Tải bản đầy đủ - 0 (trang)
Chapter 55. Differential Equations and Stability

Chapter 55. Differential Equations and Stability

Tải bản đầy đủ - 0trang

55-2



Handbook of Linear Algebra



A homogeneous system has f(t) ≡ 0; otherwise the system is inhomogeneous.

A system of linear differential-algebraic equations of order k for an unknown function x : R → Cn

has the form Ak x(k) + · · · + A0 x = f, where A0 , . . . , Ak ∈ Cm×n , Ak = 0 and f : R → Cm .

A continuously differentiable function x : R → Cn is a solution of E x˙ = Ax + f with a sufficiently

often differentiable function f if it satisfies the equation pointwise.

A solution of E x˙ = Ax + f that also satisfies an initial condition x(t0 ) = x0 with t0 ∈ R and x0 ∈ Cn is

a solution of the initial value problem.

If the initial value problem E x˙ = Ax + f, x(t0 ) = x0 has a solution, then the initial condition is called

consistent.

Facts:

1. [Cam80, p. 33] Let E , A ∈ Cn×n . If E is nonsingular, then the system E x˙ = Ax + f is equivalent

to the system of ordinary differential equations x˙ = E −1 Ax + E −1 f.

2. [Arn92, p. 105] The k-th order system of differential-algebraic equations

Ak y(k) + · · · + A0 y = g

can be rewritten as a first-order system E x˙ = Ax + f, where f = [ 0, . . . , 0, gT ]T and











E =⎢











I

..



















⎥, A = ⎢











.

I



0



I



..

.



..



.



..



0



···



0



−A0



Ak



0

.

I



· · · −Ak−2



−Ak−1











y









⎢ . ⎥



⎢ . ⎥



⎢ . ⎥

⎥, x =⎢

⎥.



⎢ (k−2) ⎥



⎣y



y(k−1)



Applications:

1. Consider a mass–spring–damper model as shown in Figure 55.1. This model is described by the

equation

măx + d x˙ + kx = 0,

where m is a mass, k is a spring constant, and d is a damping parameter. Since m = 0, we obtain

the following first-order system of ordinary differential equations







=



0



1



x



−k/m



−d/m



v



,



where the velocity is denoted by v.



x

m

k



d



FIGURE 55.1 A mass–spring–damper model.



55-3



Differential Equations and Stability



2. Consider a one-dimensional heat equation



∂2

T (t, ξ ) = c 2 T (t, ξ ),

∂t

∂ξ



(t, ξ ) ∈ (0, te ) × (0, l ),



together with an initial condition T (0, ξ ) = g (ξ ) and Cauchy boundary conditions



T (t, 0) = u(t),

∂n



β1 T (t, l ) + β2 T (t, l ) = v(t).

∂n



α1 T (t, 0) + α2



Here T (t, ξ ) is the temperature field in a thin beam of length l , c > 0 is the heat conductivity of the



denotes the derivative in the direction of

material, g (ξ ), u(t), and v(t) are given functions, and ∂n

the outward normal. A spatial discretization by a finite difference method with n + 1 equidistant

grid points leads to the initial value problem x˙ = Aa,b x + f, x(0) = x0 , where

x0 = [ g (h), g (2h), . . . , g (nh) ]T ,



x(t) = [ T (t, h), T (t, 2h), . . . , T (t, nh) ]T ,



f(t) = [c u(t)/(h 2 α1 − hα2 ), 0, . . . , 0, c v(t)/(h 2 β1 + hβ2 ) ]T ,

and







Aa,b



−a







1



⎢ 1



c ⎢



= 2⎢

h ⎢







−2



1



..



..



.



.



1



..



.



−2

1











⎥ ∈ Rn×n





1⎦

−b



with h = l /(n + 1), a = (2hα1 − α2 )/(hα1 − α2 ), and b = (2hβ1 + β2 )/(hβ1 + β2 ).

3. A simple pendulum as shown in Figure 55.2 describes the movement of a mass point with mass m

and Cartesian coordinates (x, y) under the influence of gravity in a distance l around the origin.



y



x

l



φ



m

mg

FIGURE 55.2 A simple pendulum.



