Vectors, matrices, eigenvalues and eigenvectors 1 1 2 0 . 5 2 - - PowerPoint PPT Presentation

Vectors, matrices, eigenvalues and eigenvectors

1

Scaling a vector: 0.5 V = 0.5 2 1

• =

0.5 × 2 0.5 × 1

• =

1 0.5

• Adding two vectors:

V + W = 2 1

• +

1 3

• =

2 + 1 1 + 3

• =

3 4

• A scalar times a matrix: λ

a b c d

• =

λa λb λc λd

• A matrix plus a matrix:

a b c d

• +

x y z w

• =

a + x b + y c + z d + w

• A matrix times a matrix:

a b c d x y z w

• =

ax + bz ay + bw cx + dz cy + dw

• 2

Hence the product of a matrix times a vector: a b c d x y

• =

ax + by cx + dy

• This matrix transforms the vector into another vector:

3

4

A system of linear equations: x − 2y = −5 2x + y = 10 Can be written as: 1 −2 2 1 x y

• =

−5 10

• From the first we obtain x = 2y − 5, which gives in the second:

2(2y−5)+y = 10

• r

4y−10+y = 10

• r

5y = 20 i.e. y = 4 , and hence x = 3. Finally, for the matrix A = a b c d

• define

det[A] = ad − bc and tr[A] = a + d for the determinant and the trace.

5

Linear systems only have solutions when det[A] = 0. Consider the following general linear system ax + by = p cx + dy = q . Solve x = p

a − by a from first equation, then:

cp a − cby a + dy = q

• r

cp − cby + ady = qa

• r

y(ad − cb) = qa − cp

• r

y = qa − cp ad − cb , which only has a finite solution when the denominator ad−cb is not equal to zero.

6

Forest succession: Gray Birch Blackgum Red Maple Beech Gray Birch 0.05 0.01 Blackgum 0.36 0.57 0.14 0.01 Red Maple 0.5 0.25 0.55 0.03 Beech 0.09 0.17 0.31 0.96 For example, the fraction of Red Maple trees after 50 years would be 0.5× the fraction of Gray Birch trees, plus 0.25× the fraction of Blackgum trees, plus 0.55× the fraction of Red Maples, plus 0.03× the fraction of Beech trees.

7

Write table as a matrix: A =     0.05 0.01 0.36 0.57 0.14 0.01 0.5 0.25 0.55 0.03 0.09 0.17 0.31 0.96     and define the current state of the forest as a vector, e.g.,

• V0 =
• 1 0 0 0
• After 50 years the next state of the forest is defined by:
• V50 = A

V0 =

• 0.05 0.36 0.5 0.09
• which is a forest with 5% Gray Birch, 36% Blackgum, 50% Red

Maple, and 9% Beech trees.

8

The next state of the forest is

• V100 = A

V50 =

• 0.0061 0.2941 0.3927 0.3071
• After 100 intervals of 50 years, the state is
• V5000 = A100

V0, where A100 =     0.005 0.005 0.005 0.005 0.048 0.048 0.048 0.048 0.085 0.085 0.085 0.085 0.866 0.866 0.866 0.866     which is a matrix with identical columns.

9

Now consider an arbitrary vector V = (x y z w), where w = 1 − x − y − z, and notice that A100 V =     0.005 0.005 0.005 0.005 0.048 0.048 0.048 0.048 0.085 0.085 0.085 0.085 0.866 0.866 0.866 0.866         x y z w     =     0.005(x + y + z + w) 0.048(x + y + z + w) 0.085(x + y + z + w) 0.866(x + y + z + w)     = (0.005 0.048 0.085 0.866), the succession converges into climax state. This climax vector is an eigenvector of the matrix A!

10

• 200

400 600 800 1000 1200 0.0 0.2 0.4 0.6 0.8 1.0 Time in years Frequency

• Gray birch

Blackgum Red maple Beech

(a) 5 10 15 5 10 15 x y

• (0,1)

(2,1) (4,5) (14,13) (b)

R-scripts: horn.R and eigen.R

11

Eigenvalue problem: Av = a b c d x y

• = λ

x y

• ax + by = λx

cx + dy = λy

• r

(a − λ)x + by = 0 cx + (d − λ)y = 0 Multiply first with (d − λ), and second with b: (d − λ)[(a − λ)x + by] = 0 b[cx + (d − λ)y] = 0 Subtract second from first: [(d − λ)(a − λ) − bc]x = 0 Because x = 0: (d − λ)(a − λ) − bc = 0

