Semipositone problems with falling zeros on exterior domains

J. Math. Anal. Appl. 401 (2013) 146–153

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Semipositone problems with falling zeros on exterior domains Lakshmi Sankar a , Sarath Sasi a , R. Shivaji b,∗ a

Department of Mathematics & Statistics, Mississippi State University, MS 39762, USA

b

Department of Mathematics & Statistics, University of North Carolina at Greensboro, NC 27412, USA

article

abstract

info

Article history: Received 15 September 2012 Available online 3 December 2012 Submitted by Junping Shi

We study boundary value problems of the form

Keywords: Existence Uniqueness Exterior domains Semipositone problems Falling zeros

where λ is a positive parameter, ∆u = div (∇ u) is the Laplacian of u, Ω = {x ∈ Rn ; n > 2, |x| > r0 }, K belongs to a class of C 1 functions such that limr →∞ K (r ) = 0, and f belongs to a class of C 1 functions which are negative at the origin and have falling zeros. We discuss the existence and uniqueness of nonnegative radial solutions when λ is large. © 2012 Elsevier Inc. All rights reserved.

 −∆u = λK (|x|)f (u), x ∈ Ω u = 0 if |x| = r0 u → 0 as |x| → ∞,

1. Introduction We study nonnegative radial solutions to the problem

 −1u = λK (|x|)f (u), x ∈ Ω u = 0 if |x| = r0 u → 0 as |x| → ∞,

(1)

where λ is a positive parameter, 1u = div ∇ u is the Laplacian of u, Ω = {x ∈ Rn , n > 2| |x| > r0 }, f : [0, ∞) → R is a C 1 function such that f (0) < 0 and satisfies the following.





(H1 ) There exist ρ1 , ρ2 such that 0 < ρ1 < ρ2 , f (ρ1 ) = f (ρ2 ) = 0 and f > 0 in (ρ1 , ρ2 ). ρ (H2 ) t 2 f (s)ds > 0 for every t ∈ [0, ρ2 ). And K ∈ C 1 ([r0 , ∞), (0, ∞)) satisfies the following.

(H3∗ ) K (r ) ≤ (H4 ) ∗

K (x−1 )

x2(n−1)

1 r n+ρ

for r ≫ 1 and for some ρ such that 0 < ρ < n − 2.

is decreasing for x >

1 . r0

A simple example of K satisfying our hypotheses is K (r ) = r n1+ρ , ρ < n − 2. Using the transformations r = |x| , s = ( rr )2−n , we can reduce (see the appendix of [13]) Eq. (1) to the boundary value 0 problem



∗

−u′′ (s) = λh(s)f (u(s)), u(0) = u(1) = 0,

0
Corresponding author. E-mail addresses: [email protected] (L. Sankar), [email protected] (S. Sasi), [email protected] (R. Shivaji).

0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2012.11.031

(2)

L. Sankar et al. / J. Math. Anal. Appl. 401 (2013) 146–153

where h(s) =

r02

(2−n)2

s

−2(n−1) n−2 K

147

1

(r0 s 2−n ). If K satisfies (H3∗ ), then h ∈ C 1 ((0, 1], (0, ∞)), is singular at 0, and hˆ =

inft ∈(0,1) h(t ) > 0 and satisfies the following.

(H3 ) There exist ϵ1 > 0 and a constant c > 0 such that h(t ) ≤

c tα

for all t ∈ (0, ϵ1 ), where α =

(n−2)−ρ . n −2

Further, by (H4∗ ), we have the following.

