MAS271 Methods for differential equations Dr. R. Jain

STABILITY

Phase portraits and local stability

We are interested in system of ordinary differential equations of the form

x = f(x, y), y = g(x, y), (1)

where f and g do not depend explicitly upon t i.e. the system is autonomous.

Example 1. Predator-prey model (foxes and rabbits)

x = ax− bx2 − cxy, y = −py + qxy, (2)

where a, b, c, p, q are positive constants. x(t) and y(t) are rabbit and fox population respectively.

Example 2. Simple Pendulum If θ is the angle of deviation of an undamped pendulumof length l whose bob has mass m, then equation of motion is:

mlθ = −mg sin θ

θ + ω2 sin θ = 0, (3)

where ω2 = g/l. Putting

x = θ, y = θ,

equation (3) may be written as

x = y, y = −ω2 sin x. (4)

1

The xy- plane is called the phase plane and the solution curves of (1) are x = x(t), y = y(t)the orbits, trajectories or paths. The phase plane together with the orbits is the phaseportrait.

The point (x, y) such that

f(x, y) = 0, g(x, y) = 0 (5)

is called an equilibrium point, critical point or stationary point.An isocline is a curve in the phase plane connecting points at which the gradients of the

orbits are the same. For (1)

dy

dx=

y

x=

g(x, y)

f(x, y)

and so an isocline is a curve along which f(x, y)= constant ×g(x, y). In practice the two mostimportant isoclines are f(x, y) = 0, g(x, y) = 0.

The usual procedure to construct a phase portrait is:

(1) find the equilibrium points

(2) determine the nature of each one

(3) draw the isoclines, especially

x = 0 → dy

dxinfinite; y = 0 → dy

dx= 0.

(4) use artistic imagination along with any additional information!

Classification of equilibrium points

(xy

)= A

(xy

); A =

(a11 a12

a21 a22

)

Find eigenvalues λ1 and λ2:

|A− λI| = 0, I =(

1 00 1

)

2

(1) λ1 < 0, λ2 < 0 (stable node)

3

(2) λ1 > 0, λ2 > 0 (unstable node)

4

(3) λ1 = λ2 = λ, A− λI 6= 0

(a) λ > 0 (unstable node)

5

(b) λ < 0 (stable node)

6

(4) λ1 = λ2 = λ, A− λI = 0

(a) λ > 0, unstable improper node

7

(b) λ < 0, stable improper node

8

(5) λ1 · λ2 < 0 saddle (always unstable)

9

(6) λ1,2 = ±iω centre (stable)

10

(7) λ1,2 = α± iω focus or spiral

(a) α > 0, unstable

11

(b) α < 0, stable

12

Here are some more examples of various types of equilibrium points.

13

Notes:

(1). If the real parts of both λ1 and λ2 are negative, then limt→∞(x(t), y(t)) = (0, 0), forall choices of constants, and (0,0) is called an asymptotically stable equilibrium.

(2). If λ1 and λ2 are conjugate and purely imaginary (λ1 = iq, λ2 = −iq, q 6= 0), thensolutions are periodic and (0,0) is a stable equilibrium.

(3). If either λ1 or λ2 has real part, then (0,0) is an unstable equilibrium.

(4). If either or both of λ1 and λ2 is zero, all solution trajectories are confined to straightlines and may even reduce to single points.

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Example Sketch the phase portrait of

x + ω2x = 0 (∗)where ω is a constant.

Solution Putting y = x the above equation (∗) is equivalent to

x = y, y = −ω2x

or (xy

)=

(0 1−ω2 0

) (xy

)= A

(xy

)

(i) The equilibrium point is (0,0)

(ii) |A− λI| =∣∣∣∣∣−λ 1−ω2 −λ

∣∣∣∣∣ = λ2 + ω2

→ λ = ±iω (purely imaginary)

→ (0,0) is a centre

(iii) dydx

= yx

y = 0 on x = 0 ⇒ dydx

= 0 on y-axis (y 6= 0).

x > 0 for y > 0 ⇒ x increases as t increases

x < 0 for y < 0 ⇒ x decreases as t increases

x = 0 on y = 0 ⇒ dydx

infinite on x-axis

y > 0 for x < 0 ⇒ y increases as t increases

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y < 0 for x > 0 ⇒ y decreases as t increases

(iv) dydx

= −ω2xy

→ ∫y dy = −ω2

∫x dx

⇒ 12y2 + 1

2ω2x2 = constant (∗∗)

⇒ y2 + ω2x2 = C

(**) is a first integral of (*).

