Untangling Complex Systems, page 19
∂ Y
[ ]
ss
ss
In the system [3.171], the second and higher order terms of the Taylor expansion are neglected,
because the perturbations are assumed to be small. The terms f ([ X ] , Y
[ ] ) and g([ X ] , Y[ ] )
ss
ss
ss
ss
are null. The derivatives are estimated in correspondence of the stationary points, and they are con-
stant. Therefore, the system [3.171] reduces to a system of linear differential equations. Solutions of
these linear equations are exponentials of the form x
it
=ν λ
it
i
1, e
, y =ν
λ
i
2, e
, and any linear combina-
tion of them. This result is evident if we consider, at first, two linear differential equations that are
uncoupled, like:
dx = λ1 x
dt
[3.172]
dy = λ2 y
dt
They are uncoupled because there is no y in the x-equation and vice-versa. In this simple situation,
the two differential equations can be solved separately. The solutions are x x e t
=
1
λ
2
λ
0
and y y e t
= 0
.
The x- and y- axes represent the directions of the trajectories for the x and y variables, respectively, as time goes to infinity ( t → ±∞). In other words, the x- and y- axes contain the straight-line trajectories: a trajectory starting from one point of the two axes, stays on that axis forever. For the general
case, we want to find the analog of these straight-line trajectories.
Introducing the solutions x
it
=ν λ
it
i e
,1
and y =ν
λ
i,2 e
into equation [3.171] and dividing by the non-
zero scalar factor e iλ t, we achieve
f∂
f
∂
ν
∂
[ X ]
∂ Y
[ ] ν i,1
λ
i,1
ss
ss
i
[3.173]
ν
=
i,2 g
∂
∂ ν i
g
,2
∂
Y
[ X ]
∂
ss
[ ] ss
In matrix form, equation [3.173] becomes
λν ii = ν
J i [3.174]
where J is the Jacobian matrix whose terms are the derivatives calculated at the fixed point; ν i is the 2 × 1 vector whose terms are ν i,1 and ν i,2. This vector represents the direction of the phase space,
which has trajectories with straight-lines, and is called the eigenvector of J. Linear equations such
as [3.174] have nontrivial solutions (i.e., solutions different from that with ν i,1 and ν i,2 both null)
78
Untangling Complex Systems
only when λ is an eigenvalue of the matrix J. The eigenvalues of J can be determined by solving the characteristic equation
f∂
f
λ
∂
−
∂
[ X ]
∂ Y
[ ]
det ( J λ I ) = det
ss
ss
−
= 0
[3.175]
g
∂
∂ g
λ
∂
−
Y
[ X ]
∂
ss
[ ]
ss
wherein I is the 2 × 2 identity matrix. If we expand the determinant, we have
λ2 − λ tr ( J ) + det ( J ) = 0 [3.176]
where tr( J) and det( J) are the trace (that is the sum of its diagonal elements) and the determinant of the Jacobian, 21 respectively. Equation [3.176] is a quadratic equation in λ having two solutions that are functions of the elements of the Jacobian:
2
tr J
( tr( J )) −4
( )
det ( J )
λ1 =
+
[3.177]
2
2
2
tr J
( tr( J )) −4
( )
det( J )
λ2 =
−
[3.178]
2
2
For the two eigenvalues λ and , we will have two eigenvectors, ν
1
λ 2
1 and ν2, and the general solutions
will be of the type
x( t ) = c
1
λ
2
λ
1ν1, e t
1
+ c 2ν2, e t
1
[3.179]
y( t ) = c
1
λ
2
λ
1ν1,2 e t + c 2ν 2,2 e t
wherein the values of the parameters c and depend on the initial conditions.22 The eigenvalues 1
c 2
λ 1
and λ can be either real or pure imaginary or complex numbers. The eigenvalues are related with
2
the trace and the determinant of J through the following equations:
λ1 + λ = tr ( J )
2
[3.180]
λ λ
1 2 = det ( J )
All the possible combinations of solutions for the eigenvalues can be expressed as functions of tr( J) and det( J) as shown in Table 3.2, Figure 3.18 and explained as follows.
