Research Article

# Positive blow-up solutions of nonlinear models from real world dynamics

Jürgen Gschwindl1, Irena Rachůnková2*, Svatoslav Staněk2 and Ewa B Weinmüller1

Author Affiliations

1 Department for Analysis and Scientific Computing, Vienna University of Technology, Wiedner Hauptstraße 8-10, Wien, A-1040, Austria

2 Department of Mathematics, Faculty of Science, Palacký University, 17. listopadu 12, Olomouc, 77146, Czech Republic

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Boundary Value Problems 2014, 2014:121  doi:10.1186/1687-2770-2014-121

Dedicated to Professor Ivan Kiguradze for his merits in mathematical sciences

 Received: 13 December 2013 Accepted: 5 May 2014 Published: 16 May 2014

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

### Abstract

In this paper, we investigate the structure and properties of the set of positive blow-up solutions of the differential equation ( t k v ( t ) ) = t k h ( t , v ( t ) ) , t ( 0 , T ] R , where k ( 1 , ) . The differential equation is studied together with the boundary conditions lim t 0 + v ( t ) = , v ( T ) = 0 . 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

### 1 Introduction

In this paper, we investigate the structure and properties of the set of positive blow-up solutions of the differential equation

( t k v ( t ) ) = t k h ( t , v ( t ) ) , t ( 0 , T ] R , (1)

where k ( 1 , ) .

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 [6], in relativistic cosmology for particles which can be treated as domains in the universe [7], and in the nonlinear field theory, in particular, in context of bubbles generated by scalar fields of the Higgs type in Minkowski spaces [8].

Here, we assume that h is positive and satisfies the Carathéodory conditions on [ 0 , T ] × [ 0 , ) . We define a positive solution of (1) as a function v which satisfies (1) for a.e. t [ 0 , T ] , is positive on ( 0 , T ) , and has absolutely continuous first derivative on each compact subinterval in ( 0 , T ] .

According to Lemma 4, if v is a positive solution of (1), then either

lim t 0 + v ( t ) [ 0 , ) , lim t 0 + v ( t ) = 0 ,

or

lim t 0 + v ( t ) = , lim t 0 + v ( t ) = .

In the literature, bounded solutions of (1) have been widely investigated; see e.g.[9-13]. Such solutions are characterized by the initial condition lim t 0 + v ( t ) = 0 . In contrast to this, some real problems lead to the investigation of unbounded solutions which are characterized by the condition lim t t v ( t ) = for some t [ 0 , T ] and which are called blow-up solutions. We refer to [14-16]. Here, we are interested in blow-up solutions of (1), where t = 0 . In particular, (1) will be considered together with the boundary conditions

lim t 0 + v ( t ) = , v ( T ) = 0 . (2)

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 R = { v Z : t k v  is bounded on  ( 0 , T ] } and R c = { v R : v ( T ) = c } for c 0 . Since v ( T ) 0 for each v Z , it is obvious that R = c 0 R c .

Our main goal is to find conditions for the data function h in (1), which guarantee that the set R c is nonempty for each c 0 and then to investigate the properties of this set. For example, we prove that the difference of any two functions in R c , c 0 , retains its sign on [ 0 , T ) , and that there exist minimal and maximal solutions v c , min , v c , max R c for each c 0 , cf. Theorem 5. If the interior of the set { ( t , x ) R 2 : 0 t T , v c , min ( t ) x v c , max ( t ) } is nonempty, we show that this interior is fully covered by ordered graphs of other functions belonging to R c for each c 0 ; see Theorem 6. Finally, in Theorem 7, the existence of a positive constant α 0 such that α 0 = lim t 0 + t k 1 v 0 , min ( t ) is shown and all functions v from ℛ are uniquely characterized by the condition

lim t 0 + t k 1 v ( t ) = α , α [ α 0 , ) .

If we denote such v by v α and define w α ( t ) = t k 1 v ( t ) for t ( 0 , T ] , and w α ( 0 ) = α , then we find that the graphs of these functions do not intersect, cf. Theorem 8, and that for each α > α 0 , the set { w α : α 0 α α } is compact in C [ 0 , T ] ; see Theorem 9.

The study of a structure of positive solutions to other types of ordinary differential equations can be found for example in [17-19].

#### Notation

Let us denote by C [ 0 , T ] the Banach space of functions continuous on [ 0 , T ] equipped with the maximum norm

x = max { | x ( t ) | : t [ 0 , T ] } .

Similarly, C 1 [ 0 , T ] means the Banach space of functions having a continuous first derivative on [ 0 , T ] with the corresponding maximum norm x + x . By L 1 [ 0 , T ] we denote the set of functions which are Lebesgue integrable on [ 0 , T ] . Moreover, A C 1 [ 0 , T ] is the set of functions whose first derivative is absolutely continuous on [ 0 , T ] , while A C loc 1 ( 0 , T ] is the set of functions having absolutely continuous first derivative on each compact subinterval of ( 0 , T ] . We say that h ( t , x ) satisfies the Carathéodory conditions on [ 0 , T ] × [ 0 , ) , if the following three conditions hold:

(i) The function h ( , x ) : [ 0 , T ] R is measurable for all x [ 0 , ) .

