1 Introduction

We consider the nonlinear elliptic boundary value problem

Δ u = λ K 1 ( | x | ) f ( v ) in  Ω e , Δ v = λ K 2 ( | x | ) g ( u ) in  Ω e , u ( x ) = v ( x ) = 0 if  | x | = r 0 ( > 0 ) , u ( x ) 0 , v ( x ) 0 as  | x | , }
(1.1)

where λ>0 is a parameter, Δu=div(u) is the Laplace operator, and Ω e ={x R n |x|> r 0 ,n>2} is an exterior domain. Here the nonlinearities f,g:[0,)R are C 1 functions which satisfy:

(H1):f(0)<0 and g(0)<0 (semipositone).

(H2): For i=1,2 there exist b i >0 and q i >1 such that lim s f ( s ) s q 1 = b 1 , and lim s g ( s ) s q 2 = b 2 .

Further, for i=1,2, the weight functions K i C 1 ([ r 0 ,),(0,)) are such that K i (|x|)0 as |x|. In particular, we are interested in the challenging case, where K i do not decay too fast. Namely, we assume

(H3): There exist d 1 ˜ >0, d 2 ˜ >0, ρ(0,n2) such that for i=1,2

d 1 ˜ | x | n + ρ K i ( | x | ) d 2 ˜ | x | n + ρ for |x|1.

We then establish the following.

Theorem 1.1

Let (H1)-(H3) hold. Then (1.1) has a positive radial solution(u,v) (u>0, v>0in Ω e ) when λ is small, and u , v asλ0.

We prove this result via the Leray-Schauder degree theory, by arguments similar to those used in [1] and [2]. The study of such eigenvalue problems with semipositone structure has been documented to be mathematically challenging (see [3], [4]), yet a rich history is developing starting from the 1980s (see [5]–[7]) until recently (see [8]–[12]). In [1], [2] the authors studied such superlinear semipositone problems on bounded domains. In particular, in [12] the authors studied the system

Δ u = λ f ( v ) in  Ω , Δ v = λ g ( u ) in  Ω , u = v = 0 on  Ω , }

where Ω is a bounded domain in R n , n1, and establish an existence result when λ is small. The main motivation of this paper is to extend this study in the case of exterior domains (see Theorem 1.1).

We also discuss a non-existence result for the single equation model:

Δ u = λ K 1 ( | x | ) f ˜ ( u ) in  Ω e , u ( x ) = 0 if  | x | = r 0 ( > 0 ) , u ( x ) 0 as  | x | , }
(1.2)

for large values of λ, when f ˜ , K 1 satisfy the following hypotheses:

(H4): f ˜ C 1 ([0,),R), f ˜ (z)>0 for all z>0, f ˜ (0)<0, and there exists m 0 >0 such that lim z f ˜ ( z ) z m 0 .

(H5): The weight function K 1 C 1 ([ r 0 ,),(0,)) is such that s 2 ( n 1 ) n 2 K 1 ( r 0 s 1 2 n ) is decreasing for s(0,1].

We establish the following.

Theorem 1.2

Let (H3)-(H5) hold. Then (1.2) has no nonnegative radial solution forλ1.

We establish Theorem 1.2 by recalling various useful properties of solutions established in [13], where the authors prove a uniqueness result for λ1 for such an equation in the case when f ˜ is sublinear at ∞. However, the properties we recall from [13] are independent of the growth behavior of f ˜ at ∞. Non-existence results for such superlinear semipositone problems on bounded domain also have a considerable history starting from the work in the 1980s in [14] leading to the recent work in [15]. Here we discuss such a result for the first time on exterior domains.

Finally, we note that the study of radial solutions (u(r),v(r)) (with r=|x|) of (1.1) corresponds to studying

( r n 1 u ( r ) ) = λ r n 1 K 1 ( r ) f ( v ( r ) ) for  r > r 0 , ( r n 1 v ( r ) ) = λ r n 1 K 2 ( r ) g ( u ( r ) ) for  r > r 0 , u ( r ) = v ( r ) = 0 if  r = r 0 ( > 0 ) , u ( r ) 0 , v ( r ) 0 as  r , }

which can be reduced to the study of solutions (u(s),v(s)); s[0,1] to the singular system:

u ( s ) = λ h 1 ( s ) f ( v ( s ) ) , 0 < s < 1 , v ( s ) = λ h 2 ( s ) g ( u ( s ) ) , 0 < s < 1 , u ( 0 ) = u ( 1 ) = 0 , v ( 0 ) = v ( 1 ) = 0 , }
(1.3)

via the Kelvin transformation s= ( r r 0 ) 2 n , where h i (s)= r 0 2 ( n 2 ) 2 s 2 ( n 1 ) ( n 2 ) K i ( r 0 s 1 2 n ), i=1,2 (see [16]).