The curves ω2x2 = y2 = C or x2

C/ω2 + y2

C= 1 are ellipses (ω 6= 1)

Notes

1. Each orbit corresponds to a different value of C.

2. Each orbit is closed ⇒ periodic solution

3. For physical systems (**) is an energy equation.

Linear approximation of non-linear system near the equilibrium point (x, y)

To obtain the linear approximation near (x, y) set

x = x + X(t), y = y + Y (t) (6)

and ignore the non-linear terms in f and g.

Example Find the linear approximation near (π, 0) of

x = y, y = −ω2 sin x

Solution Putting x = π + X, y = Y we get

X = Y, Y = −ω2 sin(π + X)

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Now sin(π + X) = sin π cos X + cos π sin X = − sin X ≈ −X for small X.

The linear approximation is therefore

(XY

)=

(0 1ω2 0

) (xy

).

Here we used first principles to find the linear approximation. It can also be found using theJacobian matrix

J =(

∂f/∂x ∂f/∂y∂g/∂x ∂g/∂y

).

The associated linear system near (x, y) is, using (6)

(XY

)= A

(XY

), (7)

where A = J evaluated at (x, y).

In the example f(x, y) = y, g(x, y) = −ω2 sin xso that

J =(

∂f/∂x ∂f/∂y∂g/∂x ∂g/∂y

)=

(0 1

−ω2 cos x 0

)

and

A = J(π,0) =(

0 1−ω2 cos π 0

)=

(0 1ω2 0

)

as before.Why is this linear approximation useful?

An important theorem, due to Poincare, states that provided the eigenvalues of A are eithernon-zero, real and distinct or have non-zero real parts the equilibrium point of the non-linearsystem (1) and the linear approximation (7) are of the same type.

This result enables us to deduce the nature of the equilibrium points of (1) by consideringthe linear approximation (7).

If the equilibrium point is asymptotically stable for the linear system then it is asymptoti-cally stable for the non-linear system.

In other cases more work is necessary.

λ1 = λ2 and real: system (1) can have either a node or a spiral

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λ1, λ2, purely imaginary: system (1) can have either a centre or a spiral.

Example Sketch the phase portrait of

(A) x = x(3− x− 2y), y = y(x− 1) (predator-prey)

(B) x = y, y = −ω2 sin x (simple pendulum)

Solution

A (i) the equilibrium points are given by

x(3− x− 2y) = 0 (a)

y(x− 1) = 0 (b)

(b) is satisfied by y = 0 or x = 1.

If y = 0 (a) ⇒ x(3− x) = 0 ⇒ x = 0 or x = 3

⇒ (0, 0) and (3, 0)

If x = 1 (a) ⇒ 2− 2y = 0 ⇒ y = 1 ⇒ (1, 1)

The equilibrium points are (0,0), (3,0) and (1,1).

(ii) Here f(x, y) = 3x− x2 − 2xy, g(x, y) = xy − y

so that

J =(

∂f/∂x ∂f/∂y∂g/∂x ∂g/∂y

)=

(3− 2x− 2y −2x

y x− 1

)

At (0,0)

A = J(0,0) =(

3 00 −1

)

and the eigenvalues are 3 and -1 so that (0,0) is an unstable saddle for the linear approximationand hence for the non-linear system (by Poincare).At (3,0)

A = J(3,0) =(−3 −6

0 2

)

and the eigenvalues are -3 and 2 so that (3,0) is an unstable saddle for linear approximationand hence for the non-linear system (by Poincare).At (1,1)

A = J(1,1) =(−1 −2

1 0

)

and the eigenvalues are given by

0 = |A− λI| =∣∣∣∣∣−1− λ −2

1 −λ

∣∣∣∣∣ = λ(1 + λ) + 2 = λ2 + λ + 2

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⇒ λ =1

2(−1±

√7i)

⇒ (1,1) is a stable focus for the linear approximation and hence for the non-linear system (byPoincare)

(iii) dydx

= yx

= y(x−1)x(3−x−2y)

y = 0 ⇒ either y = 0 x = x(3− x)

or x = 1, x = 2(1− y)

x = 0 ⇒ either x = 0, y = −y

or x = 3− 2y, y = 2y(1− y)

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(B) x = y, y = −ω2 sin x

(i) Equilibrium points are given by

y = 0, sin x = 0 ⇒ x = nπ, n ∈ Z, (n = 0,±1,±2, ...)

