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Uniqueness of positive radial solutions for infinite semipositone p-Laplacian problems in exterior domains
Accepted Manuscript Uniqueness of positive radial solutions for infinite semipositone p-Laplacian problems in exterior domains
K.D. Chu, D.D. Hai, R. Shivaji
PII: DOI: Reference:
S0022-247X(18)30975-2 https://doi.org/10.1016/j.jmaa.2018.11.037 YJMAA 22723
To appear in:
Journal of Mathematical Analysis and Applications
Received date:
26 June 2018
Please cite this article in press as: K.D. Chu et al., Uniqueness of positive radial solutions for infinite semipositone p-Laplacian problems in exterior domains, J. Math. Anal. Appl. (2018), https://doi.org/10.1016/j.jmaa.2018.11.037
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Uniqueness of positive radial solutions for infinite semipositone p-Laplacian problems in exterior domains K. D. Chu Faculty of Mathematics and Statistics Ton Duc Thang University Ho Chi Minh City, Vietnam Email: [email protected] D. D. Hai Department of Mathematics and Statistics Mississippi State University Mississippi State, MS 39762, USA Email:[email protected] R. Shivaji Department of Mathematics and Statisitics University of North Carolina at Greensboro Greensboro, NC 27402, USA Email:[email protected] November 13, 2018 Abstract We prove uniqueness and asymptotic behavior of positive radial solutions to the p-Laplacian problem −Δp u = λK(|x|)f (u) in |x| > r0 , u = 0 on |x| = r0 , u(x) → 0 as |x| → ∞.
1
where Ω = {x ∈ Rn : |x| > r0 > 0}, n > p, f : (0, ∞) → R is continuous, f (u) ∼ uq at ∞ for some q ∈ [0, p − 1) with possible infinite semipositone structure at 0, and λ is a large parameter.
1. INTRODUCTION Consider the boundary value problem ⎧ ⎨ −Δp u = λK(|x|)f (u) in Ω, u = 0 on |x| = r0 , ⎩ u(x) → 0 as |x| → ∞,
(1.1)
where Δp u = div(|∇u|p−2 ∇u), 1 < p < n, Ω = {x ∈ Rn : |x| > r0 > 0}, K : [r0 , ∞) → (0, ∞), f : (0, ∞) → R are continuous, and λ is a positive parameter. n−p By letting r = |x| and t = (r/r0 ) 1−p , we see that radial solutions of (1.1) satisfy the two-point boundary value problem −(φ(u )) = λh(t)f (u), t ∈ (0, 1), (1.2) u(0) = u(1) = 0, 1−p p p p(1−n) p−1 p−2 n−p where φ(s) = |s| s and h(t) = n−p r0 t K r0 t n−p . We are motivated by the uniqueness results in [1,27] in the finite semi(u) positone, sublinear case i.e. f (0) < 0 and limu→∞ ufp−1 = 0. In [1], Castro et al. obtained uniqueness of non-negative solutions to (1.2) for λ large when p = 2, f > 0 on (0, ∞), f is concave with limu→∞ f (u) = ∞, and h is positive with h < 0 and tα h(t) bounded near 0 for some α ∈ (0, 1). Uniqueness for positive solutions to (1.2) in the general case p > 1 for λ large was established in [23] when f > 0 on (0, ∞), limu→∞ f (u) = ∞, and fu(u) is nonincreasing q for u large for some q ∈ (0, p − 1). Note that the proofs in [1,27] depend heavily on the fact that f (0) is finite and can not be applied to the infinite semipositone case i.e. limu→0+ f (u) = −∞. (see figure 1.1). In this paper, we will prove uniqueness and asymptotic behavior of positive solutions to (1.2) when λ is large. In fact, this is the first paper in the literature dealing with this challenging uniqueness analysis for infinite semipositone problems. (We 2010 Mathematics Subject Classification: Primary 34B16; Secondary 34B15. Key words and phrases: infinite semipositone, positive solutions, uniqueness
2
Figure 1.1 Figure 1: * note that when h(t) ≡ constant, the problem can be analyzed by time-map analysis, but here we do not assume h is a constant). We focus on reaction terms f that satisfy f (u) ∼ uq at ∞ for some q ∈ [0, p − 1), which allows f to be bounded and we do not require any monotonicity assumption on f as in [1,27]. In particular, our results when applied to the model case 1 λ 1 q 1+u −(φ(u )) = tη − uα + u e , t ∈ (0, 1), (1.3) u(0) = u(1) = 0, where 0 ≤ η, α, η + α < 1, 0 ≤ q < p − 1, give the uniqueness of positive solutions to (1.3) when λ is large. Furthermore, the unique solution uλ of uλ (1.3) satisfies lim = 1 uniformly for t ∈ (0, 1), where wq denotes 1 λ→∞ λ p−1−q wq
the unique positive solution of −(φ(w )) = t−η wq , t ∈ (0, 1), w(0) = w(1) = 0. It should be noted that our proof can be applied to include cases when f (u) do not behave like powers of u (see Remark 1.1 (iii)). We refer to [2,7] for uniqueness results to the sublinear semilinear problem −Δu = λf (u) in Ω, (1.4) u = 0 on ∂Ω, 3
when λ is large in the semipositone case, and to [3,6,17,18,24,26,30] to the positone case. Uniqueness results to related quasilinear Dirichlet problems were discussed in [9-11,13] for the positone case on a bounded domain, and in [15] for the semipositone and the infinite semipositone case on a ball. Uniqueness of positive solutions for quasilinear semipositone problems on a bounded domain is still an open question. For existence results for λ >> 1 in the semipositone and the infinite semipositone case, see [15,16,22,23]. Related results to (1.4) when f is singular at 0 can be found in [4,5,8,12,19,20,21,28,29]. We shall make the following assumptions: (A1) h : (0, 1] → (0, ∞) is differentiable, nonincreasing, and there exists a constant η ∈ [0, 1) such that lim sup tη h(t) < ∞. t→0+
(A2) f : (0, ∞) → R is differentiable and there exists a constant β > 0 such that (z − β)f (z) > 0 for z = β. (A3) There exist numbers L > 0 and 0 ≤ q < p − 1 such that lim
z→∞
and lim sup z→∞
f (z) =L zq
z|f (z)| < p − 1. f (z)
(A4) There exist constants l < 0 and 0 ≤ α < 1 − η such that lim z α f (z) = l and lim sup z α+1 |f (z)| < ∞.
z→0+
z→0+
By a solution of (1.2), we mean a function u ∈ C 1 [0, 1] such that φ(u ) is differentiable on (0, 1] and satisfies (1.2). Our main result is Theorem 1.1. Let (A1)-(A4) hold. Then there exists a constant λ0 > 0 such that problem (1.2) has a unique positive solution uλ for λ > λ0 . Furthermore, uλ (t) lim =1 1 λ→∞ (λL) p−1−q w (t) q 4
uniformly for t ∈ (0, 1), where wq is the unique solution of −(φ(w )) = h(t)wq in (0, 1), w(0) = w(1) = 0. Remark 1.1. (i) Note that (A4) is satisfied if limz→0+ z α+1 f (z) = l1 ∈ (0, ∞). (ii) A key ingredient of the proof of Theorem 1.1 is to obtain the estimate 1 u(t) ≥ C0 λ p−1−q min(t, 1 − t) for λ large in Lemma 2.5 below. This estimate requires the assumptions (A2),(A4), and the fact that lim inf f (z) > 0 (could z→∞