Points in the region (a):
2
tr ( J ) < , det ( J ) > and the discriminant ∆ = ( tr ( J )) − det ( J )
0
0
4
> 0
21 The trace of the Jacobian is the sum of its diagonal elements, i.e., tr ( J
f
∂
g
) = (
∂
. The determinant of J is
∂[ ] )
+ ( ∂[ ] )
X
ss
Y
ss
det ( J
f
∂
g
∂
f
∂
g
) = (
∂
.
∂[ ] ) ( ∂[ ] )
− ( ∂[ ] ) ( ∂[ ] )
X
ss
Y
ss
Y
ss
X
ss
22 You can look up any text on Linear Algebra to refresh your memory on the solutions of systems of linear equations, their existence, and uniqueness.
Out-of-Equilibrium Thermodynamics
79
TABLE 3.2
List of All the Possible Solutions of Linear Differential Equations
Region
tr( J)
det( J)
Δa
Roots
Name
(a)
<0
>0
>0
λ
Stable node
1,2 < 0; λ 1 ≠ λ 2
(b)
<0
>0
=0
λ
Stable star
1,2 < 0; λ 1 = λ 2
(c)
<0
>0
<0
λ 1,2 = a ± ib; a < 0
Stable spiral
(d)
=0
>0
<0
λ = ± i
det ( J )
Center
,
4
1 2
(e)
>0
>0
<0
λ 1,2 = a ± ib; a > 0
Unstable spiral
(f)
>0
>0
=0
λ
0;
Unstable star
1,2 >
λ 1 = λ 2
(g)
>0
>0
>0
λ
0;
Unstable node
1,2 >
λ 1 ≠ λ 2
(h)
<0
>0
λ 0;
1 >
λ 2 < 0
Saddle node
(l )
1
>0
=0
=[ tr( J)]2
λ 0;
1 =
λ 2 > 0
Lines of unstable fixed points
(l
0;
2 = 0)
=0
=0
=0
λ 1 = λ 2 = 0
Plane of fixed points
(l )
3
<0
=0
=[ tr( J)]2
λ 0;
1 =
λ 2 < 0
Lines of stable fixed points
a Δ is the discriminant.
tr( J)
y
y
(g)
(l)
(f)
unstable
y
unstable
x
nodes
stars
(h)
x
y
x
saddle
y
points
(e)
unstable
y
spirals
x
(o)
(d)
x
centres
(c) y
det(J)
saddle
x
spirals
y
x
(a)
y
saddle
(b)
y
nodes
stable
x
stars
x
x
FIGURE 3.18 All the possible solutions of the quadratic equations [3.176] grouped in different regions and
curves.
For any point in this region the roots, λ and , are two different negative real numbers. This means
1
λ 2
that the stationary states represented by the points in the region (a) are stable. When a system is per-
turbed, it will restore the initial state with exponential speed. The relative stationary state is termed
a stable node. Points on the branch (b) of the curve are tr ( J) = 4 det( J) :
2
tr ( J ) < 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
=
80
Untangling Complex Systems
For any point along the branch (b) of the parabola, we have two equal roots that are negative real
numbers. This result means that after the perturbations, x and y decrease exponentially with the
same speed. The stationary state is stable and is defined as a stable star.
Points in the region (c):
2
tr ( J ) < 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
<
For the points in the region (c), the two roots are two complex conjugates of the form λ1,2 = a ± ib
with the real term a being negative. Reminding that ea± ib = ea (cos( b) ± isen( b)), it is evident that after the perturbation, the system will restore the initial stationary state after oscillations. These
kinds of stationary states are called stable spirals.
The points of the regions (a), (b), and (c) are attractors.
