Quiz
EE-602
Fall Discussion 09
1. Using a forward Euler approximation with h = 0.1,
!
x (t k ) =
to the differential equation,
x (t k + h ) ! x (t k )
h
dx !1
=
,!!! x (0 ) = 0.1, compute an
dt 2 x
estimate for x(0.1).
2. Using separation of variables, solve the diff. equation,
dx !1
=
,!!! x (0 ) = 0.1, to compute an exact expression for x(t) and
dt 2 x
evaluate it at t = 0.1. Compare with your answer to the Euler
approximation of part 1.
EE-602,Chap 7, Fall 08
7- 2
R. A. DeCarlo
CHAPTER 7: EXISTENCE AND UNIQUENESS OF STATE
TRAJECTORIES
I. INTRODUCTION:
1. QUESTION 7.1: Precisely, what is meant by a solution to
!
x = f ( x, t ), x (t 0 ) = x0 ?
ANSWER: ! (", t 0 ; x0 ) denotes a solution if and only if
i. ! (t 0 , t 0 ; x0 ) = x0 (satisfies IC), and
!
ii. ! (t , t 0 ; x0 ) = f (! (t , t 0 ; x0 ), t ) (satisfies DE)
2. QUESTION 7.2: What does uniqueness of a solution mean?
ANSWER: A UNIQUE ! (t , t 0 ; x0 ) means that if !1 and !2 both
satisfy the differential equation:
!
!i = f (!i , t )
and the same IC
!1 (t 0 , t 0 ; x0 ) = !2 (t 0 , t 0 ; x0 )
THEN for all t in R + = {t !|! t ! 0} .
!1 (t , t 0 ; x0 ) = !2 (t , t 0 ; x0 )
EE-602,Chap 7, Fall 08
7- 3
R. A. DeCarlo
3. QUESTION 7.3: When does there exist a solution to the nonlinear
state dynamics:
!
x (t ) = f [ x (t ), t ], x (t 0 ) = x0
where x (t ) ! R n and f (!, !) : R n " R # R n .
ANSWER: Momentarily postponed.
3. QUESTION 7.4: Why is EXISTENCE and UNIQUENESS of a
solution important?
ANSWER VIA MOTIVATIONAL EXAMPLE:
OBJECTIVE: Numerically, using forward Euler, and analytically
compute solution to
!
x=
!1
2x
x (0 ) = x0 = 0.1
at t = 0.1.
Part 1: Numerical solution via Euler at t = 0.1:
See Quiz
Part 2: Analytic solution via separation of variables:
EE-602,Chap 7, Fall 08
7- 4
R. A. DeCarlo
II. MATHEMATICAL PRELIMINARIES
PART I: NORMS AND NORM PROPERTIES
1. Vector Norms
(a) EUCLIDEAN VECTOR NORM:
v = ! v1 v2 ! vn #
"
$
T
has 2-norm (Euclidean)
2
2
2
= v1 + v2 + ! + vn
v
(b) -norm:
2. Proposition: v
v
!
2
!
= max vi
"v
i
2
" nv
!
Why are NORMS important?
ANSWER: They measure different properties of a function or
matrix, like the color of eyes, the color of hair, the height, the weight,
the shoe size of the matrix or function.
EE-602,Chap 7, Fall 08
7- 5
R. A. DeCarlo
3. THE FUNCTION L OR SUP-NORM
(a) Let a(!) : R " R . The L! or sup-norm is
a(.)
!
= sup a(t )
t
(b) Restricted to a domain, D:
a(.)
!, D
= sup a(t )
t "D
(c) The vector L! -norm: Let a(!) : R " R n , then
a(.)
!, D
= max ai (t )
i
!, D
QUESTION: Why is this norm important?
ANSWER: It measures the "highest absolute value" a function can
achieve over R or over some domain D.
EE-602,Chap 7, Fall 08
7- 6
R. A. DeCarlo
4. MATRIX NORMS:
(a) Matrix spectral or 2-norm: for a constant matrix A
A
2
= max
x!0
Ax
x
2
= max Ax
2
x =1
2
= "1
where ! 1 is the largest singular value of A.
(b) Matrix L-norm of constant A
A
!
#n
= max % " aij
i % j =1
$
&
(
(
'
(c) Matrix L-norm of A(!) = " aij (!) $ : R & R n' n restricted to
#
%
a domain D is:
A(.)
