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On the solvability of a Neumann boundary value problem for the differential equation f ( t , x , x , x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq1_HTML.gif

Boundary Value Problems20122012:77

DOI: 10.1186/1687-2770-2012-77

Received: 5 July 2012

Accepted: 6 July 2012

Published: 23 July 2012

Abstract

Using barrier strip arguments, we investigate the existence of C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq2_HTML.gif-solutions to the Neumann boundary value problem f ( t , x , x , x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq1_HTML.gif, x ( 0 ) = a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq3_HTML.gif, x ( 1 ) = b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq4_HTML.gif.

MSC:34B15.

Keywords

boundary value problem equation unsolved with respect to the second derivative Neumann boundary conditions existence

1 Introduction

The purpose of this paper is to establish the existence of C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq5_HTML.gif-solutions to the scalar Neumann boundary value problem (BVP)
{ f ( t , x , x , x ) = 0 , t [ 0 , 1 ] , x ( 0 ) = a , x ( 1 ) = b , a b , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ1_HTML.gif
(N)

where the function f ( t , x , p , q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq6_HTML.gif and its first derivatives are continuous only on suitable subsets of the set [ 0 , 1 ] × R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq7_HTML.gif.

The literature devoted to the solvability of singular and nonsingular Neumann BVPs for second order ordinary differential equations whose main nonlinearities do not depend on the second derivative is vast. We quote here only [15] for results and references.

The solvability of the homogeneous Neumann problem for the equation ( p ( t ) x ) + f ( t , x , x , x ) = y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq8_HTML.gif, under appropriate conditions on f, has been studied in [68]. Results, concerning the existence of solutions to the homogeneous and nonhomogeneous Neumann problem for the equation x = f ( t , x , x , x ) y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq9_HTML.gif can be found in [9] and [10] respectively. BVPs for the same equation with various linear boundary conditions have been studied in [9, 1113]. The results of [14] guarantee the solvability of BVPs for the equation x = f ( t , x , x , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq10_HTML.gif with fully linear boundary conditions. BVPs for the equation f ( t , x , x , x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq11_HTML.gif with fully nonlinear boundary conditions have been studied in [15]. For results, which guarantee the solvability of the Dirichlet BVP for the same equation, in the scalar and in the vector cases, see [12] and [16] respectively.

Concerning the kind of the nonlinearity of the function f ( t , x , p , q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq12_HTML.gif, we note that it is assumed sublinear in [6], semilinear in [11] and linear with respect to x, p and q in [8, 12]. Finally, in [9] and [17]f is a linear function with respect to q, while with respect to p, it is a quadratic function or satisfies Nagumo type growth conditions respectively.

As in [10, 15, 18, 19], we use sign conditions to establish a priori bounds for x, x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq13_HTML.gif and x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq14_HTML.gif, where x ( t ) C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq15_HTML.gif is a solution to a suitable family of BVPs similar to that in [10, 19]. Using these a priori bounds and applying the topological transversality theorem from [20], we prove our main existence result.

2 Basic hypotheses

To formulate our hypotheses, we use the sets
J x = [ min { 0 , a + b 2 , a 2 2 ( a b ) } , max { 0 , a + b 2 , a 2 2 ( a b ) } ] and J p = [ min { a , b } , max { a , b } ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equa_HTML.gif
So, we assume that there are positive constants K, M and a sufficiently small ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq16_HTML.gif such that: H1.
f ( t , x , p , q )  is continuous with respect to  x R  for each  ( t , p , q ) [ 0 , 1 ] × R 2 , f ( t , x , p , q )  is continuous with respect to  q R  for each  ( t , x , p ) [ 0 , 1 ] × R 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equb_HTML.gif
there are constants K x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq17_HTML.gif and K q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq18_HTML.gif such that
f x ( t , x , p , q ) K x > 0 for  ( t , x , p , q ) [ 0 , 1 ] × R 3 , f q ( t , x , p , q ) K q < 0 for  ( t , x , p , q ) [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × R 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equc_HTML.gif
where
M 0 = max { e e 2 1 ( | a b e | + | a e b | ) , L min { K , K x , K q } + max { | a + b | 2 , a 2 2 | a b | } } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equd_HTML.gif
f ( t , x , p , b a ( 1 λ ) x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq19_HTML.gif is bounded for ( λ , t , x , p ) [ 0 , 1 ] 2 × J x × J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq20_HTML.gif and L = max { sup | f ( t , x , p , b a ( 1 λ ) x ) | , max K | b a ( 1 λ ) x | } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq21_HTML.gif for ( λ , t , x , p ) [ 0 , 1 ] 2 × J x × J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq22_HTML.gif.H2.
f ( t , x , p , q ) + K q 0 for  ( t , x , p , q ) [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × R × ( , M ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Eque_HTML.gif
and
f ( t , x , p , q ) + K q 0 for  ( t , x , p , q ) [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × R × ( M , ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equf_HTML.gif

