Sliding Modes in Filippov Systems.

in Filippov Systems. School of Mathematics, Georgia Tech Joint work with Luciano Lopez (Univ. Bari) Geneva Conference In honor of Ernst Hairer 60th birthday

References Filippov [DEs with discontinuous RHS, 1988] Utkin [Sliding modes in control & optimization, 1991] Yakubovich Leonov Gelig [Stability of stationary sets in control systems with discontinuous nonlinearities, 2004] di Bernardo-Budd-Champneys-Kowalczyk [P.wise smooth dynamical systems, 2008] Acary-Brogliato [Numerical methods for nonsmooth DynSys, 2008] Hairer-Norsett-Wanner [v.1, 1992, pp. 196 & ss]... and many more in specialized literature... Control (Bartolini, Usai, Fridman, Johansson,...), in Optimization (Cortes, Glazos-Hui-Zak, Venets,...), in Dynamical Systems (Leine et al., Y. Kuznetsov, Mueller, Zou-Kupper-Beyn, Dercole, Rinaldi, Galvanetto,...), and Numerical: D.Stewart, Mannshardt, Shampine-Thompson, Gear, Enright, DAE,... [D. Lopez].

Outline of Talk 1. Which non-smooth? Filippov theory, sliding modes.

Outline of Talk 1. Which non-smooth? Filippov theory, sliding modes. 2. Entry and exit.

Outline of Talk 1. Which non-smooth? Filippov theory, sliding modes. 2. Entry and exit. 3. Projections and sliding vector field.

Outline of Talk 1. Which non-smooth? Filippov theory, sliding modes. 2. Entry and exit. 3. Projections and sliding vector field. 4. Sliding on intersection of p 2 surfaces.

Outline of Talk 1. Which non-smooth? Filippov theory, sliding modes. 2. Entry and exit. 3. Projections and sliding vector field. 4. Sliding on intersection of p 2 surfaces. (a) Case of p = 2: [DiLo] vector and [Alexander-Seidman] choice.

Which non-smooth? Focus on piecewise smooth systems (switched systems). One surface case. x (t) = f(x(t)) = { f1 (x(t)), x S 1, f 2 (x(t)), x S 2, x(0) = x 0 R n, R n = S 1 Σ S 2, Σ = {x R n h(x) = 0}, h : R n R, S 1 = {x R n h(x) < 0}, S 2 = {x R n h(x) > 0}. Assume f 1 smooth on S 1 Σ, f 2 smooth on S 2 Σ, h is smooth on Σ and h(x) 0 for all x Σ.

... Which non-smooth?... S 2 n(x) S 2 n(x) x T T x x x S 1 Σ Σ S1

Filippov theory Issue: f(x) not defined if x Σ!

Filippov theory Issue: f(x) not defined if x Σ! Work with differential inclusion: f 1 (x(t)) x S 1 x (t) F(x(t)) = {(1 α)f 1 + αf 2, α [0, 1]} x Σ. f 2 (x(t)) x S 2

Filippov theory Issue: f(x) not defined if x Σ! Work with differential inclusion: f 1 (x(t)) x S 1 x (t) F(x(t)) = {(1 α)f 1 + αf 2, α [0, 1]} x Σ. f 2 (x(t)) x S 2 Filippov Solution. An absolutely continuous function x : [0, τ) R n is a solution if for almost all t [0, τ) it holds that x (t) F(x(t)).

... Filippov... Theorem Let f 1,2 be C 1, f 1 f 2 also C 1 (on Σ), and h be C 2. If, at any point x Σ we have that at least one of n T (x)f 1 (x) > 0 and n T (x)f 2 (x) < 0 holds, then there exists a unique Filippov solution from each initial condition.

Crossing and Sliding Transversal Intersection [n T (x)f 1 (x)] [n T (x)f 2 (x)] > 0. Leave Σ to S 1, if n T (x)f 1 (x) < 0, or S 2, if n T (x)f 1 (x) > 0.

Crossing and Sliding Transversal Intersection [n T (x)f 1 (x)] [n T (x)f 2 (x)] > 0. Leave Σ to S 1, if n T (x)f 1 (x) < 0, or S 2, if n T (x)f 1 (x) > 0. Sliding Mode [n T (x)f 1 (x)] [n T (x)f 2 (x)] < 0. Attracting. Have existence and uniqueness: [n T (x)f 1(x)] > 0 and [n T (x)f 2(x)] < 0.

Sliding f 2 S 2 n(x) n(x) S 2 f 2 x T x x T x Σ f 1 S 1 Σ f 1 S 1

Sliding f 2 S 2 n(x) n(x) S 2 f 2 x T x x T x Σ f 1 S 1 Σ f 1 S 1 How is attracting sliding motion taking place on Σ?

