Introduction to the Modeling and Analysis of Complex Systems

18.3. SYNCHRONIZABILITY 411The first terms on both sides cancel out each other because x s is the solution of dx/dt =R(x) by definition. But what about those annoying H(x s )’s included in the vector in thelast term? Is there any way to eliminate them? Well, the answer is that we don’t have todo anything, because the Laplacian matrix will eat them all. Remember that a Laplacianmatrix always satisfies Lh = 0. In this case, those H(x s )’s constitute a homogeneousvector H(x s )h altogether. Therefore, L(H(x s )h) = H(x s )Lh vanishes immediately, andwe obtaind∆x idt⎛= R ′ (x s )∆x i − αH ′ (x s )L ⎜⎝⎞∆x 1∆x 2⎟.∆x nor, by collecting all the ∆x i ’s into a new perturbation vector ∆x,d∆xdt⎠ , (18.11)= (R ′ (x s )I − αH ′ (x s )L) ∆x, (18.12)as the final result of linearization. Note that x s still changes over time, so in order forthis trajectory to be stable, all the eigenvalues of this rather complicated coefficient matrix(R ′ (x s )I − αH ′ (x s )L) should always indicate stability at any point in time.We can go even further. It is known that the eigenvalues of a matrix aX +bI are aλ i +b,where λ i are the eigenvalues of X. So, the eigenvalues of (R ′ (x s )I − αH ′ (x s )L) are−αλ i H ′ (x s ) + R ′ (x s ), (18.13)where λ i are L’s eigenvalues. The eigenvalue that corresponds to the smallest eigenvalueof L, 0, is just R ′ (x s ), which is determined solely by the inherent dynamics of R(x) (andthus the nature of x s (t)), so we can’t do anything about that. But all the other n − 1eigenvalues must be negative all the time, in order for the target trajectory x s (t) to bestable. So, if we represent the second smallest eigenvalue (the spectral gap for connectednetworks) and the largest eigenvalue of L by λ 2 and λ n , respectively, then the stabilitycriteria can be written asαλ 2 H ′ (x s (t)) > R ′ (x s (t)) for all t, and (18.14)αλ n H ′ (x s (t)) > R ′ (x s (t)) for all t, (18.15)because all other intermediate eigenvalues are “sandwiched” by λ 2 and λ n . These inequalitiesprovide us with a nice intuitive interpretation of the stability condition: the influenceof diffusion of node outputs (left hand side) should be stronger than the internaldynamical drive (right hand side) all the time.