12

Characteristic equation: λ2 − (a + d)λ + (ad − bc) = 0 Since A = a b c d

• , this can be written as:

λ2 − trλ + det = 0 Hence: λ1,2 = tr ±

• tr2 − 4 det

2

13

Numerical example: Av = 1 2 2 1 x y

• = λ

x y

• tr[A] = 2

and det[A] = 1 − 4 = −3 Characteristic equation: λ1,2 = tr ±

• tr2 − 4 det

2 = 2 ± √4 + 12 2 = 2 ± 4 2 Hence: λ1 = 3 and λ2 = −1

14

Corresponding eigenvectors: (a − λ)x + by = 0 cx + (d − λ)y = 0

• r
• y = λ−a

b

x x = λ−d

c

y First eigenvector λ1 = 3, a = 1, b = 2, c = 2, d = 1: y = 3 − 1 2 x = x and x = 3 − 1 2 y = y hence v1 = 1 1

• Second eigenvector λ2 = −1, a = 1, b = 2, c = 2, d = 1:

y = −1 − 1 2 x = −x and x = −1 − 1 2 y = −y hence v2 = −1 1

• We only need one of the two equations!

15

Indeed, general case for eigenvectors: (a − λ)x + by = 0 cx + (d − λ)y = 0 First equation delivers: x y

• =

−b a − λ

• Indeed, second equals zero (delivers characteristic equation):

−bc + (d − λ)(a − λ) = 0 Thus, λ1 = 3, a = 1, b = 2: v1 = −2 −2

• r

v1 = 1 1

• for λ2 = −1, a = 1, b = 2: v2 =
• −2

1 − −1

• r

v2 = −1 1

• 16

Special case, diagonal matrix: A = a 0 0 d

• Characteristic equation:

(a − λ)(d − λ) − 0 = 0 → λ1 = a and λ2 = d Eigenvectors: (a − λ)x + 0y = 0 0x + (d − λ)y = 0 λ1 = a gives (d − a)y = 0 or y = 0, i.e., v1 = 1

• λ2 = d gives (a − d)x = 0 or x = 0, i.e., v2 =

1

• 17

Linear differential equations The solution of dx(t)/dt = ax(t) is x(t) = Ceat, where C = x(0). Check this: ∂t Ceat = aCeat = ax(t) Now two-dimensional systems: dx/dt = f(x, y) dy/dt = g(x, y) where x(t) and y(t) are unknown functions of time t, and f and g are functions of x and y.

18

An example: dx/dt = ax + by dy/dt = cx + dy and dx/dt = −2x + y dy/dt = x − 2y where x and y decay at a rate −1, and are converted into one another at a rate 1. In matrix notation: dx/dt dy/dt

• =

a b c d x y

• 19

We claim that dx/dt dy/dt

• =

a b c d x y

• has as a general solution:

x(t) = C1x1eλ1t + C2x2eλ2t y(t) = C1y1eλ1t + C2y2eλ2t

• r

x(t) y(t)

• = C1

x1 y1

• eλ1t + C2

x2 y2

• eλ2t

where λ1,2 are eigenvalues and (xi yi) are the corresponding eigen- vectors of the matrix given above. Like x(t) = Ceat, this has only one steady state: (x, y) = (0, 0).

20

Notice that the solutions x(t) y(t)

• = C1

x1 y1

• eλ1t + C2

x2 y2

• eλ2t

are a linear combination of the growth along the eigenvectors. Since x(t) and y(t) grow when λ1,2 > 0 we obtain:

• a stable node when both λ1,2 < 0
• an unstable node when both λ1,2 > 0
• an (unstable) saddle point when λ1 > 0 and λ2 < 0 (or vice versa)

When λ1,2 are complex, i.e., λ1,2 = α ± iβ, we obtain

• a stable spiral when the real part α < 0
• an unstable spiral when the real part α > 0
• a neutrally stable center point when the real part α = 0

21

Example: dx/dt dy/dt

• =

a b c d x y

• =

−2 1 1 −2 x y

• Since tr = −4 and det = 4 − 1 = 3 we obtain:

λ1,2 = −4 ± √16 − 12 2 = −2 ± 1 so λ1 = −1 and λ2 = −3. Hence solutions tend to zero and (x, y) = (0, 0) is a stable node. To find the eigenvector v1 we write: v1 = −b a − λ1

• =

−1 −1

• r

v1 1 1

• 22

For v2 we write v2 = −b a − λ2

• =

−1 1

• In combination this gives

x(t) y(t)

• = C1

1 1

• e−t + C2

−1 1

• e−3t
• r

x(t) = C1e−t − C2e−3t y(t) = C1e−t + C2e−3t The integration constants C1 and C2 can be solved from the initial condition: i.e., x(0) = C1 − C2 and y(0) = C1 + C2.