Remark 1.3

The assumption (H3) implies that lim s 0 + h i (s)=, for i=1,2, h ˆ = inf t ( 0 , 1 ) { h 1 (t), h 2 (t)}>0, and there exist d>0, η(0,1) such that h i (s) d s η for s(0,1], and for i=1,2. When in addition (H5) is satisfied, h 1 is decreasing in (0,1].

We will prove Theorem 1.1 in Section 2 by studying the singular system (1.3), and Theorem 1.2 in Section 3 by studying the corresponding single equation

u ( s ) = λ h 1 ( s ) f ˜ ( u ( s ) ) , 0 < s < 1 , u ( 0 ) = u ( 1 ) = 0 . }
(1.4)

2 Existence result

We first establish some useful results for solutions to the system

u ( s ) = b 1 h 1 ( s ) | v ( s ) + l | q 1 , 0 < s < 1 , v ( s ) = b 2 h 2 ( s ) | u ( s ) + l | q 2 , 0 < s < 1 , u ( 0 ) = u ( 1 ) = 0 , v ( 0 ) = v ( 1 ) = 0 , }
(2.1)

where l0 is a parameter. (Clearly, any solution ( u l , v l ) of (2.1) for l>0 must satisfy u l (s)>0, v l (s)>0 for s(0,1). This is also true for any nontrivial solution when l=0.) We prove the following.

Lemma 2.1

  1. (i)

    There exists l 0 >0 such that 2.1 has no solution if l l 0 .

  2. (ii)

    For each l[0, l 0 ), there exists M>0 (independent of l) such that if ( u l , v l ) is a solution of (2.1), then max{ u l , v l }M.

Proof of (i)

Let λ 1 := π 2 , ϕ 1 :=sin(πs). Here λ 1 is the principal eigenvalue and ϕ 1 a corresponding eigenfunction of ϕ (s)=λϕ(s) in (0,1) with ϕ(0)=0=ϕ(1). Let a> λ 1 b 1 b 2 h ˆ , c>0 be such that ( s + l ) q i asc for all s0 and for i=1,2. Now let ( u l , v l ) be a solution of (2.1). Multiplying (2.1) by ϕ 1 and integrating, we obtain

λ 1 0 1 u l ϕ 1 ds= b 1 0 1 h 1 (s) ( v l + l ) q 1 ϕ 1 ds b 1 0 1 h 1 (s)(a v l c) ϕ 1 ds

and

λ 1 0 1 v l ϕ 1 ds= b 2 0 1 h 2 (s) ( u l + l ) q 2 ϕ 1 ds b 2 0 1 h 2 (s)(a u l c) ϕ 1 ds.

By Remark 1.3, h ˆ = inf t ( 0 , 1 ) { h 1 (t), h 2 (t)}>0, and h i 1 := 0 1 h i (s)ds< for i=1,2. Then from the above inequalities we obtain

0 1 v l ϕ 1 ds 1 a b 1 h ˆ ( λ 1 0 1 u l ϕ 1 d s + b 1 c h 1 1 )

and

0 1 u l ϕ 1 ds 1 a b 2 h ˆ ( λ 1 0 1 v l ϕ 1 d s + b 2 c h 2 1 ) .

Hence we deduce that

0 1 u l ϕ 1 ds m 1 m := m 2 ,

where m:=(a b 2 h ˆ λ 1 2 a b 1 h ˆ ), and m 1 := λ 1 c h 1 1 a h ˆ + b 2 c h 2 1 . This implies

0 1 ( v l + l ) q 1 ϕ 1 ds λ 1 m 2 b 1 h ˆ := m 3 .