(ii) The function h ( t , ) : [ 0 , ) R is continuous for a.e. t [ 0 , T ] .

(iii) For each compact set U [ 0 , ) there exists a function m U L 1 [ 0 , T ] such that

| h ( t , x ) | m U ( t ) for a.e.  t [ 0 , T ]  and all  x U .

For functions satisfying above conditions, we use the notation h Car ( [ 0 , T ] × [ 0 , ) ) .

#### 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 [20] 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,

u ( t ) + a t u ( t ) a t 2 u ( t ) = f ( t , u ( t ) ) , (3)

where a < 1 . For k : = a and v ( t ) : = t k u ( t ) , (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:

(H1) f Car ( [ 0 , T ] × [ 0 , ) ) ,

(H2) 0 < f ( t , x ) for a.e. t [ 0 , T ] and all x [ 0 , ) ,

(H3) f ( t , x ) is increasing in x for a.e. t [ 0 , T ] and

lim x 1 x 0 T f ( t , x ) d t = 0 .

We now study (3) subject to the boundary conditions

u ( 0 ) = 0 , u ( T ) = 0 , (4)

and require that

u ( T ) = c , c 0 , (5)

holds. We call a function u : [ 0 , T ] R a positive solution of the Dirichlet problem (3), (4) if u A C 1 [ 0 , T ] , u > 0 on ( 0 , T ) , u satisfies the boundary conditions (4), and (3) holds for a.e. t [ 0 , T ] . Clearly, for each positive solution u of (3), (4), there exists c 0 such that (5) is satisfied.

We now denote by the set of all positive solutions of problem (3), (4), and S c = { u S : u ( T ) = c } , where c 0 .

In the following lemma we cite those results from [20] which will be used in the analysis of problem (1), (2).

Lemma 1Let (H1)-(H3) hold. Then the following statements hold:

(a) For each c 0 the set S c is nonempty and there exist functions u c , min , u c , max S c such that u c , min ( t ) u ( t ) u c , max ( t ) for t [ 0 , T ] and u S c .

(b) If c 1 > c 2 0 , u i S c i , i = 1 , 2 , then u 1 ( t ) > u 2 ( t ) for t ( 0 , T ) .

(c) If c 0 , u i S c , i = 1 , 2 , and u 1 ( t 0 ) > u 2 ( t 0 ) for some t 0 ( 0 , T ) , then either u 1 ( t ) > u 2 ( t ) for t ( 0 , T ) or there exists t ( t 0 , T ) such that u 1 ( t ) > u 2 ( t ) for t ( 0 , t ) and u 1 ( t ) = u 2 ( t ) for t [ t , T ] .

(d) For each t 0 ( 0 , T ) and each A > u 0 , max ( t 0 ) there exists u S satisfying u ( t 0 ) = A .

(e) S c is a one-point set for each c [ 0 , ) Γ , where Γ [ 0 , ) is at most countable.

(f) For each 0 K Q < , the set K c Q S c is compact in C 1 [ 0 , T ] .

(g) If c 0 , then u S c if and only if it is a solution of the equation

u ( t ) = t c T a + 1 | a + 1 | ( T a 1 t a 1 ) + t t T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s (6)

in the set C 1 [ 0 , T ] .

We now formulate new results which complete those from [20]. We first establish a relation between S 0 and the set { ( t , x ) R 2 : 0 t T , u 0 , min ( t ) x u 0 , max ( t ) } if its interior is nonempty. This question was a short time ago an open problem [[20], Remark 4.4]. We note that the relation between S c and the set { ( t , x ) R 2 : 0 t T , u c , min ( t ) x u c , max ( t ) } with c > 0 having nonempty interior is described in Lemma 1(d).

Theorem 1Let (H1)-(H3) hold. Let us assume that there exists t 0 ( 0 , T ) such that u 0 , min ( t 0 ) < u 0 , max ( t 0 ) . Then for any Q ( u 0 , min ( t 0 ) , u 0 , max ( t 0 ) ) there exists u S 0 satisfying u ( t 0 ) = Q .

ProofStep 1. Auxiliary Dirichlet problem.

Choose Q ( u 0 , min ( t 0 ) , u 0 , max ( t 0 ) ) . Consider (3) subject to the Dirichlet conditions

u ( t 0 ) = Q , u ( T ) = 0 . (7)

We claim that there exists a solution v to problem (3), (7) such that

u 0 , min ( t ) v ( t ) u 0 , max ( t ) for  t [ t 0 , T ] . (8)

We show this result utilizing the method of lower and upper functions. It follows from (H1) that there exists φ L 1 [ t 0 , T ] such that

| a t 2 x + f ( t , x ) | φ ( t ) for a.e.  t [ t 0 , T ]  and  u 0 , min ( t ) x u 0 , max ( t ) . (9)

Let

W = t 0 T φ ( t ) d t , S = max { u 0 , min , u 0 , max , ( T t 0 ) | a | W } + 1 ,

and let χ : R [ 0 , 1 ] be given as

χ ( x ) = { 1 if  | x | S , 2 | x | S if  S < | x | < 2 S , 0 if  | x | 2 S .

