Mathematical Modelling and Simulation

Department of Applied Mathematics

Adama Science and Technology University

May 19, 2021

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Bifurcation

Bifurcation is a sudden qualitative or topological change in the behaviour

of a dynamical system when there is a small smooth change to the system
parameter values (the bifurcation parameter). Generally, at a bifurcation,
the local stability properties of equilibria, periodic orbits or other invariant
sets changes. The point at which bifurcation occurs is known as the
bifurcation point.
Bifurcation means a structural change in the orbit of a system. The study
of bifurcation is concerned with how the structural change occurs when
the parameter(s) of the system are changing. The structural change and
the transition behaviour of a system are the central part of dynamical
evolution.

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Bifurcation

The dynamics of a continuous one dimensional system

dx
= f (x, µ)
dt
depends on the parameter µ ∈ R. It is often found that as µ crosses a
critical value, the properties of dynamical evolution, for example, its
stability, fixed points (equilibrium points), periodicity etc may change.
Moreover, a completely new orbit may be created.
Bifurcations associated with a single parameter are called codimension-1
bifurcations. On the other hand, bifurcations connected with two
parameters are known as codimension-2 bifurcations.

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Bifurcation Type

There are different types of bifurcations

1 Saddle node bifurcation
2 Trans critical bifurcations
3 Pitchfork bifurcation
4 Hope bifurcation

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Bifurcation

Consider a one-dimensional continuous system

dx
= f (x, µ) x, µ ∈ R
dt
which depends on the parameter µ, where f : R → R is a smooth function
of x and µ. Then all the equilibrium points of the system dx dt = f (x, µ)
depend on the parameter µ, and there may be changes in stabilities
and(or) existence of equilibrium points as µ varies. Thus, bifurcations of a
one-dimensional system are associated with the existence and(or)
stabilities of its equilibrium points. Such bifurcations are known as local
bifurcations as they occur in the neighbourhood of the equilibrium points.

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Saddle-node Bifurcation

Consider the one-dimensional system

dx
= x2 + µ x ∈R
dt
with µ as the system parameter. Then the equilibrium points of the
√
system are obtained by solving the equation x 2 + µ = 0. Hence x = −µ
√
and x = − −µ are the equilibrium points of the system provided that
µ ≤ 0. The the system will have one equilibrium point if µ = 0 and two
equilibrium points if µ < 0.

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Saddle-node Bifurcation

dx
Figure: Saddle-node bifurcation for the system dt = x2 + µ

As µ approaches 0 the two equilibrium points approach zero and at µ = 0

they collied and disappear.
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Saddle-node Bifurcation

Consider also the one-dimensional system

dx
= µ − x2 x ∈R
dt
√ √
where the equilibrium points of the system are x = µ and x = − µ.
The system has two equilibrium points if µ > 0 and one equilibrium point
for µ = 0, that is at µ = 0 the equilibrium points collide and disappear for
µ < 0. This is an example of a supercritical saddle-node bifurcation, since
the equilibrium points exist for values of µ above the bifurcation point
µ = 0. The saddle-node bifurcation in a one-dimensional system is usually
connected with appearance and disappearance (vice versa) of the fixed
points of the system as the parameter exceeds the critical value.

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Pitchfork Bifurcation

Pitchfork bifurcation in a one-dimensional system which appears when the

system has symmetry between left and right directions. In such a
system,the fixed points tend to appear and disappear in symmetrical pair.
For example,consider the one-dimensional system
dx
= µx − x 3 , x, µ ∈ R.
dt
The system invariant under the transformation xto − x. The equilibrium
√ √
points of the system are x = 0, x = µ and x = − µ.

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Pitchfork Bifurcation

Figure: phase diagram for µ < 0, for µ = 0 and µ > 0.

When µ = 0 the system has only one equilibrium point and when µ > 0
the system has three equilibrium points. The equilibrium point x ∗ = 0 is
√ √
unstabe and the other two equilibrium points x = µ and x = − µ are
stable. For µ < 0, the system has only one stable equilibrium point at the
origin.

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Pitchfork Bifurcation

The bifurcation for this vector field is called a supercritical pitchfork

bifurcation, in which a stable equilibrium bifurcates into two stable
equilibria. Consider the system
dx
= µx + x 3
dt
which is obtained by transforming (x, µ) into (−x, −µ). The system has
√
equilibrium points x ∗ = 0, and x ∗ = ± −µ provided tat µ ≤ 0 in which
x ∗ = 0 is stable and the other two are unstable. For µ > 0 the system has
only one equilibrium point x ∗ = 0 which is unstable.