Points on the straight line (d)
2
tr ( J ) = 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
<
The solutions are λ
i
det ( J )
1,2 = ±
4
, i.e., they are purely imaginary. In these cases, when the sys-
tem is removed from the stationary state, it enters a limit cycle or periodic orbit: it starts to oscillate.
The corresponding fixed points are named as centers.
Points in the region (e):
2
tr ( J ) > 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
<
In this region, the roots are two complex numbers of the type λ1,2 = a ± ib with a positive real a. In this case, the perturbation grows exponentially by oscillating. The fixed points of the region (e) are
termed unstable spirals.
Points on the branch (f) of the curve are tr ( J ) = 4 det ( J ):
2
tr ( J ) > 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
=
For any point along the branch (f) of the parabola, we have two equal roots that are positive real
numbers. This result means that the effects of the perturbations x and y grow exponentially with the
same speed. The initial stationary state is unstable and is defined as an unstable star.
Points in the region (g):
2
tr ( J ) > 0, det ( J ) > 0 and the discriminant ∆ = ( tr ( J )) − 4 det ( J ) > 0
For any point in this region the roots, λ and , are two different positive real numbers. This result
1
λ 2
means that the many stationary states represented by the points in the region (g) are unstable. When
a system is perturbed, it will go away from the initial state. These stationary states are termed
unstable nodes.
The points of the regions (e), (f), and (g) are repellers.
Points in the region (h)
2
det ( J ) < 0, ∆ = ( tr( J )) − 4 det ( J ) 0, and ∆ tr( J )
>
>
Out-of-Equilibrium Thermodynamics
81
The roots λ and are two real numbers, one positive and the other negative. The stationary state is
1
λ 2
stable if the perturbation involves just the eigenvector with the negative eigenvalue λ , whereas it is
i
unstable if it involves the other eigenvector with the positive value of the eigenvalue. A stationary state
of this type is called a saddle point because its vector field in the phase space resembles a riding saddle.
Points on the straight line (l)
2
det ( J ) = 0 and ∆ = ( tr ( J )) 0
>
If we exclude the origin (o), for the stationary states represented by the points along the straight line
(l), an eigenvalue (let us say λ ) is null, whereas the other ( ) is a positive (for the portion of the line
1
λ 2
above the origin (o)) or a negative (for the portion of the line below (o)) real number. This situation
means that there is a line of fixed points that are stable when λ is negative, but unstable when λ is 2
2
positive. In the origin (o) of the graph, both eigenvalues are null, and we have a plane of fixed points.
Now, the most curious readers would ask themselves: “is it always correct to overlook the qua-
dratic and the higher order terms in the expansion [3.171]?” The answer is yes, as long as we deal
with fixed points that are not those of the borderline cases, such as stars, centers, degenerate nodes,
and non-isolated fixed points. The latter can be altered by small nonlinear terms. However, stars,
degenerate nodes, and non-isolated fixed points do not change their stability. For instance, an unsta-
ble star due to non-linear terms may transform into an unstable node or an unstable spiral, but
not into a stable spiral. This possibility is plausible if we retake a look at the stability diagram
(Figure 3.18). The centers are much more delicate because they lie at the edge between the stability and the instability regions. Small non-linear terms may transform a center into a stable or unstable spiral.
TRY EXERCISES 3.14 AND 3.15
3.8.3 The mulTi-dimensional case
The linear stability analysis presented in the mono- and bi-dimensional cases can be generalized to
systems with more than two variables
dx
i = x
( 1, 2, , )
i = fi x x … xn with i = 1,2 …
, , n [3.181]
dt
The character of a steady state is determined by the eigenvalues of the n × n Jacobian matrix, whose
elements are of the type
f
J
i
ij =
∂
[3.182]
x
∂
j ss
If all the eigenvalues have negative real parts, the stationary state is stable; if just a single eigenvalue
has a positive real part, the stationary state has an intrinsic instability.