We now consider the auxiliary differential equation

v ( t ) = χ ( v ( t ) ) ( a t v ( t ) + a t 2 v ( t ) + f ( t , v ( t ) ) ) . (10)

It is not difficult to verify that χ ( y ) | y | S for y R . Hence, cf. (9),

χ ( y ) | a t y + a t 2 x + f ( t , x ) | χ ( y ) | a y | t + χ ( y ) φ ( t ) | a | S t + φ ( t ) (11)

for a.e. t [ t 0 , T ] and all u 0 , min ( t ) x u 0 , max ( t ) , y R . Since χ ( u 0 , min ( t ) ) = 1 and χ ( u 0 , max ( t ) ) = 1 for t [ t 0 , T ] , u 0 , min ( T ) = u 0 , max ( T ) = 0 , u 0 , min ( t 0 ) < Q < u 0 , max ( t 0 ) , and u 0 , min , u 0 , max solve (3) on [ 0 , T ] , we conclude that u 0 , min and u 0 , max are lower and upper functions of problem (3), (7) (see e.g.[21] or [22]). This fact together with (11) implies the existence of a solution v to problem (3), (7) satisfying (8), cf. [[21], Lemma 3.7]. Moreover,

u 0 , min ( t ) T t v ( t ) T t u 0 , max ( t ) T t , t [ t 0 , T ) ,

and taking the limit t T we obtaina u 0 , min ( T ) v ( T ) u 0 , max ( T ) , which together with u 0 , min ( T ) = u 0 , max ( T ) = 0 gives v ( T ) = 0 .

We now prove that | v | < S on [ t 0 , T ] . The proof is indirect. Let us assume that there exists ξ [ t 0 , T ) such that | v ( ξ ) | = S and | v ( t ) | < S for t ( ξ , T ] . Then χ ( v ( t ) ) = 1 on this interval, and therefore

v ( t ) = a t v ( t ) + a t 2 v ( t ) + f ( t , v ( t ) ) for a.e.  t [ ξ , T ] .

From (8) and (9) we conclude that

| a t 2 v ( t ) + f ( t , v ( t ) ) | φ ( t ) for a.e.  t [ ξ , T ] .

Since

v ( t ) = v ( T ) v ( t ) = t T v ( s ) d s = a t T v ( s ) s d s + t T ( a s 2 v ( s ) + f ( s , v ( s ) ) ) d s ,

we have

| v ( t ) | | a | t T | v ( s ) | s d s + t T φ ( s ) d s .

In particular,

| v ( t ) | | a | t T | v ( s ) | s d s + W for  t [ ξ , T ] .

By the Gronwall lemma,

| v ( t ) | W exp ( | a | t T d s s ) = W ( T t ) | a | , t [ ξ , T ] .

Therefore

S = | v ( ξ ) | W ( T ξ ) | a | ,

W ( T ξ ) | a | W ( T t 0 ) | a | S 1 .

Consequently, | v ( t ) | < S for t [ t 0 , T ] , and so χ ( v ) = 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 [ t 0 , T ] , v ( T ) = 0 and u 0 , min ( t 0 ) < v ( t 0 ) < u 0 , max ( t 0 ) because v ( t 0 ) = Q ( u 0 , min ( t 0 ) , u 0 , max ( t 0 ) ) . It is easy to verify that the equality

( t a + 2 ( v ( t ) t ) ) = t a + 1 f ( t , v ( t ) ) (12)

is satisfied for a.e. t [ t 0 , T ] . We now integrate the last equality two times over [ t , T ] [ t 0 , T ] and have (note that v ( T ) = v ( T ) = 0 )

v ( t ) = t t T s a 2 ( s T ξ a + 1 f ( ξ , v ( ξ ) ) d ξ ) d s , t [ t 0 , T ] .

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 γ = inf { t : t J } . Then 0 γ < t 0 and (12) with v replaced by u holds for a.e. t J . The integration now yields

u ( t ) = t t T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s , t J , (13)

and we claim that

u 0 , min ( t ) < u ( t ) < u 0 , max ( t ) for  t ( γ , t 0 ] . (14)

We show inequality (14) indirectly. Let us assume that (14) does not hold. Then there exists ν ( γ , t 0 ) such that

u 0 , min ( t ) < u ( t ) < u 0 , max ( t ) for  t ( ν , t 0 ]

and either u ( ν ) = u 0 , max ( ν ) or u ( ν ) = u 0 , min ( ν ) . Assume that u ( ν ) = u 0 , max ( ν ) .

Since, by (H3), f ( t , u ( t ) ) < f ( t , u 0 , max ( t ) ) for a.e. t [ ν , t 0 ] and f ( t , u ( t ) ) f ( t , u 0 , max ( t ) ) for a.e. t [ t 0 , T ] , we have

u ( ν ) = ν ν T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s < ν ν T s a 2 ( s T ξ a + 1 f ( ξ , u 0 , max ( ξ ) ) d ξ ) d s = u 0 , max ( ν ) ,

which is not possible. The case u ( ν ) = u 0 , min ( ν ) can be discussed analogously. Hence, (14) holds.

