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Existence of positive solutions of the Cauchy problem for a second-order differential equation

Abstract

In this paper we consider the equation u (t)=f(t,u(t), u (t)) and prove the unique solvability of the Cauchy problem u(0)=0, u (0)=λ with λ>0.

1 Introduction

In [1], Knežević-Miljanović considered the Cauchy problem

{ u ( t ) = P ( t ) t a u ( t ) σ , t ( 0 , 1 ] , u ( 0 ) = 0 , u ( 0 ) = λ ,
(1)

where P is continuous, a,σ,λR with σ<0 and λ>0, and 0 1 |P(t)| t a + σ dt<. Moreover, in [2], Kawasaki and Toyoda considered the Cauchy problem

{ u ( t ) = f ( t , u ( t ) ) , for almost every  t [ 0 , 1 ] , u ( 0 ) = 0 , u ( 0 ) = λ ,
(2)

where f is a mapping from [0,1]×(0,) into and λR with λ>0. They proved the unique solvability of Cauchy problem (2) using the Banach fixed point theorem. The theorem in [2] is as follows.

Theorem Suppose that a mapping f from [0,1]×[0,) into satisfies the following.

  1. (a)

    The mapping tf(t,u) is measurable for any u(0,), and the mapping uf(t,u) is continuous for almost every t[0,1].

  2. (b)

    |f(t, u 1 )||f(t, u 2 )| for almost every t[0,1] and for any u 1 , u 2 [0,) with u 1 u 2 .

  3. (c)

    There exists αR with 0<α<λ such that

    0 1 |f(t,αt)|dt<.
  4. (d)

    There exists βR with β>0 such that

    | f u (t,u)| β | f ( t , u ) | u

for almost every t[0,1] and for any u(0,).

Then there exists hR with 0<h1 such that Cauchy problem (2) has a unique solution in X, where X is a subset

X= { u | u C [ 0 , h ] , u ( 0 ) = 0 , u ( 0 ) = λ and  α t u ( t )  for any  t [ 0 , h ] }

of C[0,h], which is the class of continuous mappings from [0,h] into .

The case that f(t,u(t))=P(t) t a u ( t ) σ in the above theorem is the theorem of Knežević-Miljanović [1].

In this paper, we consider the Cauchy problem

{ u ( t ) = f ( t , u ( t ) , u ( t ) ) , for almost every  t [ 0 , 1 ] , u ( 0 ) = 0 , u ( 0 ) = λ ,

where f is a mapping from [0,1]×(0,)×R into and λR with λ>0. We prove the unique solvability of this Cauchy problem using the Banach fixed point theorem.

In Section 2, we consider the following four cases for u and v.

  1. (I)

    Decreasing for u in f(t,u,v) (b1) and decreasing for v in f(t,u,v) (b3).

  2. (II)

    Decreasing for u in f(t,u,v) (b1) and increasing for v in f(t,u,v) (b4).

  3. (III)

    Increasing for u in f(t,u,v) (b2) and decreasing for v in f(t,u,v) (b3).

  4. (IV)

    Increasing for u in f(t,u,v) (b2) and increasing for v in f(t,u,v) (b4).

Theorems 2.1, 2.2, 2.3 and 2.4 are the cases of (I), (II), (III) and (IV), respectively.

2 Main results

In this section, we consider the Cauchy problem

{ u ( t ) = f ( t , u ( t ) , u ( t ) ) , for almost every  t [ 0 , 1 ] , u ( 0 ) = 0 , u ( 0 ) = λ ,
(3)

where f is a mapping from [0,1]×(0,)×R into and λR with λ>0.

First, we consider the case of (I).