3.9 KEY QUESTIONS
• Which equation defines the entropy change in the thermodynamics of out-of-equilibrium
[ ]
ss
ss
In the system [3.171], the second and higher order terms of the Taylor expansion are neglected,
because the perturbations are assumed to be small. The terms f ([ X ] , Y
[ ] ) and g([ X ] , Y[ ] )
ss
ss
ss
ss
are null. The derivatives are estimated in correspondence of the stationary points, and they are con-
stant. Therefore, the system [3.171] reduces to a system of linear differential equations. Solutions of
these linear equations are exponentials of the form x
it
=ν λ
it
i
1, e
, y =ν
λ
i
2, e
, and any linear combina-
tion of them. This result is evident if we consider, at first, two linear differential equations that are
uncoupled, like:
dx = λ1 x
dt
[3.172]
dy = λ2 y
dt
They are uncoupled because there is no y in the x-equation and vice-versa. In this simple situation,
the two differential equations can be solved separately. The solutions are x x e t
=
1
λ
2
λ
0
and y y e t
= 0
.
The x- and y- axes represent the directions of the trajectories for the x and y variables, respectively, as time goes to infinity ( t → ±∞). In other words, the x- and y- axes contain the straight-line trajectories: a trajectory starting from one point of the two axes, stays on that axis forever. For the general
case, we want to find the analog of these straight-line trajectories.
Introducing the solutions x
it
=ν λ
it
i e
,1
and y =ν
λ
i,2 e
into equation [3.171] and dividing by the non-
zero scalar factor e iλ t, we achieve
f∂
f
∂
ν
∂
[ X ]
∂ Y
[ ] ν i,1
λ
i,1
ss
ss
i
[3.173]
ν
=
i,2 g
∂
∂ ν i
g
,2
∂
Y
[ X ]
∂
ss
[ ] ss
In matrix form, equation [3.173] becomes
λν ii = ν
J i [3.174]
where J is the Jacobian matrix whose terms are the derivatives calculated at the fixed point; ν i is the 2 × 1 vector whose terms are ν i,1 and ν i,2. This vector represents the direction of the phase space,
which has trajectories with straight-lines, and is called the eigenvector of J. Linear equations such
as [3.174] have nontrivial solutions (i.e., solutions different from that with ν i,1 and ν i,2 both null)
78
Untangling Complex Systems
only when λ is an eigenvalue of the matrix J. The eigenvalues of J can be determined by solving the characteristic equation
f∂
f
λ
∂
−
∂
[ X ]
∂ Y
[ ]
det ( J λ I ) = det
ss
ss
−
= 0
[3.175]
g
∂
∂ g
λ
∂
−
Y
[ X ]
∂
ss
[ ]
ss
wherein I is the 2 × 2 identity matrix. If we expand the determinant, we have
λ2 − λ tr ( J ) + det ( J ) = 0 [3.176]
where tr( J) and det( J) are the trace (that is the sum of its diagonal elements) and the determinant of the Jacobian, 21 respectively. Equation [3.176] is a quadratic equation in λ having two solutions that are functions of the elements of the Jacobian:
2
tr J
( tr( J )) −4
( )
det ( J )
λ1 =
+
[3.177]
2
2
2
tr J
( tr( J )) −4
( )
det( J )
λ2 =
−
[3.178]
2
2
For the two eigenvalues λ and , we will have two eigenvectors, ν
1
λ 2
1 and ν2, and the general solutions
will be of the type
x( t ) = c
1
λ
2
λ
1ν1, e t
1
+ c 2ν2, e t
1
[3.179]
y( t ) = c
1
λ
2
λ
1ν1,2 e t + c 2ν 2,2 e t
wherein the values of the parameters c and depend on the initial conditions.22 The eigenvalues 1
c 2
λ 1
and λ can be either real or pure imaginary or complex numbers. The eigenvalues are related with
2
the trace and the determinant of J through the following equations:
λ1 + λ = tr ( J )
2
[3.180]
λ λ
1 2 = det ( J )
All the possible combinations of solutions for the eigenvalues can be expressed as functions of tr( J) and det( J) as shown in Table 3.2, Figure 3.18 and explained as follows.