55-4



Handbook of Linear Algebra



The equations of motion have the form

măx + 2x = 0,

măy + 2yλ + mg = 0,

x 2 + y 2 − l 2 = 0,

where λ is a Lagrange multiplier. Transformation of this system into the first-order form by introducing new variables v = x˙ and w = y˙ and linearization at the equilibrium xe = 0, ye = −l ,

v e = 0, w e = 0, and λe = mg /(2l ) yields the homogeneous first-order linear differential-algebraic

system



⎤⎡ ˙ ⎤ ⎡

x

0



⎥ ⎢ y˙ ⎥ ⎢

⎢0 1 0 0 0⎥ ⎢ ⎥ ⎢ 0



⎥⎢ ⎥ ⎢

⎢0 0 m 0 0⎥ ⎢ v˙ ⎥ = ⎢−2λe



⎥⎢ ⎥ ⎢



⎥⎢ ⎥ ⎢

⎣0 0 0 m 0⎦ ⎣w˙ ⎦ ⎣ 0

˙

0 0 0 0 0

0

λ





1



0



0



0



0



0



1



0



0



0



1



0



0



0



−2λe



0



0



−2l



0



0



⎤⎡ ⎤

x

⎥⎢ ⎥

0⎥⎢ y ⎥

⎥⎢ ⎥

⎢ ⎥

0⎥

⎥ ⎢ v ⎥,

⎥⎢ ⎥

2l ⎦ ⎣w ⎦

0

λ

0



where x = x − xe , y = y − ye , v = v − v e , w = w − w e , and λ = λ − λe .

The motion of the pendulum can also be described by the ordinary differential equation

ă = 2 sin(),



where is an angle between the vertical axis and the pendulum and ω = g /l is an angular

frequency of the motion. By introducing a new variable ψ = ϕ˙ and linearization at the equilibrium

ϕe = 0 and ψe = 0, we obtain the first-order homogeneous system

ϕ˙

˙

ψ



0



=



−ω



2



1



ϕ



0



ψ



.



4. Consider a simple RLC electrical circuit as shown in Figure 55.3. Using Kirchoff ’s and Ohm’s laws,

the circuit can be described by the system E x˙ = Ax + f with







L



⎢0



⎣0

0



E =⎢



0



0



0



1



0

0



0











0



1



0



0











i











0







0



0



⎢1/C

0⎥





⎥, A = ⎢

⎣ −R

0⎦



0



⎢v ⎥

⎢ 0 ⎥

0⎥



⎢ L⎥

⎢ ⎥

⎥, x = ⎢ ⎥, f = ⎢ ⎥.

⎣v C ⎦

⎣ 0 ⎦

0 1⎦



0



0



1



1



0



0



1



vR



−v



Here R, L , and C are the resistance, inductance, and capacitance, respectively; v R , v L , and v C are the

corresponding voltage drops, i is the current, and v is the voltage source. From the last two equations



R



v



L



i



FIGURE 55.3 A simple RLC circuit.



C



55-5



Differential Equations and Stability



in the system, we find v R = Ri and v L = v − v C − Ri . Substituting v L in the first equation and

introducing a new variable w C = i /C , we obtain the system of ordinary differential equations

v˙ C

w˙ C



=



0



1



vC



−1/(L C )



−R/L



wC



0



+



.



v/(L C )



This shows the relationship to the mass–spring–damper model as in Application 1.



55.2



Linear Ordinary Differential Equations



Facts:

The following facts can be found in [Gan59a, pp. 116–124, 153–154].

1. Let J A = T −1 AT be in Jordan canonical form. Then e At = T e J A t T −1 . (See Chapter 6 and

Chapter 11 for more information on the Jordan canonical form and the matrix exponential.)