!, D
= max
i
" aij (.)
j
!, D
EE-602,Chap 7, Fall 08
7- 7
R. A. DeCarlo
EXAMPLE 7.1. Compute the L! -norm of
# e! t
! e! t
A(t ) = %
% ! cos(t ) 2 sin("t )
$
Step 1: Compute L! -norm of row 1:
Step 2: Compute L! -norm of row 2:
Step 3: Choose largest:
&
(1+ (t )
(
'
EE-602,Chap 7, Fall 08
7- 8
R. A. DeCarlo
5. MATRIX NORM PROPERTIES:
(a)
1
A
n
!
"A
2
" nA
!
(b) Ax 2 ! A 2 x 2
(c) A USEFUL LEMMA:
A(!) x
2, D j
" A(!) 2, D x
j
2
" n A(!) #, D x
j
2
EE-602,Chap 7, Fall 08
7- 9
R. A. DeCarlo
PART 2. CONTINUITY
1. CONTINUITY IN A DOMAIN: f (!, !) : R n " R # R n is
continuous on some domain D ! R n " R if it is continuous at each
point of D. Pictorially,
2. LEFT AND RIGHT HAND LIMITS:
EE-602,Chap 7, Fall 08
7-10
3. PIECEWISE CONTINUOUS
R. A. DeCarlo
EE-602,Chap 7, Fall 08
7-11
R. A. DeCarlo
III. GLOBAL EXISTENCE AND UNIQUENESS
1. PROBLEM STATEMENT: Determine sufficient conditions for
the existence of a unique solution to
!
x = f ( x, t ), x (0 ) = x0
where
x (t ) ! R n
and
t ! R + = [ 0,! " )
REMARK: A simple change of variable makes the forthcoming
conditions valid over an arbitrary interval [t 0 ,! ! ) .
2. What ASSUMPTIONS are pertinent?
i. There is a set S contained in R + containing at most a finite
number of points per unit interval. S will depend on f ( x, t ) and will
denote possible discontinuity points.
! 1$
REMARK: The points, t, 0 ! t ! 1, for which sin # & = 0 could not
"t%
be in S.
EE-602,Chap 7, Fall 08
7-12
R. A. DeCarlo
ii. for each x ! R n , f ( x, t ) is continuous for t ! S .
iii. for each x ! R n and for each ti ! S , f ( x, t ) has finite left
and right hand limits at t = ti .
iv. (!, f !) : R n " R # R n satisfies a global Lipschitz condition-i.e. there exists a piecewise continuous function k (!) : R + " R + such
that
f ( x1, t ) ! f ( x 2 , t )
2
" k (t ) x1 ! x 2
for all t ! R + and all points x1, x 2 ! R n .
2
EE-602,Chap 7, Fall 08
7-13
R. A. DeCarlo
3. WHY ARE THESE ASSUMPTIONS PERTINENT?
i. For each fixed t , f ( x, t ) is continuous at x ;
ii. For each x , the points of discontinuity of f ( x, t ) lie in S;
iii. If x (t ) is continuous, f ( x (t ), t ) is piecewise continuous in t .
iv. From (i) - (iii), f ( x (t ), t ) is integrable with respect to t and
the derivative of its integral equals f ( x (t ), t ) except possibly at t ! S
by the fundamental theorem of calculus.
v. They allow us to state and prove an existence and uniqueness
theorem.
EE-602,Chap 7, Fall 08
7-14
R. A. DeCarlo
4. EXISTENCE AND UNIQUENESS THEOREM: Given
assumptions (i) through (iv), for each x0 ! R n and t 0 ! R + , there
exists a unique continuous function
! (., t 0 ; x0 ) :R+ ---> Rn
such that
!
i. ! (t , t 0 ; x0 ) = f [! (t , t 0 ; x0 ), t ] , and
ii. ! (t 0 , t 0 ; x0 ) = x0
for all t ! R + and t S.
5. EXAMPLE 7.2. Determine if the state equations for the inverted
pendulum have a unique solution. The state equations are
! x1
!
#
!
# x2
"
$!
x2
&=#
& # q sin( x1 )
%"
$
& ' f ( x, t ) = f ( x )
&
%
The only essential part of the proof is to show that f ( x, t ) is
Lipschitz.
