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. If F : B → B is a continuous mapping such that for each countable
subset V of B the following implication holds
V ⊂ 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
Adam Mickiewicz University
Umultowska 87
61-614 Poznań
Poland
szukala@amu.edu.pl
(Received 30 01 2007)
(Revised 08 09 2008)