The expression for the averaging equation for magnitude is:

x(t,ε) = x0+O(ε)

Here, x is the position and x0 is the initial position of the system.

The expression of initial displacement x0 is:

x0 = r(T)cos(ϕ(T)+τ)

Here, r(T) is the radius of the polar coordinate system, and ϕ is the angle formed by the radius.

The expression of the amplitude of any limit cycle for the original system is r=r0+O(ε).

The expression of the frequency of any limit cycle for the original system is:

ω = 1+εϕ'

Here, ϕ' is the Taylor series expansion of the function ϕ(T).

Taylor series expansion of x(t,ε) is :

x(t,ε)∼x(t,T)+O(ε)

Here, O(ε) are the higher order terms of the Taylor series expansion.

(a)

The first derivative of r(t) and ϕ(t) is expressed as follows:

The system equation is:

x¨+x+εh(x,x˙,t) = 0

And,

x(t) = r(t)cos(t+ϕ(t))x˙(t) = -r(t)sin(t+ϕ(t))

Differentiate function x(t) with respect to time,

ddtx(t) = ddtr(t)cos(t+ϕ(t))x˙(t) = r˙cos(t+ϕ)-r(1+ϕ˙)sin(t+ϕ)

Substitute -rsin(t+ϕ) for x˙(t) in the above,

-rsin(t+ϕ)=r˙cos(t+ϕ)-r(1+ϕ˙)sin(t+ϕ)-rsin(t+ϕ) = r˙cos(t+ϕ)-rsin(t+ϕ)-rϕ˙sin(t+ϕ)r˙cos(t+ϕ)−rϕ˙sin(t+ϕ) = 0

Further differentiate the function x˙(t) from the above expression,

dx˙(t)dt = d(-r(t)sin(t+ϕ(t)))dtx¨(t) = -r˙sin(t+ϕ(t))-r(1+ϕ˙)cos(t+ϕ(t))

Substitute -r˙sin(t+ϕ(t))-r(1+ϕ˙)cos(t+ϕ(t)) for x¨(t) and r(t)cos(t+ϕ(t)) for x(t) in the above expression of x¨+x+εh(x,x˙,t)

-r˙sin(t+ϕ(t))-r(1+ϕ˙)cos(t+ϕ(t))+r(t)cos(t+ϕ(t))+εh = 0-r˙sin(t+ϕ(t))-rϕ˙cos(t+ϕ(t))+εh = 0r˙sin(t+ϕ(t))+rϕ˙cos(t+ϕ(t)) = εh

From the above calculation of r(t) and ϕ(t), there are two expressions obtained,

r˙cos(t+ϕ)-rϕ˙sin(t+ϕ) = 0r˙sin(t+ϕ(t))+rϕ˙cos(t+ϕ(t)) = εh

Multiply cos(t+ϕ) in equation r˙cos(t+ϕ)-rϕ˙sin(t+ϕ) = 0 and sin(t+ϕ) in equation r˙sin(t+ϕ(t))+rϕ˙cos(t+ϕ(t)) = εh

Therefore, the equation is:

r˙cos2(t+ϕ(t))-rϕ˙sin(t+ϕ(t))cos(t+ϕ(t)) = 0r˙sin2(t+ϕ(t))+rϕ˙cos(t+ϕ(t))sin(t+ϕ(t)) = εhsin(t+ϕ(t)) 

sin2θ+cos2θ=1

Add the above two expression,

r˙cos2(t+ϕ(t)) - rϕ˙sin(t+ϕ(t))cos(t+ϕ(t))+r˙sin2(t+ϕ(t))+rϕ˙cos(t+ϕ(t))sin(t+ϕ(t))=εhsin(t + ϕ(t)) From the trigonometric expression,

sin2θ+cos2θ=1

r˙ = εhsin(t + ϕ(t))

Multiply sin(t+ϕ) in equation r˙cos(t+ϕ)- rϕ˙sin(t+ϕ) = 0 and cos(t+ϕ) in equation r˙sin(t+ϕ(t))+rϕ˙cos(t+ϕ(t))=εh

r˙cos(t+ϕ(t))sin(t+ϕ(t))-rϕ˙sin2(t+ϕ(t))=0r˙sin(t+ϕ(t))cos(t+ϕ(t))+rϕ˙cos2(t+ϕ(t))sin(t+ϕ(t)) = εhcos(t+ϕ(t))

Adding the above expressions,

rϕ˙ = εhcos(t+ϕ(t))

Hence, expression of r˙ and rϕ˙ is rϕ˙=εh cos(t+ϕ(t))r˙=εh sin(t+ϕ(t)).

