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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equa_HTML.gif

where a ( 1 , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq1_HTML.gif. The nonlinearity f ( t , x , y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq2_HTML.gif may be singular for the space variables x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq3_HTML.gif and/or y = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq4_HTML.gif. Moreover, since a 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq5_HTML.gif, the differential operator on the left-hand side of the differential equation is singular at t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq6_HTML.gif. 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq7_HTML.gif, a ( 1 , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq1_HTML.gif and f satisfies the local Carathéodory conditions on [ 0 , T ] × D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq8_HTML.gif, where D : = R + × R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq9_HTML.gif and R + : = ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq10_HTML.gif, R 0 : = R { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq11_HTML.gif. Let us recall that a function h : [ 0 , T ] × A R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq12_HTML.gif, A R × R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq13_HTML.gif, satisfies the local Carathéodory conditions on [ 0 , T ] × A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq14_HTML.gif if
  1. (i)

    h ( , x , y ) : [ 0 , T ] R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq15_HTML.gif is measurable for all ( x , y ) A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq16_HTML.gif,

     
  2. (ii)

    h ( t , , ) : A R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq17_HTML.gif is continuous for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq18_HTML.gif,

     
  3. (iii)
    for each compact set U A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq19_HTML.gif, there exists a function m U L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq20_HTML.gif such that
    | h ( t , x , y ) | m U ( t ) for a.e.  t [ 0 , T ]  and all  ( x , y ) U . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equb_HTML.gif
     
For such functions, we use the notation h Car ( [ 0 , T ] × A ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq21_HTML.gif. Moreover, f ( t , x , y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq2_HTML.gif may become singular when the space variables x and/or y vanish, which means that f ( t , x , y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq2_HTML.gif may become unbounded for x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq3_HTML.gif and a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq22_HTML.gif and all y R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq23_HTML.gif, and/or it may be unbounded for y = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq4_HTML.gif and a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq22_HTML.gif and all x R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq24_HTML.gif. Finally, since a 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq5_HTML.gif, Eq. (1a) has a singularity of the first kind at the time variable t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq6_HTML.gif because
0 T 1 t d t = , 0 T 1 t 2 d t = . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equc_HTML.gif

The differential operator on the left-hand side of Eq. (1a) can be equivalently written as ( t a ( t a u ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq25_HTML.gif and, after the substitution x ( t ) = t a u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq26_HTML.gif, it takes the form ( t a x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq27_HTML.gif, 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq28_HTML.gif which is fundamentally different from the case a ( , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq29_HTML.gif. 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq28_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq4_HTML.gif. 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq2_HTML.gif has a singularity at y = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq4_HTML.gif. Then there exists t 0 ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq30_HTML.gif such that u ( t 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq31_HTML.gif, u ( t 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq32_HTML.gif and hence f is unbounded in a neighborhood of the point ( t 0 , u ( t 0 ) , u ( t 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq33_HTML.gif. Unfortunately, we do not know the exact position of t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq34_HTML.gif and therefore, it is not possible to construct a universal Lebesgue integrable majorant for all functions f ( t , u n ( t ) , u n ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq35_HTML.gif, where u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq36_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq37_HTML.gif, where D = R + × R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq38_HTML.gif.

(H2) There exists an ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq39_HTML.gif such that
f ( t , x , y ) ε for a.e.  t [ 0 , T ]  and all  ( x , y ) D . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equd_HTML.gif
(H3) For a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif and all ( x , y ) D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq41_HTML.gif, the estimate
f ( t , x , y ) φ ( t ) h ( x , | y | ) + g ( x ) + r ( | y | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Eque_HTML.gif
holds, where φ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq42_HTML.gif, h C ( [ 0 , ) × [ 0 , ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq43_HTML.gif, g , r C ( R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq44_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equg_HTML.gif

we denote the norms in C [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq45_HTML.gif and L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq46_HTML.gif, respectively. A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq47_HTML.gif denotes the set of functions whose first derivative is absolutely continuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif, while A C loc 1 ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq49_HTML.gif is the set of functions having absolutely continuous first derivative on each compact subinterval of ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq50_HTML.gif. We use the symbol meas ( M ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq51_HTML.gif to denote the Lebesgue measure of .