Theorem 2.1 Let λ be a real number with λ>0. Suppose that a mapping f from [0,1]×(0,)×R into satisfies the following:

  1. (a)

    The mapping tf(t,u,v) is measurable for any (u,v)(0,)×R, and the mapping (u,v)f(t,u,v) is continuous for almost every t[0,1];

  2. (b1)

    |f(t, u 1 ,v)||f(t, u 2 ,v)| for almost every t[0,1], for any u 1 , u 2 (0,) with u 1 u 2 and for any vR;

  3. (b3)

    |f(t,u, v 1 )||f(t,u, v 2 )| for almost every t[0,1], for any u(0,) and for any v 1 , v 2 R with v 1 v 2 ;

  4. (c1)

    There exist α 1 R with 0< α 1 <λ and α 2 R with α 2 <λ such that

    0 1 |f(t, α 1 t, α 2 )|dt<;
  5. (d1)

    There exists β 1 R with β 1 >0 such that

    | f u (t,u,v)| β 1 | f ( t , u , v ) | u

    for almost every t[0,1], for any u(0,) and for any vR;

  6. (d2)

    There exists β 2 R with β 2 >0 such that

    | f v (t,u,v)| β 2 |f(t,u,v)|

    for almost every t[0,1], for any u(0,) and for any vR;

  7. (e)

    There exists the limit

    lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds

    for any continuously differentiable mapping u from [0,1] into [0,);

  8. (f1)

    For α 1 and α 2 ,

    lim t 0 + 1 t 2 0 t s|f(s, α 1 s, α 2 )|ds=0.

Then there exists hR with 0<h1 such that Cauchy problem (3) has a unique solution in X, where X is a subset

X= { u | u C 1 [ 0 , h ] , u ( 0 ) = 0 , u ( 0 ) = λ , α 1 t u ( t )  and  α 2 u ( t )  for any  t [ 0 , h ] and there exists the limit  lim t 0 + t u ( t ) u ( t ) t 2 }

of C 1 [0,h], which is the class of continuously differentiable mappings from [0,h] into .

Proof It is noted that C 1 [0,h] is a Banach space by the maximum norm

u=max { max { | u ( t ) | t [ 0 , h ] } , max { | u ( t ) | t [ 0 , h ] } } .

Instead of Cauchy problem (3), we consider the integral equation

u(t)=λt+ 0 t (ts)f ( s , u ( s ) , u ( s ) ) ds.

By condition (c1), there exists h 1 R with 0< h 1 1 such that

0 h 1 |f(t, α 1 t, α 2 )|dt<min { λ α 1 , λ α 2 , ( β 1 α 1 + 2 β 2 ) 1 } .

By condition (f1), there exists hR with 0<h h 1 such that

sup t ( 0 , h ] 1 t 2 0 t s|f(s, α 1 s, α 2 )|ds 0 h 1 |f(t, α 1 t, α 2 )|dt.

Let A be an operator from X into C 1 [0,h] defined by

Au(t)=λt+ 0 t (ts)f ( s , u ( s ) , u ( s ) ) ds.

Since a mapping tλt belongs to X, X. Moreover, we have A(X)X. Indeed, by condition (a), Au C 1 [0,h], Au(0)=0 and

( A u ) (0)= [ λ + 0 t f ( s , u ( s ) , u ( s ) ) d s ] t = 0 =λ.

By conditions (b1) and (b3), we obtain that

A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s λ t t 0 h | f ( s , u ( s ) , u ( s ) ) | d s λ t t 0 h | f ( s , α 1 s , α 2 ) | d s α 1 t

and

( A u ) ( t ) = λ + 0 t f ( s , u ( s ) , u ( s ) ) d s λ 0 h | f ( s , u ( s ) , u ( s ) ) | d s λ 0 h | f ( s , α 1 s , α 2 ) | d s α 2

for any t[0,h]. Moreover, by condition (e), there exists the limit

lim t 0 + t ( A u ) ( t ) A u ( t ) t 2 = lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds.

We will find a fixed point of A. Let φ be an operator from X into C 1 [0,h] defined by

φ[u](t)= { u ( t ) t if  t ( 0 , h ] , λ if  t = 0 .

Let φ[X] be a subset defined by

φ[X]= { φ [ u ] u X } .

