Agallery of Solution Curves of Linear Ystems Systems of Dimension N

Geometrically, this means that all solution curves given by (4) with both c1 and c2 nonzero have two asymptotes, namely the lines 1 and 12passing through the origin and parallel to the eigenvectors v1 and ,v2 respectively; the solution curves approach 1 as t —* —ooand 12 as t —s. +00. Indeed, as Fig. 5.3.1 illustrates, the lines 1 and 12effectively divide the plane into four “quadrants” within which all solution curves flow from the asymptote l to the asymptote l2 as t increases. (The eigenvectors shown in Fig. 5.3.1—and in other figures—are scaled so as to have equal length.) The particular quadrant in which a solution curve lies is determined by the signs of the coefficients c1 and .c2 If c1 and c2 are both positive, for example, then the corresponding solution curve extends asymptotically in the direction of the eigenvector v1 as t — —00, and asymptotically in the direction of v2 as t —+ oo. If instead c1 > 0 but c2 < 0, then the corresponding solution curve still extends asymptotically in the direction of v1 as t —÷ —00, but extends asymptotically in the direction opposite v2 as t —* +00 (because the negative coefficient c2 causes the vector c2v to point “backwards” from v2). If c1 or c2 equals zero, then the solution curve remains confined to one of the lines i and 12. For example, if c1 0 but c2 = 0, then the solution (4) becomes = clvleIt, xl xQ) which means that the corresponding solution curve lies along the line l. It approaches the origin as t —÷ +00, because 1A < 0, and recedes farther FIGURE 5.3.1. Solutioncurves and farther from the origin as t —t —00, either in the direction of v1 (if c1 > 0) or x(t) = civte1t eA2t for the +c the direction opposite v1 (if c1 < 0). Similarly, if c1 = 0 and c2 0, then because system x’ = Ax when2v the eigenvalues Ai, 2A of A are real with 2A > 0, the solution curve flows along the line 12away from the origin as t —* +00 A <0 1 2surface (like z = xy), we call the origin a saddle point for the system x’ = Ax.

Example 1 The solutioncurvesin Fig. 5.3.1correspondto the choice

A=[ 2] (8)

in the system x’ = Ax; as you can verify, the eigenvalues of A are 1A —2and 2A 5 (thus 1A < 0 < 2),A with associated eigenvectors r...il ru vi=[ and 6] v21].=[ According to Eq. (4), the resulting general solution is

x(t) ci e2 + c2 e5 (9) [] [], or, in scalar form, xi(t) = 2t—cie + c2eSt, (10) x2(t) 216cie + c2eSt,

OurgalleryFig. 5.3.16at the endof this sectionshows a more complete set of solution curves, togetherwith a directionfield,for the systemx’ = Ax with A given by Eq. (8). (In Problem 29 we explore“Cartesian”equationsfor the solutioncurves(10)relativeto the “axes”defined by the lines 1 and 12, whichform a naturalframe of reference for the solutioncurves.) •

hus

all ii—

blem

es e

fined

rVeS,

A 1

system x(t)

A 1

FIGURE

x’

Sinks

tangent

(4)

trajectory

solution

every vector

the

NEGATIVE

that

parallel

means

pattern

then

from

both

in =

the

say

c 2

from

gives

Ax. systems.

origin;

every is

eigenvectors

Fig.

1 v 1 e

that

the

5.3.2,

situation

the

x(t)

the

through the

that

trajectory

negative.)

the

to and

cl)Livie()I2)t

Fig. the

Fig.

The to

trajectories

5.3

that as

the

tangent

5.3.2), the

curve

the v 2 .

coefficients

straight

the

origin

trajectory

appearance

other

origin

approaches

vector

the

likewise,

then

the

5.3.2

eigenvalues

the

5.3.2, which

which

increases.

+ the

origin, tangent

EIGENVALUES:

following In

Sources

Vi=[

lines

line

the

c 22 v,eA2t

origin

trajectories Gallery

general,

“goes

eigenvector

solution

hand,

x(t)

then

correspond

approaches

and

vector

Fig.

Thus, v 2

lines;

12; all

origin

1—11

will

illustrates

then

are

A=[j

—

for as

trajectories—apart

—

and

2] the

5.3.6,

then

off

the

Indeed,

the

conditions

+00

tangent clvle1t

x’(t)

tangent

indeed,

the

—

and +00;

curve

—+

and

the

phase

origin result

proper origin

c 2 ) 2 v 2 ,

useful

c 2

Solution

origin

the

are

and

call —00,

v 2

+00

system

the

origin infinity.”

divide c 2 .

parallel

Fig.

the

typical

A 1

solution

as which

the

there

portraits

the

When

origin

a 0

V2L 1

(More

both

(because

choice

]

both

solution

5.3.2

[ciivie12)t

node

are

proper

which

called the

—

the origin

c 2 v 2 eA2t

call

—14

solution

x’ the

then

some

Curves

solution

the

e)1t

satisfied:

now

direction

= )i

Moreover,

same

as curve

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provided

source.

