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# Second-order initial value problems with singularities

## Abstract

Using barrier strip arguments, we investigate the existence of $C[0,T]∩ C 2 (0,T]$-solutions to the initial value problem $x ″ =f(t,x, x ′ )$, $x(0)=A$, $lim t → 0 + x ′ (t)=B$, which may be singular at $x=A$ and $x ′ =B$.

MSC: 34B15, 34B16, 34B18.

## 1 Introduction

In this paper we study the solvability of initial value problems (IVPs) of the form

$x ″ =f ( t , x , x ′ ) ,$
(1.1)
$x(0)=A, lim t → 0 + x ′ (t)=B,B>0.$
(1.2)

Here the scalar function $f(t,x,p)$ is defined on a set of the form $( D t × D x × D p )∖( S A ∪ S B )$, where $D t , D x , D p ⊆R$, $S A = T 1 ×{A}×P$, $S B = T 2 ×X×{B}$, $T i ⊆ D t$, $i=1,2$, $X⊆ D x$, $P⊆ D p$, and so it may be singular at $x=A$ and $p=B$.

IVPs of the form

$( φ ( t ) x ′ ( t ) ) ′ = φ ( t ) f ( x ( t ) ) , x ( 0 ) = A , x ′ ( 0 ) = 0 ,$

have been investigated by Rachůnková and Tomeček –. For example in , the authors have discussed the set of all solutions to this problem with a singularity at $t=0$. Here $A<0$, $φ∈C[0,∞)∩ C 1 (0,∞)$ with $φ(0)=0$, $φ ′ (t)>0$ for $t∈(0,∞)$ and $lim t → ∞ φ ′ ( t ) φ ( t ) =0$, f is locally Lipschitz on $(−∞,L]$ with the properties $f(L)=0$ and $xf(x)<0$ for $x∈(−∞,0)∪(0,L)$, where $L>0$ is a suitable constant.

Agarwal and O’Regan  have studied the problem

$x ″ = φ ( t ) f ( t , x , x ′ ) , t ∈ ( 0 , T ] , x ( 0 ) = x ′ ( 0 ) = 0 ,$

where $f(t,x,p)$ may be singular at $x=0$ and/or $p=0$. The obtained results give a positive $C 1 [0,T]∩ C 2 (0,T]$-solution under the assumptions that $φ∈C[0,T]$, $φ(t)>0$ for $t∈(0,T]$, $f:[0,T]× ( 0 , ∞ ) 2 →(0,∞)$ is continuous and

where g, h, r, and w are suitable functions.

IVPs of the form

$x ″ ( t ) = f ( t , x ( t ) , x ′ ( t ) ) , 0 < t < 1 , x ( 0 ) = x ′ ( 0 ) = 0 ,$

where $f(t,x,p)∈C((0,1)× ( 0 , ∞ ) 2 )$, maybe singular at $t=0$, $t=1$, $x=0$ or $p=0$, have been studied by Yang , . The solvability in $C 1 [0,1]$ and $C[0,1]∩ C 2 (0,1)$ is established in these works, respectively, under the assumption that

where k, F, and G are suitable functions.

The solvability of various IVPs has been studied also by Bobisud and O’Regan , Bobisud and Lee , Cabada and Heikkilä , Cabada et al. , , Cid , Maagli and Masmoudi , and Zhao . Existence results for problem (1.1), (1.2) with a singularity at the initial value of $x ′$ have been reported in Kelevedjiev-Popivanov .

Here, as usual, we use regularization and sequential techniques. Namely, we proceed as follows. First, by means of the topological transversality theorem , we prove an existence result guaranteeing $C 2 [a,T]$-solutions to the nonsingular IVP for equations of the form (1.1) with boundary conditions

$x(a)=A, x ′ (a)=B.$

Moreover, we establish the needed a priori bounds by the barrier strips technique. Further, the obtained existence theorem assures $C 2 [0,T]$-solutions for each nonsingular IVP included in the family

$x ″ = f ( t , x , x ′ ) , x ( 0 ) = A + n − 1 , x ′ ( 0 ) = B − n − 1 ,$
(1.3)

where $n∈N$ is suitable. Finally, we apply the Arzela-Ascoli theorem on the sequence ${ x n }$ of $C 2 [0,T]$-solutions thus constructed to (1.3) to extract a uniformly convergent subsequence and show that its limit is a $C[0,T]∩ C 2 (0,T]$-solution to singular problem (1.1), (1.2). In the case $A≥0$, $B≥0$ we establish $C[0,T]∩ C 2 (0,T]$-solutions with important properties - monotony and positivity.

