In order to prove the solvability of problem (7a) and (7b), we first have to investigate the problem
depending on the parameter λ. Here, is from (H2) and .
The following result shows that the solvability of problem (17a) and (17b) is equivalent to the solvability of an integral equation in the set .
Lemma 7 Let (H1) hold. Then u is a solution of problem (17a) and (17b) if and only if u is a solution of the integral equation
(18)
in the set .
Proof Let u be a solution of Eq. (17a). Then , and by Corollary 1 (with replaced by ), there exists such that the equation
holds for . Hence, and if and only if . Consequently, if u is a solution of problem (17a) and (17b), then u is a solution of Eq. (18) in .
Let u be a solution of Eq. (18) in . Then . Hence, Lemma 3(ii) (with ρ replaced by ) guarantees that and u is a solution of Eq. (17a). Moreover, . Consequently, u is a solution of problem (17a) and (17b) which completes the proof. □
The following results provide bounds for solutions of problem (17a) and (17b).
Lemma 8 Let (H1)-(H3) hold. Then there exists a positive constant S (independent of and ) such that for all solutions u of problem (17a) and (17b), the estimates
hold. Moreover, for any solution u of problem (17a) and (17b), there exists such that
(21)
Proof Let u be a solution of problem (17a) and (17b). Then by Lemma 7, equality (18) holds for . Since by (5), for a.e. , the relation
follows from (18). Hence, for because g is nonincreasing on . Due to Remark 1, , which means that
(22)
It is clear that L is independent of the choice of solution u to problem (17a) and (17b) and independent of , .
We now show that inequality (21) holds for some . Differentiation of (18) gives
(23)
Since , it follows from (5) and (23) that
Hence, is decreasing on , and therefore vanishes at a unique point due to . The inequality (21) now follows from the relations
Hence, on , and
In particular,
(24)
Let . Taking into account (6), (9), (18), (22), (24), and Lemma 5, we obtain
It follows from for ,
(25)
and therefore, we have
(26)
where and . By (H3),
Consequently, there exists such that for . Now, due to (26), , and therefore, by (25), . □
We are now in the position to prove the existence result for problem (7a) and (7b).
Lemma 9 Let (H1)-(H3) hold. Then for each , problem (7a) and (7b) has a solution u satisfying inequalities (19)-(21), where S is a positive constant independent of n.
Proof Let S be a positive constant in Lemma 8 and let us define
Then Ω is an open and bounded subset of the Banach space . Keeping in mind Lemma 3, define an operator by the formula
(27)
By Lemma 7, any fixed point of the operator is a solution of problem (7a) and (7b). In order to show the existence of a fixed point of , we apply Lemma 1 with , , and . Especially, we show that
-
(i)
is a compact operator, and
-
(ii)
for each and .
We first verify that is a continuous operator. To this end, let be a convergent sequence, and let in . Let
It follows from Lemma 5 and (9) that
for . Here . In particular, for ,
(28)
Since for a.e. and there exists such that
we have by the Lebesgue dominated convergence theorem. Hence, by (28), is a continuous operator. We now show that the set is relatively compact in . It follows from and bounded in that there exists such that
Then by Lemma 5 and (9), the inequalities
are satisfied for and , and therefore, the set is bounded in . Moreover, the relation
holds for a.e. and all (cf. (9)). Consequently, the set is equicontinuous on . Hence, the set is relatively compact in by the Arzelà-Ascoli theorem. As a result, is a compact operator and the condition (i) follows.
Due to the fact that by Lemma 7 any fixed point u of the operator is a solution of problem (17a) and (17b), Lemma 8 guarantees that u satisfies inequality (20). Therefore, has property (ii). Consequently, by Lemmas 1 and 8, for each , problem (7a) and (7b) has a solution u satisfying estimates (19)-(21). □
Let be a solution of problem (7a) and (7b) for . The following property of the sequence is an important prerequisite for solving problem (1a) and (1b).
Lemma 10 Let (H1)-(H3) hold. Let be a solution of problem (7a) and (7b) for . Then the sequence is uniformly integrable on .
Proof
By Lemma 9, the inequalities
hold, where S is a positive constant and . Hence, by (3) and (4),
(32)
for a.e. , where , see Remark 1. Since the sequence is uniformly integrable on (cf. [[20], criterion A.4], [21, 22]), it follows from (32) that is uniformly integrable on and the result follows. □