23

Let’s check this solution: x(t) = C1e−t − C2e−3t y(t) = C1e−t + C2e−3t

• r

dx dt = −C1e−t + 3C2e−3t dy dt = −C1e−t − 3C2e−3t which should be equal to dx dt = −2x+y = −2(C1e−t−C2e−3t)+C1e−t+C2e−3t = −C1e−t+3C2e−3t dy dt = x−2y = C1e−t−C2e−3t−2(C1e−t+C2e−3t) = −C1e−t−3C2e−3t

24

Linear approximations Derivative: f′(¯ x) = lim

x→¯ x

f(x) − f(¯ x) x − ¯ x

• r

f′(¯ x) = lim

h→0

f(¯ x + h) − f(¯ x) h , Rewrite this into: f(x) ≃ f(¯ x) + f′(¯ x) (x − ¯ x)

• r

f(x) ≃ f(¯ x) + f′(¯ x) h , Example: f(x) = ax2 + b → ∂xf(x) = 2ax a = 2 , b = 1 , x = 3 → f(3) = 2×9+1 = 19 , ∂xf(3) = 2×2×3 = 12 f(3.1) = 20.22

• r

f(3.1) ≃ f(3)+∂xf(3)×0.1 = 19+12×0.1 = 20.2

25

f(x) ≃ f(¯ x) + ∂xf(¯ x) (x − ¯ x) ¯ x x f(¯ x) f(x) ∂xf(¯ x)

The function f(x, y) = 3x − x2 − 2xy: ∂xf(x, y) = 3 − 2x − 2y and ∂yf(x, y) = −2x and in the point f(1, 1) = 0: ∂x f(x, y) = −1 and ∂y f(x, y) = −2

Generally f(x, y) ≃ f(¯ x, ¯ y) + ∂xf (x − ¯ x) + ∂yf (y − ¯ y) Or, after defining hx = x − ¯ x and hy = y − ¯ y: f(x, y) = f(¯ x + hx, ¯ y + hy) ≃ f(¯ x, ¯ y) + ∂xf hx + ∂yf hy Example: f(x, y) = 3x − x2 − 2xy , f(1, 1) = 0 , ∂x = −1, ∂y = −2 f(1.25, 1.25) = 3.75 − 1.5625 − 3.125 = −0.9375 f(1.25, 1.25) ≃ 0 − 1 × 0.25 − 2 × 0.25 = −0.75

28

Consider dx/dt = f(x, y) dy/dt = g(x, y) close an equilibrium point at (¯ x, ¯ y), i.e., f(¯ x, ¯ y) = g(¯ x, ¯ y) = 0 Linear approximation of f(x, y) close to the equilibrium: f(x, y) ≃ f(¯ x, ¯ y) + ∂xf (x − ¯ x) + ∂yf (y − ¯ y) As f(¯ x, ¯ y) = 0 we obtain f(x, y) ≃ ∂xf (x − ¯ x) + ∂yf (y − ¯ y) For g(x, y) this yields: g(x, y) ≃ ∂xg (x − ¯ x) + ∂yg (y − ¯ y)

29

dx/dt = f(x, y) dy/dt = g(x, y) became dx/dt ≃ ∂xf (x − ¯ x) + ∂yf (y − ¯ y) dy/dt ≃ ∂xg (x − ¯ x) + ∂yg (y − ¯ y) Since the partial derivatives are merely the slopes of f(x, y) and g(x, y) at the point (¯ x, ¯ y), they are constants that we can write as a = ∂xf, b = ∂yf, c = ∂xg, d = ∂yg Steady states ¯ x and ¯ y are also constants, with derivatives zero: dx dt = dx dt − d¯ x dt = d(x − ¯ x) dt and dy dt = dy dt − d¯ y dt = d(y − ¯ y) dt Hence d(x − ¯ x)/dt = a(x − ¯ x) + b(y − ¯ y) d(y − ¯ y)/dt = c(x − ¯ x) + d(y − ¯ y)

30

Changing variables to the distances hx = x − ¯ x and hy = y − ¯ y: dhx/dt = ahx + bhy dhy/dt = chx + dhy having the solution hx(t) hy(t)

• = C1

x1 y1

• eλ1t + C2

x2 y2

• eλ2t

where λ1,2 and (xi yi) are the eigenvalues and corresponding eigen- vectors of the Jacobi matrix J = ∂xf ∂yf ∂xg ∂yg

• =

a b c d

• 31

Knowing the two eigenvalues of J = ∂xf ∂yf ∂xg ∂yg

• =

a b c d

• the steady state will be stable when λ1 < 0 and λ2 < 0.

