Open Access

On positive solutions to equations involving the one-dimensional p-Laplacian

Boundary Value Problems20132013:125

DOI: 10.1186/1687-2770-2013-125

Received: 15 October 2012

Accepted: 29 April 2013

Published: 15 May 2013

Abstract

We consider equations involving the one-dimensional p-Laplacian

( | u ( t ) | p 2 u ( t ) ) + λ f ( u ( t ) ) = 0 , t ( 0 , 1 )

with the Dirichlet boundary conditions. By using time map methods, we show how changes of the sign of f ( ) lead to multiple positive solutions of the problem for sufficiently large λ.

MSC:34B10, 34B18.

Keywords

positive solutions one-dimensional p-Laplacian uniqueness time map

1 Introduction

Let f : [ 0 , ) R be continuous and change its sign. Let Ω be an open subset of R N with smooth boundary Ω. The semi-positone problems and their special cases
Δ u + λ f ( u ) = 0 in  Ω , u = 0 on  Ω
(1.1)
and
u ( t ) + λ f ( u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = 0
(1.2)

(and their finite difference analogues) have been extensively studied since early 1980s. Several different approaches such as variational methods, bifurcation theory, lower and upper solutions method and quadrature arguments have been successfully applied to show the existence of multiple solutions. See Brown and Budin [1], Peitgen et al. [2], Peitgen and Schmitt [3], Hess [4], Ambrosetti and Hess [5], Cosner and Schmitt [6], Dancer and Schmitt [7], Espinoza [8], Anuradha and Shivaji [9], Anuradha et al. [10], de Figueiredo [11], Lin and Pai [12], Clément and Sweers [13] and the references therein.

Very recently, Loc and Schmitt [14] considered the problem
Δ p u + λ f ( u ) = 0 in  Ω , u = 0 on  Ω ,
(1.3)
where Δ p is the p-Laplace operator for p ( 1 , ) . They assumed that the nonlinearity f is a continuous function on , f ( 0 ) 0 , and there exist 0 < a 1 < b 1 < a 2 < b 2 < < b m 1 < a m such that f 0 on ( a k , b k ) and f 0 on ( b k , a k + 1 ) for every k = 1 , , m 1 . They proved that, for λ sufficiently large, if
a k a k + 1 f ( s ) d s > 0 for all  k { 1 , , m 1 } ,
(1.4)
then the problem (1.3) has at least m 1 positive bounded solutions u 1 , , u m 1 which belong to the Sobolev space W 0 1 , p ( Ω ) and are such that u ( a k , a k + 1 ] for each k { 1 , , m 1 } , where
u = max { | u ( x ) | x Ω ¯ } .

In the special case that p = 2 and N = 1 , Brown and Budin [1] applied the quadrature arguments to get the following more detailed results.

Theorem A [[1], Theorem 3.8]

Assume that

(H1) f C 1 [ 0 , ) ;

(H2) f ( 0 ) > 0 ;

(H3) There exists a 1 , , a n R such that 0 < a 1 < a 2 < < a n and f ( a i ) 0 for i = 1 , 2 , , n ;

(H4) If F ( u ) = 0 u f ( s ) d s , there exist b 1 , , b n 1 R with a 1 < b 1 < a 2 < b 2 < < a n 1 < b n 1 < a n such that f ( b i ) > 0 and F ( b i ) > F ( u ) for 0 u b i , i = 1 , 2 , , n 1 .

Then:
  1. (a)

    For all λ > 0 , there exists a solution ( λ , u ) of (1.2).

     
  2. (b)
    If λ > inf { λ ( ρ ) : ρ ( α i , β i ) } , there exist at least two solutions ( λ , u ) of (1.2) such that
    α i < u < β i , i = 1 , 2 , , n 1 ,
     
where
β i = inf { u > b i f ( u ) = 0 } , α i = inf { u ( u , β i ) S }
(1.5)
and
S = { u u > 0 , f ( u ) > 0 , F ( u ) > F ( s ) for all s : 0 s < u } .
(1.6)
  1. (c)
    If ( λ , u ) is any solution of (1.2) such that α i < u < β i , then
    λ > 4 α i k 1 ,
     

where k = sup { | f ( u ) | : 0 u β i } .

Of course the natural question is whether or not the similar results still hold for the corresponding problem involving the one-dimensional p-Laplacian
( | u ( t ) | p 2 u ( t ) ) + λ f ( u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = 0 .
(1.7)

We shall answer these questions in the affirmative if p ( 1 , 2 ] . More precisely, we get the following theorem.