Then we have

φ[X]= { v | v C 1 [ 0 , h ] , v ( 0 ) = λ , α 1 v ( t )  and  α 2 v ( t ) + t v ( t )  for any  t [ 0 , h ] }

and φ[X] is a closed subset of C 1 [0,h]. Hence it is a complete metric space. Let Φ be an operator from φ[X] into φ[X] defined by

Φφ[u]=φ[Au].

By the mean value theorem, for any u 1 , u 2 X, there exist mappings ξ, η such that

f ( t , u 1 ( t ) , u 1 ( t ) ) f ( t , u 2 ( t ) , u 2 ( t ) ) = f u ( t , ξ ( t ) , u 1 ( t ) ) ( u 1 ( t ) u 2 ( t ) ) + f v ( t , u 2 ( t ) , η ( t ) ) ( u 1 ( t ) u 2 ( t ) ) = ( t f u ( t , ξ ( t ) , u 1 ( t ) ) + f v ( t , u 2 ( t ) , η ( t ) ) ) ( φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) ) + t f v ( t , u 2 ( t ) , η ( t ) ) ( φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) ) , min { u 1 ( t ) , u 2 ( t ) } ξ ( t ) max { u 1 ( t ) , u 2 ( t ) }

and

min { u 1 ( t ) , u 2 ( t ) } η(t)max { u 1 ( t ) , u 2 ( t ) }

for almost every t[0,h]. Therefore, by conditions (b1), (b3), (d1) and (d2), we obtain that

| f ( t , u 1 ( t ) , u 1 ( t ) ) f ( t , u 2 ( t ) , u 2 ( t ) ) | = | ( t f u ( t , ξ ( t ) , u 1 ( t ) ) + f v ( t , u 2 ( t ) , η ( t ) ) ) ( φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) ) + t f v ( t , u 2 ( t ) , η ( t ) ) ( φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) ) | ( t | f u ( t , ξ ( t ) , u 1 ( t ) ) | + | f v ( t , u 2 ( t ) , η ( t ) ) | ) | φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) | + t | f v ( t , u 2 ( t ) , η ( t ) ) | | ( φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) ) | ( β 1 α 1 + β 2 ) | f ( t , α 1 t , α 2 ) | | φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) | + β 2 t | f ( t , α 1 t , α 2 ) | | ( φ [ u 1 ] ( t ) φ [ u 2 ] ( t ) ) |

for almost every t[0,h]. Therefore we have

| Φ φ [ u 1 ] ( t ) Φ φ [ u 2 ] ( t ) | = | 1 t 0 t ( t s ) ( f ( s , u 1 ( s ) , u 1 ( s ) ) f ( s , u 2 ( s ) , u 2 ( s ) ) ) d s | 0 t | f ( s , u 1 ( s ) , u 1 ( s ) ) f ( s , u 2 ( s ) , u 2 ( s ) ) | d s 0 t [ ( β 1 α 1 + β 2 ) | f ( s , α 1 s , α 2 ) | | φ [ u 1 ] ( s ) φ [ u 2 ] ( s ) | + β 2 s | f ( s , α 1 s , α 2 ) | | ( φ [ u 1 ] ( s ) φ [ u 2 ] ( s ) ) | ] ] d s ( β 1 α 1 + 2 β 2 ) 0 h | f ( s , α 1 s , α 2 ) | d s φ [ u 1 ] φ [ u 2 ]

for any t[0,h]. Moreover, we have

| ( Φ φ [ u 1 ] ) ( t ) ( Φ φ [ u 2 ] ) ( t ) | = | 1 t 2 0 t s ( f ( s , u 1 ( s ) , u 1 ( s ) ) f ( s , u 2 ( s ) , u 2 ( s ) ) ) d s | 1 t 2 0 t s | f ( s , u 1 ( s ) , u 1 ( s ) ) f ( s , u 2 ( s ) , u 2 ( s ) ) | d s 1 t 2 0 t s [ ( β 1 α 1 + β 2 ) | f ( s , α 1 s , α 2 ) | | φ [ u 1 ] ( s ) φ [ u 2 ] ( s ) | + β 2 s | f ( s , α 1 s , α 2 ) | | ( φ [ u 1 ] ( s ) φ [ u 2 ] ( s ) ) | ] ] d s [ ( β 1 α 1 + β 2 ) 0 h 1 | f ( s , α 1 s , α 2 ) | d s + β 2 0 h | f ( s , α 1 s , α 2 ) | d s ] φ [ u 1 ] φ [ u 2 ]

