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 L^{p}-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, d_{1}] × ⋯ × [0, d_{m}] and
D(t) = {s = (s_{1}, …, s_{m}) ∈ R^{m} : 0 ⩽ s_{i} ⩽ t_{i}, i = 1, …, m}

for t = (t_{1}, …, t_{m}) ∈ D. Denote by L^{p}(D, E) (p > 1) the space of all strongly measurable functions x : D ↦ E with ∫_{D} ∥x(t)∥^{p}dt < ∞, provided with the norm

∥x∥_{p} = (∫_{D} ∥x(t)∥^{p}dt)^{1/p}.

We consider the following functional–integral equation of Volterra type

(1)  x(t) = ϕ(t, ∫_{D(t)} K(t, s) g(s, x(s)) ds)

with the kernel K(t, s) = \frac{A(t,s)}{|t−s|^{r}}, 0 < r < n (t, s ∈ D, t ≠ s). We give sufficient conditions for the existence of a solution x ∈ L^{p}(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 L^{1}-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 L^{p}-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, L^{p}-solutions, measure of noncompactness.

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)∥ ⩽ |a_{1}(t) − a_{1}(τ)| + b_{1}∥x − y∥ for t, τ ∈ D and x, y ∈ E,

where a_{1} ∈ L^{p}(D, R) and b_{1} ⩾ 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)∥ ⩽ a_{2}(t) + b_{2}∥x∥ for s ∈ D and x ∈ E, where a_{2} ∈ L^{p}(D, R) and

b_{2} ⩾ 0. In what follows we shall need the following lemmas:

Lemma 1. The linear integral operator

(Sx)(t) = ∫_{D} K(t, s) x(s) ds (x ∈ L^{p}(D, E), t ∈ D)

maps L^{p}(D, E) into itself continuously. Moreover,

∥S∥ ⩽ aQ, where a = sup{∥A(t, s)∥ : t, s ∈ D}

and

(2)  \frac{2π^{n/2}(diam D)^{n−r}}{(n − r)Γ(n/2)} = Q ⩾ ∫_{D} \frac{ds}{|t − s|^{r}} for all t ∈ D.

Lemma 2. Put G(x)(t) = g(t, x(t)) for x ∈ L^{p}(D, E) and t ∈ D. Then G is a continuous mapping of L^{p}(D, E) into itself.

For the proofs we refer for example to [15].

Denote by α and α_{1} the Kuratowski measures of noncompactness in E and L^{1}(D, E), respectively. For any set V of functions belonging to L^{1}(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 α_{1}.

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

α({∫_{D} x(t) dt : x ∈ V }) ⩽ 2 ∫_{D} v(t) dt.

If, in addition lim_{h→∞} sup_{x∈V} ∫_{D} ∥x(t + h) − x(t)∥ dt = 0, then

α_{1}(V) ⩽ 2 ∫_{D} v(t) dt.

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
(3)  α(g(s, X)) ⩽ H(s)α(X)

for any s ∈ D and for any bounded subset X of E, then the equation (1) has a solution x ∈ L^{p}(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

(4)  ∫_{0}^{δ} \frac{1}{s} [\frac{s}{ω(s)}]^{\frac{1}{1−r}} ds = ∞ (δ > 0).  (cf. [12])

Let 1^{∘}– 3^{∘} hold, 0 < r < 1 and J = [0, d] be a compact interval in R. If

(5)  α(g(s, X)) ⩽ ω(α(X))

for any s ∈ J and for any bounded subset X of E, then the equation (1) has a solution x ∈ L^{p}(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) = a_{1}(t) + b_{1} ∫_{D(t)} ∥K(t, s)∥a_{2}(s) ds + b_{1}b_{2} ∫_{D(t)} ∥K(t, s)∥u(s) ds.

More precisely, as the spectral radius r(K) of the Volterra integral operator

(6)  Ku(t) = ∫_{D(t)} ∥K(t, s)∥u(s) ds

is equal to 0, by Theorem 2.2 from [10] the sequence of successive approximations u_{n}(t) for (6) is convergent; obviously all u_{n}(t) are nonnegative.

Put B = {x ∈ L^{p}(D, E) : ∥x(t)∥ ⩽ u(t) for a.e. t ∈ D}. Define F : B ↦ L^{p}(D, E) by

(F x)(t) = ϕ(t, ∫_{D(t)} K(t, s)g(s, x(s)) ds) for x ∈ B and t ∈ D.

Since

∥(F x)(t)∥ = ∥ϕ(t, SGx(t))∥ ⩽ a_{1}(t) + b_{1}∥SGx(t)∥

⩽ a_{1}(t) + b_{1}∥ ∫_{D(t)} K(t, s) g(s, x(s)) ds∥

⩽ a_{1}(t) + b_{1} ∫_{D(t)} ∥K(t, s)∥(a_{2}(s) + b_{2}∥x(s)∥) ds

⩽ a_{1}(t) + b_{1} ∫_{D(t)} ∥K(t, s)∥ a_{2}(s) ds + b_{1}b_{2} ∫_{D(t)} ∥K(t, s)∥ u(s) ds = u(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 L^{p}(D, E) are extended to R^{n} 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

d(t, h) = \begin{cases}u(t) & if t ∈ D and t + h ∉ D \\ ∥a_{1}(t + h) − a_{1}(t)∥ \\ + b_{1}∫_{D}∥K(t + h, s) − K(t, s)∥(a_{2}(s) + b_{2}u(s)) ds & if t, t + h ∈ D. \end{cases}

From (2) it follows that for each z ∈ L^{1}(D, R) we have

(7)  ∫∫_{D×D} \frac{|z(s)|}{|t − s|^{r}} ds dt = ∫_{D} (∫_{D} \frac{dt}{|t − s|^{r}} )|z(s)| ds ⩽ Q ∫_{D} |z(s)| ds.