(H4 ) h(s) decreases for s ∈ (0, 1]. Note that, if ρ ≥ n − 2, then the corresponding h is nonsingular at zero. We concentrate on the more difficult case ρ < n − 2, which forces h to be singular at zero. The rest of the paper we will be focusing on studying nonnegative solutions of (2) as this is equivalent to studying nonnegative radial solutions of (1). We establish the following. Theorem 1.1. Assume that (H1 )–(H4 ) hold. Then (2) has a nonnegative solution for λ ≫ 1. Next, under the additional assumption

(H5 ) f is concave and f ′ (s) < 0 in (ρ2 − τ , ρ2 ] for some τ > 0, we establish the following uniqueness result. Theorem 1.2. Assume that (H1 )–(H5 ) hold. Then (2) has a unique nonnegative solution for λ ≫ 1. Remark. The solution of (2) we obtain in Theorem 1.1 is strictly positive in (0, 1). Also, in our proof of Theorem 1.2, we establish that all nonnegative solutions for (2) are strictly positive when λ is large. We note here that the fact that h is strictly decreasing plays a role in our proof of this positivity result. See [8] where, when h(t ) ≡ 1, the authors establish a sequence {λn } and nonnegative solutions {un } with λn → ∞ as n → ∞ and the un have exactly n interior zeros. Problems of the form



−1u = λf (u), x ∈ Ω u = 0, x ∈ ∂ Ω ,

(3)

where f (0) < 0, are known in the literature as semipositone problems. When Ω is a smooth bounded domain and f : [0, ∞) → R is a C 1 function, the existence and uniqueness of positive solutions of (3) have been studied extensively in the past (see [1–6,8–11,14]). When f satisfies (H1 ) and (H2 ), the existence of a positive solution of (3) was established in [14] when the parameter λ is large. The uniqueness of such a solution was proved in [10] when Ω is a ball, and later in [11] when Ω is a bounded domain in RN . In our paper, we extend this study to the case when Ω is an exterior domain. One can refer to [7,13] for some recent existence and uniqueness results on semipositone problems on exterior domains when f is unbounded and satisfies a sublinear growth condition at ∞. We use the method of subsolutions and supersolutions and the sweeping principle to establish our existence result. The main idea of our existence result is adapted from [14]. Necessary modifications are made to overcome the difficulty caused  by the singularity of h at the origin. By a subsolution of (2) we mean a function ψ ∈ W 1,2 (0, 1) C [0, 1] that satisfies

 

1

−ψφ ≤ λ ′′

0



ψ(0) ≤ 0,

1



h(t )f (ψ)φ,

for every φ ∈ V ,

0

ψ(1) ≤ 0,

and by a supersolution we mean a function Z ∈ W 1,2 (0, 1)

 

1

−Z φ ′′ ≥ λ 0





C [0, 1] that satisfies

1



h(t )f (Z )φ,

for every φ ∈ V ,

0

Z (0) ≥ 0,

Z (1) ≥ 0,

where V = {ζ ∈ C0 (0, 1) : ζ ≥ 0 in (0, 1)}. Then we have the following lemma (see [12]). ∞

Lemma 1.3. Let ψ be a subsolution and Z be a supersolution such that ψ ≤ Z in (0, 1). Then (2) has a solution u ∈ C 2 ((0, 1)) ∩ C 1 ([0, 1]) such that ψ ≤ u ≤ Z in (0, 1). Next, we state a sweeping principle for singular boundary value problems of the form (2). This is adapted from the sweeping principle discussed in [14,15] and will be used to establish the positivity of our solution. Lemma 1.4. Let u be a solution of (2), B be a connected set, and let A = {wt : t ∈ B} be a family of subsolutions satisfying

wt (x) < 0 at x = 0, 1 for all t ∈ B. If • t → wt is continuous with respect to ∥ · ∥∞ and • wt0 ≤ u in [0, 1] for some t0 ∈ B, then wt ≤ u for all t ∈ B.

See the Appendix for the proof. We prove Theorem 1.1 in Section 2 and Theorem 1.2 in Section 3. Our uniqueness result is obtained by analyzing the properties of the solution.

148

L. Sankar et al. / J. Math. Anal. Appl. 401 (2013) 146–153

a

b

Fig. 1. Graphs of f˜ (u) and F˜ (u).

2. Proof of Theorem 1.1 We first establish two useful results for such nonlinear eigenvalue problems when the nonlinearities are zero at the origin. Namely, we consider f˜ ∈ C 1 ((0, ∞), R) such that f˜ (0) = 0 and satisfies the following.

