STABILITY Phase Portraits and Local Stability

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 pendulum of 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 phase portrait. 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 y˙ g(x, y) = = dx x˙ f(x, y)

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

dy dy x˙ = 0 → inﬁnite;y ˙ = 0 → = 0. dx dx

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

Classiﬁcation of equilibrium points

µ ¶ µ ¶ µ ¶ x˙ x a a = A ; A = 11 12 y˙ y a21 a22

Find eigenvalues λ1 and λ2: µ ¶ 1 0 |A − λI| = 0,I = 0 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), for all 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), then solutions 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 straight lines and may even reduce to single points.

14 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 µ ¶ µ ¶ µ ¶ µ ¶ x˙ 0 1 x x = = A y˙ −ω2 0 y y

(i) The equilibrium point is (0,0) ¯ ¯ ¯ ¯ ¯ −λ 1 ¯ (ii) |A − λI| = ¯ ¯ = λ2 + ω2 ¯ −ω2 −λ ¯

→ λ = ±iω (purely imaginary)

→ (0,0) is a centre

dy y˙ dy (iii) dx = x˙ y˙ = 0 on x = 0 ⇒ dx = 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

dy x˙ = 0 on y = 0 ⇒ dx inﬁnite on x-axis y˙ > 0 for x < 0 ⇒ y increases as t increases

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

dy −ω2x R 2 R (iv) dx = y → y dy = −ω x dx

1 2 1 2 2 ⇒ 2 y + 2 ω x = constant (∗∗)

⇒ y2 + ω2x2 = C

(**) is a ﬁrst integral of (*).

2 2 2 x2 y2 The curves ω x = y = C or C/ω2 + 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)

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

The linear approximation is therefore

µ ¶ µ ¶ µ ¶ X˙ 0 1 x = . Y˙ ω2 0 y

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

µ ¶ ∂f/∂x ∂f/∂y J = . ∂g/∂x ∂g/∂y

The associated linear system near (x, y) is, using (6) µ ¶ µ ¶ X˙ X = A , (7) Y˙ Y

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

In the example f(x, y) = y, g(x, y) = −ω2 sin x so that µ ¶ µ ¶ ∂f/∂x ∂f/∂y 0 1 J = = ∂g/∂x ∂g/∂y −ω2 cos x 0

and µ ¶ µ ¶ 0 1 0 1 A = J = = (π,0) −ω2 cos π 0 ω2 0 as before. Why is this linear approximation useful?

An important theorem, due to Poincar´e,states that provided the eigenvalues of A are either non-zero, real and distinct or have non-zero real parts the equilibrium point of the non-linear system (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 considering the linear approximation (7).

If the equilibrium point is asymptotically stable for the linear system then it is asymptotically 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

17 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 µ ¶ µ ¶ ∂f/∂x ∂f/∂y 3 − 2x − 2y −2x J = = ∂g/∂x ∂g/∂y y x − 1

At (0,0) µ ¶ 3 0 A = J = (0,0) 0 −1 and the eigenvalues are 3 and -1 so that (0,0) is an unstable saddle for the linear approximation and hence for the non-linear system (by Poincar´e). At (3,0) µ ¶ −3 −6 A = J = (3,0) 0 2 and the eigenvalues are -3 and 2 so that (3,0) is an unstable saddle for linear approximation and hence for the non-linear system (by Poincar´e). At (1,1) µ ¶ −1 −2 A = J = (1,1) 1 0 and the eigenvalues are given by ¯ ¯ ¯ ¯ ¯ −1 − λ −2 ¯ 0 = |A − λI| = ¯ ¯ = λ(1 + λ) + 2 = λ2 + λ + 2 ¯ 1 −λ ¯

18 1 √ ⇒ λ = (−1 ± 7i) 2 ⇒ (1,1) is a stable focus for the linear approximation and hence for the non-linear system (by Poincar´e)

dy y˙ y(x−1) (iii) dx = x˙ = x(3−x−2y) y˙ = 0 ⇒ either y = 0x ˙ = 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)