Definition 1 We say that a function u A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq52_HTML.gif is a positive solution of problem (1a) and (1b) if u > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq53_HTML.gif on ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq54_HTML.gif, u satisfies the boundary conditions (1b) and (1a) holds for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq18_HTML.gif.

Remark 1 Let a function g have the properties specified in (H3). Then for each b , c R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq55_HTML.gif, 0 b g ( c s ) d s < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq56_HTML.gif, 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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equh_HTML.gif
that
0 T g ( c t ( T t ) ) d t < for each  c R + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ3_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq57_HTML.gif we define functions f n : [ 0 , T ] × ( R × R 0 ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq58_HTML.gif and f n : [ 0 , T ] × R 2 R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq59_HTML.gif by
f n ( t , x , y ) = { f ( t , x , y ) if  x 1 n , f ( t , 1 n , y ) if  x < 1 n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equi_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equj_HTML.gif
respectively. Then it follows from (H1) that f n Car ( [ 0 , T ] × R 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq60_HTML.gif 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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ6_HTML.gif
(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 ] . } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ7_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq61_HTML.gif. Assume that F : U ¯ E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq62_HTML.gif is a compact operator. Then either

A1: has a fixed point in U ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq63_HTML.gif, or

A2: there exists an element u U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq64_HTML.gif and λ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq65_HTML.gif with u = λ F ( u ) + ( 1 λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq66_HTML.gif.

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

Lemma 2 Let { ρ n } L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq67_HTML.gif and let lim n ρ n ( t ) = ρ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq68_HTML.gif for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif. Then the following statements are equivalent:
  1. (i)

    ρ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq69_HTML.gif and lim n ρ n ρ 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq70_HTML.gif,

     
  2. (ii)

    the sequence { ρ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq71_HTML.gif is uniformly integrable on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif.

     
We recall that a sequence { ρ n } L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq72_HTML.gif is called uniformly integrable on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif if for any ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq39_HTML.gif there exists δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq73_HTML.gif such that if M [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq74_HTML.gif and meas ( M ) < δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq75_HTML.gif, then
M | ρ n ( t ) | d t < ε , n N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equk_HTML.gif

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq1_HTML.gif.

2 Preliminaries

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

Lemma 3 Let ρ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq69_HTML.gif 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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif with r C [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq76_HTML.gif and r ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq77_HTML.gif,

     
  2. (ii)
    H can be extended on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif with H A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq78_HTML.gif, and the equality
    H ( t ) + a t H ( t ) a t 2 H ( t ) = ρ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ10_HTML.gif
    (8)
     

holds for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif.

Proof (i) It is clear that r C ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq79_HTML.gif. Since
| 1 t a + 1 0 t s a + 1 ρ ( s ) d s | 0 t | ρ ( s ) | d s for  t ( 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ11_HTML.gif
(9)
we have lim t 0 + r ( t ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq80_HTML.gif. Setting r ( 0 ) : = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq81_HTML.gif, r C [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq82_HTML.gif 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equm_HTML.gif
     
Then p C ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq83_HTML.gif and H ( t ) = t p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq84_HTML.gif for t ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq85_HTML.gif. We now show that p can be extended on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif in such a way that p C [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq86_HTML.gif. 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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq85_HTML.gif. Hence lim t 0 + p ( t ) = A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq87_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equn_HTML.gif
Let p ( 0 ) : = A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq88_HTML.gif. Then p C [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq86_HTML.gif. Since H ( t ) = p ( t ) + t p ( t ) = p ( t ) r ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq89_HTML.gif for t ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq85_HTML.gif, we see that H can be extended on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif with H C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq90_HTML.gif. 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equo_HTML.gif
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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ13_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equp_HTML.gif

and therefore, H L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq91_HTML.gif. Consequently, H A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq78_HTML.gif. Finally, it follows from H ( t ) = p ( t ) r ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq92_HTML.gif and p ( t ) = H ( t ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq93_HTML.gif that r ( t ) = H ( t ) t H ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq94_HTML.gif. Since, by (11), a t r ( t ) = H ( t ) + ρ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq95_HTML.gif, we see that equality (8) is satisfied for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif which completes the proof. □

