2. Electron response in the presence of a hole

In the frame moving with the electrons, assume there exists a (Sagdeev Reference Sagdeev1966) soliton or (Schamel Reference Schamel1972) electron hole generating potential moving at speed V of the form

(2.1) \begin{align}\phi =\phi _{0}\, \mathrm{sec}h^{2}{x}=\phi _{0}/\mathrm{cos}h^{2}{x},\end{align}

where ${x}={k}({z}-{Vt})$ is the phase, ${k}\gt 0$ is the wavenumber and $\phi _{0}\gt 0$ . If the speed of the potential is greater than the electron thermal speed, ${V}\gt {v}_{{e}}=(2{T}/{m})^{1/2}$ , there is the possibility of a long-lived electron island containing bound or reflecting or librating electrons in a weakly collisional plasma in the presence of a finite electric field ${E}=-\partial \phi /\partial {x}$ , where T and m are the electron temperature and mass. The electron hole requires $\partial ^{2}\phi /\partial {x}^{2}\lt 0$ in a small region $x^{2}\lesssim 1$ about x = 0, implying there is a deficit of electrons associated with the hole (Hutchinson Reference Hutchinson2017, Reference Hutchinson2024). A nearly steady state solution to Poisson’s equation, $\partial {E}/\partial {x}=4\pi {e}({Zn}_{{i}}-{n})$ , must be found to sustain a hole for a very long time, where n and ${n}_{{i}}$ are the electron and ion densities, e is the proton charge and Z the ion charge number.

As Zakharov & Karpman (Reference Zakharov and Karpman1963) and, more recently, Catto (Reference Catto2025a , Reference Cattob ), have shown, there is no collisionless limit for a finite amplitude plasma wave when the Coulomb logarithm is large. As a result, Landau damping is actually a collisional process even though the collision frequency does not appear in the damping rate. The absence of the explicit appearance of the collision frequency in what is a collisional calculation is often referred to as a plateau regime (as in neoclassical theory of a large aspect ratio tokamak), and when there is a well defined resonance, it is often referred to as a resonant plateau process. As Landau damping would be expected to result in rather strong damping of an electron hole and require a collisional velocity space boundary layer wider than a velocity space island width it is not considered in detail here. Instead, the calculation here is performed in the extremely small collisionality limit by assuming a finite amplitude, electron hole is established by some perturbing event and its associated velocity space island width is much larger than any collisional boundary layer width. Once the hole exists the question of how long it continues to exist determines whether it has the desired long lifetime property of an electron hole. The calculation performed here evaluates the damping rate of a phase space island that allows electron hole formation, to demonstrate that an electron hole is able to persist for a very long time in the weak collisionality, finite amplitude limit, and that collisions provide a meaningful estimate of its lifetime.

Assuming the lowest order electron distribution function is Maxwellian

(2.2) \begin{align}{f}_{0}={n}_{0}({m}/2\pi {T})^{3/2}{e}^{-{m}{{v}^{2}}/2{T}}={n}_{0}\pi ^{-3/2}{{v}}_{{e}}^{-3}{e}^{-{{v^{2}}}/{{v}}_{{e}}^{2}},\end{align}

with ${n}_{0}$ the constant unperturbed electron density, and taking ${f}={f}_{0}+{f}_{1}$ with ${f}_{1}\ll {f}_{0}$ , but allowing $\partial {f}_{1}/\partial {v}_{| | }\sim \partial {f}_{0}/\partial {v}_{| | }$ , then ${f}_{1}$ must satisfy the nonlinear kinetic equation

(2.3) \begin{align}\partial {f}_{1}/\partial {t}+{v}_{| | }\partial {f}_{1}/\partial {z}+({e}/{m})(\partial \phi /\partial {z})\partial ({f}_{0}+{f}_{1})/\partial {v}_{| | }={C}\{{f}_{1}\}={\nu {v}}_{\bot {z}}^{2}\partial ^{2}{f}_{1}/\partial {{v}}_{| | }^{2},\end{align}

where $\nu =\nu _{{e}}{{v}}_{{e}}^{3}/2{v}^{3}$ , $\nu _{{e}}=3\sqrt{\pi }({Z}+1)\nu _{{ee}}/4$ , $\nu _{{ee}}=4\sqrt{2\pi }{e}^{4}{n}_{0}\ell{n}{\Lambda} /3{m}^{1/2}{T}^{3/2}=\nu _{{ei}}/{Z}$ and ${{v}}_{\bot {z}}^{2}={{v}}_{\bot }^{2}+{{v}}_{{e}}^{2}/({Z}+1)$ . The velocity ${v}_{| | }$ is in the direction of z, with ${{v}}_{\bot }^{2}={v}^{2}{-{v}}_{| | }^{2}$ . The precise form of the diffusive collision operator is unimportant until the dissipation is evaluated. At that point only pitch angle and energy scattering by the ions and electrons as in (2.3) will matter. The weak collisions occur in a very narrow collisional boundary layer and act to oppose any phase space vortex structure formation inside the localised phase space island.

