Vlasov equation in Fourier space with $k=0$ I'm trying to write the Vlasov equation (1.1):

in Fourier space using (1.2):
.
And (1.3):

First I want to write this for $\vec k = 0$. The solution I should get is (1.4):

however I'm having trouble with two terms. I've substituted $f$ and $\varphi$ in (1.2) into (1.1), and I get this:
$\frac {\partial f_0} {\partial t} + \frac{\partial \delta f_{\vec k=0}}{\partial t}\sum_{\vec k}e^{i\vec k \cdot \vec r} + \vec v \cdot (\sum_{\vec k} i \vec k e^{i\vec k \cdot \vec r} \delta f_{\vec k}) = \frac{q}{m} (\sum_{\vec k} i \vec k e^{i\vec k \cdot \vec r} \varphi_{\vec k}) \cdot \frac {\partial \delta f_{\vec k}}{\partial \vec v}\sum_{\vec k}e^{i\vec k \cdot \vec r}$
Now my guess is that the 3rd term on the LHS goes to 0, and I can cancel out the exponentials on the RHS by writing $\varphi_{-\vec k}$ and get a minus sign. However I am confused about the notation $\varphi_{-\vec k}$: does it mean I simply replace $\vec k$ with $-\vec k$ in $\varphi_{\vec k}$, and if so, why would the sign of the exponential and $i \vec k$ change? Or does this have something to do with $\varphi(\vec r, t)$ being a real field? I've also assumed that $\frac {\partial f_0} {\partial \vec v} = \frac {\partial <f>} {\partial \vec v}$ is zero, but in the derivation of the Vlasov equation for $\vec k \neq 0$ that doesn't seem to be true.
And as for the LHS, why would the dot product go to zero, and what happens to the exponential in the second term? I understand that since $f_{\vec k}$ depends on $\vec v$, the dot product of $\vec v$ with $f_{\vec k}$ should be zero, but since there's the exponential term I'm unsure if I can use that relation.
Naturally if I just wrote $\vec k = 0$, I would be able to make the exponentials disappear, but then I figure then RHS would go to zero as $i\vec k = 0$. Using dot product properties the dot product term on the LHS would then vanish since $i\vec k \cdot \vec v \delta f_{\vec k}$ would be zero.
Any help is much appreciated as there is clearly some error in my thinking.
 A: I think I have some answers to your questions. To begin with, note that, according to the text you have to take the spatial average of the equation you are referring to. Namely, for some volume $V$
$$\frac{1}{V}\int d^3\vec{x} \bigg(\frac{\partial{f_a}}{\partial t}+\vec{\upsilon}\cdot \vec{\nabla}f_a-\frac{q_a}{m_a}\vec{\nabla}\phi\cdot \frac{\partial f_a}{\partial \vec{\upsilon}}\bigg)=0$$
And now I think everything will become clear. I list the contribution associated with each term and I leave it to you to workout the details. If you have trouble with something, please let me know:

*

*The first term
$$\frac{1}{V}\int d^3\vec{x} \frac{\partial f_a}{\partial t}=
\frac{1}{V}\int d^3\vec{x}\bigg(\frac{\partial f_0}{\partial t}+\sum_ke^{i\vec{k}\cdot\vec{r}}\frac{\partial \delta f_k}{\partial t}\bigg)=
\frac{\partial f_0}{\partial t}+\frac{\partial \delta f_a}{\partial t}\bigg|_{k=0}$$
yields these two contributions, after exploiting the fact that $$
\frac{1}{V}\int d^3\vec{x}e^{i\vec{k}\cdot{r}}=\delta_{k,0}\Rightarrow
\frac{1}{V}\sum_k\int d^3\vec{x}e^{i\vec{k}\cdot{r}}=\sum_k\delta_{k,0}=1$$

*The second term yields a term that is proportional to $\delta_{k,0}\times \vec{k}$, which is zero always!

*The non vanishing contribution from the third term is actually
$$\frac{1}{V}\int d^3\vec{x}
\frac{q_a}{m_a}\bigg[\sum_k (i\vec{k})e^{i\vec{k}\cdot\vec{r}}\phi_k\bigg]
\cdot
\bigg[\sum_{k'} e^{i\vec{k}'\cdot\vec{r}}\frac{\partial\delta f_{k'}}{\partial \vec{\upsilon}}\bigg]=
\sum_{k,k'}\delta_{k,-k'}\frac{q_a}{m_a}i\vec{k}\cdot\frac{\partial\delta f_a}{\partial \vec{\upsilon}}\phi_k\\=
-i\frac{q_a}{m_a}\sum_k 
\vec{k} \cdot\frac{\partial\delta f_k}{\partial \vec{\upsilon}}
\phi_{-k}$$
so the minus k comes from the Kronecker $\delta$ symbol. Upon requiring the field $\phi$ to be real, then you can further say that $\phi_k^*=\phi_{-k}$ and this comes from $\phi^*(x)=\phi(x)$ (try it!)

If there are more questions, please do not hesitate!
