In the present work, we analyze the existence of positive solutions to the singular Dirichlet BVP,

Here, we assume that T > 0 , a∈(−1,0) and f satisfies the local Carathéodory conditions on [ 0 , T ] × D , where D : = R + × R 0 and R + : = ( 0 , ∞ ) , R 0 : = R ∖ { 0 } . Let us recall that a function h : [ 0 , T ] × A → R , A ⊂ R × R , satisfies the local Carathéodory conditions on [ 0 , T ] × A if

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

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

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

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

For such functions, we use the notation h ∈ Car ( [ 0 , T ] × A ) . Moreover, f(t,x,y) may become singular when the space variables x and/or y vanish, which means that f(t,x,y) may become unbounded for x=0 and a.e. t ∈ [ 0 , T ] and all y ∈ R 0 , and/or it may be unbounded for y=0 and a.e. t∈[0,T] and all x ∈ R + . Finally, since a≠0, Eq. (1a) has a singularity of the first kind at the time variable t=0 because

∫ 0 T 1 t d t = ∞ , ∫ 0 T 1 t 2 d t = ∞ .

The differential operator on the left-hand side of Eq. (1a) can be equivalently written as ( t − a ( t a u ) ′ ) ′ and, after the substitution x ( t ) = t a u ( t ) , it takes the form ( t − a x ′ ) ′ , which arises in numerous important applications. Operators of such type were studied in phase transitions of Van der Waals fluids 1234, in population genetics, especially in models for the spatial distribution of the genetic composition of a population 56, in the homogeneous nucleation theory 7, in relativistic cosmology for description of particles which can be treated as domains in the universe 8, and in the nonlinear field theory 9, in particular, when describing bubbles generated by scalar fields of Higgs type in the Minkowski spaces 10.

The aim of this paper is to study the case a ∈ ( − 1 , 0 ) which is fundamentally different from the case a ∈ ( − ∞ , − 1 ) . The latter setting was studied in 1112, where the structure and properties of the set of all positive solutions to (1a) and (1b) were investigated (the cardinality of this set is a continuum).

In the sequel, we introduce the basic notation and state the preliminary results required in the analysis of problem (1a) and (1b). Here, we focus our attention on the case a∈(−1,0) and prove the existence of at least one positive solution of (1a) and (1b). In contrast to 1112, we consider the more general situation in which f may be also singular at y=0. This means that we have to deal with the following additional difficulties.

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

The investigation of singular Dirichlet BVPs has a long history and a lot of methods for their analysis are available. One of the most important ones is the topological degree method providing various fixed point theorems and existence alternative theorems; see, e.g., Lemma 1. For more information on the topological degree method and its application to numerous BVPs, including Dirichlet problems, we refer the reader to the monographs by Mawhin 131415.

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

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

(H₂) There exists an ε > 0 such that

− f ( t , x , y ) ≥ ε for a.e.  t ∈ [ 0 , T ]  and all  ( x , y ) ∈ D .

(H₃) For a.e. t ∈ [ 0 , T ] and all ( x , y ) ∈ D , the estimate

− f ( t , x , y ) ≤ φ ( t ) h ( x , | y | ) + g ( x ) + r ( | y | )

holds, where φ ∈ L 1 [ 0 , T ] , h ∈ C ( [ 0 , ∞ ) × [ 0 , ∞ ) ) , g , r ∈ C ( R + ) are positive, h is nondecreasing in both its arguments, g and r are nonincreasing, and

By

∥ x ∥ ∞ = max { | x ( t ) | : t ∈ [ 0 , T ] } , ∥ x ∥ 1 = ∫ 0 T | x ( t ) | d t

we denote the norms in C [ 0 , T ] and L 1 [ 0 , T ] , respectively. A C 1 [ 0 , T ] denotes the set of functions whose first derivative is absolutely continuous on [ 0 , T ] , while A C loc 1 ( 0 , T ] is the set of functions having absolutely continuous first derivative on each compact subinterval of ( 0 , T ] . We use the symbol meas ( M ) to denote the Lebesgue measure of ℳ.

