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

Irena Rachůnková; Alexander Spielauer; Svatoslav Staněk; Ewa B Weinmüller

doi:10.1186/1687-2770-2013-6

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Open Access
Published: 16 January 2013

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

Irena Rachůnková¹,
Alexander Spielauer²,
Svatoslav Staněk¹ &
Ewa B Weinmüller²

Boundary Value Problems volume 2013, Article number: 6 (2013) Cite this article

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Abstract

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

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

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

MSC:34B18, 34B16, 34A12.

1 Introduction

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

(1a)

(1b)

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

(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 [1–4], in population genetics, especially in models for the spatial distribution of the genetic composition of a population [5, 6], in the homogeneous nucleation theory [7], in relativistic cosmology for description of particles which can be treated as domains in the universe [8], and in the nonlinear field theory [9], in particular, when describing bubbles generated by scalar fields of Higgs type in the Minkowski spaces [10].

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

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

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

The investigation of singular Dirichlet BVPs has a long history and a lot of methods for their analysis are available. One of the most important ones is the topological degree method providing various fixed point theorems and existence alternative theorems; see, e.g., Lemma 1. For more information on the topological degree method and its application to numerous BVPs, including Dirichlet problems, we refer the reader to the monographs by Mawhin [13–15].

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 + .

(2)

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

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

and

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

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

(3)

(4)

Hence,

ε ≤ − λ f n (t, x, y) + (1 − λ) ε for a.e. t ∈ [0, T] and all x, y ∈ R, λ ∈ [0, 1],

(5)

and

− λ f n (t, x, y) + (1 − λ) ε ≤ φ (t) h (1 + | x |, 1 + | y |) + g (| x |) + r (| y |) for a.e. t ∈ [0, T] and all x, y ∈ R 0, λ ∈ [0, 1] .}

(6)

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

(7a)

(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 $ℓ ∈ U$ . 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. [17–19].

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 $a ∈ (− 1, 0)$ .

2 Preliminaries

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

Lemma 3 Let $ρ ∈ L 1 [0, T]$ 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
$H ″ (t) + a t H ′ (t) − a t 2 H (t) = − ρ (t)$
(8)

holds for a.e. $t ∈ [0, T]$ .

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

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

(9)

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

(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

(10)

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

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

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

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

In particular,

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

(11)

Hence, cf. (10),

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

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

Lemma 4 Let ${ρ n} ⊂ L 1 [0, T]$ be a uniformly integrable sequence on $[0, T]$ and let $lim n → ∞ ρ n (t) = ρ (t)$ for a.e. $t ∈ [0, T]$ . Then the sequence

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

(12)

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

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

(13)

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

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

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

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

we conclude from the relation

that

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

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

Lemma 5 Let $ρ ∈ L 1 [0, T]$ . Then

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

(14)

Proof Since (cf. (10))

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

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

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

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

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

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

(15)

holds for $t ∈ [0, T]$ .

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

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

and

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

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

We now discuss the linear Euler differential equation

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

(16)

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

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

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

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

4 Auxiliary regular problems

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

(17a)

(17b)

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

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

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

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

(18)

in the set $C 1 [0, T]$ .

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

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

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

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

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

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

(19)

(20)

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

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

(21)

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

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

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

∫ 0 T g (u (t)) d t ≤ L .

(22)

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

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

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

(23)

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

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

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

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

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

In particular,

∥ r (| u ′ (t) |) ∥ 1 < V .

(24)

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

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

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

∥ u ∥ ∞ ≤ T ∥ u ′ ∥ ∞,

(25)

and therefore, we have

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

(26)

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

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

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

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

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

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

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

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

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

(27)

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

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

It follows from Lemma 5 and (9) that

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

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

(28)

Since $lim m → ∞ f n (t, x m (t), x m ′ (t)) = f n (t, x (t), x ′ (t))$ for a.e. $t ∈ [0, T]$ and there exists $ρ ∈ L 1 [0, T]$ such that

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

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

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

Then by Lemma 5 and (9), the inequalities

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

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

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

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

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

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

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

Proof

By Lemma 9, the inequalities

(29)

(30)

(31)

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

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

(32)

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

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

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

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

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

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

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

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

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

and

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

(33)

follows. By Lemma 4 (for $ρ n (t) = f n (t, u n (t), u n ′ (t))$ ), the sequence ${r n}$ is equicontinuous on $[0, T]$ . Since the sequence ${| f n (t, u n (t), u n ′ (t)) |}$ is uniformly integrable on $[0, T]$ , the sequence ${∫ 0 t f n (s, u (s), u ′ (s)) d s}$ is equicontinuous on $[0, T]$ . Hence, it follows from (33), that ${u n ′}$ is equicontinuous on $[0, T]$ . We summarize: ${u n}$ is bounded in $C 1 [0, T]$ and ${u n ′}$ is equicontinuous on $[0, T]$ . Also, ${ξ n} ⊂ (0, T)$ . Using appropriate subsequences, if necessary, we can assume, by the Arzelà-Ascoli theorem and the Bolzano-Weierstrass theorem, that ${u n}$ is convergent in $C 1 [0, T]$ and ${ξ n}$ is convergent in ℝ. Let $lim n → ∞ u n = : u$ and $lim n → ∞ ξ n = : ξ$ . With $n → ∞$ in (29)-(31), we conclude