EE-602,Chap 7, Fall 08
7-15
R. A. DeCarlo
EE-602,Chap 7, Fall 08
7-16
R. A. DeCarlo
IV. EXISTENCE AND UNIQUENESS OF LINEAR STATE
DYNAMICS
1. Linear State Dynamics
!
x (t ) = f ( x (t ), t ) = A(t ) x (t )
where A(!) = " aij (!) $ has piecewise continuous entries over the
#
%
interval [t 0 , ! ] .
2. PROPOSITION: If A(!) has piecewise continuous entries on R +
then f ( x, t ) = A(t ) x satisfies a global Lipschitz condition on R n ! R + .
PROOF: Step 1: Define D j = [ j ! 1, j ) for j = 1,2,3,....
!
Step 2. Observe that
! D j = R+
j =1
EE-602,Chap 7, Fall 08
7-17
R. A. DeCarlo
Step 3. For each t ! D j and all points x1, x 2 ! R n . it follows that
f ( x1, t ) ! f ( x 2 , t )
2
= A(t ) " x1 ! x 2 $
#
%
2
& A(t ) 2 x1 ! x 2
2
It follows that
A(t ) " x1 ! x 2 $
#
%
2, D j
& n A(t ) ', D x1 ! x 2
j
2
Step 4. Define
#
%
k (t ) = $
%
&
n A(t ) !, D
0
j
t "Dj
otherwise
Thus
A(t ) " x1 ! x 2 $
#
%
2
& k (t ) x1 ! x 2
2
means that the right side of our differential equation satisfies a global
Lipschitz condition.
EE-602,Chap 7, Fall 08
7-18
R. A. DeCarlo
3. THEOREM: A solution ! (", t 0 , x0 , u ) to
!
x (t ) = A(t ) x (t ) + B(t ) u (t ) , x (t 0 ) = x0
exists and is unique over [t 0 , ! ) provided A(!) and B(!)u (!) are
piecewise continuous over [t 0 , ! ) 1.
Corollary: If A(!) is piecewise continuous, a solution, designated
!
! (", t 0 , x0 ) , to x (t ) = A(t ) x (t ) , exists and is unique for each initial
conditon x0 .
1
For our work, the solution will be unique even when B(.)u(.) is not piecewise continuous for example when u(t) = (t).
EE-602,Chap 7, Fall 08
7-19
R. A. DeCarlo
Quiz Questions:
Part 1. True-false.
Question 1: The differential equation
!
x = f ( x, u ) = sin(t )(1 ! 2 x )u (t )
has a unique solution solvable by separation of variables for each
initial condition x(0) 0.5 and t 0.
Question 2: Consider the state dynamics,
!
x (t ) = A(t ) x (t ) + B(t ) u (t ) ! f ( x, t ) .
If there exists a unique solution, then f ( x, t ) satisfies a Lipschitz
condition. _______________
Question 2: If A(t ) and B(t ) have piecewise continuous entries,
then there exists a unique solution. _______________
EE-602,Chap 7, Fall 08
7-20
R. A. DeCarlo
Question 3: The state dynamics
" 1 1 + cos(t ) ! sin(t )
!
x (t ) = $
$0
exp(! t )1+ (t )
#
%
' x (t )
'
&
has a unique solution. _______________
!
Problem 1. (a) Suppose x = f(x,t). Precisely state the definition of
a Lipschitz condition for f(x,t).
(b) Prove or disprove: the state dynamics
! x1
!
#
!
# x2
"
$ ! t cos(t ) sin( x2 )
&=#
& # q cos( x1 ) sin( x1 )
%"
$ ! cos(t )
&+#
&# 1
%"
has a unique solution where q is a non-zero constant.
$
& u (t )
&
%
EE-602,Chap 7, Fall 08
7-21
R. A. DeCarlo
Problem 2. Consider the state dynamics
!
x = A(t ) f ( x ) + B( x, t )u (t )
(a) State the Lipschitz condition, precisely.
(b) Develop conditions on A(t ) , f ( x ) , B( x, t ) , and u (t ) which are
sufficient for the existence of a unique solution. Explain your
reasoning.
EE-602,Chap 7, Fall 08
7-22
OTHER MATHEMATICAL PROPERTIES
1. OTHER VECTOR NORMS
n
(a) 1-norm:
(b) p-norm:
v
v
1
p
= ! vi
i =1
"n
= $ ! vi
$ i =1
#
p %1/ p
'
'
&
2. VECTOR NORM PROPERTIES
(a)
v
(b)
v
2
!v
1
! nv
!
"v
1
"n v
2
!
R. A. DeCarlo