PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 77(91) (2005), 87–92

KNESER’S THEOREM FOR WEAK SOLUTIONS OF AN INTEGRAL EQUATION WITH WEAKLY SINGULAR KERNEL

Aldona Dutkiewicz and Stanisław Szufla

Abstract. We prove that the set of all weak solutions of the Volterra integral equation (1) is nonempty, compact and connected.

Assume that D = [0, a] is a compact interval in ℝ, E is a sequentially weakly complete Banach space, B = {x ∈ E : ∥x∥ ⩽ b}. We prove the existence of a weak solution of the integral equation

(1)  x(t) = ∫_{0}^{t} K(t, s)f(s, x(s)) ds,

where

1^{∘} f : D × B ↦ E is a weakly-weakly continuous function such that ∥f(t, x)∥ ⩽ M for (t, x) ∈ D × B;

2^{∘} K(t, s) = \frac{H(t, s)}{(t − s)^{r}}, 0 < r < 1, where H is a real continuous function.

Moreover, we study the topological structure of the set of all weak solutions of (1). In what follows we shall need the following result of W. Mydlarczyk given in [6].

Theorem 1. Let α > 0 and let g : ℝ_{+} ↦ ℝ_{+} be a nondecreasing function such that g(0) = 0, g(t) > 0 for t > 0. Then the equation

u(t) = ∫_{0}^{t}(t − s)^{α−1}g(u(s)) ds (t ⩾ 0)

has a nontrivial continuous solution if and only if

∫_{0}^{δ} \frac{1}{s} [\frac{s}{g(s)}]^{1/α} ds < ∞ (δ > 0).

Let c = max_{t,s∈D} |H(t, s)|. Choose a positive number d such that d ⩽ a and

M · c · \frac{d^{1−r}}{1 − r} < b. Denote by L = M · c · \frac{d^{1−r}}{1 − r}. Hence L < b.

2000 Mathematics Subject Classification: Primary 45N05.

Let J = [0, d]. Denote by C_{w}(J, E) the space of weakly continuous functions J ↦ E endowed with the topology of weak uniform convergence. Moreover, denote by β the measure of weak noncompactness introduced by De Blasi [2].

Let us recall that for any nonvoid, bounded subset A of a Banach space E, β(A) = inf{ε > 0 : there exists a weakly compact set K such that A ⊂ K + εB}, where B is the norm unit ball. Recall that β has the following properties:

1^{∘} A ⊂ B ⇒ β(A) ⩽ β(B);

2^{∘} β(Ā^{w}) = β(A), where Ā^{w} denotes the weak closure of A;
3^{∘} β(A) = 0 ⇔ Ā^{w} is weakly compact;
4^{∘} β(A ∪ B) = max(β(A), β(B));
5^{∘} β(convA) = β(A);
6^{∘} β(A + B) ⩽ β(A) + β(B);
7^{∘} β(λA) = |λ|β(A), (λ ∈ ℝ);
8^{∘} β( ⋃_{|λ|⩽h} λA) = hβ(A).

Let V be a subset of C_{w}(J, E). Put V (t) = {u(t) : u ∈ V } and V (T) = {u(t) : u ∈ V, t ∈ T }. Let us recall the well known Ambrosetti type

Lemma 1. If the set V is strongly equicontinuous and uniformly bounded, then (a) the function t ↦ β(V (t)) is continuous on J;

(b) for each compact subset T of J one has β(V (T)) = sup{β(V (t)) : t ∈ T }.

Let B̃ denote the set of all weakly continuous functions J ↦ B. We shall consider B̃ as a topological subspace of C_{w}(J, E). Put

F (x)(t) = ∫_{0}^{t} K(t, s)f(s, x(s)) ds, (x ∈ B̃, t ∈ J).

Arguing similarly as in [4, p. 132–133] we can prove that the set F (B̃) is strongly equicontinuous. On the other hand, from the following Krasnoselskiĭ type

Lemma 2. For any φ ∈ E^{∗}, ε > 0 and z ∈ B̃ there exists a weak neighbourhood U of 0 in E such that | φ(f(t, z(t)) − f(t, w(t))) |⩽ ε for t ∈ J and w ∈ B̃ such that w(s) − z(s) ∈ U for all s ∈ J. [8]

It follows that F is a continuous mapping from B̃ into C_{w}(J, E).