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Pitchfork Bifurcation

Figure: Pitchfork bifurcation diagram for the one-dimensional system

dx 3
dt = µx − x

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Trans-critical Bifurcation

The trans-critical bifurcation is one such type of bifurcation in which the

stability characters of the fixed points are changed for varying values of
the parameters. Consider the one-dimensional system
dx
= µx − x 2 , x ∈R
dt
where µ is the system parameter. The equilibrium points of this system are
x ∗ = 0 and x = µ. When µ = 0 the system has only one equilibrium point
x ∗ = 0, which is non hyperbolic equilibrium points. For µ 6= 0 the system
two distinct equilibrium points x ∗ = 0 and x ∗ µ, in which the equilibrium
point x ∗ = 0 is unstable for µ > 0 and it is a stable equilibrium point for
µ < 0. The other equilibrium point x ∗ = µ is unstable if µ < 0 and stable
for µ > 0. The phase portrait for these three cases is shown in fig1.4.

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Trans-critical Bifurcation

dx
Figure: Phase portraits of the system dt = µx − x 2

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Trans-critical Bifurcation
Trans-critical bifurcation is an exchange of stabilities between the two
fixed points of the system. The bifurcation diagram is presented in the
following figure.

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Trans-critical Bifurcation

Consider a general one-dimensional system

dx
= f (x, µ), x, µ ∈ R
dt
where f : R → R is a continuously differentiable function. Let µ0 be the
bifurcation point and x0 be the corresponding equilibrium point of the
system, then x0 is non hyperbolic if
∂f
(x0 , µ0 ) = 0.
∂x

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Saddle-node Bifurcation

Theorem
Suppose the system
dx
= f (x, µ), x, µ ∈ R
dt
has an equilibrium point x = x0 at µ = µ0 satisfying the conditions
∂f
f (x0 , µ0 ) = 0, (x0 , µ0 ) = 0.
∂x
If
∂f ∂2f
(x0 , µ0 ) 6= 0, (x0 , µ0 ) 6= 0
∂x ∂x 2
then the system has a saddle-node bifurcation at (x0 , µ0 ).

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Trans-critical and pitchfork bifurcations

Theorem
Suppose the system
dx
= f (x, µ) x, µ ∈ R
dt
has an equilibrium point x = x0 at µ = µ0 satisfying the conditions
∂f
f (x0 , µ0 ) = 0 (x0 , µ0 ) = 0.
∂x
if
a)

∂f ∂2f ∂2f
(x0 , µ0 ) = 0, (x0 , µ0 ) 6= 0, and (x0 , µ0 ) 6= 0,
∂µ ∂x 2 ∂x∂µ

then the system has a trans-critical bifurcation at (x0 , µ0 ).

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Trans-critical and pitchfork bifurcations

Theorem
b)

∂f ∂2f
(x0 , µ0 ) = 0, (x0 , µ0 ) = 0,
∂µ ∂x 2
∂2f ∂3f
(x0 , µ0 ) 6= 0, and (x0 , µ0 ) 6= 0,
∂x∂µ ∂x 3

then the system has a pitchfork bifurcation at (x0 , µ0 ).

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Two-Dimensional Systems Bifurcation

Dynamics of two-dimension systems are vast and their qualitative

behaviours are determined by the nature of equilibrium points, periodic
orbits, limit cycles, etc. The parameters and their critical values for
bifurcations are highly associated with the evolution of system and have
physical significances. The critical parameter value is a deciding factor for
a system to undergo bifurcation.

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cont...
Consider a two-dimensional system
dx dy
= µ − x 2, = −y , µ ∈ R.
dt dt
√ √
Then the equilibrium points of the system are ( µ, 0) and (− µ, 0) for
µ > 0. When µ = 0 the two equilibrium points collide we have one
equilibrium point (0, 0) and for µ < 0 the two equilibrium points vanish.
To determine the stability of the equilibrium points we find the Jacobian
matrices at the equilibrium points. The the Jacobian matrix of the system
is
 
