Constant-Coefficient Systems

Case III. Complex conjugate roots are of minor practical importance, and we discuss the derivation of real solutions from complex ones just in terms of a typical example

Case 2. Damped Forced Oscillations

4.3 Constant-Coefficient Systems

T H E O R E M 1 General Solution

If the constant matrix A in the system (1) has a linearly independent set of n eigenvectors, then the corresponding solutions in (4)form a basis of solutions of (1), and the corresponding general solution is

(5)

How to Graph Solutions in the Phase Plane

We shall now concentrate on systems (1) with constant coefficients consisting of two ODEs

(6) in components,

Of course, we can graph solutions of (6), (7)

as two curves over the t-axis, one for each component of y(t). (Figure 80a in Sec. 4.1 shows an example.) But we can also graph (7) as a single curve in the -plane. This is a parametric representation (parametric equation) with parameter t. (See Fig. 80b for an example. Many more follow. Parametric equations also occur in calculus.) Such a curve is called a trajectory (or sometimes an orbitor path) of (6). The -plane is called the phase plane.¹If we fill the phase plane with trajectories of (6), we obtain the so-called phase portraitof (6).

Studies of solutions in the phase plane have become quite important, along with advances in computer graphics, because a phase portrait gives a good general qualitative impression of the entire family of solutions. Consider the following example, in which we develop such a phase portrait.

E X A M P L E 1 Trajectories in the Phase Plane (Phase Portrait) Find and graph solutions of the system.

In order to see what is going on, let us find and graph solutions of the system

(8) thus

y1r_{⫽ ⫺3y1⫹ y2} y2r_{⫽ y1⫺3y2.} yr_⫽Ay⫽ c^{⫺3 1}

1 ⫺3d ^y, y₁y₂

y₁y₂ y(t)⫽

c

^y1(t)

y₂(t)

d

^,

y₁

r

_{⫽a11y1⫹a12y2}

y₂

r

_{⫽a21y1⫹a22y2.}

yⴕ⫽Ay;

y⫽c₁x⁽¹⁾e^l1t⫹ Á ⫹c_nx⁽ⁿ⁾e^lnt. y⁽¹⁾,Á, y⁽ⁿ⁾

SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 141

1A name that comes from physics, where it is the y-(mv)-plane, used to plot a motion in terms of position y and velocity y⬘⫽v(m⫽mass); but the name is now used quite generally for the y₁y₂-plane.

The use of the phase plane is a qualitative method, a method of obtaining general qualitative information on solutions without actually solving an ODE or a system. This method was created by HENRI POINCARÉ (1854–1912), a great French mathematician, whose work was also fundamental in complex analysis, divergent series, topology, and astronomy.

Solution. By substituting and and dropping the exponential function we get The characteristic equation is

This gives the eigenvalues and . Eigenvectors are then obtained from

For this is . Hence we can take . For this becomes

and an eigenvector is . This gives the general solution

Figure 82 shows a phase portrait of some of the trajectories (to which more trajectories could be added if so desired). The two straight trajectories correspond to and and the others to other choices of

The method of the phase plane is particularly valuable in the frequent cases when solving an ODE or a system is inconvenient of impossible.

Critical Points of the System (6)

The point in Fig. 82 seems to be a common point of all trajectories, and we want to explore the reason for this remarkable observation. The answer will follow by calculus.

Indeed, from (6) we obtain (9)

This associates with every point a unique tangent direction of the trajectory passing through P, except for the point , where the right side of (9) becomes . This point , at which becomes undetermined, is called a critical pointof (6).

Five Types of Critical Points

There are five types of critical points depending on the geometric shape of the trajectories near them. They are called improper nodes, proper nodes, saddle points, centers, and spiral points. We define and illustrate them in Examples 1–5.

E X A M P L E 1 (Continued) Improper Node (Fig. 82)

An improper node is a critical point at which all the trajectories, except for two of them, have the same limiting direction of the tangent. The two exceptional trajectories also have a limiting direction of the tangent at which, however, is different.

