2.1 The simplest case: 1-D geometry, Maxwellian $f_{0}$

Let us consider a particular situation, when (for whatever reason) the third term on the left-hand side of equation (2.12) disappears, i.e. let us briefly consider

(2.16) $$\begin{eqnarray}(\boldsymbol{v}\times \boldsymbol{B}_{0})\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}\,f^{(1)}=0,\end{eqnarray}$$

which according to (2.10) implies that $f^{(1)}$ is gyrotropic (it does not depend on the angle $\unicode[STIX]{x1D719}$ ). Let us also consider the even more special case in which $f_{0}$ is isotropic. In such a case, that is a specific case of (2.16), the direction of $\boldsymbol{B}^{(1)}$ does not matter at all for $f_{0}$ and naturally

(2.17) $$\begin{eqnarray}(\boldsymbol{v}\times \boldsymbol{B}^{(1)})\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}\,f_{0}=0.\end{eqnarray}$$

To quickly double check the correctness of the above expression, for isotropic $f_{0}(v)$ the velocity gradient is given by $\unicode[STIX]{x2202}f_{0}/\unicode[STIX]{x2202}v_{i}=(\unicode[STIX]{x2202}f_{0}/\unicode[STIX]{x2202}v)(\unicode[STIX]{x2202}v/\unicode[STIX]{x2202}v_{i})=f_{0}^{\prime }v_{i}/v$ and the velocity gradient $\unicode[STIX]{x1D735}_{v}\,f_{0}=f_{0}^{\prime }\hat{\boldsymbol{v}}$ is in the direction of velocity $\boldsymbol{v}$ . The result (2.17) then immediately follows since $\unicode[STIX]{x1D716}_{ijk}v_{j}B_{k}^{(1)}v_{i}=0$ . The equation (2.12) therefore reduces to

(2.18) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}f^{(1)}}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x1D735}f^{(1)}=-\frac{q_{r}}{m_{r}}\boldsymbol{E}^{(1)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}\,f_{0}.\end{eqnarray}$$

Fourier transforming the first-order quantities and $\unicode[STIX]{x2202}/\unicode[STIX]{x2202}t\rightarrow -i\unicode[STIX]{x1D714}$ , $\unicode[STIX]{x1D735}\rightarrow i\boldsymbol{k}$ yields

(2.19) $$\begin{eqnarray}(-i\unicode[STIX]{x1D714}+i\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{k})f^{(1)}=-\frac{q_{r}}{m_{r}}\boldsymbol{E}^{(1)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}\,f_{0},\end{eqnarray}$$

which allows us to obtain an expression for $f^{(1)}$ in the form

(2.20) $$\begin{eqnarray}f^{(1)}=-i\frac{q_{r}}{m_{r}}\;\frac{\boldsymbol{E}^{(1)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}\,f_{0}}{\unicode[STIX]{x1D714}-\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{k}}.\end{eqnarray}$$

Even though it is not necessary, it is useful to express the (electrostatic) electric field through the scalar potential $\boldsymbol{E}^{(1)}=-\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}$ , which in Fourier space reads $\boldsymbol{E}^{(1)}=-i\boldsymbol{k}\unicode[STIX]{x1D6F7}$ , yielding

(2.21) $$\begin{eqnarray}f^{(1)}=-\frac{q_{r}}{m_{r}}\unicode[STIX]{x1D6F7}\;\frac{\boldsymbol{k}\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}\,f_{0}}{\unicode[STIX]{x1D714}-\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{k}}.\end{eqnarray}$$

Now, we want to integrate $f^{(1)}$ , and obtain the linear ‘kinetic’ moments for density, velocity (current), pressure (temperature), heat flux and the fourth-order moment $r$ (or the correction $\tilde{r}$ ). To continue, we have to prescribe some distribution function $f_{0}$ .

The 3-D (isotropic) Maxwellian distribution is

(2.22) $$\begin{eqnarray}f_{0r}=n_{0r}\left(\frac{\unicode[STIX]{x1D6FC}_{r}}{\unicode[STIX]{x03C0}}\right)^{3/2}e^{-\unicode[STIX]{x1D6FC}_{r}v^{2}},\end{eqnarray}$$

where the isotropic $v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}$ and $\unicode[STIX]{x1D6FC}_{r}=m_{r}/(2T_{r}^{(0)})=1/v_{\text{th}r}^{2}$ . For simplicity, let us drop the species index $r$ , except for the charge $q_{r}$ . The velocity gradient

(2.23) $$\begin{eqnarray}\displaystyle {\displaystyle \frac{\unicode[STIX]{x2202}f_{0}}{\unicode[STIX]{x2202}v_{i}}} & = & \displaystyle n_{0}\left({\displaystyle \frac{\unicode[STIX]{x1D6FC}}{\unicode[STIX]{x03C0}}}\right)^{3/2}(-\unicode[STIX]{x1D6FC})2v_{i}e^{-\unicode[STIX]{x1D6FC}v^{2}}=-2\unicode[STIX]{x1D6FC}v_{i}f_{0}=-{\displaystyle \frac{m}{T^{(0)}}}v_{i}f_{0},\nonumber\\ \displaystyle \unicode[STIX]{x1D735}_{v}\,f_{0} & = & \displaystyle -{\displaystyle \frac{m}{T^{(0)}}}\boldsymbol{v}f_{0}.\end{eqnarray}$$

Therefore, for a Maxwellian

(2.24) $$\begin{eqnarray}f^{(1)}=+\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\;\frac{\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}}{\unicode[STIX]{x1D714}-\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{k}}f_{0}.\end{eqnarray}$$

Before continuing, let us slightly rearrange the above expression for $f^{(1)}$ and add $0=\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}$ to the numerator, otherwise we will have to do this each time, when calculating the higher-order moments. The rearrangement yields

(2.25) $$\begin{eqnarray}\displaystyle f^{(1)}=+\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\;\frac{\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}-\unicode[STIX]{x1D714}+\unicode[STIX]{x1D714}}{\unicode[STIX]{x1D714}-\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{k}}f_{0}=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\;\left(1+\frac{\unicode[STIX]{x1D714}}{\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{ k}-\unicode[STIX]{x1D714}}\right)f_{0}. & & \displaystyle\end{eqnarray}$$

For clarity, let us simplify even further and discuss the simplest possible 1-D case, for a 1-D Maxwellian distribution

(2.26) $$\begin{eqnarray}f_{0}=n_{0}\sqrt{\frac{\unicode[STIX]{x1D6FC}}{\unicode[STIX]{x03C0}}}e^{-\unicode[STIX]{x1D6FC}v^{2}};\quad \text{where }\unicode[STIX]{x1D6FC}\equiv \frac{m}{2T^{(0)}}=\frac{1}{v_{\text{th}}^{2}}.\end{eqnarray}$$

Here we consider fluctuations along the magnetic field $B_{0}$ and the wavenumber is therefore denoted as $k_{\Vert }$ . Note that the case is strictly one-dimensional, and the velocity fluctuations are along the $B_{0}$ as well. For example from the MHD perspective, we are therefore considering the parallel propagating ion-acoustic mode. The $f^{(1)}$ for a Maxwellian $f_{0}$ is expressed as (dropping all the species indices ‘ $r$ ’ except for the charge $q_{r}$ )

(2.27) $$\begin{eqnarray}\displaystyle & \displaystyle f^{(1)}=i\frac{q_{r}}{T^{(0)}}E^{(1)}\frac{v}{\unicode[STIX]{x1D714}-vk_{\Vert }}f_{0}, & \displaystyle\end{eqnarray}$$

(2.28) $$\begin{eqnarray}\displaystyle & \displaystyle f^{(1)}=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(1+\frac{{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}\right)f_{0}. & \displaystyle\end{eqnarray}$$

Now, we are ready to calculate the velocity integrals. Let us start with the density $n^{(1)}$ , by integrating

(2.29) $$\begin{eqnarray}n^{(1)}=\int _{-\infty }^{\infty }f^{(1)}\,\text{d}v=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(\underbrace{\int _{-\infty }^{\infty }f_{0}\,\text{d}v}_{=n_{0}}+\int _{-\infty }^{\infty }\frac{{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}f_{0}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}\,\text{d}v\right).\end{eqnarray}$$

By using the prescribed Maxwellian $f_{0}$ , the second integral is rewritten as

(2.30) $$\begin{eqnarray}\displaystyle \int _{-\infty }^{\infty }\frac{{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}f_{0}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}\,\text{d}v & = & \displaystyle n_{0}\sqrt{\frac{\unicode[STIX]{x1D6FC}}{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\int _{-\infty }^{\infty }\frac{e^{-\unicode[STIX]{x1D6FC}v^{2}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}\,\text{d}v\nonumber\\ \displaystyle & = & \displaystyle \left[\begin{array}{@{}c@{}}\sqrt{\unicode[STIX]{x1D6FC}}v=x\\ \sqrt{\unicode[STIX]{x1D6FC}}\text{d}v=\text{d}x\end{array}\right]=n_{0}\sqrt{\frac{\unicode[STIX]{x1D6FC}}{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}\sqrt{\unicode[STIX]{x1D6FC}}}\,\text{d}x=\left[\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\sqrt{\unicode[STIX]{x1D6FC}}\equiv x_{0}\right]\nonumber\\ \displaystyle & = & \displaystyle \frac{n_{0}}{\sqrt{\unicode[STIX]{x03C0}}}x_{0}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x.\end{eqnarray}$$

The notation $[\cdots \,]$ just indicates change of a variable. We purposely wrote the integral with

(2.31) $$\begin{eqnarray}x_{0}\equiv \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\sqrt{\unicode[STIX]{x1D6FC}}=\frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}},\end{eqnarray}$$

instead of the usual $\unicode[STIX]{x1D701}$ , since we want to define $\unicode[STIX]{x1D701}$ slightly differently. The integral is related to the famous plasma dispersion function $Z(\unicode[STIX]{x1D701})$ , that is responsible for the famous Landau damping. Each plasma physics book devotes many pages to the discussion of Landau damping, that was first correctly described by Landau (Reference Landau1946), by considering an initial value problem and using Laplace transforms. It was later shown by van Kampen (Reference van Kampen1955), that the Landau damping can be indeed obtained by using Fourier analysis. We refer the reader for example to books by Swanson, Stix, Akhiezer, Gary, Gurnett and Bhattacharjee, Fitzpatrick, etc. Let us call the integral (2.30) the ‘Landau integral’. Nevertheless, the very-well-known secret is that, even if one is armed with all these excellent books, the Landau damping effect can still be very confusing (even at the linear level). We did not find any secret recipe that explains the Landau damping in a simplified and different way, and the reader is referred to the thick plasma physics books. Here, we want to concentrate only how to express the integral (2.30) through the plasma dispersion function.

Since the Landau integral can be very confusing and boring to explain, to increase the ‘pedagogical’ value of this text, let us talk a bit more freely on the next few pages. The plasma dispersion function can be defined with a short definition

(2.32) $$\begin{eqnarray}Z(\unicode[STIX]{x1D701})\equiv \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x,\quad \text{for }\text{Im}(\unicode[STIX]{x1D701})>0.\end{eqnarray}$$

In the definition of $x_{0}$ , the thermal speed $v_{\text{th}}$ is always a positive real number, and we do not have to worry about it. Now, considering the specific case $k_{\Vert }>0$ and $\text{Im}(\unicode[STIX]{x1D714})>0$ , where we indeed have $\text{Im}(x_{0})>0$ , we can directly use the plasma dispersion function and the result of the Landau integral (2.30) is $n_{0}x_{0}Z(x_{0})$ . For this case, we are done. Really? Yes, there is nothing else we can do for this case, we calculated the Landau integral. Reeeaallyy?? Yes, because the Landau integral cannot be analytically ‘calculated’, the integral cannot be expressed through elementary functions, unless the $Z(\unicode[STIX]{x1D701})$ function is somehow simplified, for example by expansion for cases $|\unicode[STIX]{x1D701}|\ll 1$ or $|\unicode[STIX]{x1D701}|\gg 1$ , or by considering the weak damping limit when $\text{Im}(x_{0})$ is small (see plasma physics books). We are not interested in these limits and the $Z(\unicode[STIX]{x1D701})$ function has to be calculated numerically or looked up in the table. We are really done here!Footnote ¹ So why is the Landau integral so confusing for the other cases? It is exactly because of that – basically nothing gets ‘really calculated’.

2.2 The dreadful Landau integral $\int e^{-x^{2}}/(x-x_{0})\,\text{d}x$

There are many reasons why the ‘Landau integral’ (2.30) can be so confusing. The first reason is, (i) that the integral (2.30) cannot be expressed by using only elementary functions. If we did not arrive at this integral in the middle of a thick plasma physics book, but instead encountered it during our undergraduate studies of complex analysis, we would perhaps not have such a respect to this integral, and immediately attempt to calculate it by using the residue theorem. The integral appears to be so simple. Instead of calculating $\int _{-\infty }^{\infty }$ , we would calculate a different integral over a closed contour in the complex plane $\oint _{C}$ . That integral can be calculated by using the residue theorem, that states that $\oint _{C}=2\unicode[STIX]{x03C0}i\sum \text{Res}$ , if the big path that encircles all the poles is counter-clockwise.Footnote ² An equivalent statement is that the integral is equal to $\oint _{C}=-2\unicode[STIX]{x03C0}i\sum \text{Res}$ , if the big path that encircles all the poles is clockwise. In our case, there is always just one pole, at $x=x_{0}$ , and the residue of $e^{-x^{2}}/(x-x_{0})$ evaluated at $x=x_{0}$ is actually very simple, it is always

(2.33) $$\begin{eqnarray}\underset{x=x_{0}}{\text{Res}}\;\frac{e^{-x^{2}}}{x-x_{0}}=e^{-x_{0}^{2}},\end{eqnarray}$$

regardless of the value of $x_{0}$ , since for a general function $f(x)$ , the residue $\underset{x=x_{0}}{\text{Res}}\;f(x)/(x-x_{0})=f(x_{0})$ .

However, to make the result $\oint _{C}$ useful for the calculation of our integral on the real axis $\int _{-\infty }^{\infty }$ , we need to separate the closed contour integral to $\oint _{C}=\int _{-\infty }^{\infty }+\int _{\text{arc}}$ , where the $\int _{\text{arc}}$ represents the big half-circle at infinitely large radius. To preserve the direction of integration along the real axis $\int _{-\infty }^{\infty }$ , if the pole is in the upper complex plane, i.e. if $\text{Im}(x_{0})>0$ , we need to close the big arc contour in the upper half of the complex plane counter-clockwise. Similarly, if the pole is in the lower complex plane, i.e. if $\text{Im}(x_{0})<0$ , we need to close the big arc contour clockwise. Importantly, in contrast to typical examples presented in basic complex analysis classes, the arc integral $\int _{\text{arc}}$ does not disappear. The problem is, that the function $f(x)=e^{-z^{2}}$ is a very strongly decaying function on the real axis (for $z=x\rightarrow \pm \infty$ ), however, this is not true at all in the complex plane. Considering the purely Imaginary axis $z=\pm iy$ , the function $e^{-z^{2}}=e^{+y^{2}}$ is a very strongly diverging function as $y$ increases, and the arc integral $\int _{\text{arc}}$ cannot be neglected! This is a very sad news, since now we clearly see, that with $\int _{\text{arc}}\neq 0$ , we will not be able to use the complex analysis to actually ‘calculate’ the Landau integral (2.30).

We note that the well-known Gaussian integral $I=\int _{-\infty }^{\infty }e^{-x^{2}}\,\text{d}x=\sqrt{\unicode[STIX]{x03C0}}$ , is typically calculated in the real plane by means of a trick which consists in evaluating $I^{2}$ in polar coordinates, $I^{2}=\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}\,\text{d}x\,\text{d}y=2\unicode[STIX]{x03C0}\int _{0}^{\infty }e^{-r^{2}}r\,\text{d}r$ . The Gaussian integral can still be calculated in the complex plane by using the residue theorem, even though quite sophisticated tricks are required.Footnote ³

The second reason why Landau damping is confusing is, (ii) the necessity of analytic continuation. The third reason is very closely related to the second and it is (iii) The analytic continuation has to be done differently for $k_{\Vert }>0$ and for $k_{\Vert }<0$ . The big result of Landau (Reference Landau1946) can be summarized as follows: if $k_{\Vert }>0$ , the path of integration always has to pass below the pole $x=x_{0}$ . Therefore, starting with the basic case in the upper complex plane $\text{Im}(x_{0})>0$ , nothing has to be done and the integration is just along the real axis. Now, if the pole is moved to the real axis, so $\text{Im}(x_{0})=0$ , one needs to go around that pole with a tiny half-circle from below. This creates a contribution of $1/2$ times $2\unicode[STIX]{x03C0}i$ times the residue at that pole, so the contribution is $\unicode[STIX]{x03C0}ie^{-x_{0}^{2}}$ . If the pole $x_{0}$ is moved further down to the lower complex plane, a full circle around the pole is required to enclose it from below, which yields a contribution of $2\unicode[STIX]{x03C0}ie^{-x_{0}^{2}}$ . The situation is demonstrated in figure 1(a). The integral (2.30) for $k_{\Vert }>0$ is therefore ‘calculated’ as

(2.34) $$\begin{eqnarray}\int _{C}\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x\overset{k_{\Vert }>0}{=}\left\{\begin{array}{@{}ll@{}}\displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x,\quad & \text{Im}(\unicode[STIX]{x1D714})>0;\text{Im}(x_{0})>0;\\ \displaystyle V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x+\unicode[STIX]{x03C0}ie^{-x_{0}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})=0;\text{Im}(x_{0})=0;\\ \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x+2\unicode[STIX]{x03C0}ie^{-x_{0}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})<0;\text{Im}(x_{0})<0.\end{array}\right.\end{eqnarray}$$

For the Cauchy principal value, we prefer the original French pronunciation ‘Valeur Principale’, abbreviated as $V.P$ .

Figure 1. (a) Landau contours for $k_{\Vert }>0$ . (b) Landau contours for $k_{\Vert }<0$ .

The above result is completely consistent with the definition of the plasma dispersion function, since the plasma dispersion function was developed exactly to describe this integral. One starts with the definition in the upper complex plane (2.32), and analytically continues this function to a lower complex plane, according to

(2.35) $$\begin{eqnarray}Z(\unicode[STIX]{x1D701})\equiv \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{C}\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x=\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\left\{\begin{array}{@{}ll@{}}\displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x,\quad & \text{Im}(\unicode[STIX]{x1D701})>0;\\ \displaystyle V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+\unicode[STIX]{x03C0}ie^{-\unicode[STIX]{x1D701}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D701})=0;\\ \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+2\unicode[STIX]{x03C0}ie^{-\unicode[STIX]{x1D701}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D701})<0.\end{array}\right.\end{eqnarray}$$

To save space in scientific papers and plasma physics books, the definition of $Z(\unicode[STIX]{x1D701})$ is often abbreviated as (2.32), i.e. only as a first line of (2.35), with a powerful statement that for $\text{Im}(\unicode[STIX]{x1D701})<0$ the function is analytically continued. That statement indeed completely defines $Z(\unicode[STIX]{x1D701})$ , since the powerful complex analysis tells us that an analytic continuation of a function, if it exists, is unique. Another abbreviated definition is by essentially writing down only the second (middle) line of (2.35). This is the most useful 1-line abbreviation because one can immediately recognize how the $\text{sign}(k_{\Vert })$ was treated (as we will see soon). However, such a definition of $Z(\unicode[STIX]{x1D701})$ , only specifying it for $\text{Im}(\unicode[STIX]{x1D701})=0$ , would not be a complete definition of that function, and no powerful statement regarding how the function is extended above and below from the $x$ -axis is available. So plasma physicists found a very smart workaround, how not to write the $\text{Im}(\unicode[STIX]{x1D701})=0$ restriction in the second line of (2.35) and how to completely define the $Z(\unicode[STIX]{x1D701})$ with this 1-line statement. Let us still consider the case $k_{\Vert }>0$ , where our $x_{0}$ and $\unicode[STIX]{x1D701}$ are equivalent. It is often stated (e.g. Stix, bottom of page 190), that ‘the principal value of an integral through an isolated singular point may be considered the average of the two integrals that pass just above and just below the point’. For example, for a specific situation when $x_{0}$ lies on the real $x$ -axis, integrating along a horizontal line below the $x$ -axis yields the first line of (2.35), and integration along a horizontal line above the $x$ -axis yields the third line of (2.35) since when the pole is encountered we have to pass it from below. An average of the first line and third line of (2.35) yields the second line. The idea can now be generalized to an entire complex plane, for all values of $\text{Im}(x_{0})$ , where two integrals are done; one integral along a horizontal line that passes below the $x_{0}$ point (where nothing has to be done) and one integral along a horizontal line that passes above the $x_{0}$ point (and where a deformation that passes below the point has to be performed, accounting for the full residuum). The average of these two integrals yields an abbreviated $Z(\unicode[STIX]{x1D701})$ definition for all values of $\unicode[STIX]{x1D701}$ in the form

(2.36) $$\begin{eqnarray}Z(\unicode[STIX]{x1D701})=\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+i\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}},\quad \text{for }\forall \;\text{Im}(\unicode[STIX]{x1D701}),\end{eqnarray}$$

where the integration is said to go through the pole. Of course, no integration can be really done ‘through’ a singular point, and what the wording means is that the integration is done along the horizontal axis that goes through $x_{0}$ , i.e. the integration is along horizontal axis $\text{Im}(x_{0})$ .

It is possible to look at it from another (perhaps more illuminating) perspective. Consider the situation in which $x_{0}$ is somewhere in the upper half of the complex plane. One can perform the integral along the real axis, so that the first line of (2.35) applies. Let us call this result $c_{1}$ . Alternatively, one can perform the integral along the horizontal line that passes through $\text{Im}(x_{0})$ (with the required tiny half-circle passing below $x_{0}$ ), and (2.36) applies. Let us call this result $c_{2}$ . These two different integrals must be equal. Why? Because one can plot two vertical lines (passing through $\text{Re}(x_{0})=\pm \infty$ ) that, together with the two horizontal integration lines, enclose an area that does not contain any pole, and integration around all four lines (in a circular direction, let us say counter-clockwise) must yield zero. The two integrals along the vertical lines cancel each other, yielding that $c_{1}-c_{2}=0$ , the minus sign in front of $c_{2}$ appears since the integration along $c_{2}$ was now done in the opposite direction. Even though it is perhaps a bit confusing when seen at first, the definition (2.36) is very useful, and when encountered, it should be just interpreted as an abbreviated definition of (2.35).

Unfortunately, the plasma dispersion function was obviously developed only with the case $k_{\Vert }>0$ in mind. The Landau result requires that for $k_{\Vert }<0$ , the path of integration always encircles the pole from above, see figure 1(b). For $k_{\Vert }<0$ , the Landau integral is defined as

(2.37) $$\begin{eqnarray}\int _{C}\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x\overset{k_{\Vert }<0}{=}\left\{\begin{array}{@{}ll@{}}\displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x,\quad & \text{Im}(\unicode[STIX]{x1D714})>0;\;\text{Im}(x_{0})<0;\\ \displaystyle V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x-\unicode[STIX]{x03C0}ie^{-x_{0}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})=0;\;\text{Im}(x_{0})=0;\\ \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x-2\unicode[STIX]{x03C0}ie^{-x_{0}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})<0;\;\text{Im}(x_{0})>0.\end{array}\right.\end{eqnarray}$$

The two different cases for $k_{\Vert }>0$ and $k_{\Vert }<0$ can be easily combined together by using the sign of the wavenumber $k_{\Vert }$ function, that is equal to $+1$ for $k_{\Vert }>0$ , and equal to $-1$ for $k_{\Vert }<0$ . However, one needs to forget the sign of $\text{Im}(x_{0})$ , and arrange the results only with respect to the sign of $\text{Im}(\unicode[STIX]{x1D714})$ . The Landau integral with $x_{0}=\unicode[STIX]{x1D714}/(k_{\Vert }v_{\text{th}})$ therefore reads

(2.38) $$\begin{eqnarray}\int _{C}\frac{e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}}}\,\text{d}x\overset{\forall k_{\Vert }}{=}\left\{\begin{array}{@{}ll@{}}\displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x,\quad & \text{Im}(\unicode[STIX]{x1D714})>0;\\ \displaystyle V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x+\text{sign}(k_{\Vert })\unicode[STIX]{x03C0}ie^{-x_{0}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})=0;\\ \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x+\text{sign}(k_{\Vert })2\unicode[STIX]{x03C0}ie^{-x_{0}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})<0.\end{array}\right.\end{eqnarray}$$

Obviously, it is the sign of $\text{Im}(\unicode[STIX]{x1D714})$ , and not the sign of $\text{Im}(x_{0})$ , that is the ‘natural language’ of the Landau integral. However, the connection to the plasma dispersion function $Z(\unicode[STIX]{x1D701})$ is unnecessarily difficult. Sometimes, the definition of the plasma dispersion function is then altered so that the above expression is satisfied. Stix, for example, uses, in addition to the usual $Z(\unicode[STIX]{x1D701})$ , also a different function $Z_{0}(\unicode[STIX]{x1D701})$ that can be defined with respect to the sign of $\text{Im}(\unicode[STIX]{x1D714})$ instead of the sign of $\text{Im}(\unicode[STIX]{x1D701})$ , where as noted on page 202, $Z_{0}(\unicode[STIX]{x1D701})=Z(\unicode[STIX]{x1D701})$ for $k_{\Vert }>0$ , and, $Z_{0}(\unicode[STIX]{x1D701})=-Z(-\unicode[STIX]{x1D701})$ for $k_{\Vert }<0$ . With $\unicode[STIX]{x1D701}=\unicode[STIX]{x1D714}/(k_{\Vert }v_{\text{th}})$ , the function $Z_{0}(\unicode[STIX]{x1D701})$ is defined according to

(2.39) $$\begin{eqnarray}Z_{0}(\unicode[STIX]{x1D701})=\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\left\{\begin{array}{@{}ll@{}}\displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x,\quad & \text{Im}(\unicode[STIX]{x1D714})>0;\\ \displaystyle V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+\text{sign}(k_{\Vert })\unicode[STIX]{x03C0}ie^{-\unicode[STIX]{x1D701}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})=0;\\ \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+\text{sign}(k_{\Vert })2\unicode[STIX]{x03C0}ie^{-\unicode[STIX]{x1D701}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D714})<0.\end{array}\right.\end{eqnarray}$$

Again, the function $Z_{0}(\unicode[STIX]{x1D701})$ can be defined in an abbreviated form as the first line of (2.39), with analytic continuation for $\text{Im}(\unicode[STIX]{x1D714})\leqslant 0$ (Stix, page 206, equation (91)). The second possible abbreviated definition of $Z_{0}(\unicode[STIX]{x1D701})$ , valid for all values of $\text{Im}(\unicode[STIX]{x1D714})$ , is the trick with the principal value (Stix, page 206, equation (92))