2. Every solution of the homogeneous system x˙ = Ax has the form x(t) = e At v with v ∈ Cn .

3. The initial value problem x˙ = Ax, x(t0 ) = x0 has the unique solution x(t) = e A(t−t0 ) x0 .

4. The initial value problem x˙ = Ax + f, x(t0 ) = x0 has a unique solution for every initial vector x0

and every continuous inhomogeneity f. This solution is given by

t



x(t) = e A(t−t0 ) x0 +



e A(t−τ ) f(τ ) dτ.

t0



Examples:

1. Let











3



3



1



A=⎣ 0



0



0⎦,



−1



−1







For







⎡ ⎤

1

⎢ ⎥

x0 = ⎣2⎦,







1



−1



T =⎣ 0



0



1⎦



−1



1



0











we have









e At



1



0



T −1 = ⎣1



1



1⎦ ,



0



1



0



1



1



1



0



J A = T −1 AT = ⎣0



2



0⎦.



0



0



0



1



0



−1



⎤⎡



e 2t



te 2t











2









⎦.







1







and



2t







t2 + t − 1







0



Then







−3t 2 − 3t



f(t) = ⎣



3



1















0



⎤⎡



0









⎥⎢

⎥⎢



e 2t

0⎦ ⎣1 1 1⎦

1⎦ ⎣ 0

⎣ 0 0

−1 1

0

0 1 0

0

0

1





(1 + t)e 2t (1 + t)e 2t − 1

te 2t





= ⎣

0

1

0

⎦.

2t

2t

2t

−te

−te

(1 − t)e

=



55-6



Handbook of Linear Algebra



Every solution of the homogeneous system x˙ = Ax has the form









((1 + t)v 1 + (1 + t)v 2 + tv 3 )e 2t − v 2



x(t) = e At v = ⎣



v2











(−tv 1 − tv 2 + (1 − t)v 3 )e 2t

with v = [v 1 , v 2 , v 3 ]T . The solution of the initial value problem x˙ = Ax, x(0) = x0 has the form









(3 + 6t)e 2t − 2



x(t) = e At x0 = ⎣



2







⎦.



(3 − 6t)e 2t

The initial value problem x˙ = Ax + f, x(0) = x0 has the solution









(3 + 6t)e 2t + t − 2



x(t) = ⎣



t2 + 2









⎦.



(3 − 6t)e 2t + 1



Applications:

1. Consider the matrix A from the mass–spring–damper example

A=



0



1



−k/m



−d/m



.



The Jordan canonical form of A is given by J A = T −1 AT = diag(λ1 , λ2 ), where

T=



1



1



λ1



λ2



and

λ1 =



−d −



,



T −1 =





d 2 − 4km

,

2m



λ2

1

λ2 − λ1 −λ1



λ2 =



−d +



−1

1





d 2 − 4km

2m



are the eigenvalues of A. We have

e At = T diag(e λ1 t , e λ2 t )T −1 =



λ2 e λ1 t − λ1 e λ2 t

1

λ2 − λ1 λ1 λ2 (e λ1 t − e λ2 t )



e λ2 t − e λ1 t

λ2 e λ2 t − λ1 e λ1 t



.



The solution of the mass–spring–damper model with the initial conditions x(0) = x0 and v(0) = 0

is given by

x(t) =



x0

λ2 e λ1 t − λ1 e λ2 t ,

λ2 − λ1



v(t) =



λ1 λ2 x0 λ1 t

e − e λ2 t .

λ2 − λ1



2. Since the matrix Aa,b in the semidiscretized heat equation is symmetric, there exists an orthogonal

matrix U such that U T Aa,b U = diag(λ1 , . . . , λn ), where λ1 , . . . , λn ∈ R are the eigenvalues of Aa,b .

(See Chapter 7.2 and Chapter 45.) In this case e Aa,b t = U diag(e λ1 t , . . . , e λn t )U T .



55-7



Differential Equations and Stability



55.3



Linear Differential-Algebraic Equations



Definitions:

A Drazin inverse A D of a matrix A ∈ Cn×n is defined as the unique solution of the system of matrix

equations

A D AA D = A D ,



AA D = A D A,



Ak+1 A D = Ak ,



where k is a smallest nonnegative integer such that rank(Ak+1 ) = rank(Ak ).