Sliding f 2 S 2 n(x) n(x) S 2 f 2 x T x x T x Σ f 1 S 1 Σ f 1 S 1 How is attracting sliding motion taking place on Σ? Filippov vector field: f F (x) = (1 α(x))f 1 (x) + α(x)f 2 (x) α(x) such that n T (x)f F (x) = 0: α(x) = n T (x)f 1 (x) n T (x)(f 1 (x) f 2 (x)).

Entry and Exit As we reach Σ, or are sliding on it, we can have several distinct possible behaviors: Transversal crossing, attractive Sliding, smooth exits. And these can occur at first order, or higher order. As far as sliding motion is concerned, entries are less troublesome than exits. Still, for the sake of this presentation, we will tacitly assume that (insofar as sliding motion) entries and exits happen in the generic way. This means that: (a) Entries are non-tangential (we hit Σ); (b) Exits are tangential. That is, we leave the surface with the sliding vector field being (aligned with) one of the vector fields f i s we have outside of the surface.

Projections and tangents: Revisit sliding vector field Beginning from the end... sliding must keep us on Σ.

Projections and tangents: Revisit sliding vector field Beginning from the end... sliding must keep us on Σ. Projection onto T Σ (x), π Σ = [I nn T ]. Define admissible sliding vector field: f S (x) = (1 β(x))(π Σ f 1 )(x) + β(x)(π Σ f 2 )(x), 0 β 1. Choose β of the form β(x) = a 1n T (x)f 1 (x) + b 1 n T (x)f 2 (x) + c 1 a 2 n T (x)f 1 (x) + b 2 n T (x)f 2 (x) + c 2 and find coefficients by imposing 1st Order Exit Conditions: { β(x) = 0 when n T (x)f 1 (x) = 0, β(x) = 1 when n T (x)f 2 (x) = 0, β(x) = n T (x)f 1(x) n T (x)(f 1(x) af 2(x)), a > 0 is free; a = 1 is Filippov.

Two and more sliding surfaces Consider now the case of two surfaces of discontinuity: Σ 1 := {x R n : h 1 (x) = 0}, Σ 2 := {x R n : h 2 (x) = 0}. Assume h 1 (x) 0, for all x Σ 1, and h 2 (x) 0, for all x Σ 2. Assume n 1 (x) and n 2 (x) linearly independent for all x Σ 1 Σ 2. Issue: To define sliding vector field on Σ 1 Σ 2.

... Two or more... f 3 Σ 2 n 1 (x) f 4 T x x n (x) 2 f 1 T x Σ 1 01 f 2 Figure: Two surfaces

... Two or more... Attractive sliding... w i (x) = [ ] w 1 i (x) wi 2(x) = [ ] n T 1 f i (x) n T 2 f = N i(x) T f i (x), i = 1, 2, 3, 4.

... Two or more... Attractive sliding... w i (x) = [ ] w 1 i (x) wi 2(x) = Attractive sliding motion means ( ) ( ) > 0 > 0 w 1 =, w > 0 2 =, w < 0 3 = [ ] n T 1 f i (x) n T 2 f = N i(x) T f i (x), i = 1, 2, 3, 4. ( ) < 0, w > 0 4 = ( ) < 0. < 0 f 3 Σ 2 n 1 (x) f 4 T x x n (x) 2 f 1 T x Σ 1 01 01 f 2

... Two or more... Filippov vector field, ambiguity: 4 f F (x) = α i f i (x), i=1 4 α i = 1, 0 α i 1. i=1 Impose: One relation too few. n T 1 (x)f F (x) = 0, n T 2 (x)f F (x) = 0.

... Two or more... Filippov vector field, ambiguity: 4 f F (x) = α i f i (x), i=1 4 α i = 1, 0 α i 1. i=1 Impose: One relation too few. n T 1 (x)f F (x) = 0, n T 2 (x)f F (x) = 0. Special cases treated in [Filippov], [Piiroinen Kuznetsov], [Utkin], [Stewart],....

... Two or more... Geometric approach: Use tangent vectors. Let Π(x) be projection onto T x, x Σ 1 Σ 2 : Π = I N(N T N) 1 N T.