for any t[0,h]. Hence we obtain that

Φ φ [ u 1 ] Φ φ [ u 2 ] ( β 1 α 1 + 2 β 2 ) 0 h 1 | f ( s , α 1 s , α 2 ) | d s φ [ u 1 ] φ [ u 2 ] .

By the Banach fixed point theorem, there exists a unique mapping φ[u]φ[X] such that Φφ[u]=φ[u]. Then Au=u. u is a solution of (3). □

Next, we consider the case of (II).

Theorem 2.2 Let λ be a real number with λ>0. Suppose that a mapping f from [0,1]×(0,)×R into satisfies the following:

  1. (a)

    The mapping tf(t,u,v) is measurable for any (u,v)(0,)×R, and the mapping (u,v)f(t,u,v) is continuous for almost every t[0,1];

  2. (b1)

    |f(t, u 1 ,v)||f(t, u 2 ,v)| for almost every t[0,1], for any u 1 , u 2 (0,) with u 1 u 2 and for any vR;

  3. (b4)

    |f(t,u, v 1 )||f(t,u, v 2 )| for almost every t[0,1], for any u(0,) and for any v 1 , v 2 R with v 1 v 2 ;

  4. (c2)

    There exist α 1 R with 0< α 1 <λ and α 2 R with α 2 >λ such that

    0 1 |f(t, α 1 t, α 2 )|dt<;
  5. (d1)

    There exists β 1 R with β 1 >0 such that

    | f u (t,u,v)| β 1 | f ( t , u , v ) | u

    for almost every t[0,1], for any u(0,) and for any vR;

  6. (d2)

    There exists β 2 R with β 2 >0 such that

    | f v (t,u,v)| β 2 |f(t,u,v)|

    for almost every t[0,1], for any u(0,) and for any vR;

  7. (e)

    There exists the limit

    lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds

    for any continuously differentiable mapping u from [0,1] into [0,);

  8. (f1)

    For α 1 and α 2 ,

    lim t 0 + 1 t 2 0 t s|f(s, α 1 s, α 2 )|ds=0.

Then there exists hR with 0<h1 such that Cauchy problem (3) has a unique solution in X, where X is a subset

X= { u | u C 1 [ 0 , h ] , u ( 0 ) = 0 , u ( 0 ) = λ , α 1 t u ( t )  and  u ( t ) α 2  for any  t [ 0 , h ] and there exists the limit  lim t 0 + t u ( t ) u ( t ) t 2 }

of C 1 [0,h].

Proof By condition (c2), there exists h 1 R with 0< h 1 1 such that

0 h 1 |f(t, α 1 t, α 2 )|dt<min { λ α 1 , α 2 λ , ( β 1 α 1 + 2 β 2 ) 1 } .

By condition (f1), there exists hR with 0<h h 1 such that

sup t ( 0 , h ] 1 t 2 0 t s|f(s, α 1 s, α 2 )|ds 0 h 1 |f(t, α 1 t, α 2 )|dt.

Let A be an operator from X into C 1 [0,h] defined by

Au(t)=λt+ 0 t (ts)f ( s , u ( s ) , u ( s ) ) ds.

Since a mapping tλt belongs to X, X. Moreover, we have A(X)X. Indeed, by condition (a), Au C 1 [0,h], Au(0)=0 and

( A u ) (0)= [ λ + 0 t f ( s , u ( s ) , u ( s ) ) d s ] t = 0 =λ.

