The motivating principal of linear systems is that even if a system isn t linear, it is at least locally linear.

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1 Linear Systems Notes 1 1 Lecture 1 Introduction Consider the following linear system ẋ = Ax + Bx (1) where x R n, which describes the dynamics of x, given by two components: Ax the position of x (A L(R n, R n ), a linear mapping), and Bu, where u R m is called the control (if u = 0, the system has no control and its dynamics are determined only by the state) and B L(R m, R n ). There s another element of the system: y = Cx + Du (2) where y represents some measurable state/output (in real systems this would correspond to components with sensors on them). Think of these two equations as a the inner workings of a black box into which you put a control u and get output y. The motivating principal of linear systems is that even if a system isn t linear, it is at least locally linear. Consider a pendulum, with rod length l, mass m, reference angle θ which lives in the unit circle, hence not linear, and finally some actuator which can be controlled (u). If θ = 0 corresponds to the direction of the acceleration of gravity, the mechanics which govern this system are given by ml θ + mg sin(θ) = u (3) Recall that for higher order systems, we introduce a new variable, e.g. θ = v and v = g l sin(θ) + u ml. Note that even though the location coordinates are nonlinear, velocity v is always tangent and therefore linear. The picture is as follows: on the one hand we have a circle and another a line (so composed a cylinder). Let s unwrap the cylinder so that we have a plane in R 2 with θ as (l θ) 2 the x-axis and v as the y axis. Conservation of energy implies that 2 + gl cos(θ) = a constant, i.e. v 2 /2 + g/l cos(θ) = c. These level sets describe the dynamics for various conditions. E.g. if the pendulum starts vertical then the pendulum will approach the vertical position again at, see figure. Two equilibrium: θ = 0, v = 0. Given a small perturbation, θ = 0 + ɛx 1 and v = 0 + ɛx 2, after substitution we get ɛẋ 1 = ɛx 2 (4) ɛẋ 2 = g/l sin(ɛx 1 ) = g/l(ɛx 1 ɛ3 x which can be simplified by killing higher order terms as +...) (5) ẋ 1 = x 2 (6) ẋ 2 = g/lx 1 (7)

2 Linear Systems Notes 2 whose level sets look like elipsoids instead, see figure 2. This linearized system simplifies the mathematics, but in so doing loses accurate representation of the system. Nevertheless, locally (near equilibrium) it satisfactorily describes how the system works. Again, given θ = v (8) v = g/l sin(θ) (9) with equilibrium points θ = π + ɛx 1 and v = 0 + ɛx 2, we get by substitution If ɛẋ 1 = ɛx 2 (10) ɛx 2 = g/l sin(π + ɛx 1 ) = g/l(ɛx 1 + O(3)) (11) ẋ 1 = x 2 (12) ẋ 2 = g/lx 2 (13) see figure 3. Apparently, this point is unstable, but this linearization shows how the system behaves locally. This system is given by linear components (i.e. x 1 and x 2 ) and constant coefficients. Now consider θ = v (14) v = g/l sin(θ) (15) where θ = θ(t) and v = v(t); then we consider a slight perturbation, θ θ + ɛx 1 and v v + ɛx 2. Substituting results in θ + ɛẋ 1 = v + ɛx 2 (16) i.e. v + ɛẋ 2 = g/l sin(θ + ɛx 1 ) = g/l sin(θ) g/lɛx 1 cos(θ) + O(2) (17) ẋ 1 = x 2 (18) ẋ 2 = g/l cos(θ(t))x 1 (19) Linearization about an equilibrium point gives us a linear time invariant system, and linearization about general trajectory gives a linear time variant system. Linear Algebra Definition 1. A field (F, +, ) consists of a set F together with two binary operations, + : F F satisfying the following: 1. (F, +) is an abelian group. 2. (F \ 0, ) is an abelian group and interact by distribution: a (b + c) = a b + a c for all a, b, c F. ( ) a b Consider the set of matrices of the form = a b a commutative. ( ) b 0 1 ( ) 0 1 which is clearly 1 0