We have used variants of the approach described above for various boundary value problems (BVPs); see Grammatikopoulos et al. , Kelevedjiev and Popivanov  and Palamides et al. . For example in , we have established the existence of positive solutions to the BVP

$g ( t , x , x ′ , x ″ ) = 0 , t ∈ ( 0 , 1 ) , x ( 0 ) = 0 , x ′ ( 1 ) = B , B > 0 ,$

which may be singular at $x=0$. Note that despite the more general equation of this problem, the conditions imposed here as well as the results obtained are not consequences of those in .

## 2 Topological transversality theorem

In this short section we state our main tools - the topological transversality theorem and a theorem giving an important property of the constant maps.

So, let X be a metric space and Y be a convex subset of a Banach space E. Let $U⊂Y$ be open in Y. The compact map $F: U ¯ →Y$ is called admissible if it is fixed point free on ∂U. We denote the set of all such maps by $L ∂ u ( U ¯ ,Y)$.

A map F in $L ∂ u ( U ¯ ,Y)$ is essential if every map G in $L ∂ u ( U ¯ ,Y)$ such that $G|∂U=F|∂U$ has a fixed point in U. It is clear, in particular, every essential map has a fixed point in U.

### Theorem 2.1

(, Chapter I, Theorem 2.2])

Let$p∈U$be fixed and$F∈ L ∂ u ( U ¯ ,Y)$be the constant map$F(x)=p$for$x∈ U ¯$. Then F is essential.

We say that the homotopy${ H λ :X→Y}$, $0≤λ≤1$, is compact if the map$H(x,λ):X×[0,1]→Y$given by$H(x,λ)≡ H λ (x)$for$(x,λ)∈X×[0,1]$is compact.

### Theorem 2.2

(, Chapter I, Theorem 2.6])

Let Y be a convex subset of a Banach space E and$U⊂Y$be open. Suppose:

1. (i)

$F,G: U ¯ →Y$ are compact maps.

2. (ii)

$G∈ L ∂ U ( U ¯ ,Y)$ is essential.

3. (iii)

$H(x,λ)$, $λ∈[0,1]$, is a compact homotopy joining F and G, i.e.

$H(x,1)=F(x)andH(x,0)=G(x).$
1. (iv)

$H(x,λ)$, $λ∈[0,1]$, is fixed point free on ∂U.

Then$H(x,λ)$, $λ∈[0,1]$, has at least one fixed point in U and in particular there is a$x 0 ∈U$such that$x 0 =F( x 0 )$.

## 3 Nonsingular problem

Consider the IVP

${ x ″ = f ( t , x , x ′ ) , x ( a ) = A , x ′ ( a ) = B , B ≥ 0 ,$
(3.1)

where $f: D t × D x × D p →R$, $D t , D x , D p ⊆R$.

We include this problem into the following family of regular IVPs constructed for $λ∈[0,1]$

${ x ″ = λ f ( t , x , x ′ ) , x ( a ) = A , x ′ ( a ) = B ,$
(3.2)

and suppose the following.

1. (R)

There exist constants $T>a$, $m 1$, $m ¯ 1$, $M 1$, $M ¯ 1$, and a sufficiently small $τ>0$ such that

$m 1 ≥ 0 , M ¯ 1 − τ ≥ M 1 ≥ B ≥ m 1 ≥ m ¯ 1 + τ , [ a , T ] ⊆ D t , [ A − τ , M 0 + τ ] ⊆ D x , [ m ¯ 1 , M ¯ 1 ] ⊆ D p ,$

where $M 0 =A+ M 1 (T−a)$,

(3.3)

where $D M 0 = D x ∩(−∞, M 0 ]$.

Our first result ensures bounds for the eventual $C 2$-solutions to (3.2). We need them to prepare the application of the topological transversality theorem.

