INTRODUCTION 3 In particular (0.14) tv (λt, 1, d) = 2λvλ(t, λ2, d). Remark 3. Using the contraction mapping principle it can be shown that for every (λ, d) there exists σ(λ, d) such (0.12) has a unique positive solution defined on the maximal interval [0, σ(λ, d)). If σ(λ, d) +∞ one has limt→σ(λ,d) v(t, λ, d) = 0 and v(t, λ, d) 0 on [0, σ(λ, d)). For the sake of simplicity in the notation we will write (0.15) v(t, 1, d) ≡ v(t, d) and σ(1, d) = σ(d). The steady state solutions to (0.9) subject to (0.7) or to (0.8) are the solutions to (0.12) satisfying (0.16) v(1, λ, d) = 1, or (0.17) v (1, λ, d) = 0, respectively. When p ≤ q, it follows by a phase plane analysis that v ≡ 1 is the only solution of (0.12) satisfying v(1, λ, d) = 1 or v (1, λ, d) = 0 respectively. Therefore, we will assume in the rest of this paper that p q. Throughout this paper (0.18) 0 μ1 2 μ2 2 · · · → ∞, and (0.19) 0 = ν2 1 ν2 2 · · · → ∞ will denote the radial eigenvalues of the negative Laplacian operator with zero Dirichlet and Neumann boundary data on the unit ball in RN respectively. That is, for each k = 1, . . . there exists radial functions wk = 0, zk = 0 satisfying Δwk + μkwk = 0 and the boundary condition (0.7), and Δzk + νkzk = 0 and the boundary condition (0.8). For future reference we note that the μks, νks are given by the Bessel function J defined as the solution to (0.20) J (t) + N − 1 t J (t) + J(t) = 0 t ≥ 0, J(0) = 1, J (0) = 0. In fact, the μks are the zeroes of J and the νks are the zeroes of J . These numbers also satisfy (0.21) νk μk νk+1 for all k = 1, 2 . . .. Our main results are the next theorems. Theorem 1. Let q p, and Σ = {(λ, d) ∈ (0, ∞) × (0, ∞) v(1, λ, d) = 1}. The set of positive solutions to ( 0.9) subject to ( 0.7) is homeomorphic to Σ and is connected. Moreover, for each positive integer j there exist 0 ≤ dj 1 and Dj 1 and a continuous function ξj : (dj, Dj) → (0, ∞) such that v(·, λ, d), d = 1, is a solution to ( 0.9) subject to ( 0.7) with v(·, λ, d) − 1 having j − 1 zeroes in (0, 1) if and only if λ = ξj(d). In addition, ξj(1) = μj p−q .

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