Mathematical Systems Theory: Advanced Course Exercise Session 5. 1 Accessibility of a nonlinear system

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1 Mathematical Systems Theory: dvanced Course Exercise Session 5 1 ccessibility of a nonlinear system Consider an affine nonlinear control system: [ ẋ = f(x)+g(x)u, x() = x, G(x) = g 1 (x) g m (x) ], where x N R n, N is an open set and u R m. We will discuss the accessibility of this system, which is a weaker concept than the controllability. Definition The system is called locally strongly accessible from x if for any initial point in the neighborhood of x, the set of reachable points with appropriate u contains a non-empty open set for any sufficiently small final time T. Proposition If dim R c (x )=n, then the system is locally strongly accessible from x, where R c (x) isthestrong accessibility distribution (see page 66 in the lecture note). Procedure to compute R c (x) Step 1. Take Set k =. Step 2. Compute Lie brackets and take R (x) =span{g 1 (x),,g m (x)}. [f,d], [g i,d], d(x) R k (x), R k+1 (x) =R k (x)+span{lie brackets which are not in R k (x)}. Step 3. Stop and set R c (x) =R k+1 (x)ifr k+1 (x) =R k (x), or dim R k+1 (x) = n, x N. Otherwise, return to Step 2 with k = k +1. Note: There is no guarantee that the process will end up. 1

2 Example Consider the angular motion of a spacecraft. Here we assume there are only two controls (two pairs of boosters) available. The model for angular velocities around the three principal axes is as follows: ẋ 1 = a 2 a 3 a 1 x 2 x 3 ẋ 2 = a 3 a 1 a 2 x 1 x 3 + u 1 ẋ 3 = a 1 a 2 a 3 x 2 x 1 + u 2 a 1 >, a 2 >, a 3 >. Let us compute the strong accessibility distribution R c (x) and check the accessibility of the system. In this case, αx 2 x 3 f(x) := βx 3 x 1, g 1 (x) := 1, g 2 (x) :=. γx 1 x 2 1 where α := (a 2 a 3 )/a 1, β =(a 3 a 1 )/a 2 and γ =(a 1 a 2 )/a 3. Step 1. R (x) =span{g 1 (x),g 2 (x)} =span{e 2,e 3 }. Step 2. Lie brackets are computed as follows: [f,g 1 ]= e αx 2 f 3 f(x) x x e 2 = γx 1 [f,g 2 ]= e αx 3 f 2 f(x) x x e 3 = βx 1 [g 1,g 2 ]=. =: g 3 (x) =: g 4 (x) Thus, R 1 (x) =span{g i (x),i=1,...,4}. Step 3. If α = (i.e. a 2 = a 3 ), then R 1 (x) =R (x). So, R c (x) =R (x) = span {e 2,e 3 }. If α,thenr 1 (x) R (x) and dim R 1 (x) =2< 3 for x 2 = x 3 =. Hence, go back to Step 2. 2

3 Step 2-2. R 2 (x) =R 1 (x)+span{[f,g i ], [g i,g j ], i,j =1, 2, 3, 4} Since [g 1,g 4 ]= g 4 x e 2 e 2 x g 4(x) = α, (α ) R 2 (x) =R 3 (whole space). Step 3-2 Since dim R 2 (x) = 3 for any x, R c (x) =R 3. Therefore, if a 2 a 3, then the system is locally strongly accessible from any point in R 3. 2 Stability for linear systems We will discuss the stability of the linear system 2.1 symptotic stability ẋ(t) =x(t). The matrix is stable (i.e., all the eigenvalues of have negative real parts) if and only if for any N<, there exists a unique solution P > forthe Lyapunov equation T P + P = N. In this case, if we define the Lyapunov function with the solution P,then V (x) :=x T Px, V (x(t)) = ẋ T (t)px(t)+x T (t)p ẋ(t) =x T (t)( T P + P)x(t) < for x(t). 3

