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R O C Z N IK I PO LSK IEG O T O W A R Z Y S T W A M A T E M A T Y C ZN E G O Séria I: PRACE M A T E M A T Y C ZN E XX IX (1989)

D. D. Ba i n o v, S. I. Ko s t a d i n o v and A. I. Za h a r i e v (Plovdiv)

Abstract Volterra type integral equations

Abstract. The aim of the present paper is to study some properties of an abstract nonlinear analogue of Volterra equation. Sufficient conditions have been obtained guaranteeing the existence of solutions in the homogeneous and nonhomogeneous case.

The main results are due to some ideas from [1 ].

The present paper studies an abstract nonlinear analogue of the Volterra equation. The main results are due to some ideas of [1].

Let Q be a metric space with metric q and Borel measure /t while Ф: x\—>MX is a map associating every element x e Q with a closed subset Mx of Q.

We say that conditions (A) hold when the following set of suppositions is fulfilled:

(Al) The set Q is compact.

(A2) For any e > 0 and x e Q , there exists a number Ô = S (e, x) > 0 such that for every element y e Q for which q(x, y) < <5

ia({Mx\My} и {My\Mx}) < e.

(A3) (Transitivity) For any x e Q and y e M x one has My ç Mx.

(A4) For every choice of e > 0 and x e Q there exists a number S > 0 such that for every element y e Q for which ^(x, y) < S the inclusion Mx £ U(e, My) holds, where U(s, My) denotes the ^neighbourhood of the set My.

(A5) There exists a point x0eQ for which ц{Мхо) = 0.

Let В be a Banach space with norm || ||B and let C(Q, В) denote the linear space of all continuous maps / : Q-+B.

Remark 1. The space C(Q, B) is a Banach space with the norm ll/ll = sup ||/(x)||B.

x e Q

Let us denote by со the measure of Q and let M be the family of all sets Mx, x eQ.

We say that the operator A: Qx C(Q, B)->C(Q, В) fulfils conditions (B) when:

(Bl) The operator A is continuous.

3 — Comment. Math. 29.1

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(B2) For every function f e C ( Q , B ) and for every x e Q the following inequality holds

\\A(x,f)\\^Q(fy\\f\\,

where the function Q{ f ) is bounded over every bounded subset of C{Q, B).

Consider the equation

(1) f + * f = P ,

where/, peC(Q, B) and the operator К is defined by means of the equation

(2) (Kf)(x)= J A(x,f){y)djxy.

m x

Since (Bl) holds, the integral in (2) exists, see e.g. [3].

Remark 2. Condition (A3) allows for every point aeQ to consider the restriction Ka: C(Ma, B)-+C(Ma, B), where the operator Ka is defined by (2) and C(Ma, В) is the space of the continuous functions from Ma to B. In this case the restriction/Ma of the solution of the equation f + K f = p, f e C(Q, B) is a solution of the equation ф + Каф = p\Ma, феС(Ма, В).

Le m m a 1. Let conditions (Al), (A2) and (Bl) hold. Then the operator К defined by equality (2) maps continuously C(Q, B) into C(Q, B).

Proof. Let x0eO be an arbitrary element and let {x„}*=1 be a sequence of points convergent to x0. Then, if/eC(£2, B) is an arbitrary fixed element, we have

(3) ||(tf/)(xJ-(tf/)(x0)||B = || f A(xn,f)(y)dpy- J A(x0J)(y)dpy\\B

MXn Afx0

= 11 J lA(xn, f ) ( y ) - A { x 0,f){y)']dpy+ J A(xn,f){y)dpy

M x n n M XQ M x n \ M x0

- J А(Хо’ Л(у№у\\в

M x0\ M x n

^ j \\A(xn, f ) ( y ) - A { x 0,f)(y)\\Bdpy+ j \\A{xn, f){y)WBdpy

M x „ r \ M x0 M x n \ M x0

+ j \\A(x0J)(y)\\BdVy

M x0\ M x „

Let e > 0 be an arbitrary number. Condition (Bl) implies that a number n0 = n0(e, x 0, f ) exists such that for n ^ n0 and y e M Xo the following inequality holds

(4) M(x„, f){y) — A (x0, f){y) || B < s/3p(MXo).