(2.40) $$\begin{eqnarray}Z_{0}(\unicode[STIX]{x1D701})=\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+\text{sign}(k_{\Vert })i\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}},\quad \text{for }\forall \;\text{Im}(\unicode[STIX]{x1D714}),\end{eqnarray}$$

where the integration path goes ‘through’ the pole, i.e. the integration is done along the horizontal line $\text{Im}(\unicode[STIX]{x1D701})$ . With the use of this new function $Z_{0}(\unicode[STIX]{x1D701})$ of Stix, we can therefore express the dreadful Landau integral for all values of $k_{\Vert }$ as

(2.41) $$\begin{eqnarray}\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{C}\frac{e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}}}\,\text{d}x\overset{\forall k_{\Vert }}{=}Z_{0}(\unicode[STIX]{x1D701});\quad \text{where }\unicode[STIX]{x1D701}=\frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}.\end{eqnarray}$$

However, we do not like this formulation with $Z_{0}$ . Here, we insist on using the original plasma dispersion function $Z$ . In our opinion, the most elegant solution is the one that is used for example in the book by Peter Gary and in some Landau fluid papers, and that is to use $|k_{\Vert }|$ in the definition of $\unicode[STIX]{x1D701}$ , by defining

(2.42) $$\begin{eqnarray}\unicode[STIX]{x1D701}\equiv \frac{\unicode[STIX]{x1D714}}{|k_{\Vert }|v_{\text{th}}}.\end{eqnarray}$$

This amazingly convenient definition simplifies the expressions and represents the ‘natural language’ of the plasma dispersion function. We note that $|k_{\Vert }|=\text{sign}(k_{\Vert })k_{\Vert }$ , and also $k_{\Vert }=\text{sign}(k_{\Vert })|k_{\Vert }|$ . With the new definition of $\unicode[STIX]{x1D701}$ , for $k_{\Vert }>0$ obviously nothing is changed since $|k_{\Vert }|=k_{\Vert }$ . However, for $k_{\Vert }<0$ ,

(2.43) $$\begin{eqnarray}\displaystyle \int _{-\infty }^{\infty }{\displaystyle \frac{e^{-x^{2}}}{x-\frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}}}\,\text{d}x & \overset{k_{\Vert }<0}{=} & \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x+\unicode[STIX]{x1D701}}\,\text{d}x=\left[\begin{array}{@{}c@{}}x=-y\\ \text{d}x=-\text{d}y\end{array}\right]\nonumber\\ \displaystyle & = & \displaystyle \int _{\infty }^{-\infty }\frac{e^{-y^{2}}}{-y+\unicode[STIX]{x1D701}}(-\text{d}y)=\left[\begin{array}{@{}c@{}}\text{rename}\\ y\rightarrow x\end{array}\right]=-\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x\nonumber\\ \displaystyle & = & \displaystyle \text{sign}(k_{\Vert })\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x,\end{eqnarray}$$

where in the last step we used $-1=\text{sign}(k_{\Vert })$ . By examining the first and the last expressions, the result is also obviously valid for $k_{\Vert }>0$ , and therefore for all $k_{\Vert }$ . Or alternatively, (perhaps more confusingly, but keeping an exact track of the $\text{sign}(k_{\Vert })$ ), for all values of $k_{\Vert }$

(2.44) $$\begin{eqnarray}\displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}}}\,\text{d}x & \overset{\forall k_{\Vert }}{=} & \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\text{sign}(k_{\Vert })\unicode[STIX]{x1D701}}\,\text{d}x\nonumber\\ \displaystyle & = & \displaystyle \text{sign}(k_{\Vert })\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{\text{sign}(k_{\Vert })x-\unicode[STIX]{x1D701}}\,\text{d}x=\left[\begin{array}{@{}c@{}}y=\text{sign}(k_{\Vert })x\\ \text{d}y=\text{sign}(k_{\Vert })\,\text{d}x\end{array}\right]\nonumber\\ \displaystyle & = & \displaystyle \text{sign}(k_{\Vert })\int _{-\infty \text{sign}(k_{\Vert })}^{+\infty \text{sign}(k_{\Vert })}\frac{e^{-y^{2}}}{y-\unicode[STIX]{x1D701}}\left(\frac{\text{d}y}{\text{sign}(k_{\Vert })}\right)\nonumber\\ \displaystyle & = & \displaystyle \int _{-\infty \text{sign}(k_{\Vert })}^{+\infty \text{sign}(k_{\Vert })}\frac{e^{-y^{2}}}{y-\unicode[STIX]{x1D701}}\,\text{d}y=\text{sign}(k_{\Vert })\int _{-\infty }^{+\infty }\frac{e^{-y^{2}}}{y-\unicode[STIX]{x1D701}}\,\text{d}y\nonumber\\ \displaystyle & = & \displaystyle \text{sign}(k_{\Vert })\int _{-\infty }^{+\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x,\end{eqnarray}$$

which is the same result as the one obtained above. The definition of $\unicode[STIX]{x1D701}$ (2.42) therefore yields

(2.45) $$\begin{eqnarray}\int _{C}\frac{e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}}}\,\text{d}x\overset{\forall k_{\Vert }}{=}\text{sign}(k_{\Vert })\left\{\begin{array}{@{}ll@{}}\displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x,\quad & \text{Im}(\unicode[STIX]{x1D701})>0;\\ \displaystyle V.P.\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+\unicode[STIX]{x03C0}ie^{-\unicode[STIX]{x1D701}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D701})=0;\\ \displaystyle \int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-\unicode[STIX]{x1D701}}\,\text{d}x+2\unicode[STIX]{x03C0}ie^{-\unicode[STIX]{x1D701}^{2}},\quad & \text{Im}(\unicode[STIX]{x1D701})<0.\end{array}\right.\end{eqnarray}$$

This result allows us to use the original plasma dispersion function definition (2.35), and express the dreadful Landau integral for all $k_{\Vert }$ simply as

(2.46) $$\begin{eqnarray}\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{C}\frac{e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}}}\,\text{d}x\overset{\forall k_{\Vert }}{=}\text{sign}(k_{\Vert })Z(\unicode[STIX]{x1D701});\quad \text{where }\unicode[STIX]{x1D701}\equiv \frac{\unicode[STIX]{x1D714}}{|k_{\Vert }|v_{\text{th}}}.\end{eqnarray}$$

Now we have calculated the Landau integral to our satisfaction, and we can continue with the calculation of the linear kinetic hierarchy. Wait. We had basically the same result several pages back! For the case $k_{\Vert }>0$ and $\text{Im}(\unicode[STIX]{x1D714})>0$ . The Landau integral was just expressed through the plasma dispersion function, basically the same result as is done now, there is just one $\text{sign}(k_{\Vert })$ in front of the integral and one in the definition of $\unicode[STIX]{x1D701}$ . Are we suggesting that all these calculations, contour drawings and discussions, were done just to get a sign right? Affirmative. The Landau integral is all about chasing minus signs, but to get the correct signs is very important. This is exactly the reason why the Landau damping is so confusing, and why it needed the genius of Landau to correctly figure it out. Nevertheless, that the Landau damping (Landau Reference Landau1946) is indeed very confusing can be understood from the fact that the effect was questioned for almost 20 years before it was experimentally verified by Malmberg & Wharton (Reference Malmberg and Wharton1966).

To conclude, and to summarize the differences between the plasma physics books of Stix and Peter Gary, we have two equivalent recipes to ‘calculate’ the Landau integral, that can be written as

(2.47)

where the $A.C.$ stands for analytic continuation. It is important to emphasize that some plasma books, such as for example that by Gurnett and Bhattacharjee, take a different approach and call the function $Z_{0}(\unicode[STIX]{x1D701})$ simply $Z(\unicode[STIX]{x1D701})$ , as is obvious from their expressions for $Z(\unicode[STIX]{x1D701})$ (pages 347–348) that contain the $\text{sign}(k_{\Vert })$ . Of course, this approach is fully kosher, however, one needs to be extra careful when adopting a numerical routine for the plasma dispersion function. The second choice in (2.47) appears inconvenient, however, it is not, since the expression (2.30) contains

(2.48) $$\begin{eqnarray}\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}\int _{C}\frac{e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}}}\,\text{d}x=\left\{\begin{array}{@{}ll@{}}\displaystyle \unicode[STIX]{x1D701}Z_{0}(\unicode[STIX]{x1D701}),\quad & \unicode[STIX]{x1D701}={\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}};\\ \displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }v_{\text{th}}}\text{sign}(k_{\Vert })Z(\unicode[STIX]{x1D701})=\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701}),\quad & \unicode[STIX]{x1D701}={\displaystyle \frac{\unicode[STIX]{x1D714}}{|k_{\Vert }|v_{\text{th}}}}.\end{array}\right.\end{eqnarray}$$

The book by Peter Gary, and many Landau fluid papers prefer the second choice, since this small trick with redefining $\unicode[STIX]{x1D701}$ allows the use of the original plasma dispersion function $Z(\unicode[STIX]{x1D701})$ , that was tabulated by Fried & Conte (Reference Fried and Conte1961).Footnote ⁴ We prefer it too, and therefore, the integral that we will use frequently in the kinetic hierarchy is

(2.49)

and obviously $x_{0}=\text{sign}(k_{\Vert })\unicode[STIX]{x1D701}$ . Now we are able to finish the calculation of the density $n^{(1)}$ , equation (2.29), that yields

(2.50) $$\begin{eqnarray}n^{(1)}=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(n_{0}+n_{0}\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})\right)=-\frac{q_{r}n_{0}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(1+\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})\right).\end{eqnarray}$$

The result can also be expressed by using the derivative $Z^{\prime }(\unicode[STIX]{x1D701})=-2(1+\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701}))$ . The quantity $1+\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})$ appears very frequently in kinetic calculations with Maxwellian distribution and it is called the plasma response function

(2.51)

For a different (general) distribution function $f_{0}$ , the plasma response function $R(\unicode[STIX]{x1D701})$ can be defined according to what is obtained after calculating the density $n^{(1)}=-(q_{r}n_{0}/T^{(0)})\unicode[STIX]{x1D6F7}R(\unicode[STIX]{x1D701})$ . The name is very appropriate, since the $R(\unicode[STIX]{x1D701})$ describes, how a plasma with some distribution function ‘responds’ to an applied electric field (or a scalar potential).

2.3 Short afterthoughts, after the Landau integral

Why does some ‘analytic continuation’ have to be done? Even though we did not manage to express the Landau integral (2.30) through elementary functions, the integral appears to be well defined in both upper and lower halves of the complex plane, regardless of where the $x_{0}$ is. And it indeed is. So why the analytic continuation? The very-deep reason why the analytic continuation is necessary, is that the integral is not continuous when crossing the real axis $\text{Im}(x_{0})=0$ in the complex plane.Footnote ⁵ If a function is not continuous, it is not analytic (a fancy well-defined language that says that the function is not infinitely differentiable, basically meaning that it matters from what direction that point is approached in the complex plane, very similarly to a derivative of function $|x|$ on the real axis). And if a function is analytic in some area, and not analytic outside of that area, we can sometimes push/extend the area of where the function is analytic, to/through the area where the function is not analytic, therefore the term ‘analytic continuation’.

Why is the analytic continuation so important? Why is it a big problem that the integral is not continuous when crossing the real axis? Because it directly relates to the causality principle, that is, if something happens, then the response to this incident must come after, and not before, the time in which that incident happened. This can be perhaps more intuitively addressed by performing the Laplace transforms in time (instead of the Fourier transforms), and considering an initial value problem, as was done by Landau (Reference Landau1946). For more information, see plasma physics books, for example Stix (Reference Stix1992), chapter 3 on causality etc.

The necessity of analytic continuation and the definition of the plasma dispersion function can be nicely clarified by a formula from a higher complex analysis, known as the Plemelj formula (Plemelj Reference Plemelj1908), which can be written in the following convenient form

(2.52) $$\begin{eqnarray}\lim _{\unicode[STIX]{x1D716}\rightarrow 0^{+}}\frac{1}{x-x_{0}\pm i\unicode[STIX]{x1D716}}=V.P.\frac{1}{x-x_{0}}\mp i\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FF}(x-x_{0}).\end{eqnarray}$$

The formula (2.52) is meant to be applied on a function $f(x)$ and integrated ‘through’ the pole, i.e. along the horizontal line $\text{Im}(x_{0})$ . The easiest is to consider $x_{0}=0$ (or $\text{Im}(x_{0})=0$ ) with integration along the real axis. The Dirac delta function $\unicode[STIX]{x1D6FF}(x-x_{0})$ in (2.52) represents contributions of the Landau residue. If the Landau residue is neglected, i.e. if only the $V.P.$ part in (2.52) is considered, as done by Vlasov (1945), yields that there is no damping present. In § 3.3, we will construct Padé approximants of $Z(\unicode[STIX]{x1D701})$ and $R(\unicode[STIX]{x1D701})$ . One can easily check that, by neglecting the Landau residue in the power-series expansions (the residue will be neglected in the asymptotic-series expansions), yields no collisionless damping. Therefore, as shown by van Kampen (Reference van Kampen1955), it is indeed possible to derive Landau damping by using Fourier analysis (an approach adopted here), provided the Landau residue in (2.52) is retained. The formula (2.52) is often attributed only to Plemelj (Reference Plemelj1908), for his rigorous proof. Sometimes it is called the Sokhotski–Plemelj formula, because it is argued the formula was derived in the doctoral thesis of Y. V. Sokhotski in 1873, with a proof that can be viewed as sufficiently rigorous for mathematical standards that existed at those times, i.e. 35 years before the rigorous proof of Plemelj. The Sokhotski–Plemelj formula is used in many areas of physics, from the theory of elasticity to quantum field theory.

2.4 Easy Landau integrals $\int x^{n}e^{-x^{2}}/(x-x_{0})\,\text{d}x$

We want to calculate moments in velocity space all the way up to the fourth-order moment $r$ , and (including the 3-D geometry) we will need integrals only up to $n=5$ . To this aim, we will use frequently (2.49), where we find it convenient to use $x_{0}$ and $\unicode[STIX]{x1D701}$ instead of chasing the $\text{sign}(k_{\Vert })$ , and in the end we will just use the definition $x_{0}=\text{sign}(k_{\Vert })\unicode[STIX]{x1D701}$ . We already saw that the zeroth-order moment was

(2.53) $$\begin{eqnarray}\displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x=\text{sign}(k_{\Vert })Z(\unicode[STIX]{x1D701}). & & \displaystyle\end{eqnarray}$$

Let us now calculate the higher-order moments. Since we talked so much in the last pages, we will remain silent for a moment and we will just enjoy the calculation,

(2.54) $$\begin{eqnarray}\displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{xe^{-x^{2}}}{x-x_{0}}\,\text{d}x & = & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x-x_{0}+x_{0}}{x-x_{0}}e^{-x^{2}}\,\text{d}x\nonumber\\ \displaystyle & = & \displaystyle \underbrace{\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }e^{-x^{2}}\,\text{d}x}_{=1}+\underbrace{\frac{x_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x}_{=\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})}\nonumber\\ \displaystyle & = & \displaystyle 1+\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})=R(\unicode[STIX]{x1D701}).\end{eqnarray}$$

(2.55) $$\begin{eqnarray}\displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{2}e^{-x^{2}}}{x-x_{0}}\,\text{d}x & = & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{2}-x_{0}^{2}+x_{0}^{2}}{x-x_{0}}e^{-x^{2}}\,\text{d}x\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }(x+x_{0})e^{-x^{2}}\,\text{d}x+\underbrace{\frac{x_{0}^{2}}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x}_{x_{0}\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})}\nonumber\\ \displaystyle & = & \displaystyle \underbrace{\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }xe^{-x^{2}}\,\text{d}x}_{=0}+\underbrace{\frac{x_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }e^{-x^{2}}\,\text{d}x}_{=x_{0}}+x_{0}\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})=x_{0}(1+\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701}))\nonumber\\ \displaystyle & = & \displaystyle \text{sign}(k_{\Vert })\unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701}).\end{eqnarray}$$

(2.56) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{3}e^{-x^{2}}}{x-x_{0}}\,\text{d}x=\underbrace{\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{3}-x_{0}^{3}}{x-x_{0}}e^{-x^{2}}\,\text{d}x}_{=(1/2)+x_{0}^{2}}+\underbrace{\frac{x_{0}^{3}}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x}_{=\unicode[STIX]{x1D701}^{3}Z(\unicode[STIX]{x1D701})}=\frac{1}{2}+\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701}). & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(2.57) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{4}e^{-x^{2}}}{x-x_{0}}\,\text{d}x=\text{sign}(k_{\Vert })\left(\frac{1}{2}\unicode[STIX]{x1D701}+\unicode[STIX]{x1D701}^{3}R(\unicode[STIX]{x1D701})\right). & \displaystyle\end{eqnarray}$$

(2.58) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{5}e^{-x^{2}}}{x-x_{0}}\,\text{d}x=\frac{3}{4}+\frac{\unicode[STIX]{x1D701}^{2}}{2}+\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701}). & \displaystyle\end{eqnarray}$$

That was easy! If we ever need a higher order, we will just blindly calculate

(2.59) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{n}e^{-x^{2}}}{x-x_{0}}\,\text{d}x=\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{x^{n}-x_{0}^{n}}{x-x_{0}}e^{-x^{2}}\,\text{d}x+\underbrace{\frac{x_{0}^{n}}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{e^{-x^{2}}}{x-x_{0}}\,\text{d}x}_{=x_{0}^{n}\text{sign}(k_{\Vert })Z(\unicode[STIX]{x1D701})}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }(x^{n-1}+x^{n-2}x_{0}+x^{n-3}x_{0}^{2}+\cdots +xx_{0}^{n-2}+x_{0}^{n-1})e^{-x^{2}}\,\text{d}x\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\text{sign}(k_{\Vert })^{n+1}\unicode[STIX]{x1D701}^{n}Z(\unicode[STIX]{x1D701}),\end{eqnarray}$$

and we do not worry right now if this general case can be expressed in some smarter way. Now we know how to calculate the kinetic Landau integrals, so let us use this knowledge to calculate the first few integrals of the linear ‘kinetic hierarchy’.

3.1 Kinetic moments for Maxwellian $f_{0}$

With the previous integrals already calculated, the calculation of the linear kinetic hierarchy is an easy process. However, it is important to emphasize that the hierarchy is linear, and must be calculated as such. Again, as emphasized before, the kinetic velocity $v$ is an independent quantity, and is not linearized. The total density $n=\int f\,\text{d}v$ , and at the first order of course $n_{0}=\int f_{0}\,\text{d}v$ . The expansion $n_{0}+n^{(1)}=\int (f_{0}+f^{(1)})\,\text{d}v$ implies $n^{(1)}=\int f^{(1)}\,\text{d}v$ . The density $n^{(1)}$ , already calculated in (2.50), was therefore calculated correctly, and using the plasma response function

(3.1) $$\begin{eqnarray}\frac{n^{(1)}}{n_{0}}=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}R(\unicode[STIX]{x1D701}).\end{eqnarray}$$

The velocity moment is $nu=\int vf\,\text{d}v$ and at the first order $n_{0}u_{0}=\int vf_{0}\,\text{d}v$ . In our specific case, because we do not consider any drifts in the distribution function, $u_{0}=0$ . Expanding $(n_{0}+n^{(1)})(u_{0}+u^{(1)})=\int v(f_{0}+f^{(1)})\,\text{d}v$ and neglecting the nonlinear quantity $n^{(1)}u^{(1)}$ , yields $n_{0}u^{(1)}=\int vf^{(1)}\,\text{d}v$ . The velocity moment calculates as

(3.2) $$\begin{eqnarray}\displaystyle n_{0}u^{(1)} & = & \displaystyle \int vf^{(1)}\,\text{d}v=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\int v\left(1+\frac{{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}\right)f_{0}\,\text{d}v\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(\underbrace{\int vf_{0}\,\text{d}v}_{=n_{0}u_{0}=0}+\int \frac{v{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}f_{0}\,\text{d}v\right)\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}n_{0}\sqrt{\frac{\unicode[STIX]{x1D6FC}}{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\int \frac{ve^{-\unicode[STIX]{x1D6FC}v^{2}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}\,\text{d}v=\left[\begin{array}{@{}c@{}}\sqrt{\unicode[STIX]{x1D6FC}}v=x\\ \sqrt{\unicode[STIX]{x1D6FC}}\,\text{d}v=\,\text{d}x\end{array}\right]\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\frac{n_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\frac{\sqrt{\unicode[STIX]{x1D6FC}}}{\sqrt{\unicode[STIX]{x1D6FC}}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\int \frac{xe^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}\sqrt{\unicode[STIX]{x1D6FC}}}\,\text{d}x\nonumber\\ \displaystyle & = & \displaystyle \left[x_{0}=\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\sqrt{\unicode[STIX]{x1D6FC}}\right]=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\frac{n_{0}}{\sqrt{\unicode[STIX]{x1D6FC}}}\frac{x_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\int _{-\infty }^{\infty }\frac{xe^{-x^{2}}}{x-x_{0}}\,\text{d}x\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\frac{n_{0}}{\sqrt{\unicode[STIX]{x1D6FC}}}\text{sign}(k_{\Vert })\unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701}),\end{eqnarray}$$

and cancelling $n_{0}$ and using $1/\sqrt{\unicode[STIX]{x1D6FC}}=v_{\text{th}}=\sqrt{2T^{(0)}/m}$ yields

(3.3) $$\begin{eqnarray}u^{(1)}=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\sqrt{\frac{2T^{(0)}}{m}}\text{sign}(k_{\Vert })\unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701}).\end{eqnarray}$$

The definition of the scalar pressure is $p=m\int (v-u)^{2}f\,\text{d}v$ and at the first order $p_{0}=m\int v^{2}f_{0}\,\text{d}v$ , because again $u_{0}=0$ . The quantity $(v-u)^{2}=v^{2}-2vu+u^{2}$ is linearized as $v^{2}-2vu^{(1)}$ , and expanding $p_{0}+p^{(1)}=m\int (v^{2}-2vu^{(1)})(f_{0}+f^{(1)})\,\text{d}v$ , further linearizing by neglecting $u^{(1)}f^{(1)}$ and using $u_{0}=0$ yields $p^{(1)}=m\int v^{2}f^{(1)}\,\text{d}v$ . The pressure calculates as

(3.4) $$\begin{eqnarray}\displaystyle p^{(1)} & = & \displaystyle m\int v^{2}f^{(1)}\,\text{d}v=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(\underbrace{m\int v^{2}f_{0}\,\text{d}v}_{=p_{0}}+m\int \frac{v^{2}{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}f_{0}\,\text{d}v\right)=\left[\begin{array}{@{}c@{}}\sqrt{\unicode[STIX]{x1D6FC}}v=x\\ \sqrt{\unicode[STIX]{x1D6FC}}\,\text{d}v=\,\text{d}x\end{array}\right]\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(p_{0}+m\frac{n_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\frac{\sqrt{\unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x1D6FC}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\int \frac{x^{2}e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}\sqrt{\unicode[STIX]{x1D6FC}}}\,\text{d}x\right)=\left[x_{0}=\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\sqrt{\unicode[STIX]{x1D6FC}}\right]\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(p_{0}+m\frac{n_{0}}{\unicode[STIX]{x1D6FC}}\frac{x_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\int \frac{x^{2}e^{-x^{2}}}{x-x_{0}}\,\text{d}x\right)=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(p_{0}+m\frac{n_{0}}{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701})\right),\end{eqnarray}$$

and dividing by $p_{0}$ and using $p_{0}=n_{0}T^{(0)}$ to calculate $mn_{0}/(p_{0}\unicode[STIX]{x1D6FC})=2$ , the pressure moment reads

(3.5) $$\begin{eqnarray}\frac{p^{(1)}}{p_{0}}=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(1+2\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701})\right).\end{eqnarray}$$

We will also need the temperature $T^{(1)}$ . The general temperature is defined as $T=p/n$ , i.e. the definition is nonlinear. The process of linearization is essentially like doing a derivative

(3.6) $$\begin{eqnarray}\displaystyle T=\frac{p}{n}\overset{\text{lin.}}{\rightarrow }T^{(1)}=\frac{p^{(1)}}{n_{0}}-\frac{p_{0}}{n_{0}^{2}}n^{(1)}, & & \displaystyle\end{eqnarray}$$

and dividing by $T^{(0)}=p_{0}/n_{0}$ yields

(3.7) $$\begin{eqnarray}\frac{T^{(1)}}{T^{(0)}}=\frac{p^{(1)}}{p_{0}}-\frac{n^{(1)}}{n_{0}}.\end{eqnarray}$$

If one does not like the ‘derivative’, the same result is obtained by writing $p=Tn$ instead, and linearizing $(p_{0}+p^{(1)})=(T^{(0)}+T^{(1)})(n_{0}+n^{(1)})$ . Which after subtracting $p_{0}=T^{(0)}n_{0}$ , neglecting $T^{(1)}n^{(1)}$ , yields $p^{(1)}=T^{(1)}n_{0}+T^{(0)}n^{(1)}$ , which after dividing by $p_{0}$ yields (3.7). The temperature is therefore easily calculated as

(3.8) $$\begin{eqnarray}\frac{T^{(1)}}{T^{(0)}}=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(1+2\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701})-R(\unicode[STIX]{x1D701})\right).\end{eqnarray}$$

The scalar heat flux is defined as $q=m\int (v-u)^{3}f\,\text{d}v$ and at the first order $q_{0}=m\int v^{3}f_{0}\,\text{d}v$ , and for our case $q_{0}=0$ . The quantity $(v-u)^{3}=v^{3}-3v^{2}u+3vu^{2}-u^{3}$ is linearized as $v^{3}-3v^{2}u^{(1)}$ . Expanding $q_{0}+q^{(1)}=m\int (v^{3}-3v^{2}u^{(1)})(f_{0}+f^{(1)})\,\text{d}v$ , neglecting $u^{(1)}f^{(1)}$ , yields one contribution that is very easy to overlook, and that is of the same order as the expected $m\int v^{3}f^{(1)}\,\text{d}v$ , and that is proportional to $m\int v^{2}f_{0}\,\text{d}v=p_{0}$ . Therefore, the linearized heat flux $q^{(1)}$ must be correctly calculated according to

(3.9) $$\begin{eqnarray}q^{(1)}=m\int v^{3}f^{(1)}\,\text{d}v-3p_{0}u^{(1)}.\end{eqnarray}$$