If so the return time is defined by the dominant eigenvalue: TR = −1 max(λ1, λ2)

32

Example: dx dt = f(x, y) = a − bx − cxy and dy dt = g(x, y) = dxy − ey , with ¯ x = a

b when y = 0, and ¯

x = e

d and ¯

y = ad

ce − b c

J = ∂xf ∂yf ∂xg ∂yg

• =

−b − c¯ y −c¯ x d¯ y d¯ x − e

• 33

J = ∂xf ∂yf ∂xg ∂yg

• =

−b − c¯ y −c¯ x d¯ y d¯ x − e

• Fill in ¯

x = a

b and ¯

y = 0, J1 = −b −ca

b da b − e

• Since this matrix is in a diagonal form we know that the diagonal

elements provide the eigenvalues, i.e., λ1 = −b and λ2 = da

b − e.

Hence this state is stable whenever λ2 < 0, i.e., a

b < e d.

34

J = ∂xf ∂yf ∂xg ∂yg

• =

−b − c¯ y −c¯ x d¯ y d¯ x − e

• Now consider ¯

x = e

d and ¯

y = ad

ce − b c and first fill in ¯

x: J2 = −b − c¯ y −ce

d

d¯ y

• When ¯

y > 0 the signs of this matrix are given by J3 = −α −β γ

• with

trJ3 = −α < 0 and det J3 = βγ > 0 , such that λ1,2 = tr ±

• tr2 − 4 det

2 = −α ±

• α2 − 4βγ

2 < 0 , Since λ1,2 < 0 the non-trivial steady state is stable.

Having J = ∂xf ∂yf ∂xg ∂yg

• =

a b c d

• we know that

λ1,2 = tr ± √ D 2 where D = tr2 − 4 det Observing that λ1 + λ2 = tr[J] and λ1 × λ2 = det[J] , the latter because 1 4(tr + √ D)(tr − √ D) = 1 4(tr2 − D) = 1 4(tr2 − tr2 + 4 det) = det we can classify steady states by just the trace and determinant of their Jacobi matrix.

36

center saddle node stable non−stable node non−stable spiral stable spiral

tr det 5 1 2 3 4 6 D=0

λ1,2 = tr ± √ D 2 D = tr2 − 4 det λ1 + λ2 = tr λ1 × λ2 = det

• 1. if det < 0 then D > 0: λ1,2 are real with unequal sign: saddle
• 2. if det > 0, tr > 0 and D > 0 then λ1,2 > 0: unstable node.
• 3. if det > 0, tr < 0 and D > 0 then λ1,2 < 0: stable node.
• 4. if det > 0, tr > 0 and D < 0 then λ1,2 > 0: unstable spiral.
• 5. if det > 0, tr < 0 and D < 0 then λ1,2 > 0: stable spiral.

Graphical Jacobian: use the signs only J =     ∂xf ≃ f(¯ x + h, ¯ y) h ∂yf ≃ f(¯ x, ¯ y + h) h ∂xg ≃ g(¯ x + h, ¯ y) h ∂yg ≃ g(¯ x, ¯ y + h) h     =     α β γ δ     with tr[J] = α + δ and det[J] = αδ − βγ. If tr < 0 and det > 0 the state will be stable.

38

x y

(x,y)

a b

x y

(x,y) (x+h,y) (x,y+h)

x y

(x,y)

c

J =     ∂xf ≃ f(¯ x + h, ¯ y) h ∂yf ≃ f(¯ x, ¯ y + h) h ∂xg ≃ g(¯ x + h, ¯ y) h ∂yg ≃ g(¯ x, ¯ y + h) h     =     − − + −    

39

The Graphical Jacobian of the Lotka Volterra model: R N

e d a b a c

−α −β γ

• tr(J) = −α < 0

and det(J) = βγ > 0

Finally: The full Jacobian of the Lotka-Volterra model: dR dt = aR − bR2 − cRN dN dt = dRN − eN Nullclines: R = 0 , N = a − bR c and N = 0 , R = e d Steady states: (R, N) = (0, 0) , (R, N) = (a/b, 0) and (R, N) = e d, da − eb dc

