Open Access

Positive solutions of nonlinear Dirichlet BVPs in ODEs with time and space singularities

  • Irena Rachůnková1Email author,
  • Alexander Spielauer2,
  • Svatoslav Staněk1 and
  • Ewa B Weinmüller2
Boundary Value Problems20132013:6

DOI: 10.1186/1687-2770-2013-6

Received: 17 October 2012

Accepted: 21 December 2012

Published: 16 January 2013

Abstract

In this paper, we discuss the existence of positive solutions to the singular Dirichlet boundary value problems (BVPs) for ordinary differential equations (ODEs) of the form

u ( t ) + a t u ( t ) a t 2 u ( t ) = f ( t , u ( t ) , u ( t ) ) , u ( 0 ) = 0 , u ( T ) = 0 ,

where a ( 1 , 0 ) . The nonlinearity f ( t , x , y ) may be singular for the space variables x = 0 and/or y = 0 . Moreover, since a 0 , the differential operator on the left-hand side of the differential equation is singular at t = 0 . Sufficient conditions for the existence of positive solutions of the above BVPs are formulated and asymptotic properties of solutions are specified. The theory is illustrated by numerical experiments computed using the open domain MATLAB code bvpsuite, based on polynomial collocation.

MSC:34B18, 34B16, 34A12.

Keywords

singular ordinary differential equation of the second order time singularities space singularities positive solutions existence of solutions polynomial collocation

1 Introduction

In the present work, we analyze the existence of positive solutions to the singular Dirichlet BVP,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ1_HTML.gif
(1a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ2_HTML.gif
(1b)
Here, we assume that T > 0 , a ( 1 , 0 ) and f satisfies the local Carathéodory conditions on [ 0 , T ] × D , where D : = R + × R 0 and R + : = ( 0 , ) , R 0 : = R { 0 } . Let us recall that a function h : [ 0 , T ] × A R , A R × R , satisfies the local Carathéodory conditions on [ 0 , T ] × A if
  1. (i)

    h ( , x , y ) : [ 0 , T ] R is measurable for all ( x , y ) A ,

     
  2. (ii)

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

     
  3. (iii)
    for each compact set U A , there exists a function m U L 1 [ 0 , T ] such that
    | h ( t , x , y ) | m U ( t ) for a.e.  t [ 0 , T ]  and all  ( x , y ) U .
     
For such functions, we use the notation h Car ( [ 0 , T ] × A ) . Moreover, f ( t , x , y ) may become singular when the space variables x and/or y vanish, which means that f ( t , x , y ) may become unbounded for x = 0 and a.e. t [ 0 , T ] and all y R 0 , and/or it may be unbounded for y = 0 and a.e. t [ 0 , T ] and all x R + . Finally, since a 0 , Eq. (1a) has a singularity of the first kind at the time variable t = 0 because
0 T 1 t d t = , 0 T 1 t 2 d t = .

The differential operator on the left-hand side of Eq. (1a) can be equivalently written as ( t a ( t a u ) ) and, after the substitution x ( t ) = t a u ( t ) , it takes the form ( t a x ) , which arises in numerous important applications. Operators of such type were studied in phase transitions of Van der Waals fluids [14], in population genetics, especially in models for the spatial distribution of the genetic composition of a population [5, 6], in the homogeneous nucleation theory [7], in relativistic cosmology for description of particles which can be treated as domains in the universe [8], and in the nonlinear field theory [9], in particular, when describing bubbles generated by scalar fields of Higgs type in the Minkowski spaces [10].

The aim of this paper is to study the case a ( 1 , 0 ) which is fundamentally different from the case a ( , 1 ) . The latter setting was studied in [11, 12], where the structure and properties of the set of all positive solutions to (1a) and (1b) were investigated (the cardinality of this set is a continuum).

In the sequel, we introduce the basic notation and state the preliminary results required in the analysis of problem (1a) and (1b). Here, we focus our attention on the case a ( 1 , 0 ) and prove the existence of at least one positive solution of (1a) and (1b). In contrast to [11, 12], we consider the more general situation in which f may be also singular at y = 0 . This means that we have to deal with the following additional difficulties.

Let u be a positive solution of problem (1a) and (1b), where f ( t , x , y ) has a singularity at y = 0 . Then there exists t 0 ( 0 , T ) such that u ( t 0 ) > 0 , u ( t 0 ) = 0 and hence f is unbounded in a neighborhood of the point ( t 0 , u ( t 0 ) , u ( t 0 ) ) . Unfortunately, we do not know the exact position of t 0 and therefore, it is not possible to construct a universal Lebesgue integrable majorant for all functions f ( t , u n ( t ) , u n ( t ) ) , where u n are positive solutions of a sequence of auxiliary regular problems. Consequently, the Lebesgue dominated convergence theorem is not applicable and we have to use arguments based on the Vitali convergence theorem instead; see Lemma 2. Another tool used in the proofs is a combination of regularization and sequential techniques with the Leray-Schauder nonlinear alternative.

The investigation of singular Dirichlet BVPs has a long history and a lot of methods for their analysis are available. One of the most important ones is the topological degree method providing various fixed point theorems and existence alternative theorems; see, e.g., Lemma 1. For more information on the topological degree method and its application to numerous BVPs, including Dirichlet problems, we refer the reader to the monographs by Mawhin [1315].

Throughout this paper, we work with the following conditions on the function f in (1a):

(H1) f Car ( [ 0 , T ] × D ) , where D = R + × R 0 .

(H2) There exists an ε > 0 such that
f ( t , x , y ) ε for a.e.  t [ 0 , T ]  and all  ( x , y ) D .
(H3) For a.e. t [ 0 , T ] and all ( x , y ) D , the estimate
f ( t , x , y ) φ ( t ) h ( x , | y | ) + g ( x ) + r ( | y | )
holds, where φ L 1 [ 0 , T ] , h C ( [ 0 , ) × [ 0 , ) ) , g , r C ( R + ) are positive, h is nondecreasing in both its arguments, g and r are nonincreasing, and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equf_HTML.gif
By
x = max { | x ( t ) | : t [ 0 , T ] } , x 1 = 0 T | x ( t ) | d t

we denote the norms in C [ 0 , T ] and L 1 [ 0 , T ] , respectively. A C 1 [ 0 , T ] denotes 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 use the symbol meas ( M ) to denote the Lebesgue measure of .

Definition 1 We say that a function u A C 1 [ 0 , T ] is a positive solution of problem (1a) and (1b) if u > 0 on ( 0 , T ) , u satisfies the boundary conditions (1b) and (1a) holds for a.e. t [ 0 , T ] .

