In reaction diffusion processes, steady states define the long term dynamics. Here we consider a steady state reaction diffusion equation on an exterior domain with a nonlinear boundary condition on the interior boundary. Namely, we study positive radial solutions to:

{ − Δ u = λ k 1 ( | x | ) f 1 ( u , v )   on   Ω , − Δ v = λ k 2 ( | x | ) f 2 ( u , v )   on   Ω , u ( x ) = v ( x ) = 0   on   | x | → ∞ , ∂ u ∂ η + c ˜ 1 ( u ) u = 0   on   | x | = r 0 , ∂ v ∂ η + c ˜ 2 ( v ) v = 0 ,   on   | x | = r 0 . (1.1)

where Δ u = d i v ( ∇ u ) and Δ v = d i v ( ∇ v ) are the Laplacian of u, λ is a positive parameter, Ω = { x ∈ ℝ n : N > 2 , | x | > r 0 , r 0 > 0 } , let i = [ 1 , 2 ] then K i : [ r 0 , ∞ ) → ( 0, ∞ ) is a continuous function such that lim r → ∞ k i ( r ) = 0 and

∂ ∂ η is the outward normal derivative, and c ˜ i : [ 0, ∞ ) → ( 0, ∞ ) is a is an non

decreasing (increasing) function. Here the reaction term f i : [ 0, ∞ ) × [ 0, ∞ ) → R is a C 1 function. The case when f i < 0 (see [1] [2], that the study of positive solutions to such problems is considerably more challenging than in the case f i > 0 (positone problems). For a rich history on semipositone problems with Dirichlet boundary conditions on bounded domains, (see [3] - [8], and on domains exterior to a ball, see [9] [10] [11]. Such nonlinear boundary conditions occur very naturally in applications see [12] for a detailed description in a model arising in combustion theory. Recently, the existence of a radial positive solution for (1.1) when λ ≫ 1 has been established in [13], via the method of subsuper solutions. Here we discuss the uniqueness of this radial solution when some additional assumptions hold. In [14], the authors study such a uniqueness result for the case of Dirichlet boundary condition on | x | = r 0 . Our focus in this paper is to consider the uniqueness result for semipositone problem when a class of nonlinear boundary condition is satisfied at | x | = r 0 . The fact that we have no longer a fixed value of u on | x | = r 0 results in quite a challenge in extending the results in [15].

Namely, we need to establish a detailed behavior of u at | x | = r 0 to achieve our goal. Instead of working directly with (1), we note that the change of

variables r = | x | and s = ( r r 0 ) 2 − N transforms (1) into the following boundary value problem:

{ − u ″ ( t ) = λ a ˜ 1 ( t ) f 1 ( u ( t ) , v ( t ) )   t ∈ [ 0 , 1 ] , − v ″ ( t ) = λ a ˜ 2 ( t ) f 2 ( u ( t ) , v ( t ) )   t ∈ [ 0 , 1 ] , N − 2 r 0 u ′ + c ˜ 1 ( u ( 1 ) ) u ( 1 ) = 0 , N − 2 r 0 v ′ + c ˜ 2 ( v ( 1 ) ) v ( 1 ) = 0 , u ( 0 ) = v ( 0 ) = 0. (1.2)

where a ˜ i = r 0 2 ( 2 − N ) 2 t − 2 ( N − 1 ) N − 2 k i ( r 0 t 1 2 − N ) . We will only assume k i ≤ 1 r N + μ for r ≫ 1 and for some μ ∈ ( 0, N − 2 ) . then a ˜ i ∈ ( ( 0,1 ] , ( 0, ∞ ) ) could be singular at 0.if μ ≥ N − 2 , a ˜ i will be nonsingular at 0 and it will be an easier case to study. Note that a ˜ i = inf t ∈ ( 0 , 1 ] a ˜ i ( t ) > 0 and there exists a constant d ˜ > 0 such

that a ˜ i ≤ d ˜ t α for all t ∈ ( 0,1 ] where α ˜ = ( N − 2 ) − μ N − 2 . Motivated by the above discussion, in this paper, we will study positive solutions in C 2 ( 0,1 ) ∩ C 1 [ 0,1 ] to the following boundary value problems:

{ − u ″ ( t ) = λ a 1 ( t ) f 1 ( u ( t ) , v ( t ) )   t ∈ [ 0 , 1 ] , − v ″ ( t ) = λ a 2 ( t ) f 2 ( u ( t ) , v ( t ) )   t ∈ [ 0 , 1 ] , u ′ ( 1 ) + c 1 ( u ( 1 ) ) u ( 1 ) = 0 , v ′ ( 1 ) + c 2 ( v ( 1 ) ) v ( 1 ) = 0 , u ( 0 ) = v ( 0 ) = 0. (1.3)

where c i : [ 0, ∞ ) → ( 0, ∞ ) is a continuous function and a i ∈ C ( ( 0,1 ] , ( 0, ∞ ) ) is such that:

(H1) a i = inf t ∈ ( 0 , 1 ] a i ( t ) > 0 ;

(H2) there exists a constant d > 0 such that a i ( t ) = d t α for all t ∈ ( 0, ε ] where a ∈ ( 0,1 ) and ε ≈ 0

(H3) a i is decreasing. We consider various C 1 classes of the reaction term f i : [ 0, ∞ ) × [ 0, ∞ ) → R satisfying the following:

(F1) f i < 0 and lim s → ∞ f i ( s ) s = 0 ; i = 1 , 2

(F2) f i is increasing and lim s → ∞ f i ( s ) = ∞ ; i = 1 , 2

(F3) f i is concave on [ 0, ∞ ) . i = 1 , 2

Theorem 1.1. Assume (H1) - (H3) and (F1) - (F3). Then (1.3) has a unique positive solution for all λ sufficiently large.

In Section two we will establish important a priori estimates. We will first recall some important results from [8] where the authors studied the case of Dirichlet boundary condition, or equivalently (1.3) with the boundary condition t = 1 replaced by u ( 1 ) = v ( 1 ) = 0 . These results do not depend on the boundary condition at t = 1 and hence it is also true for solutions of (1.3). In view of the readers convenience we include the proofs of these results. In Section three, we prove Theorem 1.1.

Let F ( s ) = ∫ 0 s   f i ( t ) d s . Note that there exist unique positive numbers β , θ such that f i ( β ) = 0 and F ( θ ) = 0 and β < θ

Theorem 2.1. (See [8].) Let u , v are a positive solution of (1.3). Then u and v has only one interior maximum in ( 0,1 ) , say at t m , depending on λ , and u ( t m ) > θ , v ( t m ) > θ .

