Nonlinear Singular BVP of Limit Circle Type and the Presence of Reverse-Ordered Upper and Lower Solutions
We consider the following class of nonlinear singular differential equation subject to the Neumann boundary condition . Conditions on and ensure that is a singular point of limit circle type. A simple approximation scheme which is iterative in nature is considered. The initial iterates are upper and lower solutions which can be ordered in one way or the other .
The upper and lower solution technique is the most promising technique as far as singular boundary value problems, are concerned . Recently, lot of activities are there regarding upper and lower solutions technique (see [2, 3] and the references therein). To see the application of the similar kind of problems, one should see the references of . In most of the results, upper and lower solutions are well ordered, that is, . As far as reverse-ordered upper and lower solutions are considered, that is, , the literature is not that rich. Though references are there for nonsingular boundary value problem, but singular boundary value problems require further exploration. The details of the work done for the nonsingular problem when upper and lower solutions are in reverse order can be seen in [4, 5]. To fill this gap in the present paper, we consider the following singular BVP:(A1)Let satisfy the following conditions. (i) and in . (ii). (A2) Let satisfy the following conditions. (i) in and . (ii). (iii). (iv).
In this paper, we consider a computationally simple iterative scheme defined by Starting with upper and lower solutions, we generate monotone sequences. To generate these monotonic sequences, we need the existence of some differential inequalities. To prove these differential inequalities, we analyze the corresponding singular IVP and extract properties of the solutions and their derivative.
We have arranged the paper in four sections. In Section 2, we discuss some elementary results, for example, maximum principles and existence of two differential inequalities. Then using these elementary results, we establish existence results for well-ordered upper and lower solutions in Section 3 and for reverse-ordered upper and lower solutions in Section 4. In Section 5, we conclude this paper with some remarks.
Let , and let , let and let . Now, consider the following class of linear singular problems: The corresponding homogeneous system (eigenvalue problem) is given by The solution of the nonhomogeneous problem (2.1)-(2.2) can be written as follows: where is the solution of is the solution of and . By replacing with in (2.6), it is easy to verify that for both positive and negative values of .
Remark 2.2. Let be a Hilbert space with inner product defined by From (A2) (iv), it can easily be verified that is a singular point of limit circle type (see [6, Remark (i) page 434]) in . Thus, we have pure point spectrum [7, page 125]. It is easy to show that the eigenvalues are real, simple, and negative.
Remark 2.3. Since and are two linearly independent solutions of (2.3), the eigenvalues of the eigenvalue problem (2.3)-(2.4) will be the zeros of . Since is an analytic function of so its zeros will be isolated and they all will be negative. Let them be , where for . Now, we have as the first negative zero of or in other words first negative eigenvalue of (2.3)-(2.4).
Since does not change its sign for and ; therefore, for all and for all .
Remark 2.4. Using (2.6), , it is easy to prove that if then for all , , and and for all , we have and .
Remark 2.6. Let and let . If , then are nonnegative (or nonpositive).
Remark 2.7. Let and let . If , then are nonpositive (or nonnegative).
Now, we derive conditions on which will help us to prove the monotonicity of the solutions generated by the iterative scheme (1.2).
Lemma 2.10. Let and . If is such that then for all ,
Lemma 2.11. Let and . If is such that and then for all ,
Note 2. In Lemma 2.11, we arrive at the integral which is an improper integral, and it should be convergent. Using the assumption (A2) (iv) and Remark and at [6, page 434] its convergence can be established.
3. Well-Ordered Upper and Lower Solutions
Let us define upper and lower solutions.
Definition 3.1. A function is an upper solution of (1.1) if
In this section, we show that for the proposed scheme (1.2) a good choice of is possible so that the solutions generated by the approximation scheme converge monotonically to solutions of (1.1). We require a number of results.
Proof. Let , then and using Proposition 2.8, we have .
Proposition 3.4. Assume that there exist upper solution and lower solution in such that for all ,the function is continuous onthere exists such that for all ,there exist such that for all , Let be such that . Then the functions defined recursively by (1.2) are such that, for all , is an upper solution of (1.1). .
Proof. We prove the claims by the principle of mathematical induction. Since is an upper solution and by Lemma 3.3 ; therefore, both the claims are true for .
