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Second-order initial value problems with singularities
Boundary Value Problems volume 2014, Article number: 161 (2014)
Using barrier strip arguments, we investigate the existence of -solutions to the initial value problem , , , which may be singular at and .
MSC: 34B15, 34B16, 34B18.
In this paper we study the solvability of initial value problems (IVPs) of the form
Here the scalar function is defined on a set of the form , where , , , , , , , and so it may be singular at and .
IVPs of the form
have been investigated by Rachůnková and Tomeček –. For example in , the authors have discussed the set of all solutions to this problem with a singularity at . Here , with , for and , f is locally Lipschitz on with the properties and for , where is a suitable constant.
Agarwal and O’Regan  have studied the problem
where may be singular at and/or . The obtained results give a positive -solution under the assumptions that , for , is continuous and
where g, h, r, and w are suitable functions.
IVPs of the form
where k, F, and G are suitable functions.
The solvability of various IVPs has been studied also by Bobisud and O’Regan , Bobisud and Lee , Cabada and Heikkilä , Cabada et al. , , Cid , Maagli and Masmoudi , and Zhao . Existence results for problem (1.1), (1.2) with a singularity at the initial value of have been reported in Kelevedjiev-Popivanov .
Here, as usual, we use regularization and sequential techniques. Namely, we proceed as follows. First, by means of the topological transversality theorem , we prove an existence result guaranteeing -solutions to the nonsingular IVP for equations of the form (1.1) with boundary conditions
Moreover, we establish the needed a priori bounds by the barrier strips technique. Further, the obtained existence theorem assures -solutions for each nonsingular IVP included in the family
where is suitable. Finally, we apply the Arzela-Ascoli theorem on the sequence of -solutions thus constructed to (1.3) to extract a uniformly convergent subsequence and show that its limit is a -solution to singular problem (1.1), (1.2). In the case , we establish -solutions with important properties - monotony and positivity.
We have used variants of the approach described above for various boundary value problems (BVPs); see Grammatikopoulos et al. , Kelevedjiev and Popivanov  and Palamides et al. . For example in , we have established the existence of positive solutions to the BVP
which may be singular at . Note that despite the more general equation of this problem, the conditions imposed here as well as the results obtained are not consequences of those in .
2 Topological transversality theorem
In this short section we state our main tools - the topological transversality theorem and a theorem giving an important property of the constant maps.
So, let X be a metric space and Y be a convex subset of a Banach space E. Let be open in Y. The compact map is called admissible if it is fixed point free on ∂U. We denote the set of all such maps by .
A map F in is essential if every map G in such that has a fixed point in U. It is clear, in particular, every essential map has a fixed point in U.
(, Chapter I, Theorem 2.2])
Letbe fixed andbe the constant mapfor. Then F is essential.
We say that the homotopy, , is compact if the mapgiven byforis compact.
(, Chapter I, Theorem 2.6])
Let Y be a convex subset of a Banach space E andbe open. Suppose:
are compact maps.
, , is a compact homotopy joining F and G, i.e.
, , is fixed point free on ∂U.
Then, , has at least one fixed point in U and in particular there is asuch that.
3 Nonsingular problem
Consider the IVP
where , .
We include this problem into the following family of regular IVPs constructed for
and suppose the following.
There exist constants , , , , , and a sufficiently small such that
Our first result ensures bounds for the eventual -solutions to (3.2). We need them to prepare the application of the topological transversality theorem.
Let (R) hold. Then each solutionto the family (3.2) λ , , satisfies the bounds
Suppose that the set
is not empty. Then
imply that there exists an interval such that
This inequality and the continuity of guarantee the existence of some for which
Since , , is a solution of the differential equation, we have . In particular for γ we have
Thus, we apply (R) to conclude that
which contradicts the inequality . This has been established above. Thus, is empty and as a result
Now, by the mean value theorem for each there exists a such that
This allows us to use (3.3) to show similarly to above that the set
is empty. Hence,
To estimate , we observe firstly that (R) implies in particular
which yield and . Multiplying both sides of the inequality by and , we get, respectively, and . On the other hand, we have established
and each and so
Let us mention that some analogous results have been obtained in Kelevedjiev . For completeness of our explanations, we present the full proofs here.
Now we prove an existence result guaranteeing the solvability of IVP (3.1).
Let (R) hold. Then nonsingular problem (3.1) has at least one non-decreasing solution in.
Preparing the application of Theorem 2.2, we define first the set
where . It is important to notice that according to Lemma 3.1 all -solutions to family (3.2) are interior points of U. Further, we introduce the continuous maps
and for and the map
Clearly, the map Φ is also continuous since, by assumption, the function is continuous on if
In addition we verify that exists and is also continuous. To this aim we introduce the linear map
defined by , where . It is one-to-one because each function has a unique image, and each function has a unique inverse image which is the unique solution to the IVP
It is not hard to see that W is bounded and so, by the bounded inverse theorem, the map exists and is linear and bounded. Thus, it is continuous. Now, using , we define
where is the unique solution of the problem
Clearly, is continuous since is continuous.
We already can introduce a homotopy
It is well known that j is completely continuous, that is, j maps each bounded subset of into a compact subset of . Thus, the image of the bounded set U is compact. Now, from the continuity of Φ and it follows that the sets and are also compact. In summary, we have established that the homotopy is compact. On the other hand, for its fixed points we have
which is the operator form of family (3.2). So, each fixed point of is a solution to (3.2), which, according to Lemma 3.1, lies in U. Consequently, the homotopy is fixed point free on ∂U.