⇒ (nπ, 0)

(ii) Here

J =(

0 1−ω2 cos x 0

)

and

A = J(nπ,0) =(

0 1−ω2 cos nπ 0

)=

(0 1

−ω2(−1)n 0

)

so that the eigenvalues are given by

0 = |A− λI| =

∣∣∣∣∣−λ 1

−ω2(−1)n −λ

∣∣∣∣∣ = λ2 + (−1)nω2

n even: λ = ±iω i.e. purely imaginary ⇒ centre (linear approximation but care neededfor non-linear system)

n odd: λ = ±ω ⇒ unstable saddle (non-linear system)

(iii) dydx

= yx

y = 0 on x = nπ

⇒ dydx

= 0 on x = nπ, y 6= 0

x = 0 on y = 0 ⇒ dydx

infinite on x-axis except at x = nπ.

Also, since ω2 sin x is bounded dydx→ 0 as |y| → ∞ for every x.

(iv) Since replacing x by x + 2π gives the same equations the portrait repeats itself everyinterval of length 2π. Consider −π ≤ x ≤ π.Also

dy

dx=

−ω2 sin x

y

which is separable and gives∫

y dy = −ω2∫

sin x dx

y2

2= ω2 cos x + C (∗)

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Phase portrait is therefore symmetric about x- and y- axes (x → −x or y → −y does notalter the equation) and

y = ±√

2(ω2 cos x + C)

For real y, we require ω2 cos x + C ≥ 0 ⇒ C ≥ −ω2.Equation (∗) is the first integral of the equation of motion θ = −ω2 sin θ and expresses con-servation of mechanical energy. [For the equation x = f(x) similar approach gives 1

2y2 =∫

f(x)dx + C]

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simple pendulum phase portrait - physical interpretation

Each path corresponds to a different value of C e.g. the path passing through (π, 0) correspondsto C = ω2 and crosses the y-axis when y = ±2ω. The other closed paths cross the x-axisat x 6= nπ. When y = 0, ω2 cos x + C = 0 and since | cos x < 1| for this to be positive

−ω2 < C < ω2. For these solutions y = θ = 0 for some θ at which point the pendulum reversesdirection and periodic oscillations occur (closed paths ⇒ periodic solutions). These occur when

|y| = |θ| < 2ω at θ = 0 and the initial speed l|θ| is not great enough for complete revolutionsto occur.

For values of C > ω2, y = θ = 0 is impossible, the paths are not closed and the solutionsare not periodic. When θ = 0, |θ| > 2ω and the initial speed is large enough for completerevolutions to occur in finite time or ”whirling” motion. The path corresponding to C = ω2

separates these two types of motion and is called the separatrix. The initial speed is justsufficient to give complete revolutions but they take an infinite time to complete.Equilibrium points:(0,0) mass hanging at rest vertically below P (stable).(π, 0) mass vertically above P (unstable).

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For this range of C the paths also cross the y-axis since when x = 0

y = ±√

2(ω2 + C)

and ω2 + C > 0.

Since the phase portrait is symmetric about x- and y- axes and for −ω2 ≤ C ≤ ω2 the pathscross the x- and y-axes the points x = nπ for n even are centres for the non-linear system.

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Liapunov’s direct method for stability

For systems in which linearisation fails to give useful information, an alternative approachis to consider the non-linear system

x = f(x, y), y = g(x, y) (8)

and use Liapunov’s direct method.In essence this method seeks a scalar function of x and y (which can be regarded as a

measure of the ”energy” of system (1)) and then seeks to demonstrate that this function eitherdecreases as t →∞ indicating stability or it increases indicating instability.

