Flashcards in Phase Portraits and Linearisation Deck (30)

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1

##
Linearisation

Description

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-close to a fixed point it is quite accurate to approximate a function using the linear terms of a Taylor expansion:

f(x) = f(x1) + f'(x1)(x-x1)

-the f(x1) will equal 0 if x1 is a fixed point since f=x'

-although adding more terms makes the Taylor expansion more accurate it would also be much more difficult or even impossible to solve

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##
Nullcline

Definition

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-exist in all dimensions but this definition is for the 2D case:

-lines where the vector field is either horizontal or vertical

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##
Isocline

Definition

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exist in all dimensions but this definition is for the 2D case:

-lines where the vector field has a constant gradient

4

##
Phase Portrait

Description

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-given a 2D system:

x' = f(x,y)

y' = g(x,y)

-the phase portrait plots y against x with trajectories drawn on

5

## How to go about drawing a phase portrait

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-given a 2D system

x' = f(x,y)

y' = g(x,y)

-start with horizontal isoclines and vertical isoclines

-plot other isoclines e.g. g/f=-1

-by looking at the plotted vectors draw trajectories onto the phase portrait

-use local analysis for the regions surrounding fixed points

*********

6

## Horizontal Nullclines

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-given a 2D system

x' = f(x,y)

y' = g(x,y)

-horizontal isoclines are lines where g=0

-so set y'=0 to find the equations of the lines where this is true

-evaluate f at different regions on these lines

-if f is positive the vector field points to the right

-if f is negative the vector field points to the left

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## Vertical Nullclines

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-given a 2D system

x' = f(x,y)

y' = g(x,y)

-vertical isoclines are lines where f=0

-so set x'=0 to find the equations of the lines where this is true

-evaluate g at different regions on these lines

-if g is positive the vector field points p

-if g is negative the vector field points down

8

##
Linearisation

N=2`

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x' = f(x,y)

y' = g(x,y)

-to linearise the system near the fixed point, compute the Jacobi matrix:

J(x,y) = 2x2 matrix, top row: ∂f/∂x, ∂f/∂y, second row: ∂g/∂x, ∂g/∂y

-evaluate at the fixed point:

A = J(xo,yo)

-sub into

(ζ', η') = A (ζ,η)

where the row vectors should be 2x1 column vectors

-where

ζ = x-xo

η = y-yo

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##
Linear Systems With Constant Coefficients

N=1

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x' = a*x , x(0)=xo

-assume solution of the form x(t)=A*e^(λt)

-differentiate

x' = λA*e^(λt)

-sub in:

A*e^(λt) (λ-a) = 0

=> λ=a

=> x = A*e^(at)

-sub in initial condition

xo = A

=> x(t) = xo*e^(at)

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##
Linear Systems With Constant Coefficients

N≥1 , General Equation

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|x' = A |x

-where A is an NxN matrix

-and |x ϵ R^N

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##
Linear Systems With Constant Coefficients

N=2 General Equation

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|x' = (x' , y')

x' = ax + by

y' = cx+dy

-in this case |x' = A |x, where |x=(x,y) and A is a 2x2 matrix with entries a, b, c, d

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##
Linear Systems With Constant Coefficients

N=2 Eigenvalue Problem

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|x' = A |x , |x(to) = |xo

-assume a solution of the form:

|x(t) = |V*exp(λt)

-where |V is a constant complex vector and λ is a non-complex constant

-differentiate:

|x' = λ*|V*exp(λt)

-sub in

=> λ|V exp(λt) = A|V exp(λt)

-since exp(λt)≠0, we arrive at the eigenvalue problem, to tind a non-zero vector |V such that:

λ |V = A |V

13

##
Linear Systems With Constant Coefficients

N=2 Eigenvalues and Eigenvectors

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-to solve the system;

|x' = A |x , |x(to) = |xo

-we are looking for solutions to the eigenvalue problem;

λ |V = A |V

=> (A-λI) |V = 0

-there is a non-trivial solution only if (A-λI) is not invertible i.e. if:

det(A - λI) = 0

-this determinant is a polynomial of degree N in λ and thus there are exactly N solutions λ1, λ2, ... , λn

-once these eigenvalues are found we sub back in for the eigenvectors Vn

14

##
Linear Systems With Constant Coefficients

N=2 Characteristic Equation

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-in the case N=2, the characteristic equation can be written generally in the form:

λ² - Tλ + D = 0

-where T is the trace (sum of diagonal elements) of matrix A

-and D is the determinant of matrix A

-thus eigenvalues are solutions of form:

λ1 = T+√ (T²-4D)/2

λ2 = T-√ (T²-4D)/2

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##
Linear Systems With Constant Coefficients

N=2 Case 1 : λ1≠λ2

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λ1≠λ2, this happens when T²≠4D

-in this case we have two distinct Eigenvalues λ1,2 and the corresponding eigenvectors |V1,2

-a general solution of the system in this case is:

|x(t) = C1|V1exp(λ1t) + C2|V2exp(λ2t)

16

##
Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real

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-the eigenvalues λ1,2 re real only if T²>4D

-in this case, the eigenvectors |v1,2 are real and the constants C1,2 should be taken as real as well

-there are also subcases involving all combinations of λ1,2 being positive, zero or negative

17

##
Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)*

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-the eigenvalues λ1 and λ2 are complex conjugated λ2=(λ1)*, i.e. T²<4D

=>

λ1 = γ + iω

λ2 = γ - iω

-where γ=T/2 and ω = 1/2*√(4D-T²)