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

Remark 1 Let a function g have the properties specified in (H₃). Then for each b , c ∈ R + , ∫ 0 b g ( c s ) d s < ∞ , and it follows from the inequality

t ( T − t ) ≥ { T 2 t for  t ∈ [ 0 , T 2 ] , T 2 ( T − t ) for  t ∈ [ T 2 , T ] ,

that

∫ 0 T g ( c t ( T − t ) ) d t < ∞ for each  c ∈ R + .

In order to prove that the singular problem (1a) and (1b) has a positive solution, we use regularization and sequential techniques. To this end, for n ∈ N we define functions f n ∗ : [ 0 , T ] × ( R × R 0 ) → R and f n : [ 0 , T ] × R 2 → R by

f n ∗ ( t , x , y ) = { f ( t , x , y ) if  x ≥ 1 n , f ( t , 1 n , y ) if  x < 1 n ,

and

f n ( t , x , y ) = { f n ∗ ( t , x , y ) if  | y | ≥ 1 n , n 2 [ f n ∗ ( t , x , 1 n ) ( y + 1 n ) − f n ∗ ( t , x , − 1 n ) ( y − 1 n ) ] if  | y | < 1 n ,

respectively. Then it follows from (H₁) that f n ∈ Car ( [ 0 , T ] × R 2 ) and (H₂) and (H₃) yield

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 ] ,

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 ] . }

As a first step in the analysis, we investigate auxiliary regular BVPs of the form

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 ℓ ∈ U . Assume that F : U ¯ → E is a compact operator. Then either

A1: ℱ has a fixed point in U ¯ , or

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

In limit processes, we apply the following Vitali convergence theorem; cf. 171819.

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

(i) ρ ∈ L 1 [ 0 , T ] and lim n → ∞ ∥ ρ n − ρ ∥ 1 = 0 ,

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

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

∫ M | ρ n ( t ) | d t < ε , n ∈ N .

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

Throughout the paper a∈(−1,0).

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

Lemma 3 Let ρ∈L1[0,T] and

Then

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

(ii) H can be extended on [0,T] with H ∈ A C 1 [ 0 , T ] , and the equality

H ″ ( t ) + a t H ′ ( t ) − a t 2 H ( t ) = − ρ ( t )

holds for a.e. t∈[0,T].

Proof (i) It is clear that r ∈ C ( 0 , T ] . Since

| 1 t a + 1 ∫ 0 t s a + 1 ρ ( s ) d s | ≤ ∫ 0 t | ρ ( s ) | d s for  t ∈ ( 0 , T ] ,

we have lim t → 0 + r ( t ) = 0 . Setting r ( 0 ) : = 0 , r ∈ C [ 0 , T ] follows.

(ii) Let

p ( t ) = ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s for  t ∈ ( 0 , T ] .

Then p ∈ C ( 0 , T ] and H ( t ) = t p ( t ) for t ∈ ( 0 , T ] . We now show that p can be extended on [0,T] in such a way that p ∈ C [ 0 , T ] . Integrating by parts yields

for t∈(0,T]. Hence lim t → 0 + p ( t ) = A , where

A = − 1 ( a + 1 ) T a + 1 ∫ 0 T s a + 1 ρ ( s ) d s + 1 a + 1 ∫ 0 T ρ ( s ) d s .

Let p ( 0 ) : = A . Then p∈C[0,T]. Since H ′ ( t ) = p ( t ) + t p ′ ( t ) = p ( t ) − r ( t ) for t∈(0,T], we see that H can be extended on [0,T] with H ∈ C 1 [ 0 , T ] . Moreover,

H ″ ( t ) = p ′ ( t ) − r ′ ( t ) = − 1 t a + 2 ∫ 0 t s a + 1 ρ ( s ) d s + a + 1 t a + 2 ∫ 0 t s a + 1 ρ ( s ) d s − ρ ( t ) = a t a + 2 ∫ 0 t s a + 1 ρ ( s ) d s − ρ ( t ) .