In view of (7) the function (t, s) ↦ W (t, s) = K(t, s)(a_{2}(s) + b_{2}u(s)) is integrable on D × D. Therefore

lim_{h→0} ∫_{D} d(t, h) dt = lim_{h→0} ∫_{D} (∫_{D} ∥K(t + h, s) − K(t, s)∥ (a_{2}(s) + b_{2}u(s)) ds) dt

= lim_{h→0} ∫_{D} ∫_{D} ∥W (t + h, s) − W (t, s)∥ ds dt = 0

for t ∈ D. Hence

(8)  lim_{h→0} sup_{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 ⊂ \overline{conv}(F (V) ∪ \{0\}).

Then V (t) ⊂ \overline{conv}(F (V)(t) ∪ \{0\}) for a.e. t ∈ D, so that

(10)  α(V (t)) ⩽ α(F (V)(t)) for a.e. t ∈ D.

Put v(t) = α(V (t)) for t ∈ D. From (8) and (9) we deduce that

lim_{h→0} sup_{x∈V} ∫_{D} ∥x(t + h) − x(t)∥ dt = 0.

Moreover, ∥x(t)∥ ⩽ u(t) for all x ∈ V and a.e. t ∈ D. Consequently, by Lemma 3, v ∈ L^{p}(D, R) and

(11)  α_{1}(V) ⩽ 2 ∫_{D} v(t) dt.

According to 1^{∘}(ii), we have ∥ϕ(t, x) − ϕ(t, y)∥ ⩽ b_{1}∥x − y∥ for t ∈ D and x, y ∈ E. Then α(ϕ(t, X)) ⩽ b_{1}α(X) for any bounded subset X of E.

From (7) it is clear that

(12)  ∫_{D} \frac{a_{2}(s) + b_{2}u(s)}{|t − s|^{r}} ds < ∞ for a.e. t ∈ D.

Fix t ∈ D such that the integral (12) is finite. Next, we have

∥K(t, s)g(s, x(s))∥ ⩽ a\frac{a_{2}(s) + b_{2}u(s)}{|t − s|^{r}} for x ∈ B and s ∈ D.

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)))

⩽ b_{1}α({ ∫_{D(t)} K(t, s) g(s, x(s)) ds : x ∈ V })

⩽ 2b_{1} ∫_{D(t)} α({K(t, s) g(s, x(s)) ds : x ∈ V }) ds

⩽ 2b_{1} ∫_{D(t)} ∥K(t, s)∥ α(g(s, V (s)) ds ⩽ 2b_{1} ∫_{D(t)} ∥K(t, s)∥ H(s) α(V (s)) ds

i.e.

v(t) ⩽ 2b_{1} ∫_{D(t)} ∥K(t, s)∥ H(s) v(s) ds.

Putting

w(t) = 2b_{1}c ∫_{D(t)} \frac{v(s)}{|t − s|^{r}} ds,

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

(13)  w(t) ⩽ 2b_{1}c ∫_{D(t)} \frac{w(s)}{|t − s|^{r}} ds.

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)))

⩽ b_{1}α({∫_{0}^{t} K(t, s) g(s, x(s)) ds : x ∈ V })

⩽ 2b_{1} ∫_{0}^{t} α({K(t, s) g(s, x(s)) ds : x ∈ V }) ds

⩽ 2b_{1} ∫_{0}^{t} ∥K(t, s)∥ α(g(s, V (s))) ds ⩽ 2b_{1} ∫_{0}^{t} ∥K(t, s)∥ ω(α(V (s))) ds,

i.e.

v(t) ⩽ 2b_{1}a ∫_{0}^{t} \frac{ω(v(s))}{(t − s)^{r}} ds for t ∈ J.

Putting

w(t) = 2b_{1}a ∫_{0}^{t} \frac{ω(v(s))}{(t − s)^{r}} ds for t ∈ J

we see that w is a continuous function such that v(t) ⩽ w(t) for t ∈ J. Hence

(14)  w(t) ⩽ 2b_{1}a ∫_{0}^{t} \frac{ω(w(s))}{(t − s)^{r}} ds for t ∈ J.

By the Mydlarczyk theorem [12, Th. 3.1] and assumption (4), the integral equation

z(t) = 2b_{1}a ∫_{0}^{t} \frac{ω(z(s))}{(t − s)^{r}} ds 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 α_{1}(V) = 0, so that V is relatively compact in L^{1}(D, E). On the other hand, the set B has equiabsolutely continuous norms in L^{p}(D, E) and V ⊂ B. Consequently, V is relatively compact in L^{p}(D, E).

Applying now the following Mönch fixed point theorem [11]:

Theorem 3. Let B be a closed, convex, and bounded subset of a Banach space such that 0 ∈ B. If F : B ↦ B is a continuous mapping such that for each countable subset V of B the following implication holds

V ⊂ \overline{conv}(F (V) ∪ 0) ⟹ V is relatively compact,

then F has a fixed point.

we conclude that there exists x ∈ B such that x = F (x). Obviously x is a solution of (1). □

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Faculty of Mathematics and Computer Science (Received 30 01 2007) Adam Mickiewicz University (Revised 08 09 2008) Umultowska 87 61-614 Poznań Poland

szukala@amu.edu.pl