(H˜ 1 ) There exist ρ˜1 , ρ˜2 such that 0 < ρ˜1 < ρ˜2 , f˜ (ρ˜1 ) = f˜ (ρ˜2 ) = 0 and f˜ > 0 in (ρ˜1 , ρ˜2 ).  ρ˜ (H˜ 2 ) t 2 f˜ (s)ds > 0 for every t ∈ [0, ρ˜2 ). See Fig. 1. We study the following boundary value problem:

 ′′ −u (s) = λh(s)f˜ (u(s)), u(0) = u(1) = 0,

0
(4)

where h(s) is as before. First, we establish the following.

˜ 1 ), (H˜ 2 ) and (H3 ) hold and that f˜ (0) = 0. Then (4) has a positive solution uλ for λ ≫ 1 with Lemma 2.1. Assume that (H max uλ ∈ (ρ˜1 , ρ˜2 ]. Proof. First, modify f˜ outside [0, ρ˜2 ] as f˜ (s) = 0 if s > ρ˜2 and f˜ (s) = −f˜ (−s) for s < 0. Define I (u, λ) =

1

(u′ (x))2 dx − λ 0 h(x)F˜ (u(x))dx in (0, 1), where F˜ (u) = 0 f˜ (s)ds. Since h is integrable and F˜ is bounded, it is easy to see that I (u, λ) is bounded below, weakly lower semi-continuous, and coercive for λ > 0. Also, since F˜ is an even function and h(x) > 0, I (|u|, λ) ≤ I (u, λ), for all λ > 0. Hence I (u, λ) has a nonnegative minimizer, say uλ . We now prove that ∥uλ ∥∞ ≤ ρ˜2 . Suppose that ∥uλ ∥∞ > ρ˜2 and let t1 ∗ be such that uλ (t1 ∗ ) = ∥uλ ∥∞ . Then there exists a t0 ∗ < t1 ∗ such that uλ (t0 ∗ ) = ρ˜2 and uλ is nondecreasing in [t0 ∗ , t1 ∗ ]. Integrating (4) from t to t1 ∗ , where t0 ∗ < t < t1 ∗ , 1,2 W0

1

u

1 2

0

we see that u′ (t ) = λ

t1 ∗



h(s)f˜ (uλ (s))ds = 0 (since f˜ (s) = 0 for s > ρ˜2 ),

t

which is a contradiction. Next, we prove that ∥uλ ∥∞ > ρ˜1 for λ ≫ 1. Suppose that ∥uλ ∥∞ ≤ ρ˜1 for all positive λ. We choose a w ∈ C0∞ (0, 1) such that 0 ≤ w ≤ ρ˜2 in [0, δ] ∪ [1 − δ, 1] and w = ρ˜2 in (δ, 1 − δ), where δ will be chosen later. Then 1

 1 ((w′ )2 − (u′λ )2 )dx − λ h(x)(F˜ (w) − F˜ (uλ ))dx 2 0 0    δ 1 1 1 ′ 2 (w ) dx − λ h(x)F˜ (ρ˜2 )dx + ≤ h(x)(F˜ (w) − F˜ (ρ˜2 ))dx

I (w, λ) − I (uλ , λ) =

1



2

0

0

h(x)(F˜ (w) − F˜ (ρ˜2 ))dx −

+ 1−δ

≤

1 2

(w ′ )2 dx − λ 0



1



h(x)F˜ (uλ )dx 0

1



−λ

0

1



δ



h(x)(F˜ (w) − F˜ (ρ˜2 ))dx

0



1

1−δ

h(x)(F˜ (w) − F˜ (ρ˜2 ))dx − λ

1



h(x) 0



ρ˜2 uλ

f˜ (s)dsdx.

L. Sankar et al. / J. Math. Anal. Appl. 401 (2013) 146–153

149

Fig. 2. A solution with more than one maximum.

