Theorem 3.1.

If the assumptions and hold, then for any , (2.1) has at least a solution defined on with the initial value .

Proof.

By the assumption , it is easy to get that : is an upper semicontinuous set-valued map with nonempty, compact, and convex values. Hence, by Definition 2.1, the local existence of a solution for (2.1) on , , with , is obvious [17].

Set . Since is a continuous -periodic vector function, is bounded, that is, there exists a constant such that , . By the assumption , we have

By , we can choose a constant , such that when ,

By (2.4), (3.1), (3.2), and the Cauchy inequality, when ,

Therefore, let , then, by (3.3), it follows that on . This means that the local solution is bounded. Thus, (2.1) has at least a solution with the initial value on . This completes the proof.

Theorem 3.1 shows the existence of solutions of (2.1). In the following, we will prove that (2.1) has an -periodic solution.

Let for all , then is a linear operator.

Proposition 3.2.

is bounded, one to one and surjective.

Proof.

For any , we have

this implies that is bounded.

Let . If , then

By the assumption ,

Noting , we have

By (3.6),

Hence . It follows . This shows that is one to one.

Let . In order to verify that is surjective, in the following, we will prove that there exists such that

that is, we will prove that there exists a solution for the differential equation

Consider Cauchy problem

It is easily checked that

is the solution of (3.11). By (3.12), we want , then

that is,

By the assumption , is a nonsingular matrix, where is a unit matrix. Thus, by (3.14), if we take as

in (3.12), then (3.12) is the solution of (3.10). This shows that is surjective. This completes the proof.

By the Banach inverse operator theorem, is a bounded linear operator.

For any , define the set-valued map as

Then has the following properties.

Proposition 3.3.

has nonempty closed convex values in and is also upper semicontinuous from into endowed with the weak topology.

Proof.

The closedness and convexity of values of are clear. Next, we verify the nonemptiness. In fact, for any , there exists a sequence of step functions such that and a.e. on . By the assumption (1) and the continuity of , we can get that is graph measurable. Hence, for any , admits a measurable selector . By the assumption (2), is uniformly integrable. So by Dunford-Pettis theorem, there exists a subsequence such that weakly in . Hence, from [21, Theorem 3.1], we have

Noting that is an upper semicontinuous set-valued map with nonempty closed convex values on for a.e. , . Therefore, . This shows that is nonempty.

At last we will prove that is upper semicontinuous from into . Let be a nonempty and weakly closed subset of , then we need only to prove that the set

is closed. Let and assume in , then there exists a subsequence such that a.e. on . Take , , then By the assumption (2) and Dunford-Pettis theorem, there exists a subsequence such that weakly in . As before we have

This implies , that is, is closed in . The proof is complete.

Theorem 3.4.

Under the assumptions and , there exists a solution for the boundary-value problem (2.5), that is, the neural network (2.1) has an -periodic solution.

Proof.

Consider the set-valued map . Since is continuous and is upper semicontinuous, the set-valued map is upper semicontinuous. Let be a bounded set, then

is a bounded subset of . Since is a bounded linear operator, is a bounded subset of . Noting that is compactly embedded in , is a relatively compact subset of . Hence by Proposition 3.3, is the upper semicontinuous set-valued map which maps bounded sets into relatively compact sets.

For any , when , by (3.1) and the Cauchy inequality,

Arguing as (3.2), we can choose a constant , such that when ,

Therefore, when , by (3.21),

Set

In the following, we will prove that is a bounded subset of . Let , then , that is, . By the definition of , there exists a measurable selection , such that

By (3.23) and (3.25), . Otherwise, . By , we have . Since is continuous, we can choose , such that

Thus, there exists a constant , such that when , . By (3.23) and (3.25),

This is a contradiction. Thus, for any . Furthermore, we have

This shows that is a bounded subset of .

By Lemma 2.3, the set-valued map has a fixed point, that is, there exists such that , . Hence there exists a measurable selection , such that

By the definition of , . Moreover, by Definition 2.1 and (3.29), is a solution of the boundary-value problem (2.5), that is, the neural network (2.1) has an -periodic solution. The proof is completed.