1.4 the fundamental theorem for linear...

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1.4 The Fundamental Theorem for Linear Systems

• In the last section, we have seen how to define the matrix expo-

nential eAt through the Taylor series

eAt =∞∑

k=0

1

k!Aktk.

• The purpose of this section is to verify the identity

d

dteAt = AeAt

and solve the linear system x = Ax in terms of the matrix ex-

ponential eAt:

x(t) = eAtx(0).

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The Fundamental Theorem

• We first compute the derivative of eAt in the following lemma.

• Lemma 13 Let A be a square matrix, then

d

dteAt = AeAt.

• Now we are ready to formulate the main theorem.

• Theorem 14 Let A be an n×n matrix. Then for a given x0 ∈

Rn, the initial value problem

x = Ax, x(0) = x0,

has a unique solution given by

x(t) = eAtx0 .

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Example

• Solve the initial value problem

x =

[

−2 −1

1 −2

]

x, x(0) =

[

1

0

]

,

and sketch the solution curves in the phase plane R2.

• By applying the fundamental theorem and the results from the

last section, the solution is easily found to be

x(t) = e−2t

[

cos t

sin t

]

.

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Example (Cont’d)

• The phase portrait of the above system is shown below.

Figure 5: The phase portrait of the example.

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1.5 Linear Systems in R2

• Now we have learned how to solve the linear system x = Ax in

terms of the matrix exponential eAt:

x(t) = eAtx(0),

and the only task that remains is to compute eAt for arbitrary

square matrices A.

• Before moving on, we shall take a short discursion and discuss

in this section the various phase portraits that are possible for

linear systems in R2, i.e. systems with 2 × 2 matrices A.

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Reduction to Jordan Form

• Based on the results summarized at the end of Section 1.3, for

any 2 × 2 matrix A, there is an invertible 2 × 2 matrix P such

that the matrix B = P−1AP has one of the following forms:

B =

[

λ 0

0 µ

]

, B =

[

λ 1

0 λ

]

, or B =

[

a −b

b a

]

.

• It suffices to describe the phase portraits for the linear system

y = By since the phase portrait for x = Ax can be obtained

from that for x = Bx under the linear transformation x = Py.

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The Solution

• It follows from the fundamental theorem in Section 1.4 and the

form of the matrix eBt computed in Section 1.3 that the solution

of y = By with y(0) = y0 is given by

y(t) =

[

eλt 0

0 eµt

]

y0, y(t) = eλt

[

1 t

0 1

]

y0,

or y(t) = eat

[

cos b − sin b

sin b cos b

]

y0.

• We now list the various phase portraits that result from these

solutions.

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Case I: Saddle

• Consider the case B =

[

λ 0

0 µ

]

with λ < 0 < µ. The system

y = By has a saddle at the origin in this case.

Figure 6: A saddle at the origin.

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Case II: Stable/Unstable Node


[

λ 0

0 µ

]

with λ ≤ µ < 0 or B =

[

λ 1

0 λ

]

with λ < 0. The system y = By has a stable node

at the origin in these cases. If λ ≥ µ > 0 or λ > 0, the node is

called an unstable node.

Figure 7: A stable node at the origin.

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Case III: Stable/Unstable Focus


[

a −b

b a

]

with a < 0. The system

y = By has a stable focus at the origin in this case. If a > 0,

the focus is called an unstable focus.

Figure 8: A stable focus at the origin.

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Case IV: Center


[

0 −b

b 0

]

. The system y = By has a

center at the origin in this case.

Figure 9: A center at the origin.

• Remark. If one (or both) of the eigenvalues of B is zero, i.e., if

detB = 0, the origin is called a degenerate equilibrium point of

the system y = By. �

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Example

• As an example, consider the linear system

x = Ax =

[

0 −4

1 0

]

x, x(0) = c.

• It can be reduced to the Jordan form

y = By =

[

0 −2

2 0

]

y,

and its solution can be found to be

x(t) =

[

cos 2t −2 sin 2t

1

2sin 2t cos 2t

]

c.

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Example (Cont’d)

• The phase portrait of the above system consists of concentric

ellipses as shown in the following figure.


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Classification of Linear Systems in R2

• Now we may classify the linear systems x = Ax in R2.

• Definition 15 The linear system x = Ax is said to have a sad-

dle, a node, a focus or a center at the origin if the matrix A

is similar to one of the matrices B in Cases I, II, III or IV

respectively.