3 Linear Systems Notes 3 2 Lecture 4 Definition 2. A vector space (V, F ) is a set V over a field F such that (V, +) is an abelian group such that F acts on V as follows: 1. av V for all a F and v V. 2. (a + b)v = av + bv for all a, b F, v V. In particular, 0 F v = 0 V for all v V. 3. a(v + w) = av + aw for all a F, v, w V. 4. (ab)v = a(bv) for all a, b F, v V. The most important vector space is F un(a, F ), the set of functions f : A F. Define f +g in F un as (f + g)(a) = f(a) + g(a). The standard euclidean vector space is R n = F un({1,..., n}, R) defined by F (1,..., n) = (v 1,..., v n ).? (Can think of (0,..., 1,..., 0) as e.g. the function δ k ). If u R n, can write u = (a 1,..., a n ) = n a i δ i. Cannot always represent vectors in this i=1 fashion, e.g. if the space is infinite dimensional. Nevertheless, we can always consider the set of elements in a linear space spanned by only a finite number of vectors. This is also a vector space. (For sum of two sums is also finite.) Definition 3. Given a vector space V, and W V, then W is a subspace if W is also a vector space (on its own). Let V = R 1, then Z R is not a subspace. On the other hand, let B A and consider the set of all functions f on A for which B ker(f). Another example of linear subspace. Let V be a vector space, and consider the set of linear functions f : V F (i.e. where f(av + w) = af(v) + f(w)), called linear functionals (or covectors), when the range is contained in R; call this set V ; the dual space of V. For example, if f V = C([0, 1], R), then 1 0 f(x)dx is a linear functional. f(0) = fδ 0 dx, evaluation at zero, with dirac function. Then can consider the set of functions which evaluated at zero are zero. R n = F un({1,..., n}, R). If u R n = u k δ k. Often there are canonical bases given by {δ k }, but there are always different ones. Definition 4. Given a collection of vectors (e i ) n i V, we say that (e i ) are linearly independent if n a i e i = 0 a i = 0 F for i = 1,..., n. i=0 Fact: Consider V, and consider f 1,..., f n V ; consider the set {f k (e j ) : j, k = 1,..., n}. Then f 1 (e 1 ) f 1 (e n ) det (20) f n (e 1 ) f n (e n ) and hence f k are linearly independent.

4 Linear Systems Notes 4 Definition 5. A basis is a maximal linear independent set of vectors. Proposition 1. Every vector space has a basis. Proof. Finite case: keep adding vectors not in the span until you can t go anymore. Infinite case: Zorn s lemma. Given a basis, there is a unique representation of each vector by its basis elements. Proof. Assume a i v i = b i v i ; then (a i b i )v i = 0 a i = b i by linear independence of basis vectors. Given u = c k e k, can think of the coefficients as [linear] functions of u, i.e. u = c k (u)e k = ( e k c k (u))u, which means that the sum e k c k is the identity. Now suppose u = n a i e i in the basis (e i ), and in another basis (e i ), u = i=1 n b i e i; u is the same vector, but working with it depends on understanding how to navigate the bases. There is a standard procedure: write each e i in terms of e i, i.e. e i = n n c i e i. So u = a k e k ( ẽ j e j) = a k (ẽ j(e k )) e j = j j k k=1 (P jk a k )f j, where P jk = ẽ j (e k ), and ẽ is the covector of e. k Example Consider the set of functions on {0, 1,..., 5} represented by the basis of δ k functions. Another basis: f 0 = z 0 = (1, 1, 1, 1, 1, 1), f 1 = z 1 = (0, 1, 2, 3, 4, 5), and in like manner the set of monomials of degree less than or equal to 5. i=1 j=1

