PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 83(97) (2008), 57–63 DOI: 10.2298/PIM0897057D ON THE FUNCTIONAL–INTEGRAL EQUATION OF VOLTERRA TYPE WITH WEAKLY SINGULAR KERNEL Aldona Dutkiewicz Communicated by Stevan Pilipović Abstract. We give sufficient conditions for the existence of Lp -solution of a Volterra functional–integral equation in a Banach space. Our assumptions and proofs are expressed in terms of measures of noncompactness. 1. Introduction Let E, F be Banach spaces, D = [0, d1 ] × · · · × [0, dm ] and D(t) = {s = (s1 , . . . , sm ) ∈ Rm : 0 si ti , i = 1, . . . , m} p for t = (t1 , . . . , tm ) ∈ D. Denote by L (D, E) p(p > 1) the space of all strongly measurable functions x : D → E with D x(t) dt < ∞, provided with the norm 1/p xp = D x(t)p dt . We consider the following functional–integral equation of Volterra type K(t, s) g(s, x(s)) ds (1) x(t) = φ t, D(t) A(t,s) |t−s|r , with the kernel K(t, s) = 0 < r < n (t, s ∈ D, t = s). We give sufficient conditions for the existence of a solution x ∈ Lp (D, E) of (1). Moreover, for r < 1 we present one-dimensional result involving a generalized Osgood condition. Our considerations are inspirated by a paper of Darwish [5] concerning the functional– integral equation of Hammerstein type. The existence of L1 -solution of functional– integral equation of Hammerstein type was studied in [4] and when g(s, x) = x we get an equation considered in [3]. In [15] Szufla has established the existence of Lp -solution of Hammerstein integral equation with weakly singular kernel. Throughout this paper we shall assume that: 2000 Mathematics Subject Classification: 45N05. Key words and phrases: Functional–integral equation, Lp -solutions, measure of noncompactness. 57 58 DUTKIEWICZ 1◦ (t, x) → φ(t, x) is a function from D × E into E such that (i) φ is strongly measurable in t and continuous in x; (ii) φ(t, x) − φ(τ, y) |a1 (t) − a1 (τ )| + b1 x − y for t, τ ∈ D and x, y ∈ E, where a1 ∈ Lp (D, R) and b1 0; (iii) φ(0, 0) = 0; ◦ 2 A is a bounded strongly measurable function from D × D into the space of continuous linear mappings F → E; 3◦ (t, x) → g(t, x) is a function from D × E into F such that (i) g is strongly measurable in t and continuous in x; (ii) g(t, x) a2 (t) + b2 x for s ∈ D and x ∈ E, where a2 ∈ Lp (D, R) and b2 0. In what follows we shall need the following lemmas: Lemma 1. The linear integral operator (Sx)(t) = K(t, s) x(s) ds (x ∈ Lp (D, E), t ∈ D) D maps Lp (D, E) into itself continuously. Moreover, S aQ, where a = sup{A(t, s) : t, s ∈ D} and (2) 2π n/2 (diam D)n−r =Q (n − r)Γ(n/2) D ds |t − s|r for all t ∈ D. Lemma 2. Put G(x)(t) = g(t, x(t)) for x ∈ Lp (D, E) and t ∈ D. Then G is a continuous mapping of Lp (D, E) into itself. For the proofs we refer for example to [15]. Denote by α and α1 the Kuratowski measures of noncompactness in E and L1 (D, E), respectively. For any set V of functions belonging to L1 (D, E) denote by v the function defined