Chapter 2 Flashcards Preview

22 > Chapter 2 > Flashcards

Flashcards in Chapter 2 Deck (17):
1

First order ODE: qualitative analysis

Examine first order odes with general form: du/dt = f(t,u)

•direction fields
•phase line
•planar odes...

2

Direction fields

For first order ode: du/ft=f(t,u)
•select a point (t,u)
•draw short vector with gradient f(t,u)
•map out region of interest using table of values which f(t,u) are referred to as tangent slopes

*solution curves don't intersect usually, if they go can solve with initial value

Examples that are translationally invariant in t: a direction field is translationally invariant in t if the rate of T doesn't depend on t. Ie du/dt = F(u)
*if translationally invariant in t then slopes of tangent lines/ vectors are the same

*translationally invariant in t usually t doesn't appear or...at least doesn't affect g(u)

3

Autonomous

Differential equation doesn't depender explicitly on the independent variable ie invariant also

du/ft = f(t,u) = g(u)

4

Phase line

If autonomous first order ODE
Phase line:
•finding equilibrium points du/dt=0
These correspond to solutions u that remain constant for all t , rate of change is 0
•for values in between EPS determine sign of function du/dt. Only change here. DRAW FUNCTION
•marking point on phase line and arrowhead for sign
STABLE if points towards, UNSTABLE if either side outwards
**0 will always be an equilibrium solution
**phase line tells us how solutions that start with different initial conditions behave.

Eg decays to -infinity if initial 0
**if the initial condition belongs to one region then sol satisfying this initial condition stays in the region eg if initial condition in certain region decays to value in region or end point

5

Planar autonomous first order

Form du/dt =f(u,v) and dv/dt =g(u,v)
System at time t can be described by point (u,v)
Autonomous may depend only on one independent
Usually converting 2nd order ODEs into planar first order by setting y=dx/dt
Where dx/dt =y is one of the two.

We use a phase plane.

6

Phase plane

Used for first order planar system
•finding EPs (u_e, v_e) st both functions f(u_e, v_e) = g(u_e, v_e) =0
•if the system not at EP then as the value of the independent variable t changes (u(t), v(t)) moves around the plane tracing a curve in (u,v)-plane called THE TRAJECTORY OF THE SYSTEM
**this is called the curve of particular solution by using the tangent vectors

• we construct the table for values of (x•, y•) evaluated at range of points for both x and y. Eg -1 to 3/2 for y and -pi to 2pi for x. Note some may be invariant in one variable some may not.
•we then draw these vectors and can draw phase portrait for EPs....
• usually trajectories cross perpendicular at coordinate axes up down up down etc

7

Nullclines

Nullclines are drawn on phase portraits.
u-nullclines: C_u = {(u,v) : f(u,v) =0} ie curve for u• =0 du/dt=0


v-nullclines: C_v = {(u,v) : g(u,v) =0} ie curve for v• =0 dv/dt=0

System trajectories cross these: C_u vertically as u•=0 hence rate of change in v only and C_v horizontally...
•they also cross only at EPs as both for v•=0 and u•=0
• nullclines split the plane into 4 regions in each direction of trajectory
*** we pick 4 points one in each region and find values of u• and v• assigning NE NW SE and SW, help see if spiral direction

8

Linear systems- linear planar systems

These are systems that can be represented in matrix form

du/dt = a_11 u + a_12 v dv/dt = a_21 u + a_22 v


Nullclines: straight lines passing through the origin. non zero constants

v= -(a_11/a_12)u

V=(-a_21/a_22)u
If we have that a_11a_22=/=a_12a_21 then nullclines only at (0,0)

At eps Ax =0 vector

9

Linear systems:
In matrix form d(vector(S))/dt = A vector(S)

Vector (S) = (u,v)

A= [a_11 a_12]
[a_21 a_22]

Significance of eigenvalues

Seeking directions vector(w) in (u,v) space such that

d(vector(w))/dt = Aw = λw
(As vectors)
Ie eigenvectors and eigenvalues
This equation will be a linear ODE with solution:
Found by separating variables
Vector(w(t)) = vector(w_0) e^(λt)
Where w_0 = w(0)


Therefore if the system has an eigenvector we know it's exact solution at all times. We can determine it's trajectory in terms of λ.