Remark 1 Let a function g have the properties specified in (H3). Then for each b , c R + , 0 b g ( c s ) d s < , and it follows from the inequality
t ( T t ) { T 2 t for  t [ 0 , T 2 ] , T 2 ( T t ) for  t [ T 2 , T ] ,
that
0 T g ( c t ( T t ) ) d t < for each  c R + .
(2)
In order to prove that the singular problem (1a) and (1b) has a positive solution, we use regularization and sequential techniques. To this end, for n N we define functions f n : [ 0 , T ] × ( R × R 0 ) R and f n : [ 0 , T ] × R 2 R by
f n ( t , x , y ) = { f ( t , x , y ) if  x 1 n , f ( t , 1 n , y ) if  x < 1 n ,
and
f n ( t , x , y ) = { f n ( t , x , y ) if  | y | 1 n , n 2 [ f n ( t , x , 1 n ) ( y + 1 n ) f n ( t , x , 1 n ) ( y 1 n ) ] if  | y | < 1 n ,
respectively. Then it follows from (H1) that f n Car ( [ 0 , T ] × R 2 ) and (H2) and (H3) yield
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ4_HTML.gif
(3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ5_HTML.gif
(4)
Hence,
ε λ f n ( t , x , y ) + ( 1 λ ) ε for a.e.  t [ 0 , T ]  and all  x , y R , λ [ 0 , 1 ] ,
(5)
and
λ f n ( t , x , y ) + ( 1 λ ) ε φ ( t ) h ( 1 + | x | , 1 + | y | ) + g ( | x | ) + r ( | y | ) for a.e.  t [ 0 , T ]  and all  x , y R 0 , λ [ 0 , 1 ] . }
(6)
As a first step in the analysis, we investigate auxiliary regular BVPs of the form
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ8_HTML.gif
(7a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ9_HTML.gif
(7b)

To show the solvability of problem (7a) and (7b), we use the following alternative of Leray-Schauder type which follows from [[16], Theorem 5.1].

Lemma 1 Let E be a Banach space, U be an open subset of E and U . Assume that F : U ¯ E is a compact operator. Then either

A1: has a fixed point in U ¯ , or

A2: there exists an element u U and λ ( 0 , 1 ) with u = λ F ( u ) + ( 1 λ ) .

In limit processes, we apply the following Vitali convergence theorem; cf. [1719].

Lemma 2 Let { ρ n } L 1 [ 0 , T ] and let lim n ρ n ( t ) = ρ ( t ) for a.e. t [ 0 , T ] . Then the following statements are equivalent:
  1. (i)

    ρ L 1 [ 0 , T ] and lim n ρ n ρ 1 = 0 ,

     
  2. (ii)

    the sequence { ρ n } is uniformly integrable on [ 0 , T ] .

     
We recall that a sequence { ρ n } L 1 [ 0 , T ] is called uniformly integrable on [ 0 , T ] if for any ε > 0 there exists δ > 0 such that if M [ 0 , T ] and meas ( M ) < δ , then
M | ρ n ( t ) | d t < ε , n N .

The paper is organized as follows. In Section 2, we collect auxiliary results used in the subsequent analysis. Section 3 is devoted to the study of limit properties of solutions to Eq. (7a). In Section 4, we investigate auxiliary regular problems associated with the singular problem (1a) and (1b). We show their solvability and describe properties of their solutions. An existence result for the singular problem (1a) and (1b) is given in Section 5. Finally, in Section 6, we illustrate the theoretical findings by means of numerical experiments.

Throughout the paper a ( 1 , 0 ) .

2 Preliminaries

In this section, auxiliary statements necessary for the subsequent analysis are formulated.

Lemma 3 Let ρ L 1 [ 0 , T ] and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equl_HTML.gif
Then
  1. (i)

    r can be extended on [ 0 , T ] with r C [ 0 , T ] and r ( 0 ) = 0 ,

     
  2. (ii)
    H can be extended on [ 0 , T ] with H A C 1 [ 0 , T ] , and the equality
    H ( t ) + a t H ( t ) a t 2 H ( t ) = ρ ( t )
    (8)
     

holds for a.e. t [ 0 , T ] .

Proof (i) It is clear that r C ( 0 , T ] . Since
| 1 t a + 1 0 t s a + 1 ρ ( s ) d s | 0 t | ρ ( s ) | d s for  t ( 0 , T ] ,
(9)
we have lim t 0 + r ( t ) = 0 . Setting r ( 0 ) : = 0 , r C [ 0 , T ] follows.
  1. (ii)
    Let
    p ( t ) = t T 1 s a + 2 ( 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s for  t ( 0 , T ] .
     
Then p C ( 0 , T ] and H ( t ) = t p ( t ) for t ( 0 , T ] . We now show that p can be extended on [ 0 , T ] in such a way that p C [ 0 , T ] . Integrating by parts yields
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ12_HTML.gif
(10)
for t ( 0 , T ] . Hence lim t 0 + p ( t ) = A , where
A = 1 ( a + 1 ) T a + 1 0 T s a + 1 ρ ( s ) d s + 1 a + 1 0 T ρ ( s ) d s .
Let p ( 0 ) : = A . Then p C [ 0 , T ] . Since H ( t ) = p ( t ) + t p ( t ) = p ( t ) r ( t ) for t ( 0 , T ] , we see that H can be extended on [ 0 , T ] with H C 1 [ 0 , T ] . Moreover,
H ( t ) = p ( t ) r ( t ) = 1 t a + 2 0 t s a + 1 ρ ( s ) d s + a + 1 t a + 2 0 t s a + 1 ρ ( s ) d s ρ ( t ) = a t a + 2 0 t s a + 1 ρ ( s ) d s ρ ( t ) .
In particular,
H ( t ) = a t a + 2 0 t s a + 1 ρ ( s ) d s ρ ( t ) for a.e.  t [ 0 , T ] .
(11)
Hence, cf. (10),
0 T H ( s ) d s = a 0 T 1 s a + 2 ( 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s 0 T ρ ( t ) d t = a A 0 T ρ ( t ) d t ,

and therefore, H L 1 [ 0 , T ] . Consequently, H A C 1 [ 0 , T ] . Finally, it follows from H ( t ) = p ( t ) r ( t ) and p ( t ) = H ( t ) t that r ( t ) = H ( t ) t H ( t ) . Since, by (11), a t r ( t ) = H ( t ) + ρ ( t ) , we see that equality (8) is satisfied for a.e. t [ 0 , T ] which completes the proof. □