by v(t) = α(V (t)) for t ∈ D (under the convention that α(X) = ∞ if X is unbounded), where V (t) = {x(t) : x ∈ V }. The next lemma clarifies the relation between α and α1 . Lemma 3. ([7, Th.2.1]; and [16, Th.1]) Assume that V is a countable set of strongly measurable functions D → E and there exists an integrable function µ such that x(t) µ(t) for all x ∈ V and t ∈ D. Then the corresponding function v is integrable on D and x(t) dt : x ∈ V 2 v(t) dt. α If, in addition lim sup h→∞ x∈V D D D x(t + h) − x(t) dt = 0, then α1 (V ) 2 v(t) dt. D FUNCTIONAL–INTEGRAL EQUATION OF VOLTERRA TYPE 59 2. The main results Let H : D → R+ be a measurable function such that the function (t, s) → A(t, s)H(s) is bounded on D × D. Theorem 1. Let 1◦ − 3◦ hold and 0 < r < n. If α(g(s, X)) H(s)α(X) (3) for any s ∈ D and for any bounded subset X of E, then the equation (1) has a solution x ∈ Lp (D, E). In the case, when r < 1, we can apply the famous Mydlarczyk theorem [12, Th.3.1], and consequently we obtain a stronger theorem if we replace (3) by the condition (5) given below. Theorem 2. Let ω : R+ → R+ be a continuous nondecreasing function such that ω(0) = 0, ω(t) > 0 for t > 0 and δ (4) 0 1 1 s 1−r ds = ∞ s ω(s) (δ > 0). (cf. [12]) Let 1◦ – 3◦ hold, 0 < r < 1 and J = [0, d] be a compact interval in R. If α(g(s, X)) ω(α(X)) (5) for any s ∈ J and for any bounded subset X of E, then the equation (1) has a solution x ∈ Lp (J, E). Proof. By the theory of scalar linear Volterra integral equations it follows that there exists a nonnegative solution u(t) of the equation u(t) = a1 (t) + b1 K(t, s)a2 (s) ds + b1 b2 K(t, s)u(s) ds. D(t) D(t) More precisely, as the spectral radius r(K) of the Volterra integral operator (6) Ku(t) = K(t, s)u(s) ds D(t) is equal to 0, by Theorem 2.2 from [10] the sequence of successive approximations un (t) for (6) is convergent; obviously all un (t) are nonnegative. Put B = {x ∈ Lp (D, E) : x(t) u(t) for a.e. t ∈ D}. Define F : B → p L (D, E) by (F x)(t) = φ t, D(t) K(t, s)g(s, x(s)) ds for x ∈ B and t ∈ D. 60 DUTKIEWICZ Since (F x)(t) = φ(t, SGx(t)) a1 (t) + b1 SGx(t) K(t, s) g(s, x(s)) ds a1 (t) + b1 D(t) a1 (t) + b1 K(t, s) a2 (s) + b2 x(s) ds D(t) a1 (t) + b1 K(t, s) a2 (s) ds + b1 b2 D(t) K(t, s) u(s) ds = u(t) D(t) for x ∈ B and t ∈ D, Lemmas 1 and 2 prove that F is a continuous mapping B → B. Without loss of generality we shall always assume that all functions from Lp (D, E) are extended to Rn by putting x(t) = 0 outside D. Moreover, by 1◦ (ii) we obtain F (x)(t + h) − F (x)(t) d(t, h) for x ∈ B, t ∈ D and small |h|, where ⎧ u(t) if t ∈ D and t + h ∈ /D ⎪ ⎪ ⎨ d(t, h) = a1 (t + h) − a1 (t) ⎪ ⎪ ⎩ + b1 K(t + h, s) − K(t, s) a2 (s) + b2 u(s) ds if t, t + h ∈ D. D From (2) it follows that for each z ∈ L1 (D, R) we have |z(s)| dt (7) |z(s)| ds Q ds dt = |z(s)| ds. |t − s|r |t − s|r D×D D D D In view of (7) the function (t, s) → W (t, s) = K(t, s)(a2 (s) + b2 u(s)) is integrable on D × D. Therefore K(t + h, s) − K(t, s) a2 (s) + b2 u(s) ds dt lim d(t, h) dt = lim h→0 D h→0 D D W (t + h, s) − W (t, s) ds dt = 0 = lim h→0 D D for t ∈ D. Hence (8) lim sup h→0 x∈B D(t) (F x)(t + h) − (F x)(t) dt = 0. Next, let V be a countable subset of B such that (9) V ⊂ conv(F (V ) ∪ {0}). FUNCTIONAL–INTEGRAL EQUATION OF VOLTERRA TYPE 61 Then V (t) ⊂ conv F (V )(t) ∪ {0} for a.e. t ∈ D, so that α(V (t)) α(F (V )(t)) for a.e. t ∈ D. (10) Put v(t) = α(V (t)) for t ∈ D. From (8) and (9) we deduce that lim sup x(t + h) − x(t) dt = 0. h→0 x∈V D Moreover, x(t) u(t) for all x ∈ V and a.e. t ∈ D. Consequently, by Lemma 3, v ∈ Lp (D, R) and (11) α1 (V ) 2 v(t) dt. D ◦ According to 1 (ii), we have φ(t, x) − φ(t, y) b1 x − y for t ∈ D and x, y ∈ E. Then α(φ(t, X)) b1 α(X) for any bounded subset X of E. From (7) it is clear that a2 (s) + b2 u(s) (12) ds < ∞ for a.e. t ∈ D. |t − s|r D Fix t ∈ D such that the integral (12) is finite. Next, we have K(t, s)g(s, x(s)) a a2 (s) + b2 u(s) for x ∈ B and s ∈ D. |t − s|r Case 1. Suppose that the assumptions of Theorem 1 hold. Thus, by (10), (3) and Lemma 3, we get α(V (t)) α((F V )(t)) = α(φ(t, SGV (t))) b1 α K(t, s) g(s, x(s)) ds : x ∈ V D(t) 2b1 α {K(t, s) g(s, x(s)) ds : x ∈ V } ds D(t) 2b1 K(t, s) α(g(s, V (s)) ds 2b1 D(t) K(t, s) H(s) α(V (s)) ds D(t) i.e. v(t) 2b1 K(t, s) H(s) v(s) ds. D(t) Putting w(t) = 2b1 c D(t) v(s) ds, |t − s|r where c = sup A(t, s) H(s) : t, s ∈ D , we see that w(t) is a continuous function such that v(t) w(t) for t ∈ D. Hence 62 DUTKIEWICZ w(t) 2b1 c (13) D(t) w(s) ds. |t − s|r Arguing similarly as in [8; p. 134–135] we can prove that w(t) = 0 for t ∈ D. Since v(t) w(t), we have v(t) = 0 for t ∈ D. Case 2. Suppose that the assumptions of Theorem 2 hold. Thus, by (10), (5) and Lemma 3, we get α(V (t)) α((F V )(t)) = α φ(t, SGV (t)) t b1 α K(t, s) g(s, x(s)) ds : x ∈ V 0 t 2b1 α {K(t, s) g(s, x(s)) ds : x ∈ V } ds 0 t 2b1 K(t, s) α g(s, V (s)) ds 2b1 0 t K(t, s) ω α(V (s)) ds, 0 i.e. t v(t) 2b1 a 0 Putting t w(t) = 2b1 a 0 ω(v(s)) ds for t ∈ J. (t − s)r ω(v(s)) ds (t − s)r for t ∈ J we see that w is a continuous function such that v(t) w(t) for t ∈ J. Hence t (14) w(t) 2b1 a 0 ω(w(s)) ds (t − s)r for t ∈ J. By the Mydlarczyk theorem [12, Th. 3.1] and assumption (4), the integral equation t z(t) = 2b1 a 0 ω(z(s)) ds (t − s)r for ∈ J has the unique continuous solution z(t) ≡ 0. Applying now theorem on integral inequalities [1, Th. 2], from (14) we deduce that w(t) ≡ 0. Thus v(t) = 0 for t ∈ J. In view of (11) this shows that α1 (V ) = 0, so that V is relatively compact in L1 (D, E). On the other hand, the set B has equiabsolutely continuous norms in Lp (D, E) and V ⊂ B. Consequently, V is relatively compact in Lp (D, E). Applying now the following Mönch fixed point theorem [11]: FUNCTIONAL–INTEGRAL EQUATION OF VOLTERRA TYPE 63 Theorem 3. Let B be a closed, convex, and bounded subset of a Banach space such that 0 ∈ B. 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