5 Linear Systems Notes 5 3 Lecture 3 Let V = span({e 1,..., e n }) where any v V can be uniquely represented as v = n c k e k, c k R. Given another basis {f 1,..., f n }, to write the same vector v in terms of f i, just represent each e k in terms of f i, i.e. e i = n n p k,i f k, so v = c i p k,i f k = ( p k,i f k ). k i=1 k k i Recall Definition 6. The dual space V = L(V, F), where V is a linear space. Remarks: recall that f(av + w) = af(v) + f(w). Field action on the space is explicitly connected to the space you re in! So e.g. a W f(v) whereas in the domain a V v. The space of linear operators is itself a linear (vector space) defined for f, g : V W by (f + g)(a) = f(a) + g(a) where the latter sum is taken in W and the former in L(V, W ). Definition 7. Given a linear operator A L(V, W ), Null(A) = Ker(A) = {v V : Av = 0 W } and Ran(A) = Im(A) = {w W : Av = w some v V }. Example Consider the set of linear functions on F, L(F, F), with a L defined by a : v a v. Proposition 2. dim(n(a)) + dim(r(a)) = dim(v ). Consider the space V = F un(s, C) where S = {0, 1,..., 5}. Then A(f(0),..., f(5)) = (0, f(1),..., 5 f(5)); i.e. A is multiplication by z. Then N(A) = span({δ 0 }) and R(A) = {δ 1,..., δ 5 }. (Must verify that these elements actually are in the range, but that s easy since Aδ k = δ k.) In the standard δ basis A can be represented as A =. 0.. (21) Definition 8. The collection of all subspaces of a given space is called the Grassmanian ring, and it is not linear, denoted G(n, k) with dimension (n k)k Consider a linear mapping A L(V, W ), where V has basis (e i ) i m and W has basis (f j ) j n, and A represented as an n m matrix, Ae l = A(k, l)f k. With new bases (e ) and (f ), find a new matrix for the same operator. Represent old basis in terms of new basis: e k = P k,i e i. As i it stands A : V W by (e) A(f). Want a mapping (e ) (e) (f) (f ). So just compose representations: A(e k ) = A( P k,i e i), and also f j = Q j,l f l. Messy to expand, but conceptually i l simple: just iterate the process for each change of basis/transformation One special case: if V = W and (e ) = (f ), then A : (e) (f) becomes à = P AP 1 where P is the change of basis matrix from e e. The mapping A à defines an equivalence relation. i

6 Linear Systems Notes 6 Proof. A A by P = I. If A B, then A = P BP 1 B = P 1 AP = P 1 A(P 1 ) 1. Finally, if A B and B C, then A = P BP 1 = P QCQ 1 P 1 = P QC(P Q) 1. Example Consider the following operator. Let n N. Let f : n R (represented by n people who have, say, some amount of stuff), and the operation which takes what person i has, splits it in two and gives it to i 1 and i + 1. Can be represented as the following matrix: 0 1/2 1/2 1/ (22). 1/2 0 0 Definition 9. Let A : V V ; v 0 is an eigenvector if A(v) = λv for some λ F. If the eigenvalues form a basis of the range space, then the operator A : V W can be represented diagonally. Similarly if A D D(n, F), then the eigenvectors of A form a basis. Over C there s always at least one eigenvalue (follows from FTA).

7 Linear Systems Notes 7 4 Lecture 4 Let V, W be vector spaces over some field F, and A : V W (from dimension n to m), where B V = {v 1,..., v n } is a basis for V and B W = {w 1,..., w m } is a basis for W. In order to find the image, need to find rank(a) = r; start by finding the kernel of A, ker(a) V, and a basis for it, namely some B ker(a) {v j1,..., v jr } B V, then to find the image, see where the remaining basis elements B v \ B ker(a) map to in W. Now consider A : V V. Can split the image space into the following: the set of things which map to zero (null space), the set of things which map to the image, then the remainder of the range space which isn t mapped to by elements in the domain. Suppose that each individual basis element is mapped to a scaled factor of another basis element, then the matrix representing this linear transformation is given by a diagonal matrix. Unfortunately, not every linear mapping has a diagonal representation. For example, consider V = {p F[x] : dim(p) k}, given by basis B V = {x j : j = 0,..., k}. Take the linear mapping D : V V given by action on the standard basis element Dx j = jx j 1 (where x 1 = 0 by convention). The matrix of this transformation is represented by D = (23) This operator is called nilpotent, because there is an n N (namely d = k) such that D n 0 on L(V, V ). e.g = 0 M 3 (R) (24) Obviously a nilpotent matrix has a nontrivial kernel. 4.1 Jordan Normal Form Let A : V V where V has dimension V. It is possible to find a canonical Jordan representation of A consisting of Jordan blocks, which is a direct sum of a identity operator and nilpotent part, i.e. A = λi k N (25) where k indicates the size of the Jordan block. For example λ 1 0 A = 0 λ 1 (26) 0 0 λ

