⊂ D¯ and every 0 s T , the family {U (·, s)y : [s, T ] → X, y ∈ B} is an equicontinuous family of functions from [s, T ] into X (cf. [14, p. 43]).
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3. Main results We begin with the following definition. Definition 3.1. A function u : [−r, T ] → X is said to be an integral solution of problem (1.1) if u(t) = 0 (t) for every t ∈ [−r, 0] and u; [0, T ] → X is an integral solution of problem (2.2) with G(t, ut , Lt u) and 0 (0) in place of f (t) and u0 , respectively. t We put ut,p = ( −r u()p d)1/p for every u ∈ Lp (−r, t; X) with t ∈ [0, T ] and consider the initial-value problem (1.1) under the following assumptions as in [7]: (L) For every t ∈ [0, T ], Lt : Lp (−r, t; X) → X satisfies (L1) Lt u − Lt v a(t)u − vt,p , (L2) Lt u − Ls u r1 (uT ,p )|t − s| for every t, s ∈ [0, T ] and u, v ∈ Lp (−r, T ; X) where a : [0, T ] → [0, ∞) is continuous and r1 : [0, ∞) → [0, ∞) is nondecreasing. (G) G : [0, T ] × Lp (−r, 0; X) × X → X satisfies (G1) G(t, , x) − G(t, , y) b(t)( − 0,p + x − y), (G2) G(t, , x) − G(s, , x) r2 (0,p , x)|t − s|, for every t, s ∈ [0, T ], , ∈ Lp (−r, 0; X) and x, y ∈ X, where b : [0, T ] → [0, ∞) is continuous and r2 : [0, ∞) × [0, ∞) → [0, ∞) is nondecreasing for both variables. Assume that (L) and (G) are satisfied. We put aT =maxt∈[0,T ] a(t) and bT =maxt∈[0,T ] b(t). Let u ∈ Lp (−r, T ; X). From (L), for every t ∈ [0, T ], Lt u Lt u − Lt 0 + Lt 0 − L0 0 + L0 0 a(t)ut,p + r1 (0)t + L0 0 at uT ,p + r1 (0)T + L0 0 = aT uT ,p + cT , where cT = r1 (0)T + L0 0 and hence from (G), for every t ∈ [0, T ], G(t, ut , Lt u) G(t, ut , Lt u) − G(t, 0, 0) + G(t, 0, 0) − G(0, 0, 0) + G(0, 0, 0) b(t)(ut 0,p + Lt u) + r2 (0, 0)t + G(0, 0, 0) bT (uT ,p + aT uT ,p + cT ) + r2 (0, 0)T + G(0, 0, 0) = (1 + aT )bT uT ,p + bT cT + r2 (0, 0)T + G(0, 0, 0). Hence we have, for every u ∈ Lp (−r, T ; X), 0
T
G(t, ut , Lt u) dt {(1 + aT )bT uT ,p + bT cT + r2 (0, 0)T + G(0, 0, 0)}T .
(3.1)
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Now we establish local existence results for nonlinear evolution Eq. (1.1) with delay condition. Theorem 3.2. Let A(t) satisfy (C1) and (C2). Assume that the evolution operator U (t, s) generated by {A(t)} via f ormula (2.1), 0 s t T , is equicontinuous and that (L) and ¯ Then there exists T ∈ (0, T0 ] (G) are satisfied. Let 0 ∈ Lp (−r, 0; X) with 0 (0) ∈ D. such that problem (1.1) has a local integral solution. 1/p
Proof. Choose 0 < T0 T , M > 0 and > 0 such that T0
T0
0
MT0 and
G(t, ut , Lt u) dt MT0
(3.2)
T for every u ∈ Lp (−r, T0 ; X) with ( 0 0 u(t) − U (t, 0)0 (0)p dt)1/p and u(t) = 0 (t) a.e. t ∈ [−r, 0], where
p 1/p 1/p T0 p MT0 = (1 + aT0 )bT0 0 0,p + U (t, 0)0 (0)p dt + 0 + bT0 cT0 + r2 (0, 0)T0 + G(0, 0, 0) T0 p
from (3.1) and uT0 ,p = (0 0,p + K=
u ∈ Lp (0, T0 ; X) :
0
T0
T0
0
u(t)p dt)1/p . Set
u(t) − U (t, 0)0 (0)p dt
1/p
and u(t) = 0 (t) a.e. t ∈ [−r, 0] . Then K is a nonempty closed convex subset of Lp (−r, T0 ; X). We set K0 = {u|[0,T0 ] : u ∈ K}, where u|[0,T0 ] is the restriction of u on [0, T0 ]. Then K0 is a nonempty closed convex subset of Lp (0, T0 ; X). Consider the initial-value problem dv(t) + A(t)v(t) G(t, ut , Lt u), dt
0 t T0 ,
v(0) = 0 (0).
(3.3)
Then it follows from 0 (0) ∈ D¯ and (3.2) that there is a unique integral solution v to problem (3.3) on [0, T0 ]. Define an operator Q : K0 → Lp (0, T0 ; X) by Qu|[0,T0 ] = v for every u|[0,T0 ] ∈ K0 . We show that Q has a fixed point. To do this, we shall make use of Schauder’s fixed point theorem.
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First, we prove that Q : K0 → K0 is a continuous operator. Let u|[0,T0 ] ∈ K0 . Then u ∈ K and u ∈ Lp (−r, T0 ; X) with u(t) = 0 (t) a.e. t ∈ [−r, 0]. Since (Qu|[0,T0 ] )(t) − U (t, 0)0 (0) we have
T0
0
T0
G(, u , L u) d,
0
1/p
p
(Qu|[0,T0 ] )(t) − U (t, 0)0 (0) dt
T0
0
T0
0
1/p
p G(, u , L u) d
dt
1/p
T0
MT0 .
We put w(t) =
(Qu|[0,T0 ] )(t), 0 t T0 , a.e. t ∈ [−r, 0]. 0 (t)
Then w ∈ Lp (−r, T0 ; X) and so w ∈ K. This implies that Qu|[0,T0 ] = w|[0,T0 ] ∈ K0 . and hence Q(K0 ) ⊂ K0 . Let w|[0,T0 ] , z|[0,T0 ] ∈ K0 . Then w, z ∈ K and hence w, z ∈ Lp (−r, T0 ; X) with w(t) = z(t)= 0 (t) a.e. t ∈ [−r, 0]. Since it follows from (2.4) that (Qw|[0,T0 ] )(t) − (Qz|[0,T0 ] )(t)
0
T0
G(, w , L w) − G(, z , L z) d,
we have, from (L1) and (G1), Qw|[0,T0 ] − Qz|[0,T0 ] Lp (0,T0 ;X) T0 1/p p = (Qw |[0,T0 ] )(t) − (Qz|[0,T0 ] )(t) dt 0 T0 1/p T0 G(, w , L w) − G(, z , L z) d 0 T0 1/p (w − z 0,p + aT0 w − z,p ) d bT0 T0 0 (1/p)+1
bT0 T0
(1 + aT0 )w − zT0 ,p T0 1/p (1/p)+1 = bT0 T0 (1 + aT0 ) w|[0,T0 ] (t) − z|[0,T0 ] (t)p dt .