Proof. Let

{ E 1 ( t ) = λ F ( u ( t ) ) a 1 ( t ) + | u ′ ( t ) | 2 2 , t ∈ ( 0 , 1 ) , E 2 ( t ) = λ F ( v ( t ) ) a 2 ( t ) + | v ′ ( t ) | 2 2 , t ∈ ( 0 , 1 ) (1.4)

then

{ E ′ 1 ( t ) = λ F ( u ( t ) ) a ′ 1 ( t ) , t ∈ ( 0 , 1 ) , E ′ 2 ( t ) = λ F ( v ( t ) ) a ′ 2 ( t ) , t ∈ ( 0 , 1 ) (1.5)

Note that by (H3), a ′ 1 ( t ) < 0 and a ′ 2 ( t ) < 0 for all t ∈ ( 0,1 ] . Hence, E 1 ( t ) and E 2 ( t ) are increases when u ( t ) < θ , v ( t ) < θ and decreases when u ( t ) > θ , v ( t ) > θ .

Let t m ∈ ( 0,1 ) be the first point at which u has a local maximum and assume that u ( t ) ≤ θ and v ( t ) ≤ θ for all t ∈ [ 0, t m ] . Then E 1 ( t ) and E 2 ( t ) are increases in [ 0, t m ] . Now integrating (1.3) from t to t m , for t < t m

{ u ′ ( t ) = ∫ t t m   λ a 1 ( s ) f 1 ( u ( s ) ) d s ≤ λ f 1 ( θ ) ∫ t t m c 1 s α d s ≤ λ c 1 f 1 ( θ ) 1 − α v ′ ( t ) = ∫ t t m   λ a 2 ( s ) f 2 ( v ( s ) ) d s ≤ λ f 2 ( θ ) ∫ t t m c 2 s α d s ≤ λ c 2 f 2 ( θ ) 1 − α (1.6)

where c i > d are such that a i ( t ) ≤ c i t α for all t ∈ ( 0,1 ) using (H2). Integrating again (1.6) from 0 to t, t ≤ t m

{ u ( t ) ≤ ∫ 0 t     λ c 1 f 1 ( θ ) 1 − α d s ,   C 0 = c 1 f 1 ( θ ) 1 − α V ( t ) ≤ ∫ 0 t     λ c 2 f 2 ( θ ) 1 − α d s ,   C 0 = c 2 f 2 ( θ ) 1 − α (1.7)

Since f i are continuous, there exists K 0 > 0 such that | F ( u ) | ≤ K 0 u and | F ( v ) | ≤ K 0 v for all ( u , v ) ∈ [ 0, θ ] . Hence

lim t → 0 + λ | F ( u ( t ) ) | a 1 ( t ) ≤ lim t → 0 + λ K 0 u ( t ) a 1 ( t ) ≤ lim t → 0 + λ 2 K 0 C 0 c 1 d t 1 − α = 0

lim t → 0 + λ | F ( v ( t ) ) | a 2 ( t ) ≤ lim t → 0 + λ K 0 v ( t ) a 2 ( t ) ≤ lim t → 0 + λ 2 K 0 C 0 c 2 d t 1 − α = 0

Hence lim t → 0 + E i ( t ) ≥ 0 . Since E i ( t ) increases on [ 0, t m ] , E 1 ( t m ) = λ F ( u ( t m ) ) a 1 ( t m ) > 0 and E 2 ( t m ) = λ F ( v ( t m ) ) a 2 ( t m ) > 0 and which is a contra-diction if u ( t m ) ≤ θ and v ( t m ) ≤ θ . Suppose that u and v has two interior maxima. Then there exists t ˜ ∈ ( t ˜ ,1 ) such that u ′ ( t ˜ ) = 0 , v ′ ( t ˜ ) = 0 and u ″ ( t ˜ ) ≥ 0 , v ″ ( t ˜ ) ≥ 0 . Since u ″ ( t ˜ ) = λ a 1 ( t ˜ ) f 1 ( u ( t ˜ ) ) ≥ 0 , and

v ″ ( t ˜ ) = λ a 2 ( t ˜ ) f 2 ( v ( t ˜ ) ) ≥ 0 we have f 1 ( u ( t ˜ ) ) ≤ 0 and f 2 ( v ( t ˜ ) ) ≤ 0 which implies that u ( t ˜ ) ≤ β and u ( t ˜ ) ≤ β . Thus E i ( t ˜ ) < 0 . Let t θ ( t m , t ˜ ) such that

u ( t θ ) = θ and v ( t θ ) = θ . Then E 1 ( t θ ) = | u ′ ( t θ ) | 2 2 ≥ 0 , E 2 ( t θ ) = | v ′ ( t θ ) | 2 2 ≥ 0

and E i increases in ( t θ , t ˜ ) since u ( t ) < θ and since v ( t ) < θ in ( t θ , t ˜ ) . Hence E i ( t ˜ ) > 0 , which is a contradiction. Therefore, we have only one interior maximum and that maximum value is larger than θ .

Theorem 2.2. (See [8].) Let u and v be a positive solution of (1.3) and let t β ∈ ( 0, t m ) such that u ( t β ) = β and v ( t β ) = β . Then t β ≤   ○ ( λ − 1 2 ) as λ → ∞ .

Proof. Let t β 2 ∈ ( 0, t β ) be the point such that u ( t β 2 ) = β 2 and v ( t β 2 ) = β 2 from Integrating (1.3) from 0 to t for some t < t β 2

u ′ ( t ) = u ′ ( 0 ) − λ ∫ 0 t     a 1 ( s ) f 1 ( u ( s ) ) d s ≥ λ a 1 _ ( − f 1 ( β 2 ) ) t

v ′ ( t ) = v ′ ( 0 ) − λ ∫ 0 t     a 2 ( s ) f 2 ( v ( s ) ) d s ≥ λ a 2 _ ( − f 2 ( β 2 ) ) t ,

Integrating the above again from 0 to t β

t β 2 ≤ c ˜ 1 λ − 1 2 ,     t β 2 ≤ c ˜ 2 λ − 1 2       where     { c ˜ 1 = ( − β a 1 _ ( − f 1 ( β 2 ) ) 1 2 ) > 0 c ˜ 2 = ( − β a 2 _ ( − f 2 ( β 2 ) ) 1 2 ) > 0 (1.8)