To clarify the limit of interest, notice there are two velocity space scales in a weakly collisional plasma when ${f}_{1}({z},{v}_{| | },{t})={f}_{1}({x},{v}_{| | })$ is assumed. The first is a linear regime in which nonlinearities are unimportant and collisions resolve the singularity associated with the resonance in the kinetic equation. Balancing $\partial {f}_{1}/\partial {t}+$ ${v}_{| | }\partial {f}_{1}/\partial {z}={k}({v}_{| | }-{V})\partial {f}_{1}/\partial {x} \sim {kf}_{1}\Delta {v}_{| | }$ with collisions, by letting $(\Delta {v}_{| | })_{\nu }={v}_{| | }-{V}$ and estimating $\partial ^{2}{f}_{1}/\partial {{v}}_{| | }^{2}\sim {f}_{1}/(\Delta {v}_{| | }{)}_{\nu }^{2}$ and $\partial {f}_{1}/\partial {x}\sim {f}_{1}$ , gives a normalised collisional boundary layer velocity space width of

(2.4) \begin{align}(\Delta {v}_{| | })_{\nu }/{v}_{{e}}\approx (\nu /{kv}_{{e}})^{1/3}\ll 1,\end{align}

with ${{v}}_{\bot {z}}^{2}\sim {{v}}_{{e}}^{2}$ . The effective collision frequency is then

(2.5) \begin{align}\nu _{{eff}}\approx \nu {{v}}_{{e}}^{2}/(\Delta {v}_{| | }{)}_{\nu }^{2}\gg \nu.\end{align}

The second velocity space scale is defined by letting $(\Delta {v}_{| | })_{{is}}={v}_{| | }-{V}$ and balancing the nonlinear term, $e{k}\phi _{0}{f}_{1}/{m}(\Delta {v}_{| | })_{{is}}$ , with streaming to obtain the normalised island width

(2.6) \begin{align}(\Delta {v}_{| | })_{{is}}/{v}_{{e}}\approx ({e}\phi _{0}/{T})^{1/2}\ll 1.\end{align}

To allow a localised island to exist the treatment presented here focuses on the very weakly collisional limit in which the island width is much larger than the collisional boundary layer

(2.7) \begin{align}1\gg ({e}\phi _{0}/{T})^{1/2}\gg (\nu /{kv}_{{e}})^{1/3}.\end{align}

In this limit the separatrix enclosing the phase space island is surrounded by a narrow collisional boundary layer and no fine scale velocity space structure is allowed for the bound electrons inside the island separatrix.

Far from the island centred at ${v}_{| | }={V}$ the outer unbound response is found from

(2.8) \begin{align}{u}\partial {f}_{1}/\partial {x}+({e}/{m})(\partial \phi /\partial {x})\partial {f}_{0}/\partial {v}_{| | }=0,\end{align}

where ${u}={v}_{| | }-{V}$ , with the flow speed larger than the electron thermal speed, ${V}\gt {v}_{{e}}$ . The ${V}\gt {v}_{{e}}$ ordering allows a non-Maxwell Boltzmann electron response (not unlike a plasma wave). Solving using ${f}_{1}={{f}}_{1}^{{out}}+{{f}}_{2}^{{out}}+\cdots$ gives the outer solution away from the island to be

(2.9) \begin{align}{{f}}_{1}^{{out}} & =-\frac{{e}\phi }{{mu}}\frac{\partial {f}_{0}}{\partial {v}_{| | }}=-\frac{{e}\phi _{0}\,\mathrm{sec}h^{2}{x}}{{m}({v}_{| | }-{V})}\frac{\partial {f}_{0}}{\partial {v}_{| | }}\nonumber\\& =\frac{{e}\phi {v}_{| | }{f}_{0}}{{T}({v}_{| | }-{V})}\approx -\frac{{e}\phi {v}_{| | }{f}_{0}}{{TV}}\left[1+\frac{{v}_{| | }}{{V}}+\left(\frac{{v}_{| | }}{{V}}\right)^{2}+\left(\frac{{v}_{| | }}{{V}}\right)^{3}+\cdots \right],\end{align}

where here $\partial {f}_{0}/\partial {v}_{| | }=-{mv}_{| | }{f}_{0}/{T}$ . The next order outer response for ${v}_{{e}}\sim {v}_{| | }\lt {V}$ from

(2.10) \begin{align}\quad {u}\partial {{f}}_{2}^{{out}}/\partial {x}+({e}/{m})(\partial \phi /\partial {x})\partial {{f}}_{1}^{{out}}/\partial {v}_{| | }\approx {u}\partial /\partial {x}[{{f}}_{2}^{{out}}-{V\kern-4pt f}_{0}{e}^{2}\phi ^{2}/2{mT}u^{3}]\approx 0\end{align}

yields the lowest order result

(2.11) \begin{align}{{f}}_{2}^{{out}}\approx {V\kern-4pt f}_{0}({e}\phi )^{2}/2{mT\kern-2pt u}^{3}\approx -{f}_{0}({v}_{{e}}{e}\phi /{TV})^{2}/4,\end{align}

which needs to be retained in Poisson’s equation to balance the $\mathrm{sec}h^{4}{x}$ term from $(\partial ^{2}/\partial {x}^{2})\,\mathrm{sec}h^{2}{x}=-6\,\mathrm{sec}h^{4}{x}+4\,\mathrm{sec}h^{2}{x}$ .