8 Linear Systems Notes 8 For example, e λx p k (x) where p k (x) F[x] is a polynomial of degree k; this is called a quasipolynomial, which has the property that its closed under differentiation, i.e. the space V = {e λx p(x) : p(x) V }, (V from above), then Ee l = λe l + e l 1. Theorem 1. Let (V, C) be a vector space, and A L(V, V ); then there is a basis B w.r.t. which A has a matrix representation in Jordan Normal Form. Take away: all transformations look like generalized differentiation of quasi polynomials. 4.2 Other Canonical Forms Let A : V V. Definition 10. let A L(V, V ). A vector v V is called cyclic in A if the set {A k v : k = 0,...} spans of V. Note that if v is cyclic, then {v,..., A n 1 v} suffices to span the space. Let A L(V, V ), can write A mat M n (R), and consider the sequence (I = A 0, A, A 2,...); these matrices are linearly dependent, and moreover the relation can be easily described: Theorem 2 (Cayley-Hamilton). A n + k=1 n Aa k A n k = 0 where a k are coefficients of the characteristic polynomial of A, i.e. det(λi A) = λ n + a k λ n k. Proof. Easy if A is diagonalizable. Let char A (t) be the characteristic polynomial of A in the n indeterminate variable t. If A is diagonalizable, then char A (t) = (t λ k ) where λ k are the diagonal values of A (i.e. the eigenvalues of A); then char A (A), i.e. evaluation of char A (t) at A. k=1 4.3 Functions of Matrices n Given a polynomial p F[x], can evaluate at A, P (A) = a k A k. On the other hand, we can express functions as formal power series, namely f(t) = if it is convergent for all t. k=1 f k t k /k!, f k A k Definition 11. Can do so similarly for A, f(a) =. In particular if A D n (R), then k! f(a) = (f(a kk ) n k=1 ). k=1 k=1

9 Linear Systems Notes 9 Definition 12 (Matrix Exponential). ϕ(t) = e ta is well defined, as e ta = I + ta + t 2 A 2 / satisfying various properties: 1. e (t+s)a = e ta e sa 2. ϕ (t) = Aϕ(t). Suppose x 0 V, and we have a differential equation of the form ẋ = Ax, x(0) = x 0. Then the above tells us that if x(t) = ϕ(t)x 0, then ẋ = Aϕ(t)x 0 = Ax by definition.

10 Linear Systems Notes 10 5 Lecture 5 For ẋ = Ax where A M n (R) is called a linear time invariant system. The solution x(t) = ϕ(t)x 0, with ϕ(0) = x 0. Recall that ϕ = ɛ ta = I + ta + t2 A 2 2! +... Note that e ta does in fact converge. Take some matrix A, and consider A = max i,j a ij, then A 2 i,j na2, and similarly A k i,j nk 1 a k. So e ta i,j = δ i,j + ta i,j + t 2 A 2 i,j /2! tk (A k ) i,j /k! +... k! = 1 k k+1/2 2π e k (1 + o(1)). How to compute the matrix exponential? It s not good to use the power series expansion to compute the exponential. There are several better alternatives. 5.1 Computing the Matrix Exponential First method: diagonalization. Matrices behave well w.r.t. change of basis. Given à = P 1 AP, then e A = e P 1 AP = I + tp AP 1 + t 2 P A 2 P 1 /2! +... = P (1 + ta +...)P 1 = P e At P 1. This implies that if we can find a basis w.r.t. which matrix multiplication is easy to compute (e.g. diagonal matrices), then we should transform to that. Example Let A = ( A k fk 1 f = k f k f k+1 ( ) 0 1, A = ( ) 1 1, A = ), where f k is the k-th Fibonacci number. ( ) 1 2, A = ( ) 2 3. In general 3 5 ( ) Since σ(a ) = { 0.618, 1.618} are distinct, A is diagonalizable, with P =, ( ) ( ) e hence A = P A P 1 =, and e A 0.618t 0 t = 0 e 1.618t. Therefore, e A t = e d 1 0 P 1 e At P. (Note, we skipped steps: e D = because (D k ) i,j = (D i,j ) k.) 0 e dn Now consider Jordan matrices: J = λi + N where N is nilpotent. Then e tj = e tλi+tn = e tλi e N = (e tλ ) (I + tn + λn t k 1 N k 1 /(k 1)!), since N k = 0. Therefore, 1 t t k 1 /(k 1)! e tn 0 1 t k 2 /(k 2)! = (27) Another method: e ta = L 1 [(si A) 1 ]. This requires inverting the matrix (si A) and computing inverse Laplace component wise. e.g. ( ) 1 ( ) s 1 1 s 1 1 = 1 s 1 s 2 (28) s 1 1 s