Let E , A ∈ Cm×n . A pencil of the form

λE − A = diag Ln1 , . . . , Ln p , Mm1 , . . . , Mmq , Jk , Ns

is called pencil in Kronecker canonical form if the block entries have the following properties: every entry

Ln j = λL n j − Rn j is a bidiagonal block of size n j × (n j + 1), n j ∈ N, where









1



Lnj = ⎢





0

..



.



..



.



1













⎥,





Rn j = ⎢









0







1

..



.



0



..





⎥;





.



0



1



T

− RmT j is a bidiagonal block of size (m j +1)×m j , m j ∈ N; the entry Jk = λIk − Ak

every entry Mm j = λL m

j

is a block of size k × k, k ∈ N, where Ak is in Jordan canonical form; the entry Ns = λNs − Is is a block

of size s × s , s ∈ N, where Ns = diag(Ns 1 , . . . , Ns r ); and











Ns j = ⎢







0







1

..



.



..



.



..



.













1⎦

0



is a nilpotent Jordan block with index of nilpotency s j .

The numbers n1 , . . . , n p are called the right Kronecker indices of the pencil λE − A.

The numbers m1 , . . . , mq are called the left Kronecker indices of the pencil λE − A.

The number ν = max1≤ j ≤r s j is called the index of the pencil λE − A.

A matrix pencil λE − A with E , A ∈ Cm×n is called regular, if m = n and det(λE − A) = 0 for some

λ ∈ C. Otherwise, the pencil is called singular.

Let E , A ∈ Cm,n . Subspaces Wl ⊂ Cm and Wr ⊂ Cn are called left and right reducing subspaces of the

pencil λE − A if Wl = E Wr + AWr and dim(Wl ) = dim(Wr ) − p, where p is the number of Ln j blocks

in the Kronecker canonical form.

Let λE − A be a regular pencil. Subspaces Wl , Wr ⊂ Cn are called left and right deflating subspaces

of λE − A if Wl = E Wr + AWr and dim(Wl ) = dim(Wr ).

Let W1 , W2 ⊂ Cn be subspaces such that W1 ∩ W2 = {0} and W1 + W2 = Cn . A matrix P ∈ Cn,n is

called a projection onto W1 along W2 if P 2 = P , range(P ) = W1 , and ker(P ) = W2 .

Let λE − A be a regular pencil. If Tl , Tr ∈ Cn×n are nonsingular matrices such that Tl −1 (λE − A)Tr is

in Kronecker canonical form, then

Pl = Tl



Ik



0



0



0



Tl −1 ,



Pr = Tr



Ik



0



0



0



Tr−1



55-8



Handbook of Linear Algebra



are the spectral projections onto the left and right deflating subspaces of λE − A corresponding to the

finite eigenvalues along the left and right deflating subspaces corresponding to the eigenvalue at infinity.

Facts:

1. [Cam80, p. 8] If A ∈ Cn×n is nonsingular, then A D = A−1 .

2. [Cam80, p. 8] Let

J A = T −1 AT =



A1



0



0



A0



be in Jordan canonical form, where A1 contains all the Jordan blocks associated with the nonzero

eigenvalues, and A0 contains all the Jordan blocks associated with the eigenvalue 0. Then

AD = T



A−1

1



0



0



0



T −1 .



3. [Gan59b, pp. 29–37] For every matrix pencil λE − A with E , A ∈ Cm×n there exist nonsingular

matrices Tl ∈ Cm×m and Tr ∈ Cn×n such that Tl −1 (λE − A)Tr is in Kronecker canonical form.

The Kronecker canonical form is unique up to permutation of the diagonal blocks, i.e., the kind,

size, and number of the blocks are characteristic for the pencil λE − A. (For more information on

matrix pencils, see Section 43.1.)

4. [Gan59b, p. 47] If f(t) = [ f 1 (t), . . . , f n j (t)]T is an n j -times continuously differentiable vectorvalued function and g (t) is an arbitrary (n j + 1)-times continuously differentiable function, then

the system L n j x˙ = Rn j x + f has a continuously differentiable solution of the form







x1 (t)















g (t)







g (1) (t) − f 1 (t)

⎢ x (t) ⎥ ⎢



⎢ 2

⎥ ⎢



.