(3.4)
0
Thus Q is continuous in Lp (0, T0 ; X). Now we show that Q is compact. To this end, we first note that K0 is a bounded closed convex subset of Lp (0, T0 ; X) and hence Q(K0 ) is a bounded subset of Lp (0, T0 ; X). Let u|[0,T0 ] ∈ K0 . Then u ∈ K and so u ∈ Lp (−r, T0 ; X) with u(t) = 0 (t) a.e. t ∈ [−r, 0]. Since U (t + h + , t + h) is nonexpansive and U (t + h + , t)(Qu|[0,T0 ] )(t) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) converges uniformly to zero as h → 0, by (2.8) in Lemma 2.1,
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we have U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − U (t + h + , t)(Qu|[0,T0 ] )(t) + U (t + h + , t)(Qu|[0,T0 ] )(t) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) (Qu|[0,T0 ] )(t + h) − U (t + h, t)(Qu|[0,T0 ] )(t) + (h) t+h G(, u , L u) d + (h), t
where (h) → 0 as h → 0. Hence we have (Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t) (Qu|[0,T0 ] )(t + h) − U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) + U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) + U (t + h + , t + h)(Qu|[0,T0 ] )(t) − (Qu|[0,T0 ] )(t) U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t + h) t+h G(, u , L u) d + (h) + t
+ U (t + h + , t + h)(Qu|[0,T0 ] )(t) − (Qu|[0,T0 ] )(t). Since U (t, s) is equicontinuous, by letting → 0, we obtain t+h G(, u , L u) d + (h). (Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t) t
Thus we have T0 1/p (Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t)p dt 0
T0
0 1/p
cp−1
lim
h→0
0
T0
1/p
+ T0
where c = supu∈K
t+h t
0
ch
T0
t+h t
0
T0
T0 0
G(, u , L u) d + (h)
dt
p 1/p G(, u , L u) d dt + t+h
t
1/p
p
1/p
T0 0
1/p
G(, u , L u) d dt
(h)p dt
+ T0
(h)
(h),
G(t, ut , Lt u) dt, and hence p
(Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t) dt
1/p =0
1/p
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
uniformly for u|[0,T0 ] ∈ K0 . From the same lines as those in the proof of Theorem A.1 (Proposition 2.5) in [5], it follows that {Qu|[0,T0 ] ; u|[0,T0 ] ∈ K0 } is relatively compact in Lp (0, T0 ; X), and hence Q is compact. Therefore, by Schauder’s fixed point theorem, there exists a fixed point u|[0,T0 ] ∈ K0 of Q. We define (u|[0,T0 ] )(t), 0 t T0 , u(t) ¯ = a.e. t ∈ [−r, 0]. 0 (t) It is clear that u¯ is a local integral solution of problem (1.1). This completes the proof. Theorem 3.3. Let {A(t)} satisfy (C1) and (C2). Assume that the evolution operator U (t, s) generated by {A(t)}, t ∈ [0, T ], is equicontinuous and that (L) and (G) are satisfied. Let ¯ Then problem (1.1) has a local integral solution. 0 ∈ L∞ (−r, 0; X) with 0 (0) ∈ D. Proof. Let > 0. Let u ∈ L∞ (−r, T ; X) with ess sup0 T u() and u(t) = 0 (t) a.e. t ∈ [−r, 0]. Since uT ,p =
T −r
p
1/p
u(t) dt
p
(0 0,p + p T )1/p
for every u ∈ Lp (−r, T ; X), from (3.1), we have, for 0 < h < T ,
h
0
p
G(t, ut , Lt u) dt {(1 + ah )bh (0 0,p + p T )1/p + bh ch + r2 (0, 0)h + G(0, 0, 0)}h
and hence for every > 0,
h
lim
h→0+
0
G(t, ut , Lt u) dt = 0
(3.5)
uniformly for u ∈ L∞ (−r, T ; X) with ess sup0 T u() and u(t) = 0 (t) a.e. ¯ From (3.5), there exist > 0 and 0 < T0 T such that t ∈ [−r, 0]. Let 0 (0) ∈ D.
T0
0
G(t, ut , Lt u) dt
(3.6)
uniformly for u ∈ L∞ (0, T0 ; X) with ess sup0 T0 u() , where = +max0 t T0 U (t, 0)0 (0). We define K=
u ∈ L∞ (−r, T0 ; X) : ess
sup
0 t T0
u(t) − U (t, 0)0 (0)
and u(t) = 0 (t) a.e. t ∈ [−r, 0] .
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Then, K is a nonempty closed convex subset of L∞ (−r, T0 ; X) with respect to · T0 .p . We set K0 = {u|[0,T0 ] : u ∈ K}. Then, K0 is a nonempty closed convex subset of L∞ (0, T0 ; X) with respect to ·Lp (0,T0 ;X) . As in the proof of Theorem 3.2, we consider the initial-value problem dv(t) + A(t)v(t) G(t, ut , Lt u), dt
0 t T0 ,
v(0) = 0 (0)
(3.7)
and define an operator Q : K0 → Lp (0, T0 ; X) by Qu|[0,T0 ] = v for every u|[0,T0 ] ∈ K0 , where v is a unique integral solution of problem (3.7). We shall show that Q : K0 → K0 is a continuous in Lp (0, T0 ; X). Let u|[0,T0 ] ∈ K0 . Then u ∈ K and ess sup0 t T0 u . Since it follows from (2.4) and (3.6) that T0 G(t, ut , Lt u) dt (Qu|[0,T0 ] )(t) − U (t, 0)0 (0) 0
and thus ess
sup
0 t T0
(Qu|[0,T0 ] )(t) − U (t, 0)0 (0) ,
this implies that Qu|[0,T0 ] ∈ K0 and so Q(K0 ) ⊂ K0 as in the proof of Theorem 3.2. From (3.4), Q is also continuous in Lp (0, T0 ; X). From the same lines as those in the proof of Theorem 3.2, it follows that Q is also compact. Hence, by Schauder’s fixed point theorem, there exists a fixed point u|[0,T0 ] ∈ K0 of Q. Hence, as in the proof of Theorem 3.2, problem (1.1) has a local integral solution. This completes the proof. Now we denote the integral solution u of problem (2.2) by uf , in order to exhibit the dependence of u on f ∈ L1 (0, T ; X). The following result is a time-dependent version of Theorem 4 in [17]. Lemma 3.4. Let {A(t)} satisfy (C1) and (C2) and U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1). Assume that the resolvent operator J (t)=(I +A(t))−1 from X into D(A(t)) is compact for every > 0. Let T > 0 and B be a bounded subset of L1 (0, T ; X) such that T −h f (t + h) − f (t) dt = 0 lim h→0
0
uniformly for f ∈ B and h lim f (t) dt = 0 h→0
0
uniformly for f ∈ B. Then {uf : f ∈ B} is relatively compact in Lp (0, T ; X) for every 1 p < ∞ and it is bounded in L∞ (0, T ; X).
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Proof. Let 1 p < ∞. Note that {uf (t) : f ∈ B, 0 t T } is a bounded subset of X. First we show that {J (r)uf : f ∈ B} is relatively compact in Lp (0, T ; X) for every > 0 and r ∈ [0, T ]. Since J (r) is compact for every > 0 and r ∈ [0, T ], {J (r)uf (t) : f ∈B, 0 t T } is a bounded subset in X and hence we have
T
J (r)uf (t)p dt
sup
f ∈B
0
1/p = L<∞
for every > 0 and r ∈ [0, T ]. Let f ∈ B, > 0, and r ∈ [0, T ]. Since, by (2.9) in Lemma 2.2,
T −h
0
f
p
f
1/p
J (r)u (t + h) − J (r)u (t) dt
T −h
0
T −h
uf (t + h) − uf (t)p dt
1/p
h
U (h, 0)u0 − u0 +
f () d
0
0
+
T −h
p 1/p f ( + h) − f () d dt
0
(T −h)1/p U (h, 0)u0 −u0 +
h
T −h
f ()d+
0
f (+h)−f ()d
0
and U (t, 0) is continuous at (0, 0), we have
T −h
lim
h→0
0
J (r)uf (t + h) − J (r)uf (t)p dt
1/p =0
uniformly for f ∈ B. Thus, from the same lines as those in the proof of Theorem A.1 (Proposition 2.5) in [5], it follows that {J (r)uf : f ∈ B} is relatively compact in Lp (0, T ; X) for every > 0 and r ∈ [0, T ]. Now, using (2.5), we have J (0)uf (t) − uf (t) 2 t U (, 0)uf (t) − uf (t) d + U (t, 0)uf (t) − uf (t) t 0 t C t + q() − q(s) d t 0
(3.8)
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
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for f ∈ B, t ∈ [0, T ) and > 0 with t + ∈ [0, T ], where C is the constant in (2.5). At this point, we observe that U (, t)uf (t) is the unique integral solution of the problem dv () + A()v() 0, d
v (t) = uf (t)
t t + ,
for 0 < t t + . Then, by (2.8) in Lemma 2.2, f
f
U (t + , t)u (t) − u (t + )
t
t+
f () d.
We also observe that f
f
u (t) − U (t + , t)u (t)
t+ t
f () d + uf (t + ) − uf (t).
(3.9)
From (3.8) and (3.9), we obtain for = J (t)uf (t) − uf (t) 2 t+ U (, t)uf (t) − U (t + , t)uf (t) d t
t+ 2 t+ 2 t+ f + f () d d + u (t + ) − uf (t) d t t t t+ t+ f f + f () d + u (t + ) − u (t) + C q() − q(t) d t
t
for f ∈ B, t ∈ [0, T ) and t + ∈ [0, T ]. Hence, by (2.9) in Lemma 2.2, we have J (t)uf (t) − uf (t) 2 t+ U (, t)uf (t) − U (t + , t)uf (t) d t
t+ 2 t+ 2 t+ + f () d d + U (, 0)u0 − u0 d t t t
t 2 t+ 2 t+ + f () d d + f ( + ) − f () d d t t 0 0 t+ + f () d + U (, 0)u0 − u0 + f () d t 0 t t+ + f ( + ) − f () d + C q() − q(t) d. (3.10) 0
t
Since (t, s, x) → U (t, s)x is continuous and q is continuous, there exists a nondecreasing function t : [0, ∞) → [0, ∞), which depends on t with limh↓0 t (h) = 0 and satisfies
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
the following: U (, t)uf (t) − U (t + h, t)uf (t) t (h), t+h f () d t (h), U (h, 0)u0 − u0 t (h), t t t+h f ( + h) − f () t (h), C q() − q(t) d t (h) t
0
(3.11)
for each f ∈ B, t ∈ [0, T ), > 0 with t + ∈ [0, T ] and h ∈ [0, ] and ∈ [t, t + h]. Thus, from (3.10) and (3.11), we deduce that J (t)uf (t) − uf (t) 15 t (). So, we have
T −
0
1/p f
f
p
T
15
J (t)u (t) − u (t) dt
0
t ()p dt
1/p .
