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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 [1]–[3]. For example in [1], the authors have discussed the set of all solutions to this problem with a singularity at t=0. Here A<0, φC[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 [4] 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 φC[0,T], φ(t)>0 for t(0,T], f:[0,T]× ( 0 , ) 2 (0,) is continuous and

f(t,x,p) [ g ( x ) + h ( x ) ] [ r ( p ) + w ( p ) ] for (t,x,p)[0,T]× ( 0 , ) 2 ,

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 [5], [6]. 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

0<f(t,x,p)k(t)F(x)G(y)for (t,x,p)(0,1)× ( 0 , ) 2 ,

where k, F, and G are suitable functions.

The solvability of various IVPs has been studied also by Bobisud and O’Regan [7], Bobisud and Lee [8], Cabada and Heikkilä [9], Cabada et al. [10], [11], Cid [12], Maagli and Masmoudi [13], and Zhao [14]. Existence results for problem (1.1), (1.2) with a singularity at the initial value of x have been reported in Kelevedjiev-Popivanov [15].

Here, as usual, we use regularization and sequential techniques. Namely, we proceed as follows. First, by means of the topological transversality theorem [16], 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 nN 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 A0, B0 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. [17], Kelevedjiev and Popivanov [18] and Palamides et al. [19]. For example in [17], 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 [17].

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 UY 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

([16], Chapter I, Theorem 2.2])

LetpUbe fixed andF L u ( U ¯ ,Y)be the constant mapF(x)=pforx U ¯ . Then F is essential.

We say that the homotopy{ H λ :XY}, 0λ1, is compact if the mapH(x,λ):X×[0,1]Ygiven byH(x,λ) H λ (x)for(x,λ)X×[0,1]is compact.

Theorem 2.2

([16], Chapter I, Theorem 2.6])

Let Y be a convex subset of a Banach space E andUYbe 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.

ThenH(x,λ), λ[0,1], has at least one fixed point in U and in particular there is a x 0 Usuch 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 (Ta),

f ( t , x , p ) C ( [ a , T ] × [ A τ , M 0 + τ ] × [ m 1 τ , M 1 + τ ] ) , f ( t , x , p ) 0 for  ( t , x , p ) [ a , T ] × D x × [ M 1 , M ¯ 1 ] , f ( t , x , p ) 0 for  ( t , x , p ) [ a , T ] × D M 0 × [ m ¯ 1 , m 1 ] ,
(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 solutionx C 2 [a,T]to the family (3.2) λ , λ[0,1], satisfies the bounds

Ax(t) M 0 , m 1 x (t) M 1 , m 2 x (t) M 2 for t[a,T],

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

x (t) M 1 for t[a,T].

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)(ta),

which yields

x(t) M 0 for t[a,T].

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,

0 m 1 x (t)for t[a,T]

and so

Ax(t)for t[a,T].

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

f(t,x, M 1 )0for (t,x)[a,T]×[A, M 0 ]

and

f(t,x, m 1 )0for (t,x)[a,T]×[A, M 0 ],

which yield m 2 0 and M 2 0. Multiplying both sides of the inequality λ1 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

x(t)[A, M 0 ]and x (t)[ m 1 , M 1 ]for t[a,T].

Thus,

m 2 λ m 2 λf ( t , x ( t ) , x ( t ) ) λ M 2 M 2 for t[a,T]

and each λ[0,1] and so

x (t)[ m 2 , M 2 ]for t[a,T].

 □

Let us mention that some analogous results have been obtained in Kelevedjiev [20]. 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

j : C I 2 [ a , T ] C 1 [ a , T ] by  j x = x , V : C I 2 [ a , T ] C [ a , T ] by  V x = x ,

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

Φ: C 1 [a,T]C[a,T]by (Φx)(t)=f ( t , x ( t ) , x ( t ) ) .

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

x(t)[ m 0 τ, M 0 +τ]and x (t)[ m 1 τ, M 1 +τ]for t[a,T].

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 yC[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

V 1 :C[a,T] C I 2 [a,T]by  ( V 1 y ) (t)=(t)+ ( W 1 y ) (t),

where (t)=B(ta)+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)+(1λ).

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)+(1λ)=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 (t). Thus, according to Theorem 2.1, H 0 (x)= 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

x (t) m 1 0for t[a,T],

from which its monotony follows. □

The validity of the following results follows similarly.

Theorem 3.3

LetB>0and 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

LetA>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

LetA0, B>0and 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 fort(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)
f(t,x,p)0for (t,x,p) ( [ 0 , T ] × D x × [ m 1 , B ) ) S A
(4.2)

and

f(t,x,p)0for (t,x,p) ( [ 0 , T ] × D M ˜ 0 × [ m ¯ 1 , m 1 ] ) S A ,

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

f(t,x,p)k<0for (t,x,p)[0,α]×(A, M ˜ 0 ]×[μ,B),

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 >μkα>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

A< x n (t) M ˜ 0 and m 1 x n (t)<Bfor t[0,T].

