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As $t\to +\infty$, the displayed limit holds in distribution theory. Indeed $\sin (tx)/x=\frac{1}{2}{\int }_{-t}^t{e}^{iux}du$ and the Fourier representation of the Dirac delta function gives the limit on smooth rapidly decaying test functions. For a regular detuning $\Delta (\mathbf{J})$, the coarea formula reduces its use under an integral over action variables to this one-dimensional identity. The kernel grows like $t$ at exact resonance, so this is not pointwise convergence.

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.

In the sense of distribution theory, $H^{\prime }(t)=\delta (t)$. Since $\sin (\alpha t)$ vanishes at zero,and therefore

Let $G_{\mathrm{ret}}(\mathbf{x},t;\mathbf{y})$ solveThe spatial Fourier transform turns this into the preceding oscillator equation with $\alpha =ck$, soInverting with the supplied radial integral gives the three-dimensional retarded Green function

For zero initial displacement and initial velocity $f$, convolution with this kernel gives the Kirchhoff formulaThus the value at $(\mathbf{x},t)$ depends only on the initial data on the sphere of radius $ct$, not on its interior. This is the Huygens principle and describes propagation at the finite wave speed $c$.

Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

Use the Fourier transform conventionSince $\mathcal{F}[e^{i\Omega t}]=2\pi \delta (\omega -\Omega )$ in distribution theory, the carrier has transformThe modulation property of the Fourier transform translates the message spectrum to the two carrier frequencies. ThereforeThe first term represents the unmodulated carrier wave; the other terms are the amplitude modulation sidebands.

Green's second identity, with the outward normal derivative, isSet $u=\Phi$ and $v=G(\cdot ;{\mathbf{r}}^{\prime })$, interpreting the logarithmic singularity by a small excised disc or distribution theory. The volume integral is $\Phi ({\mathbf{r}}^{\prime })$ because $\Delta \Phi =0$ and $\Delta G=\delta$. On the boundary $G=0$ and $\Phi =f$, leavingFor the exterior problem below, apply the identity first in a large truncated region. For bounded harmonic functions the extra circle at infinity contributes zero: $G$ tends to a constant, its radial derivative is $O(r^{-2})$, and a bounded exterior harmonic function also has radial derivative $O(r^{-2})$. This is the condition under which the same boundary representation applies there.