By the mean value theorem, there exists a t ˜ ∈ [ 0, t β 2 ] such that u ( t β 2 ) − u ( 0 ) = u ′ ( t ˜ ) t β 2 , v ( t β 2 ) − v ( 0 ) = v ′ ( t ˜ ) t β 2 and by (2.5) u ′ ( t ˜ ) ≥ β 2 c ˜ 1 λ 1 2 and v ′ ( t ˜ ) ≥ β 2 c ˜ 2 λ 1 2 . Since u ′ and v ′ are increases in [ 0, t β ] ,

{ u ′ ( t ) ≥ β 2 c ˜ 1 λ 1 2 ,     t ∈ [ t β 2 , t β ] v ′ ( t ) ≥ β 2 c ˜ 2 λ 1 2 ,     t ∈ [ t β 2 , t β ] (1.9)

Integrating (2.6) from t β 2 to t β , we have that ( t β − t β 2 ) ≤ c ˜ 1 λ − 1 2 and ( t β − t β 2 ) ≤ c ˜ 1 λ − 1 2 . This implies t β ≤   ○ ( λ − 1 2 ) by (2.5). □

Lemma 2.3. Let u and v be a positive solution of (1.3). Then u ( 1 ) → ∞ and v ( 1 ) → ∞ as λ → ∞

Proof. We first claim that u ( 1 ) ≥ β + θ 2 and v ( 1 ) ≥ β + θ 2 for λ ≫ 1 . Assume that u ( 1 ) < β + θ 2 and v ( 1 ) < β + θ 2 Then there exists a t ˜ θ ∈ ( t m ,1 )

such that u ( t ˜ θ ) = θ and v ( t ˜ θ ) = θ where t m is the point at which u , v achieves are maximum, and u ( t m ) > θ , v ( t m ) > θ by Lemma 2.1.

From (2.1) and (2.2), E 1 ( t ˜ θ ) = | u ′ ( t ˜ θ ) | 2 > 0 , E 2 ( t ˜ θ ) = | v ′ ( t ˜ θ ) | 2 > 0 and E i ( t ) ≥ 0 on [ t ˜ θ ,1 ] since u ( t ) ≤ θ and v ( t ) ≤ θ in [ t ˜ θ ,1 ] . Hence we obtain that

E 1 ( 1 ) = λ F ( u ( 1 ) ) a 1 ( 1 ) + | u ′ ( 1 ) | 2 > 0 ,

E 2 ( 1 ) = λ F ( v ( 1 ) ) a 2 ( 1 ) + | v ′ ( 1 ) | 2 > 0

and from (1.3), we have

{ c 1 ( u ( 1 ) ) u ( 1 ) = − u ′ ( 1 ) ≥ − 2 λ F ( u ( 1 ) ) a 1 ( 1 ) , c 2 ( v ( 1 ) ) v ( 1 ) = − v ′ ( 1 ) ≥ − 2 λ F ( v ( 1 ) ) a 2 ( 1 ) . (1.10)

This cannot hold unless u ( 1 ) → 0 and v ( 1 ) → 0 as λ → ∞ . However, rewriting (1.10) as

{ c 1 ( u ( 1 ) ) u ( 1 ) 1 2 ≥ − 2 λ F ( u ( 1 ) ) u ( 1 ) a 1 ( 1 ) , c 2 ( v ( 1 ) ) v ( 1 ) 1 2 ≥ − 2 λ F ( v ( 1 ) ) v ( 1 ) a 2 ( 1 ) . (1.11)

we obtain a contradiction when λ ≫ 1 since F ( u ( 1 ) ) u ( 1 ) → f 1 ( 0 ) , F ( v ( 1 ) ) v ( 1 ) → f 2 ( 0 ) if u ( 1 ) → 0 and v ( 1 ) → 0 as λ → ∞ . Hence,

u ( 1 ) ≥ β + θ 2     and     v ( 1 ) ≥ β + θ 2   for   λ ≫ 1 (1.12)

Next, we claim that u ( t m ) = ‖ u ‖ ∞ → ∞ and v ( t m ) = ‖ v ‖ ∞ → ∞ as λ → ∞ . Let

h : = u − β and w : = v − β then h > 0 , w > 0 in ( t β ,1 ] and satisfies

{ − h ″ = λ a 1 ( t ) f 1 ( u ) u − β h   ( t β , 1 ) − w ″ = λ a 2 ( t ) f 2 ( v ) v − β w   ( t β , 1 ) h ( t β ) = w ( t β ) = 0 , h ( 1 ) = u ( 1 ) − β > 0 w ( 1 ) = v ( 1 ) − β > 0 (1.13)

Let ψ = : sin ( π ( t − t β ) t − t β ) . Then ψ satisfies:

{ − ψ ″ = π 2 ( t − t β ) 2 ψ ,   ( t β , 1 ) ψ ( t β ) = ψ ( 1 ) = 0 (1.14)

Multiplying (1.13) by ψ and (1.14) by h,and w integrating both from t β to 1 and subtracting, we have

∫ t β 1 ( h ″ ψ − ψ ″ h ) d t = ∫ t β 1 ( π 2 ( 1 − t β ) 2 − λ f 1 ( u ) u − β a 1 ( t ) ) h ψ d t ,

∫ t β 1 ( w ″ ψ − ψ ″ w ) d t = ∫ t β 1 ( π 2 ( 1 − t β ) 2 − λ f 2 ( v ) v − β a 2 ( t ) ) w ψ d t

Since ∫ t β 1 ( h ″ ψ − ψ ″ h ) d t = − ψ ′ ( 1 ) h ( 1 ) ( > 0 ) and ∫ t β 1 ( w ″ ψ − ψ ″ w ) d t = − ψ ′ ( 1 ) w ( 1 ) ( > 0 ) by integration by parts, we can see that

π 2 ( 1 − t β ) 2 > λ f 1 ( u ) u − β a 1 ( t )     and     π 2 ( 1 − t β ) 2 > λ f 2 ( v ) v − β a 2 ( t )     for   some     t ∈ ( t β , 1 ) (1.15)