In particular,

H ″ ( t ) = a t a + 2 ∫ 0 t s a + 1 ρ ( s ) d s − ρ ( t ) for a.e.  t ∈ [ 0 , T ] .

Hence, cf. (10),

∫ 0 T H ″ ( s ) d s = a ∫ 0 T 1 s a + 2 ( ∫ 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s − ∫ 0 T ρ ( t ) d t = a A − ∫ 0 T ρ ( t ) d t ,

and therefore, H ″ ∈ L 1 [ 0 , T ] . Consequently, H∈AC1[0,T]. Finally, it follows from H ′ ( t ) = p ( t ) − r ( t ) and p ( t ) = H ( t ) t that r ( t ) = H ( t ) t − H ′ ( t ) . Since, by (11), a t r ( t ) = H ″ ( t ) + ρ ( t ) , we see that equality (8) is satisfied for a.e. t∈[0,T] which completes the proof. □

Lemma 4 Let {ρn}⊂L1[0,T] be a uniformly integrable sequence on [0,T] and let lim n → ∞ ρ n ( t ) = ρ ( t ) for a.e. t∈[0,T]. Then the sequence

{ 1 t a + 1 ∫ 0 t s a + 1 ρ n ( s ) d s } is equicontinuous on   [ 0 , T ] .

Proof It follows from Lemma 2 that ∥ ρ n ∥ 1 ≤ L for n∈N, where L is a positive constant. Recall that by Lemma 3(i), { 1 t a + 1 ∫ 0 t s a + 1 ρ n ( s ) d s } ⊂ C [ 0 , T ] . Let us assume that (12) does not hold. Then there exist ε>0, { k n } ⊂ N and { ξ n } , { η n } ⊂ [ 0 , T ] such that lim n → ∞ k n = ∞ , lim n → ∞ ( ξ n − η n ) = 0 and

| 1 ξ n a + 1 ∫ 0 ξ n s a + 1 ρ k n ( s ) d s − 1 η n a + 1 ∫ 0 η n s a + 1 ρ k n ( s ) d s | ≥ ε for  n ∈ N .

Since { ξ n } and { η n } are bounded sequences, we may assume that they are convergent, and lim n → ∞ ξ n = τ = lim n → ∞ η n . If τ = 0 , then (cf. (9))

lim n → ∞ 1 ξ n a + 1 ∫ 0 ξ n s a + 1 ρ k n ( s ) d s = 0 , lim n → ∞ 1 η n a + 1 ∫ 0 η n s a + 1 ρ k n ( s ) d s = 0 ,

which contradicts (13). Let τ ∈ ( 0 , T ] . Since lim n → ∞ ( 1 ξ n a + 1 − 1 η n a + 1 ) = 0 and since the uniform integrability of the sequence {ρn} on [0,T] results in

lim n → ∞ | ∫ η n ξ n | ρ k n ( t ) | d t | = 0 ,

we conclude from the relation

that

lim n → ∞ | 1 ξ n a + 1 ∫ 0 ξ n s a + 1 ρ k n ( s ) d s − 1 η n a + 1 ∫ 0 η n s a + 1 ρ k n ( s ) d s | = 0 .

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

Lemma 5 Let ρ∈L1[0,T]. Then

| ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s | ≤ 2 ∥ ρ ∥ 1 a + 1 for   t ∈ [ 0 , T ] .

Proof Since (cf. (10))

| ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 ρ ( ξ ) d ξ ) d s | ≤ 1 a + 1 ( ∫ 0 t | ρ ( s ) | d s + ∫ t T | ρ ( s ) | d s + ∥ ρ ∥ 1 ) = 2 ∥ ρ ∥ 1 a + 1

for t∈[0,T], estimate (14) holds. □

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

Lemma 6 Let (H₁) hold. Let u ∈ A C loc 1 ( 0 , T ] satisfy Eq. (7a) for a.e. t∈[0,T] and L : = sup { | u ( t ) | + | u ′ ( t ) | : t ∈ ( 0 , T ] } < ∞ . Then u can be extended on [0,T] with u∈AC1[0,T], and there exists c ∈ R such that the integral equation

u ( t ) = t ( c − ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 f n ( ξ , u ( ξ ) , u ′ ( ξ ) ) d ξ ) d s )

holds for t∈[0,T].

