5. Initial Value Problems

Question 1

First Order IVPs

General Form

Answer 1

x’ = f(x(t),t)

-with x(0)=xo

Question 2

Second Order IVPs

General Form

Answer 2

x’’ = f(x’(t),x(t),t)

-with x(0)=xo and x’(0)=xo’

Question 3

First Order IVPs - Standard Expansion

Steps

Answer 3

1) suppose that:
x ~ xo(t) + ε*x1(t) + ε²x2(t)+...
as ε->0
2) sub into equation
3) consider O(1) terms, then O(ε), then O(ε²) etc.

Question 4

Second Order IVPs

Most Fundamental Form

Answer 4

x’’ + ωo²x = 0

-with initial conditions

Question 5

Second Order IVPs

Fundamental Form - General Solution

Answer 5

x = a cos(ωot) + b sin(ωot)
OR
x = Ae^(iωot) + Be^(-iωot)
= Ae^(iωot) + A* e^(-iωot)
-since x(t) needs to be real
-to fit the initial conditions, use a and b i the first equation or Re(A) and Im(A) in the second

Question 6

Second Order IVPs

Fundamental Form - Period and Frequency

Answer 6

• solution is oscillatory and periodic in time, once cycle corresponds to ωot increasing by 2π
• so the period is, T=2π/ωo
• and ωo is the frequency

Question 7

Second Order IVPs

Forced Systems

Answer 7

x’’ + ωo²x = α cosωt + β sinωt

-where ωo is the natural frequency and ω is the forcing frequency

Question 8

Second Order IVPs

Forced Systems - Complimentary Function

Answer 8

xcf = a cosωot + b sinωot

-determine a and b with initial conditions

Question 9

Second Order IVPs

Forced Systems - Particular Integral

Answer 9

xpi = C cosωot + D sinωot
-where:
C = α / (ωo²-ω²)
-and
D = β / (ωo² - ω²)

Question 10

Second Order IVPs

Forced Systems - Resonant Response

Answer 10

-there is a resonant response when ω->ωo

Question 11

Second Order IVPs

Forced Systems - Forcing at Resonant Frequency

Answer 11

-governing equation:
x’’ + ωo²x = α cosωot + β sinωot
=>
x(t) = a cosωot + b sinωot - β/2ωo t cos(ωot) + α/2ωo t sin(ωot)

Question 12

Motion of a Non-Linear Pendulum

Equation

Answer 12

d²θ/dq² + g/L sinθ = 0
-with:
θ(0)=θo
dθ(0)/dq = 0
-where θo>0

Question 13

Motion of a Non-Linear Pendulum

Non-Dimensionalising

Answer 13

-introduce a non-dimensional time, t, with t∝q where q has dimensions of seconds:
t = √[g/L] q

Question 14

Motion of a Non-Linear Pendulum

Conservation of Energy

Answer 14

• the non-linear pendulum system conserves energy

- prove this by multiplying by dθ/dt

Question 15

Motion of a Non-Linear Pendulum

Solution

Answer 15

-make the substitution x(t) = θ/θo
-apply a Taylor expansion to sine
=>
x’’ + x = εx³/6 - ε²x^5/120 + O(ε³)
-gather terms of like order and apply the boundary conditions

Question 16

Motion of a Non-Linear Pendulum

Time Period at Leading Order

Answer 16

-at leading order:
x(t) ~ cost
=>
T = 2π

Question 17

Motion of a Non-Linear Pendulum

Time Period at O(ε)

Answer 17

-we anticipate:
T = 2π + 4α1ε + O(ε²)
-we know that x=0 when t=T/4=π/2 + α1ε + O(ε²)
-sub in to the O(ε) expression for x(t)

Question 18

Motion of a Non-Linear Pendulum

Secular Term

Answer 18

• the order ε expansion for x(t) contains a secular term ∝ tsint which grown in time
• the expansion is valid when t=O(1) but the secular term makes it invalid when εt=Os(1)
• to improve this, alternative techniques are needed

Question 19

Method of Multiple Scales

Aim

Answer 19

-to develop an asymptotic expansion that works for both t=O(1) and t=Os(1/ε), as ε->0

Question 20

Method of Multiple Scales

Time

Answer 20

-define fast and slow timescales:
τ = t and T = εt
-so if t=O(1) then τ=O(1) but T is small
-if t=Os(1/ε) then τ is large but T is O(1)
-as the name suggests, multiple of these timescales can be defined

Question 21

Method of Multiple Scales

Steps

Answer 21

1) define time scales e.g. τ=t and T=εt
2) we then seek solutions x(τ ,T) rather than x(t), so rewrite the governing equation in terms of τ and T
3) assume that x ~ xo + εx1 + ε²x2 +…
4) consider O(1) terms from the governing equation, then O(ε) terms etc.
5) make choices about the unknown constants which will eliminate the secular terms
6) convert back to original timescale, t

Question 22

Linstedt-Poincare Technique

Aim

Answer 22

-we use LPT when the initial value problem with ε≠0 (the perturbed problem) has a frequency shift from the unperturbed problem

Question 23

Linstedt-Poincare Technique

Strained Coordinate

Answer 23

-replace t by a strained coordinate:
τ = (1 + εω1 + ε²ω2 + …)t
-where we choose ω1,ω2,… to remove secularities

Question 24

Linstedt-Poincare Technique

Steps

Answer 24

1) introduce strained coordinate τ = (1 + εω1 + ε²ω2 + …)t
2) rewrite governing equation and boundary conditions in terms of τ instead of t
3) let x ~ xo + ε x1 + ε² x2 + ….
4) consider O(1) terms, choose ω’s to remove forcing at resonant frequency (i.e. to avoid secularities)
5) then do the same for O(ε) terms and so on

Question 25

Renormalisation

Aim

Answer 25

-similar technique to LPT
-starts from the results of a completely standard asymptotic expansion using:
x(t) ~ xo(t) + εx1(t) + ε²x2(t) + …
-the results of which is known to be secular

Question 26

Renormalisation

Straining Functions

Answer 26

-introduce:
t = τ + ε f1(τ) + ε² f2(τ) + …
-where the straining functions fj(τ) will be chosen to eliminate secularities

Question 27

Renormalisation

Steps

Answer 27

1) introduce strained functions, t = τ + ε f1(τ) + ε² f2(τ) + …
2) take the standard expansion with secularities that need removing and rewrite in terms of τ
3) choose fj(τ) to eliminate the secularities
4) truncate series at desired order