11 Linear Systems Notes 11 This method works for small matrices, but it is computationally intensive. Finally, we can also use Cayley Hamilton: recall that for for any A M n (R), char A (A) 0 M n (R). Therefore A n can be expressed as a linear combination of matrices of lower power (similarly, A n+1 = A n A can be too, and so on). Then t j A j /j! = β k (t)a k. Though this k<n makes the sum of matrices finite, β k (t) will be a formal power series in t. j=0 5.2 Using Matrix Exponential in System Given ẋ = Ax + Bu and y = Cx + Du, a solution is given by x(t) = ϕ(x)x 0 + t 0 ϕ(t s)bu(s)ds; this is the solution for linear LTI systems; y(t) = Cϕ(t)x 0 + t 0 Cϕ(t s)du(s)ds + Du(t). ϕ should satisfy several properties: 1. ϕ(t s)ϕ(s r) = ϕ(t r). 2. ϕ(t t) = id. 3. ϕ(t s)ϕ(s t) = id.

12 Linear Systems Notes 12 6 Lecture 6 Now consider ẋ = A(t)x + B(t)u and y = C(t)x + D(t)u, the solution given by x(t) = ϕ(t s)x(s). The solution is found by integrating: x(t) = ϕ(t s)b(s)ds. Thus x(t + ) x(t) ϕ(t + t)ϕ(t + )x(s) ϕ(t s)x(s) ϕ(t + ) I A(t)x(t) = lim = lim = lim ϕ(t + t) ϕ(t) from which it follows that lim = A(t)ϕ(t s) = 0 ϕ(t s). t Question: how to solve ϕ t (t s) = A(t)ϕ(t s). If A(t) A is linear time invariant, then ϕ(t s) = e (t s)a. On the other hand, consider the ordinary differential equation, ẋ = a(t)x, the solution is given by x(t) = e a(s)ds x(0). But in higher dimension, this approach may not work. However, if [A(t), A(s)] = 0 for all t, s R (where [, ] denotes the standard Lie bracket in gl n (V )), then ϕ(t s) = e t s A(u)du (30) The way to verify this is to use power series expansion from the definition of the matrix exponential. Two operators commute iff they can be diagonalized simultaneously, i.e. we can represent the basis for each as a direct sum of one-dimensional spaces (in other words, if there is a diagonal basis for both of them). (29) x(t) 6.1 Uniqueness of Differential Equations Let x V = R n, and consider We want to consider conditions which guarantee uniqueness. ẋ = f(x, t) (31) Example ẋ = x, x = 0, and x = t 2 /4 are two different solutions to this system. But f x = 1 and explodes as x 0 which is why the solution is not unique. 2 x The statement: if f(x, t) is continuously differentiable in x, then the solution exists and is unique. Proof Idea. Proof uses Picard approximation. See [Arnold] for details. Work in steps: start with 0 order approximation and follow the trajectory traced out by velocities recorded along the guessed solution, to get the first approximation: x 1 (t) = x 0 + t 0 f(x 0(x), s)ds, then iterate this procedure to get the next approximated solution, x 2 (t) = x 0 + t 0 f(x 1(s), s)ds. Etc. For example, let ẋ = x with x(0) = 1 and let x 0 (t) = 1, then x 1 (t) = x 0 + t 0 x 0(s)ds = 1 t, then x 2 (t) = x 0 + t 0 (1 s)ds = 1 t + t2 /2. Then x 3 (t) = 1 + t 0 (1 s + s2 /2)ds = 1 t + t 2 /2 t 3 /6, continuing in this way it is clear that the solution is x (t) = ( 1) j t j /j!, and each finite sum is just the Taylor approximation. j=0