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On the other hand, condition (A2) yields that there exists a number 3 = <5(e, x0) > 0 so that for any x from the ball q(x, x0) < 3 one has (5) /x({Mm\MJ u {M,\MxJ < ф Л ,

where X = sup \\A(x, f)(y)l\B.

x,yeQ

Besides, there exists a number nl = n^e) such that for n ^ пг there holds g(x„, x0) < 3. Then, if n ^ max(nl5 n0) from (5) we get

(6) j \\MXr,,f)(y)\\BdHy+ J \\Ax0,f)(y)\\BdVy < 2Ф -

M Xn\MXQ M X0\MXn

For n^max(nl9 n0), using (5), (4) and (6), we finally obtain

\\(K f)(X n)-(K f)(x0)\\B < 8 .

We will prove that the operator К is continuous. Let {/„}*= i be a convergent sequence from C(Q, B) and let/ 0 be its limit. For e > 0, in view of (Bl) it follows that there exists a number n0 = n0(e,/0) such that for n0 the following inequality holds

\\A(x, f n) — A(x, /o)|| < e/со, xeQ.

Then \\Kfn — Kf0\\ < e provided n0.

This completes the proof of Lemma 1.

Le m m a 2. Suppose that conditions (Al), (A2) and (B) are fulfilled. Then the set K(B(0, R)) is equicontinuous for every central ball B{0, R).

Proof. Let B(0, R) be an arbitrary ball centered at zero with radius R.

Then, for geB(0,R) and x, yeQ,

(7) \\(Kg)(x)-(Kg)(y)\\ ^ j \\A(x, g)(z)-A(y, g){z)\\Bdpz

M x c\My

+ J \\A(x, g){z)\\Bdpz+ j \\A(y, g){z)\\Bdgz.

M x \My M y\M x

In view of (Bl), for any fixed x eQ and e > 0 there exists a number <5 =

<5(e, x) > 0 such that for g(x, y) < 3 and z e Q the following inequality holds (8) ||A(x, g)(z)-A(y, g)(z)\\B < e/2m.

Moreover, there exists a.number 5*e(0, 3) such that for g(x, y) < 3* we have

g({Mx\My} и {My\Mx}) < B/4QR,

where Q = sup Q{g).

g e B t O . R )

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Then from (7), (8) and (9), we obtain that for q(x, y) < (5*

\\(Kg)(x)-(K9m B < 8,

i.e. the set K(B(0, R}) is equicontinuous at any point x e Q , which is to be proved.

Observe that (B2) implies that the set K(B(0, R)) is uniformly bounded.

Remark 3. The set K(B{0, R)) is relatively compact if and only if the sets (9) H = { J A(x0J)(y)dnsI f e B ( 0, R)}

лО

are relatively compact for x0e£2 (see [3]).

Remark 4. Let the conditions of Lemma 2 hold supposing that Q{ f ) in condition (B2) does not depend on /, and let coQ < 1. Then the equation

(10) Kf = f

has the trivial solution / = 0 only.

Remark 4 implies that the condition Qco ^ 1 is necessary for (10) to have a nonzero solution. Moreover, this condition is necessary for the existence of a nonzero element f e C ( Q , B) with the property ||/|| = ||üf/||.

Theorem 1. Suppose that the following conditions hold:

1. Conditions (A) and (B) hold.

2. The space Q is connected.

3. For every x0eQ and f eC(Q, В) the following inequality holds sup \\A(x,f){y)|)e ^ Qfif, x0) sup \]f{y)\\B, x e M X0,

yeMx yeMx

where Q^{f, x0) is continuous with respect to x0 for fixed f.

Then the equation

(11) f = AKf

has the trivial solution f = 0 for XeC only.