In the island vicinity the full electron kinetic equation must be solved. To begin let

(2.12) \begin{align}{f}_{1}={g}({u},{x})-[{u}-\sigma ({e}\phi _{0}/{m})^{1/2}\alpha ]\partial {f}_{0}/\partial {v}_{| | }| _{u=0},\end{align}

where $\sigma ={u}/| {u}| =\pm 1$ or 0, $\alpha$ is a constant to be determined and $\partial {f}_{0}/\partial {v}_{| | }| _{u=0}\approx -{mV}\kern-3pt\bar{{f}}_{0}/{T}$ with $\bar{{f}}_{0}={f}_{0}({v}_{| | }={V})={n}_{0}\pi ^{-3/2}{{v}}_{{e}}^{-3}{e}^{-{({{v}}_{\bot }^{2}+{V}^{2}})/{{v}}_{{e}}^{2}}$ a constant to lowest order in the highly localised inner velocity space region containing the island having a width much bigger than that of the collisional boundary layer surrounding its separatrix. Substituting (2.12) into (2.3) leads to

(2.13) \begin{align}{ku}\ \partial {g}/\partial {x}+({ek}/{m})(\partial \phi /\partial {x})\partial {g}/\partial {u}={\nu {v}}_{\bot {z}}^{2}\partial ^{2}{g}/\partial {u}^{2}.\end{align}

Far from the island ${f}_{1}$ must go to (2.9), implying the inner solution for g must match to

(2.14) \begin{align}{g}\rightarrow \left[{u}-\sigma \alpha \left(\frac{{e}\phi _{0}}{{m}}\right)^{1/2}-\frac{{e}\phi _{0}}{{mu}}\,\mathrm{sec}h^{2}{x}\right]\frac{\partial {f}_{0}}{\partial {v}_{| | }}| _{u=0}.\end{align}

Defining

(2.15) \begin{align}{j}=({m}/{e}\phi _{0})^{1/2}{u}\end{align}

and

(2.16) \begin{align}\varDelta =\nu {k}^{-1}{{v}}_{\bot {z}}^{2}({m}/{e}\phi _{0})^{3/2}\approx (\Delta {v}_{| | }{)}_{\nu }^{3}/(\Delta {v}_{| | })_{{is}}^{3}\end{align}

then (2.3) becomes a similar equation to the one solved numerically by Hamilton et al. (Reference Hamilton, Tolman, Arzamasskiy and Duarte2023) for the motion associated with a galactic bar or spiral arm galaxy, namely

(2.17) \begin{align}{j}\frac{\partial {g}}{\partial {x}}+\frac{\partial \phi }{\partial {x}}\frac{\partial {g}}{\partial {j}}=\varDelta \frac{\partial ^{2}{g}}{\partial {j}^{2}},\end{align}

where $\varDelta \gt 0$ since ${{v}}_{\bot }^{2}\approx {v}^{2}-{V}^{2}\gt 0$ in ${{v}}_{\bot {z}}^{2}$ . Here, however, a localised non-periodic solution is essential as a solitary electron hole potential is being considered.

The steady state solutions of interest are skew symmetric so satisfy

(2.18) \begin{align}{g}({j},{x})=-{g}({-}{j},-{x}).\end{align}

An analytic treatment is possible for weak collisions satisfying $\varDelta \ll 1$ by taking advantage of the lowest order collisionless trajectories. Employing the reduced constant of the motion

(2.19) \begin{align}{h}({j},{x})={j}^{2}/2-\mathrm{sec}h^{2}{x}={j}^{2}/2-1/\mathrm{cos}h^{2}{x},\end{align}

leads to an island with a separatrix at ${h}=0$ in the phase space j and x, when $-1\leq {h}\lt 0$ and ${h}\gt 0$ are considered. The solution (2.14) far from the island becomes

(2.20) \begin{align}{g}\rightarrow \sigma \big(\sqrt{2{h}}-\alpha \big)\left({e}\phi _{0}/{m}\right)^{1/2}\partial {f}_{0}/\partial {v}_{| | }| _{u=0}.\end{align}

For the unbound,

(2.21) \begin{align}{j}=\pm \sqrt{2({h}+\mathrm{sec}h^{2}{x})},\end{align}

with ${h}+\mathrm{sec}h^{2}{x}\gt {h}\gt 0$ , while for the bound, a real ${j}$ requires $-1\leq {h}\lt 0$ . The separatrix at h = 0 is enclosed by a narrow collisional boundary layer separating the bound ( $-1 \lt h \lt 0$ ) and the unbound ( $h \gt 0$ ) electrons.

Using the constant of the motion h and x as variables, the kinetic equation becomes

(2.22) \begin{align}\left.\frac{\partial {g}}{\partial {x}}\right| _{{h}}=\varDelta \left.\frac{\partial }{\partial {h}}\right| _{{x}}\left({j}\left.\frac{\partial {g}}{\partial {h}}\right| _{{x}}\right).\end{align}

The weak collisionality $\varDelta \ll 1$ limit allows a solution of the form ${g}={g}_{1}({h})+{g}_{2}({h},{x})+\cdots$ to be found. To lowest order $\partial {g}_{1}/\partial {x}| _{{h}}= 0$ , while to next order

(2.23) \begin{align}\left.\frac{\partial {g}_{2}}{\partial {x}}\right| _{{h}}=\varDelta \left.\frac{\partial }{\partial {h}}\right| _{{x}}\left({j}\left.\frac{\partial {g}_{1}}{\partial {h}}\right| _{{x}}\right).\end{align}

Further progress is made by defining the x integral of j at fixed h as

(2.24) \begin{align}\left\langle {j}\right\rangle _{{h}}=\oint _{{h}}{\mathrm{d}xj},\end{align}

where for the unbound x is from $-\infty$ to $\infty$ , but for the bound it is over a complete bounce.