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from

with

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node

approaches

a t curve

straight e(A1

an <)2

and

eigenvector

node

A 2 and

—

sink;

both

line

curve

improper

straight

approaches x(t)

those

Fig.

into

curves increasing

eA2t

+00

might

2 )t

<0,

xQ)

Linear

opposite

Thus

—7

improper.

e1t

that

flows

line

Chapter

differentiation

5.3.2,

x(t)

four

grow

passing

approaches

the

that

(and

note

instead

the

becomes

are

the

corresponding

and

be line c 2 2v7]

through

every

system v 1 .

Systems

nodal is

similarly

“quadrants”

flow

thus

factors

without

straight

called

that

two

the

fixed

describe

e2t

through

6, Once

the

through

v 2 ,

scalar

every

A 1

trajectory

along

origin introduce

when

different

sink.

x’ approach

solution

the

because zero).

nonzero

ternatively, traverse follows We

field, gallery

functions

note

shows

increases) 0

< scalar

x(t)

system

The

than

Differential

Fig.

a form,

x’

POSITIVE

curves that

then

—A 1

the

system

x’

sink,

short,

instead

common

gives

x(t)

other

the

each

solution

“run

5.3.16

along

5.3.3

system

have

the

gives

<—A 2

verification

distinct

for

same

then

Time

and

with

x’

i(t)

system point

(t)

backwards”

under

Fig.

has

illustrates

shows

the

each

the

x’

may

thus

(t)

EIGEN

solution

<0 x(r)

that

Equations

—tin

direction

analyzing 5.3.3

x(t)=c 1

the

A=_[ general

system positive

Reversal

same

the

you

“time

nonzero

solution

x(—t)

x’

but

are

say

—x’Q);

x 1 (t)

x 2 (t)

their

same

the

correspond

is 2-dimensional

VALUES:

with

can

the

applying

negatives

that

typical

solution

the

solution

curve

x’

righthand —Ax

reversal,”

[—i]

routine

common

two eigenvalues

[]e14t

complete

trajectories,

verify

same

= A

the

values

curve

since

negative

2c 1 e 141

—cie’ 4

the

independently,

given

Ax,

solution x’=Ax

has

_]=[_

increases

vector-valued

solutions =

e’ 41

solution

solutions

the Linear

eigenvectors

curve

matter

5Q)

opposite

(Problem

-Ai

reversed.

and

side

the

solution

since

set

principle

each

linear

+c 2 []e7t.

the

improper

Eq.

mirrors

and

choice

the of

except

(or C2

3c 2 e_ 7 t,

curves

Systems

of for

(13)

coefficients

matrix

the

solution

other.

call (12).

[

x(—t)

matrix

directions

checking

image) system

the

Thus

31),

curve

of get

one

the

system

leads

]

v 1

the

]

that

functions

with

e 7

time

nodal

other.

given

Therefore

the

systems

represent

and

solution

curves,

can

the

origin

the

has

Example

reversal

matrix

solutions

the

distinct

signs

v 2 .

origin

c 1

velocity

sink

rely

the

positive

direction

and

together

The

x(t)

(1)

and

increases—or,

plane.

at the

the (Problem

improper

—A

c 2 .

negative

and

the

preceding

now

the

vectors same

two

the

decreases and

Therefore

has

eigenvalues with

civ 1 e1t

one

origin.

(14)

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following

However,

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negative Q)

point,

source,

30).

motion

system

eigen

nodal

corre

(14)

direc

case

(1)

(16)

(15)

(13)

But

the

for

in it 5.3 A Gallery of Solution Curves of Linear Systems 301 Of course, we could also have “started from scratch” by finding the eigenvalues ,1A 2A and eigenvectors,v1 v2 of A. These can be found from the definitionof eigenvalue, but it is easier to note (see Problem 31 again) that because A is the negative of the matrix in Eq. (12), 1A and 2A are likewise the negatives of their values in Example 2, whereas we can take Vj and 2V to be the same as in Example 2. By either means we find that 1A = 14and 2A = 7 (so that o < 2A 1),

From Eq. (4), then, the general solution is

xl x(t) = c t4el + cz te7 FIGURE 5.3.3. Solutioncurves [—] [] x(t) = civlelt forthe +ceA2t (in agreement with Eq. (16)), or, in scalar form, system x’ = Ax when2v7 the eigenvalues A 2A of A are real with ,1 14t 71 o2