Lemma 4 Let { ρ n } L 1 [ 0 , T ] be a uniformly integrable sequence on [ 0 , T ] and let lim n ρ n ( t ) = ρ ( t ) for a.e. t [ 0 , T ] . Then the sequence
{ 1 t a + 1 0 t s a + 1 ρ n ( s ) d s } is equicontinuous on [ 0 , T ] .
(12)
Proof It follows from Lemma 2 that ρ n 1 L for n N , where L is a positive constant. Recall that by Lemma 3(i), { 1 t a + 1 0 t s a + 1 ρ n ( s ) d s } C [ 0 , T ] . Let us assume that (12) does not hold. Then there exist ε > 0 , { k n } N and { ξ n } , { η n } [ 0 , T ] such that lim n k n = , lim n ( ξ n η n ) = 0 and
| 1 ξ n a + 1 0 ξ n s a + 1 ρ k n ( s ) d s 1 η n a + 1 0 η n s a + 1 ρ k n ( s ) d s | ε for  n N .
(13)
Since { ξ n } and { η n } are bounded sequences, we may assume that they are convergent, and lim n ξ n = τ = lim n η n . If τ = 0 , then (cf. (9))
lim n 1 ξ n a + 1 0 ξ n s a + 1 ρ k n ( s ) d s = 0 , lim n 1 η n a + 1 0 η n s a + 1 ρ k n ( s ) d s = 0 ,
which contradicts (13). Let τ ( 0 , T ] . Since lim n ( 1 ξ n a + 1 1 η n a + 1 ) = 0 and since the uniform integrability of the sequence { ρ n } on [ 0 , T ] results in
lim n | η n ξ n | ρ k n ( t ) | d t | = 0 ,
we conclude from the relation
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equs_HTML.gif
that
lim n | 1 ξ n a + 1 0 ξ n s a + 1 ρ k n ( s ) d s 1 η n a + 1 0 η n s a + 1 ρ k n ( s ) d s | = 0 .

The last equality contradicts (13). Consequently, (12) holds and the result follows. □

Lemma 5 Let ρ L 1 [ 0 , T ] . Then
| t T 1 s a + 2 ( 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s | 2 ρ 1 a + 1 for t [ 0 , T ] .
(14)
Proof Since (cf. (10))
| t T 1 s a + 2 ( 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s | 1 a + 1 ( 0 t | ρ ( s ) | d s + t T | ρ ( s ) | d s + ρ 1 ) = 2 ρ 1 a + 1

for t [ 0 , T ] , estimate (14) holds. □

3 Limit properties of solutions to Eq. (7a)

Here, we investigate asymptotic properties of solutions of (7a). We also provide a related integral equation this solution satisfies.

Lemma 6 Let (H1) hold. Let u A C loc 1 ( 0 , T ] satisfy Eq. (7a) for a.e. t [ 0 , T ] and L : = sup { | u ( t ) | + | u ( t ) | : t ( 0 , T ] } < . Then u can be extended on [ 0 , T ] with u A C 1 [ 0 , T ] , and there exists c R such that the integral equation
u ( t ) = t ( c t T 1 s a + 2 ( 0 s ξ a + 1 f n ( ξ , u ( ξ ) , u ( ξ ) ) d ξ ) d s )
(15)

holds for t [ 0 , T ] .

Proof Choose n N and denote by ρ ( t ) = f n ( t , u ( t ) , u ( t ) ) for a.e. t [ 0 , T ] . In order to prove that ρ L 1 [ 0 , T ] , define for m N , 1 m < T ,
v m ( t ) : = { u ( t ) if  t ( 1 m , T ] , u ( 1 m ) if  t [ 0 , 1 m ] , w m ( t ) : = { u ( t ) if  t ( 1 m , T ] , u ( 1 m ) if  t [ 0 , 1 m ] ,
and
ρ m ( t ) : = f n ( t , v m ( t ) , w m ( t ) ) for a.e.  t [ 0 , T ] .

Then ρ m L 1 [ 0 , T ] and lim m ρ m ( t ) = ρ ( t ) for a.e. t [ 0 , T ] . Moreover, | ρ m ( t ) | μ ( t ) for a.e. t [ 0 , T ] and all m N , where μ ( t ) = sup { | f n ( t , x , y ) | : | x | L , | y | L } L 1 [ 0 , T ] . Consequently, by the Lebesgue dominated convergence theorem, ρ L 1 [ 0 , T ] .

We now discuss the linear Euler differential equation
v ( t ) + a t v ( t ) a t 2 v ( t ) = ρ ( t ) .
(16)

Let H be the function given in Lemma 3. By Lemma 3(ii), H can be extended on [ 0 , T ] with H A C 1 [ 0 , T ] and −H satisfies (16) for a.e. t [ 0 , T ] . Therefore, each function v A C loc 1 ( 0 , T ] which satisfies Eq. (16) a.e. on [ 0 , T ] has the form v ( t ) = c t + d t a H ( t ) for t ( 0 , T ] , with some c , d R . By assumption we know that u A C loc 1 ( 0 , T ] satisfies (16) a.e. on [ 0 , T ] , and therefore there exist c , d R such that u ( t ) = c t + d t a H ( t ) , t ( 0 , T ] . Since by assumption | u | L on ( 0 , T ] , we have d = 0 . Consequently, the function u can be extended on the interval [ 0 , T ] in the class A C 1 [ 0 , T ] and (15) holds on [ 0 , T ] . □

Corollary 1 Let (H1) hold. Let u A C 1 [ 0 , T ] be a solution of Eq. (7a). Then there exists a constant c R such that equality (15) is satisfied for t [ 0 , T ] .

Proof The result holds by Lemma 6 with sup { | u ( t ) | + | u ( t ) | : t [ 0 , T ] } < . □

Remark 2 Corollary 1 says that the set of all solutions u A C 1 [ 0 , T ] of Eq. (7a) depends on one parameter c R and u ( 0 ) = 0 .

4 Auxiliary regular problems

In order to prove the solvability of problem (7a) and (7b), we first have to investigate the problem
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ19_HTML.gif
(17a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ20_HTML.gif
(17b)

depending on the parameter λ. Here, ε > 0 is from (H2) and n N .

The following result shows that the solvability of problem (17a) and (17b) is equivalent to the solvability of an integral equation in the set C 1 [ 0 , T ] .

Lemma 7 Let (H1) hold. Then u is a solution of problem (17a) and (17b) if and only if u is a solution of the integral equation
u ( t ) = t t T 1 s a + 2 ( 0 s ξ a + 1 [ λ f n ( ξ , u ( ξ ) , u ( ξ ) ) ( 1 λ ) ε ] d ξ ) d s
(18)

in the set C 1 [ 0 , T ] .

Proof Let u be a solution of Eq. (17a). Then u A C 1 [ 0 , T ] , and by Corollary 1 (with f n replaced by λ f n ( 1 λ ) ε ), there exists c R such that the equation
u ( t ) = t ( c t T 1 s a + 2 ( 0 s ξ a + 1 [ λ f n ( ξ , u ( ξ ) , u ( ξ ) ) ( 1 λ ) ε ] d ξ ) d s )

holds for t [ 0 , T ] . Hence, u ( 0 ) = 0 and u ( T ) = 0 if and only if c = 0 . Consequently, if u is a solution of problem (17a) and (17b), then u is a solution of Eq. (18) in C 1 [ 0 , T ] .

