ANNALES SOCIETATIS MATHEMATICAE POLONAE Series 1: COMMENTATIONES MATHEMATICAE XXX (1990) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO
Séria I: PRACE MATEMATYCZNE XXX (1990)
Ja n u s z Ja n u s z e w s k i (Bydgoszcz)
On the existence of continuous solutions of nonlinear integral equations in Banach spaces
1. Introduction. Let £ be a Banach space and let D = [0, d] be a compact interval in R. In this paper we shall prove some existence theorems for the Urysohn integral equation
(1) x(t) = p(t) + À f f ( t , s, x(s))ds,
D
and for the Volterra integral equation
(2) x(t) = p(t) + J /( t, s, x(s))ds, о
where f , p and x are functions with values in E.
Our fundamental tool are measures of noncompactness. There have appeared a lot of papers using the Kuratowski measure of noncompactness in proving existence theorems for ordinary differential equations [1, 5, 6, 8, 10]
and integral equations in Banach spaces [9, 11]. In [3] Banas and Goebel have introduced an axiomatic approach to the notion of a measure of noncompact
ness. The usefulness of their axiomatics was illustrated by an existence theorem for the Cauchy problem [3, Th. 13.3.1]. In our paper we shall show that axiomatic measures of noncompactness are useful for integral equations too.
Our notation is the same as that in [3] or [2]. Let ME denote the family of all nonempty and bounded subsets of E and NE the family of all nonempty and relatively compact sets in E.
De f in it io n 1. A function p: M£ -+[0, + oo) is said to be a measure of noncompactness if it satisfies the following conditions:
1° the family ker/x — { X e M E\ p{X) = 0} is nonempty and ker/x c NE, 2° if X c Y, then p(X) ^ p(Y),
3° p(X) = p(X), 4° p(convX) = p{X),
5° /х(лЛг+ (1 -л )Г )^ л /х (Х ) + (1-/)/х(У) for Ae[0, 1].
For a given measure p in the space E set Ец = {xeE: {xjeker^}.
86 J. Ja n u s z e ws k i
In what follows we shall need the following lemmas:
Lemma 1 [2]. I f ||X|| = sup{||x||: x e l } < 1, then li{X+Y)<n(Y)+\\X\\n(K(Y, 1)), where K(Y, 1) = {xeE: dist(x, 7) ^ 1}.
Lemma 2 [2]. Let tx, t 2, . . . , t n be nonnegative real numbers such that
£ " =1£г^ 1 and let р е Е ц. Then I
Kp+ t iix è ^ i t M p + x i)-
i = 1 i = 1
Lemma 3. I f X is a bounded equicontinuous subset of C(D, E), peE^ and 0 < Я mes D ^ 1, then
p(pA Я j X(s)ds) ^ Я j p(p + X(s))ds,
D D
where X(t) = {x(t): x e l } , jDX(s)ds = {Ji>x(s)ds: x e X ).
Lemma 3 is a modification of Lemma 7.2.4 from [2].
Moreover, in this paper we shall apply some modification of Sadovskii’s fixed point theorem [1 1].
Theorem 1. Let В be a bounded, closed and convex subset of a Banach space, ре В and let G be a continuous mapping of В into itself I f the implication (3) F c conv(G(K) и {p}) => V is relatively compact
holds for every subset V of B, then G has a fixed point.
2. The Urysohn integral equation. We consider the integral equation (1).
Assume that
I. p: D-*E is a continuous function;
II. (t, s, x) —>/(t, s, x) is a function from D2 x E into E which satisfies the following conditions:
(i) / is continuous in x and strongly measurable in (t, s);
(ii) for any h > 0 there exists a measurable function mh: D2^ R + such that ||f ( t , s, x)|| ^ mh(t, s,)(t, seD, ||x|| ^ h) and §Dmh(t, s)ds ^ a(h) < oo;
(iii) for any h > 0 there is a function dh: D3->R+ such that ||/(t, s, x)
—/ (t, s, x)|| ^ dh(x, t, s) (t, t, s e D , ||xII ^ h) and limt_t §Ddh(T, t, s)ds = 0.
Theorem 2. Assume, in addition to I and II, that the function (s, x)-*f(t, s, x) is uniformly continuous and р(7)е£д for teD. Moreover, assume that there exists an integrable function к: D2 —► R + such that- (4) p({p(t)} u (p(t) +f{t, s, X))) ^ k{t, s)p{X)
for any s, te D and for any bounded subset X of E. Then there exists q > 0 such that for any Я, 0 ^ Я < q, the equation (1) has at least one continuous solution.