Points in the region (a):
2
tr ( J ) < , det ( J ) > and the discriminant ∆ = ( tr ( J )) − det ( J )
0
0
4
> 0
21 The trace of the Jacobian is the sum of its diagonal elements, i.e., tr ( J
f
∂
g
) = (
∂
. The determinant of J is
∂[ ] )
+ ( ∂[ ] )
X
ss
Y
ss
det ( J
f
∂
g
∂
f
∂
g
) = (
∂
.
∂[ ] ) ( ∂[ ] )
− ( ∂[ ] ) ( ∂[ ] )
X
ss
Y
ss
Y
ss
X
ss
22 You can look up any text on Linear Algebra to refresh your memory on the solutions of systems of linear equations, their existence, and uniqueness.
Out-of-Equilibrium Thermodynamics
79
TABLE 3.2
List of All the Possible Solutions of Linear Differential Equations
Region
tr( J)
det( J)
Δa
Roots
Name
(a)
<0
>0
>0
λ
Stable node
1,2 < 0; λ 1 ≠ λ 2
(b)
<0
>0
=0
λ
Stable star
1,2 < 0; λ 1 = λ 2
(c)
<0
>0
<0
λ 1,2 = a ± ib; a < 0
Stable spiral
(d)
=0
>0
<0
λ = ± i
det ( J )
Center
,
4
1 2
(e)
>0
>0
<0
λ 1,2 = a ± ib; a > 0
Unstable spiral
(f)
>0
>0
=0
λ
0;
Unstable star
1,2 >
λ 1 = λ 2
(g)
>0
>0
>0
λ
0;
Unstable node
1,2 >
λ 1 ≠ λ 2
(h)
<0
>0
λ 0;
1 >
λ 2 < 0
Saddle node
(l )
1
>0
=0
=[ tr( J)]2
λ 0;
1 =
λ 2 > 0
Lines of unstable fixed points
(l
0;
2 = 0)
=0
=0
=0
λ 1 = λ 2 = 0
Plane of fixed points
(l )
3
<0
=0
=[ tr( J)]2
λ 0;
1 =
λ 2 < 0
Lines of stable fixed points
a Δ is the discriminant.
tr( J)
y
y
(g)
(l)
(f)
unstable
y
unstable
x
nodes
stars
(h)
x
y
x
saddle
y
points
(e)
unstable
y
spirals
x
(o)
(d)
x
centres
(c) y
det(J)
saddle
x
spirals
y
x
(a)
y
saddle
(b)
y
nodes
stable
x
stars
x
x
FIGURE 3.18 All the possible solutions of the quadratic equations [3.176] grouped in different regions and
curves.
For any point in this region the roots, λ and , are two different negative real numbers. This means
1
λ 2
that the stationary states represented by the points in the region (a) are stable. When a system is per-
turbed, it will restore the initial state with exponential speed. The relative stationary state is termed
a stable node. Points on the branch (b) of the curve are tr ( J) = 4 det( J) :
2
tr ( J ) < 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
=
80
Untangling Complex Systems
For any point along the branch (b) of the parabola, we have two equal roots that are negative real
numbers. This result means that after the perturbations, x and y decrease exponentially with the
same speed. The stationary state is stable and is defined as a stable star.
Points in the region (c):
2
tr ( J ) < 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
<
For the points in the region (c), the two roots are two complex conjugates of the form λ1,2 = a ± ib
with the real term a being negative. Reminding that ea± ib = ea (cos( b) ± isen( b)), it is evident that after the perturbation, the system will restore the initial stationary state after oscillations. These
kinds of stationary states are called stable spirals.
The points of the regions (a), (b), and (c) are attractors.
Points on the straight line (d)
2
tr ( J ) = 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
<
The solutions are λ
i
det ( J )
1,2 = ±
4
, i.e., they are purely imaginary. In these cases, when the sys-
tem is removed from the stationary state, it enters a limit cycle or periodic orbit: it starts to oscillate.