Let u be a solution of Eq. (18) in C 1 [ 0 , T ] . Then f n ( t , u ( t ) , u ( t ) ) L 1 [ 0 , T ] . Hence, Lemma 3(ii) (with ρ replaced by λ f n + ( 1 λ ) ε ) guarantees that u A C 1 [ 0 , T ] and u is a solution of Eq. (17a). Moreover, u ( 0 ) = u ( T ) = 0 . Consequently, u is a solution of problem (17a) and (17b) which completes the proof. □

The following results provide bounds for solutions of problem (17a) and (17b).

Lemma 8 Let (H1)-(H3) hold. Then there exists a positive constant S (independent of n N and λ [ 0 , 1 ] ) such that for all solutions u of problem (17a) and (17b), the estimates
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ22_HTML.gif
(19)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ23_HTML.gif
(20)
hold. Moreover, for any solution u of problem (17a) and (17b), there exists ξ ( 0 , T ) such that
| u ( t ) | 2 ε a + 2 | t ξ | for t [ 0 , T ] .
(21)
Proof Let u be a solution of problem (17a) and (17b). Then by Lemma 7, equality (18) holds for t [ 0 , T ] . Since by (5), λ f n ( t , u ( t ) , u ( t ) ) ( 1 λ ) ε ε for a.e. t [ 0 , T ] , the relation
u ( t ) ε t t T 1 s a + 2 ( 0 s ξ a + 1 d ξ ) d s = ε a + 2 t ( T t ) for  t [ 0 , T ]
follows from (18). Hence, g ( u ( t ) ) g ( ε a + 2 t ( T t ) ) for t ( 0 , T ) because g is nonincreasing on R + . Due to Remark 1, L = g ( ε a + 2 t ( T t ) ) 1 < , which means that
0 T g ( u ( t ) ) d t L .
(22)

It is clear that L is independent of the choice of solution u to problem (17a) and (17b) and independent of n N , λ [ 0 , 1 ] .

We now show that inequality (21) holds for some ξ ( 0 , T ) . Differentiation of (18) gives
u ( t ) = a t a + 2 0 t s a + 1 [ λ f n ( s , u ( s ) , u ( s ) ) ( 1 λ ) ε ] d s + λ f n ( t , u ( t ) , u ( t ) ) ( 1 λ ) ε for a.e.  t [ 0 , T ] .
(23)
Since a < 0 , it follows from (5) and (23) that
u ( t ) a ε t a + 2 0 t s a + 1 d s ε = 2 ε a + 2 for a.e.  t [ 0 , T ] .
Hence, u is decreasing on [ 0 , T ] , and therefore u vanishes at a unique point ξ ( 0 , T ) due to u ( 0 ) = u ( T ) = 0 . The inequality (21) now follows from the relations
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equaa_HTML.gif
Hence, r ( | u ( t ) | ) r ( 2 ε a + 2 | t ξ | ) on [ 0 , T ] { ξ } , and
0 T ( r ( | u ( t ) | ) ) d t 0 ξ ( r ( 2 ε a + 2 ( ξ t ) ) ) d t + ξ T ( r ( 2 ε a + 2 ( t ξ ) ) ) d t < a + 2 ε 0 ( 2 ε T ) / ( a + 2 ) r ( s ) d s = V for  t [ 0 , T ] .
In particular,
r ( | u ( t ) | ) 1 < V .
(24)
Let W = a + 3 a + 1 . Taking into account (6), (9), (18), (22), (24), and Lemma 5, we obtain
| u ( t ) | = | t T 1 s a + 2 ( 0 s ξ a + 1 [ λ f n ( ξ , u ( ξ ) , u ( ξ ) ) ( 1 λ ) ε ] d ξ ) d s + 1 t a + 1 0 t s a + 1 [ λ f n ( s , u ( s ) , u ( s ) ) ( 1 λ ) ε ] d s | W 0 T | λ f n ( t , u ( t ) , u ( t ) ) ( 1 λ ) ε | d t W 0 T ( φ ( t ) h ( 1 + u , 1 + u ) + g ( u ( t ) ) + r ( | u ( t ) | ) ) d t W ( h ( 1 + u , 1 + u ) φ 1 + L + V ) , t [ 0 , T ] .
It follows from u ( t ) = 0 t u ( s ) d s for t [ 0 , T ] ,
u T u ,
(25)
and therefore, we have
u K h ( 1 + T u , 1 + u ) + M ,
(26)
where K = W φ 1 and M = W ( L + V ) . By (H3),
lim z ( 1 / z ) ( K h ( 1 + T z , 1 + z ) + M ) = 0 .

Consequently, there exists S > 0 such that K h ( 1 + T z , 1 + z ) + M < z for z S . Now, due to (26), u < S , and therefore, by (25), u < S T . □

We are now in the position to prove the existence result for problem (7a) and (7b).

Lemma 9 Let (H1)-(H3) hold. Then for each n N , problem (7a) and (7b) has a solution u satisfying inequalities (19)-(21), where S is a positive constant independent of n.

Proof Let S be a positive constant in Lemma 8 and let us define
Ω : = { x C 1 [ 0 , T ] : x < S T , x < S } .
Then Ω is an open and bounded subset of the Banach space C 1 [ 0 , T ] . Keeping in mind Lemma 3, define an operator K : [ 0 , T ] × Ω ¯ C 1 [ 0 , T ] by the formula
K ( λ , x ) ( t ) = t t T 1 s a + 2 ( 0 s ξ a + 1 [ λ f n ( ξ , x ( ξ ) , x ( ξ ) ) ( 1 λ ) ε ] d ξ ) d s .
(27)
By Lemma 7, any fixed point of the operator K ( 1 , ) is a solution of problem (7a) and (7b). In order to show the existence of a fixed point of K ( 1 , ) , we apply Lemma 1 with E = C 1 [ 0 , T ] , U = Ω , F = K ( 1 , ) and = ε a + 2 t ( T t ) . Especially, we show that
  1. (i)

    K ( 1 , ) : Ω ¯ C 1 [ 0 , T ] is a compact operator, and

     
  2. (ii)

    K ( λ , x ) x for each λ ( 0 , 1 ] and x Ω .

