Introduction of Linear and Nonlinear Dynamical Systems 1. Dynamical Systems 1.1 Ordinary Differential Equation (ODE) In this lecture, we will consider dynamical systems that can be described by the following ordinary differential equation.

( , ; )x f x t µ= (1)

where the dynamical variable x may be an element of n-dimensional Euclidian space n . The dynamical variable x may also be called as the state point of the system. The

dot represents differentiation with respect to time t. The right–hand side of the

equation f that smoothly maps n × to n , referred to as the vector field. When the vector field f includes time t explicitly as its argument, that dynamical system is referred to as non-autonomous, while when it does not as autonomous. When f is a linear function that can be represented by n n× matrix, Eq. (1) is a linear dynamical system. When f is a nonlinear function, it is a nonlinear dynamical system. Question: Explain why f is called as the vector field? In particular, consider graphically the case with n=2 (a two dimensional dynamical system). Answer: Let me explain the reason using the following dynamical system as an example that is called FitzHugh-Nagumo (FHN) equation (FitzHugh 1961, Nagumo et al. 1962) and is very famous in the field of computational neuroscience.

( )

3

31

dx xc x y zdtdy x by adt c

= − − +

= − +

where the parameters are set specifically to the following values; 0.7, 0.8, 3.0, 0.0a b c z= = = = . In this case, the vector filed may be defined as

( )

1

2

3

( , )( , )

( , )

/ 3

1 ( )

f x yf x y

f x y

c x x y z

x by ac

= − − + =

− +

We could display this vector-valued function on a two dimensional plane with its x-y coordinate as shown in Figure 1. That is, for any given point (x,y) on the plane, we could calculate a single two dimensional vector as f(x,y). The obtained vector is plotted on the

plane with its origin at the point (x,y).

Figure 1: The vector field of FitzHugh-Nagumo equation with its nullclines and equilibrium

solution.

It is important to notice that the state point of the system at the point (x,y) at some time instant t will move to the direction defined by this vector. The moving “velocity” of the state point will be fast if the length of this vector is large, and the state point moves slowly if it is small. After some small amount of time t∆ from the time t, the system’s state will be at

1

2

( , )( , )

f x yxt

f x yy

+ ∆

.

One could obtain a curve called “solution” or “trajectory” of the dynamical system by repeating this process. The vectors are distributed anywhere on the plane, and the direction and length of the vectors depend on the location of the origin of each vector. The figure looks like a rice field when the rice ears are waving in the wind. However, since this example is for the case with the autonomous dynamical system, the waving rice era should not change in time. In other cases with non-autonomous dynamical systems, the corresponding waving era will change as time since the function defining the vector field changes as time. Usually, the vector field f may be parameterized smoothly by k-dimensional vector

parameterµ . This means that the vector field may show a different form depending on the value of µ . The vector field of FHN equation will be different from the one

displayed in Fig. 1 for different values of a,b,c, and z. Example from biology: The Hodgkin-Huxley Equation

Hodgkin and Huxley (Hodgkin and Huxley, 1952) proposed that the membrane potential of the squid giant axon can be described by the differential equation

3 4ext Na Na K K L L

d ( ) ( ) ( )dVC I g m h V V g n V V g V Vt= − − − − − −

where V (mV) is the membrane potential, t is time in millisecond, C (1 µ F/cm2) is the membrane capacitance, NaV (115 mV) is the sodium equilibrium potential, KV ( 12− mV) is the potassium equilibrium potential, LV ( 10.613 mV) is the leakage equilibrium potential, Nag (120 mS/cm2) is the maximum sodium conductance, Kg (36 mS/cm2) is the maximum potassium conductance, Lg ( 0.3 mS/cm2) is the leak conductance and extI ( µ A/cm2) is the externally applied current. m , h and n are

also the dynamical variables which may be interpreted as probabilities that satisfy

d ( )[ (1 ) ]d m mm m mt

φ θ α β= − −

d ( )[ (1 ) ]d h hh h ht

φ θ α β= − −

d ( )[ (1 ) ]d n nn n nt

φ θ α β= − −

where ( 6.3) /10( ) 3 θφ θ −= expresses their dependence on temperature θ ( C). By fitting experimental data, Hodgkin and Huxley represented the rate functions in the equations by the following formulae

