linear system theory 2 e sol

8/15/2019 Linear System Theory 2 e Sol

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Solutions Manual

LINEAR SYSTEM THEORY, 2/E

Wilson J. Rugh

Department of Electrical and Computer Engineering

Johns Hopkins University


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PREFACE

With some lingering ambivalence about the merits of the undertaking, but with a bit more dedication than

the first time around, I prepared this Solutions Manual for the second edition of Linear System Theory. Roughly

40% of the exercises are addressed, including all exercises in Chapter 1 and all others used in developments in the

text. This coverage complements the 60% of those in an unscientific survey who wanted a solutions manual, andperhaps does not overly upset the 40% who voted no. (The main contention between the two groups involved the

inevitable appearance of pirated student copies and the view that an available solution spoils the exercise.)

I expect that a number of my solutions could be improved, and that some could be improved using only

techniques from the text. Also the press of time and my flagging enthusiasm for text processing impeded the

crafting of economical solutions—some solutions may contain too many steps or too many words. However I

hope that the error rate in these pages is low and that the value of this manual is greater than the price paid.

Please send comments and corrections to the author at [email protected] or ECE Department, Johns Hopkins

University, Baltimore, MD 21218 USA.


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CHAPTER 1

Solution 1.1

(a) For k = 2, ( A + B)2

= A2

+ AB + BA + B2

. If AB = BA, then ( A + B)2

= A2

+ 2 AB + B2

. In general if AB = BA, then the k -fold product ( A + B)k can be written as a sum of terms of the form A j Bk − j, j = 0, . . . , k . The

number of terms that can be written as A j Bk − j is given by the binomial coefficient

jk

. Therefore AB = BA

implies

( A + B)k = j=0Σk

jk

A j Bk − j

(b) Write

det [λ I − A (t )] = λn + an−1(t )λn−1 + . . . + a1(t )λ + a 0(t )

where invertibility of A (t ) implies a 0(t ) ≠ 0. The Cayley-Hamilton theorem implies

An(t ) + an−1(t ) An−1(t ) + . . . + a0(t ) I = 0

for all t . Multiplying through by A−1(t ) yields

A−1(t ) =a0(t )

−a1(t ) I − . . . − an−1(t ) An−2(t ) − An−1(t )_________________________________

for all t . Since a 0(t ) = det [− A (t )], a 0(t ) = det A (t ). Assume ε > 0 is such that det A (t ) ≥ ε for all t . Since A (t ) ≤ α we have aij (t ) ≤ α, and thus there exists a γ such that a j(t ) ≤ γ for all t . Then, for all t ,

A−1(t ) =det A (t )

a 1(t ) I + . . . + An−1(t )______________________

≤ε

γ + γ α + . . . + αn−1_________________=∆ β

Solution 1.2(a) If λ is an eigenvalue of A, then recursive use of Ap = λ p shows that λk is an eigenvalue of A k . However toshow multiplicities are preserved is more difficult, and apparently requires Jordan form, or at least results on

similarity to upper triangular form.

(b) If λ is an eigenvalue of invertible A, then λ is nonzero and Ap = λ p implies A−1 p = (1 / λ) p. As in (a),addressing preservation of multiplicities is more difficult.

(c) AT has eigenvalues λ1 , . . . , λn since det (λ I − AT ) = det (λ I − A)T = det (λ I − A).(d) A H has eigenvalues λ1

__, . . . , λn

__using (c) and the fact that the determinant (sum of products) of a conjugate is

the conjugate of the determinant. That is

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Linear System Theory, 2/E Solutions Manual

det (λ I − A H ) = det (λ_ I − A) H = det (λ

_ I − A)

__________

(e) α A has eigenvalues α λ1 , . . . , αλn since Ap = λ p implies (α A) p = (αλ) p.(f) Eigenvalues of AT A are not nicely related to eigenvalues of A. Consider the example

A =

00

0α

, AT A =

00

α0

where the eigenvalues of A are both zero, and the eigenvalues of AT A are 0, α. (If A is symmetric, then (a)applies.)

Solution 1.3(a) If the eigenvalues of A are all zero, then det (λ I − A) = λn and the Cayley-Hamilton theorem shows that A isnilpotent. On the other hand if one eigenvalue, say λ1 is nonzero, let p be a corresponding eigenvector. Then Ak p = λ1

k p ≠ 0 for all k ≥ 0, and A cannot be nilpotent.(b) Suppose Q is real and symmetric, and λ is an eigenvalue of Q. Then λ

_also is an eigenvalue. From the

eigenvalue/eigenvector equation Qp = λ p we get p H Qp = λ p H p. Also Qp_

= λ

_

p

_, and transposing gives

p H Qp = λ_ p H p. Subtracting the two results gives (λ − λ

_) p H p = 0. Since p ≠ 0, this gives λ = λ

_, that is, λ is real.

(c) If A is upper triangular, then λ I − A is upper triangular. Recursive Laplace expansion of the determinant aboutthe first column gives

det (λ I − A) = (λ − a11) . . . (λ − ann)

which implies the eigenvalues of A are the diagonal entries a 11 , . . . , ann .

Solution 1.4(a)

A =

10

00

implies AT A =

01

00

implies A = 1

(b)

A =

13

31

implies AT A =

610

106

Then

det (λ I − AT A) = (λ − 16)(λ − 4)

which implies A = 4.(c)

A =

0

1−i1+i0

implies A H A =

0

(1+i)(1−i)(1−i)(1+i)

0

=

02

20

This gives A = √ 2 .

Solution 1.5 Let

A =

0

1 / α1 / αα

, α > 1

Then the eigenvalues are 1 / α and, using an inequality on text page 7,

A ≥1 ≤ i, j ≤ 2max aij = α

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Solution 1.6 By definition of the spectral norm, for any α ≠ 0 we can write

A = x = 1

max A x = x = 1

max x

A x_______

=α x = 1

maxα x

Aα x________=

x = 1 / αmax

α xα A x__________

Since this holds for any α ≠ 0,

A = x ≠ 0max

x

A x_______=

x ≠ 0max

x

A x_______

Therefore

A ≥ x

A x_______

for any x ≠ 0, which gives

A x ≤ A x

Solution 1.7 By definition of the spectral norm,

AB = x = 1max ( AB) x =

x = 1max A ( Bx)

≤ x = 1max { A Bx} , by Exercise 1.6

= A x = 1max Bx = A B

If A is invertible, then A A−1 = I and the obvious I = 1 give

1 = A A−1 ≤ A A−1

Therefore

A−1 ≥ A

1_____

Solution 1.8 We use the following easily verified facts about partitioned vectors:

x 2

x 1

≥ x 1, x 2 ;

0

x1

= x 1 ,

x2

0

= x 2

Write

Ax =

A21 A11

A22 A12

x2 x1

=

A21 x1 + A22 x 2 A11 x1 + A12 x 2

Then for A11 , for example,

A = x = 1max A x ≥

x = 1max A11 x 1 + A12 x 2

≥ x1 = 1

max A11 x 1 = A11

The other partitions are handled similarly. The last part is easy from the definition of induced norm. For example

if

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A =

0

0

0

A12

then partitioning the vector x similarly we see that

x = 1max A x =

x 2 = 1max A12 x 2 = A12

Solution 1.9 By the Cauchy-Schwarz inequality, and x T = x,

x T A x ≤ x T A x = AT x x

≤ AT x2 = A x2

This immediately gives

x T A x ≥ − A x2

If λ is an eigenvalue of A and x is a corresponding unity-norm eigenvector, then

λ = λ x = λ x = A x ≤ A x = A

Solution 1.10 Since Q = QT , QT Q = Q2 , and the eigenvalues of Q2 are λ12 , . . . , λn

2 . Therefore

Q = √ λmax(Q2) =1 ≤ i ≤ nmax λi

For the other equality Cauchy-Schwarz gives

x T Qx | ≤ xT Q x = Qx x

≤ Q x2 =

[ 1 ≤ i ≤ nmax λ

i

] x T x

Therefore | x T Qx | ≤ Q for all unity-norm x. Choosing xa as a unity-norm eigenvector of Q corresponding tothe eigenvalue that yields

1 ≤ i ≤ nmax λi gives

xaT Qxa = xa

T [1 ≤ i ≤ nmax λi ] xa =

1 ≤ i ≤ nmax λi

Thus x = 1max x T Qx = Q.

Solution 1.11 Since A x = √ ( A x)T ( A x) = √ x T AT A x ,

A = x = 1max √ x T AT A x

=

x = 1max x T AT A x

1 / 2

The Rayleigh-Ritz inequality gives, for all unity-norm x,

x T AT A x ≤ λmax( AT A) xT x = λmax( AT A)

and since A T A ≥ 0, λmax( AT A) ≥ 0. Choosing xa to be a unity-norm eigenvector corresponding to λmax( AT A) gives

xaT AT A xa = λmax( A

T A)

Thus

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x = 1max x T AT A x = λmax( A

T A)

so we have A = √

λmax( AT

A) .