Suppose that γ > 0 . Then J = ( γ , T ] and since u is bounded on ( γ , T ] , it follows that lim sup t γ + | u ( t ) | = . The integration of the equality

u ( t ) = a t u ( t ) + a t 2 u ( t ) + f ( t , u ( t ) ) for a.e.  t ( γ , T ]

over [ t , T ] ( γ , T ] gives

u ( t ) = a t T u ( s ) s d s + t T ( a s 2 u ( s ) + f ( s , u ( s ) ) ) d s , t ( γ , T ] .

Since | a t 2 u ( t ) + f ( t , u ( t ) ) | p ( t ) for a.e. t [ γ , T ] , where p L 1 [ γ , T ] , we have

| u ( t ) | | a | t T | u ( s ) | s d s + γ T p ( s ) d s , t ( γ , T ] .

Using the Gronwall lemma, we deduce that for t ( γ , T ] ,

| u ( t ) | ( T t ) | a | γ T p ( s ) d s ( T γ ) | a | γ T p ( s ) d s

Therefore γ = 0 , and then (14) yields lim t 0 + u ( t ) = 0 . Consequently J = [ 0 , T ] , 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 t 0 ( 0 , T ) and each A u 0 , min ( t 0 ) there exists u S satisfying u ( t 0 ) = A .

Proof The result follows immediately from Lemma 1(d) and Theorem 1. □

Remark 1 Corollary 1 says that the set U = { ( t , x ) R 2 : t [ 0 , T ] , x u 0 , min ( t ) } is covered by the graphs of functions from , that is,

U = { ( t , u ( t ) ) : t [ 0 , T ] , u S } .

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 t = T only in the case that S c is a singleton set for each c 0 . Since we cannot uniquely determine all functions from via their derivatives at t = T , see Lemma 1(e), we discuss their derivatives at the singular point t = 0 .

Lemma 2Let (H1)-(H3) hold. Let u S c , c 0 . Then

u ( 0 ) = c | a + 1 | + 0 T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s . (15)

Proof It follows from Lemma 1(g) that (6) holds for t [ 0 , T ] . Since u ( 0 ) = 0 , we have (note that a + 1 < 0 )

u ( 0 ) = lim t 0 + u ( t ) u ( 0 ) t = lim t 0 + ( c T a + 1 | a + 1 | ( T a 1 t a 1 ) + t T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s ) = c | a + 1 | + 0 T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s ,

and (15) follows. □

Corollary 2Let u S . Then

u ( 0 ) = u ( T ) | a + 1 | + 0 T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s .

Proof The result follows from Lemma 2, since u S u ( T ) . □

Corollary 3Let u , v S , u v . Then u ( 0 ) v ( 0 ) .

Proof Since u v , there exists t 0 ( 0 , T ) such that u ( t 0 ) v ( t 0 ) . We can assume that for instance u ( t 0 ) < v ( t 0 ) . Then u < v on ( 0 , t 0 ] and u v on ( t 0 , T ] by Lemma 1(b)(c). Hence, u ( T ) v ( T ) and

0 T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s < 0 T s a 2 ( s T ξ a + 1 f ( ξ , v ( ξ ) ) d ξ ) d s ,

which together with Corollary 2 gives u ( 0 ) < v ( 0 ) . □

Corollary 4Let u S c , c 0 . Then

u ( 0 ) > c | a + 1 | . (16)

In particular, u 0 , min ( 0 ) > 0 .

Proof Inequality (16) follows from Lemma 2 and the fact that f ( t , u ( t ) ) > 0 for a.e. t [ 0 , T ] by (H2). □

Corollary 5Let u S . Then

0 T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s = 1 | a + 1 | 0 T f ( s , u ( s ) ) d s .

Proof The integration of (3) over [ ε , T ] ( 0 , T ] gives

u ( T ) u ( ε ) + a ε T u ( t ) t d t a ε T u ( t ) t 2 d t = ε T f ( t , u ( t ) ) d t .

Using integration by parts we obtain

ε T u ( t ) t 2 d t = u ( ε ) ε + ε T u ( t ) t d t ,

and, therefore,

u ( T ) u ( ε ) a u ( ε ) ε = ε T f ( t , u ( t ) ) d t .

Taking the limit ε 0 , we have

u ( T ) + | a + 1 | u ( 0 ) = 0 T f ( t , u ( t ) ) d t .

On the other hand, it follows from Corollary 2 that

u ( T ) + | a + 1 | u ( 0 ) = | a + 1 | 0 T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s .

Combining the above two equalities yields the result. □

Lemma 3Let (H1)-(H3) hold. Let c 0 and S c be not a singleton set. Then for each α [ u c , min ( 0 ) , u c , max ( 0 ) ] there exists a unique u S c such that u ( 0 ) = α . Consequently, [ u c , min ( 0 ) , u c , max ( 0 ) ] = { u ( 0 ) : u S c } .

Proof Let u S c , u c , min u u c , max . Then it follows from Lemma 1(a) and (H3) that

0 T s a 2 ( s T ξ a + 1 f ( ξ , u c , min ( ξ ) ) d ξ ) d s < 0 T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s < 0 T s a 2 ( s T ξ a + 1 f ( ξ , u c , max ( ξ ) ) d ξ ) d s .