### Lemma 3.1

Let (R) hold. Then each solution$x∈ C 2 [a,T]$to the family (3.2) λ , $λ∈[0,1]$, satisfies the bounds

where

$m 2 = min { f ( t , x , p ) : ( t , x , p ) ∈ [ a , T ] × [ A , M 0 ] × [ m 1 , M 1 ] } , M 2 = max { f ( t , x , p ) : ( t , x , p ) ∈ [ a , T ] × [ A , M 0 ] × [ m 1 , M 1 ] } .$

### Proof

Suppose that the set

$S − = { t ∈ [ a , T ] : M 1 < x ′ ( t ) ≤ M ¯ 1 }$

is not empty. Then

$x ′ (a)=B≤ M 1 and x ′ ∈C[a,T]$

imply that there exists an interval $[α,β]⊂ S −$ such that

$x ′ (α)< x ′ (β).$

This inequality and the continuity of $x ′ (t)$ guarantee the existence of some $γ∈[α,β]$ for which

$x ″ (γ)>0.$

Since $x(t)$, $t∈[a,T]$, is a solution of the differential equation, we have $(t,x(t), x ′ (t))∈[a,T]× D x × D p$. In particular for γ we have

$( γ , x ( γ ) , x ′ ( γ ) ) ∈ S − × D x ×( M 1 , M ¯ 1 ].$

Thus, we apply (R) to conclude that

$x ″ (γ)=λf ( γ , x ( γ ) , x ′ ( γ ) ) ≤0,$

which contradicts the inequality $x ″ (γ)>0$. This has been established above. Thus, $S −$ is empty and as a result

Now, by the mean value theorem for each $t∈(a,T]$ there exists a $ξ∈(a,t)$ such that

$x(t)−x(a)= x ′ (ξ)(t−a),$

which yields

This allows us to use (3.3) to show similarly to above that the set

$S + = { t ∈ [ a , T ] : m ¯ 1 ≤ x ′ ( t ) < m 1 }$

is empty. Hence,

and so

To estimate $x ″ (t)$, we observe firstly that (R) implies in particular

and

which yield $m 2 ≤0$ and $M 2 ≥0$. Multiplying both sides of the inequality $λ≤1$ by $m 2$ and $M 2$, we get, respectively, $m 2 ≤λ m 2$ and $λ M 2 ≤ M 2$. On the other hand, we have established

Thus,

and each $λ∈[0,1]$ and so

□

Let us mention that some analogous results have been obtained in Kelevedjiev . For completeness of our explanations, we present the full proofs here.

Now we prove an existence result guaranteeing the solvability of IVP (3.1).

### Theorem 3.2

Let (R) hold. Then nonsingular problem (3.1) has at least one non-decreasing solution in$C 2 [a,T]$.

### Proof

Preparing the application of Theorem 2.2, we define first the set

$U= { x ∈ C I 2 [ a , T ] : A − τ < x < M 0 + τ , m 1 − τ < x ′ < M 1 + τ , m 2 − τ < x ″ < M 2 + τ } ,$

where $C I 2 [a,T]={x∈ C 2 [a,T]:x(a)=A, x ′ (a)=B}$. It is important to notice that according to Lemma 3.1 all $C 2 [a,T]$-solutions to family (3.2) are interior points of U. Further, we introduce the continuous maps

and for $t∈[a,T]$ and $x(t)∈j( U ¯ )$ the map

Clearly, the map Φ is also continuous since, by assumption, the function $f(t,x(t), x ′ (t))$ is continuous on $[a,T]$ if

In addition we verify that $V − 1$ exists and is also continuous. To this aim we introduce the linear map

$W: C I 0 2 [a,T]→C[a,T],$

defined by $Wx= x ″$, where $C I 0 2 [a,T]={x∈ C 2 [a,T]:x(a)=0, x ′ (a)=0}$. It is one-to-one because each function $x∈ C I 0 2 [a,T]$ has a unique image, and each function $y∈C[a,T]$ has a unique inverse image which is the unique solution to the IVP

$x ″ =y,x(a)=0, x ′ (a)=0.$

It is not hard to see that W is bounded and so, by the bounded inverse theorem, the map $W − 1$ exists and is linear and bounded. Thus, it is continuous. Now, using $W − 1$, we define

where $ℓ(t)=B(t−a)+A$ is the unique solution of the problem

$x ″ =0,x(a)=A, x ′ (a)=B.$

Clearly, $V − 1$ is continuous since $W − 1$ is continuous.

