Open Access

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

Journal of Inequalities and Applications20132013:465

https://doi.org/10.1186/1029-242X-2013-465

Received: 1 June 2013

Accepted: 20 September 2013

Published: 7 November 2013

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 , λ R with σ < 0 and λ > 0 , and 0 1 | P ( t ) | t a + σ d t < . 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 λ R 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 t f ( t , u ) is measurable for any u ( 0 , ) , and the mapping u f ( 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 α R with 0 < α < λ such that
    0 1 | f ( t , α t ) | d t < .
     
  4. (d)
    There exists β R 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 h R with 0 < h 1 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 λ R 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 λ R 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 t f ( 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 v R ;

     
  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 ) | d t < ;
     
  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 v R ;

     
  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 v R ;

     
  7. (e)
    There exists the limit
    lim t 0 + 1 t 2 0 t s f ( s , u ( s ) , u ( s ) ) d s

    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 ) | d s = 0 .
     
Then there exists h R with 0 < h 1 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 ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s .
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 ) | d t < min { λ α 1 , λ α 2 , ( β 1 α 1 + 2 β 2 ) 1 } .
By condition (f1), there exists h R with 0 < h h 1 such that
sup t ( 0 , h ] 1 t 2 0 t s | f ( s , α 1 s , α 2 ) | d s 0 h 1 | f ( t , α 1 t , α 2 ) | d t .
Let A be an operator from X into C 1 [ 0 , h ] defined by
A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s .
Since a mapping t λ t belongs to X, X . Moreover, we have A ( X ) X . Indeed, by condition (a), A u C 1 [ 0 , h ] , A u ( 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 s f ( s , u ( s ) , u ( s ) ) d s .
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
Φ φ [ u ] = φ [ A u ] .
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 Φ φ [ u ] = φ [ u ] . Then A u = 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 t f ( 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 v R ;

     
  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 ) | d t < ;
     
  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 v R ;

     
  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 v R ;

     
  7. (e)
    There exists the limit
    lim t 0 + 1 t 2 0 t s f ( s , u ( s ) , u ( s ) ) d s

    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 ) | d s = 0 .
     
Then there exists h R with 0 < h 1 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 ) | d t < min { λ α 1 , α 2 λ , ( β 1 α 1 + 2 β 2 ) 1 } .
By condition (f1), there exists h R with 0 < h h 1 such that
sup t ( 0 , h ] 1 t 2 0 t s | f ( s , α 1 s , α 2 ) | d s 0 h 1 | f ( t , α 1 t , α 2 ) | d t .
Let A be an operator from X into C 1 [ 0 , h ] defined by
A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s .
Since a mapping t λ t belongs to X, X . Moreover, we have A ( X ) X . Indeed, by condition (a), A u C 1 [ 0 , h ] , A u ( 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 s f ( s , u ( s ) , u ( s ) ) d s .
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
Φ φ [ u ] = φ [ A u ] .

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 Φ φ [ u ] = φ [ u ] and hence A u = 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 t f ( 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 v R ;

     
  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 ) | d t < ;
     
  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 v R ;

     
  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 v R ;

     
  7. (e)
    There exists the limit
    lim t 0 + 1 t 2 0 t s f ( s , u ( s ) , u ( s ) ) d s

    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 ) | d s = 0 .
     
Then there exists h R with 0 < h 1 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 ) | d t < min { λ α 1 , λ α 2 , ( β 1 α 1 + 2 β 2 ) 1 } .
By condition (f2), there exists h R 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 ) | d s 0 h 1 | f ( t , ( 2 λ α 1 ) t , α 2 ) | d t .
Let A be an operator from X into C 1 [ 0 , h ] defined by
A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s .
Since a mapping t λ t belongs to X, X . Moreover, A ( X ) X . Indeed, by condition (a), A u C 1 [ 0 , h ] , A u ( 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 s f ( s , u ( s ) , u ( s ) ) d s .
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
Φ φ [ u ] = φ [ A u ] .

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 Φ φ [ u ] = φ [ u ] and hence A u = 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 t f ( 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 v R ;

     
  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 ) | d t < ;
     
  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 v R ;

     
  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 v R ;

     
  7. (e)
    There exists the limit
    lim t 0 + 1 t 2 0 t s f ( s , u ( s ) , u ( s ) ) d s

    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 ) | d s = 0 .
     
Then there exists h R with 0 < h 1 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 ) | d t < min { λ α 1 , α 2 λ , ( β 1 α 1 + 2 β 2 ) 1 } .
By condition (f2), there exists h R 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 ) | d s 0 h 1 | f ( t , ( 2 λ α 1 ) t , α 2 ) | d t .
Let A be an operator from X into C 1 [ 0 , h ] defined by
A u ( t ) = λ t + 0 t ( t s ) f ( s , u ( s ) , u ( s ) ) d s .
Since a mapping t λ t belongs to X, X . Moreover, A ( X ) X . Indeed, by condition (a), A u C 1 [ 0 , h ] , A u ( 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 s f ( s , u ( s ) , u ( s ) ) d s .
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
Φ φ [ u ] = φ [ A u ] .

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 Φ φ [ u ] = φ [ u ] and hence A u = u . □

Declarations

Authors’ Affiliations

(1)
College of Engineering, Nihon University
(2)
Marine Faculty of Engineering, Tamagawa University

References

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Copyright

© Kawasaki and Toyoda; licensee Springer. 2013

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