The first term calculates as

(3.10) $$\begin{eqnarray}\displaystyle m\int v^{3}f^{(1)}\,\text{d}v & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(\underbrace{m\int v^{3}f_{0}\,\text{d}v}_{=q_{0}=0}+m\int \frac{v^{3}{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}f_{0}\,\text{d}v\right)=\left[\begin{array}{@{}c@{}}\sqrt{\unicode[STIX]{x1D6FC}}v=x\\ \sqrt{\unicode[STIX]{x1D6FC}}\,\text{d}v=\,\text{d}x\end{array}\right]\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}m\frac{n_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\frac{\sqrt{\unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x1D6FC}^{3/2}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\int \frac{x^{3}e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}\sqrt{\unicode[STIX]{x1D6FC}}}\,\text{d}x=\left[x_{0}=\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\sqrt{\unicode[STIX]{x1D6FC}}\right]\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}m\frac{n_{0}}{\unicode[STIX]{x1D6FC}^{3/2}}\frac{x_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\int \frac{x^{3}e^{-x^{2}}}{x-x_{0}}\,\text{d}x\nonumber\\ \displaystyle & = & \displaystyle -q_{r}n_{0}\unicode[STIX]{x1D6F7}\sqrt{\frac{2T^{(0)}}{m}}\text{sign}(k_{\Vert })\left(\unicode[STIX]{x1D701}+2\unicode[STIX]{x1D701}^{3}R(\unicode[STIX]{x1D701})\right),\end{eqnarray}$$

where we used $\unicode[STIX]{x1D6FC}^{-3/2}=(2T^{(0)}/m)\sqrt{2T^{(0)}/m}$ . And the entire heat flux (3.9) then reads

(3.11) $$\begin{eqnarray}\displaystyle q^{(1)}=-q_{r}n_{0}\unicode[STIX]{x1D6F7}\sqrt{\frac{2T^{(0)}}{m}}\text{sign}(k_{\Vert })\left(\unicode[STIX]{x1D701}+2\unicode[STIX]{x1D701}^{3}R(\unicode[STIX]{x1D701})-3\unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701})\right). & & \displaystyle\end{eqnarray}$$

The scalar fourth-order moment is defined as $r=m\int (v-u)^{4}f\,\text{d}v$ and at the first order of course $r_{0}=m\int v^{4}f_{0}\,\text{d}v$ , since again $u_{0}=0$ . Also, $r_{0}=3p_{0}^{2}/\unicode[STIX]{x1D70C}_{0}$ , where $\unicode[STIX]{x1D70C}_{0}=mn_{0}$ . The quantity $(v-u)^{4}$ is linearized as $v^{4}-4v^{3}u^{(1)}$ . Expanding $r_{0}+r^{(1)}=m\int (v^{4}-4v^{3}u^{(1)})(f_{0}+f^{(1)})\,\text{d}v$ , the quantity $m\int v^{3}f_{0}=q_{0}=0$ , which yields a simple $r^{(1)}=m\int v^{4}f^{(1)}\,\text{d}v$ . The fourth-order moment calculates as

(3.12) $$\begin{eqnarray}\displaystyle r^{(1)} & = & \displaystyle m\int v^{4}f^{(1)}\,\text{d}v=-\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(\underbrace{m\int v^{4}f_{0}\,\text{d}v}_{=r_{0}}+m\int \frac{v^{4}{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}{v-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}}f_{0}\,\text{d}v\right)=\left[\begin{array}{@{}c@{}}\sqrt{\unicode[STIX]{x1D6FC}}v=x\\ \sqrt{\unicode[STIX]{x1D6FC}}\,\text{d}v=\,\text{d}x\end{array}\right]\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(r_{0}+m\frac{n_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\frac{\sqrt{\unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x1D6FC}^{2}}\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\int \frac{x^{4}e^{-x^{2}}}{x-{\displaystyle \frac{\unicode[STIX]{x1D714}}{k_{\Vert }}}\sqrt{\unicode[STIX]{x1D6FC}}}\,\text{d}x\right)=\left[x_{0}=\frac{\unicode[STIX]{x1D714}}{k_{\Vert }}\sqrt{\unicode[STIX]{x1D6FC}}\right]\nonumber\\ \displaystyle & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\left(r_{0}+m\frac{n_{0}}{\unicode[STIX]{x1D6FC}^{2}}\frac{x_{0}}{\sqrt{\unicode[STIX]{x03C0}}}\int \frac{x^{4}e^{-x^{2}}}{x-x_{0}}\,\text{d}x\right)=-\frac{q_{r}p_{0}}{m}\unicode[STIX]{x1D6F7}\left(3+2\unicode[STIX]{x1D701}^{2}+4\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701})\right),\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where we have used $1/\unicode[STIX]{x1D6FC}^{2}=4T^{(0)2}/m^{2}$ , and $mn_{0}/\unicode[STIX]{x1D6FC}^{2}=4p_{0}^{2}/\unicode[STIX]{x1D70C}_{0}$ .

The entire nonlinear $r$ is decomposed as $r=3p^{2}/\unicode[STIX]{x1D70C}+\widetilde{r}$ . The first term can be linearized in a number of ways, and of course, all techniques must yield the same result, since linearization must be unique. For example, by using the derivative

(3.13) $$\begin{eqnarray}\left(\frac{p^{2}}{\unicode[STIX]{x1D70C}}\right)^{\prime }=\frac{1}{\unicode[STIX]{x1D70C}}2pp^{\prime }-\frac{p^{2}}{\unicode[STIX]{x1D70C}^{2}}\unicode[STIX]{x1D70C}^{\prime }=\frac{p^{2}}{\unicode[STIX]{x1D70C}}\left(\frac{2p^{\prime }}{p}-\frac{\unicode[STIX]{x1D70C}^{\prime }}{\unicode[STIX]{x1D70C}}\right),\end{eqnarray}$$

the term is easily linearized as

(3.14) $$\begin{eqnarray}\frac{p^{2}}{\unicode[STIX]{x1D70C}}\overset{\text{lin.}}{=}\frac{p_{0}^{2}}{\unicode[STIX]{x1D70C}_{0}}\left(\frac{2p^{(1)}}{p_{0}}-\frac{\unicode[STIX]{x1D70C}^{(1)}}{\unicode[STIX]{x1D70C}_{0}}\right)=\frac{p_{0}^{2}}{\unicode[STIX]{x1D70C}_{0}}\left(\frac{2p^{(1)}}{p_{0}}-\frac{n^{(1)}}{n_{0}}\right),\end{eqnarray}$$

and by further using (3.7), also alternatively as

(3.15) $$\begin{eqnarray}\frac{p^{2}}{\unicode[STIX]{x1D70C}}\overset{\text{lin.}}{=}\frac{p_{0}^{2}}{\unicode[STIX]{x1D70C}_{0}}\left(\frac{2T^{(1)}}{T^{(0)}}+\frac{n^{(1)}}{n_{0}}\right).\end{eqnarray}$$

Using $r_{0}=3p_{0}^{2}/\unicode[STIX]{x1D70C}_{0}$ therefore yields useful relations (valid for a Maxwellian)

(3.16) $$\begin{eqnarray}\frac{r^{(1)}}{r_{0}}=2\frac{p^{(1)}}{p_{0}}-\frac{n^{(1)}}{n_{0}}+\frac{\widetilde{r}^{(1)}}{r_{0}}=2\frac{T^{(1)}}{T^{(0)}}+\frac{n^{(1)}}{n_{0}}+\frac{\widetilde{r}^{(1)}}{r_{0}}.\end{eqnarray}$$

Another possibility (to double check the linearization), is to rewrite $p^{2}/\unicode[STIX]{x1D70C}=(1/m)pT$ , so that $r=(3/m)pT+\widetilde{r}$ , and to linearize that one instead. Expanding that expression into $r_{0}+r^{(1)}=(3/m)((p_{0}+p^{(1)})(T^{(0)}+T^{(1)}))+\widetilde{r}^{(1)}$ (where by definition/construction $\widetilde{r}^{(0)}=0$ ), after subtracting $r_{0}=(3/m)p_{0}T^{(0)}$ , and neglecting $p^{(1)}T^{(1)}$ , yields $r^{(1)}=(3/m)(p^{(1)}T^{(0)}+p_{0}T^{(1)})+\widetilde{r}^{(1)}$ . Dividing this expression by $r_{0}$ yields

(3.17) $$\begin{eqnarray}\frac{r^{(1)}}{r_{0}}=\frac{p^{(1)}}{p_{0}}+\frac{T^{(1)}}{T^{(0)}}+\frac{\widetilde{r}^{(1)}}{r_{0}},\end{eqnarray}$$

which when used with (3.7), is equivalent to (3.16). Now we can easily calculate the $\widetilde{r}^{(1)}$ component as

(3.18) $$\begin{eqnarray}\widetilde{r}^{(1)}=r^{(1)}-3\frac{p_{0}^{2}}{\unicode[STIX]{x1D70C}_{0}}\left(\frac{2p^{(1)}}{p_{0}}-\frac{n^{(1)}}{n_{0}}\right),\end{eqnarray}$$

that directly yields

(3.19) $$\begin{eqnarray}\widetilde{r}^{(1)}=-\frac{q_{r}p_{0}}{m}\unicode[STIX]{x1D6F7}\left(2\unicode[STIX]{x1D701}^{2}+4\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701})+3R(\unicode[STIX]{x1D701})-3-12\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701})\right).\end{eqnarray}$$

Now we are ready to explore the possible closures.

3.2 Exploring possibilities of a closure

Let us summarize the obtained linear hierarchy so that we can directly see the similarities. Let us also for a moment introduce back the species index $r$ , so that we are completely clear

(3.20) $$\begin{eqnarray}\displaystyle \frac{n_{r}^{(1)}}{n_{0r}} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}R(\unicode[STIX]{x1D701}_{r});\end{eqnarray}$$

(3.21) $$\begin{eqnarray}\displaystyle u_{r}^{(1)} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}\sqrt{\frac{2T_{r}^{(0)}}{m_{r}}}\text{sign}(k_{\Vert })\unicode[STIX]{x1D701}_{r}R(\unicode[STIX]{x1D701}_{r});\end{eqnarray}$$

(3.22) $$\begin{eqnarray}\displaystyle \frac{p_{r}^{(1)}}{p_{0r}} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}(1+2\unicode[STIX]{x1D701}_{r}^{2}R(\unicode[STIX]{x1D701}_{r}));\end{eqnarray}$$

(3.23) $$\begin{eqnarray}\displaystyle \frac{T_{r}^{(1)}}{T_{r}^{(0)}} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}(1+2\unicode[STIX]{x1D701}_{r}^{2}R(\unicode[STIX]{x1D701}_{r})-R(\unicode[STIX]{x1D701}_{r}));\end{eqnarray}$$

(3.24) $$\begin{eqnarray}\displaystyle q_{r}^{(1)} & = & \displaystyle -q_{r}n_{0r}\unicode[STIX]{x1D6F7}\sqrt{\frac{2T_{r}^{(0)}}{m_{r}}}\text{sign}(k_{\Vert })(\unicode[STIX]{x1D701}_{r}+2\unicode[STIX]{x1D701}_{r}^{3}R(\unicode[STIX]{x1D701}_{r})-3\unicode[STIX]{x1D701}_{r}R(\unicode[STIX]{x1D701}_{r}));\end{eqnarray}$$

(3.25) $$\begin{eqnarray}\displaystyle r_{r}^{(1)} & = & \displaystyle -\frac{q_{r}p_{0r}}{m_{r}}\unicode[STIX]{x1D6F7}(3+2\unicode[STIX]{x1D701}_{r}^{2}+4\unicode[STIX]{x1D701}_{r}^{4}R(\unicode[STIX]{x1D701}_{r}));\end{eqnarray}$$

(3.26) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle -\frac{q_{r}p_{0r}}{m_{r}}\unicode[STIX]{x1D6F7}(2\unicode[STIX]{x1D701}_{r}^{2}+4\unicode[STIX]{x1D701}_{r}^{4}R(\unicode[STIX]{x1D701}_{r})+3R(\unicode[STIX]{x1D701}_{r})-3-12\unicode[STIX]{x1D701}_{r}^{2}R(\unicode[STIX]{x1D701}_{r})),\end{eqnarray}$$

with an emphasis that the charge $q_{r}$ should not be confused with the heat flux $q_{r}^{(1)}$ . The $\unicode[STIX]{x1D701}_{r}=\unicode[STIX]{x1D714}/(|k_{\Vert }|v_{\text{th}r})$ and the thermal speed $v_{\text{th}r}=\sqrt{2T_{r}^{(0)}/m_{r}}$ . Note the presence of $\text{sign}(k_{\Vert })$ in the expressions for $u_{r}^{(1)}$ and $q_{r}^{(1)}$ . The presence of $\text{sign}(k_{\Vert })$ can be verified a posteriori, for example by considering the simplest situation when the Landau damping is neglected, and the $R(\unicode[STIX]{x1D701}_{r})$ function yields only real numbers for real valued $\unicode[STIX]{x1D701}_{r}$ (i.e. the $R(\unicode[STIX]{x1D701}_{r})$ function can be approximated with Padé approximants that contain only powers of $\unicode[STIX]{x1D701}_{r}^{2}$ ). Simultaneously changing the signs of $k_{\Vert }$ and $\unicode[STIX]{x1D714}$ in a Fourier mode should give its complex conjugate, i.e. the real part of expressions (3.20)–(3.26) cannot change its sign in that transformation. This is indeed true because the expressions for $u_{r}^{(1)}$ and $q_{r}^{(1)}$ contain $\text{sign}(k_{\Vert })\unicode[STIX]{x1D701}_{r}=\unicode[STIX]{x1D714}/(k_{\Vert }v_{\text{th}r})$ .

To better understand what is meant by ‘a closure’, let us first examine what is not a closure. Let us examine the density $n^{(1)}$ equation. Since in this specific example we used the electrostatic electric field $\boldsymbol{E}^{(1)}=-\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}$ , the only Maxwell equation left is $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{E}^{(1)}=4\unicode[STIX]{x03C0}\sum _{r}q_{r}n_{r}$ , where $q_{r}$ is the charge and $n_{r}$ is the total density. Linearization of this equation, and using the natural charge neutrality that must be satisfied at the zeroth order $\sum _{r}q_{r}n_{0r}=0$ , yields $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{E}^{(1)}=4\unicode[STIX]{x03C0}\sum _{r}q_{r}n_{r}^{(1)}$ , or written with the scalar potential $-\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D6F7}=4\unicode[STIX]{x03C0}\sum _{r}q_{r}n_{r}^{(1)}$ , and transformed to Fourier space, $k^{2}\unicode[STIX]{x1D6F7}=4\unicode[STIX]{x03C0}\sum _{r}q_{r}n_{r}^{(1)}$ . We consider 1-D propagation parallel to $\boldsymbol{B}_{0}$ with wavenumber $k_{\Vert }$ , and to be consistent, we therefore continue with $k_{\Vert }$ and

(3.27) $$\begin{eqnarray}k_{\Vert }^{2}\unicode[STIX]{x1D6F7}=4\unicode[STIX]{x03C0}\mathop{\sum }_{r}q_{r}n_{r}^{(1)}=4\unicode[STIX]{x03C0}\mathop{\sum }_{r}q_{r}n_{0r}\frac{n_{r}^{(1)}}{n_{0r}}=-4\unicode[STIX]{x03C0}\mathop{\sum }_{r}n_{0r}\frac{q_{r}^{2}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}R(\unicode[STIX]{x1D701}_{r}),\end{eqnarray}$$

which can be rewritten as

(3.28) $$\begin{eqnarray}\left(k_{\Vert }^{2}+4\unicode[STIX]{x03C0}\mathop{\sum }_{r}n_{0r}\frac{q_{r}^{2}}{T_{r}^{(0)}}R(\unicode[STIX]{x1D701}_{r})\right)\unicode[STIX]{x1D6F7}=0.\end{eqnarray}$$

Even though the system is now ‘closed’, the equation (3.28) does not represent a fluid closure, and should be viewed only as a kinetic ‘dispersion relation’. To have a non-trivial solution for the potential $\unicode[STIX]{x1D6F7}$ , the expression inside of the bracket must be equal to zero. By declaring that $k_{\Vert }\neq 0$ (the case $k_{\Vert }=0$ is trivial since we need some wavenumber), we can divide by $k_{\Vert }^{2}$ . By using the definition of the Debye length of $r$ -species $\unicode[STIX]{x1D706}_{Dr}=1/k_{Dr}$ , where $k_{Dr}^{2}=4\unicode[STIX]{x03C0}n_{0r}q_{r}^{2}/T_{r}^{(0)}$ , one obtains a dispersion relationFootnote ⁶

(3.29) $$\begin{eqnarray}1+\mathop{\sum }_{r}\frac{1}{k_{\Vert }^{2}\unicode[STIX]{x1D706}_{Dr}^{2}}R(\unicode[STIX]{x1D701}_{r})=0.\end{eqnarray}$$

If one replaces here $k_{\Vert }\rightarrow k$ , the expression is actually equivalent to a multi-species dispersion relation, usually found in plasma physics books under the electrostatic waves in hot unmagnetized plasmas, with Maxwellian $f_{0r}$ . See for example Gurnett & Bhattachrjee, page 353, equation (9.4.18). We are not interested here in studying unmagnetized plasmas, and instead, we will just remember (3.29) as the dispersion relation of the parallel propagating (to $\boldsymbol{B}_{0}$ ) electrostatic mode in a magnetized plasma, since this mode indeed does not contain any magnetic field fluctuations.

Let us consider only the proton and electron species, $r=p,e$ , so that

(3.30) $$\begin{eqnarray}1+\frac{1}{k_{\Vert }^{2}\unicode[STIX]{x1D706}_{De}^{2}}\left[\frac{T_{e}^{(0)}}{T_{p}^{(0)}}R(\unicode[STIX]{x1D701}_{p})+R(\unicode[STIX]{x1D701}_{e})\right]=0,\end{eqnarray}$$

where the proton Debye length was rewritten with the electron Debye length $\unicode[STIX]{x1D706}_{De}=\unicode[STIX]{x1D706}_{Dp}\sqrt{T_{e}^{(0)}/T_{p}^{(0)}}$ . For a general case, the dispersion relation has to be solved numerically, and again, cannot be much simplified, unless one wants to consider the long-wavelength limit $k_{\Vert }\unicode[STIX]{x1D706}_{De}\ll 1$ , where only the expression inside of the big brackets can be used. The solution contains the usual Langmuir waves, that are obtained by neglecting the ion term (by making the ions immobile) and by expanding the $R(\unicode[STIX]{x1D701}_{e})$ in the limit $|\unicode[STIX]{x1D701}_{e}|\gg 1$ , i.e. in the limit when the wave phase speed $\unicode[STIX]{x1D714}/k$ is much larger than the electron thermal speed $v_{\text{th}e}$ . Langmuir waves propagate with speeds that are higher than the electron plasma frequency $\unicode[STIX]{x1D714}_{pe}=\sqrt{4\unicode[STIX]{x03C0}n_{e0}e^{2}/m_{e}}$ , which for us are extremely high frequencies. The solution also contains the ‘ion-acoustic mode’, which in plasma books is obtained in the limit $|\unicode[STIX]{x1D701}_{p}|\gg 1$ and $|\unicode[STIX]{x1D701}_{e}|\ll 1$ , i.e. in the limit where the wave phase speed is much larger than the proton thermal speed, $\unicode[STIX]{x1D714}/k\gg v_{\text{th}p}$ , but also where the phase speed is much smaller than the electron thermal speed, $\unicode[STIX]{x1D714}/k\ll v_{\text{th}e}$ , for the result see for example Gurnett and Bhattacharjee, page 356, equation (9.4.28-29).

So what about the limit $|\unicode[STIX]{x1D701}_{p}|\ll 1$ , when the phase speed is much smaller than the proton thermal speed $\unicode[STIX]{x1D714}/k\ll v_{\text{th}p}$ ? Does the ion-acoustic mode not exist in this limit? Unfortunately, in the classical long-wavelength limit, the phase speeds do not become smaller and smaller, the phase speeds $\unicode[STIX]{x1D714}/k$ just become non-dispersive and constant. In the CGL description (with cold electrons), the parallel propagating ion-acoustic mode has a phase speed $\unicode[STIX]{x1D714}/k_{\Vert }=\pm C_{\Vert }$ , where the parallel sound speed $C_{\Vert }^{2}=3p_{\Vert p}^{(0)}/\unicode[STIX]{x1D70C}_{0}=3T_{\Vert p}^{(0)}/m_{p}$ . The limit $|\unicode[STIX]{x1D701}_{p}|\ll 1$ is never satisfied, because $C_{\Vert }\ll v_{\text{th}\Vert p}$ means $\sqrt{3T_{\Vert p}^{(0)}/m_{p}}\ll \sqrt{2T_{\Vert p}^{(0)}/m_{p}}$ , which is never true. One can estimate the lowest possible value of $|\unicode[STIX]{x1D701}_{p}|$ to be roughly in the neighbourhood of $|\unicode[STIX]{x1D701}_{p}|_{\text{min}}\approx C_{\Vert }^{\text{CGL}}/v_{\text{th}\Vert p}=\sqrt{3/2}$ , or in another words $|\unicode[STIX]{x1D701}_{p}|_{\text{min}}\approx 1$ . There is no expansion of the $Z(\unicode[STIX]{x1D701})$ for $|\unicode[STIX]{x1D701}|\approx 1$ and the result has to be found numerically.

So what constitutes a Landau fluid closure? We will use the following definition: express the last retained moment through lower-order moments in such a way, that the kinetic $R(\unicode[STIX]{x1D701})$ function is eliminated (for example by using Padé approximation), so that the closure is expressed only through fluid variables and it is prescribed for all $\unicode[STIX]{x1D701}$ values.

3.2.1 Preliminary closures for $|\unicode[STIX]{x1D701}|\ll 1$

As explained above, the limit $|\unicode[STIX]{x1D701}|\ll 1$ is actually a bit unphysical for the proton species in the electrostatic limit, and is physically plausible only for the electron species. Nevertheless, briefly exploring the linear kinetic hierarchy in this limit allows us to explore what kind of closures might be possible. In this limit, the plasma dispersion function can be expanded as

(3.31) $$\begin{eqnarray}\displaystyle Z(\unicode[STIX]{x1D701}) & = & \displaystyle i\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}-2\unicode[STIX]{x1D701}\left[1-\frac{2}{3}\unicode[STIX]{x1D701}^{2}+\frac{4}{15}\unicode[STIX]{x1D701}^{4}-\frac{8}{105}\unicode[STIX]{x1D701}^{6}+\cdots \,\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{(-2)^{n}\unicode[STIX]{x1D701}^{2n}}{(2n+1)!!}+\cdots \,\right];\quad |\unicode[STIX]{x1D701}|\ll 1,\end{eqnarray}$$

(3.32) $$\begin{eqnarray}\displaystyle Z(\unicode[STIX]{x1D701})=i\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}-2\unicode[STIX]{x1D701}+\frac{4}{3}\unicode[STIX]{x1D701}^{3}-\frac{8}{15}\unicode[STIX]{x1D701}^{5}+\frac{16}{105}\unicode[STIX]{x1D701}^{7}+\cdots \,; & & \displaystyle\end{eqnarray}$$

and the plasma response function as

(3.33) $$\begin{eqnarray}\displaystyle R(\unicode[STIX]{x1D701}) & = & \displaystyle 1+i\unicode[STIX]{x1D701}\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}+\left[-2\unicode[STIX]{x1D701}^{2}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}-\frac{8}{15}\unicode[STIX]{x1D701}^{6}+\frac{16}{105}\unicode[STIX]{x1D701}^{8}+\cdots \,\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{(-2)^{n+1}\unicode[STIX]{x1D701}^{2n+2}}{(2n+1)!!}+\cdots \,\right];\quad |\unicode[STIX]{x1D701}|\ll 1,\end{eqnarray}$$

and where for small $\unicode[STIX]{x1D701}$ , $e^{-\unicode[STIX]{x1D701}^{2}}$ is naturally expanded as

(3.34) $$\begin{eqnarray}e^{-\unicode[STIX]{x1D701}^{2}}=1-\unicode[STIX]{x1D701}^{2}+\frac{\unicode[STIX]{x1D701}^{4}}{2!}-\frac{\unicode[STIX]{x1D701}^{6}}{3!}+\cdots +\frac{(-1)^{n}\unicode[STIX]{x1D701}^{2n}}{n!}+\cdots \,;\quad |\unicode[STIX]{x1D701}|\ll 1,\end{eqnarray}$$

yielding

(3.35) $$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D701}|\ll 1:\quad Z(\unicode[STIX]{x1D701}) & = & \displaystyle i\sqrt{\unicode[STIX]{x03C0}}-2\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}+\frac{4}{3}\unicode[STIX]{x1D701}^{3}+i\frac{\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}^{4}\nonumber\\ \displaystyle & & \displaystyle -\,\frac{8}{15}\unicode[STIX]{x1D701}^{5}-i\frac{\sqrt{\unicode[STIX]{x03C0}}}{6}\unicode[STIX]{x1D701}^{6}+\frac{16}{105}\unicode[STIX]{x1D701}^{7}+\cdots \,;\end{eqnarray}$$

(3.36) $$\begin{eqnarray}\displaystyle R(\unicode[STIX]{x1D701}) & = & \displaystyle 1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}+i\frac{\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}^{5}\nonumber\\ \displaystyle & & \displaystyle -\,\frac{8}{15}\unicode[STIX]{x1D701}^{6}-i\frac{\sqrt{\unicode[STIX]{x03C0}}}{6}\unicode[STIX]{x1D701}^{7}+\frac{16}{105}\unicode[STIX]{x1D701}^{8}+\cdots\end{eqnarray}$$

For our purposes it is sufficient to keep the series only up to $\unicode[STIX]{x1D701}^{3}$ , i.e. to work with the precision $o(\unicode[STIX]{x1D701}^{3})$ . The expressions entering the kinetic hierarchy in (3.20)–(3.26) are

(3.37) $$\begin{eqnarray}\displaystyle R(\unicode[STIX]{x1D701}) & = & \displaystyle 1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\cdots \,;\quad\end{eqnarray}$$

(3.38) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701}) & = & \displaystyle \unicode[STIX]{x1D701}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}-2\unicode[STIX]{x1D701}^{3};\end{eqnarray}$$

(3.39) $$\begin{eqnarray}\displaystyle 1+2\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701}) & = & \displaystyle 1+2\unicode[STIX]{x1D701}^{2}+2i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3};\end{eqnarray}$$

(3.40) $$\begin{eqnarray}\displaystyle 1-R(\unicode[STIX]{x1D701})+2\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701}) & = & \displaystyle -i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}+4\unicode[STIX]{x1D701}^{2}+3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3};\end{eqnarray}$$

(3.41) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}+2\unicode[STIX]{x1D701}^{3}R(\unicode[STIX]{x1D701})-3\unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701}) & = & \displaystyle -2\unicode[STIX]{x1D701}-3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}+8\unicode[STIX]{x1D701}^{3};\end{eqnarray}$$

(3.42) $$\begin{eqnarray}\displaystyle 3+2\unicode[STIX]{x1D701}^{2}+4\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701}) & = & \displaystyle 3+2\unicode[STIX]{x1D701}^{2}+4\unicode[STIX]{x1D701}^{4};\end{eqnarray}$$

(3.43) $$\begin{eqnarray}\displaystyle 2\unicode[STIX]{x1D701}^{2}+4\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701})+3R(\unicode[STIX]{x1D701})-3-12\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701}) & = & \displaystyle 3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-16\unicode[STIX]{x1D701}^{2}-15i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}.\end{eqnarray}$$

An interesting observation is that for small $\unicode[STIX]{x1D701}$ , moments $n^{(1)}$ , $p^{(1)}$ and $r^{(1)}$ are finite, and moments $u^{(1)}$ , $T^{(1)}$ , $q^{(1)}$ and $\widetilde{r}^{(1)}$ are proportional to $\unicode[STIX]{x1D701}$ and therefore small. We want to make a simple closure for the heat flux $q^{(1)}$ or the fourth-order correction $\widetilde{r}^{(1)}$ , and thus, let us concentrate on the moments that are small. To clarify how the closure is performed, let us write them down only up to the precision $o(\unicode[STIX]{x1D701}_{r}^{2})$ , so

(3.44) $$\begin{eqnarray}\displaystyle u_{r}^{(1)} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}\sqrt{\frac{2T_{r}^{(0)}}{m_{r}}}\text{sign}(k_{\Vert })(\unicode[STIX]{x1D701}_{r}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}_{r}^{2});\end{eqnarray}$$

(3.45) $$\begin{eqnarray}\displaystyle T_{r}^{(1)} & = & \displaystyle -q_{r}\unicode[STIX]{x1D6F7}\left(-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}_{r}+4\unicode[STIX]{x1D701}_{r}^{2}\right);\end{eqnarray}$$

(3.46) $$\begin{eqnarray}\displaystyle q_{r}^{(1)} & = & \displaystyle -q_{r}n_{0r}\unicode[STIX]{x1D6F7}\sqrt{\frac{2T_{r}^{(0)}}{m_{r}}}\text{sign}(k_{\Vert })(-2\unicode[STIX]{x1D701}_{r}-3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}_{r}^{2});\end{eqnarray}$$

(3.47) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle -\frac{q_{r}p_{0r}}{m_{r}}\unicode[STIX]{x1D6F7}(3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}_{r}-16\unicode[STIX]{x1D701}_{r}^{2}).\end{eqnarray}$$

If we further restrict ourselves to only precision $o(\unicode[STIX]{x1D701}_{r})$ and neglect the $\unicode[STIX]{x1D701}_{r}^{2}$ terms, we can find an amazing result that we can express the heat flux $q_{r}^{(1)}$ with respect to temperature $T_{r}^{(1)}$ according to

(3.48)

The above result is of upmost importance, because it emphasizes the major difference between collisionless and collisional systems. At this point, the result is derived only with the assumption $|\unicode[STIX]{x1D701}|\ll 1$ , even though we will see later that the result is not restricted to this limit, and the result has a much wider applicability. The result is the famous expression for collisionless heat flux, that here reads $q\sim -i\,\text{sign}(k_{\Vert })T$ , which is in strong contrast to the usual collisional heat flux $q\sim -\unicode[STIX]{x1D735}_{\Vert }T$ that in Fourier space reads $q\sim -ik_{\Vert }T$ . We will come to this expression later.