For both the bound and unbound, integration over x leads to the solubility constraint

(2.25) \begin{align}\left.\frac{\partial }{\partial {h}}\right| _{{x}}\left(\left\langle {j}\right\rangle _{{h}}\left.\frac{\partial {g}_{1}}{\partial {h}}\right| _{{x}}\right)=0,\end{align}

since

(2.26) \begin{align}{\int }_{-{\infty }}^{{\infty }}{dx}\left.\frac{\partial {g}_{2}}{\partial {x}}\right| _{{h}}=0.\end{align}

For the bound, the preceding is clearly satisfied since an integration over a full bounce returns to the same ${g}_{2}$ . For the unbound, skew symmetry is employed to find

(2.27) \begin{align}{g}_{2}\big({j} & =\sigma \sqrt{2{h}},{\infty }\big)-{g}_{2}\big({j}=\sigma \sqrt{2{h}},-{\infty }\big)={g}_{2}\big({j}=\sigma \sqrt{2{h}},{\infty }\big)\nonumber\\& \quad +{g}_{2}\big({-}{j}=-\sigma \sqrt{2{h}},{\infty }\big)=0.\end{align}

By requiring (2.27) to vanish, the unbound solution odd in ${j}$ avoids a jump in ${g}_{2}$ as | ${x}| \rightarrow {\infty }$ .

Of course, the inner solution ${g}_{1}$ must have the desired skew symmetry to match the outer solution far from the island where ${f}_{1}\propto 1/{u}\rightarrow 0$ . The solution for ${g}_{1}$ unconstrained by solubility would just be a version of a BGK mode (Bernstein et al. Reference Bernstein, Greene and Kruskal1957). But the constrained solution found next is no longer a typical piecewise continuous BGK mode because its form is determined by a collisional solubility constraint, whereas Bernstein et al. (Reference Bernstein, Greene and Kruskal1957) were careful to note that their treatment ignored collisions.

Integrating the solubility constraint once more gives

(2.28) \begin{align}\left\langle {j}\right\rangle _{{h}}\frac{\partial {g}_{1}}{\partial {h}}=\mathrm{constant},\end{align}

where the constant must be independent of h. Using integrals no. 2.261 and no. 2.266 from Gradshteyn & Ryzhik (Reference Gradshteyn and Ryzhik2000) the integrals in $\langle {j}\rangle _{{h}}$ are performed by letting ${y}=\mathrm{sec}h^{2}{x}$ .

Integrating from ${y}=-$ h to 0 to find $\langle {j}\rangle _{{h}\lt 0}=4\pi (\sqrt{2}-\sqrt{-2{h}})$ for the bound, leads to

(2.29) \begin{align} \left. 4\pi \big(\sqrt{2}-\sqrt{-2{h}}\big)\frac{\partial {g}_{1}}{\partial {h}}\right| _{{h}\lt 0}=\mathrm{constant},\end{align}

requiring the constant to vanish to avoid a singularity at h = –1, and implying ${g}_{1}$ is a constant at most. But at x = 0, ${g}({j},{x}=0)=-{g}({-}{j},-{x}=0)$ and a constant ${g}_{1}$ is not odd in j. Hence,

(2.30) \begin{align}{g}_{1}({h}\lt 0)=0,\end{align}

along with $\partial {g}_{1}/\partial {h}| _{{h}\lt 0}=0$ . In the steady state the skew symmetric vortex structure of the bound solution is dissipated by collisions. Phase mixing is not required to regularise the moments of ${f}_{1}$ by integrating over velocity space. The full bound solution is

(2.31) \begin{align}{f}_{1}\left({h}\lt 0\right)=-{u}\partial {f}_{0}/\partial {v}_{| | }| _{u=0}={m\kern-1pt V\kern-2pt u}\bar{{f}}_{0}/{T}.\end{align}

For the unbound, skew symmetry implies only x $\geq 0$ and $\sigma =+1$ need be considered. Again letting ${y}=\mathrm{sec}h^{2}{x}$ to perform the integral, but this time the range of interest is from x = 0 to a large ${x}={X}\gg 1$ at which ${y}={Y}\approx 4{e}^{-2{X}}\ll 1$ . Again, using no. 2.261 and no. 2.266 from Gradshteyn & Ryzhik (Reference Gradshteyn and Ryzhik2000)

(2.32) \begin{align}\left\langle {j}\right\rangle _{{h}\geq 0} & =\sqrt{2{h}} \left[2{X}+\ell{n}\left(\frac{{h}}{1+{h}}\right)\right]+\sqrt{2}\mathrm{sin}^{-1}\left(\frac{1-{h}}{1+{h}}\right)\nonumber\\& \quad +\frac{\pi }{\sqrt{2}}\rightarrow \begin{cases} \pi \sqrt{2}\\[3pt] 2{X}\sqrt{2{h}}+2/\sqrt{2{h}} \end{cases} \begin{array}{c} {h}=0,\\[3pt] {h}\gg 1. \end{array}\end{align}