where M 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq23_HTML.gif is as in H1.

H3. The functions f ( t , x , p , q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq6_HTML.gif and f q ( t , x , p , q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq24_HTML.gif are continuous for ( t , x , p , q ) [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × [ M 1 ε , M 1 + ε ] × [ M 2 ε , M 2 + ε ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq25_HTML.gif, where M 1 = min { | a | , | b | } + M 0 + M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq26_HTML.gif, M 2 = M 0 + M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq27_HTML.gif and M 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq23_HTML.gif is as in H1.

3 Auxiliary lemmas

In order to obtain our main existence results, we use the constant K from the hypotheses to construct the family of BVPs
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equg_HTML.gif

where λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq28_HTML.gif and prove the following three auxiliary results.

Lemma 3.1 Let H 1 hold and x ( t ) C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq15_HTML.gif be a solution to (3.1) λ , λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq28_HTML.gif. Then
| x ( t ) | M 0 , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equh_HTML.gif
Proof For λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq29_HTML.gif, problem (3.1)0 is of the form
x x = 0 , x ( 0 ) = a , x ( 1 ) = b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equi_HTML.gif
The unique solution to this BVP satisfies the bound
| x ( t ) | e e 2 1 ( | a b e | + | a e b | ) , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equj_HTML.gif
Let now λ ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq30_HTML.gif. Then the function
y ( t ) = x ( t ) s ( t ) , t [ 0 , 1 ] ,  where  s ( t ) = 1 2 ( b a ) t 2 + a t , t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equk_HTML.gif
is a solution to the homogeneous boundary value problem
K ( y + b a ( 1 λ ) ( y + s ) ) = λ ( K ( y + b a ( 1 λ ) ( y + s ) ) + f ( t , y + s , y + s , y + b a ( 1 λ ) ( y + s ) ) ) , y ( 0 ) = 0 , y ( 1 ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equl_HTML.gif
The equation is equivalent to the following one
( 1 λ ) K y = ( 1 λ ) 2 K y ( 1 λ ) K ( b a ( 1 λ ) s ) + λ f ( t , y + s , y + s , y + b a ( 1 λ ) ( y + s ) ) λ f ( t , s , y + s , y + b a ( 1 λ ) ( y + s ) ) + λ f ( t , s , y + s , y + b a ( 1 λ ) ( y + s ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equm_HTML.gif
Hence, by the intermediate value theorem, we obtain consecutively
( 1 λ ) K y = ( 1 λ ) 2 K y ( 1 λ ) K ( b a ( 1 λ ) s ) + λ f x ( t , s + θ 1 y , y + s , y + b a ( 1 λ ) ( y + s ) ) y + λ f ( t , s , y + s , y + b a ( 1 λ ) ( y + s ) ) λ f ( t , s , y + s , y + b a ( 1 λ ) s ) + λ f ( t , s , y + s , y + b a ( 1 λ ) s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equn_HTML.gif
