The following paper
https://core.ac.uk/reader/82037870
Oscillators with nonlinear elastic and damping forces
L.Cveticanin
studies the general problem
$$ \ddot{x} + 2 b_k \, \dot{x} \, |\dot{x}|^k + \omega^2 x = 0 $$
The author introduces slow degrees of freedoms and integrates out (averages over) the fast ones, using the ansatz (17)
$$ x(t) = A(t) \, \sin(\omega t + \theta) $$
My question is about the slow amplitude $A(t)$.
Using $k=2$ and (55) one finds a solution
$$ A(t) = \frac{A_0}{1 + A_0 \xi t} $$
$$ A_0 = A(0) $$
This agrees with a straight forward approach using Krylov–Bogoliubov averaging.
First question: Is there any reason why the author restricts this solution to $k < 1$? I used a different approach for $k=2$ and found exactly the same expression.
Then the author claims that "in Eq (55), the first two terms represent the series expansion of an exponential function and Eq.(55) can be rewritten as"
$$ A(t) = A_0 \exp\left(- A_0 \xi t\right) $$
Second question: Is there any further justification for this in this 'improvement'?
I have doubts.
For (55) one can immediately show that
$$ A(t) = \frac{A_0}{1 + A_0 \xi t} \stackrel{!}{=} \frac{A_1 }{1 + A_1 \xi (t-t_1)} $$
$$ A_1 = A(t_1) = \frac{A_0 }{1 + A_0 \xi t_1} $$
which means that one shifts the initial condition from $t_0=0$ to some $t_1$.
However, this fails for the exponential approach
$$ A(t) = A_0 \exp\left(- A_0 \xi t\right) \stackrel{!}{=} A_1 \exp\left(- A_1 \xi (t-t_1)\right) $$
$$ A_1 = A(t_1) = A_0 \exp\left(- A_0 \xi t_1\right) $$