Lemma 4 Let { ρ n } L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq67_HTML.gif be a uniformly integrable sequence on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif and let lim n ρ n ( t ) = ρ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq96_HTML.gif for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif. Then the sequence
{ 1 t a + 1 0 t s a + 1 ρ n ( s ) d s } is equicontinuous on [ 0 , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ14_HTML.gif
(12)
Proof It follows from Lemma 2 that ρ n 1 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq97_HTML.gif for n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq57_HTML.gif, 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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq98_HTML.gif. Let us assume that (12) does not hold. Then there exist ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq39_HTML.gif, { k n } N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq99_HTML.gif and { ξ n } , { η n } [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq100_HTML.gif such that lim n k n = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq101_HTML.gif, lim n ( ξ n η n ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq102_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ15_HTML.gif
(13)
Since { ξ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq103_HTML.gif and { η n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq104_HTML.gif are bounded sequences, we may assume that they are convergent, and lim n ξ n = τ = lim n η n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq105_HTML.gif. If τ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq106_HTML.gif, 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equq_HTML.gif
which contradicts (13). Let τ ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq107_HTML.gif. Since lim n ( 1 ξ n a + 1 1 η n a + 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq108_HTML.gif and since the uniform integrability of the sequence { ρ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq71_HTML.gif on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif results in
lim n | η n ξ n | ρ k n ( t ) | d t | = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equr_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equt_HTML.gif

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

Lemma 5 Let ρ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq69_HTML.gif. 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ16_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equu_HTML.gif

for t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif, 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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq109_HTML.gif satisfy Eq. (7a) for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif and L : = sup { | u ( t ) | + | u ( t ) | : t ( 0 , T ] } < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq110_HTML.gif. Then u can be extended on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif with u A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq52_HTML.gif, and there exists c R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq111_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ17_HTML.gif
(15)

holds for t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif.

Proof Choose n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq112_HTML.gif and denote by ρ ( t ) = f n ( t , u ( t ) , u ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq113_HTML.gif for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq18_HTML.gif. In order to prove that ρ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq69_HTML.gif, define for m N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq114_HTML.gif, 1 m < T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq115_HTML.gif,
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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equv_HTML.gif
and
ρ m ( t ) : = f n ( t , v m ( t ) , w m ( t ) ) for a.e.  t [ 0 , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equw_HTML.gif

Then ρ m L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq116_HTML.gif and lim m ρ m ( t ) = ρ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq117_HTML.gif for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif. Moreover, | ρ m ( t ) | μ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq118_HTML.gif for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif and all m N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq119_HTML.gif, where μ ( t ) = sup { | f n ( t , x , y ) | : | x | L , | y | L } L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq120_HTML.gif. Consequently, by the Lebesgue dominated convergence theorem, ρ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq121_HTML.gif.

We now discuss the linear Euler differential equation
v ( t ) + a t v ( t ) a t 2 v ( t ) = ρ ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ18_HTML.gif
(16)

Let H be the function given in Lemma 3. By Lemma 3(ii), H can be extended on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif with H A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq78_HTML.gif and −H satisfies (16) for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif. Therefore, each function v A C loc 1 ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq122_HTML.gif which satisfies Eq. (16) a.e. on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif has the form v ( t ) = c t + d t a H ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq123_HTML.gif for t ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq85_HTML.gif, with some c , d R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq124_HTML.gif. By assumption we know that u A C loc 1 ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq109_HTML.gif satisfies (16) a.e. on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif, and therefore there exist c , d R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq125_HTML.gif such that u ( t ) = c t + d t a H ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq126_HTML.gif, t ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq85_HTML.gif. Since by assumption | u | L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq127_HTML.gif on ( 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq50_HTML.gif, we have d = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq128_HTML.gif. Consequently, the function u can be extended on the interval [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif in the class A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq47_HTML.gif and (15) holds on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. □

Corollary 1 Let (H1) hold. Let u A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq52_HTML.gif be a solution of Eq. (7a). Then there exists a constant c R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq129_HTML.gif such that equality (15) is satisfied for t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif.

