Modeling bone regeneration around endosseous implants

32 Chapter 2. Linear stability analysis2. If D γ = 0, and• for ∀k n ∈ (0, ∞): c(k 2 n) > 0 and λ 1 (k 2 n), λ 2 (k 2 n) are eitherreal and negative, or complex with a negative real part;• c(0) = 0 and λ 1 (0) < 0 and λ 2 (0) = 0.(a) if γ 1 ≤ 0, then ∃κ 1 = κ 2 =√− γ 12γ 2≥ 0 , such that• c(κ 2 1 ) = 0, and λ 1(κ 2 1 ) < 0 and λ 2(κ 2 1 ) = 0;• for k n ∈ [0, ∞)/{κ 2 1 }: c(k2 n) > 0 and λ 1 (k 2 n), λ 2 (k 2 n) are eitherreal and negative, or complex with a negative real part;(b) if γ 1 > 0, then for ∀k n ∈ [0, ∞): c(k 2 n) > 0 and λ 1 (k 2 n), λ 2 (k 2 n) areeither real and negative, or complex with a negative real part.3. If D γ < 0, then for ∀k n ∈ [0, ∞): c(k 2 n) > 0 and λ 1 (k 2 n), λ 2 (k 2 n) areeither real and negative, or complex with a negative real part.Therefore, if D γ > 0, and γ 1 < 0, then λ 2 (kn) 2 > 0, if k n ∈ (κ 1 , κ 2 ), andthe magnitude of the perturbation modes having wavenumbers k n ∈ (κ 1 , κ 2 )grows monotonically after a certain period of time. Hence, the constant solution˜z + = (m + , s 2+ ) is unstable with respect to these perturbation modes.Otherwise, ∀k n ∈ [0, ∞) the eigenvalues of the matrix Ãk nare eitherreal non-positive numbers (the matrix Ãk ncan not have more than onezero eigenvalue) or complex numbers with a negative real part. Hence,initially small perturbations remain small during any period of time, oreven disappear when t → ∞, and the constant solution ˜z + is stable in thiscase.The parameters γ 1 and D γ , can be written in terms of the model parametersasγ 1 = (D m A s2 + D s2 α m − χm + B m2 ) + D s2 α m0 m 0 , (2.54)β m2 + s 2+s 2+(s) 2+2D γ = (D m A s2 + D s2 α m − χm + B m2 ) + D s2 α m0 m 0β m2 + s(2+s 2+s 2+− 4D m D s2 A s2 α m0 m 0 + α m (2 − m + )β m2 + s 2+ β m2 + s 2+)+ α m (m + − 1) .(2.55)