Solution 1.12 Since A T A > 0 we have λi( AT A) > 0, i = 1, . . . , n, and ( AT A)−1 > 0. Then by Exercise 1.11,

A−12 = λmax(( AT A)−1) =

λmin( AT A)

1_________

=λmin( A

T A) . det ( AT A)

i =1Πn

λi( AT A)

__________________ ≤(det A)2

[λmax( AT A)]n −1______________

=(det A)2

A2(n−1)_________

Therefore

A−1 ≤det A

An−1________

Solution 1.13 Assume A ≠ 0, for the zero case is trivial. For any unity-norm x and y,

yT A x ≤ y T A x

≤ y A x = A

Therefore

x, y = 1max yT

A x ≤ A

Now let unity-norm xa be such that A xa = A, and let

ya = A

Axa_____

Then ya = 1 and

yaT A xa =

A

xaT AT A xa___________

= A

A xa2

________=

A

A2______= A

Therefore

x, y = 1max yT A x = A

Solution 1.14 The coefficients of the characteristic polynomial of a matrix are continuous functions of matrixentries, since determinant is a continuous function of the entries (sum of products). Also the roots of a

polynomial are continuous functions of the coefficients. (A proof is given in Appendix A.4 of E.D. Sontag,

Mathematical Control Theory, Springer-Verlag, New York, 1990.) Since a composition of continuous functions

is a continuous function, the pointwise-in-t eigenvalues of A (t ) are continuous in t .

This argument gives that the (nonnegative) eigenvalues of A T (t ) A (t ) are continuous in t . Then the maximum at

each t is continuous in t — plot two eigenvalues and consider their pointwise maximum to see this. Finally since

square root is a continuous function of nonnegative arguments, we conclude A (t ) is continuous in t .However for continuously-differentiable A (t ), A (t ) need not be continuously differentiable in t . Consider the

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Solution 1.17 Using the product rule to differentiate A (t ) A−1(t ) = I yields

A.(t ) A−1(t ) + A (t )

dt

d ___ A−1(t ) = 0

which gives

dt

d ___ A−1(t ) = − A−1(t ) A

.(t ) A−1(t )

Solution 1.18 Assuming differentiability of both x (t ) and x (t ), and using the chain rule for scalarfunctions,

dt

d ___ x (t )2 = 2 x (t )dt

d ___ x (t )

= 2 x (t )dt

d ___ x (t )

Also we can write, using the product rule and the Cauchy-Schwarz inequality,

dt

d ___ x (t )2 =

dt

d ___ x T (t ) x (t ) = x

. T (t ) x (t ) + x T (t ) x

.(t ) = 2 x T (t ) x

.(t )

≤ 2 x (t ) x.(t )

For t such that x (t ) ≠ 0, comparing these expressions gives

dt

d ___ x (t ) ≤ x.(t )

If x (t ) = 0 on a closed interval, then on that interval the result is trivial. If x (t ) = 0 at an isolated point, then

continuity arguments show that the result is valid. Note that for the differentiable function x (t ) = t , x (t ) = t is not differentiable at t = 0. Thus we must make the assumption that x (t ) is differentiable. (While thisinequality is not explicitly used in the book, the added differentiability hypothesis explains why we always

differentiate x (t )2 = xT (t ) x (t ) instead of x (t ).)

Solution 1.19 To prove the contrapositive claim, suppose for each i, j there is a constant βij such that

0

∫ t

f ij (σ) d σ ≤ βij , t ≥ 0

Then by the inequality on page 7, noting thati, j

max f ij (t ) is a continuous function of t and taking the pointwise-

in-t maximum,

0∫ t

F (σ) d σ ≤0∫ t

√ mni, j

max f ij (σ) d σ

≤ √ mn0

∫ t

i =1Σm

j=1Σn

| f ij (σ) d σ

≤ √ mni=1Σn

j=1Σm

βij


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Solution 1.20 If λ(t ), p (t ) are a pointwise-in-t eigenvalue/eigenvector pair for A−1(t ), then

A−1(t ) p (t ) = λ(t ) p (t ) = λ(t ) p (t )

Therefore, for every t ,

λ(t ) = p (t )

A−1(t ) p (t )_____________ ≤ p (t )

A−1(t ) p (t )_______________ ≤ α

Since this holds for any eigenvalue/eigenvector pair,

det A (t ) =det A −1(t )

1___________=

λ1(t ) . . . λn(t )1_________________ ≥

αn1___

> 0

for all t .

Solution 1.21 Using Exercise 1.10 and the assumptions Q (t ) ≥ 0, t b ≥ t a,

t a

∫ t b

Q (σ) d σ =t a

∫ t b

λmax[Q (σ)] d σ ≤t a

∫ t b

tr [Q (σ)] d σ = trt a

∫ t b

Q (σ) d σ

Note that

t a

∫ t b

Q (σ) d σ ≥ 0

since for every x

xT

t a

∫ t b

Q (σ) d σ x =t a

∫ t b

xT Q (σ) x d σ ≥ 0

Thus, using a property of the trace on page 8 of Chapter 1, we have

t a

∫ t b

Q (σ) d σ ≤ trt a

∫ t b

Q (σ) d σ ≤ nt a

∫ t b

Q (σ) d σ

Finally,

t a

∫ t b

Q (σ) d σ ≤ ε I

implies, using Rayleigh-Ritz,

t a

∫ t b

Q (σ) d σ ≤ ε

Therefore

t a

∫ t b

Q (σ) d σ ≤ nε

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CHAPTER 2

Solution 2.3 The nominal solution for ũ (t ) = sin (3t ) is ỹ (t ) = sin t . Let x 1(t ) = y (t ), x 2(t ) = y.(t ) to write the

state equation

x.(t ) =

−(4 / 3) x13 (t ) − (1 / 3)u (t ) x 2(t )

Computing the Jacobians and evaluating gives the linearized state equation

x δ.

(t ) =

−4 sin2t 0

01

x δ(t ) +

−1 / 30

u δ(t )

y δ(t ) = 1 0

x δ(t )

where

xδ(t ) = x (t ) −

cos t

sin t

, uδ(t ) = u (t ) − sin (3t ) , y

δ(t ) = y (t ) − sin t , x

δ(0) = x (0) −

1

0

Solution 2.5 For ũ = 0 constant nominal solutions are solutions of

0 = x̃2 − 2 x̃ 1 x̃ 2 = x̃ 2(1−2 x̃ 1)

0 = − x̃1 + x̃12

+ x̃22

= x̃1( x̃ 1−1) + x̃22

Evidently there are 4 possible solutions:

x̃a =

00

, x̃b =

01

, x̃c =

1 / 21 / 2

, x̃d =

−1 / 21 / 2

Since

∂ x∂ f ___

=

−1+2 x1−2 x 2

2 x 2

1−2 x1

,∂u∂ f ___

=

1

0

evaluating at each of the constant nominals gives the corresponding 4 linearized state equations.

Solution 2.7 Clearly x̃ is a constant nominal if and only if

0 = A x̃ + bũ

that is, if and only if A x̃ = −bũ. There exists such an x̃ if and only if b ∈ Im [ A ], in other words

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rank A = rank [ A b ].Also, x̃ is a constant nominal with c x̃ = 0 if and only if

0 = A x̃ + bũ

0 = c x̃

that is, if and only if

c A

x̃ =

0−bũ

As above, this holds if and only if

rank

c A

= rank

c A

0b

Finally, x̃ is a constant nominal with c x̃ = ũ if and only if

0 = A x̃ + bũ = ( A + bc ) x̃

and this holds if and only if

x̃ ∈ Ker [ A + bc ]

(If A is invertible, we can be more explicit. For any ũ the unique constant nominal is x̃ = − A−1bũ . Then ỹ = 0 for ũ ≠ 0 if and only if c A−1b = 0 , and ỹ = ũ if and only if c A−1b = −1.)

Solution 2.8(a) Since

C A

0 B

is invertible, for any K

C A + BK

0 B

=

C A

0 B

K I

I 0

is invertible. Let

C

A + BK

0

B

R3

R1 R4

R2

=

0

I

I

0

Then the 1, 2-block gives R 2 = −( A + BK )−1 BR4 and the 2, 2-block gives CR 2 = I , that is, I = −C ( A + BK )−1 BR4Thus [ C ( A + BK )−1 B ]−1 exists and is given by − R4 .

(b) We need to show that there exists N such that

0 = ( A + BK ) x̃ + BNũ

ũ = Cx̃

The first equation gives

x̃ = −( A + BK )−1 BN ũ

Thus we need to choose N such that

−C ( A + BK )−1 BN ũ = ũ

From part (a) we take N = [−C ( A + BK )−1 B ]−1 = R4 .

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Solution 2.10 For u (t ) = ũ, x̃ is a constant nominal if and only if

0 = ( A + Dũ) x̃ + bũ

This holds if and only if bũ ∈ Im [ A + Dũ], that is, if and only if

rank ( A + Dũ ) = rank

A+Dũ bũ

If A +Dũ is invertible, then

x̃ = −( A + Dũ)−1bũ (+)

If A is invertible, then by continuity of the determinant det ( A + Dũ) ≠ 0 for all ũ such that ũ is sufficientlysmall, and (+) defines a corresponding constant nominal. The corresponding linearized state equation is

x.

δ(t ) = ( A + Dũ) x δ(t ) + [ b − D ( A + Dũ)−1bũ ] u δ(t )

y δ(t ) = C x δ(t )

Solution 2.12 For the given nominal input, nominal output, and nominal initial state, the nominal solutionsatisfies

x̃

.