With the precision $o(\unicode[STIX]{x1D701}_{r})$ , other obvious possibilities are to express $q_{r}^{(1)}$ with respect to velocity $u_{r}^{(1)}$ , or to express $\widetilde{r}^{(1)}$ through $u_{r}^{(1)}$ , $T_{r}^{(1)}$ , $q_{r}^{(1)}$ according to

(3.49) $$\begin{eqnarray}\displaystyle o(\unicode[STIX]{x1D701}_{r}):\quad q_{r}^{(1)} & = & \displaystyle -2n_{0r}T_{r}^{(0)}u_{r}^{(1)}=-2p_{0r}u_{r}^{(1)};\end{eqnarray}$$

(3.50) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle i{\textstyle \frac{3}{2}}\sqrt{\unicode[STIX]{x03C0}}v_{\text{th}r}p_{0r}\text{sign}(k_{\Vert })u_{r}^{(1)};\end{eqnarray}$$

(3.51) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle -{\textstyle \frac{3}{2}}v_{\text{th}r}^{2}n_{0r}T_{r}^{(1)};\end{eqnarray}$$

(3.52) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle -i{\textstyle \frac{3}{4}}\sqrt{\unicode[STIX]{x03C0}}v_{\text{th}r}\text{sign}(k_{\Vert })q_{r}^{(1)}.\end{eqnarray}$$

However, if we did so much work that we consider the fourth-order moment, it would be a shame not to increase the precision to $o(\unicode[STIX]{x1D701}_{r}^{2})$ . Obviously, we need to use a combination of at least 2 different lower-order moments. For example, by trying

(3.53) $$\begin{eqnarray}o(\unicode[STIX]{x1D701}_{r}^{2}):\quad \widetilde{r}_{r}^{(1)}=\unicode[STIX]{x1D6FC}_{q}q_{r}^{(1)}+\unicode[STIX]{x1D6FC}_{T}T_{r}^{(1)}.\end{eqnarray}$$

The proportionality constants $\unicode[STIX]{x1D6FC}_{q}$ , $\unicode[STIX]{x1D6FC}_{T}$ are easily obtained by separation to two equations for $\unicode[STIX]{x1D701}_{r}$ and $\unicode[STIX]{x1D701}_{r}^{2}$ that must be satisfied

(3.54) $$\begin{eqnarray}\displaystyle & \displaystyle -\frac{p_{0r}}{m_{r}}3i\sqrt{\unicode[STIX]{x03C0}}=+2\unicode[STIX]{x1D6FC}_{q}n_{0r}v_{\text{th}r}\text{sign}(k_{\Vert })+i\unicode[STIX]{x1D6FC}_{T}\sqrt{\unicode[STIX]{x03C0}}; & \displaystyle\end{eqnarray}$$

(3.55) $$\begin{eqnarray}\displaystyle & \displaystyle 16\frac{p_{0r}}{m_{r}}=+\unicode[STIX]{x1D6FC}_{q}n_{0r}v_{\text{th}r}\text{sign}(k_{\Vert })3i\sqrt{\unicode[STIX]{x03C0}}-4\unicode[STIX]{x1D6FC}_{T}. & \displaystyle\end{eqnarray}$$

Playing with the algebra little bit (for example $p_{0r}/m_{r}=T_{r}^{(0)}n_{0r}/m_{r}=v_{\text{th}r}^{2}n_{0r}/2$ ), the two equations can be solved easily for the unknown quantities $\unicode[STIX]{x1D6FC}_{q}$ , $\unicode[STIX]{x1D6FC}_{T}$ , and the final result is

(3.56)

There are naturally other possibilities and with the precision $o(\unicode[STIX]{x1D701}_{r}^{2})$ , one can search for closures

(3.57) $$\begin{eqnarray}\displaystyle o(\unicode[STIX]{x1D701}_{r}^{2}):\quad q_{r}^{(1)} & = & \displaystyle \unicode[STIX]{x1D6FC}_{T}T_{r}^{(1)}+\unicode[STIX]{x1D6FC}_{u}u_{r}^{(1)};\end{eqnarray}$$

(3.58) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle \unicode[STIX]{x1D6FC}_{q}q_{r}^{(1)}+\unicode[STIX]{x1D6FC}_{u}u_{r}^{(1)};\end{eqnarray}$$

(3.59) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle \unicode[STIX]{x1D6FC}_{T}T_{r}^{(1)}+\unicode[STIX]{x1D6FC}_{u}u_{r}^{(1)},\end{eqnarray}$$

where the first choice yields a closure

(3.60)

and the other two choices yield

(3.61) $$\begin{eqnarray}\displaystyle o(\unicode[STIX]{x1D701}_{r}^{2}):\quad \widetilde{r}_{r}^{(1)} & = & \displaystyle -i\frac{16-3\unicode[STIX]{x03C0}}{2\sqrt{\unicode[STIX]{x03C0}}}v_{\text{th}r}\text{sign}(k_{\Vert })q_{r}^{(1)}-i\frac{32-9\unicode[STIX]{x03C0}}{2\sqrt{\unicode[STIX]{x03C0}}}v_{\text{th}r}n_{0r}T_{r}^{(0)}\text{sign}(k_{\Vert })u_{r}^{(1)};\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.62) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle -\frac{16-3\unicode[STIX]{x03C0}}{8-2\unicode[STIX]{x03C0}}v_{\text{th}r}^{2}n_{0r}T_{r}^{(1)}+i\frac{2\sqrt{\unicode[STIX]{x03C0}}}{\unicode[STIX]{x03C0}-4}v_{\text{th}r}n_{0r}T_{r}^{(0)}\text{sign}(k_{\Vert })u_{r}^{(1)}.\end{eqnarray}$$

For completeness, one can easily find a closure for $\widetilde{r}_{r}^{(1)}$ with precision $o(\unicode[STIX]{x1D701}_{r}^{3})$ (after updating (3.44)–(3.47) to precision $o(\unicode[STIX]{x1D701}_{r}^{3})$ ) by searching for a solution

(3.63) $$\begin{eqnarray}o(\unicode[STIX]{x1D701}_{r}^{3}):\quad \widetilde{r}_{r}^{(1)}=\unicode[STIX]{x1D6FC}_{q}q_{r}^{(1)}+\unicode[STIX]{x1D6FC}_{T}T_{r}^{(1)}+\unicode[STIX]{x1D6FC}_{u}u_{r}^{(1)},\end{eqnarray}$$

and the solution reads

(3.64)

We purposely kept the species index $r$ in the calculations, to clearly show that the closures are performed for each species separately, and no Maxwell equations or other physical principles are used. The equations would be perhaps easier to read without the index $r$ .

3.2.2 Exploring the case $|\unicode[STIX]{x1D701}|\gg 1$

For large value of $|\unicode[STIX]{x1D701}|$ , we need to use an asymptotic expansion of the plasma dispersion function that reads

(3.65) $$\begin{eqnarray}\displaystyle Z(\unicode[STIX]{x1D701}) & = & \displaystyle i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}-\frac{1}{\unicode[STIX]{x1D701}}\left[1+\frac{1}{2\unicode[STIX]{x1D701}^{2}}+\frac{3}{4\unicode[STIX]{x1D701}^{4}}+\frac{15}{8\unicode[STIX]{x1D701}^{6}}+\frac{105}{16\unicode[STIX]{x1D701}^{8}}\cdots \,\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{(2n-1)!!}{(2\unicode[STIX]{x1D701}^{2})^{n}}+\cdots \,\right];\quad |\unicode[STIX]{x1D701}|\gg 1,\end{eqnarray}$$

where

(3.66) $$\begin{eqnarray}\unicode[STIX]{x1D70E}=\left\{\begin{array}{@{}ll@{}}0,\quad & \text{Im}(\unicode[STIX]{x1D701})>0;\\ 1,\quad & \text{Im}(\unicode[STIX]{x1D701})=0;\\ 2,\quad & \text{Im}(\unicode[STIX]{x1D701})<0.\end{array}\right.\end{eqnarray}$$

The term with $\unicode[STIX]{x1D70E}$ comes directly from the definition of $Z(\unicode[STIX]{x1D701})$ and there is not much one can do further about it, since there is no further asymptotic expansion for $\exp (-\unicode[STIX]{x1D701}^{2})$ when $\unicode[STIX]{x1D701}$ is large. The term is zero in the upper half of complex plane ( $\unicode[STIX]{x1D70E}=0$ ). When very close to the real axis, i.e. when $\unicode[STIX]{x1D70E}=1$ , the term mainly contributes to the imaginary part of $Z(\unicode[STIX]{x1D701})$ (even though only very weakly) and for the real part of $Z(\unicode[STIX]{x1D701})$ , its contribution can be neglected. However, when deeply down in the lower half of complex plane, the term can become very large (for example if $\unicode[STIX]{x1D701}=-iy$ , $\exp (-\unicode[STIX]{x1D701}^{2})=\exp (y^{2})$ and if $y$ is large the term obviously explodes). Deeply down in the lower complex plane the term is a real trouble, and even some kinetic solvers such as WHAMP (Rönnmark Reference Rönnmark1982) have trouble with calculations when the damping is too large.

We will see shortly, that for our purposes the term can be completely neglected, but let us keep it for a moment. The expansion of the Maxwellian plasma response function therefore reads

(3.67) $$\begin{eqnarray}R(\unicode[STIX]{x1D701})=i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}e^{-\unicode[STIX]{x1D701}^{2}}-\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}-\frac{15}{8\unicode[STIX]{x1D701}^{6}}-\frac{105}{16\unicode[STIX]{x1D701}^{8}}-\frac{945}{32\unicode[STIX]{x1D701}^{10}}\cdots \,;\quad |\unicode[STIX]{x1D701}|\gg 1.\end{eqnarray}$$

Let us calculate the kinetic hierarchy, at least up to $1/\unicode[STIX]{x1D701}^{4}$ . After a short inspection, one immediately sees that the hierarchy calculates a bit differently than in the previous case, and to get the fourth-order moments with the precision $o(1/\unicode[STIX]{x1D701}^{4})$ , it is important to keep all the terms up to ${\sim}1/\unicode[STIX]{x1D701}^{8}$ in the $R(\unicode[STIX]{x1D701})$ expression, since the fourth-order moments contain $\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701})$ terms. The expressions entering the kinetic hierarchy dully calculate as

(3.68) $$\begin{eqnarray}\displaystyle & \displaystyle n^{(1)}\sim R(\unicode[STIX]{x1D701})=i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}e^{-\unicode[STIX]{x1D701}^{2}}-\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}+o\left(\frac{1}{\unicode[STIX]{x1D701}^{4}}\right); & \displaystyle\end{eqnarray}$$

(3.69) $$\begin{eqnarray}\displaystyle & \displaystyle u^{(1)}\sim \unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701})=i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}e^{-\unicode[STIX]{x1D701}^{2}}-\frac{1}{2\unicode[STIX]{x1D701}}-\frac{3}{4\unicode[STIX]{x1D701}^{3}}-\frac{15}{8\unicode[STIX]{x1D701}^{5}}; & \displaystyle\end{eqnarray}$$

(3.70) $$\begin{eqnarray}\displaystyle & \displaystyle p^{(1)}\sim 1+2\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701})=2i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}e^{-\unicode[STIX]{x1D701}^{2}}-\frac{3}{2\unicode[STIX]{x1D701}^{2}}-\frac{15}{4\unicode[STIX]{x1D701}^{4}}; & \displaystyle\end{eqnarray}$$

(3.71) $$\begin{eqnarray}\displaystyle & \displaystyle T^{(1)}\sim 1-R(\unicode[STIX]{x1D701})+2\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701})=i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}(2\unicode[STIX]{x1D701}^{3}-\unicode[STIX]{x1D701})-\frac{1}{\unicode[STIX]{x1D701}^{2}}-\frac{3}{\unicode[STIX]{x1D701}^{4}}; & \displaystyle\end{eqnarray}$$

(3.72) $$\begin{eqnarray}\displaystyle & \displaystyle q^{(1)}\sim \unicode[STIX]{x1D701}+2\unicode[STIX]{x1D701}^{3}R(\unicode[STIX]{x1D701})-3\unicode[STIX]{x1D701}R(\unicode[STIX]{x1D701})=i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}(2\unicode[STIX]{x1D701}^{4}-3\unicode[STIX]{x1D701}^{2})-\frac{3}{2\unicode[STIX]{x1D701}^{3}}-\frac{15}{2\unicode[STIX]{x1D701}^{5}}; & \displaystyle\end{eqnarray}$$

(3.73) $$\begin{eqnarray}\displaystyle & \displaystyle r^{(1)}\sim 3+2\unicode[STIX]{x1D701}^{2}+4\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701})=4i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{5}e^{-\unicode[STIX]{x1D701}^{2}}-\frac{15}{2\unicode[STIX]{x1D701}^{2}}-\frac{105}{4\unicode[STIX]{x1D701}^{4}}; & \displaystyle\end{eqnarray}$$

(3.74) $$\begin{eqnarray}\widetilde{r}^{(1)}\sim 2\unicode[STIX]{x1D701}^{2}+4\unicode[STIX]{x1D701}^{4}R(\unicode[STIX]{x1D701})+3R(\unicode[STIX]{x1D701})-3-12\unicode[STIX]{x1D701}^{2}R(\unicode[STIX]{x1D701})=i\unicode[STIX]{x1D70E}\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}(4\unicode[STIX]{x1D701}^{5}-12\unicode[STIX]{x1D701}^{3}+3\unicode[STIX]{x1D701})-\frac{6}{\unicode[STIX]{x1D701}^{4}},\end{eqnarray}$$

where for brevity we suppressed the proportionality constants, including the $\text{sign}(k_{\Vert })$ . Interestingly, the velocity $u^{(1)}$ decreases the slowest, only as $1/\unicode[STIX]{x1D701}$ . The $n^{(1)}$ , $p^{(1)}$ , $r^{(1)}$ and also the temperature $T^{(1)}$ , decrease as $1/\unicode[STIX]{x1D701}^{2}$ . The heat flux $q^{(1)}$ decreases as $1/\unicode[STIX]{x1D701}^{3}$ and the cumulant $\widetilde{r}^{(1)}$ decreases the fastest, as $1/\unicode[STIX]{x1D701}^{4}$ . This is not good news, since it is obvious that the direct closures that were easily obtained for the small $\unicode[STIX]{x1D701}$ case, cannot be easily done here.

To understand how the terms contribute to the real frequency and damping, it is useful to separate $\unicode[STIX]{x1D701}=x+iy$ and calculate expressions with $y$ being small, i.e. the weak growth-rate (actually weak damping) approximation. The exponential term entering (3.67) can be approximated as

(3.75) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}^{2} & = & \displaystyle (x+iy)^{2}=(x^{2}-y^{2})+2ixy\approx x^{2}+2ixy;\nonumber\\ \displaystyle e^{-\unicode[STIX]{x1D701}^{2}} & {\approx} & \displaystyle e^{-x^{2}}e^{-2ixy};\nonumber\\ \displaystyle i\unicode[STIX]{x1D701}e^{-\unicode[STIX]{x1D701}^{2}} & = & \displaystyle i(x+iy)e^{-(x+iy)^{2}}\approx (-y+ix)e^{-x^{2}}e^{-2ixy},\end{eqnarray}$$

and the fractions of $\unicode[STIX]{x1D701}$ are approximately

(3.76) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{\unicode[STIX]{x1D701}}=\frac{1}{x+iy}=\frac{1}{x\left(1+i{\displaystyle \frac{y}{x}}\right)}\approx \frac{1}{x}\left(1-i\frac{y}{x}\right)=\frac{1}{x}-i\frac{y}{x^{2}}; & \displaystyle\end{eqnarray}$$

(3.77) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{\unicode[STIX]{x1D701}^{2}}=\frac{1}{x^{2}\left(1+i{\displaystyle \frac{y}{x}}\right)^{2}}\approx \frac{1}{x^{2}}\left(1-2i\frac{y}{x}\right)=\frac{1}{x^{2}}-2i\frac{y}{x^{3}}; & \displaystyle\end{eqnarray}$$

(3.78) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{\unicode[STIX]{x1D701}^{3}}\approx \frac{1}{x^{3}}-3i\frac{y}{x^{4}}; & \displaystyle\end{eqnarray}$$

(3.79) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{\unicode[STIX]{x1D701}^{4}}\approx \frac{1}{x^{4}}-4i\frac{y}{x^{5}}, & \displaystyle\end{eqnarray}$$

etc. For large $x$ , the exponential term (3.75) is strongly suppressed as $e^{-x^{2}}$ (with oscillations $e^{i2xy}$ ). Additionally, the real part of (3.75) is proportional to $y$ , which is also small, and its contribution to the real part of $R(\unicode[STIX]{x1D701})$ can be therefore completely neglected. The imaginary part of the exponential term (3.75) has to be kept, if one wants to match the approximate kinetic dispersion relations from plasma books (usually calculated in the weak growth rate/damping approximation), for example for the damping of the Langmuir mode or the ion-acoustic mode. However, even smart plasma physics books have trouble analytically reproducing the full kinetic dispersion relations that have to be solved numerically, see for example figures in Gurnett and Bhattacharjee on pages 341 and 355, that compare the analytic and full solutions for the Langmuir mode and the ion-acoustic mode. The trouble is that the damping can become large, and the entire approach with the weak damping invalid. If kinetic plasma books have trouble analytically reproducing the damping with full accuracy under these conditions, we would be naive to think that we can do better with a fluid model and we know we cannot be analytically exact for $|\unicode[STIX]{x1D701}|\gg 1$ if the damping is too large. If the damping is far too large, and the imaginary frequency starts to be comparable to real frequency, the mode will be damped away very quickly.

In fact, even the well-known kinetic solver WHAMP, neglects this term in calculation of $Z(\unicode[STIX]{x1D701})$ for large $\unicode[STIX]{x1D701}$ values, as can be verified in the WHAMP full manual (Rönnmark Reference Rönnmark1982) from the asymptotic expansion of $Z(\unicode[STIX]{x1D701})$ , equation (III-6) on page 10, and the discussion of numerical errors on page 13. The WHAMP solver uses an eight-pole Padé approximant of $Z(\unicode[STIX]{x1D701})$ , which is a very precise approximant, and imprecision starts to show up only if the damping become too large. For example, in the very damped regime when the $\text{Im}(\unicode[STIX]{x1D701})=-\text{Re}(\unicode[STIX]{x1D701})/2$ , the error in real and imaginary values of $Z(\unicode[STIX]{x1D701})$ is still less than 2 %–3 %, where the calculation should be stopped (in a less damped regime, the precision is much higher).

If a full kinetic solver can neglect the exponential term for large $\unicode[STIX]{x1D701}$ values, we can surely neglect it as well. It should be emphasized that the term is neglected only for large $\unicode[STIX]{x1D701}$ values (i.e. in the asymptotic expansion), the exponential term is otherwise fully retained and enters the Padé approximation through the power series expansion for small $\unicode[STIX]{x1D701}$ . To summarize, the ‘ideal’ large $\unicode[STIX]{x1D701}$ asymptotic behaviour that we would like to obtain reads

(3.80) $$\begin{eqnarray}\displaystyle \frac{n_{r}^{(1)}}{n_{0r}} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}\left[-\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}-\cdots \,\right];\end{eqnarray}$$

(3.81) $$\begin{eqnarray}\displaystyle u_{r}^{(1)} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}v_{\text{th}r}\text{sign}(k_{\Vert })\left[-\frac{1}{2\unicode[STIX]{x1D701}}-\frac{3}{4\unicode[STIX]{x1D701}^{3}}-\cdots \,\right];\end{eqnarray}$$

(3.82) $$\begin{eqnarray}\displaystyle \frac{p_{r}^{(1)}}{p_{0r}} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}\left[-\frac{3}{2\unicode[STIX]{x1D701}^{2}}-\frac{15}{4\unicode[STIX]{x1D701}^{4}}-\cdots \,\right];\end{eqnarray}$$

(3.83) $$\begin{eqnarray}\displaystyle \frac{T_{r}^{(1)}}{T_{r}^{(0)}} & = & \displaystyle -\frac{q_{r}}{T_{r}^{(0)}}\unicode[STIX]{x1D6F7}\left[-\frac{1}{\unicode[STIX]{x1D701}^{2}}-\frac{3}{\unicode[STIX]{x1D701}^{4}}-\cdots \,\right];\end{eqnarray}$$

(3.84) $$\begin{eqnarray}\displaystyle q_{r}^{(1)} & = & \displaystyle -q_{r}n_{0r}\unicode[STIX]{x1D6F7}v_{\text{th}r}\text{sign}(k_{\Vert })\left[-\frac{3}{2\unicode[STIX]{x1D701}^{3}}-\frac{15}{2\unicode[STIX]{x1D701}^{5}}-\cdots \,\right];\end{eqnarray}$$

(3.85) $$\begin{eqnarray}\displaystyle r_{r}^{(1)} & = & \displaystyle -\frac{q_{r}p_{0r}}{m_{r}}\unicode[STIX]{x1D6F7}\left[-\frac{15}{2\unicode[STIX]{x1D701}^{2}}-\frac{105}{4\unicode[STIX]{x1D701}^{4}}-\cdots \,\right];\end{eqnarray}$$

(3.86) $$\begin{eqnarray}\displaystyle \widetilde{r}_{r}^{(1)} & = & \displaystyle -\frac{q_{r}p_{0r}}{m_{r}}\unicode[STIX]{x1D6F7}\left[-\frac{6}{\unicode[STIX]{x1D701}^{4}}-\cdots \,\right].\end{eqnarray}$$

3.3 A brief introduction to Padé approximants

Padé approximants, i.e. Padé series approximation/expansion, is a very powerful mathematical technique, comparable to the usual Taylor series and the Laurent series. Nevertheless, for some unknown reason, Padé series seems to somehow disappear from the modern educational system that a typical physicist encounters. The lack of Padé series in classes is even more surprising, if one realizes that the technique is in fact very simple, and anybody can fully grasp it in very short time. We therefore make a quick introduction to the technique here.