Note that inf ( 0 , 1 ] a 1 ( t ) > 0 , inf ( 0 , 1 ] a 2 ( t ) > 0 and from Lemma 2.2 we can assume ( 1 − t β ) > 1 2 . Thus (2.11) is only true if f 1 ( u ) u − β → 0 and f 2 ( v ) v − β → 0

when λ ≫ 1 . By (F1) (F2), we conclude that u ( t m ) = ‖ u ‖ ∞ and v ( t m ) = ‖ v ‖ ∞ as λ → ∞ . Notice that since u ″ < 0 and v ″ < 0 in ( t β ,1 ] , we have

u ( t ) ≥ u ( t m ) − β t m − t β ( t − t β ) + β ,   t ∈ [ t β , t m ]

v ( t ) ≥ v ( t m ) − β t m − t β ( t − t β ) + β ,   t ∈ [ t β , t m ]

u ( t ) ≥ u ( t m ) − β + θ 2 1 − t m ( 1 − t ) + β + θ 2 ,   t ∈ [ t m , 1 ]

v ( t ) ≥ v ( t m ) − β + θ 2 1 − t m ( 1 − t ) + β + θ 2 ,   t ∈ [ t m , 1 ]

Since u ( t m ) → ∞ , v ( t m ) → ∞ and t β → 0 as λ → ∞ , it is all true that

v ( t ) ≥ β + θ 2     and     u ( t ) ≥ β + θ 2 ,     in   [ 1 4 , 1 ]   for   λ ≫ 1 (1.16)

Now we show that u ( 1 ) → ∞ and v ( 1 ) → ∞ as λ → ∞ . Since u , v is a solution of (1.3), u and v can be written as: (see Appendix 8.2 in [5] for details)

u ( t ) = λ ∫ 0 1     G ( t , s ) a 1 ( s ) f 1 ( u ( s ) ) d s − c 1 ( u ( 1 ) ) u ( 1 ) t (1.17)

v ( t ) = λ ∫ 0 1     G ( t , s ) a 2 ( s ) f 2 ( v ( s ) ) d s − c 2 ( v ( 1 ) ) v ( 1 ) t (1.18)

where

G ( t , s ) = { s ,   0 ≤ s ≤ t ≤ 1 t ,   0 ≤ t ≤ s ≤ 1

Let t = 1 . Then from (1.17) and (1.18), we have

[ 1 + c 1 ( u ( 1 ) ) ] u ( 1 ) = λ [ ∫ 0 t β     G ( 1 , s ) a 1 ( s ) f 1 ( u ( s ) ) d s + ∫ t β 1     G ( 1 , s ) a 1 ( s ) f 1 ( u ( s ) ) d s ] (1.19)

[ 1 + c 2 ( v ( 1 ) ) ] v ( 1 ) = λ [ ∫ 0 t β     G ( 1 , s ) a 2 ( s ) f 2 ( v ( s ) ) d s + ∫ t β 1     G ( 1 , s ) a 2 ( s ) f 2 ( v ( s ) ) d s ] (1.20)

Then using the fact G ( 1 , s ) = s and t β → ∞ as λ → ∞ , for λ large we obtain

[ 1 + c 1 ( u ( 1 ) ) ] u ( 1 ) = λ ( ∫ 0 t β     s a 1 ( s ) f 1 ( u ( s ) ) d s + ∫ t β 1     s a 1 ( s ) f 1 ( u ( s ) ) d s )

[ 1 + c 2 ( v ( 1 ) ) ] v ( 1 ) = λ ( ∫ 0 t β     s a 2 ( s ) f 2 ( v ( s ) ) d s + ∫ t β 1     s a 2 ( s ) f 2 ( v ( s ) ) d s )

≥ λ ( ∫ 0 t β     s a 1 ( s ) f 1 ( u ( s ) ) d s + ∫ 1 4 1     s a 1 ( s ) f 1 ( u ( s ) ) d s )

≥ λ ( ∫ 0 t β     s a 2 ( s ) f 2 ( v ( s ) ) d s + ∫ 1 4 1     s a 2 ( s ) f 2 ( v ( s ) ) d s )

≥ λ 2 f 1 ( β + θ 2 ) ∫ 1 4 1     s a 1 ( s ) d s

≥ λ 2 f 2 ( β + θ 2 ) ∫ 1 4 1     s a 2 ( s ) d s

where the last inequality is obtained by (1.16). Hence, we have

[ 1 + c 1 ( u ( 1 ) ) ] u ( 1 ) ≥ λ K     and     [ 1 + c 2 ( v ( 1 ) ) ] v ( 1 ) ≥ λ K , (1.21)

where K = 1 2 f 1 ( β + θ 2 ) ∫ 1 4 1     s a 1 ( s ) d s > 0 and K = 1 2 f 2 ( β + θ 2 ) ∫ 1 4 1     s a 2 ( s ) d s > 0 .

Now, from (1.21), clearly u ( 1 ) → ∞ , v ( 1 ) → ∞ as λ → ∞ . □

Lemma 2.4. Let u and v be a positive solution of (1.3). Then there exists [ α , μ ] ⊂ [ 1 4 ,1 ] , α ≠ μ , both independent of λ , such that inf [ α , μ ] u ( t ) → ∞ and inf [ α , μ ] v ( t ) → ∞ as λ → ∞

Proof. As λ → ∞ , t m may converge to 1 or to any other point in ( 0,1 ) . First we consider the case t m ↛ 1 as λ → ∞ . Since u ( 1 ) < u ( t m ) and v ( 1 ) < v ( t m ) clearly there exists α < 1 such that inf [ α ,1 ] u ( t ) → ∞ and inf [ α ,1 ] v ( t ) → ∞ as λ → ∞ by Lemma 2.3.

Now, let us consider the case when t m → 1 as λ → ∞ . By differentiating (1.17) and (1.18) (or integrating (1.3)), we obtain

u ′ ( t ) = λ ∫ t 1     a 1 ( s ) f 1 ( u ( s ) ) d s − c 1 ( u ( 1 ) ) u ( 1 ) ,   t ∈ [ 0 , 1 ] ,

v ′ ( t ) = λ ∫ t 1     a 2 ( s ) f 2 ( v ( s ) ) d s − c 2 ( v ( 1 ) ) v ( 1 ) ,   t ∈ [ 0 , 1 ]

which gives us that

u ′ ( t ) = λ ∫ t 1     a 1 ( s ) f 1 ( u ( s ) ) d s − c 1 ( u ( 1 ) ) u ( 1 ) = 0 , (1.22)

v ′ ( t ) = λ ∫ t m 1     a 2 ( s ) f 2 ( v ( s ) ) d s − c 2 ( v ( 1 ) ) v ( 1 ) = 0 (1.23)

Now we rewrite (1.17) and (1.18) by using (1.22), (1.23) as:

u ( t ) = λ ∫ 0 1     G ( t , s ) a 1 ( s ) f 1 ( u ( s ) ) d s − λ ( ∫ t m 1     a 1 ( s ) f 1 ( u ( s ) ) d s ) t ,

v ( t ) = λ ∫ 0 1     G ( t , s ) a 2 ( s ) f 2 ( v ( s ) ) d s − λ ( ∫ t m 1     a 2 ( s ) f 2 ( v ( s ) ) d s ) t

= λ ∫ 0 t β     G ( s , t ) a 1 ( s ) f 1 ( u ( s ) ) d s + λ ∫ t β t m     G ( t , s ) a 1 ( s ) f 1 ( u ( s ) ) d s     + λ ∫ t m 1 [ G ( t , s ) − t ] a 1 ( s ) f 1 ( u ( s ) ) d s .