Proof Choose n ∈ N and denote by ρ ( t ) = f n ( t , u ( t ) , u ′ ( t ) ) for a.e. t∈[0,T]. In order to prove that ρ∈L1[0,T], define for m ∈ N , 1 m < T ,

v m ( t ) : = { u ( t ) if  t ∈ ( 1 m , T ] , u ( 1 m ) if  t ∈ [ 0 , 1 m ] , w m ( t ) : = { u ′ ( t ) if  t ∈ ( 1 m , T ] , u ′ ( 1 m ) if  t ∈ [ 0 , 1 m ] ,

and

ρ m ( t ) : = f n ( t , v m ( t ) , w m ( t ) ) for a.e.  t ∈ [ 0 , T ] .

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

We now discuss the linear Euler differential equation

v ″ ( t ) + a t v ′ ( t ) − a t 2 v ( t ) = ρ ( t ) .

Let H be the function given in Lemma 3. By Lemma 3(ii), H can be extended on [0,T] with H∈AC1[0,T] and −H satisfies (16) for a.e. t∈[0,T]. Therefore, each function v ∈ A C loc 1 ( 0 , T ] which satisfies Eq. (16) a.e. on [0,T] has the form v ( t ) = c ∗ t + d ∗ t − a − H ( t ) for t∈(0,T], with some c ∗ , d ∗ ∈ R . By assumption we know that u∈ACloc1(0,T] satisfies (16) a.e. on [0,T], and therefore there exist c , d ∈ R such that u ( t ) = c t + d t − a − H ( t ) , t∈(0,T]. Since by assumption | u ′ | ≤ L on (0,T], we have d = 0 . Consequently, the function u can be extended on the interval [0,T] in the class AC1[0,T] and (15) holds on [0,T]. □

Corollary 1 Let (H₁) hold. Let u∈AC1[0,T] be a solution of Eq. (7a). Then there exists a constant c ∈ R such that equality (15) is satisfied for t∈[0,T].

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

Remark 2 Corollary 1 says that the set of all solutions u∈AC1[0,T] of Eq. (7a) depends on one parameter c∈R and u ( 0 ) = 0 .

In order to prove the solvability of problem (7a) and (7b), we first have to investigate the problem

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

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

Lemma 7 Let (H₁) hold. Then u is a solution of problem (17a) and (17b) if and only if u is a solution of the integral equation

u ( t ) = − t ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 [ λ f n ( ξ , u ( ξ ) , u ′ ( ξ ) ) − ( 1 − λ ) ε ] d ξ ) d s

in the set C1[0,T].

Proof Let u be a solution of Eq. (17a). Then u∈AC1[0,T], and by Corollary 1 (with f n replaced by λ f n − ( 1 − λ ) ε ), there exists c∈R such that the equation

u ( t ) = t ( c − ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 [ λ f n ( ξ , u ( ξ ) , u ′ ( ξ ) ) − ( 1 − λ ) ε ] d ξ ) d s )

holds for t∈[0,T]. Hence, u(0)=0 and u ( T ) = 0 if and only if c = 0 . Consequently, if u is a solution of problem (17a) and (17b), then u is a solution of Eq. (18) in C1[0,T].

Let u be a solution of Eq. (18) in C1[0,T]. Then f n ( t , u ( t ) , u ′ ( t ) ) ∈ L 1 [ 0 , T ] . Hence, Lemma 3(ii) (with ρ replaced by − λ f n + ( 1 − λ ) ε ) guarantees that u∈AC1[0,T] and u is a solution of Eq. (17a). Moreover, u ( 0 ) = u ( T ) = 0 . Consequently, u is a solution of problem (17a) and (17b) which completes the proof. □

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

Lemma 8 Let (H₁)-(H₃) hold. Then there exists a positive constant S (independent of n∈N and λ ∈ [ 0 , 1 ] ) such that for all solutions u of problem (17a) and (17b), the estimates

hold. Moreover, for any solution u of problem (17a) and (17b), there exists ξ ∈ ( 0 , T ) such that

| u ′ ( t ) | ≥ 2 ε a + 2 | t − ξ |   for   t ∈ [ 0 , T ] .