Therefore
T −
lim
→0
0
1/p f
f
p
J (t)u (t) − u (t) dt
=0
uniformly for f ∈ B. This implies that {uf : f ∈ B} is relatively compact in Lp (0, T ; X), which completes the proof. Theorem 3.5. Let A(t) satisfy (C1) and (C2) and U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1). Assume that the resolvent operator J (t)=(I +A(t))−1 is compact for every > 0 and that (L) and (G) are satisfied. Let 0 ∈ L∞ (−r, 0; X) with 0 (0) ∈ D(A). Let B=
u ∈ L∞ (−r, T0 ; X) with ess
sup
0 t T0
u(t)
and u(t) = 0 (t) a.e. t ∈ [−r, 0] , where 0 T0 T and > 0 are as in the proof of Theorem 3.3. Assume that lim
h→0+
T −h −r
u(t + h) − u(t) dt = 0,
uniformly for u ∈ B. Then problem (1.1) has a local integral solution.
(3.12)
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
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¯ Define K as in the proof of Theorem 3.3. Then for every h > 0 with Proof. Let 0 (0) ∈ D. t + h T0 , we have, from (L2) and (G), G(t + h, ut+h , Lt+h u) − G(t, ut , Lt u) G(t + h, ut+h , Lt+h u) − G(t + h, ut , Lt u) + G(t + h, ut , Lt u) − G(t, ut , Lt u) bT0 (ut+h − ut 0,1 + Lt+h u − Lt u) + hr 2 (ut 0,1 , Lt u) T0 −h u( + h) − u() d + hr 1 (uT0 ,1 ) bT0 −r
+ hr 2 (0 0,1 + T0 , aT0 (0 0,1 + T0 ) + cT0 ) and hence 0
T0 −h
G(t + h, ut+h , Lt+h u) − G(t, ut , Lt u) dt T0 −h bT0 T0 u(t + h) − u(t) dt + T0 ,h ,
(3.13)
−r
where
T0 ,h = T0 {bT0 hr 1 (0 0,1 + T0 ) + hr 2 (0 0,1 + T0 , aT0 (0 0,1 + T0 ) + cT0 )} and limh→0+ T0 ,h = 0. From (3.12) and (3.13),
T0 −h
lim
h→0+
0
G(t + h, ut+h , Lt+h u) − G(t, ut , Lt u) dt = 0
(3.14)
uniformly for u ∈ B. Choose 0 < T1 T0 satisfying 1 − bT1 T12 > 0. We define K1 =
T1 −h
u∈K:
p
u(t + h) − u(t) dt
0
1/p
(h)
for every 0 < h < T1 , where 1/p
(h) =
T1
1 − bT1 T12
h
sup
u∈K
0
+ bT1 T1 T1 + T1 ,h
G(t, ut , Lt u) dt + U (h, 0)0 (0) − 0 (0)
(3.15)
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
and T1 = 30 0,1 + T1 . Then K1 is a nonempty closed convex subset of L∞ (−r, T1 ; X) with respect to · T1 ,p We set K0 = {u|[0,T1 ] ; u ∈ K1 }. Then, K0 is a nonempty closed convex subset of L∞ (0, T0 ; X) with respect to ·Lp (0,T1 ;X) . As in the proof of Theorem 3.2, we consider the initial-value problem dv(t) + A(t)v(t) G(t, ut , Lt u), dt
0 t T1 ,
v(0) = 0 (0)
(3.16)
and define an operator Q : K0 → Lp (0, T1 ; X) by Qu|[0,T1 ] = v for every u|[0,T1 ] ∈ K0 , where v is a unique integral solution of problem (3.16). We shall show that Q : K0 → K0 is a continuous in Lp (0, T1 ; X). To this end, let u|[0,T1 ] ∈ K0 . Then, u ∈ K1 and u ∈ K. Since, by (2.9) in Lemma 2.2 with u0 = 0 (0), (Qu|[0,T1 ] )(t + h) − (Qu|[0,T1 ] )(t) h U (h, 0)0 (0) − 0 (0) + G(, u , L u) d 0 T1 −h G( + h, u+h , L+h u) − G(, u , L u) d + 0 h U (h, 0)0 (0) − 0 (0) + sup G(, u , L u) d u∈K1
0
T1 −h
u( + h) − u() d + T1 ,h h U (h, 0)0 (0) − 0 (0) + sup G(, u , L u) d + bT1 T1
−r
u∈K1
+ bT1 T1 T1 +
T1 −h 0
0
u( + h) − u() d + T1 ,h ,
where
T1 −h
−r
=
u( + h) − u() d −h
−r
0 ( + h) − 0 () d + T1 −h
+ 0
T1 +
−h
u( + h) − u() d
T1 −h 0
0
u( + h) − u() d
u( + h) − 0 () d
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
1235
for 0 < h < T1 with t + h T1 , we have T1 −h 1/p p (Qu|[0,T1 ] )(t + h) − (Qu|[0,T1 ] )(t) dt 0 T1 −h
0
u∈K1
+ bT1 T1 T1 + 1/p
T1 −h 0
+ bT1 T1 T1 +
1/p T1
u∈K1
T1 −h 0
+ bT1 T1 1/p T1
G(, u , L u) d
G(, u , L u) d
u( + h) − u() d + T1 ,h u∈K1
1/q T1 + T1
h 0
U (h, 0)0 (0) − 0 (0) + sup
0
p 1/p u( + h) − u() d + T1 ,h dt
U (h, 0)0 (0) − 0 (0) + sup
T1
h
U (h, 0)0 (0) − 0 (0) + sup
T1 −h
h
G(, u , L u) d
0
u( + h) − u()p d
0
U (h, 0)0 (0) − 0 (0) + sup
u∈K1 0
h
1/p
+ T1 ,h
G(, u , L u) d
1/q + bT1 T1 (T1 + T1 (h)) + T1 ,h
= (h),
where 1/p + 1/q = 1. We put (Qu|[0,T1 ] )(t), 0 t T1 w(t) = 0 (t) a.e. t ∈ [−r, 0]. Then, w ∈ L∞ (−r, T1 ; X) and from the proof of Theorem 3.3 it follows that w ∈ K. Thus w ∈ K1 . This implies that Qu|[0,T1 ] = w|[0,T1 ] ∈ K0 , and hence Q(K0 ) ⊂ K0 . From (3.4), Q : K0 → K0 is continuous in Lp (0, T1 ; X). Also it follows from Lemma 3.4 that {Qu|[0,T1 ] : u|[0,T1 ] ∈ K0 } is relatively compact in Lp (0, T1 ; X) and hence Q : K0 → K0 is compact. Thus, by Schauder’s fixed point theorem, there exists a fixed point u|[0,T1 ] ∈ K0 . Therefore, as in the proof of Theorem 3.2, problem (1.1) has a local integral solution. This completes the proof. Remark. (1) It is known by Calvert and Kartsatos [4] and Pavel [16] that the evolution operator U (t, s) generated by {A(t)} via f ormula (2.1) is compact for every 0 s < t T if and only if the following two conditions below hold: (i) For each t ∈ [0, T ] and > 0, the resolvent operator J (t) = (I + A(t))−1 of A(t) is compact.