4 Example 1 Check if the matrix = is stable, without computing the 2 3 eigenvalues. Set N = I 2 and solve the Lyapunov equation (you can use lyap.m): 2 p1 p 2 p1 p = 1 3 p 2 p 3 p 2 p }{{}}{{}}{{}}{{}}{{} T P P N P = The matrix P is positive definite, and therefore, is stable. 2.2 Critical stability Suppose that the matrix does not have eigenvalues with positive real part and has some eigenvalues on the imaginary axis. Such a case is called a critical case. In critical cases, the system is stable if and only if algebraic multiplicities of the eigenvalues on the imaginary axis equal geometric multiplicities. Example First, consider the system 1 ẋ = x, }{{} where has two eigenvalues at the origin (algebraic multiplicity is two). T For the two eigenvalues, there is only one eigenvector 1 (geometric multiplicity is one). Hence, the system is unstable. Indeed, { ẋ1 = x 2 ẋ 2 = = { x1 (t) =x 2 t + x 1 x 2 (t) =x 2 and x 2 (t) divergesifx 2. 4

5 Next, consider the system 1 ẋ = x, 1 }{{} where has eigenvalues at ±i, with algebraic multiplicity one. Since each eigenvalue corresponds to one eigenvector, algebraic and geometric multiplicities are the same for each eigenvalue. Hence, this system is stable. Indeed, { { ẋ1 = x 2 x1 (t) =r sin(t + φ) = ẋ 2 = x 1 x 2 (t) =r cos(t + φ) and the trajectory {(x 1 (t),x 2 (t))} t forms a circle with radius r. 3 Stability for nonlinear systems 3.1 Principle of stability in the first approximation Consider a nonlinear system ẋ = x + g(x), where g(x) indicates higher order terms than order one (i.e., g may include x 2 1, x 1x 2, x 2 2 etc.). Denote the set of all the eigenvalues of by σ(). Then, x =is exponentially stable if σ() C.(C is the open left half-plane.) unstable if σ() C +. (C + is the open right half-plane.) If has no eigenvalue in the open right half-plane but has at least one eigenvalue on the imaginary axis, then we need nonlinear stability theory, such as center manifold theory, to determine the stability of x =. Next, consider a nonlinear system with a control ẋ = x + g(x)+bu, where g isthesameasabove. If(, B) is stabilizable, then x = of the nonlinear system can be exponentially stable by using a state feedback u = Fx,whereF is chosen so that + BF is stable. 5

6 Example Consider a nonlinear system 1 ẋ = x + g(x), α β }{{} where g is a higher order term than order one. If α = 1 and β = 2: has two eigenvalues at 1, and hence x =is exponentially stable. (In fact, if α and β are negative, then is a stable matrix and x = is exponentially stable.) If α =and β =1: has 1 eigenvalue at 1, and hence x = is unstable. If α = β =: Since has eigenvalues only on the imaginary axis, we cannot determine the stability of x = by Principle of stability in the first approximation. 3.2 Stability for a special but important nonlinear system Consider a scalar nonlinear system ẋ = ax n, x() = x, where a is a real constant and n is a positive integer. Study the stability of this system. We consider several cases. If n =1: The system is linear. x(t) =e at x. If a < : Since x(t) ast, x =isasymptotically stable. If a =: Since x(t) =x for all t, x =is(critically) stable. If a > : Since x(t) as t, x = is unstable. If n>1: We solve the differential equation. ẋ = ax n x n dx = adt 1 1 n x1 n = at n x1 n, (since x() = x ) x(t) n 1 1 = (1 n)at + x 1 n 6

7 If a =: Since x(t) =x for all t, x =iscritically stable. If a : Since (1 n)at + x 1 n as t, x(t) ast. But the question is if x(t) at some time t (, ) for some x. By setting the denominator of x(t) n 1 equal zero, t := x1 n a(n 1) = 1 ax n 1 (n 1). Fact If a< and n is odd: Since x n 1 > for all nonzero x, t < and hence x(t) and x = is asymptotically stable. Otherwise: We can always choose x such that In summary, x =is 1 t = ax n 1 (n 1) >. Thus, x =isunstable. asymptotically stable if a < andn is odd, critically stable if a =, unstable otherwise. The stability of the system ẋ = ax n + O( x n+1 ) is the same as the stability of ẋ = ax n. (This fact will be useful when you learn center manifold theory.) 7