Padé series consist of approximating a function as a ratio of two polynomials. If a power series (e.g. Taylor series) of a function $f(x)$ is known around some point with coefficients $c_{n}$ , the goal is to express it as a ratio of two polynomials

(3.87) $$\begin{eqnarray}c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+\cdots =\frac{a_{0}+a_{1}x+a_{2}x^{2}+\cdots }{1+b_{1}x+b_{2}x^{2}+\cdots }\end{eqnarray}$$

The choice of $b_{0}=1$ is an ad hoc choice and the entire decomposition can be done without it, leading to the same results at the end. Multiplying the left-hand side by the denominator $1+b_{1}x+b_{2}x^{2}+\cdots \,$ , and grouping the $x^{n}$ contributions together, that must be satisfied independently, leads to the system of equations

(3.88) $$\begin{eqnarray}\displaystyle a_{0} & = & \displaystyle c_{0};\nonumber\\ \displaystyle a_{1} & = & \displaystyle c_{1}+c_{0}b_{1};\nonumber\\ \displaystyle a_{2} & = & \displaystyle c_{2}+c_{1}b_{1}+c_{0}b_{2};\nonumber\\ \displaystyle a_{3} & = & \displaystyle c_{3}+c_{2}b_{1}+c_{1}b_{2}+c_{0}b_{3};\nonumber\\ \displaystyle a_{4} & = & \displaystyle c_{4}+c_{3}b_{1}+c_{2}b_{2}+c_{1}b_{3}+c_{0}b_{4};\nonumber\\ \displaystyle a_{5} & = & \displaystyle c_{5}+c_{4}b_{1}+c_{3}b_{2}+c_{2}b_{3}+c_{1}b_{4}+c_{0}b_{5};\nonumber\\ \displaystyle a_{6} & = & \displaystyle c_{6}+c_{5}b_{1}+c_{4}b_{2}+c_{3}b_{3}+c_{2}b_{4}+c_{1}b_{5}+c_{0}b_{6},\end{eqnarray}$$

etc. The necessary condition for the system being solvable, is that the number of variables is equivalent to the number of equations. Therefore, if we want to approximate function $f(x)$ with a ratio of two polynomials $P_{m}/Q_{n}$ , of degrees $m$ and $n$ , we will need the Taylor series on the left-hand side of (3.87) up to the order $m+n$ . The Padé approximation is sometimes denoted as $R_{m,n}$ or using a function $f(x)$ that is being approximated as $f(x)_{m,n}$ or $[f(x)]_{m,n}$ . If the Padé approximation exists, it is unique.

For example, the function $e^{x}$ has a Taylor series around the point $x=0$

(3.89) $$\begin{eqnarray}e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots\end{eqnarray}$$

Let us say we want to approximate $e^{x}$ as a ratio of two polynomials of zeroth and first order $e^{x}\approx a_{0}/(1+b_{1}x)$ , i.e. we want to find the Padé approximant $[e^{x}]_{0,1}$ . Respecting the $n+m$ rule, the approximation therefore consists of equating

(3.90) $$\begin{eqnarray}\underbrace{c_{0}}_{=1}+\underbrace{c_{1}}_{=1}x=\frac{a_{0}}{1+b_{1}x},\end{eqnarray}$$

that leads to the system of equations

(3.91) $$\begin{eqnarray}\displaystyle a_{0} & = & \displaystyle c_{0}=1;\nonumber\\ \displaystyle a_{1}=0 & = & \displaystyle c_{1}+b_{1}\quad \Rightarrow \quad b_{1}=-c_{1}=-1,\end{eqnarray}$$

yielding the Padé approximation

(3.92) $$\begin{eqnarray}[e^{x}]_{0,1}=\frac{1}{1-x}.\end{eqnarray}$$

To feel confident with the Padé approximations, let us find another approximant of $e^{x}$ , for example $[e^{x}]_{1,2}$ . The system is written as

(3.93) $$\begin{eqnarray}\underbrace{1}_{=c_{0}}+\underbrace{1}_{=c_{1}}x+\underbrace{\frac{1}{2}}_{=c_{2}}x^{2}+\underbrace{\frac{1}{6}}_{=c_{3}}x^{3}=\frac{a_{0}+a_{1}x}{1+b_{1}x+b_{2}x^{2}},\end{eqnarray}$$

and yields a system of equations

(3.94) $$\begin{eqnarray}\displaystyle a_{0} & = & \displaystyle 1;\nonumber\\ \displaystyle a_{1} & = & \displaystyle 1+b_{1};\nonumber\\ \displaystyle a_{2}=0 & = & \displaystyle {\textstyle \frac{1}{2}}+b_{1}+b_{2};\nonumber\\ \displaystyle a_{3}=0 & = & \displaystyle {\textstyle \frac{1}{6}}+{\textstyle \frac{1}{2}}b_{1}+b_{2}+0,\end{eqnarray}$$

which have a solution $b_{1}=-2/3$ , $b_{2}=1/6$ , $a_{1}=1/3$ and the Padé approximant

(3.95) $$\begin{eqnarray}[e^{x}]_{1,2}=\frac{1+\frac{1}{3}x}{1-\frac{2}{3}x+\frac{1}{6}x^{2}}=\frac{6+2x}{6-4x+x^{2}}.\end{eqnarray}$$

It is just a straightforward algebraic exercise to find other Padé approximations, for example

(3.96) $$\begin{eqnarray}\displaystyle \displaystyle [e^{x}]_{1,1} & = & \displaystyle \frac{1+\frac{1}{2}x}{1-\frac{1}{2}x}=\frac{2+x}{2-x};\nonumber\\ \displaystyle \displaystyle [e^{x}]_{2,1} & = & \displaystyle \frac{1+\frac{2}{3}x+\frac{1}{6}x^{2}}{1-\frac{1}{3}x}=\frac{6+4x+x^{2}}{6-2x};\nonumber\\ \displaystyle \displaystyle [e^{x}]_{3,1} & = & \displaystyle \frac{1+\frac{3}{4}x+\frac{1}{4}x^{2}+\frac{1}{24}x^{3}}{1-\frac{1}{4}x}=\frac{24+18x+6x^{2}+x^{3}}{24-6x},\end{eqnarray}$$

etc. Similarly, it is easy to find Padé approximations to a function $e^{-x}$ , and for example (obviously)

(3.97) $$\begin{eqnarray}[e^{-x}]_{1,2}=\frac{6-2x}{6+4x+x^{2}};\quad [e^{-x}]_{1,1}=\frac{2-x}{2+x};\quad [e^{-x}]_{2,1}=\frac{6-4x+x^{2}}{6+2x}.\end{eqnarray}$$

The approximations were derived from Taylor expansion of $e^{-x}$ around $x=0$ , and all 3 choices naturally have the correct limit $\lim _{x\rightarrow 0}e^{-x}=1$ . However, we can see that, by choosing the degree of the Padé approximation, we can also control what the function is doing for large values of $x$ . For example, for large values of $x$ the Padé approximations (3.97) go to $0$ , $-1$ and $+\infty$ . Obviously, the smart choice is $[e^{-x}]_{1,2}$ which approximately reproduces the behaviour of $e^{-x}$ also for large $x$ . The usefulness of Padé approximation becomes especially apparent when considering analytically difficult functions, for example the $e^{-x^{2}}$ , where the ‘smart’ lowest Padé approximants are

(3.98) $$\begin{eqnarray}[e^{-x^{2}}]_{0,2}=\frac{1}{1+x^{2}};\quad [e^{-x^{2}}]_{0,4}=\frac{1}{1+x^{2}+\frac{1}{2}x^{4}};\quad [e^{-x^{2}}]_{2,4}=\frac{6-2x^{2}}{6+4x^{2}+x^{4}}.\end{eqnarray}$$

Therefore, depending on the required precision of a physical problem, instead of working with $e^{-x^{2}}$ (which for example does not have an indefinite integral that can be expressed in elementary functions), one can approximate the function $e^{-x^{2}}$ for all x, as $1/(1+x^{2})$ , that is much easier to work with. Curiously, the reader might recognize that the $1/(1+x^{2})$ is the Cauchy distribution function, often used in plasma physics books to get a better understanding of the complicated Landau damping. The Cauchy distribution therefore can be thought of as the simplest Padé approximation of the Maxwellian distribution.

Now we are ready to use the Padé approximation for the plasma dispersion function $Z(\unicode[STIX]{x1D701})$ or the plasma response function $R(\unicode[STIX]{x1D701})$ . We do not have to explore all the possibilities, and we can immediately pick up only the smart choices. For large $\unicode[STIX]{x1D701}$ (by neglecting the exponential term as discussed in the previous section), at the first order $Z(\unicode[STIX]{x1D701})\sim 1/\unicode[STIX]{x1D701}$ and $R(\unicode[STIX]{x1D701})\sim 1/\unicode[STIX]{x1D701}^{2}$ , and both functions approach zero as $\unicode[STIX]{x1D701}$ increases. Obviously, a smart choice worth exploring will always be a Padé approximant $[\;\;]_{m,n}$ where $n>m$ . In fact, we can be even more specific. We know the asymptotic behaviour for large $\unicode[STIX]{x1D701}$ , and obviously, even smarter choice is to concentrate only on approximants $[Z(\unicode[STIX]{x1D701})]_{n-1,n}$ and $[R(\unicode[STIX]{x1D701})]_{n-2,n}$ , since such a choice will naturally lead to the correct asymptotic behaviour

(3.99) $$\begin{eqnarray}\unicode[STIX]{x1D701}\gg 1:\quad [Z(\unicode[STIX]{x1D701})]_{n-1,n}\sim \frac{1}{\unicode[STIX]{x1D701}};\quad [R(\unicode[STIX]{x1D701})]_{n-2,n}\sim \frac{1}{\unicode[STIX]{x1D701}^{2}}.\end{eqnarray}$$

Any other choice is not really interesting and therefore, the usual 2-digit notation of the Padé approximation becomes redundant. We can just use 1-digit notation with ‘ $n$ ’, that represents the degree of a chosen polynomial in the denominator, and we can omit writing the $(n-1)$ and $(n-2)$ , since this will always be the case (except for the $R_{1}(\unicode[STIX]{x1D701})$ ). The ‘ $n$ ’ represents the number of poles, and we therefore talk about an ‘ $n$ -pole Padé approximation’ of $Z(\unicode[STIX]{x1D701})$ or $R(\unicode[STIX]{x1D701})$ , and

(3.100) $$\begin{eqnarray}Z_{n}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+\cdots +a_{n-1}\unicode[STIX]{x1D701}^{n-1}}{1+b_{1}\unicode[STIX]{x1D701}+\cdots +b_{n}\unicode[STIX]{x1D701}^{n}};\quad R_{n}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+\cdots +a_{n-2}\unicode[STIX]{x1D701}^{n-2}}{1+b_{1}\unicode[STIX]{x1D701}+\cdots +b_{n}\unicode[STIX]{x1D701}^{n}}.\end{eqnarray}$$

Note that one can directly work with Padé approximants for both $Z_{n}(\unicode[STIX]{x1D701})$ and $R_{n}(\unicode[STIX]{x1D701})$ , and that in general, according to definitions (3.100), the approximants are not automatically equivalent. The difference is as if one does approximations to a function $f(x)$ or its derivative $f^{\prime }(x)$ . Usually in papers, the approximant $Z_{n}(\unicode[STIX]{x1D701})$ is calculated, and $R_{n}(\unicode[STIX]{x1D701})$ is just defined according to $R_{n}(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z_{n}(\unicode[STIX]{x1D701})$ . One can choose another (and better approach in our opinion) and to calculate directly approximants $R_{n}(\unicode[STIX]{x1D701})$ , and if really required (which should not be the case), obtain $Z_{n}(\unicode[STIX]{x1D701})$ approximants as $Z_{n}(\unicode[STIX]{x1D701})=(R_{n}(\unicode[STIX]{x1D701})-1)/\unicode[STIX]{x1D701}$ .

Moreover, we can do even better than (3.100). We shall not be satisfied just by approximating the asymptotic trend ${\sim}1/\unicode[STIX]{x1D701}$ for $\unicode[STIX]{x1D701}\gg 1$ , and hope for the best. For large $\unicode[STIX]{x1D701}$ , the correct asymptotic expansions are $Z(\unicode[STIX]{x1D701})\rightarrow -1/\unicode[STIX]{x1D701}$ and $R(\unicode[STIX]{x1D701})\rightarrow -1/(2\unicode[STIX]{x1D701}^{2})$ . By prescribing $a_{n-1}/b_{n}=-1$ for $Z(\unicode[STIX]{x1D701})$ , and $a_{n-2}/b_{n}=-1/2$ for $R(\unicode[STIX]{x1D701})$ , we will obtain correct asymptotic behaviour of these functions, at least at the first order. By doing this, we are not ‘destroying’ the Padé approximation, since it is easy to argue that if an $n$ -pole approximation is determined to be sufficient for small $\unicode[STIX]{x1D701}$ values, we can just add one more pole and use that one to control the asymptotic behaviour for large $\unicode[STIX]{x1D701}$ values. Of course, we will always use at least the first term in the expansion for $\unicode[STIX]{x1D701}\ll 1$ , that yields $a_{0}=i\sqrt{\unicode[STIX]{x03C0}}$ for $Z_{n}(\unicode[STIX]{x1D701})$ and $a_{0}=1$ for $R_{n}(\unicode[STIX]{x1D701})$ , otherwise the functions will have incorrect values at $\unicode[STIX]{x1D701}=0$ . The ‘smart’ choices worth considering therefore can be summarized as

(3.101) $$\begin{eqnarray}\displaystyle \displaystyle Z_{n}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}+a_{1}\unicode[STIX]{x1D701}+\cdots +a_{n-1}\unicode[STIX]{x1D701}^{n-1}}{1+b_{1}\unicode[STIX]{x1D701}+\cdots +b_{n-1}\unicode[STIX]{x1D701}^{n-1}-a_{n-1}\unicode[STIX]{x1D701}^{n}};\nonumber\\ \displaystyle \displaystyle R_{n}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1+a_{1}\unicode[STIX]{x1D701}+\cdots +a_{n-2}\unicode[STIX]{x1D701}^{n-2}}{1+b_{1}\unicode[STIX]{x1D701}+\cdots +b_{n-1}\unicode[STIX]{x1D701}^{n-1}-2a_{n-2}\unicode[STIX]{x1D701}^{n}},\end{eqnarray}$$

and have a property to correctly match the $Z$ and $R$ functions at $\unicode[STIX]{x1D701}=0$ and, have the correct first-order asymptotic expansion at $\unicode[STIX]{x1D701}\gg 1$ . The one-pole approximant $R_{1}(\unicode[STIX]{x1D701})$ is an exception, and can be defined only as $R_{1}(\unicode[STIX]{x1D701})=1/(1+b_{1}\unicode[STIX]{x1D701})$ . This function obviously cannot have correct asymptotic expansion ${\sim}1/\unicode[STIX]{x1D701}^{2}$ and the only possibility is to use $\unicode[STIX]{x1D701}\ll 1$ expansion $R_{1}(\unicode[STIX]{x1D701})=1/(1+b_{1}\unicode[STIX]{x1D701})=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}$ , which yields $b_{1}=-i\sqrt{\unicode[STIX]{x03C0}}$ and

(3.102) $$\begin{eqnarray}R_{1}(\unicode[STIX]{x1D701})=\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}.\end{eqnarray}$$

The one-pole approximant $Z_{1}(\unicode[STIX]{x1D701})$ can be obtained directly from the definition (3.101), that yields

(3.103) $$\begin{eqnarray}Z_{1}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}},\end{eqnarray}$$

and that has correct asymptotic behaviour $Z_{1}(\unicode[STIX]{x1D701})\rightarrow -1/\unicode[STIX]{x1D701}$ for large $\unicode[STIX]{x1D701}$ values, even though it has only precision $o(\unicode[STIX]{x1D701}^{0})$ for small $\unicode[STIX]{x1D701}$ values. Perhaps curiously, in this case $R_{1}(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z_{1}(\unicode[STIX]{x1D701})$ exactly. Alternatively, if the precision for small $\unicode[STIX]{x1D701}$ is more important than the exact asymptotic expansion for large $\unicode[STIX]{x1D701}$ , it is possible to increase the $Z_{1}$ precision to $o(\unicode[STIX]{x1D701}^{1})$ and write $Z_{1}(\unicode[STIX]{x1D701})=i\sqrt{\unicode[STIX]{x03C0}}/(1-2i\unicode[STIX]{x1D701}/\sqrt{\unicode[STIX]{x03C0}})$ . In this case $R_{1}(\unicode[STIX]{x1D701})\neq 1+\unicode[STIX]{x1D701}Z_{1}(\unicode[STIX]{x1D701})$ exactly, and the functions are equal only for small $\unicode[STIX]{x1D701}$ and only with precision $o(\unicode[STIX]{x1D701}^{1})$ .

Right now, in the definition (3.101), we just used 1 pole for the asymptotic series of $Z(\unicode[STIX]{x1D701})$ and $R(\unicode[STIX]{x1D701})$ , but one can naturally use more poles. By opening the possibility of increasing the number of matching asymptotic points in the $n$ -pole Padé approximation (3.101), the number of possible approximants for a given $n$ naturally increases. To keep track of all the possibilities, we obviously need some kind of classification scheme. It is useful to modify the usual 1-index Padé series notation for $Z_{n}(\unicode[STIX]{x1D701})$ and $R_{n}(\unicode[STIX]{x1D701})$ functions (that only specify the number of poles), to a two index notation $Z_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ , $R_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ . Now we have a wide range of possibilities how to define $n,n^{\prime }$ and there is no clear ‘natural’ winner.

There are two different existing notations (likely more), introduced by Martín et al. (Reference Martín, Donoso and Zamudio-Cristi1980) and by Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992), that consider $Z_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ Padé approximants. The first reference defines $n=\text{number}$ of points (equations) used in the power-series expansion, and $n^{\prime }=\text{number}$ of points (equations) used in the asymptotic series expansion. Even though perhaps clear, for example three-pole approximants in this notation are expressed as $Z_{5,1},Z_{4,2},Z_{3,3}$ etc., to get the number of poles (which is the most important information), one has to calculate $(n+n^{\prime })/2$ . When using a lot of different approximants, this notation is a bit confusing and is rarely used.

The notation of Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992) can be interpreted as defining $Z_{n,n^{\prime }}$ with $n=\text{ number}$ of poles (which we like), and $n^{\prime }=\text{number}$ of additional poles in the asymptotic expansion that is used, compared to some ‘minimally interesting’ or ‘basic’ definition $Z_{n}$ , that can be denoted as $Z_{n,0}$ (which we like too). The problem with the notation of Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992) is with the definition of the ‘basic’ $Z_{n,0}$ , since the number of asymptotic points used in the definition of $Z_{n,0}$ keeps changing with $n$ (and is actually equal to $n$ ). The notation is physically motivated, but the motivation is difficult to follow. The $Z_{2,0}$ is defined with 2 asymptotic points, $Z_{3,0}$ with 3 asymptotic points and so on. This can be easily deduced from their definitions of $Z_{2}-Z_{5}$ , as we will discuss later. We find this notation confusing.

Importantly, both mentioned notations consider the Padé approximants to $Z(\unicode[STIX]{x1D701})$ . We do not really care about $Z(\unicode[STIX]{x1D701})$ , since all the kinetic moments are formulated with $R(\unicode[STIX]{x1D701})$ at this stage. We want to calculate direct Padé approximants to $R(\unicode[STIX]{x1D701})$ , which is actually slightly less analytically complicated for a given $n$ . Here we define the 2-index Padé approximation to the plasma response function $R(\unicode[STIX]{x1D701})$ simply as

(3.104)

i.e. as having asymptote $-1/(2\unicode[STIX]{x1D701}^{2})$ for large $\unicode[STIX]{x1D701}$ , and notation $R_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ means that $n^{\prime }$ additional asymptotic points are used compared to the basic definition $R_{n,0}(\unicode[STIX]{x1D701})$ . The notation feels natural, and the $n^{\prime }=0$ index helps us to orient in the hierarchy of many possible $R(\unicode[STIX]{x1D701})$ approximants. It is easy to remember that this asymptotic profile is the minimum ‘desired’ profile that correctly captures the zeroth-order (density) moment, and any profile with less asymptotic points should be avoided if possible. The $R_{n,0}(\unicode[STIX]{x1D701})$ has power-series precision $o(\unicode[STIX]{x1D701}^{2n-3})$ and asymptotic-series precision $o(\unicode[STIX]{x1D701}^{-2})$ , so $R_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ has precision $o(\unicode[STIX]{x1D701}^{2n-3-n^{\prime }})$ and $o(\unicode[STIX]{x1D701}^{-2-n^{\prime }})$ .

Of course, we want to make the $R_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ and $Z_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ definitions fully consistent, and $Z_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ is defined so that

(3.105) $$\begin{eqnarray}R_{n,n^{\prime }}(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z_{n,n^{\prime }}(\unicode[STIX]{x1D701}),\end{eqnarray}$$

is satisfied. This dictates that in comparison to $Z_{n}(\unicode[STIX]{x1D701})$ definition (3.101), two additional asymptotic points must be used to define the $Z_{n,0}(\unicode[STIX]{x1D701})$ . We have no other choice and when calculating the $Z_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ , we have to start counting from $n^{\prime }=-2$ , and we define

(3.106) $$\begin{eqnarray}Z_{n,-2}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+a_{1}\unicode[STIX]{x1D701}+\cdots +a_{n-1}\unicode[STIX]{x1D701}^{n-1}}{1+b_{1}\unicode[STIX]{x1D701}+\cdots +b_{n-1}\unicode[STIX]{x1D701}^{n-1}-a_{n-1}\unicode[STIX]{x1D701}^{n}}.\end{eqnarray}$$

When calculating the hierarchy of plasma dispersion functions $Z(\unicode[STIX]{x1D701})$ , the $-2$ index is actually a nice reminder that we are two asymptotic points short of the ‘desired’ profile (3.104) for the plasma response function $R(\unicode[STIX]{x1D701})$ . We want to feel fully confident that we understand both Padé approximants $R(\unicode[STIX]{x1D701})$ and $Z(\unicode[STIX]{x1D701})$ , and we will calculate two-pole and three-pole approximants for both functions. For four-pole approximants and above, we will only work with $R(\unicode[STIX]{x1D701})$ .

Padé approximants were also used for other interesting physical problems, such as developing analytic models for the Rayleigh–Taylor and Richtmyer–Meshkov instabilities (Zhou Reference Zhou2017a ,Reference Zhou b ).

3.3.1 Two-pole approximants of $R(\unicode[STIX]{x1D701})$ and $Z(\unicode[STIX]{x1D701})$

Let us be patient and go slowly. A general two-pole Padé approximant to $R(\unicode[STIX]{x1D701})$ is

(3.107) $$\begin{eqnarray}R_{2}(\unicode[STIX]{x1D701})=\frac{a_{0}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}},\end{eqnarray}$$

where $a_{0}=1$ . The asymptotic expansion for large $\unicode[STIX]{x1D701}$ values calculates as

(3.108) $$\begin{eqnarray}\displaystyle \frac{a_{0}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}} & = & \displaystyle \frac{a_{0}}{b_{2}\unicode[STIX]{x1D701}^{2}\left({\displaystyle \frac{1}{b_{2}\unicode[STIX]{x1D701}^{2}}}+{\displaystyle \frac{b_{1}}{b_{2}\unicode[STIX]{x1D701}}}+1\right)}\nonumber\\ \displaystyle & = & \displaystyle \frac{a_{0}}{b_{2}\unicode[STIX]{x1D701}^{2}}\left[1-\left(\frac{b_{1}}{b_{2}\unicode[STIX]{x1D701}}+\frac{1}{b_{2}\unicode[STIX]{x1D701}^{2}}\right)+\left(\frac{b_{1}}{b_{2}\unicode[STIX]{x1D701}}+\frac{1}{b_{2}\unicode[STIX]{x1D701}^{2}}\right)^{2}+\cdots \,\right]\nonumber\\ \displaystyle & = & \displaystyle \frac{a_{0}}{b_{2}\unicode[STIX]{x1D701}^{2}}-a_{0}\frac{b_{1}}{b_{2}^{2}\unicode[STIX]{x1D701}^{3}}+a_{0}\frac{b_{1}^{2}-b_{2}}{b_{2}^{3}\unicode[STIX]{x1D701}^{4}}+\cdots \,;\quad \unicode[STIX]{x1D701}\gg 1,\end{eqnarray}$$

and must be matched with the asymptotic expansion (3.67)

(3.109) $$\begin{eqnarray}R(\unicode[STIX]{x1D701})=-\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{0}{\unicode[STIX]{x1D701}^{3}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}+\cdots \,;\quad \unicode[STIX]{x1D701}\gg 1.\end{eqnarray}$$

Matching the first point implies $b_{2}=-2a_{0}$ , and this is how $R_{2,0}(\unicode[STIX]{x1D701})$ is defined. Then, matching with 2 equations for the small $\unicode[STIX]{x1D701}$ expansion, equation (3.36), the classical Padé approach yields

(3.110) $$\begin{eqnarray}R_{2,0}(\unicode[STIX]{x1D701})=\frac{a_{0}}{1+b_{1}\unicode[STIX]{x1D701}-2a_{0}\unicode[STIX]{x1D701}^{2}}=\underbrace{1}_{=c_{0}}+\underbrace{i\sqrt{\unicode[STIX]{x03C0}}}_{=c_{1}}\unicode[STIX]{x1D701};\quad \Rightarrow \quad R_{2,0}(\unicode[STIX]{x1D701})=\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}}.\end{eqnarray}$$

To match additional asymptotic point (and to potentially find $R_{2,1}(\unicode[STIX]{x1D701})$ ), dictates that $b_{1}=0$ . However, the resulting function $R_{2,1}(\unicode[STIX]{x1D701})=1/(1-2\unicode[STIX]{x1D701}^{2})$ does not have any imaginary part for real valued $\unicode[STIX]{x1D701}$ , since it uses too many asymptotic points and the Landau residue is not accounted for. Therefore, the $R_{2,1}(\unicode[STIX]{x1D701})$ does not represent a valuable approximation of $R(\unicode[STIX]{x1D701})$ , and this approximant is eliminated.

Let us now explore possible two-pole approximations of $Z(\unicode[STIX]{x1D701})$ . A general two-pole approximant is defined as

(3.111) $$\begin{eqnarray}\displaystyle Z_{2}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}}, & & \displaystyle\end{eqnarray}$$

and has the following asymptotic expansion for large $\unicode[STIX]{x1D701}$ values

(3.112) $$\begin{eqnarray}\displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}} & = & \displaystyle \left(\frac{a_{1}}{b_{2}}\right)\frac{1}{\unicode[STIX]{x1D701}}+\left(\frac{a_{0}}{b_{2}}-\frac{a_{1}b_{1}}{b_{2}^{2}}\right)\frac{1}{\unicode[STIX]{x1D701}^{2}}\nonumber\\ \displaystyle & & \displaystyle +\,\left(-\frac{a_{0}b_{1}}{b_{2}^{2}}+\frac{a_{1}(b_{1}^{2}-b_{2})}{b_{2}^{3}}\right)\frac{1}{\unicode[STIX]{x1D701}^{3}}+\cdots \,;\quad \unicode[STIX]{x1D701}\gg 1.\end{eqnarray}$$

The $Z(\unicode[STIX]{x1D701})$ has asymptotic expansion

(3.113) $$\begin{eqnarray}Z(\unicode[STIX]{x1D701})=-\frac{1}{\unicode[STIX]{x1D701}}-\frac{0}{\unicode[STIX]{x1D701}^{2}}-\frac{1}{2\unicode[STIX]{x1D701}^{3}}+\cdots \,;\quad \unicode[STIX]{x1D701}\gg 1,\end{eqnarray}$$

so by matching with $1/\unicode[STIX]{x1D701}$ implies $b_{2}=-a_{1}$ (as already used previously) that defines $Z_{2,-2}$ (remember, we are starting to count with $n^{\prime }=-2$ ). By further matching with $1/\unicode[STIX]{x1D701}^{2}$ implies $b_{1}=-a_{0}$ , that defines $Z_{2,-1}$ , and by further matching with $1/\unicode[STIX]{x1D701}^{3}$ implies $a_{1}=2$ , that defines $Z_{2,0}$ .