 ρ˜2

Let β = min{ ρ f˜ (s)ds; 0 ≤ ρ ≤ ρ˜1 }. By our assumption, β > 0. Also, F˜ (uλ ) ≤ m for some m > 0 and h(s) ≤ t ∈ (0, 1); thus I (w, λ) − I (uλ , λ) ≤

1 2



1

(w ′ )2 dx +

0

2λmdδ 1−α 1−α

+

2λmd(1 − (1 − δ)1−α ) 1−α

−

d tα

for all

λβ d . 1−α

We now choose δ ≈ 0. Then it follows that I (w, λ) < I (uλ , λ) for λ ≫ 1, a contradiction. Thus ∥uλ ∥∞ > ρ˜1 for λ ≫ 1. Next, we prove that uλ > 0 in (0, 1). Suppose that uλ (tˆ) = 0 for some tˆ ∈ (0, 1). Then uλ satisfies the initial value problem



−u′′λ (s) = λh(s)f˜ (uλ (s)), u′λ (tˆ) = uλ (tˆ) = 0.

But f˜ (0) = 0, and hence, by the uniqueness result of Picard, uλ ≡ 0, which is a contradiction. Hence uλ > 0 in (0, 1).



Next, we prove that the solution uλ has only one interior maximum.

˜ 1 ), (H˜ 2 ) and (H3 )–(H4 ) hold, and let uλ be the solution of (4) for λ ≫ 1. Then uλ has only one Lemma 2.2. Assume that (H interior maximum. Proof (See also Lemma 2.1 in [7]). Let E˜ (t ) := λF˜ (uλ (t ))h(t ) +

[u′λ (t )]2

, t ∈ (0, 1). Hence E˜ ′ (t ) = λF˜ (uλ (t ))h′ (t ). Note that, by (H4 ), h(s) decreases for s > 0. Let θ˜ be such that ρ˜1 < θ˜ < ρ˜2 and F˜ (θ˜ ) = 0. Then E˜ (t ) increases when uλ (t ) < θ˜ and decreases when uλ (t ) > θ˜ . Let t ∗ ∈ (0, 1) be the first point at which uλ has a local maximum, and assume that uλ (t ) ≤ θ˜ , ∀t ≤ t ∗ . Integrating (2) from t to t ∗ , t < t ∗ , and using (H3 ),  t∗ df˜ (θ˜ ) ∗ 1−α df˜ (θ˜ ) ′ uλ ( t ) = λ (t − t 1−α ) ≤ λ , (5) h(s)f˜ (uλ (s))ds ≤ λ 1−α 1−α t where d ≥ c is such that h(t ) ≤ where M0 =

˜ df˜ (θ) . 1−α

d tα

2

for all t ∈ (0, 1) and α ∈ (0, 1). Integrating (5) again from 0 to t, t < t ∗ , uλ (t ) ≤ λM0 t,

˜ . Hence Since f˜ is continuous, there exists k0 > 0 such that |F˜ (uλ )| ≤ k0 uλ for uλ ∈ [0, θ]

lim λ|F˜ (uλ (t ))|h(t ) ≤ lim k0 λM0 dt 1−α = 0,

t →0+

t →0+

which implies that limt →0+ E˜ (t ) ≥ 0. Since E˜ (t ) increases on [0, t ∗ ], E˜ (t ∗ ) = λF˜ (uλ (t ∗ ))h(t ∗ ) > 0, which is a contradiction if uλ (t ∗ ) ≤ θ˜ . Hence uλ (t ∗ ) > θ˜ .  Now, suppose that uλ has two interior maxima. Let t˜ ∈ (t ∗ , 1) be such that u′λ (t˜) = 0 and u′′λ (t˜) ≥ 0 (as in Fig. 2). Since

u′′λ (t˜)