250.1 , 4exp( /18),25exp 1

10

m mV VV

α β− += = −

− + −

10.07exp( / 20), ,30exp 1

10

h hVV

α β= − =− + +

100.01 , 0.125exp( /80).10exp 1

10

n nV VV

α β− += = −

− + −

HOME WORK: Plot (may be numerically by computer) the form of these functions, , , , , ,m m h h n nα β α β α β , as the function of the membrane potential V, may be for a range

from -100 mV to +100 mV. We will use these function form later. A basic understanding of the Hodgkin-Huxley equation is relatively easy if the variables m, h, and n are assumed to be constant parameters. Let us do so by defining the following parameters.

3

4Na Na

K K

L L

G g m h

G g nG g

=

==

Then, the first equation of the Hodgkin-Huxley equation becomes as

ext Na Na K K L Ld ( ) ( ) ( )dVC I G V V G V V G V Vt= − − − − − −

The electrical circuit of this equation should be easily illustrated on your notebook using a single capacitor, three resisters and three butteries, with external current sources with the intensity of extI .

Ex. Do that illustration. The Hodgkin-Huxley equation with dynamic variables m, h, and n has no closed analytical solutions, but it is easily solved numerically, for example, by Euler or Runge-Kutta method. The Hodgkin-Huxley equation can be rewritten in the vector form as in Eq. (1).

( , , , ; , , , , )( , ; )( , ; )( , ; )

Na K lV F V m h n C V V Vm M V mdh H V hdtn N V n

θθθ

=

.

Figure 2: Example waveforms of the Hodgkin-Huxley equation, exhibiting membrane

action potentials.

1.2 Equilibrium Solutions, Steady State Solutions Let us look at Figure 1 for the vector field of FHN equation. At the lower left of the plane marked by the open circle, one can see that the length of the corresponding vector

is zero. This sort of special point is referred to as the equilibrium point of the dynamical system. Equivalently, it is also called as the steady state of the system. The equilibrium point if an autonomous dynamical system

( ; )x f x µ= is defined as follows. That is, a particular state point x satisfying

( ; ) 0f x µ =

is called the equilibrium point of the system. The system’s state point located exactly at the equilibrium point will not move and keep staying at the point since the vector field, i.e., the velocity of change in the state point position x there, is zero. Usually, because of the continuation argument of the function ( ; )f x µ with respect to x, the vector field

near the equilibrium point is small, implying that the state point near the equilibrium point may move slowly. The equilibrium solutions of the FHN equation The equilibrium solution of the FHN equation may be obtained from the following set of algebraic equations

( )

3

03

1 0

xc x y z

x by ac

− − + =

− + =

by setting dx/dt=dy/dt=0. From these equations, we have 3

31

xy x z

ay xb b

= − +

= +

The first equation, referred to as the x-nullcline (isocline), represents a cubic curve on the x-y plane. The second one, referred to as the y-nullcline (isocline) is just a straight line. The equilibrium point of the FHN shown in Fig. 1 is an intersection point of the two nullclines. One should note that, on the x-nullcline, the arrows representing vector field are parallel to the y axis, i.e., their x components are zero. In the same way, on the y-nullcline, the arrows representing vector field are parallel to the x axis, i.e., their y components are zero. It is worthwhile to note that the shapes and locations of these isoclines, thus of the equilibrium point, may change depending on the parameter values of the model. We consider this property below using a simple model and in detail later in this lecture.

The equilibrium solutions of Euler Buckling equation

Figure 3: Finite element analogue of Euler buckling

Figure 3 illustrates a simple physical system representing a finite element analogue of the Euler column. The system consists of two rigid rods of unit length connected by friction-less pins which permit rotation in a horizontal plane. It is subjected to a compressive force λ . This force is resisted by a torsional spring of unit length. The state of the system is described by the angle x measuring the deviation of the rods from the horizontal. The potential energy of this system is

2

( ; ) 2 (cos 1)2xV x xλ λ= + − .