By conditions (b1) and (b4), we obtain that

A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s λ t t 0 h | f ( s , u ( s ) , u ( s ) ) | d s λ t t 0 h | f ( s , α 1 s , α 2 ) | d s α 1 t

and

( A u ) ( t ) = λ + 0 t f ( s , u ( s ) , u ( s ) ) d s λ + 0 h | f ( s , u ( s ) , u ( s ) ) | d s λ + 0 h | f ( s , α 1 s , α 2 ) | d s α 2

for any t[0,h]. Moreover, by condition (e), there exists the limit

lim t 0 + t ( A u ) ( t ) A u ( t ) t 2 = lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds.

We will find a fixed point of A. Let φ be an operator from X into C 1 [0,h] defined by

φ[u](t)= { u ( t ) t if  t ( 0 , h ] , λ if  t = 0 ,

and

φ [ X ] = { φ [ u ] u X } = { v | v C 1 [ 0 , h ] , v ( 0 ) = λ , α 1 v ( t )  and  v ( t ) + t v ( t ) α 2  for any  t [ 0 , h ] } .

Then φ[X] is a closed subset of C 1 [0,h] and hence it is a complete metric space. Let Φ be an operator from φ[X] into φ[X] defined by

Φφ[u]=φ[Au].

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping φ[u]φ[X] such that Φφ[u]=φ[u] and hence Au=u. □

Next, we consider the case of (III).

Theorem 2.3 Let λ be a real number with λ>0. Suppose that a mapping f from [0,1]×(0,)×R into satisfies the following:

  1. (a)

    The mapping tf(t,u,v) is measurable for any (u,v)(0,)×R, and the mapping (u,v)f(t,u,v) is continuous for almost every t[0,1];

  2. (b2)

    |f(t, u 1 ,v)||f(t, u 2 ,v)| for almost every t[0,1], for any u 1 , u 2 (0,) with u 1 u 2 and for any vR;

  3. (b3)

    |f(t,u, v 1 )||f(t,u, v 2 )| for almost every t[0,1], for any u(0,) and for any v 1 , v 2 R with v 1 v 2 ;

  4. (c3)

    (c3) There exist α 1 R with 0< α 1 <λ and α 2 R with α 2 <λ such that

    0 1 |f ( t , ( 2 λ α 1 ) t , α 2 ) |dt<;
  5. (d1)

    There exists β 1 R with β 1 >0 such that

    | f u (t,u,v)| β 1 | f ( t , u , v ) | u

    for almost every t[0,1], for any u(0,) and for any vR;

  6. (d2)

    There exists β 2 R with β 2 >0 such that

    | f v (t,u,v)| β 2 |f(t,u,v)|

    for almost every t[0,1], for any u(0,) and for any vR;

  7. (e)

    There exists the limit

    lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds

    for any continuously differentiable mapping u from [0,1] into [0,);

  8. (f2)

    For α 1 and α 2 ,

    lim t 0 + 1 t 2 0 t s|f ( s , ( 2 λ α 1 ) s , α 2 ) |ds=0.

Then there exists hR with 0<h1 such that Cauchy problem (3) has a unique solution in X, where X is a subset

X= { u | u C 1 [ 0 , h ] , u ( 0 ) = 0 , u ( 0 ) = λ , α 1 t u ( t ) ( 2 λ α 1 ) t  and  α 2 u ( t )  for any  t [ 0 , h ] and there exists the limit  lim t 0 + t u ( t ) u ( t ) t 2 }

of C 1 [0,h].

Proof By condition (c3), there exists h 1 R with 0< h 1 1 such that

0 h 1 |f ( t , ( 2 λ α 1 ) t , α 2 ) |dt<min { λ α 1 , λ α 2 , ( β 1 α 1 + 2 β 2 ) 1 } .

By condition (f2), there exists hR with 0<h h 1 such that

sup t ( 0 , h ] 1 t 2 0 t s|f ( s , ( 2 λ α 1 ) s , α 2 ) |ds 0 h 1 |f ( t , ( 2 λ α 1 ) t , α 2 ) |dt.