Zero Eigenvalues and Straight-Line Solutions ONE ZERO AND ONE NEGATIVE EIGEN VALUE: When ) 2 0, for example, then civieAlt extends in the direction of ,v1 approaching the origin as t increases, and receding from the origin as t decreases. If instead Cl < 0 , then civieit extends in the direction opposite v1 while still approaching the origin as t increases. Loosely speaking, we can visualize the flow of the term civiet taken alone as a pair of arrows opposing is each other head-to-head at the origin. The solution curve xQ) in Eq. (17) is simply this same trajectory CivleAit, then, shifted (or offset) by the constant vector c2V Thus in this case the phase portrait of the system x’ = Ax consists of all .lines it FIGURE 5.3.4. Solutioncurves parallel to the eigenvector v1 where along each such line the solution flows (from in x(t) =c1vie11 forthe , +c both directions) toward the line 12 passing through the origin and parallel to v1 systemx’ Ax whenthe eigenvalues . A A of A are real2v with Figure 5.3.4 illustrates typical solution curves corresponding to nonzero values of al ,1 2 1A

guaranteed by Theorem 1 of Section 4.1. Note that this situation is in marked con trast with the other eigenvalue cases we have considered so far, in which x(t) 0 302 Chapter 5 Linear Systems of Differential Equations ii is theonly constant solution of the system x’ = Ax. (In Problem 32 we explore the general circumstances under which the system x’ Ax has constant solutions other than x(t) 0.)

Example 4 The solution curves in Fig. 53.4 correspondto the choice A=[3 ] (18) in the system x’ Ax. The eigenvalues of A are A = —35and 2A = 0, with associated eigenvectors

F 61 F 1V=[ and Based on Eq. (17), the generalj solution is x(t) Cl e351 + C2 (19) [_ ] [ or, in scalar form, x (t) e16c35 + C2, = _cle_ x2(t) t35 — 6c2. Our gallery Fig. 5.3.16 shows a more completet set of solution curves, together with a direc 12 tion field, for the system x’ = Ax with A given by Eq. (18).

-_7-c>O,c,0. c2>0 of the system x’ = Ax is again given by ,

-- 1V x(t) = civieAt 0

= A = [3 (20) — [ fl in the systemx’ = Ax; thus A is the negativeof the matrixin Example4. Once againwe can solve the systemusing the principle of lime reversal:Replacing t with —t in the right-hand side of the solutionin Eq. (19) of Example4 leadsto

= c [] e351 + C2 []. (21) Alternatively, directly finding the eigenvalues and eigenvectors of A leads to 1A 35 and 2A = 0, with associated eigenvectors

F 61 r i j1Vi=[ and 26V=[

S I Equation pendent one tion Our vector pendent (in of portraits, Therefore REPEATED scalar The Repeated or We independent free general to trajectories multiple A the in agreement field, gallery repeated could general to with vector With Either A same is take times eigenvectors, solution eigenvectors. (17) for of we an A the also Fig. two x(t) the of format, result the > v 1 solution POSITIVE way, eigenvector gives with eigenvectors, will Eigenvalues; eigenvalue, system the 0. the 5.3.16 have system = fixed independent = Eq. consider (4): our identity the as [1 system 5.3 civieAt arrived (21)), Eq. shows of general vector x’ solution then x’ Because Because A 0 Eq. EIGENVALUE: = = of then (25): ]T Gallery x(t)__ci[]e35t+c 2 [] (1) matrix them each or, Ax it Ax A_AH a + (23) A, [c 1 and solution at is more in — are c 2 v 2 eAt eigenvectors: A with Proper in becomes x 2 (t) xi(t) easy Eq. from associated these all scalar separately. may is x(t) v 2 Eq. of half-lines, x(t) x’ 1 (t) complete o c 2 x 2 (t) x 1 (t) of A nonzero (25) = to = = order of 1 T given (25) which or Solution two form, = —cie 35t 6c 1 e 35 t show [0 the olD = = ij[o (in eAt and by = = may e Thus shows two, possibilities system Ax 2 Q). Axi(t). As with by c 2 et, c 1 e set scalar We

starting, [n]. 1 (civi it vectors (Problem or First, not Eq. we +c2, Improper of follows — apart rays, that Curves let the 6C2. that solution as x’ have (20). noted form) + if A eigenvalue = is, A a 0 from c 2 v 2 ) are A denote as emanating xQ) representative Ax 33) two that lead does in earlier, of eigenvectors curves, as is the that previous = associated Linear Nodes A to the always have eAt case must A. in quite repeated together if from fact [Ci Then two the Systems c 1 a cases, be different pair positive every

the ] = linearly linearly matrix of Eq. equal with c 2 eigenvalue of A, origin from (4) nonzero = linearly a we A to phase scalar 0, direc inde inde leads 303 (23) (24) (25) has and 22 our the are the

is A 5.3 A Gallery of Solution Curves of Linear Systems 305

the fixed nonzero multiple 2v1Ac of the eigenvector v1 as t — + or as t — —oo. In this case it follows that as t gets larger and larger numerically (in either direction), the tangent line to the solution curve at the point xQ)—since it is parallel to the tan gent vector x’(t) which approaches Ac—becomes more and more nearly parallel to the eigenvector .v1 In short, we might say that as t increases numerically, the point x(t) on the solution curve moves2v1 in a direction that is more and more nearly parallel to the vector ,v1 or still more briefly, that near xQ) the solution curve itself is virtually parallel to .v1 We conclude that if c2 0, then as t —* — the point x(t) approaches the ori gin along the solution curve which is tangent there to the vector .v1 But as t —÷ +00 and the point x(t) recedes further and further from the origin, the tangent line to the trajectory at this point tends to differ (in direction) less and less from the (moving) line through xQ) that is parallel to the (fixed) vector .v1 Speaking loosely but sug gestively, we might therefore say that at points sufficiently far from the origin, all trajectories are essentially parallel to the single vector .v1 If instead c2 = 0, then our solution (7) becomes

x(t) = clviet, (31)