for any θ 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq31_HTML.gif depending on λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq32_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq33_HTML.gif and y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq34_HTML.gif,
( 1 λ ) K y = ( 1 λ ) 2 K y ( 1 λ ) K ( b a ( 1 λ ) s ) + λ f x ( t , s + θ 1 y , y + s , y + b a ( 1 λ ) ( y + s ) ) y + λ f q ( t , s , y + s , y + b a ( 1 λ ) s θ 2 ( 1 λ ) y ) ( ( λ 1 ) y ) + λ f ( t , s , y + s , y + b a ( 1 λ ) s ) λ f ( t , s , y + s , b a ( 1 λ ) s ) + λ f ( t , s , y + s , b a ( 1 λ ) s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equo_HTML.gif
and
( 1 λ ) K y = ( 1 λ ) 2 K y ( 1 λ ) K ( b a ( 1 λ ) s ) + λ f x ( t , s + θ 1 y , y + s , y + b a ( 1 λ ) ( y + s ) ) y + λ f q ( t , s , y + s , y + b a ( 1 λ ) s θ 2 ( 1 λ ) y ) ( ( 1 λ ) y ) + λ f q ( t , s , y + s , b a ( 1 λ ) s + θ 3 y ) y + λ f ( t , s , y + s , b a ( 1 λ ) s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equp_HTML.gif
for any θ 2 , θ 3 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq35_HTML.gif depending on λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq36_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq33_HTML.gif and y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq34_HTML.gif,
{ ( ( 1 λ ) K λ f q ( t , s , y + s , b a ( 1 λ ) s + θ 3 y ) ) y = { ( 1 λ ) 2 K + λ f x ( t , s + θ 1 y , y + s , y + b a ( 1 λ ) ( y + s ) ) λ ( 1 λ ) f q ( t , s , y + s , y + b a ( 1 λ ) s θ 2 ( 1 λ ) y ) } y + λ f ( t , s , y + s , b a ( 1 λ ) s ) ( 1 λ ) K ( b a ( 1 λ ) s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ2_HTML.gif
(3.2)
Next, suppose that | y ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq37_HTML.gif achieves its maximum at t 0 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq38_HTML.gif. Then the function z = y 2 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq39_HTML.gif has also a maximum at t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq40_HTML.gif. Consequently, we have
0 z ( t 0 ) = 2 y ( t 0 ) y ( t 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ3_HTML.gif
(3.3)
Using the fact that y ( t 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq41_HTML.gif, from (3.2) we obtain
{ { ( 1 λ ) K λ f q ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 + θ 3 , 0 y 0 ) } y 0 = { ( 1 λ ) { ( 1 λ ) K λ f q ( t 0 , s 0 , s 0 , y 0 + b a ( 1 λ ) s 0 θ 2 , 0 ( 1 λ ) y 0 ) } + λ f x ( t 0 , s 0 + θ 1 , 0 y 0 , s 0 , y 0 + b a ( 1 λ ) ( y 0 + s 0 ) ) } y 0 + λ f ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 ) ( 1 λ ) K ( b a ( 1 λ ) s 0 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ4_HTML.gif
(3.4)