be finite). It will then be used to obtain sharp upper and lower bound estimates for positive solutions of (1.2) when f (u) ∼ uq at ∞ for some q ∈ [0, p − 1). Next, we establish the existence of a positive constant c independent of u such that cv ≤ u ≤ c−1 v for any two positive solutions of (1.2), and then show that c = 1. (iii) The steps in (ii) can be used with suitable modifications to obtain the uniqueness result in Theorem 1.1 under the assumptions (A2),(A4), and (A5) There exist constants a > 0 and q ∈ [0, p − 1) such that f (z) is nondecreasing and fz(z) is nonincreasing on [a, ∞). q This allows the case when f (u) do not behave like powers of u at ∞ e.g. f (u) = − u1α + uq ln(1 + u). 2. PRELIMINARY RESULTS z We shall denote the norm in Lp (0, 1) by ||.||p . Let F (z) = 0 f (s)ds. By (A2) and (A3), there exists a number θ > β such that F (z) < 0 for 0 < z < θ and F (z) > 0 for z > θ. We first extend [27, Lemma 2.1] to include the infinite semipositone case with h not necessarily strictly decreasing. Lemma 2.1. Let u be a positive solution of (1.2). Then u has only one interior maximum at t0 and u(t0 ) ≥ θ. Proof. Using the equation in (1.2) and (A2), we see that φ(u ) is increasing for t near 0. Hence φ(u (t)) > φ(u (0)) ≥ 0 i.e. u (t) > 0 for t near 0. Let t0 ∈ (0, 1) be the first zero of u . Note that by (A4), lim+ uα−1 (t)F (u(t)) = t→0
l(1−α)−1 , and since |u(t)| ≤ ||u ||∞ t, it follows from (A1) that tη+α−1 h(t)F (u) is bounded near t = 0. Since η + α < 1, lim+ h(t)F (u(t)) = 0. Suppose t→0
u(t0 ) < θ. Then h F (u) ≥ 0 on [0, t0 ]. Hence, by multiplying the equation in
5
(1.2) by u and integrating on [0, t0 ], we obtain
t0 1 p |u (0)| = λh(t0 )F (u(t0 )) − λ h (t)F (u)dt 0≤ 1− p 0 ≤ λh(t0 )F (u(t0 )) < 0, a contradiction. Thus u(t0 ) ≥ θ. Next, we claim that u is decreasing on (t0 , 1). Suppose to the contrary that there exist t1 , t2 ∈ (t0 , 1) with t1 < t2 such that u(t1 ) ≤ u(t2 ). Since (φ(u )) (t0 ) = −λh(t0 )f (u(t0 )) < 0, it follows that u is decreasing for t near t0 , t > t0 . Hence u has a minimum value on [t0 , t2 ] at some t∗ ∈ (t0 , t2 ). This implies λh(t∗ )f (u(t∗ )) = −(φ(u )) (t∗ ) ≤ 0 i.e. 0 < u(t∗ ) ≤ β and so F (u(t∗ )) < 0. Since F (u(t0 )) ≥ 0, there exists tˆ ∈ [t0 , t∗ ) such that F (u) < 0 on (tˆ, t∗ ] and F (u(tˆ)) = 0. Multiplying the equation in (1.2) by u and integrating on [tˆ, t∗ ] gives
t∗ 1 ˆ p ∗ ∗ |u (t)| = λh(t )F (u(t )) − λ h (t)F (u)dt < 0, 0≤ 1− p tˆ a contradiction. Hence u is increasing on (0, t0 ) and decreasing on (t0 , 1), which completes the proof. Let ρ = β+θ . By Lemma 2.1, for each μ ∈ (0, θ] there exist tμ ≤ t0 ≤ t˜μ 2 such that u(tμ ) = u(t˜μ ) = μ. Lemma 2.2. There exists a constant cρ > 0 such that any positive solution u of (1.2) satisfies ⎧ p t 1/p p−1+α ⎪ ⎪ h if t ∈ [0, tρ ], ⎪ 0 ⎨ u(t) ≥ 1 ⎪ p ⎪ cρ λ p−1+α 1 1/p p−1+α ⎪ ⎩ h if t ∈ [t˜ρ , 1]. t In particular, there exists a constant c˜ρ > 0 independent of u and λ such 1 that max{tρ , 1 − t˜ρ } ≤ c˜ρ λ− p . Proof. Note that h F (u) ≥ 0 on [0, tθ ] ∪ [t˜θ , 1]. Let s ∈ [0, tθ ]. Multiplying the equation in (1.2) by u and integrating on [0, s] gives
s 1 1 p p (u (s)) = 1 − (u (0)) − λh(s)F (u) + λ h (τ )F (u)dτ 1− p p 0 6
≥ −λh(s)F (u). Hence
u (s) 1/p
≥ (λh(s))1/p ,
(−F (u)) which implies upon integrating on [0, t], t ∈ [0, tθ ], that t u(t) dz 1/p ≥λ h1/p . 1/p (−F (z)) 0 0
(2.1)
By (A4), there exists a constant k > 0 such that −F (z) ≥ kz 1−α for z ∈ [0, ρ]. Hence (2.1) implies t p−1+α p−1 1/p p u (λk) (t) ≥ h1/p , p 0 p t 1 p−1+α for t ∈ [0, tρ ] follows, where cρ = from which u(t) ≥ cρ λ p−1+α 0 h1/p p p−1+α p−1 1/p k . In particular, since h is nonincreasing, p 1
p
ρ = u(tρ ) ≥ cρ (λh(1)) p−1+α tρp−1+α , p−1+α 1 p which implies tρ ≤ c˜ρ λ−1/p , where c˜ρ = ρ/(cρ h p−1+α (1)) . Next, let s ∈ [t˜θ , 1]. Since F (θ) = 0, it follows from multiplying the equation in (1.2) by u and integrating on [t˜θ , s] that
1 1− |u (s)|p ≥ −λh(s)F (u). p Hence
|u (s)| (−F (u))
1/p
≥ (λh(s))1/p ,
and since u ≤ 0 on [t˜θ , 1], it follows upon integrating on [t, 1], t ∈ [t˜θ , 1], that 1 u(t) dz 1/p ≥λ h1/p 1/p (−F (z)) 0 t p p−1+α 1 1 and as in the above argument, we get u(t) ≥ cρ λ p−1+α t h1/p for t ∈ [t˜ρ , 1] follows. By choosing t = t˜ρ , we conclude as in the above that 1 − t˜ρ ≤ c˜ρ λ−1/p , which completes the proof. 7
The next result establishes a comparison principle when the forcing terms are close in L1 −norm, but not necessarily ordered. Lemma 2.3. Let M > 0 and hi ∈ L1 (0, 1) with ||hi ||1 < M for i = 1, 2. Let u˜i ∈ C 1 [0, 1], i = 1, 2, satisfy −(φ(˜ u1 )) ≥ h1 , t ∈ (0, 1), −(φ(˜ u2 )) ≤ h2 , t ∈ (0, 1), , u˜1 (0) = u˜1 (1) = 0, u˜2 (0) = u˜2 (1) = 0. Then (i) If h2 ≥ 0, h2 ≡ 0 then for a given ε0 > 0, there exists a constant δ > 0 such that u˜1 ≥ (1 − ε0 )˜ u2 on [0, 1], provided that ||h1 − h2 ||1 < δ. (ii) Let tˇ < 1/2 < tˆ . Suppose u˜2 defined above satisfies −(φ(˜ u2 )) = h2 on (0, 1) and there exists a constant C > 0 such that |˜ u2 | > C on [0, tˇ] ∪ ˜ [tˆ, 1]. Then there exist constants δC , kC > 0 such that if ||h1 − h2 ||1 < δ˜C then u˜1 (t) ≥ u˜2 (t) − kC ||h1 − h2 ||1 p(t) for t ∈ [0, tˇ] ∪ [tˆ, 1], where p(t) = min(t, 1 − t). Proof. Let ui ∈ C 1 [0, 1], i = 1, 2, satisfy −(φ(ui )) = hi , t ∈ (0, 1), ui (0) = ui (1) = 0. Then u˜1 ≥ u1 and u˜2 ≤ u2 on [0, 1] by the weak comparison principle (see [25, Lemma A.2]). Let ε > 0. We first show that there exists a constant δ > 0 such that ||h1 − h2 ||1 < δ =⇒ ||u1 − u2 ||∞ < ε. (2.2) By integrating, we obtain ui (t)
−1
=φ
Ci −
t 0
hi ,
1 s where Ci are constants satisfying 0 φ−1 Ci − 0 hi ds = 0, i = 1, 2. Using the fact that φ−1 is increasing with φ−1 (0) = 0, we deduce that |Ci | ≤ ||hi ||1 and |C1 − C2 | ≤ ||h1 − h2 ||1 . 8
(2.3)
Since φ−1 is uniformly continuous on [−2M, 2M ], there exists δ0 > 0 such that x, y ∈ [−2M, 2M ], |x − y| < δ0 =⇒ |φ−1 (x) − φ−1 (y)| < ε. Suppose ||h1 − h2 ||1 < δ0 /2. Then in view of (2.3), t < δ0 , C1 − C2 − (h − h ) 1 2 0
t and since Ci − 0 hi ≤ 2M, it follows that |u1 (t) − u2 (t)| < ε
(2.4)
for all t ∈ [0, 1] i.e. (2.2) holds with δ = δ0 /2. (i) If h2 ≥ 0, h2 ≡ 0 then u2 (t) ≥ ||u2 ||∞ p(t) for t ∈ [0, 1] (see e.g. [14, Lemma 3.4]. Let ε < ε0 ||u2 ||∞ . Then there exists δ > 0 such that (2.4) holds provided that ||h1 − h2 ||1 < δ. Since ui (0) = ui (1) = 0, the Mean Value Theorem gives u1 (t) ≥ u2 (t) − εp(t) ≥ (1 − ε0 )u2 (t) for t ∈ [0, 1], and therefore u˜1 ≥ (1 − ε0 )˜ u2 on [0, 1]. (ii) Suppose u˜2 satisfies −(φ(˜ u2 )) = h2 on (0, 1) and |˜ u2 | > C > 0 on I ∪ J, where I = [0, tˇ], J = [tˆ, 1]. Then u˜2 ≡ u2 and, in view of (2.4), there exists 0 < δ˜ < 12 min(φ(C/2), δ0 ) such that |u1 | > C/2 on [0, tˇ] ∪ [tˆ, 1] ˜ which we shall assume. Hence provided that ||h1 − h2 ||1 < δ, t Ci − hi = φ(|ui (t)|) > φ(C/2) > 0 (2.5) 0
for t ∈ I ∪ J and Ci − Theorem, |u1 (t)
−
t
u2 (t)|
0
hi are of the same sign, i = 1, 2. By the Mean Value
t t −1 −1 C1 − C2 − = φ h1 − φ h2 0
2−p
=
|ξ(t)| p−1
t C1 − C2 − 0 (h1 − h2 ) p−1 9
0
,
(2.6)
t t where ξ(t) is between C1 − 0 h1 and C2 − 0 h2 . By (2.3) and (2.5), φ(C/2) ≤ |ξ(t)| ≤ 2M, from which (2.3) and (2.6) gives |u1 (t) − u2 (t)| ≤ kC ||h1 − h2 ||1 , 2−p
2−p
2 for t ∈ I, where kC = p−1 max{(2M ) p−1 , φ p−1 (C/2)}. Since u1 (0) = 0 and u2 (0) = 0, it follows that
u˜1 (t) ≥ u1 (t) ≥ u2 (t) − kC ||h1 − h2 ||1 t for t ∈ I. Similarly, u˜1 (t) ≥ u1 (t) ≥ u2 (t) − kC ||h1 − h2 ||1 (1 − t) for t ∈ J, which completes the proof.
