Modeling bone regeneration around endosseous implants

34 Chapter 2. Linear stability analysisThen, the characteristic polynomials of the matrices A kn and Ãk n, whichare evaluated at the constant solutions (m + , s 2+ , b + ) and (m + , s 2+ ), respectively,are denoted by a cubic polynomial P 3 (λ) and a quadratic polynomialP 2 (λ) with respect to λ:P 3 (λ) = det(A kn (m + , s 2+ , b + )−λI 3 ); P 2 (λ) = det(Ãk n(m + , s 2+ , b + )−λI 2 ).Equation (2.58) can be written as:whereP 3 (λ) = ( − A b − λ ) P 2 (λ) − C(k 2 n)λ, (2.59)C(kn) 2 = α p0AA kn(1,2)(m + , s 2+ , b + )A kn(2,1)(m + , s 2+ , b + )b= α ()p0 α m2 s 2+ αm β mA b β m2 + s 2+ (β m + s 2+ ) 2 m +(1 − m + ) + knB 2 m2 m + .(2.60)If s 2+ > 0, it follows from (2.20) that m + > 0, and equation (2.40) yields:m + = 1 −< 1. Thus,α p0+A mα m0 + αms 2+βm+s 2+s 2+ > 0 ⇒ 0 < m + < 1 ⇒ C(k 2 n) > 0, ∀k 2 n ∈ [0, ∞). (2.61)Lemma 2.4. Suppose, that for the chosen parameter values m 0 defined in(2.18) is positive, and that there exists a real positive s 2+ defined in (2.19).If the matrix Ãk n(m + , s 2+ ) has one real negative eigenvalue ˜λ 1 < 0 andone real positive eigenvalue ˜λ 2 > 0, then A kn (m + , s 2+ , b + ) has one realpositive eigenvalue and either two real negative eigenvalues, or two complexconjugated eigenvalues with a negative real part.Proof. From the assumption of the lemma and from (2.61) it follows, thatC(k 2 n) > 0. Let Ãk n(m + , s 2+ ) have one real negative eigenvalue ˜λ 1 < 0 andone real positive eigenvalue ˜λ 2 > 0. The characteristic polynomial can bewritten as P 2 (λ) = (λ − ˜λ 1 )(λ − ˜λ 2 ). Then from equation (2.59)P 3 (λ) = ( − A b − λ ) (λ − ˜λ 1 )(λ − ˜λ 2 ) − C(k 2 n)λ = −λ 3 + (˜λ 1 + ˜λ 2 − A b )λ 2From (2.62) it follows:+ (−˜λ 1˜λ2 + A b (˜λ 1 + ˜λ 2 ) − C(k 2 n))λ − A b˜λ1˜λ2 . (2.62)P 3 (0) = −˜λ 1˜λ2 A b > 0 and P 3 (˜λ 2 ) = −˜λ 2 C(k 2 n) < 0. (2.63)Since P 3 (λ) is continuous, it follows from relation (2.63), that the polynomialP 3 (λ) has at least one real positive root λ 1 on the interval (0, ˜λ 2 ).