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
(ii) For each t0 ∈ (s, T ], the family {t → U (t, s)x : x ∈ Y } is equicontinuous at t0 on ¯ bounded subsets Y ∈ D. This result is an extension of Brézis’s result [3]. Brézis studied that the nonlinear semigroup {S(t) : D(A) → D(A), t 0} generated by constant −A = −A(t) (independent of t) is compact if and only if the resolvent operator J = (I + A)−1 is compact for every > 0 and for each bounded subset Y of X, {S(·) : [0, ∞) → X, x ∈ Y } is equicontinuous at each t > 0. Thus our results are improvements of Ha and Shin’s results in [6] as well the time-dependent version of them. In fact, using the methods of Baras [1] and Shioji [17] in Lp (0, T ; X), Ha and Shin [6] considered problem (1.1) in the case when {S(t) : D(A) → D(A), t 0} and resolvent J are compact. (2) As in [6,7,9], our existence results can be adapted to handle the nonlinear integrodifferential equations with finite delay. Acknowledgements The author is indebted to the referee for his/her valuable comments on this manuscript and the paper [4] in references. References [1] P. Baras, Compacité de l’opérateur f → u solution d’une equation non linéaire du/dt + Au f , C. R. Acad. Sci. Paris 286 (1978) 1113–1116. [2] Ph. Bénilan, Solutions integrales d’équation d’évolution dans un espace de Banach, C. R. Acad. Sci. Paris 224 (1972) 47–50. [3] H. Brézis, New results concerning monotone operators and nonlinear semigroups, in: Proceedings of the RIMS (on “Analysis of Nonlinear Problems”), Kyoto University, 1975. [4] B. Calvert, A.G. Kartsatos, On the compactness of the nonlinear evolution operator in a Banach space, Bull. London Math. Soc. 19 (6) (1987) 551–558. [5] S. Gutman, Compact perturbations of m-accretive operators in general Banach spaces, SIAM J. Math. Anal. 13 (1982) 789–800. [6] K.S. Ha, K. Shin, Local existence of integral solutions via compactness methods for nonlinear functional differential equations in Lp spaces, Nonlinear Anal. 45 (2001) 447–458. [7] K.S. Ha, K. Shin, B.J. Jin, Existence of solutions of functional integrodifferential equations in Banach spaces, Differential Integral Equations 8 (1995) 553–566. [8] A.G. Kartsatos, A compact evolution operator generated by a nonlinear time-dependent m-accretive operator in a Banach space, Math. Ann. 302 (1995) 477–487. [9] A.G. Kartsatos, X. Liu, On the construction and the convergence of the method of lines for quasi-nonlinear functional evolutions in general Banach spaces, Nonlinear Anal. 29 (1997) 385–414. [10] A.G. Kartsatos, K. Shin, Solvability of functional evolution via compactness methods in general Banach spaces, Nonlinear Anal. 21 (1993) 517–535. [11] E. Mitidieri, I.I. Vrabie, Existence for nonlinear functional differential equations, Hiroshima Math. J. 17 (1987) 627–649. [12] E. Mitidieri, I.I.Vrabie, Differential inclusions governed by nonconvex perturbations of m-accretive operators, Differential Integral Equations 2 (1989) 525–531. [13] N.H. Pavel, Nonlinear evolution equations governed f-quasi-dissipative operators, Nonlinear Anal. 5 (1981) 449–468. [14] N.H. Pavel, Nonlinear Evolution Operators and Semigroups, Lecture Notes in Mathematics, vol. 1260, Springer, Berlin, 1987.
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[15] N.H. Pavel, Differential equations associated with compact evolution generators, Differential Integral Equations 1 (1988) 117–123. [16] N.H. Pavel, Compact evolution operators, Differential Integral Equations 2 (1989) 57–62. [17] N. Shioji, Local existence theorems for nonlinear differential equations and compactness of integral solutions in Lp (0, T ; X), Nonlinear Anal. 26 (1996) 799–811. [18] I.I. Vrabie, Compactness methods for an abstract nonlinear Volterra integrodifferential equation, Nonlinear Anal. 5 (1981) 365–371. [19] I.I. Vrabie, Compactness Methods for Nonlinear Evolutions, Pitman Monographs Surveys, Pure and Applied Mathematics, vol. 32, Longman Scientific and Technical, 1987.
Existence results for nonlinear functional evolution equations with delay conditions夡 Jong Soo Jung Department of Mathematics, Dong-A University, Pusan 604-714, Republic of Korea Received 24 September 2002; accepted 12 October 2004
Abstract Let X be a real Banach space, A(t) : D(A(t)) ⊂ X → 2X be an m-accretive operator, G : [0, T ] × Lp (−r, 0; X) × X → X be a mapping, Lt : Lp (−r, t; X) → X be a mapping, ut : [−r, 0] → X ¯ satisfy ut (s)=u(t +s) for every s ∈ [−r, 0], and 0 ∈ Lp (−r, 0; X) for 1 p < ∞ with 0 (0) ∈ D, ¯ where D=D(A(t)) (independent of t). The local existence of integral solutions of nonlinear functional evolution equation with delay condition du(t) + A(t)u(t) G(t, ut , Lt u), 0 t T , dt u(0) = 0 (t), −r t 0 is established in the case when the evolution operator {U (t, s)} generated by {A(t)} is equicontinuous. 䉷 2004 Elsevier Ltd. All rights reserved. Keywords: Delay condition; Equicontinuity; m-Accretive operator; Evolution operator; Compact operator; Resolvent
1. Introduction Let X be a real Banach space. We consider the abstract nonlinear functional evolution equation with delay condition du(t) + A(t)u(t) G(t, ut , Lt u), 0 t T , dt (1.1) u(0) = 0 (t), −r t 0. 夡 This
work was supported by Korea Research Foundation Grant (KRF-2001-041-D00016). E-mail addresses: [email protected], [email protected] (J.S. Jung).
0362-546X/$ - see front matter 䉷 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.10.015
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
Here u : [0, T ] → X is an unknown function, A(t) : D(A(t)) ⊂ X → 2X is an m-accretive operator, G : [0, T ] × Lp (−r, 0; X) × X → X is a mapping, Lt : Lp (−r, t; X) → X is a mapping, ut : [−r, 0] → X satisfy ut (s) = u(t + s) for every s ∈ [−r, 0], and 0 ∈ ¯ where D¯ = D(A(t)) (independent of t). Lp (−r, 0; X) for 1 p < ∞ with 0 (0) ∈ D, The existence problems for this kind have been studied under various different conditions by many authors, for instance, Ha et al. [7], Ha and Shin [6], Kartsatos and Liu [9], Kartsatos and Shin [10], Mitidieri and Vrabie [11,12]. In particular, Kartsatos and Shin [10] studied nonlinear abstract functional evolution equation du(t) + A(t)u(t) G(t, ut ), dt
0t T ,
u(0) = 0 (t),
−r t 0
to prove the existence of local solutions by the compactness methods with continuity of G : [0, T ] × C([−r, 0]; X) → X and 0 ∈ C([−r, 0]; X). They utilized Schauder’s fixed point theorem in C([0, T ]; X) as in others (for instance, Mitidieri and Vrabie [11,12] in the case when A = A(t) is independent of t). Recently, using the theory of compact nonlinear semigroups and fixed point technique (Schauder’s fixed point theorem in Lp (0, T ; X) as in [17]), Ha and Shin [6] proved the existence of local solutions to problem (1.1) with Lipschitz continuity of G and Lt in the case when A = A(t) is independent of t. Their results improved those of the quasi-nonlinear t case in [7,9] when Lt u = 0 k(t, s, us ) ds for every t ∈ [0, T ] to the fully nonlinear case. In this paper, by using Schauder’s fixed point theorem in Lp (0, T ; X) for 1 p ∞, we establish the local existence of integral solutions of problem (1.1) in the case when the evolution operator U (t, s) generated by {A(t)} is equicontinuous. Further, by utilizing the relative compactness of {uf : f ∈ B} in Lp (0, T ; X) for 1 p < ∞ under the assumption that the resolvent operator J (t) = (I + A(t))−1 is compact for every > 0 and B is a bounded subset of L1 (0, T ; X) satisfying the condition of equiintegrable type, we show the existence of local solution of problem (1.1) in the case when the resolvent operator J (t) = (I + A(t))−1 is compact for every > 0. Our results improve the main results of Ha and Shin [6]. Also our work may be viewed as an attempt to develop a general existence theory for abstract delay problems of general form (1.1) under various compactness assumptions.
2. Preliminaries Let X be a real Banach space with norm and X ∗ be the dual space of X. A set-valued operator A in X with domain D(A) and range R(A) is said to be accretive if x1 −x2 x1 − x2 + (y1 − y2 ) for all > 0 and yi ∈ Ax i , i = 1, 2, . . . , A is called m-accretive if it is accretive and R(I + A) = X for all > 0. (Here I stands for the identity on X.) For each T 0, let {A(t) : 0 t T } be a family of nonlinear (possibly multivalued) operators A(t) : D(A(t)) ⊂ X → 2X with D(A(t)) = D¯ (independent of t) satisfying the following assumptions, which have been considered by Pavel [13–16]: (C1) {A(t) : 0 t T } is a family of m-accretive operators on X.
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
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(C2) There exist a continuous function q : [0, T ] → X, and a nondecreasing bounded (on bounded subsets) function L : [0, ∞) → [0, ∞) such that y1 − y2 , x1 − x2 s − q(t) − q(s)x1 − x2 L(max{x1 , x2 }) for all 0 s t T , y1 ∈ A(t)x1 and y2 ∈ A(s)x2 . Here the function ·, ·s : X×X → R is defined by y, xs =sup{x ∗ (y) : x ∗ ∈ J (x)}, where J : X → X∗ is the duality mapping of X. Recall that {A(t)} which satisfies conditions (C1) and (C2) generates an evolution operator U (t, s) via the following formula: U (t, s)x ≡ lim
n→∞
n
I+
j =1
−1 t −s t −s A s+j x n n
(2.1)
for all x ∈ D¯ and all 0 s t T and U (t, s) is an evolution operator in the following sense: (i) (ii) (iii) (iv)
¯ U (s, s) = I, 0 s t T , U (t, s) : D¯ → D, U (t, s)U (s, r) = U (t, r), 0 r s t T , the function (t, s, x) → U (t, s)x is continuous, ¯ U (t, s)x − U (t, s)y x − y, 0 s t, x, y ∈ D.