(t ) =

x̃ 2(t ) − 2 x̃3(t ) x̃ 1(t ) − x̃ 3(t )

1

, x̃(0) =

−2−30

1 = x̃2(t ) − 2 x̃3(t )

Integrating for x̃ 1(t ) and then x̃ 3(t ) easily gives the nominal solution x̃1(t ) = t , x̃ 2(t ) = 2 t − 3, and x̃ 3(t ) = t − 2.The corresponding linearized state equation is specified by

A =

01

0

10

0

−2−10

, B (t )=

0t

0

, C =

0 1 −2

It is unusual that the nominal input and nominal output are constants, but the linearization is time varying.

Solution 2.14 Compute

z.(t ) = x

.(t ) − q

.(t ) = A x (t ) + Bu (t ) + A−1 Bu

.(t )

= A x (t ) − A[− A−1 Bu (t )] + A−1 Bu.(t )

= A z (t ) + A−1 Bu.(t )

If at any value of t a > 0 we have x (t a) = q (t a), that is z (t a) = 0, and u.(t ) = 0 for t ≥ t a, that is u (t ) = u (t a) for

t ≥ t a , then z (t ) = 0 for t ≥ t a . Thus x (t ) = q (t a) for t ≥ t a , and q (t ) represents what could be called an‘instantaneous constant nominal.’

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CHAPTER 3

Solution 3.2 Differentiating term k +1 of the Peano-Baker series using Leibniz rule gives

∂τ∂___

τ∫ t

A (σ1)τ∫

σ1

A (σ2)τ∫

σ2. . .

τ∫

σk

A (σk +1) d σk +1 . . . d σ1

=

A (t )τ∫ t

A (σ2)τ∫

σ2. . .

τ∫ σk

A (σk +1) d σk +1 . . . d σ2

d τd ___

t −

A (τ)τ∫ τ

A (σ2)τ∫ τ

. . . d σk +1 . . . d σ2

d τd ___ τ

+τ∫ t

A (σ1) ∂τ∂___

τ∫

σ1

A (σ2)τ∫

σ2. . .

τ∫

σk

A (σk +1)

d σk +1 . . . d σ1

=τ∫ t

A (σ1) ∂τ∂___

τ∫

σ1

A (σ2)τ∫

σ2

. . .τ∫

σk

A (σk +1)

d σk +1 . . . d σ1

Repeating this process k times gives

∂τ∂___

τ∫ t

A (σ1)τ∫

σ1

A (σ2)τ∫

σ2. . .

τ∫

σk

A (σk +1) d σk +1 . . . d σ1

=τ∫ t

A (σ1)τ∫

σ1. . .

τ∫

σk −1

A (σk ) ∂τ∂___

τ∫ σk

A (σk +1) d σk +1

d σk . . . d σ1

=τ∫

t

A (σ1)τ∫

σ1

. . .τ∫

σk −1

A (σk )

0 − A (τ) + τ∫

σk

0 d σk +1

d σk . . . d σ1

=τ∫ t

A (σ1)τ∫

σ1

A (σ2)τ∫

σ2. . .

τ∫

σk −1

A (σk ) d σk . . . d σ1

− A (τ)

Recognizing this as term k of the uniformly convergent series for −Φ(t , τ) A (τ) gives

∂τ∂___ Φ(t , τ) = −Φ(t , τ) A (τ)

(Of course it is simpler to use the formula for the derivative of an inverse matrix given in Exercise 1.17 .)

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Solution 3.6 Writing the state equation as a pair of scalar equations, the first one is

x.

1

(t ) =1 + t 2

−t ______ x 1(t )

and an easy computation gives

x 1(t ) =(1 + t 2)1 / 2

x1o_________

Then the second scalar equation then becomes

x.

2(t ) =1 + t 2

−4t ______ x 2(t ) +

(1 + t 2)1 / 2

x1o_________

The complete solution formula gives, with some help from Mathematica,

x.

2(t ) =

(1 + t

2

)

2

1________ x 2o +

0

∫ t

(1 + t 2)2(1 + σ2)3 / 2_________

d σ x 1o

=(1 + t 2)2

1________ x 2o +

(1 + t 2)2√ 1+t 2 (t 3 / 4+5t / 8)+(3 / 8) sinh−1(t )_____________________________

x1o

If x 1o = 1, then as t →∞, x2(t ) → 1 / 4, not zero.

Solution 3.7 From the hint, letting

r (t ) =t o

∫ t

v (σ)φ(σ) d σ

we have r .(t ) = v (t )φ(t ), and

φ(t ) ≤ ψ (t ) + r (t ) (*)

Multiplying (*) through by the nonnegative v (t ) gives

v (t )φ(t ) ≤ v (t )ψ (t ) + v (t )r (t )

or

r .(t ) − v (t )r (t ) ≤ v (t )ψ (t )

Multiply both sides by the positive quantity

e

−t o

∫ t

v (τ) d τ

to obtain

dt

d ___

r (t )e

−t o

∫ t

v (τ) d τ

≤ v (t )ψ (t )e−

t o

∫ t

v (τ) d τ

Integrating both sides from t o to t , and using r (t o) = 0 gives

r (t )e

−t o

∫ t

v (τ) d τ

≤t o

∫ t

v (σ)ψ (σ)e−

t o

∫ σ

v (τ) d τ

d σ

Multiplying through by the positive quantity

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et o

∫ t

v (τ) d τ

gives

r (t ) ≤t o

∫ t

v (σ)ψ (σ)e σ∫ t

v (τ) d τ

d σ

and using (*) yields the desired inequality.

Solution 3.10 Multiply the state equation by 2 zT (t ) to obtain

2 zT (t ) z.(t ) =

dt

d ___ z (t )2

=i=1Σ

n

j=1Σ

n

2 zi(t )aij (t ) z j(t )

≤i=1Σn

j =1Σn

2aij (t ) zi(t ) z j(t ) , t ≥ t o

At each t ≥ t o let

a (t ) = 2n21 ≤ i, j ≤ n

max aij (t )

Note a (t ) is a continuous function of t , as a quick sample sketch indicates. Then, since zi(t ) ≤ z (t ),

dt

d ___ z (t )2 ≤ a (t ) z (t )2 , t ≤ t o

Multiplying through by the positive quantity

e

−t o

∫ t

a (σ) d σ

gives

dt

d ___

e

−t o

∫ t

a (σ) d σ

z (t )2

≤ 0 , t ≤ t o

Integrating both sides from t o to t and using z (t o) = 0 gives

z (t ) = 0 , t ≥ t o

which implies z (t ) = 0 for t ≥ t o .

Solution 3.11 The vector function x (t ) satisfies the given state equation if and only if it satisfies

x (t ) = xo +t o

∫ t

A (σ) x(σ) d σ +t o

∫ t

t o

∫ τ

E (τ, σ) x(σ) d σd τ +t o

∫ t

B (σ)u (σ) d σ

Assuming there are two solutions, their difference z (t ) satisfies

z (t ) =t o

∫ t

A (σ) z(σ) d σ +t o

∫ t

t o

∫ τ

E (τ, σ) z(σ) d σd τ

Interchanging the order of integration in the double integral (Dirichlet’s formula) gives

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z (t ) =t o

∫ t

A (σ) z(σ) d σ +t o

∫ t

σ∫ t

E (τ, σ) d τ z(σ) d σ

=t o

∫ t

A (σ) +σ∫ t

E (τ, σ) d τ

z(σ) d σ

=∆

t o

∫ t

Â(t , σ) z (σ) d σ

Thus

z (t ) = t o

∫ t

Â(t , σ) z (σ) d σ ≤t o

∫ t

Â(t , σ) z (σ) d σ

By continuity, given T > 0 there exists a finite constant α such that Â(t , σ) ≤ α for t o ≤ σ ≤ t ≤ t o + T . Thus

z (t ) ≤t o∫ t

α z (t ) d σ , t ∈ [t o , t o+T ]

and the Gronwall-Bellman inequality gives z (t ) = 0 for t ∈ [t o, t o+T ], implying that there can be no morethan one solution.

Solution 3.13 From the Peano-Baker series,

Φ(t , τ) −

I +τ∫ t

A (σ1) d σ1 + . . . +τ∫ t

A (σ1)τ∫

σ1. . .

τ∫

σk −1

A (σk ) d σk . . . d σ1

= j=k +1Σ

∞

τ∫

t

A (σ1)τ∫

σ1

. . .τ∫

σ j−1

A (σ j) d σ j . . . d σ1

For any fixed T > 0 there is a finite constant α such that A (t ) ≤ α for t ∈ [−T , T ], by continuity. Therefore

j=k +1Σ∞

τ∫ t

A (σ1)τ∫

σ1. . .

τ∫

σ j−1

A (σ j) d σ j . . . d σ1 ≤ j=k +1Σ∞

τ∫ t

A (σ1)τ∫

σ1. . .

τ∫

σ j−1

A (σ j) d σ j . . . d σ1

≤ j=k +1Σ∞

τ∫ t

A (σ1)τ∫

σ1. . .

τ∫

σ j−1

A (σ j) d σ j . . . d σ1

.

.

.

≤ j=k +1Σ∞ α j

τ∫ t

. . .