The first term is the stored energy in the torsional spring and the second is the work done by the external force. The equation of motion of the mass with unit weight attached at the right side of the rod becomes as

( 2 sin )Vx x xx

λ∂= − = − −

∂.

If one desire representing this equation with the form defined by Eq. (1), it can be done as follows.

( 2 sin )x yy x xλ

= − −

The equilibrium solution of this system can be calculated from the single algebraic equation as

( , ) 2 sin 0g x x xλ λ= − = . For some reason we will see later in this lecture, we need to look for a point ( , )x λ that

satisfied the following relations.

( , ) ( , ) 0g x g xx

λ λ∂= =∂

One can confirm that 0 0( , ) (0,1/ 2)x λ = is such a point. Let us assume that the angle x

and the force λ is close to this point. That is, we are interested in only a case with small angle deviation for force near 1/2. For this case, ( , )g x λ can be Taylor expanded with respect to 0 0( , ) (0,1/ 2)x λ = . The equation for the equilibrium point becomes

3 12( ) 0

6 2x xλ− − = (2)

by neglecting some higher order terms. Ex. Plot solutions of Eq. (2) as the function of λ , to notice that, depending on the value of λ , the number of equilibrium solution may change. That is, in one case ( 1/ 2λ < ), there is only one equilibrium, in another case ( 1/ 2λ > ), there are three. The equilibrium solution of the Hodgkin-Huxley equation For a steady current extI , the Hodgkin-Huxley equation has a steady state solution, V , satisfying

3 4ext Na Na K K L L( ) ( ) ( )I g m h V V g n V V g V V= − + − + −

where the functions m , h and n take their steady state values

( )( ) ,

( ) ( )( )

( ) ,( ) ( )

( )( ) ,

( ) ( )

m

m m

h

h h

n

n n

Vm m V

V VV

h h VV V

Vn n V

V V

αα β

αα β

αα β

∞

∞

∞

= =+

= =+

= =+

which can be derived by setting the vector field of the equation is equal to zero. 2. Linearization As mentioned for the Hodgkin-Huxley equation, in may cases, it is not possible to obtain analytical expression of solutions for a given nonlinear dynamical system which is described by an ODE. This means that we should usually investigate dynamics of a given nonlinear dynamical system using numerical integration (computer simulation) of the ODE. However, there are several ways for us to understand some aspects of the ODE dynamics without numerical integration of the ODE. The most common way to do this is to restrict ourselves to consider the dynamical behavior of the system only when its state point is close to an equilibrium of the system. There are two points that we have to clarify before getting into detail. The first point may be summarized as the following question. Q1: How can we analyze the dynamical behavior of the system near the equilibrium point? A1: It can be done by obtaining a linearized dynamical system of the original nonlinear dynamical system. Once we have such a linear dynamical system, it is relatively easy to analyze its dynamics analytically.

Q2: Is there any relationship between dynamical behavior of the system near the equilibrium point of the original nonlinear dynamical system and those of the linearized dynamical system? A2: Yes, there is. There may be one-to-one correspondence between these dynamics if some simple condition on the linearized dynamical system is satisfied. Let us consider the following nonlinear autonomous dynamical system.

( )x f x= and its equilibrium point x that satisfies ( ) 0f x = .

Let us consider time-dependent changes (dynamics) of a state point x close to this equilibrium point. Such a state point may be represented as x x η= +

where

1

n

ii

η η=

= ∑

is small enough. That is, η is a small deviation vector from the equilibrium point. The

vector field near the equilibrium point may be Taylor expanded as

2( ) ( ) ( ) ( )f x f x Df xη η η+ = + +Ο .

Calculation of derivatives for a vector-valued function Let us consider a function f

1

1 22

1 2

( , )( )

( , )f x x

f xf x x

=

that maps 2 to 2 as an example. Let us denote an equilibrium point of this function as

1

2

xx

x

=

.

