Nonlinear Dynamics Lec 3 (but actually 2) Flashcards
What nonlinearity does perturbation theory consider
stiffness of k3y^3 where k3 is equal to epsilon which is very small
Why do we use pertubation theory
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
Can we solve nonlinear systems
Very few exact mathematical solutions for nonlinear equations
What is the idea of perturbation theory
nonlinearity is small so the solution will be close to the linear system
For perturbation theory what do we treat as variables
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 + …
If we simplify the taylor expansion of y(t,eps) to say epsilon is zero what do we get
y(t) = y0(t) i.e the linear term which makes sense
How do we simplify the taylor expansion in perturbation theory
As said epsilon is very small, assume all terms with eps^2 are 0
What equation does the perturbation work on
Free, undamped duffing equation i.e
y.. + y + eps*y^3 = 0
What trial solution do we sub into duffings equation and what does this simplify to
simplified taylor exanpsion y = y0 + eps*y1
simplifies to epsy1.. + epsy1 + eps*y0^3 = 0
What system of equations do we get from simplified duffings equation y0.. + epsy1.. + y0 + epsy1 + eps*y0^3 = 0
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
From the perturbation system of equations y0.. + y0 = 0 and y..1 + y1 = -y0^3 what can we determine
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
What is de moivres theorem
cos(t) = 1/2 * (e^it + e^-it) sin(t) = 1/2i * (e^it - e^-it)
How do we subsitue y0 = Y cost into y0^3
Use demoivres theorem
How do we solve the y1 equation
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
How do we finalise the solution for y1 in perturbation theory
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)
What is the final trick in perturbation theory to finalise the solution
Have to share out the initial among yi(0) such that y(0) = Y. Usally assign it all to y0 term
What are the steps in basic perturbation theory
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
Whats the problem with basic perturbation theory solution
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
How do we resolve the secular term in the perturbation
Logically Y increases with natural frequency
Replace original equation with
y.. + w^2(eps)*y + eps(y^3) = 0 with w^2 = 0
What is the new taylor series expansion for the updated perturbation method
w^2(eps) = 1 + epsalpha1 + eps^2alpha2+…
How do the y0 and y1 equations change in the updated perturbation method
y0 stays the same
y1.. + y1 = -alpha1*y0 - y0^3
What is the benefit of the update perturbation method
Recover amplitude dependence of the frequency
What is the full duffing equation
my.. + cy. + ky + k3y^3 = x(t)
Derive the scaled duffing equation
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)
What is the critical damping ratio equal to
c / critical damping
critical damping = 2 * sqrt(k*m)
