(A1) are lower and upper solutions of (1.1) respectively such that for with ;
(A2) is convex, that is, exists, continuous with for each , and , where ;
(A3) is convex, that is, exists, continuous with for each , and , where ;
(A4) g is continuous on ℝ such that , exist and , .
Then there exist monotone sequences and that converge uniformly and quadratically in the space of continuous functions on to a unique solution of (1.1).
Proof Using the generalized mean value theorem together with (A2), (A3), and (A4), we obtain
Now, we set
Consider the singular BVP
Using assumption (A1), (3.1) and (3.2), we obtain
Further, we note that and using assumption (A4), for , we find that
which implies that . Thus, it follows that and are respectively lower and upper solutions of (3.4). Since (), therefore it follows by Lemmas 2.6 and 2.7 that there exists a unique solution of (3.4) satisfying
Observe that the hypotheses of Lemma 2.7 hold in view of the conditions , , and , demanded in (A2), (A3), and (A4).
Next, consider the singular BVP
Using the definition of the lower and upper solutions, (3.1) and (3.2), we get
By assumption (A4), we have as
and . In view of the above inequalities, we find that and are respectively lower and upper solutions of (3.5). Therefore it follows by Lemmas 2.6 and 2.7 that there exists a unique solution of (3.5) satisfying
Now we show that . Using (3.1), (3.2), and assumption (A1), we obtain
and in view of (A4), we get , which implies that . Clearly, the above inequalities and singular boundary value problem (3.5) satisfy the hypotheses of Lemma 2.7. Therefore, by the conclusion of Lemma 2.7, we have , . Consequently, we obtain
As a next step, we prove that
for . For that, we consider the following SBVP:
Using (3.1), (3.2), (3.3), and the inequality , we get
Since , are increasing in v by assumptions (A2) and (A3), therefore, , for . Consequently, in view of (3.1), (3.2), (3.3), we obtain the inequality
and by virtue of the inequality
we get . Thus, as argued earlier, there exists a unique solution of (3.6) such that
Now let us consider the following SBVP:
Again, using assumptions (A2)-(A4), (3.1), (3.2), (3.3), and the inequality
Applying the earlier arguments, it follows that there exists a unique solution of (3.7) such that
Following the procedure employed to prove , it can be shown that . Hence we have
As the monotone sequences and are both bounded, therefore, they converge to the limit functions and pointwise respectively.
Now we show that the convergence of the sequences and to the limit functions and respectively is indeed uniform. Using SBVP (3.6), integral equation (2.1), and relation (3.8), we observe that , , , are uniformly bounded sequences. Thus, by the Arzela-Ascoli theorem, the sequences , have uniformly convergent subsequences. Hence, by the monotonicity of the sequence in ℰ, it follows that the sequence converges uniformly to the limit function in ℰ. In a similar manner, the sequence in ℰ converges uniformly to the limit function in ℰ.
Finally, we show that the convergence of the sequences is quadratic. We only prove the quadratic convergence for the sequence as that of the sequence follows a similar procedure.
Let us define
Then, for , , , , we have
Setting , , where and provide bounds for and respectively and , , the above inequality takes the form
Given M, N and , there exists such that
By a comparison theorem , it follows from (3.9) and (3.10) that
which implies that
Now, we consider
where . Letting on and , we find that
From (3.11) and (3.12), we conclude that the sequence converges to the unique solution of SBVP (1.1) quadratically. Similarly, we can prove the quadratic convergence of the sequence . This completes the proof. □
Example Consider the following singular nonlocal boundary value problem:
where , , , , and A, B, C are suitable positive constants. Let and be respectively lower and upper solutions of (3.13). Clearly, and are not the solutions of (3.13) and , . Moreover, assumptions (A1), (A2), (A3), and (A4) of Theorem 3.1 are satisfied. Thus, the conclusion of Theorem 3.1 applies to problem (3.13).