τ∫

σ j−1

1 d σ j . . . d σ1

≤ j=k +1Σ∞

α j j !

t − τ j_______

≤ j=k +1Σ∞

j !

(α2T ) j______, t , τ ∈ [−T , T ]

We need to show that given ε > 0 there exists K such that

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j=K +1Σ∞

j !

(α2T ) j_______ α2T , then

j=k +1Σ∞

j !

(α2T ) j_______ ≤(k +1)!

(α2T )k +1________ .1 − α2T/k

1_________=

(k −1)!(k +1)(k −α2T )(α2T )k +1__________________

Because of the factorial in the denominator, given ε > 0 there exists a K > α2T such that (*) holds.

Solution 3.15 Writing the complete solution of the state equation at t f , we need to satisfy

H o xo + H f

Φ(t f , t o) xo +t o

∫ t f

Φ(t f , σ) f (σ) d σ

= h (+)

Thus there exists a solution that satisfies the boundary conditions if and only if

h − H f t o

∫ t f

Φ(t f , σ) f (σ) d σ ∈ Im[ H o + H f Φ(t f , t o) ]

There exists a unique solution that satisfies the boundary conditions if H o + H f Φ(t f , t o) is invertible. To computea solution x (t ) satisfying the boundary conditions:

(1) Compute Φ(t , t o) for t ∈ [t o, t f ]

(2) Compute H o + H f Φ(t f , t o)

(3) Computet o

∫ t f

Φ(t f , σ) f (σ) d σ

(4) Solve (+) for xo

(5) Set x (t ) = Φ(t , t o) xo +t o

∫ t

Φ(t , σ) f (σ) d σ, t ∈ [t o, t f ]

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CHAPTER 4

Solution 4.1 An easy way to compute A (t ) is to use A (t ) = Φ.

(t , 0)Φ(0, t ). This gives

A (t ) =

1−2t −2t −

1

This A (t ) commutes with its integral, so we can write Φ(t , τ) as the matrix exponential

Φ(t , τ) = exp

τ∫ t

A (σ) d σ

= exp

(t −τ)−(t −τ)2

−(t −τ)2−(t −τ)

Solution 4.4 A linear state equation corresponding to the n th-order differential equation is

x

.

(t ) =

−a0(t )0

.

.

.

00

−a1(t )0

.

.

.

01

. . .

. . .

.

..

. . .

. . .

−an−1(t )1

.

..

00

x (t )

The corresponding adjoint state equation is

z.(t ) =

0

0

.

.

.

−10

. . .

. . .

.

.

.

. . .

. . .

−10

.

.

.

0

0

an−1(t )

an−2(t )

.

.

.

a 1(t )

a 0(t )

z (t )

To put this in the form of an n th-order differential equation, start with

z.n(t ) = − zn−1(t ) + an−1(t ) zn(t )

z.n−1(t ) = − zn−2(t ) + an−2(t ) zn(t )

These give

z..

n(t ) = − z.n−1(t ) +

dt

d ___[ an−1(t ) zn(t ) ]

= zn−2(t ) − an−2(t ) zn(t ) +dt

d ___[ an−1(t ) zn(t ) ]

Next,

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z.n−2(t ) = − zn−3(t ) + an−3(t ) zn(t )

gives

dt 3d 3____

zn(t ) = z.n−2(t ) −

dt

d ___[ an−2(t ) zn(t ) ] +

dt 2d 2____

[ an−1(t ) zn(t ) ]

= − zn−3(t ) + an−3(t ) zn(t ) −dt

d ___[ an−2(t ) zn(t ) ] +

dt 2d 2____

[ an−1(t ) zn(t ) ]

Continuing gives the n th-order differential equation

dt nd n____

zn(t ) =dt n−1d n−1_____

[ an−1(t ) zn(t ) ] −dt n−2d n−2_____

[ an−2(t ) zn(t ) ]

+ . . . + (−1)ndt

d ___[ a 1(t ) zn(t ) ] + (−1)n +1a0(t ) zn(t )

Solution 4.6 For the first matrix differential equation, write the transpose of the equation as (transpose anddifferentiation commute)

X . T

(t ) = AT (t ) X T (t ) , X T (t o) = X oT

This has the unique solution X T (t ) = Φ AT (t )(t , t o) X oT , so that

X (t ) = X oΦ AT (t )T (t , t o)

In the second matrix differential equation, let Φk (t , τ) be the transition matrix for Ak (t ), k = 1, 2. Then it is easyto verify (Leibniz rule) that a solution is

X (t ) = Φ1(t , t o) X oΦ2T (t , t o) +t o

∫ t

Φ1(t , σ)F (σ)Φ2T (t , σ) d σ

Or, one can generate this expression by using the obvious integrating factors on the left and right sides of the

differential equation. (To show this is a unique solution, show that the difference Z (t ) between any two solutions

satisfies Z .(t ) = A1(t ) Z (t ) + Z (t ) A2

T (t ), with Z (t o) = 0. Integrate both sides and apply the Bellman-Gronwall

inequality to show Z (t ) is identically zero.)

Solution 4.9 Clearly A (t ) commutes with its integral. Thus we compute

exp

−10

01

τ

and then replace τ by0

∫ t

a (σ) d σ. From the power series for the exponential,

exp

−10

01

τ

=k =0Σ∞

k !

1___ −1

001

k

τk

=k =0Σ∞

(2k )!

1_____ −1

001

2k

τ2k +k =0Σ∞

(2k +1)!

1________ −1

001

2k +1

τ2k +1

=k =0Σ∞

(2k )!

1_____

0(−1)k

(−1)k 0

τ2k +k =0Σ∞

(2k +1)!

1________

(−1)k +10

0(−1)k

τ2k +1

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21/106


=

0cos τ

cos τ0

+

−sin τ0

0sin τ

=

−sin τcos τ

cos τsin τ

Replacing τ as noted above gives Φ(t , 0).

Solution 4.10 For sufficiency, suppose Φ x(t , 0) = T (t )e Rt . Then T (0) = I and T (t ) is continuouslydifferentiable. Let z (t ) = T −1(t ) x (t ) so that

Φ z(t , 0) = T −1(t )Φ x(t , 0)T (0) = T −1(t )T (t )e Rt = e Rt

Thus z.(t ) = R z (t ).

For necessity, suppose P (t ) is a variable change that gives

z.(t ) = Ra z (t )

Then

Φ z(t , 0) = e Ra t = P−1(t )Φ x(t , 0)P (0)

that is,

Φ x(t , 0) = P (t )e Ra t P−1(0)

Let T (t ) = P (t )P−1(0) and R = P (0) RaP−1(0). Then

Φ x(t , 0) = T (t )P (0) e P−1(0) RP (0)t P−1(0)

= T (t )P (0) [ P−1(0)e Rt P (0) ] P−1(0)

= T (t )e Rt

Solution 4.11 Suppose

Φ(t , 0) = e A1t e A2t

Then

Φ.

(t , 0) =dt

d ___

e A1t e

A2t

= e A1t ( A1+A2 ) e

A2t

= e A1t ( A1+A2 ) e

− A1t . e A1t e

A 2t

This implies A (t ) = e A1t [ A1+A2 ] e

− A 1t . Therefore A (0) = A1+A2 is clear, and

A.(t ) = A1e

A1t ( A1+A2 ) e− A1t + e

A1t ( A1+A2 ) e− A1t (− A1)

= A1 A (t ) − A (t ) A1

Conversely, assume A1 and A2 are such that

A.(t ) = A1 A (t ) − A (t ) A1 , A (0) = A1 + A2

This matrix differential equation has a unique solution (by rewriting it as a linear vector differential equation), and

from the calculation above this solution is

A (t ) = e A 1t ( A1 + A2 ) e

− A1t

Since

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22/106


dt

d ___

e A1t e

A2t

= A (t )e A1t e

A2t , e A10e

A20 = I

we have that Φ(t , 0) = e A1t

e A2t

.

Solution 4.13 Writing

∂t ∂___ Φ A(t , τ) = A (t )Φ A(t , τ) , Φ(τ, τ) = I

in partitioned form shows that

∂t ∂___ Φ21(t , τ) = A22(t )Φ21(t , τ) , Φ21(τ, τ) = 0

Thus Φ21(t , τ) is identically zero. But then

∂t

∂___ Φii

(t , τ) = Aii

(t )Φii

(t , τ) , Φii

(τ, τ) = I

for i = 1, 2, and

∂t ∂___ Φ12(t , τ) = A11(t )Φ12(t , τ) + A12(t )Φ22(t , τ) , Φ12(τ, τ) = 0

Using Exercise 4.6 with F (t ) = A12(t ) Φ22(t , τ) gives

Φ12(t , τ) =τ∫ t

Φ11(t , σ) A12(σ) Φ22(σ, τ) d σ

Solution 4.17 We need to compute a continuously-differentiable, invertible P (t ) such that

1t

t

1

= P−1(t )

2−t 20

2 t 1

P (t ) − P−1(t )P. (t )

Multiplying on the left by P (t ), the result can be written as a dimension-4 linear state equation. Choosing the

initial condition corresponding to P (0) = I , some clever guessing gives

P (t ) =

t 1

10

Solution 4.23 Using the formula for the derivative of an inverse matrix given in Exercise 1.17,

∂t ∂___ Φ A(−τ, −t ) = ∂t

∂___ Φ A−1(−t , −τ) = −Φ A−1(−t , −τ)

∂t ∂___ Φ A(−t , −τ)

Φ A−1(−t , −τ)

= −Φ A−1(−t , −τ)

−∂(−t )

∂_____ Φ A(−t , −τ)

Φ A−1(−t , −τ)

= −Φ A−1(−t , −τ)

− A (−t )Φ A(−t , −τ)

Φ A−1(−t , −τ)

= Φ A−1(−t , −τ) A (−t ) = Φ A(−τ, −t ) A (−t )

Transposing gives

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23/106


∂t ∂___ Φ AT (−τ, −t ) = AT (−t )Φ AT (−τ, −t )

Since Φ(−τ, −τ) = I , we have F (t ) = AT

(−t ).