Hence, by Lemma 2,

u c , min ( 0 ) < u ( 0 ) < u c , max ( 0 ) for  u S c , u c , min u u c , max .

Since S c is a compact set in C 1 [ 0 , T ] by Lemma 1(f), the set I = { u ( 0 ) : u S c } [ u c , min ( 0 ) , u c , max ( 0 ) ] is closed. In fact, let { β n } I , β n = u n ( 0 ) , where u n S c , and let lim n β n = β . Then there exist a subsequence { β k n } of { β n } and u S c such that lim n u k n = u in C 1 [ 0 , T ] . In particular, β k n = u k n ( 0 ) u ( 0 ) as n . Therefore, β = u ( 0 ) I .

It remains to prove that I = [ u c , min ( 0 ) , u c , max ( 0 ) ] . Assume that the equality does not hold. Then, from the structure of bounded and closed subsets of ℝ the existence of an open interval ( α , β ) [ u c , min ( 0 ) , u c , max ( 0 ) ] I , α , β I , follows. Let α = u α ( 0 ) and β = u β ( 0 ) , where u α , u β S c . Due to Lemma 1(c), there exists t 0 ( 0 , T ) such that u α < u β on ( 0 , t 0 ] , u α u β on ( t 0 , T ] . Choose τ ( u α ( t 0 ) , u β ( t 0 ) ) . By Corollary 1, there exists v S c such that v ( t 0 ) = τ . Then u α < v < u β on ( 0 , t 0 ] and u α v u β on ( t 0 , T ] . Consequently, u α ( 0 ) < v ( 0 ) < u β ( 0 ) on ( t 0 , T ] , that is, v ( 0 ) ( α , β ) , which is not possible. □

Since functions from belong to A C 1 [ 0 , T ] , we have u ( 0 ) < for each u S . Corollary 4 yields u 0 , min ( 0 ) > 0 . Let us denote

J : = [ u 0 , min ( 0 ) , ) ( 0 , ) . (17)

Lemma 3 implies that functions from can be uniquely determined by the values of their derivatives at the singular point t = 0 ; see Theorem 2.

Theorem 2Let (H1)-(H3) hold. Then there exists a unique u S satisfying u ( 0 ) = α if and only if α J .

Proof We first show

{ u ( 0 ) : u S } = J . (18)

It follows from Lemma 3 that

{ u ( 0 ) : u 0 c K S c } = [ u 0 , min ( 0 ) , u K , max ( 0 ) ]

for each K 0 , where we set u c , min = u = u c , max if S c is a singleton set and S c = { u } . In view of Corollary 4, u ( 0 ) > c / | a + 1 | for u S c . Consequently, (18) follows.

Let us now choose α J . Then there exists u S satisfying u ( 0 ) = α . The uniqueness of u follows from Corollary 3. □

For α J , we denote by u α the unique element of satisfying u α ( 0 ) = α .

Theorem 3Let (H1)-(H3) hold. Assume that { α n } J is a convergent sequence and lim n α n = α . Then lim n u α n = u α in C 1 [ 0 , T ] .

Proof Let K = sup { u α n ( T ) : n N } . Then K < because α n = u α n ( 0 ) > u α n ( T ) / | a + 1 | for n N by Corollary 4. Since { u α n } 0 c K S c and 0 c K S c is compact in C 1 [ 0 , T ] by Lemma 1(f), the sequence { u α n } is relatively compact in C 1 [ 0 , T ] . Let { u α n } be a subsequence of { u α n } which is convergent in C 1 [ 0 , T ] , and let lim n u α n = u . Then u S c 0 for a c 0 [ 0 , K ] and u ( 0 ) = lim n u α n ( 0 ) = lim n α n = α . Therefore, u = u α and hence any subsequence { u α n } of { u α n } converging in C 1 [ 0 , T ] has the same limit u α . Consequently, { u α n } is convergent in C 1 [ 0 , T ] and u α 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 t = 0 .

Lemma 4Let k ( 1 , ) and let the function h Car ( [ 0 , T ] × ( 0 , ) ) be positive. Letvbe a positive solution of (1). Then either

lim t 0 + v ( t ) [ 0 , ) , lim t 0 + v ( t ) = 0 , (19)

or

lim t 0 + v ( t ) = , lim t 0 + v ( t ) = . (20)

Proof By Corollary 3.5 in [12], if sup { | v ( t ) | + | v ( t ) | : t ( 0 , T ] } < , then v satisfies (19). Now, assume that

sup { | v ( t ) | + | v ( t ) | : t ( 0 , T ] } = . (21)

Then v has to satisfy

sup { | v ( t ) | : t ( 0 , T ] } = , (22)

because otherwise, if (22) was not true, then sup { | v ( t ) | : t ( 0 , T ] } < , contradicting (21). Since h is positive, (1) indicates that the function t k v ( t ) is increasing on ( 0 , T ] and hence there exists lim t 0 + t k v ( t ) . The next part of the proof is divided into three cases.