The corresponding fixed points are named as centers.
Points in the region (e):
2
tr ( J ) > 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
<
In this region, the roots are two complex numbers of the type λ1,2 = a ± ib with a positive real a. In this case, the perturbation grows exponentially by oscillating. The fixed points of the region (e) are
termed unstable spirals.
Points on the branch (f) of the curve are tr ( J ) = 4 det ( J ):
2
tr ( J ) > 0, det ( J ) > 0 and ∆ = ( tr( J )) − 4 det ( J ) 0
=
For any point along the branch (f) of the parabola, we have two equal roots that are positive real
numbers. This result means that the effects of the perturbations x and y grow exponentially with the
same speed. The initial stationary state is unstable and is defined as an unstable star.
Points in the region (g):
2
tr ( J ) > 0, det ( J ) > 0 and the discriminant ∆ = ( tr ( J )) − 4 det ( J ) > 0
For any point in this region the roots, λ and , are two different positive real numbers. This result
1
λ 2
means that the many stationary states represented by the points in the region (g) are unstable. When
a system is perturbed, it will go away from the initial state. These stationary states are termed
unstable nodes.
The points of the regions (e), (f), and (g) are repellers.
Points in the region (h)
2
det ( J ) < 0, ∆ = ( tr( J )) − 4 det ( J ) 0, and ∆ tr( J )
>
>
Out-of-Equilibrium Thermodynamics
81
The roots λ and are two real numbers, one positive and the other negative. The stationary state is
1
λ 2
stable if the perturbation involves just the eigenvector with the negative eigenvalue λ , whereas it is
i
unstable if it involves the other eigenvector with the positive value of the eigenvalue. A stationary state
of this type is called a saddle point because its vector field in the phase space resembles a riding saddle.
Points on the straight line (l)
2
det ( J ) = 0 and ∆ = ( tr ( J )) 0
>
If we exclude the origin (o), for the stationary states represented by the points along the straight line
(l), an eigenvalue (let us say λ ) is null, whereas the other ( ) is a positive (for the portion of the line
1
λ 2
above the origin (o)) or a negative (for the portion of the line below (o)) real number. This situation
means that there is a line of fixed points that are stable when λ is negative, but unstable when λ is 2
2
positive. In the origin (o) of the graph, both eigenvalues are null, and we have a plane of fixed points.
Now, the most curious readers would ask themselves: “is it always correct to overlook the qua-
dratic and the higher order terms in the expansion [3.171]?” The answer is yes, as long as we deal
with fixed points that are not those of the borderline cases, such as stars, centers, degenerate nodes,
and non-isolated fixed points. The latter can be altered by small nonlinear terms. However, stars,
degenerate nodes, and non-isolated fixed points do not change their stability. For instance, an unsta-
ble star due to non-linear terms may transform into an unstable node or an unstable spiral, but
not into a stable spiral. This possibility is plausible if we retake a look at the stability diagram
(Figure 3.18). The centers are much more delicate because they lie at the edge between the stability and the instability regions. Small non-linear terms may transform a center into a stable or unstable spiral.
TRY EXERCISES 3.14 AND 3.15
3.8.3 The mulTi-dimensional case
The linear stability analysis presented in the mono- and bi-dimensional cases can be generalized to
systems with more than two variables
dx
i = x
( 1, 2, , )
i = fi x x … xn with i = 1,2 …
, , n [3.181]
dt
The character of a steady state is determined by the eigenvalues of the n × n Jacobian matrix, whose
elements are of the type
f
J
i
ij =
∂
[3.182]
x
∂
j ss
If all the eigenvalues have negative real parts, the stationary state is stable; if just a single eigenvalue
has a positive real part, the stationary state has an intrinsic instability.
3.9 KEY QUESTIONS
• Which equation defines the entropy change in the thermodynamics of out-of-equilibrium
