Nonlinear Dynamics Lec 3 (but actually 2)

Question 1

What nonlinearity does perturbation theory consider

Answer 1

stiffness of k3y^3 where k3 is equal to epsilon which is very small

Question 2

Why do we use pertubation theory

Answer 2

In the complete range of systems linear systems are a single point (very few real ones compared to all possible systems) perturbation allows us to expand the systems we can get solutions to. Very close to linear but not quite

Question 3

Can we solve nonlinear systems

Answer 3

Very few exact mathematical solutions for nonlinear equations

Question 4

What is the idea of perturbation theory

Answer 4

nonlinearity is small so the solution will be close to the linear system

Question 5

For perturbation theory what do we treat as variables

Answer 5

we say y is a function of t and epsilon, this means we can compute the taylor series expansion of y
y(t,eps) = y0(t) + epsy1(t) + eps^2 * y2(t) +…
y = yo + epsy1 + eps^2 * y2 + …

Question 6

If we simplify the taylor expansion of y(t,eps) to say epsilon is zero what do we get

Answer 6

y(t) = y0(t) i.e the linear term which makes sense

Question 7

How do we simplify the taylor expansion in perturbation theory

Answer 7

As said epsilon is very small, assume all terms with eps^2 are 0

Question 8

What equation does the perturbation work on

Answer 8

Free, undamped duffing equation i.e

y.. + y + eps*y^3 = 0

Question 9

What trial solution do we sub into duffings equation and what does this simplify to

Answer 9

simplified taylor exanpsion y = y0 + eps*y1

simplifies to epsy1.. + epsy1 + eps*y0^3 = 0

Question 10

What system of equations do we get from simplified duffings equation y0.. + epsy1.. + y0 + epsy1 + eps*y0^3 = 0

Answer 10

Equate powers of epsilon as equation must be true for all values of epsilon therefore
y0.. + y0 = 0
y..1 + y1 = -y0^3
with similar equations for higher order terms if the exapnsion were allowed to continue

Question 11

From the perturbation system of equations y0.. + y0 = 0 and y..1 + y1 = -y0^3 what can we determine

Answer 11

First equation is linear, and once first equation is fixed, second equation is also linear
y0 = Y cost (t) when y0(0) = Y and y.(0) = 0

Question 12

What is de moivres theorem

Answer 12

cos(t) = 1/2 * (e^it + e^-it)
sin(t) = 1/2i * (e^it - e^-it)

Question 13

How do we subsitue y0 = Y cost into y0^3

Answer 13

Use demoivres theorem

Question 14

How do we solve the y1 equation

Answer 14

Equation is linear but inhomogenous, has complementary function and particular integral
CF (when equation equal to zero) simply y1 = Ycost again
PI more complicated assume y1 = Ct cos(t) + D cos(3t) (attempt first then add in a phase to Ct cos(t) term

Question 15

How do we finalise the solution for y1 in perturbation theory

Answer 15

Equate cosine terms to find constants then define y1(t) as equal to complimentary function + particular integral and then y(t) = y0(t) + eps*y1(t)

Question 16

What is the final trick in perturbation theory to finalise the solution

Answer 16

Have to share out the initial among yi(0) such that y(0) = Y. Usally assign it all to y0 term

Question 17

What are the steps in basic perturbation theory

Answer 17

1) Taylor series expansion of y(t, eps) and simplify
2) Sub simplified taylor series into undamped free duffing equation y.. + y + epsy^3
3) Equate powers of epsilon
4) Guess at solutions for y0 and y1 remember pi/2!
5) Simplify with De Moivres Theorem and equate terms
6) Find PI and CF solution for y1
7) Bring together for y = y0 + epsy1 remember to sort out initial conditions

Question 18

Whats the problem with basic perturbation theory solution

Answer 18

t sin(t) term is not periodic it grow without bond as t increases, it is a secular term and an artefact of the perturbation method (would cancel with a term or terms later in series expansion)
Want a valid solution at any given level of expansion

Question 19

How do we resolve the secular term in the perturbation

Answer 19

Logically Y increases with natural frequency
Replace original equation with
y.. + w^2(eps)*y + eps(y^3) = 0 with w^2 = 0

Question 20

What is the new taylor series expansion for the updated perturbation method

Answer 20

w^2(eps) = 1 + epsalpha1 + eps^2alpha2+…

Question 21

How do the y0 and y1 equations change in the updated perturbation method

Answer 21

y0 stays the same

y1.. + y1 = -alpha1*y0 - y0^3

Question 22

What is the benefit of the update perturbation method

Answer 22

Recover amplitude dependence of the frequency

Question 23

What is the full duffing equation

Answer 23

my.. + cy. + ky + k3y^3 = x(t)

Question 24

Derive the scaled duffing equation

Answer 24

See presentation
start - my.. + cy. + ky + k3y^3 = x(t)
use - u = alphay p = betax tau=gamma*t
solution beta = 1, alpha = k, gamma = sqrt (k/m)

Question 25

What is the critical damping ratio equal to

Answer 25

c / critical damping

critical damping = 2 * sqrt(k*m)