We say that the evolution operator U is compact if U (t, s) maps bounded subsets of D¯ into relatively compact subsets of D¯ for all 0 s t T . We refer to [8] for an example. Next, we consider the initial value problem du(t) + A(t)u(t) f (t), dt
0t T ,
u(0) = u0 ,
(2.2)
¯ where A(t) is m-accretive in X, f ∈ L1 (0, T ; X) and u0 ∈ D. ¯ is said to be an integral solution Definition 2.1 (Pavel [13]). A function u ∈ C([0, T ]; D) of problem (2.2) if u(0) = u0 and the following inequalities are satisfied: t (f () − y, u() − xs u(t) − x2 − u(s) − x2 2 s
+ Cu() − xq() − q(r)) d
(2.3)
for all 0 s t T , r ∈ [0, T ], x ∈ D(A(r)) and y ∈ A(r)x, q and L as in (C2), and C = L(max{u, x}), where u = sup0 t T u(t). It can be shown that if (C1) and (C2) are fulfilled, then for each u0 ∈ D¯ and f ∈ problem (2.2) has a unique integral solution. Moreover if v is the integral solution of the problem L1 (0, T ; X),
dv(t) + A(t)v(t) g(t), dt
v(0) = v0
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
¯ then we refer to for g ∈ L1 (0, T ; X) and v0 ∈ D, t f () − g() d u(t) − v(t) u(s) − v(s) + s
(2.4)
for all 0 s t T as “Bénilan’s inequality” [2]. And the function t → U (t, s)u0 is an integral solution to the problem u (t) + A(t)u(t) 0, u(0) = u0 , 0 s t T . For these facts, we refer the reader to [13] or [14, Chapter 1]. Now we mention the relation between the evolution operator U (t, s) and the resolvent operator J (t). To this end, let U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1) and J (t) = (I + A(t))−1 be the operator from X into D¯ for each > 0. It is known that the following inequalities: t 2 U (t, s)x − x U (, s)x − x d + J (s)x − x t −s t −s s t C + q() − q(s) d (2.5) t −s s and
t −s J (s)x − x + (t − s)L(J (s)x) U (t, s)x − x 2 + × sup{q(t) − q(s) : t, s ∈ [0, T ], |t − s| T }
(2.6)
¯ 0 s t T and > 0, where C = C(x, s) is a positive constant hold for every x ∈ D, with property that C = C(x, s) is bounded on x-bounded set, and L and q are as in (C2). For proof of this result, known as the extension of Brézis’s inequalities [3], we refer to [14, Chapter 1; 16]. We prepare the following lemma, which improves that in [18] (cf. [19]). Lemma 2.2. Assume that {A(t)} satisfies conditions (C1) and (C2). Let U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1), f ∈ L1 (0, T ; X), and u0 ∈ ¯ Let u be the unique integral solution of (2.2) on [0, T ] corresponding to f and u0 . Then D. for all t ∈ (0, T ], l ∈ [0, T ) and h > 0 with t − h ∈ [0, T ], l + h ∈ [0, T ], t f () d (2.7) U (t, t − h)u(t − h) − u(t) t−h
and
U (l + h, l)u(l) − u(l + h)
l+h l
f () d
hold. Furthermore, the following inequalities hold: h f () d u(l) − u(l + h) U (h, 0)u0 − u0 + 0 l f ( + h) − f () d + 0
(2.8)
(2.9)
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
and
1223
h
f () d u(t) − u(t − h) U (h, 0)u0 − u0 + 0 t + f ( − h) − f () d. h
(2.10)
Proof. We prove only (2.7) since (2.8) follows in a similar way. Let v h : [t − h, t] → D¯ be a function defined by v h = U (, t − h)u(t − h), Then
vh
t − h t.
is the unique integral solution of the problem
dv h () + A()v h () 0, t − h t, v h (t − h) = u(t − h). d Applying (2.4) to v h and to u on [t − h, t], we obtain t f () d, U (t, t − h)u(t − h) − u(t) t−h
which proves (2.7). Next we prove (2.9). Eq. (2.10) follows in the same method. Let w : [0, T − h] → X be the function defined by w(l) = u(l + h),
0 l T − h.
Then w is the integral solution of problem dw(l) + A(l)w(l) f (l + h), dl From (2.4) we deduce
0 l T − h,
l
u(l) − w(l) u(0) − w(0) +
w(0) = u(h).
f ( + h) − f () d.
0
Since u(0) − w(0) = u0 − u(h), we obtain u(l) − u(l + h) U (h, 0)u0 − u0 + u(h) − U (h, 0)u0 l + f ( + h) − f () d. 0
Applying (2.8) to dominate u(h) − U (h, 0)u0 , we have (2.9), which completes the proof. Finally, let U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1). We recall that U (t, s) is said to be equicontinuous if for every bounded subset B ⊂ D¯ and every 0 s T , the family {U (·, s)y : [s, T ] → X, y ∈ B} is an equicontinuous family of functions from [s, T ] into X (cf. [14, p. 43]).
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
3. Main results We begin with the following definition. Definition 3.1. A function u : [−r, T ] → X is said to be an integral solution of problem (1.1) if u(t) = 0 (t) for every t ∈ [−r, 0] and u; [0, T ] → X is an integral solution of problem (2.2) with G(t, ut , Lt u) and 0 (0) in place of f (t) and u0 , respectively. t We put ut,p = ( −r u()p d)1/p for every u ∈ Lp (−r, t; X) with t ∈ [0, T ] and consider the initial-value problem (1.1) under the following assumptions as in [7]: (L) For every t ∈ [0, T ], Lt : Lp (−r, t; X) → X satisfies (L1) Lt u − Lt v a(t)u − vt,p , (L2) Lt u − Ls u r1 (uT ,p )|t − s| for every t, s ∈ [0, T ] and u, v ∈ Lp (−r, T ; X) where a : [0, T ] → [0, ∞) is continuous and r1 : [0, ∞) → [0, ∞) is nondecreasing. (G) G : [0, T ] × Lp (−r, 0; X) × X → X satisfies (G1) G(t, , x) − G(t, , y) b(t)( − 0,p + x − y), (G2) G(t, , x) − G(s, , x) r2 (0,p , x)|t − s|, for every t, s ∈ [0, T ], , ∈ Lp (−r, 0; X) and x, y ∈ X, where b : [0, T ] → [0, ∞) is continuous and r2 : [0, ∞) × [0, ∞) → [0, ∞) is nondecreasing for both variables. Assume that (L) and (G) are satisfied. We put aT =maxt∈[0,T ] a(t) and bT =maxt∈[0,T ] b(t). Let u ∈ Lp (−r, T ; X). From (L), for every t ∈ [0, T ], Lt u Lt u − Lt 0 + Lt 0 − L0 0 + L0 0 a(t)ut,p + r1 (0)t + L0 0 at uT ,p + r1 (0)T + L0 0 = aT uT ,p + cT , where cT = r1 (0)T + L0 0 and hence from (G), for every t ∈ [0, T ], G(t, ut , Lt u) G(t, ut , Lt u) − G(t, 0, 0) + G(t, 0, 0) − G(0, 0, 0) + G(0, 0, 0) b(t)(ut 0,p + Lt u) + r2 (0, 0)t + G(0, 0, 0) bT (uT ,p + aT uT ,p + cT ) + r2 (0, 0)T + G(0, 0, 0) = (1 + aT )bT uT ,p + bT cT + r2 (0, 0)T + G(0, 0, 0). Hence we have, for every u ∈ Lp (−r, T ; X), 0
T
G(t, ut , Lt u) dt {(1 + aT )bT uT ,p + bT cT + r2 (0, 0)T + G(0, 0, 0)}T .