For given ε > 0 denote by S_{ε} the set of all z ∈ B̃ such that ∥z(t)−F (z)(t)∥ < ε for all t ∈ J.

Lemma 3. For each ε, 0 < ε < b − L, the set S_{ε} is nonempty and connected in C_{w}(J, E).

Proof. For any positive integer n we define F_{n}(x)(t) = F (x)(r_{n}(t)) (x ∈ B̃, t ∈ J), where

r_{n}(t) = \begin{cases}0, & if 0 ⩽ t ⩽ d/n \\ t − d/n, & if d/n ⩽ t ⩽ d. \end{cases}

Put

(2)  sup_{t∈J, x∈B̃} ∥F (x)(t) − F (x)(r_{n}(t))∥ = w_{n}.

KNESER’S THEOREM FOR WEAK SOLUTIONS OF AN INTEGRAL EQUATION 89

Because the set F (B̃) is equicontinuous, we have w_{n} ↦ 0 as n ↦ ∞. Moreover, there exists a unique z_{n} ∈ B̃ such that z_{n} = F_{n}(z_{n}). It is clear from (2) that z_{n} ∈ S_{ε} for sufficiently large n. Fix u_{0}, u_{1} ∈ S_{ε}. Put

η = max ( sup_{t∈J} ∥u_{0}(t) − F (u_{0})(t)∥, sup_{t∈J} ∥u_{1}(t) − F (u_{1})(t)∥)

and δ = ε − η. Fix a positive integer n such that 2w_{n} < δ. Let

a_{λ} = λ(u_{1} − F_{n}(u_{1})) + (1 − λ)(u_{0} − F_{n}(u_{0})) for 0 ⩽ λ ⩽ 1.

It follows from (2) that

∥u_{i}(t) − F_{n}(u_{i})(t)∥ ⩽ ∥u_{i}(t) − F (u_{i})(t)∥ + ∥F (u_{i})(t) − F_{n}(u_{i})(t)∥ ⩽ η + w_{n}

(i = 0, 1).

Hence

(3)  ∥a_{λ}(t)∥ ⩽ η + w_{n} for t ∈ J and 0 ⩽ λ ⩽ 1.

Arguing similarly as in [8, p. 122] we can prove that for each λ ∈ [0, 1] there exists a unique u_{λ} such that u_{λ} = a_{λ} + F_{n}(u_{λ}) and u_{λ} depends continuosly on λ. Since

∫_{0}^{t} \frac{ds}{(t − s)^{r}} = \frac{t^{1−r}}{1 − r} ,

we have

∥F (x)(t)∥ = ∥∫_{0}^{t} K(t, s)f(s, x(s)) ds∥ ⩽ c · M · \frac{d^{1−r}}{1 − r} = L for x ∈ B̃, t ∈ J.

From this and inequalities (2)–(3) we obtain

∥u_{λ}(t)∥ ⩽ ∥a_{λ}(t)∥ + ∥F (u_{λ})(r_{n}(t))∥ ⩽ η + w_{n} + L < η + δ + L = ε + L < b

and

∥u_{λ}(t) − F (u_{λ})(t)∥ = ∥a_{λ}(t) + F_{n}(u_{λ})(t) − F (u_{λ})(t)∥
⩽ ∥a_{λ}(t)∥ + ∥F_{n}(u_{λ})(t) − F (u_{λ})(t)∥
⩽ η + 2w_{n} < η + δ = ε (t ∈ J, 0 ⩽ λ ⩽ 1),

so that u_{λ} ∈ S_{ε}. From this we conclude that for any u_{0}, u_{1} ∈ S_{ε} there exists a continuous curve in S_{ε} connecting u_{0} and u_{1}, which proves that S_{ε} is arcwise connected. □

The main result of the paper is the following

Theorem 2. Let g : ℝ_{+} ↦ ℝ_{+} be a continuous nondecreasing function such that g(0) = 0, g(t) > 0 for t > 0 and

(4)  ∫_{0}^{δ} \frac{1}{s} [\frac{s}{g(s)}]^{1/(1−r)} ds = ∞ (δ > 0).
If 1^{∘} and 2^{∘} hold and
(5)  β(f(J × X)) ⩽ g(β(X)) for X ⊂ B,

then the set S of all weak solutions of (1) defined on J is nonempty, compact and connected in C_{w}(J, E).