⎥.

x(t) = ⎢

..

..

⎥=⎢





⎣ . ⎦ ⎢

nj





(n j −i )

(n j )

fi

(t)

g (t) −

xn j +1 (t)

i =1



A consistent initial condition has to satisfy this defining equation at t0 .

T ˙

5. [Gan59b, p. 47] A system of differential-algebraic equations L m

x = RmT j x + f with a vectorj

T

valued function f(t) = [ f 1 (t), . . . , f m j +1 (t)] has a unique solution if and only if f is m j -times

m j +1



continuously differentiable and

i =1



f i(i −1) (t) ≡ 0. If this holds, then the solution is given by















mj



x1 (t)



⎢ . ⎥ ⎢

i =1







x(t) = ⎣ .. ⎦ = ⎢



xm j (t)



f i(i+1−1) (t)

..

.









⎥.





− f m j +1 (t)



A consistent initial condition has to satisfy this defining equation at t0 .

6. [Gan59b, p. 48] A system Ns j x˙ = x + f has a unique continuously differentiable solution x if f is

s j -times continuously differentiable. This solution is given by

s j −1



x(t) = −



Nsi j f(i ) (t).

i =0



A consistent initial condition has to satisfy this defining equation at t0 .



55-9



Differential Equations and Stability



7. [Cam80, pp. 37–39] If the pencil λE − A is regular of index ν, then for every ν-times differentiable

inhomogeneity f there exists a solution of the differential-algebraic system E x˙ = Ax + f. Every

solution of this system has the form

ˆ D A(t−t

ˆ

0)



x(t) = e E



t



Eˆ DEˆ v +



ˆ D A(t−τ

ˆ

)



eE



ν−1



ˆ ) dτ − (I − Eˆ DEˆ )

Eˆ D f(τ



t0



ˆ D)j A

ˆ D fˆ( j ) (t),

( Eˆ A

j =0



ˆ = (λ0 E − A)−1A, and fˆ = (λ0 E − A)−1 f for some λ0 ∈ C

where v ∈ Cn , Eˆ = (λ0 E − A)−1 E , A

such that λ0 E − A is nonsingular.

8. [Cam80, pp. 37–39] If the pencil λE − A is regular of index ν and if f is ν-times differentiable, then

the initial value problem E x˙ = Ax + f, x(t0 ) = x0 possesses a solution if and only if there exists

v ∈ Cn that satisfies

ν−1



ˆ D)j A

ˆ D fˆ( j ) (t0 ).

( Eˆ A



x0 = Eˆ D Eˆ v − (I − Eˆ DEˆ )

j =0



If such a v exists, then the solution is unique.

9. [KM06, p. 21] The existence of a unique solution of E x˙ = Ax + f, x(t0 ) = x0 does not imply that

the pencil λE − A is regular.

10. [Cam80, pp. 41–44] If the pencil λE − A is singular, then the initial value problem E x˙ = Ax + f,

x(t0 ) = x0 may have no solutions or the solution, if it exists, may not be unique.

11. [Sty02, pp. 23–26] Let the pencil λE − A be regular of index ν. If

Tl −1 (λE − A)Tr =



λI − Ak



0



0



λNs − I



is in Kronecker canonical form, then the solution of the initial value problem E x˙ = Ax + f,

x(t0 ) = x0 can be represented as

ν−1



t



x(t) = F(t − t0 )E x0 +



F(t − τ )f(τ ) dτ +

t0



F − j −1 f( j ) (t),

j =0



where

F(t) = Tr



e Ak t



0



0



0



Tl −1 ,



F − j = Tr



0



0



0



−Nsj −1



Tl −1 .



Examples:

1. The system

1



0



0



0



x˙ =



1



0



0



0



x+



0

g (t)



x(0) =



,



1

0



has no solution if g (t) ≡ 0. For g (t) ≡ 0, this system has the solution x(t) = [ e t , φ(t) ]T , where

φ(t) is a differentiable function such that φ(0) = 0.