Proof The result holds by Lemma 6 with sup { | u ( t ) | + | u ( t ) | : t [ 0 , T ] } < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq130_HTML.gif. □

Remark 2 Corollary 1 says that the set of all solutions u A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq52_HTML.gif of Eq. (7a) depends on one parameter c R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq129_HTML.gif and u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif.

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq39_HTML.gif is from (H2) and n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq112_HTML.gif.

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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif.

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ21_HTML.gif
(18)

in the set C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif.

Proof Let u be a solution of Eq. (17a). Then u A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq52_HTML.gif, and by Corollary 1 (with f n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq133_HTML.gif replaced by λ f n ( 1 λ ) ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq134_HTML.gif), there exists c R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq111_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equx_HTML.gif

holds for t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif. Hence, u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif and u ( T ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq135_HTML.gif if and only if c = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq136_HTML.gif. Consequently, if u is a solution of problem (17a) and (17b), then u is a solution of Eq. (18) in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif.

Let u be a solution of Eq. (18) in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif. Then f n ( t , u ( t ) , u ( t ) ) L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq137_HTML.gif. Hence, Lemma 3(ii) (with ρ replaced by λ f n + ( 1 λ ) ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq138_HTML.gif) guarantees that u A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq52_HTML.gif and u is a solution of Eq. (17a). Moreover, u ( 0 ) = u ( T ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq139_HTML.gif. 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq57_HTML.gif and λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq140_HTML.gif) 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq141_HTML.gif such that
| u ( t ) | 2 ε a + 2 | t ξ | for t [ 0 , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ24_HTML.gif
(21)
Proof Let u be a solution of problem (17a) and (17b). Then by Lemma 7, equality (18) holds for t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif. Since by (5), λ f n ( t , u ( t ) , u ( t ) ) ( 1 λ ) ε ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq142_HTML.gif for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif, 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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equy_HTML.gif
follows from (18). Hence, g ( u ( t ) ) g ( ε a + 2 t ( T t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq143_HTML.gif for t ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq144_HTML.gif because g is nonincreasing on R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq145_HTML.gif. Due to Remark 1, L = g ( ε a + 2 t ( T t ) ) 1 < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq146_HTML.gif, which means that
0 T g ( u ( t ) ) d t L . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ25_HTML.gif
(22)

It is clear that L is independent of the choice of solution u to problem (17a) and (17b) and independent of n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq57_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq140_HTML.gif.

We now show that inequality (21) holds for some ξ ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq147_HTML.gif. 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ26_HTML.gif
(23)
Since a < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq148_HTML.gif, 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equz_HTML.gif
Hence, u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq149_HTML.gif is decreasing on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif, and therefore u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq149_HTML.gif vanishes at a unique point ξ ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq141_HTML.gif due to u ( 0 ) = u ( T ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq139_HTML.gif. 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 | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq150_HTML.gif on [ 0 , T ] { ξ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq151_HTML.gif, 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equab_HTML.gif
In particular,
r ( | u ( t ) | ) 1 < V . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ27_HTML.gif
(24)
Let W = a + 3 a + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq152_HTML.gif. 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equac_HTML.gif
It follows from u ( t ) = 0 t u ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq153_HTML.gif for t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif,
u T u , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ28_HTML.gif
(25)
and therefore, we have
u K h ( 1 + T u , 1 + u ) + M , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ29_HTML.gif
(26)
where K = W φ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq154_HTML.gif and M = W ( L + V ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq155_HTML.gif. By (H3),
lim z ( 1 / z ) ( K h ( 1 + T z , 1 + z ) + M ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equad_HTML.gif

Consequently, there exists S > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq156_HTML.gif such that K h ( 1 + T z , 1 + z ) + M < z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq157_HTML.gif for z S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq158_HTML.gif. Now, due to (26), u < S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq159_HTML.gif, and therefore, by (25), u < S T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq160_HTML.gif. □