Existence o f solutions o f nonlinear integral equations 87
P roof. Denote by C the Banach space of all continuous functions и: D-+E with the usual supremum norm || • ||c. Let r(K) be the spectral radius of the integral operator К defined by
Put
Ku(t) = J k(t, s)u(s)ds (u e C , teD).
D
q = min sup---- ’h~\\p\\
,й > О Ф ) r{K)’ mes D J For fixed XeR, 0 X < q, choose b > 0 in such a way that
(5) \\p\\c + Xa(b) ^ b.
Put В = {xeC: ||x||c < b}. Consider the operator G defined by G(x)(t) = p(t) + Xj f ( t , s, x(s))ds (x eB , t ED).
D
The assumptions I, II and (5) imply that G is a continuous mapping B-+B and the set G(B) is equicontinuous (see [7] or [9]).
Let F be a subset of В such that F c conv(G(F) u {p}). Obviously V(t) c: conv(G(F)(£) и {p(0}) f°r tED.
Let us fix teD . In virtue of the uniform continuity of/ the transformation (Jx)(s) = f(t , s, x(s)) maps equicontinuous sets into equicontinuous sets. Hence
p(F(0) ^ p(conv(G(F)(0 u {p(t)})) = p(G(V)(t) u {p(t)})
= p({p(t)} u (p(t) + X j (fV)(s) ds))
D
= p(p{t) + x \ (/V u {9})(s)ds),
D
by Lemma 3 (F c G (В) is equicontinuous) it follows that p(p{t) + X${JVv{0}){s)ds) < X\ p(p{t) + {JVu {9})(s))ds
D D
= Я f n({p{t)} U (p(t)+f(t, s, V(s))))ds.
Thus, by (4), we obtain D
p(V{t)) ^ X J k(t, s)p(V(s))ds.
D
Because this inequality holds for every î e D and Xr(K) < 1, by applying the theorem on integral inequalities, we conclude that n(V(tj) = 0 for tED.
Hence Ascoli’s theorem proves that the set F is relatively compact.
Consequently, by Theorem 1, G has a fixed point.
R em ark. If the measure p has the property: p{{a) u A) = p(A) for any then the assumption (4) has the form p(p(t)+f(t, s, X)) < k(t, s)p(X).
In the next theorem we shall assume additionally that the measure p has the maximum property:
6° p(A u B) = max(p(A), p(B)) for A, BeM e.
Then arguing similarly to [1], we can prove the following Ambrosetti-type
J. J a n us z e ws k i
Lemma 4. Let V be a bounded equicontinuous subset of C(D, E). Then for each compact subset T of D
p(F(T)) = supp(V(t)), where V(T) = {x(t): x eV, te T } .
t e T
Theorem 3. Let p fulfil conditions \ ° ~ 6 ° and pit)eE^ for tED. Assume, in addition, that there exists an integrable function k : D2-+R+ such that for every tED, e > 0 and for every bounded subset X of E there exists a closed subset De of D such that mes (D\De) < e and
(6) p(p(t)+f(t, T x X ) ) ^ sup/c(L s)g(X)
s e T
for any compact subset T of De.
Then there exists g > 0 such that for each a, 0 ^ Я < g, there exists at least one continuous solution of (1).
P ro o f. Choose g, b, B, G as in the preceding proof. Let F be a subset of В such that F c conv(G(F) u {p}). Since F is equicontinuous, the function t-*v(t) = p(F(f)) is continuous on D.
Fix tED and e > 0. By (6) and the Lusin theorem there exists a compact subset Ds of D such that mes(D\De) < s and p ( p ( t ) + f ( t , T x X ) ) ^ supseT k{t, s)p(X) for any compact subset T of Ds, while the function s-*/c(t, s) is continuous.
We divide the interval D = [0, d] into n parts 0 = d0 < dx < ... < dn — d in such a way that \k(t, s)v{r) — k(t, u)v(z)\ < г for s, r, u, z eTl = Dt r\ DE, where Dt = [di-i, df\ (i = 1, ..., n). Set Ц = {u(s): u eV, s eD(}. By Lemma 1 (7) p(p(t) + X $f(t, s, V(s))ds) ^ p(p(t) + X j f{t, s, V{s))ds)
D De
+ || J f ( t , s , V(s))ds\\p(K(B(D), 1)).