10

Linear system: as a matrix product. Significance of trace and determinant

The eigenvalues for the matrix are given from solutions:

λ^2 -Tλ + D =0

Where T=Tr(A) = a_11 + a_22
*is also the sum for a symmetric
D=Det(A)= a_11a_22 - a_12a_21
*is also the product for a symmetric

11

Linear systems: as represented by matrix, finding eigenvalues and classification of stationary points

Roots for two possible λ values:

λ±= 0.5( T ± sqrt(T^2 -4D))

Classification: 9 cases

•STABLE NODE: λ± both real and negative
D>0, T^2>4D and T<0
•UNSTABLE NODE:λ± both real and positive
D>0, T^2>4D and T>0
•STABLE SPIRAL: λ± are complex and Re(λ±) are both negative.
D>0,T^2<4D and T<0
•UNSTABLE SPIRAL: λ± are complex and Re(λ±) are both negative
D>0, T^2<4D and T>0
•SADDLE: λ± are both real and λ+>0 and λ-<0
D<0
•CENTRE: λ± are both imaginary
D>0 and T=0
•STABLE DEGENERATE NODE: only one eigenvalue- real and negative
D>0, T^2=4D and T<0
•UNSTABLE DEGENERATE NODE: only one eigenvector- real and positive.
D>0, T^2 =4D and T>0
•LINE OF FIXED POINTS: λ+ =T and λ-=0
D=0

12

Classification of stationary points

GRAPH

Graph:

TD graph:
Left hand stable side: stable spiral above T^2 =4D and stable node under.
On T^2 =4D degenerate node.
Right hand side: unstable, unstable spiral above unstable node below.


D
|
CENTRE
|
---------->T
|
SADDLE
|

Saddle when D is negative, any T
Centre when D is positive and T is 0
Degenerate node on the T^2 =4D graph, D positive and T affects stable or unstable
Line of fixed points when D=0. One eigenvalue is T and the other 0

13

Theorem: considering a linear system with two distinct eigen values and corresponding eigenvectors

For the given linear system (defined previously) if A has two distinct eigenvalues with corresponding eigenvectors w.

Then the equation of a trajectory is given by

Vector(s(t))
= α_0e^(λ₁t) vector(w₁) + β_0e^(λ₂t) vector(w₂)

where vector(s(0)) = αvector(w₁)+βvector(w₂) is the starting point for the trajectory at time t=0

That is the trajectory can be described as a linear combination of the two vectors.

14

Vector(s(t))
= α_0e^(λ₁t) vector(w₁) + β_0e^(λ₂t) vector(w₂)

where vector(s(0)) = αvector(w₁)+βvector(w₂) is the starting point for the trajectory at time t=0

CASES IN THE CLASSIFICATION:
Eigenvalues are real

• both the eigenvalues are real:
If the entries of A are real then so are the eigenvectors and so the vector(w_i) specify directions in the phase plane. If one positive the contribution of its eigenvector to the state increases with time, if one negative its eigenvector decreases the state decreases with time.
••• cases of these:
• both negative do eigenvector coefficients decrease with time so trajectories move towards the equilibrium point, STABLE NODE
•both positive so eigenvectors increase in time and so all trajectories move away from the equilibrium point, UNSTABLE NODE
• if one pos one neg trajectories close to either move down and away either sides, SADDLE

Note that if initial state is exactly on one eigenvector then it moves along the direction the whole time, this no trajectories cross the eigenvector at a non equilibrium point.

The eigenvectors also split the phase space into four distinct sections and trajectories must therefore stay within these sectors

The larger the magnitude of eigenvalue the faster contribution of eigenvector grows or shrinks therefore for a stable node trajectories off the eigenvectors are curves towards the eigenvector with the largest eigenvalue

If negative arrow inwards sink (stable)
If positive eigenvalue arrow outwards source ( unstable)

15

Vector(s(t))
= α_0e^(λ₁t) vector(w₁) + β_0e^(λ₂t) vector(w₂)

where vector(s(0)) = αvector(w₁)+βvector(w₂) is the starting point for the trajectory at time t=0

CASES IN THE CLASSIFICATION:
Eigenvalues are not real

In this case they're always complex conjugated and the eigenvectors are complex, written as:

λ₁ = σ +iω
λ₂= /λ₁ = σ −iω

Vector(w_1) = (1,z)
Vector(w_2) = conjugate(w_1) = (1, conjugate(z))

Where z = (σ + iω - a_11)/ (a_12)

Combining with trajectory equation gives that for vector(s(t)) to be real we require that alpha_0 equals beta_0.

We can find exact solutions:
.....

16

...

....

17

Linearisation of a non linear planar system

Mea