We investigate the existence of positive solutions of singular problem , , , . Here, and the Carathéodory function may be singular in all its space variables . The results are proved by regularization and sequential techniques. In limit processes, the Vitali convergence theorem is used.
Let be a positive constant, and , , . We consider the singular complementary Lidstone boundary value problem
where satisfies the local Carathéodory function on () with
The function is positive and may be singular at the value zero of all its space variables .
Let . We say that is singular at the value zero of its space variable if for a.e. and all , , such that , the relation
A function (i.e., has absolutely continuous th derivative on ) is a positive solution of problem (1.1), (1.2) if for , satisfies the boundary conditions (1.2) and (1.1) holds a.e. on .
The regular complementary Lidstone problem
was discussed in . Here, is continuous at least in the interior of the domain of interest. Existence and uniqueness criteria for problem (1.5) are proved by the complementary Lidstone interpolating polynomial of degree . No contributions exist, as far as we know, concerning the existence of positive solutions of singular complementary Lidstone problems.
We observe that differential equations in complementary Lidstone problems as well as derivatives in boundary conditions are odd orders, in contrast to the Lidstone problem
The aim of this paper is to give the conditions on the function in (1.1) which guarantee that the singular problem (1.1), (1.2) has a solution. The existence results are proved by regularization and sequential techniques, and in limit processes, the Vitali convergence theorem [16, 17] is applied.
Throughout the paper, and , stands for the norm in and , respectively. denotes the set of functions (Lebesgue) integrable on and meas the Lebesgue measure of .
We work with the following conditions on the function in (1.1).
() and there exists such that
for a.e. and each .
()For a.e. and for all , the inequality
is fulfilled, where is positive and nondecreasing in the second variable, is nonincreasing, ,
The paper is organized as follows. In Section 2, we construct a sequence of auxiliary regular differential equations associated with (1.1). Section 3 is devoted to the study of auxiliary regular complementary Lidstone problems. We show that the solvability of these problems is reduced to the existence of a fixed point of an operator . The existence of a fixed point of is proved by a fixed point theorem of cone compression type according to Guo-Krasnosel'skii [18, 19]. The properties of solutions to auxiliary problems are also investigated here. In Section 4, applying the results of Section 3, the existence of a positive solution of the singular problem (1.1), (1.2) is proved.
Let be from (1.1). For , define , , and by the formulas
Let . Chose and put
for . Now, define an auxiliary function by means of the following recurrence formulas:
for , and
Then, under condition (), and
Condition () gives
We investigate the regular differential equation
If a function satisfies (2.8) for a.e. , then is called a solution of (2.8).
3. Auxiliary Regular Problems
Let and denote by the Green function of the problem
Lemma 3.1 (see [10, Lemmas 2.1 and 2.3]).
For and , the inequalities
Let and let be a solution of the differential equation
satisfying the Lidstone boundary conditions
It follows from the definition of the Green function that
It is easy to check that is a solution of problem (2.8), (1.2) if and only if , and its derivative is a solution of a problem involving the functional differential equation
and the Lidstone boundary conditions (3.8). From (3.9) (for ), we see that is a solution of problem (3.10), (3.8) exactly if it is a solution of the equation
in the set . Consequently, is a solution of problem (2.8), (1.2) if and only if it is a solution of the equation
in the set . It means that is a solution of problem (2.8), (1.2) if is a fixed point of the operator defined as
Let be a Banach space, and let be a cone in . Let be bounded open balls of centered at the origin with . Suppose that is completely continuous operator such that
holds. Then, has a fixed point in .
We are now in the position to prove that problem (2.8), (1.2) has a solution.
Let () and () hold. Then, problem (2.8), (1.2) has a solution.
Let the operator be given in (3.13), and let
Then, is a cone in and since for by (3.4) and satisfies (2.5), we see that . The fact that is a completely continuous operator follows from , from Lebesgue dominated convergence theorem, and from the Arzelà-Ascoli theorem.