Then the following bound is satisfied for eachn n α , μ :

x n (t)< ϕ α (t)<Bfor t(0,T],

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 >μkα>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

x n (t)=f ( t , x n ( t ) , x n ( t ) ) kfor t[0,α],

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

x n (t)kt+B n 1 for t[0,α].

Now m 1 x n (t)<B, t[0,T], and (4.2) imply

x n (t)=f ( t , x n ( t ) , x n ( t ) ) 0for t[0,T].

In particular x n (t)0 for t[α,T], thus

x n (t) x n (α)kα+B n 1 for t(α,T].

Case 2. As in the first case, we derive

x n (t)kt+B n 1 for t[0,β].

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

x n (t)=f ( t , x n ( t ) , x n ( t ) ) 0for t[β,T],

which yields

x n (t) x n (β)=μ<kα+B n 1 kt+B n 1 for t[β,α]

and

x n (t)<kα+B n 1 for t(α,T].

So, as a result of the considered cases we get

x n (t) { k t + B n 1 , t [ 0 , α ] , k α + B n 1 , t ( α , T ] < ϕ α (t)for t[0,T] and n n α , μ ,

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 inC[0,T] C 2 (0,T]such that

m 1 t+Ax(t)Bt+Afor t[0,T], m 1 x (t)<Bfor t(0,T].

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 <B. Besides,

f ( t , x , p ) 0 for  ( t , x , p ) ( [ 0 , T ] × D x × [ M 1 , M ¯ 1 ] ) S A , f ( t , x , p ) 0 for  ( t , x , p ) ( [ 0 , T ] × ( D x × ( , M 0 ] ) × [ m ¯ 1 , m 1 ] ) S A

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<A+ n 1 x n (t) M 0 < M ˜ 0
(4.4)

and

m 1 x n (t)B n 1 <B.

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

m 1 t+A+ n 1 x n (t)<Bt+A+ n 1 for t[0,T]
(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

x n (t) m 1 >0for t[α,T],

which together with (4.6) gives

x n (t) x n (α) m 1 α+A+ n 1 > A 1 >Afor t[α,T],

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

A 1 < x n (t)< M ˜ 0 for t[α,T],n n α , μ .
(4.7)

From (4.5) we have in addition

m 1 x n (t)< ϕ α (α)=B+kαfor t[α,T],n n α , μ .
(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

x n (t)=f ( t , x n ( t ) , x n ( t ) ) for t[α,T],

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

| x n (t)| M 2 for t[α,T] and n n α , μ .

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 }, kN, n k n α , μ , of { x n } and a function x α C 1 [α,T] such that

x n k x α 1 0on the interval [α,T],

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

A 1 x α ( t ) M ˜ 0 for  t [ α , T ] , m 1 x α ( t ) ϕ α ( α ) < B for  t [ α , T ] .

Clearly, the functions x n k (t), kN, 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

m 1 t+A x α (t)Bt+Afor t[α,T].

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), iN, with the properties

α i + 1 < α i for iN and  lim i α i =0.

For each iN we consider sequences

{ x i , n k }, n k n i + 1 , μ ,kN, 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 }, kN, 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,

x i + 1 , n k x α i + 1 1 0on [ α i + 1 ,T],
(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

m 1 t + A x α i + 1 ( t ) B t + A for  t [ α i + 1 , T ] , m 1 x α i + 1 ( t ) ϕ α ( α i + 1 ) < B for  t [ α i + 1 , T ] , x α i + 1 ( t ) = x α i ( t ) and x α i + 1 ( t ) = x α i ( t ) for  t [ α i , T ] .

The properties of the functions from { x α i }, iN, 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

m 1 t+A x 0 (t)Bt+Afor t(0,T],

hence lim t 0 + x 0 (t)=A,

m 1 x 0 ( t ) ϕ α ( t ) < B for  t ( 0 , T ] , x 0 ( t ) = x α i ( t ) for  t [ α i , T ]  and  i N , x 0 ( t ) = x α i ( t ) for  t [ α i , T ]  and  i N .
(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)<Bε.

In other words, assume that for every sequence { δ j }(0,T], jN, 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 )<Bε.
(4.12)

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

x i , n k x 0 0on [ α i , δ j )
(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 jN there is a t j [ α i , δ j ) where i> i j . In summary, for every sufficiently large jN, 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 )<Bε,

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

LetA0and let (S1) and (S2) hold. Then the singular IVP (1.1), (1.2) has at least one strictly increasing solution inC[0,T] C 2 (0,T]with positive values fort(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), k2, has simple zeroes p 1 and p 2 such that

0< p 1 <B< p 2 .

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 +θB 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

P k (p)>0for p[ p 1 θ, p 1 )and P k (p)<0for p( p 2 , p 2 +θ];

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 [Aτ, 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/2band p 2 T 0 +θ/2b,

which yield θ(0,2b] and T 2 b θ 2 p 2 . Besides, [0,T](c,c) yields T<C. Thus, 0<T<min{c, 2 b θ 2 p 2 }. 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<min{c, 2 b θ 2 p 2 }.

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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Kelevedjiev, P., Popivanov, N. Second-order initial value problems with singularities. Bound Value Probl 2014, 161 (2014). https://doi.org/10.1186/s13661-014-0161-z

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