Now, we present the necessary definitions from the theory of cones in Banach spaces.
Definition 1 Let E be a real Banach space. A nonempty convex closed set is said to be a cone provided that
for all and all , and
Note that every cone induces an ordering in E given by if .
Definition 2 A map Φ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if is continuous and
for all and .
Similarly, we say the map φ is a nonnegative continuous convex functional on a cone P of a real Banach space E if is continuous and
for all and .
Definition 3 An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P. Then, for positive numbers a, b, c, d, we define the following sets:
We will use the following fixed point theorem of Avery and Peterson to establish multiple positive solutions to problem (1).
Theorem 1 (see )
Let P be a cone in a real Banach space E. Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P satisfying for such that for some positive numbers and d,
for all . Suppose
is completely continuous and there exist positive numbers a, b, c with such that
(S1): and for ,
(S2): for with ,
(S3): and for with .
has at least three fixed points
We apply Theorem 1 with the cone K instead of P and let . Now, we define the nonnegative continuous concave functional Φ on K by
Note that . Put .
Now, we can formulate the main result of this section.
Theorem 2 Let the assumptions H1-H4 hold with , . Let , . In addition, we assume that there exist positive constants a, b, c, d, M, and such that
(A1): for ,
(A2): for ,
(A3): for .
Then problem (1) has at least three nonnegative solutions , , satisfying , ,
Proof Basing on the definitions of T, we see that is equicontinuous on J, so T is completely continuous.
Let , so . By Lemma 1, , so , . Assumption (A1) implies .
Moreover, in view of (7),
Combining it, we have
This proves that .
Now, we need to show that condition (S1) is satisfied. Take
Then , , and
for , . Moreover,
This proves that
Let for . Then for , so , . Assumption (A2) implies . Hence,
This proves that condition (S1) holds.
Now, we need to prove that condition (S2) is satisfied. Take and . Then
so condition (S2) holds.
Indeed, , so . Suppose that with . Note that , . Then
This shows that condition (S3) is satisfied.
Since all the conditions of Theorem 1 are satisfied, problem (1) has at least three nonnegative solutions , , such that for , and
This ends the proof. □
Example Consider the following problem:
with . For example, we can take , on J with fixed . Indeed, , , , , and
Note that , so the measure changes the sign on J. Moreover,
so the assumption H4 holds; see Remark 4. Next,
Put , , , then , , . Let , . Then
All the assumptions of Theorem 2 hold, so problem (8) has at least three positive solutions.
Remark 5 We can also construct an example in which, for example, to use the results of Remark 4. Note that also this measure changes the sign.