−2x 0
J(x, y ) = .
0 −1
√
Thus the Jcobian matrix at the equilibrium point ( µ, 0) for µ > 0 is
given by
 √ 
√ −2 µ 0
J( µ, 0) = .
0 −1
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cont...
√ √
Then the eigenvalues of the Jacobian matrix, J( µ, 0) are λ1 = −2 µ
and λ2 = −1 which are negative and therefore, the equilibrium point
√
( µ, 0) is locally asymptotically stable. Similarly the Jacobian matrix at
√
the equilibrium point (− µ, 0) is given by
 √ 
√ 2 µ 0
J(− µ, 0) = .
0 −1
√
and then the corresponding eigenvalues are λ1 = 2 µ and λ2 = −1. Thus
√
the equilibrium point (− µ, 0) is a saddle point.
When µ = 0 the system has only one equilibrium point (0, 0) and the
Jacobian matrix at the point (0, 0) is given by
 
0 0
J(0, 0) = .
0 −1

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cont...
Then the eigenvalues are λ1 = 0 and λ2 = −1 and hence the equilibrium
point (0, 0) is semi stable. As µ decreases to zero the two equilibrium
points approach each other and collide at µ = 0 and disappear for µ < 0.
For different values of the system parameter µ the phase portrait is given
in the figure below.

Figure: Phase portrait of the system for different values of the parameter µ

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Cont...

Saddle-node Bifurcation
Therefore, the system experiences a saddle node bifurcation at µ = 0 (the
bifurcation point) since as the value of µ crosses the origin the system has
a change in the stability of the equilibrium points.

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Trans-critical Bifurcation

Consider a two dimensional system

dx
= µx − x 2
dt
dy
= −y , µ ∈ R.
dt
Then the system has tow equilibrium points (0, 0) and (µ, 0) for µ 6= 0.
When the value of µ approaches zero the two equilibrium points approach
each other and collide at µ = 0. The Jacobian matrix of the system is give
by
 
µ − 2x 0
J(x, y ) = .
0 −1

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Trans-critical Bifurcation
Thus the eigenvalues of the system at the point (0, 0) are λ1 = µ and
λ2 = −1 and therefore, the equilibrium point is stable for µ < 0, a saddle
node for µ > 0 and for µ = 0 the equilibrium point is semi stable.
The eigenvalues of the system at the equilibrium point (µ, 0) are λ1 = −µ
and λ2 = −1 and hence the equilibrium point is stable for µ > 0 and a
saddle point for µ < 0. The phase diagram for different values of the
system is given in figure blow.

Figure: Phase portrait of the system for different values of the bifurcation
parameter µ.
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Cont ...

Trans-critical Bifurcation
The behaviour of the system changes when the parameter µ passes
through the origin. As the value of the parameter crosses the critical value
µ = 0 the saddle becomes a stable node and the stable node becomes a
saddle. That is, when µ passes through the origin from left, the equilibrium
point (0, 0) changes to a saddle from a stable node and the fixed point
(µ, 0) changes from a saddle to a stable node. This type of bifurcation is
known as trans-critical bifurcation and µ = 0 is the bifurcation point.

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Cont ...

There are two types of pitchfork bifurcations, supercritical and subcritical

pitchfork bifurcations. Consider a two-dimensional system
dx
= µx − x 3
dt
dy
= −y , µ ∈ R.
dt
√
For µ > 0, the system has three equilibrium points (0, 0) , ( µ, 0) and
√
(− µ, 0). The Jacobian matrix of the system is given by

µ − 3x 2 0
 
J(x, y ) = .
0 −1

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Cont ...
Then the eigenvalues at point (0, 0) and λ1 = µ and λ2 = −1 and hence
the equilibrium point (0, 0) is saddle point for µ > 0, stable for µ < 0 and
√
semi stable for µ = 0. Similarly, the eigenvalues at the points (± µ, 0)
√
are λ1 = −2µ and λ2 = −1. Hence the equilibrium points (± µ, 0) are
stable node for µ > 0 and a saddle point for µ < 0. The phase diagrams
for different values of the bifurcation parameter µ are presented in figure
below.

Figure: Phase portrait of the system for different parameter values

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Cont ...

Trans-critical Bifurcation
Therefore, as the values of parameter µ crosses the bifurcation point
µ = 0, the equilibrium point (0, 0) bifurcates into a saddle point from a
stable node. In this situation, it also gives birth to two stable nodes at the
√
points (± µ, 0). This type of bifurcation is known as supercritical
pitchfork bifurcation.