Lemma 2.4. (i)
1
(ii)
1
0
h(t) αp dt < ∞. t 1/p p−1+α h 0
1/2
h(t) αp dt < ∞. 1 1/p p−1+α h t
˜ = t−γ h. Then η + γ < p and Proof. (i) Let γ = p − 1 + α and define h 1/p 1 ˜ ∈ L (0, 1) in view of (A1). Note that therefore h t t 1/p γ/p ˜ 1/p . h ≤t h 0
Since h is nonincreasing, h(t)
p−1 p
0
p−1
t −1 1/p ≤ t h . 0
Hence 0
1
h(t) αp dt ≤ t 1/p p−1+α h 0
1
h1/p (t)
t 0
h1/p tp−1
0
10
αp p−1− p−1+α
dt
≤
1
= 0
˜ 1/p h
1
γ˜
t h
0
t 0
1/p
t
˜ 1/p h
t
γ/p 0
˜ 1/p h
αp p−1− p−1+α
αp p−1− p−1+α
dt = Cα,p
1 0
t1−p dt
˜ 1/p h
p(p−1) p−1+α
<∞
by the choice of γ, where Cα,p is a positive constant depending on α, p. (ii) Since h is nonincreasing,
1 1/2
p−1 h(t) dt ≤ (h(1/2)) p αp p−1+α 1 1/p h t
= C˜α,p
1 1/2
h
1/p
1 1/2
αp 1− p−1+α
h1/p (t) αp dt 1 1/p p−1+α h t
< ∞,
where C˜α,p is a positive constant depending on α, p. ˜ > 0 such that for λ > λ, ˜ any Lemma 2.5. There exist constants C0 , λ positive solution u of (1.2) satisfies 1
u(t) ≥ λ p−1 C0 p(t) for t ∈ [0, 1], where p(t) = min(t, 1 − t). Proof. By (A2), (A4), and continuity of z α f (z), there exists a constant a > 0 such that a f (z) ≥ − α (2.7) z for all z > 0. Since u ≥ ρ on [tρ , t˜ρ ], where tρ , t˜ρ are defined immediately preceding Lemma 2.2, it follows that f (u) ≥ δρ > 0 on [tρ , t˜ρ ] for some 1 δρ > 0. Then assuming cρ λ p−1+α > 1, (2.7) and Lemma 2.2 give ⎧ ah(t) ⎪ ⎪ − if t ∈ [0, tρ ] αp ⎪ ⎪ t 1 p−1+α ⎪ ⎪ hp 0 ⎨ ⎪ 1 − p−1 if t ∈ (tρ , t˜ρ ) ≡ h1 , δρ h(t) − φ λ u ≥ (2.8) ⎪ ⎪ ah(t) ⎪ ⎪ − if t ∈ [t˜ρ , 1] ⎪ αp ⎪ 1 1 p−1+α ⎪ ⎩ p h t
11
Defining the right-hand side of (2.8) as h1 . Then h1 ∈ L1 (0, 1) by Lemma 2.4. Let ω ∈ C 1 [0, 1] satisfy −(φ(ω )) = δρ h(t) ≡ h2 , t ∈ (0, 1), (2.9) ω(0) = ω(1) = 0. Then ω(t) ≥ ||ω||∞ p(t) for t ∈ [0, 1] (see [14, Lemma 3.4]). Note that 1 tρ tρ h(t) h+ h +a dt ||h1 − h2 ||1 = δρ αp 1 p−1+α t 0 0 t˜ρ p h 0 +a
1 t˜ρ
h(t) αp dt → 0 as λ → ∞ 1 1 p−1+α p h t
since max{tρ , 1 − t˜ρ } → 0 as λ → ∞ by Lemma 2.2. Hence it follows by 1 applying Lemma 2.3 (i) with u˜1 = λ− p−1 u, u˜2 = w, and ε0 = 1/2 that for λ large enough, 1 λ− p−1 u ≥ ω/2 ≥ (||ω||∞ /2)p(t), which completes the proof. Lemma 2.6. Let 0 < q < p − 1. Then there exists a unique positive solution wq ∈ C 1 [0, 1] to the problem −(φ(w )) = h(t)wq , t ∈ (0, 1), (2.10) w(0) = w(1) = 0. Proof. Let ω0 , ω1 ∈ C 1 [0, 1] satisfy −(φ(ω2 )) = h(t), t ∈ (0, 1), −(φ(ω1 )) = ω1q , t ∈ (0, 1), , . ω1 (0) = ω1 (1) = 0, ω2 (0) = ω2 (1) = 0. whose existence follow from [14,Theorem 2.2 (ii)] and [14, Lemma 3.1]. Then it is easy to see that εω1 is a subsolution of (2.10) if ε is small enough, and M ω2 is a supersolution of (2.10) if M is large enough. By decreasing ε and increasing M if necessary, we can assume that εω1 ≤ M ω2 on [0, 1]. Hence (2.10) has a solution wq with wq ≥ εω1 in [0, 1]. To show uniqueness, let w1 , w2 i (t) i (t) be two positive solutions of (2.10). Since sup wp(t) < ∞ and inf wp(t) > t∈(0,1)
12
t∈(0,1)
0 for i = 1, 2, there exists a largest constant c > 0 such that w1 ≥ cw2 in [0, 1]. Suppose c < 1. Then −(φ(w1 )) = h(t)w1q ≥ cq h(t)w2q q
for t ∈ (0, 1), and the weak comparison principle gives w1 ≥ c p−1 w2 in [0, 1], a contradiction with the maximality of c. Thus c ≥ 1 i.e. w1 ≥ w2 , which completes the proof. The asymptotic behavior of the solutions to (1.2) as λ → ∞ is given in the next result. Lemma 2.7. Let ε ∈ (0, 1). Let wq be the unique solution of (2.10) given ¯ > 0 such that for λ > λ, ¯ any by Lemma 2.6. Then there exists a constant λ positive solution u of (1.2) satisfies p
p
1
1
(1 − ε) p−1−q (λL) p−1−q wq ≤ u ≤ (1 + ε) p−1−q (λL) p−1−q wq on [0, 1], where L is defined in (A3). In other words, lim
λ→∞
u(t) 1
(λL) p−1−q wq (t)
=1
uniformly in t ∈ (0, 1). Proof. Let ε ∈ (0, 1). By (A3), there exists a constant A > 0 such that (1 − ε)Lz q < f (z) < (1 + ε)Lz q ˜ where λ ˜ is defined in Lemma 2.5. Since wq (t) ≤ for z > A. Suppose λ > λ, ||wq ||∞ p(t) for t ∈ [0, 1], it follows from Lemma 2.5 that u ≥ d0 wq in [0, 1], 1 ˜ p−1 where d0 = λ C0 ||wq ||−1 ∞ . Let dλ be the largest number such that u ≥ dλ wq in [0, 1]. By (A4), there exists a constant b > 0 such that |f (z)| ≤
b for z ∈ (0, A]. zα
1
Assuming λ p−1 C0 > 1 then Lemma 2.5 gives ⎧ (1 − ε)Ldλq h(t)wqq ⎪ ⎪ ⎨ −(φ(u )) ≥ λ bh(t) ⎪ ⎪ ⎩ − α p (t) 13
if u(t) > A, if u(t) ≤ A.
This implies − φ
u
≥
1
(λ(1 − ε)Ldqλ ) p−1
⎧ h(t)wqq ⎪ ⎪ ⎨
if u(t) > A,
b0 h(t) ⎪ ⎪ ⎩ − α p (t)
if u(t) ≤ A,
≡ h1
b where b0 = (1−ε)Ld q and h1 is defined to be the right-hand side of the previous 0 equation. Since −(φ(wq )) = h(t)wqq ≡ h2 in (0, 1), 1 − p−1 −1 C0 and the measure of {t : u(t) ≤ A} ≤ the measure of t : p(t) ≤ λ → 0 as λ → ∞ by Lemma 2.5, it follows that
b0 q ||h1 − h2 ||1 = dt → 0 h(t) wq + α p (t) u≤A
as λ → ∞. Hence if λ is large enough, Lemma 2.3(i) gives u 1
(λ(1 − ε)Ldλq ) p−1 p
≥ (1 − ε)wq
on [0, 1] p
1
1
i.e. u ≥ (1−ε) p−1 (λLdqλ ) p−1 wq in [0, 1]. Consequently, dλ ≥ (1−ε) p−1 (λLdqλ ) p−1 , and therefore p 1 u ≥ (1 − ε) p−1−q (λL) p−1−q wq on [0, 1]. Next, since sup t∈(0,1)
u(t) p(t)
wq (t) t∈(0,1) p(t)
< ∞ and inf
> 0, there exists a constant d1 ≥ 0
(depending on u) such that u ≤ d1 wq in [0, 1]. Let d˜λ be the smallest number such that u ≤ d˜λ wq in [0, 1]. Then d˜λ ≥ d0 and ⎧ (1 + ε)Ld˜qλ h(t)wqq if u(t) > A, ⎪ ⎪ ⎨ −(φ(u )) ≤ λ bh(t) ⎪ ⎪ ⎩ α if u(t) ≤ A. p (t) Hence − φ
u
1
(λ(1 + ε)Ld˜λq ) p−1
≤
⎧ h(t)wqq ⎪ ⎪ ⎨
if u(t) > A,
b1 h(t) ⎪ ⎪ ⎩ α p (t)
if u(t) ≤ A,
14
˜ 1, ≡h
where b1 = Since
˜ 1 be the righ-hand side of the previous equation. Let h
b 1 q ˜ 1 − h2 ||1 ≤ ||h →0 h(t) wq + α p (t) u≤A
b . (1+ε)Ldq0
p
1 gives u ≤ (1 + ε) p−1 (λLd˜λq ) p−1 wq in p 1 [0, 1] if λ is large enough. Consequently, d˜λ ≤ (1 + ε) p−1 (λLd˜λq ) p−1 , which implies p 1 u ≤ (1 + ε) p−1−q (λL) p−1−q wq in [0, 1],
as λ → ∞, Lemma 2.3(i) with ε0 =
ε ε+1
which completes the proof. Lemma 2.8. There exist positive constants K0 , tˇ , tˆ with tˇ < 1/2 < tˆ ¯ any positive solution u of (1.2) satisfies such that for λ > λ, 1
|u (t)| ≥ K0 λ p−1−q for t ∈ [0, tˇ] ∪ [tˆ, 1]. Proof. By Lemma 2.7, there exist positive constants D1 , D2 independent of u and λ such that 1
1
D1 λ p−1−q wq (t) ≤ u(t) ≤ D2 λ p−1−q wq (t)
(2.11)
¯ Consequently, there exists a constant for t ∈ [0, 1], provided that λ > λ. 1 D3 > 0 depending on D1 , D2 and wq such that u (0) ≥ D3 λ p−1−q and u (1) ≤ 1 −D3 λ p−1−q . By (A2) and (A3), there exists a constant m > 0 such that f (z) ≤ mz q for all z > 0. This, together with (2.11), gives t φ(u (t)) = φ(u (0)) − λ h(s)f (u)ds 0
p−1
≥ D3p−1 λ p−1−q − λm ≥λ
p−1 p−1−q
D3p−1
−
mD2q
t 0
t 0
h(s)uq ds
h(s)wqq ds
15
p−1
λ p−1−q D3p−1 ≥ 2
tˇ for t < tˇ, where tˇ < 1/2 is such that 0 h(s)wqq ds < D3p−1 /(2mD2q ). Hence 1 1 u (t) ≥ K0 λ p−1−q for t ∈ [0, tˇ], where K0 = D3 /2 p−1 . Similarly,
1
φ(u (t)) = φ(u (1)) + λ
h(s)f (u)ds t
≤ ≤λ
p−1 p−1−q
p−1 −D3p−1 λ p−1−q
+
mD2q
for t > tˆ, where tˆ > 1/2 is such that 1
h(s)uq ds ≤
t
−D3p−1
1
+ λm 1
q
h(s)w ds t
1 tˆ
p−1
λ p−1−q D3p−1 ≤− 2
h(s)wqq ds < D3p−1 /(2mD2q ).
