Modeling bone regeneration around endosseous implants

30 Chapter 2. Linear stability analysiswhere χ is defined in (2.22). Next, it is proved, that inequality (2.48) holds.From equation (2.19) and assumption β m2 = β m it follows, thats 2+ + β m = −a 1 + √ a 2 1 − 4a 2a 02a 2+ β m ≥ − a 12a 2+ β m= − (α m0 + α m )(β m A s2 − χm 0 ) + α m χ(m 0 − 1) + α m0 β m A s22A s2 (α m0 + α m )+ 2β mA s2 (α m0 + α m )2A s2 (α m0 + α m )= α mβ m A s2 + χ(α m + α m0 m 0 ), (2.49)2A s2 (α m0 + α m )where a 2 , a 1 , a 0 are defined in (2.21). Since χ = α m2 (1 + α p0 /A b ) > 0 andm 0 is supposed to be positive, then from (2.26) it is derived that:(χ m 0 + α )√mα m≥ −A s2 β m + η α mχ. (2.50)α m0 α m0 α m0where η is defined in (2.25). Thus from (2.49) and (2.50) it follows:β m + s 2+ ≥ α mβ m A s2 + χ(α m + α m0 m 0 )2A s2 (α m0 + α m )≥ α mβ m A s2 − α m β m A s2 + √ ηα m α m0 χ2A s2 (α m0 + α m )√√ηαm α m0 χ=2A s2 (α m0 + α m ) = α m β m χA s2 (α m0 + α m )Thus inequality (2.48) holds, and consequently γ 0 ≥ 0.Remark 2.9. From the proof of Lemma 2.2 it follows, that γ 0 = 0, if andonly if a 2 1 − 4a 2a 0 = 0 which is equivalent for m 0 > 0 to(χ m 0 + α )√mα m= −A s2 β m + η α mχ. (2.51)α m0 α m0 α m0where η is defined in (2.25). In this case two constant solutions ˜z − and ˜z +coincide, since s 2− = s 2+ = − a 12a 0.It should be mentioned here, that under the assumptions of Lemma 2.2,c(0) = γ 0 ≥ 0. Then, Lemma 2.3 follows from Lemma 2.1.Lemma 2.3. If, for the chosen parameter values, m 0 defined in (2.18) ispositive, β m2 = β m and there exists a real positive s 2+ defined in (2.19),then for zero wavenumber k 0 , the matrix Ãk n(m + , s 2+ ) has either one zeroeigenvalue and one negative, or two negative eigenvalues, or two complexeigenvalues with a negative real part; and the constant solution (m + , s 2+ ) ofsystem (2.38) is stable against the purely temporal perturbations.