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Cont...
Now consider a parameter-dependent two-dimensional system
dx dy
= µx + x 3 , = −y , µ ∈ R.
dt dt
√
Hence the equilibrium points of the system are (0, 0) and (± −µ, 0) for
µ < 0.
Pitchfork Bifurcation
The eigenvalues of the Jacobian matrix at point (0, 0) are λ1 = µ and
λ2 = −1 and hence the point (0, 0) is a saddle point for µ > 0 and a
stable node for µ < 0. The eigenvalues of the Jacobian at the points
√
(± −µ, 0) are λ1 = −4µ and λ2 = −1. Therefore, the equilibrium points
√
(± −µ, 0) are saddle points for µ < 0. But for µ > 0 the equilibrium
points do not exist. Therefore, as the value of the parameter crosses the
bifurcation point µ = 0, the stable node at the origin coincides with the
saddles and then bifurcates into a saddle. This type of bifurcation is
known as subcritical pitchfork bifurcation.
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Sotomayor Theorem

Theorem
Consider the system
dx
= f (x, µ), (x, µ) ∈ Rn xR.
dt
Suppose that f (x0 , µ0 ) = 0 and that the n x n matrix A = Df (x0 , µ0 ) has a
simple eigenvalue λ = 0 with eigenvector v and that AT has an eigenvector
w corresponding to the eigenvalue λ = 0. Furthermore, suppose that A
has k eigenvalues with negative real parts and (n − k − 1) eigenvalues with
positive real parts and that the following conditions are satisfied,
w T fµ (x0 , µ0 ) 6= 0
w T [D 2 f (x0 , µ0 )(v , v )] 6= 0.

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Cont ...

Theorem
Then there is a smooth curve of equilibrium points of the system in Rn xR
passing through the point (x0 , µ0 ) and tangent to the hyper plane
Rn x{µ0 }. Depending on the sign of the expression in the above equation
there are no equilibrium points of the system near x0 when when
µ < µ0 (or µ > µ0 ) and there are two equilibrium points of the system near
x0 when µ > µ0 (or when µ < µ0 ). The two equilibrium points of the
system near x0 are hyperbolic and have stable manifolds of dimension k
and k + 1 respectively, that is the system experiences a saddle node
bifurcation at the equilibrium point x0 as the parameter µ passes through
the bifurcation value µ = µ0 .

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Cont ...

Theorem ...
If
w T fµ (x0 , µ0 ) = 0
w T [Dfµ (x0 , µ0 )v ] 6= 0
w T [Dfµ (x0 , µ0 )(v , v )] 6= 0,
then the system experiences a trans-critical bifurcation at the equilibrium
point x0 as the parameter µ varies through the bifurcation value µ = µ0 .

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Cont ...

Theorem ...
If
w T fµ (x0 , µ0 ) = 0
w T [Dfµ (x0 , µ0 )v ] 6= 0
w T [d 2 f (x0 , µ0 )(v , v )] = 0
w T [D 3 f (x0 , µ0 )(v , v , v )] 6= 0,
then the system experiences pitchfork bifurcation at the equilibrium point
x0 as the parameter µ varies through the bifurcation value µ = µ0 .

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Hopf Bifurcation

Hopf bifurcation is an interesting periodic bifurcation phenomenon for a

two-dimensional system where the eigenvalues are complex. This type of
bifurcating phenomenon in two-dimensional or higher dimensional systems
was studied by the German Scientist Eberhard Hopf 1902 - 1983. This
type of bifurcation was also recognized by Henri Poincare and later by
A.D. Andronov in 1930.
Hopf bifurcation occurs when a stable equilibrium point losses its stability
and gives birth to a limit cycle and vice versa. There are two types of Hopf
bifurcations,
Supercritical: when stable limit cycles are created for an unstable
equilibrium point, then the bifurcation is called a supercritical Hopf
bifurcation,
Subcritical: when an unstable limit cycle is created for a stable
equilibrium point, then the bifurcation is called a subcritical Hopf
bifurcation.

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Cont ...

Consider a two-dimensional non linear system

dx
= µx − y − x(x 2 + y 2 )
dt
dx
= x + µy − y (x 2 + y 2 ), µ∈R
dt
Then the system has the unique equilibrium point (0, 0). In the polar
coordinate x = r cos θ, y = r sin θ, the system can be written as
dr
= µr − r 3
dt
dθ
= 1
dt
which are decoupled system of equations and easy to analyze. The phase
portraits of the system for µ > 0 and µ < 0 are shown in figure below.

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Cont ...

Figure: Phase portraits of the system for µ = −0.5 and µ = 0.5 respectively

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Cont ...