2. The system







1



0











0



0











− sin(t)



















⎣0 1⎦ x˙ = ⎣1 0⎦ x + ⎣− cos(t) ⎦,

0 0

0 1

0



x(0) =



1

0



has a unique solution x(t) = [ cos(t), 0 ]T , but the pencil λE − A is singular.



55-10



Handbook of Linear Algebra



Applications:

1. The pencil in the linearized pendulum example has the Kronecker canonical form



⎤ ⎡ √

1 0

0 0 0

−i g /l

0

0 0





⎥ ⎢

0

i g /l

0 0 0⎥ ⎢

0 0

⎢0 1





diag(J2 , N3 ) = λ ⎢

⎢0 0



⎣0 0

0



⎥ ⎢



0 1 0⎥

⎥−⎢

⎥ ⎢

0 0 1⎦ ⎣

0 0 0



0



0



0



0



0



0



0



0











1 0 0⎥

⎥.



0 1 0⎦

0 0 1

0⎥



This pencil is regular of index 3. Since the linearized pendulum system is homogeneous, it has

a unique solution for every consistent initial condition.

2. The pencil of the circuit equation has the Kronecker canonical form







1



0



⎢0 1



⎣0 0

0 0



diag(J2 , N2 ) = λ ⎢



0



0











⎢0



⎥−⎢

0 0⎦ ⎣ 0

0

0 0

0



0⎥





λ1



0







0



0



λ2



0



0



1



0⎥



⎥,

0⎦



0



0



1



with

λ1 = −



R



2L



R2

1



,

4L 2

LC



λ2 = −



R

+

2L



R2

1



.

4L 2

LC



This pencil is regular of index 1. Hence, there exists a unique continuous solution for every continuous voltage source v(t) and for every consistent initial condition.



55.4



Stability of Linear Ordinary Differential Equations



The notion of stability is used to study the behavior of dynamical systems under initial perturbations

around equilibrium points. In this section, we consider the stability of linear homogeneous ordinary

differential equations with constant coefficients only. For extensions of this concept to general nonlinear

systems, see, e.g., [Ces63] and [Hah67].

Definitions:

The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is called stable in the sense of Lyapunov, or simply

stable, if for every ε > 0 there exists a δ = δ(ε) > 0 such that any solution x of x˙ = Ax, x(t0 ) = x0 with

x0 2 < δ satisfies x(t) 2 < ε for all t ≥ t0 .

The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is called asymptotically stable if it is stable and

lim x(t) = 0 for any solution x of x˙ = Ax.

t→∞



The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is called unstable if it is not stable.

The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is called exponentially stable if there exist α > 0 and

β > 0 such that the solution x of x˙ = Ax, x(t0 ) = x0 satisfies x(t) 2 ≤ α e −β(t−t0 ) x0 2 for all t ≥ t0 .

Facts:

1. [Gan59a, pp. 125–129] The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is stable if and only if all

the eigenvalues of A have nonpositive real part and those with zero real part have the same algebraic

and geometric multiplicities. If at least one of these conditions is violated, then the equilibrium

xe (t) ≡ 0 of x˙ = Ax is unstable.

2. [Gan59a, pp. 125–129] The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is asymptotically stable if

and only if all the eigenvalues of A have negative real part.



55-11



Differential Equations and Stability



3. [Ces63, p. 22] Let p A (λ) = det(λI − A) = λn + a1 λn−1 + · · · + an be the characteristic polynomial

of A ∈ Rn,n . If the equilibrium xe (t) ≡ 0 of the system x˙ = Ax is asymptotically stable, then a j > 0

for j = 1, . . . , n.

4. [Gan59b, pp. 185–189] The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is asymptotically stable

if and only if the Lyapunov equation A∗ X + X A = −Q has a unique Hermitian, positive definite

solution X for every Hermitian, positive definite matrix Q.

5. [God97] Let H be a Hermitian, positive definite solution of the Lyapunov equation

A∗ H + H A = −I

and let x be a solution of the initial value problem x˙ = Ax, x(0) = x0 . Then in terms of the original

data,

x(t)



2



κ(A) e −t







A 2 /κ(A)



x0 2 ,



where κ(A) = 2 A 2 H 2 .