Theorem 1.1 Let p ( 1 , 2 ] and let (H1), (H3), (H4) hold. Assume that

(H2′) either f ( 0 ) > 0 or f ( 0 ) = 0 and
f 0 = lim s 0 + f ( s ) s p 1 > 0 .
(1.8)
Then:
  1. (a)
    For all λ > λ 1 f 0 , there exists a solution ( λ , u ) of (1.7), and λ 1 is the least eigenvalue of BVP
    ( | u ( t ) | p 2 u ( t ) ) + λ | u ( t ) | p 2 u ( t ) = 0 , t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = 0 .
    (1.9)
     
  2. (b)
    If λ > inf { λ ( ρ ) : ρ ( α i , β i ) } , there exist at least two solutions ( λ , u ) of (1.7) such that
    α i < u < β i , i = 1 , 2 , , n 1 .
     
  3. (c)
    If ( λ , u ) is any solution of (1.7) such that α i < u < β i , then
    λ > ( α i C ) p 1 ,
     
where
C = p 1 p ( 1 2 ) p p 1 ( sup s [ 0 , β i ] | f ( s ) | ) 1 p 1 .
(1.10)

We shall apply the time map method to show how changes of the sign of f ( ) lead to multiple positive solutions of (1.7) for sufficiently large λ.

In the following, we extend f so that f ( u ) > 0 for all u < 0 , then all the solutions of (1.7) are positive on ( 0 , 1 ) .

2 Preliminaries

To prove our main results, we use the uniqueness results due to Reichel and Walter [15] on the initial value problem
( | u ( t ) | p 2 u ( t ) ) + λ f ( u ( t ) ) = 0 , t ( 0 , 1 ) , u ( a ) = b , u ( a ) = d ,
(2.1)

where a [ 0 , 1 ] and b , d R .

Lemma 2.1 Let (H1) hold. If a ( 0 , 1 ] and d 0 , then the initial value problem (2.1) has a unique local solution. The extension u ( t ) remains unique as long as u ( t ) 0 .

Proof It is an immediate consequence of Reichel and Walter [[15], Theorem 2]. □

Lemma 2.2 Let (H1) hold. Let a ( 0 , 1 ) , and let ρ ( 0 , ) be such that
f ( ρ ) 0 .
Then the initial value problem
( | u ( t ) | p 2 u ( t ) ) + λ f ( u ( t ) ) = 0 , t ( 0 , 1 ) , u ( a ) = ρ , u ( a ) = 0
(2.2)

has a unique local solution.

Proof (H1) implies that f is locally Lipschitzian. This together with the assumption f ( ρ ) 0 and using [[15], (iii) and (v) in the case (β) of Theorem 4] yields that (2.2) has a unique solution in some neighborhood of a. □

Lemma 2.3 Let g : R R be continuous. Let u be a solution of the equation
( | u ( t ) | p 2 u ( t ) ) + g ( u ( t ) ) = 0 , t ( 0 , 1 )
(2.3)
with u = ρ S . Let x 0 ( 0 , 1 ) be such that u ( x 0 ) = 0 . Then
u ( x 0 t ) u ( x 0 + t ) , t ( 0 , min { x 0 , 1 x 0 } ) .
(2.4)
Proof Since g is independent of t, both u ( x 0 t ) and u ( x 0 + t ) satisfy the initial value problem
{ ( | w ( t ) | p 2 w ( t ) ) + g ( w ( t ) ) = 0 , t ( 0 , min { x 0 , 1 x 0 } ) , w ( 0 ) = u ( x 0 ) , w ( 0 ) = 0 .
(2.5)

By Lemmas 2.1 and 2.2, (2.5) has a unique solution defined on t ( 0 , min { x 0 , 1 x 0 } ) . Therefore, (2.4) is true. □

Lemma 2.4 Let ( λ , u ) be a positive solution of the problem
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-125/MediaObjects/13661_2012_Article_370_Equ16_HTML.gif
(2.6)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-125/MediaObjects/13661_2012_Article_370_Equ17_HTML.gif
(2.7)
with u = ρ S and λ > 0 . Let x 0 ( 0 , 1 ) be such that u ( x 0 ) = 0 . Then
  1. (a)

    x 0 = 1 2 ;

     
  2. (b)

    x 0 is the unique point on which u attains its maximum;

     
  3. (c)

    u ( t ) > 0 , t ( 0 , 1 2 ) .