(b)

The expression for 〈r〉(t) is given in the question is:

〈r〉(t) = 12π∫t-πt+πr(τ)dτ

From Leibniz’s integral rule, considered the function r˙(t) is continuous for τ∈[t-π, t+π].

Differentiate the above expression,

d〈r〉dt = ddt(12π∫t-πt+πr(t)dτ)d〈r〉dt = 12π(r(t+π)(ddt(t+π))-r(t-π)(ddt(t-π))∫t-πt+π∂∂tr(τ)dτ)d〈r〉dt = 12π(r(t+π)(1)-r(t-π)(1)+∫t-πt+π0 dτ)d〈r〉dt = 12π(r(t+π)-r(t-π))

d〈r〉dt = 12π∫t-πt+πddτr(τ)dτd〈r〉dt = 〈drdt〉

Hence, expression d〈r〉dt = 〈drdt〉 is proved.

(c)

d〈r〉dt = 〈drdt〉d〈r〉dt = 〈r˙(t)〉

Substitute ∈h(x,x˙,t)sin(t+ϕ) for r˙(t) in above,

〈r˙(t)〉 = 〈∈h(x,x˙,t)sin(t+ϕ)〉

Substitute -r(t)sin(t+ϕ(t)) for x˙(t) and r(t)cos(t+ϕ(t)) for x(t) in the above expression,

〈r˙(t)〉 = 〈∈h(r(t)cos(t+ϕ(t)), - rsin(t+ϕ(t)),t)sin(t+ϕ)〉

Therefore the expression of 〈r˙(t)〉 is

〈r˙(t)〉 = 〈∈h(r(t)cos(t+ϕ(t)), - rsin(t+ϕ(t)),t)sin(t+ϕ)〉

(d)

The magnitude and phase expression for the system equation is calculated as:

Apply Taylor series expansion of r(t) and ϕ(t),

r(t) = r¯ + O(ε)ϕ(t) = ϕ¯+ O(ε)

The above expression is true only because the average value of one oscillation and the rate of change over that cycle,

r = r¯+〈r˙〉r = r¯+ε〈hsin(t+ϕ)〉r(t) = r¯+O(∈)

ϕ= ϕ¯+〈ϕ˙〉ϕ= ϕ¯+∈r¯〈hcos(t+ϕ)〉ϕ(t) = ϕ¯(t)+O(∈)

Therefore, from above expression for 〈r˙(t)〉 and ϕ˙(t),

dr¯dt= 〈r˙〉dr¯dt= ε〈(h(r¯ cos(t + ϕ¯)), - r¯ sin(t + ϕ¯), t)sin(t + ϕ)〉dr¯dt= ε〈(h(r¯ cos(t + ϕ¯)), - r¯ sin(t + ϕ¯), t)sin(t + ϕ) + O(ε)〉dr¯dt= ε〈(h(r¯ cos(t + ϕ¯)), - r¯ sin(t + ϕ¯), t)sin(t + ϕ) + O(ε2)〉

r¯dϕ¯dt = 〈rϕ˙〉r¯dϕ¯dt= ε〈(h(r(t) cos(t + ϕ¯)), - r(t) sin(t + ϕ¯), t)cos(t + ϕ¯)〉r¯dϕ¯dt= ε〈(h(r¯ cos(t + ϕ¯)), - r¯ sin(t + ϕ¯), t)cos(t + ϕ¯) + O(ε)〉r¯dϕ¯dt= ε〈(h(r¯ cos(t + ϕ¯)), - r¯ sin(t + ϕ¯), t)cos(t + ϕ¯) + O(ε2)〉

Therefore, the expression of dr¯dt and r¯dϕ¯dt is

dr¯dt = ε〈(h(r¯ cos(t + ϕ¯)), - r¯ sin(t + ϕ¯), t)sin(t + ϕ¯) + O(ε2)〉r¯dϕ¯dt= ε〈(h(r¯ cos(t + ϕ¯)), - r¯ sin(t + ϕ¯), t)cos(t + ϕ¯) + O(ε2)〉.