Further, let the claims be true for , that is, is an upper solution and . Now, we are required to prove that is an upper solution and . To prove this, let , then we have Thus, to prove that is an upper solution, we are required to prove that Now, since satisfies from Proposition 2.8, we have for . Now, putting the value of from (2.5) in (3.8), and in view of , we deduce that to prove (3.8) it is sufficient to prove that for all . Since , using Remark 2.6, the above inequalities will be true if for all we have Which is true (Lemma 2.10). Therefore, (3.8) holds, and hence is an upper solution.
Now applying Lemma 3.3, we deduce that . This completes the proof.
In the next result, we prove that upper solution is larger than lower solution for all .
Proposition 3.7. Assume that , , , and hold, and let such that and for all Then for all , the functions and defined recursively by (1.2) satisfy .
Proof. We define a function
It is easy to see that for all , satisfies the following differential equation:
Now to prove this proposition again, we use the principle of mathematical induction. For , we have , and is the solution of (2.1)-(2.2) with and . Using Proposition 2.8, we deduce that , that is, .
Now, let , let , and let , then we are required to prove that and . First, we show that for all the function is nonnegative. Indeed, we have Here is a solution of (2.1) with , , and . Arguments similar to Proposition 3.4 can be used to prove that . Now, we have , , and , thus from Proposition 2.8, we deduce that , that is, .
Lemma 3.8. If satisfies , , and for all , , where is continuous and satisfies. then there exists such that any solution of with , for all , satisfies .
Proof. Consider an interval such that Now using , we have and after integrating it from to and using , we have Similarly for the interval , we have Thus In the same way, we can prove the following result for lower solutions.
Lemma 3.9. If satisfies , , and , then there exists such that any solution of with , for all , satisfies .
Now we are in a situation to prove our final result for the case when upper and lower solutions are well ordered.
Proof. Using Lemma 3.3 to Lemma 3.9 and Proposition 3.4 to Proposition 3.7, we deduce that the sequences and are monotonic and are bounded by and in , and by Dini's theorem, they converge uniformly to and (say). We can also deduce that the sequences and are uniformly bounded and equicontinuous in , and by Arzela-Ascoli theorem, there exists uniformly convergent subsequences and in . It is easy to observe that and imply and .
Solution of (1.2) is given by (2.5) where . Since the sequences are uniformly convergent taking limit as , we get and as the solutions of the nonlinear boundary value problem (1.1). Any solution in plays the role of . Hence, . Similarly .
Remark 3.11. When the source function is derivative independent, that is, , in this case we can choose .
4. Upper and Lower Solutions in Reverse Order
In this section, we consider the case when the upper and lower solutions are in reverse order, that is, For this, we require opposite one-sided Lipschitz condition, and we assume that there exists upper solution () and lower solution () in such that for all ,the function is continuous onthere exists such that for all ,there exist such that for all , Here again we define the approximation scheme by (1.2) and use the Antimaximum principle. We make a good choice of so that the sequences thus generated converge to the solution of the nonlinear problem. Similar to Section 3, we require the following lemmas and propositions.
Proof. Let , then and using Proposition 2.8, we have .
Proof. Using Remarks 2.5 and 2.7, Lemmas 2.11 and 4.1, and on the lines of the proof of Proposition 3.4, this proposition can be deduced easily.
In the same way, we can prove the following results for the lower solutions.
In the next result, we prove that lower solution is larger than upper solution for all .
Proposition 4.5. Assume that , , , and hold. Let be such that and and for all , Then, for all , the functions and defined recursively by (1.2) satisfy .
Lemma 4.6. If satisfies , , and for all , , where is continuous and satisfies then there exists such that any solution of with , for all , satisfies .
Lemma 4.7. If satisfies , , and , then there exists such that any solution of with , for all , satisfies .
Finally we arrive at the theorem similar to Theorem 3.10.
Theorem 4.8. Assume , , , , and are true. Let be such that and , and for all ,
Then the sequences and defined by (1.2) converge monotonically to solutions and of (1.1). Any solution of (1.1) in satisfies
Remark 4.9. When the source function is derivative independent, that is, , in this case we can choose .
We establish some existence results under quite general conditions on , , and . We prove some fundamental differential inequalities which enables us to prove the monotonicity of the sequences and . For this we have analyzed the singular differential equation and derived properties of the solutions and their derivatives. This work generalizes our previous work . Lot of exploration is still left. For example, one can consider different type of boundary conditions, and one can also try to remove the Lipschitz condition.
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