Proof. Let А ф 0, AeC be an arbitrary fixed complex number and let f ф 0 be a solution of equation (11). Consider the set

JTf = {x| x eQ, f(y) = 0, уeMx} .

We will prove that JTj ф 0 . (A5) implies that there exists a point x0eQ with the property p{MX0) = 0. Let us assume that MXo Ф0 . (The case Mxo = 0 is trivial.) Then, in view of (A3), it follows that for any x eMXo, Mx Ç Mxo, and hence p(Mx) = 0, whence with the help of (11) we obtain that / (x) = 0. Thus, we proved that x0e Л7-

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The set JTf is closed. Let {x„}£L i be any sequence of elements of JVf and let x* = lim xneQ be its limit. We will prove that

П~> 00

It is sufficient to consider the case pt(Mx*) Ф 0 only, then, infinitely many terms of the sequence {x„}*=1 exist for which ц(МХп) # 0. Let z e M x* be an arbitrary fixed point, while e > 0 is a number. Then there exists a number

= <5i(e) > 0 such that for x e Q and £>(x, z) < 31 the following inequality holds

||/(x)-/(z)||B < £.

If we denote q(z, MXn) = inf q(z, q), n = 1 ,2 ,..., then for any n there exists

qeMXri

an element zne M Xn for which

Q ( Z , M x n) + 2^ l > Q ( Z , Z „ ) .

Condition (A4) implies that there exists a number 32 = ^2(^1) > 0 sucb that if у е й and @(y, x*) < 32, then Mx* ^ U(j31, My). Let the number n0 be so large that the following g(xn, x*) < b2 holds for n0. Then Mx*

^ U(^3l , MXn), and hence zeU(^3l , MXn), j.e., q(z, MXn) < j 3 t whence we obtain that <5j > q(z, MXn) + ^3l > g(z, z„). Taking into account the continuity of /, we get

l l / ( 0 - / ( * ) I U < e .

Since zne M Xn, we have f(z„) = 0, and hence ||/(z)||B<£, i.e./(z) = 0.

Thus, we proved that x*eJ^f , i.e., jVf is closed. We will show that is an open set as well. Let a be an element of and let e > 0 be a number satisfying the condition

a) < i

Condition (A2) implies that there exists a number 3 = <5(e, a) > 0 such that for x e Q and g {a, x) < 3 the following inequality holds:

There exists a number «^ (O , <5) such that for any x for which g (a, x) < <Sj one has

\QAf, a ) - Q , { f , x ) \ < l/4|A|e.

Let beQ, g (a, b) < and let tp(x) denote the restriction of /(x ) over Mb, i.e., tp(x) =f(x)\Mb. Then from conditions (A) and from condition 3 of

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Theorem 1 the following inequalities are fulfilled sup |M*)||B = |2| sup || J A(x, (p)(y)dpy\\B

хеМь хеМъ Mx

= Щ sup J \\А(х, (p)(y)))Bdpy

хеМь Мх\Ма

< \MQi(f, b) sup p{Mx\Ma) sup ||<р(х)||в

хеМь хеМъ

< z\M(Qt(f, a)+ l/4|/l|e) sup ||<p(x)||B < i sup ||<j£>(x)||B.

хеМъ хеМь

Hence (p(x) = 0 for x e M b, i.e.,/(x) = 0 for x e M b. Thus we proved that beJV'j-, and hence jVf is an open set.

Taking into account that О is a connected set we conclude that JTf = Û, i.e., for every ÀeC equation (11) possesses the trivial solution only.

Remark 5. We note that in the proof of the closeness of the set jVf the compactness of the space Q is not used.

Remark 6. It is not difficult to see that the assertion of Theorem 1 remains valid replacing (Al) by the following condition.

(AT) For x 6 Q the sets Mx are compact, connected, (J Mx = Q and every

xeQ

set Mx contains a point y = y(x) for which fi(My) = 0.