= λ ∫ 0 t β     G ( s , t ) a 2 ( s ) f 2 ( v ( s ) ) d s + λ ∫ t β t m     G ( t , s ) a 2 ( s ) f 2 ( v ( s ) ) d s     + λ ∫ t m 1 [ G ( t , s ) − t ] a 2 ( s ) f 2 ( v ( s ) ) d s .

Note that if t ∈ [ 0, t m ] , then

∫ t m 1 [ G ( t , s ) − t ] a 1 ( s ) f 1 ( u ( s ) ) d s = 0

and

∫ t m 1 [ G ( t , s ) − t ] a 2 ( s ) f 2 ( v ( s ) ) d s = 0

since t ≤ t m ≤ s ≤ 1 implies G ( t , s ) = t . Now t β → 0 and t m → 1 as λ → ∞ . Hence, for t ∈ [ 1 4 , 3 4 ] and λ large we obtain

u ( t ) = λ ( ∫ 0 t β     G ( t , s ) a 1 ( s ) f 1 ( u ( s ) ) d s + ∫ t β t m     G ( t , s ) a 1 ( s ) f 1 ( u ( s ) ) d s ) ,

v ( t ) = λ ( ∫ 0 t β     G ( t , s ) a 2 ( s ) f 2 ( v ( s ) ) d s + ∫ t β t m     G ( t , s ) a 2 ( s ) f 2 ( v ( s ) ) d s )

≥ λ ( ∫ 0 t β     G ( t , s ) a 1 ( s ) f 1 ( u ( s ) ) d s + ∫ 1 4 3 4     G ( t , s ) a 1 ( s ) f 1 ( u ( s ) ) d s ) ,

≥ λ ( ∫ 0 t β     G ( t , s ) a 2 ( s ) f 2 ( v ( s ) ) d s + ∫ 1 4 3 4     G ( t , s ) a 2 ( s ) f 2 ( v ( s ) ) d s )

≥ λ a 1 _ 2 f 1 ( β + θ 2 ) ∫ 1 4 3 4     G ( t , s ) d s ,

≥ λ a 2 _ 2 f 2 ( β + θ 2 ) ∫ 1 4 3 4     G ( t , s ) d s .

Thus,

u ( t ) ≥ λ a 1 _ 2 f 1 ( β + θ 2 ) inf [ 1 4 , 3 4 ] ∫ 1 4 3 4     G ( t , s ) d s

and

v ( t ) ≥ λ a 2 _ 2 f 2 ( β + θ 2 ) inf [ 1 4 , 3 4 ] ∫ 1 4 3 4   G ( t , s ) d s

on [ 1 4 , 3 4 ] , which means that u ( t ) → ∞ and v ( t ) → ∞ for all t ∈ [ 1 4 , 3 4 ] as λ → ∞ .

Lemma 2.5. Let u and v be a positive solution of (1.3). Then there exists λ ˜ such that if λ > λ ˜ , then

u ( t ) ≥ λ C d ( t , ∂ Ω )     and     v ( t ) ≥ λ C d ( t , ∂ Ω ) (1.24)

for some positive constant C, independent of λ . Here Ω = ( 0,1 ) .

Proof. Let ϕ i be the unique solution of the problems

{ − ϕ ″ i = ω α , μ a i ( t ) ,   ( t β , 1 ) ψ i ( t β ) = ψ i ( 1 ) = 0 ,     and     i = 1 , 2 (1.25)

where ω is the characteristic function. By the Hopf maximum principle there exists c ˜ i > 0 such that ϕ i ( t ) > c ˜ i e i ( t ) for all t ∈ [ 0,1 ] , where e i are a solution of

{ − e ″ i = ω α , μ a i ( t ) ,   ( t β , 1 ) e i ( 0 ) = e i ( 1 ) = 0 ,     and     i = 1 , 2 (1.26)

Let H > 0 be such that D : = c ˜ i f i ( H ) + f i ( 0 ) > 0 , and this is possible by (F2). Let u 1 , v 1 and u 2 , v 2 satisfy

− u ″ 1 = λ f 1 ( H ) ω [ α , μ ] a 1 ( t ) ,   t ∈ ( 0 , 1 ) ,   u 1 ( 0 ) = 0 = u 1 ( 1 )

− v ″ 1 = λ f 2 ( H ) ω [ α , μ ] a 2 ( t ) ,   t ∈ ( 0 , 1 ) ,   v 1 ( 0 ) = 0 = v 1 ( 1 )

and

− u ″ 2 = λ f 1 ( 0 ) ω [ α , μ ] a 1 ( t ) ,   t ∈ ( 0 , 1 ) ,   u 2 ( 0 ) = 0 = u 2 ( 1 )

− v ″ 2 = λ f 2 ( 0 ) ω [ α , μ ] a 2 ( t ) ,   t ∈ ( 0 , 1 ) ,   v 2 ( 0 ) = 0 = v 2 ( 1 )

Now by Lemma 2.4, there exists λ ˜ > 0 such that if λ > λ ˜ , then

u ( t ) ≥ H     and     v ( t ) ≥ H     on   [ α , μ ] . (1.27)

Hence, by (1.27), for λ ≫ 1 we have that for t ∈ ( 0,1 )

− u ″ = λ f 1 ( u ) a 1 ( t ) ≥ λ f 1 ( u ) a 1 ( t ) ω [ 0 , t B ] + λ f 1 ( u ) a 1 ( t ) ω [ α , μ ] ,

− v ″ = λ f 2 ( v ) a 2 ( t ) ≥ λ f 2 ( v ) a 2 ( t ) ω [ 0 , t B ] + λ f 2 ( v ) a 2 ( t ) ω [ α , μ ]