Here we confine our attention to the case when the equilibrium point is the origin (0,0).Let V (x, y) be a scalar function defined and continuous with continuous partial derivatives

in some neighbourhood (nhd) of the origin 0 and such that V (0, 0) = 0.

Definition 1. V (x, y) is positive (negative) definite in a nhd N of O if

(i)V (0, 0) = 0

(ii)V (x, y) > 0 (< 0) for all (x, y) 6= (0, 0) in N

2. V (x, y) is positive (negative) semi-definite in N if

(i) V (0, 0) = 0

(ii)V (x, y) ≥ 0 (≤ 0) for all (x, y) 6= (0, 0) in N .

Examples: x2 + y2 is positive definite (pd).

−x2 − 2exy6 is negative definite (nd).

−x2 is negative semi-definite (nsd) since it is 0 at points (0, y)

(x− y)2 is positive semidefinite (psd) since it is zero at points where x = y.

y2(x2 − 1) is nsd on the strip |x| < 1 it is zero at points (x, 0).

On a solution curve x = x(t), y = y(t) of system (1) V (x, y) = V (x(t), y(t)) = V (t) andthe rate of change of V along this curve is

V

(=

dV

dt

)=

∂V

∂xx +

∂V

∂yy = f

∂V

∂x+ g

∂V

∂y

Theorem Let V (x, y) be p.d. If V is n.s.d (n.d) in N then (0,0) is stable (asymptoti-cally stable).

A p.d function V (x, y) such that V ≤ 0 (V < 0) is called a weak (strong) Liapunov function.

Example 1. x = y, y = −ω2x (simple harmonic motion)

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x = −ω2xTypical mass-spring systemT = kx when spring has extension x

mx = −kx

⇒ x + ω2x = 0 (ω2 = k/m)

total energy = kinetic energy + elastic energy

=1

2mx2 +

1

2kx2

= m(1

2y2 +

1

2ω2x2)

Motivated by above working take

V (x, y) = y2 + ω2x2

then

V =∂V

∂xx +

∂V

∂yy = 2ω2x(y) + 2y(−ω2x) = 0

Since V is p.d. it follows that V is a weak Liapunov function ⇒ (0,0) is stable.

Example 2: x = y, y = −cy − ω2x (c > 0) (damped oscillations)

Take V (x, y) = y2 + ω2x2

then

V =∂V

∂xx +

∂V

∂yy = 2ω2x(y) + 2y(−cy − ω2x) = −2cy2 ≤ 0

V is a weak Liapunov function.

⇒ (0, 0) is stable.

N.B. In fact (0,0) is asymptotically stable

Ex. Show this using the linearisation method but this Liapunov function does not detectthis.

Exercise For the simple pendulum use

v(x, y) =1

2y2 + ω2(1− cos x)

to show that (0,0) is stable.

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[y = x y = −ω2 sin x , V = ω2 sin x(y) + y(−ω2 sin x) = 0

Since V is p.d. ⇒ is a weak Liapunov function. (0,0) is stable. ]

In the above examples V has been associated with the energy of the mechanical system. How-ever finding the Liapunov function can be difficult. A quadratic expression can be helpful.

Example Find necessary and sufficient conditions on the constants a, b, c for

F (x, y) = ax2 + 2bxy + cy2

to be (a) p.d. ; (b) n.d.Solution First note that F (0, 0) = 0.

Need F (x, 0) > 0. But

F (x, 0) = ax2 ⇒ a > 0⇒ a > 0 is necessary.Also

F (x, y) = a

{x2 +

2b

axy +

c

ay2

}

= a

{(x +

b

ay)2 +

(c

a− b2

a2

)y2

}

= a

(x +

b

ay

)2

+

(ac− b2

a

)y2

Sufficient conditions for F (x, y) > 0 for every (x, y) 6= (0, 0) are

a > 0 and ac− b2 > 0.

These are also necessary since a > 0 necessary from above and

F

(− b

ay, y

)=

ac− b2

ay2 > 0 for every y 6= 0

⇒ ac− b2 > 0

⇒ Necessary and Sufficient Condition (NSC) for F (x, y) to be p.d. are

a > 0 and ac− b2 > 0,(or c > 0 ac− b2 > 0).