Recall, far from the island the unbound solution is ${g}_{1}\rightarrow -{mV}({e}\phi _{0}/{m})^{1/2}(\sqrt{2{h}}-\alpha )\bar{{f}}_{0}/{T}$ , thereby giving $\left.\partial {g}_{1}/\partial {h}\right| _{{h}\gg 1}=-{mV}\left({e}\phi _{0}/{m}\right)^{1/2}\bar{{f}}_{0}/{T}\sqrt{2{h}}$ . Therefore,

(2.33) \begin{align}\left\langle {j}\right\rangle _{{h}}\left.\frac{\partial {g}_{1}}{\partial {h}}\right| _{{h}\gt 0}=\pi \sqrt{2}\left.\frac{\partial {g}_{1}}{\partial {h}}\right| _{{x},{h}=0}=-2{X}\frac{{mV}}{{T}}\left(\frac{{e}\phi _{0}}{{m}}\right)^{1/2}\bar{{f}}_{0}.\end{align}

As a result, for ${h}\geq 0$ , integrating from h = 0, where ${g}_{1}({h}=0)=0$ to match the bound, gives

(2.34) \begin{align}{g}_{1}({h})=-\sigma \frac{{mV}\kern-3pt\bar{{f}}_{0}}{{T}}\left(\frac{{e}\phi _{0}}{{m}}\right)^{1/2}\left[2{X}{\int }_{0}^{{h}}\frac{{{\rm d}h}}{\left\langle {j}\right\rangle _{{h}\geq 0}}\right]\underset{h\gg 1}{\rightarrow }-\sigma \frac{{mV}\kern-3pt\bar{{f}}_{0}}{{T}}\left(\frac{{e}\phi _{0}}{{m}}\right)^{1/2}[\sqrt{2{h}}-\alpha ({X})],\end{align}

where $\sigma$ is restored to display the proper skew symmetric behaviour of the solution, and

(2.35) \begin{align}\alpha ({X}) & ={\int }_{\kern-3pt 0}^{{\infty }}{{\rm d}h}\left(\frac{1}{\sqrt{2{h}}}-\frac{2{X}}{\left\langle {j}\right\rangle _{{h}\geq 0}}\right)\nonumber\\[5pt]& ={\int }_{\kern-3pt 0}^{{\infty }}{{\rm d}h}\frac{\sqrt{2{h}}{\mathrm{ln}}[{{h}}/({1+{h}})]+\sqrt{2}{\sin}^{-1}[({1-{h}})/({1+{h}})]+({\pi }/{\sqrt{2}})}{\sqrt{2{h}}\left\langle {j}\right\rangle _{{h}\geq 0}}\underset{{X}\rightarrow {\infty }}{\longrightarrow }0.\end{align}

The constant $\alpha$ is a function of X since as ${x}={X}\rightarrow {\infty }$ , $\alpha$ must vanish to allow g to become continuous. It is interesting to realise that the value of $\alpha ({X})$ is unimportant since it is only required to insure the proper behaviour at ${h}\gg 1$ . The dependence of $\alpha$ on ${X}$ is an artefact of the hole potential $\phi \propto \mathrm{sec}h^{2}{x}$ extending to infinity. An analytic solution allows ${X}\rightarrow {\infty }$ and $\alpha \rightarrow 0$ . However, a numerical solution has to pick an ${X}\gg 1$ and therefore has a finite $\alpha ({X})$ .

Using ${g}_{1}({h})$ from (2.35), the full unbound solution for ${f}_{1}$ about the island is

(2.36) \begin{align}{f}_{1}({h}\geq 0) & =\frac{{mV}\kern-3pt\bar{{f}}_{0}}{{T}}\left\{{u}-\sigma \left(\frac{{e}\phi _{0}}{{m}}\right)^{1/2}\left[\alpha ({X})+2{X}{\int }_{0}^{{h}}\frac{{{\rm d}h}}{\left\langle {j}\right\rangle _{{h}\geq 0}}\right]\right\}\underset{h\gg 1}{\longrightarrow }\nonumber\\[3pt]& \quad \frac{{mV}\kern-3pt\bar{{f}}_{0}}{{T}} \left[{u}-\sigma \left(\frac{{e}\phi _{0}}{{m}}\right)^{1/2}\!\!\sqrt{2{h}}\right]=\frac{{mV}\kern-3pt\bar{{f}}_{0}}{{T}}\big[{u}-{u}\sqrt{1-\big(2\,\mathrm{sec}h^{2}{x}/{j}^{2}\big)}\big]\approx \frac{{e}\phi {V}\bar{{f}}_{0}}{{Tu}}.\end{align}

Steps in ${f}_{1}$ (for finite X) and $\partial {g}_{1}/\partial {h}$ across the separatrix are allowed because it is enclosed by a thin collisional boundary layer that provides smooth matching. Piecewise continuous matching is adequate for the purposes here because the collisional solubility constraints (2.29) and (2.33) are satisfied – just as it is in the banana regime of neoclassical transport in a tokamak for which there are magnetically trapped and passing particles.