− v ″ ≥ λ f 2 ( 0 ) a 2 ( t ) + λ f 2 ( H ) a 2 ( t ) ω [ α , μ ]

− u ″ ≥ λ f 1 ( 0 ) a 1 ( t ) + λ f 1 ( H ) a 1 ( t ) ω [ α , μ ]

= − ( u 1 − u 2 ) ′ ​ ′ ( t )

= − ( v 1 − v 2 ) ′ ​ ′ ( t )

u ( 0 ) − ( u 1 − u 2 ) ( 0 ) = 0 , v ( 0 ) − ( v 1 − v 2 ) ( 0 ) = 0 and u ( 1 ) − ( u 1 − u 2 ) ( 1 ) = u ( 1 ) > 0 , v ( 1 ) − ( v 1 − v 2 ) ( 1 ) = v ( 1 ) > 0 . By the maximum principle, u ( t ) = u 1 ( t ) − u 2 ( t ) = λ f 1 ( H ) ϕ 1 ( t ) + λ f 1 ( 0 ) e 1 ( t ) and v ( t ) = v 1 ( t ) − v 2 ( t ) = λ f 2 ( H ) ϕ 2 ( t ) + λ f 2 ( 0 ) e 2 ( t ) in [ 0,1 ] . Hence

u ( t ) ≥ f 1 ( H ) c ˜ 1 e 1 ( t ) + λ f 1 ( 0 ) e 1 ( t ) = λ D e 1 ( t )

and

v ( t ) ≥ f 2 ( H ) c ˜ 2 e 2 ( t ) + λ f 2 ( 0 ) e 2 ( t ) = λ D e 2 ( t )

for all t ∈ [ 0,1 ] .

Note that there exists L > 0 such that e 1 ( t ) ≥ L d ( t , ∂ Ω ) and e 2 ( t ) ≥ L d ( t , ∂ Ω ) for all t ∈ [ 0,1 ] . Hence, for λ large u ( t ) ≥ λ C d ( t , ∂ Ω ) and v ( t ) ≥ λ C d ( t , ∂ Ω ) for all t ∈ [ 0,1 ] , where C : = D L > 0 .

Lemma 2.6. Let u and v be a positive solution of (1.3). Then there exists H λ such that ‖ u ‖ ∞ ≤ H λ and ‖ v ‖ ∞ ≤ H λ .

Proof.

Let B = ∫ 0 1     a i ( s ) d s . Then B < ∞ since a i ( t ) ≤ c i t α for all t ∈ ( 0,1 ) for some c i > 0 . Now for each given λ > 0 , there exists W λ > 0 such that if

W > W λ , then f i ( W ) W ≤ 1 2 λ B due to the hypothesis (F1). Also since f i ∈ C 1 ( [ 0, ∞ ) , R ) , there exists K λ > 0 such that f i ( W ) ≤ K λ on [ 0, W λ ] . Hence,

f i ( W ) ≤ W 2 λ B + K λ ,   W ∈ [ 0, ∞ ) . (1.28)

Now by Lemma 2.1 and (1.28), we have

‖ u ‖ ∞ = u ( t m ) = λ ∫ 0 1     G ( t m , s ) a 1 ( s ) f 1 ( u ( s ) ) d s − c 1 ( u ( 1 ) ) u ( 1 ) t m

‖ v ‖ ∞ = v ( t m ) = λ ∫ 0 1     G ( t m , s ) a 2 ( s ) f 2 ( v ( s ) ) d s − c 2 ( u ( 1 ) ) u ( 1 ) t m

≤ λ ∫ 0 1     G ( t m , s ) a 1 ( s ) f 1 ( u ( s ) ) d s

≤ λ ∫ 0 1     G ( t m , s ) a 2 ( s ) f 2 ( v ( s ) ) d s

≤ λ ∫ 0 1     G ( t m , s ) a 1 ( s ) [ u ( t m ) 2 λ B + K λ ] d s

≤ λ ∫ 0 1     G ( t m , s ) a 2 ( s ) [ v ( t m ) 2 λ B + K λ ] d s

≤ λ ∫ 0 1     a 1 ( s ) [ 1 2 λ B u ( t m ) + K λ ] d s

≤ λ ∫ 0 1     a 2 ( s ) [ 1 2 λ B v ( t m ) + K λ ] d s   ( since     G ( t , s ) ≤ 1     in     [ 0 , 1 ] × [ 0 , 1 ] )

= 1 2 u ( t m ) + λ B K λ

= 1 2 v ( t m ) + λ B K λ

Hence, for each λ > 0 , ‖ u ‖ ∞ ≤ H λ and ‖ v ‖ ∞ ≤ H λ , where H λ = 2 λ B K λ .

We first claim that (1.3) has a maximal positive solutions u ˜ , v ˜ for λ ≫ 1 . Let φ i be the solutions of the problems

{ − φ ″ 1 = a i ( t ) ,   t ∈ ( 0 , 1 ) φ ′ i ( 1 ) = c i ( φ i ( 1 ) ) φ i ( 1 ) = 0 , φ i ( 0 ) = 0   and   i = 1 , 2 (1.29)

Note that (1.29) has the unique solution since e i , φ 0 are sub solutions and super solutions of (1.29), respectively, where e i is defined in (1.26) and φ 0 is the solutions of the linear boundary condition problems

{ − φ ″ 0 = a i ( t ) ,   t ∈ ( 0 , 1 ) φ ′ 0 ( 1 ) = c i ( φ 0 ( 1 ) ) φ 0 ( 1 ) = 0 , φ 0 ( 0 ) = 0   and   i = 1 , 2 (1.30)

Since f i satisfies (F1), given λ > 0 , we can choose Z λ ≥ 1 such that Z λ > λ f i ( Z λ ‖ φ i ‖ ∞ ) and Z λ > λ f i ( H λ ) where H λ is as in Lemma 2.6. Then, Z λ φ i are a super solutions of (1.3) since

{ − ( Z λ φ i ) ′ ​ ′ = Z λ a i ( t ) ≥ λ a i f i ( Z λ ‖ φ i ‖ λ ) ≥ λ a i f i ( Z λ φ i ) ,   t ∈ ( 0 , 1 ) ( Z λ φ ′ i ( 1 ) ) + c i ( Z λ ( φ i ( 1 ) ) ) Z λ φ i ( 1 ) = Z λ ( φ ′ i ( 1 ) + c i ( Z λ ( φ i ( 1 ) ) ) φ i ( 1 ) ) ≥ Z λ ( φ ′ i ( 1 ) + c i ( φ i ( 1 ) ) φ i ( 1 ) ) = 0 , Z λ φ i ( 0 ) = 0   and   i = 1 , 2