To find the eigenvalues of the system, we evaluate the Jacobian matrix of

the system at the equilibrium point (0, 0). The Jacobian matrix of the
system is given by
 
µ −1
J(0, 0) = .
1 µ

Thus, the eigenvalues of the system are λ1,2 = µ ± i. Hence the origin is a
stable spiral when µ < 0 and an unstable spiral when µ > 0. Therefore,
the eigenvalues cross the imaginary axis from left to right as the parameter
changes from negative to positive values and hence a supercritical Hopf
bifurcation occurs when a stable spiral changes into an unstable spiral
surrounded by a limit cycle.

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Cont ...

Consider a two-dimensional system represented by

dx
= µx − y + x(x 2 + y 2 ) − x(x 2 + y 2 )2
dt
dx
= x + µy + y (x 2 + y 2 ) − y (x 2 + y 2 )2 , µ∈R
dt
In the polar coordinate, the system can be transformed in to the form
dr
= µr + r 3 − r 5
dt
dθ
= 1.
dt
The system has unique equilibrium point (0, 0) and the phase portraite for
µ > 0 and for µ < 0 is shown in figure below

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Cont ...

Figure: Phase portraits of the system for µ = −0.1 (a) and for µ + 0.5 (b)

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Cont ...
It is clear that when µ > 0, the equilibrium point (0, 0) (or r = 0) is a
stable spiral and all trajectories are attracted to it in anti-clockwise
direction, and for µ < 0, it is an unstable spiral. The diagram exhibits that
the system has two limits cycles when µ < 0, one of which is stable and
other is unstable. For µ > 0, it has only a stable limit cycle. All these
cycles can be determined from the equation µr + r 2 − r 5 = 0. For µ < 0,
the system has two limit cycles at
√
2 1 ± 1 + 4µ
r = ,
2
and for µ > 0, the unique limit cycle occurs at
√
2 1 + 1 + 4µ
r = .
2
√
1+ 1+4µ
One can clearly see that when µ < 0, the limit cycle at r 2 = 2 is
√
1− 1+4µ
stable while the limit cycle at r2 = 2 is unstable. For µ > 0, the
√
1+ 1+4µ
limit cycle at r2 = 2 is stable.
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Cont ...

Theorem
Let (x0 , y0 ) be an equilibrium point of the planar autonomous system

dr
= f (x, y , µ)
dt
dθ
= g (x, y , µ)
dt
which depends on some parameter µ ∈ R and let J be the Jacobian matrix
of the system evalueted at the equilibrium point (x0 , y0 ) has purely
imaginary eigenvalues λ1 (µ) = iω and λ2 (µ) = −iω, ω 6= 0 at µ = µ0 . If
d 1
i) dµ (Reλ(µ)) = 2 > 0, at µ = 0,
ii) (fµx + gµy ) = 1 6= 0, and,
iii) a = − µ8 6= 0 for µ 6= 0

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Cont ...

theorem ...
where the constant is given by
1 1
a = (fxxx + gxxy + fxyy + gyyy ) + (fxy (fxx + fyy )
16 16ω
−gxy (gxx + gyy ) − fxx gxx + fyy gyy )

evaluated at the equilibrium point (x0 , y0 ) then a periodic solution

bifurcates from the equilibrium point (x0 , y0 ) into µ < µ0 if
a(fµx + gµy ) > 0 or into µ > µ0 if a(fµx + gµy ) < 0.
Also, the equilibrium point is stable for µ > µ0 (respectively µ < µ0 ) and
unstable for µ < µ0 (respectively µ > µ0 ) if (fµx + gµy ) < 0 (respectively
> 0).
In both the cases, the periodic solution is stable (respectively unstable) if
the equilibrium point is unstable (respectively stable) on the side of
µ = µ0 for which the periodic solutions exist.
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Limit Cycle

Definition
Consider a non-autonomous system
dx
= P(x, y ),
dt
dy
= Q(x, y ).
dt
A closed path C of the system which is approached spirally from either the
inside or outside by a non-closed path C1 of the system either as t → ∞ or
t → −∞ is called a limit cycle of the system.

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Limit Cycle Example

Example
Consider the system
dx
= y + x(1 − x 2 − y 2 )
dt
dy
= −x + y (1 − x 2 − y 2 ).
dt
Using the polar coordinate system x = r cos θ and y = r sin θ we have
dx dy dr
x +y = r
dt dt dt
dy dx 2 dθ
x −y = r
dt dt dt

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Cont ...

Example
Thus
dx dy
x +y = (1 − x 2 − y 2 )(x 2 + y 2 ) = r 2 (1 − r 2 )
dt dt
dy dx
x −y = x2 + y2 = r2
dt dt
Therefore,in polar coordinate system the system will become
dr
= r (1 − r 2 )
dt
dθ
= −1.
dt

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Cont ...