(i) Let us assume that

lim t 0 + t k v ( t ) = 0 . (23)

Since t k v ( t ) is increasing on ( 0 , T ] , we deduce t k v ( t ) > 0 and v ( t ) > 0 for t ( 0 , T ] . Consequently, v is increasing on ( 0 , T ] . This together with v > 0 and v A C loc 1 ( 0 , T ] yields sup { v ( t ) : t ( 0 , T ] } < . Therefore, h ( t , v ( t ) ) L 1 [ 0 , T ] . Integrating (1) and using (23), we obtain

t k v ( t ) = 0 t s k h ( s , v ( s ) ) d s , t ( 0 , T ] , 0 < v ( t ) = 1 t k 0 t s k h ( s , v ( s ) ) d s < 0 T h ( s , v ( s ) ) d s < , t ( 0 , T ] ,

which is a contradiction to (22).

(ii) Assume that

lim t 0 + t k v ( t ) = > 0 .

Then there exists t 0 ( 0 , T ) such that

v ( t ) > / 2 t k for  t ( 0 , t 0 ] . (24)

Hence, v is increasing on ( 0 , t 0 ] and there exists lim t 0 + v ( t ) 0 .

By integration, we obtain from (24)

v ( t 0 ) v ( t ) > ( t 0 1 k t 1 k ) 2 ( 1 k ) , t ( 0 , t 0 ]

and, by virtue of k > 1 and > 0 , we arrive at

lim t 0 + v ( t ) v ( t 0 ) lim t 0 + ( t 0 1 k t 1 k ) 2 ( 1 k ) = ,

(iii) Assume that

lim t 0 + t k v ( t ) = < 0 . (25)

Then there exists t 0 ( 0 , T ) such that

v ( t ) < / 2 t k for  t ( 0 , t 0 ] . (26)

Hence, v is decreasing on ( 0 , t 0 ] and there exists lim t 0 + v ( t ) ( 0 , ] . By integration, we obtain from (26)

v ( t 0 ) v ( t ) < ( t 0 1 k t 1 k ) 2 ( 1 k ) , t ( 0 , t 0 ]

and, since k > 1 and < 0 , we have

lim t 0 + v ( t ) v ( t 0 ) lim t 0 + ( t 0 1 k t 1 k ) 2 ( 1 k ) = .

In addition (25) yields lim t 0 + v ( t ) = 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

h ( t , x ) = ψ ( t ) + g ( t , x ) .

In particular, we study the equation

v ( t ) + k t v ( t ) = ψ ( t ) + g ( t , v ( t ) ) , (27)

where k ( 1 , ) and ψ, g satisfy the conditions

( H 1 0 ) ψ L 1 [ 0 , T ] and ψ > 0 for a.e. t [ 0 , T ] ,

( H 2 0 ) g Car ( [ 0 , T ] × [ 0 , ) ) and g ( t , x ) is increasing in x for a.e. t [ 0 , T ] ,

( H 3 0 ) 0 g ( t , x ) ϕ ( x ) for a.e. t [ 0 , T ] and all x [ 0 , ) , where ϕ C [ 0 , ) , ϕ is increasing and

lim x ϕ ( x ) x = 0 .

Together with (27) we consider the conditions

lim t 0 v ( t ) = , v ( T ) = 0 , (28)

and

v ( T ) = c . (29)

We define a positive solution to problem (27), (28) as a function v A C loc 1 ( 0 , T ] such that v > 0 on ( 0 , T ) , v satisfies the boundary conditions (28), and (27) holds for a.e. t [ 0 , T ] .

Define a set by

Z = { v A C loc 1 ( 0 , T ] : v  is a positive solution of (27), (28) } . (30)

Clearly, for each v Z there exists c 0 such that (29) holds.

Lemma 5Let ( H 1 0 )-( H 3 0 ) hold. Let v Z . Then v < 0 on ( 0 , T ) .

Proof Since ( t k v ) = t k ( v + ( k / t ) v ) , the integration of (27) over the subinterval [ t , T ] ( 0 , T ] gives

T k v ( T ) t k v ( t ) = t T s k ( ψ ( s ) + g ( s , v ( s ) ) ) d s .

Assume that v ( t 0 ) = 0 for some t 0 ( 0 , T ) . Then

T k v ( T ) = t 0 T s k ( ψ ( s ) + g ( s , v ( s ) ) ) d s ,

which is not possible, since v ( T ) 0 and t 0 T s k ( ψ ( s ) + g ( s , v ( s ) ) ) d s > 0 . □

Moreover, let us define sets

R = { v Z : t k v  is bounded on  ( 0 , T ] } (31)

and

R c = { v R : v ( T ) = c } for  c 0 . (32)

It is obvious that R = c 0 R c . Using the substitution

a = k , u ( t ) = t a v ( t ) = t k v ( t ) for  t ( 0 , T ] , (33)

we can rewrite (27) and obtain the form

u ( t ) + a t u ( t ) a t 2 u ( t ) = t a ( ψ ( t ) + g ( t , t a x ) ) . (34)

We now introduce a function f,

f ( t , x ) = t a ( ψ ( t ) + g ( t , t a x ) ) for a.e.  t [ 0 , T ]  and all  x [ 0 , ) .