Hence u (t) ≤ −K0 λ p−1−q for t ∈ [tˆ, 1], which completes the proof. 3. PROOF OF THE MAIN RESULT In what follows, we shall assume that λ > 1 is large enough so that Lemmas 2.5 and 2.7 apply. Proof of Theorem 1.1. By (A3), there exist positive constants A0 , δ, ε0 , δ1 with δ(1 + ε0 ) < δ1 < p − 1 and A0 > β such that |f (z)| ≤
and (1 − ε0 )
z1 z2
q
δf (z) z
f (z1 ) ≤ (1 + ε0 ) ≤ f (z2 )
(3.1)
z1 z2
q (3.2)
for z, z1 , z2 > A0 . Let u, v be positive solutions of (1.2). Then Lemma 2.7 with ε small gives 1 1 1 3 (λL) p−1−q wq ≤ u, v ≤ (λL) p−1−q wq 2 2
(3.3)
in [0, 1], which implies (1/3)v ≤ u ≤ 3v on [0, 1]. Let c be the largest number such that cv ≤ u ≤ c−1 v on [0, 1] and suppose that c < 1. We claim that there exists a constant L0 > 0 such that f (u) − cp−1 f (v) ≥ L0 (1 − c) 16
(3.4)
for v > A, where A = 3A0 . For v > A, we have u > A0 . Hence (3.1) and the Mean Value Theorem gives |f (u) − f (v)| = |u − v||f (ψ)| ≤
δ(1 − c)vf (ψ) cψ
δ(1 − c)f (ψ) , c2 where ψ is between u and v. Here we have used the fact that cv ≤ ψ ≤ c−1 v. By (3.2),
q ψ f (ψ) ≤ (1 + ε0 ) ≤ (1 + ε0 )c−q , f (v) v ≤
and so |f (u) − f (v)| ≤ for v > A. Since
lim
z→1−
δ1 (1 − c) δ(1 + ε0 )(1 − c) f (v) ≤ f (v) q+2 c cq+2
1 − z p−1 δ1 − 2+q 1−z z
(3.5)
= p − 1 − δ1 ≡ 2k0 > 0,
there exists c˜ ∈ (0, 1) such that 1 − z p−1 δ1 − 2+q ≥ k0 1−z z
(3.6)
for z ≥ c˜. Combining (3.5) and (3.6), we obtain f (u) − cp−1 f (v) = f (u) − f (v) + (1 − cp−1 )f (v)
δ1 1 − cp−1 − q+2 f (v) ≥ k0 (1 − c)f (v) ≥ (1 − c) 1−c c
(3.7)
for v > A if c ≥ c˜. Suppose next that c < c˜. Since c ≥ 1/3 and q < p − 1, there exists a constant m0 > 0 such that cq − cp−1 ≥ m0 (1 − c) > m0 (1 − c˜), from which (3.2) gives
f (u) − c
p−1
f (v) =
f (u) p−1 −c f (v) ≥ (1 − ε0 )cq − cp−1 f (v) f (v) 17
≥ (c − c q
p−1
− ε0 )f (v) ≥
m0 (1 − c˜) 2
f (v) ≥ B(1 − c)
(3.8)
if ε0 is decreased further, where B = (m0 (1 − c˜)/2) inf f. From (3.7), (3.8) (A,∞)
and inf f > 0, we see that (3.4) holds. (A,∞)
For v ≤ A, we have u ≤ 3A, and it follows from (A4) that there exists a constants KA > 0 such that |f (u) − f (v)| = |u − v||f (ξ)| ≤ ≤
KA (1 − c)v c|ξ|α+1
KA (1 − c) KA (1 − c) ≤ , α+2 α c v wqα
and
KA KA ≤ α α w wq
|f (w)| ≤
(3.9)
for w ∈ {u, v}, where we have assumed that λ is large enough (independent of u, v) so that (3.3) gives w ≥ wq and c2+α wα ≥ wqα on (0, 1). Consequently, |f (u) − cp−1 f (v)| ≤ |f (u) − f (v)| + (1 − cp−1 )|f (v)| ≤
˜ A (1 − c) K , wqα
(3.10)
˜ A = 2KA max{1, p − 1}. By (3.3) and (A3), there exists a constant where K L1 > 0 depending on L such that q
f (v) ≤ mv q ≤ L1 λ p−1−q
(3.11)
on [0, 1] for some m > 0. Hence (3.4) gives
L0 (1 − c) p−1 −(φ(u )) ≥ λh(t) c f (v) + q L1 λ p−1−q for v > A, which implies ⎧ q − p−1−q ⎪ λ h(t) cp−1 + ⎨ − (φ (˜ u)) ≥ ⎪ q ⎩ − p−1−q λ h(t)f (u)
L2 (1−c) q λ p−1−q
18
) f (v) if v > A, if v < A.
≡ h1 , (3.12)
1
where u˜ = λ− p−1−q u and L2 = L0 /L1 . Define h1 to be the right-hand side of (3.12). 1 p−1 q 1 Let v˜ = λ− p−1−q cp−1 + L2 (1 − c)λ− p−1−q v. Then
− (φ(˜ v )) = λ
q − p−1−q
p−1
h(t) c
+
L2 (1 − c)
f (v) ≡ h2 .
q
λ p−1−q
(3.13)
in [0, 1]. Note that, in view of (3.11), there exists a constant M0 independent of u, v, λ, such that |hi (t)| ≤ M0 h(t) for t ∈ (0, 1), i = 1, 2. By (3.9) and (3.10), q (1 − c) L 2 − p−1−q ||h1 − h2 ||1 = λ h(t) f (u) − cp−1 f (v) − f (v) dt q λ p−1−q v
≤λ
(1 − c)L3 v
h dt, wqα
(3.14)
˜ A + L2 KA . Let tˇ , tˆ be given by Lemma 2.8. Then |˜ where L3 = K v (t)| ≥ cK0 ≥ K0 /3 ≡ C for t ∈ 0, tˇ ∪ [tˆ, 1]. Since ||h1 − h2 ||1 → 0 as λ → ∞, it follows from (3,12), (3.13) and Lemma 2.3(ii) that for λ sufficiently large, u˜(t) ≥ v˜(t) − kC ||h1 − h2 ||1 p(t) for t ∈ [0, tˇ] ∪ [tˆ, 1]. This, together with (2.11), gives 1 p−1 q 1 u(t) ≥ cp−1 + L2 (1 − c)λ− p−1−q v − λ p−1−q kC ||h1 − h2 ||1 p(t)
≥
p−1
c
q − p−1−q
+ L2 (1 − c)λ
1 p−1
˜ − kC ||h1 − h2 ||1 v,
for t ∈ [0, tˇ] ∪ [tˆ, 1], where k˜C = (kC /D1 ) sup(p/wq ). Since (0,1)
q
(1/3)p−1 ≤ cp−1 , cp−1 + L2 (1 − c)λ− p−1−q ≤ 1 + L2 ,
19
(3.15)
it follows from the Mean Value Theorem that 1 p−1 1 q q − p−1−q p−1 + L2 (1 − c)λ − cp−1 p−1 ≥ L4 (1 − c)λ− p−1−q , c
(3.16)
where L4 is a positive constant depending only on p and L2 . Since v
h L4 (1 − c)v dt v ≥ q α wq 2λ p−1−q
L4 − k˜C L3 q λ p−1−q v 0 such that ≥
u ≥ (c + γ)v
on [0, tˇ] ∪ [tˆ, 1]. (3.17) In particular, u(τ ) ≥ cv(τ )+γ0 for τ ∈ tˇ , tˆ , where γ0 = γ min(v(tˇ), v(tˆ)) > 0. By (2.11), 1 v(t) ≥ D1 λ p−1−q min wq > A [tˇ,tˆ]
for t ∈ [tˇ, tˆ], provided that λ is large enough. Hence (3.4) gives −(φ(u )) = λh(t)f (u) ≥ λcp−1 h(t)f (v) on (tˇ, tˆ). Since
−(φ(cv + γ0 ) ) = λcp−1 h(t)f (v) on (tˇ, tˆ), and u(τ ) ≥ cv(τ ) + γ0 for τ ∈ tˇ , tˆ , the weak comparison principle implies u ≥ cv + γ0
on [tˇ, tˆ].
This, together with (3.17), gives the existence of a number γ1 > 0 such that u ≥ (c + γ1 )v in [0, 1]. By symmetry, there exists a constant γ2 > 0 such that v ≤ (c + γ2 )u in [0, 1]. Hence (c + γ˜ )v ≤ u ≤ (c + γ˜ )−1 v in [0, 1], a contradiction with the definition of c. Thus c ≥ 1 i.e. u = v in [0, 1]. The asymptotic behavior of the solution as λ → ∞ is given in Lemma 2.7. This completes the proof of Theorem 1.1. Ackowledgements. The authors thank the referee for carefully reading the manuscript and providing helpful suggestions. 20
References [1] A. Castro, L. Sankar, and R. Shivaji, Uniqueness of non-negative solutions for semipositone problems on exterior domains, J. Math. Anal. Appl. 394 (2012), 432-437. [2] A. Castro, M. Hassanpour, and R. Shivaji, Uniqueness of non-negative solutions for a semipositone problem with concave nonlinearities, Comm. Partial Differential Equations 20 (1995), 1927-1936. [3] A. Castro and R. Shivaji, Uniqueness of positive solutions for a class of elliptic boundary value problems, Proc. Roy. Soc. Edinburgh A 98A (1984), 267-269. [4] M. M. Coclite and G. Palmieri, On a singular nonlinear Dirichlet problem, Comm. Partial Differential Equations 14 (1989), 1315-1327. [5] M. G. Crandall, P. H. Rabinowitz, and L. Tartar, On a Dirichlet problem with a singular nonlinearity, Comm. Partial Differential Equations 2 (1977), 193-222. [6] E. N. Dancer, On the number of positive solutions of semilinear elliptic systems, Proc. London Math. Soc. 53 (1986), 429-452. [7] E. N. Dancer and J. Shi, Uniqueness and nonexistence of positive solutions to semipositone problems, Bull. London Math. Soc. 38 (2006), no. 6, 1033–1044. [8] J. I. Diaz, J. M. Morel, and L. Oswald, An elliptic equation with singular nonlinearity, Comm. Partial Differential Equations 2 (1997), 193-222. [9] J. I. Diaz and J. E. Saa, Existence et unicite de solutions positives pur certaines equations elliptiques quasilineares, Compte Rendus Acad. Sci. Paris 305 (1987), 521-524. [10] P. Drabek and J. Hernandez, Existence and uniqueness of positive solutions for some quasilinear elliptic problems, Nonlinear Anal. 27 (1996), 229-247.