= −λh(t˜)f˜ (uλ (t˜)) ≥ 0, we see that uλ (t˜) ≤ ρ˜1 , and thus E˜ (t˜) < 0. Let t ∈ (t ∗ , t˜) be such that uλ (t ) = θ˜ . Since E˜ (t ) ≥ 0 and E˜ increases in (t , t˜), E˜ (t˜) > 0, which is a contradiction. Hence uλ can have only one interior maximum, and that maximum value is bigger than θ˜ . Now, we prove Theorem 1.1. ρ First, modify f in R \ [0, ρ2 ] as follows. Let f (s) ≤ 0 for s ∈ (ρ2 , ∞), f (s) = 0 for s ∈ (−∞, −1], and t 2 f (s)ds > 0 for t ∈ [−1, 0) such that f ∈ C 1 . By Lemma 2.1,



−u′′ (s) = µh(s)f (u(s) − 1), u(0) = u(1) = 0

0
150

L. Sankar et al. / J. Math. Anal. Appl. 401 (2013) 146–153

Fig. 3. Graphs of v(t ) and w(t ).

b

a

Fig. 4. Graphs of f (u) and F (u).

has a positive solution w for some µ large enough with max w ∈ (ρ1 + 1, ρ2 + 1]. Define v(t ) = w(t ) − 1 for all t ∈ (0, 1) (see Fig. 3). By Lemma 2.2, v has only two zeros, say α1 , α2 and v > 0 in (α1 , α2 ). Extend v in (1, ∞) such that v(t ) ≤ −1 and v ′′ (t ) = 0 for all t ∈ (1, ∞). Also extend h(t ) as h(t ) = h(1) for all t ∈ (1, ∞).Then v(t ) satisfies 

−v ′′ (s) = µh(s)f (v), in (0, ∞) with max v ∈ (ρ1 , ρ2 ]. Now, for a fixed y0 ∈ (0, 1), define ψy0 (λ, x) = v ( µλ ) 2 |x−y0 |+t ∗ , ˜ = (0, 1) and d(y0 , ∂ Ω ˜ ) denote the distance from y0 to the boundary where t ∗ is a point at which v has maximum. Let Ω ˜ . If λ > µ(α2 − t ∗ )2 d(y0 , Ω ˜ )−2 = λ∗ , ψy0 < 0 on ∂ Ω ˜ . Thus ψy0 is a subsolution of (2) for λ > λ∗ . Clearly Z = ρ2 is a of Ω supersolution of (2). Also, the subsolution ψy0 ≤ ρ2 for all λ. Thus (2) has a solution uy0 ∈ [ψy0 , ρ2 ] if λ > λ∗ .   µ 1 µ 1 Next we will show that uy0 > 0, using the sweeping principle. For y ∈ Iλ = (α2 − t ∗ )( λ ) 2 , 1 − (α2 − t ∗ )( λ ) 2 , define   1 ψy (λ, x) = v ( µλ ) 2 |x − y| + t ∗ . Then {ψy , y ∈ Iλ } is a family of subsolutions to problem (2) with ψy < 0 on ∂ Ω and 1

• y → ψy (λ, x) is continuous with respect to ∥ · ∥∞ and • y0 ∈ Iλ and uy0 ≥ ψy0 in [0, 1]. Thus, by the sweeping principle, uy0 (x) > ψy (λ, x) for all y ∈ Iλ . For x ∈ Iλ , by choosing y = x, we see that uy0 (x) ≥

v(t ∗ ) > 0. For x ∈ (0, 1) − Iλ , we choose y ∈ Iλ such that d(x, y) < ( µλ ) 2 (α2 − t ∗ ). Since t ∗ > α1 , and by the choice of y, 1

α1 < ( µλ ) 2 |x − y| + t ∗ < α2 , which implies that ψy (λ, x) > 0 for x ∈ (0, 1) − Iλ . Hence uy0 (x) > 0 for all x ∈ (0, 1). 1

3. Proof of Theorem 1.2

s

Let F (s) = 0 f (t )dt. Note that there exists a positive real number θ such that ρ1 < θ < ρ2 and F (θ ) = 0 (see Fig. 4). Let u denote a nonnegative solution of problem (2) for λ ≫ 1 under assumptions (H1 )–(H5 ). We first establish some properties of u, namely, Lemmas 3.1–3.4, which will help us to prove Theorem 1.2. Lemma 3.1. u has only one interior maximum, say at t0 , and u(t0 ) > θ . Proof. Follows by similar arguments as in the proof of Lemma 2.2.