where θ 1 , 0 = θ 1 ( t 0 , s 0 , y 0 + b a ( 1 λ ) ( y 0 + s 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq42_HTML.gif, θ 2 , 0 = θ 2 ( t 0 , s 0 , s 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq43_HTML.gif, θ 3 , 0 = θ 3 ( t 0 , s 0 , s 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq44_HTML.gif, and s 0 = s ( t 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq45_HTML.gif, s 0 = s ( t 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq46_HTML.gif, y 0 = y ( t 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq47_HTML.gif, y 0 = y ( t 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq48_HTML.gif.

In view of H1, from (3.4) we have
{ ( 1 λ ) { ( 1 λ ) K λ f ¯ q } + λ f ¯ x min { ( 1 λ ) K λ f ¯ q , f ¯ x } ( 1 λ ) ( ( 1 λ ) K λ f ¯ q ) + λ f ¯ x min { K , f ¯ q , f ¯ x } min { K , K x , K q } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ5_HTML.gif
(3.5)
where
f ¯ q = f q ( t 0 , s 0 , s 0 , y 0 + b a ( 1 λ ) s 0 θ 2 , 0 ( 1 λ ) y 0 ) , f ¯ x = f x ( t 0 , s 0 + θ 1 , 0 y 0 , s 0 , y 0 + b a ( 1 λ ) ( y 0 + s 0 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equq_HTML.gif
Suppose now that | y ( t 0 ) | > L ( min { K , K x , K q } ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq49_HTML.gif. Then, from (3.4) and (3.5) it follows that
{ { ( 1 λ ) K λ f q ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 + θ 3 , 0 y 0 ) } y 0 min { K , K x , K q } y ( t 0 ) + λ f ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 ) ( 1 λ ) K ( b a ( 1 λ ) s 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ6_HTML.gif
(3.6)
if y ( t 0 ) > L ( min { K , K x , K q } ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq50_HTML.gif or
{ { ( 1 λ ) K λ f q ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 + θ 3 , 0 y 0 ) } y 0 min { K , K x , K q } y ( t 0 ) + λ f ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 ) ( 1 λ ) K ( b a ( 1 λ ) s 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ7_HTML.gif
(3.7)
if y ( t 0 ) < L ( min { K , K x , K q } ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq51_HTML.gif. Multiplying (3.6) and (3.7) by y ( t 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq52_HTML.gif, we obtain
{ ( 1 λ ) K λ f q ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 + θ 3 , 0 y 0 ) } y 0 y 0 y 0 ( min { K , K x , K q } y 0 L ) > 0 , { ( 1 λ ) K λ f q ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 + θ 3 , 0 y 0 ) } y 0 y 0 | y 0 | ( min { K , K x , K q } | y 0 | L ) > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equr_HTML.gif
respectively. Finally, since ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 + θ 3 , 0 y 0 ) [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × R 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq53_HTML.gif we have f q ( t 0 , s 0 , s 0 , b a ( 1 λ ) s 0 + θ 3 , 0 y 0 ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq54_HTML.gif. So
y 0 y 0 > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equs_HTML.gif
which contradicts (3.3). Thus, we infer that if | y ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq37_HTML.gif achieves its maximum on ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq55_HTML.gif, then
| y ( t ) | L min { K , K x , K q } for  t [ 0 , 1 ]  and  λ ( 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equt_HTML.gif
Let | y ( 1 ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq56_HTML.gif be the maximum of | y ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq57_HTML.gif and suppose that | y ( 1 ) | > L ( min { K , K x , K q } ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq58_HTML.gif. Following the above reasoning and using the fact that y ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq59_HTML.gif, we obtain
y ( 1 ) y ( 1 ) > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equu_HTML.gif
If y ( 1 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq60_HTML.gif, then y ( 1 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq61_HTML.gif and so y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq62_HTML.gif is a strictly increasing function for t U 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq63_HTML.gif, where U 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq64_HTML.gif is a sufficiently small neighbourhood of t = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq65_HTML.gif. So, we see that
y ( t ) < y ( 1 ) = 0 for  t U 1 { 1 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equv_HTML.gif
i.e., y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq66_HTML.gif is a strictly decreasing function for t U 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq63_HTML.gif. Therefore, y ( 1 ) = | y ( 1 ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq67_HTML.gif can not be the maximum of | y ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq57_HTML.gif on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq68_HTML.gif, which is a contradiction. Assume next that y ( 1 ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq69_HTML.gif. Then similar to the above arguments lead again to a contradiction. Thus, we see that
| y ( 1 ) | L min { K , K x , K q } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equw_HTML.gif
The inequality
| y ( 0 ) | L min { K , K x , K q } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equx_HTML.gif
can be obtained in the same manner. Consequently, the eventual solutions of (3.1) λ , λ ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq70_HTML.gif satisfy the bound
| x ( t ) | | y ( t ) | + | s ( t ) | L min { K , K x , K q } + max { a 2 2 | a b | , | a + b | 2 } , t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equy_HTML.gif

and the proof of the lemma is completed. □

Lemma 3.2 Let H 1 and H 2 hold and x ( t ) C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq15_HTML.gif be a solution to (3.1) λ , λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif. Then:
  1. (a)

    | x ( t ) ( 1 λ ) x ( t ) | M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq72_HTML.gif, | x ( t ) | M 0 + M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq73_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq33_HTML.gif.

     
  2. (b)

    | x ( t ) | min { | a | , | b | } + M 0 + M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq74_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq33_HTML.gif.