Proof. 1. First we shall show that the set S is nonempty. By Lemma 3 there exists a sequence (u_{n}) such that u_{n} ∈ B̃ and

(6)  lim_{n→∞} sup_{t∈J} ∥u_{n}(t) − F (u_{n})(t)∥ = 0.

Let V = {u_{n} : n ∈ N}. Since

V ⊂ {u_{n} − F (u_{n}) : n ∈ N} + F (V)
V (t) ⊂ {u_{n}(t) − F (u_{n})(t) : n ∈ N} + F (V)(t),
F (V)(t) ⊂ V (t) − {u_{n}(t) − F (u_{n})(t) : n ∈ N},

it follows from (6) that the set V is strongly equicontinuous and

(7)  β(V (t)) = β(F (V)(t)) for t ∈ J.

Hence, by Lemma 1, the function t ↦ v(t) = β(V (t)) is continuous on J. Fix t ∈ J and ε > 0, and choose η > 0 such that

∥∫_{t−η}^{t} \frac{H(t, s)}{(t − s)^{r}} f(s, x(s)) ds∥ ⩽ ∫_{t−η}^{t} \frac{|H(t, s)|}{(t − s)^{r}}M ds < ε for all x ∈ B̃.

From the continuity of the function \frac{H(t, s)}{(t − s)^{r}} g(v(s)) on [0, t−η] it follows that there exists δ > 0 such that

(8)  |\frac{H(t, τ)}{(t − τ)^{r}} g(v(q)) − \frac{H(t, s)}{(t − s)^{r}} g(v(s))| < ε

if |τ − s| < δ, |q − s| < δ, q, s, τ ∈ [0, t − η]. Divide the interval [0, t − η] into n

OK? parts 0 = t_{0} < t_{1} < … < t_{n} = t − η so that Δt_{i} = t_{i} − t_{i−1} < δ (i = 1, …, n). Put T_{i} = [t_{i−1}, t_{i}]. By Lemma 1 for each i there exists s_{i} ∈ T_{i} such that β(V (T_{i})) = v(s_{i}) (i = 1, …, n). Put

∫_{T} \frac{H(t, s)}{(t − s)^{r}} f(s, V (s)) ds = \{∫_{T} \frac{H(t, s)}{(t − s)^{r}} f(s, x(s)) ds : x ∈ V \} .

Because

∫_{0}^{t−η} \frac{H(t, s)}{(t − s)^{r}} f(s, V (s)) ds ⊂ ∑_{i=1}^{n} ∫_{t_{i−1}}^{t_{i}} \frac{H(t, s)}{(t − s)^{r}} f(s, V (s)) ds

⊂ ∑_{i=1}^{n} Δt_{i}\overline{conv} \{\frac{H(t, s)}{(t − s)^{r}} f(s, x(s)) : s ∈ T_{i}, x ∈ V \} ,

KNESER’S THEOREM FOR WEAK SOLUTIONS OF AN INTEGRAL EQUATION 91

we get

β (∫_{0}^{t−η} \frac{H(t, s)}{(t − s)^{r}} f(s, V (s)) ds)

⩽ ∑_{i=1}^{n} Δt_{i}β (\overline{conv} \{\frac{H(t, s)}{(t − s)^{r}} f(s, x(s)) : s ∈ T_{i}, x ∈ V \})

= ∑_{i=1}^{n} Δt_{i}β (\{\frac{H(t, s)}{(t − s)^{r}} f(s, x(s)) : s ∈ T_{i}, x ∈ V \})

= ∑_{i=1}^{n} Δt_{i} max_{s∈T_{i}} \frac{|H(t, s)|}{(t − s)^{r}} β(f(J × V (T_{i}))

⩽ ∑_{i=1}^{n} Δt_{i} \frac{|H(t, τ_{i})|}{(t − τ_{i})^{r}} g(β(V (T_{i}))) ⩽ ∑_{i=1}^{n} Δt_{i} \frac{|H(t, τ_{i})|}{(t − τ_{i})^{r}} g(v(s_{i})).

