In this paper, we investigate the structure and properties of the set of positive blow-up solutions of the differential equation , , where . The differential equation is studied together with the boundary conditions , . We specify conditions for the data function h which guarantee that the set of all positive solutions to the above boundary value problem is nonempty. Further properties of the solutions are discussed and results of numerical simulations are presented.
MSC: 34B18, 34B16, 34A12.
Keywords:singular ordinary differential equation of the second order; time singularities; blow-up, positive solutions; existence of solutions; polynomial collocation
In this paper, we investigate the structure and properties of the set of positive blow-up solutions of the differential equation
Models in the form of (1) arise in many applications. Among others, they occur in the study of phase transitions of Van der Waals fluids [1-3], in population genetics, where they characterize the spatial distribution of the genetic composition of a population [4,5], in the homogeneous nucleation theory , in relativistic cosmology for particles which can be treated as domains in the universe , and in the nonlinear field theory, in particular, in context of bubbles generated by scalar fields of the Higgs type in Minkowski spaces .
Here, we assume that h is positive and satisfies the Carathéodory conditions on . We define a positive solution of (1) as a function v which satisfies (1) for a.e. , is positive on , and has absolutely continuous first derivative on each compact subinterval in .
According to Lemma 4, if v is a positive solution of (1), then either
In the literature, bounded solutions of (1) have been widely investigated; see e.g.[9-13]. Such solutions are characterized by the initial condition . In contrast to this, some real problems lead to the investigation of unbounded solutions which are characterized by the condition for some and which are called blow-up solutions. We refer to [14-16]. Here, we are interested in blow-up solutions of (1), where . In particular, (1) will be considered together with the boundary conditions
In this case, we speak about a positive solution of problem (1), (2). Let us denote by the set of all positive solutions to (1), (2). Moreover, let and for . Since for each , it is obvious that .
Our main goal is to find conditions for the data function h in (1), which guarantee that the set is nonempty for each and then to investigate the properties of this set. For example, we prove that the difference of any two functions in , , retains its sign on , and that there exist minimal and maximal solutions for each , cf. Theorem 5. If the interior of the set is nonempty, we show that this interior is fully covered by ordered graphs of other functions belonging to for each ; see Theorem 6. Finally, in Theorem 7, the existence of a positive constant such that is shown and all functions v from ℛ are uniquely characterized by the condition
If we denote such v by and define for , and , then we find that the graphs of these functions do not intersect, cf. Theorem 8, and that for each , the set is compact in ; see Theorem 9.
Let us denote by the Banach space of functions continuous on equipped with the maximum norm
Similarly, means the Banach space of functions having a continuous first derivative on with the corresponding maximum norm . By we denote the set of functions which are Lebesgue integrable on . Moreover, is the set of functions whose first derivative is absolutely continuous on , while is the set of functions having absolutely continuous first derivative on each compact subinterval of . We say that satisfies the Carathéodory conditions on , if the following three conditions hold:
(i) The function is measurable for all .
(ii) The function is continuous for a.e. .
(iii) For each compact set there exists a function such that
For functions satisfying above conditions, we use the notation .
Structure of the paper
The paper is organized as follows. In Section 2 we discuss properties of the solutions of the auxiliary Dirichlet problem (3), (4). We recapitulate previous results from  and also present new results in Theorems 1, 2, and 3. The main results of the paper can be found in Section 3, where we describe a relation between solutions of problem (3), (4) and blow-up solutions of problem (1), (2); see Theorem 4. Using this relation and the results of Section 2, we obtain various interesting properties of blow-up solutions; see Theorems 5 to 9. Section 4 contains three examples illustrating the theoretical findings. Final remarks and open problems are formulated in Section 5.