     
Proof (a) Suppose there exists a ( λ 0 , t 0 ) [ 0 , 1 ] 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq75_HTML.gif or a ( λ 1 , t 1 ) [ 0 , 1 ] 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq76_HTML.gif such that
x ( t 0 ) ( 1 λ 0 ) x ( t 0 ) < M or x ( t 1 ) ( 1 λ 1 ) x ( t 1 ) > M . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equz_HTML.gif
By Lemma 3.1, we have
| x ( t ) | M 0 for  t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ8_HTML.gif
(3.8)
In particular, (3.8) holds for t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq40_HTML.gif and t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq77_HTML.gif. Thus, in view of H2, we have
0 > K ( x ( t 0 ) ( 1 λ 0 ) x ( t 0 ) ) = λ 0 { K ( x ( t 0 ) ( 1 λ 0 ) x ( t 0 ) ) + f ( t 0 , x ( t 0 ) , x ( t 0 ) , ( x ( t 0 ) ( 1 λ 0 ) x ( t 0 ) ) ) } 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equaa_HTML.gif
or
0 < K ( x ( t 1 ) ( 1 λ 1 ) x ( t 1 ) ) = λ 1 { K ( x ( t 1 ) ( 1 λ 1 ) x ( t 1 ) ) + f ( t 1 , x ( t 1 ) , x ( t 1 ) , ( x ( t 1 ) ( 1 λ 1 ) x ( t 1 ) ) ) } 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equab_HTML.gif
respectively. The obtained contradictions show that
M x ( t ) ( 1 λ ) x ( t ) M for  t [ 0 , 1 ]  and  λ [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equac_HTML.gif
and therefore
( M 0 + M ) x ( t ) M 0 + M for  t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equad_HTML.gif
which proves (a).
  1. (b)
    By the mean value theorem, for each t ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq78_HTML.gif there is a ξ ( 0 , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq79_HTML.gif such that
    x ( t ) x ( 0 ) = x ( ξ ) t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equae_HTML.gif
     
Since, in view of (a), we have | x ( ξ ) | M 0 + M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq80_HTML.gif, from the last formula we find that
| x ( t ) | | x ( 0 ) | + | x ( ξ ) | min { | a | , | b | } + M 0 + M , t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equaf_HTML.gif

which proves (b) and completes the proof of the lemma. □

Lemma 3.3 Let H 1, H 2 and H 3 hold. Then there exists a function G ( λ , t , x , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq81_HTML.gif continuous for ( λ , t , x , p ) [ 0 , 1 ] 2 × [ M 0 ε , M 0 + ε ] × [ M 1 ε , M 1 + ε ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq82_HTML.gif and such that
  1. (a)
    the BVP
    x ( 1 λ ) x = G ( λ , t , x , x ) , t [ 0 , 1 ] , x ( 0 ) = a , x ( 1 ) = b , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equag_HTML.gif
     
is equivalent to BVP (3.1) λ .
  1. (b)

    G ( 0 , t , x , p ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq83_HTML.gif for ( t , x , p ) Π q [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × [ M 1 ε , M 1 + ε ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq84_HTML.gif.

     
Proof (a) We write the differential equation from (3.1) λ as
λ f ( t , x , x , ( x ( 1 λ ) x ) ) ( 1 λ ) K ( x ( 1 λ ) x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ9_HTML.gif
(3.9)
and consider the function
F ( λ , t , x , p , q ) = λ f ( t , x , p , q ) ( 1 λ ) K q for  ( λ , t , x , p , q ) [ 0 , 1 ] × Π , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equah_HTML.gif
where Π = [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × [ M 1 ε , M 1 + ε ] × [ M 2 ε , M 2 + ε ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq85_HTML.gif. Since M 2 ε < M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq86_HTML.gif and M 2 + ε > M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq87_HTML.gif, we can use H2 to conclude that
F ( λ , t , x , p , M 2 ε ) F ( λ , t , x , p , M 2 + ε ) < 0 for  ( λ , t , x , p ) [ 0 , 1 ] × Π q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ10_HTML.gif
(3.10)
On the other hand, for ( λ , t , x , p , q ) [ 0 , 1 ] × Π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq88_HTML.gif we have
F q ( λ , t , x , p , q ) = λ f q ( t , x , p , q ) ( 1 λ ) K max { K , K q } < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ11_HTML.gif
(3.11)
Finally, from H3 we have that
F ( λ , t , x , p , q )  and  F q ( λ , t , x , p , q )  are continuous for  ( λ , t , x , p , q ) [ 0 , 1 ] × Π . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equ12_HTML.gif
(3.12)
So, (3.10), (3.11) and (3.12) allow us to apply a well-known theorem to conclude that there is a unique function G ( λ , t , x , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq89_HTML.gif which is continuous for ( λ , t , x , p ) [ 0 , 1 ] × Π q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq90_HTML.gif and such that the equations
q = G ( λ , t , x , p ) , ( λ , t , x , p ) [ 0 , 1 ] × Π q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equai_HTML.gif
and
F ( λ , t , x , p , q ) = 0 , ( λ , t , x , p , q ) [ 0 , 1 ] × Π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equaj_HTML.gif
are equivalent. Now from Lemma 3.1 we have
M 0 ε x ( t ) M 0 + ε for  t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equak_HTML.gif
and Lemma 3.2 yields
M 1 ε x ( t ) M 1 + ε and M 2 ε < M x ( t ) ( 1 λ ) x ( t ) M < M 2 + ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equal_HTML.gif
for t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq33_HTML.gif and λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif. Consequently, equation (3.9) is equivalent to the equation
x ( 1 λ ) x = G ( λ , t , x , x ) , t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equam_HTML.gif
which yields the first assertion.
  1. (b)

