EG4321/EG7040. Nonlinear Control. Dr. Matt Turner

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1 EG4321/EG7040 Nonlinear Control Dr. Matt Turner

2 EG4321/EG7040 [An introduction to] Nonlinear Control Dr. Matt Turner

3 EG4321/EG7040 [An introduction to] Nonlinear [System Analysis] and Control Dr. Matt Turner

4 Motivation What we know: Lyapunov functions can be used for proving stability of nonlinear systems (without requiring solution of the differential equations) The choice of a Lyapunov function is (in general) not straightforward Aims of Lecture 1. To look at quadratic Lyapunov functions as initial choices for Lyapunov functions 2. To look at ways of removing indefinite terms in expressions for V(x)

5 Quadratic Lyapunov functions Given a (symmetric) positive definite matrix, P > 0, a quadratic Lyapunov function candidate is Automatically satisfies 1. Positive definiteness: 2. Radial unboundedness: V(x) = x Px V(x) = x Px > 0 x 0 x V(x) Therefore to check global asymptotic stability of the origin of ẋ = f(x), simply need to verify The trick is in the choice of V(x) V(x) = V(x) f(x) < 0 x 0 x

6 Quadratic Lyapunov functions - linear systems Consider the linear system ẋ = Ax Then the following statements are equivalent The system is globally asymptotically stable The eigenvalues of A have strictly negative real part There exists a positive definite symmetric matrix P such that A P +PA < 0 For any (symmetric) Q > 0 there exists a positive definite symmetric matrix P such that A P +PA = Q Implication: forlinear systems the Lyapunov function V(x) = x Px is entirely adequate

7 Quadratic Lyapunov functions - nonlinear systems Main difficulty with Lyapunov s method for nonlinear systems is Choosing an appropriate Lyapunov function Potentially limitless variety of Lyapunov function candidates Quadratic Lyapunov functions are not guaranteed to work However a good starting point... Many systems permit following partition ẋ = f(x) = Ax }{{} linear + g(x) }{{} nonlinear V(x) = x Px Positive definite Radially unbounded Only check time-derivative! If A is stable, quadratic Lyapunov functions useful

8 Use of quadratic Lyapunov functions - Process Assume ẋ = f(x) = Ax +g(x) R{λ i (A)} < 0 i Implication: can choose a Q > 0 such that P > 0 A P +PA = Q Hence, with V(x) = x Px V(x) = V(x) f(x) x = 2x P(Ax +g(x)) = x (A P +PA)x +2x Pg(x) = x Qx +2x Pg(x) Thus for V(x) < 0 1. x Pg(x) 0 x 0 We re done! either: 2. x Pg(x) 0 x 0...But if x Q x > x Pg(x) x 0 We re happy

9 Example Consider the nonlinear system: ẋ 1 = x 2 ẋ 2 = 2x 1 3x 2 +x 2 sinx 2 In this example, clearly [ ] [ ] ẋ1 0 1 = ẋ }{{} A [ x1 x 2 ] [ ] x 2 sinx 2 } {{ } g(x) Because A is stable, we know for our choice of Q > 0 there will exists a positive definite matrix P > 0 such that With Q = I (identity matrix) A P +PA = Q P = [ 5/4 1/4 1/4 1/4 ]

10 Example Thus choose V(x) = x Px and evaluate V(x) V(x) = x (A P +PA)x +2x Pg(x) [ ] [ ][ ] = x x1 P11 P x x x 2 P 12 P sinx 2 = x1 2 x2 2 +2[ ] [ ] P x 1 x 12 2 x P 2 sinx 2 22 = x 2 1 x x 1 P 12 x 2 sinx 2 +2x 2 P 22 x 2 sinx 2 Now because P 22 > 0 and because sinx 2 < 1 x 2 V(x) x 2 1 x x 1 P 12 x 2 sinx 2 +2P 22 x 2 2 = x 2 1 (1 2P 22)x x 1P 12 x 2 sinx 2

11 Example Next consider Young s Inequality 2z y 1 γ z 2 +γ y 2 γ > 0 z,y R m Applying Young s Inequality to yields 2x 1 P }{{ 12 x } 2 sinx 2 }{{} z y 2x 1 P 12 x 2 sinx 2 1 γ P2 12x 2 1 +γx 2 2 sinx γ P2 12x 2 1 +γx 2 2 Thus V(x) x 2 1 (1 2P 22)x γ P2 12 x2 1 +γx2 2 = (1 1 γ P2 12 )x2 1 (1 2P 22 γ)x 2 2

12 Example Bound on our Lyapunov function derivative is a quadratic function V (1 1 γ P2 12 )x2 1 (1 2P 22 γ)x 2 2 For this to be negative definite (and thus imply V(x) certainly negative definite) must have 1 1 γ P 12 > 0 1 2P 22 γ > 0 This translates into two inequalities on γ In our case P 12 = P 22 = 1/4 so γ > P 2 12 γ < 1 2P < γ < 1 2 Because Young s inequality holds for any γ we can always choose γ appropriately We have proved global asymptotic stability of the origin (using a quadratic Lyapunov function)

13 Another approach - conceptually more difficult Assume, as before, ẋ = f(x) = Ax +g(x) R{λ i (A)} < 0 i Again let P > 0 solve the Lyapunov equation A P +PA = Q and choose V(x) = x Px Now assume that where V(x) = x (A P +PA)x +2x Pg(x) = x Qx +2x Pg(x) x Qx +2x Pg(x) η 1 (x)+ǫ(x) η1(x) is positive definite ǫ(x) is indefinite...but η1(x) ǫ(x) To ensure V(x) < 0 we need to knock out the term ǫ(x) - how?

14 Another approach Assume there exists another positive definite function V 2 (x) (not necessarily radially unbounded) Assume further that where η 2 (x) 0 x V 2 (x) f(x) η 2 (x) ǫ(x) x Then, with the augmented Lyapunov function Ṽ(x) = V(x)+V 2(x) Ṽ(x) = V(x) f(x)+ V 2 x x f(x) = η 1 (x)+ǫ(x) η 2 (x) ǫ(x) = η 1 (x) η 2 (x) < 0 x 0 Note that 1. Ṽ(x) is positive definite 2. Ṽ(x) it radially unbounded because V(x) is radially unbounded

3 rd Tutorial on EG4321/EG7040 Nonlinear Control

3 rd Tutorial on EG4321/EG7040 Nonlinear Control Lyapunov Stability Dr Angeliki Lekka 1 1 Control Systems Research Group Department of Engineering, University of Leicester arch 9, 2017 Dr Angeliki Lekka