When the collisionality constraint is not imposed only $\partial {g}/\partial {x}| _{{h}}=0$ need be satisfied. Then, only unconstrained solutions for ${g}$ depending on h need to be considered. Taking $\alpha =0$ in the relation between ${f}_{1}$ and ${g}$ reduces (2.12) to

(2.37) \begin{align}{f}_{1}={g}({h})-{u}\partial {f}_{0}/\partial {v}_{| | }| _{{u}=0}={g}({h})+{MuV }\bar{{f}}_{0}/{T},\end{align}

with $\bar{{f}}_{0}={f}_{0}({v}_{| | }={V})$ . As a result, an allowed unconstrained piecewise continuous at the separatrix at ${h}=0$ that matches to the inner limit of the outer solution is

(2.38) \begin{align}{g}={g}({h})=\begin{cases} \qquad\quad\qquad 0 & \mathrm{bound }\ ( -1\lt {h}\lt 0),\\[4pt] \sigma \sqrt{2{h}}\left(\dfrac{{e}\phi _{0}}{{m}}\right)^{1/2}\dfrac{\partial {f}_{0}}{\partial {v}_{| | }}| _{{u}=0} & \mathrm{unbound}{\kern.5pt} \left({h}\gt 0\right). \end{cases}\end{align}

This leads to the unconstrained unbound and bound solutions

(2.39) \begin{align}{f}_{1}=-{u}\frac{\partial {f}_{0}}{\partial {v}_{| | }}| _{{u}=0}\begin{cases} \qquad\qquad\quad 1 & \mathrm{bound},\\ 1-\big(1-2\,\mathrm{sec}h^{2}{x}/{j}^{2}\big)^{1/2} & \mathrm{unbound}. \end{cases}\end{align}

The unbound unconstrained solution recovers the ${h}\gg 1$ limit of the collisionally constrained solution as in (2.36), but differs from the collisionally constrained solution as ${h}\rightarrow 0$ , that is, in the vicinity of the separatrix for the barely unbound. Its singular behaviour, $\partial ^{2}\sqrt{2{h}}/\partial {h}^{2}=-1/(2{h})^{3/2}$ , means collisions can never be neglected when the Coulomb logarithm is large.

3. Electron and ion response and Poisson’s equation

To determine the conditions under which the electron response self-consistently satisfies Poisson’s equation, both the electron and ion density are required. The skew symmetry of the inner solution means it does not perturb the density to all orders for both the correct, collisionally constrained solution, and the unphysical unconstrained solution (2.37)–(2.39). The electron density is only perturbed by the outer or away from the island electron response (that is, in the outer region, including the matching region, where the inner limit of the outer solution equals the outer limit of the inner solution). As the inner or island solution does not perturb the density, the electron density is simply

(3.1) \begin{align}{n}=\int {d}^{3}{v}\left({f}_{0}+{{f}}_{1}^{{out}}+{{f}}_{2}^{{out}}\right)={n}_{0}\left[1-\frac{{e}\phi }{{T}}\left(\frac{{v}_{{e}}}{{V}}\right)^{2}\left(1+\frac{3{{v}}_{{e}}^{2}}{2{V}^{2}}+\cdots \right)+\frac{1}{4}\left(\frac{{v}_{{e}}{e}\phi }{{VT}}\right)^{2}+\cdots \right],\nonumber\\[5pt]\end{align}

where the outer solution ${{f}}_{1}^{{out}}+{{f}}_{2}^{{out}}$ is valid away from the island and $\phi ^{2}$ terms are needed, unlike the standard sinusoidal island chain case considered by Zakharov & Karpman (Reference Zakharov and Karpman1963) and Catto (Reference Catto2025). The Poisson equation solution allows ${e}\phi _{0}/{T}\ll 1$ for a parameter range, thereby implying the self-consistency of the velocity space localised island model considered. The deficit of electrons needed to form a hole is not directly due to the island, but rather is from the matching region and beyond. The presence of the island allows a non-Maxwell Boltzmann electron response permitting an electron deficit, and thereby an electron hole, as will be shown in the remainder of this section. Numerical simulations (see figure 2 in Hutchinson Reference Hutchinson2017) form finite islands having a depleted electron population, but are unlikely to properly resolve the very narrow collisional boundary layer surrounding the separatrix to obtain the correct solubility constraint (2.25) and associated damping rate.

The ions are assumed to be stationary in the electron frame. Their lowest order distribution function is also assumed to be Maxwellian

(3.2) \begin{align}{f}_{0{i}}={n}_{{i}0}({M}/2\pi {T}_{{i}})^{3/2}{e}^{-{M}{{v}^{2}}/2{{T}_{{i}}}},\end{align}

to make the ion–ion collision operator vanish to lowest order, and with no significant resonant interactions occurring at higher order because ${V}\gt {v}_{{e}}\gg {v}_{{i}}=(2{T}_{{i}}/{M})^{1/2}$ . The constants ${n}_{{i}0}$ , ${T}_{{i}}$ and M are the ion density, temperature and mass. In the absence of a resonance, ion collisions are unimportant when calculating the corrections to ${f}_{0{i}}$ . Therefore, the lowest order perturbed ion response is the solution to

(3.3) \begin{align}{v}_{| | }[\partial {f}_{1{i}}/\partial {x}+({Ze}/{T}_{{i}})(\partial \phi /\partial {x}){f}_{0{i}}]=0.\end{align}