The calculation is continued by matching with the power series for small $\unicode[STIX]{x1D701}$ values, i.e. by using the classical Padé approach, that is described as

(3.114) $$\begin{eqnarray}\displaystyle Z_{2,-2}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}-a_{1}\unicode[STIX]{x1D701}^{2}}=\underbrace{i\sqrt{\unicode[STIX]{x03C0}}}_{=c_{0}}\underbrace{-2}_{=c_{1}}\unicode[STIX]{x1D701}\underbrace{-i\sqrt{\unicode[STIX]{x03C0}}}_{=c_{2}}\unicode[STIX]{x1D701}^{2}, & & \displaystyle\end{eqnarray}$$

and the solution is

(3.115) $$\begin{eqnarray}\displaystyle Z_{2,-2}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\unicode[STIX]{x03C0}-2}}\unicode[STIX]{x1D701}}{1-{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}}{\unicode[STIX]{x03C0}-2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\unicode[STIX]{x03C0}-2}}\unicode[STIX]{x1D701}^{2}}. & & \displaystyle\end{eqnarray}$$

Continuing with $Z_{2,-1}(\unicode[STIX]{x1D701})$ , i.e. by using one more additional asymptotic term that dictates $b_{1}=-a_{0}$ , the matching with the power series yields

(3.116) $$\begin{eqnarray}Z_{2,-1}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1-a_{0}\unicode[STIX]{x1D701}-a_{1}\unicode[STIX]{x1D701}^{2}}=i\sqrt{\unicode[STIX]{x03C0}}-2\unicode[STIX]{x1D701};\quad \Rightarrow \quad Z_{2,-1}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+(\unicode[STIX]{x03C0}-2)\unicode[STIX]{x1D701}}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-(\unicode[STIX]{x03C0}-2)\unicode[STIX]{x1D701}^{2}}.\end{eqnarray}$$

Similarly, considering $Z_{2,0}(\unicode[STIX]{x1D701})$ yields

(3.117) $$\begin{eqnarray}Z_{2,0}(\unicode[STIX]{x1D701})=\frac{a_{0}+2\unicode[STIX]{x1D701}}{1-a_{0}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}}=i\sqrt{\unicode[STIX]{x03C0}};\quad \Rightarrow \quad Z_{2,0}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+2\unicode[STIX]{x1D701}}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}}.\end{eqnarray}$$

Obviously, $R_{2,0}(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z_{2,0}(\unicode[STIX]{x1D701})$ exactly.

3.3.2 Three-pole approximants of $R(\unicode[STIX]{x1D701})$ and $Z(\unicode[STIX]{x1D701})$

A general three-pole approximant of $R(\unicode[STIX]{x1D701})$ is

(3.118) $$\begin{eqnarray}\displaystyle R_{3}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}}. & & \displaystyle\end{eqnarray}$$

The asymptotic expansion calculates as

(3.119) $$\begin{eqnarray}\displaystyle \frac{1}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}} & = & \displaystyle \frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}\left({\displaystyle \frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}}}+{\displaystyle \frac{b_{1}}{b_{3}\unicode[STIX]{x1D701}^{2}}}+{\displaystyle \frac{b_{2}}{b_{3}\unicode[STIX]{x1D701}}}+1\right)}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}}\left[1-\left(\frac{b_{2}}{b_{3}\unicode[STIX]{x1D701}}+\frac{b_{1}}{b_{3}\unicode[STIX]{x1D701}^{2}}+\frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}}\right)\vphantom{\left(\frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}}\right)^{2}}\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\left(\frac{b_{2}}{b_{3}\unicode[STIX]{x1D701}}+\frac{b_{1}}{b_{3}\unicode[STIX]{x1D701}^{2}}+\frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}}\right)^{2}+\cdots \,\right]\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}}\left[1-\frac{1}{\unicode[STIX]{x1D701}}\frac{b_{2}}{b_{3}}+\frac{1}{\unicode[STIX]{x1D701}^{2}}\left(-\frac{b_{1}}{b_{3}}+\frac{b_{2}^{2}}{b_{3}^{2}}\right)+\cdots \,\right]\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{b_{3}\unicode[STIX]{x1D701}^{3}}-\frac{b_{2}}{b_{3}^{2}\unicode[STIX]{x1D701}^{4}}+\frac{b_{2}^{2}-b_{1}b_{3}}{b_{3}^{3}\unicode[STIX]{x1D701}^{5}}+\cdots \,,\end{eqnarray}$$

so that

(3.120) $$\begin{eqnarray}\displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}} & = & \displaystyle \frac{a_{1}}{b_{3}\unicode[STIX]{x1D701}^{2}}+\frac{1}{\unicode[STIX]{x1D701}^{3}}\left(\frac{a_{0}}{b_{3}}-\frac{a_{1}b_{2}}{b_{3}^{2}}\right)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{\unicode[STIX]{x1D701}^{4}}\left(-\frac{a_{0}b_{2}}{b_{3}^{2}}+a_{1}\frac{b_{2}^{2}-b_{1}b_{3}}{b_{3}^{3}}\right)+\cdots\end{eqnarray}$$

For $R_{3,0}(\unicode[STIX]{x1D701})$ this implies $b_{3}=-2a_{1}$ , for $R_{3,1}(\unicode[STIX]{x1D701})$ additionally $b_{2}=-2a_{0}$ and for $R_{3,2}(\unicode[STIX]{x1D701})$ also $b_{1}=3a_{1}$ . The asymptotic expansions (3.120) can become very long for higher orders of $\unicode[STIX]{x1D701}$ , especially when more poles are considered. It is beneficial to write down the following scheme, where in each line, we advance the matching with one more asymptotic point,

(3.121) $$\begin{eqnarray}\displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}}=\underbrace{\frac{a_{1}}{b_{3}}}_{=-1/2}\frac{1}{\unicode[STIX]{x1D701}^{2}}+\cdots \quad \Rightarrow \quad b_{3}=-2a_{1}; & & \displaystyle\end{eqnarray}$$

(3.122) $$\begin{eqnarray}\displaystyle R_{3,0}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\underbrace{\frac{a_{0}+{\displaystyle \frac{b_{2}}{2}}}{2a_{1}}}_{=0}\frac{1}{\unicode[STIX]{x1D701}^{3}}+\cdots \quad \Rightarrow \quad b_{2}=-2a_{0};\end{eqnarray}$$

(3.123) $$\begin{eqnarray}\displaystyle R_{3,1}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}-2a_{0}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\underbrace{\frac{b_{1}}{4a_{1}}}_{=3/4}\frac{1}{\unicode[STIX]{x1D701}^{4}}+\cdots \quad \Rightarrow \quad b_{1}=3a_{1};\end{eqnarray}$$

(3.124) $$\begin{eqnarray}\displaystyle R_{3,2}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+3a_{1}\unicode[STIX]{x1D701}-2a_{0}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}-\underbrace{\frac{1-3a_{0}}{4a_{1}}}_{=0}\frac{1}{\unicode[STIX]{x1D701}^{5}}+\cdots \quad \Rightarrow \quad a_{1}\rightarrow \infty .\end{eqnarray}$$

In the last expression the $a_{1}\rightarrow \infty$ since $a_{0}=1$ , implying the $R_{3,3}(\unicode[STIX]{x1D701})$ does not make sense and it is not defined. The scheme can be very quickly verified by using Maple (or Mathematica) software, by using command $asympt(expression(\unicode[STIX]{x1D701}),\unicode[STIX]{x1D701},n)$ , where $\unicode[STIX]{x1D701}$ is the variable, and $n$ prescribes the precision of the expansion that is calculated up to the $o(\unicode[STIX]{x1D701}^{-n})$ order. Now by matching with the power series for small $\unicode[STIX]{x1D701}$ values

(3.125) $$\begin{eqnarray}\displaystyle R_{3,0}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}}=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3};\end{eqnarray}$$

(3.126) $$\begin{eqnarray}\displaystyle R_{3,1}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+b_{1}\unicode[STIX]{x1D701}-2a_{0}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}}=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2};\end{eqnarray}$$

(3.127) $$\begin{eqnarray}\displaystyle R_{3,2}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}}{1+3a_{1}\unicode[STIX]{x1D701}-2a_{0}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}}=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701};\end{eqnarray}$$

and the solutions are

(3.128) $$\begin{eqnarray}\displaystyle R_{3,0}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1-i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{\unicode[STIX]{x03C0}-3}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}}{1-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}-{\displaystyle \frac{3\unicode[STIX]{x03C0}-8}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}^{2}+2i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{\unicode[STIX]{x03C0}-3}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}^{3}};\end{eqnarray}$$

(3.129) $$\begin{eqnarray}\displaystyle R_{3,1}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1-i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}}{1-{\displaystyle \frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{3}};\end{eqnarray}$$

(3.130) $$\begin{eqnarray}\displaystyle R_{3,2}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1-{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}}{1-{\displaystyle \frac{3i\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}}.\end{eqnarray}$$

A general three-pole approximant of $Z(\unicode[STIX]{x1D701})$ is

(3.131) $$\begin{eqnarray}Z_{3}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}},\end{eqnarray}$$

and has the following asymptotic expansion

(3.132) $$\begin{eqnarray}\displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}} & = & \displaystyle \frac{a_{2}}{b_{3}\unicode[STIX]{x1D701}}+\left(\frac{a_{1}}{b_{3}}-\frac{a_{2}b_{2}}{b_{3}^{2}}\right)\frac{1}{\unicode[STIX]{x1D701}^{2}}\nonumber\\ \displaystyle & & \displaystyle +\,\left(\frac{a_{0}}{b_{3}}-\frac{a_{1}b_{2}}{b_{3}^{2}}+\frac{a_{2}(b_{2}^{2}-b_{1}b_{3})}{b_{3}^{3}}\right)\frac{1}{\unicode[STIX]{x1D701}^{3}}+\cdots\end{eqnarray}$$

Matching the first asymptotic term implies $b_{3}=-a_{2}$ , which defines $Z_{3,-2}(\unicode[STIX]{x1D701})$ . For $Z_{3,-1}(\unicode[STIX]{x1D701})$ the second term is matched as well and $b_{2}=-a_{1}$ . For $Z_{3,0}(\unicode[STIX]{x1D701})$ the third term is also matched and $b_{1}=-a_{0}+a_{2}/2$ . To go higher requires higher-order expansion (3.132). It is again easier to write down the asymptotic expansion scheme step by step

(3.133) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}}=\underbrace{\frac{a_{2}}{b_{3}}}_{=-1}\frac{1}{\unicode[STIX]{x1D701}}+\cdots \,;\quad \Rightarrow \quad b_{3}=-a_{2}; & \displaystyle\end{eqnarray}$$

(3.134) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,-2}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}=-\frac{1}{\unicode[STIX]{x1D701}}-\underbrace{\frac{a_{1}+b_{2}}{a_{2}}}_{=0}\frac{1}{\unicode[STIX]{x1D701}^{2}}+\cdots \quad \Rightarrow \quad b_{2}=-a_{1}; & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.135) $$\begin{eqnarray}\displaystyle Z_{3,-1}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}-a_{1}\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{\unicode[STIX]{x1D701}}-\frac{0}{\unicode[STIX]{x1D701}^{2}}-\underbrace{\frac{a_{0}+b_{1}}{a_{2}}}_{=1/2}\frac{1}{\unicode[STIX]{x1D701}^{3}}+\cdots \quad \Rightarrow \quad b_{1}=\frac{a_{2}}{2}-a_{0};\end{eqnarray}$$

(3.136) $$\begin{eqnarray}\displaystyle Z_{3,0}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+\left({\displaystyle \frac{a_{2}}{2}}-a_{0}\right)\unicode[STIX]{x1D701}-a_{1}\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{\unicode[STIX]{x1D701}}-\frac{1}{2\unicode[STIX]{x1D701}^{3}}-\underbrace{\frac{2-a_{1}}{2a_{2}}}_{=0}\frac{1}{\unicode[STIX]{x1D701}^{4}}+\cdots \quad \Rightarrow \quad a_{1}=2;\end{eqnarray}$$

(3.137) $$\begin{eqnarray}\displaystyle Z_{3,1}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{a_{0}+2\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+\left({\displaystyle \frac{a_{2}}{2}}-a_{0}\right)\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{\unicode[STIX]{x1D701}}-\frac{1}{2\unicode[STIX]{x1D701}^{3}}-\underbrace{\frac{a_{2}-2a_{0}}{4a_{2}}}_{=3/4}\frac{1}{\unicode[STIX]{x1D701}^{5}}+\cdots \quad \Rightarrow \quad a_{2}=-a_{0};\end{eqnarray}$$

(3.138) $$\begin{eqnarray}\displaystyle Z_{3,2}(\unicode[STIX]{x1D701})=\frac{a_{0}+2\unicode[STIX]{x1D701}-a_{0}\unicode[STIX]{x1D701}^{2}}{1-\frac{3}{2}a_{0}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+a_{0}\unicode[STIX]{x1D701}^{3}}. & & \displaystyle\end{eqnarray}$$

Matching these results with an expansion for small $\unicode[STIX]{x1D701}$ values is done according to

(3.139) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,-2}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}=i\sqrt{\unicode[STIX]{x03C0}}-2\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}+\frac{4}{3}\unicode[STIX]{x1D701}^{3}+i\frac{\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}^{4}; & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.140) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,-1}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}-a_{1}\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}=i\sqrt{\unicode[STIX]{x03C0}}-2\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}+\frac{4}{3}\unicode[STIX]{x1D701}^{3}; & \displaystyle\end{eqnarray}$$

(3.141) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,0}(\unicode[STIX]{x1D701})=\frac{a_{0}+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+\left(-a_{0}+{\displaystyle \frac{a_{2}}{2}}\right)\unicode[STIX]{x1D701}-a_{1}\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}=i\sqrt{\unicode[STIX]{x03C0}}-2\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}; & \displaystyle\end{eqnarray}$$

(3.142) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,1}(\unicode[STIX]{x1D701})=\frac{a_{0}+2\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+\left({\displaystyle \frac{a_{2}}{2}}-a_{0}\right)\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}^{3}}=i\sqrt{\unicode[STIX]{x03C0}}-2\unicode[STIX]{x1D701}; & \displaystyle\end{eqnarray}$$

(3.143) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,2}(\unicode[STIX]{x1D701})=\frac{a_{0}+2\unicode[STIX]{x1D701}-a_{0}\unicode[STIX]{x1D701}^{2}}{1-\frac{3}{2}a_{0}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+a_{0}\unicode[STIX]{x1D701}^{3}}=i\sqrt{\unicode[STIX]{x03C0}}, & \displaystyle\end{eqnarray}$$

and the solutions are

(3.144) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,-2}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+{\displaystyle \frac{3\unicode[STIX]{x03C0}^{2}-30\unicode[STIX]{x03C0}+64}{2(5\unicode[STIX]{x03C0}-16)}}\unicode[STIX]{x1D701}+{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}(9\unicode[STIX]{x03C0}-28)}{6(5\unicode[STIX]{x03C0}-16)}}\unicode[STIX]{x1D701}^{2}}{1-{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}(3\unicode[STIX]{x03C0}-10)}{2(5\unicode[STIX]{x03C0}-16)}}\unicode[STIX]{x1D701}+{\displaystyle \frac{21\unicode[STIX]{x03C0}-64}{6(5\unicode[STIX]{x03C0}-16)}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}(9\unicode[STIX]{x03C0}-28)}{6(5\unicode[STIX]{x03C0}-16)}}\unicode[STIX]{x1D701}^{3}}; & \displaystyle\end{eqnarray}$$

(3.145) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,-1}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+{\displaystyle \frac{10-3\unicode[STIX]{x03C0}}{3(\unicode[STIX]{x03C0}-3)}}\unicode[STIX]{x1D701}+{\displaystyle \frac{i(5\unicode[STIX]{x03C0}-16)}{3\sqrt{\unicode[STIX]{x03C0}}(\unicode[STIX]{x03C0}-3)}}\unicode[STIX]{x1D701}^{2}}{1-{\displaystyle \frac{i(3\unicode[STIX]{x03C0}-8)}{3\sqrt{\unicode[STIX]{x03C0}}(\unicode[STIX]{x03C0}-3)}}\unicode[STIX]{x1D701}-{\displaystyle \frac{10-3\unicode[STIX]{x03C0}}{3(\unicode[STIX]{x03C0}-3)}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{i(5\unicode[STIX]{x03C0}-16)}{3\sqrt{\unicode[STIX]{x03C0}}(\unicode[STIX]{x03C0}-3)}}\unicode[STIX]{x1D701}^{3}}; & \displaystyle\end{eqnarray}$$

(3.146) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,0}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+{\displaystyle \frac{3\unicode[STIX]{x03C0}-8}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}-2i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{\unicode[STIX]{x03C0}-3}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}^{2}}{1-{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}-{\displaystyle \frac{3\unicode[STIX]{x03C0}-8}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}^{2}+2i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{\unicode[STIX]{x03C0}-3}{4-\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}^{3}}. & \displaystyle\end{eqnarray}$$

(3.147) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,1}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+2\unicode[STIX]{x1D701}-2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{2}}{1-i{\displaystyle \frac{4}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{3}}; & \displaystyle\end{eqnarray}$$

(3.148) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3,2}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+2\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}}{1-\frac{3}{2}i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}}. & \displaystyle\end{eqnarray}$$

Of course, the following relations now hold exactly

(3.149) $$\begin{eqnarray}\displaystyle & \displaystyle R_{3,0}(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z_{3,0}(\unicode[STIX]{x1D701}); & \displaystyle\end{eqnarray}$$

(3.150) $$\begin{eqnarray}\displaystyle & \displaystyle R_{3,1}(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z_{3,1}(\unicode[STIX]{x1D701}); & \displaystyle\end{eqnarray}$$

(3.151) $$\begin{eqnarray}\displaystyle & \displaystyle R_{3,2}(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z_{3,2}(\unicode[STIX]{x1D701}). & \displaystyle\end{eqnarray}$$

3.3.3 Four-pole approximants of $R(\unicode[STIX]{x1D701})$ and $Z(\unicode[STIX]{x1D701})$

As before, the procedure of matching with asymptotic expansion yields (for simplicity already assuming $a_{0}=1$ )

(3.152) $$\begin{eqnarray}\frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}+b_{4}\unicode[STIX]{x1D701}^{4}}=\underbrace{\frac{a_{2}}{b_{4}}}_{=-1/2}\frac{1}{\unicode[STIX]{x1D701}^{2}}+\cdots \quad \Rightarrow \quad b_{4}=-2a_{2};\end{eqnarray}$$

(3.153) $$\begin{eqnarray}\displaystyle R_{4,0}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\underbrace{\frac{a_{1}+{\displaystyle \frac{b_{3}}{2}}}{2a_{2}}}_{=0}\frac{1}{\unicode[STIX]{x1D701}^{3}}+\cdots \quad \Rightarrow \quad b_{3}=-2a_{1};\end{eqnarray}$$

(3.154) $$\begin{eqnarray}\displaystyle R_{4,1}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\underbrace{\frac{1+{\displaystyle \frac{b_{2}}{2}}}{2a_{2}}}_{=3/4}\frac{1}{\unicode[STIX]{x1D701}^{4}}+\cdots \quad \Rightarrow \quad b_{2}=3a_{2}-2;\end{eqnarray}$$

(3.155) $$\begin{eqnarray}\displaystyle R_{4,2}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+(3a_{2}-2)\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}-\underbrace{\frac{b_{1}-3a_{1}}{4a_{2}}}_{=0}\frac{1}{\unicode[STIX]{x1D701}^{5}}+\cdots \quad \Rightarrow \quad b_{1}=3a_{1};\end{eqnarray}$$

(3.156) $$\begin{eqnarray}\displaystyle R_{4,3}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+3a_{1}\unicode[STIX]{x1D701}+(3a_{2}-2)\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}-\underbrace{\frac{\frac{9}{2}a_{2}-2}{4a_{2}}}_{=15/8}\frac{1}{\unicode[STIX]{x1D701}^{6}}+\cdots \quad \Rightarrow \quad a_{2}=-\frac{2}{3};\end{eqnarray}$$

(3.157) $$\begin{eqnarray}\displaystyle R_{4,4}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1+a_{1}\unicode[STIX]{x1D701}-\frac{2}{3}\unicode[STIX]{x1D701}^{2}}{1+3a_{1}\unicode[STIX]{x1D701}-4\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{2\unicode[STIX]{x1D701}^{2}}-\frac{3}{4\unicode[STIX]{x1D701}^{4}}-\frac{15}{8\unicode[STIX]{x1D701}^{6}}-\underbrace{\frac{9a_{1}}{8}}_{=0}\frac{1}{\unicode[STIX]{x1D701}^{7}}\quad \Rightarrow \quad a_{1}=0,\end{eqnarray}$$

where the last relation implies a possible approximant $R_{4,5}(\unicode[STIX]{x1D701})=(1-\frac{2}{3}\unicode[STIX]{x1D701}^{2})/(1-4\unicode[STIX]{x1D701}^{2}+\frac{4}{3}\unicode[STIX]{x1D701}^{4})$ . However, such an approximant is not well behaved (it has zero imaginary part for real valued $\unicode[STIX]{x1D701}$ ) and the $R_{4,5}(\unicode[STIX]{x1D701})$ is eliminated. Matching with the power series is performed according to

(3.158) $$\begin{eqnarray}\displaystyle R_{4,0}(\unicode[STIX]{x1D701}) & = & \displaystyle \frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}\nonumber\\ \displaystyle & = & \displaystyle 1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}+i\frac{\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}^{5};\end{eqnarray}$$

(3.159) $$\begin{eqnarray}R_{4,1}(\unicode[STIX]{x1D701})=\frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4};\end{eqnarray}$$

(3.160) $$\begin{eqnarray}R_{4,2}(\unicode[STIX]{x1D701})=\frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+(3a_{2}-2)\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3};\end{eqnarray}$$

(3.161) $$\begin{eqnarray}\displaystyle & \displaystyle R_{4,3}(\unicode[STIX]{x1D701})=\frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+3a_{1}\unicode[STIX]{x1D701}+(3a_{2}-2)\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}-2a_{2}\unicode[STIX]{x1D701}^{4}}=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}; & \displaystyle\end{eqnarray}$$

(3.162) $$\begin{eqnarray}\displaystyle & \displaystyle R_{4,4}(\unicode[STIX]{x1D701})=\frac{1+a_{1}\unicode[STIX]{x1D701}-\frac{2}{3}\unicode[STIX]{x1D701}^{2}}{1+3a_{1}\unicode[STIX]{x1D701}-4\unicode[STIX]{x1D701}^{2}-2a_{1}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}}=1+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}, & \displaystyle\end{eqnarray}$$

and the results are

(3.163) $$\begin{eqnarray}\displaystyle & & \displaystyle R_{4,0}(\unicode[STIX]{x1D701})=\frac{1+i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}}{2}}{\displaystyle \frac{(12\unicode[STIX]{x03C0}^{2}-67\unicode[STIX]{x03C0}+92)}{(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(9\unicode[STIX]{x03C0}^{2}-69\unicode[STIX]{x03C0}+128)}{6(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{2}}{1-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}}{2}}{\displaystyle \frac{(9\unicode[STIX]{x03C0}-28)}{(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}+{\displaystyle \frac{(36\unicode[STIX]{x03C0}^{2}-195\unicode[STIX]{x03C0}+256)}{6(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{2}-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}(33\unicode[STIX]{x03C0}-104)}{6(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{(9\unicode[STIX]{x03C0}^{2}-69\unicode[STIX]{x03C0}+128)}{3(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{4}};\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.164) $$\begin{eqnarray}\displaystyle & & \displaystyle R_{4,1}(\unicode[STIX]{x1D701})=\frac{1-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}}{3}}{\displaystyle \frac{(9\unicode[STIX]{x03C0}-28)}{(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}{3(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{2}}{1-i{\displaystyle \frac{2\sqrt{\unicode[STIX]{x03C0}}}{3}}{\displaystyle \frac{(10-3\unicode[STIX]{x03C0})}{(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(21\unicode[STIX]{x03C0}-64)}{3(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{2}+i{\displaystyle \frac{2\sqrt{\unicode[STIX]{x03C0}}}{3}}{\displaystyle \frac{(9\unicode[STIX]{x03C0}-28)}{(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{2(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}{3(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{4}};\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.165) $$\begin{eqnarray}R_{4,2}(\unicode[STIX]{x1D701})=\frac{1-i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{(10-3\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(16-5\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{2}}{1-i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{2}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(32-9\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{2(10-3\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{2(16-5\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{4}}\end{eqnarray}$$

(3.166) $$\begin{eqnarray}\displaystyle & \displaystyle R_{4,3}(\unicode[STIX]{x1D701})=\frac{1-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(3\unicode[STIX]{x03C0}-8)}{4}}\unicode[STIX]{x1D701}^{2}}{1-i{\displaystyle \frac{3\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(9\unicode[STIX]{x03C0}-16)}{4}}\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{(3\unicode[STIX]{x03C0}-8)}{2}}\unicode[STIX]{x1D701}^{4}}; & \displaystyle\end{eqnarray}$$

(3.167) $$\begin{eqnarray}\displaystyle & \displaystyle R_{4,4}(\unicode[STIX]{x1D701})=\frac{1-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{2}{3}}\unicode[STIX]{x1D701}^{2}}{1-i{\displaystyle \frac{3\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}-4\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}}. & \displaystyle\end{eqnarray}$$

Of the four-pole approximants, perhaps the most well-known one is $R_{4,3}(\unicode[STIX]{x1D701})$ used for example by Hammett & Perkins (Reference Hammett and Perkins1990), Passot & Sulem (Reference Passot and Sulem2007) etc., and which can be written in a convenient form

(3.168)

Here we do not double check the derivation of the $Z_{4}(\unicode[STIX]{x1D701})$ approximants ‘from scratch’, and for a given $R_{4}$ coefficients, the $Z_{4}$ coefficients are of course easily obtained by

