Iterative method for solving equation of motion in wave field

Consider the motion of a test particle moving in a longitudinal wave,

$$\mathbf{E}= \hat{\mathbf{z}} E \cos (k z - \omega t),$$ (12)

then the equation of motion is given by

$$m \frac{d v}{d t} = q E \cos (k z - \omega t),$$ (13)

and

$$\frac{d z}{d t} = v,$$ (14)

with initial condition $v (0) = v_0$ and $z (0) = z_0$, where $v \equiv \mathbf{v} \cdot \hat{\mathbf{z}}$ and $\mathbf{v}$ is the velocity of the particle. Equations (13) and (14) are nonlinear system, for which exact solutions are hard to be found. Here, we consider the amplitude of the electrical field, $E$, as a small perturbation, and use the iterative method[2] to solve Eqs. (13) and (14) approximately. The initial guess of the solution is obtained by setting $E = 0$, which gives

$$v^{(0)} = v_0,$$ (15)

and

$$z^{(0)} = z_0 + v_0 t.$$ (16)

Substituting this solution back into the right-hand side of Eq. (13), we obtain

$$m \frac{d v^{(1)}}{d t} = q E \cos (k z_0 + k v_0 t - \omega t),$$ (17)

which can be integrated over time to give

$$v^{(1)} = v_0 + \frac{q E}{m} \frac{\sin (k z_0 + \alpha t) - \sin (k z_0)}{\alpha},$$ (18)

where $\alpha = k v_0 - \omega$ and use has been made of the initial condition $v^{(1)} (0) = v_0$. Substituting this solution for the velocity, Eq. (14) is written

$$\frac{d z^{(1)}}{d t} = v_0 + \frac{q E}{m} \frac{\sin (k z_0 + \alpha t) - \sin (k z_0)}{\alpha},$$ (19)

which can be integrated over time, giving

$$z^{(1)} = z_0 + v_0 t + \frac{q E}{m} \int_0^t \frac{\sin (k z_0 + \alpha t) - \sin (k z_0)}{\alpha} d t,$$

$$= z_0 + v_0 t + \frac{q E}{m} \int_0^t \frac{\sin (k z_0 + \alpha t) - \sin (k z_0)}{\alpha^2} d (k z_0 + \alpha t),$$

$$= z_0 + v_0 t + \frac{q E}{m} \left[ \frac{- \cos (k z_0 + \alpha t) + \cos (k z_0)}{\alpha^2} - \frac{\sin (k z_0)}{\alpha} t \right],$$ (20)

where use has been made of the initial condition $z^{(1)} (0) = z_0$. Substituting the solution in Eq. (20) back into the right-hand side of Eq. (13), we obtain

$$m \frac{d v^{(2)}}{d t} = q E \cos (k z^{(1)} - \omega t)$$

$$= q E \cos \left\{ k z_0 + k v_0 t + k \frac{q E}{m} \left[ \frac{-... ... (k z_0)}{\alpha^2} - \frac{t \sin (k z_0)}{\alpha} \right] - \omega t \right\}$$

$$= q E \cos \left\{ k z_0 + \alpha t + k \frac{q E}{m} \left[ \frac{... ...t) + \cos (k z_0)}{\alpha^2} - \frac{\sin (k z_0)}{\alpha} t \right] \right\} .$$ (21)

Since $E$ is considered to be a small parameter, the term proportional to $E$ can be considered to be small when compared with $k z_0 + \alpha t$. Therefore, we expand the first cosine function in the vicinity of $k z_0 + \alpha t$. Thus the above equation is written approximately as

$$m \frac{d v^{(2)}}{d t} \approx q E \cos (k z_0 + \alpha t) - \si... ...+ \alpha t) + \cos (k z_0)}{\alpha^2} - \frac{\sin (k z_0)}{\alpha} t \right] .$$ (22)

Next, calculate the time change rate of the kinetic energy of the particle, which is written as

$$\frac{d}{d t} \left( \frac{1}{2} m v^2 \right) = m v \frac{d v}{d t}$$

$$\approx m v^{(1)} \frac{d v^{(2)}}{d t}$$ (23)

Using Eq. (18) for $v^{(1)}$ and Eq. (22) for $m d v^{(2)} / d t$, Eq. (23) is written

$$\frac{d}{d t} \left( \frac{1}{2} m v^2 \right) \approx \left[ v_0 + \frac{q E}{m} \frac{\sin (k z_0 + \alpha t) - \sin (k z_0)}{\alpha} \right]$$

$$\times \left\{ q E \cos (k z_0 + \alpha t) - \sin (k z_0 + \alpha t) k \... ...a t) + \cos (k z_0)}{\alpha^2} - \frac{\sin (k z_0)}{\alpha} t \right] \right\}$$

$$\approx v_0 q E \cos (k z_0 + \alpha t)$$

$$- v_0 \sin (k z_0 + \alpha t) k \frac{(q E)^2}{m} \left[ \frac{- \c... ...0 + \alpha t) + \cos (k z_0)}{\alpha^2} - \frac{\sin (k z_0)}{\alpha} t \right]$$

$$+ \frac{(q E)^2}{m} \frac{\sin (k z_0 + \alpha t) - \sin (k z_0)}{\alpha} \cos (k z_0 + \alpha t),$$ (24)

where the terms of order $E^3$ have been neglected (if terms of order $E^3$ and higher are included, then the result will correspond to nonlinear Landau damping, is this correct?). Equations (24) agrees with Eq. (8) in Chapter 8 of Stix's book[4] (however Stix's formula misses, by mistakes, the first term of the above equation).