Let us observe that DS'De
Я J f ( t , s, V(s))ds c= £ Я j f(t , s, V(s))ds -
Dc i = l . Ti
Hence
Я £ mes 7]conv/(f, 7] x V f i= 1
n
р(р(0 + Я j f(t , s, V(s))ds) ^ р(р(0 + Я £ mes f com f i t , 7) x Vf)
De i = l
n
< я
£ mes 7>(p(f) + conv/(t, f x Vf)i = 1
< я
£ mes 7]p(conv {pit) + fit, Tt xVf)) i = 1^ Я £ mes 7] sup ^it, s)p(^)
i = 1 s e T i n
= Я £ mes T f i t, q M s f i= 1
Existence o f solutions of nonlinear integral equations 89
where qte 7], s.eD,. Moreover, as \k(t, s)v{s) — k{t, < e for seTJ, we have
mes Ttk(t, g , ) ^ ) ^ J k(t, s)v(s)ds + s mes 7].
Ti
Thus Л
р(Я J f(t, s, V(s))ds + p(t)) ^ Я £ (J k(t, s)v(s)ds + ernes 7])
DE i = l Ti
= Я J /c(r, s)t>(s)<is + A£mesD£.
As г is arbitrarily small, from this and (7) we deduce that р(р(0 + Я{ /(C s, F(s))ds) ^ Я J /c(r, s)v(s)ds,
i> D
and therefore fi{V{t)) ^ y.JD/c(t, 5 )15 (5 ) ds. Further we argue as in the final part of the proof of Th. 2.
Now we assume additionally that the measure p has the property 7° if Х я eM E, X n = X„, X n +1 c for n = 1, 2, ... and if ü m , ^ p{Xn) = 0, then X x = O?=1X n * 0 .
Theorem 4. Let p fulfil conditions l°-5° and 7°, let the function (s, x)-» f(t, s, x) be uniformly continuous and pitfeE^for any teD. Assume, in addition, that there exists an integrable function к : D2-+R+ such that for any bounded subset X of E
(8) p(p(t)+f(t, s, X)) ^ k(t, s)p(X).
Then there exists q > 0 such that for every Я, 0 < Я < q, the equation (1) has at least one continuous solution x(t) such that x(t)eЕц for teD.
P roof. Choose q, b, B, G as in the proof of Th. 2. Denote by X 0 the set of all functions x e B such that
IIx(t) x(s)|| ^ ||p(r)-p(s)|| + j d b(s, t, T)di for t, s eD.
D
The set X 0 is bounded, closed, convex and equicontinuous. Set X n+ i = conv GXn (n = 1 ,2 ,...). Since G{B) cz B, therefore all these sets are of the same type as X 0 and X n+1c : X n. Put un(t) = p(Xn(t)). Obviously 0 ^ un + 1(t) ^ un(t) (n — 0, 1, ...). The functions щ are continuous. Therefore the sequence un(t) converges uniformly to a function u^f). We have
*V+i(0 = p(con\{GXn)(tj) = p(p(t) + X J ( f X n)(s)ds)
D
< Я j p(p(t) + { f X n)(s})ds ^ Я j k(t, s)un(s)ds,
D D
which implies u^it) ^ ЯJDk(t, s)uo0{s)ds for tED. Since Xr(K) < 1, we get uoo(0 = 0 for teD.
90 J. J a n u s z e w s k i
Set цс(Х) — max {p(X(t)): teD} for X equicontinuous. Then pc is a mea
sure of noncompactness (cf. [2], p. 79). Since lim max{n„(t): teD} = 0 , n-> 00
therefore pc(Xn)-> 0 as n-> oo.
Hence the set X œ = P)“=1 X n is nonempty, convex, closed and Х ж c= £ д.
Now we can apply the Schauder fixed point theorem to the mapping G|Xoo, which yields the existence of x e X ^ such that x = G(x).
R em ark. The proofs of Theorems 3 and 4 are suggested by the corresponding proofs from [10] and [6] for differential equations.
3. The Volterra integral equation. Consider now the integral equation (2) assuming that p and / satisfy I and II. Choose b > 0 in such a way that b > 2sup(6D ||p(f)||. From II(ii) it follows that there is a number a1, 0 < ^ d, such that jo mb(t, s)ds ^ b/2 for 0 < t < a^.
Let J = [0, a], where a = min(al5 1). Put В = {u e C ( J , E): ||w||c ^ b} and t
F(x)(t) = p(t) + J / ( t , s, x{s))ds for x e B , te J . о
Similarly to the Urysohn integral equation, we can show that F is a continuous mapping B-+B and the set F(B) is equiuniformly continuous.