In particular, this implies 1 4 3 4 l q 1 ds m 3 inf [ 1 4 , 3 4 ] ϕ 1 . Since m 3 is independent of l, clearly this is a contradiction for l1, and hence there must exists an l 0 >0 such that for l l 0 , (2.1) has no solution.

Proof of (ii)

Assume the contrary. Then without loss of generality we can assume there exists { l n }(0, l 0 ) such that u l n as n. Clearly u l n (s)<0, and v l n (s)<0 for all s(0,1). Let s 1 ( l n ) (0,1), s 2 ( l n ) (0,1) be the points at which u l n and v l n attain their maximums. Now since u l n (s)<0 for all s(0,1), we have

u l n (s) { s u l n ( s 1 ( l n ) ) s 1 ( l n ) for  s ( 0 , s 1 ( l n ) ) , ( 1 s ) u l n ( s 1 ( l n ) ) 1 s 1 ( l n ) for  s ( s 1 ( l n ) , 1 ) .

Hence u l n (s)min{ s u l n s 1 ( l n ) , ( 1 s ) u l n 1 s 1 ( l n ) }, and in particular, for s[ 1 4 , 3 4 ],

u l n (s)min { 1 4 u l n , 1 4 u l n } = 1 4 u l n .

Let s l n ˜ , s l n ¯ [ 1 4 , 3 4 ] be such that min [ 1 4 , 3 4 ] u l n (s)= u l n ( s l n ˜ ), and min [ 1 4 , 3 4 ] v l n (s)= v l n ( s l n ¯ ). Now for s[ 1 4 , 3 4 ],

v l n (s) b 2 h ˆ m ˜ 1 4 3 4 | u l n ( t ) + l | q 2 dt,

where m ˜ := min [ 1 4 , 3 4 ] × [ 1 4 , 3 4 ] G(s,t) (>0), and G is the Green’s function of Z with Z(0)=0=Z(1). In particular, v l n ( s l n ¯ ) b 2 h ˆ m ˜ 2 ( u l n ( s l n ˜ ) ) q 2 . Similarly u l n ( s l n ˜ ) b 1 h ˆ m 2 ( v l n ( s l n ¯ ) ) q 1 . Hence, there exists a constant A>0 such that

u l n ( s l n ˜ )A ( u l n ( s l n ˜ ) ) q 1 q 2 .

This is a contradiction since q 1 q 2 >1 and u l n ( s l n ˜ ) 1 4 u l n as n. Thus (ii) holds. □

Proof of Theorem 1.1

We first extend f and g as even functions on ℝ by setting f(s)=f(s) and g(s)=g(s). Then we use the rescaling, λ= γ δ , w 1 =γu, and w 2 = γ θ v with γ>0, θ= q 2 + 1 q 1 + 1 , and δ= q 1 q 2 1 q 1 + 1 . With this rescaling, (1.3) reduces to

w 1 ( s ) = F ( s , γ , w 2 ) , 0 < s < 1 , w 2 ( s ) = G ( s , γ , w 1 ) , 0 < s < 1 , w 1 ( 0 ) = w 1 ( 1 ) = 0 , w 2 ( 0 ) = w 2 ( 1 ) = 0 , }
(2.2)

where

F ( s , γ , w 2 ) : = γ 1 + δ h 1 ( s ) ( f ( w 2 γ θ ) b 1 | w 2 γ θ | q 1 ) + b 1 | w 2 | q 1 h 1 ( s ) , and G ( s , γ , w 1 ) : = γ θ + δ h 2 ( s ) ( g ( w 1 γ ) b 2 | w 1 γ | q 2 ) + b 2 | w 1 | q 2 h 2 ( s ) .

Note that by our hypothesis (H2), F(s,γ, w 2 ) b 1 | w 2 | q 1 h 1 (s) and G(s,γ, w 1 ) b 2 | w 1 | q 2 h 2 (s) as γ0. Hence we can continuously extend F(s,γ, w 2 ) and G(s,γ, w 1 ) to F(s,0, w 2 )= b 1 | w 2 | q 1 h 1 (s) and G(s,0, w 1 )= b 2 | w 1 | q 2 h 2 (s), respectively. Note that proving (1.3) has a positive solution for λ small is equivalent to proving (2.2) has a solution ( w 1 , w 2 ) with w 1 >0, w 2 >0 in (0,1) for small γ>0. We will achieve this by establishing that the limiting equation (when γ=0)

w 1 ( s ) = F ( s , 0 , w 2 ) = b 1 h 1 ( s ) | w 2 | q 1 , 0 < s < 1 , w 2 ( s ) = G ( s , 0 , w 1 ) = b 2 h 2 ( s ) | w 1 | q 2 , 0 < s < 1 , w 1 ( 0 ) = w 1 ( 1 ) = 0 , w 2 ( 0 ) = w 2 ( 1 ) = 0 }
(2.3)