We already can introduce a homotopy

$H: U ¯ ×[0,1]→ C I 2 [a,T],$

defined by

$H(x,λ)≡ H λ (x)≡λ V − 1 Φj(x)+(1−λ)ℓ.$

It is well known that j is completely continuous, that is, j maps each bounded subset of $C I 2 [a,T]$ into a compact subset of $C 1 [a,T]$. Thus, the image $j( U ¯ )$ of the bounded set U is compact. Now, from the continuity of Φ and $V − 1$ it follows that the sets $Φ(j( U ¯ ))$ and $V − 1 (Φ(j( U ¯ )))$ are also compact. In summary, we have established that the homotopy is compact. On the other hand, for its fixed points we have

$λ V − 1 Φj(x)+(1−λ)ℓ=x$

and

$Vx=λΦj(x),$

which is the operator form of family (3.2). So, each fixed point of $H λ$ is a solution to (3.2), which, according to Lemma 3.1, lies in U. Consequently, the homotopy is fixed point free on ∂U.

Finally, $H 0 (x)$ is a constant map mapping each function $x∈ U ¯$ to $ℓ(t)$. Thus, according to Theorem 2.1, $H 0 (x)=ℓ$ is essential.

So, all assumptions of Theorem 2.2 are fulfilled. Hence $H 1 (x)$ has a fixed point in U which means that the IVP of (3.2) obtained for $λ=1$ (i.e. (3.1)) has at least one solution $x(t)$ in $C 2 [a,T]$. From Lemma 3.1 we know that

from which its monotony follows. □

The validity of the following results follows similarly.

### Theorem 3.3

Let$B>0$and let (R) hold for$m 1 >0$. Then problem (3.1) has at least one strictly increasing solution in$C 2 [a,T]$.

### Theorem 3.4

Let$A>0$ ($A=0$) and let (R) hold for$m 1 =0$. Then problem (3.1) has at least one positive (nonnegative) non-decreasing solution in$C 2 [a,T]$.

### Theorem 3.5

Let$A≥0$, $B>0$and let (R) hold for$m 1 >0$. Then problem (3.1) has at least one strictly increasing solution in$C 2 [a,T]$with positive values for$t∈(a,T]$.

## 4 A problem singular at x and $x ′$

In this section we study the solvability of singular IVP (1.1), (1.2) under the following assumptions.

(S1): There are constants $T>0$, $m 1$, $m ¯ 1$ and a sufficiently small $ν>0$ such that

$m 1 > 0 , B > m 1 ≥ m ¯ 1 + ν , [ 0 , T ] ⊆ D t , ( A , M ˜ 0 + ν ] ⊆ D x , [ m ¯ 1 , B ) ⊆ D p ,$

where $M ˜ 0 =A+BT+1$,

$f(t,x,p)∈C ( [ 0 , T ] × ( A , M ˜ 0 + ν ] × [ m 1 − ν , B ) ) ,$
(4.1)
(4.2)

and

where $D M ˜ 0 =(−∞, M ˜ 0 ]∩ D x$.

(S2): For some $α∈(0,T]$ and $μ∈( m 1 ,B)$ there exists a constant $k<0$ such that $kα+B>μ$ and

where T, $m 1$ and $M ˜ 0$ are as in (S1).

Now, for $n≥ n α , μ$, where $n α , μ >max{ α − 1 , ( B + k α − μ ) − 1 }$, and α, μ, and k are as in (S2), we construct the following family of regular IVPs:

${ x ″ = f ( t , x , x ′ ) , x ( 0 ) = A + n − 1 , x ′ ( 0 ) = B − n − 1 .$
(4.3)

Notice, for $n≥ n α , μ$, that we have $B− n − 1 >μ−kα>μ> m 1 >0$.

### Lemma 4.1

Let (S1) and (S2) hold and let$x n ∈ C 2 [0,T]$, $n≥ n α , μ$, be a solution to (4.3) such that

Then the following bound is satisfied for each$n≥ n α , μ$:

where $ϕ α (t)= { k t + B , t ∈ [ 0 , α ] , k α + B , t ∈ ( α , T ] .$

### Proof

Since for each $n≥ n α , μ$ we have

$x n ′ (0)=B− n − 1 >μ−kα>μ,$

we will consider the proof for an arbitrary fixed $n≥ n α , μ$, considering two cases. Namely, $x n ′ (t)>μ$ for $t∈[0,α]$ is the first case and the second one is $x n ′ (t)>μ$ for $t∈[0,β)$ with $x n ′ (β)=μ$ for some $β∈(0,α]$.