While proving the solvability of equation (1) we will employ the following theorem:

Th e o r e m 2 [ 4 ] . Let D be a bounded open subset of the real Banach space

££ and let К: be a compact operator. If the point p&D is such that f+ tK f Ф pfor f e d D and te[0, 1], then the equation f + K f = p has at least one

solution in D.

It will be assumed further that В is a real Banach space.

Let peC{Q, В), p Ф 0 and let f eC(Q, В) be such an element for which the inequality ||p—f\\ < ||p|| holds.

De f i n i t i o n 1. We say that the element / is well dislocated to the ball B(p, Rf ),

Rf e(\\p-f\\, llpll),

if for any element ge dB(p, Rf ) there exist a point x = x(f, g)eQ and a number ос = <x(/, g)e(0, p(Mx)) for which the following inequality holds

(12) max {sup ||A(x,/)(y)||B, sup \\A(x, g)(y)\\B,

yeM- yeM-

sup ||A( x,f)(y)~A(x, 0)(y)||B} <ü\\f(x)-gffî\\B‘

yeM-

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De f i n i t i o n 2. The element peC(Q, В), p Ф 0, is called regular for the operator A if there exist a radius Ro e(0, ||p||) and a number ee(0, ||p|| —R0) such that every element fedB(p, R0) is well dislocated with respect to some ball B(p, Rj), where Rf e(R0 + e, ||p||).

Th e o r e m 3. Let conditions (Al), (A2) and (B) be fulfilled and besides, let the set K(B(p, ||p К)) be relatively compact. Then for every regular point peC(Q, В) equation (1) has at least one solution f ( x ) ^ 0 in the ball B(p, ||p||).

Proof. Condition (B2) implies that f{ x) = 0 is not a solution of equation (1).

Let p be a regular point of the operator A, and let {B(p, JR)}, R e ( 0, Ы1) be a family of closed balls. Assume that the condition of Theorem 2 is violated on the boundary of one of these balls.

Then for every ball B(p, R) there exist a number tR = [0, 1] and an element f ReôB(p, R) for which the following equality holds

(13) /я + tR KfR = p.

Then (13) implies that # 0 for R > 0. Moreover, one can consider tR Ф 1 for every R since the opposite means that the desired solution is found.

Let R0e(0, ||p||) and ee(0, ||p|| — R0) be the radius and the number from Definition 2, and let f 0eôB(p, R0), foe(0, 1), satisfy (13).

The regularity of p implies that we can find a ball B(p, R/o), R/oe(R0+e, ||p II) so that /о should be well dislocated with respect to B(p, Rfo). Then, if for g* edB(p, Rfo) and t* e(0, 1) equality (13) holds, then there exist a point x — x ( f0, g*)eQ and a number â = â (/0, p*)e(0,

such that (12) implies the estimate

sup ||t0A( x, f0)(y)-t*A(x, д*){у)\\в < а|[/0(х)-р*(х)||в.

уеМ-

The above inequality and (13) yield

\\f0(x)-g*(x)\\B = \\t0(Kf0)(x)-t*(Kg*)(x)\\B

^ p(Ms) sup \\t0A{ x,f 0)(y)-t*A(x, g*){y)||B

yeM-

< II/o(x) - 0 * ( * ) IIb>

which is a contradiction.

This completes the proof of Theorem 3.

We give an example for an operator A: Q x C(Q, B)->C(Q, B) for which every element peC(Q, В), p ф 0, is regular.

Suppose that the following conditions are fulfilled:

1. A(x, 0) = 0 for x eQ.

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2. F o r /, gGC(Q,B) there exists a point x = x { f , g ) e Q so that the following inequality holds

sup \\A(x,f)(y)-A{x, 0)00 II в ^ T(f, g)\\f{x)-g(x)\\B,

y e M -

where T {f ,g )e (0, 1/2p{Mx)).