Or we can use the result of Exercise 3.2 to compute:

∂t ∂___ Φ A(−τ, −t ) =− ∂(−t )

∂_____ Φ A(−τ, −t ) = Φ A(−τ, −t ) A (−t )

This implies

∂t ∂___ Φ AT (−τ, −t ) = AT (−t )Φ A(−τ, −t )

Since Φ(−τ, −τ) = I , we have F (t ) = AT (−t ).

Solution 4.25 We can write

Φ(t + σ, σ) = I +σ∫

t+σ

A (τ) d τ +k =2Σ∞

σ∫

t+σ

A (τ1)σ∫ τ1

A (τ2) . . .σ∫

τk −1

A (τk ) d τk . . . d τ1

and

e A

__

t (σ)t = I + A_

t (σ)t +k =2Σ∞

k !

1___ A_

t

k (σ)t k

Then

R (t , σ) = Φ(t + σ, σ) − e A__

t (σ)t

= k =2

Σ

∞

σ∫

t +σ

A (τ1)σ∫ τ1

A (τ2) . . .

σ∫

τk −1

A (τk ) d τk . . . d τ1 −k !

1___ A_

t

k (σ)t k

From A (t ) ≤ α and the triangle inequality,

R (t , σ) ≤ 2k =2Σ∞

αk k !

t k ___= α2t 2

k =2Σ∞

k !

2___ αk −2 t k −2

Using

k !

2___ ≤(k −2)!

1______, k ≥ 2

gives

R (t , σ) ≤ α2t 2k =2Σ∞

(k −2)!1______ αk −2 t k −2

= α2t 2e α t

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CHAPTER 5

Solution 5.3 Using the series definition, which involves talent in series recognition,

A2k +1 =

10

01

, A2k =

01

10

, k = 0, 1, . . .

gives

e At = I +

t

0

0

t

+2!

1___

0

t 2

t 20

+3!

1___

t 30

0

t 3

+ . . .

=

(e t −e −t ) / 2(e t +e−t ) / 2

(e t +e −t ) / 2

(e t −e−t ) / 2

=

sinh t

cosh t

cosh t

sinh t

Using the Laplace transform method,

(sI − A)−1

=

−1s

s

−1

−1

=

s2−1s_____

s2−11_____

s2−11_____

s2−1s_____

which gives again

e At =

sinh t cosh t

cosh t sinh t

Using the diagonalization method, computing eigenvectors for A and letting

P =

11

−11

gives

P−1 AP =

01

−10

Then

e At = P

0

et

e −t 0

P−1 =

sinh t

cosh t

cosh t

sinh t

Solution 5.4 Since

A (t ) =

1t

t 1

commutes with its integral,

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25/106


Φ(t , 0) = e 0∫ t

A (σ) d σ

= exp

t t 2 / 2

t 2 / 2t

And since

0

t 2 / 2

t 2 / 2

0

,

t

0

0

t

commute,

Φ(t , 0) = exp

01

10

t 2 / 2

. exp

10

01

t

Using Exercise 5.3 gives

Φ(t , 0) =

0

e t 2 / 2

e

t 2 / 2

0

sinh t

cosh t

cosh t

sinh t

=

e

t 2 / 2

sinh t

e t 2 / 2 cosh t

e

t 2 / 2

cosh t

e t 2 / 2 sinh t

Solution 5.7 To verify that

A0

∫ t

e Aσ d σ = e At − I

note that the two sides agree at t = 0, and the derivatives of the two sides with respect to t are identical.

If A is invertible and all its eigenvalues have negative real parts, then lim t → ∞ e At = 0. This gives

A0

∫ ∞

e Aσ d σ = − I

that is,

A−1 = −0

∫ ∞

e Aσ d σ =∞∫ 0

e Aσ d σ

Solution 5.9 Evaluating the given expression at t = 0 gives x (0) = 0. Using Leibniz rule to differentiate theexpression gives

x.(t ) =

dt

d ___

0

∫ t

e A (t −σ)e D

σ∫ t

u (τ) d τ

bu (σ) d σ

= bu (t ) +0

∫ t

∂t ∂___

e A (t −σ)e D

σ∫

t

u (τ) d τbu (σ)

d σ

Using the product rule and differentiating the power series for e D

σ∫ t

u (τ) d τ

gives

x.(t ) = bu (t ) +

0

∫ t

Ae A (t −σ)e D

σ∫ t

u (τ) d τ

bu (σ) + e A (t −σ) Du (t )e D

σ∫ t

u (τ) d τ

bu (σ)

d σ

If we assume that AD = DA, then e A (t −σ) D = De A (t −σ) and

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x.(t ) = bu (t ) + A

0

∫ t

e A (t −σ)e D

σ∫ t

u (τ) d τ

bu (σ) d σ + Du (t )

0

∫ t

e A (t −σ)e D

σ∫ t

u (τ) d τ

bu (σ) d σ

= A x (t ) + Dx (t )u (t ) + bu (t )

Solution 5.12 We will show how to define β0(t ), . . . , βn−1(t ) such that

k =0Σn−1

β.k (t )Pk =

k =0Σn−1

βk (t ) APk ,k =0Σn−1

βk (0)Pk = I (*)

which then gives the desired expression by Property 5.1. From the definitions,

P1 = AP 0 − λ1 I , P2 = AP1 − λ2P1 , . . . , Pn−1 = APn−2 − λn−1Pn−2

Also Pn = ( A−λn I )Pn−1 = 0 by the Cayley-Hamilton theorem, so APn−1 = λnPn−1. Now we equate coefficients of like Pk ’s in (*), rewritten as

k =0Σn−1

β.

k (t )Pk =k =0Σn−1

βk (t )[Pk+1 + λk +1Pk ]

to get equations for the desired βk (t )’s:

P0 : β.

0(t ) = λ1β0(t )

P1 : β.

1(t ) = β0(t ) + λ2β1(t )...

Pn−1 : β.

n−1(t ) = βn−2(t ) + λnβn−1(t )

that is,

β.

n−1(t )

.

.

.

β.

1(t )

β.

0(t )

=

0

0

.

.

.

1

λ1

0

0

.

.

.

λ2

0

. . .

. . .

.

.

.

. . .

. . .

1

λn−1

.

.

.

0

0

λn

0

.

.

.

0

0

βn−1(t )

.

.

.

β1(t )β0(t )

With the initial condition provided by β0(0) = 1, βk (0) = 0, k = 1, . . . , n−1, the analytic solution of this stateequation provides a solution for (*). (The resulting expression for e At is sometimes called Putzer’s formula.)

Solution 5.17 Write, by Property 5.11,Φ(t , t o) = P−1(t )e

R (t −t o)P (t o)

where P (t ) is continuous, T -periodic, and invertible at each t . Let

S = P −1(t o) RP (t o) , Q (t , t o) = P−1(t )P (t o)

Then Q (t , t o) is continuous and invertible at each t , and satisfies

Q (t +T , t o) = P−1(t +T )P (t o) = P

−1(t )P (t o) = Q (t , t o)

with Q (t o , t o) = I . Also,

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27/106


Φ(t , t o) = P−1(t ) eP (t o)SP

−1(t o) (t −t o)P (t o) = P−1(t )P (t o) e

S (t −t o)P−1(t o)P (t o)

= Q (t , t o)eS (t −t o)

Solution 5.19 From the Floquet decomposition and Property 4.9,

det Φ(T , 0) = det e RT = e 0∫ T

tr [ A (σ)] d σ

Because the integral in the exponent is positive, the product of eigenvalues of Φ(T , 0) is greater than unity, whichimplies that at least one eigenvalue of Φ(T , 0) has magnitude greater than unity.Thus by the argument followingExample 5.12 there exist unbounded solutions.

Solution 5.20 Following the hint, define a real matrix S by

eS 2T = Φ2(T , 0)

and set

Q (t ) = Φ(t , 0)e −St

Clearly Q (t ) is real and continuous, and

Q (t +2T ) = Φ(t +2T , 0)e−S (t+2T ) = Φ(t +2T , T )Φ(T , 0)e−S 2T e −St

= Φ(t +T , 0)Φ(T , 0)e−S 2T e −St = Φ(t +T , T )Φ2(T , 0)e −S 2T e−St

= Φ(t +T , T )e −St = Φ(t , 0)e−St

= Q (t )

That is, Q (t ) is 2T -periodic. (For a proof of the hint, see Chapter 8 of D.L. Lukes, Differential Equations:Classical to Controlled , Academic Press, 1982.)

