Central Configurations for Newtonian N + 2 p + 1 - Body Problems

and Applied Analysis 3 −λ (0, 0, rp) = − N 󵄨󵄨󵄨󵄨󵄨1 + r2 p󵄨󵄨󵄨󵄨󵄨3/2 (0, 0, rp) − (0, 0, 1 r2 p)M0 − ∑ 1≤i j; a0,0 = ai,0 = 1 when i = 1, . . . , p; b0 = β + M0, bi = N/|1 + r2 i |3/2 + M0/r3 i , where i = 1, . . . , p. Equations (9)–(11) can be written as a linear system of the form AX = b given by (1 a0,1 a0,2 a0,3 ⋅ ⋅ ⋅ a0,p 1 a1,1 a1,2 a1,3 ⋅ ⋅ ⋅ a1,p 1 a2,1 a2,2 a2,3 ⋅ ⋅ ⋅ a2,p .. .. .. . . . .. .. 1 ap,1 ap,2 ap,3 ⋅ ⋅ ⋅ ap,p) (( ( λ M1 M2 M3 .. MP )) ) = (( ( b0 b1 b2 b3 .. bp )) ) . (12) The column vector is given by the variables X = (λ,M1, M2, . . . ,Mp)T. Since the aij is function of r1, r2, . . . , rp, we write the coefficient matrix as Ap(r1, r2, . . . , rp). 2.2. For p = 1. We need the next lemma. Lemma 3 (see [12]). Assuming m1 = ⋅ ⋅ ⋅ = mN = 1, mN+1 = mN+2 = M1, there is a nonempty interval I ⊂ R, M0(r1) and M1(r1), such that for each r1 ∈ I, (q1, . . . , qN, qN+1, qN+2, qN+3) forms a central configuration of theN + 2 + 1-body problem. For p = 1, system (12) becomes (1 a0,1 1 a1,1)( λ M1) = (b0 b1) , (13) 󵄨󵄨󵄨󵄨A1 (r1)󵄨󵄨󵄨󵄨 = a1,1 − a0,1 = − 1 4r3 1 + 2 󵄨󵄨󵄨󵄨1 + r2 1 󵄨󵄨󵄨󵄨3/2 . (14) If we consider |A1(r1)| as a function of r1, then |A1(r1)| is an analytic function and nonconstant. By Lemma 3, there exists a ř1 ∈ I such that (13) has a unique solution (λ,M1) satisfying λ > 0 andM1 > 0. 2.3. For All p>1. The proof for p ≥ 1 is done by induction. We claim that there exists 0 < r1 < r2 < ⋅ ⋅ ⋅ < rp such that system (12) has a unique solution λ = λ(r1, . . . , rp) > 0,Mi = Mi(r1, . . . , rp) > 0 for i = 1, . . . , p. We have seen that the claim is true for p = 1. We assume the claim is true for p − 1 and we will prove it for p. Assume by induction hypothesis that there exists 0 < ?̃?1 < ?̃?2 < ⋅ ⋅ ⋅ < ?̃?p−1 such that system (12) has a unique solution ?̃? = λ(?̃?1, ?̃?2, . . . , ?̃?p−1) > 0 and ?̃?i = Mi(?̃?1, ?̃?2, . . . , ?̃?p−1) > 0 for i = 1, 2, . . . , p − 1. We need the next lemma. Lemma 4. There exists ?̃?p > ?̃?p−1 such that ?̃? = λ(?̃?1, ?̃?2, . . . , ?̃?p−1), ?̃?i = Mi(?̃?1, ?̃?2, . . . , ?̃?p−1) for i = 1, 2, . . . , p − 1 and ?̃?p = 0 is a solution of (12). Proof. Since Mp = ?̃?p = 0, we have that the first p − 1 equation of (12) is satisfied when λ = ?̃?, Mi = ?̃?i for i = 1, 2, . . . , p − 1 and Mp = ?̃?p = 0. Substituting this solution into the last equation of (12), we let f (rp) = ?̃? + ap,1?̃?1 + ⋅ ⋅ ⋅ + ap,p−1?̃?p−1 − N 󵄨󵄨󵄨󵄨󵄨1 + r2 p󵄨󵄨󵄨󵄨󵄨3/2 − M0 r3 p .