(b) Similar arguments shows that NSC for F (x, y) to be n.d. are

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a < 0 and ac > b2, (or c < 0 and ac− b2 > 0)

Example For the system

x = −2xy2 − x3, y = −y + x2y

show that (0,0) is asymptotically stable.

Solution For a Liapunov function try

V (x, y) = αx2 + y2

for suitable α. By previous example for V to be positive definite α > 0.Now

V =∂V

∂xx +

∂V

∂yy

= 2αx(−2xy2 − x3) + 2y(−y + x2y)

= −2αx4 + 2(1− 2α)x2y2 − 2y2

Taking α = 1/2 gives

V = −x4 − 2y2 < 0 for all (x, y) 6= (0, 0)

Since V (0, 0) = 0 it now follows that

V =1

2x2 + y2 is a strong Liapunov function

as V is positive definite and V is negative definite⇒ (0,0) is asymptotically stable.

Note 1. Linear approximation is(

xy

)=

(0 0) −1

) (xy

)

giving det A = 0 and previous theory does not apply.

2. (0, 0) is the only equilibrium point.Since V →∞ as |x| → ∞ all orbits whatever their starting point tend to (0,0) as t →∞⇒ (0, 0) is globally asymptotically stable.

Example: x = −3x3 − y, y = x5 − 2y3

V (x, y) = αx2m + βy2n

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Solution V (0, 0) = 0 and provided m, n are positive integers and α > 0, β > 0V (x, y) > 0 for all (x, y) 6= (0, 0) and so V (x, y) is positive definite. Now

V =∂V

∂xx +

∂V

∂yy

= 2mαx2m−1(−3x3 − y) + 2nβy2n−1(x5 − 2y3)

= −6mαx2m+2 + (2nβx5y2n−1 − 2mαx2m−1y) − 4nβy2n+2

The choice m = 3, n = 1 gives

V = −18αx8 + (2β − 6α)x5y − 4βy4

Taking α = 1, β = 3 gives

V = −18x8 − 12y4 < 0

Since V (0, 0) = 0 V is negative definite.

⇒ V is a strong Liapunov function.

⇒ is asymptotically stable.

Theorem If, in some region Σ ⊂ Rn that contains the origin, there exists a scalar func-tion V (x, y) such that V (0, 0) = 0 and V (x, y) is either positive definite or negative definite,and if, in every neighbourhood N of the origin with N ⊂ Σ, there exists at least one point(xa, ya) such that V (xa, ya) has the same sign as V (xa, ya) then the origin is unstable.

Example Consider the system

x = x2 − y2 y = −2xy

then (0,0) is the equilibrium point.

J =(

2x −2y−2y −2x

)⇒ J(0,0) =

(0 00 0

)= A

|A− λI| =∣∣∣∣∣−λ 00 −λ

∣∣∣∣∣ = 0 ⇒ λ2 = 0 i.e.λ = 0, 0

So, both eigenvalues are zero. Linearisation fails.

Let us take the Liapunov function

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V (x, y) = αxy2 − x3

V (0, 0) = 0

V =dV

dt=

(αy2 − 3x2

)(x2 − y2)− 4αx2y2 = 3(1− α)x2y2 − αy4 − 3x4

We can choose α = 1, so that V = −y4 − 3x4 which is negative definite.For α = 1, V = x(y2− x2) = 0 for x = 0 or y = ±x so changes sign six times on the unit circlearound the origin.

Now consider

V = −y4 − 3x4

This is negative everywhere. So, in every neighbourhood of the origin there is at least one pointwhere V has the same sign as V and hence the origin is unstable.

Bifurcation Theory

Consider the equation

x = µ− x2

where µ is a parameter.

For µ > 0, the equilibrium points are x = ±√µ.

For µ < 0, there are no real equilibrium points.

For µ = 0, x = −x2 and x = 0 which is unstable.

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If we plot a graph of x versus µ, we can see that there is a region of stable equilibrium and thereis a region of unstable equilibrium which meet at x = 0, µ = 0. the diagram is called bifurcationdiagram and the point x = 0, µ = 0 is the bifurcation point. The bifurcation associated withx = µ− x2 is called a saddle-node bifurcation.