Proof Let u be a solution of problem (17a) and (17b). Then by Lemma 7, equality (18) holds for t∈[0,T]. Since by (5), λ f n ( t , u ( t ) , u ′ ( t ) ) − ( 1 − λ ) ε ≤ − ε for a.e. t∈[0,T], the relation

u ( t ) ≥ ε t ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 d ξ ) d s = ε a + 2 t ( T − t ) for  t ∈ [ 0 , T ]

follows from (18). Hence, g ( u ( t ) ) ≤ g ( ε a + 2 t ( T − t ) ) for t ∈ ( 0 , T ) because g is nonincreasing on R + . Due to Remark 1, L = ∥ g ( ε a + 2 t ( T − t ) ) ∥ 1 < ∞ , which means that

∫ 0 T g ( u ( t ) ) d t ≤ L .

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

We now show that inequality (21) holds for some ξ ∈ ( 0 , T ) . Differentiation of (18) gives

u ″ ( t ) = − a t a + 2 ∫ 0 t s a + 1 [ λ f n ( s , u ( s ) , u ′ ( s ) ) − ( 1 − λ ) ε ] d s + λ f n ( t , u ( t ) , u ′ ( t ) ) − ( 1 − λ ) ε for a.e.  t ∈ [ 0 , T ] .

Since a < 0 , it follows from (5) and (23) that

u ″ ( t ) ≤ a ε t a + 2 ∫ 0 t s a + 1 d s − ε = − 2 ε a + 2 for a.e.  t ∈ [ 0 , T ] .

Hence, u ′ is decreasing on [0,T], and therefore u′ vanishes at a unique point ξ∈(0,T) due to u(0)=u(T)=0. The inequality (21) now follows from the relations

Hence, r ( | u ′ ( t ) | ) ≤ r ( 2 ε a + 2 | t − ξ | ) on [ 0 , T ] ∖ { ξ } , and

∫ 0 T ( r ( | u ′ ( t ) | ) ) d t ≤ ∫ 0 ξ ( r ( 2 ε a + 2 ( ξ − t ) ) ) d t + ∫ ξ T ( r ( 2 ε a + 2 ( t − ξ ) ) ) d t < a + 2 ε ∫ 0 ( 2 ε T ) / ( a + 2 ) r ( s ) d s = V for  t ∈ [ 0 , T ] .

In particular,

∥ r ( | u ′ ( t ) | ) ∥ 1 < V .

Let W = a + 3 a + 1 . Taking into account (6), (9), (18), (22), (24), and Lemma 5, we obtain

| u ′ ( t ) | = | − ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 [ λ f n ( ξ , u ( ξ ) , u ′ ( ξ ) ) − ( 1 − λ ) ε ] d ξ ) d s + 1 t a + 1 ∫ 0 t s a + 1 [ λ f n ( s , u ( s ) , u ′ ( s ) ) − ( 1 − λ ) ε ] d s | ≤ W ∫ 0 T | λ f n ( t , u ( t ) , u ′ ( t ) ) − ( 1 − λ ) ε | d t ≤ W ∫ 0 T ( φ ( t ) h ( 1 + ∥ u ∥ ∞ , 1 + ∥ u ′ ∥ ∞ ) + g ( u ( t ) ) + r ( | u ′ ( t ) | ) ) d t ≤ W ( h ( 1 + ∥ u ∥ ∞ , 1 + ∥ u ′ ∥ ∞ ) ∥ φ ∥ 1 + L + V ) , t ∈ [ 0 , T ] .

It follows from u ( t ) = ∫ 0 t u ′ ( s ) d s for t∈[0,T],

∥ u ∥ ∞ ≤ T ∥ u ′ ∥ ∞ ,

and therefore, we have

∥ u ′ ∥ ∞ ≤ K h ( 1 + T ∥ u ′ ∥ ∞ , 1 + ∥ u ′ ∥ ∞ ) + M ,

where K = W ∥ φ ∥ 1 and M = W ( L + V ) . By (H₃),

lim z → ∞ ( 1 / z ) ( K h ( 1 + T z , 1 + z ) + M ) = 0 .