The function can be rewritten as

1

1 1 2 22

1 1 2 2

( , )( )

( , )f x x

f xf x x

η ηη

η η + +

+ = + +

Each of two components, say 1 1 2 2( , )if x xη η+ + , can be Taylor expanded as

1 1 2 2 1 2 1 2 1 1 2 21 2

2 2 232 2

1 2 1 1 2 1 2 1 2 22 21 1 2 2

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( )

i ii i

i i i

f ff x x f x x x x x xx x

f f fx x x x x xx x x x

η η η η

η ηη η η

∂ ∂+ + = + + +

∂ ∂

∂ ∂ ∂+ + +Ο

∂ ∂ ∂ ∂

Using this equation and the deviation vector that can also rewritten as heη = ,

where e is the unit vector parallel to η and h η= , one can observe that

0

1 11 2 1

1 2 1 2 1 1 2 2 1 21 22 2

2 2 21 2 1 2 1 1 2 2 1 2

1 2

1 1

1 2 1 1 2 21 22

( ) ( )( ) lim

( , ) ( , ) ( , ) ( ) ( , )1

( , ) ( , ) ( , ) ( ) ( , )

( , ) ( , )

h

f x he f xDf x eh

f ff x x x x he x x he h f x xx x

h f ff x x x x he x x he h f x xx x

f fx x e x x ex xfx

→

+ −=

∂ ∂+ + +Ο − ∂ ∂ =

∂ ∂+ + +Ο − ∂ ∂

∂ ∂+

∂ ∂=

∂∂

2

1 2 1 1 2 21 2

1 1

1 2 1 21 2 12 2

21 2 1 2

1 2

( )( , ) ( , )

( , ) ( , )

( , ) ( , )

hfx x e x x ex

f fx x x xx x e

ef fx x x xx x

+Ο ∂

+ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂

This means that

1 1

1 2 1 21 22 2

1 2 1 21 2

( , ) ( , )( )

( , ) ( , )

f fx x x xx x

Df xf fx x x xx x

∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂

which is independent of the unit vector e and only the function of the equilibrium point. This matrix is usually called as Jacobian. Ex. Consider FitzHugh-Nagumo equation.

( )

3

31

dx xc x y zdtdy x by adt c

= − − +

= − +

Let us denote its equilibrium point as ( , )Tx y . Show that Jacobian of FHN equation with respect to ( , )Tx y may be expressed as

2(1 )1

c x cb

c c

− − −

.

Ex. Show that Jacobian of the Hodgkin-Huxley equation with respect to its equilibrium point ( , , , )TV m h n may be

3 4 2 3 3Na K L Na Na Na Na K K( ) 3 ( ) ( ) 4 ( )

( ) ( ) 0 0

( ) 0 ( ) 0

( ) 0 0 ( )

m m mm m

h h hh h

n n nn n

g m h g n g g m h V V g m V V g n V Vd d d

mdV dV dV

J d d dh

dV dV dVd d d

ndV dV dV

α α βα β

α α βα β

α α βα β

− + + − − − − − − − + − + = − + − + − + − +

Using a Jacobian matrix with respect to an equilibrium point x , we can obtain a linear or linearized dynamical system that may represent dynamical behavior of the original nonlinear dynamical system when its state point is close to the equilibrium point as follows,

2

0

( ) ( )

0 ( ) ( ) ( )

( )

d x f xdt

f x Df x

Df x

η η

η η η

η η=

+ = +

+ = + +Ο

=

by neglecting the higher order terms than two. ( )Df x is just a n n× constant matrix,

and this type linear differential equation can be easily and analytically solved. Indeed, the solution of the linear differential equation Aη η= where we rewrite ( )A Df x= , with an initial condition at time t=0 as (0)η may be

given by ( ) exp( ) (0)t Atη η= .

Remind the definition of exponential function of a matrix A .

2 31 12! 3!

Ae I A A A= + + + +

and

2 3

2 3

2! 3!At t te I At A A= + + + +

Thus,

22 3

22

2!

2!