Lemma 3.2. Let t2 and tˆ2 ∈ (0, 1) be such that t2 < tˆ2 and u(t2 ) = u(tˆ2 ) =

ρ1 +θ 2

1

; then t2 , 1 − tˆ2 ≤ O(λ− 2 ) (see Fig. 5).

L. Sankar et al. / J. Math. Anal. Appl. 401 (2013) 146–153

151

Fig. 5. Graph of a solution.

Proof. Let t1 be the first point in (0, 1) such that u(t1 ) = u′ (t ) = u′ (0) − λ

t





ˆ −f h(s)f (u(s))ds ≥ λht

ρ1 2

. Integrating (2) from 0 to t, t < t1 ,

 ρ 

0 1

1

2

Integrating (6) from 0 to t1 yields t1 ≤ c˜ λ− 2 , where c˜ =



.

(6)

ρ1 ρ hˆ (−f ( 21 ))

 12

. Now, let E (t ) := λF (u)h(t ) +

[u′ (t )]2 2

, t ∈ (0, 1). As

in the discussion in Lemma 2.2, limt →0+ E (t ) ≥ 0 and E is increasing if u(t ) < θ . Hence we have E (t ) ≥ 0 for all t ∈ (0, 1). This implies that

(u′ (t ))2 2

≥ λ(−F (u))h(t ) for all t ∈ (0, 1).

√ 1 For t ∈ (t1 , t2 ), u′ (t ) ≥ λ 2 k1 , where k1 = mint ∈(t1 ,t2 ) −2F (u)h(t ) > 0. Integrating this from t1 to t2 , we obtain 1

1

1

1

(t2 − t1 ) ≤ O(λ− 2 ). Since t1 ≤ c˜ λ− 2 , this implies that t2 ≤ O(λ− 2 ). Similarly 1 − tˆ2 ≤ O(λ− 2 ).



Lemma 3.3. ∥u∥∞ → ρ2 as λ → ∞. Proof. Suppose that there exists ϵ > 0 such that ∥u∥∞ < ρ2 − ϵ , for all λ > 0. Let G(t , s) denote the Green’s function of the operator −u′′ with boundary condition u(0) = 0 = u(1). Then, for t ∈ (0, 1), we have u( t ) = λ

≥λ

1



G(t , s)h(s)f (u(s))ds

0

t2

3 4



G(t , s)h(s)f (u(s))ds +

1 4

0

G(t , s)h(s)f (u(s))ds +



1

 G(t , s)h(s)f (u(s))ds .

tˆ2

Since h is integrable, by Lemma 3.2, we have u( t ) ≥

1 2

λ

3 4

 1 4

G(t , s)h(s)f (u(s))ds for λ ≫ 1.

By our assumption (∥u∥∞ < ρ2 − ϵ ), there exists k2 > 0 such that f (u(s)) > k2 in [ 41 , 34 ]. Then, for all t ∈ [ 41 , 34 ], u( t ) ≥

1 2

ˆ 2 inf λhk

t ∈[ 14 , 34 ]

3 4

 1 4

G(t , s)ds,

which is a contradiction, since all positive solutions of (2) are bounded above by ρ2 . Hence ∥u∥∞ → ρ2 as λ → ∞.