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[11] Z. M. Guo and J. R. L. Webb, Uniqueness of positive solutions for quasilinear elliptic equations when a parameter is large, Proc. Roy. Soc. Edinburgh 124 A (1994), 189-198. [12] W. Fulks and J. S. Maybee, A singular nonlinear equation, Osaka J. Math 12 (1960), 1-19. [13] D. D. Hai, Uniqueness of positive solutions for a class of quasilinear problems, Nonlinear Anal. 69 (2008), 2720-2732. [14] D. D. Hai, On singular Sturm-Liouville boundary value problems, Proc. Roy. Soc. Edinburgh 140A (2010), 49-63. [15] D. D. Hai and R. Shivaji, Existence and uniqueness for a class of quasilinear elliptic boundary value problems, J. Differential Equations 193 (2003), 500-510. [16] D. D. Hai and R. Shivaji, An existence result on positive solutions for a class of p-Laplacian systems, Nonlinear Anal. 56 (2004), 1007-1010. [17] D. D. Hai and R. C. Smith, Uniqueness for singular semilinear elliptic boundary value problems, Glasgow Math. J. 55 (2013), 399-409. [18] D. D. Hai and R. C. Smith, On uniqueness for a class of nonlinear boundary value problems, Proc. Roy. Soc. Edinburgh A 136A (2006), 779-784. [19] J. Hernandez, J. Karatson, and P. L. Simon, Multiplicity for semilinear elliptic equations involving singular nonlinearity, Nonlinear Anal. 65 (2006), 265-283. [20] J. Hernandez and F. J. Mancebo, Singular elliptic and parabolic equations, in Handbook of Differential Equations: Stationary Partial Differential Equations, (2006), 317-400. [21] S. Karlin and L. Nirenberg, On a theorem of P. Nowosad, J. Math. Anal. Appl. 17 (1967), 61-67. [22] E. K. Lee, L. Sankar, and R. Shivaji, Positive solutions for infinite semipositone problems on exterior domains, Differential Integral Equations 24 (2011), 861-875. 22
[23] E. K. Lee, R. Shivaji, and J. Ye, Classes of infinite semipositone systems, Proc. Roy. Soc. Edinburgh A 139A (2009), 853-865. [24] S. S. Lin, On the number of positive solutions for nonlinear elliptic equations when a parameter is large, Nonlinear Anal. 16 (1991), 283297. [25] S. Sakaguchi, Concavity properties of solutions to some degenerate quasilinear elliptic Dirichlet problems, Ann. Scuola Norm. Sup. Pisa Cl. Sci (4) 14 (1987), 403-421. [26] R. Shivaji, Uniqueness results for a class of positone problems, Nonlinear Anal. 7 (1983), 223-230. [27] R. Shivaji, I. Sim, and B. Son, A uniqueness result for a semipositone p-Laplacian problem on the exterior of a ball, J. Math. Anal. Appl. 445 (2017), 459-475. [28] C. A. Stuart, Existence and approximation of solutions of nonlinear elliptic equations, Math. Z. 147 (1976), 53-63. [29] C. A. Stuart, Existence theorems for a class of nonlinear integral equations, Math. Z. 137 (1974), 49-66. [30] M. Wiegner, A uniqueness theorem for some nonlinear boundary value problems with a large parameter, Math. Ann. 270 (1985), 401-402.
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K.D. Chu, D.D. Hai, R. Shivaji
PII: DOI: Reference:
S0022-247X(18)30975-2 https://doi.org/10.1016/j.jmaa.2018.11.037 YJMAA 22723
To appear in:
Journal of Mathematical Analysis and Applications
Received date:
26 June 2018
Please cite this article in press as: K.D. Chu et al., Uniqueness of positive radial solutions for infinite semipositone p-Laplacian problems in exterior domains, J. Math. Anal. Appl. (2018), https://doi.org/10.1016/j.jmaa.2018.11.037
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Uniqueness of positive radial solutions for infinite semipositone p-Laplacian problems in exterior domains K. D. Chu Faculty of Mathematics and Statistics Ton Duc Thang University Ho Chi Minh City, Vietnam Email: [email protected] D. D. Hai Department of Mathematics and Statistics Mississippi State University Mississippi State, MS 39762, USA Email:[email protected] R. Shivaji Department of Mathematics and Statisitics University of North Carolina at Greensboro Greensboro, NC 27402, USA Email:[email protected] November 13, 2018 Abstract We prove uniqueness and asymptotic behavior of positive radial solutions to the p-Laplacian problem −Δp u = λK(|x|)f (u) in |x| > r0 , u = 0 on |x| = r0 , u(x) → 0 as |x| → ∞.
1
where Ω = {x ∈ Rn : |x| > r0 > 0}, n > p, f : (0, ∞) → R is continuous, f (u) ∼ uq at ∞ for some q ∈ [0, p − 1) with possible infinite semipositone structure at 0, and λ is a large parameter.
1. INTRODUCTION Consider the boundary value problem ⎧ ⎨ −Δp u = λK(|x|)f (u) in Ω, u = 0 on |x| = r0 , ⎩ u(x) → 0 as |x| → ∞,
(1.1)
where Δp u = div(|∇u|p−2 ∇u), 1 < p < n, Ω = {x ∈ Rn : |x| > r0 > 0}, K : [r0 , ∞) → (0, ∞), f : (0, ∞) → R are continuous, and λ is a positive parameter. n−p By letting r = |x| and t = (r/r0 ) 1−p , we see that radial solutions of (1.1) satisfy the two-point boundary value problem −(φ(u )) = λh(t)f (u), t ∈ (0, 1), (1.2) u(0) = u(1) = 0, 1−p p p p(1−n) p−1 p−2 n−p where φ(s) = |s| s and h(t) = n−p r0 t K r0 t n−p . We are motivated by the uniqueness results in [1,27] in the finite semi(u) positone, sublinear case i.e. f (0) < 0 and limu→∞ ufp−1 = 0. In [1], Castro et al. obtained uniqueness of non-negative solutions to (1.2) for λ large when p = 2, f > 0 on (0, ∞), f is concave with limu→∞ f (u) = ∞, and h is positive with h < 0 and tα h(t) bounded near 0 for some α ∈ (0, 1). Uniqueness for positive solutions to (1.2) in the general case p > 1 for λ large was established in [23] when f > 0 on (0, ∞), limu→∞ f (u) = ∞, and fu(u) is nonincreasing q for u large for some q ∈ (0, p − 1). Note that the proofs in [1,27] depend heavily on the fact that f (0) is finite and can not be applied to the infinite semipositone case i.e. limu→0+ f (u) = −∞. (see figure 1.1). In this paper, we will prove uniqueness and asymptotic behavior of positive solutions to (1.2) when λ is large. In fact, this is the first paper in the literature dealing with this challenging uniqueness analysis for infinite semipositone problems. (We 2010 Mathematics Subject Classification: Primary 34B16; Secondary 34B15. Key words and phrases: infinite semipositone, positive solutions, uniqueness
2
Figure 1.1 Figure 1: * note that when h(t) ≡ constant, the problem can be analyzed by time-map analysis, but here we do not assume h is a constant). We focus on reaction terms f that satisfy f (u) ∼ uq at ∞ for some q ∈ [0, p − 1), which allows f to be bounded and we do not require any monotonicity assumption on f as in [1,27]. In particular, our results when applied to the model case 1 λ 1 q 1+u −(φ(u )) = tη − uα + u e , t ∈ (0, 1), (1.3) u(0) = u(1) = 0, where 0 ≤ η, α, η + α < 1, 0 ≤ q < p − 1, give the uniqueness of positive solutions to (1.3) when λ is large. Furthermore, the unique solution uλ of uλ (1.3) satisfies lim = 1 uniformly for t ∈ (0, 1), where wq denotes 1 λ→∞ λ p−1−q wq
the unique positive solution of −(φ(w )) = t−η wq , t ∈ (0, 1), w(0) = w(1) = 0. It should be noted that our proof can be applied to include cases when f (u) do not behave like powers of u (see Remark 1.1 (iii)). We refer to [2,7] for uniqueness results to the sublinear semilinear problem −Δu = λf (u) in Ω, (1.4) u = 0 on ∂Ω, 3
when λ is large in the semipositone case, and to [3,6,17,18,24,26,30] to the positone case. Uniqueness results to related quasilinear Dirichlet problems were discussed in [9-11,13] for the positone case on a bounded domain, and in [15] for the semipositone and the infinite semipositone case on a ball. Uniqueness of positive solutions for quasilinear semipositone problems on a bounded domain is still an open question. For existence results for λ >> 1 in the semipositone and the infinite semipositone case, see [15,16,22,23]. Related results to (1.4) when f is singular at 0 can be found in [4,5,8,12,19,20,21,28,29]. We shall make the following assumptions: (A1) h : (0, 1] → (0, ∞) is differentiable, nonincreasing, and there exists a constant η ∈ [0, 1) such that lim sup tη h(t) < ∞. t→0+
(A2) f : (0, ∞) → R is differentiable and there exists a constant β > 0 such that (z − β)f (z) > 0 for z = β. (A3) There exist numbers L > 0 and 0 ≤ q < p − 1 such that lim
z→∞
and lim sup z→∞
f (z) =L zq
z|f (z)| < p − 1. f (z)
(A4) There exist constants l < 0 and 0 ≤ α < 1 − η such that lim z α f (z) = l and lim sup z α+1 |f (z)| < ∞.
