Modeling bone regeneration around endosseous implants

24 Chapter 2. Linear stability analysiswhere(A kn(1,1) = α m0 + α ms ′ )2β m + s ′ (1 − 2m ′ ) − (α p0 + A m ) − knD 2 m ,2A kn(1,2) =α mβ m(β m + s ′ 2 )2 m′ (1 − m ′ ) + knB 2 m2 m ′ ,A kn(2,2) =α m2β m2 ( α p0 )1 + m ′(β m2 + s ′ − A s2 − k2 )2 AnD 2 s2 .bThen from (2.30):⃗C n (t) = e A kn t ⃗ C0n , n = 0, 1, . . . , (2.33)where C ⃗ n 0 define the perturbations imposed on the constant solution of thesystem initially at time t = 0:⎡ ⎤m p (x, 0) ∞∑⎣ s 2p (x, 0) ⎦ = ⃗C nφ 0 n (x).b p (x, 0) n=0Thus the solution of (2.28) is written as:⎡ ⎤m p (x, t) ∞∑⎣ s 2p (x, t) ⎦ = e A kn t C ⃗ 0n φ n (x). (2.34)b p (x, t) n=0The magnitude of perturbations ‖ C ⃗ n (t)‖ = ‖e A kn t C ⃗ 0n ‖ of mode n, willgrow in time, if at least one of the eigenvalues of the matrix A kn is a positivereal number or a complex number with a positive real part. And ‖ C ⃗ n (t)‖ willconverge to zero, if all the eigenvalues of A kn are real negative, or complexnumbers with the real part less than zero. If the matrix A kn has preciselyone zero eigenvalue, and other eigenvalues are real negative or complex witha negative real part, then small perturbations remain small for infinite timeperiod.It is not complicated to find expressions for the eigenvalues of A kn ,evaluated at the ‘chronic non healing state’ z t = (0, 0, 0) and ‘low densitystate’ z 0 = (m 0 , 0, b 0 ). For the constant solution z t eigenvalues of A kn are:λ 1t (k 2 n) = α m0 m 0 − k 2 nD m > 0, if 0 ≤ k 2 n < α m0m 0D m,λ 2t (k 2 n) = −A s2 − k 2 nD s2 < 0, λ 3t (k 2 n) = −A b < 0.(2.35)Therefore, if m 0 is positive, constant solution z t is unstable against purelytemporal √ perturbations and perturbations with a small wavenumber 0