I and thus runs along the line l passing through the origin and parallel to the eigen t vector .v1 Because A > 0, x(t) flows away from the origin as t increases; the flow is in the direction of v1 if c1 > 0, and opposite v1 if c1 <0. S We can further see the influence of the coefficient c2 by writing Eq. (7) in yet a different way:

x(t) = civieAt )eAt t)vieAt eAt. + 2C(vit + v2 = 1(c + c2 + c2v (32) It follows from Eq. (32) that if c2 0, then the solution curve x(t) does not cross the line 11. Indeed, if c2 > 0, then Eq. (32) shows that for all t, the solution curve xQ) lies on the same side of l as ,v2 whereas if c2 <0, then xQ) lies on the opposite side of l. To see the overall picture, then, suppose for example that the coefficient c2 > 0. Starting from a large negative value oft, Eq. (30) shows that as t increases, the direction in which the solution curve x(t) initially proceeds from the origin is roughly that of the vector tetAc the teAtAc vi. Since scalar 2 is negative (because <0 and 2Ac > 0), the direction2 of the trajectory is opposite that of .v1 For large positive values of t, on the other hand, the scalar 2tettAc is positive, and so x(t) flows in nearly the same direction as .v1 Thus, as t increases from —no to +no, FIGURE 5.3.7. Solutioncurves the solution curve leaves the origin flowing in the direction opposite v1 makes a = civieAt )eAt , x(t) + c,(vit + v7 “U-turn” as it moves away from the origin, and ultimately flows in the direction of for the systemx’ = Ax whenA has vi. one repeated positiveeigenvalueA with associated eigenvectorv and Because all nonzero trajectories are tangent at the origin to the line l, the ori “generalizedeigenvector”V2. gin represents an improper nodal source. Figure 5.3.7 illustrates typical solution curves given by x(t) = civieAt + c21(v + v21)e’ for the system x’ = Ax when A has a repeated eigenvalue but does tnot have ’two linearly independent eigenvectors. Example 7 The solution curves in Fig. 5.3.7 correspond to the choice I=[ 1] (33) in the system x’ = Ax. In Examples 2 and 3 of Section 5.5 we will see that A has the repeated eigenvalue A = 4 with associated eigenvector and generalized eigenvector given by Li 1—31 Ii es vi=[ 3] and (34)

“generalized for FIGURE

in 11 e e n H S r 1 S

ng ‘is

r L Example 9 Eq. in The together You Alternatively, the with However, We can in Note or, that solution The Eq. that vector two set xiQ) Therefore REPEATED A vectors simply where simply = the the in of general could (34), apply (38). is, solution the can 0. v 1 that linearly scalar solution system right-hand and Special v 2 If c 1 is is two the with in v. that verify replacing the it also as now is an and x 2 Q) Eq. It solution the form, does, fixed solutions the curves the x’ principle is, a ZERO eigenvector follows because curves, arrive given c 2 independent direction (37). that system negative methods side are Case

then are points Ax. C2 in of A 5.3 by at EIGENVALUE: Our each of are together with arbitrary Fig. the that A of Thus x(t) has (using an x(t) Eq. x’ field, of is time indeed A gallery of (c 1 , of system equivalent constant 5.3.9 of —c2 A the = = given the (35) Gallery A = Section A=z_[ x 2 (t) xi(t) a eigenvectors is Cl for c 2 ) A, reversal Ax is with Problem Cl repeated generalized in constants, Repeated correspond the equivalent. yields

the [1 Fig. x’ the that by in the reduces = = —3 a negative functions. zero 5.5 the direction Eq. system A=[ form (3c 2 t (—3c 2 t x(t) 5.3.16 of

solution ] to Vi=[ Ax is, 1 e 41 will eigenvalue 33 e Once the phase Solution Dl

(22) —]=[: —41 matrix that as = of + eigenvector to and once to Our F—3 solution x’ shows — associated show, of + 3ci)e_ 4 1.

the [], + with field, Zero (39) the x(t) Av 3ci again = C2 the plane. 3 Thus C2 the

gallery ]. solution more) F—i of Ax choice — = F yields matrix a A a Curves A for “trajectories” order 3t found c2)e 4 1, generalized more = with the = 0 Eigenvalue —3t the = the in + Fig. xQ) —2

we _fl v —2 with the matrix Eq. in 11 general in A 1] system two, complete in

= , conclude Eq. 5.3.16 Example given with of solution e Eq. Eq. (34). 0 —41 = the Linear that for (39) eigenvector A (35): (25)