As ${v}_{| | }\neq 0$ , the ion response is Maxwell Boltzmann

(3.4) \begin{align}{f}_{1{i}}=-({Ze}\phi /{T}_{{i}}){f}_{0{i}}=-({Ze}\phi _{0}/{T}_{{i}}){f}_{0{i}}\,\mathrm{sec}h^{2}{x}.\end{align}

Solving to next order by noticing ${f}_{1{i}}\propto {Ze}\phi /{T}\ll 1$ , then

(3.5) \begin{align}{v}_{| | }[\partial {f}_{2{i}}/\partial {x}+({Ze}/{T}_{{i}})(\partial \phi /\partial {x}){f}_{1{i}}]=0\end{align}

gives the next order Maxwell Boltzmann response

(3.6) \begin{align}{f}_{2{i}}=-({Ze}\phi /2{T}_{{i}}){f}_{1{i}}=({Ze}\phi /{T}_{{i}})^{2}{f}_{0{i}}/2=({Ze}\phi _{0}/{T}_{{i}})^{2}{f}_{0{i}}\,\mathrm{sec}h^{4}{x}/2.\end{align}

Consequently, the ions only experience the Maxwell Boltzmann response

(3.7) \begin{align}{n}_{{i}}({x})=\int {d}^{3}{v}({f}_{0{i}}+{f}_{1{i}}+{f}_{2{i}})={n}_{{i}0}\left[1-\frac{{Ze}\phi }{{T}_{{i}}}+\frac{1}{2}\left(\frac{{Ze}\phi }{{T}_{{i}}}\right)^{2}\right].\end{align}

As a check, notice that the ion distribution function can only depend on the constant of the motion ${Mv}^{2}/2+{Ze}\phi$ , requiring the Maxwell Boltzmann response ${n}_{{i}}={n}_{{i}0}{e}^{-{Ze}\phi /{{T}_{{i}}}}$ . Recall that keeping the $({Ze}\phi /{T}_{{i}})^{2}$ term is necessary to obtain the assumed potential (2.1) because the Laplacian acting on $\mathrm{sec}h^{2}{x}$ gives a $\mathrm{sec}h^{4}{x}$ term.

Taking the background plasma as quasineutral ( ${Zn}_{{i}0}={n}_{0}$ ), the lowest order Poisson equation only requires the non-resonant electron and ion responses (and is valid for both the collisionally constrained and unconstrained cases), and becomes

(3.8) \begin{align}\frac{\partial ^{2}\phi }{\partial {z}^{2}} & =-4\pi \rho =4\pi {e}\left({n}-{Zn}_{{i}}\right)\approx -\frac{\phi }{{\lambda }_{{D}}^{2}}\left[\frac{{{v}}_{{e}}^{2}}{{V}^{2}}\left(1+\frac{3{{v}}_{{e}}^{2}}{2{V}^{2}}\right)-\frac{{ZT}}{{T}_{{i}}}\right]\nonumber\\& \quad -\frac{{e}\phi ^{2}}{{2{T}\lambda }_{{D}}^{2}}\left(\frac{{Z}^{2}{T}^{2}}{{{T}}_{{i}}^{2}}+\frac{{{v}}_{{e}}^{2}}{2{V}^{2}}\right),\end{align}

with the Debye length defined by ${\lambda }_{{D}}^{2}={T}/4\pi {e}^{2}{n}_{0}$ . The last term need not be smaller than the first term on the right side. Inserting $\phi$ and using $(\partial ^{2}/\partial {x}^{2})\,\mathrm{sec}h^{2}{x}=-6\,\mathrm{sec}h^{4}{x}+4\,\mathrm{sec}h^{2}{x}=-2\,\mathrm{sec}h^{2}{x}(1-3\,\mathrm{tan}h^{2}{x})\underset{x\rightarrow 0}{\longrightarrow }-2$ , requires satisfying two conditions from

(3.9) \begin{align}\left[6{k}^{2}{\lambda }_{{D}}^{2}-\frac{{e}\phi _{0}}{2{T}}\left(\frac{{Z}^{2}{T}^{2}}{{{T}}_{{i}}^{2}}+\frac{{{v}}_{{e}}^{2}}{2{V}^{2}}\right)\right]\mathrm{sec}h^{4}{x}=\left[4{k}^{2}{\lambda }_{{D}}^{2}+\frac{{{v}}_{{e}}^{2}}{{V}^{2}}\left(1+\frac{3{{v}}_{{e}}^{2}}{2{V}^{2}}\right)-\frac{{ZT}}{{T}_{{i}}}\right]\mathrm{sec}h^{2}{x}.\end{align}

Self-consistency requires both sides vanish separately giving

(3.10) \begin{align}4{k}^{2}{\lambda }_{{D}}^{2}={ZT}/{T}_{{i}}-\big({{v}}_{{e}}^{2}/{V}^{2}\big)\big(1+{3{v}}_{{e}}^{2}/2{V}^{2}\big)\gt 0\end{align}

and

(3.11) \begin{align}1\gg \frac{{e}\phi _{0}}{{T}}=\frac{12{k}^{2}{\lambda }_{{D}}^{2}}{{Z}^{2}{T}^{2}/{{T}}_{{i}}^{2}+{{v}}_{{e}}^{2}/2{V}^{2}}.\end{align}

