Nonlinear Control Theory: Introduction To Dynamical Systems

Nonlinear Control Theory
Introduction to Dynamical Systems

Harry G. Kwatny
Department of Mechanical Engineering & Mechanics Drexel University
Outline
Ordinary Differential Equations

Existence & uniqueness Continuous dependence on parameters Invariant sets, nonwandering sets, limit sets Autonomous systems Basic stability theorems Stable, unstable & center manifolds
Lyapunov Stability

Basics of Nonlinear ODEs
Dynamical Systems
d x ( t ) = f ( x ( t ) , t ) , x R n , t R non-autonomous dt d autonomous x (t ) = f ( x (t )) , x Rn , t R dt A solution on a time interval t [t0 , t1 ] is a function x(t ) :[t0 , t1 ] R n that satisfies the ode.
Vector Fields and Flow

We can visualize an individual solution as a graph x(t ) : t R n . For autonomous systems it is convenient to think of f ( x ) as a vector field on R n - f ( x) assigns a vector to each point in R n . As t varies, a solution x(t ) traces a path through R n tangent to the field f ( x ) . These curves are often called trajectories or orbits. The collection of all trajectories in R n is called the flow of the vector field f ( x).
Auto at Constant Speed

Vs = V
d = f ( , , V , ) dt
u
Body Frame
y v
m, J
Y
b
Fl
Fr
Space Frame
Notice the three equilibria.
Van der Pol

x2 1 x = x 2 2 0.8 (1 x1 ) x2 x1
Damped Pendulum
1 x2 x = x x / 2 sin x 1 2 2
Lipschitz Condition
The existence and uniqueness of solutions depend on properties of the function f . In many applications f ( x, t ) has continuous derivatives in x. We relax this - we require that f is Lipschitz in x. Def : f : R n R n is locally Lipschitz on an open subset D R n if each point x0 D has a neighborhood U 0 such that f ( x ) f ( x0 ) L x x0 for some constant L and all x U 0 Note: C 0 (continuous) functions need not be Lipschitz, C1 functions always are.
Lipschitz Condition
Intuitively, a Lipschitz continuous function is limited in how fast it can change; a line joining any two points on the graph of this function will never have a slope steeper than its Lipschitz constant L. The mean value theorem can be used to prove that any differentiable function with bounded derivative is Lipschitz continuous, with the Lipschitz constant being the largest magnitude of the derivative.
Examples: Lipschitz
f ( x ) = x 2 , x [ 3, 7 ]
L = 14
L =1
f ( x ) = x 2 + 1, x R
L =1
f ( x ) = x 1 , x [ 5,5]
Fails
f ( x ) = x , x [ 0, 4]
Local Existence & Uniqueness

Proposition (Local Existence and Uniqueness) Let f ( x, t ) be piece-wise continuous in t and satisfy the Lipschitz condition f ( x, t ) f ( y , t ) L x y for all x, y Br = x R n x x0 < r and all t [t0 , t1 ]. Then there exists > 0 such that the differential equation with initial condition = f ( x, t ) , x ( t0 ) = x0 Br x has a unique solution over [t0 , t0 + ].
The Flow of a Vector Field

= f ( x ), x ( t0 ) = x0 x ( x0 , t ) this notation indicates x 'the solution of the ode that passes through x0 at t = 0' More generally, let ( x, t ) denote the solution that passes through x at t = 0. The function : R n R R n satisfies ( x, t ) = f ( ( x, t ) ) , ( x,0 ) = x t is called the flow or flow function of the vector field f
Example: Flow of a Linear Vector Field

= Ax x Example: x R3 , x1 cos ( t ) + x2 sin ( t ) 0 1 0 , x, t = x cos t x sin t A= 1 0 0 ( ) 2 ( ) 1 ( ) t e x3 0 0 1 x2
1 0.5 0 -0.5 -1 0.1 x3 0
( x, t ) t
= A ( x, t ) ( x, t ) = e At x
-0.1 -1 -0.5 0 0.5 1 x1
Invariant Set
A set of points S R n is invariant with respect to the vector field f if trajectories beginning in S remain in S both forward and backward in time. Examples of invariant sets: any entire trajectory (equilibrium points, limit cycles) collections of entire trajectories
Example: Invariant Set

x2 1
x1 0 1 0 x1 d x 1 0 0 = x 2 2 dt x3 0 0 1 x3
x3
0.5 0 -0.5 -1 0.1 0
each of the three trajectories shown are invariant sets the x1-x2 plane is an invariant set
-0.1 -1 -0.5 0 0.5 1 x1
Limit Points & Sets

