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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 207352, 12 pages
Singular Initial Value Problem for Certain Classes of Systems of Ordinary Differential Equations
1Department of Mathematics and Descriptive Geometry, Faculty of Civil Engineering, Brno University of Technology, Veveří 331/95, 602 00 Brno, Czech Republic
2Brno University of Technology, Department of Mathematics, Faculty of Electrical Engineering and Communication, Technická 8, 61600 Brno, Czech Republic
3CEITEC BUT, Brno University of Technology, Technická 3058/10, 61600 Brno, Czech Republic
Received 11 August 2013; Accepted 16 October 2013
Academic Editor: Yuriy Rogovchenko
Copyright © 2013 Josef Diblík et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The paper is devoted to the study of the solvability of a singular initial value problem for systems of ordinary differential equations. The main results give sufficient conditions for the existence of solutions in the right-hand neighbourhood of a singular point. In addition, the dimension of the set of initial data generating such solutions is estimated. An asymptotic behavior of solutions is determined as well and relevant asymptotic formulas are derived. The method of functions defined implicitly and the topological method (Ważewski's method) are used in the proofs. The results generalize some previous ones on singular initial value problems for differential equations.
Let and be given constants. Define auxiliary set of points
In the paper, we consider a system of ordinary differential equations in the form
where and functions can satisfy
for some indices or
for some indices , where is the -dimensional zero vector. Together with system (2) we consider the initial problem
Definition 1. Denote by a class of vector-functions having the following properties: (1)is continuously differentiable on ;(2), ;(3) for , , ;(4), .
For , define an auxiliary vector-function
are derived, where are the coordinates of a function and the symbol is the well-known Landau order symbol.
There are numerous papers and books dealing with singular initial value problems (see, e.g., [1–16] and the references therein). Among others, we should mention pioneering results on the solvability of singular problems for ordinary differential equations achieved by Chechyk  and Kiguradze . The results of the paper generalize previous investigation of the first author on the solvability of singular problems [5–7]. The main differences are as follows. In , a scalar singular differential equation was studied for the case that a function similar to the function above does not change the sign for . In , system (2) is investigated under the assumption that the th right-hand side of the system is bounded by the product of two functions, with the first depending only on the variable while the second one only depends on the variable , . In comparison with the results of , we cannot expect that a first approximation of system (2) consists of equations with separable variables.
The structure of the paper is the following. In Section 2, auxiliary results on implicit functions are given. We refer to Corollary 4 where formula (27) is crucial for the proofs of the asymptotic behavior of solutions. The main results of the paper are formulated in Section 3. New results are proved and a progress is achieved by implicit construction of funnels, where solutions of the singular problem are expected. To prove the existence of such solutions, the topological method of Ważewski (see, e.g., [17–19]) is used. A simple illustrative example is shown here as well. A generalization of the results derived is discussed in Section 4.
2. Auxiliary Results on Implicit Functions
First, we give some properties of implicit functions used in the following proofs.
Lemma 2. Assume that a function satisfies the following conditions:(1) is continuously differentiable with respect to and ;(2);(3)for every , there exists a finite limit , for every , there exists a finite limit , and ;(4), where .
defines a unique implicit function on some interval , such that .
Proof. Analysing assumptions (1)–(4), we deduce that only the following two cases can occur: either
while the remaining two cases
are in contradiction with assumptions and . The rest of the proof is analogous to the proofs of the well-known implicit-function theorems and, therefore, we leave it out.
To formulate the second lemma, we need some auxiliary notions. Define, for a given and satisfying the inequality , the set
Moreover, for a given continuously differentiable function , let
where argument is assumed to be positive. In addition, we define
provided that the limits exist and are finite.
Lemma 3. Assume that functions
satisfy the following conditions: (1) is continuously differentiable and ;(2) is continuous with respect to and continuously differentiable with respect to ;(3).
Then, for an arbitrary and , where is a positive and sufficiently small constant, there exists a unique continuous solution
of the equation
and is a sufficiently small positive constant, .