(which is the same as (2.1) with l=0) has a positive solution w 1 >0, w 2 >0 in (0,1) that persists for small γ>0.

Let X= C 0 [0,1]× C 0 [0,1] be the Banach space equipped with w ̲ X = ( w 1 , w 2 ) X =max{ w 1 , w 2 }, where denotes the usual supremum norm in C 0 ([0,1]). Then for fixed γ0, we define the map S(γ,):XX by

S(γ, w ̲ ):= w ̲ ( K ( F ( s , γ , w 2 ) ) , K ( G ( s , γ , w 1 ) ) ) ,

where K(H(s,γ,Z(s)))= 0 1 G(t,s)H(t,γ,Z(t))dt. Note that F(s,γ,),G(s,γ,): C 0 ([0,1]) L 1 (0,1) are continuous and K: L 1 (0,1) C 0 1 ([0,1]) is compact. Hence S(γ,) is a compact perturbation of the identity. Clearly for γ>0, if S(γ, w ̲ )= 0 ̲ , then w ̲ =( w 1 , w 2 ) is a solution of (2.2), and if S(0, w ̲ )= 0 ̲ , then w ̲ =( w 1 , w 2 ) is a solution of (2.3).

We first establish the following.

Lemma 2.2

There existsR>0such thatS(0, w ̲ ) 0 ̲ for all w ̲ =( w 1 , w 2 )Xwith w ̲ X =Randdeg(S(0,), B R ( 0 ̲ ), 0 ̲ )=0.

Proof

Define S l (0, w ̲ ):XX by

S l (0, w ̲ ):= w ̲ ( K ( b 1 h 1 ( s ) | w 2 + l | q 1 ) , K ( b 2 h 2 ( s ) | w 1 + l | q 2 ) )

for l0. (Note S 0 (0, w ̲ )=S(0, w ̲ ).) By Lemma 2.1, if l l 0 then S l (0, w ̲ ) 0 ̲ and if S l (0, w ̲ )= 0 ̲ for l[0, l 0 ), then w X M. This implies that there exists R1 such that S l (0, w ̲ ) 0 ̲ for w ̲ B R ( 0 ̲ ) for any l0. Also, since (2.1) has no solution for l l 0 , deg( S l 0 (0,), B R ( 0 ̲ ), 0 ̲ )=0. Hence, using the homotopy invariance of degree with the parameter l[0, l 0 ] we get

deg ( S ( 0 , ) , B R ( 0 ̲ ) , 0 ̲ ) =deg ( S l 0 ( 0 , ) , B R ( 0 ̲ ) , 0 ̲ ) =0.

 □

Next we establish the following.

Lemma 2.3

There existsr(0,R)small enough such thatS(0, w ̲ ) 0 ̲ for all w ̲ =( w 1 , w 2 )Xwith w ̲ X =randdeg(S(0,), B r ( 0 ̲ ), 0 ̲ )=1.

Proof

Define T τ (0, w ̲ ):XX by

T τ (0, w ̲ ):= w ̲ ( K ( τ b 1 h 1 ( s ) | w 2 | q 1 ) , K ( τ b 2 h 2 ( s ) | w 1 | q 2 ) )

for τ[0,1]. Clearly T 1 (0, w ̲ )=S(0, w ̲ ), and T 0 (0, w ̲ )=I is the identity operator. Note that T τ (0, w ̲ )=0 if w ̲ =( w 1 , w 2 ) is a solution of

w 1 ( s ) = τ b 1 h 1 ( s ) | w 2 | q 1 , 0 < s < 1 , w 2 ( s ) = τ b 2 h 2 ( s ) | w 1 | q 2 , 0 < s < 1 , w 1 ( 0 ) = w 1 ( 1 ) = 0 , w 2 ( 0 ) = w 2 ( 1 ) = 0 , }
(2.4)

and for τ=1, (2.4) coincides with (2.3). Assume to the contrary that (2.4) has a solution w ̲ =( w 1 , w 2 ) with w ̲ X = r ˜ >0. Without loss of generality assume w 1 = r ˜ . Now,

w 1 (s)=τ 0 1 G(s,t) b 1 h 1 (s) | w 2 | q 1 ds.

