Introduction to the Modeling and Analysis of Complex Systems

286 CHAPTER 14. CONTINUOUS FIELD MODELS II: ANALYSISthe Jacobian matrices, and the stability analysis is just calculating a Jacobian matrix andthen investigating its eigenvalues. Everything is so mechanistic and automatic, comparedto what we went through in the previous section. You may wonder, aren’t there any easiershortcuts in analyzing the stability of continuous field models?Well, if you feel that way, you will become a big fan of the reaction-diffusion systems wediscussed in Section 13.6. Their linear stability analysis is much easier, because of theclear separation of local reaction dynamics and spatial diffusion dynamics. To be morespecific, you can bring the Jacobian matrix back to the analysis! Here is how and why itworks.Consider conducting a linear stability analysis to the following standard reaction-diffusionsystem:∂f 1∂t = R 1(f 1 , f 2 , . . . , f n ) + D 1 ∇ 2 f 1 (14.80)∂f 2∂t = R 2(f 1 , f 2 , . . . , f n ) + D 2 ∇ 2 f 2 (14.81).∂f n∂t = R n(f 1 , f 2 , . . . , f n ) + D n ∇ 2 f n (14.82)The homogeneous equilibrium state of this system, (f 1eq , f 2eq , . . . , f neq ), is a solution ofthe following equations:0 = R 1 (f 1eq , f 2eq , . . . , f neq ) (14.83)0 = R 2 (f 1eq , f 2eq , . . . , f neq ) (14.84).0 = R n (f 1eq , f 2eq , . . . , f neq ) (14.85)To conduct a linear stability analysis, we replace the original state variables as follows:f i (x, t) ⇒ f ieq + ∆f i (x, t) = f ieq + sin(ωx + φ)∆f i (t) for all i (14.86)This replacement turns the dynamical equations into the following form:S ∂∆f 1∂tS ∂∆f 2∂t= R 1 (f 1eq + S∆f 1 , f 2eq + S∆f 2 , . . . , f neq + S∆f n ) − D 1 ω 2 S∆f 1 (14.87)= R 2 (f 1eq + S∆f 1 , f 2eq + S∆f 2 , . . . , f neq + S∆f n ) − D 2 ω 2 S∆f 2 (14.88)S ∂∆f n∂t.= R n (f 1eq + S∆f 1 , f 2eq + S∆f 2 , . . . , f neq + S∆f n ) − D n ω 2 S∆f n (14.89)