... Two or more... Geometric approach: Use tangent vectors. Let Π(x) be projection onto T x, x Σ 1 Σ 2 : Π = I N(N T N) 1 N T. Define f = 4 λ i v i (x), i=1 with λ i 0, and 4 i=1 λ i = 1. v i (x) = Πf i (x)

... Two or more... Geometric approach: Use tangent vectors. Let Π(x) be projection onto T x, x Σ 1 Σ 2 : Π = I N(N T N) 1 N T. Define f = 4 λ i v i (x), i=1 with λ i 0, and 4 i=1 λ i = 1. To determine λ i... v i (x) = Πf i (x)

... Two or more... We want to satisfy exit conditions v i = f i λ i = 1, i = 1,...,4.

... Two or more... We want to satisfy exit conditions v i = f i λ i = 1, i = 1,...,4. We end up proposing to take µ i λ i = 4 i=1 µ, i = 1, 2, 3, 4, i µ i = [ 4 ] 1/3 j=1,j i at j w j [ 4 j=1,j i at j w j [ 1 a 1 =, a 1] 2 = [ 1 1 ] 1/3 a T i w i i = 1,...,4 ] [ [ 1 1, a 3 =, a 1] 4 = 1], Notice: a T 1 y 1 > 0, a T 2 y 2 < 0, a T 3 y 3 < 0, a T 4 y 4 < 0 and we satisfy 1st order exit conditions: v i = f i λ i = 1, i = 1,...,4.

... Two or more... [Alexander-Seidman] interpolation approach. ([AS] call it sigmoid blending).

... Two or more... [Alexander-Seidman] interpolation approach. ([AS] call it sigmoid blending). The idea is to obtain a vector field on the intersection, call it f B, by forming a bilinear interpolant amongst the four vector fields and then imposing the orthogonality conditions: f B (x) = (1 α(x))(1 β(x))f 1 (x) + (1 α(x))β(x)f 2 (x)+ α(x)(1 β(x))f 3 (x) + α(x)β(x)f 4 (x), where α, β [0, 1] [0, 1] must be found by solving the nonlinear system n T 1 (x)f B (x) = 0, nt 2 (x)f B (x) = 0.

... Two or more... On the intersection of p surfaces (n p + 1)... Let us generalize the geometric approach of [DiLo].

... Two or more... On the intersection of p surfaces (n p + 1)... Let us generalize the geometric approach of [DiLo]. Now, have p surfaces in R n, Σ 1, Σ 2,..., Σ p, with n p + 1, and the vectors h i (x) 0 are linearly independent for x p i=1 Σ i. Locally, we have 2 p vector fields f i, i = 1,...,2 p. Set N = [n 1, n 2,...,n p ] and as before let the projection onto N be Π = I N(N T N) 1 N T. Let w i = N T f i, for i = 1,...,2 p. Attractive sliding means that the components of the vectors w i, i = 1,...,2 p, take all signs patterns in the 2 p combinations of [±1, ±1,..., ±1]. E.g., for appropriate ordering of the vector fields, hence of the w i s, we can assume that all components of w 1 are positive, the last component of w 2 is negative all the others being positive, etc..

... Two or more...

... Two or more... Propose to take the sliding vector field on the intersection as: f S (x) = 2 p i=1 For the µ i s we take µ i = λ i v i (x), λ i = [ 2 p j=1,j i at j w j] m [ 2 p j=1,j i at j w j] m a T i w i µ i 2 p j=1 µ, i = 1,..., 2 p. j, i = 1,...,2 p, m = 1 2 p 1, and a i = [±1, ±1,..., ±1] T, with the signs chosen so that sgn(a i ) j = sgn(w i ) j < 0 for i = 2,...,2 p, j = 1,...,p, and the signs of the entries of a 1 being the same as the signs of the entries of w 1.

... Two or more... Extension of [AS] to the case of attractive sliding motion on the intersection of p = 3 surfaces: ẋ =(1 α)(1 β)(1 γ)f 1 (x) + (1 α)(1 β)(1 + γ)f 2 (x) +(1 α)(1 + β)(1 γ)f 3 (x) + (1 α)(1 + β)(1 + γ)f 4 (x) +(1 + α)(1 β)(1 γ)f 5 (x) + (1 + α)(1 β)(1 + γ)f 6 (x) +(1 + α)(1 + β)(1 γ)f 7 (x) + (1 + α)(1 + β)(1 + γ)f 8 (x)

... Two or more... Not at all clear that this nonlinear system is solvable.

... Two or more... Not at all clear that this nonlinear system is solvable. However, we recently proved that: System has a solution (α, β, γ) [0, 1] 3.

... Two or more... Not at all clear that this nonlinear system is solvable. However, we recently proved that: System has a solution (α, β, γ) [0, 1] 3. Solution is isolated (locally unique).