0, . eigenvector x’ by solution shows Equation given may repeated all set which = is, in Eq. (40), 7, that leads Replacing Ax the of two-dimensional Systems and or (36). a by with thus solution every following more may of is directly once (7) v 2 Eq. v 1 eigenvalue to x’ confirming A associated then not t complete nonzero given again say (41) = given with curves, Ax to 307 gives have way. (41) (38) (40) that (39) are the we —t by by is

.5.3 A Gallery of Solution Curves of Linear Systems 309 Pure Imaginary Eigenvalues: Centers and Elliptical Orbits

PURE IMAGINARY EIGENVALUES: Here we assume that the eigenvalues of the matrix A are given by 1A A = ±iq with q 0. Taking p = 0 in Eq. (5) gives the ) 2 general solution

x(t) = ci(acosqt —bsinqt) + (bcosqt + asinqt) (45) r c2 V for the system x’ = Ax. Rather than directly analyze the trajectories given by 5 Eq. (45), as we have done in the previous cases, we begin instead with an exam El ple that will shed light on the nature of these solution curves.

Example 11 Solve the initial value problem a e 16 —171 x(O)= r41 h = [ _6]X [2] (46) The coefficient matrix S Solution 0 A=[ ] (47) has characteristic equation 8AA_AIl=[6 7]A2+loo_o and hence has the complex conjugate eigenvalues Ai, 2A = ±lOi. If v = [a b is an eigenvector associated with A lOi, then the eigenvector equation (A — AI)v 0 yields

[A—lOi.I]v= 16—lOi —17 lEal 10 [ 8 _6_lOi][b] = [o Upon division of the second row by 2, this gives the two scalar equations

(6— lOi)a — 17b = 0, (48) 4a — (3 + 5i)b = 0,

each of which is satisfied by a = 3 + Si and b = 4. Thus the desired eigenvector is v = ]E’, [3 + 5i with real and imaginary parts r31 r51 and (49)

respectively. Taking q = 10 in Eq. (45) therefore gives the general solution of the system x’ Ax:

x(t) = ci ([]cos10t_ []sinl0t +c2 ([]cos lot + []sin lot) (50) (3 cos lOt — 5 3 — [ c1 sin lOt) + C2(5 cos lOt + sin lOt) cos lOt sin lOt — [ 4Ci + 4C2 To solve the given initial value problem it remains only to determine values of the coefficients ]T le Cl and c2. The initial condition x(O) = [4 2 readily yields c = C2 = 4,and with these of values Eq. (50) becomes (in scalar form)

xi(t) = 4cos lOt — sin lot, (51) FIGURE 5.3.11. Solutioncurve x2(t) = 2cos lOt + 2sin lOt. xi(t) =4coslOt—sinlOt, x2(t) = 2cos lOt + 2sin 101 for the Figure 5.3.11 shows the trajectory given by Eq. (51) together with the initial initialvalueproblemin Eq. (46). point (4, 2). 310 Chapter 5 Linear Systems of Differential Equations

This solution curve appears to be an ellipse rotated counterclockwise by the angle 9 = arctan 0.4636. We can verify this by finding the equations of the solution curve relative to the rotated u- and v-axes shown in Fig. 5.3.11. By a standard formula from analytic geometry, these new equations are given by

2 1 u = x1 cos 9 + x2 sin 9 = —xi + 2—x (52) 1 , 2 v = —xisin 9 + x2 cos 9 = ———xi+ .2—x In Problem 34 we ask you to substitute the expressions for x1 and x2 from Eq. (51) into Eq. (52), leading (after simplification) to

a, = 2’../cos lOt, v = /sin lOt. (53)

Equation (53) not only confirms that the solution curve in Eq. (51) is indeed an ellipse rotated by the angle 9 , but it also shows that the lengths of the semi-major and semi-minor axes of the ellipse are 2/ and respectively. Furthermore, we can demonstrate that any choice of initial point (apart from the origin) leads to a solution curve that is an ellipse rotated by the same angle 9 and “concentric” (in an obvious sense) with the trajectory in Fig. 5.3.11 (see Problems 35—37).All these concentric rotated ellipses are centered at the origin (0, 0), which is therefore called a center for the system x’ = Ax whose coefficient matrix A has pure imaginary eigenvalues. Our gallery Fig. 5.3.16 shows a more complete set of solution curves, together with a direction field, for the system x’ = Ax with A given by Eq. (47). Further investigation: Geometric significance of the eigenvector. Our general solution in Eq. (50) was based upon the vectors a and b in Eq. (49), that is, the real and imaginary parts of the complex eigenvector v = [3 + 5i ]“ of the matrix A. We might therefore expect a and b to have some clear geometric connection to the solution curve in Fig. 5.3.11. For example, we might guess that a and b would be parallel to the major and minor axes of the elliptical trajectory. However, it is clear from Fig. 5.3.12—which shows the vectors a and b together with the solution curve given by Eq. (51)—that this is not the case. Do the eigenvectors of A, then, play FIGURE 5.3.12. Solution curve for any geometric role in the phase portrait of the system x’ = Ax? the initial value problem in Eq. (46) The (affirmative) answer lies in the fact that any nonzero real or complex showing the vectors a, b, , and b. multiple of a complex eigenvector of the matrix A is still an eigenvector of A as sociated with that eigenvalue. Perhaps, then, if we multiply the eigenvector v = 3 5i { + 4]” by a suitable nonzero complex constant z, the resulting eigenvector will have real and imaginary parts a and b that can be readily identified with ge ometric features of the ellipse. To this end, let us multiply v by the complex scalar z = -(1 + i). (The reason for this particular choice will become clear shortly.) The resulting new complex eigenvector of the matrix A is

