Math 415

One-Dimensional System

Fixed Point

Consider the one-dimensional system:

x˙=f(x)

The fixed point found by setting x˙=0. For the fixed point x. We will use it for determining the stability.

Example

For equation x˙=x21, we have it's plot here:

---
title: x' = x^2-1
xLabel: x
yLabel: x'
bounds: [-2,2,-2,2]
disableZoom: false
grid: true
---
f(x)=x^2-1
g(x)=0

The left fixed point x=(1,0) is a stable point, while the right FP(1,0) is unstable.

Linear Stability Analysis

In order to know the behavior around fixed point linearization is a great method to analyze.

Let x be a fixed point, and let η(t)=x(t)x be a small perturbation away from x. To see whether the perturbation grows or decays, we derive a differential equation for η. Differentiation yields

η˙=ddt(xx)=x˙,

since x is constant. Thus η˙=x˙=f(x)=f(x+η). Now using Taylor’s expansion we obtain

f(x+η)=f(x)+ηf(x)+O(η2),

where O(η2) denotes quadratically small terms in η. Finally, note that f(x)=0 since x is a fixed point. Hence

η˙=ηf(x)+O(η2).

Now if f(x)0, the O(η2) terms are negligible and we may write the approximation

η˙ηf(x).

This is a linear equation in η, and is called the linearization about x*. It shows that the perturbation:

η(t)={Grows Exponentiallyf(x)>0decays Exponentiallyf(x)<0

Potentials

There’s another way to visualize the dynamics of the first-order system: potential V(x) is defined by

f(x)=dVdx

Consider the relation between potential and time V(x(t)), by using chain rule we have

dVdt=dVdxdxdt

since x˙=f(x), we obtain

dVdt=(dVdx)20.

The equilibrium point happens at dV/dx=0, the V remains constant. Since dV/dx=0 implies x˙=0.

Example

Graph the potential for the system x˙=xx3, and identify all equilibrium points.

Solution: Solving dV/dx=xx3 yields

V=12x2+14x4+C.

Once again we set C=0. Figure shows the graph of V. The local minima at x=±1 correspond to stable equilibria, and the local maximum at x=0 corresponds to an unstable equilibrium.

---
title: V(x)-Potential
xLabel: x
yLabel: V(x)
bounds: [-2,2,-0.5,1]
disableZoom: false
grid: true
---
f(x)=-0.5x^2+0.25x^4

Bifurcations

Saddle-Node

Typical Form:

x˙=r±x2 image/svg+xml Master slide Slide Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing

Splitting into two branches.

Transcritical Bifurcation

Normal Form:

x˙=rxx2 image/svg+xml

Some people say that an exchange of stabilities has taken place between the two fixed points.

Pitchfork Bifurcation

This bifurcation is common in physical problems that have a symmetry.

Supercritical Pitchfork Bifurcation

Normal Form:

x˙=rxx3

Note that this equation is invariant under the change of variables xx.

image/svg+xml Master slide Slide Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing Drawing

One stable splitting two stable and one unstable.

Subcritical Pitchfork Bifurcation

Normal Form:

x˙=rx+x3

One stable and two unstable to one stable.

Dimensional Analysis

Non-dimensionalization

Differential equations that show up in modeling “real world situations” usually have many constants in them. Often one can reduce the number of constants in a problem by choosing the right units for the various quantities in the problem. In the other words, we can convert the variables to ratio that has no unit to analyze multiple different coefficient problems.

We define a dimensionless time τ by

τ=tT

where τ is dimensionless time, t is dimensional time, T is characteristic time scale. We need to choose T very carefully to do the non-dimensionalization.

Timescale

λ is the rate of growth. 1/|λ| is timescales of growth.

Example 1

Suppose that a quantity x(t) changes in time according to the ODE

(1)dxdt=a+bx+cx2.

The coefficients a, b, and c must all have different units, otherwise we could not add the terms a, bx, and cx2.

To simplify the equation we choose a constant value for x, let's say x0, and we let this value be our unit. The ratio

ξ=xx0

has no units. In the same way we can pick a unit of time t0 and introduce the quantity

τ=tt0

which also has no units.