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq112_HTML.gif, 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 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equae_HTML.gif
Then Ω is an open and bounded subset of the Banach space C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif. Keeping in mind Lemma 3, define an operator K : [ 0 , T ] × Ω ¯ C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq161_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ30_HTML.gif
(27)
By Lemma 7, any fixed point of the operator K ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq162_HTML.gif is a solution of problem (7a) and (7b). In order to show the existence of a fixed point of K ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq162_HTML.gif, we apply Lemma 1 with E = C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq163_HTML.gif, U = Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq164_HTML.gif, F = K ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq165_HTML.gif and = ε a + 2 t ( T t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq166_HTML.gif. Especially, we show that
  1. (i)

    K ( 1 , ) : Ω ¯ C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq167_HTML.gif is a compact operator, and

     
  2. (ii)

    K ( λ , x ) x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq168_HTML.gif for each λ ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq169_HTML.gif and x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq170_HTML.gif.

     
We first verify that K ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq162_HTML.gif is a continuous operator. To this end, let { x m } Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq171_HTML.gif be a convergent sequence, and let lim m x m = x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq172_HTML.gif in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif. 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equaf_HTML.gif
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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif. Here K ( 1 , x ) = d d t K ( 1 , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq173_HTML.gif. In particular, for m N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq114_HTML.gif,
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ31_HTML.gif
(28)
Since lim m f n ( t , x m ( t ) , x m ( t ) ) = f n ( t , x ( t ) , x ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq174_HTML.gif for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif and there exists ρ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq69_HTML.gif such that
| f n ( t , x m ( t ) , x m ( t ) ) | ρ ( t ) for a.e.  t [ 0 , T ]  and all  m N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equah_HTML.gif
we have lim m r m 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq175_HTML.gif by the Lebesgue dominated convergence theorem. Hence, by (28), K ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq162_HTML.gif is a continuous operator. We now show that the set K ( 1 , Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq176_HTML.gif is relatively compact in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif. It follows from f n Car ( [ 0 , T ] × R 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq177_HTML.gif and Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq178_HTML.gif bounded in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif that there exists μ L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq179_HTML.gif such that
| f n ( t , x ( t ) , x ( t ) ) | μ ( t ) for a.e.  t [ 0 , T ]  and all  x Ω ¯ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equai_HTML.gif
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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equaj_HTML.gif
are satisfied for t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif and x Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq180_HTML.gif, and therefore, the set K ( 1 , Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq181_HTML.gif is bounded in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif. 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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equak_HTML.gif

holds for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif and all x Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq180_HTML.gif (cf. (9)). Consequently, the set { K ( 1 , x ) : x Ω ¯ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq182_HTML.gif is equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. Hence, the set K ( 1 , Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq176_HTML.gif is relatively compact in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif by the Arzelà-Ascoli theorem. As a result, K ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq162_HTML.gif 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 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq183_HTML.gif is a solution of problem (17a) and (17b), Lemma 8 guarantees that u satisfies inequality (20). Therefore, K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq184_HTML.gif has property (ii). Consequently, by Lemmas 1 and 8, for each n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq112_HTML.gif, problem (7a) and (7b) has a solution u satisfying estimates (19)-(21). □

Let u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq36_HTML.gif be a solution of problem (7a) and (7b) for n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq112_HTML.gif. The following property of the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq185_HTML.gif is an important prerequisite for solving problem (1a) and (1b).

Lemma 10 Let (H1)-(H3) hold. Let u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq36_HTML.gif be a solution of problem (7a) and (7b) for n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq112_HTML.gif. Then the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq185_HTML.gif is uniformly integrable on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq186_HTML.gif. 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 | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ35_HTML.gif
(32)