Lemma 3.4. Let ρ˜ ∈ (ρ2 − τ , ρ2 ] and tλ , tˆλ be such that u(tλ ) = u(tˆλ ) = ρ˜ with tλ < tˆλ . Then tλ , (1 − tˆλ ) → 0 as λ → ∞. Proof. By (H5 ), f ′ (s) < 0 for s ∈ [ρ, ˜ ρ2 ], and, by Lemma 3.3, there exists tλ and tˆλ such that u(tλ ) = u(tˆλ ) = ρ˜ when λ is large. We first prove tλ → 0 as λ → ∞. Suppose that there exists γ1 > 0 such that tλ > γ1 > 0 for all λ > 0. Then u( t ) = λ

1



G(t , s)h(s)f (u(s))ds 0

≥λ

t2

 0

G(t , s)h(s)f (u(s))ds +



γ1 γ0

G(t , s)h(s)f (u(s))ds +



1 tˆ2



G(t , s)h(s)f (u(s))ds ,

152

L. Sankar et al. / J. Math. Anal. Appl. 401 (2013) 146–153

where γ0 is such that 0 < t2 < γ0 < γ1 . Now, using Lemma 3.2, for t in, say, [ 14 , 34 ], u(t ) ≥

1 2

hˆ λk3



γ1

G(t , s)ds,

inf  γ0 t ∈ 14 , 34

where k3 > 0 is such that f (u(s)) > k3 in [γ0 , γ1 ]. This again contradicts the fact that solutions of (2) are bounded. Hence tλ → 0 as λ → ∞. Similarly, (1 − tˆλ ) → 0 as λ → ∞.  Now, we prove Theorem 1.2. By Theorem 1.1, (2) has a positive solution for λ ≫ 1. Note that (2) has a maximal solution, u¯ , since all positive solutions of (2) are bounded above by ρ2 , which is also a supersolution. To prove the uniqueness of the positive solution, u for λ ≫ 1, we will show that u ≡ u¯ . Since u¯ and u satisfy (2),

  −(¯u − u)′′ (t ) = λh(t ) f (¯u(t )) − f (u(t )) ,

0
(7)

(¯u − u)(0) = (¯u − u)(1) = 0. By the mean value theorem there exists ξ such that u ≤ ξ ≤ u¯ in [0, 1] and

−(¯u − u)′′ (t ) = λh(t )f ′ (ξ )(¯u(t ) − u(t )), (¯u − u)(0) = (¯u − u)(1) = 0.

0
(8)

Multiplying (2) by (¯u − u), (8) by u, and integrating,

λ

 1



f (u) − f ′ (u)u h(s)(¯u − u)ds ≤ 0.

(9)

0

˜ + = (tλ , tˆλ ) and Ω ˜ − = (0, 1) − Ω ˜ + , where tλ is as in Lemma 3.4. Since f ′ (s) ≤ 0 Here we also used the concavity of f . Let Ω ˜ + . Also, since f is concave, f (z ) − f ′ (z )z ≥ f (0) for for s > ρ˜ , there exists a constant a > 0 such that f (z ) − f ′ (z )z > a in Ω all z > 0. Thus  ˜+ Ω

ah(s)(¯u − u)ds +

 ˜− Ω

f (0)h(s)(¯u − u)ds ≤ 0.

(10)

˜ − | → 0 as λ → ∞. Also, using the facts that (¯u − u) is bounded and h(s) is positive and integrable, we By Lemma 3.4, |Ω see that (10) is true only if (¯u − u) ≡ 0. Appendix In this appendix we provide the proof of Lemma 1.4. Set I = {t ∈ B : wt ≤ u in [0, 1]}. I is nonempty as wt0 ≤ u in [0, 1]. We will show that I is both closed and open. Then the connectedness of B would imply that I = B. Clearly, I is closed, since t → wt is continuous with respect to ∥ · ∥∞ . In order to show  1 that I is open, we will prove that every point in I is an interior point. Let t ∈ I be given. Then 1 ′′ −(w − u )φ ≤ λ h(x)[f (wt ) − f (u)]φ, for every φ ∈ V and wt (x) − u(x) < −ξt for some ξt > 0 at x = 0, 1. Define t 0 0

 f (wt (x)) − f (u(x))   ;  wt (x) − u(x) g (x) =    ∂ f (wt (x)); ∂x

wt (x) ̸= u(x) wt (x) = u(x).