Finally, is a constant map mapping each function to . Thus, according to Theorem 2.1, is essential.
So, all assumptions of Theorem 2.2 are fulfilled. Hence has a fixed point in U which means that the IVP of (3.2) obtained for (i.e. (3.1)) has at least one solution in . From Lemma 3.1 we know that
from which its monotony follows. □
The validity of the following results follows similarly.
Letand let (R) hold for. Then problem (3.1) has at least one strictly increasing solution in.
Let () and let (R) hold for. Then problem (3.1) has at least one positive (nonnegative) non-decreasing solution in.
Let, and let (R) hold for. Then problem (3.1) has at least one strictly increasing solution inwith positive values for.
4 A problem singular at x and
In this section we study the solvability of singular IVP (1.1), (1.2) under the following assumptions.
(S1): There are constants , , and a sufficiently small such that
(S2): For some and there exists a constant such that and
where T, and are as in (S1).
Now, for , where , and α, μ, and k are as in (S2), we construct the following family of regular IVPs:
Notice, for , that we have .
Let (S1) and (S2) hold and let, , be a solution to (4.3) such that
Then the following bound is satisfied for each:
Since for each we have
we will consider the proof for an arbitrary fixed , considering two cases. Namely, for is the first case and the second one is for with for some .
Case 1. From , , and (S2) we have
i.e. for . Integrating the last inequality from 0 to t we get
Now , , and (4.2) imply
In particular for , thus
Case 2. As in the first case, we derive
On the other hand, since for , again from (4.2) it follows that
So, as a result of the considered cases we get
from which the assertion follows immediately. □
Having this lemma, we prove the basic result of this section.
Let (S1) and (S2) hold. Then singular IVP (1.1), (1.2) has at least one strictly increasing solution insuch that
For each fixed introduce ,
having the properties
and since . Besides,
and, in view of (4.1),
All this implies that for each the corresponding IVP of family (4.3) satisfies (R). Thus, we apply Theorem 3.2 to conclude that (4.3) has a solution for each . We can use also Lemma 3.1 to conclude that for each and we have
Now, these bounds allow the application of Lemma 4.1 from which one infers that for each and the bounds
hold. For later use, integrating the least inequality from 0 to t, , we get
We consider firstly the sequence of -solutions of (4.3) only for each . Clearly, for each we have in particular
which together with (4.6) gives
where . On combining the last inequality and (4.4) we obtain
From (4.5) we have in addition
Now, using the fact that (4.1) implies continuity of on the compact set and keeping in mind that for each
we conclude that there is a constant , independent of n, such that
Using the obtained a priori bounds for , and on the interval , we apply the Arzela-Ascoli theorem to conclude that there exists a subsequence , , , of and a function such that
i.e., the sequences and converge uniformly on the interval to and , respectively. Obviously, (4.7) and (4.8) are valid in particular for the elements of and , respectively, from which, letting , one finds
Clearly, the functions , , , satisfy integral equations of the form
Now, since is uniformly continuous on the compact set , from the uniform convergence of it follows that the sequence , is uniformly convergent on to the function , which means
for each . Returning to the integral equation and letting yield
which implies that is a -solution to the differential equation on . Besides, (4.6) implies
Further, we observe that if the condition (S2) holds for some , then it is true also for an arbitrary . We will use this fact considering a sequence , , with the properties
For each we consider sequences
on the interval . Thus, we establish that each sequence has a subsequence , , , converging uniformly on the interval to any function , , that is,
which is a -solution to the differential equation on and
The properties of the functions from , , imply that there exists a function which is a -solution to the equation on the interval and is such that
We have to show also that
Reasoning by contradiction, assume that there exists a sufficiently small such that for every there is a such that
In other words, assume that for every sequence , , with , there exists a sequence having the properties , and
It is clear that every interval , , contains a subsequence of converging to 0. Besides, from (4.9) and (4.10) it follows that for every there are such that and
for all and all . Moreover, since the accumulation point of is 0, for each sufficiently large there is a where . In summary, for every sufficiently large , that is, for every sufficiently small , there are such that for all and from (4.12) and (4.13) we have
which contradicts to the fact that and . This contradiction proves that (4.11) is true.
Now, it is easy to verify that the function
is a -solution to (1.1), (1.2). This function is strictly increasing because for , and the bounds for and follows immediately from the corresponding bounds for and . □
The following results provide information about the presence of other useful properties of the assured solutions. Their correctness follows directly from Theorem 4.2.
Letand let (S1) and (S2) hold. Then the singular IVP (1.1), (1.2) has at least one strictly increasing solution inwith positive values for.
Consider the IVP
where , and the polynomial , , has simple zeroes and such that
Let us note that here , and .
Clearly, there is a sufficiently small such that
and for .
We will show that all assumptions of Theorem 3.2 are fulfilled in the case
the other cases as regards the sign of around and may be treated similarly. For this case choose , and . Next, using the requirement , i.e., we get the following conditions for θ and T:
which yield and . Besides, yields . Thus, . Now, choosing
we really can apply Theorem 3.2 to conclude that the considered problem has a strictly increasing solution with on for each .
Consider the IVP
Notice that here
It is easy to check that (S1) holds, for example, for , , , and an arbitrary fixed , moreover, . Besides, for , and , for example, we have
and on , which means that (S2) also holds. By Theorem 4.3, the considered IVP has at least one positive strictly increasing solution in .
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The work is partially supported by the Sofia University Grant 158/2013 and by the Bulgarian NSF under Grant DCVP - 02/1/2009.
The authors declare that they have no competing interests.
All authors read and approved the final manuscript.