Here τ_{i} ∈ T_{i} is a number such that \frac{|H(t, τ_{i})|}{(t − τ_{i})^{r}} = max_{s∈T_{i}} \frac{|H(t, s)|}{(t − s)^{r}}. Furthermore, from inequality (8) we infer that

(9) ∑_{i=1}^{n} Δt_{i} \frac{|H(t, τ_{i})|}{(t − τ_{i})^{r}} g(v(s_{i})) ⩽ ∫_{0}^{t−η} \frac{|H(t, s)|}{(t − s)^{r}} g(v(s)) ds + ε(t − η).

Since

F (V)(t) ⊂ ∫_{0}^{t−η} \frac{H(t, s)}{(t − s)^{r}} f(s, V (s)) ds + ∫_{t−η}^{t} \frac{H(t, s)}{(t − s)^{r}} f(s, V (s)) ds

and

β (\{∫_{t−η}^{t} \frac{H(t, s)}{(t − s)^{r}} f(s, x(s)) ds : x ∈ V \}) ⩽ 2ε,

from inequalities (8) and (9) it follows that

β(F (V)(t)) ⩽ ∫_{0}^{t−η} \frac{|H(t, s)|}{(t − s)^{r}} g(v(s)) ds + ε(t − η) + 2ε

⩽ ∫_{0}^{t} \frac{|H(t, s)|}{(t − s)^{r}} g(v(s))ds + εt + 2ε.

As the last inequality is satisfied for every ε > 0, we get

β(F (V)(t)) ⩽ ∫_{0}^{t} \frac{|H(t, s)|}{(t − s)^{r}} g(v(s)) ds.

Therefore, by (7),

β(V (t)) ⩽ ∫_{0}^{t} \frac{|H(t, s)|}{(t − s)^{r}} g(v(s)) ds

i.e.,

v(t) ⩽ ∫_{0}^{t} \frac{|H(t, s)|}{(t − s)^{r}}g(v(s)) ds for t ∈ J.

Applying Theorem 1 with α = 1 − r and theorem on integral inequalities [3, Lemma 1] from this we deduce that v(t) = 0 for t ∈ J. Therefore, by Lemma 1

β(V (J)) = sup{β(V (t)) : t ∈ J} = 0

i.e., V is relatively compact in C_{w}(J, E). Hence we can find a subsequence (u_{n_{k}}) of (u_{n}) which converges in C_{w}(J, E) to a limit u. As F is continuous, from this and (6) we conclude that u = F (u). This proves that the set S is nonempty.

2. Further, since F is continuous, S is closed in C_{w}(J, E). As S = F (S), we have β(S(t)) = β(F (S)(t)) for t ∈ J. Arguing similarly as in 1, we can show that S is compact in C_{w}(J, E).

Now we shall prove that S is connected. Suppose that S is not connected in C_{w}(J, E). As S is compact, there are nonempty compact sets S_{1}, S_{2} such that S = S_{1} ∪ S_{2} and S_{1} ∩ S_{2} = ∅, and there are two disjoint open sets U_{1}, U_{2} such that S_{1} ⊂ U_{1}, S_{2} ⊂ U_{2}. Let U = U_{1} ∪ U_{2}. We choose n_{0} such that 1/n_{0} < b − L. Suppose that for each n ⩾ n_{0} there exists u_{n} ∈ S_{1/n} ∖ U. Put V = {u_{n} : n ∈ N}. Because lim_{n→∞} sup_{t∈J} ∥u_{n}(t)−F (u_{n})(t)∥ = 0, by repeating the argument from 1 we can

prove that there exists u_{0} ∈ \overline{V} such that u_{0} = F (u_{0}), i.e., u_{0} ∈ S. Furthermore, \overline{V} ⊂ C_{w}(J, E) ∖ U, as U is open, so that u_{0} ∈ S ∖ U, a contradiction. Therefore there exists k ∈ N such that S_{1/k} ⊂ U. Since U_{1} ∩S_{1/k} ≠ ∅ ≠ U_{2} ∩S_{1/k}, this shows that S_{1/k} is not connected, which contradicts Lemma 3. Hence S is connected. □

The authors express their appreciation to the referee for his suggestions to amend an earlier version of the paper.

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Faculty of Mathematics and Computer Science (Received 20 02 2004) Adam Mickiewicz University (Revised 20 01 2005) Umultowska 87 61-614 Poznań Poland