    It follows immediately from F ( 0 , t , x , p , 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq91_HTML.gif for ( t , x , p ) Π q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq92_HTML.gif. □

     

4 The main result

Our main result is the following existence theorem, the proof of which is based on the lemmas of the previous sections and the Topological transversality theorem [20].

Theorem 4.1 Let H 1, H 2 and H 3 hold. Then problem (N) has at least one solution in C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq93_HTML.gif.

Proof First, we observe that according to Lemma 3.3, the family of boundary value problems
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equan_HTML.gif
is equivalent to the family (3.1) λ for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif. Next define the set
U = { x C B 2 [ 0 , 1 ] : | x | < M 0 + ε , | x | < M 1 + ε , | x | < M 2 + ε } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equao_HTML.gif
where C B 2 [ 0 , 1 ] = { x ( t ) C 2 [ 0 , 1 ] : x ( 0 ) = a , x ( 1 ) = b } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq94_HTML.gif, and the maps
j : C B 2 [ 0 , 1 ] C 1 [ 0 , 1 ] by  j x = x , G λ : C 1 [ 0 , 1 ] C [ 0 , 1 ] by  ( G λ x ) ( t ) = G ( λ , t , x ( t ) , x ( t ) ) x ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equap_HTML.gif
where t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq33_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif, x ( t ) j ( U ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq95_HTML.gif and
L λ : C B 2 [ 0 , 1 ] C [ 0 , 1 ] by  L λ x = x ( 2 λ ) x , λ [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equaq_HTML.gif
Since L λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq96_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif, is a continuous, linear, one-to-one map of C B 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq97_HTML.gif onto C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq98_HTML.gif, the map L λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq99_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq28_HTML.gif exists and is continuous. In addition, G λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq100_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif, is a continuous and j is a completely continuous embedding. Since j ( U ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq101_HTML.gif is a compact subset of C 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq102_HTML.gif, and G λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq100_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq36_HTML.gif, and L λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq99_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif, are continuous on j ( U ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq101_HTML.gif and G λ ( j ( U ¯ ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq103_HTML.gif respectively, the homotopy
H : U ¯ × [ 0 , 1 ] C 2 [ 0 , 1 ] defined by  H ( x , λ ) H λ ( x ) L λ 1 G λ j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equar_HTML.gif
is compact. Besides, the equation
L λ 1 G λ j ( x ) = x for  x U ¯ yields L λ x = G λ j ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equas_HTML.gif
which coincides with BVP (3.13) λ . Thus, the fixed points of H λ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq104_HTML.gif are solutions to (3.13) λ . But, from Lemma 3.1 and Lemma 3.2 it follows that the solutions to (3.13) λ are elements of U. Consequently, H λ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq105_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif, is a fixed point free on ∂U, i.e., H λ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq105_HTML.gif is an admissible map for all λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq71_HTML.gif. Finally, we see that the map H 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq106_HTML.gif is a constant map, i.e., H 0 ( x ) l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq107_HTML.gif, where l is the unique solution to the BVP
x 2 x = x , x ( 0 ) = a , x ( 1 ) = b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equat_HTML.gif

From the fact that l U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq108_HTML.gif, it follows that H 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq109_HTML.gif is an essential map (see, [20]). By the Topological transversality theorem (see, [20]), H 1 = L 1 1 G 1 j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq110_HTML.gif is also essential, i.e., problem (3.13)1 has a C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq93_HTML.gif-solution. It is also a solution to (3.1)1, by Lemma 3.3. To complete the proof, remark that problem (3.1)1 coincides with the problem (N). □

We conclude with the following example, which illustrates our main result.