The quantities ξ and τ are nondimensionalized versions of our original variables x and t. The point of nondimensionalization is that we can now derive a differential equation for ξ and τ, and then afterwards figure out which choice of the units x0 and t0 simplifies things most.

In this example we substitute

x(t)=x0ξ(τ),andt=t0τ

which leads to

dxdt=x0t0dξdτ

by the chain rule, and

a+bx+cx2=a+(bx0)ξ+(cx02)ξ2

by direct substitution. The differential equation (1) for x and t is therefore equivalent with

x0t0dξdτ=a+(bx0)ξ+(cx02)ξ2,

and thus

dξdτ=t0ax0+(t0b)ξ+(t0x0c)ξ2.

At this point we choose x0 and t0. We can try to make the constant term and the coefficient of ξ both equal to 1. If a0 and b0 then this is possible provided we choose

t0=1b,x0=at0=ab.

The coefficient of ξ2 then becomes

α=t0x02c=a2cb3,

and we get the following differential equation for ξ as a function of τ:

(2)dξdτ=1+ξ+αξ2.

This is a nondimensionalized version of equation (1). Note that instead of three undetermined parameters (a,b,c) it only has one parameter, namely α.

Example 2

The Logistic equation:

N˙=rN(1NK)

where N is population, t is time, r growth rate, K carrying capacity.

The non-dimensionalize term:

τ=tTandx=NN0

Thus, we have

dNdt=N0Tdxdτdxdτ=Trx(1N0Kx)=x(1x)

when T=1/r and N0=K.

Two-Dimensional System

General form

2D system:

{x˙=f(x,y)y˙=g(x,y)

Example

mcϕ¨=bϕ˙mgsinϕ+mcsinϕcosϕ

Let x=ϕ, y=ϕ˙, then we have

{x˙=yy˙=bmcygcsinx+sinxcosx

Linear form

{x˙=ax+byy˙=cx+dy

where a,b,c,d are R. Let x=(xy) , A=(abcd), so we have x˙=Ax.
The fixed points: when x˙=y˙=0 or x˙=0.

Uncoupled System

Consider the system

x˙=(a001)x

Notice {x˙=axy˙=y = {x=x0eaty=y0et the trajectory is different and depends on a:

Created by potrace 1.10, written by Peter Selinger 2001-2011

Classified these cases: (a) Stable node, (b) Stable star,
(c) stable node, (d) line of fixed points, (e) saddle point

Stability Technology

When x is a fixed point:

  1. Attracting but not Lyapunov Stable
Created by potrace 1.10, written by Peter Selinger 2001-2011
  1. Lyapunov Stable but not attracting
Created by potrace 1.10, written by Peter Selinger 2001-2011
  1. Attracting and Lyapunov is Stable Start (b) case.

General Linear System

Consider following form

x˙=(abcd)x

Note: The straight-line trajectories is if x(0)=(x00) then x(t)=(x(t)0) which is stays on x-axis.

Now let's generalize this idea. Let's guess x(t)=eλtv , hence λ is the growth rate and v is the direction of growth (the vector).

x(t)=eλtvx˙(t)=λeλtv&Ax=eλtAv

So, x˙(t)=Ax implies λeλtv=eλtAv and

Av=λv

So, v is an eigenvector and λ is an eigenvalue. x(t)=eλtv is an eigensolution.

2D Eigenvalues

Definition
(AλI)v=0

Then the this singular matrix has determinant zero:

det(AλI)=0λ2τλ+Δ=0λ2(a+d)λ+(adbc)=0

Thus, λ=τ2±τ24Δ2, τ=a+d, and Δ=adbc.
The eigenvectors are when you plug the eigenvalues back to the original form.

Note: τ=λ1+λ2, Δ=λ1λ2

System form

When λ1λ2 the general form of the system is

x(t)=c1eλ1tv1+c2eλ2tv2

Example

For linear system:

{x˙=x+yy˙=4x2ysubject to{x(0)=2y(0)=3

we have A=(1142) and x0=(23) . Then the τ=12=1, Δ=24=6, so λ2+λ6=0 gives λ1=2 and λ2=3.
So, by plugging λ1,2 backs to determent equation, we have

λ1=2:(11)andλ2=3:(14)

By set t=0 I get c1=c2=1. Therefore the general solution is