Example
Solving the system we get

θ(t) = −t + t0

for some arbitrary constant t0 . Solving the first equation we get

c0 e 2t
r2 =
1 + c0 e 2t
or
1
r=√
1 + ce −2t
1
where c = c0 .

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Cont ...

Example
Thus, the solution of the system is
1
r (t) = √
1 + ce −2t
θ(t) = −t + t0

where c and t0 are arbitrary constants. If t0 = 0 the θ(t) = −t. Hence

cos t
x(t) = √
1 + ce −2t
sin t
y (t) = − √
1 + ce −2t

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Cont ...

Example
1 If c = 0 then the path defined by the system is the circle x 2 + y 2 = 1
described in the clockwise direction.
2 If c 6= 0 then the path defined by the system
cos t sin t
x(t) = √ , y (t) = − √
1 + ce −2t 1 + ce −2t
are not closed paths but rather paths having a spiral behavior.
If c > 0 the paths are spiral lying inside the circle x 2 + y 2 = 1. As
t → ∞ the paths approach this circle while as t → −∞ they approach
the critical point (0, 0) of the system.
If c < 0 the paths lie outside of the circle x 2 + y 2p
= 1. These outer
paths approach the circle as t → ±∞. As t → ln |c| both |x| and |y |
becomes infinite.

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Cont ...

Theorem (Bendixson’s non-existence theorem)

Let D be a domain in the xy-plane. Consider the autonomous system
dx
= P(x, y )
dt
dy
= Q(x, y ).
dt
where P and Q are functions with continuous first order partial derivatives
in D. Suppose that
∂P(x, y ) ∂Q(x, y )
+
∂x ∂y
has the same sign throughout D. Then the system has no closed path in D.

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Proof
Proof.
Let C be a closed curve in D and R be the region bounded by C. By
Green’s theorem
Z Z Z
∂P ∂Q
(P(x, y )dy − Q(x, y )dx) = ( + )ds
C R ∂x ∂y

where the integral is taken in the positive sense. Now assume that C is
closed path of the system and let x = f(t) and y = g(t) be arbitrary
solution of the system defining C parametrically and let T be the period of
this solution. Then
df (t)
= P(f (t), g (t))
dt
dg (t)
= Q(f (t), g (t))
dt
along C and
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Cont ...

Proof.
Z Z T
dg (t) df (t)
(Pdy − Qdx) = (P(f (t), g (t)) − Q(f (t), g (t) )dt
C 0 dt dt
= P(f (t), g (t))Q(f (t), g (t)) − Q(f (t), g (t))P(f (t), g (t)
= 0.

Thus Z Z
∂P ∂Q
( + )ds = 0.
R ∂x ∂y
∂Q
But the double integral can be zero only if ∂P
∂x + ∂y changes sign which is
a contradiction. Thus C is not a closed path.
Exercise:

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Exercise!!

Determine whether the following system has a limit cycle or not.

1

dx
= x + 7y 2 + 2x 3
dt
dy
= −x + 3y + yx 2
dt
2

dx
= x − xy 2 + y 3
dt
dy
= 3y − yx 2 + x 3
dt

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Theorem

Theorem (Poincare-Bendixson Theorem)

Consider a non-linear autonomous system
dx
= P(x, y )
dt
dy
= Q(x, y ).
dt
where P(x, y) and Q(x, y) are continuous and have continuous first order
partial derivatives. Let R be a bounded region of the phase plane together
with its boundary and assume that R does not contain any critical point of
the system. If C = (x(t) , y(t)) is a path of the system that lies in R for
some t0 and remains in R for t ≥ t0 then C is either a closed path or it
spirals towards a closed path as t → ∞. Thus in either case the system
has a closed path in R.

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Exercise!!
1 Consider the Van der Pol equation
d 2x dx
2
+ µ(x 2 − 1) +x =0
dt dt
where µ is assumed to be a positive constant. Show that the DE has
unique closed solution that is approached spirally by every other path.
2 Consider the autonomous system
dx
= 4x − 4y − x(x 2 + y 2 )
dt
dy
= 4x + 4y − y (x 2 + y 2 ).
dt

a) Transform the system to polar coordinate form.

b) Using the Poincare-Bendixson theorem show that there exists a limit
cycle between the circles x 2 + y 2 = 14 and x 2 + y 2 = 16.
c) Find the explicit solutions (x(t), y(t)) of the original system.
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