Proof. Define an auxiliary function
and consider implicit equation (18) in the form
with respect to . In what follows, we will assume that is a parameter. Since
is continuous with respect to all , , and and continuously differentiable with respect to , and
for arbitrary . Hence we can apply the classical implicit-function theorem. As a result, we state that (18) is uniquely solvable with respect to . Thus
where is a continuous function with respect to both and and is a sufficiently small positive number. The sign of the function can be specified. In particular, since is a decreasing function, the function is decreasing with respect to ,
and is a sufficiently small positive constant satisfying .
3. Main Results
Using , define the sets
where . To formulate the results we need auxiliary functions
defined as follows:
Theorem 5. Let and , , be continuous functions. Let, moreover, for a function , the following conditions be true:
where is a sufficiently small constant, , , and , , , are constants such that
there is an integer such that(a) if and ;(b) if and ;(c) if and ;(d) if and .
Here, if , conditions , are omitted and, if , conditions , are omitted.
Proof. The proof is divided into two parts. First, implicit curves are constructed and their properties are derived. Then, Ważewski’s method is applied to special domains having the shape of funnels with sides constructed using implicitly defined hypersurfaces. In this construction, we use implicit curves from the first part of the proof.
Implicitly Defined Curves and Their Properties. Let be fixed. Define auxiliary functions
We prove that
define unique implicit functions
on the interval . Observe that the function is a solution of
on the interval as well. Therefore, we consider the latter equation and investigate its solvability using Lemma 2. Set
where and . We show that the function satisfies all assumptions (1)–(4) of Lemma 2, where, instead of the region , we assume the region with sufficiently small , such that , .
(a) It is easy to see (in view of the above assumptions) that the function
defined by (39) is continuously differentiable with respect to and and assumption of Lemma 2, holds.
(b) Compute the limit . We get
if the last two limits exist and are finite. Substituting into the first integral of the last expression and using condition , we get
and assumption of Lemma 2 holds.
(c) Now we consider the existence of the product for , and determine its sign. We get
Substituting into the first integrals in the square brackets, using conditions and of Theorem 5 and the property , and assuming , we have
(d) Determine the sign of . We get
From condition , it follows that
Because of (a)–(d) all assumptions (1)–(4) of Lemma 2 on hold and (39) defines an implicit function
on some interval , where .
Now we turn to (37) and show that its solution given by formula (49) can be extended beyond .
We show that . On the contrary, assume . Then, (after a proper transformation of variables) we can apply Lemma 2 to the point again and, by well-known procedure, implicit function can be continued up to the boundary of the region .
If , then . Moreover, we have
since and by condition . This is a contradiction and . Therefore, the implicit function can be continued on the whole interval . Similarly we will show that the inequality
holds. We have
because, by condition , holds. It is obvious that
Because the function is monotonously increasing and
we get . Hence inequality (51) is proved.
Now we will investigate the behavior of implicit curves in a neighborhood of the function . Since
we have (by condition (8))
Thus, (in the first integral we substitute )
where and . Consequently, we deduce that
Since functions , increase with respect to their second co-ordinates and
on , we get
for each . Finally, we recall that
Application of Ważewski’s Method to an Implicitly Defined Domain. In the next part of the proof we will apply the topological method of Ważewski. We use the above mentioned functions given implicitly to define an open set
Now we start to investigate the behavior of the integral curves of system (2) with respect to the boundary of the set , that is, on the sets
First, we calculate the full derivative of the function along trajectories of system (2) on the set , . It is clear that . Further, for , we have
On the set , as it follows from the condition , we have either or .
(1) Let . Then, we can see that
The derivative of the function can be calculated using the well-known rules for differentiation of implicit functions given by identities
Using that relation, we have
Since, by (60), and , assumption of the theorem yields
In view of assumption (4b) ( when ), we get
(2) Let . Then, by similar calculations and using assumption (4a) ( when ), we obtain
Hence for all .