Next, we show that this super solution Z λ φ i is larger than any positive solutions of (1.3). Let θ i be any positive solutions of (1.3). From Lemma 2.6, we have

− ( Z λ φ i − θ i ) ′ ​ ′ = Z λ a i ( t ) − λ a i f i ( θ i ) = a i [ Z λ − λ f i ( θ i ) ]                                       ≥ a i [ Z λ − λ f i ( H λ ) ] > 0 ,   t ∈ ( 0 , 1 )

by the choice of Z λ . Note that ( Z λ φ i − θ i ) ( 0 ) = 0 . Now we show that ( Z λ φ i − θ i ) ( 1 ) ≥ 0 . Indeed, since Z λ φ ′ i ( 1 ) + c i ( Z λ ( φ i ( 1 ) ) ) Z λ φ i ( 1 ) ≥ 0 = θ ′ i ( 1 ) + c i ( θ i ( 1 ) ) θ i ( 1 ) , we have

Z λ φ ′ i ( 1 ) − θ ′ i ( 1 ) + c i ( Z λ ( φ i ( 1 ) ) ) Z λ φ i ( 1 ) − c i ( θ i ( 1 ) ) θ i ( 1 ) ≥ 0 (1.31)

If we assume that Z λ φ i ( 1 ) − θ i ( 1 ) < 0 , then c i ( Z λ ( φ i ( 1 ) ) ) Z λ φ i ( 1 ) − c i ( θ i ( 1 ) ) θ i ( 1 ) < 0 since c i increases. Hence from (3.3) we obtain Z λ φ ′ i ( 1 ) − θ ′ i ( 1 ) > 0 . However, − ( Z λ φ i − θ i ) ′ ​ ′ > 0 in ( 0,1 ) , ( Z λ φ i − θ i ) ( 0 ) = 0 and ( Z λ φ i − θ i ) ( 1 ) < 0 implies that ( Z λ φ i − θ i ) ′ ( 1 ) > 0 , which is a contradiction. Hence Z λ φ i ( 1 ) − θ i ( 1 ) ≥ 0 . Therefore, by the maximum principle, Z λ φ i ≥ φ i in [ 0,1 ] . Therefore, (1.3) has a maximal positive solutions u ˜ , v ˜ . Now, let u and v be any other positive solutions of (1.3). To establish our theorem, we will show that u ≡ u ˜ and v ≡ v ˜ for λ ≫ 1 . Since u , v and u ˜ , v ˜ are solutions of (1.3), we obtain

{ − ( u ˜ − u ) ′ ​ ′ ( t ) = λ a 1 ( t ) ( f 1 ( u ˜ ( t ) ) − f 1 ( u ( t ) ) ) ,   t ∈ ( 0 , 1 ) − ( v ˜ − v ) ′ ​ ′ ( t ) = λ a 2 ( t ) ( f 2 ( v ˜ ( t ) ) − f 2 ( v ( t ) ) ) ,   t ∈ ( 0 , 1 ) ( u ˜ − u ) ( 0 ) = ( v ˜ − v ) ( 0 ) = 0 , ( u ˜ − u ) ′ ( 1 ) + c 1 ( u ˜ ( 1 ) ) u ˜ ( 1 ) − c 1 ( u ( 1 ) ) u ( 1 ) = 0 , ( v ˜ − v ) ′ ( 1 ) + c 2 ( v ˜ ( 1 ) ) v ˜ ( 1 ) − c 2 ( v ( 1 ) ) v ( 1 ) = 0

By the mean value theorem, there exists ξ such that u ≤ ξ ≤ u ˜ and v ≤ ξ ≤ v ˜ quadin [ 0,1 ] and

{ − ( u ˜ − u ) ′ ​ ′ ( t ) = λ a 1 ( t ) f ′ 1 ( ξ ) ( u ˜ ( t ) − u ( t ) ) ,   t ∈ ( 0 , 1 ) − ( v ˜ − v ) ′ ​ ′ ( t ) = λ a 2 ( t ) f ′ 2 ( ξ ) ( v ˜ ( t ) − v ( t ) ) ,   t ∈ ( 0 , 1 ) ( u ˜ − u ) ( 0 ) = ( v ˜ − v ) ( 0 ) = 0 , ( u ˜ − u ) ′ ( 1 ) + c 1 ( u ˜ ( 1 ) ) u ˜ ( 1 ) − c 1 ( u ( 1 ) ) u ( 1 ) = 0 , ( v ˜ − v ) ′ ( 1 ) + c 2 ( v ˜ ( 1 ) ) v ˜ ( 1 ) − c 2 ( v ( 1 ) ) v ( 1 ) = 0 (1.32)

By multiplying (1.3) by ( u ˜ − u ) , ( v ˜ − v ) and (1.32) by u , v and integrating, we first obtain, using integration by parts,

∫ 0 1 [ ( u ˜ − u ) ′ ​ ′ u − ( u ˜ − u ) u ″ ] d t = u ˜ ′ ( 1 ) u ( 1 ) − u ′ ( 1 ) u ˜ ( 1 ) = u ( 1 ) [ − c 1 ( u ˜ ( 1 ) ) u ˜ ( 1 ) ] + u ˜ ( 1 ) [ c 1 ( u ( 1 ) ) u ( 1 ) ] = u ( 1 ) u ˜ ( 1 ) [ c 1 ( u ( 1 ) ) − c 1 ( u ˜ ( 1 ) ) ] ≤ 0

∫ 0 1 [ ( v ˜ − v ) ′ ​ ′ v − ( v ˜ − v ) v ″ ] d t = v ˜ ′ ( 1 ) v ( 1 ) − v ′ ( 1 ) v ˜ ( 1 ) = v ( 1 ) [ − c 2 ( v ˜ ( 1 ) ) v ˜ ( 1 ) ] + v ˜ ( 1 ) [ c 2 ( v ( 1 ) ) v ( 1 ) ] = v ( 1 ) v ˜ ( 1 ) [ c 2 ( v ( 1 ) ) − c 2 ( v ˜ ( 1 ) ) ] ≤ 0

since c i are increasing. Using that f i is concave, we also have

∫ 0 1 [ ( u ˜ − u ) ′ ​ ′ u − ( u ˜ − u ) u ″ ] d t = λ ∫ 0 1     a 1 ( t ) [ f 1 ( u ) − f ′ 1 ( ξ ) u ] ( u ˜ − u ) d t ≥ λ ∫ 0 1     a 1 ( t ) [ f 1 ( u ) − f ′ 1 ( u ) u ] ( u ˜ − u ) d t (1.33)