     
Proof (a) Suppose on the contrary that x 0 1 2 , say x 0 > 1 2 , then
0 = u ( 1 ) = u ( 1 2 x 0 ) .
However, this is impossible since 1 2 x 0 ( 0 , 1 ) and u > 0 in ( 0 , 1 ) . Therefore x 0 = 1 2 .
  1. (b)
    Suppose on the contrary that there exists x 1 ( 0 , 1 ) with x 1 x 0 and
    u ( x 1 ) = u ( x 0 ) = : ρ .
     

We may assume that x 1 < x 0 . The other case can be treated in a similar way.

If u ( t ) u ( x 0 ) in the interval ( x 1 , x 0 ) , then Lemma 2.3 yields that
u ( t ) u ( x 0 ) = ρ > 0 , t ( 0 , 1 ) .

This contradicts the boundary conditions u ( 0 ) = u ( 1 ) = 0 . Therefore, u ( t ) u ( x 0 ) in any subinterval of ( 0 , 1 ) .

So, there exists x ( x 1 , x 0 ) , such that
u ( x ) = min { u ( t ) t ( x 1 , x 0 ) } .
Obviously,
0 < u ( x ) < ρ , u ( x ) = 0 .
Multiplying both sides of the equation in (2.6) by u and integrating from t to x 0 , we get that
| u ( t ) | p = λ p p 1 [ F ( ρ ) F ( u ( t ) ) ] , t [ 0 , 1 2 ]
(2.8)
and subsequently,
0 = | u ( x ) | p = λ p p 1 [ F ( ρ ) F ( u ( x ) ) ] .
This contradicts the facts that ρ S and u ( x ) < ρ . Therefore,
u ( 1 2 ) > u ( t ) , t [ 0 , 1 2 ) .
Similarly, we can prove that
u ( 1 2 ) > u ( t ) , t ( 1 2 , 1 ] .
  1. (c)
    Suppose on the contrary that there exists x ˆ ( 0 , 1 2 ) with u ( x ˆ ) = 0 . Then
    u ( x ˆ ) < ρ .
     
This together with (2.8) implies that
0 = | u ( x ˆ ) | p = λ p p 1 [ F ( ρ ) F ( u ( x ˆ ) ) ] .

This contradicts the facts that ρ S and u ( x ˆ ) < ρ . □

3 Proof of the main results

To prove Theorem 1.1, we need the following preliminary results.

Lemma 3.1 For any ρ S , there exists a unique λ > 0 such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-125/MediaObjects/13661_2012_Article_370_Equ19_HTML.gif
(3.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-125/MediaObjects/13661_2012_Article_370_Equ20_HTML.gif
(3.2)

has a positive solution ( λ , u ) with u = ρ . Moreover, ρ λ ( ρ ) is a continuous function on S .

Proof By Lemma 2.4, ( λ , u ) is a positive solution of (3.1), (3.2) if and only if ( λ , u ) is a positive solution of
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-125/MediaObjects/13661_2012_Article_370_Equ21_HTML.gif
(3.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-125/MediaObjects/13661_2012_Article_370_Equ22_HTML.gif
(3.4)
Suppose that ( λ , u ) is a solution of (3.3), (3.4) with u = ρ . Then
| u ( t ) | p = λ p p 1 ( F ( ρ ) F ( u ( t ) ) ) , t [ 0 , 1 2 ]
and so
t ( p p 1 λ ) 1 / p = 0 u ( t ) ( F ( ρ ) F ( s ) ) 1 / p d s , t [ 0 , 1 2 ] .
(3.5)
Putting t = 1 2 , we obtain
λ 1 / p = 2 ( p 1 p ) 1 / p 0 ρ ( F ( ρ ) F ( s ) ) 1 / p d s .
(3.6)

Hence λ (if exists) is uniquely determined by ρ.

If ρ S , we define λ ( ρ ) by (3.6) and u ( t ) by (3.5). It is straightforward to verify that u is twice differentiable, u satisfies (3.3), (3.4), u > 0 in ( 0 , 1 ) and u ( 1 / 2 ) = ρ . The continuity of λ ( ) is implied by (3.6) and this completes the proof. □

Let
r = inf { u > 0 : f ( u ) = 0 } .

Then ( 0 , r ) S .

Lemma 3.2 Let (H1) and (H2′) hold, and let p ( 1 , ) . Then
lim ρ 0 λ ( ρ ) = λ 1 f 0 , lim ρ r λ ( ρ ) = ,

where λ 1 is the least eigenvalue of (1.9).

Proof We only deal with lim ρ 0 λ ( ρ ) = λ 1 f 0 . The other one can be treated by the same method.

To this end, we divide the proof into two cases.