2 Auxiliary results
In this section we consider the auxiliary singular differential equation,
where . For and , (3) becomes a special case of (1) and therefore, results obtained for (3) apply for (1). For the further analysis, we assume that f satisfies the following conditions:
(H2) for a.e. and all ,
(H3) is increasing in x for a.e. and
We now study (3) subject to the boundary conditions
and require that
holds. We call a function a positive solution of the Dirichlet problem (3), (4) if , on , u satisfies the boundary conditions (4), and (3) holds for a.e. . Clearly, for each positive solution u of (3), (4), there exists such that (5) is satisfied.
We now denote by the set of all positive solutions of problem (3), (4), and , where .
In the following lemma we cite those results from  which will be used in the analysis of problem (1), (2).
Lemma 1Let (H1)-(H3) hold. Then the following statements hold:
(a) For each the set is nonempty and there exist functions such that for and .
(b) If , , , then for .
(c) If , , , and for some , then either for or there exists such that for and for .
(d) For each and each there exists satisfying .
(e) is a one-point set for each , where is at most countable.
(f) For each , the set is compact in .
(g) If , then if and only if it is a solution of the equation
in the set .
We now formulate new results which complete those from . We first establish a relation between and the set if its interior is nonempty. This question was a short time ago an open problem [, Remark 4.4]. We note that the relation between and the set with having nonempty interior is described in Lemma 1(d).
Theorem 1Let (H1)-(H3) hold. Let us assume that there exists such that . Then for any there exists satisfying .
ProofStep 1. Auxiliary Dirichlet problem.
Choose . Consider (3) subject to the Dirichlet conditions
We claim that there exists a solution v to problem (3), (7) such that
We show this result utilizing the method of lower and upper functions. It follows from (H1) that there exists such that
and let be given as
We now consider the auxiliary differential equation
It is not difficult to verify that for . Hence, cf. (9),
for a.e. and all , . Since and for , , , and , solve (3) on , we conclude that and are lower and upper functions of problem (3), (7) (see e.g. or ). This fact together with (11) implies the existence of a solution v to problem (3), (7) satisfying (8), cf. [, Lemma 3.7]. Moreover,
and taking the limit we obtaina , which together with gives .
We now prove that on . The proof is indirect. Let us assume that there exists such that and for . Then on this interval, and therefore
From (8) and (9) we conclude that
By the Gronwall lemma,
Consequently, for , and so = 1 on this interval. Thus, v is a solution of problem (3), (7).
Step 2. Continuation of the solutionv.
It follows from the arguments given in Step 1 that v is a solution of problem (3), (7) on , and because . It is easy to verify that the equality
is satisfied for a.e. . We now integrate the last equality two times over and have (note that )
Let u be a solution of problem (3), (7) on an interval J which is a left-continuation of v. Let us assume that u is not continuable. Let . Then and (12) with v replaced by u holds for a.e. . The integration now yields
and we claim that
We show inequality (14) indirectly. Let us assume that (14) does not hold. Then there exists such that
and either or . Assume that .
Since, by (H3), for a.e. and for a.e. , we have
which is not possible. The case can be discussed analogously. Hence, (14) holds.
Suppose that . Then and since u is bounded on , it follows that . The integration of the equality
Since for a.e. , where , we have
Using the Gronwall lemma, we deduce that for ,
holds. This is a contradiction.
Therefore , and then (14) yields . Consequently , and the assertion of the theorem follows from (13). □
In the next corollary we extend the statement (d) from Lemma 1 to a large set of A values.
Corollary 1For each and each there exists satisfying .
Proof The result follows immediately from Lemma 1(d) and Theorem 1. □
Remark 1 Corollary 1 says that the set is covered by the graphs of functions from , that is,
By Lemma 1(b), (c) we know that functions from are uniquely determined by the values −c of their derivatives at the right end point only in the case that is a singleton set for each . Since we cannot uniquely determine all functions from via their derivatives at , see Lemma 1(e), we discuss their derivatives at the singular point .
Lemma 2Let (H1)-(H3) hold. Let , . Then
Proof It follows from Lemma 1(g) that (6) holds for . Since , we have (note that )
and (15) follows. □
Corollary 2Let . Then
Proof The result follows from Lemma 2, since . □
Corollary 3Let , . Then .