19 20 (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 µ ¶ 0 1 J = −ω2 cos x 0 and µ ¶ µ ¶ 0 1 0 1 A = J = = (nπ,0) −ω2 cos nπ 0 −ω2(−1)n 0 so that the eigenvalues are given by ¯ ¯ ¯ ¯ ¯ −λ 1 ¯ 0 = |A − λI| = ¯ ¯ = λ2 + (−1)nω2 ¯ −ω2(−1)n −λ ¯ n even: λ = ±iω i.e. purely imaginary ⇒ centre (linear approximation but care needed for non-linear system) n odd: λ = ±ω ⇒ unstable saddle (non-linear system)

dy y˙ (iii) dx = x˙ y˙ = 0 on x = nπ

dy ⇒ dx = 0 on x = nπ, y 6= 0

dy x˙ = 0 on y = 0 ⇒ dx inﬁnite on x-axis except at x = nπ. 2 dy Also, since ω sin x is bounded dx → 0 as |y| → ∞ for every x.

(iv) Since replacing x by x + 2π gives the same equations the portrait repeats itself every interval of length 2π. Consider −π ≤ x ≤ π. Also dy −ω2 sin x = dx y

which is separable and gives Z Z y dy = −ω2 sin x dx

y2 = ω2 cos x + C (∗) 2

21 Phase portrait is therefore symmetric about x- and y- axes (x → −x or y → −y does not alter the equation) and q 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 equationx ¨ = f(x) similar approach gives 1 y2 = R 2 f(x)dx + C]

22 simple pendulum phase portrait - physical interpretation

Each path corresponds to a different value of C e.g. the path passing through (π, 0) corresponds to C = ω2 and crosses the y-axis when y = ±2ω. The other closed paths cross the x-axis at 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 reverses direction 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 revolutions to occur. For values of C > ω2, y = θ˙ = 0 is impossible, the paths are not closed and the solutions are not periodic. When θ = 0, |θ˙| > 2ω and the initial speed is large enough for complete revolutions 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 just sufficient 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).

23 For this range of C the paths also cross the y-axis since when x = 0 q 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 paths cross the x- and y-axes the points x = nπ for n even are centres for the non-linear system.

24 Liapunov’s direct method for stability

For systems in which linearisation fails to give useful information, an alternative approach is 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 either decreases 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) and the rate of change of V along this curve is Ã ! dV ∂V ∂V ∂V ∂V V˙ = = x˙ + y˙ = f + g dt ∂x ∂y ∂x ∂y

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

25 x¨ = −ω2x Typical mass-spring system T = kx when spring has extension x

mx¨ = −kx ⇒ x¨ + ω2x = 0 (ω2 = k/m) total energy = kinetic energy + elastic energy 1 1 = mx˙ 2 + kx2 2 2 1 1 = m( y2 + ω2x2) 2 2

Motivated by above working take V (x, y) = y2 + ω2x2 then ∂V ∂V V˙ = x˙ + y˙ = 2ω2x(y) + 2y(−ω2x) = 0 ∂x ∂y

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 V˙ = x˙ + y˙ = 2ω2x(y) + 2y(−cy − ω2x) = −2cy2 ≤ 0 ∂x ∂y

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 detect this.

Exercise For the simple pendulum use 1 v(x, y) = y2 + ω2(1 − cos x) 2 to show that (0,0) is stable.

