Phase Portraits and Linearisation

Question 1

Linearisation

Description

Answer 1

-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

Question 2

Nullcline

Definition

Answer 2

• exist in all dimensions but this definition is for the 2D case:
• lines where the vector field is either horizontal or vertical

Question 3

Isocline

Definition

Answer 3

exist in all dimensions but this definition is for the 2D case:
-lines where the vector field has a constant gradient

Question 4

Phase Portrait

Description

Answer 4

-given a 2D system:
x’ = f(x,y)
y’ = g(x,y)
-the phase portrait plots y against x with trajectories drawn on

Question 5

How to go about drawing a phase portrait

Answer 5

-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
*******

Question 6

Horizontal Nullclines

Answer 6

-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

Question 7

Vertical Nullclines

Answer 7

-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

Question 8

Linearisation

N=2`

Answer 8

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

Question 9

Linear Systems With Constant Coefficients

N=1

Answer 9

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)

Question 10

Linear Systems With Constant Coefficients

N≥1 , General Equation

Answer 10

|x’ = A |x

• where A is an NxN matrix
• and |x ϵ R^N

Question 11

Linear Systems With Constant Coefficients

N=2 General Equation

Answer 11

|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

Question 12

Linear Systems With Constant Coefficients

N=2 Eigenvalue Problem

Answer 12

|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

Question 13

Linear Systems With Constant Coefficients

N=2 Eigenvalues and Eigenvectors

Answer 13

-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

Question 14

Linear Systems With Constant Coefficients

N=2 Characteristic Equation

Answer 14

-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

Question 15

Linear Systems With Constant Coefficients

N=2 Case 1 : λ1≠λ2

Answer 15

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)

Question 16

Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real

Answer 16

• 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

Question 17

Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)*

Answer 17

-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

Question 18

Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (+,+)

Answer 18

• 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

Question 19

Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (-,-)

Answer 19

• 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

Question 20

Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (+,-)

Answer 20

-if two eigenvalues have different sign: λ2<0

Question 21

Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (0,+)

Answer 21

• 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

Question 22

Linear Systems With Constant Coefficients

N=2 Case 1a : λ1,2 real (0,-)

Answer 22

• 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

Question 23

Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)* γ=0

Answer 23

• 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

Question 24

Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)* γ>0

Answer 24

• 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

Question 25

Linear Systems With Constant Coefficients

N=2 Case 1b : λ2 = (λ1)* γ<0

Answer 25

• 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

Question 26

Linear Systems With Constant Coefficients

N=2 Case 2 : λ1=λ2

Answer 26

• if T²=4D, then λ1=λ2
• there are two subcases:
  2a) matrix A is diagonal and thus of the form A=λ1I
  2b) matrix A is not diagonal and thus A-λ1I≠0

Question 27

Linear Systems With Constant Coefficients

N=2 Case 2a : λ1=λ2 and A diagonal

Answer 27

-in this case λ1=λ2 and A=λ1I
-equations are of the form:
x’ = λ1x
y’ = λ1y
-trajectories satisfy the equation dy/dx = y/x which has general solution y=Γx 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) = c1exp(λ1t)
y(t) = c2exp(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

Question 28

Linear Systems With Constant Coefficients

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

Answer 28

-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(λ1t) + |Wexp(λ1t)
-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|V1exp(λ1t)
+ C2[t|V1exp(λ1t) + |Wexp(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

Question 29

Linear Systems With Constant Coefficients

N=2 Case 2a : λ1=λ2=0 A=λ1*I

Answer 29

-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

Question 30

Linear Systems With Constant Coefficients

N=2 Case 2b : λ1=λ2=0 A≠λ1*I

Answer 30

-in this case the general solution is given by:
|x(t) = C1*|V1*exp(λ1*t) 
\+ C2*[t*|V1*exp(λ1*t) + |W*exp(λ1*t)]
=>
|x(t) = C1*|V1 + C2[t*|V1 + |W]
-i.e. it is a linear function of t
-there is a line of fixed points (C1|V1)
-every trajectory is parallel to this line of fixed points