Proof Since , there exists such that . We can assume that for instance . Then on and on by Lemma 1(b)(c). Hence, and
which together with Corollary 2 gives . □
Corollary 4Let , . Then
In particular, .
Proof Inequality (16) follows from Lemma 2 and the fact that for a.e. by (H2). □
Corollary 5Let . Then
Proof The integration of (3) over gives
Using integration by parts we obtain
Taking the limit , we have
On the other hand, it follows from Corollary 2 that
Combining the above two equalities yields the result. □
Lemma 3Let (H1)-(H3) hold. Let and be not a singleton set. Then for each there exists a unique such that . Consequently, .
Proof Let , . Then it follows from Lemma 1(a) and (H3) that
Hence, by Lemma 2,
Since is a compact set in by Lemma 1(f), the set is closed. In fact, let , , where , and let . Then there exist a subsequence of and such that in . In particular, as . Therefore, .
It remains to prove that . Assume that the equality does not hold. Then, from the structure of bounded and closed subsets of ℝ the existence of an open interval , , follows. Let and , where . Due to Lemma 1(c), there exists such that on , on . Choose . By Corollary 1, there exists such that . Then on and on . Consequently, on , that is, , which is not possible. □
Since functions from belong to , we have for each . Corollary 4 yields . Let us denote
Lemma 3 implies that functions from can be uniquely determined by the values of their derivatives at the singular point ; see Theorem 2.
Theorem 2Let (H1)-(H3) hold. Then there exists a unique satisfying if and only if .
Proof We first show
It follows from Lemma 3 that
for each , where we set if is a singleton set and . In view of Corollary 4, for . Consequently, (18) follows.
Let us now choose . Then there exists satisfying . The uniqueness of u follows from Corollary 3. □
For , we denote by the unique element of satisfying .
Theorem 3Let (H1)-(H3) hold. Assume that is a convergent sequence and . Then in .
Proof Let . Then because for by Corollary 4. Since and is compact in by Lemma 1(f), the sequence is relatively compact in . Let be a subsequence of which is convergent in , and let . Then for a and . Therefore, and hence any subsequence of converging in has the same limit . Consequently, is convergent in and is its limit. □
3 Blow-up solutions and their properties
This section contains the main results of the paper. First, we present a lemma which describes how positive solutions of (1) may behave at the singular point .
Lemma 4Let and let the function be positive. Letvbe a positive solution of (1). Then either
Proof By Corollary 3.5 in , if , then v satisfies (19). Now, assume that
Then v has to satisfy
because otherwise, if (22) was not true, then , contradicting (21). Since h is positive, (1) indicates that the function is increasing on and hence there exists . The next part of the proof is divided into three cases.
(i) Let us assume that
Since is increasing on , we deduce and for . Consequently, v is increasing on . This together with and yields . Therefore, . Integrating (1) and using (23), we obtain
which is a contradiction to (22).
(ii) Assume that
Then there exists such that
Hence, v is increasing on and there exists .
By integration, we obtain from (24)
and, by virtue of and , we arrive at
which again is a contradiction.
(iii) Assume that
Then there exists such that
Hence, v is decreasing on and there exists . By integration, we obtain from (26)
and, since and , we have
In addition (25) yields and (20) follows. □
Now, we investigate the existence and properties of blow-up solutions of (1), for the case that the function h has the form
In particular, we study the equation
where and ψ, g satisfy the conditions
( ) and for a.e. ,
( ) and is increasing in x for a.e. ,
( ) for a.e. and all , where , ϕ is increasing and
Together with (27) we consider the conditions
We define a positive solution to problem (27), (28) as a function such that on , v satisfies the boundary conditions (28), and (27) holds for a.e. .
Define a set by
Clearly, for each there exists such that (29) holds.
Lemma 5Let ( )-( ) hold. Let . Then on .