In addition, $u (0) = 0$ , $u (T) = 0$ . Since

lim n → ∞ f n (t, u n (t), u n ′ (t)) = f (t, u (t), u ′ (t)) for a.e. t ∈ [0, T],

(34)

it follows from Lemma 2 that

lim n → ∞ ∥ f n (t, u n (t), u n ′ (t)) − f (t, u (t), u ′ (t)) ∥ 1 = 0

and $f (t, u (t), u ′ (t)) ∈ L 1 [0, T]$ . We now deduce from the inequality (cf. Lemma 5)

that

Taking the limit $n → ∞$ in

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

we have

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

(35)

Hence,

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

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

6 Numerical simulations

For the numerical simulation, we choose $T = 1$ and use an alternative formulation of problem (1a) and (1b),

(36a)

(36b)

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

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

In what follows, we work with $f (t, x, y) = q (t) + h (x, y)$ for a.e. $t ∈ [0, 1]$ and all $x ∈ R +$ , $y ∈ R ∖ {0}$ and, according to the next numerical approach (see Section 6.2), we consider Eq. (36a), where $h ≡ 0$ , that is,

u ″ (t) + a t u ′ (t) − a t 2 u (t) = q (t) .

(37)

By [23], problem (37), (36b) is well posed and therefore it is suitable for the numerical treatment. To see this, we need to look at a general solution of the homogeneous equation

u ″ (t) + a t u ′ (t) − a t 2 u (t) = 0, t ∈ [0, 1] .

(38)

If we set $u (t) = t λ$ , we arrive at the characteristic polynomial of (38),

λ 2 − (1 − a) λ − a = 0

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

6.1 MATLAB Code bvpsuite

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

6.2 Preliminaries

Before dealing with two nonlinear models specified in Sections 6.3 and 6.4, we have to compute numerical solutions for a simpler linear^c model of the form

(39a)

(39b)

where a was chosen as $a = − 0.1, − 0.5, − 0.9$ . Since in this case the exact solution is given, $u (t) = 2 t (1 − t) a + 1.5$ , the value $u ′ (1)$ is available, $u ′ (1) = − 0.72, − 1, − 1.67$ , respectively. In Figure 1, the numerical solutions of BVPs (39a) and (39b) are shown. They will be used as starting values for the numerical solution of Examples 1 and 2; see Sections 6.3 and 6.4, respectively. All numerical results have been obtained using collocation with five Gaussian collocation points on an equidistant grid (justified by a very simple solution structure) with the step size 0.01.

6.3 Example 1

We first investigate the following problem:

(40a)

(40b)

The nonlinearity f in (40a) has the form

f (t, x) = − 1 t − x 13

(41)

and it satisfies (H₁)-(H₃) with $ε = 1$ , $φ (t) = 1 t$ for $t ∈ (0, 1]$ and

h (x, y) = 1, g (x) = x 13, r (y) = 0 for x, y ∈ R + .

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

Let u be a solution of problem (1a) and (1b) with f from (41). Then by (35), we obtain

| u (t) | ≤ t ∫ t 1 1 s a + 2 (∫ 0 s ξ a + 1 (1 ξ + ∥ u ∥ ∞ 13) d ξ) d s < 4 2 a + 3 + ∥ u ∥ ∞ 13 4 (a + 2), t ∈ [0, 1] .

Therefore,

∥ u ∥ ∞ < 4 2 a + 3 + ∥ u ∥ ∞ 13 4 (a + 2) .

(42)

Let $K > 0$ satisfy

K = 4 2 a + 3 + K 13 4 (a + 2) .

(43)

Then (42) implies $∥ u ∥ ∞ < K$ and due to (35) and (41),

u ′ (1) = ∫ 01 t a + 1 f (t, u (t), u ′ (t)) d t = ∫ 01 t a + 1 (− 1 t − u 13 (t)) d t ≥ − ∫ 01 (t a + 12 + K 13 t a + 1) d t = − 2 2 a + 3 − K 13 1 a + 2 .

Consequently,

c ∗ = 2 2 a + 3 + K 13 1 a + 2 .

(44)

In order to solve the nonlinear problem (40a) and (40b), we first have to solve a series of auxiliary problems for parameter-dependent differential equations

u ″ (t) + a t u ′ (t) − a t 2 u (t) = − 1 t − δ u 13 (t), t ∈ (0, 1], δ > 0 .

(45)

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

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

1.
Numerical approximation of BVP (39a) and (39b) is used as an initial guess for ODE (45) with $δ = 0$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 0.5$ .
2.
Use the above approximation as an initial guess for ODE (45) with $δ = 0.01$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 1.0$ .
3.
Use the above approximation as an initial guess for ODE (45) with $δ = 0.1$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 1.0$ .
4.
Use the above approximation as an initial guess for ODE (45) with $δ = 1.0$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 1.0$ .