Solution 5.22 The solution will be T -periodic for initial state xo if and only if xo satisfies (see text equation(32))

[ Φ−1(t o+T , t o) − I ] xo =t o

∫ t o+T

Φ(t o , σ) f (σ) d σ

This linear equation has a solution for xo if and only if

zoT

t o

∫ t o+T

Φ(t o , σ) f (σ) d σ = 0 (*)

for every nonzero vector zo that satisfies

[ Φ−1(t o+T , t o) − I ]T zo = 0 (**)

The solution of the adjoint state equation can be written as

z (t ) = [ Φ−1(t , t o) ]T zo

Then by Lemma 5.14, (**) is precisely the condition that z (t ) be T -periodic. Thus writing (*) in the form

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28/106


0 =t o

∫ t o+T

zoT Φ(t o , σ) f (σ) d σ =

t o

∫ t o+T

zT (σ) f (σ) d σ

completes the proof.

Solution 5.24 Note A = − AT , and from Example 5.9,

e At =

−sin t cos t

cos t sin t

Therefore all solutions of the adjoint equation are periodic, with period of the form k 2π, where k is a positiveinteger. The forcing term has period T = 2π / ω, where we assume ω > 0. The rest of the analysis breaks downinto 3 cases.

Case 1: If ω ≠ 1, 1 / 2, 1 / 3, . . . then the adjoint equation has no T -periodic solution, so the condition (Exercise5.22)

0∫ T

zT (σ) f (σ) d σ = 0 (+)

holds vacuously. Thus there will exist corresponding periodic solutions.

Case 2: If ω = 1, then

0

∫ T

zT (σ) f (σ) d σ =0

∫ T

zoT e Aσ f (σ) d σ

= − zo 10

∫ T

sin2(σ) d σ + zo 20

∫ T

cos σ sin σ d σ

≠ 0

so there is no periodic solution.

Case 3: If ω = 1 /k , k = 2, 3, . . . , then since

0

∫ T

cos σ sin (σ /k ) d σ =0

∫ T

sin σ sin (σ /k ) d σ = 0

the condition (+) will hold, and there exist periodic solutions.

In summary, there exist periodic solutions for all ω > 0 except ω = 1.

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CHAPTER 6

Solution 6.1 If the state equation is uniformly stable, then there exists a positive γ such that for any t o and xo

the corresponding solution satisfies

x (t ) ≤ γ xo , t ≥ t o

Given a positive ε, take δ = ε / γ . Then, regardless of t o , xo ≤ δ implies

x (t ) ≤ γ δ = ε , t ≥ t o

Conversely, given a positive ε suppose positive δ is such that, regardless of t o , xo ≤ δ implies x (t ) ≤ ε,t ≥ t o . For any t a ≥ t o let xa be such that

xa = 1 , Φ(t a , t o) xa = Φ(t a , t o)

Then xo = δ xa satisfies xo = δ, and the corresponding solution at t = t a satisfies

x (t a) = Φ(t a, t o) xo = δΦ(t a, t o) ≤ ε

Therefore

Φ(t a , t o) ≤ ε / δ

Such an xa can be selected for any t a , t o such that t a ≥ t o . Therefore

Φ(t , t o) ≤ ε / δ

for all t and t o with t ≥ t o , and we can take γ = ε / δ to obtain

x (t ) = Φ(t , t o) xo ≤ Φ(t , t o) xo ≤ γ xo , t ≥ t o

This implies uniform stability.

Solution 6.4 Using the fact that A (t ) commutes with its integral,

Φ(t , τ) = e τ∫ t

A (σ) d σ

= I +

e −(t −τ)t −τ

t −τ−e −(t −τ)

+2!

1___

e−(t −τ)t −τ

t −τ−e−(t −τ)

2

+ . . .

For any fixed τ, φ11(t , τ) clearly grows without bound as t → ∞, and thus the state equation is not uniformlystable.

Solution 6.6 Using elementary properties of the norm,

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Φ(t , τ) = I +τ∫ t

A (σ) d σ +τ∫ t

A (σ1)τ∫

σ1

A (σ2) d σ2d σ1 + . . .

= I + τ∫ t

A (σ) d σ + τ∫ t

A (σ1)τ∫

σ1

A (σ2) d σ2d σ1 + . . .

= 1 + τ∫ t

A (σ) d σ + τ∫ t

A (σ1) τ∫

σ1

A (σ2) d σ2d σ1 + . . .

(Be careful of t 0 by assumption, so that

t eλt

= t e−ηt

, t ≥ 0A simple maximization argument (setting the derivative to zero) gives

t e −ηt ≤η e1___

=∆ β , t ≥ 0

so that

t e λt ≤ β , t ≥ 0

Using this bound we can write

t eλt = t e−ηt = t e−(η / 2)t e −(η / 2)t ≤η e2___

e−(η / 2)t , t ≥ 0

Similarly,

t 2 eλt = t 2 e −ηt ≤η e2___

t e−(η / 2)t =η e2___

t e −(η / 4)t e−(η / 4)t ≤η e2___ .

η e4___

e−(η / 4)t , t ≥ 0

and continuing we get, for any j ≥ 0,

t j e λt ≤(η e) j

2 j+( j −1)+ . . . +1

____________e−(η / 2

j)t , t ≥ 0

Therefore

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0

∫ ∞

t j e λt dt ≤(η e) j

2 j +( j−1)+ . . . +1

____________

0

∫ ∞

e−(η / 2 j)t dt

≤(η e) j

2 j+( j−1)+ . . . +1____________ .η2 j___

=e j Re [λ] j +122 j+( j−1)+

. . . +1_____________

Solution 6.12 By Theorem 6.4 uniform stability is equivalent to existence of a finite constant γ such thate At ≤ γ for all t ≥ 0. Writing

e At =k =1Σm

j=1Σσk

W kj( j−1)!

t j−1______e

λk t

where λ1, . . . , λm are the distinct eigenvalues of A, supposeRe[λk ] ≤ 0 , k = 1, . . . , m (*)

Re[λk ] = 0 implies σk = 1

Since t j−1eλk t is bounded if Re[λk ] 0 for some k , the proof of Theorem 6.2 shows that e At grows without bound as t → ∞. The gap

between this necessary condition and the sufficient condition is illustrated by the two cases

A =

0

0

0

0

, A =

0

0

0

1

Both satisfy the necessary condition, neither satisfy the sufficient condition, and the first case is uniformly stable

while the second case is not (unbounded solutions exist, as shown by easy computation of the transition matrix).

(It can be shown that a necessary and sufficient condition for uniform stability is that each eigenvalue of A has

nonpositive real part and any eigenvalue of A with zero real part has algebraic multiplicity equal to its geometric

multiplicity.)

Solution 6.14 Suppose γ , λ > 0 are such that

Φ(t , t o) ≤ γ e−λ(t −t o)

for all t , t o such that t ≥ t o . Then given any xo , t o , the corresponding solution at t ≥ t o satisfies

x (t ) = Φ(t , t o) xo ≤ Φ(t , t o) xo ≤ γ e−λ(t −t o)

xoand the state equation is uniformly exponentially stable.

Now suppose the state equation is uniformly exponentially stable, so that there exist γ , λ > 0 such that

x (t ) ≤ γ e−λ(t −t o) xo , t ≥ t o

for any xo and t o . Given any t o and t a ≥ t o , choose xa such that

Φ(t a , t o) xa = Φ(t a , t o) , xa = 1

Then with xo = xa the corresponding solution at t a satisfies

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x (t a) = Φ(t a , t o) xa = Φ(t a , t o) ≤ γ e−λ(t a−t o)

Since such an xa can be selected for any t o and t a > t o, we have

Φ(t , τ) ≤ γ e−λ(t −τ)

for all t , τ such that t ≥ τ, and the proof is complete.

Solution 6.18 The variable change z (t ) = P−1(t ) x (t ) yields z.(t ) = 0 if and only if

P−1(t ) A (t )P (t ) − P−1(t )P.(t ) = 0

for all t . This clearly is equivalent to P.(t ) = A (t )P (t ), which is equivalent to Φ A(t , τ) = P (t )P−1(τ). Now, if P (t )

is a Lyapunov transformation, that is P (t ) ≤ ρ 0 for all t , then

Φ A(t , τ) ≤ P (t )P−1(τ) ≤ P (t )det P (τ)P (τ)n−1__________

≤ ρn / η =∆ γ

for all t and τ.Conversely, suppose Φ A(t , τ) ≤ γ for all t and τ. Let P (t ) = Φ A(t , 0). Then P (t ) ≤ γ and

P (t ) ≤det P−1(t )

P−1(t )n−1___________= P−1(t )n−1det P (t )

for all t . Using P (t ) ≥ 1 / P−1(t ) gives

det P (t ) ≥P−1(t )n

1__________

and since P−1(t ) = Φ A(0, t ) ≤ γ ,

det P (t ) ≥ γ n1___

Thus P (t ) is a Lyapunov transformation, and clearly

P−1(t ) A (t )P (t ) − P−1(t )P.(t ) = 0

for all t .