(3.1)
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
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Now we establish local existence results for nonlinear evolution Eq. (1.1) with delay condition. Theorem 3.2. Let A(t) satisfy (C1) and (C2). Assume that the evolution operator U (t, s) generated by {A(t)} via f ormula (2.1), 0 s t T , is equicontinuous and that (L) and ¯ Then there exists T ∈ (0, T0 ] (G) are satisfied. Let 0 ∈ Lp (−r, 0; X) with 0 (0) ∈ D. such that problem (1.1) has a local integral solution. 1/p
Proof. Choose 0 < T0 T , M > 0 and > 0 such that T0
T0
0
MT0 and
G(t, ut , Lt u) dt MT0
(3.2)
T for every u ∈ Lp (−r, T0 ; X) with ( 0 0 u(t) − U (t, 0)0 (0)p dt)1/p and u(t) = 0 (t) a.e. t ∈ [−r, 0], where
p 1/p 1/p T0 p MT0 = (1 + aT0 )bT0 0 0,p + U (t, 0)0 (0)p dt + 0 + bT0 cT0 + r2 (0, 0)T0 + G(0, 0, 0) T0 p
from (3.1) and uT0 ,p = (0 0,p + K=
u ∈ Lp (0, T0 ; X) :
0
T0
T0
0
u(t)p dt)1/p . Set
u(t) − U (t, 0)0 (0)p dt
1/p
and u(t) = 0 (t) a.e. t ∈ [−r, 0] . Then K is a nonempty closed convex subset of Lp (−r, T0 ; X). We set K0 = {u|[0,T0 ] : u ∈ K}, where u|[0,T0 ] is the restriction of u on [0, T0 ]. Then K0 is a nonempty closed convex subset of Lp (0, T0 ; X). Consider the initial-value problem dv(t) + A(t)v(t) G(t, ut , Lt u), dt
0 t T0 ,
v(0) = 0 (0).
(3.3)
Then it follows from 0 (0) ∈ D¯ and (3.2) that there is a unique integral solution v to problem (3.3) on [0, T0 ]. Define an operator Q : K0 → Lp (0, T0 ; X) by Qu|[0,T0 ] = v for every u|[0,T0 ] ∈ K0 . We show that Q has a fixed point. To do this, we shall make use of Schauder’s fixed point theorem.
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First, we prove that Q : K0 → K0 is a continuous operator. Let u|[0,T0 ] ∈ K0 . Then u ∈ K and u ∈ Lp (−r, T0 ; X) with u(t) = 0 (t) a.e. t ∈ [−r, 0]. Since (Qu|[0,T0 ] )(t) − U (t, 0)0 (0) we have
T0
0
T0
G(, u , L u) d,
0
1/p
p
(Qu|[0,T0 ] )(t) − U (t, 0)0 (0) dt
T0
0
T0
0
1/p
p G(, u , L u) d
dt
1/p
T0
MT0 .
We put w(t) =
(Qu|[0,T0 ] )(t), 0 t T0 , a.e. t ∈ [−r, 0]. 0 (t)
Then w ∈ Lp (−r, T0 ; X) and so w ∈ K. This implies that Qu|[0,T0 ] = w|[0,T0 ] ∈ K0 . and hence Q(K0 ) ⊂ K0 . Let w|[0,T0 ] , z|[0,T0 ] ∈ K0 . Then w, z ∈ K and hence w, z ∈ Lp (−r, T0 ; X) with w(t) = z(t)= 0 (t) a.e. t ∈ [−r, 0]. Since it follows from (2.4) that (Qw|[0,T0 ] )(t) − (Qz|[0,T0 ] )(t)
0
T0
G(, w , L w) − G(, z , L z) d,
we have, from (L1) and (G1), Qw|[0,T0 ] − Qz|[0,T0 ] Lp (0,T0 ;X) T0 1/p p = (Qw |[0,T0 ] )(t) − (Qz|[0,T0 ] )(t) dt 0 T0 1/p T0 G(, w , L w) − G(, z , L z) d 0 T0 1/p (w − z 0,p + aT0 w − z,p ) d bT0 T0 0 (1/p)+1
bT0 T0
(1 + aT0 )w − zT0 ,p T0 1/p (1/p)+1 = bT0 T0 (1 + aT0 ) w|[0,T0 ] (t) − z|[0,T0 ] (t)p dt .
(3.4)
0
Thus Q is continuous in Lp (0, T0 ; X). Now we show that Q is compact. To this end, we first note that K0 is a bounded closed convex subset of Lp (0, T0 ; X) and hence Q(K0 ) is a bounded subset of Lp (0, T0 ; X). Let u|[0,T0 ] ∈ K0 . Then u ∈ K and so u ∈ Lp (−r, T0 ; X) with u(t) = 0 (t) a.e. t ∈ [−r, 0]. Since U (t + h + , t + h) is nonexpansive and U (t + h + , t)(Qu|[0,T0 ] )(t) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) converges uniformly to zero as h → 0, by (2.8) in Lemma 2.1,
J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
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we have U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − U (t + h + , t)(Qu|[0,T0 ] )(t) + U (t + h + , t)(Qu|[0,T0 ] )(t) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) (Qu|[0,T0 ] )(t + h) − U (t + h, t)(Qu|[0,T0 ] )(t) + (h) t+h G(, u , L u) d + (h), t
where (h) → 0 as h → 0. Hence we have (Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t) (Qu|[0,T0 ] )(t + h) − U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) + U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − U (t + h + , t + h)(Qu|[0,T0 ] )(t) + U (t + h + , t + h)(Qu|[0,T0 ] )(t) − (Qu|[0,T0 ] )(t) U (t + h + , t + h)(Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t + h) t+h G(, u , L u) d + (h) + t
+ U (t + h + , t + h)(Qu|[0,T0 ] )(t) − (Qu|[0,T0 ] )(t). Since U (t, s) is equicontinuous, by letting → 0, we obtain t+h G(, u , L u) d + (h). (Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t) t
Thus we have T0 1/p (Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t)p dt 0
T0
0 1/p
cp−1
lim
h→0
0
T0
1/p
+ T0
where c = supu∈K
t+h t
0
ch
T0
t+h t
0
T0
T0 0
G(, u , L u) d + (h)
dt
p 1/p G(, u , L u) d dt + t+h
t
1/p
p
1/p
T0 0
1/p
G(, u , L u) d dt
(h)p dt
+ T0
(h)
(h),
G(t, ut , Lt u) dt, and hence p
(Qu|[0,T0 ] )(t + h) − (Qu|[0,T0 ] )(t) dt
1/p =0
1/p
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
uniformly for u|[0,T0 ] ∈ K0 . From the same lines as those in the proof of Theorem A.1 (Proposition 2.5) in [5], it follows that {Qu|[0,T0 ] ; u|[0,T0 ] ∈ K0 } is relatively compact in Lp (0, T0 ; X), and hence Q is compact. Therefore, by Schauder’s fixed point theorem, there exists a fixed point u|[0,T0 ] ∈ K0 of Q. We define (u|[0,T0 ] )(t), 0 t T0 , u(t) ¯ = a.e. t ∈ [−r, 0]. 0 (t) It is clear that u¯ is a local integral solution of problem (1.1). This completes the proof. Theorem 3.3. Let {A(t)} satisfy (C1) and (C2). Assume that the evolution operator U (t, s) generated by {A(t)}, t ∈ [0, T ], is equicontinuous and that (L) and (G) are satisfied. Let ¯ Then problem (1.1) has a local integral solution. 0 ∈ L∞ (−r, 0; X) with 0 (0) ∈ D. Proof. Let > 0. Let u ∈ L∞ (−r, T ; X) with ess sup0 T u() and u(t) = 0 (t) a.e. t ∈ [−r, 0]. Since uT ,p =
T −r
p
1/p
u(t) dt
p
(0 0,p + p T )1/p
for every u ∈ Lp (−r, T ; X), from (3.1), we have, for 0 < h < T ,
h
0
p
G(t, ut , Lt u) dt {(1 + ah )bh (0 0,p + p T )1/p + bh ch + r2 (0, 0)h + G(0, 0, 0)}h
and hence for every > 0,
h
lim
h→0+
0
G(t, ut , Lt u) dt = 0
(3.5)
uniformly for u ∈ L∞ (−r, T ; X) with ess sup0 T u() and u(t) = 0 (t) a.e. ¯ From (3.5), there exist > 0 and 0 < T0 T such that t ∈ [−r, 0]. Let 0 (0) ∈ D.
T0
0
G(t, ut , Lt u) dt
(3.6)
uniformly for u ∈ L∞ (0, T0 ; X) with ess sup0 T0 u() , where = +max0 t T0 U (t, 0)0 (0). We define K=
u ∈ L∞ (−r, T0 ; X) : ess
sup
0 t T0
u(t) − U (t, 0)0 (0)
and u(t) = 0 (t) a.e. t ∈ [−r, 0] .
