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Systems, Structure and Control 2012 Part 10

Tham khảo tài liệu 'systems, structure and control 2012 part 10', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Almost Global Synchronization of Symmetric Kuramoto Coupled Oscillators 173 6i Sin 6 -_1 -ỡi sin 6i 1 -di i 1 . N Here the configuration is circular we identify 67 with 61 and 60 with 66. Consider the equilibrium point showed in Figure 3. Using an approach that will be presented later it can be shown that this configuration is locally attractive. n 3 Figure 3. Stable non-consensus equilibrium for the Kuramoto model of Example 4.1 We thus see that guaranteeing almost global asymptotical consensus is more involved. We will analyze the stability of the equilibrium points using Jacobian linearization. A first order approximation of the system at an equilibrium point 6 takes the form 86 A.86 with 86 6-6 and A the symmetric matrix N X N with entries aii - y cos 6k -6i -ai z J keN _lcos6 -6i h e N ahi 1 Io h e Ni i with ai defined as in Proposition 2.1. The matrix A can be written as A -B.diag cos bt 6 .Bt 5 and can be seen as a weighted Laplacian since A -L -B.BT at a consensus equilibrium. Two facts must be remarked. First of all A is symmetric reflecting the bidirectional influence of the agents. This implies that it is a diagonalizable matrix with real eigenvalues. Note also that A.1N 0. Hence A always has the zero eigenvalue with associated eigenvector 1n . We will analyze the transversal stability of the consensus set Khalil 1996 that is the convergence to the consensus set. The following results are true for general graph topologies. Their were originally introduced in Monzón et al. 2005 Monzón 2006 and Monzón et al. 2006 . Lemma 4.2 Let 6 be an equilibrium point of 3 such that at least one a 0 . Then 6 is unstable. 174 Systems Structure and Control Proof The numbers a appear at the diagonal of the matrix symmetric A. So a negative a implies that A has a positive eigenvalue. Then 0 is unstable. Lemma 4.3 Let 0 be an equilibrium point of 3 such that cosịớ c di 0 for every k e Nị i 1 . N . Then 0 is stable. Proof Since the underlying graph G is connected 0 is a .