First, we investigate attractive solutions of the problem (2.1). In this way, suppose that for . We assume that is Lebesgue measurable with respect to t on and is continuous with respect to on for . Note that the problem (2.1) is equivalent to the system of equations
or
for . Define the operator by
where
for . It is easy to check that is a solution of the problem (2.1) if and only if is a fixed point of the operator T.
Theorem 3.1
Suppose that for each
there exist
and
such that
for all and . Then the problem (2.1) has at least one attractive solution such that for all .
Proof Consider the set
where is a constant. It is easy to check that is a closed, bounded, and convex subset of
. We show that the operator T has a fixed point in . This implies that the problem (2.1) has a solution. Note that for all and so . Now, we show that T is continuous. Let for all and for all . Then, we have for all and . Let be given. Choose such that implies that . Let and note that for . Also, we have
for . Thus, for all . Also, we have
for . Hence, for . This implies that is continuous for and so T is continuous. Now, we show that the set is equi-continuous. Let . Since for , there is a such that for all and . Let and . If , then
and so . If , then
Now, let . Since
we get in all cases. This implies that the set is equi-continuous. Since is uniformly bounded, is relatively compact. Now by using Theorem 2.1, T has a fixed point in which is a solution of the problem (2.1). Since , . Thus, is an attractive solution for the problem (2.1). □
Theorem 3.2 Suppose that for each there exist , and such that and
for all , and . Then the problem (2.1) has at least one attractive solution such that for all .
Proof It is sufficient we consider the set
where is a constant. By using a similar techniques and proof in Theorem 3.1, one can show that , T is continuous and is relatively compact. Now by using Theorem 2.1, T has a fixed point in which is a solution of the problem (2.1). Since , . Thus, is an attractive solution for the problem (2.1). □
Theorem 3.3
Suppose that for each
there exists
such that
for all . Then the problem (2.1) has at least one attractive solution such that for all .
Proof It is sufficient we consider the set
where is a constant. By using a similar techniques and proof in Theorem 3.1, one can show that , T is continuous and is relatively compact. Now, by using Theorem 2.1, T has a fixed point in which is a solution of the problem (2.1). Since , . Thus, is an attractive solution for the problem (2.1). □
Here, we are going to investigate global attractivity of solutions of the problem (2.2). We assume that is Lebesgue measurable with respect to t on and there exists a constant such that and is continuous with respect to on for all . Note that the problem (2.2) is equivalent to the system of equations
for all and . Define the operator by
where
for all . Now, define
where for all . Finally, define
where for all . It is easy to check that is a solution of the problem (2.2) if and only if it is a fixed point of the operator T. Note that A is a contraction with constant 0.
Theorem 3.4 Suppose that for each there exist and such that for all and . Then the zero solution of the problem (2.2) is globally attractive.
Proof Consider the set
where and is chosen such that for all . First, we show that B maps into . It is easy to check that is a closed, bounded, and convex subset of . Note that
and for all and . Thus,
for all and . Hence, . Now, we show that B is continuous on . Let for all and . Then, one can get for all . Let be given. Choose such that for all . Let for . Then, we have
for all . Hence, for all . Also,
for all . Thus, for all . This implies that is continuous on for and so B is continuous on . Now, we show that is equi-continuous. Let be given. Since , there exists such that for . Let and . If , then we have
and so . If , then
If , then we have
and so . Thus, is equi-continuous. Since is uniformly bounded, is relatively compact. Now, suppose that , and . Then,
for all . Since for , we get
Thus, for all and . This implies that for all . Therefore, by using Theorem 2.2 T has a fixed point in which is a solution of the problem (2.2). Since all elements of the set tend to 0 as , the zero solution of the problem (2.2) is globally attractive. □
Theorem 3.5 Suppose that for each there exist and such that for all and . Then the zero solution of the problem (2.2) is globally attractive.
Proof It is sufficient to consider the set
where and is chosen such that for all . Similar to the proof of Theorem 3.4, one can show that is a closed, bounded, and convex set, B maps into , is relatively compact, and B is continuous on . Now, suppose that , and . Then,
for all . Since for , we get
Thus, for all and . This implies that , for . Since all elements of the set tend to 0 as , the zero solution of the problem (2.2) is globally attractive. □