     
We first verify that K ( 1 , ) is a continuous operator. To this end, let { x m } Ω ¯ be a convergent sequence, and let lim m x m = x in C 1 [ 0 , T ] . Let
r m ( t ) : = f n ( t , x m ( t ) , x m ( t ) ) f n ( t , x ( t ) , x ( t ) ) for a.e.  t [ 0 , T ] .
It follows from Lemma 5 and (9) that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equag_HTML.gif
for t [ 0 , T ] . Here K ( 1 , x ) = d d t K ( 1 , x ) . In particular, for m N ,
K ( 1 , x m ) K ( 1 , x ) 2 T r m 1 a + 1 , K ( 1 , x m ) K ( 1 , x ) ( a + 3 ) r m 1 a + 1 .
(28)
Since lim m f n ( t , x m ( t ) , x m ( t ) ) = f n ( t , x ( t ) , x ( t ) ) for a.e. t [ 0 , T ] and there exists ρ L 1 [ 0 , T ] such that
| f n ( t , x m ( t ) , x m ( t ) ) | ρ ( t ) for a.e.  t [ 0 , T ]  and all  m N ,
we have lim m r m 1 = 0 by the Lebesgue dominated convergence theorem. Hence, by (28), K ( 1 , ) is a continuous operator. We now show that the set K ( 1 , Ω ¯ ) is relatively compact in C 1 [ 0 , T ] . It follows from f n Car ( [ 0 , T ] × R 2 ) and Ω ¯ bounded in C 1 [ 0 , T ] that there exists μ L 1 [ 0 , T ] such that
| f n ( t , x ( t ) , x ( t ) ) | μ ( t ) for a.e.  t [ 0 , T ]  and all  x Ω ¯ .
Then by Lemma 5 and (9), the inequalities
| K ( 1 , x ) ( t ) | 2 T μ 1 a + 1 , | K ( 1 , x ) ( t ) | ( a + 3 ) μ 1 a + 1
are satisfied for t [ 0 , T ] and x Ω ¯ , and therefore, the set K ( 1 , Ω ¯ ) is bounded in C 1 [ 0 , T ] . Moreover, the relation
| K ( 1 , x ) ( t ) | = | a t a + 2 0 t s a + 1 f n ( s , x ( s ) , x ( s ) ) d s + f n ( t , x ( t ) , x ( t ) ) | | a | t a + 2 0 t s a + 1 μ ( s ) d s + μ ( t ) L 1 [ 0 , T ]

holds for a.e. t [ 0 , T ] and all x Ω ¯ (cf. (9)). Consequently, the set { K ( 1 , x ) : x Ω ¯ } is equicontinuous on [ 0 , T ] . Hence, the set K ( 1 , Ω ¯ ) is relatively compact in C 1 [ 0 , T ] by the Arzelà-Ascoli theorem. As a result, K ( 1 , ) is a compact operator and the condition (i) follows.

Due to the fact that by Lemma 7 any fixed point u of the operator K ( λ , ) is a solution of problem (17a) and (17b), Lemma 8 guarantees that u satisfies inequality (20). Therefore, K has property (ii). Consequently, by Lemmas 1 and 8, for each n N , problem (7a) and (7b) has a solution u satisfying estimates (19)-(21). □

Let u n be a solution of problem (7a) and (7b) for n N . The following property of the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } is an important prerequisite for solving problem (1a) and (1b).

Lemma 10 Let (H1)-(H3) hold. Let u n be a solution of problem (7a) and (7b) for n N . Then the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } is uniformly integrable on [ 0 , T ] .

Proof

By Lemma 9, the inequalities
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ32_HTML.gif
(29)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ33_HTML.gif
(30)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ34_HTML.gif
(31)
hold, where S is a positive constant and ξ n ( 0 , T ) . Hence, by (3) and (4),
0 < f n ( t , u n ( t ) , u n ( t ) ) φ ( t ) h ( 1 + S T , 1 + S ) + g ( t ) + r ( 2 ε a + 2 | t ξ n | )
(32)

for a.e. t [ 0 , T ] , where g ( t ) = g ( ε a + 2 t ( T t ) ) L 1 [ 0 , T ] , see Remark 1. Since the sequence { r ( 2 ε a + 2 | t ξ n | ) } is uniformly integrable on [ 0 , T ] (cf. [[20], criterion A.4], [21, 22]), it follows from (32) that { | f n ( t , u n ( t ) , u n ( t ) ) | } is uniformly integrable on [ 0 , T ] and the result follows. □

5 The existence result for BVP (1a) and (1b)

This section is devoted to the main result on the existence of positive solutions to the original BVP (1a) and (1b).

Theorem 1 Let (H1)-(H3) hold. Then problem (1a) and (1b) has at least one positive solution.

Proof By Lemma 9, for each n N , problem (7a) and (7b) has a solution u n satisfying inequalities (29)-(31), where S is a positive constant and ξ n ( 0 , T ) . Moreover, by Lemma 10, the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } is uniformly integrable on [ 0 , T ] . We now prove that { u n } is equicontinuous on [ 0 , T ] . Since u n is a fixed point of the operator K ( 1 , ) given in (27), the equality
u n ( t ) = a t a + 2 0 t s a + 1 f n ( s , u n ( s ) , u n ( s ) ) d s + f n ( t , u n ( t ) , u n ( t ) )
holds for a.e. t [ 0 , T ] and all n N . Let 0 t 1 < t 2 T . Then
| u n ( t 2 ) u n ( t 1 ) | = | a t 1 t 2 1 s a + 2 ( 0 s ξ a + 1 f n ( ξ , u n ( ξ ) , u n ( ξ ) ) d ξ ) d s + t 1 t 2 f n ( s , u n ( s ) , u n ( s ) ) d s | .
Let r n ( t ) = 1 t a + 1 0 t s a + 1 f n ( s , u n ( s ) , u n ( s ) ) d s . By Lemma 3(i), { r n } C [ 0 , T ] and r n ( 0 ) = 0 . Integrating by parts yields
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equan_HTML.gif
and
| u n ( t 2 ) u n ( t 1 ) | = | a a + 1 ( r n ( t 2 ) r n ( t 1 ) ) + 1 a + 1 t 1 t 2 f n ( s , u ( s ) , u ( s ) ) d s |
(33)
follows. By Lemma 4 (for ρ n ( t ) = f n ( t , u n ( t ) , u n ( t ) ) ), the sequence { r n } is equicontinuous on [ 0 , T ] . Since the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } is uniformly integrable on [ 0 , T ] , the sequence { 0 t f n ( s , u ( s ) , u ( s ) ) d s } is equicontinuous on [ 0 , T ] . Hence, it follows from (33), that { u n } is equicontinuous on [ 0 , T ] . We summarize: { u n } is bounded in C 1 [ 0 , T ] and { u n } is equicontinuous on [ 0 , T ] . Also, { ξ n } ( 0 , T ) . Using appropriate subsequences, if necessary, we can assume, by the Arzelà-Ascoli theorem and the Bolzano-Weierstrass theorem, that { u n } is convergent in C 1 [ 0 , T ] and { ξ n } is convergent in . Let lim n u n = : u and lim n ξ n = : ξ . With n in (29)-(31), we conclude
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equao_HTML.gif
In addition, u ( 0 ) = 0 , u ( T ) = 0 . Since
lim n f n ( t , u n ( t ) , u n ( t ) ) = f ( t , u ( t ) , u ( t ) ) for a.e.  t [ 0 , T ] ,
(34)
it follows from Lemma 2 that
lim n f n ( t , u n ( t ) , u n ( t ) ) f ( t , u ( t ) , u ( t ) ) 1 = 0
and f ( t , u ( t ) , u ( t ) ) L 1 [ 0 , T ] . We now deduce from the inequality (cf. Lemma 5)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equaq_HTML.gif
that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equar_HTML.gif
Taking the limit n in
u n ( t ) = t t T 1 s a + 2 ( 0 s ξ a + 1 f n ( ξ , u n ( ξ ) , u n ( ξ ) ) d ξ ) d s ,
we have
u ( t ) = t t T 1 s a + 2 ( 0 s ξ a + 1 f ( ξ , u ( ξ ) , u ( ξ ) ) d ξ ) d s for  t [ 0 , T ] .
(35)
Hence,
u ( t ) + a t u ( t ) a t 2 u ( t ) = f ( t , u ( t ) , u ( t ) ) for a.e.  t [ 0 , T ] ,

and u A C 1 [ 0 , T ] by Lemma 3(ii). This means that u is a positive solution of problem (1a) and (1b) and the result follows. □