Case 1. From $μ< x n ′ (t)≤B$, $t∈[0,α]$, and (S2) we have

i.e.$x n ″ (t)≤k$ for $t∈[0,α]$. Integrating the last inequality from 0 to t we get

$x n ′ (t)− x n ′ (0)≤kt,t∈[0,α],$

which yields

Now $m 1 ≤ x n ′ (t), $t∈[0,T]$, and (4.2) imply

In particular $x n ″ (t)≤0$ for $t∈[α,T]$, thus

Case 2. As in the first case, we derive

On the other hand, since $m 1 ≤ x n ′ (t) for $t∈[β,T]$, again from (4.2) it follows that

which yields

and

So, as a result of the considered cases we get

from which the assertion follows immediately. □

Having this lemma, we prove the basic result of this section.

### Theorem 4.2

Let (S1) and (S2) hold. Then singular IVP (1.1), (1.2) has at least one strictly increasing solution in$C[0,T]∩ C 2 (0,T]$such that

### Proof

For each fixed $n≥ n α , μ$ introduce $τ=min{ ( 2 n ) − 1 ,ν}$,

$M 1 =B− n − 1 , M ¯ 1 =B− ( 2 n ) − 1 and M 0 = ( B − n − 1 ) T+A+1< M ˜ 0$

having the properties

$M ¯ 1 − τ > M 1 = B − n − 1 > μ − k α > μ > m 1 ≥ m ¯ 1 + τ , [ 0 , T ] ⊆ D t , [ A + n − 1 − τ , M 0 + τ ] ⊆ ( A , M ˜ 0 + τ ] ⊆ D x$

and $[ m ¯ 1 , M ¯ 1 ]⊆ D p$ since $M ¯ 1 =B− ( 2 n ) − 1 . Besides,

and, in view of (4.1),

$f(t,x,p)∈C ( [ 0 , T ] × [ A + n − 1 − τ , M 0 + ν ] × [ m 1 − τ , M 1 + τ ] ) .$

All this implies that for each $n≥ n α , μ$ the corresponding IVP of family (4.3) satisfies (R). Thus, we apply Theorem 3.2 to conclude that (4.3) has a solution $x n ∈ C 2 [0,T]$ for each $n≥ n α , μ$. We can use also Lemma 3.1 to conclude that for each $n≥ n α , μ$ and $t∈[0,T]$ we have

$A
(4.4)

and

$m 1 ≤ x n ′ (t)≤B− n − 1

Now, these bounds allow the application of Lemma 4.1 from which one infers that for each $n≥ n α , μ$ and $t∈[0,T]$ the bounds

$m 1 ≤ x n ′ (t)< ϕ α (t)≤B$
(4.5)

hold. For later use, integrating the least inequality from 0 to t, $t∈(0,T]$, we get

(4.6)

and $n≥ n α , μ$.

We consider firstly the sequence ${ x n }$ of $C 2 [0,T]$-solutions of (4.3) only for each $n≥ n α , μ$. Clearly, for each $n≥ n α , μ$ we have in particular

which together with (4.6) gives

where $A 1 = m 1 α+A$. On combining the last inequality and (4.4) we obtain

(4.7)

From (4.5) we have in addition

(4.8)

Now, using the fact that (4.1) implies continuity of $f(t,x,p)$ on the compact set $[α,T]×[ A 1 , M ˜ 0 ]×[ m 1 , ϕ α (α)]$ and keeping in mind that for each $n≥ n α , μ$

we conclude that there is a constant $M 2$, independent of n, such that

Using the obtained a priori bounds for $x n (t)$, $x n ′ (t)$ and $x n ″ (t)$ on the interval $[α,T]$, we apply the Arzela-Ascoli theorem to conclude that there exists a subsequence ${ x n k }$, $k∈N$, $n k ≥ n α , μ$, of ${ x n }$ and a function $x α ∈ C 1 [α,T]$ such that

i.e., the sequences ${ x n k }$ and ${ x n k ′ }$ converge uniformly on the interval $[α,T]$ to $x α$ and $x α ′$, respectively. Obviously, (4.7) and (4.8) are valid in particular for the elements of ${ x n k }$ and ${ x n k ′ }$, respectively, from which, letting $k→∞$, one finds