Proof Since , the integration of (27) over the subinterval gives
Assume that for some . Then
which is not possible, since and . □
Moreover, let us define sets
It is obvious that . Using the substitution
we can rewrite (27) and obtain the form
We now introduce a function f,
Under conditions ( )-( ), the function f satisfies (H1)-(H3); see the proof of [, Theorem 5.1]. Therefore, the results of Section 2 hold for problem (34), (4). As in Section 2, we define the sets and for solutions of (34).
The following result is the key-stone to the analysis of the structure of the set ℛ and describes the relation between the sets and ℛ.
Theorem 4Let ( )-( ) hold and . Let us assume that (33) holds. Then if and only if and .
Proof (⇒) Let . Then (28), (32), and (33) provide the following properties: , , and u is bounded on . It follows from the equality
We now argue as in the proof of [, Lemma 3.3] and arrive at
for . Since u is bounded on , we have . Then, by [, Lemma 2.1],
for , and hence, . Now, [, Corollary 1] guarantees that u can be extended on with such that the equality (35) holds for . Consequently, .
(⇐) Let and . Then it follows from (4), (5), and (33) that and , . Hence, is bounded on and
yields , that is,
Since , the equality (35) holds on , and consequently,
Hence, note that , , and so . As a result, we have . □
Now, we are in the position to provide results on the solvability of problem (27), (28) and formulate the properties of its solutions.
Theorem 5Let ( )-( ) hold. Then the following statements hold:
(a) For each the set is nonempty and there exist such that for and .
(b) If , , , then for .
(c) If , , , and for some , then either for or there exists such that for and for .
(d) is a singleton set for each , where is at most countable.
Proof The result follows by combining results from Lemma 1(a), (b), (c), and (e) with those from Theorem 4. □
If , , and if (33) holds, then Theorems 4 and 5 yield
The next theorem shows that the set
is covered by graphs of the functions from ℛ.
Theorem 6Let ( )-( ) hold. Then, for each and each , there exists satisfying . In particular, if for some and the inequality holds, then for each there exist satisfying .
Proof Choose and . Define . Then, using (33) and (36), we deduce . Therefore, by Corollary 1, there exists such that . Consequently, (33) and Theorem 4 give and . The last statement follows from Theorem 5(c). □
By Corollary 4, there exists such that
and hence, by (33) and (36) with ,
Using constants from the interval , we can uniquely determine all functions in ℛ.
Theorem 7Let ( )-( ) hold and let be as in (38). Then the following statements hold:
(a) For each there exists such that
(b) For each there exists a unique satisfying (39).
Proof (a) Choose . Using (33), (37), and Theorem 4, we have and
By Lemma 1(c), . Hence, if we denote , the first condition in (39) follows. Moreover,
and therefore the second condition in (39) holds.
(b) Choose . Theorem 2 guarantees that there exists a unique satisfying . Using (33) and Theorem 4, we conclude . Then v satisfies (40) and (41) which results in (39). It remains to prove that v is unique. Assume that there exists a function , , such that w satisfies (39). Let . Then and . Consequently, by Theorem 2, we arrive at and , which is a contradiction. □
Let be given in (38) and choose . Keeping Theorem 7(b) in mind, there exists a unique satisfying (39). We denote such v by and define a function as
Then . We now specify further properties of functions .
Theorem 8Let ( )-( ) hold and let be from (38). Let and . Then either for or there exists such that for and for .
Proof According to (42), there exist , , and such that , for . Also, and, since , it follows that in a right neighborhood of . Therefore, there exists such that and hence, . (Note that if , then Theorem 5(b) yields for , which is not possible.) The result now follows from Theorem 5(b) for and from Theorem 5(c) for . □
Corollary 6Let be a convergent sequence and let us denote . Then in .
Proof The proof is indirect. Assume that the statement of the corollary does not hold. Then there exist ε, and a subsequence of , we denote it again by , such that
Let and . Let be countable. Then there exists a decreasing subsequence of . By Theorem 8, the sequence is not increasing on and on this interval. Using (33) and (42), we obtain , . Since in it follows from Theorem 3, note that , that and that for . This fact together with the monotonicity of and gives