Under conditions ( H 1 0 )-( H 3 0 ), the function f satisfies (H1)-(H3); see the proof of [[20], Theorem 5.1]. Therefore, the results of Section 2 hold for problem (34), (4). As in Section 2, we define the sets and S c 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 ( H 1 0 )-( H 3 0 ) hold and c 0 . Let us assume that (33) holds. Then v R c if and only if u S c 0 and c 0 = T k c .

Proof (⇒) Let v R c . Then (28), (32), and (33) provide the following properties: u A C loc 1 ( 0 , T ] , u ( T ) = c 0 = T k c , and u is bounded on ( 0 , T ] . It follows from the equality

( t k v ( t ) ) = t k ( ψ ( t ) + g ( t , v ( t ) ) ) for a.e.  t [ 0 , T ]

that

( t a ( t a u ( t ) ) ) = t a ( ψ ( t ) + g ( t , t a u ( t ) ) ) = f ( t , u ( t ) ) for a.e.  t [ 0 , T ] .

Hence,

u ( t ) + a t u ( t ) a t 2 u ( t ) = f ( t , u ( t ) ) for a.e.  t [ 0 , T ] .

We now argue as in the proof of [[20], Lemma 3.3] and arrive at

u ( t ) = t c 0 T a + 1 | a + 1 | ( T a 1 t a 1 ) + t t T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s (35)

for t ( 0 , T ] . Since u is bounded on ( 0 , T ] , we have f ( t , u ( t ) ) L 1 [ 0 , T ] . Then, by [[20], Lemma 2.1],

0 t a 1 t T s a + 1 f ( s , u ( s ) ) d s 0 T f ( s , u ( s ) ) d s , 0 t T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s 1 | a + 1 | 0 T f ( s , u ( s ) ) d s ,

for t ( 0 , T ] , and hence, sup { | u ( t ) | + | u ( t ) | : t ( 0 , T ] } < . Now, [[23], Corollary 1] guarantees that u can be extended on [ 0 , T ] with u A C 1 [ 0 , T ] such that the equality (35) holds for t [ 0 , T ] . Consequently, u S c 0 .

(⇐) Let u S c 0 and c 0 = T k c . Then it follows from (4), (5), and (33) that v A C loc 1 ( 0 , T ] and v ( T ) = c , lim t 0 + t k v ( t ) = u ( 0 ) = 0 . Hence, t k v is bounded on ( 0 , T ] and

( t a ( t a u ( t ) ) ) = f ( t , u ( t ) ) for a.e.  t [ 0 , T ]

yields ( t k v ( t ) ) = f ( t , t k v ( t ) ) , that is,

v ( t ) + k t v ( t ) = t k f ( t , t k v ( t ) ) = ψ ( t ) + g ( t , v ( t ) ) for a.e.  t [ 0 , T ] .

Since u S c 0 , the equality (35) holds on [ 0 , T ] , and consequently,

t a u ( t ) t a + 1 t T s a 2 ( s T ξ a + 1 f ( ξ , u ( ξ ) ) d ξ ) d s , t ( 0 , T ] .

Hence, note that a + 1 < 0 , lim t 0 t a u ( t ) = , and so lim t 0 v ( t ) = . As a result, we have v R c . □

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 ( H 1 0 )-( H 3 0 ) hold. Then the following statements hold:

(a) For each c 0 the set R c is nonempty and there exist v c , min , v c , max R c such that v c , min ( t ) v ( t ) v c , max ( t ) for t ( 0 , T ] and v R c .

(b) If c 1 > c 2 0 , v i R c i , i = 1 , 2 , then v 1 ( t ) > v 2 ( t ) for t ( 0 , T ) .

(c) If c 0 , v i R c , i = 1 , 2 , and v 1 ( t 0 ) > v 2 ( t 0 ) for some t 0 ( 0 , T ) , then either v 1 ( t ) > v 2 ( t ) for t ( 0 , T ) or there exists t ( t 0 , T ) such that v 1 ( t ) > v 2 ( t ) for t ( 0 , t ) and v 1 ( t ) = v 2 ( t ) for t [ t , T ] .

(d) R c is a singleton set for each c [ 0 , ) Γ , where Γ [ 0 , ) 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 c 0 , c 0 = T k c , and if (33) holds, then Theorems 4 and 5 yield

v c , min ( t ) = t k u c 0 , min ( t ) , v c , max ( t ) = t k u c 0 , max ( t ) , t ( 0 , T ] . (36)

The next theorem shows that the set

V = { ( t , x ) R 2 : t ( 0 , T ] , x v 0 , min ( t ) }

is covered by graphs of the functions from ℛ.

Theorem 6Let ( H 1 0 )-( H 3 0 ) hold. Then, for each t 0 ( 0 , T ) and each Q v 0 , min ( t 0 ) , there exists v R satisfying v ( t 0 ) = Q . In particular, if for some c 0 and t ˜ ( 0 , T ) the inequality v c , min ( t ˜ ) < v c , max ( t ˜ ) holds, then for each Q ˜ ( v c , min ( t ˜ ) , v c , max ( t ˜ ) ) there exist v R c satisfying v ( t ˜ ) = Q ˜ .