Case 1. We show that f 0 = implies lim ρ 0 λ ( ρ ) = 0 .

In this case, for any M > 0 , there is a positive number R such that f ( w ) > M w p 1 for 0 w R . Thus, if ρ < R , then
F ( ρ ) F ( w ) = w ρ f ( v ) d v M p ( ρ p w p )
for 0 w ρ . From (3.6), we have that for any ρ R ,
[ λ ( ρ ) ] 1 p = 2 ( p 1 p ) 1 p 0 ρ d w [ F ( ρ ) F ( w ) ] 1 / p 2 ( p 1 p ) 1 / p ( p M ) 1 / p 0 ρ d w [ ρ p w p ] 1 / p 2 ( p 1 M ) 1 / p 0 1 d w ρ [ 1 ( w ρ ) p ] 1 / p 2 ( p 1 M ) 1 / p ( 1 p 1 ) 1 / p 0 ( p 1 ) 1 / p d s [ 1 s p p 1 ] 1 / p 2 ( 1 M ) 1 / p 0 ( p 1 ) 1 / p d s [ 1 s p p 1 ] 1 / p = ( 1 M ) 1 / p π p ,
where
π p : = 2 0 ( p 1 ) 1 / p d s [ 1 s p p 1 ] 1 / p ,
see Zhang [16]. Hence
lim ρ 0 λ ( ρ ) = 0 .
Case 2. We show that f 0 = m for some m ( 0 , ) implies that
lim ρ 0 λ ( ρ ) = p 1 p m τ p p = ( π p ) p f 0 ,
(3.7)
where
τ p = 2 0 1 [ p 1 v p ] 1 / p d v , p > 1 .
(3.8)
In fact, (3.6) yields
[ m λ ( ρ ) ] 1 / p = 2 [ m ( p 1 ) p ] 1 p 0 ρ d w [ F ( ρ ) F ( w ) ] 1 / p = 2 ( p 1 p ) 1 / p 0 1 [ p 1 v p ] 1 / p d v 2 ( p 1 p ) 1 / p 0 1 [ p 1 v p ] 1 / p [ 1 + γ ( ρ , v ) ] 1 / p 1 [ 1 + γ ( ρ , v ) ] 1 / p d v
(3.9)
for p > 1 , where
γ ( ρ , v ) = p m ρ v ρ [ f ( w ) m w p 1 ] d w ρ p ( 1 v p ) .

We will show that the last integral in (3.9) converges to zeros as ρ 0 .

For 0 v 1 2 , using l’Hospital’s rule, it follows that as ρ 0 ,
| γ ( ρ , v ) | = p m ρ v ρ | f ( w ) m w p 1 | d w ρ p ( 1 v p ) p m 0 ρ | f ( w ) m w p 1 | d w ρ p ( 1 v p ) p m | f ( ρ ) m ρ p 1 | p ρ p 1 ( 1 v p ) = p m | f ( ρ ) ρ p 1 m | p ( 1 v p ) 0 .
For 1 2 v 1 ,
lim ρ 0 | γ ( ρ , v ) | = lim sup ρ 0 { | f ( w ) m w p 1 w p 1 | : 1 2 ρ w ρ } p m ρ p ( 1 v p ) ρ v ρ w p 1 d w = 0
uniformly in v. Therefore, (3.9) implies
lim ρ 0 [ m λ ( ρ ) ] 1 / p = 2 ( p 1 p ) 1 p 0 1 [ p 1 v p ] 1 / p d v .
(3.10)

Therefore, (3.7) holds. □

From the definitions of α i and β i , we have that a i α i < β i a i + 1 and ( α i , β i ) S for i = 1 , 2 , , n 1 . Moreover, we have the following.

Lemma 3.3 Let p ( 1 , 2 ] . Then
  1. (i)

    lim ρ α i + λ ( ρ ) = ;

     
  2. (ii)

    lim ρ β i λ ( ρ ) = .