Then w 1 C ˜ w 2 q 1 for some constant C ˜ >0 independent of τ[0,1]. Similarly w 2 C ˆ w 1 q 2 for some constant C ˆ >0. This implies that

r ˜ = w 1 C w 1 q 1 q 2 =C r ˜ q 1 q 2

for some constant C>0. But q 1 q 2 >1, and hence this is a contradiction if r ˜ >0 is small. Thus there exists small r>0 such that (2.4) has no solution w ̲ with w ̲ X =r for all τ[0,1]. Now using the homotopy invariance of degree with the parameter τ[0,1], in particular using the values τ=1 and τ=0, we obtain

deg ( S ( 0 , ) , B r ( 0 ̲ ) , 0 ̲ ) =deg ( T 1 ( 0 , ) , B r ( 0 ̲ ) , 0 ̲ ) =deg ( T 0 ( 0 , ) , B r ( 0 ̲ ) , 0 ̲ ) =1.

 □

By Lemma 2.2 and Lemma 2.3, with 0<r<R, we conclude that

deg ( S ( 0 , ) , B R ( 0 ̲ ) B r ¯ ( 0 ̲ ) , 0 ̲ ) =1,

and hence (2.3) has a solution w ̲ =( w 1 , w 2 ) with w 1 >0, w 2 >0 in (0,1), and r< w X <R. Now we show that the solution obtained above (when γ=0) persists for small γ>0 and remains positive componentwise.

Lemma 2.4

Let R, r be as in Lemmas 2.2, 2.3, respectively. Then there exists γ 0 >0such that:

  1. (i)

    deg(S(γ,), B R ( 0 ̲ ) B r ¯ ( 0 ̲ ), 0 ̲ )=1 for all γ[0, γ 0 ].

  2. (ii)

    If S(γ, w ̲ )= 0 ̲ for γ[0, γ 0 ] with r< w ̲ X <R, then w 1 >0, w 2 >0 in (0,1).

Proof of (i)

We first show that there exists γ 0 >0 such that S(γ, w ̲ ) 0 ̲ for all w ̲ =( w 1 , w 2 )X with w ̲ X {R,r}, for all γ[0, γ 0 ]. Suppose to the contrary that there exists { γ n } with γ n 0, S( γ n , w n ̲ )= 0 ̲ and w n ̲ X {r,R}. Since K ̲ =(K,K): L 1 (0,1)× L 1 (0,1) C 0 1 ([0,1])× C 0 1 ([0,1]) is compact, and {F(s, γ n , w 2 n ),G(s, γ n , w 1 n )} are bounded in L 1 (0,1)× L 1 (0,1), w ̲ n Z ̲ =( Z 1 , Z 2 ) C 0 1 ([0,1])× C 0 1 ([0,1]) (up to a subsequence) with Z ̲ X =R or r and S(0, Z ̲ )= 0 ̲ . This is a contradiction to Lemma 2.2 or 2.3 and hence there exists a small γ 0 >0 satisfying the assertions. Now, by the homotopy invariance of degree with respect to γ[0, γ 0 ],

deg ( S ( γ , ) , B R ( 0 ̲ ) B r ¯ ( 0 ̲ ) , 0 ̲ ) =deg ( S ( 0 , ) , B R ( 0 ̲ ) B r ¯ ( 0 ̲ ) , 0 ̲ ) =1

for all γ[0, γ 0 ].

Proof of (ii)

Assume to the contrary that there exists γ n 0 and a corresponding solution w n ̲ =( w 1 n , w 2 n ) such that r< w n ̲ X <R and

Ω n := { x ( 0 , 1 ) w 1 n ( x ) 0  or  w 2 n ( x ) 0 } .