6 Numerical simulations

For the numerical simulation, we choose T = 1 and use an alternative formulation of problem (1a) and (1b),
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ39_HTML.gif
(36a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ40_HTML.gif
(36b)

where c > 0 is a parameter. We can use the above formulation because problem (1a) and (1b) is solvable for f satisfying the assumptions of Theorem 1 and, therefore, solutions of problem (1a) and (1b) can be computed as solutions of problem (36a) and (36b) using the proper value c ( 0 , c ) depending on f. The values c are provided for given f in Examples 1 and 2, below.

The reason for changing the boundary conditions from (1b) to (36b) is that the differential equation (36a) subject to (1b) is not well posed; see [23]. However, to enable successful numerical treatment, well-posedness of the model is crucial. This property means that Eq. (36a) subject to proper boundary conditions has at least a locally unique solution,a and this solution depends continuously on the problem data. The well-posedness of the problem is important for two reasons. First of all, it allows to express errors in the solution of the analytical problem in terms of modeling errors and data errors (all measured via appropriate norms). Therefore, when the errors in the data become smaller due to more precise modeling or smaller measurement inaccuracies, the errors in the solution will decrease. The second reason is that the well-posedness decides if the numerical simulation will be at all successful. If the analytical problem is ill-posed, then the inevitable round-off errors can become extremely magnified and fully spoil the accuracy of the approximation.

In what follows, we work with f ( t , x , y ) = q ( t ) + h ( x , y ) for a.e. t [ 0 , 1 ] and all x R + , y R { 0 } and, according to the next numerical approach (see Section 6.2), we consider Eq. (36a), where h 0 , that is,
u ( t ) + a t u ( t ) a t 2 u ( t ) = q ( t ) .
(37)
By [23], problem (37), (36b) is well posed and therefore it is suitable for the numerical treatment. To see this, we need to look at a general solution of the homogeneous equation
u ( t ) + a t u ( t ) a t 2 u ( t ) = 0 , t [ 0 , 1 ] .
(38)
If we set u ( t ) = t λ , we arrive at the characteristic polynomial of (38),
λ 2 ( 1 a ) λ a = 0

whose roots λ 1 = 1 and λ 2 = a are positive. Therefore, conditions for u and u can be prescribed at t = 1 as it is done in (36b).

6.1 MATLAB Code bvpsuite

To illustrate the analytical results discussed in the previous section, we solved numerically examples of the form (36a) and (36b) using a MATLAB™ software package bvpsuite designed to solve BVPs in ODEs and differential algebraic equations. The solver routine is based on a class of collocation methods whose orders may vary from two to eight. Collocation has been investigated in the context of singular differential equations of first and second order in [24, 25], respectively. This method could be shown to be robust with respect to singularities in time and retains its high convergence order in the case that the analytical solution is appropriately smooth. The code also provides an asymptotically correct estimate for the global error of the numerical approximation. To enhance the efficiency of the method, a mesh adaptation strategy is implemented, which attempts to choose grids related to the solution behavior in such a way that the tolerance is satisfied with the least possible effort. Error estimate procedure and the mesh adaptation work dependably provided that the solution of the problem and its global error are appropriately smooth.b The code and the manual can be downloaded from http://www.math.tuwien.ac.at/~ewa. For further information, see [26]. This software proved useful for the approximation of numerous singular BVPs important for applications; see, e.g., [3, 9, 27, 28].

6.2 Preliminaries

Before dealing with two nonlinear models specified in Sections 6.3 and 6.4, we have to compute numerical solutions for a simpler linearc model of the form
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ43_HTML.gif
(39a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ44_HTML.gif
(39b)
where a was chosen as a = 0.1 , 0.5 , 0.9 . Since in this case the exact solution is given, u ( t ) = 2 t ( 1 t ) a + 1.5 , the value u ( 1 ) is available, u ( 1 ) = 0.72 , 1 , 1.67 , respectively. In Figure 1, the numerical solutions of BVPs (39a) and (39b) are shown. They will be used as starting values for the numerical solution of Examples 1 and 2; see Sections 6.3 and 6.4, respectively. All numerical results have been obtained using collocation with five Gaussian collocation points on an equidistant grid (justified by a very simple solution structure) with the step size 0.01.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Fig1_HTML.jpg
Figure 1

Problem ( 39a ) and ( 39b ): Numerical solutions for different values of a .

6.3 Example 1

We first investigate the following problem:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ45_HTML.gif
(40a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ46_HTML.gif
(40b)
The nonlinearity f in (40a) has the form
f ( t , x ) = 1 t x 1 3
(41)
and it satisfies (H1)-(H3) with ε = 1 , φ ( t ) = 1 t for t ( 0 , 1 ] and
h ( x , y ) = 1 , g ( x ) = x 1 3 , r ( y ) = 0 for  x , y R + .

It follows from Theorem 1 that there exists at least one value of c > 0 such that the related solution u of problem (40a) and (40b) with u ( 1 ) = c is positive on ( 0 , 1 ) with u ( 0 ) = 0 . Using formula (35), we now determine an interval ( 0 , c ) ( 0 , ) containing all admissible values of c.