Example 4.2 Consider the boundary value problem
1 ( 1.5 + t ) x t x 5 cos x + x = 0 , x ( 0 ) = 0 , x ( 1 ) = 10 4 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equau_HTML.gif
It is clear that for ( t , x , p , q ) [ 0 , 1 ] × R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq111_HTML.gif the function
f ( t , x , p , q ) = 1 ( 1.5 + t ) q t q 5 cos p + x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equav_HTML.gif

is continuous and f x ( t , x , p , q ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq112_HTML.gif and f q ( t , x , p , q ) = 1.5 t 5 t q 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq113_HTML.gif. Thus H1 holds for K x = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq114_HTML.gif and K q = 1.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq115_HTML.gif.

To verify H2 we choose, for example, K = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq116_HTML.gif, M = 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq117_HTML.gif and ε = 3 10 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq118_HTML.gif. Next we need the constants L and M 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq23_HTML.gif. Having in mind that J x = [ 0 , 5 10 5 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq119_HTML.gif and J p = [ 0 , 10 4 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq120_HTML.gif, from
5 10 5 10 4 ( 1 λ ) x 10 4 for  ( λ , x ) [ 0 , 1 ] × J x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equaw_HTML.gif
it follows that max K | b a ( 1 λ ) x | = 0.5 max ( 10 4 ( 1 λ ) x ) = 5 10 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq121_HTML.gif. On the other hand, from
2 , 5 10 4 10 20 ( 1 , 5 + t ) ( 10 4 ( 1 λ ) x ) t ( 10 4 ( 1 λ ) x ) 5 7 , 5 10 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equax_HTML.gif
for ( λ , t , x ) [ 0 , 1 ] 2 × J x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq122_HTML.gif and
0 1 cos p 5 10 9 for  p J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equay_HTML.gif
we have
26 10 5 < 1 ( 1 , 5 + t ) ( 10 4 ( 1 λ ) x ) t ( 10 4 ( 1 λ ) x ) 5 cos p + x 2 , 5 10 5 + 5 10 9 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equaz_HTML.gif
for ( λ , t , x , p ) [ 0 , 1 ] 2 × J x × J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq22_HTML.gif, which means that for ( λ , t , x , p ) [ 0 , 1 ] 2 × J x × J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq22_HTML.gif
max | f ( t , x , p , b a ( 1 λ ) x ) | = = max | 1 ( 1 , 5 + t ) ( 10 4 ( 1 λ ) x ) t ( 10 4 ( 1 λ ) x ) 5 cos p + x | 26 10 5 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equba_HTML.gif
So, L = max { 26 10 5 , 5 10 5 } = 26 10 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq123_HTML.gif. Then
M 0 = max { e e 2 1 ( 10 4 e + 10 4 ) , 26 10 5 min { 0.5 , 1 , 1.5 } + 5 10 5 } = 57 10 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equbb_HTML.gif
and we see that for ( t , x , p , q ) [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × R × ( , M ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq124_HTML.gif
f ( t , x , p , q ) + K q = ( 1 + t ) q t q 5 + 1 cos p + x > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equbc_HTML.gif
and
f ( t , x , p , q ) + K q < 0 for  ( t , x , p , q ) [ 0 , 1 ] × [ M 0 ε , M 0 + ε ] × R × ( M , ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_Equbd_HTML.gif

Thus, H2 also holds.

Finally, H3 holds since f ( t , x , p , q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq12_HTML.gif and f q ( t , x , p , q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq24_HTML.gif are continuous for ( t , x , p , q ) [ 0 , 1 ] × R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq111_HTML.gif.

Thus, we can apply Theorem 4.1 to conclude that the considered problem has a solution in C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-77/MediaObjects/13661_2012_Article_196_IEq93_HTML.gif.

Declarations

Acknowledgements

In memory of Professor Myron K. Grammatikopoulos, 1938-2007.

This research was partially supported by Sofia University Grant N350/2012. The research of N. Popivanov was partially supported by the Bulgarian NSF under Grants DO 02-75/2008 and DO 02-115/2008.

Authors’ Affiliations

(1)
Naval Academy of Greece
(2)
Department of Mathematics, Technical University of Sliven
(3)
Faculty of Mathematics and Informatics, ‘St. Kl. Ohridski’ University of Sofia

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© Palamides et al.; licensee Springer 2012

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