Let A be an operator from X into C 1 [0,h] defined by

Au(t)=λt+ 0 t (ts)f ( s , u ( s ) , u ( s ) ) ds.

Since a mapping tλt belongs to X, X. Moreover, A(X)X. Indeed, by condition (a), Au C 1 [0,h], Au(0)=0,

( A u ) (0)= [ λ + 0 t f ( s , u ( s ) , u ( s ) ) d s ] t = 0 =λ,

by conditions (b2) and (b3),

A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s A u ( t ) λ t t 0 h | f ( s , u ( s ) , u ( s ) ) | d s A u ( t ) λ t t 0 h | f ( s , ( 2 λ α 1 ) s , α 2 ) | d s A u ( t ) α 1 t , A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s A u ( t ) λ t + t 0 h | f ( s , u ( s ) , u ( s ) ) | d s A u ( t ) λ t + t 0 h | f ( s , ( 2 λ α 1 ) s , α 2 ) | d s A u ( t ) ( 2 λ α 1 ) t , ( A u ) ( t ) = λ + 0 t f ( s , u ( s ) , u ( s ) ) d s ( A u ) ( t ) λ 0 h | f ( s , u ( s ) , u ( s ) ) | d s ( A u ) ( t ) λ 0 h | f ( s , ( 2 λ α 1 ) s , α 2 ) | d s ( A u ) ( t ) α 2

for any t[0,h], and by condition (e), there exists the limit

lim t 0 + t ( A u ) ( t ) A u ( t ) t 2 = lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds.

We will find a fixed point of A. Let φ be an operator from X into C 1 [0,h] defined by

φ[u](t)= { u ( t ) t if  t ( 0 , h ] , λ if  t = 0 ,

and

φ [ X ] = { φ [ u ] u X } = { v | v C 1 [ 0 , h ] , v ( 0 ) = λ , α 1 v ( t ) 2 λ α 1  and  α 2 v ( t ) + t v ( t )  for any  t [ 0 , h ] } .

Then φ[X] is a closed subset of C 1 [0,h], and hence it is a complete metric space. Let Φ be an operator from φ[X] into φ[X] defined by

Φφ[u]=φ[Au].

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping φ[u]φ[X] such that Φφ[u]=φ[u] and hence Au=u. □

Finally, we consider the case of (IV).

Theorem 2.4 Let λ be a real number with λ>0. Suppose that a mapping f from [0,1]×(0,)×R into satisfies the following:

  1. (a)

    The mapping tf(t,u,v) is measurable for any (u,v)(0,)×R, and the mapping (u,v)f(t,u,v) is continuous for almost every t[0,1];

  2. (b2)

    |f(t, u 1 ,v)||f(t, u 2 ,v)| for almost every t[0,1], for any u 1 , u 2 (0,) with u 1 u 2 and for any vR;

  3. (b4)

    |f(t,u, v 1 )||f(t,u, v 2 )| for almost every t[0,1], for any u(0,) and for any v 1 , v 2 R with v 1 v 2 ;

  4. (c4)

    There exist α 1 R with 0< α 1 <λ and α 2 R with α 2 >λ such that

    0 1 |f ( t , ( 2 λ α 1 ) t , α 2 ) |dt<;
  5. (d1)

    There exists β 1 R with β 1 >0 such that

    | f u (t,u,v)| β 1 | f ( t , u , v ) | u

    for almost every t[0,1], for any u(0,) and for any vR;

  6. (d2)

    There exists β 2 R with β 2 >0 such that

    | f v (t,u,v)| β 2 |f(t,u,v)|

    for almost every t[0,1], for any u(0,) and for any vR;

  7. (e)

    There exists the limit

    lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds

    for any continuously differentiable mapping u from [0,1] into [0,);

  8. (f2)

    For α 1 and α 2 ,

    lim t 0 + 1 t 2 0 t s|f ( s , ( 2 λ α 1 ) s , α 2 ) |ds=0.