The system (8) has an improper node at 0, as its phase portrait Fig. 82 shows. The common limiting direction at 0is that of the eigenvector because goes to zero faster than as tincreases. The two exceptional limiting tangent directions are those of x⁽²⁾⫽[1 ⫺1]^Tand ⫺x⁽²⁾⫽[⫺1 1]^T. 䊏

e^ⴚ2t e^ⴚ4t

x⁽¹⁾⫽[1 1]^T P0

P0

dy2>dy1

P0

0>0

P⫽P0: (0, 0)

dy2>dy1

P: (y1, y2) dy2

dy1 ⫽y2

r

_dt

y1

r

_dt ^⫽

y2

r

y1

r

^⫽

a21y1⫹a22y2

a11y1⫹a12y2

. y⫽0

䊏 c₁, c₂.

c₂⫽0 c₁⫽0

y⫽c^y1

y2d^{⫽c1y(1)⫹c2y(2)⫽c1} c¹

1d ^eⴚ2t⫹c2 c ¹

⫺1d ^eⴚ4t.

x⁽²⁾⫽[1 ⫺1]^T

x₁⫹x₂⫽0, l₂⫽ ⫺4

x⁽¹⁾⫽[1 1]^T

⫺x₁⫹x₂⫽0 l₁⫽ ⫺2

(⫺3⫺l)x1⫹x2⫽0.

l₂⫽ ⫺4 l₁⫽ ⫺2

⫽l²⫹6l⫹8⫽0.

det (A⫺lI)⫽2^{⫺3⫺l 1}

1 ⫺3⫺l

2

Ax⫽lx.

yr_⫽lxe^lt y⫽xe^lt

E X A M P L E 2 Proper Node (Fig. 83)

A proper node is a critical point at which every trajectory has a definite limiting direction and for any given direction dat there is a trajectory having das its limiting direction.

The system

(10)

has a proper node at the origin (see Fig. 83). Indeed, the matrix is the unit matrix. Its characteristic equation has the root . Any is an eigenvector, and we can take and . Hence a general solution is

䊏 y⫽c₁ c¹₀d ^et⫹c2 c⁰₁d ^{et or y}_y^1⫽c1et

2⫽c₂e^t or c₁y₂⫽c₂y₁.

[0 1]^T [1 0]^T

x⫽0 l⫽1

(1⫺l)²⫽0

yr_⫽ c¹₀ ⁰₁d ^{y, thus y}_y^1r⫽y1

2r_⫽y2 P₀

P₀

SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 143

y₂

y₁ y⁽¹⁾(t)

y⁽²⁾(t)

Fig. 82. Trajectories of the system (8) (Improper node)

y₂

y₁

Fig. 83. Trajectories of the system (10) (Proper node)

E X A M P L E 3 Saddle Point (Fig. 84)

A saddle point is a critical point at which there are two incoming trajectories, two outgoing trajectories, and all the other trajectories in a neighborhood of bypass .

The system

(11)

has a saddle point at the origin. Its characteristic equation has the roots and . For an eigenvector is obtained from the second row of that is,

. For the first row gives . Hence a general solution is

This is a family of hyperbolas (and the coordinate axes); see Fig. 84. 䊏 y⫽c1 c¹

0d ^et⫹c2 c⁰

1d ^{eⴚt or y1⫽c1et}

y2⫽c2e^ⴚt or y1y2⫽const.

[0 1]^T l₂⫽ ⫺1

0x1⫹(⫺1⫺1)x2⫽0

(A⫺lI)x⫽0, [1 0]^T

l⫽1 l₂⫽ ⫺1

l₁⫽1 (1⫺l)(⫺1⫺l)⫽0

yr_⫽ c^{1 0}

0 ⫺1d ^{y, thus y1r⫽ y1}

y1r_{⫽ ⫺y2} P0

P0

P0

E X A M P L E 4 Center (Fig. 85)

A center is a critical point that is enclosed by infinitely many closed trajectories.

The system

(12)

has a center at the origin. The characteristic equation gives the eigenvalues 2i and . For 2ian

eigenvector follows from the first equation of , say, . For that

equation is and gives, say, . Hence a complex general solution is

(12 )

A real solution is obtained from (12 ) by the Euler formula or directly from (12) by a trick. (Remember the trick and call it a method when you apply it again.) Namely, the left side of (a) times the right side of (b) is

. This must equal the left side of (b) times the right side of (a). Thus,

. By integration, .

This is a family of ellipses (see Fig. 85) enclosing the center at the origin. 䊏 2y₁²⫹¹2y₂²⫽const

⫺4y₁y₁r_⫽y2y2r

⫺4y₁y₁r

*

y⫽c₁ c_2i¹d ^e2it⫹c2 c_⫺¹_2id ^{eⴚ2it, thus y}_y^{1⫽ c1e2it⫹ c2eⴚ2it}

2⫽2ic₁e^2it⫺2ic₂e^ⴚ2it.