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Then, K is a nonempty closed convex subset of L∞ (−r, T0 ; X) with respect to · T0 .p . We set K0 = {u|[0,T0 ] : u ∈ K}. Then, K0 is a nonempty closed convex subset of L∞ (0, T0 ; X) with respect to ·Lp (0,T0 ;X) . As in the proof of Theorem 3.2, we consider the initial-value problem dv(t) + A(t)v(t) G(t, ut , Lt u), dt
0 t T0 ,
v(0) = 0 (0)
(3.7)
and define an operator Q : K0 → Lp (0, T0 ; X) by Qu|[0,T0 ] = v for every u|[0,T0 ] ∈ K0 , where v is a unique integral solution of problem (3.7). We shall show that Q : K0 → K0 is a continuous in Lp (0, T0 ; X). Let u|[0,T0 ] ∈ K0 . Then u ∈ K and ess sup0 t T0 u . Since it follows from (2.4) and (3.6) that T0 G(t, ut , Lt u) dt (Qu|[0,T0 ] )(t) − U (t, 0)0 (0) 0
and thus ess
sup
0 t T0
(Qu|[0,T0 ] )(t) − U (t, 0)0 (0) ,
this implies that Qu|[0,T0 ] ∈ K0 and so Q(K0 ) ⊂ K0 as in the proof of Theorem 3.2. From (3.4), Q is also continuous in Lp (0, T0 ; X). From the same lines as those in the proof of Theorem 3.2, it follows that Q is also compact. Hence, by Schauder’s fixed point theorem, there exists a fixed point u|[0,T0 ] ∈ K0 of Q. Hence, as in the proof of Theorem 3.2, problem (1.1) has a local integral solution. This completes the proof. Now we denote the integral solution u of problem (2.2) by uf , in order to exhibit the dependence of u on f ∈ L1 (0, T ; X). The following result is a time-dependent version of Theorem 4 in [17]. Lemma 3.4. Let {A(t)} satisfy (C1) and (C2) and U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1). Assume that the resolvent operator J (t)=(I +A(t))−1 from X into D(A(t)) is compact for every > 0. Let T > 0 and B be a bounded subset of L1 (0, T ; X) such that T −h f (t + h) − f (t) dt = 0 lim h→0
0
uniformly for f ∈ B and h lim f (t) dt = 0 h→0
0
uniformly for f ∈ B. Then {uf : f ∈ B} is relatively compact in Lp (0, T ; X) for every 1 p < ∞ and it is bounded in L∞ (0, T ; X).
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Proof. Let 1 p < ∞. Note that {uf (t) : f ∈ B, 0 t T } is a bounded subset of X. First we show that {J (r)uf : f ∈ B} is relatively compact in Lp (0, T ; X) for every > 0 and r ∈ [0, T ]. Since J (r) is compact for every > 0 and r ∈ [0, T ], {J (r)uf (t) : f ∈B, 0 t T } is a bounded subset in X and hence we have
T
J (r)uf (t)p dt
sup
f ∈B
0
1/p = L<∞
for every > 0 and r ∈ [0, T ]. Let f ∈ B, > 0, and r ∈ [0, T ]. Since, by (2.9) in Lemma 2.2,
T −h
0
f
p
f
1/p
J (r)u (t + h) − J (r)u (t) dt
T −h
0
T −h
uf (t + h) − uf (t)p dt
1/p
h
U (h, 0)u0 − u0 +
f () d
0
0
+
T −h
p 1/p f ( + h) − f () d dt
0
(T −h)1/p U (h, 0)u0 −u0 +
h
T −h
f ()d+
0
f (+h)−f ()d
0
and U (t, 0) is continuous at (0, 0), we have
T −h
lim
h→0
0
J (r)uf (t + h) − J (r)uf (t)p dt
1/p =0
uniformly for f ∈ B. Thus, from the same lines as those in the proof of Theorem A.1 (Proposition 2.5) in [5], it follows that {J (r)uf : f ∈ B} is relatively compact in Lp (0, T ; X) for every > 0 and r ∈ [0, T ]. Now, using (2.5), we have J (0)uf (t) − uf (t) 2 t U (, 0)uf (t) − uf (t) d + U (t, 0)uf (t) − uf (t) t 0 t C t + q() − q(s) d t 0
(3.8)
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for f ∈ B, t ∈ [0, T ) and > 0 with t + ∈ [0, T ], where C is the constant in (2.5). At this point, we observe that U (, t)uf (t) is the unique integral solution of the problem dv () + A()v() 0, d
v (t) = uf (t)
t t + ,
for 0 < t t + . Then, by (2.8) in Lemma 2.2, f
f
U (t + , t)u (t) − u (t + )
t
t+
f () d.
We also observe that f
f
u (t) − U (t + , t)u (t)
t+ t
f () d + uf (t + ) − uf (t).
(3.9)
From (3.8) and (3.9), we obtain for = J (t)uf (t) − uf (t) 2 t+ U (, t)uf (t) − U (t + , t)uf (t) d t
t+ 2 t+ 2 t+ f + f () d d + u (t + ) − uf (t) d t t t t+ t+ f f + f () d + u (t + ) − u (t) + C q() − q(t) d t
t
for f ∈ B, t ∈ [0, T ) and t + ∈ [0, T ]. Hence, by (2.9) in Lemma 2.2, we have J (t)uf (t) − uf (t) 2 t+ U (, t)uf (t) − U (t + , t)uf (t) d t
t+ 2 t+ 2 t+ + f () d d + U (, 0)u0 − u0 d t t t
t 2 t+ 2 t+ + f () d d + f ( + ) − f () d d t t 0 0 t+ + f () d + U (, 0)u0 − u0 + f () d t 0 t t+ + f ( + ) − f () d + C q() − q(t) d. (3.10) 0
t
Since (t, s, x) → U (t, s)x is continuous and q is continuous, there exists a nondecreasing function t : [0, ∞) → [0, ∞), which depends on t with limh↓0 t (h) = 0 and satisfies
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
the following: U (, t)uf (t) − U (t + h, t)uf (t) t (h), t+h f () d t (h), U (h, 0)u0 − u0 t (h), t t t+h f ( + h) − f () t (h), C q() − q(t) d t (h) t
0
(3.11)
for each f ∈ B, t ∈ [0, T ), > 0 with t + ∈ [0, T ] and h ∈ [0, ] and ∈ [t, t + h]. Thus, from (3.10) and (3.11), we deduce that J (t)uf (t) − uf (t) 15 t (). So, we have
T −
0
1/p f
f
p
T
15
J (t)u (t) − u (t) dt
0
t ()p dt
1/p .
Therefore
T −
lim
→0
0
1/p f
f
p
J (t)u (t) − u (t) dt
=0
uniformly for f ∈ B. This implies that {uf : f ∈ B} is relatively compact in Lp (0, T ; X), which completes the proof. Theorem 3.5. Let A(t) satisfy (C1) and (C2) and U (t, s) be the evolution operator generated by {A(t)} via f ormula (2.1). Assume that the resolvent operator J (t)=(I +A(t))−1 is compact for every > 0 and that (L) and (G) are satisfied. Let 0 ∈ L∞ (−r, 0; X) with 0 (0) ∈ D(A). Let B=
u ∈ L∞ (−r, T0 ; X) with ess
sup
0 t T0
u(t)
and u(t) = 0 (t) a.e. t ∈ [−r, 0] , where 0 T0 T and > 0 are as in the proof of Theorem 3.3. Assume that lim
h→0+
T −h −r
u(t + h) − u(t) dt = 0,
uniformly for u ∈ B. Then problem (1.1) has a local integral solution.