-the corresponding eigenvectors |V1 and |V2 can be chosen to be complex conjugated:

|V2 = (|V1)*

-there are three subcases of this case, γ=0, γ>0, γ<0

18

##
Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (+,+)

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-both eigenvalues are distinct and positive

-both exponents, e^(λ1,2 t) are growing with t and the solution goes to infinity as t->∞

-the fixed point at the origin is unstable, an unstable node

-there are two trajectories represented by straight lines corresponding to c1≠0,c2=0 and c1=0,c2≠0

-the lines go through the origin and are in the directions of |v1 and |v2 respectively

-eigenvalue λ1 with the smallest value is called the slow eigenvalue and |V1 the slow eigenvector

-the bigger value λ2 is the fast eigenvalue and |V2 the fast eigenvector

-at the origin trajectories are tangent to the slow eigenvector, as t goes to infinity the slope of a generic trajectory approaches the fast direction

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##
Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (-,-)

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-if both eigenvalues are negative, λ2∞

-the corresponding fixed point at the origin is called a stable node

-as for (+,+) λ1 is still the slow eigenvalue as it has the smallest absolute value, while λ2 has the larger absolute value and is the fast eigenvalue

-similar to (-,-) but lines point towards the stable node

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##
Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (+,-)

### -if two eigenvalues have different sign: λ2<0

21

##
Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (0,+)

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-if one eigenvalue is zero, λ1=0 and the other positive, λ2>0, we have a case with a line of equilibrium (a line filled with fixed points)

-in this case there are infinitely many time-independent solutions of the form x(t)=c1|V1

-every point on the line n|Vi where n is a real number is a point of equilibrium

-this line is unstable

-any solution with initial conditions outside of this line is growing exponentially with time

-all trajectories are straight lines parallel to |V2 and going to infinity

22

##
Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (0,-)

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-similar to (0,+), there is a line of equilibria, λ1=1, λ2<0

-but this line of equilibrium is a stable one

-every solution with initial conditions outside of the line will tend to the line as t->∞

-all trajectories are straight lines parallel to |V2 are going towards the line of fixed points

23

##
Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)* γ=0

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-if γ=0, λ1=iω, λ2= -iω

-then the solution is expressed in terms of sin and cos and thus is periodic with period T=2π/ω

-all trajectories are closed loops around the origin

-direction depends on the value of b in the original A=(abcd) matrix

-if b>0 then they are clockwise, if b<0 then they are anticlockwise

-in this case solutions are stable but not asymptotically stable

-a fixed point of a linear system with pure imaginary eigenvalues is called a centre

24

##
Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)* γ>0

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-if γ>0, then there is an exponential term e^(γt) in the solution so solutions will are unstable and will grow as t->∞

-trajectories of such solutions on the phase plane are unfolding spirals

-direction depends on b from original matrix A, b>0 clockwise, b<0 anticlockwise

-the fixed point is an unstable node

25

##
Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)* γ<0

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-if γ<0, then the exponential term e^(γt) is vanishing as t->∞

-therefore solutions are asymptotically stable

-trajectories of such solutions in the phase plane are spirals winding towards the origin

-direction depends on b from original matrix A, b>0 clockwise, b<0 anticlockwise

-the fixed point is a stable node

26

##
Linear Systems With Constant Coefficients

N=2 Case 2 : λ1=λ2

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-if T²=4D, then λ1=λ2

-there are two subcases:

2a) matrix A is diagonal and thus of the form A=λ1*I

2b) matrix A is not diagonal and thus A-λ1*I≠0

27

##
Linear Systems With Constant Coefficients

N=2 Case 2a : λ1=λ2 and A diagonal

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-in this case λ1=λ2 and A=λ1*I

-equations are of the form:

x' = λ1*x

y' = λ1*y

-trajectories satisfy the equation dy/dx = y/x which has general solution y=Γ*x where Γ is an arbitrary constant

-so trajectories are straight lines passing through the origin

-any two linearly independent vectors |V1 and |V2 are eigenvectors of the matrix A corresponding to eigenvalue λ1

-the general solution is:

x(t) = c1*exp(λ1t)

y(t) = c2*exp(λ1t)

-if λ1>0 then the fixed point is unstable and is called an unstable star

-if λ1<0 the fixed point is stable and called a stable star

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##
Linear Systems With Constant Coefficients

N=2 Case 2b : λ1=λ2 and A not diagonal

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-in this case we have only one eigenvector |V1

-it provides us with only one solution:

|x1(t) = |V1 * exp(λ1t)

-of the system |x'=A|x

-we seek a second linearly independent solution of the form:

|x2(t) = t*|V1exp(λ1*t) + |Wexp(λ1*t)

-where |W is a constant vector to be found

-differentiate and sub into the original system, to get

A|W = λ1*|W + |V1 the vector |W satisfying this is a generalised eigenvector

-general solution:

|x(t) = C1*|V1*exp(λ1*t)

+ C2*[t*|V1*exp(λ1*t) + |W*exp(λ1*t)]

-if λ1<0, the fixed point is a stable degenerate/ stable improper node

-if λ1>0, the fixed point is unstable degenerate/ stable improper node

29

##
Linear Systems With Constant Coefficients

N=2 Case 2a : λ1=λ2=0 A=λ1*I

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-this is the most trivial case of the system

-in this case x'=0 and y'=0

-general solution:

|x(t) = |xo

-i.e. a constant vector

-every point on the phase plane is a fixed point

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