6. [Hah67, pp. 113–117] The equilibrium xe (t) ≡ 0 of the system x˙ = Ax is exponentially stable if

and only if it is asymptotically stable.

7. [Hah67, p. 16] If the equilibrium xe (t) ≡ 0 of the homogeneous system x˙ = Ax is asymptotically stable, then all the solutions of the inhomogeneous system x˙ = Ax + f with a bounded

inhomogeneity f are bounded.

Examples:

1. Consider the linear system x˙ = Ax with

A=



−1



0



0



2



.



For the initial condition x(0) = [ 1, 0 ]T , this system has the solution x(t) = [ e −t , 0 ]T that is

bounded for all t ≥ 0. However, this does not mean that the equilibrium xe (t) ≡ 0 is stable. For

linear systems with constant coefficients, stability means that the solution x(t) remains bounded for

all time and for all initial conditions, but not just for some specific initial condition. If we can find

at least one initial condition that causes one of the states to approach infinity with time, then the

equilibrium is unstable. For the above system, we can choose, for example, x(0) = [ 1, 1 ]T . In this

case x(t) = [ e −t , e 2t ]T is unbounded, which proves that the equilibrium xe (t) ≡ 0 is unstable.

2. Consider the linear system x˙ = Ax with

A=



−0.1



−1



1



−0.1



.



The eigenvalues of A are −0.1 ± i , and, hence, the equilibrium xe (t) ≡ 0 is asymptotically stable.

Indeed, the solution of this system is given by x(t) = e At x(0), which can be written in the real form as

x1 (t) = e −0.1t (x1 (0) cos(t) − x2 (0) sin(t)),

x2 (t) = e −0.1t (x1 (0) sin(t) + x2 (0) cos(t)).

(See also Chapter 56.) Thus, for all initial conditions x1 (0) and x2 (0), the solution tends to zero as

t → ∞. The phase portrait for x1 (0) = 1 and x2 (0) = 0 is presented in Figure 55.4.

3. Consider the linear system x˙ = Ax with

A=



0



−1



1



0



.



55-12



Handbook of Linear Algebra



1

0.8

0.6

0.4



x2



0.2

0

−0.2

−0.4

−0.6

−0.8

−0.8



−0.6



−0.4



−0.2



0



0.2



0.4



0.6



0.8



1



x1

FIGURE 55.4 Asymptotic stability.



The matrix A has the eigenvalues ±i . The solution of this system x(t) = e At x(0) can be written in

the real form as

x1 (t) = x1 (0) cos(t) − x2 (0) sin(t),

x2 (t) = x1 (0) sin(t) + x2 (0) cos(t).

It remains bounded for all initial values x1 (0) and x2 (0), and, hence, the equilibrium xe (t) ≡ 0 is

stable. The phase portrait for x1 (0) = 1 and x2 (0) = 0 is given in Figure 55.5.

4. Consider the linear system x˙ = Ax with

A=



0.1



−1



1



0.1



.



The eigenvalues of A are 0.1 ± i . The solution of this system in the real form is given by

x1 (t) = e 0.1t (x1 (0) cos(t) − x2 (0) sin(t)),

x2 (t) = e 0.1t (x1 (0) sin(t) + x2 (0) cos(t)).

It is unbounded for all nontrivial initial conditions. Thus, the equilibrium xe (t) ≡ 0 is unstable.

The phase portrait of the solution with x1 (0) = 1 and x2 (0) = 0 is shown in Figure 55.6.

5. Consider the linear system x˙ = Ax with

A=



a



b



c



d



∈ R2×2 .



The characteristic polynomial of the matrix A is given by p A (λ) = λ2 − (a + d)λ + (ad − bc ) and

the eigenvalues of A have the form





a +d

a +d

(a + d)2 − 4(ad − bc )

(a + d)2 − 4(ad − bc )

+

,

λ2 =



.

λ1 =

2

2

2

2



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Chapter 55. Differential Equations and Stability

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