v=Z.v=—(l+i).[- I r3+5i1 F—l+4i

and has real and imaginary parts

and I

5.3 A Gallery of Solution Curves of Linear Systems 311

It is clear that the vector b is parallel to the major axis of our elliptical trajectory. Further, you can easily check that a b = 0, which means that a is perpendicular to b, and hence is parallel to the minor axis of the ellipse, as Fig. 5.3.12 illustrates. Moreover, the length of b is twice that of a, reflecting the fact that the lengths of the major and minor axes of the ellipse are in this same ratio. Thus for a matrix A with pure imaginary eigenvalues, the complex eigenvector of A used in the general solution (45)—if suitably chosen—is indeed of great significance to the geometry of the elliptical solution curves of the system x’ = Ax. How was the value (l + i) chosen for the scalar z? In order that the real and imaginary parts a and b of z v be parallel to the axes of the ellipse, at a minimum a and b must be perpendicular to each other. In Problem 38 we ask you to show that this condition is satisfied if and only if z is of the form r(l ± i), where r is a nonzero real number, and that if z is chosen in this way, then a and b are in fact parallel to the axes of the ellipse. The value r = then aligns the lengths of a and b with those of the semi-minor and -major axes of the elliptical trajectory. More generally, we can show that given any eigenvector v of a matrix A with pure imaginary eigenvalues, there exists a constant z such that the real and imaginary

parts a and b of the eigenvector = z . vare parallel to the axes of the (elliptical) trajectories of the system x’ = Ax. Further investigation: Direction of flow. Figs. 5.3.11 and 5.3.12 suggest that the solution curve in Eq. (51) flows in a counterclockwise direction with increasing t. However, you can check that the matrix

_A_F_6 17 6

has the same eigenvalues and eigenvectors as the matrix A in Eq. (47) itself, and yet the principle of time reversal) the trajectories of the system x’ —Axare (by = iden tical to those of x’ = Ax while flowing in the opposite direction, that is, clockwise. Clearly, mere knowledge of the eigenvalues and eigenvectors of the matrix A is not e sufficient to predict the direction of flow of the elliptical trajectories of the system = Ax as t increases. How then can we detennine this direction of flow? One simple approach is to use the tangent vector x’ to monitor the direction in e which the solution curves flow as they cross the positive -axis. Ifs is any positive y (so x1 number that the point (s, 0) lies on the positive xi-axis), and if the matrix A is given by b LC d then any trajectory for the system x’ = Ax passing through (s, 0) satisfies X!=AX=[a ][]=[a5]=s[a]

at the point (s, 0). Therefore, at this point the direction of flow of the solution curve is a positive scalar multiple of the vector [a c Since c cannot be zero (see Problem 39), this vector either points “upward” into the first quadrant of the phase plane (if c > 0), or “downward” into the fourth quadrant (if c <0). If upward, then the flow of the solution curve is counterclockwise; if downward,1T then clockwise. For the matrix A in Eq. (47), the vector [a c = [6 8 points into the first quadrant because c = 8 > 0, thus indicating a counterclockwise direction of flow (as Figs. 5.3.11 and 5.3.12 suggest). 312 Chapter 5 Linear Systems of Differential Equations Complex Eigenvalues: Spiral Sinks and Sources COMPLEX EIGENVALUES WITH NEGATIVE REAL PART: Now we assume that the eigenvalues of the matrix A are given by )i, 2A = p ± Iq with q 0 and p <0. In this case the general solution of the system x’ = Axis given directly by Eq. (5): x(t) = ciePt(acosqt (bcosqt—bsinqt)+c +asinqt), (5) where the vectors a and b have theire” usual meaning. Once again we begin with an example to gain an understanding2 of these solution curves. Example 12 Solve the initial value problem

/ Es —171 E41 x, x(0) (54) = [8 t 7 = [2 j j Solution The coefficient matrix A=[ —;] (55) has characteristic equation 8AlAAIl=5 A=(A++100=0, and hence has the complexconjugateeigenvaluesA, 2A —1± lOi. If v = [a b is an eigenvectorassociatedwith A —1 101, + 7 then the eigenvectorequation(A — AI)v = 0 yields the same system(48) of equationsfound in Example 11:

(6—lOi)a-— l7b=O, (48) 4a —(3 + 5i)b = 0.