Then there exists hR with 0<h1 such that Cauchy problem (3) has a unique solution in X, where X is a subset

X= { u | u C 1 [ 0 , h ] , u ( 0 ) = 0 , u ( 0 ) = λ , α 1 t u ( t ) ( 2 λ α 1 ) t  and  u ( t ) α 2  for any  t [ 0 , h ] and there exists the limit  lim t 0 + t u ( t ) u ( t ) t 2 }

of C 1 [0,h].

Proof By condition (c4), there exists h 1 R with 0< h 1 1 such that

0 h 1 |f ( t , ( 2 λ α 1 ) t , α 2 ) |dt<min { λ α 1 , α 2 λ , ( β 1 α 1 + 2 β 2 ) 1 } .

By condition (f2), there exists hR with 0<h h 1 such that

sup t ( 0 , h ] 1 t 2 0 t s|f ( s , ( 2 λ α 1 ) s , α 2 ) |ds 0 h 1 |f ( t , ( 2 λ α 1 ) t , α 2 ) |dt.

Let A be an operator from X into C 1 [0,h] defined by

Au(t)=λt+ 0 t (ts)f ( s , u ( s ) , u ( s ) ) ds.

Since a mapping tλt belongs to X, X. Moreover, A(X)X. Indeed, by condition (a), Au C 1 [0,h], Au(0)=0,

( A u ) (0)= [ λ + 0 t f ( s , u ( s ) , u ( s ) ) d s ] t = 0 =λ,

by conditions (b2) and (b4),

A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s A u ( t ) λ t t 0 h | f ( s , u ( s ) , u ( s ) ) | d s A u ( t ) λ t t 0 h | f ( s , ( 2 λ α 1 ) s , α 2 ) | d s A u ( t ) α 1 t , A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s A u ( t ) λ t + t 0 h | f ( s , u ( s ) , u ( s ) ) | d s A u ( t ) λ t + t 0 h | f ( s , ( 2 λ α 1 ) s , α 2 ) | d s A u ( t ) ( 2 λ α 1 ) t , ( A u ) ( t ) = λ + 0 t f ( s , u ( s ) , u ( s ) ) d s ( A u ) ( t ) λ + 0 h | f ( s , u ( s ) , u ( s ) ) | d s ( A u ) ( t ) λ + 0 h | f ( s , α 1 s , α 2 ) | d s ( A u ) ( t ) α 2

for any t[0,h], and by condition (e), there exists the limit

lim t 0 + t ( A u ) ( t ) A u ( t ) t 2 = lim t 0 + 1 t 2 0 t sf ( s , u ( s ) , u ( s ) ) ds.

We will find a fixed point of A. Let φ be an operator from X into C 1 [0,h] defined by

φ[u](t)= { u ( t ) t if  t ( 0 , h ] , λ if  t = 0 ,

and

φ [ X ] = { φ [ u ] u X } = { v | v C 1 [ 0 , h ] , v ( 0 ) = λ , α 1 v ( t ) 2 λ α 1  and  v ( t ) + t v ( t ) α 2  for any  t [ 0 , h ] } .

Then φ[X] is a closed subset of C 1 [0,h] and hence it is a complete metric space. Let Φ be an operator from φ[X] into φ[X] defined by

Φφ[u]=φ[Au].

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping φ[u]φ[X] such that Φφ[u]=φ[u] and hence Au=u. □

References

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Correspondence to Toshiharu Kawasaki.

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TK wrote first draft. MT wrote final manuscript. All authors read and approved the final manuscript.

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Kawasaki, T., Toyoda, M. Existence of positive solutions of the Cauchy problem for a second-order differential equation. J Inequal Appl 2013, 465 (2013). https://doi.org/10.1186/1029-242X-2013-465

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Keywords

  • Differential Equation
  • Banach Space
  • Continuous Mapping
  • Real Number
  • Integral Equation