At

At

d te A A t Adt

tA I At A

Ae

= + + +

= + + +

=

which shows that

( ) ( )( ) ( ) ( ) ( )At Atd dt e t A e t A tdt dtη η η η= = =

3. Stability Stability of an equilibrium of a given linear or linearized dynamical system may be classified into three, stable, unstable, neutral. The state point of the system close to a stable equilibrium point may asymptote the equilibrium point as time evolves, while the state point of the system close to an unstable equilibrium point may be away from the equilibrium point as time evolves. Let us consider the stability of a n dimensional linear dynamical system x Ax= where A is a n n× constant matrix. As we have seen above, a linearized system of a nonlinear dynamical system with respect to its equilibrium point has the same form as this. That is, ( )A Df x= .

A vector field of this linear dynamical system can be obtained by the matrix A which maps a given state point x to a n dimensional vector Ax. The equilibrium point of the system x Ax= is located at the origin. Stability of the system may be determined by eigenvalues of the matrix A, say λ , satisfying the following eigen equation; det( ) 0A Iλ− = .

The equilibrium point x=0 of the linear or the linearized dynamical system x Ax= is stable if real part of every n eigenvalue λ is negative. If at least one of them is positive, the equilibrium point is unstable. Ex. Using the fact that the solution of x Ax= is given by

( ) exp( ) (0)x t At x= ,

show that the equilibrium point x=0 of the linear or the linearized dynamical system

x Ax= is stable if real part of every n eigenvalue λ is negative. Ex. Jacobin of FHN equation with respect to ( , )Tx y is

2(1 )1

c x cA b

c c

− − = −

as we showed above. Show that the eigen equation of this system is given by 2 0λ λ−Γ + ∆ = where

2

2

(1 ) /1 (1 )c x b c

b xΓ = − −

∆ = − −

Assume that 0Γ < is true for a range of our interest. Let us consider a case in which the parameter z in the FHN equation is controllable and the others a, b, and c are fixed constants. It will be our point of interest to clarify the stability of the equilibrium point ( , )Tx y as the function of the value of the control parameter z. Indeed, as we see later in this lecture, the parameter z represents an intensity of the direct current injection to the model membrane. Since the eigen values (the solutions of the above second order algebraic equation) is

2 4

2λ Γ ± Γ − ∆= ,

a critical parameter value of Cz at which the stability of the equilibrium point changes

may be associated with the following equation;

2(1 ( )) / 0Cc x z b cΓ = − − =

Here, the control parameter z dependency of the x-component of the equilibrium point x is expressed explicitly. Using the fact that

3

03

0

xx y z

x by a

− − + =

− + =

calculate Cz that satisfies 0Γ = . This point in the one dimensional parameter space

may be called as the Hopf bifurcation point at which non-trivial periodic solution (limit cycle) appears as we will see later.

Figure 4: Bifurcation diagram of the FHN equation (a=0.8, b=0.7, c=3.0).

The x-component of the equilibrium point is shown as the function of the parameter value z. The periodic orbit branch emerging when the equilibrium point is destabilized

is also shown. XPPAUT was used to draw this diagram.

Figure 5: Oscillation period of FHN equation as the function of z.

Figure 6: Magnification of a part of Figure 4. Notice bistability of the equilibrium point

and a stable oscillation within a small range of z value.

Figure 7: Two trajectories of FHN with the same parameter values (z=0.3375) but

starting from different initial conditions, showing the bistability. One asymptotes to the stable limit cycle and the other to the equilibrium point.

4. Phase Portrait and Topological Classification of Dynamical System Under construction

2. The Hodgkin-Huxley equation and its generalization A Landscape of membrane dynamics

At resting state (potassium ion)

At resting state (sodium)

At resting state (chloride)

At the onset of brief current injection (sodium)

When the membrane voltage increases (potassium)

At the offset of the action potential (sodium)

Membrane voltage dependency of alpha and beta functions

Membrane voltage dependency of infinity and tau functions

Response to pulse stimulus Example 1

Response to pulse stimulus Example 2

Response to pulse stimulus Example 3

Response to pulse stimulus Example 4