     
Proof (i) Suppose firstly that f ( α i ) > 0 . Since S is open, α i S and so there exists k : 0 < k < α i such that
F ( α i ) = F ( k ) .
Clearly k must be a local maximum for F and so f ( k ) = 0 . If M = max { | f ( u ) | : 0 u b i } , then
f ( u ) M | u k | , 0 u b i .
Let
N = max { | f ( u ) | : 0 u b i } .
Then if α i < ρ < b i ,
F ( ρ ) F ( u ) = F ( ρ ) F ( α i ) + F ( k ) F ( u ) = ( ρ α i ) f ( ξ ) + ( k u ) f ( η ) , where  ξ ( α i , ρ )  and  η ( k , u ) N ( ρ α i ) + M ( k u ) 2 .
Hence
( λ ( ρ ) ) 1 / p = 2 ( p 1 p ) 1 / p 0 ρ ( F ( ρ ) F ( s ) ) 1 / p d s 2 ( p 1 p ) 1 / p 0 α i ( N ( ρ α i ) + M ( k u ) 2 ) 1 / p d u = 0 α i H ρ ( u ) d u .
As ρ α i + , H ρ ( u ) = 2 ( p 1 p ) 1 / p ( N ( ρ α i ) + M ( k u ) 2 ) 1 / p is a nondecreasing sequence of measurable functions. Therefore, by the monotone convergence theorem and the assumption p 2 , it follows that
lim ρ α i + [ λ ( ρ ) ] 1 / p lim ρ α i + 0 α i H ρ ( u ) d u = 0 α i 2 ( p 1 p ) 1 / p M 1 / p [ k u ] 2 / p d u =

since k ( 0 , α i ) .

Suppose next that f ( α i ) = 0 . Then F ( α i ) = 0 .

Since
F ( α i ) F ( u ) = f ( η ) ( α i u ) , where  η ( u , α i ) , | f ( u ) | = | f ( u ) f ( α i ) | M | u α i | .
Thus
0 α i [ F ( α i ) F ( u ) ] 1 / p d u 0 α i [ M | α i u | 2 ] 1 / p d u = 0 α i M 1 / p | α i u | 2 / p d u = .
  1. (ii)
    Let K 1 = max { | f ( u ) | : 0 u β i } and K 2 = max { | f ( u ) | : 0 u β i } . Since f ( β i ) = 0 ,
    f ( u ) K 2 | u β i | , 0 u < β i .
    (3.11)
     
Hence, if 0 u ρ < β i , then it follows from (3.11) that
F ( ρ ) F ( u ) = F ( ρ ) F ( β i ) + F ( β i ) F ( u ) = ( ρ β i ) f ( ξ ) + ( β i u ) f ( η ) , where  ξ ( ρ , β i ) , η ( u , β i ) K 1 ( β i ρ ) + K 2 ( β i u ) 2 .
Hence, if 0 < ρ < β i ,
( λ ( ρ ) ) 1 / p 2 ( p 1 p ) 1 / p 0 ρ ( K 1 ( β i ρ ) + K 2 ( β i u ) 2 ) 1 / p d u = 0 β i G ρ ( u ) d u ,
where G ρ ( u ) = 2 ( p 1 p ) 1 / p ( K 1 ( β i ρ ) + K 2 ( β i u ) 2 ) 1 / p χ [ 0 , ρ ] and χ [ 0 , ρ ] denotes the characteristic function of [ 0 , ρ ] . As G ρ is a nondecreasing sequence of measurable functions, by the monotone convergence theorem
lim ρ β i [ λ ( ρ ) ] 1 / p lim ρ β i 0 β i G ρ ( u ) d u = 0 β i 2 ( p 1 p ) 1 / p K 2 1 / p | β i u | 2 / p d u = .

 □

Proof of Theorem 1.1 (a) follows from the continuity of ρ λ ( ρ ) and Lemma 3.2.
  1. (b)

    follows from the continuity of ρ λ ( ρ ) and Lemma 3.3.

     
  2. (c)
    ( λ , u ) is any solution of (3.1), (3.2) if and only if
    u ( t ) = 0 t ( τ 1 / 2 λ f ( u ( s ) ) d s ) 1 p 1 d τ , t [ 0 , 1 2 ] .
     
Hence
| u ( t ) | λ 1 p 1 0 1 / 2 ( τ 1 / 2 | f ( u ( s ) ) | d s ) 1 p 1 d τ λ 1 p 1 p 1 p ( 1 2 ) p p 1 ( sup y [ 0 , 1 ] | f ( u ( y ) ) | ) 1 p 1 .
Now, if α i < u < β i , then
α i λ 1 p 1 p 1 p ( 1 2 ) p p 1 ( sup s [ 0 , β i ] | f ( s ) | ) 1 p 1
and so
λ > ( α i C ) p 1 .

 □

Declarations

Acknowledgements

The authors are very grateful to the anonymous referees for their valuable suggestions. This work was supported by the NSFC (No. 11061030), NSFC (No. 11126296), SRFDP (No. 20126203110004) and Gansu Provincial National Science Foundation of China (No. 1208RJZA258).

Authors’ Affiliations

(1)
Department of Mathematics, Northwest Normal University

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