• 41

Steady states of the Lotka-Volterra model R N

e d a b a c

(0,0) a

b, 0

• e

d, da−eb dc

• N = a−bR

c

Note that da > eb

42

Lotka Volterra model: dR dt = aR−bR2−cRN = f(R, N) and dN dt = dRN−eN = g(R, N) New variables hR and hN define the distance to the steady state: dR dt = d( ¯ R + hR) dt = dhR dt and dN dt = d( ¯ N + hN) dt = dhN dt dhR dt = f( ¯ R+hR, ¯ N+hN) ≈ f( ¯ R, ¯ N)+∂Rf

• ( ¯

R, ¯ N)hR+∂Nf

• ( ¯

R, ¯ N)hN

dhN dt = g( ¯ R+hR, ¯ N+hN) ≈ g( ¯ R, ¯ N)+∂Rg

• ( ¯

R, ¯ N)hR+∂Ng

• ( ¯

R, ¯ N)hN

43

Because f( ¯ R, ¯ N) = 0 and g( ¯ R, ¯ N) = 0: dhR dt = ∂Rf

• ( ¯

R, ¯ N)hR + ∂Nf

• ( ¯

R, ¯ N)hN

dhN dt = ∂Rg

• ( ¯

R, ¯ N)hR + ∂Ng

• ( ¯

R, ¯ N)hN

J = a − 2b ¯ R − c ¯ N −c ¯ R d ¯ N d ¯ R − e

• = J =

? −β +γ ?

• The solution of the linear system has the form

hR(t) hN(t)

• = C1

R1 N1

• eλ1t + C2

R2 N2

• eλ2t

where λ1,2 are the eigenvalues of J and (R1 N1) and (R2 N2) the corresponding eigenvectors.

For ( ¯ R, ¯ N) = (0, 0) one finds: J = a 0 0 −e

• with λ1 = a > 0 and λ2 = −e < 0, i.e., a saddle point.

For ( ¯ R, ¯ N) = (a/b, 0) one finds: J = −a −ac

b da−eb b

• The eigenvalues are λ1 = −a < 0 and λ2 = (da − eb)/b.

Since da > eb this is also a saddle point.

45

For ( ¯ R, ¯ N) =

• e

d, da−eb dc

• ne obtains:

J =

• −be

d

−ce

d da−eb c

• =

−b ¯ R −c ¯ R d ¯ N

• =

−α −β γ

• Hence

trJ = −b ¯ R < 0 and det J = cd ¯ R ¯ N > 0 which tells us that the non-trivial steady state is stable. If D = tr2 − 4 det = (b ¯ R)2 − 4cd ¯ R < 0 this is a stable spiral point.

46

The graphical Jacobian has the same signs: R N

e d a b a c

−α −β γ

• tr(J) = −α < 0

and det(J) = βγ > 0

Complex numbers Quadratic equation: aλ2 + bλ + c = 0 , with roots λ1,2 = −b ± √ b2 − 4ac 2a = −b ± √ D 2a where D = b2 − 4ac What if D < 0? Define: i2 = −1

• r equivalently

i = √ −1 Solve λ2 = −3 by using i2 = −1: λ2 = i2 × 3

• r

λ1,2 = ±i √ 3

48

So if D < 0 write: λ1,2 = −b ± i √ −D 2a Solve the equation λ2 + 2λ + 10 = 0: λ1,2 = −2 ± √4 − 4 × 10 2 = −2 ± √−36 2 = −2 ± 6i 2 In other words, λ1 = −1 + 3i and λ2 = −1 − 3i. A complex number z is written as z = α + iβ, where α is called the real part and iβ is called the imaginary part.

These two solutions are complex conjugates: z1 = a + ib and z2 = a − ib

49

Argand diagram: complex number as a vector:

complex number z

The complex plane

real part imaginary part

Addition of two complex numbers: adding their real parts, and add their imaginary parts. With z1 = 3 + 10i and z2 = −5 + 4i: z1 + z2 = (3 + 10i) + (−5 + 4i) = 3 − 5 + 10i + 4i = −2 + 14i .

50

Multiplication works like (a + bx)(c + dx): z1 × z2 = (3 + 10i)(−5 + 4i) = 3(−5) + 3 × 4i + 10i(−5) + 10i4i = −15 + 12i − 50i + 40i2 = −15 − 38i − 40 = −55 − 38i . Note: (a + ib)(a − ib) = a2 + b2 If z = a+ib, its modulus |z| = √ a2 + b2 (magnitude, length vector). Hence z¯ z = |z|2. (Used for division).