for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif, where g ( t ) = g ( ε a + 2 t ( T t ) ) L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq187_HTML.gif, see Remark 1. Since the sequence { r ( 2 ε a + 2 | t ξ n | ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq188_HTML.gif is uniformly integrable on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif (cf. [[20], criterion A.4], [21, 22]), it follows from (32) that { | f n ( t , u n ( t ) , u n ( t ) ) | } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq185_HTML.gif is uniformly integrable on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq57_HTML.gif, problem (7a) and (7b) has a solution u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq36_HTML.gif satisfying inequalities (29)-(31), where S is a positive constant and ξ n ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq186_HTML.gif. Moreover, by Lemma 10, the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq189_HTML.gif is uniformly integrable on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. We now prove that { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq190_HTML.gif is equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. Since u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq36_HTML.gif is a fixed point of the operator K ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq162_HTML.gif 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 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equal_HTML.gif
holds for a.e. t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq40_HTML.gif and all n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq57_HTML.gif. Let 0 t 1 < t 2 T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq191_HTML.gif. 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 | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equam_HTML.gif
Let r n ( t ) = 1 t a + 1 0 t s a + 1 f n ( s , u n ( s ) , u n ( s ) ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq192_HTML.gif. By Lemma 3(i), { r n } C [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq193_HTML.gif and r n ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq194_HTML.gif. 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 | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ36_HTML.gif
(33)
follows. By Lemma 4 (for ρ n ( t ) = f n ( t , u n ( t ) , u n ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq195_HTML.gif), the sequence { r n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq196_HTML.gif is equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. Since the sequence { | f n ( t , u n ( t ) , u n ( t ) ) | } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq185_HTML.gif is uniformly integrable on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif, the sequence { 0 t f n ( s , u ( s ) , u ( s ) ) d s } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq197_HTML.gif is equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. Hence, it follows from (33), that { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq198_HTML.gif is equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. We summarize: { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq199_HTML.gif is bounded in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif and { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq200_HTML.gif is equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq48_HTML.gif. Also, { ξ n } ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq201_HTML.gif. Using appropriate subsequences, if necessary, we can assume, by the Arzelà-Ascoli theorem and the Bolzano-Weierstrass theorem, that { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq199_HTML.gif is convergent in C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq132_HTML.gif and { ξ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq103_HTML.gif is convergent in . Let lim n u n = : u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq202_HTML.gif and lim n ξ n = : ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq203_HTML.gif. With n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq204_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif, u ( T ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq135_HTML.gif. Since
lim n f n ( t , u n ( t ) , u n ( t ) ) = f ( t , u ( t ) , u ( t ) ) for a.e.  t [ 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ37_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equap_HTML.gif
and f ( t , u ( t ) , u ( t ) ) L 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq205_HTML.gif. 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq204_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equas_HTML.gif
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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ38_HTML.gif
(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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equat_HTML.gif

and u A C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq52_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq206_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq207_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq208_HTML.gif depending on f. The values c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq209_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq210_HTML.gif for a.e. t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq211_HTML.gif and all x R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq24_HTML.gif, y R { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq212_HTML.gif and, according to the next numerical approach (see Section 6.2), we consider Eq. (36a), where h 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq213_HTML.gif, that is,
u ( t ) + a t u ( t ) a t 2 u ( t ) = q ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ41_HTML.gif
(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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ42_HTML.gif
(38)
If we set u ( t ) = t λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq214_HTML.gif, we arrive at the characteristic polynomial of (38),
λ 2 ( 1 a ) λ a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equau_HTML.gif

whose roots λ 1 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq215_HTML.gif and λ 2 = a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq216_HTML.gif are positive. Therefore, conditions for u and u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq149_HTML.gif can be prescribed at t = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq217_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq218_HTML.gif. Since in this case the exact solution is given, u ( t ) = 2 t ( 1 t ) a + 1.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq219_HTML.gif, the value u ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq220_HTML.gif is available, u ( 1 ) = 0.72 , 1 , 1.67 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq221_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ47_HTML.gif
(41)
and it satisfies (H1)-(H3) with ε = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq222_HTML.gif, φ ( t ) = 1 t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq223_HTML.gif for t ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq224_HTML.gif and
h ( x , y ) = 1 , g ( x ) = x 1 3 , r ( y ) = 0 for  x , y R + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equav_HTML.gif

It follows from Theorem 1 that there exists at least one value of c > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq225_HTML.gif such that the related solution u of problem (40a) and (40b) with u ( 1 ) = c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq226_HTML.gif is positive on ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq227_HTML.gif with u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif. Using formula (35), we now determine an interval ( 0 , c ) ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq228_HTML.gif 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equaw_HTML.gif
Therefore,
u < 4 2 a + 3 + u 1 3 4 ( a + 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ48_HTML.gif
(42)
Let K > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq229_HTML.gif satisfy
K = 4 2 a + 3 + K 1 3 4 ( a + 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ49_HTML.gif
(43)
Then (42) implies u < K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq230_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equax_HTML.gif
Consequently,
c = 2 2 a + 3 + K 1 3 1 a + 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ50_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ51_HTML.gif
(45)

We begin the calculations with δ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq231_HTML.gif and increase its value gradually until we arrive at δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq232_HTML.gif; 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq220_HTML.gif before solving the BVP (40a) and (40b), i.e., find the final value of c ( 0 , c ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq233_HTML.gif such that u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif.