After the last step, we have solved problem (40a) and (40b) subject to boundary conditions $u (1) = 0$ , $u ′ (1) = − 1.0$ . In this case, the value of $u (0)$ was not small enough to consider it a reasonable approximation for $u (0) = 0$ . Therefore, we use a shooting idea combined with a bisection strategy to find a better value for $c = − u ′ (1)$ . The complete numerical results for Example 1 can be found in Table 1 and Figure 2.

Table 1 Problem ( 40a ) and ( 40b ): Complete data of the numerical simulation for different values of a

Full size table

6.4 Example 2

The above approach has been also accordingly applied for Example 2. Here, we consider the problem

(46a)

(46b)

The right-hand side f in Eq. (46a) now reads

f (t, x) = − 1 t − x − 13

(47)

and has a singularity at $x = 0$ . The function f satisfies conditions (H₁)-(H₃) with $ε = 1$ , $φ (t) = 1 t$ for $t ∈ (0, 1]$ and

h (x, y) = 1, g (x) = x − 13, r (y) = 0 for x, y ∈ R + .

Theorem 1 guarantees the existence of at least one $c > 0$ such that a solution u of problem (46a) and (46b) is positive on $(0, 1)$ and $u (0) = 0$ holds. We now again determine an interval $(0, c ∗) ⊂ (0, ∞)$ containing all such values of c. Let u be a solution of problem (1a) and (1b) with f given in (47). Inequality (19) yields

u (t) ≥ 1 a + 2 t (1 − t), t ∈ [0, 1],

and hence by (35),

u ′ (1) = ∫ 01 t a + 1 f (t, u (t), u ′ (t)) d t = ∫ 01 t a + 1 (− 1 t − u − 13 (t)) d t ≥ ∫ 01 t a + 1 (− 1 t − (1 a + 2 t (1 − t)) − 13) d t .

Consequently,

c ∗ = ∫ 01 t a + 1 (1 t + (a + 2) 1 / 3 t 1 / 3 (1 − t) 1 / 3) d t .

(48)

For Example 2, the auxiliary ODE is constructed using ODE (39a),

u ″ (t) + a t u ′ (t) − a t 2 u (t) = − 1 t − δ u − 13, t ∈ [0, 1], δ > 0 .

(49)

For all values of a, we choose $u ′ (1) ∈ (− c ∗, 0)$ and analogously carry out the path-following in δ first. The related chain for $a = − 0.1$ is as follows.

1.
Numerical approximation of BVP (39a) and (39b) is used as an initial guess for ODE (49) with $δ = 0$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 0.8$ .
2.
Use the above approximation as an initial guess for ODE (49) with $δ = 0.01$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 0.8$ .
3.
Use the above approximation as an initial guess for ODE (49) with $δ = 0.1$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 1.2$ .
4.
Use the above approximation as an initial guess for ODE (49) with $δ = 0.5$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 1.3$ .
5.
Use the above approximation as an initial guess for ODE (49) with $δ = 1.0$ subject to terminal conditions $u (1) = 0$ , $u ′ (1) = − 1.7$ .

After the last step, we have solved BVP (46a) and (46b) subject to boundary conditions $u (1) = 0$ , $u ′ (1) = − 1.7$ , but also, in this case, the value of $u (0)$ is too large and we have to find a better value for $c = − u ′ (1)$ . The complete numerical results for Example 2 can be found in Table 2 and Figure 3.

Table 2 Problem ( 46a ) and ( 46b ): Complete data of the numerical simulation for different values of a

Full size table

7 Conclusions

In the present article, we deal with the existence of positive solutions to the singular Dirichlet problem of the form

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

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

u ″ (t) + a t u ′ (t) − a t 2 u (t) = f (t, u (t), u ′ (t)), u (T) = 0, u ′ (T) = − c .

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

Endnotes

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

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

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

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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.

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Affiliations

Department of Mathematical Analysis, Faculty of Science, Palacký University, 17. listopadu 12, Olomouc, CZ-771 46, Czech Republic
- Irena Rachůnková
- & Svatoslav Staněk
Department of Analysis and Scientific Computing, Vienna University of Technology, Wiedner Hauptstraße 8-10, Wien, A-1040, Austria
- Alexander Spielauer
- & Ewa B Weinmüller

Authors

Irena Rachůnková
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Alexander Spielauer
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Svatoslav Staněk
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Ewa B Weinmüller
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Correspondence to Irena Rachůnková.

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The authors declare that they have no competing interests.

Authors’ contributions

IR and SS contributed to the analytical part of the work and AS and EBW contributed to its numerical part. All authors read and approved the final version of the manuscript.

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Rachůnková, I., Spielauer, A., Staněk, S. et al. Positive solutions of nonlinear Dirichlet BVPs in ODEs with time and space singularities. Bound Value Probl 2013, 6 (2013). https://doi.org/10.1186/1687-2770-2013-6

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Received: 17 October 2012
Accepted: 21 December 2012
Published: 16 January 2013
DOI: https://doi.org/10.1186/1687-2770-2013-6

Keywords

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