As in Example 11, each of these equations is satisfiedby a = 3 + 51 and b = 4. Thus the desired eigenvector, associated with 1A = —1+ lOi, is once againv = [3 + 51 4 ]‘, with real and imaginaryparts a=[] and b=[] (56)

respectively. Taking p = —1and q = lOin Eq. (5) thereforegivesthe generalsolutionof the system x’ = Ax:

x(t) = cle_t cos lOt sin lot) + 2e_r cos lOt + sin lOt) ([] — [] c ([] [] — —Fcie(3 cos lOt 5sin lOt) + c2e (5cos lOt + 3sin lot) 4cie_ cos lOt + 4c2e’ Sill lOt —L t t 1T The initial conditionx(0) = [4 2 gives c = C2 = once again, and with these values Eq. (57) becomes (in scalar form) FIGURE 5.3.13. Solution curve = x1 (t) et (4 cos lOt —sin lOt), x (t) = e_t(4cos lOt — sin lot), x2(t) = e_t(2cos lOt +2sinlOt) (58) for the initial value problem in x2(t) = e_t(2cos lOt + 2sin lOt). Eq. (54). The dashed and solid portions of the curve correspond to negative and Figure 5.3.13 shows the trajectory given by Eq. (58) together with the initial positive values of t, respectively. point (4, 2). It is noteworthy to compare this spiral trajectory with the elliptical trajectory in Eq. (51). The equations for xi(t) and x2Q) in (58) are obtained by multiplying their counterparts in (51) by the common factor e1 which is positive and , decreasing with increasing t. Thus for positive values oft, the spiral trajectory is generated, so to speak, by standing at the origin and “reeling in” the point on the elliptical trajectory (51) as it is traced out. When t is negative, the picture is rather

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FIGURE 5.3.15. Three-dimensional trajectories for the system x’ Ax with the matrix A given by Eq. (62).

The matrix A has the single real eigenvalue —l with the single (real) eigen T vector [0 0 1 and the complex conjugate eigenvalues —l + 51. The negative real eigenvalue corresponds to trajectories that lie on the x3-axis and approach the origin as —÷ 0 (as illustrated by the beads on the vertical axis of the figure). Thus the origin (0, 0, 0) is a sink that “attracts” all the trajectories of the system. The complex conjugate eigenvalues with negative real part correspond to tra jectories in the horizontal xix-plane that spiral around the origin while approaching it. Any other trajectory—one2 which starts at a point lying neither on the z-axis nor in the -plane—combines the preceding behaviors by spiraling around the surface of ax12cone while approaching the origin at its vertex. _

T 5.3 A Gallery of Solution Curves of Linear Systems 315

Gallery of Typical Phase Portraits for the System x’ = Ax: Nodes / // / / - //// \\ ‘ N — S

- —/ - — / / / i / / I ,/j////I\\/ //I\”’\\\ \\ \ Proper Nodal Source: A repeated posi Proper Nodal Sink: A repeated negative tive real eigenvalue with two linearly in real eigenvalue with two linearly indepen dependent eigenvectors. dent eigenvectors.

ie is Improper Nodal Source: Distinct positive real eigenvalues (left) or a repeated positive real a eigenvalue without two linearly independent eigenvectors (right). in

Improper Nodal Sink: Distinct negative real eigenvalues (left) or a repeated negative real eigenvalue without two linearly independent eigenvectors (right).

FIGURE 5.3.16. Galleryof typicalphaseplaneportraitsfor thesystemx’ = Ax. 316 Chapter 5 Linear Systems of Differential Equations

Gallery of Typical Phase Portraits for the System x’ = Ax: Saddles, Centers, Spirals, and Parallel Lines

Saddle Point: Real eigenvalues of oppo Center: Pure imaginary eigenvalues. site sign.

/ / /1 / /

Spiral Source: Complex conjugate Spiral Sink: Complex conjugate eigen eigenvalues with positive real part. values with negative real part. :=E\ : \

Parallel Lines: One zero and one neg Parallel Lines: A repeated zero eigen ative real eigenvalue. (If the nonzero value without two linearly independent eigenvalue is positive, then the trajecto eigenvectors. ries flow away from the dotted line.) FIGURE 5.3.16. (Continued)

I 1

5.3 A Gallery of Solution Curves of Linear Systems 31 7 Problems For each of the systems in Problems 1 through 16 in Section 20. - 5.2, categorize the eigenvalues and eigenvectors of the coeffi cient matrix A according to Fig. 5.3.16 and sketch the phase portrait of the system by hand. Then use a computer system or graphing calculator to check your answet:

The phase portraits in Problem 17 through 28 corre spond to linear systems ofthe form x’ = Ax in which the matrix A has two linearly independent eigenvectors. Determine the nature of the eigenvalues and eigenvectors of each system. For example, you may discern that the system has pure imaginary eigenvalues, or that it has real eigenvalues of opposite sign; that an eigenvector associated with the positive eigenvalue is T —1 , roughly [2 ] etc. 21.

17.

22. lx.

19. 23. 27. 26. 25.