Let p eC( Q,B) be an arbitrary element {p Ф 0). For all functions /, geC(Q, B) we set x(f,g) = x* where x* is a point for which \\f— g\\

= ||/(* * ) — g(x*)\\B. Choose R0 so small that

T(f, g) < 1 IIpII Ы М Х*)

for suitable choice of epe(0, ||p|| — R0). Then the following inequality is fulfilled:

sup \\A(x*,f){y)\\B ^ T{f, 0)||/(x*)||B < Л у ^ -7^7

уемх* IIpII 2р(Мд

<

<

_ £ P\\f(x*)-g(x*) llpll 2g(Mx* ) " p(Mx<) g(Mx*)\\f—g\\

1

p(Mx*)II / ( x * ) -0(x?

Analogically, we get

sup И (х*, g)(y)||B < —j - — \\f(x*)-g(x*)\\B.

y e Mx *

It is not difficult to see that Remark 5 remains valid for Theorem 3 as well if equation (1) has unique solution for every regular peC(Q,B).

We present two corollaries of Theorem 3 for the case where equation (1) has unique solution:

Corollary 1. Suppose that the following conditions are fulfilled:

1. The conditions of Theorem 3 hold, where condition (Al) is replaced with

(АГ).

2. For x e Q there exists a number L(x)e(0, l/p(Mx)), such that sup IIA ( x , f )( y )- A (x , 0)(y)|| ^ L(x) sup ||/OO-0()OII*,

y e M x y e M x

for all f, geC(Q, B).

Then, for every regular element peC(Q,B), equation (1) has a unique solution.

Definition 3. We say that the function peC(Q, B) is weakly regular if the restriction p\Mx over every Mx, x eQ , is a regular element in the sense of Definition 2 for the Banach space C(MX, В).

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Corollary 2. Suppose that the following conditions are fulfilled:

1. Conditions (АГ), (A2)-(A5), (B) hold.

OO

2. There exists a sequence MXn+1 => MXn such that (J MXn = Q.

n = 1

3. 77iere exists a continuous function L: Q-> [0, oo) such that for x e Q and for f g eC (Mx, В) the following inequality holds

IIA ( x , f ){ z) - A{ x , g)(z)\\B ^ L(x) \\f (z) — g(z)\\B, z e M x.

4. The sets Kx(B(p\Mx, ||p|MJ|)) are relatively compact in C(MX, B), xeQ.

Then equation (1) has unique solution for every weakly regular element peC(Q, B).

Proof. Let x e Q be arbitrary point. Then from Theorem 3 it follows that there exists / eC( Mx, В) for which f + K xf = p\Mx. Suppose that there exists another function g eC ( Mx, В) for which g + Kxg = p\Mx. According to con­

dition 3 of Corollary 2 we have

ll/(y)-^(T)lle< f L(y)\\f(z)~g(z)\\Bdpz, y e M x.

M y

Consider the equation

Ф(У)= S L ( y ) 0 W n z, y e M x,

M y

where Ф: Mx-+R.

Using Theorem 3 from [1] we conclude that for y e M x the following inequality holds

\\f(y)-g(y)\\B^4>(y)- Taking into account Theorem 1, we obtain

II/0 0 - 0 0 0 IIВ = 0, y e M x, which contradicts our supposition.

The assertion of Corollary 2 follows from condition 2 of Corollary 2, condition (Al) and the proved uniqueness of the solution of the equation f + K xf = p .

Acknowledgment. The authors wishes to thank Professor A. D. Myshkis and Professor G. S. Skordev for their kindness in discussing these results.

References

[1 ] D . D. B a i n o v , A . D. M y s h k i s and A. I. Z a h a r i e v , On an abstract analogue o f the Bellm an-Gronwall inequality, Publications o f the Research Institute for Mathematical Sciences, K yo to University 20 (1984).

[2 ] R. E. E d w a r d s , Functional analysis, H olt, Reinhart and Winston (1965), 1070 pp.

[3 ] E. H i l l e and R. P h i l l i p s , Functional analysis and semi-groups, Amer. M ath. Soc., Colloq.

Publ., Providence 1957.

[4 ] V . C. L. H u t s o n and J. S. P y m , Applications o f functional analysis and operator theory, Academic Press, 1980, 431 pp.

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