(3.169) $$\begin{eqnarray}R_{4}(\unicode[STIX]{x1D701})=\frac{1+a_{1}\unicode[STIX]{x1D701}+a_{2}\unicode[STIX]{x1D701}^{2}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}+b_{4}\unicode[STIX]{x1D701}^{4}}\quad \Rightarrow \quad Z_{4}(\unicode[STIX]{x1D701})=\frac{(a_{1}-b_{1})+(a_{2}-b_{2})\unicode[STIX]{x1D701}-b_{3}\unicode[STIX]{x1D701}^{2}-b_{4}\unicode[STIX]{x1D701}^{3}}{1+b_{1}\unicode[STIX]{x1D701}+b_{2}\unicode[STIX]{x1D701}^{2}+b_{3}\unicode[STIX]{x1D701}^{3}+b_{4}\unicode[STIX]{x1D701}^{4}}.\end{eqnarray}$$

For completeness, the corresponding results are

(3.170) $$\begin{eqnarray}\displaystyle Z_{4,0}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}-{\displaystyle \frac{(15\unicode[STIX]{x03C0}^{2}-88\unicode[STIX]{x03C0}+128)}{2(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}+i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}(33\unicode[STIX]{x03C0}-104)}{6(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{(9\unicode[STIX]{x03C0}^{2}-69\unicode[STIX]{x03C0}+128)}{3(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{3}}{1-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}}{2}}{\displaystyle \frac{(9\unicode[STIX]{x03C0}-28)}{(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}+{\displaystyle \frac{(36\unicode[STIX]{x03C0}^{2}-195\unicode[STIX]{x03C0}+256)}{6(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{2}-i{\displaystyle \frac{\sqrt{\unicode[STIX]{x03C0}}(33\unicode[STIX]{x03C0}-104)}{6(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{(9\unicode[STIX]{x03C0}^{2}-69\unicode[STIX]{x03C0}+128)}{3(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}}\unicode[STIX]{x1D701}^{4}}; & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.171) $$\begin{eqnarray}\displaystyle Z_{4,1}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}-{\displaystyle \frac{2(3\unicode[STIX]{x03C0}^{2}-25\unicode[STIX]{x03C0}+48)}{3(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}-i{\displaystyle \frac{2\sqrt{\unicode[STIX]{x03C0}}}{3}}{\displaystyle \frac{(9\unicode[STIX]{x03C0}-28)}{(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{2(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}{3(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{3}}{1-i{\displaystyle \frac{2\sqrt{\unicode[STIX]{x03C0}}}{3}}{\displaystyle \frac{(10-3\unicode[STIX]{x03C0})}{(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(21\unicode[STIX]{x03C0}-64)}{3(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{2}+i{\displaystyle \frac{2\sqrt{\unicode[STIX]{x03C0}}}{3}}{\displaystyle \frac{(9\unicode[STIX]{x03C0}-28)}{(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{2(6\unicode[STIX]{x03C0}^{2}-29\unicode[STIX]{x03C0}+32)}{3(16-5\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{4}}; & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.172) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{4,2}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+{\displaystyle \frac{4(4-\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{2(10-3\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{2(16-5\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{3}}{1-i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{2}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(32-9\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}{\displaystyle \frac{2(10-3\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{2(16-5\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}}\unicode[STIX]{x1D701}^{4}} & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.173) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{4,3}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+{\displaystyle \frac{3\unicode[STIX]{x03C0}-4}{2}}\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{(3\unicode[STIX]{x03C0}-8)}{2}}\unicode[STIX]{x1D701}^{3}}{1-i{\displaystyle \frac{3\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{(9\unicode[STIX]{x03C0}-16)}{4}}\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{(3\unicode[STIX]{x03C0}-8)}{2}}\unicode[STIX]{x1D701}^{4}}; & \displaystyle\end{eqnarray}$$

(3.174) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{4,4}(\unicode[STIX]{x1D701})=\frac{i\sqrt{\unicode[STIX]{x03C0}}+\frac{10}{3}\unicode[STIX]{x1D701}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}-\frac{4}{3}\unicode[STIX]{x1D701}^{3}}{1-i{\displaystyle \frac{3\sqrt{\unicode[STIX]{x03C0}}}{2}}\unicode[STIX]{x1D701}-4\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}}. & \displaystyle\end{eqnarray}$$

3.4 Conversion of our 2-index $R_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ notation to other notations

For clarity, we provide conversion tables of Padé approximants in the notation of Martín et al. (Reference Martín, Donoso and Zamudio-Cristi1980) and Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992) to our notation. Comparing our analytic results to those of Martín et al. (Reference Martín, Donoso and Zamudio-Cristi1980) (introducing superscript $M$ ), can be done easily according to

(3.175) $$\begin{eqnarray}Z_{3,1}^{M}=Z_{2,-2};\quad Z_{2,2}^{M}=Z_{2,-1};\quad Z_{1,3}^{M}=Z_{2,0};\end{eqnarray}$$

(3.176) $$\begin{eqnarray}Z_{5,1}^{M}=Z_{3,-2};\quad Z_{4,2}^{M}=Z_{3,-1};\quad Z_{3,3}^{M}=Z_{3,0};\end{eqnarray}$$

(3.177) $$\begin{eqnarray}Z_{5,3}^{M}=Z_{4,0},\end{eqnarray}$$

and the general conversion can be written as

(3.178) $$\begin{eqnarray}Z_{n,n^{\prime }}^{M}=Z_{n+n^{\prime }/2,n^{\prime }-3}.\end{eqnarray}$$

The table 1 of Martín et al. (Reference Martín, Donoso and Zamudio-Cristi1980) can be now easily verified, which reveals a small obvious typo in their $Z_{4,2}^{M}$ , where the coefficient $p_{2}$ is missing the imaginary i number.

To compare our results to those of Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992), it is useful to calculate asymptotic expansions of their $Z_{n}$ definitions (that is defined as $Z_{n,0}$ ), that calculate as

(3.179) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{2}^{HL}=-\frac{a_{1}-\unicode[STIX]{x1D701}}{a_{0}+a_{1}\unicode[STIX]{x1D701}-\unicode[STIX]{x1D701}^{2}}=-\frac{1}{\unicode[STIX]{x1D701}}-\frac{0}{\unicode[STIX]{x1D701}^{2}}+o\left(\frac{1}{\unicode[STIX]{x1D701}^{2}}\right); & \displaystyle\end{eqnarray}$$

(3.180) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{3}^{HL}=-\frac{\unicode[STIX]{x1D701}^{2}-a_{2}\unicode[STIX]{x1D701}-a_{1}+\frac{1}{2}}{\unicode[STIX]{x1D701}^{3}-a_{2}\unicode[STIX]{x1D701}^{2}-a_{1}\unicode[STIX]{x1D701}-a_{0}}=-\frac{1}{\unicode[STIX]{x1D701}}-\frac{1}{2\unicode[STIX]{x1D701}^{3}}+o\left(\frac{1}{\unicode[STIX]{x1D701}^{3}}\right); & \displaystyle\end{eqnarray}$$

(3.181) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{4}^{HL}=-\frac{\unicode[STIX]{x1D701}^{3}-a_{3}\unicode[STIX]{x1D701}^{2}-(a_{2}-\frac{1}{2}\unicode[STIX]{x1D701})-(a_{1}+a_{3}/2)}{\unicode[STIX]{x1D701}^{4}-a_{3}\unicode[STIX]{x1D701}^{3}-a_{2}\unicode[STIX]{x1D701}^{2}-a_{1}\unicode[STIX]{x1D701}-a_{0}}=-\frac{1}{\unicode[STIX]{x1D701}}-\frac{1}{2\unicode[STIX]{x1D701}^{3}}-\frac{0}{\unicode[STIX]{x1D701}^{4}}+o\left(\frac{1}{\unicode[STIX]{x1D701}^{4}}\right); & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.182) $$\begin{eqnarray}\displaystyle Z_{5}^{HL} & = & \displaystyle -\frac{\unicode[STIX]{x1D701}^{4}-a_{4}\unicode[STIX]{x1D701}^{3}-(a_{3}-\frac{1}{2})\unicode[STIX]{x1D701}^{2}-(a_{2}+a_{4}/2)\unicode[STIX]{x1D701}-(a_{1}+a_{3}/2-\frac{3}{4})}{\unicode[STIX]{x1D701}^{5}-a_{4}\unicode[STIX]{x1D701}^{4}-a_{3}\unicode[STIX]{x1D701}^{3}-a_{2}\unicode[STIX]{x1D701}^{2}-a_{1}\unicode[STIX]{x1D701}-a_{0}}\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{\unicode[STIX]{x1D701}}-\frac{1}{2\unicode[STIX]{x1D701}^{3}}-\frac{3}{4\unicode[STIX]{x1D701}^{5}}+o\left(\frac{1}{\unicode[STIX]{x1D701}^{5}}\right),\end{eqnarray}$$

where ‘ $HL$ ’ stands for Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992). As one can see, the number of asymptotic points used in their basic definition of $Z_{n}$ , increases with the number of poles $n$ . Compared to our definition, their $Z_{2,0}$ is defined as having another asymptotic point (for a total of 2), $Z_{3,0}$ has another asymptotic point (for a total of 3), $Z_{4,0}$ another one (for a total of 4), and so on. Essentially, in their notation the basic $Z_{n,0}$ is defined as having ‘ $n$ ’ asymptotic points, and asymptotic precision $o(1/\unicode[STIX]{x1D701}^{n})$ . The conversion between their and our notation is easy, and

(3.183) $$\begin{eqnarray}\displaystyle Z_{2,0}^{HL} & = & \displaystyle Z_{2,-1};\quad Z_{2,1}^{HL}=Z_{2,0};\end{eqnarray}$$

(3.184) $$\begin{eqnarray}\displaystyle Z_{3,1}^{HL} & = & \displaystyle Z_{3,1};\quad Z_{3,2}^{HL}=Z_{3,2};\end{eqnarray}$$

(3.185) $$\begin{eqnarray}\displaystyle Z_{4,1}^{HL} & = & \displaystyle Z_{4,2};\quad Z_{4,2}^{HL}=Z_{4,3};\quad Z_{4,3}^{HL}=Z_{4,4};\end{eqnarray}$$

(3.186) $$\begin{eqnarray}\displaystyle Z_{5,1}^{HL} & = & \displaystyle Z_{5,3};\quad Z_{5,2}^{HL}=Z_{5,4};\quad Z_{5,3}^{HL}=Z_{5,5};\quad Z_{5,4}^{HL}=Z_{5,6},\end{eqnarray}$$

or the general conversion can be written as

(3.187) $$\begin{eqnarray}Z_{n,n^{\prime }}^{HL}=Z_{n,n^{\prime }+n-3}.\end{eqnarray}$$

We checked the table 1 of Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992) that provides coefficients for the Padé approximants (3.183)–(3.186) and we can confirm that the table is essentially correct, except for one coefficient.Footnote ⁷ The coefficient where a simple typo is suspected is the coefficient $a_{1}$ in $Z_{3,1}$ . Rewriting our three-pole approximant $R_{3}(\unicode[STIX]{x1D701})$ to the form used by Passot & Sulem (Reference Passot and Sulem2007) and Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992) (that corresponds to the $Z_{3}^{HL}$ as written in (3.180)) yields

(3.188) $$\begin{eqnarray}R_{3}(\unicode[STIX]{x1D701})=\frac{-\frac{1}{2}\unicode[STIX]{x1D701}-a_{0}}{\unicode[STIX]{x1D701}^{3}-a_{2}\unicode[STIX]{x1D701}^{2}-a_{1}\unicode[STIX]{x1D701}-a_{0}},\end{eqnarray}$$

which further yields

(3.189) $$\begin{eqnarray}\displaystyle & \displaystyle R_{3,1}(\unicode[STIX]{x1D701})=\frac{-{\displaystyle \frac{1}{2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}}{2(4-\unicode[STIX]{x03C0})}}}{\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}}{(4-\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{2}{(4-\unicode[STIX]{x03C0})}}\unicode[STIX]{x1D701}-{\displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}}{2(4-\unicode[STIX]{x03C0})}}}; & \displaystyle\end{eqnarray}$$

(3.190) $$\begin{eqnarray}\displaystyle & \displaystyle R_{3,2}(\unicode[STIX]{x1D701})=\frac{-{\displaystyle \frac{1}{2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{i}{\sqrt{\unicode[STIX]{x03C0}}}}}{\unicode[STIX]{x1D701}^{3}+{\displaystyle \frac{2i}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{2}-{\displaystyle \frac{3}{2}}\unicode[STIX]{x1D701}-{\displaystyle \frac{i}{\sqrt{\unicode[STIX]{x03C0}}}}}, & \displaystyle\end{eqnarray}$$

and our approximants are

(3.191) $$\begin{eqnarray}\displaystyle R_{3,1}(\unicode[STIX]{x1D701}):\quad a_{0} & = & \displaystyle \frac{i\sqrt{\unicode[STIX]{x03C0}}}{2(4-\unicode[STIX]{x03C0})}=1.03241i;\quad a_{1}=\frac{2}{4-\unicode[STIX]{x03C0}}=2.32990;\nonumber\\ \displaystyle a_{2} & = & \displaystyle -\frac{i\sqrt{\unicode[STIX]{x03C0}}}{4-\unicode[STIX]{x03C0}}=-2.06482i;\end{eqnarray}$$

(3.192) $$\begin{eqnarray}\displaystyle R_{3,2}(\unicode[STIX]{x1D701}):\quad a_{0} & = & \displaystyle \frac{i}{\sqrt{\unicode[STIX]{x03C0}}}=0.56419i;\quad a_{1}=\frac{3}{2};\nonumber\\ \displaystyle a_{2} & = & \displaystyle -\frac{2i}{\sqrt{\unicode[STIX]{x03C0}}}=-1.12838i.\end{eqnarray}$$

For the $a_{1}$ coefficient in $R_{3,1}$ , both Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992) and Passot & Sulem (Reference Passot and Sulem2007) use $a_{1}=2.23990$ instead of the correct $a_{1}=2.32990$ . The differences are of course small. Nevertheless, the new correct value explains the observation made by Passot & Sulem (Reference Passot and Sulem2007), in the paragraph below their figure 1, where they write: ‘It is conspicuous that $R_{3,2}$ provides a fit that is slightly better for small $\unicode[STIX]{x1D701}$ , but turns out to be globally less accurate than $R_{3,1}$ ’. The authors obviously noticed that something is not right, since for small $\unicode[STIX]{x1D701}$ , the $R_{3,1}$ has precision $o(\unicode[STIX]{x1D701}^{2})$ and $R_{3,2}$ only $o(\unicode[STIX]{x1D701})$ , so the $R_{3,1}$ should be more precise. And it indeed is, the authors were just misguided by the wrong value of $a_{1}$ introduced by Hedrick & Leboeuf (Reference Hedrick and Leboeuf1992).

3.5 Precision of $R(\unicode[STIX]{x1D701})$ approximants

It is useful to compare the Padé approximants to the exact $R(\unicode[STIX]{x1D701})=1+\unicode[STIX]{x1D701}Z(\unicode[STIX]{x1D701})$ , where the plasma dispersion function can be conveniently calculated (for example in Maple) according to

(3.193) $$\begin{eqnarray}\displaystyle Z(\unicode[STIX]{x1D701})=i\sqrt{\unicode[STIX]{x03C0}}e^{-\unicode[STIX]{x1D701}^{2}}\left(1+\text{erf}(i\unicode[STIX]{x1D701})\right), & & \displaystyle\end{eqnarray}$$

where $\text{erf}(z)=(2/\sqrt{\unicode[STIX]{x03C0}})\int _{0}^{z}e^{-t^{2}}\,\text{d}t$ is the well-known error function, defined for any complex $z$ . We plot only approximants for which we were able to obtain closures. The exact $R(\unicode[STIX]{x1D701})$ is plotted as a black solid line in all the figures. Figure 2 top shows one-pole and two-pole approximants $R_{1}(\unicode[STIX]{x1D701})$ (red dashed line) and $R_{2,0}(\unicode[STIX]{x1D701})$ (blue dot-dashed line). Figure 2 bottom shows three-pole approximants $R_{3,0}(\unicode[STIX]{x1D701})$ (red dashed line), $R_{3,1}(\unicode[STIX]{x1D701})$ (green dotted line) and $R_{3,2}(\unicode[STIX]{x1D701})$ (blue dot-dashed line). Figures in the left column show the imaginary part and figures in the right column show the real part. The input variable $\unicode[STIX]{x1D701}$ plotted on the $x$ -axis is prescribed to be real, i.e. states in the weak growth-rate/damping approximation are explored (one might as well prescribe $\text{Im}(\unicode[STIX]{x1D701})=\pm 0.01\text{Re}(\unicode[STIX]{x1D701})$ and plot essentially the same graphs, with only small differences in solutions).

Figure 2. One-pole, two-pole (a,b) and three-pole (c,d) Padé approximants of $R(\unicode[STIX]{x1D701})$ . (a,c)  $\text{Im}R(\unicode[STIX]{x1D701})$ , (b,d)  $\text{Re}R(\unicode[STIX]{x1D701})$ , for $\unicode[STIX]{x1D701}$ being real.

Figure 3. Four-pole (a,b) and five-pole (c,d) Padé approximants of $R(\unicode[STIX]{x1D701})$ .

As expected, the very simple approximant $R_{1}(\unicode[STIX]{x1D701})$ is unprecise for larger values of $\unicode[STIX]{x1D701}$ , and above $\unicode[STIX]{x1D701}>1$ , the $\text{Re}R_{1}(\unicode[STIX]{x1D701})$ even has a wrong sign. Nevertheless, the approximant is still a good approximant for small $\unicode[STIX]{x1D701}\ll 1$ values, and it is also very valuable from a theoretical perspective, since it is the only approximant that provides a quasi-static closure for the perpendicular heat flux $q_{\bot }$ (see the 3-D geometry § 4, closure (4.108)). This has one important implication:

If one renders the approximant $R_{1}$ as not satisfactory (which is true unless $\unicode[STIX]{x1D701}\ll 1$ or at least $\unicode[STIX]{x1D701}<1$ ), 3-D simulations with fluid models that contain Landau damping can be only performed with time-dependent heat flux equations. All other approximants in figure 2 perform reasonably well, and the most precise is $R_{3,0}(\unicode[STIX]{x1D701})$ , followed by $R_{3,1}(\unicode[STIX]{x1D701})$ .

Figure 3 shows selected four-pole and five-pole approximants for which we were able to obtain closures. Unfortunately, approximants $R_{4,4}(\unicode[STIX]{x1D701})$ and $R_{5,6}(\unicode[STIX]{x1D701})$ show a bit unpleasant behaviour, and the associated closures obtained with these approximants are therefore difficult to recommend, unless the considered domain is $\unicode[STIX]{x1D701}\ll 1$ or $\unicode[STIX]{x1D701}\gg 1$ , or more specifically, at least $\unicode[STIX]{x1D701}<0.5$ or $\unicode[STIX]{x1D701}>2$ . The behaviour is not surprising, since approximants $R_{4,4}(\unicode[STIX]{x1D701})$ and $R_{5,6}(\unicode[STIX]{x1D701})$ have the maximum available number of poles devoted to the asymptotic expansion $\unicode[STIX]{x1D701}\gg 1$ , without being ill posed. The closures are therefore specifically suitable for $\unicode[STIX]{x1D701}\gg 1$ regime, for example in the low-temperature limit, or, in the high-frequency (actually high phase speed) limit (since $\unicode[STIX]{x1D701}=\unicode[STIX]{x1D714}/(|k_{\Vert }|v_{\text{th}\Vert })$ ). For $R_{4,4}(\unicode[STIX]{x1D701})$ , the corresponding closures are the quasi-static closure (3.268) and time-dependent closures (3.295), (3.297), (3.299). For $R_{5,6}(\unicode[STIX]{x1D701})$ , the corresponding closure is time dependent (3.325) and naturally, this is the most precise closure in the $\unicode[STIX]{x1D701}\gg 1$ regime, with precision $o(\unicode[STIX]{x1D701}^{-8})$ . Noticeably, the asymptotic precision is even better than the $R_{8,3}(\unicode[STIX]{x1D701})$ approximant used in the WHAMP code, which has a precision $o(\unicode[STIX]{x1D701}^{-5})$ .

All other approximants in figure 3 are very precise in the entire considered range of $\unicode[STIX]{x1D701}$ . To clearly see the precision, it is useful to calculate the maximum relative errors

(3.194) $$\begin{eqnarray}\text{Im}\left(\frac{R_{n,n^{\prime }}(\unicode[STIX]{x1D701})-R(\unicode[STIX]{x1D701})}{R(\unicode[STIX]{x1D701})}\right)100\,\%;\quad \text{Re}\left(\frac{R_{n,n^{\prime }}(\unicode[STIX]{x1D701})-R(\unicode[STIX]{x1D701})}{R(\unicode[STIX]{x1D701})}\right)100\,\%,\end{eqnarray}$$

which we define this way instead of for example $\text{Re}(R_{n,n^{\prime }}(\unicode[STIX]{x1D701})-R(\unicode[STIX]{x1D701}))/\text{Re}R(\unicode[STIX]{x1D701})$ , since the real part of $R(\unicode[STIX]{x1D701})$ is going through zero. The maximum relative errors typically appear for $\unicode[STIX]{x1D701}\in (0,4)$ , even though some reported values are outside of this range. The $R_{1}(\unicode[STIX]{x1D701})$ approximant is excluded from the table since its relative error of the imaginary part increases with $\unicode[STIX]{x1D701}$ . We omit if errors are positive or negative and the results are:

The numbers of course do not reveal the entire story, since the maximum error can occur for different $\unicode[STIX]{x1D701}$ values. For example, from the plots of $\text{Im}R(\unicode[STIX]{x1D701})$ in figure 2, the approximant $R_{3,0}(\unicode[STIX]{x1D701})$ captures the maximum (the peak around $\unicode[STIX]{x1D701}\sim 1$ ) with much better accuracy than the approximant $R_{3,1}(\unicode[STIX]{x1D701})$ . However, according to the above table, the $R_{3,1}(\unicode[STIX]{x1D701})$ appears to be more precise globally. The discrepancy is easily understood from figure 4, where $\%$ errors of both approximants are plotted with respect to $\unicode[STIX]{x1D701}$ . A similar table and figures can be created for the heavily damped regime, for example for $\unicode[STIX]{x1D701}$ with the imaginary part $\text{Im}(\unicode[STIX]{x1D701})=-\text{Re}(\unicode[STIX]{x1D701})/2$ , where the Padé approximants are less precise.

3.6 Landau fluid closures – fascinating closures for all $\unicode[STIX]{x1D701}$

Now, let us use various Padé approximations of the plasma response function $R(\unicode[STIX]{x1D701})$ , and calculate the kinetic moments. Let us start with the simplest choice of replacing the exact $R(\unicode[STIX]{x1D701})$ with approximant $R_{1}(\unicode[STIX]{x1D701})=1/(1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701})$ . Let us drop the index $r$ . The linear kinetic moments (3.20)–(3.26) calculate as

(3.195) $$\begin{eqnarray}\displaystyle R_{1}(\unicode[STIX]{x1D701}):\quad \frac{n^{(1)}}{n_{0}} & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}\left[1\right];\end{eqnarray}$$

(3.196) $$\begin{eqnarray}\displaystyle u^{(1)} & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}v_{\text{th}}\text{sign}(k_{\Vert })\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}\left[\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.197) $$\begin{eqnarray}\displaystyle \frac{p^{(1)}}{p_{0}} & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}\left[2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}+1\right];\end{eqnarray}$$

(3.198) $$\begin{eqnarray}\displaystyle T^{(1)} & = & \displaystyle -q_{r}\unicode[STIX]{x1D6F7}\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}\left[2\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.199) $$\begin{eqnarray}\displaystyle q^{(1)} & = & \displaystyle -q_{r}n_{0}\unicode[STIX]{x1D6F7}v_{\text{th}}\text{sign}(k_{\Vert })\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}\left[2\unicode[STIX]{x1D701}^{3}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}-2\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.200) $$\begin{eqnarray}\displaystyle r^{(1)} & = & \displaystyle -q_{r}\frac{n_{0}}{2}v_{\text{th}}^{2}\unicode[STIX]{x1D6F7}\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}\left[4\unicode[STIX]{x1D701}^{4}-2i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+2\unicode[STIX]{x1D701}^{2}-3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}+3\right];\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.201) $$\begin{eqnarray}\displaystyle \widetilde{r}^{(1)} & = & \displaystyle -q_{r}\frac{n_{0}}{2}v_{\text{th}}^{2}\unicode[STIX]{x1D6F7}\frac{1}{1-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}}\left[4\unicode[STIX]{x1D701}^{4}-2i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}-10\unicode[STIX]{x1D701}^{2}+3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right].\end{eqnarray}$$

We are looking for a closure, and we want to express either $q^{(1)}$ or $\widetilde{r}^{(1)}$ , as a linear combination of lower-order moments. To immediately see possible closures, it is always useful to pull the denominator of the Padé approximant out (as done above), and concentrate only on the expressions inside the big brackets. Also, similarly to the closures explored for small $\unicode[STIX]{x1D701}$ , it is useful to forget the $n^{(1)}$ , $p^{(1)}$ and $r^{(1)}$ moments, and just concentrate at the $u^{(1)}$ , $T^{(1)},q^{(1)}$ and $\widetilde{r}^{(1)}$ moments. Nevertheless, we will keep the $n^{(1)}$ moment, since it helps us to understand the expressions and to somehow ‘maintain touch with reality’.

Figure 4. The $\%$ error of the imaginary parts of $R_{3,0}(\unicode[STIX]{x1D701})$ (red line), and $R_{3,1}(\unicode[STIX]{x1D701})$ (green line).

By exploring the expressions inside of the brackets, it is obvious that it is impossible to express $q^{(1)}$ or $\widetilde{r}^{(1)}$ as a linear combination of lower-order moments that eliminate the $\unicode[STIX]{x1D701}$ dependence. Moreover, for large $\unicode[STIX]{x1D701}$ , the moments $q^{(1)}\sim \unicode[STIX]{x1D701}^{2}$ and $\widetilde{r}^{(1)}\sim \unicode[STIX]{x1D701}^{3}$ , which does not make physical sense, since these quantities should converge to zero with increasing $\unicode[STIX]{x1D701}$ , as explored in the $\unicode[STIX]{x1D701}\gg 1$ limit, see equations (3.80)–(3.86). The $R_{1}(\unicode[STIX]{x1D701})$ approximant therefore does not yield a closure. The same conclusion is obtained by using the $R_{2,0}(\unicode[STIX]{x1D701})$ approximant, where no closure for $q^{(1)}$ or $\widetilde{r}^{(1)}$ is possible. We note that the approximant $R_{2,1}(\unicode[STIX]{x1D701})$ , that was eliminated because it is not a good approximant of $R(\unicode[STIX]{x1D701})$ yields a closure $q^{(1)}=-2p_{0}u^{(1)}$ , which is equivalent to the closure (3.49), that was obtained for small $\unicode[STIX]{x1D701}$ with the precision $o(\unicode[STIX]{x1D701})$ . This closure is therefore disregarded.