Clearly, the functions $x n k (t)$, $k∈N$, $n k ≥ n α , μ$, satisfy integral equations of the form

$x n k ′ (t)= x n k ′ (α)+ ∫ α t f ( s , x n k ( s ) , x n k ′ ( s ) ) ds,t∈(α,T].$

Now, since $f(t,x,p)$ is uniformly continuous on the compact set $[α,T]×[ A 1 , M ˜ 0 ]×[ m 1 , ϕ α (α)]$, from the uniform convergence of ${ x n k }$ it follows that the sequence ${f(s, x n k (s), x n k ′ (s))}$, $n k ≥ n α , μ$ is uniformly convergent on $[α,T]$ to the function $f(s, x α (s), x α ′ (s))$, which means

$lim k → ∞ ∫ α t f ( s , x n k ( s ) , x n k ′ ( s ) ) ds= ∫ α t f ( s , x α ( s ) , x α ′ ( s ) ) ds$

for each $t∈(α,T]$. Returning to the integral equation and letting $k→∞$ yield

$x α ′ (t)= x α ′ (α)+ ∫ α t f ( s , x α ( s ) , x α ′ ( s ) ) ds,t∈(α,T],$

which implies that $x α (t)$ is a $C 2 (α,T]$-solution to the differential equation $x ″ =f(t,x, x ′ )$ on $(α,T]$. Besides, (4.6) implies

Further, we observe that if the condition (S2) holds for some $α>0$, then it is true also for an arbitrary $α 0 ∈(0,α)$. We will use this fact considering a sequence ${ α i }⊂(0,α)$, $i∈N$, with the properties

For each $i∈N$ we consider sequences

${ x i , n k }, n k ≥ n i + 1 , μ ,k∈N, n i + 1 , μ >max { α i + 1 − 1 , ( B + k α i + 1 − μ ) − 1 } ,$

on the interval $[ α i + 1 ,T]$. Thus, we establish that each sequence ${ x i , n k }$ has a subsequence ${ x i + 1 , n k }$, $k∈N$, $n k ≥ n i + 1 , μ$, converging uniformly on the interval $[ α i + 1 ,T]$ to any function $x α i + 1 (t)$, $t∈[ α i + 1 ,T]$, that is,

(4.9)

which is a $C 2 ( α i + 1 ,T]$-solution to the differential equation $x ″ (t)=f(t,x(t), x ′ (t))$ on $( α i + 1 ,T]$ and

The properties of the functions from ${ x α i }$, $i∈N$, imply that there exists a function $x 0 (t)$ which is a $C 2 (0,T]$-solution to the equation $x ″ =f(t,x, x ′ )$ on the interval $(0,T]$ and is such that

hence $lim t → 0 + x 0 (t)=A$,

(4.10)

We have to show also that

$lim t → 0 + x 0 ′ (t)=B.$
(4.11)

Reasoning by contradiction, assume that there exists a sufficiently small $ε>0$ such that for every $δ>0$ there is a $t∈(0,δ)$ such that

$x 0 ′ (t)

In other words, assume that for every sequence ${ δ j }⊂(0,T]$, $j∈N$, with $lim j → ∞ δ j =0$, there exists a sequence ${ t j }$ having the properties $t j ∈(0, δ j )$, $lim j → ∞ t j =0$ and

$x 0 ′ ( t j )
(4.12)

It is clear that every interval $(0, δ j )$, $j∈N$, contains a subsequence of ${ t j }$ converging to 0. Besides, from (4.9) and (4.10) it follows that for every $j∈N$ there are $i j , n j ∈N$ such that $α i j < δ j$ and

(4.13)

for all $i> i j$ and all $n k ≥max{ n i , μ , n j }$. Moreover, since the accumulation point of ${ t j }$ is 0, for each sufficiently large $j∈N$ there is a $t j ∈[ α i , δ j )$ where $i> i j$. In summary, for every sufficiently large $j∈N$, that is, for every sufficiently small $δ j >0$, there are $i j , n j ∈N$ such that for all $i> i j$ and $n k ≥max{ n i , μ , n j }$ from (4.12) and (4.13) we have

$x i , n k ′ ( t j )

which contradicts to the fact that $x i , n k ′ (0)=B− n k − 1$ and $x i , n k ′ ∈C[0,T]$. This contradiction proves that (4.11) is true.

