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)〉.