Non-negative Matrices and Distributed Control

Transcription

1 Non-negatve Matrces an Dstrbute Control Yln Mo July 2, 2015 We moel a network compose of m agents as a graph G = {V, E}. V = {1, 2,..., m} s the set of vertces representng the agents. E V V s the set of eges. (, ) E f an only f can sen nformaton to. The graph can be recte. Defne the neghbors N of agent as the set of agents who can sen nformaton to,.e., N { : (, ) E, }. Suppose each agent has a state x (t). The agent upate the state base on the followng upate equaton: Contnuous tme: Dscrete tme: t x (t) = a x (t) + x (t + 1) = a x (t) + N a x (t). N a x (t). We can wrte everythng n matrx form: Contnuous tme: Dscrete tme: x(t) = Ax(t). t x(t + 1) = Ax(t). Queston: s the system stable? Can we know the answer n a strbute fashon? 1 Some Defntons Let R n m + an R n m ++ be convex cones efne as R n m + {M R n m : M 0,, } R n m ++ {M R n m : M > 0,, } 1

2 Hence, we can efne X Y X Y R n m +, X > Y X Y R n m ++ Moreover, we efne Q R n +\{0}. A matrx A s calle postve f A > 0. It s calle non-negatve f A 0. It s calle a Metzler matrx f all the off-agonal entres are non-negatve,.e., A = B si, where B s non-negatve. Some observatons: If A 0 an x 0, then Ax 0. On the contrary, f for all x 0 an Ax 0, then A 0. Smlarly, f A > 0 an x Q, then Ax > 0. On the contrary, f for all x Q an Ax > 0, then A > 0. If A s Metzler, then exp(at) s non-negatve. A matrx A s calle Hurwtz f all ts egenvalues have strctly negatve real part. A s calle stable f all ts egenvalues satsfy λ < 1. A non-negatve matrx s calle prmtve f there exsts an k, such that A k s postve. A non-negatve matrx s calle rreucble f for any,, there exsts an k, such that ( A k) s postve. In general, a matrx A s calle rreucble f A s rreucble. Defne G(A) = (V, E) aw the graph assocate wth A, where V = {1,..., n} an (, ) E f an only f a 0. Let the pero of a vertex to be the greatest common vsor of the lengths of all cycles startng from. Some observatons: If A s rreucble then A + I s prmtve. (A k ) > 0 f an only f there exsts a path of length k from to. A s rreucble s equvalent to G(A) to be strongly connecte. If G(A) s strongly connecte, then all the vertces have the same pero. A s prmtve f A s rreucble an G(A) has pero 1 (aperoc). 2 Important Propertes of Non-negatve matrces an Metzler matrces Theorem 1 (Perron Frobenus Theorem). Let A be an rreucble matrx, then the followng propostons hol: 2

3 1. Let the spectral raus of A to be ρ(a), then there exsts an egenvalue λ of A, such that λ = ρ(a). 2. λ has geometrc an algebrac multplcty of The left an rght egenvectors of λ s strctly postve. Any other egenvector has negatve entres. 4. If A s prmtve, then all the other egenvalues satsfy λ < ρ(a). 5. ρ(a) satsfes: mn a ρ(a) max a. Proof. Frst let us efne the followng functon L : Q R + : L(x) max{s : sx Ax}. Clearly L(αx) = L(x) for any α > 0. Defne P = (I + A) k, where k s large enough such that P s postve. Hence, f sx Ax, whch mples that P (sx) P Ax = AP x, L(P x) L(x). Furthermore, f L(x)x Ax, then L(P x) > L(x). Now efne: λ max{l(x) : x 2 = 1, x Q}. Suppose λ s acheve at v. Then λv = Av. (otherwse L(P v) L(v).) Hence, λ s an egenvalue of A wth a postve egenvector v. Applyng the same proceure to A T, snce the spectral raus of A s the same as A T, we can fn a strctly postve left egenvector of A. Let us enote t as w. Now let µ λ be an egenvalue of A wth egenvector y. Then w T Ay = λw T y = µw T y. Hence, w T y = 0, whch mples that y must have negatve entres. Furthermore, µ y = Ay A y. Hence, µ L( y ) λ, whch fnshes the proof of tem 1. To prove tem 2, one can conser et(λi A) λ, λ=ρ(a) an prove that t s strctly postve. The etal s omtte. Please check the reference. 3

4 If A s prmtve, then A k s postve. Clearly the egenvalues of A k s the k-th power of the egenvalues of A. Hence, wthout loss of generalty, we can assume that A s postve an ρ(a) = 1 to prove tem 4. Let y be an egenvalue of A wth corresponng egenvalue µ, where µ = 1, then Suppose that z 0, then z = A y y 0. Az > 0, whch mples that there exsts an ε > 0, such that whch s equvalent to Thus, for all k, Az εa y, A A z A z. 1 + ε ( ) k A A z A z, 1 + ε whch contracts wth the fact that ρ(a) = 1. As a result, z = 0. Thus, y = A y, an y = Ay. Hence, y s ether all non-negatve or all non-postve, whch mples that y s ust a scalar multplcaton of v. Now to prove tem 5 we have an Hence, L(1) = mn A1 max w T A1 = λw T 1 whch mples that λ max a. a λ, max a 1. a w T 1, For a general A matrx, to prove t s stable, we nee to conser a Lyapunov functon of the followng form: V (x) = x T P x, where P s postve efnte an A T P A P s negatve efnte. Snce there s no guarantee that P s agonal (or comply wth the network topology), ths crteron cannot be easly strbute. However f A s non-negatve an rreucble, then we have 4

5 Theorem 2. If A s non-negatve an rreucble, then A s stable f an only f there exsts a postve w R n an 0 < δ < 1, such that The corresponng Lyapunov functon s gven by Proof. f : (1) s equvalent to w T A < δw T. (1) V (z) = w T z. V (Az) < δv (z). only f : If A s stable, then we can choose w as the left egenvector assocate wth λ = ρ(a). We can generalze ths result to contnuous tme an conser Metzler matrx. Assumng that A s a Metzler matrx wth A = B si, where B s rreucble. Hence, A s Hurwtz f an only f ρ(b) < s, whch s equvalent to the exstence of a postve w, such that w T B < sw T w T A < 0. To see ths, let v be the rght egenvector assocate wth ρ(b), then w T Bv = ρ(b)w T v < sw T v, whch mples that ρ(b) < s. Thus, we have the followng theorem: Theorem 3. If A s Meltzer an rreucble, then A s Hurwtz f an only f there exsts a postve w R n, such that The corresponng Lyapunov functon s gven by w T A < 0. (2) V (z) = w T z. Eq (1) an (2) can be verfe n a strbute fashon. 5