*

[1 ⫺2i]^T

⫺(⫺2i)x₁⫹x₂⫽0

l⫽ ⫺2i [1 2i]^T

(A⫺lI)x⫽0

⫺2ix₁⫹x₂⫽0

⫺2i l²⫹4⫽0

yr_⫽ c_⫺⁰₄ ¹₀d ^{y, thus (a)}_(b) ^y_y^1r⫽y2

2r_{⫽ ⫺4y1}

y₂

y₁

Fig. 84. Trajectories of the system (11) (Saddle point)

y₂

y₁

Fig. 85. Trajectories of the system (12) (Center)

E X A M P L E 5 Spiral Point (Fig. 86)

Aspiral point is a critical point about which the trajectories spiral, approaching as (or tracing these spirals in the opposite sense, away from ).

The system

(13)

has a spiral point at the origin, as we shall see. The characteristic equation is . It gives the eigenvalues ⫺1⫹iand ⫺1⫺i. Corresponding eigenvectors are obtained from (⫺1⫺l)x₁⫹x₂⫽0. For

l²⫹2l⫹2⫽0 yr_⫽c^⫺_⫺¹_{1 ⫺}¹₁d ^{y, thus y}_y^{1r⫽ ⫺y1⫹y2}

2r_{⫽ ⫺y1⫺y2} P₀

t:⬁ P₀ P₀

this becomes and we can take as an eigenvector. Similarly, an eigenvector corresponding to is . This gives the complex general solution

The next step would be the transformation of this complex solution to a real general solution by the Euler formula. But, as in the last example, we just wanted to see what eigenvalues to expect in the case of a spiral point. Accordingly, we start again from the beginning and instead of that rather lengthy systematic calculation we use a shortcut. We multiply the first equation in (13) by , the second by , and add, obtaining

.

We now introduce polar coordinates r, t, where . Differentiating this with respect to tgives . Hence the previous equation can be written

, Thus, , , .

For each real c this is a spiral, as claimed (see Fig. 86). 䊏

r⫽ce^ⴚt ln ƒrƒ⫽ ⫺t⫹c*,

dr>r⫽ ⫺dt rr_{⫽ ⫺r}

rrr_{⫽ ⫺r}² 2rrr_⫽2y1y1r_⫹2y2yr₂

r²⫽y₁²⫹y₂² y₁y₁r_{⫹ y2y2}r_{⫽ ⫺(y1}²_⫹y2²₎

y₂ y₁

y⫽c₁ c¹_id ^{e(ⴚ1ⴙi)t⫹c2} c_⫺¹_id ^{e(ⴚ1ⴚi)t.}

[1 ⫺i]^T

⫺1⫺i

[1 i]^T

⫺ix₁⫹x₂⫽0 l⫽ ⫺1⫹i

SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 145

y₂

y₁

Fig. 86. Trajectories of the system (13) (Spiral point)

E X A M P L E 6 No Basis of Eigenvectors Available. Degenerate Node (Fig. 87)

This cannot happen if Ain (1) is symmetric , as in Examples 1–3) or skew-symmetric

thus . And it does not happen in many other cases (see Examples 4 and 5). Hence it suffices to explain the method to be used by an example.

Find and graph a general solution of

(14)

Solution. Ais not skew-symmetric! Its characteristic equation is

. det (A⫺lI)⫽2^{4⫺l 1}

⫺1 2⫺l2⫽l²⫺6l⫹9⫽(l⫺3)²⫽0 yr_⫽Ay⫽c_⫺⁴₁ ¹₂d ^y.

a_jj⫽0)

(a_kj⫽ ⫺a_jk, (a_kj⫽a_jk

It has a double root . Hence eigenvectors are obtained from , thus from say, and nonzero multiples of it (which do not help). The method now is to substitute

with constant into (14). (The xt-term alone, the analog of what we did in Sec. 2.2 in the case of a double root, would not be enough. Try it.) This gives

. On the right, . Hence the terms cancel, and then division by gives

, thus .

Here and , so that

, thus

A solution, linearly independent of , is . This yields the answer (Fig. 87)

The critical point at the origin is often called a degenerate node. gives the heavy straight line, with the lower part and the upper part of it. gives the right part of the heavy curve from 0 through the second, first, and—finally—fourth quadrants. ⫺y⁽²⁾gives the other part of that curve. 䊏

y⁽²⁾ c1⬍0

c1⬎0

c1y⁽¹⁾ y⫽c1y⁽¹⁾⫹c2y⁽²⁾⫽c1 c ¹

⫺1d ^{e3t⫹c2 £}c ¹

⫺1d ^t⫹c⁰

1d^{≥ e3t.}

u⫽[0 1]^T x⫽[1 ⫺1]^T

u1⫹ u2⫽1

⫺u1⫺ u2⫽ ⫺1.