13 Linear Systems Notes 13 Now given ϕ(t) = A(t)ϕ(t), with ϕ(0) = Id. Using Picard approximation, ϕ 0 (t) = Id, ϕ 1 (t) = Id + t 0 A(s)ϕ 0(s)ds = Id + t 0 A(s)ds, ϕ 2(t) = Id + t 0 A(s)ϕ 1(s)ds = Id + t 0 A(s)(Id + s 0 A(u)du)ds = Id + t 0 A(s)ds + t s 0 0 A(s)A(u)duds. Again, ϕ 3(t) = Id + 0<s 1 <t A(s)ds + 0<s 1 <s 2 <t A(s 2)A(s 1 )ds 1 ds 2 + 0<s 1 <s 2 <s 3 <t A(s 3)A(s 2 )A(s 1 )ds 1 ds 2 ds 3. Continuing in this way to infinity, can prove (analogous to proof of convergence of matrix exponential) that this series will converge, will be well defined (i.e. unique), and will be the solution. Returning to controls, x(t) = ϕ(t 0)x(0) + t 0 ϕ(t s)b(s)u(s)ds and y = C(t)x(t) + D(t)u(t). Given a purported ϕ, what are the conditions which verify that it is the fundamental matrix? Det(ϕ) cannot be negative; any ϕ with positive determinant can be a fundamental ( ) matrix 2 0 of a linear time variant system ϕ(t) = A(t)ϕ(t). For LTI, for example if ϕ(1) =, then let 0 3 ( ) log(2) 0 A =, then e 0 log(3) A works. In this case it is not enough that the determinant is positive, ( ) 2 0 but if e.g. ϕ =, then we have a system which expands and rotates, which cannot be 0 3 done time independently. Let J = ( ) ( cos(t) sin(t) e tj = sin(t) cos(t), and compute e tj. It rotates by 90. Then J 2 = Id, i.e. J = Id. Then J on [0, π] ), the operator of rotation. Let A(t) = ẋ(t) = ϕ(t)x(0), ẋ = A(t)x, l x(t) = y(t), lϕ(t)x(0) ( log(2) 0 0 log(3) ) on[π, π + 1]

14 Linear Systems Notes 14 7 Lecture 7 LTV ẋ(t) = Ax(t) (32) where x V = R n, ϕ(t s) : V V, and x(s) = x s then the solution to the above two has x(t) = ϕ(t s)x(s), y(t) = l(t) x(t) = l(t)ϕ(t s)x(s) where l(s) = l(t)ϕ(t s); ϕ : V s V t, but the l Vt = L(V t, R), so ϕ : Vt Vs. ϕ = Id + s u 1 t A(u 1)du 1 + s u 1 u 2 t A(u 2)A(u 1 )du 1 du s u 1... u A(u n t n) A(u 1 )du 1... du n. b af (b, a) = a a f(x)dx... dl/ds = l(t)(i + A +...) = l(s)a(s). Example ẋ 1 = v, v = a and ȧ = 0. This system can be represented by ẋ = Ax where A = Then ϕ(t s) = e (t s)a ; suppose the initial conditions are x(0) = x 0, v(0) = v t s (t s) 2 /2 x 0 x 0 + (t s)v 0 + (t s) 2 /2a 0 and a(0) = a 0. Then x(t) = 0 1 t s v 0 = v 0 + (t s)a a 0 a 0 Say that y(t) = ( ) x 1 (t) x 2 (t) = x(s) + (t s)v(s) + (t s) 2 /2a(s). Then l(s) = (1, (t x 3 (t) s), (t s) 2 /2). Then d/ds(l(s)) = (0, 1, (t s)) = (0, l(1), l(2)) = ( ) l 1 l 2 l Stability Consider the system ẋ 1 = x 2, ẋ 2 = 2x 1 + x 2 + u, and y = x 2 2x 1. Applying Laplace, sx 1 = X 2, Y = X 2 2X 1, sx 2 = 2X 1 + X 2 + U, s 2 X 1 = 2X 1 + sx 1 + ( U, and ) solving we obtain that X 1 = U s 2 s 2, X 2 = su s 2 s 2, Y = (s 2)U 0 1 (s 2)(s+1). In matrix form, ẋ = x. 2 1 Definition 13 (Lyapunov Stability). x 0 is Lyapunov stable if for any ball B(x 0, δ) one can find ɛ > 0 such that x(0) x 0 < ɛ x(t) x 0 δ A more familiar way to say it is that the mapping from equilibrium point to trajectory is continuous (with norm on the on trajectory s sup). Definition 14. An equilibrium point x 0 is asymptotically stable if it is Lyapunov stable and there is an ɛ > 0 such that for every trajectory satisfying x 0 x(0) < ɛ, x(t) x 0 as t.

