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

Abstract

We consider equations involving the one-dimensional p-Laplacian

( | u ( t ) | p 2 u ( t ) ) +λf ( 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.

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

Δu+λf(u)=0in Ω,u=0on Ω
(1.1)

and

u (t)+λf ( 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+λf(u)=0in Ω,u=0on 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 f0 on ( a k , b k ) and f0 on ( b k , a k + 1 ) for every k=1,,m1. They proved that, for λ sufficiently large, if

a k a k + 1 f(s)ds>0for all k{1,,m1},
(1.4)

then the problem (1.3) has at least m1 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,,m1}, 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)ds, 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 0u b i , i=1,2,,n1.

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,,n1,

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)|:0u 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,,n1.
  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,dR.

Lemma 2.1 Let (H1) hold. If a(0,1] and d0, 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:RR 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 =ρS. 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

(2.6)
(2.7)

with u =ρS 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(12 x 0 ).

However, this is impossible since 12 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 ρS 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 ρS 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 ρS, there exists a unique λ>0 such that

(3.1)
(3.2)

has a positive solution (λ,u) with u =ρ. Moreover, ρλ(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

(3.3)
(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 ds,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 ds.
(3.6)

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

If ρS, 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 0wR. Thus, if ρ<R, then

F ( ρ ) F ( w ) = w ρ f ( v ) d v M p ( ρ p w p )

for 0wρ. From (3.6), we have that for any ρR,

[ 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 dv,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 ρ0.

For 0v 1 2 , using l’Hospital’s rule, it follows that as ρ0,

| 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 v1,

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 dw=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 dv.
(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,,n1. 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)|:0u b i }, then

f(u)M|uk|,0u 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 p2, 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)|:0u β i } and K 2 =max{| f (u)|:0u β i }. Since f( β i )=0,

    f(u) K 2 |u β i |,0u< β i .
    (3.11)

Hence, if 0uρ< 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 ρλ(ρ) and Lemma 3.2.

  1. (b)

    follows from the continuity of ρλ(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 dτ,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 .

 □

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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).

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RM completed the main study, carried out the results of this article. YL drafted the manuscript. AOMA checked the proofs and verified the calculation. All the authors read and approved the final manuscript.

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Ma, R., Lu, Y. & Abubaker, A.O.M. On positive solutions to equations involving the one-dimensional p-Laplacian. Bound Value Probl 2013, 125 (2013). https://doi.org/10.1186/1687-2770-2013-125

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