z→0+
z→0+
By a solution of (1.2), we mean a function u ∈ C 1 [0, 1] such that φ(u ) is differentiable on (0, 1] and satisfies (1.2). Our main result is Theorem 1.1. Let (A1)-(A4) hold. Then there exists a constant λ0 > 0 such that problem (1.2) has a unique positive solution uλ for λ > λ0 . Furthermore, uλ (t) lim =1 1 λ→∞ (λL) p−1−q w (t) q 4
uniformly for t ∈ (0, 1), where wq is the unique solution of −(φ(w )) = h(t)wq in (0, 1), w(0) = w(1) = 0. Remark 1.1. (i) Note that (A4) is satisfied if limz→0+ z α+1 f (z) = l1 ∈ (0, ∞). (ii) A key ingredient of the proof of Theorem 1.1 is to obtain the estimate 1 u(t) ≥ C0 λ p−1−q min(t, 1 − t) for λ large in Lemma 2.5 below. This estimate requires the assumptions (A2),(A4), and the fact that lim inf f (z) > 0 (could z→∞
be finite). It will then be used to obtain sharp upper and lower bound estimates for positive solutions of (1.2) when f (u) ∼ uq at ∞ for some q ∈ [0, p − 1). Next, we establish the existence of a positive constant c independent of u such that cv ≤ u ≤ c−1 v for any two positive solutions of (1.2), and then show that c = 1. (iii) The steps in (ii) can be used with suitable modifications to obtain the uniqueness result in Theorem 1.1 under the assumptions (A2),(A4), and (A5) There exist constants a > 0 and q ∈ [0, p − 1) such that f (z) is nondecreasing and fz(z) is nonincreasing on [a, ∞). q This allows the case when f (u) do not behave like powers of u at ∞ e.g. f (u) = − u1α + uq ln(1 + u). 2. PRELIMINARY RESULTS z We shall denote the norm in Lp (0, 1) by ||.||p . Let F (z) = 0 f (s)ds. By (A2) and (A3), there exists a number θ > β such that F (z) < 0 for 0 < z < θ and F (z) > 0 for z > θ. We first extend [27, Lemma 2.1] to include the infinite semipositone case with h not necessarily strictly decreasing. Lemma 2.1. Let u be a positive solution of (1.2). Then u has only one interior maximum at t0 and u(t0 ) ≥ θ. Proof. Using the equation in (1.2) and (A2), we see that φ(u ) is increasing for t near 0. Hence φ(u (t)) > φ(u (0)) ≥ 0 i.e. u (t) > 0 for t near 0. Let t0 ∈ (0, 1) be the first zero of u . Note that by (A4), lim+ uα−1 (t)F (u(t)) = t→0
l(1−α)−1 , and since |u(t)| ≤ ||u ||∞ t, it follows from (A1) that tη+α−1 h(t)F (u) is bounded near t = 0. Since η + α < 1, lim+ h(t)F (u(t)) = 0. Suppose t→0
u(t0 ) < θ. Then h F (u) ≥ 0 on [0, t0 ]. Hence, by multiplying the equation in
5
(1.2) by u and integrating on [0, t0 ], we obtain
t0 1 p |u (0)| = λh(t0 )F (u(t0 )) − λ h (t)F (u)dt 0≤ 1− p 0 ≤ λh(t0 )F (u(t0 )) < 0, a contradiction. Thus u(t0 ) ≥ θ. Next, we claim that u is decreasing on (t0 , 1). Suppose to the contrary that there exist t1 , t2 ∈ (t0 , 1) with t1 < t2 such that u(t1 ) ≤ u(t2 ). Since (φ(u )) (t0 ) = −λh(t0 )f (u(t0 )) < 0, it follows that u is decreasing for t near t0 , t > t0 . Hence u has a minimum value on [t0 , t2 ] at some t∗ ∈ (t0 , t2 ). This implies λh(t∗ )f (u(t∗ )) = −(φ(u )) (t∗ ) ≤ 0 i.e. 0 < u(t∗ ) ≤ β and so F (u(t∗ )) < 0. Since F (u(t0 )) ≥ 0, there exists tˆ ∈ [t0 , t∗ ) such that F (u) < 0 on (tˆ, t∗ ] and F (u(tˆ)) = 0. Multiplying the equation in (1.2) by u and integrating on [tˆ, t∗ ] gives
t∗ 1 ˆ p ∗ ∗ |u (t)| = λh(t )F (u(t )) − λ h (t)F (u)dt < 0, 0≤ 1− p tˆ a contradiction. Hence u is increasing on (0, t0 ) and decreasing on (t0 , 1), which completes the proof. Let ρ = β+θ . By Lemma 2.1, for each μ ∈ (0, θ] there exist tμ ≤ t0 ≤ t˜μ 2 such that u(tμ ) = u(t˜μ ) = μ. Lemma 2.2. There exists a constant cρ > 0 such that any positive solution u of (1.2) satisfies ⎧ p t 1/p p−1+α ⎪ ⎪ h if t ∈ [0, tρ ], ⎪ 0 ⎨ u(t) ≥ 1 ⎪ p ⎪ cρ λ p−1+α 1 1/p p−1+α ⎪ ⎩ h if t ∈ [t˜ρ , 1]. t In particular, there exists a constant c˜ρ > 0 independent of u and λ such 1 that max{tρ , 1 − t˜ρ } ≤ c˜ρ λ− p . Proof. Note that h F (u) ≥ 0 on [0, tθ ] ∪ [t˜θ , 1]. Let s ∈ [0, tθ ]. Multiplying the equation in (1.2) by u and integrating on [0, s] gives
s 1 1 p p (u (s)) = 1 − (u (0)) − λh(s)F (u) + λ h (τ )F (u)dτ 1− p p 0 6
≥ −λh(s)F (u). Hence
u (s) 1/p
≥ (λh(s))1/p ,
(−F (u)) which implies upon integrating on [0, t], t ∈ [0, tθ ], that t u(t) dz 1/p ≥λ h1/p . 1/p (−F (z)) 0 0
(2.1)
By (A4), there exists a constant k > 0 such that −F (z) ≥ kz 1−α for z ∈ [0, ρ]. Hence (2.1) implies t p−1+α p−1 1/p p u (λk) (t) ≥ h1/p , p 0 p t 1 p−1+α for t ∈ [0, tρ ] follows, where cρ = from which u(t) ≥ cρ λ p−1+α 0 h1/p p p−1+α p−1 1/p k . In particular, since h is nonincreasing, p 1
p
ρ = u(tρ ) ≥ cρ (λh(1)) p−1+α tρp−1+α , p−1+α 1 p which implies tρ ≤ c˜ρ λ−1/p , where c˜ρ = ρ/(cρ h p−1+α (1)) . Next, let s ∈ [t˜θ , 1]. Since F (θ) = 0, it follows from multiplying the equation in (1.2) by u and integrating on [t˜θ , s] that
1 1− |u (s)|p ≥ −λh(s)F (u). p Hence
|u (s)| (−F (u))
1/p
≥ (λh(s))1/p ,
and since u ≤ 0 on [t˜θ , 1], it follows upon integrating on [t, 1], t ∈ [t˜θ , 1], that 1 u(t) dz 1/p ≥λ h1/p 1/p (−F (z)) 0 t p p−1+α 1 1 and as in the above argument, we get u(t) ≥ cρ λ p−1+α t h1/p for t ∈ [t˜ρ , 1] follows. By choosing t = t˜ρ , we conclude as in the above that 1 − t˜ρ ≤ c˜ρ λ−1/p , which completes the proof. 7
The next result establishes a comparison principle when the forcing terms are close in L1 −norm, but not necessarily ordered. Lemma 2.3. Let M > 0 and hi ∈ L1 (0, 1) with ||hi ||1 < M for i = 1, 2. Let u˜i ∈ C 1 [0, 1], i = 1, 2, satisfy −(φ(˜ u1 )) ≥ h1 , t ∈ (0, 1), −(φ(˜ u2 )) ≤ h2 , t ∈ (0, 1), , u˜1 (0) = u˜1 (1) = 0, u˜2 (0) = u˜2 (1) = 0. Then (i) If h2 ≥ 0, h2 ≡ 0 then for a given ε0 > 0, there exists a constant δ > 0 such that u˜1 ≥ (1 − ε0 )˜ u2 on [0, 1], provided that ||h1 − h2 ||1 < δ. (ii) Let tˇ < 1/2 < tˆ . Suppose u˜2 defined above satisfies −(φ(˜ u2 )) = h2 on (0, 1) and there exists a constant C > 0 such that |˜ u2 | > C on [0, tˇ] ∪ ˜ [tˆ, 1]. Then there exist constants δC , kC > 0 such that if ||h1 − h2 ||1 < δ˜C then u˜1 (t) ≥ u˜2 (t) − kC ||h1 − h2 ||1 p(t) for t ∈ [0, tˇ] ∪ [tˆ, 1], where p(t) = min(t, 1 − t). Proof. Let ui ∈ C 1 [0, 1], i = 1, 2, satisfy −(φ(ui )) = hi , t ∈ (0, 1), ui (0) = ui (1) = 0. Then u˜1 ≥ u1 and u˜2 ≤ u2 on [0, 1] by the weak comparison principle (see [25, Lemma A.2]). Let ε > 0. We first show that there exists a constant δ > 0 such that ||h1 − h2 ||1 < δ =⇒ ||u1 − u2 ||∞ < ε. (2.2) By integrating, we obtain ui (t)
−1
=φ
Ci −
t 0
hi ,
1 s where Ci are constants satisfying 0 φ−1 Ci − 0 hi ds = 0, i = 1, 2. Using the fact that φ−1 is increasing with φ−1 (0) = 0, we deduce that |Ci | ≤ ||hi ||1 and |C1 − C2 | ≤ ||h1 − h2 ||1 . 8
(2.3)
Since φ−1 is uniformly continuous on [−2M, 2M ], there exists δ0 > 0 such that x, y ∈ [−2M, 2M ], |x − y| < δ0 =⇒ |φ−1 (x) − φ−1 (y)| < ε. Suppose ||h1 − h2 ||1 < δ0 /2. Then in view of (2.3), t < δ0 , C1 − C2 − (h − h ) 1 2 0
t and since Ci − 0 hi ≤ 2M, it follows that |u1 (t) − u2 (t)| < ε
(2.4)
for all t ∈ [0, 1] i.e. (2.2) holds with δ = δ0 /2. (i) If h2 ≥ 0, h2 ≡ 0 then u2 (t) ≥ ||u2 ||∞ p(t) for t ∈ [0, 1] (see e.g. [14, Lemma 3.4]. Let ε < ε0 ||u2 ||∞ . Then there exists δ > 0 such that (2.4) holds provided that ||h1 − h2 ||1 < δ. Since ui (0) = ui (1) = 0, the Mean Value Theorem gives u1 (t) ≥ u2 (t) − εp(t) ≥ (1 − ε0 )u2 (t) for t ∈ [0, 1], and therefore u˜1 ≥ (1 − ε0 )˜ u2 on [0, 1]. (ii) Suppose u˜2 satisfies −(φ(˜ u2 )) = h2 on (0, 1) and |˜ u2 | > C > 0 on I ∪ J, where I = [0, tˇ], J = [tˆ, 1]. Then u˜2 ≡ u2 and, in view of (2.4), there exists 0 < δ˜ < 12 min(φ(C/2), δ0 ) such that |u1 | > C/2 on [0, tˇ] ∪ [tˆ, 1] ˜ which we shall assume. Hence provided that ||h1 − h2 ||1 < δ, t Ci − hi = φ(|ui (t)|) > φ(C/2) > 0 (2.5) 0
for t ∈ I ∪ J and Ci − Theorem, |u1 (t)
−
t
u2 (t)|
0
hi are of the same sign, i = 1, 2. By the Mean Value
t t −1 −1 C1 − C2 − = φ h1 − φ h2 0
2−p
=
|ξ(t)| p−1
t C1 − C2 − 0 (h1 − h2 ) p−1 9
0
,
(2.6)
t t where ξ(t) is between C1 − 0 h1 and C2 − 0 h2 . By (2.3) and (2.5), φ(C/2) ≤ |ξ(t)| ≤ 2M, from which (2.3) and (2.6) gives |u1 (t) − u2 (t)| ≤ kC ||h1 − h2 ||1 , 2−p
2−p
2 for t ∈ I, where kC = p−1 max{(2M ) p−1 , φ p−1 (C/2)}. Since u1 (0) = 0 and u2 (0) = 0, it follows that
u˜1 (t) ≥ u1 (t) ≥ u2 (t) − kC ||h1 − h2 ||1 t for t ∈ I. Similarly, u˜1 (t) ≥ u1 (t) ≥ u2 (t) − kC ||h1 − h2 ||1 (1 − t) for t ∈ J, which completes the proof.
