Existence of a positive solution for a -Laplacian semipositone problem
- PDF / 489,124 Bytes
- 5 Pages / 468 x 680 pts Page_size
- 30 Downloads / 213 Views
We consider the boundary value problem −∆ p u = λ f (u) in Ω satisfying u = 0 on ∂Ω, where u = 0 on ∂Ω, λ > 0 is a parameter, Ω is a bounded domain in Rn with C 2 boundary ∂Ω, and ∆ p u := div(|∇u| p−2 ∇u) for p > 1. Here, f : [0,r] → R is a C 1 nondecreasing function for some r > 0 satisfying f (0) < 0 (semipositone). We establish a range of λ for which the above problem has a positive solution when f satisfies certain additional conditions. We employ the method of subsuper solutions to obtain the result. 1. Introduction Consider the boundary value problem −∆ p u = λ f (u)
in Ω,
u > 0 in Ω,
(1.1)
u = 0 on ∂Ω, where λ > 0 is a parameter, Ω is a bounded domain in Rn with C 2 boundary ∂Ω and ∆ p u := div(|∇u| p−2 ∇u) for p > 1. We assume that f ∈ C 1 [0,r] is a nondecreasing function for some r > 0 such that f (0) < 0 and there exist β ∈ (0,r) such that f (s)(s − β) ≥ 0 for s ∈ [0,r]. To precisely state our theorem we first consider the eigenvalue problem −∆ p v = λ|v | p−2 v
in Ω,
v = 0 on ∂Ω.
(1.2)
Let φ1 ∈ C 1 (Ω) be the eigenfunction corresponding to the first eigenvalue λ1 of (1.2) such that φ1 > 0 in Ω and φ1 ∞ = 1. It can be shown that ∂φ1 /∂η < 0 on ∂Ω and hence, depending on Ω, there exist positive constants m,δ,σ such that ∇φ1 p − λ1 φ p ≥ m 1
φ1 ≥ σ where Ωδ := {x ∈ Ω | d(x,∂Ω) ≤ δ }. Copyright © 2006 Hindawi Publishing Corporation Boundary Value Problems 2005:3 (2005) 323–327 DOI: 10.1155/BVP.2005.323
on Ωδ ,
on Ω \ Ωδ ,
(1.3)
324
Positive solution for p-Laplacian semipositone problems
We will also consider the unique solution, e ∈ C 1 (Ω), of the boundary value problem −∆ p e = 1
in Ω,
(1.4)
e = 0 on ∂Ω
to discuss our result. It is known that e > 0 in Ω and ∂e/∂η < 0 on ∂Ω. Now we state our theorem. Theorem 1.1. Assume that there exist positive constants l1 ,l2 ∈ (β,r] satisfying (a) l2 ≥ kl1 , (b) | f (0)|λ1 /m f (l1 ) < 1, and p−1 p−1 (c) l2 / f (l2 ) > µ(l1 / f (l1 )), 1/(p−1) where k = k(Ω) = λ1 (p/(p − 1))σ (p−1)/ p e∞ and µ = µ(Ω) = (pe∞ /(p − 1)) p−1 (λ1 / σ p ). Then there exist λˆ < λ∗ such that (1.1) has a positive solution for λˆ ≤ λ ≤ λ∗ . Remark 1.2. A simple prototype example of a function f satisfying the above conditions is
f (s) = r (s + 1)1/2 − 2 ;
0 ≤ s ≤ r4 − 1
(1.5)
when r is large. Indeed, by taking l1 = r 2 − 1 and l2 = r 4 − 1 we see that the conditions β(= 3) < l1 < l2 and (a) are easily satisfied for r large. Since f (0) = −r, we have f (0)λ1 =
λ1 . m(r − 2)
m f l1
(1.6)
Therefore (b) will be satisfied for r large. Finally, p−1
p−1
r4 − 1 (r − 2) r 4p−3 = ∼ ∼ r 2p−3 p−1 p−1 2 r 2p l1 / f (l1 ) r2 − 1 r −1
l2 / f (12 )
(1.7)
for large r and hence (c) is satisfied when p > 3/2. Remark 1.3. Theorem 1.1 holds no matter what the growth condition of f is, for large u. Namely, f could satisfy p-superlinear, p-sublinear or p-linear growth condition at infinity. It is well documented in the literature that the study of positive solution is very challenging in the semipostone case. See [5] where positive solution i
Data Loading...