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CHAPTER 7

Solution 7.3 Let Â = FA, and take Q = F −1 , which is positive definite since F is positive definite. Then since F

is symmetric,

ÂT Q +QÂ = AT FF −1 + F −1FA = AT + A


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η ≤ a (t ) ≤ 1 / (2η)

for all t . Then

2a (t ) + 1 − η ≥ η + 1 > 1

a (t )

a (t )+1_______ − η ≥ 1 +1 / (2η)

1______= 1 + η > 1

and Q (t )−η I ≥ 0, for all t , follows easily. Similarly, with ρ = (2η+1) / η we can show ρ I −Q (t ) ≥ 0 using

ρ − 2a (t ) − 1 ≥η

2η+1______ − 22η1___ − 1 = 1

ρ −a (t )

a (t )+1_______ ≥η

2η+1______ − 1 −a (t )

1____ ≥ 1

Next consider

AT (t )Q (t ) + Q (t ) A (t ) + Q.

(t ) =

0

2a.(t )−2a(t )

−2a(t )−a2(t )

a. (t )_____

0

≤ − ν I

This gives that for uniform exponential stability we also need existence of a small, positive constant ν such that

νa2(t ) − 2a3(t ) ≤ a.(t ) ≤ a (t )− ν / 2

for all t . For example, a (t ) = 1 satisfies these conditions.

Solution 7.11 Suppose that for every symmetric, positive-definite M there exits a unique, symmetric,positive-definite Q such that

AT Q + QA + 2µQ = − M (*)

that is,

( A + µ I )T Q + Q ( A + µ I ) = − M (**)

Then by the argument above Theorem 7.11 we conclude that all eigenvalues of A +µ I have negative real parts.That is, if

0 = det [ λ I − ( A +µ I ) ] = det [ (λ − µ) I − A ]

then Re [λ] 0, this gives Re [λ − µ]


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36/106


x T e F T t e Ft x ≤ 2 ( A+µ−ε) x T Qx , t ≥ 0

which gives

eFt ≤ √ 2 ( A+µ−ε)Q , t ≥ 0

Thus the desired inequality follows from (*).

Solution 7.19 To show uniform exponential stability of A (t ), write the 1,2-entry of A (t ) as a (t ), and letQ (t ) = q (t ) I , where

q (t ) =

3 , t ≤ −1 / 2q ⁄ 1 2(t ) , −1 / 2 < t


37/106

CHAPTER 8

Solution 8.3 No. The matrix

A =

0−2

−1√ 8

has negative eigenvalues, but

A + AT =

√ 8−4

−2√ 8

has an eigenvalue at zero.

Solution 8.6 Viewing F (t ) x (t ) as a forcing term, for any t o, xo , and t ≥ t o we can write

x (t ) = Φ A +F (t , t o) xo = Φ A(t , t o) xo + t o∫

t

Φ A(t , σ)F (σ) x(σ) d σ

which gives, for suitable constants γ , λ > 0,

x (t ) ≤ γ e−λ(t −t o) xo +t o

∫ t

γ e−λ(t −σ)F (σ) x(σ) d σ

Thus

e λt x (t ) ≤ γ eλt o xo +t o

∫ t

γ F (σ) eλσ x(σ) d σ

and the Gronwall-Bellman inequality (Lemma 3.2) implies

e λt x (t ) ≤ γ eλt o xoet o∫ t

γ F (σ) d σ

Therefore

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x(t ) ≤ γ e−λ(t −t o)e t o∫ t

γ F (σ) d σ

xo

≤ γ e−λ(t −t o)e t o∫ ∞

γ F (σ) d σ

xo

≤ γ e−λ(t −t o) e γ β xo

and we conclude the desired uniform exponential stability.

Solution 8.8 We can follow the proof of Theorem 8.7 (first and last portions) to show that the solution

Q (t ) =0

∫ ∞

e AT (t )σ e A (t )σ d σ

of

AT (t )Q (t ) + Q (t ) A (t ) = − I

is continuously-differentiable and satisfies, for all t ,

η I ≤ Q (t ) ≤ ρ I

where η and ρ are positive constants. Then with

F (t ) = A (t ) − ⁄ 1 2Q−1(t )Q.

(t )

an easy calculation shows

F T (t )Q (t ) + Q (t )F (t ) + Q.

(t ) = AT (t )Q (t ) + Q (t ) A (t ) = − I

Thus x.(t ) = F (t ) x (t )

is uniformly exponentially stable by Theorem 7.4.

Solution 8.9 As in Exercise 8.8 we have, for all t ,

η I ≤ Q (t ) ≤ ρ I

which implies

Q−1(t ) ≤η1__

Also, by the middle portion of the proof of Theorem 8.7,

Q.

(t ) ≤ 2 A.(t )Q (t )2

Therefore

⁄ 1 2Q−1(t )Q.

(t ) ≤η

βρ2____

for all t . Write

x.(t ) = A (t ) x (t ) = [ A (t ) − ⁄ 1 2Q−1(t )Q

.(t ) ] x (t ) + ⁄ 1 2Q−1(t )Q

.(t ) x (t )

=∆

F (t ) x (t ) + ⁄ 1 2Q−1(t )Q.

(t ) x (t )

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Then the complete solution formula gives

x (t ) = ΦF (t , t o) xo + t o∫

t

ΦF (t , σ) ⁄ 1

2Q−1

(σ)Q

.

(σ) x(σ) d σ

and the result of Exercise 8.8 implies that there exists positive constants γ , λ such that, for any t o and t ≥ t o ,

x (t ) ≤ γ e−λ(t −t o) xo +t o

∫ t

γ e−λ(t −σ)η

βρ2____ x(σ) d σ

Therefore

e λt x (t ) ≤ γ eλt o xo +t o

∫ t

ηγβρ2_____

e λσ x(σ) d σ

and the Gronwall-Bellman inequality (Lemma 3.2) implies

e λt x (t ) ≤ γ eλt o xoet o∫

t

γβρ2 / η d σ

Thus

x (t ) ≤ γ e−(λ−γβρ2 / η)(t −t o) xo

Now, writing the left side as Φ A(t , t o) xo and for any t o and t ≥ t o choosing the appropriate unity-norm xo gives

Φ A(t , t o) ≤ γ e−(λ−γβρ2 / η)(t −t o)

For β sufficiently small this gives the desired uniform exponential stability. (Note that Theorem 8.6 also can beused to conclude that uniform exponential stability of x

.(t ) = F (t ) x (t ) implies uniform exponential stability of

x.(t ) = [ F (t ) + ⁄ 1 2Q−1(t )Q

.(t ) ] x (t ) = A (t ) x (t )

for β sufficiently small.)

Solution 8.10 With F (t ) = A (t ) + (µ / 2) I we have that F (t ) ≤ α + µ / 2, F .(t ) = A

.(t ), and the eigenvalues of

F (t ) satisfy Re [λF (t )] ≤ −µ / 2. The unique solution of

F T (t )Q (t ) + Q (t )F (t ) = − I

is

Q (t ) =0

∫ ∞

eF T (t )σ e F (t )σ d σ

As in the proof of Theorem 8.7, there is a constant ρ such that Q (t ) ≤ ρ for all t . Now, for any n × 1 vector z,

d σd ___ zT e F

T (t )σ e F (t )σ z = z T e F T (t )σ[ F T (t ) + F (t ) ] e F (t )σ z

≥ −(2α + µ) zT e F T (t )σ eF (t )σ z

Thus for any τ ≥ 0,

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− zT e F T (t )τ e F (t )τ z =

τ∫ ∞

d σd ___

zT e F T (t )σ e F (t )σ z

d σ

≥ −(2α + µ)τ∫ ∞

zT e F T (t )σ e F (t )σ z d σ

≥ −(2α + µ)0

∫ ∞

zT e F T (t )σ e F (t )σ z d σ

≥ −(2α + µ) zT Q (t ) z

Thus

e F T (t )τ e F (t )τ ≤ (2α + µ) Q (t ) , τ ≥ 0

and using

e F (t )τ = e A(t )τ e (µ / 2) τ , τ ≥ 0

gives

e A(t )τ ≤ √ (2α+µ)ρ e (−µ / 2) τ , τ ≥ 0

Solution 8.11 Write (the chain rule is valid since u (t ) is a scalar)

q.(t ) = − A−1(u (t ))

du

dA___(u (t ))u

.(t )

A−1(u (t ))b (u (t )) − A−1(u (t ))du

db___(u (t ))u

.(t )

=∆ − B̂(t )u

.(t )

Then

x.(t ) = A (u (t )) x (t ) + b (u (t ))

= A (u (t )) [ x (t ) − q (t ) ] + A (u (t ))q (t ) + b (u (t ))= A (u (t )) [ x (t ) − q (t ) ]

gives

dt

d ___[ x (t ) − q (t ) ] = A (u (t )) [ x (t ) − q (t ) ] + B̂(t )u

.(t ) (*)

Since

dt

d ___ A (u (t )) =

du

dA___(u (t ))u

.(t ) =

du

dA___(u (t ))u

.(t )

we can conclude from Theorem 8.7 that for δ sufficiently small, and u (t ) such that u.(t ) ≤ δ for all t , there exist

positive constants γ and η (depending on u (t )) such that

Φ A (u (t ))(t , σ) ≤ γ e−η (t −σ) , t ≥ σ ≥ 0

But the smoothness assumptions on A (.) and b (.) and the bounds on u (t ) also give that there exists a positive

constant β such that B̂(t ) ≤ β for t ≥ 0. Thus the solution formula for (*) gives

x (t ) − q (t ) ≤ γ x (0) − q (0) + γ βδ / η , t ≥ 0

for u (t ) as above, and the claimed result follows.