15 Linear Systems Notes 15 8 Lecture 8 Let ẋ = f(x) (33) and x 0 = 0 an equilibrium point. Then x 0 is Lyapunov stable if for every ɛ > 0 there is a δ > 0 such that x(0) x 0 < δ guarantees that x(t) x 0 < ɛ for all t R. x 0 is asymptotically stable if it is Lyapunov stable and for any x(0) x 0 < δ, x(t) x 0. Global asymptotic stability is asymptotic stability with no condition on initial condition. Exponential stability is stability with negative exponential bound, x(t) < e αt. 8.1 Stability for LTI Systems Theorem 3. The linear time invariant system ẋ = Ax (34) is Lyapunov stable iff for σ(a) = {λ C : Ax = λx}, Re(λ) 0 for every λ σ(a), and the Jordan cells of λ σ(a) with Re(λ) = 0 have size 1. Recall, for any operator A : R n R n, we can decompose A (i.e. represent it as a matrix over some basis) as J 1 0 A = (35) 0 J k where J j has the form λi nj + N nj where I nj is the identity matrix in M nj (R) and N nj is the canonical nilpotent matrix (N n j n j = 0) in M nj (R). The size of a Jordan block corresponds to a function of the multiplicity of the eigenvalue. Note: This does not mean that an eigenvalue of multiplicity 3 has a Jordan block of size 3, but that the Jordan form corresponding to this eigenvalue can have size 3, 2 and 1, or three size 1 Jordan blocks. ( ) λ 1 The reason for this is that if a Jordan block is nontrivial, e.g. J = will give 0 λ ẋ = ( ) (36) which has solution x(t) = e ta x(0) = 1 t 0 1 x(0) (37) Theorem 4. Linear time invariant systems are asymptotically stable iff they are globally explonentially stable.

16 Linear Systems Notes 16 Proof. ( ): Immediate from the definition. ( ): First of all, asymptotic stability implies global asymptotic stability for linear systems. Linearity guarantees that if something which starts close approaches the origin, then by (linear) scaling, everything else does too. (e.g. ˆX = cx, then d/dx ˆX = cd/dx(x) = cax = AcX = A ˆX). This all happens iff Re(σ(A)) < 0. (If Re(λ) < 0 for all λ σ(a), then A is defined to be Hurwitz.) 8.2 Lyapunov Theory Given ẋ = Ax for x V = R d, and let U : V R be a sensor (indicator) which is continuously differentiable, with x 0 an equilibrium point. Suppose the following are satisfied: 1. U(x 0 ) = 0 2. U > 0 (positive definite) 3. in some open neighborhood of x 0 du(x(t)) dt 0. If a function U satisfies these conditions, U is called a Lyapunov function. Theorem 5. If U is a Lyapunov function for ẋ = f(x) at x 0 Ω. Then x 0 is Lyapunov stable. Proof. Given some B(x 0, δ) Ω, take inf x Sδ U(X) which is certainly obtained since B is compact, call it U 0. Consider the boundary of the closed set where {U(x) = U 0 /2}; on that boundary, take the point closest to the origin, i.e. inf x {x:u(x)=u0 /2} x, and again this isn t the origin, let that radius be r. Within this ball, all values of U are smaller than U 0 /2. and ( ) ( ) Calculation: d(u(t)) dt = U = U x 1,..., U x n x (t) = U x 1,..., U x n f(x). Example Let ẋ 1 = x 2 + x 1 ( ɛ + x x 2 2) (38) ẋ 2 = x 1 + x 2 ( ɛ + x x 2 2) (39) Let U(x 1, x 2 ) = x 2 1 +x2 2. We verify that this function is indeed a Lyapunov function: first U(0, 0) = 0 and is clearly positive definite. Then U = ( ) 2x 1 2x 2, so d dt U(x(t)) = U f(x) = 2x 1 ( x 2 ɛx 1 + x 1 r 2 ) + 2x 2 (x 1 ɛx 2 + x 2 r 2 ) = 2ɛr 2 + r 4, where r 2 = x x2 2. So choose neighborhood inside ball with radius r 2 < ɛ, i.e. Ω = {x x2 2 < ɛ}.

17 Linear Systems Notes Relation between Linear and Nonlinear ẋ = f(x) (40) f(x 0 ) = 0, for x x 0, we say that f(x) = f(x 0 ) + Jf(x 0 )(x x 0 ) +... where the dots encapsulate nonlinear terms (Taylor). I.e. ẋ = Ax + g(x) (41) for small x, g(x) C x 2. If U is a good Lyapunov function for the linear part Ax (meaning that additionally U Ax is negative definite, and U is homogeneous); then U is also a Lyapunov function for ẋ = Ax + g(x). Definition 15. U is homogeneous of degree k if U(λx) = λ k U(x).