Choose and put for . Then, (cf. (2.5))
Since and for , the equality holds with some for . We now use the equality and have
Hence, , and so
Next, we deduce from the relation
and from (2.7) that
where . Since for , we have
The last inequality together with (3.21) gives
where is from (). Since is arbitrary, relations (3.18) and (3.21) imply that for all , inequalities (3.18) and
hold. By (), there exists such that
Then, it follows from (3.18), (3.24), and (3.26) that
The conclusion now follows from Lemma 3.2 (for and ).
The properties of solutions to problem (2.8), (1.2) are collected in the following lemma.
Let () and () be satisfied. Let be a solution of problem (2.8), (1.2). Then, for all , the following assertions hold:
(i) for , , and for a.e. ,
(ii) is increasing on , and for , is decreasing on , and there is a unique such that ,
(iii)there exists a positive constant such that
(iv)the sequence is bounded in .
Let us choose an arbitrary . By (2.5),
and it follows from the definition of the Green function that the equality
holds for and . Now, using (1.2), (3.4), (3.30), and (3.31), we see that assertion (i) is true. Hence, is decreasing on for and is increasing on this interval. Due to for , there exists a unique such that for . Consequently, assertion (ii) holds.
Next, in view of (2.5), (3.6), and (3.31),
and, by [13, Lemma 6.2],
and (cf. (3.32) for )
since on by assertion (ii). Let
Then estimate (3.29) follows from relations (3.32)–(3.37).
It remains to prove the boundedness of the sequence in . We use estimate (3.29), the properties of given in (), and the inequality
for all . Now, from the above estimates, from (2.6) and from for some , which is proved in (ii), we get
Notice that by (). Consequently,
Since for , which follows from the fact that vanishes in by (1.2) and assertion (ii), inequality (3.45) yields
where is from (). Due to the condition
in (), there exists a positive constant such that for all the inequality
is fulfilled. The last inequality together with estimate (3.46) gives for . Consequently, for , , and assertion (iv) follows.
The following result gives the important property of for applying the Vitali convergent theorem in the proof of Theorem 4.1.
Let () and () hold. Let be a solution of problem (2.8), (1.2). Then, the sequence
is uniformly integrable on , that is, for each , there exists such that if and , then
By Lemma 3.4 (iv), there exists such that for , the inequality holds. Now, we conclude from (2.5) and (2.6), from the properties of and given in , and finally from (3.29) that for and , the estimate
is fulfilled, where is a positive constant. Since the functions , , and () belong to the set by assumption (), in order to prove that is uniformly integrable on , it suffices to show that the sequences
are uniformly integrable on . Due to and for by (), this fact follows from [13, Criterion 11.10 (with and )].
4. The Main Result
The following theorem is the existence result for the singular problem (1.1), (1.2).
Let () and () hold. Then, problem (1.1), (1.2) has a positive solution and
Lemma 3.3 guarantees that problem (2.8), (1.2) has a solution . Consider the sequence . By Lemma 3.4, is bounded in ,
and fulfils estimate (3.29), where is a positive constant and . Furthermore, the sequence is uniformly integrable on by Lemma 3.5, and therefore, we deduce from the equality for a.e. that is equicontinuous on . Now, by the Arzelà-Ascoli theorem and the Bolzano-Weierstrass theorem, we may assume without loss of generality that is convergent in and is convergent in for . Let and (). Then and satisfies the boundary conditions (1.2). Letting in (3.29) and (4.2), we get (for )
Keeping in mind the definition of , we conclude from (4.3) that
Then, by the Vitali theorem, and
Letting in the equality
As a result, and is a solution of (1.1). Consequently, is a positive solution of problem (1.1), (1.2) and inequality (4.1) follows from (4.3).
Consider problem (1.1), (1.2) with
on , where , (that is, is essentially bounded and measurable on ) are nonnegative, for a.e. . If for and , for , then, by Theorem 4.1, the problem has a positive solution satisfying inequality (4.1).
This work was supported by the Council of Czech Government MSM no. 6198959214.
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