(3.12)
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¯ Define K as in the proof of Theorem 3.3. Then for every h > 0 with Proof. Let 0 (0) ∈ D. t + h T0 , we have, from (L2) and (G), G(t + h, ut+h , Lt+h u) − G(t, ut , Lt u) G(t + h, ut+h , Lt+h u) − G(t + h, ut , Lt u) + G(t + h, ut , Lt u) − G(t, ut , Lt u) bT0 (ut+h − ut 0,1 + Lt+h u − Lt u) + hr 2 (ut 0,1 , Lt u) T0 −h u( + h) − u() d + hr 1 (uT0 ,1 ) bT0 −r
+ hr 2 (0 0,1 + T0 , aT0 (0 0,1 + T0 ) + cT0 ) and hence 0
T0 −h
G(t + h, ut+h , Lt+h u) − G(t, ut , Lt u) dt T0 −h bT0 T0 u(t + h) − u(t) dt + T0 ,h ,
(3.13)
−r
where
T0 ,h = T0 {bT0 hr 1 (0 0,1 + T0 ) + hr 2 (0 0,1 + T0 , aT0 (0 0,1 + T0 ) + cT0 )} and limh→0+ T0 ,h = 0. From (3.12) and (3.13),
T0 −h
lim
h→0+
0
G(t + h, ut+h , Lt+h u) − G(t, ut , Lt u) dt = 0
(3.14)
uniformly for u ∈ B. Choose 0 < T1 T0 satisfying 1 − bT1 T12 > 0. We define K1 =
T1 −h
u∈K:
p
u(t + h) − u(t) dt
0
1/p
(h)
for every 0 < h < T1 , where 1/p
(h) =
T1
1 − bT1 T12
h
sup
u∈K
0
+ bT1 T1 T1 + T1 ,h
G(t, ut , Lt u) dt + U (h, 0)0 (0) − 0 (0)
(3.15)
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
and T1 = 30 0,1 + T1 . Then K1 is a nonempty closed convex subset of L∞ (−r, T1 ; X) with respect to · T1 ,p We set K0 = {u|[0,T1 ] ; u ∈ K1 }. Then, K0 is a nonempty closed convex subset of L∞ (0, T0 ; X) with respect to ·Lp (0,T1 ;X) . As in the proof of Theorem 3.2, we consider the initial-value problem dv(t) + A(t)v(t) G(t, ut , Lt u), dt
0 t T1 ,
v(0) = 0 (0)
(3.16)
and define an operator Q : K0 → Lp (0, T1 ; X) by Qu|[0,T1 ] = v for every u|[0,T1 ] ∈ K0 , where v is a unique integral solution of problem (3.16). We shall show that Q : K0 → K0 is a continuous in Lp (0, T1 ; X). To this end, let u|[0,T1 ] ∈ K0 . Then, u ∈ K1 and u ∈ K. Since, by (2.9) in Lemma 2.2 with u0 = 0 (0), (Qu|[0,T1 ] )(t + h) − (Qu|[0,T1 ] )(t) h U (h, 0)0 (0) − 0 (0) + G(, u , L u) d 0 T1 −h G( + h, u+h , L+h u) − G(, u , L u) d + 0 h U (h, 0)0 (0) − 0 (0) + sup G(, u , L u) d u∈K1
0
T1 −h
u( + h) − u() d + T1 ,h h U (h, 0)0 (0) − 0 (0) + sup G(, u , L u) d + bT1 T1
−r
u∈K1
+ bT1 T1 T1 +
T1 −h 0
0
u( + h) − u() d + T1 ,h ,
where
T1 −h
−r
=
u( + h) − u() d −h
−r
0 ( + h) − 0 () d + T1 −h
+ 0
T1 +
−h
u( + h) − u() d
T1 −h 0
0
u( + h) − u() d
u( + h) − 0 () d
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for 0 < h < T1 with t + h T1 , we have T1 −h 1/p p (Qu|[0,T1 ] )(t + h) − (Qu|[0,T1 ] )(t) dt 0 T1 −h
0
u∈K1
+ bT1 T1 T1 + 1/p
T1 −h 0
+ bT1 T1 T1 +
1/p T1
u∈K1
T1 −h 0
+ bT1 T1 1/p T1
G(, u , L u) d
G(, u , L u) d
u( + h) − u() d + T1 ,h u∈K1
1/q T1 + T1
h 0
U (h, 0)0 (0) − 0 (0) + sup
0
p 1/p u( + h) − u() d + T1 ,h dt
U (h, 0)0 (0) − 0 (0) + sup
T1
h
U (h, 0)0 (0) − 0 (0) + sup
T1 −h
h
G(, u , L u) d
0
u( + h) − u()p d
0
U (h, 0)0 (0) − 0 (0) + sup
u∈K1 0
h
1/p
+ T1 ,h
G(, u , L u) d
1/q + bT1 T1 (T1 + T1 (h)) + T1 ,h
= (h),
where 1/p + 1/q = 1. We put (Qu|[0,T1 ] )(t), 0 t T1 w(t) = 0 (t) a.e. t ∈ [−r, 0]. Then, w ∈ L∞ (−r, T1 ; X) and from the proof of Theorem 3.3 it follows that w ∈ K. Thus w ∈ K1 . This implies that Qu|[0,T1 ] = w|[0,T1 ] ∈ K0 , and hence Q(K0 ) ⊂ K0 . From (3.4), Q : K0 → K0 is continuous in Lp (0, T1 ; X). Also it follows from Lemma 3.4 that {Qu|[0,T1 ] : u|[0,T1 ] ∈ K0 } is relatively compact in Lp (0, T1 ; X) and hence Q : K0 → K0 is compact. Thus, by Schauder’s fixed point theorem, there exists a fixed point u|[0,T1 ] ∈ K0 . Therefore, as in the proof of Theorem 3.2, problem (1.1) has a local integral solution. This completes the proof. Remark. (1) It is known by Calvert and Kartsatos [4] and Pavel [16] that the evolution operator U (t, s) generated by {A(t)} via f ormula (2.1) is compact for every 0 s < t T if and only if the following two conditions below hold: (i) For each t ∈ [0, T ] and > 0, the resolvent operator J (t) = (I + A(t))−1 of A(t) is compact.
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J.S. Jung / Nonlinear Analysis 60 (2005) 1219 – 1237
(ii) For each t0 ∈ (s, T ], the family {t → U (t, s)x : x ∈ Y } is equicontinuous at t0 on ¯ bounded subsets Y ∈ D. This result is an extension of Brézis’s result [3]. Brézis studied that the nonlinear semigroup {S(t) : D(A) → D(A), t 0} generated by constant −A = −A(t) (independent of t) is compact if and only if the resolvent operator J = (I + A)−1 is compact for every > 0 and for each bounded subset Y of X, {S(·) : [0, ∞) → X, x ∈ Y } is equicontinuous at each t > 0. Thus our results are improvements of Ha and Shin’s results in [6] as well the time-dependent version of them. In fact, using the methods of Baras [1] and Shioji [17] in Lp (0, T ; X), Ha and Shin [6] considered problem (1.1) in the case when {S(t) : D(A) → D(A), t 0} and resolvent J are compact. (2) As in [6,7,9], our existence results can be adapted to handle the nonlinear integrodifferential equations with finite delay. Acknowledgements The author is indebted to the referee for his/her valuable comments on this manuscript and the paper [4] in references. References [1] P. Baras, Compacité de l’opérateur f → u solution d’une equation non linéaire du/dt + Au f , C. R. Acad. Sci. Paris 286 (1978) 1113–1116. [2] Ph. Bénilan, Solutions integrales d’équation d’évolution dans un espace de Banach, C. R. Acad. Sci. Paris 224 (1972) 47–50. [3] H. Brézis, New results concerning monotone operators and nonlinear semigroups, in: Proceedings of the RIMS (on “Analysis of Nonlinear Problems”), Kyoto University, 1975. [4] B. Calvert, A.G. Kartsatos, On the compactness of the nonlinear evolution operator in a Banach space, Bull. London Math. Soc. 19 (6) (1987) 551–558. [5] S. Gutman, Compact perturbations of m-accretive operators in general Banach spaces, SIAM J. Math. Anal. 13 (1982) 789–800. [6] K.S. Ha, K. Shin, Local existence of integral solutions via compactness methods for nonlinear functional differential equations in Lp spaces, Nonlinear Anal. 45 (2001) 447–458. [7] K.S. Ha, K. Shin, B.J. Jin, Existence of solutions of functional integrodifferential equations in Banach spaces, Differential Integral Equations 8 (1995) 553–566. [8] A.G. Kartsatos, A compact evolution operator generated by a nonlinear time-dependent m-accretive operator in a Banach space, Math. Ann. 302 (1995) 477–487. [9] A.G. Kartsatos, X. Liu, On the construction and the convergence of the method of lines for quasi-nonlinear functional evolutions in general Banach spaces, Nonlinear Anal. 29 (1997) 385–414. [10] A.G. Kartsatos, K. Shin, Solvability of functional evolution via compactness methods in general Banach spaces, Nonlinear Anal. 21 (1993) 517–535. [11] E. Mitidieri, I.I. Vrabie, Existence for nonlinear functional differential equations, Hiroshima Math. J. 17 (1987) 627–649. [12] E. Mitidieri, I.I.Vrabie, Differential inclusions governed by nonconvex perturbations of m-accretive operators, Differential Integral Equations 2 (1989) 525–531. [13] N.H. Pavel, Nonlinear evolution equations governed f-quasi-dissipative operators, Nonlinear Anal. 5 (1981) 449–468. [14] N.H. Pavel, Nonlinear Evolution Operators and Semigroups, Lecture Notes in Mathematics, vol. 1260, Springer, Berlin, 1987.
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[15] N.H. Pavel, Differential equations associated with compact evolution generators, Differential Integral Equations 1 (1988) 117–123. [16] N.H. Pavel, Compact evolution operators, Differential Integral Equations 2 (1989) 57–62. [17] N. Shioji, Local existence theorems for nonlinear differential equations and compactness of integral solutions in Lp (0, T ; X), Nonlinear Anal. 26 (1996) 799–811. [18] I.I. Vrabie, Compactness methods for an abstract nonlinear Volterra integrodifferential equation, Nonlinear Anal. 5 (1981) 365–371. [19] I.I. Vrabie, Compactness Methods for Nonlinear Evolutions, Pitman Monographs Surveys, Pure and Applied Mathematics, vol. 32, Longman Scientific and Technical, 1987.