A point q R n is called an -limit point of the trajectory ( t , p ) if there exists a sequence of time values tk + such that
tk
lim ( tk , p ) = q
q is said to be an -limit point of ( t , p ) if there exists a sequence of time values tk such that
tk
lim ( tk , p ) = q
The set of all -limit points of the trajectory through p is the -limit set, and the set of all -limit points is the -limit set.
For an example see invariant set
Introduction to Lyapunov Stability Analysis
Lyapunov Stability
= f ( x), x f ( 0 ) = 0, f : D R n ( locally Lipschitz ) The origin is
x0
a stable equilibrium point if for each > 0, there is a ( ) > 0 such that x ( 0 ) < x ( t ) < t > 0 unstable if it is not stable, and asymptotically stable if can be chosen such that x ( 0 ) < lim x ( t ) 0
t
Two Simple Results

The origin is asymptotically stable only if it is isolated. The origin of a linear system = Ax x is stable if and only if e At N < t > 0 It is asymptotically stable if and only if, in addition e At 0, t
Example: Non-isolated Equilibria

1 x2 x = x 2 x2 x1 x2
x2 4 2 x1 -3 -2 -1 -2 -4 1 2 3
All points on the x1 axis are equilibrium points
Positive Definite Functions

A function V : R n R is said to be positive definite if V ( 0 ) = 0 and V ( x ) > 0, x 0 positive semi-definite if V ( 0 ) = 0 and V ( x ) 0, x 0 negative (semi-) definite if V ( x ) is positive (semi-) definite radially unbounded if V ( x ) as x For a quadratic form: V ( x ) = xT Qx, Q = QT the following are equivalent V ( x ) is positive definite the eigenvalues of Q are positive the principal minors of Q are positive
Lyapunov Stability Theorem

V ( x ) is called a Lyapunov function relative to the flow of = f ( x ) if it is positive definite and nonincreasing with x respect to the flow: V ( 0 ) = 0, V ( x ) > 0 for x 0 V ( x ) V= f ( x) 0 x Theorem : If there exists a Lyapunov function on some
D
V =
V = <
V = <
neighborhood D of the origin, then the origin is stable. If V is negative definite on D then it is asymptotically stable.
Example: Rotating Rigid Body

x, y, z body axes; x , y , z angular velocities in body coord's; diag ( I x , I y , I z ) I x I y I z >0 inertia matrix Iz I y x = Ix Ix Iz y = I y I y Ix z = Iz z y = a z y = b x z x z y x = c y x
note a, b, c > 0
Rigid Body, Contd

Note that a state (x , y , z ) is an equilibrium point if any two of the angular velocity components are zero, i.e., the x , y , z axes are all equilibrium points. Consider a point (x ,0, 0 ) . Shift x x + x . x = a z y y = b ( x + x ) z z = c ( x + x ) y
z
x = 0 0
Rigid Body, Contd

Energy does not work for x 0. Obvious? So, how do we find Lyapunov function? We want V ( 0, 0, 0 ) =0, 0 V = 0,i.e., that satisfy the pde: Lets look at all functions that satisfy V V V V a z y + b ( x + x ) z ) + c ( x + x ) y = 0 ( x y z All solutions take the form: b x 2 + 2b x x + a y 2 c x 2 + 2c x x + a z 2 , f 2 2 a a 1 8b 2 c 2 x 2 2 2 2 2 + + V ( x , y , z ) = cA + bB + ( cA bB ) = c b + h.o.t. x y z 2 2 a V ( x , y , z ) > 0 if ( x , y , z ) ( 0, 0, 0 ) and ( x , y , z ) D
Rigid Body, Contd

Clearly, V ( 0 ) = 0, V > 0 on a neighborhood D of 0 =0 V spin about x-axis is stability

This is one approach to finding candidate Lyapunov functions The first order PDE usually has many solutions The method is connected to traditional first integral methods to the study of stability in mechanics Same method can be used to prove stability for spin about zaxis, but spin about y-axis is unstable why?
First Integrals
Definition : A first integral of the differential equation = f ( x, t ) x is a scalar function ( x, t ) that is constant along trajectories, i.e., ( x, t ) = ( x, t ) x f ( x, t ) + ( x, t ) t 0
= f ( x ) . Suppose 1 ( x ) is a first integral Observation : For simplicity, consider the autonomous case x and 2 ( x ) ,, n ( x ) are arbitrary independent functions on a neighborhhod of the point x0 , i.e., 1 ( x ) det 0 x n ( x ) x = x0 Then we can define coordinate transformation x z , via ( x ) = 1 = 0 z1 constant z = ( x) z f ( x ) z x 1 x = ( z ) The problem has been reduced to solving n 1 differential equations.
Chetaevs Method
Consider the system of equations = f ( x, t ) , x f ( 0, t ) = 0 We wish to study the stability of the equilibrium point x = 0. Obviously, if ( x, t ) is a first integral and it is also a positive definite function, then V ( x, t ) = ( x, t ) establishes stability. But suppose ( x, t ) is not positive definite? Suppose the system has k first integrals 1 ( x, t ) , , k ( x, t ) such that i ( 0, t ) = 0. Chetaev suggested the construction of Lyapunov functions of the form: V ( x, t ) = i =1 i i ( x, t ) + i =1 i i2 ( x, t )
k k
LaSalle Invariance Theorem