• Remark. We say the matrix A is similar to the matrix B if there

is a nonsingular matrix P such that P−1AP = B. Furthermore,

the direction of rotation of trajectories in the phase portraits for

the systems x = Ax and y = By will be the same if detP > 0

(i.e., if P is orientation preserving) and it will be opposite if

detP < 0 (i.e., if P is orientation reversing). �

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A Simple Criterion

• For detA 6= 0, the following simple criterion can be used to

determine the type of the linear system x = Ax without reducing

A to its Jordan form.

• Theorem 16 Let δ = detA and τ = traceA and consider the

linear system x = Ax.

(a) If δ < 0 then the system has a saddle at the origin.

(b) If δ > 0 and τ2 − 4δ ≥ 0 then the system has a node at the

origin; it is stable if τ < 0 and unstable if τ > 0.

(c) If δ > 0, τ2 − 4δ < 0, and τ 6= 0 then the system has a focus

at the origin; it is stable if τ < 0 and unstable if τ > 0.

(d) If δ > 0 and τ = 0 then the system has a center at the origin.

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The Bifurcation Diagram

• The above results can be summarized in a bifurcation diagram

as shown in the following figure. Note that a stable node or focus

of the linear system x = Ax is called a sink and an unstable node

or focus is called a source.

Figure 11: A bifurcation diagram for the linear system x = Ax.

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1.6 Complex Eigenvalues

• Having studied the classification of the linear system x = Ax in

R2, we return in this section to the computation of the matrix

exponential eAt for arbitrary square matrices A.

• We have learned how to compute eAt for matrices A that (1) have

real eigenvalues and (2) have a complete set of eigenvectors, in

which case A is diagonalizable (in R).

• We still need to handle the case where A (1) has complex eigen-

values and/or (2) has an incomplete set of eigenvectors, in which

case A is not diagonalizable (in R).

• In this section, we shall first handle the case where A has complex

eigenvalues. This has been done for 2 × 2 systems but we need

to generalize the results to arbitrary dimensions.

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Matrices with Distinct Complex Eigenvalues

• Let’s start with the special case where A has distinct, complex

(only) eigenvalues. The following theorem provides the basis for

the solution of the linear system x = Ax (recall that complex

eigenvalues occur in complex conjugate pairs).

• Theorem 17 If the 2n× 2n real matrix A has 2n distinct com-

plex eigenvalues λj = aj+ibj and λj = aj−ibj and corresponding

complex eigenvectors wj = uj + ivj, and wj = uj + ivj , j =

1, . . . , n, then {u1,v1, . . . ,un,vn} is a basis for R2n, the matrix

P = [v1 u1 v2 u2 · · · vn un]

is invertible and

P−1AP = diag

[

aj −bj

bj aj

]

,

a real 2n × 2n matrix with 2 × 2 blocks along the diagonal.

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Matrices with Distinct Complex Eigenvalues (Cont’d)

• Using the above theorem, we may solve the linear system x = Ax

by introducing the change of variable y = P−1x, as we have done

before.

• Corollary 18 Under the hypotheses of the above theorem, the

solution of the initial value problem

x = Ax, x(0) = x0

is given by

x(t) = P diag eajt

[

cos bjt − sin bjt

sin bjt cos bjt

]

P−1x0.

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Example


x =

1 −1 0 0

1 1 0 0

0 0 3 −2

0 0 1 1

x, x(0) = x0.

• By applying the above theorem, the solution is found to be

x(t) =

et cos t −et sin t 0 0

et sin t et cos t 0 0

0 0 e2t(cos t + sin t) −2e2t sin t

0 0 e2t sin t e2t(cos t − sin t)

x0.

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Matrices with Distinct Mixed Eigenvalues

• In case A has both real and complex eigenvalues and they are

distinct, we have the following result.

• Theorem 19 If A has distinct real eigenvalues λj and corre-

sponding eigenvectors vj , j = 1, . . . , k and distinct complex

eigenvalues λj = aj + ibj and λj = aj − ibj and corresponding

eigenvectors wj = uj + ivj and wj = uj − ivj , j = k +1, . . . , n,

then the matrix

P = [v1 · · · vk vk+1 uk+1 · · · vn un]

is invertible and

P−1AP = diag[λ1, . . . , λk, Bk+1, . . . , Bn], Bj =

[

aj −bj

bj aj

]

.

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Example


x =

−3 0 0

0 3 −2

0 1 1

x, x(0) = x0.

• By applying the above theorem, the solution is found to be

x(t) =

e−3t 0 0

0 e2t(cos t + sin t) −2e2t sin t

0 e2t sin t e2t(cos t − sin t)

x0.

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Example (Cont’d)

• The phase portrait of the above system is shown below.


1.4 the fundamental theorem for linear...

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