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

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

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

Proof Let S be a positive constant in Lemma 8 and let us define

Ω : = { x ∈ C 1 [ 0 , T ] : ∥ x ∥ < S T , ∥ x ′ ∥ < S } .

Then Ω is an open and bounded subset of the Banach space C1[0,T]. Keeping in mind Lemma 3, define an operator K : [ 0 , T ] × Ω ¯ → C 1 [ 0 , T ] by the formula

K ( λ , x ) ( t ) = − t ∫ t T 1 s a + 2 ( ∫ 0 s ξ a + 1 [ λ f n ( ξ , x ( ξ ) , x ′ ( ξ ) ) − ( 1 − λ ) ε ] d ξ ) d s .

By Lemma 7, any fixed point of the operator K ( 1 , ⋅ ) is a solution of problem (7a) and (7b). In order to show the existence of a fixed point of K(1,⋅), we apply Lemma 1 with E = C 1 [ 0 , T ] , U = Ω , F = K ( 1 , ⋅ ) and ℓ = ε a + 2 t ( T − t ) . Especially, we show that

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

(ii) K ( λ , x ) ≠ x for each λ ∈ ( 0 , 1 ] and x ∈ ∂ Ω .

We first verify that K(1,⋅) is a continuous operator. To this end, let { x m } ⊂ Ω ¯ be a convergent sequence, and let lim m → ∞ x m = x in C1[0,T]. Let

r m ( t ) : = f n ( t , x m ( t ) , x m ′ ( t ) ) − f n ( t , x ( t ) , x ′ ( t ) ) for a.e.  t ∈ [ 0 , T ] .

It follows from Lemma 5 and (9) that

for t∈[0,T]. Here K ( 1 , x ) ′ = d d t K ( 1 , x ) . In particular, for m∈N,

∥ K ( 1 , x m ) − K ( 1 , x ) ∥ ∞ ≤ 2 T ∥ r m ∥ 1 a + 1 , ∥ K ( 1 , x m ) ′ − K ( 1 , x ) ′ ∥ ∞ ≤ ( a + 3 ) ∥ r m ∥ 1 a + 1 .

Since lim m → ∞ f n ( t , x m ( t ) , x m ′ ( t ) ) = f n ( t , x ( t ) , x ′ ( t ) ) for a.e. t∈[0,T] and there exists ρ∈L1[0,T] such that

| f n ( t , x m ( t ) , x m ′ ( t ) ) | ≤ ρ ( t ) for a.e.  t ∈ [ 0 , T ]  and all  m ∈ N ,

we have lim m → ∞ ∥ r m ∥ 1 = 0 by the Lebesgue dominated convergence theorem. Hence, by (28), K(1,⋅) is a continuous operator. We now show that the set K ( 1 , Ω ¯ ) is relatively compact in C1[0,T]. It follows from f n ∈ Car ( [ 0 , T ] × R 2 ) and Ω ¯ bounded in C1[0,T] that there exists μ ∈ L 1 [ 0 , T ] such that

| f n ( t , x ( t ) , x ′ ( t ) ) | ≤ μ ( t ) for a.e.  t ∈ [ 0 , T ]  and all  x ∈ Ω ¯ .