Further, let P = {(t , s, z) eR3: 0 ^ s ^ t ^ l, |z| < c}, where l > a, c > 2b.
Assume that a nonnegative real-valued function (t, s, z)-+h(t, s, z) defined on P is a Kamke function, i.e. h satisfies the Carathéodory conditions II(i)—(iii) and
(iv) for each fixed t, s the function z->h(t, s, z) is nondecreasing, (v) for each q, 0 < q ^ /, the zero function is the unique continuous solution of the equation z(t) = ÿ0h(t, s, z(s))ds defined on [0, q).
Theorem 5. Assume that for each t e J the function (s, x )-> /(t, s, x) is uniformly continuous on {(5, x): 0 ^ s < t, ||x|| ^ b}, p i f e E ^ for t e J and (9) p({p{t)} u(p(t)+f{t, s, X)))^ h(t, s, p{X))
for 0 ^ s ^ t ^ a and for each bounded subset X of E. Then the equation (2) has at least one continuous solution on J.
The proof is similar to that of Th. 2.
Theorem 6. Let p fulfil conditions l°-6°. Assume, in addition, that for any e > 0, bounded X <= E and t e J there exists a closed subset IE o /[ 0, f] such that mes([0, t ] \ / £) < e and
(10) p(p{t)+f{t, T x X)) ^ sup h(t, s, p(X))
s e T
for each closed subset T of IE. Then the equation (2) has at least one continuous solution on J.
Existence o f solutions o f nonlinear integral equations 91
P roof. Let F be a subset of В such that V a conv(F(F)u {p}). Let us fix teJ, s > 0.
By the Scorza Dragoni theorem there exists a closed subset De of J such that mes(J\Ds) < e and the function h is uniformly continuous on Ds x [0, bf\, where = bp{K(B(J), 1)). As Fis equicontinuous, the function t-^v(t) = p(V(t)) is continuous on J . Choose Ô > 0 such that \h(t, s, r) — h(t, u, z)\ < г for s, u e De and r, z e[0 , bf\ satisfying \s — u\ < ô, \r — z\ < Ô, and choose r] such that 0 < q < Ô and |u(s) — t>(w)| < <5 for s , w e J such that \s — w| < ц.
We divide the interval [0, r] into n parts 0 = t0 < tl < ... < t n = t in such a way that |t£ — tr- il < for i = l , . . . , w. Let = [tt, ti-f\ n DE and Vt = {u(s): ueV, s e D,}.
By (10) we may choose a closed subset Je of J such that m e s(J\J£) < e and p(p(t)+f(t, T x V f ) ^ sup h(t, s, p{Vt))
s e T
for any compact subset T of JE and z = 1, 2, . . n. Put P = [0, t] n Den J e, Q = [0, t~\\P and 7] = Dt n J E. We see that
j / ( t , s, V(s))dsa j f(t , s, F(s))ds + j / ( t , s, F(s))ds.
0 P Q
Arguing similarly to the proof of Th. 3, we get
П
p(p(t) + J /(L s, V(s})ds) ^ X mes TiH** Vi’ v(si))
P i = 1
for qie T i, sieDi and further
p(V{t)) ^ /x(p(0 + j / ( t , s, V(s)) ds) ^ J h(t, s, v(s))ds
о 0
for te J .
From the property of Kamke functions and the theorem on integral inequalities, we conclude that p(V(t)) = 0 for t e J . The proof is completed as that of Th. 2.
Theorem 7. Let pt fulfil conditions l°-5° and 7°, and p(t)eE^for any te J.
Assume, in addition, that for teD the function (s, x)-+f(t, s, x) is uniformly continuous on {(s, x): 0 ^ s ^ t, ||x|| < b] and
f(p(0 + /(L s, X)) < h(t, s, fi(X))
for 0 ^ s ^ t ^ a and for each bounded subset X of E. Then the equation (2) has at least one continuous solution x(t) such that x i f e E ^ f o r te J .
The proof is similar to that of Th. 4.
92 J. Ja n u s z e w s k i
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INSTYTUT MATEMATYKI I FIZYKI, AKADEMIA TECHNICZNO-ROLNICZA INSTITUTE O F MATHEMATICS AND PHYSICS, ATR
KALISKIEGO 7, 85-790 BYDGOSZCZ, POLAND