In the case of Example 1 and a = 0.1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq234_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq235_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq237_HTML.gif.

     
  2. 2.

    Use the above approximation as an initial guess for ODE (45) with δ = 0.01 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq238_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq239_HTML.gif.

     
  3. 3.

    Use the above approximation as an initial guess for ODE (45) with δ = 0.1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq240_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq239_HTML.gif.

     
  4. 4.

    Use the above approximation as an initial guess for ODE (45) with δ = 1.0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq241_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq239_HTML.gif.

     
After the last step, we have solved problem (40a) and (40b) subject to boundary conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq239_HTML.gif. In this case, the value of u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq242_HTML.gif was not small enough to consider it a reasonable approximation for u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif. Therefore, we use a shooting idea combined with a bisection strategy to find a better value for c = u ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq243_HTML.gif. 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq244_HTML.gif .

Table 1

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

a

K

c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq245_HTML.gif

c

u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq246_HTML.gif

−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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ54_HTML.gif
(47)
and has a singularity at x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq3_HTML.gif. The function f satisfies conditions (H1)-(H3) with ε = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq222_HTML.gif, φ ( t ) = 1 t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq247_HTML.gif for t ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq224_HTML.gif and
h ( x , y ) = 1 , g ( x ) = x 1 3 , r ( y ) = 0 for  x , y R + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equay_HTML.gif
Theorem 1 guarantees the existence of at least one c > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq225_HTML.gif such that a solution u of problem (46a) and (46b) is positive on ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq227_HTML.gif and u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif holds. We now again determine an interval ( 0 , c ) ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq248_HTML.gif 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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equaz_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equba_HTML.gif
Consequently,
c = 0 1 t a + 1 ( 1 t + ( a + 2 ) 1 / 3 t 1 / 3 ( 1 t ) 1 / 3 ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ55_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equ56_HTML.gif
(49)
For all values of a, we choose u ( 1 ) ( c , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq249_HTML.gif and analogously carry out the path-following in δ first. The related chain for a = 0.1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq234_HTML.gif is as follows.
  1. 1.

    Numerical approximation of BVP (39a) and (39b) is used as an initial guess for ODE (49) with δ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq235_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 0.8 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq250_HTML.gif.

     
  2. 2.

    Use the above approximation as an initial guess for ODE (49) with δ = 0.01 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq238_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 0.8 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq250_HTML.gif.

     
  3. 3.

    Use the above approximation as an initial guess for ODE (49) with δ = 0.1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq240_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq251_HTML.gif.

     
  4. 4.

    Use the above approximation as an initial guess for ODE (49) with δ = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq252_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq253_HTML.gif.

     
  5. 5.

    Use the above approximation as an initial guess for ODE (49) with δ = 1.0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq241_HTML.gif subject to terminal conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.7 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq254_HTML.gif.

     
After the last step, we have solved BVP (46a) and (46b) subject to boundary conditions u ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq236_HTML.gif, u ( 1 ) = 1.7 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq254_HTML.gif, but also, in this case, the value of u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq242_HTML.gif is too large and we have to find a better value for c = u ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq243_HTML.gif. 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq255_HTML.gif .

Table 2

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

a

c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq245_HTML.gif

c

u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq246_HTML.gif

−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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equbb_HTML.gif
where a ( 1 , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq1_HTML.gif, and the nonlinearity f ( t , x , y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq2_HTML.gif may be singular at the space variables x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq3_HTML.gif and/or y = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq4_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_Equbc_HTML.gif

Here, it is only known that c ( 0 , c ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq256_HTML.gif, where c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq209_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-6/MediaObjects/13661_2012_Article_262_IEq131_HTML.gif.

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

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