Then by Lemma 5 and (9), the inequalities

| K ( 1 , x ) ( t ) | ≤ 2 T ∥ μ ∥ 1 a + 1 , | K ( 1 , x ) ′ ( t ) | ≤ ( a + 3 ) ∥ μ ∥ 1 a + 1

are satisfied for t∈[0,T] and x ∈ Ω ¯ , and therefore, the set K ( 1 , Ω ¯ ) is bounded in C1[0,T]. Moreover, the relation

| K ( 1 , x ) ″ ( t ) | = | − a t a + 2 ∫ 0 t s a + 1 f n ( s , x ( s ) , x ′ ( s ) ) d s + f n ( t , x ( t ) , x ′ ( t ) ) | ≤ | a | t a + 2 ∫ 0 t s a + 1 μ ( s ) d s + μ ( t ) ∈ L 1 [ 0 , T ]

holds for a.e. t∈[0,T] and all x∈Ω¯ (cf. (9)). Consequently, the set { K ( 1 , x ) ′ : x ∈ Ω ¯ } is equicontinuous on [0,T]. Hence, the set K(1,Ω¯) is relatively compact in C1[0,T] by the Arzelà-Ascoli theorem. As a result, K(1,⋅) is a compact operator and the condition (i) follows.

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

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

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

Proof

By Lemma 9, the inequalities

hold, where S is a positive constant and ξ n ∈ ( 0 , T ) . Hence, by (3) and (4),

0 < − f n ( t , u n ( t ) , u n ′ ( t ) ) ≤ φ ( t ) h ( 1 + S T , 1 + S ) + g ∗ ( t ) + r ( 2 ε a + 2 | t − ξ n | )

for a.e. t∈[0,T], where g ∗ ( t ) = g ( ε a + 2 t ( T − t ) ) ∈ L 1 [ 0 , T ] , see Remark 1. Since the sequence { r ( 2 ε a + 2 | t − ξ n | ) } is uniformly integrable on [0,T] (cf. [20, criterion A.4], 2122), it follows from (32) that {|fn(t,un(t),un′(t))|} is uniformly integrable on [0,T] and the result follows. □

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

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

Proof By Lemma 9, for each n∈N, problem (7a) and (7b) has a solution un satisfying inequalities (29)-(31), where S is a positive constant and ξn∈(0,T). Moreover, by Lemma 10, the sequence { | f n ( t , u n ( t ) , u n ′ ( t ) ) | } is uniformly integrable on [0,T]. We now prove that { u n ′ } is equicontinuous on [0,T]. Since un is a fixed point of the operator K(1,⋅) given in (27), the equality

u n ″ ( t ) = − a t a + 2 ∫ 0 t s a + 1 f n ( s , u n ( s ) , u n ′ ( s ) ) d s + f n ( t , u n ( t ) , u n ′ ( t ) )

holds for a.e. t∈[0,T] and all n∈N. Let 0 ≤ t 1 < t 2 ≤ T . Then

| u n ′ ( t 2 ) − u n ′ ( t 1 ) | = | − a ∫ t 1 t 2 1 s a + 2 ( ∫ 0 s ξ a + 1 f n ( ξ , u n ( ξ ) , u n ′ ( ξ ) ) d ξ ) d s + ∫ t 1 t 2 f n ( s , u n ( s ) , u n ′ ( s ) ) d s | .

Let r n ( t ) = 1 t a + 1 ∫ 0 t s a + 1 f n ( s , u n ( s ) , u n ′ ( s ) ) d s . By Lemma 3(i), { r n } ⊂ C [ 0 , T ] and r n ( 0 ) = 0 . Integrating by parts yields

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 |

follows. By Lemma 4 (for ρ n ( t ) = f n ( t , u n ( t ) , u n ′ ( t ) ) ), the sequence { r n } is equicontinuous on [0,T]. Since the sequence {|fn(t,un(t),un′(t))|} is uniformly integrable on [0,T], the sequence { ∫ 0 t f n ( s , u ( s ) , u ′ ( s ) ) d s } is equicontinuous on [0,T]. Hence, it follows from (33), that { u n ′ } is equicontinuous on [0,T]. We summarize: { u n } is bounded in C1[0,T] and { u n ′ } is equicontinuous on [0,T]. Also, { ξ n } ⊂ ( 0 , T ) . Using appropriate subsequences, if necessary, we can assume, by the Arzelà-Ascoli theorem and the Bolzano-Weierstrass theorem, that {un} is convergent in C1[0,T] and {ξn} is convergent in ℝ. Let lim n → ∞ u n = : u and lim n → ∞ ξ n = : ξ . With n → ∞ in (29)-(31), we conclude