Theorem : Suppose V : R n R is C1 and let c denote a component of the region
x Rn V ( x ) < c
}
c
( x ) 0. Suppose c is bounded and within c , V ( x ) = 0, Let E be the set of points within where V Let M be the largest invariant set within E. every sol'n beginning in c tends to M . as t .
Example: LaSalles Theorem

V
x2 d x1 x = m 1kx m 1cx , dt 2 1 2
x2
1 2 1 2 V ( x1 , x2 ) = mx2 + kx1 2 2 ( x , x ) = cx 2 V 1 2 2
M
c
E
x1
Lagrangian Systems
) L ( x, x ) d L ( x, x = QT dt x x x R n generalized coordinates = dx / dt generalized velocities x ) = T ( x, x ) U ( x) L : R 2 n R is the Lagrangian, L ( x, x
T ) = 1 x M ( x) x kinetic energy: T ( x, x 2
) = T ( x, x ) +U ( x) total energy: V ( x, x
Example
2 x2 x12 1 = x2 , T = , U = x 2 1 + x12
x2 1 x x = 1 x 2 cx2 2 2 2 1 + x ( ) 1 2 x2 x12 ( x ) = cx 0 for c > 0 ,V V ( x) = + 2 2 2 1 + x1 Notice that the level sets are unbounded for V ( x ) = constant 1 V ( x ) is not radially unbounded
Chetaev Instability Theorem

Let D be a neighborhood of the origin. Suppose there is a function V ( x ) : D R and a set D1 D such that 1) V ( x ) is C1 on D, 2) the origin belongs to the boundary of D1 , D1 , ( x ) > 0 on D , 3) V ( x ) > 0, V
1
4) on the boundary of D1 inside D, i.e., on D1 D, V ( x ) = 0 Then the origin is unstable.

D
r U x (t ) D1
Example, Rigid Body, Contd

Consider the rigid body with spin about the y -axis (intermediate inertia), = ( 0, y , 0 ) x = a z ( y + y ) y = b x z Shifted equations: z = c ( y + y ) x Attempts to prove stability fails. So, try to prove instability. Consider V ( x , y , z ) = x z
2 2 Let Br = ( x , y , z ) x + y + z2 < r 2 and D1 = ( x , y , z ) Br x > 0, z > 0 T
so that V > 0 on D1 and V = 0 on D1 = a 2 ( + ) + c ( + ) 2 = ( + ) ( a 2 + c 2 ) V z y y y y x y y z x

2 > 0 on D instability We can take r 2 < y in which case V 1
}
x
Br D1
Stability of Linear Systems

Consider the linear system = Ax x Choose V ( x ) = xT Px ( x ) = xT ( AT P + PA ) x := xT Qx V
a) if their exists a positive definite pair of matrices P, Q that satisfy AT P + PA = Q (Lyapunov equation) the origin is asymptotically stable. b) if P has at least one negative eigenvalue and Q > 0, the origin is unstable. c) if the origin is stable then for any Q > 0, there is a unique solution, P > 0, of the Lyapunov equation,
Second Order Systems

Consider the system + Cx + Kx = 0, M T = M > 0, C T = C > 0, K T = K > 0 Mx
T 1 T , x ) = 1 E(x x Mx + 2 2 x Kx
d , x ) = x T Mx + x T Kx = x T [Cx + Kx ] + xT Kx E(x dt d , x ) = x T Cx E(x dt The anti-symmetric terms correspond to Some interesting generalizations: circulatory forces (transfer conductances 1) C 0, 2) C T C , 3) K T K
in power systems) they are nonconservative.
The anti-symmetric terms correspond to gyroscope forces they are conservative.
Example
V 1,
1 -ja

Assume uniform damping Assume e=0 Designate Gen 1 as swing bus Eliminate internal bus 4
-jc -jb -jd e-jf V3,