Then 1



−(wt + ξ − u)φ ′′ ≤ λ

1



0

h(x)[g + − g − ][wt + ξ − u]φ − ξ 0

− ξλ

1



φ ′′ 0

1



h(x)[g + − g − ]φ,

∀φ ∈ V , x ∈ [0, 1],

0

and wt (x) + ξ − u(x) < 0 at x = 0, 1 for all ξ < ξt . Rearranging the terms, we have 1



−(wt + ξ − u)φ ′′ − 0

 0

1

h(x)g + (wt + ξ − u)φ ≤ −λ

1



h(x)g − (wt + ξ − u)φ − ξ 0

− ξλ

1



φ ′′ 0

1



h(x)[g + − g − ]φ, 0

L. Sankar et al. / J. Math. Anal. Appl. 401 (2013) 146–153

153

for all φ ∈ V , x ∈ [0, 1]. Now, for ξ small enough, we have 1



−(wt + ξ − u)φ ′′ − 0

1



h(x)g + (wt + ξ − u)φ ≤ 0, 0

∀φ ∈ V , x ∈ [0, 1] and wt (x) + ξ − u(x) < 0 at x = 0, 1. By the weak maximum principle, we have wt (x) + ξ − u(x) ≤ 0 on [0, 1]. Hence wt (x) < u(x) in [0, 1]. This implies that t is in the interior of I. Thus I is both closed and open, and hence I = B; i.e., wt ≤ u for all t ∈ B. References [1] I. Ali, A. Castro, R. Shivaji, Uniqueness and stability of positive solutions for semipositone problems in a ball, Proc. Amer. Math. Soc. 117 (1993) 775–782. [2] A. Ambrosetti, D. Arcoya, B. Biffoni, Positive solutions for some semipositone problems via bifurcation theory, Differential Integral Equations 7 (3) (1994) 655–663. [3] V. Anuradha, D.D. Hai, R. Shivaji, Existence results for superlinear semipositone boundary value problems, Proc. Amer. Math. Soc. 124 (3) (1996) 757–763. [4] A. Castro, S. Gadam, R. Shivaji, Positive solution curves of semipositone problems with concave nonlinearities, Proc. Roy. Soc. Edinburgh Sect. A 127 (5) (1997) 921–934. [5] A. Castro, J.B. Garner, R. Shivaji, Existence results for classes of sublinear semipositone problems, Results Math. 23 (1993) 214–220. [6] A. Castro, M. Hassanpour, R. Shivaji, Uniqueness of non-negative solutions for a semipositone problem with concave nonlinearity, Comm. Partial Differential Equations 20 (11–12) (1995) 1927–1936. [7] A. Castro, L. Sankar, R. Shivaji, Uniqueness of nonnegative solutions for semipositone problems on exterior domains, J. Math. Anal. Appl. 394 (1) (2012) 432–437. [8] A. Castro, R. Shivaji, Nonnegative solutions for a class of nonpositone problems, Proc. Roy. Soc. Edinburgh Sect. A 108 (1998) 291–302. [9] A. Castro, R. Shivaji, Nonnegative solutions for a class of radially symmetric nonpositone problems, Proc. Amer. Math. Soc. 106 (3) (1989) 735–740. [10] A. Castro, R. Shivaji, Positive solutions for a concave semipositone Dirichlet problem, Nonlinear Anal. 31 (1–2) (1998) 91–98. [11] E. Norman Dancer, Junping Shi, Uniqueness and nonexistence of positive solutions to semipositone problems, Bull. London Math. Soc. 38 (6) (2006) 1033–1044. [12] P. Habets, F. Zanolin, Upper an lower solutions for a generalized Emden–Fowler equation, J. Math. Anal. Appl. 181 (1994) 684–700. [13] E. Lee, L. Sankar, R. Shivaji, Positive solutions for infinite semipositone problems on exterior domains, Differential Integral Equations 24 (2011) 861–875. [14] I. Philippe Clement, Guido Sweers, Existence and multiplicity results for a semilinear elliptic eigenvalue problem, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 14 (1) (1987) 97–121. [15] J. Serrin, Nonlinear equations of second order, in: A.M.S. Sympos. Partial Differential Equations, Berkeley, August 1971.