Consider another example

y = λ− 2λy − y2

This has equilibrium points at y = −λ ± √λ2 + λ. So, there are no real equilibrium points

for −1 < λ < 0 and two equilibrium points otherwise. Therefore, we expect saddle-nodebifurcation at λ = 0, y = 0 and λ = −1, y = 1.Now, at the equilibrium points

dy

dy= −2λ− 2y = ∓2

√λ2 + λ

The equilibrium point with the larger value of y is stable, whilst the other one is unstable. Thebifurcation diagram is shown below.

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The bifurcation diagram for y = λ− 2λy − y2. Note there are two saddle-node bifurcation.

Transcritical Bifurcation

Consider

x = µx− x2

We have equilibrium points at x = 0 and x = µ.If µ 6= 0, there are two real equilibrium points.

∂f

∂x=

dx

dx= µ− 2x =

{−µ at x = µµ at x = 0

At x = µ, J = ∂f∂x

= A = −µ which is stable for µ > 0 and unstable for µ < 0.

At x = 0, A = J = ∂f∂x

= µ which is unstable for µ > 0 and stable for µ < 0.

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This is transcritical bifurcation. At the bifurcation point, the two equilibrium solutions passthrough each other and exchange stabilities, so this bifurcation is referred to as an exchange ofstabilities.

Pitchfork Bifurcation

Consider

x = µx− x3

f(x) = x(µ− x2) = 0 ⇒ x = 0, x = ±√µ

J =∂f

∂x= µ− 3x2

At x = 0, A = J0 = µ. Therefore, the eigenvalue is µ (stable if µ < 0 and unstable if µ > 0).At x =

√µ, A = J√µ = −2µ (stable if µ > 0 and unstable if µ < 0).

At x = −√µ, A = J−√µ = −2µ (stable if µ > 0 and unstable if µ < 0).

Similarly, x = µx + x3.

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Two new equilibrium points created at the bifurcation point are unstable.

Structurally unstable

If x = f(x) and x1 the equilibrium point, we can write

d(x− x1)

dt=

dx

dt= x = f(x1) + (x− x1)f

′(x1).

If x1 is the equilibrium solution, f(x1) = 0 and assuming that f ′(x1) 6= 0, we can write

x = (x− x1)f′(x1) for x << 1

or x = xf ′(x1)

x = kef ′(x1)t for some constant k.

If f ′(x1 < 0, x → 0 as t →∞ and the equilibrium point is stable.If f ′(x1) > 0, x →∞ as t →∞ and the equilibrium point is unstable.However, if f ′(x1) = 0, we need more terms in the Taylor expansion.

x ≈ (x− x1)f′′(x1)

2x << 1

Such systems where we have an equilibrium point (or points) for which f ′(x) = 0, we havestructurally unstable system.

Second order ordinary differential equation: Bifurcation

If the equilibrium point with real part of an eigenvalue is zero, then just like first order equa-tions, second order systems are structurally unstable. A slight perturbation (change) in thegoverning equations can change the qualitative nature of the phase portrait.

Consider the second order system

x = µ− x2 y = −y

If µ > 0, equilibrium points are (√

µ, 0) and (−√µ, 0).If µ < 0, there are no real equilibrium point.

J =(−2x 0

0 −1

)

J(√

µ,0) =(−2

õ 0

0 −1

)−

(λ 00 λ

)=

(−2√

µ− λ 00 −1− λ

)

J(−√µ,0) =(

2√

µ 00 −1

)saddle point

If µ = 0, we have x = −x2, y = −y and (0,0) is the equilibrium point.

J(0,0) =(

0 00 −1

)

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|A− λI| =(−λ 0

0 −1− λ

)

⇒ λ + λ2 = 0

⇒ λ(1 + λ) = 0

⇒ λ = 0,−1

For µ = 0, (0, 0) is a saddle-node point. At such point, a saddle and a node collide anddisappear.

Note that since y = −y, we will have y = ke−t. All integral paths asymptote to the x-axisas t →∞ where the dynamics are controlled by x = µ− x2.

In effect, we can ignore the dynamics in y-direction and concentrate on the dynamics on thex-axis. This shows why it was important to study bifurcation in first order systems, since theimportant dynamics of higher order system occur on a lower order.

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