26 [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 ( ) 2b c F (x, y) = a x2 + xy + y2 a a ( Ã ! ) b c b2 = a (x + y)2 + − y2 a a a2 Ã ! Ã ! b 2 ac − b2 = a x + y + y2 a a

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

Ã ! b ac − b2 F − y, y = y2 > 0 for every y 6= 0 a a ⇒ 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

27 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 V˙ = x˙ + y˙ ∂x ∂y = 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 1 V = x2 + y2 is a strong Liapunov function 2

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

Note 1. Linear approximation is µ ¶ µ ¶ µ ¶ x˙ 0 0 x = y˙ ) −1 y

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

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

µ ¶ µ ¶ 2x −2y 0 0 J = ⇒ J = = A −2y −2x (0,0) 0 0

¯ ¯ ¯ ¯ ¯ −λ 0 ¯ |A − λI| = ¯ ¯ = 0 ⇒ λ2 = 0 i.e.λ = 0, 0 ¯ 0 −λ ¯

So, both eigenvalues are zero. Linearisation fails.

Let us take the Liapunov function

29 V (x, y) = αxy2 − x3

V (0, 0) = 0

dV ³ ´ V˙ = = αy2 − 3x2 (x2 − y2) − 4αx2y2 = 3(1 − α)x2y2 − αy4 − 3x4 dt 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 circle around the origin.

Now consider

V˙ = −y4 − 3x4

This is negative everywhere. So, in every neighbourhood of the origin there is at least one point where 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.

30 If we plot a graph of x versus µ, we can see that there is a region of stable equilibrium and there is a region of unstable equilibrium which meet at x = 0, µ = 0. the diagram is called bifurcation diagram and the point x = 0, µ = 0 is the bifurcation point. The bifurcation associated with x˙ = 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-node bifurcation at λ = 0, y = 0 and λ = −1, y = 1. Now, at the equilibrium points dy˙ √ = −2λ − 2y = ∓2 λ2 + λ dy The equilibrium point with the larger value of y is stable, whilst the other one is unstable. The bifurcation diagram is shown below.

31 The bifurcation diagram fory ˙ = 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 dx˙ −µ at x = µ = = µ − 2x = ∂x dx µ at x = 0

∂f At x = µ, J = ∂x = A = −µ which is stable for µ > 0 and unstable for µ < 0. ∂f At x = 0, A = J = ∂x = µ which is unstable for µ > 0 and stable for µ < 0.

32 This is transcritical bifurcation. At the bifurcation point, the two equilibrium solutions pass through each other and exchange stabilities, so this bifurcation is referred to as an exchange of stabilities.

Pitchfork Bifurcation

Consider

x˙ = µx − x3 √ f(x) = x(µ − x2) = 0 ⇒ x = 0, x = ± µ ∂f J = = µ − 3x2 ∂x

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.

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

Structurally unstable

Ifx ˙ = f(x) and x1 the equilibrium point, we can write d(x − x ) dx 1 = = x˙ = f(x ) + (x − x )f 0(x ). dt dt 1 1 1 0 If x1 is the equilibrium solution, f(x1) = 0 and assuming that f (x1) 6= 0, we can write

0 x˙ = (x − x1)f (x1) for x << 1 0 or x˙ = xf (x1) 0 x = kef (x1)t for some constant k.

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

(x − x )f 00 (x ) x˙ ≈ 1 1 x << 1 2 Such systems where we have an equilibrium point (or points) for which f 0(x) = 0, we have structurally 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 equations, second order systems are structurally unstable. A slight perturbation (change) in the governing 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.

µ ¶ −2x 0 J = 0 −1

µ √ ¶ µ ¶ µ √ ¶ −2 µ 0 λ 0 −2 µ − λ 0 J √ = − = ( µ,0) 0 −1 0 λ 0 −1 − λ

µ √ ¶ 2 µ 0 J √ = saddle point (− µ,0) 0 −1

If µ = 0, we havex ˙ = −x2,y ˙ = −y and (0,0) is the equilibrium point. µ ¶ 0 0 J = (0,0) 0 −1

34 µ ¶ −λ 0 |A − λI| = 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 and disappear. Note that sincey ˙ = −y, we will have y = ke−t. All integral paths asymptote to the x-axis as t → ∞ where the dynamics are controlled byx ˙ = µ − x2.

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