51

Mandelbrot set: zi = z2

i−1 + c, where c = a + bi is a point in the

Argand diagram, and z0 = 0. Black points remain bounded, colored points keep growing. The color indicates the number of iterations i = 1, 2, . . . , n required to reach a size of zn. Start with c = 0.5: 0.5, 0.52 + 0.5 = 0.75, 0.752 + 0.5, . . .

52

Linear ODEs dx/dt = ax + by dy/dt = cx + dy

with λ1,2 = tr ±

√ D 2

and

(a − λi)x + by = 0 cx + (d − λi)y = 0 λ1,2 = tr ± i √ −D 2

• r

λ1,2 = α ± iβ v1 = k −b a − λ1

• = k
• −b

a − (α + iβ)

• = k

−b a − α

• − ik

β

• = kw1 − ikw2

where w1 = −b

a−α

• and w2 =

β

• correspond to the real and imag-

inary parts of the eigenvector.

Similarly v2 = k −b a − λ2

• = k
• −b

a − (α − iβ)

• = kw1 + ikw2

General solution: x(t) y(t)

• = C1(w1 − iw2)e(α+iβ)t + C2(w1 + iw1)e(α−iβ)t

where the constants k are absorbed into C1 and C2. Euler’s formula: eix = cos x + i sin x

• r

e−ix = cos x − i sin x hence eα+iβ = eαeiβ = eα(cos β + i sin β)

54

Hence from x(t) y(t)

• = C1(w1 − iw2)e(α+iβ)t + C2(w1 + iw1)e(α−iβ)t

we obtain

x(t) y(t)

• = C1(w1 − iw2)eαt(cos βt + i sin βt)

+ C2(w1 + iw2)eαt(cos βt − i sin βt) = eαt[C1(w1 − iw2)(cos βt + i sin βt) + C2(w1 + iw2)(cos βt − i sin βt)] . which dies out whenever α = tr/2 < 0.

55

Initial condition where t = 0, eαt = 1, cos βt = 1 and i sin βt = 0, x(0) y(0)

• = C1(w1 − iw2) + C2(w1 + iw2)

= w1(C1 + C2) + iw2(C2 − C1) ,

• r

x(0) = −b(C1 + C2)

and

y(0) = (a − α)(C1 + C2) + iβ(C2 − C1) from which we solve the complex pair C1 and C2. Note that C1 + C2 should be real, whereas C2 − C1 should be an imaginary number.

56

Lotka-Volterra model

(a) R dN/dt 0.5 1 1 (b) t hR, hN 10 20

• 0.05

0.05 (c) t x(t), y(t) 10 20

• 0.05

0.05

dR dt = aR − bR2 − cRN , dN dt = dRN − eN With a = b = c = d = 1, e = 0.5, ¯ R = 0.5 and ¯ N = 0.5, and hR = 0.05 and hN = 0

57

dR dt = aR−bR2−cRN , dN dt = dRN−eN , with ( ¯ R, ¯ N) = e d, da − eb dc

• J =
• −be

d

−ce

d da−eb c

• =

−b ¯ R −c ¯ R d ¯ N

• .

For a = b = c = d = 1 and e = 0.5, ¯ R = 0.5 and ¯ N = 0.5, and J = −0.5 −0.5 0.5

• with

D = −0.75 implying that λ1,2 = tr ± i √ −D 2

• r

λ1,2 = −0.5 ± i √ 0.75 2 = −0.25 ± i 0.43 . Hence α = −0.25 and β = 0.43, the nontrivial state is stable, has a return time of −1/α = 4, and a wave length proportional to 1/β.

v1 =

• 0.5

−0.25 − i0.43)

• and

v2 =

• 0.5

−0.25 + i0.43)

• .

(

x(t) y(t)) = e−0.25t [C1v1(cos 0.43t + i sin 0.43t) + C2v2(cos 0.43t − i sin 0.43t)]

x(t) = e−0.25t 0.5[(C1 + C2) cos 0.43t + (C1 − C2)i sin 0.43t] y(t) = . . .

Using t = 0, e−0.25t = 1, cos 0.43t = 1, x(0) = 0.05, y(0) = 0, and sin 0.43t = 0:

x(t) = e−0.25t[0.05 cos 0.433t − 0.0289 sin 0.433t] y(t) = e−0.25t0.0577 sin 0.433t

59