∫ 0 1 [ ( v ˜ − v ) ′ ​ ′ v − ( v ˜ − v ) v ″ ] d t = λ ∫ 0 1     a 2 ( t ) [ f 2 ( v ) − f ′ 2 ( ξ ) v ] ( v ˜ − v ) d t ≥ λ ∫ 0 1     a 2 ( t ) [ f 2 ( v ) − f ′ 2 ( v ) v ] ( v ˜ − v ) d t (1.34)

From (F1), there exist r i > 0 , b i > 0 such that f i ( s ) − f ′ ( s ) s ≥ b i whenever s ≥ r i . From (1.20), for λ ≫ 1 , u ( t ) ≥ r i and v ( t ) ≥ r i if

d ( t , ∂ Ω ) ≥ r i λ C . Let Ω + = [ r i λ C ,1 − r i λ C ] and Ω − = ( 0, r i λ C ) ∪ ( 1 − r i λ C ,1 ) . Then from (1.33), we have

0 ≥ λ ∫ Ω +     a 1 ( t ) b 1 ( u ˜ − u ) d t + λ ∫ Ω −     a 1 ( t ) f 1 ( 0 ) ( u ˜ − u ) d t

0 ≥ λ ∫ Ω +     a 2 ( t ) b 2 ( v ˜ − v ) d t + λ ∫ Ω −     a 2 ( t ) f 2 ( 0 ) ( v ˜ − v ) d t . (1.35)

since when f i is concave f i ( W ) − W f ′ i ( W ) ≥ f i ( 0 ) for all W ≥ 0 .

Next let m and h satisfy

− m ″ ( t ) = ω Ω + a 1 ( t ) ,   t ∈ ( 0 , 1 ) ,   m ( 0 ) = m ( 1 ) = 0 ,

m ″ ( t ) = ω Ω + a 2 ( t ) ,   t ∈ ( 0 , 1 ) ,   m ( 0 ) = m ( 1 ) = 0

and

− h ″ ( t ) = ω Ω − a 1 ( t ) ,   t ∈ ( 0 , 1 ) ,   h ( 0 ) = h ( 1 ) = 0 ,

− h ″ ( t ) = ω Ω − a 2 ( t ) ,   t ∈ ( 0 , 1 ) ,   h ( 0 ) = h ( 1 ) = 0

respectively. Now multiplying (1.32) by b i m + f i ( 0 ) h and integrating, we obtain

I : = ∫ 0 1 − ( u ˜ − u ) ′ ​ ′ [ b 1 m + f 1 ( 0 ) h ] d t

J : = ∫ 0 1 − ( v ˜ − v ) ′ ​ ′ [ b 2 m + f 2 ( 0 ) h ] d t

= ( u ˜ ( 1 ) − u ( 1 ) ) [ b 1 m ′ ( 1 ) + f 1 ( 0 ) h ′ ( 1 ) ] + ∫ Ω +     a 1 ( t ) b 1 ( u ˜ − u ) d t       + ∫ Ω −     a 1 ( t ) f 1 ( 0 ) ( u ˜ − u ) d t

= ( v ˜ ( 1 ) − v ( 1 ) ) [ b 2 m ′ ( 1 ) + f 2 ( 0 ) h ′ ( 1 ) ] + ∫ Ω +     a 2 ( t ) b 2 ( v ˜ − v ) d t       + ∫ Ω −     a 2 ( t ) f 2 ( 0 ) ( v ˜ − v ) d t

{ = I 1 + I 2 = J 1 + J 2 (1.36)

Note that as λ → ∞ , m → e i and h → 0 in C 1 [ 0,1 ] . Hence, for λ large, we obtain b i m + f i ( 0 ) h > 0 , in ( 0,1 ) and i = 1 , 2

b i m + f i ( 0 ) h > 0   in   ( 0 , 1 ) (1.37)

and

b i m ′ + f i ( 0 ) h ′ < 0 (1.38)

Hence for λ ≫ 1 , (1.37) implies I 1 ≤ 0 , J 1 ≤ 0 and combining with (1.34) (which implies I 2 ≤ 0 and J 2 ≤ 0 ) we have I ≤ 0 and J ≤ 0 . However, by (1.32), we also have

I : = ∫ 0 1 − ( u ˜ − u ) ′ ​ ′ [ b 1 m + f 1 ( 0 ) h ] d t             + λ ∫ 0 1     a 1 ( t ) f ′ 1 ( ξ ) ( u ˜ ( t ) − u ( t ) ) [ b 1 m + f 1 ( 0 ) h ] d t

J : = ∫ 0 1 − ( v ˜ − v ) ′ ​ ′ [ b 2 m + f 2 ( 0 ) h ] d t             + λ ∫ 0 1     a 2 ( t ) f ′ 2 ( ξ ) ( v ˜ ( t ) − v ( t ) ) [ b 2 m + f 2 ( 0 ) h ] d t

Now for λ ≫ 1 , using (1.36), a i > 0 , and f ′ i ≥ 0 we get I ≥ 0 and J ≥ 0 . Hence, we conclude that I ≡ 0 and J ≡ 0 for λ ≫ 1 , which implies that v ˜ ≡ v and u ˜ ≡ u in [ 0,1 ] . This proves that (1.3) has a unique positive solution for all λ large.

In the paper, were studied the positive radial solutions for elliptic systems to the nonlinear Boundary Value problems. And then, we presented that by the Theorem 1.1, and Theorem 2.2, we can obtain a solution of the problem (1.3). Moreover, for all λ ≫ 1 , then (1.3) has a unique positive solution.

The authors would like to thank the anonymous referees for their helpful comments.

The authors declare no conflicts of interest regarding the publication of this paper.

Mohamed, A., Abbakar, K.A., Awad, A., Khalil, O., Acyl, B.M., Youssouf, A.A. and Mousa, M. (2021) Uniqueness of Positive Radial Solutions for a Class of Semipositone Systems on the Exterior of a Ball. Applied Mathematics, 12, 131-146. https://doi.org/10.4236/am.2021.123009