18 Linear Systems Notes 18 9 Lecture Lyapunov Method Given ẋ = f(x) with x V = R d, f(0), let U : V R be positive definite with U f = 0, in some open neighborhood Ω containing 0. ẋ = f(x) = Aẋ + g(x) with g(x) C x 2 in Ω, where g is homogeneous of degree k, i.e. U(λx) = λ k U(x). Examples 1. Consider a j x j, x ij x i x j or J =k U(x) is positive definite of degree 2k > 0 only. a j x i x n where i + n = k. Homogeneous polynomials 2. U (homogeneous of degree 2k > 0), is positive definite iff U S(0,1) > 0 (where S(0, 1) denotes the unit sphere), for take x = λs S(0, 1), with λ > 0. Then U(x) = λ 2k U(s). ( ) x 1 3. U A(x) = Q(X) = U x 1,..., U x D A. where each U x j is polynomial of degree x d 2k 1, so the result Q(x) is also a polynomial of degree 2k. This implies thast if Q S1 < 0 then U f(x) < 0 in some open neighborhood of 0. So for small perturbations, U(Ax + g(x)) < 0 still holds, because (U(Ax + g(x))) = Q(x) + U g(x), with the first term hom. of degree 2k, and U g(x) U g(x). Recall that U x j C j x 2k 1, which implies that the total norm of U will be bounded by e.g. d max j {C 1,..., C d } x 2k 1, and therefore, U g(x) B x 2k+1 (since g(x) C x 2 ). Then Q(x) a (a > 0), for x = 1, then Q(x) a x 2k by homogeneity: x = x x x, so Q(x) = x 2k Q( x x ). Hence Q(x) + U(x) g(x) a x 2k+1 + B x 2k+1, and for x a a 2B, we have U f(x) 2 x 2k, which is negative definite, and so satisfies the conditions for Lyapunov stability. The point of this last example is that stability of nonlinear systems can be determined, at least locally, by inspection on the stability of the linear part. 9.2 Quadratic Forms Let V = R d, and Q : V R; we define Q(x, y) = 1 2 (Q(x + y) Q(x) Q(y)). Q is quadratic if Q is bilinear, i.e. Q(ax + x, by + y ) = abq(x, y) + aq(x, y ) + bq(x, y) + Q(x, y ). More concretely, fix a basis B. Then the quadratic functions are those with the form Q(x) = a i,j x i x j. They are important because Lyapunov functions for linear time invariant systems 1 i,j d will generally have a quadratic form. The matrix representing this will be symmetric: x T Ax = Q(x).

19 Linear Systems Notes 19 There s a natural quadratic form: C(x) = x 2 j = xt x = x. Note: these computations depend on the system of coordinates (choice of basis). So far we have two quadratic forms: one defined by A and the unit matrix which gives the norm of a vector. Theorem 6. For any any quadratic form Q : V R, there exists P Gl n (R) such that Q(P (Y )) = ±y 2 i. Thus, the quadratic form is characterized by (n +, n, n 0 ), the number of +yi 2, y2 i, and 0, respectively. For example, could have positive definite (2, 0, 0), negative definite (0, 2, 0), positive semidefinite (1, 0, 1), nothing (0, 0, 2). Example Let Q(x) = 9x + 1 x2 2, which is not in canonical form. Therefore, define x 1 = 3x 1 and x 2 = x 2. Then Q(x) = x 2. (Here we are stretching the space in one dimension). 1 + x 2 2 If however, we want to preserve the metric, then it won t be as easy to transform to the canonical form. Theorem 7. If P is distance preserving, i.e. x = P x for all x V, or also P T P = Id V, then any quadratic form Q can be transformed by an orthonormal matrix P to Q(P y) = λ j y 2. Theorem 8. With conditions as in the previous theorem, each λ j R. Proof. Ax = λx, and take for x = u j + v j i (i = 1), x Ax = a kj (u k v k i)(u j + v j i) + a j,i (u j v j i)(u k + v k i) from which after staring at this equation for a while, it becomes clear that all imaginary parts disappear. Furthermore, x x = u 2 j + v2 j, so λx x is real. Here Qx = λcx. Proposition 3 (Rayleigh s Min-Max Characterization of λ i ). In order to characterize the largest λ, take the unit sphere; it will be the one of steepest ascent. In other words, λ max = max x =1 Q(x), and λ min = min l max x <1 Q(x). Order λ 1... λ d. Then λ k = min linearsubspacesofdimensionk max x =1,x L k Q(x).