Let us try the three-pole Padé approximants. The moments with $R_{3,1}(\unicode[STIX]{x1D701})$ approximant are proportional to

(3.202) $$\begin{eqnarray}\displaystyle R_{3,1}(\unicode[STIX]{x1D701}):\quad n^{(1)} & {\sim} & \displaystyle \frac{1}{1-{\displaystyle \frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{3}}\left[1+\frac{i}{\sqrt{\unicode[STIX]{x03C0}}}(\unicode[STIX]{x03C0}-4)\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.203) $$\begin{eqnarray}\displaystyle u^{(1)} & {\sim} & \displaystyle \frac{1}{1-{\displaystyle \frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{3}}\left[\frac{i}{\sqrt{\unicode[STIX]{x03C0}}}(\unicode[STIX]{x03C0}-4)\unicode[STIX]{x1D701}^{2}+\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.204) $$\begin{eqnarray}\displaystyle T^{(1)} & {\sim} & \displaystyle \frac{1}{1-{\displaystyle \frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{3}}\left[-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.205) $$\begin{eqnarray}\displaystyle q^{(1)} & {\sim} & \displaystyle \frac{1}{1-{\displaystyle \frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{3}}\left[-\frac{i}{\sqrt{\unicode[STIX]{x03C0}}}(3\unicode[STIX]{x03C0}-8)\unicode[STIX]{x1D701}^{2}-2\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.206) $$\begin{eqnarray}\displaystyle \widetilde{r}^{(1)} & {\sim} & \displaystyle {\displaystyle \frac{1}{1-{\displaystyle \frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i{\displaystyle \frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}}\unicode[STIX]{x1D701}^{3}}}\left[-\frac{2i}{\sqrt{\unicode[STIX]{x03C0}}}(3\unicode[STIX]{x03C0}-8)\unicode[STIX]{x1D701}^{3}-4\unicode[STIX]{x1D701}^{2}+3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right],\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where we have suppressed writing all the multiplicative factors including the minus signs (it does not mean that these were neglected, full expressions are considered, we are just not writing down the full expressions, which helps in spotting the possible closures). There is a possibility of expressing $q^{(1)}$ through the combination of the lower moments $T^{(1)}$ and $u^{(1)}$ . The full expressions of these moments are

(3.207) $$\begin{eqnarray}\displaystyle R_{3,1}(\unicode[STIX]{x1D701}):\quad D & = & \displaystyle \left(1-\frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+2i\frac{4-\unicode[STIX]{x03C0}}{\sqrt{\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D701}^{3}\right);\end{eqnarray}$$

(3.208) $$\begin{eqnarray}\displaystyle u^{(1)} & = & \displaystyle -\frac{q_{r}}{T^{(0)}}\unicode[STIX]{x1D6F7}v_{\text{th}}\text{sign}(k_{\Vert })\frac{1}{D}\left[\frac{i}{\sqrt{\unicode[STIX]{x03C0}}}(\unicode[STIX]{x03C0}-4)\unicode[STIX]{x1D701}^{2}+\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.209) $$\begin{eqnarray}\displaystyle T^{(1)} & = & \displaystyle -q_{r}\unicode[STIX]{x1D6F7}\frac{1}{D}\left[-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.210) $$\begin{eqnarray}\displaystyle q^{(1)} & = & \displaystyle -q_{r}n_{0}\unicode[STIX]{x1D6F7}v_{\text{th}}\text{sign}(k_{\Vert })\frac{1}{D}\left[-\frac{i}{\sqrt{\unicode[STIX]{x03C0}}}(3\unicode[STIX]{x03C0}-8)\unicode[STIX]{x1D701}^{2}-2\unicode[STIX]{x1D701}\right],\end{eqnarray}$$

where we have used a convenient notation $D$ for the denominator of the plasma response function, and the closure is

(3.211)

which is equivalent to the (3.60) closure (which was obtained for small $\unicode[STIX]{x1D701}$ with the precision $o(\unicode[STIX]{x1D701}^{2})$ ). Continuing with the next approximant $R_{3,2}(\unicode[STIX]{x1D701})$ , the moments calculate as

(3.212) $$\begin{eqnarray}\displaystyle R_{3,2}(\unicode[STIX]{x1D701}):\quad D & = & \displaystyle \left(1-\frac{3i\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}-2\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}\right);\end{eqnarray}$$

(3.213) $$\begin{eqnarray}\displaystyle n^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[1-\frac{i\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.214) $$\begin{eqnarray}\displaystyle u^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-\frac{i\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}^{2}+\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.215) $$\begin{eqnarray}\displaystyle T^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.216) $$\begin{eqnarray}\displaystyle q^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-2\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.217) $$\begin{eqnarray}\displaystyle \widetilde{r}^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-4\unicode[STIX]{x1D701}^{2}+3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right].\end{eqnarray}$$

It is possible to express (1)  $q^{(1)}$ through $T^{(1)}$ ; (2)  $\widetilde{r}^{(1)}$ through the combination of $u^{(1)}$ and $q^{(1)}$ ; (3)  $\widetilde{r}^{(1)}$ through the combination of $u^{(1)}$ and $T^{(1)}$ . The first choice yields a closure

(3.218)

that is equivalent to the (3.48) closure obtained for small $\unicode[STIX]{x1D701}$ with the precision $o(\unicode[STIX]{x1D701})$ . This is indeed the famous simplest possible Landau fluid closure that expresses the collisionless heat flux with respect to temperature, and it equivalent to equation (7) of Hammett & Perkins (Reference Hammett and Perkins1990).Footnote ⁸ The closure is written here in Fourier space. The important part is the $i\text{sign}(k_{\Vert })$ that typically written as $ik_{\Vert }/|k_{\Vert }|$ , and that in real space rewrites as a Hilbert transform, which we will address later. The $R_{3,2}(\unicode[STIX]{x1D701})$ was obtained with $o(\unicode[STIX]{x1D701})$ power-series expansion, and $o(1/\unicode[STIX]{x1D701}^{4})$ asymptotic-series expansion. How good is this closure? By exploring expressions (3.212)–(3.216), the quantities $n^{(1)}$ , $u^{(1)}$ , $T^{(1)}$ have all correct asymptotic expansion for large $\unicode[STIX]{x1D701}$ (including the proportionality constants), however, the heat flux decreases only as $q^{(1)}\sim 1/\unicode[STIX]{x1D701}^{2}$ instead of the correct ${\sim}1/\unicode[STIX]{x1D701}^{3}$ , see (3.84). For large $\unicode[STIX]{x1D701}$ , the heat flux is therefore overestimated by this simple closure, which typically leads to an overestimation of the Landau damping in fluid models that use this simplest closure. Nevertheless, the closure is very beneficial because it clarifies the distinction between the collisional and collisionless heat fluxes.

The other two possible closures with $R_{3,2}(\unicode[STIX]{x1D701})$ are

(3.219) $$\begin{eqnarray}\displaystyle R_{3,2}(\unicode[STIX]{x1D701}):\quad \widetilde{r}^{(1)} & = & \displaystyle -\frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}v_{\text{th}}n_{0}T^{(0)}\text{sign}(k_{\Vert })u^{(1)}-i\frac{(3\unicode[STIX]{x03C0}+8)}{4\sqrt{\unicode[STIX]{x03C0}}}v_{\text{th}}\text{sign}(k_{\Vert })q^{(1)};\end{eqnarray}$$

(3.220) $$\begin{eqnarray}\displaystyle \widetilde{r}^{(1)} & = & \displaystyle -\frac{4i}{\sqrt{\unicode[STIX]{x03C0}}}v_{\text{th}}n_{0}T^{(0)}\text{sign}(k_{\Vert })u^{(1)}-\frac{(3\unicode[STIX]{x03C0}+8)}{2\unicode[STIX]{x03C0}}v_{\text{th}}^{2}n_{0}T^{(1)},\end{eqnarray}$$

and one can go from (3.219) to (3.220) by using (3.218). Obviously, it would be also possible to construct a closure $\widetilde{r}^{(1)}=\unicode[STIX]{x1D6FC}_{u}u^{(1)}+\unicode[STIX]{x1D6FC}_{q}q^{(1)}+\unicode[STIX]{x1D6FC}_{T}T^{(1)}$ , where $\unicode[STIX]{x1D6FC}_{u}=-(4i/\sqrt{\unicode[STIX]{x03C0}})v_{\text{th}}n_{0}T^{(0)}\text{sign}(k_{\Vert })$ , and where $\unicode[STIX]{x1D6FC}_{q}$ and $\unicode[STIX]{x1D6FC}_{T}$ are related by satisfying $2n_{0}v_{\text{th}}\text{sign}(k_{\Vert })\unicode[STIX]{x1D6FC}_{q}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D6FC}_{T}=-iv_{\text{th}}^{2}n_{0}((3\unicode[STIX]{x03C0}+8)/2\sqrt{\unicode[STIX]{x03C0}})$ , i.e. one could consider a closure with a free parameter, which we will not consider. Additionally, all constructed closures should be checked with respect to obtained dispersion relations, and closures (3.219), (3.220) will be later disregarded as not well behaved (see the discussion below equations (3.242), (3.382) and (3.399)).

For $R_{4,2}(\unicode[STIX]{x1D701})$ , the kinetic moments calculate as

(3.221) $$\begin{eqnarray}\displaystyle R_{4,2}(\unicode[STIX]{x1D701}):\quad D & = & \displaystyle \left(1-i\sqrt{\unicode[STIX]{x03C0}}\frac{2}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}-\frac{(32-9\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{2}\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,i\sqrt{\unicode[STIX]{x03C0}}\frac{2(10-3\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{3}+\frac{2(16-5\unicode[STIX]{x03C0})}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{4}\right);\end{eqnarray}$$

(3.222) $$\begin{eqnarray}\displaystyle n^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[\frac{(5\unicode[STIX]{x03C0}-16)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\frac{(3\unicode[STIX]{x03C0}-10)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}+1\right];\end{eqnarray}$$

(3.223) $$\begin{eqnarray}\displaystyle u^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[\frac{(5\unicode[STIX]{x03C0}-16)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{3}+i\sqrt{\unicode[STIX]{x03C0}}\frac{(3\unicode[STIX]{x03C0}-10)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{2}+\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.224) $$\begin{eqnarray}\displaystyle T^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[\frac{2(5\unicode[STIX]{x03C0}-16)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.225) $$\begin{eqnarray}\displaystyle q^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-i\sqrt{\unicode[STIX]{x03C0}}\frac{(9\unicode[STIX]{x03C0}-28)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{2}-2\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.226) $$\begin{eqnarray}\displaystyle \widetilde{r}^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-i\sqrt{\unicode[STIX]{x03C0}}\frac{2(9\unicode[STIX]{x03C0}-28)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{3}-\frac{2(21\unicode[STIX]{x03C0}-64)}{(3\unicode[STIX]{x03C0}-8)}\unicode[STIX]{x1D701}^{2}+3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right].\end{eqnarray}$$

The only possibility is to express $\widetilde{r}^{(1)}$ through a combination of $u^{(1)}$ , $T^{(1)}$ and $q^{(1)}$ , and the solution is

(3.227) $$\begin{eqnarray}\displaystyle & \displaystyle R_{4,2}(\unicode[STIX]{x1D701}):\quad \widetilde{r}^{(1)}=-i\sqrt{\unicode[STIX]{x03C0}}\frac{(10-3\unicode[STIX]{x03C0})}{(16-5\unicode[STIX]{x03C0})}v_{\text{th}}\text{sign}(k_{\Vert })q^{(1)}+\frac{(21\unicode[STIX]{x03C0}-64)}{2(16-5\unicode[STIX]{x03C0})}v_{\text{th}}^{2}n_{0}T^{(1)} & \displaystyle \nonumber\\ \displaystyle & \displaystyle \qquad \quad +\,i\sqrt{\unicode[STIX]{x03C0}}\frac{(9\unicode[STIX]{x03C0}-28)}{(16-5\unicode[STIX]{x03C0})}v_{\text{th}}T^{(0)}n_{0}\text{sign}(k_{\Vert })u^{(1)}, & \displaystyle\end{eqnarray}$$

which is equivalent to the closure (3.64), that was obtained for small $\unicode[STIX]{x1D701}$ with precision $o(\unicode[STIX]{x1D701}^{3})$ . Obviously such a closure is precise for small values of $\unicode[STIX]{x1D701}$ , however for large values of $\unicode[STIX]{x1D701}$ , the asymptotic behaviour of $q^{(1)}\sim \unicode[STIX]{x1D701}^{-2}$ and $\widetilde{r}^{(1)}\sim \unicode[STIX]{x1D701}^{-1}$ , instead of the correct $\unicode[STIX]{x1D701}^{-3},\unicode[STIX]{x1D701}^{-4}$ profiles (see (3.84), (3.86)), and these quantities will be therefore overestimated. Nevertheless, the solution is interesting and we are not aware of it being reporting in any literature.

Continuing with $R_{4,3}(\unicode[STIX]{x1D701})$ , the kinetic moments calculate as

(3.228) $$\begin{eqnarray}\displaystyle R_{4,3}(\unicode[STIX]{x1D701}):\quad D & = & \displaystyle \left(1-i\frac{3\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}-\frac{(9\unicode[STIX]{x03C0}-16)}{4}\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\frac{(3\unicode[STIX]{x03C0}-8)}{2}\unicode[STIX]{x1D701}^{4}\right);\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.229) $$\begin{eqnarray}\displaystyle n^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[\frac{(8-3\unicode[STIX]{x03C0})}{4}\unicode[STIX]{x1D701}^{2}-i\frac{\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}+1\right];\end{eqnarray}$$

(3.230) $$\begin{eqnarray}\displaystyle u^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[\frac{(8-3\unicode[STIX]{x03C0})}{4}\unicode[STIX]{x1D701}^{3}-i\frac{\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}^{2}+\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.231) $$\begin{eqnarray}\displaystyle T^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[\frac{(8-3\unicode[STIX]{x03C0})}{2}\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.232) $$\begin{eqnarray}\displaystyle q^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-2\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.233) $$\begin{eqnarray}\displaystyle \widetilde{r}^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[\frac{(9\unicode[STIX]{x03C0}-32)}{2}\unicode[STIX]{x1D701}^{2}+3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right].\end{eqnarray}$$

It is possible to express $\widetilde{r}^{(1)}$ through the combination of $q^{(1)}$ and $T^{(1)}$ and the result is

(3.234)

which is equivalent to the closure (3.56), that was obtained for small $\unicode[STIX]{x1D701}$ with the precision $o(\unicode[STIX]{x1D701}^{2})$ . The heat flux has a correct asymptotic behaviour $q^{(1)}\sim \unicode[STIX]{x1D701}^{-3}$ (even though with incorrect proportionality constant), and the quantity $\widetilde{r}^{(1)}\sim \unicode[STIX]{x1D701}^{-2}$ instead of the correct ${\sim}\unicode[STIX]{x1D701}^{-4}$ . The closure was first reported by Hammett & Perkins (Reference Hammett and Perkins1990), and is equivalent to the (non-numbered) expression between their equations (10) and (11).

Continuing with $R_{4,4}(\unicode[STIX]{x1D701})$ approximant, the kinetic moments are (let us stop writing down $n^{(1)}$ from now on since we know we can get it from $u^{(1)}$ )

(3.235) $$\begin{eqnarray}\displaystyle R_{4,4}(\unicode[STIX]{x1D701}):\quad D & = & \displaystyle \left(1-i\frac{3\sqrt{\unicode[STIX]{x03C0}}}{2}\unicode[STIX]{x1D701}-4\unicode[STIX]{x1D701}^{2}+i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{3}+\frac{4}{3}\unicode[STIX]{x1D701}^{4}\right);\end{eqnarray}$$

(3.236) $$\begin{eqnarray}\displaystyle u^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-\frac{2}{3}\unicode[STIX]{x1D701}^{3}-\frac{i}{2}\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}^{2}+\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.237) $$\begin{eqnarray}\displaystyle T^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-\frac{4}{3}\unicode[STIX]{x1D701}^{2}-i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.238) $$\begin{eqnarray}\displaystyle q^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[-2\unicode[STIX]{x1D701}\right];\end{eqnarray}$$

(3.239) $$\begin{eqnarray}\displaystyle \widetilde{r}^{(1)} & {\sim} & \displaystyle \frac{1}{D}\left[3i\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D701}\right].\end{eqnarray}$$

It is possible to express $\widetilde{r}^{(1)}$ through $q^{(1)}$ and the closure is

(3.240)

The result is equivalent to the (3.52) closure, that was obtained for small $\unicode[STIX]{x1D701}$ with precision $o(\unicode[STIX]{x1D701})$ . This very simple closure has only precision $o(\unicode[STIX]{x1D701})$ , however, it does have the correct asymptotic behaviour of the heat flux $q^{(1)}\sim -3/(2\unicode[STIX]{x1D701}^{3})$ (including the proportionality constant), and $\widetilde{r}^{(1)}\sim \unicode[STIX]{x1D701}^{-3}$ that is closer to the correct $\unicode[STIX]{x1D701}^{-4}$ than the previous closure.

3.7 Table of moments $(u_{\Vert },T_{\Vert },q_{\Vert },\widetilde{r}_{\Vert \Vert })$ for various Padé approximants

To clearly see possibilities of a closure, it is useful to create the following summarizing table, that is self-explanatory after reading the previous section, i.e. all the proportionality constants (including the minus signs) are suppressed. Even though the table here is created for a 1-D geometry, we will see that exactly the same table is constructed for a 3-D geometry, where all the quantities are given a ‘parallel’ sub-index, i.e. $u^{(1)}\rightarrow u_{\Vert }^{(1)}$ , $T^{(1)}\rightarrow T_{\Vert }^{(1)}$ , $q^{(1)}\rightarrow q_{\Vert }^{(1)}$ and $\widetilde{r}^{(1)}\rightarrow \widetilde{r}_{\Vert \Vert }^{(1)}$ . The table is therefore useful to spot all the possible closures that can be constructed in a 1-D geometry for quantities $q^{(1)},\widetilde{r}^{(1)}$ , as well as in a 3-D geometry for quantities $q_{\Vert }^{(1)}$ and $r_{\Vert \Vert }^{(1)}$ . The approximants $R_{2,1},R_{4,5},R_{6,9}$ and $R_{8,13}$ are marked with an asterisk ‘*’. These approximants are not well behaved (because the Landau residue is not accounted for) and are provided only for completeness, these approximants should be disregarded.

It is obvious by now that any higher-order Padé approximants will not help to achieve a closure. Or is it? One might still hope for ‘a miracle’ thinking that perhaps the seven-pole and eight-pole approximants with the maximum possible number of poles devoted to the asymptotic series – $R_{7,10}$ and $R_{8,12}$ – might yield a closure. However, this is unfortunately not the case, and the table for seven-pole and eight-pole approximants reads

By observing the entire table, there are 7 possible quasi-static closures (that were already addressed),

(3.241)

There are also 13 time-dependent closures (that are addressed in the next section),

(3.242)

New closures should be always checked. Later on, we will consider propagation of the ion-acoustic mode, satisfying kinetic dispersion relation (3.366). We believe that a good ‘reliable’ closure of a fluid model obtained with the $R_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ approximant, should yield a fluid dispersion relation that is equivalent to (3.366), after $R(\unicode[STIX]{x1D701})$ is replaced with $R_{n,n^{\prime }}(\unicode[STIX]{x1D701})$ (equivalent to the numerator of (3.366) once both terms are written with the common denominator). Closures that satisfy this requirement are marked with ‘ ${\checkmark}$ ’ in the above table. Closures that do not satisfy this requirement were eliminated, and can be further split to two categories. Both eliminated categories actually appear to describe the ion-acoustic mode with the same accuracy as a corresponding ‘reliable’ closure satisfying (3.366), however, the difference is in the higher-order modes. The first category of eliminated closures, marked with ‘x’, produces higher-order modes with positive growth rate, and these closures cannot be used for numerical simulations. The second category, marked with ‘!’, produces higher-order modes that are damped, and these closures can still be useful. However, there is no guarantee that these closures will behave well in different circumstances (for example when used in a 3-D geometry) and these closures were therefore eliminated.

3.8 Going back from Fourier space to real space – the Hilbert transform

The quasi-static Landau fluid closures explored in the previous section were constructed in Fourier space. For direct numerical simulations that can use Fourier transforms (that are usually restricted to periodic boundaries), or for solving dispersion relations $\unicode[STIX]{x1D714}(\boldsymbol{k})$ , this is the easiest and natural way how to implement these closures. Nevertheless, it is very beneficial to see how these advanced fluid closures translate to real space.

Provided all equations are linear (and homogeneous), transformation between real and Fourier space is usually very easy and so far we just needed

(3.243) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\rightarrow -i\unicode[STIX]{x1D714};\quad \unicode[STIX]{x1D735}\rightarrow i\boldsymbol{k};\quad f(\boldsymbol{x},t)\rightarrow \hat{f}(\boldsymbol{k},\unicode[STIX]{x1D714}),\end{eqnarray}$$

where we did not even bother to write the hat symbol on the quantities in Fourier space, since it was obvious and not necessary.

With equations encountered in simple fluid models, transformation back to real space is easy and one can usually just flip the direction of the arrow in relations (3.243). However, the constructed Landau fluid closures contain an unusual operator $i\text{sign}(k_{\Vert })=ik_{\Vert }/|k_{\Vert }|$ . How does this operator transforms to real space? Considering spatial 1-D transformation between coordinates $z\leftrightarrow k_{\Vert }$ , a general function Fourier transforms according to

(3.244) $$\begin{eqnarray}\displaystyle & \displaystyle f(z)=\frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{\infty }\hat{f}(k_{\Vert })e^{ik_{\Vert }z}\,\text{d}k_{\Vert }\equiv {\mathcal{F}}^{-1}\hat{f}(k_{\Vert }); & \displaystyle\end{eqnarray}$$

(3.245) $$\begin{eqnarray}\displaystyle & \displaystyle \hat{f}(k_{\Vert })=\int _{-\infty }^{\infty }f(z)e^{-ik_{\Vert }z}\,\text{d}z\equiv {\mathcal{F}}f(z), & \displaystyle\end{eqnarray}$$

where the first equation is the inverse/backward Fourier transform and the second equation is the forward Fourier transform. As usual, we often do not bother to write the hat symbols on quantities in Fourier space. The location of the normalization factor $1/(2\unicode[STIX]{x03C0})$ is an ad hoc choice, but one has to be consistent, especially when calculating convolutions. As a first step, we need to calculate ${\mathcal{F}}^{-1}$ of a function $\text{sign}(k_{\Vert })$ . However, if such an integral is calculated directly, one will find out that the result is not clearly defined.

It is beneficial to use a small trick, where instead of a function $\text{sign}(k_{\Vert })$ , one considers the function $\text{sign}(k_{\Vert })e^{-\unicode[STIX]{x1D6FC}|k_{\Vert }|}$ , where $\unicode[STIX]{x1D6FC}$ is some small positive constant $\unicode[STIX]{x1D6FC}>0$ . And after the calculation, one performs the limit $\unicode[STIX]{x1D6FC}\rightarrow 0$ . The considered function is

(3.246) $$\begin{eqnarray}\text{sign}(k_{\Vert })e^{-\unicode[STIX]{x1D6FC}|k_{\Vert }|}=\left\{\begin{array}{@{}ll@{}}-e^{+\unicode[STIX]{x1D6FC}k_{\Vert }};\quad & k_{\Vert }<0;\\ +e^{-\unicode[STIX]{x1D6FC}k_{\Vert }};\quad & k_{\Vert }>0,\end{array}\right.\end{eqnarray}$$

and the integral calculates as

(3.247) $$\begin{eqnarray}\displaystyle \int _{-\infty }^{\infty }\text{sign}(k_{\Vert })e^{-\unicode[STIX]{x1D6FC}|k_{\Vert }|}e^{ik_{\Vert }z}\,\text{d}k_{\Vert } & = & \displaystyle \int _{-\infty }^{0}(-1)e^{+\unicode[STIX]{x1D6FC}k_{\Vert }}e^{ik_{\Vert }z}\,\text{d}k_{\Vert }+\int _{0}^{\infty }(+1)e^{-\unicode[STIX]{x1D6FC}k_{\Vert }}e^{ik_{\Vert }z}\,\text{d}k_{\Vert }\nonumber\\ \displaystyle & = & \displaystyle -\int _{-\infty }^{0}e^{(+\unicode[STIX]{x1D6FC}+iz)k_{\Vert }}\,\text{d}k_{\Vert }+\int _{0}^{\infty }e^{(-\unicode[STIX]{x1D6FC}+iz)k_{\Vert }}\,\text{d}k_{\Vert }\nonumber\\ \displaystyle & = & \displaystyle \left.-\frac{1}{(\unicode[STIX]{x1D6FC}+iz)}e^{(+\unicode[STIX]{x1D6FC}+iz)k_{\Vert }}\right|_{k_{\Vert }=-\infty }^{0}+\left.\frac{1}{(-\unicode[STIX]{x1D6FC}+iz)}e^{(-\unicode[STIX]{x1D6FC}+iz)k_{\Vert }}\right|_{k_{\Vert }=0}^{\infty }\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{(\unicode[STIX]{x1D6FC}+iz)}-\frac{1}{(-\unicode[STIX]{x1D6FC}+iz)}\nonumber\\ \displaystyle & = & \displaystyle \frac{-(\unicode[STIX]{x1D6FC}-iz)+(\unicode[STIX]{x1D6FC}+iz)}{(\unicode[STIX]{x1D6FC}+iz)(\unicode[STIX]{x1D6FC}-iz)}=\frac{2iz}{\unicode[STIX]{x1D6FC}^{2}+z^{2}},\end{eqnarray}$$

further yielding

(3.248) $$\begin{eqnarray}{\mathcal{F}}^{-1}\left[i\,\text{sign}(k_{\Vert })e^{-\unicode[STIX]{x1D6FC}|k_{\Vert }|}\right]=\frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{\infty }i\,\text{sign}(k_{\Vert })e^{-\unicode[STIX]{x1D6FC}|k_{\Vert }|}e^{ik_{\Vert }z}\,\text{d}k_{\Vert }=-\frac{z}{\unicode[STIX]{x03C0}(\unicode[STIX]{x1D6FC}^{2}+z^{2})}.\end{eqnarray}$$