3
4 V4,
+ = P b sin ( ) b sin ( ) 1 1 1 13 1 12 1 2 + = P + b sin ( ) b sin ( ) 2 2 2 12 1 2 23 2
1 = 2 1 , 2 = 3 1 , P1 = P2 P1 , P2 = P3 P1
Example Contd
This is a Lagrangian system with

T (1 , 2 ) = 1 2 1 + 22 ) , Q = [ 1 ( 2 2 ]
U (1 , 2 ) = P 11 P 2 2 b13 cos (1 ) b12 cos (1 2 ) b23 cos ( 2 )
To study stability choose total energy as Lyapunov function

V = T (1 , 2 ) + U (1 , 2 ) = 2 2 0 V 1 2 Note: T ( 0, 0 ) = 0 and T (1 , 2 ) > 0 (1 , 2 ) 0
G. V. Arononvich and N. A. Kartvelishvili, "Application of Stability Theory to Static and Dynamic Stability Problems of Power Systems," presented at Second All-union Conference on Theoretical and Applied mechanics, Moscow, 1965.
Equilibria corresponding to U (1 , 2 ) a local minimum are stable.
Example, Contd
U (1 , 2 ) as a function on a torus U : T R
3
Since 1 < and 2 < we should consider
1 0 -1 -2 -3 0 -2
-1
-2
0 2
-2
-3 -3 -2 -1 0 1 1 2 3
P 1 = 0, P 2 = 0, b12 = 1, b13 = 1, b23 = 1
Example, Contd
3 2
2
2
0
0 2 -2 0 -2 0 2 -2
-1
-2
-3 -3 -2 -1 0 1 1 2 3
P 1 = .25, P 2 = 0, b12 = 1, b13 = 1, b23 = 1
Example Contd
3 2
2
2
0 2 -2 0 -2 0 2 -2
-1
-2
-3 -3 -2 -1 0 1 1 2 3
P 1 = /15, P 2 = 0, b12 = 1, b13 = 1, b23 = 1
Example Contd
3 2
5 2.5 0 -2.5 0 -2 0 2 -2
-3 -3 -2 -1 0 1 1 2 3 -2
2
-1
P 1 = / 5, P 2 = 1, b12 = 1, b13 = .5, b23 = 1

Nonlinear Control Theory: Introduction To Dynamical Systems

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Nonlinear Control Theory: Introduction To Dynamical Systems

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Nonlinear Control Theory

Introduction to Dynamical Systems

Ordinary Differential Equations

Basics of Nonlinear ODEs

Vector Fields and Flow

Auto at Constant Speed

Notice the three equilibria.

Van der Pol

Local Existence & Uniqueness

The Flow of a Vector Field

Example: Flow of a Linear Vector Field

-0.1 -1 -0.5 0 0.5 1 x1

Example: Invariant Set

0.5 0 -0.5 -1 0.1 0

-0.1 -1 -0.5 0 0.5 1 x1

Limit Points & Sets

Introduction to Lyapunov Stability Analysis

Two Simple Results

Example: Non-isolated Equilibria

All points on the x1 axis are equilibrium points

Positive Definite Functions

Lyapunov Stability Theorem

Example: Rotating Rigid Body

Rigid Body, Contd

Rigid Body, Contd

Rigid Body, Contd

LaSalle Invariance Theorem

Example: LaSalles Theorem

Chetaev Instability Theorem

4) on the boundary of D1 inside D, i.e., on D1 D, V ( x ) = 0 Then the origin is unstable.

Example, Rigid Body, Contd

so that V > 0 on D1 and V = 0 on D1 = a 2 ( + ) + c ( + ) 2 = ( + ) ( a 2 + c 2 ) V z y y y y x y y z x

Stability of Linear Systems

Second Order Systems

The anti-symmetric terms correspond to gyroscope forces they are conservative.

-jc -jb -jd e-jf V3,

+ = P b sin ( ) b sin ( ) 1 1 1 13 1 12 1 2 + = P + b sin ( ) b sin ( ) 2 2 2 12 1 2 23 2

This is a Lagrangian system with

U (1 , 2 ) = P 11 P 2 2 b13 cos (1 ) b12 cos (1 2 ) b23 cos ( 2 )

To study stability choose total energy as Lyapunov function

Equilibria corresponding to U (1 , 2 ) a local minimum are stable.

Since 1 < and 2 < we should consider

P 1 = 0, P 2 = 0, b12 = 1, b13 = 1, b23 = 1

P 1 = .25, P 2 = 0, b12 = 1, b13 = 1, b23 = 1

P 1 = /15, P 2 = 0, b12 = 1, b13 = 1, b23 = 1

P 1 = / 5, P 2 = 1, b12 = 1, b13 = .5, b23 = 1

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