Let u be a solution of problem (1a) and (1b) with f from (41). Then by (35), we obtain
| u ( t ) | t t 1 1 s a + 2 ( 0 s ξ a + 1 ( 1 ξ + u 1 3 ) d ξ ) d s < 4 2 a + 3 + u 1 3 4 ( a + 2 ) , t [ 0 , 1 ] .
Therefore,
u < 4 2 a + 3 + u 1 3 4 ( a + 2 ) .
(42)
Let K > 0 satisfy
K = 4 2 a + 3 + K 1 3 4 ( a + 2 ) .
(43)
Then (42) implies u < K and due to (35) and (41),
u ( 1 ) = 0 1 t a + 1 f ( t , u ( t ) , u ( t ) ) d t = 0 1 t a + 1 ( 1 t u 1 3 ( t ) ) d t 0 1 ( t a + 1 2 + K 1 3 t a + 1 ) d t = 2 2 a + 3 K 1 3 1 a + 2 .
Consequently,
c = 2 2 a + 3 + K 1 3 1 a + 2 .
(44)
In order to solve the nonlinear problem (40a) and (40b), we first have to solve a series of auxiliary problems for parameter-dependent differential equations
u ( t ) + a t u ( t ) a t 2 u ( t ) = 1 t δ u 1 3 ( t ) , t ( 0 , 1 ] , δ > 0 .
(45)

We begin the calculations with δ = 0 and increase its value gradually until we arrive at δ = 1 ; cf. (40a). In each step we use the solution of the previous problem to solve the next one. The aim is to find a good starting value for both the solution u and the value u ( 1 ) before solving the BVP (40a) and (40b), i.e., find the final value of c ( 0 , c ) such that u ( 0 ) = 0 .

In the case of Example 1 and a = 0.1 , this chain has the following structure:
  1. 1.

    Numerical approximation of BVP (39a) and (39b) is used as an initial guess for ODE (45) with δ = 0 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 0.5 .

     
  2. 2.

    Use the above approximation as an initial guess for ODE (45) with δ = 0.01 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 1.0 .

     
  3. 3.

    Use the above approximation as an initial guess for ODE (45) with δ = 0.1 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 1.0 .

     
  4. 4.

    Use the above approximation as an initial guess for ODE (45) with δ = 1.0 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 1.0 .

     
After the last step, we have solved problem (40a) and (40b) subject to boundary conditions u ( 1 ) = 0 , u ( 1 ) = 1.0 . In this case, the value of u ( 0 ) was not small enough to consider it a reasonable approximation for u ( 0 ) = 0 . Therefore, we use a shooting idea combined with a bisection strategy to find a better value for c = u ( 1 ) . The complete numerical results for Example 1 can be found in Table 1 and Figure 2.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Fig2_HTML.jpg
Figure 2

Problem ( 40a ) and ( 40b ): Numerical solutions for different values of a . Values of u ( 0 ) 10 14 .

Table 1

Problem ( 40a ) and ( 40b ): Complete data of the numerical simulation for different values of a

a

K

c

c

u ( 0 )

−0.1

1.5819

1.327538328

1.00569659944

1.40264071382347 E-14

−0.5

2.2173

1.869327784

1.41953539630

1.18953445347016 E-13

−0.9

3.6844

3.070760848

2.33615892300

4.13495149060736 E-14

6.4 Example 2

The above approach has been also accordingly applied for Example 2. Here, we consider the problem
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ52_HTML.gif
(46a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ53_HTML.gif
(46b)
The right-hand side f in Eq. (46a) now reads
f ( t , x ) = 1 t x 1 3
(47)
and has a singularity at x = 0 . The function f satisfies conditions (H1)-(H3) with ε = 1 , φ ( t ) = 1 t for t ( 0 , 1 ] and
h ( x , y ) = 1 , g ( x ) = x 1 3 , r ( y ) = 0 for  x , y R + .
Theorem 1 guarantees the existence of at least one c > 0 such that a solution u of problem (46a) and (46b) is positive on ( 0 , 1 ) and u ( 0 ) = 0 holds. We now again determine an interval ( 0 , c ) ( 0 , ) containing all such values of c. Let u be a solution of problem (1a) and (1b) with f given in (47). Inequality (19) yields
u ( t ) 1 a + 2 t ( 1 t ) , t [ 0 , 1 ] ,
and hence by (35),
u ( 1 ) = 0 1 t a + 1 f ( t , u ( t ) , u ( t ) ) d t = 0 1 t a + 1 ( 1 t u 1 3 ( t ) ) d t 0 1 t a + 1 ( 1 t ( 1 a + 2 t ( 1 t ) ) 1 3 ) d t .
Consequently,
c = 0 1 t a + 1 ( 1 t + ( a + 2 ) 1 / 3 t 1 / 3 ( 1 t ) 1 / 3 ) d t .
(48)
For Example 2, the auxiliary ODE is constructed using ODE (39a),
u ( t ) + a t u ( t ) a t 2 u ( t ) = 1 t δ u 1 3 , t [ 0 , 1 ] , δ > 0 .
(49)
For all values of a, we choose u ( 1 ) ( c , 0 ) and analogously carry out the path-following in δ first. The related chain for a = 0.1 is as follows.
  1. 1.

    Numerical approximation of BVP (39a) and (39b) is used as an initial guess for ODE (49) with δ = 0 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 0.8 .

     
  2. 2.

    Use the above approximation as an initial guess for ODE (49) with δ = 0.01 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 0.8 .

     
  3. 3.

    Use the above approximation as an initial guess for ODE (49) with δ = 0.1 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 1.2 .

     
  4. 4.

    Use the above approximation as an initial guess for ODE (49) with δ = 0.5 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 1.3 .

     
  5. 5.

    Use the above approximation as an initial guess for ODE (49) with δ = 1.0 subject to terminal conditions u ( 1 ) = 0 , u ( 1 ) = 1.7 .

     
After the last step, we have solved BVP (46a) and (46b) subject to boundary conditions u ( 1 ) = 0 , u ( 1 ) = 1.7 , but also, in this case, the value of u ( 0 ) is too large and we have to find a better value for c = u ( 1 ) . The complete numerical results for Example 2 can be found in Table 2 and Figure 3.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Fig3_HTML.jpg
Figure 3

Problem ( 46a ) and ( 46b ): Numerical solutions for different values of a . Values of u ( 0 ) 10 12 .

Table 2

Problem ( 46a ) and ( 46b ): Complete data of the numerical simulation for different values of a

a

c

c

u ( 0 )

−0.1

2.044582190

1.63000971355

4.57015034596816 e-12

−0.5

2.528760387

2.04278888650

1.82943058378521 e-13

−0.9

3.566085670

2.91010561000

1.16498324337590 e-12

7 Conclusions

In the present article, we deal with the existence of positive solutions to the singular Dirichlet problem of the form
u ( t ) + a t u ( t ) a t 2 u ( t ) = f ( t , u ( t ) , u ( t ) ) , u ( 0 ) = 0 , u ( T ) = 0 ,
where a ( 1 , 0 ) , and the nonlinearity f ( t , x , y ) may be singular at the space variables x = 0 and/or y = 0 . The main result for the existence of positive solutions of the above BVP is Theorem 1. It is illustrated by numerical simulations using the MATLAB code bvpsuite, based on polynomial collocation. For the successful numerical treatment, the above problem has to be reformulated to obtain its well-posed form
u ( t ) + a t u ( t ) a t 2 u ( t ) = f ( t , u ( t ) , u ( t ) ) , u ( T ) = 0 , u ( T ) = c .