Lemma 2.4. (i)
1
(ii)
1
0
h(t) αp dt < ∞. t 1/p p−1+α h 0
1/2
h(t) αp dt < ∞. 1 1/p p−1+α h t
˜ = t−γ h. Then η + γ < p and Proof. (i) Let γ = p − 1 + α and define h 1/p 1 ˜ ∈ L (0, 1) in view of (A1). Note that therefore h t t 1/p γ/p ˜ 1/p . h ≤t h 0
Since h is nonincreasing, h(t)
p−1 p
0
p−1
t −1 1/p ≤ t h . 0
Hence 0
1
h(t) αp dt ≤ t 1/p p−1+α h 0
1
h1/p (t)
t 0
h1/p tp−1
0
10
αp p−1− p−1+α
dt
≤
1
= 0
˜ 1/p h
1
γ˜
t h
0
t 0
1/p
t
˜ 1/p h
t
γ/p 0
˜ 1/p h
αp p−1− p−1+α
αp p−1− p−1+α
dt = Cα,p
1 0
t1−p dt
˜ 1/p h
p(p−1) p−1+α
<∞
by the choice of γ, where Cα,p is a positive constant depending on α, p. (ii) Since h is nonincreasing,
1 1/2
p−1 h(t) dt ≤ (h(1/2)) p αp p−1+α 1 1/p h t
= C˜α,p
1 1/2
h
1/p
1 1/2
αp 1− p−1+α
h1/p (t) αp dt 1 1/p p−1+α h t
< ∞,
where C˜α,p is a positive constant depending on α, p. ˜ > 0 such that for λ > λ, ˜ any Lemma 2.5. There exist constants C0 , λ positive solution u of (1.2) satisfies 1
u(t) ≥ λ p−1 C0 p(t) for t ∈ [0, 1], where p(t) = min(t, 1 − t). Proof. By (A2), (A4), and continuity of z α f (z), there exists a constant a > 0 such that a f (z) ≥ − α (2.7) z for all z > 0. Since u ≥ ρ on [tρ , t˜ρ ], where tρ , t˜ρ are defined immediately preceding Lemma 2.2, it follows that f (u) ≥ δρ > 0 on [tρ , t˜ρ ] for some 1 δρ > 0. Then assuming cρ λ p−1+α > 1, (2.7) and Lemma 2.2 give ⎧ ah(t) ⎪ ⎪ − if t ∈ [0, tρ ] αp ⎪ ⎪ t 1 p−1+α ⎪ ⎪ hp 0 ⎨ ⎪ 1 − p−1 if t ∈ (tρ , t˜ρ ) ≡ h1 , δρ h(t) − φ λ u ≥ (2.8) ⎪ ⎪ ah(t) ⎪ ⎪ − if t ∈ [t˜ρ , 1] ⎪ αp ⎪ 1 1 p−1+α ⎪ ⎩ p h t
11
Defining the right-hand side of (2.8) as h1 . Then h1 ∈ L1 (0, 1) by Lemma 2.4. Let ω ∈ C 1 [0, 1] satisfy −(φ(ω )) = δρ h(t) ≡ h2 , t ∈ (0, 1), (2.9) ω(0) = ω(1) = 0. Then ω(t) ≥ ||ω||∞ p(t) for t ∈ [0, 1] (see [14, Lemma 3.4]). Note that 1 tρ tρ h(t) h+ h +a dt ||h1 − h2 ||1 = δρ αp 1 p−1+α t 0 0 t˜ρ p h 0 +a
1 t˜ρ
h(t) αp dt → 0 as λ → ∞ 1 1 p−1+α p h t
since max{tρ , 1 − t˜ρ } → 0 as λ → ∞ by Lemma 2.2. Hence it follows by 1 applying Lemma 2.3 (i) with u˜1 = λ− p−1 u, u˜2 = w, and ε0 = 1/2 that for λ large enough, 1 λ− p−1 u ≥ ω/2 ≥ (||ω||∞ /2)p(t), which completes the proof. Lemma 2.6. Let 0 < q < p − 1. Then there exists a unique positive solution wq ∈ C 1 [0, 1] to the problem −(φ(w )) = h(t)wq , t ∈ (0, 1), (2.10) w(0) = w(1) = 0. Proof. Let ω0 , ω1 ∈ C 1 [0, 1] satisfy −(φ(ω2 )) = h(t), t ∈ (0, 1), −(φ(ω1 )) = ω1q , t ∈ (0, 1), , . ω1 (0) = ω1 (1) = 0, ω2 (0) = ω2 (1) = 0. whose existence follow from [14,Theorem 2.2 (ii)] and [14, Lemma 3.1]. Then it is easy to see that εω1 is a subsolution of (2.10) if ε is small enough, and M ω2 is a supersolution of (2.10) if M is large enough. By decreasing ε and increasing M if necessary, we can assume that εω1 ≤ M ω2 on [0, 1]. Hence (2.10) has a solution wq with wq ≥ εω1 in [0, 1]. To show uniqueness, let w1 , w2 i (t) i (t) be two positive solutions of (2.10). Since sup wp(t) < ∞ and inf wp(t) > t∈(0,1)
12
t∈(0,1)
0 for i = 1, 2, there exists a largest constant c > 0 such that w1 ≥ cw2 in [0, 1]. Suppose c < 1. Then −(φ(w1 )) = h(t)w1q ≥ cq h(t)w2q q
for t ∈ (0, 1), and the weak comparison principle gives w1 ≥ c p−1 w2 in [0, 1], a contradiction with the maximality of c. Thus c ≥ 1 i.e. w1 ≥ w2 , which completes the proof. The asymptotic behavior of the solutions to (1.2) as λ → ∞ is given in the next result. Lemma 2.7. Let ε ∈ (0, 1). Let wq be the unique solution of (2.10) given ¯ > 0 such that for λ > λ, ¯ any by Lemma 2.6. Then there exists a constant λ positive solution u of (1.2) satisfies p
p
1
1
(1 − ε) p−1−q (λL) p−1−q wq ≤ u ≤ (1 + ε) p−1−q (λL) p−1−q wq on [0, 1], where L is defined in (A3). In other words, lim
λ→∞
u(t) 1
(λL) p−1−q wq (t)
=1
uniformly in t ∈ (0, 1). Proof. Let ε ∈ (0, 1). By (A3), there exists a constant A > 0 such that (1 − ε)Lz q < f (z) < (1 + ε)Lz q ˜ where λ ˜ is defined in Lemma 2.5. Since wq (t) ≤ for z > A. Suppose λ > λ, ||wq ||∞ p(t) for t ∈ [0, 1], it follows from Lemma 2.5 that u ≥ d0 wq in [0, 1], 1 ˜ p−1 where d0 = λ C0 ||wq ||−1 ∞ . Let dλ be the largest number such that u ≥ dλ wq in [0, 1]. By (A4), there exists a constant b > 0 such that |f (z)| ≤
b for z ∈ (0, A]. zα
1
Assuming λ p−1 C0 > 1 then Lemma 2.5 gives ⎧ (1 − ε)Ldλq h(t)wqq ⎪ ⎪ ⎨ −(φ(u )) ≥ λ bh(t) ⎪ ⎪ ⎩ − α p (t) 13
if u(t) > A, if u(t) ≤ A.
This implies − φ
u
≥
1
(λ(1 − ε)Ldqλ ) p−1
⎧ h(t)wqq ⎪ ⎪ ⎨
if u(t) > A,
b0 h(t) ⎪ ⎪ ⎩ − α p (t)
if u(t) ≤ A,
≡ h1
b where b0 = (1−ε)Ld q and h1 is defined to be the right-hand side of the previous 0 equation. Since −(φ(wq )) = h(t)wqq ≡ h2 in (0, 1), 1 − p−1 −1 C0 and the measure of {t : u(t) ≤ A} ≤ the measure of t : p(t) ≤ λ → 0 as λ → ∞ by Lemma 2.5, it follows that
b0 q ||h1 − h2 ||1 = dt → 0 h(t) wq + α p (t) u≤A
as λ → ∞. Hence if λ is large enough, Lemma 2.3(i) gives u 1
(λ(1 − ε)Ldλq ) p−1 p
≥ (1 − ε)wq
on [0, 1] p
1
1
i.e. u ≥ (1−ε) p−1 (λLdqλ ) p−1 wq in [0, 1]. Consequently, dλ ≥ (1−ε) p−1 (λLdqλ ) p−1 , and therefore p 1 u ≥ (1 − ε) p−1−q (λL) p−1−q wq on [0, 1]. Next, since sup t∈(0,1)
u(t) p(t)
wq (t) t∈(0,1) p(t)
< ∞ and inf
> 0, there exists a constant d1 ≥ 0
(depending on u) such that u ≤ d1 wq in [0, 1]. Let d˜λ be the smallest number such that u ≤ d˜λ wq in [0, 1]. Then d˜λ ≥ d0 and ⎧ (1 + ε)Ld˜qλ h(t)wqq if u(t) > A, ⎪ ⎪ ⎨ −(φ(u )) ≤ λ bh(t) ⎪ ⎪ ⎩ α if u(t) ≤ A. p (t) Hence − φ
u
1
(λ(1 + ε)Ld˜λq ) p−1
≤
⎧ h(t)wqq ⎪ ⎪ ⎨
if u(t) > A,
b1 h(t) ⎪ ⎪ ⎩ α p (t)
if u(t) ≤ A,
14
˜ 1, ≡h
where b1 = Since
˜ 1 be the righ-hand side of the previous equation. Let h
b 1 q ˜ 1 − h2 ||1 ≤ ||h →0 h(t) wq + α p (t) u≤A
b . (1+ε)Ldq0
p
1 gives u ≤ (1 + ε) p−1 (λLd˜λq ) p−1 wq in p 1 [0, 1] if λ is large enough. Consequently, d˜λ ≤ (1 + ε) p−1 (λLd˜λq ) p−1 , which implies p 1 u ≤ (1 + ε) p−1−q (λL) p−1−q wq in [0, 1],
as λ → ∞, Lemma 2.3(i) with ε0 =
ε ε+1
which completes the proof. Lemma 2.8. There exist positive constants K0 , tˇ , tˆ with tˇ < 1/2 < tˆ ¯ any positive solution u of (1.2) satisfies such that for λ > λ, 1
|u (t)| ≥ K0 λ p−1−q for t ∈ [0, tˇ] ∪ [tˆ, 1]. Proof. By Lemma 2.7, there exist positive constants D1 , D2 independent of u and λ such that 1
1
D1 λ p−1−q wq (t) ≤ u(t) ≤ D2 λ p−1−q wq (t)
(2.11)
¯ Consequently, there exists a constant for t ∈ [0, 1], provided that λ > λ. 1 D3 > 0 depending on D1 , D2 and wq such that u (0) ≥ D3 λ p−1−q and u (1) ≤ 1 −D3 λ p−1−q . By (A2) and (A3), there exists a constant m > 0 such that f (z) ≤ mz q for all z > 0. This, together with (2.11), gives t φ(u (t)) = φ(u (0)) − λ h(s)f (u)ds 0
p−1
≥ D3p−1 λ p−1−q − λm ≥λ
p−1 p−1−q
D3p−1
−
mD2q
t 0
t 0
h(s)uq ds
h(s)wqq ds
15
p−1
λ p−1−q D3p−1 ≥ 2
tˇ for t < tˇ, where tˇ < 1/2 is such that 0 h(s)wqq ds < D3p−1 /(2mD2q ). Hence 1 1 u (t) ≥ K0 λ p−1−q for t ∈ [0, tˇ], where K0 = D3 /2 p−1 . Similarly,
1
φ(u (t)) = φ(u (1)) + λ
h(s)f (u)ds t
≤ ≤λ
p−1 p−1−q
p−1 −D3p−1 λ p−1−q
+
mD2q
for t > tˆ, where tˆ > 1/2 is such that 1
h(s)uq ds ≤
t
−D3p−1
1
+ λm 1
q
h(s)w ds t
1 tˆ
p−1
λ p−1−q D3p−1 ≤− 2
h(s)wqq ds < D3p−1 /(2mD2q ).