Now, it is easy to verify that the function

$x(t)= { A , t = 0 , x 0 ( t ) , t ∈ ( 0 , T ] ,$

is a $C[0,T]∩ C 2 (0,T]$-solution to (1.1), (1.2). This function is strictly increasing because $x ′ (t)= x 0 ′ (t)≥ m 1 >0$ for $t∈(0,T]$, and the bounds for $x(t)$ and $x ′ (t)$ follows immediately from the corresponding bounds for $x 0 (t)$ and $x 0 ′ (t)$. □

The following results provide information about the presence of other useful properties of the assured solutions. Their correctness follows directly from Theorem 4.2.

### Theorem 4.3

Let$A≥0$and let (S1) and (S2) hold. Then the singular IVP (1.1), (1.2) has at least one strictly increasing solution in$C[0,T]∩ C 2 (0,T]$with positive values for$t∈(0,T]$.

## 5 Examples

### Example 5.1

Consider the IVP

$x ″ = b 2 − x 2 c 2 − t 2 P k ( x ′ ) , x ( 0 ) = 0 , x ′ ( 0 ) = B , B > 0 ,$

where $b,c∈(0,∞)$, and the polynomial $P k (p)$, $k≥2$, has simple zeroes $p 1$ and $p 2$ such that

$0< p 1

Let us note that here $D t =(−c,c)$, $D x =[−b,b]$ and $D p =R$.

Clearly, there is a sufficiently small $θ>0$ such that

$0< p 1 −θ, p 1 +θ≤B≤ p 2 −θ$

and $P k (p)≠0$ for $p∈[ p 1 −θ, p 1 )∪( p 1 , p 1 +θ]∪[ p 2 −θ, p 2 )∪( p 2 , p 2 +θ]$.

We will show that all assumptions of Theorem 3.2 are fulfilled in the case

the other cases as regards the sign of $P k (p)$ around $p 1$ and $p 2$ may be treated similarly. For this case choose $τ=θ/2$, $m 1 = p 1 >0$ and $M 1 = p 2$. Next, using the requirement $[A−τ, M 0 +τ]⊆[−b,b]$, i.e.$[−θ/2, p 2 T 0 +θ/2]⊆[−b,b]$, we get the following conditions for θ and T:

$−θ/2≥−band p 2 T 0 +θ/2≤b,$

which yield $θ∈(0,2b]$ and $T≤ 2 b − θ 2 p 2$. Besides, $[0,T]⊆(−c,c)$ yields $T. Thus, $0. Now, choosing

$m ¯ 1 = p 1 −θand M ¯ 1 = p 2 +θ,$

we really can apply Theorem 3.2 to conclude that the considered problem has a strictly increasing solution $x∈ C 2 [0,T]$ with $x(t)>0$ on $t∈(0,T]$ for each $T.

### Example 5.2

Consider the IVP

$x ″ = ( x ′ − 5 ) ( 15 − x ′ ) ( x − 2 ) 2 ( x ′ − 10 ) , x ( 0 ) = 2 , lim t → 0 + x ′ ( t ) = 10 .$

Notice that here

$S A =R×{2}× ( ( − ∞ , 10 ) ∪ ( 10 , ∞ ) ) , S B =R× ( ( − ∞ , 2 ) ∪ ( 2 , ∞ ) ) ×{10}.$

It is easy to check that (S1) holds, for example, for $m ¯ 1 =4$, $m 1 =5$, $ν=0.1$, and an arbitrary fixed $T>0$, moreover, $M ˜ 0 =10T+3$. Besides, for $k=−24/ ( 10 T + 1 ) 2$, $α=T/100$ and $μ=9$, for example, we have

$kα+B=−24T/100 ( 10 T + 1 ) 2 +10>9=μ$

and $f(t,x,p)≤−24/ ( 10 T + 1 ) 2$ on $[0,T/100]×(2,10T+3]×[9,10)$, which means that (S2) also holds. By Theorem 4.3, the considered IVP has at least one positive strictly increasing solution in $C[0,T]∩ C 2 (0,T]$.

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## Acknowledgements

The work is partially supported by the Sofia University Grant 158/2013 and by the Bulgarian NSF under Grant DCVP - 02/1/2009.

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