Here, it is only known that c ( 0 , c ) , where c can be specified depending on functions f arising in Examples 1 and 2. Now, a simple shooting method combined with the bisection idea is used to find c in such a way that u ( 0 ) = 0 .

Endnotes

a This BVP can have more than one solution, but they may not lay close together.

b The required smoothness of higher derivatives is related to the order of the used collocation method.

c The nonlinear term in f has been omitted; see (40a) and (46a).

Declarations

Acknowledgements

Dedicated to Jean Mawhin on the occasion of his 70th birthday.

This research was supported by the grant Matematické modely a struktury, PrF-2012-017. The authors thank the referees for suggestions which improved the paper.

Authors’ Affiliations

(1)
Department of Mathematical Analysis, Faculty of Science, Palacký University
(2)
Department of Analysis and Scientific Computing, Vienna University of Technology

References

  1. Bongiorno V, Scriven LE, Davis HT: Molecular theory of fluid interfaces. J. Colloid Interface Sci. 1967, 57: 462-475.View ArticleGoogle Scholar
  2. Gouin H, Rotoli G: An analytical approximation of density profile and surface tension of microscopic bubbles for Van der Waals fluids. Mech. Res. Commun. 1997, 24: 255-260. 10.1016/S0093-6413(97)00022-0MATHView ArticleGoogle Scholar
  3. Kitzhofer G, Koch O, Lima P, Weinmüller E: Efficient numerical solution of the density profile equation in hydrodynamics. J. Sci. Comput. 2007, 32: 411-424. 10.1007/s10915-007-9141-0MATHMathSciNetView ArticleGoogle Scholar
  4. van der Waals JD, Kohnstamm R 1. In Lehrbuch der Thermodynamik. Salzwasser-Verlag, Leipzig; 1908.Google Scholar
  5. Fife PC Lecture Notes in Biomathematics 28. In Mathematical Aspects of Reacting and Diffusing Systems. Springer, Berlin; 1979.View ArticleGoogle Scholar
  6. Fischer RA: The wave of advance of advantageous genes. Ann. Eugen. 1937, 7: 355-369. 10.1111/j.1469-1809.1937.tb02153.xView ArticleGoogle Scholar
  7. Abraham FF: Homogeneous Nucleation Theory. Academic Press, New York; 1974.Google Scholar
  8. Linde AP: Particle Physics and Inflationary Cosmology. Harwood Academic, Chur; 1990.View ArticleGoogle Scholar
  9. Hammerling R, Koch O, Simon C, Weinmüller EB: Numerical solution of eigenvalue problems in electronic structure computations. J. Comp. Physiol. 2010, 181: 1557-1561.MATHGoogle Scholar
  10. Derrick GH: Comments on nonlinear wave equations as models for elementary particles. J. Math. Phys. 1965, 5: 1252-1254.MathSciNetView ArticleGoogle Scholar
  11. Rachůnková I, Staněk S: Properties of the set of positive solutions to Dirichlet boundary value problems with time singularities. Cent. Eur. J. Math. 2013, 11(1):112-132. 10.2478/s11533-012-0047-1MATHMathSciNetView ArticleGoogle Scholar
  12. Rachůnková, I, Spielauer, A, Staněk, S, Weinmüller, EB: The structure of a set of positive solutions to Dirichlet BVPs with time and space singularities. Georgian Math. J. (to appear)
  13. Gaines RE, Mawhin JL Lecture Notes Math. 568. In Coincidence Degree and Nonlinear Differential Equations. Springer, Berlin; 1977.Google Scholar
  14. Mawhin J Sémin. Math. Sup. 92. In Points Fixes, Points Critiques et Problèmes aux Limites. Presses Univ. Montréal, Montréal; 1985.Google Scholar
  15. Mawhin J Lecture Notes Math. 1537. In Topological Degree and Boundary Value Problems for Nonlinear Differential Equations. Springer, Berlin; 1993.View ArticleGoogle Scholar
  16. Agarwal RP, Meehan M, O’Regan D: Fixed Point Theory and Applications. Cambridge University Press, Cambridge; 2001.MATHView ArticleGoogle Scholar
  17. Bartle RG Graduate Studies in Mathematics 32. In A Modern Theory of Integration. AMS, Providence; 2001.View ArticleGoogle Scholar
  18. Hewitt E, Stromberg K: Real and Abstract Analysis. Springer, New York; 1965.MATHView ArticleGoogle Scholar
  19. Natanson IP Mathematische Lehrbücher und Monographien. In Theorie der Funktionen Einer Reelen Veränderlichen. 4th edition. Akademie Verlag, Berlin; 1975. Herausgegeben von Karl BögelGoogle Scholar
  20. Rachůnková I, Staněk S, Tvrdý M: Solvability of Nonlinear Singular Problems for Ordinary Differential Equations. Hindawi Publishing Corporation, New York; 2009.View ArticleGoogle Scholar
  21. Agarwal RP, O’Regan D, Rachůnková I, Staněk S: Two-point higher-order BVPs with singularities in phase variables. Comput. Math. Appl. 2003, 46: 1799-1826. 10.1016/S0898-1221(03)90238-0MATHMathSciNetView ArticleGoogle Scholar
  22. Rachůnková I, Staněk S: A singular boundary value problem for odd-order differential equations. J. Math. Anal. Appl. 2004, 291: 741-756. 10.1016/j.jmaa.2003.11.040MATHMathSciNetView ArticleGoogle Scholar
  23. Weinmüller EB: On the boundary value problems of ordinary second order differential equations with a singularity of the first kind. SIAM J. Math. Anal. 1984, 15: 287-307. 10.1137/0515023MATHMathSciNetView ArticleGoogle Scholar
  24. de Hoog F, Weiss R: Collocation methods for singular boundary value problems. SIAM J. Numer. Anal. 1978, 15: 198-217. 10.1137/0715013MATHMathSciNetView ArticleGoogle Scholar
  25. Weinmüller EB: Collocation for singular boundary value problems of second order. SIAM J. Numer. Anal. 1986, 23: 1062-1095. 10.1137/0723074MATHMathSciNetView ArticleGoogle Scholar
  26. Kitzhofer G, Pulverer G, Simon C, Koch O, Weinmüller E:The new MATLAB solver BVPSUITE for the solution of singular implicit BVPs. J. Numer. Anal. Ind. Appl. Math. 2010, 5: 113-134.Google Scholar
  27. Budd C, Koch O, Weinmüller E: From nonlinear PDEs to singular ODEs. Appl. Numer. Math. 2006, 56: 413-422. 10.1016/j.apnum.2005.04.012MATHMathSciNetView ArticleGoogle Scholar
  28. Rachůnková I, Pulverer G, Weinmüller E: A unified approach to singular problems arising in the membrane theory. Appl. Math. 2010, 55: 47-75. 10.1007/s10492-010-0002-zMATHMathSciNetView ArticleGoogle Scholar

Copyright

© Rachůnková et al.; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. 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 cited.