Hence u (t) ≤ −K0 λ p−1−q for t ∈ [tˆ, 1], which completes the proof. 3. PROOF OF THE MAIN RESULT In what follows, we shall assume that λ > 1 is large enough so that Lemmas 2.5 and 2.7 apply. Proof of Theorem 1.1. By (A3), there exist positive constants A0 , δ, ε0 , δ1 with δ(1 + ε0 ) < δ1 < p − 1 and A0 > β such that |f (z)| ≤
and (1 − ε0 )
z1 z2
q
δf (z) z
f (z1 ) ≤ (1 + ε0 ) ≤ f (z2 )
(3.1)
z1 z2
q (3.2)
for z, z1 , z2 > A0 . Let u, v be positive solutions of (1.2). Then Lemma 2.7 with ε small gives 1 1 1 3 (λL) p−1−q wq ≤ u, v ≤ (λL) p−1−q wq 2 2
(3.3)
in [0, 1], which implies (1/3)v ≤ u ≤ 3v on [0, 1]. Let c be the largest number such that cv ≤ u ≤ c−1 v on [0, 1] and suppose that c < 1. We claim that there exists a constant L0 > 0 such that f (u) − cp−1 f (v) ≥ L0 (1 − c) 16
(3.4)
for v > A, where A = 3A0 . For v > A, we have u > A0 . Hence (3.1) and the Mean Value Theorem gives |f (u) − f (v)| = |u − v||f (ψ)| ≤
δ(1 − c)vf (ψ) cψ
δ(1 − c)f (ψ) , c2 where ψ is between u and v. Here we have used the fact that cv ≤ ψ ≤ c−1 v. By (3.2),
q ψ f (ψ) ≤ (1 + ε0 ) ≤ (1 + ε0 )c−q , f (v) v ≤
and so |f (u) − f (v)| ≤ for v > A. Since
lim
z→1−
δ1 (1 − c) δ(1 + ε0 )(1 − c) f (v) ≤ f (v) q+2 c cq+2
1 − z p−1 δ1 − 2+q 1−z z
(3.5)
= p − 1 − δ1 ≡ 2k0 > 0,
there exists c˜ ∈ (0, 1) such that 1 − z p−1 δ1 − 2+q ≥ k0 1−z z
(3.6)
for z ≥ c˜. Combining (3.5) and (3.6), we obtain f (u) − cp−1 f (v) = f (u) − f (v) + (1 − cp−1 )f (v)
δ1 1 − cp−1 − q+2 f (v) ≥ k0 (1 − c)f (v) ≥ (1 − c) 1−c c
(3.7)
for v > A if c ≥ c˜. Suppose next that c < c˜. Since c ≥ 1/3 and q < p − 1, there exists a constant m0 > 0 such that cq − cp−1 ≥ m0 (1 − c) > m0 (1 − c˜), from which (3.2) gives
f (u) − c
p−1
f (v) =
f (u) p−1 −c f (v) ≥ (1 − ε0 )cq − cp−1 f (v) f (v) 17
≥ (c − c q
p−1
− ε0 )f (v) ≥
m0 (1 − c˜) 2
f (v) ≥ B(1 − c)
(3.8)
if ε0 is decreased further, where B = (m0 (1 − c˜)/2) inf f. From (3.7), (3.8) (A,∞)
and inf f > 0, we see that (3.4) holds. (A,∞)
For v ≤ A, we have u ≤ 3A, and it follows from (A4) that there exists a constants KA > 0 such that |f (u) − f (v)| = |u − v||f (ξ)| ≤ ≤
KA (1 − c)v c|ξ|α+1
KA (1 − c) KA (1 − c) ≤ , α+2 α c v wqα
and
KA KA ≤ α α w wq
|f (w)| ≤
(3.9)
for w ∈ {u, v}, where we have assumed that λ is large enough (independent of u, v) so that (3.3) gives w ≥ wq and c2+α wα ≥ wqα on (0, 1). Consequently, |f (u) − cp−1 f (v)| ≤ |f (u) − f (v)| + (1 − cp−1 )|f (v)| ≤
˜ A (1 − c) K , wqα
(3.10)
˜ A = 2KA max{1, p − 1}. By (3.3) and (A3), there exists a constant where K L1 > 0 depending on L such that q
f (v) ≤ mv q ≤ L1 λ p−1−q
(3.11)
on [0, 1] for some m > 0. Hence (3.4) gives
L0 (1 − c) p−1 −(φ(u )) ≥ λh(t) c f (v) + q L1 λ p−1−q for v > A, which implies ⎧ q − p−1−q ⎪ λ h(t) cp−1 + ⎨ − (φ (˜ u)) ≥ ⎪ q ⎩ − p−1−q λ h(t)f (u)
L2 (1−c) q λ p−1−q
18
) f (v) if v > A, if v < A.
≡ h1 , (3.12)
1
where u˜ = λ− p−1−q u and L2 = L0 /L1 . Define h1 to be the right-hand side of (3.12). 1 p−1 q 1 Let v˜ = λ− p−1−q cp−1 + L2 (1 − c)λ− p−1−q v. Then
− (φ(˜ v )) = λ
q − p−1−q
p−1
h(t) c
+
L2 (1 − c)
f (v) ≡ h2 .
q
λ p−1−q
(3.13)
in [0, 1]. Note that, in view of (3.11), there exists a constant M0 independent of u, v, λ, such that |hi (t)| ≤ M0 h(t) for t ∈ (0, 1), i = 1, 2. By (3.9) and (3.10), q (1 − c) L 2 − p−1−q ||h1 − h2 ||1 = λ h(t) f (u) − cp−1 f (v) − f (v) dt q λ p−1−q v
≤λ
(1 − c)L3 v
h dt, wqα
(3.14)
˜ A + L2 KA . Let tˇ , tˆ be given by Lemma 2.8. Then |˜ where L3 = K v (t)| ≥ cK0 ≥ K0 /3 ≡ C for t ∈ 0, tˇ ∪ [tˆ, 1]. Since ||h1 − h2 ||1 → 0 as λ → ∞, it follows from (3,12), (3.13) and Lemma 2.3(ii) that for λ sufficiently large, u˜(t) ≥ v˜(t) − kC ||h1 − h2 ||1 p(t) for t ∈ [0, tˇ] ∪ [tˆ, 1]. This, together with (2.11), gives 1 p−1 q 1 u(t) ≥ cp−1 + L2 (1 − c)λ− p−1−q v − λ p−1−q kC ||h1 − h2 ||1 p(t)
≥
p−1
c
q − p−1−q
+ L2 (1 − c)λ
1 p−1
˜ − kC ||h1 − h2 ||1 v,
for t ∈ [0, tˇ] ∪ [tˆ, 1], where k˜C = (kC /D1 ) sup(p/wq ). Since (0,1)
q
(1/3)p−1 ≤ cp−1 , cp−1 + L2 (1 − c)λ− p−1−q ≤ 1 + L2 ,
19
(3.15)
it follows from the Mean Value Theorem that 1 p−1 1 q q − p−1−q p−1 + L2 (1 − c)λ − cp−1 p−1 ≥ L4 (1 − c)λ− p−1−q , c
(3.16)
where L4 is a positive constant depending only on p and L2 . Since v
h L4 (1 − c)v dt v ≥ q α wq 2λ p−1−q
L4 − k˜C L3 q λ p−1−q v 0 such that ≥
u ≥ (c + γ)v
on [0, tˇ] ∪ [tˆ, 1]. (3.17) In particular, u(τ ) ≥ cv(τ )+γ0 for τ ∈ tˇ , tˆ , where γ0 = γ min(v(tˇ), v(tˆ)) > 0. By (2.11), 1 v(t) ≥ D1 λ p−1−q min wq > A [tˇ,tˆ]
for t ∈ [tˇ, tˆ], provided that λ is large enough. Hence (3.4) gives −(φ(u )) = λh(t)f (u) ≥ λcp−1 h(t)f (v) on (tˇ, tˆ). Since
−(φ(cv + γ0 ) ) = λcp−1 h(t)f (v) on (tˇ, tˆ), and u(τ ) ≥ cv(τ ) + γ0 for τ ∈ tˇ , tˆ , the weak comparison principle implies u ≥ cv + γ0
on [tˇ, tˆ].
This, together with (3.17), gives the existence of a number γ1 > 0 such that u ≥ (c + γ1 )v in [0, 1]. By symmetry, there exists a constant γ2 > 0 such that v ≤ (c + γ2 )u in [0, 1]. Hence (c + γ˜ )v ≤ u ≤ (c + γ˜ )−1 v in [0, 1], a contradiction with the definition of c. Thus c ≥ 1 i.e. u = v in [0, 1]. The asymptotic behavior of the solution as λ → ∞ is given in Lemma 2.7. This completes the proof of Theorem 1.1. Ackowledgements. The authors thank the referee for carefully reading the manuscript and providing helpful suggestions. 20
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