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CHAPTER 9

Solution 9.7 Write

B ( A−β I ) B ( A−β I )2 B . . .

=

B AB−β B A2 B−2β AB+β2 B . . .

=

B AB A 2 B . . .

.

.

.

0

0

0

I m

.

.

.

0

0

I m

−β I m

.

.

.

0

I m

−2β I mβ2 I m

.

.

.

. . .

. . .

. . .

. . .

Clearly the two controllability matrices have the same rank. (The solution is even easier using rank tests from

Chapter 13.)

Solution 9.8 Since A has negative-real-part eigenvalues,

Q =0

∫ ∞

e At BBT e AT t dt

is well defined, symmetric, and

AQ + QAT =0

∫ ∞

Ae At BBT e AT t + e At BBT e A

T t AT

dt

=0

∫ ∞

dt

d ___

e At BBT e AT t

dt

= − BBT

Also it is clear that Q is positive semidefinite. If it is not positive definite, then for some nonzero, n × 1 x,

0 = xT Qx =0

∫ ∞

x T e At BBT e AT t x dt

=0

∫ ∞

x T e At B2 dt

Thus xT e At B = 0 for all t ≥ 0, and it follows that

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0 =dt jd j___

x T e At B

t = 0= xT A j B

for j = 0, 1, 2, . . . . But this implies

x T

B AB . . . An−1 B

= 0

which contradicts the controllability hypothesis. Thus Q is positive definite.

Solution 9.9 Suppose λ is an eigenvalue of A, and p is a corresponding left eigenvector. Then p ≠ 0, and

p T A = λ pT

This implies both

p H A = λ_ p H , AT p = λ p

Now suppose Q is as claimed. Then

p H AQp + p H QAT p = λ_ p H Qp + λ p H Qp

= − p H BBT p

that is,

2 Re [λ] p H Q p = − p H BBT p (*)

This gives Re [λ] ≤ 0 since Q is positive definite. Now suppose Re [λ] = 0. Then (*) gives p H B = 0. Also, for j = 1, 2, . . . ,

p H A j B = λ_ p H A j−1 B = . . . = λ

_ j p H B = 0

Thus

p H

B AB . . . An−1 B

= 0

which contradicts the controllability assumption. Therefore Re [λ]


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Now suppose the state equation is output controllable on [t o , t f ], but that W y(t o , t f ) is not invertible. Then

there exists a p × 1 vector ya ≠ 0 such that yaT W y(t o, t f ) ya = 0. Using by now familiar arguments, this gives

yaT C (t f )Φ(t f , t ) B (t ) = 0 , t ∈ [t o , t f ]

Consider the initial state

xo = Φ(t o , t f )C T (t f )[ C (t f )C T (t f ) ]−1 ya

which is well defined and nonzero since rank C (t f ) = p. There exists an input ua(t ) such that

0 = C (t f )Φ(t f , t o) xo +t o

∫ t f

C (t f )Φ(t f , σ) B (σ)ua(σ) d σ

= ya +t o

∫ t f

C (t f )Φ(t f , σ) B (σ)ua(σ) d σ

Premultiplying by ya

T gives

0= yaT ya

This contradicts ya ≠ 0, and thus W y(t o , t f ) is invertible.The rank assumption on C (t f ) is needed in the necessity proof to guarantee that xo is well defined. For

m = p = 1, invertibility of W y(t o , t f ) is equivalent to existence of a t a ∈ (t o , t f ) such that

C (t f )Φ(t f , t a) B (t a) ≠ 0

That is, there exists a t a ∈ (t o , t f ) such that the output response at t f to an impulse input at t a is nonzero.

Solution 9.11 From Exercise 9.10, since rank C = p, the state equation is output controllable if and only if forsome fixed t f > 0,

W y =∆

0

∫ t f

Ce A (t f −t ) BBT e

A T (t f −t )C T dt

is invertible. We will show this holds if and only if

rank

CB CAB . . . CAn−1 B

= p

by showing equivalence of the negations. If W y is not invertible, there exists a nonzero p × 1 vector ya such that ya

T W y ya = 0. Thus

yaT Ce

A (t f −t ) B = 0 , t ∈ [0, t f ]

Differentiating repeatedly, and evaluating at t = t f gives

ya

T CA j B = 0 , j = 0, 1, . . .

Thus

yaT


= 0

and this implies

rank


< p

Conversely, if the rank condition fails, then there exists a nonzero ya such that yaT CA j B = 0,

j = 0, . . . , n−1. Then

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yaT Ce

A (t f −t ) B = y aT C

k =0Σn−1

αk (t f −t ) Ak B = 0 , t ∈ [0, t f ]

Therefore yaT W y ya = 0, which implies that W y is not invertible.For m = p = 1 argue as in Solution 9.10 to show that a linear state equation is output controllable if and

only if its impulse response (equivalently, transfer function) is not identically zero.

Solution 9.17 Beginning with

y (t ) = c (t ) x (t )

y.(t ) = c

.(t ) x (t ) + c (t ) x

.(t )

= [c.(t ) + c (t ) A (t )] x (t ) + c (t )b (t )u (t )

= L1(t ) x (t ) + L0(t )b (t )u (t )

it is easy to show by induction that

y (k )(t ) = Lk (t ) x (t ) + j =0Σk −1

dt k − j−1d k − j−1_______ [ L j(t )b (t )u (t ) ] , k = 1, 2, . . .

Now if

Ln(t ) M __

−1=∆

α0(t ) α1(t ) . . . αn −1(t )

then

i=0Σn −1

αi(t ) Li(t ) = α0(t ) . . . αn −1(t )

Ln −1(t )

.

.

.

L0(t )

= Ln(t )

Thus we can write

y (n)(t ) −i=0Σn −1

αi(t ) y(i)(t ) = Ln(t ) x (t ) +

j=0Σn−1

dt n − j−1d n − j−1_______ [ L j(t )b (t )u (t ) ]

−i=0Σn −1

αi(t ) Li(t ) x (t ) −i =0Σn −1

αi(t ) j=0Σi−1

dt i − j −1d i− j−1______ [ L j(t )b (t )u (t ) ]

= j =0Σn−1

dt n − j−1d n − j −1_______ [ L j(t )b (t )u (t ) ] −

i=0Σn −1

αi(t ) j =0Σi−1

dt i− j−1d i− j−1______ [ L j(t )b (t )u (t ) ]

This is in the desired form of an n th-order differential equation.

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C (t ) B (σ) = H (t )F (σ) (*)

for all t , σ, picking an appropriate t o and t f > t o ,

M x(t o , t f )W x(t o , t f ) =t o

∫ t f

C T (t ) H (t ) dt t o

∫ t f

F (σ) BT (σ) d σ (**)

where the left side is a product of invertible matrices by minimality. Therefore the two matrices on the right side

are invertible. Let

P−1 = M x−1(t o , t f )

t o

∫ t f

C T (t ) H (t ) dt

Then multiply both sides of (*) by C T (t ) and integrate with respect to t to obtain

M x(t o, t f ) B (σ) =

t o

∫ t f

C T (t ) H (t ) dt F (σ)

for all σ. That is,

B (σ) = P−1F (σ)

for all σ. Similarly, (*) gives

C (t )W x(t o , t f ) = H (t )t o

∫ t f

F (σ) BT (σ) d σ

that is,

C (t ) = H (t )t o

∫ t f

F (σ) BT (σ) d σ W x−1(t o , t f )

But (**) then gives

t o

∫ t f

F (σ) BT (σ) d σ W x−1(t o, t f ) =

t o

∫ t f

C T (t ) H (t ) dt

−1

M x(t o , t f ) = P

so we have

C (t ) = H (t )P

for all t . Noting that 0 = P−1 . 0 . P, we have that P is a change of variables relating the two zero- A minimal

realizations. Since a change of variables always can be used to obtain a zero- A realization, this shows that any

two minimal realizations of a given weighting pattern are related by a variable change.

Solution 10.11 Evaluating

X (t+σ) = X (t ) X (σ)

at σ = −t gives that X (t ) is invertible, and X −1(t ) = X (−t ) for all t . Differentiating with respect to t , and withrespect to σ, and using

∂t ∂___

X (t+σ) =∂σ∂___

X (t+σ)

gives

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dt

d ___ X (t )

X (σ) = X (t )

d σd ___

X (σ)

which implies

d σd ___

X (σ) = X (−t )

dt

d ___ X (t )

X (σ)

Integrate both sides with respect to t from a fixed t o to a fixed t f > t o to obtain

(t f − t o)d σd ___

X (σ) =t o

∫ t f

X (−t )

dt

d ___ X (t )

dt X (σ)

Now let

A =t f

−t o

1_____

t o

∫ t f

X (−t )

dt

d ___ X (t )

dt

to write

d σd ___

X (σ) = A X (σ) , X (0) = I

This implies X (σ) = e Aσ.

linear system theory 2 e sol

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