Proof Choose t 0 ( 0 , T ) and Q v 0 , min ( t 0 ) . Define A : = t 0 k Q . Then, using (33) and (36), we deduce u 0 , min ( t 0 ) = t 0 k v 0 , min ( t 0 ) A . Therefore, by Corollary 1, there exists u S such that u ( t 0 ) = A . Consequently, (33) and Theorem 4 give v R and v ( t 0 ) = Q . The last statement follows from Theorem 5(c). □

By Corollary 4, there exists α 0 > 0 such that

α 0 = u 0 , min ( 0 ) = lim t 0 + u 0 , min ( t ) t , (37)

and hence, by (33) and (36) with c = c 0 = 0 ,

α 0 = lim t 0 + t k 1 v 0 , min ( t ) . (38)

Using constants from the interval [ α 0 , ) , we can uniquely determine all functions in ℛ.

Theorem 7Let ( H 1 0 )-( H 3 0 ) hold and let α 0 be as in (38). Then the following statements hold:

(a) For each v R there exists α [ α 0 , ) such that

lim t 0 + t k 1 v ( t ) = α , lim t 0 + t k v ( t ) = ( 1 k ) α . (39)

(b) For each α [ α 0 , ) there exists a unique v R satisfying (39).

Proof (a) Choose v R . Using (33), (37), and Theorem 4, we have t k v = u S and

lim t 0 + t k 1 v ( t ) = lim t 0 u ( t ) t = u ( 0 ) . (40)

By Lemma 1(c), u ( 0 ) α 0 . Hence, if we denote u ( 0 ) = α , the first condition in (39) follows. Moreover,

lim t 0 + t k v ( t ) = lim t 0 + t k ( t k u ( t ) ) = lim t 0 + ( k u ( t ) t + u ( t ) ) = ( 1 k ) u ( 0 ) , (41)

and therefore the second condition in (39) holds.

(b) Choose α [ α 0 , ) . Theorem 2 guarantees that there exists a unique u S satisfying u ( 0 ) = α . Using (33) and Theorem 4, we conclude t k u = v R . Then v satisfies (40) and (41) which results in (39). It remains to prove that v is unique. Assume that there exists a function w R , w v , such that w satisfies (39). Let z = t k w . Then z S and z ( 0 ) = lim t 0 z ( t ) / t = lim t 0 t k 1 w ( t ) = α . Consequently, by Theorem 2, we arrive at z = u and w = v , which is a contradiction. □

Let α 0 be given in (38) and choose α [ α 0 , ) . Keeping Theorem 7(b) in mind, there exists a unique v R satisfying (39). We denote such v by v α and define a function w α : [ 0 , T ] R as

w α ( t ) : = { t k 1 v α ( t ) if  t ( 0 , T ] , α if  t = 0 . (42)

Then w α C [ 0 , T ] . We now specify further properties of functions w α .

Theorem 8Let ( H 1 0 )-( H 3 0 ) hold and let α 0 be from (38). Let α , β [ α 0 , ) and α < β . Then either w α ( t ) < w β ( t ) for t [ 0 , T ) or there exists t ( 0 , T ) such that w α ( t ) < w β ( t ) for t [ 0 , t ) and w α ( t ) = w β ( t ) for t [ t , T ] .

Proof According to (42), there exist c 1 , c 2 [ 0 , ) , v α R c 1 , and v β R c 2 such that w α ( t ) = t k 1 v α ( t ) , w β ( t ) = t k 1 v β ( t ) for t ( 0 , T ] . Also, w α , w β C [ 0 , T ] and, since w α ( 0 ) = α < β = w β ( 0 ) , it follows that w α < w β in a right neighborhood of t = 0 . Therefore, there exists t 0 ( 0 , T ) such that v α ( t 0 ) < v β ( t 0 ) and hence, c 1 c 2 . (Note that if c 1 > c 2 , then Theorem 5(b) yields v α ( t ) > v β ( t ) for t ( 0 , T ) , which is not possible.) The result now follows from Theorem 5(b) for c 1 < c 2 and from Theorem 5(c) for c 1 = c 2 . □

Corollary 6Let { α n } [ α 0 , ) be a convergent sequence and let us denote lim n α n = α . Then lim n w α n = w α in C [ 0 , T ] .

Proof The proof is indirect. Assume that the statement of the corollary does not hold. Then there exist ε, { t n } [ 0 , T ] and a subsequence of { α n } , we denote it again by { α n } , such that

| w α n ( t n ) w α ( t n ) | ε for  n N . (43)

Let N 1 = { n N : α n α } and N 2 = { n N : α n α } . Let N 1 be countable. Then there exists a decreasing subsequence { α n } n N of { α n } n N 1 . By Theorem 8, the sequence { w α n } is not increasing on [ 0 , T ] and w α n w α on this interval. Using (33) and (42), we obtain w α n ( t ) = u α n ( t ) / t , w α ( t ) = u α ( t ) / t . Since lim n u α n = u α in C 1 [ 0 , T ] it follows from Theorem 3, note that w α n ( 0 ) = α n , that w α ( 0 ) = α and that lim n w α n ( t ) = w α ( t ) for t [ 0 , T ] . This fact together with the monotonicity of { w α n } and w α C [ 0 , T ] gives li