Now we will calculate the full derivative of the function along trajectories of system (2) on the set , where . As above, we get
On the set , as it follows from the condition , we have either or .
(1) Let . Then, we get (proceeding like in the previous part of the proof)
Since, by (60), and , applying assumption of the theorem, we have
In view of assumption (4d) ( when ), we have
(2) Let . Then, by similar calculations and using assumption -(c) ( when ), we obtain
Thus, for all .
By [18, Lemma 3.1, page 281], for decreasing values of the variable , the set of all egress points , , from the set equals the set of all strict egress points , from the set (for definitions of and see, for example, [18, page 37 and page 278]); that is,
Let be a subset of defined as
where , , are fixed. Then,
We can see that the set is a subset of the boundary of the set , but it is not a retract of . The explanation is simple and is based on the well-known fact that the boundary of an -dimensional ball is not its retract , and the set is topologically equivalent to an -dimensional ball.
We show that is a retract of . Define a mapping
With respect to the behavior of functions , , , the mapping is continuous. From the definition of the mapping , we get that is a retract of and, furthermore, is a compact set.
By seeing Corollary 3.1, [18, page 282] of Ważewski’s theorem [18, Theorem 3.1, page 282] there exists a solution of system (2) with the initial conditions in the set and it is contained in for . This solution satisfies (5) since the set is contracted to the initial point for .
As we can change the constants , within the inequality , we can repeat the above-mentioned construction for every admissible fixed set . Then, there exists a class of solutions depending on parameters and lying in for . If , the assertion of theorem remains true as well. The proof is complete.
Remark 6. If is a solution of initial problem (2), (5), then
where and , and
Then, condition of Theorem 5 is satisfied and, in this case, we obtain the result on the dimension of the set of initial data generating solutions of initial problem (2), (5).
, provided that the limits exist and are finite.
Theorem 7. Let all assumptions of Theorem 5 hold and, moreover, the functions
satisfy the following:(1) is continuous with respect to and continuously differentiable with respect to ;(2).
Then arbitrary solution of initial problem (2), (5) mentioned in Theorem 5 has the asymptotic form
Proof. From the proof of Theorem 5, it follows that, for coordinates , , of the solution of initial problem (2), (5), the inequalities
are valid on an interval and we can assume that the inequalities
are valid on as well. Thus, to prove (89), it is sufficient to prove that
Applying L’Hospital’s rule to the limits (92), we do not obtain the desired result. Therefore we will apply L’Hospital’s rule to the limits
This is possible because, in view of condition of the theorem, we obviously have
Then, we use the auxiliary results for implicit functions. Applying the above-mentioned procedure, we get (for )
Hence it follows that there exists a sufficiently small constant such that the inequalities
hold on some interval , , where is a sufficiently small constant. Applying Lemma 3 and Corollary 4 (formula (27)) we show that inequalities (96) can be written as
where , , are constants depending on and such that
In Lemma 3 we set, for every fixed index ,
Then, we have , and we have the following:(a) is continuously differentiable, ;(b) is continuous with respect to and continuously differentiable with respect to ;(c) by condition of the theorem.
Hence, Lemma 3 holds. By Corollary 4, we can write inequalities (96) in the form of (97). From (97), we get
for . The last inequalities are equivalent to
The proof is complete.
Example 8. Consider the following simple particular case of initial problem (2), (5):
This problem is a singular one. To apply Theorem 5, we rewrite (102), (103) as
Set (below we omit this index since we deal with a scalar problem), , and
Without loss of generality, we assume that and in definition of are positive and sufficiently small (from Definition , property , it follows ). Condition , obviously holds. Condition , is valid as well because
Since is a solution of (104), we have and Condition holds (see Remark 6 as well). Next,
Set (i.e., conditions (4c) and (4d) are omitted). For and sufficiently small and positive, it is easy to see that if and if . Hence, conditions (4a) and (4b) hold. According to Theorem 5, problem (102), (103) has at least one-parametric class of solutions.
Now we apply Theorem 7. We get