Arguing as before, w n ̲ Z ̲ C 0 1 ([0,1])× C 0 1 ([0,1]) with S(0, Z ̲ )= 0 ̲ (up to a subsequence). Note that Z ̲ 0 ̲ since Z ̲ X r>0. By the strong maximum principle Z 1 >0, Z 2 >0, Z 1 (0)>0, Z 2 (0)>0, Z 1 (1)<0 and Z 2 (1)<0. Now suppose there exists { x n }(0,1) with { x n } Ω n and w 1 n ( x n )0. Then { x n } must have a subsequence (renamed as { x n } itself) such that x n x ˜ [0,1]. But Z 1 >0 in (0,1) implies that x ˜ {0,1}. Suppose x ˜ =0. Since w 1 n ( x n )0 and w 1 n (0)=0, there exists y n (0, x n ) such that w 1 n ( y n )0, and hence taking the limit as n we will have Z 1 (0)0, which is a contradiction since Z 1 (0)>0. A similar contradiction follows if x ˜ =1, using the fact that Z 1 (1)<0. Further, contradictions can be achieved if there exists { x n }Ω with { x n } Ω n and w 2 n ( x n )0 using the facts that Z 2 (0)>0 and Z 2 (1)<0. This completes the proof of the lemma. □

We now easily conclude the proof of Theorem 1.1. From Lemma 2.4, since w ̲ =( w 1 , w 2 ) is a positive solution of (2.2) for γ small, (u,v)=( γ 1 w 1 , γ θ w 2 ) with θ= q 2 + 1 q 1 + 1 is a positive solution of (1.3) for λ= γ δ where δ= q 1 q 2 1 q 1 + 1 . Further, since w 1 >0 and w 2 >0 in (0,1) for γ[0, γ 0 ], u and v as λ(= γ δ )0. This completes the proof of Theorem 1.1.  □

3 Non-existence result

We first recall from [13] that, when (H5) is satisfied, one can prove via an energy analysis that a nonnegative solution u of (1.4) must be positive in (0,1) and have a unique interior maximum with maximum value greater than θ, where θ is the unique positive zero of F ˜ (s)= 0 s f ˜ (y)dy. Further, for λ1 and s 1 , s 1 ˆ (0,1) such that s 1 ˆ > s 1 , u( s 1 )=u( s 1 ˆ )=β (see Figure 1), where β>0 is the unique zero of f ˜ , there exists a constant C such that s 1 C λ 1 2 and (1 s 1 ˆ )C λ 1 2 . Hence we can assume ( s 1 ˆ s 1 )> 1 2 for λ1. Now we provide the proof of Theorem 1.2.

Figure 1
figure 1

Graph of u .

Proof of Theorem 1.2

Let v:=uβ. Then v>0 in ( s 1 , s 1 ˆ ) and satisfies

v = λ h 1 ( s ) f ˜ ( u ) u β v , s 1 < s < s 1 ˆ , v ( s 1 ) = v ( s 1 ˆ ) = 0 . }

Note that ϕ(s)=(sin( π ( s s 1 ) ( s 1 ˆ s 1 ) ))>0 in ( s 1 , s 1 ˆ ), ϕ( s 1 )=ϕ( s 1 ˆ )=0, and it satisfies ϕ = π 2 ( s 1 ˆ s 1 ) 2 ϕ in ( s 1 , s 1 ˆ ). Hence using the fact that s 1 s 1 ˆ (ϕ v +v ϕ )ds=0, we obtain

s 1 s 1 ˆ ( λ f ˜ ( u ) u β h 1 ( s ) π 2 ( s 1 ˆ s 1 ) 2 ) vϕds=0.

In particular,

λ f ˜ ( u ( s λ ) ) u ( s λ ) β h 1 ( s λ )= π 2 ( s 1 ˆ s 1 ) 2 ,for some  s λ ( s 1 , s 1 ˆ ).
(3.1)

But h ˆ = inf ( 0 , 1 ) h 1 (s)>0, and ( s 1 ˆ s 1 )> 1 2 for λ1. Thus clearly (3.1) can hold when λ, only if Z=u( s λ ) with f ˜ ( u ( s λ ) ) u ( s λ ) β 0. But by (H4), this is not possible since lim Z f ˜ ( Z ) Z m 0 >0. Hence the nonnegative solution cannot exist for λ1. □