... Two or more... Not at all clear that this nonlinear system is solvable. However, we recently proved that: System has a solution (α, β, γ) [0, 1] 3. Solution is isolated (locally unique). Extend construction, and proof of existence of solution, to the case of the intersection of any number of surfaces.

Numerical Method Each particular state of the system is integrated with an appropriate numerical method, and the events (where structural changes in the system occur) are located accurately: Event driven method. Must decide how to proceed at event points (sliding or crossing). Must check exit conditions (1st and higher order). Must decide how to slide. Difficulties as we get close to the surface(s) of discontinuity, since vector fields may not extend past surfaces, Standard integration off Σ. Integration on Σ using projected RK methods. [We use Lagrange s multiplier technique and a simplified Newton iteration.] Alternative: Regularization Obvious simplifications in the theory. But, regularization may lead to losing the characteristics of the original problem.

Example: Sliding and Chattering on Intersection. A linear problem [IEEE-AC, 2002] x 1 y < 0, (t) = Ax(t) + bu(t) y(t) = c T with u = sign(y) = [ 1, 1] y = 0, x(t) 1 y > 0. 4 1 0 0 1 A = 6 0 1 0 4 0 0 1, c = 0 0, and 1 0 0 0 0 1 0 (i) b = 3β 3β 2, or (ii) b = 1 2β, β = 1/5. β 3 β 2

... Sliding and Chattering... (i) The trajectory has a portion on the sliding set of order 1, the plane x 1 = 0. The solution slides on the plane x 1 = 0, twice. Past the transient, the behavior is periodic, and the lack of differentiability when the solution hits the sliding plane is apparent. (ii) The trajectory exhibits chattering (frequent switching) about the sliding set of order 2, x 1 = x 2 = 0. The solution appears to slide on the sliding set of order 2, x 1 = x 2 = 0, but it does not.

... Sliding and Chattering... 2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5 2 1 0 1 2 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.6 4 0.4 x1 3 x2 2 0.2 1 0 0 1 0.2 2 0.4 0 2 4 6 x 10 4 3 0 2 4 6 x 10 4 3 1.5 2 x3 1 x4 1 0.5 0 1 0 2 0 2 4 6 x 10 4 0.5 0 2 4 6 x 10 4

... Sliding and Chattering... 2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5 1 0.5 0 0.5 1 0.2 0.15 0.1 0.05 0 0.05 0.1 0.15 0.2 0.2 1 0.15 x 1 x 2 0.1 0.5 0.05 0 0 0.05 0.1 0.5 0.15 0 0.5 1 1.5 2 2.5 x 10 5 1 0 1 2 3 x 10 5 x 10 3 0.06 1 blowup x 1 0.04 blowup x 2 0 0.02 0 1 0.02 0.04 2 0.06 1.8 2 2.2 2.4 2.6 2.8 x 10 4 0.08 1 1.5 2 2.5 3 3.5 x 10 4

Example: Double stick-slip [Galvanetto]. Mechanical system composed by two blocks on a moving belt. The velocity of the belt is constant, v. Each block is connected to a fixed support and to the other block by elastic springs. The surface between the blocks and the belt is rough so that the belt exerts a dry friction force on each block that sticks on the belt to the point where the elastic forces due to the springs exceed the maximum static force. At this point the blocks start slipping and the slipping motion will continue to the point where the velocity of the block will equal that of the belt and the elastic forces will be equilibrated by the static friction force.

... stick-slip... { m1 x = k 1 x + k 12 (x y) + f k1 (x v), m 2 y = k 2 y + k 12 (y x) + f k2 (y v), The kinetic force has the form f k2 (x v) = βf k1 (x v) with: 1 δ f k1 (x 1 γ(x v) + δ + η(x 1 v) 2, for v > x, v) =. (1 δ) 1 γ(x v) δ η(x 1 v)2 for v < x, Set x 1 = x, x 2 = y, x = x 3, y = x 4. Discontinuity surfaces Σ 1 and Σ 2 : h 1 (x) = x 3 v, h 2 (x) = x 4 v.

... stick-slip... Fix m 1 = m 2 = 1, k 1 = k 2 = k 12 = 1, δ = 0, γ = 3, η = 0, v = 0.295, β = 1.301. Time interval [0, 80], initial point at the origin. 0.5 0 x4 0.5 1 1.5 0.5 0 0.5 x3 1 1.5 1 0.5 0 x1 0.5 1 1.5

... stick-slip... Periodic solution 0.2 0 0.2 x4 0.4 0.6 0.8 1 0.5 0 0.5 x3 1 1.5 1 0.5 0 x1 0.5 1 1.5 Can define a monodromy, do stability analysis, etc. etc..