24. 318 Chapter 5 Linear Systems of Differential Equations 28. 29. 30. We ric both principle Use where moving axes of then from V 2 system in then measured the Example curve = can the FIGURE determined nonzero, it this are (9) [1 The system u lies indicated give chain point determined that in lrajectory

of ]. = in x(t) nv-coordinate on Eq. respectively. time the u(O) a 1 0(t) a rule 5.3.17. x(t) then simpler the by trajectory by = (63) directions reversal. and c the in for introducing are v-axis. a lies by uOe_ 2 t,

is [] “Cartesian” Fig. eigenvectors vector-valued vo The simply

description =L6 given the on

of r4 oblique functions 5.3.17, parallel (b) the eigenvectors e 2 t u(O). the by its Show u-axis, the _lj 0(t) system v distances ii vi +C2 equation uv-coordinate (a) in of = to oblique and functions = 0(t) that Cu_ 72 . the Show Vi

which [ whereas voe 5 t is V2. and v 1 general if described and ]e5t of from that uv-coordinate no = v 2 . the the to 0(t) system and if if the verify It u-and solution paramet no 00 follows of v origin by = (63) and the are the (9) v 0, 0, 5.3 A Gallery of Solution Curves of Linear Systems 319

r In Problems 31—33A represents a 2 x 2 matrix. 36. In analytic geometry it is shown that the general quadratic equation 31. Use the definitions of eigenvalue and eigenvector (Section Ax + Bx + C4 = k (65) 5.2) to prove that if A is an eigenvalue of A with associ 1x2 ated eigenvector v, then —Ais an eigenvalue of the ma represents an ellipse centered at the origin if and only if trix —Awith associated eigenvector v. Conclude that if Ak > 0 and the discriminant 2B — 4AC <0. Show that A has positive eigenvalues 0 < A < 1A with associated Eq. (64) satisfies these conditions if k < 0, and thus con eigenvectors Vi and ,v2 then —Ahas negative eigenvalues clude that all nondegenerate solution curves of the system —Ai < 2—A <0 with the same associated eigenvectors. in Example 11 are elliptical. 32. Show that the system x’ = Ax has constant solutions other 37. It can be further shown that Eq. (65) represents in general than x(t) 0 if and only if there exists a (constant) vector a conic section rotated by the angle 0 given by x 0 with Ax 0. (It is shown in linear algebra that such a vector x exists exactly when det(A) 0.) tan20= B 33. (a) Show that if A has the repeated eigenvalue Awith two A—C linearly independent associated eigenvectors, then every Show that this formula applied to Eq. (64) leads to the an nonzero vector V is an eigenvector of A. (Hint: Express V gle &= arctan found in Example 11, and thus conclude as a linear combination of the linearly independent eigen that all elliptical solution curves of the system in Exam vectors and multiply both sides by A.) (b) Conclude that ple 11 are rotated by the same angle 0. (Suggestion: You A must be given by Eq. (22). (Suggestion: In the equation i]T•) may find useful the double-angle formula for the tangent Av=AVtakev=[1 0]TanclV=[O function.) 34. Verify Eq. (53) by substituting the expressions for x (t) T 38. Let v 3 + 5m 4] be the complex elgenvector found and from Eq. (51) into Eq. (52) and simplifying. = { x2(t) in Example 11 and let z be a complex number. (a) Show that the real and imaginary parts a and b, respectively, Problems 35—37 show that all nontrivial solution curves of the d of the vector v are perpendicular if and only if system in Example 11 are ellipses rotated by the same angle i = z z = ± i) for some nonzero real number r. (b) Show as the trajectory in Fig. 5.3.11. r(l that if this is the case, then a and b are parallel to the 35. The system in Example 11 can be rewritten in scalar form axes of the elliptical trajectory found in Example 11 (as as Fig. 5.3.12 indicates). 39. Let A denote the 2 x 2 matrix x = 6x — 17x2, 4 = 8x — 6X2, b A=H d leading to the first-order differential equation H (a) Show that the characteristic equation of A (Eq. (8), 2dx — dx/dt — 8xi —6X2 2 Section 5.2) is given by 1dx — dxi/dt — 6xi — 17x2’

or, in differential form, 2A — (a + d)A + (ad — bc) = 0.

(6X2 — 8xi) 1dx + (6xi — l7x2) 2dx = 0. (b) Suppose that the eigenvalues of A are pure imaginary. Show trace T(A) a d A be zero Verify that this equation is exact with general solution that the + of must and that the determinant D(A) = ad — bc must be positive. Conclude that c 0. —4x + 6x1x2 — = k, (64) 40. Use the eigenvalue/eigenvector method to confirm the so where k is a constant. lution in Eq. (61) of the initial value problem in Eq. (59). in vs

5.3 Application Dynamic Phase Plane Graphics 0, 0, ie Using computer systems we can “bring to life” the static gallery of phase portraits in Fig. 5.3.16 by allowing initial conditions, eigenvalues, and even eigenvectors to vary in “real time.” Such dynamic phase plane graphics afford additional insight he into the relationship between the algebraic properties of the 2 x 2 matrix A and the phase plane portrait of the system x’ = Ax.