(A⫺3I)u⫽ c^{4⫺3 1}

⫺1 2⫺3d ^u⫽c ¹

⫺1d

x⫽[1 ⫺1]^T l⫽3

(A⫺lI)u⫽x x⫹lu⫽Au

e^lt lxte^lt

Ax⫽lx

y⁽²⁾r_⫽xe^lt_⫹lxte^lt_⫹lue^lt_⫽Ay⁽²⁾_⫽Axte^lt_⫹Aue^lt u⫽[u1 u2]^T

y⁽²⁾⫽xte^lt⫹ue^lt x⁽¹⁾⫽[1 ⫺1]^T

x₁⫹x₂⫽0, (4⫺l)x₁⫹x₂⫽0

l⫽3

y₂

y₁

y⁽¹⁾ y⁽²⁾

Fig. 87. Degenerate node in Example 6

We mention that for a system (1) with three or more equations and a triple eigenvalue with only one linearly independent eigenvector, one will get two solutions, as just discussed, and a third linearly independent one from

with vfrom u⫹lv⫽Av.

y⁽³⁾⫽¹2xt²e^lt⫹ute^lt⫹ve^lt

SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 147

1–9 GENERAL SOLUTION

Find a real general solution of the following systems. Show the details.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10–15 IVPs

Solve the following initial value problems.

10.

11.

12.

13.

y₁(0)⫽0, y₂(0)⫽2 y₂r_⫽y1

y₁r_⫽y2

y₁(0)⫽12, y₂(0)⫽2 y₂r_⫽¹_3y1⫹y2

y₁r_⫽y1⫹3y2

y₁(0)⫽ ⫺12, y₂(0)⫽0 y₂r_{⫽ ⫺}¹_2y1⫺ ³_2y2

y₁r_⫽2y1⫹5y2

y₁(0)⫽0, y₂(0)⫽7 y₂r_⫽5y1⫺y2

y₁r_⫽2y1⫹2y2

y₃r_{⫽ ⫺4y1⫺14y2⫺2y3}

y₂r_{⫽ ⫺10y1⫹y2⫺14y3}

y₁r_{⫽10y1⫺10y2⫺4y3}

y₂r_⫽y1⫹10y2

y₁r_⫽8y1⫺y2

y₃r_{⫽ ⫺y2}

y₂r_{⫽ ⫺y1⫹y3}

y₁r_⫽y2

y₂r_⫽2y1⫹2y2

y₁r_⫽2y1⫺2y2

y₂r_{⫽5y1⫹12.5y2}

y₁r_⫽2y1⫹5y2

y₂r_⫽2y1⫺4y2

y₁r_{⫽ ⫺8y1⫺2y2}

y₂r_⫽¹_2y1⫹y2

y₁r_⫽y1⫹2y2

y₂r_⫽y1⫹6y2

y₁r_⫽6y1⫹9y2

y₂r_⫽3y1⫺y2

y₁r_⫽y1⫹y2

14.

15.

16–17 CONVERSION

Find a general solution by conversion to a single ODE.

16. The system in Prob. 8.

17. The system in Example 5 of the text.

18. Mixing problem, Fig. 88. Each of the two tanks contains 200 gal of water, in which initially 100 lb (Tank ) and 200 lb (Tank ) of fertilizer are dissolved.

The inflow, circulation, and outflow are shown in Fig. 88. The mixture is kept uniform by stirring. Find the fertilizer contents y₁(t)in T₁and y₂(t)in .T₂

T₂ T₁

y₁(0)⫽0.5, y₂(0)⫽ ⫺0.5 y₂r_⫽2y1⫹3y2

y₁r_⫽3y1⫹2y2

y₁(0)⫽1, y₂(0)⫽0 y₂r_⫽y1⫺y2

y₁r_{⫽ ⫺y1⫺y2}

P R O B L E M S E T 4 . 3

Fig. 88. Tanks in Problem 18

4 gal/min

16 gal/min 12 gal/min

12 gal/min (Pure water)

T₁ T₂

19. Network. Show that a model for the currents and in Fig. 89 is

, .

Find a general solution, assuming that , .

C⫽1>12 F L⫽4 H,

R⫽3 ⍀ LIr_{2⫹R(I2⫺I1)⫽0}

1

C

冮

^{I1 dt⫹R(I1⫺I2)⫽0}

I₂(t)

I₁(t)

Fig. 89. Network in Problem 19

I₁ C

R

L I₂

20. CAS PROJECT. Phase Portraits. Graph some of the figures in this section, in particular Fig. 87 on the degenerate node, in which the vector depends ont.

In each figure highlight a trajectory that satisfies an initial condition of your choice.

y⁽²⁾