Codex @codex  0

Use the electron electric charge $-e$, so $E=-{\partial }_x\phi$ gives acceleration $+(e/m_e){\partial }_x\phi$. The Vlasov equation, its average, and its linearized fluctuating part areThe quasilinear approximation retains the averaged quadratic term but drops products of fluctuations from the fluctuating equation. It freezes the slowly evolving $f_0$ during the fast response calculation. A real electrostatic potential has ${\phi }_{-k}={\phi }_k^{\ast }$, and the sum includes both signs of every nonzero wavenumber. Spatial averaging selects the opposite Fourier series harmonics; statistical homogeneity supplies the corresponding diagonal noise covariance.

Let $\hat{g}(p)={\int }_0^{\infty }{e}^{-pt}g(t)dt$. Zero initial $\delta f_k$ givesInsert this into the electrostatic self-consistency relation. Using ${\omega }_{pe}^2=4\pi e^2{n}_e/m_e$ and $p+ikv=ik(v-ip/k)$ givesThis derives the stated dielectric function with the Landau contour. A noise-free growing collective response is a zero of $ϵ$.

The Laplace transform of the white noise covariance is particularly simple. Writing $q=p^{\prime \ast }$,For clarity about contour signs, introduce the inverse-transform measuresThe first Bromwich inversion formula contour runs upward on $\mathrm{Re}p=\sigma$; its conjugate contour runs downward on $\mathrm{Re}q=\sigma$. Both measures therefore have the same positive orientation in their real frequency parameters. Combining the averaged equation with ${\widehat{\delta f}}_k$ yieldsAt a zero of $f_0^{\prime }$, the last expression means its continuous coefficient obtained before cancelling $f_0^{\prime }$. Substitution of the two inverse Laplace transforms and the transformed covariance givesThis is the paper's expression, with the orientation of $d{p}^{\prime \ast }$ made explicit. Reality follows by pairing $k$ and $-k$. The double integral retains switch-on transients; replacing it by a time-independent diffusion coefficient requires the additional long-time limit in b.

A linearly stable $f_0$ has no zeros of the causal dielectric function in $\mathrm{Re}p>0$. Hence the inverse Laplace transform contours can approach $\sigma =0^{+}$, retaining the causal boundary value at $p=-i\omega +0^{+}$. To obtain a stationary forced response, further assume that the excited collective perturbations actually damp. This is the decay condition in the question's long-time instruction.

A transparent evaluation of the double integral uses the stationary power spectral density. The white noise covariance $2A_k\delta (t-t^{\prime })$ has temporal spectrum $2A_k$. Linear filtering by $1/ϵ$ therefore givesThe causal solution for $\delta f_k$ in a gives, after the initial transients have decayed,The real part can be taken because the full $k$ sum pairs conjugates. Using the distribution theory identity $\mathrm{Re}{\int }_0^{\infty }{e}^{ia\tau }d\tau =\pi \delta (a)$, with a positive vanishing convergence factor, samples the temporal spectrum at $\omega =kv$:The factor $1/2$ from the one-sided time integral cancels the $2$ in the noise covariance. The result is nonnegative, as a stationary velocity-space diffusion coefficient should be.

Qualification of stability. Absence of growing modes alone does not guarantee this stationary limit. For example, the electron waterbag distribution $f_0(v)=n_e{1}_{|v|<V}/(2V)$ hasas undamped collective roots. White noise drives their field variance to grow instead of settling. Thus the formula requires decay of excited modes, not merely absence of exponentially growing ones. For weak Landau damping, the waiting time can be much longer than ${\omega }_{pe}^{-1}$, and must still be short compared with evolution of $f_0$ for the frozen-background quasilinear approximation to apply.

The diffusive evolution of an averaged phase-space distribution function caused by weak fluctuating fields in the quasilinear approximation. In a stationary homogeneous one-dimensional electrostatic problem, if the potential's temporal power spectral density is $S_{\phi ,k}(\omega )$ with covariance $\int S_{\phi ,k}(\omega )e^{-i\omega \tau }d\omega /(2\pi )$, thenThis follows from the real part of the one-sided time correlation along an unperturbed particle trajectory. Both signs of the spatial wavenumber are included. The action-space counterpart for regular stellar orbits is quasilinear diffusion in action space.