If ${ZT}/{T}_{{i}}=1$ and ${{v}}_{{e}}^{2}/{V}^{2}=1/2$ , then ${e}\phi _{0}/{T}\approx 3/10$ and ${k}^{2}{\lambda }_{{D}}^{2}\approx 1/32$ . But for ${ZT}/{T}_{{i}}\gg 1$ , the ${Ze}\phi _{0}/{T}_{{i}}\ll 1$ assumption fails, as does ${e}\phi _{0}/{T}\ll 1$ if $1\gg {ZT}/{T}_{{i}}{\gg ({v}}_{{e}}^{2}/{V}^{2})(1+3{{v}}_{{e}}^{2}/2{V}^{2})$ . In any case, if ${ZT}/{T}_{{i}}\approx ({v}_{{e}}^{2}/{V}^{2})(1+3{{v}}_{{e}}^{2}/2{V}^{2})$ , small ${k}^{2}{\lambda }_{{D}}^{2}$ allows $({e}\phi _{0}/{T})^{2}$ terms to matter.

Consequently, the temperatures of interest are expected to be in the range

(3.12) \begin{align}1\geq \frac{{ZT}}{{T}_{{i}}}\approx \frac{{{v}}_{{e}}^{2}}{{V}^{2}}\left(1+\frac{3{{v}}_{{e}}^{2}}{2{V}^{2}}\right).\end{align}

For example, there are small ${e}\phi _{0}/{T}$ and ${k}^{2}{\lambda }_{{D}}^{2}$ solutions for ${ZT}/{T}_{{i}}=1/2$ if ${{v}}_{{e}}^{2}/{V}^{2}\lt 1/3$ , for ${ZT}/{T}_{{i}}=11/32$ if ${{v}}_{{e}}^{2}/{V}^{2}\lt 1/4$ , etc. The temperature restriction is not as rigid as it seems at first glance since there are exponentially few electrons at very large ${V}^{2}/{{v}}_{{e}}^{2}$ . Finally, the hole potential is only significant for ${x}^{2}\lesssim 1$ , implying the spatial extent of an electron hole normalised to a Debye length is $\Delta {z}/\lambda _{{D}}\sim 1/{k}\lambda _{{D}}\gg 1$ or many Debye lengths. The collisional velocity space boundary layer width as well as ${e}\phi _{0}/{T}$ are small since

(3.13) \begin{align}1\gg (\Delta {v}_{| | })_{{is}}/{v}_{{e}}\approx ({e}\phi _{0}/{T})^{1/2}\sim 12{k}^{2}{\lambda }_{{D}}^{2}/\big({Z}^{2}{T}^{2}/{{T}}_{{i}}^{2}+{{v}}_{{e}}^{2}/2{V}^{2}\big).\end{align}

4. Collisional dissipation in the boundary layer

To higher order there is some dissipation associated with the electron hole because the collisional electrons in the separatrix boundary layer heat the background plasma as the island slowly decays away, as will now be demonstrated. The power lost to the background plasma by the slowly decaying hole potential due to the dissipation in the collisional boundary layer surrounding the separatrix in a region of phase space extent $2{X}\gg 1$ is

(4.1) \begin{align}\left\langle {P}\right\rangle _{{h}}=-{e}\left\langle \frac{\partial \phi }{\partial {z}}\int {d}^{3}{vv}_{| | }{f}\right\rangle _{{h}}=-{ek}\left\langle \frac{\partial \phi }{\partial {z}}\int {d}^{3}{vv}_{| | }{f}\right\rangle _{{h}},\end{align}

where the subscript h on $\langle \ldots \rangle _{{h}}$ denotes it is held fixed to perform the x integration. The skew symmetry of the inner unbound region gives $\sigma \int {{\rm d}u}(\sqrt{2{h}}-\alpha ) =0$ and $\langle \int {{\rm d}u\ g}\rangle _{{h}}=0$ , leaving

(4.2) \begin{align}\left\langle {P}\right\rangle _{{h}}=-{ek}\left\langle \frac{\partial \phi }{\partial {x}}\int {d}^{3}{vu\ g}\right\rangle _{{h}}={ek}\left\langle \phi \int {d}^{3}{vu}\left.\frac{\partial {g}}{\partial {x}}\right| _{{h}}\right\rangle _{{h}}.\end{align}

Inserting the reduced kinetic equation yields

(4.3) \begin{align}\left\langle {P}\right\rangle _{{h}} & ={ek}\left(\frac{{e}\phi _{0}}{{m}}\right)^{{1}/{2}}\left\langle \phi \int {d}^{3}{vj}\varDelta \left.\frac{\partial }{\partial {h}}\right| _{{x}}\left({j}\left.\frac{\partial {g}}{\partial {h}}\right| _{{x}}\right)\right\rangle _{{h}}\nonumber\\[5pt]& =2\pi {ek}\frac{{e}\phi _{0}}{{m}}\left\langle \phi {\int }_{\kern-3.4pt 0}^{{\infty }}{{\rm d}v}_{\bot }{v}_{\bot }\varDelta {\int }_{\kern-4pt -{\infty }}^{{\infty }}{{\rm d}jj}\left.\frac{\partial }{\partial {h}}\right| _{{x}}\left({j}\left.\frac{\partial {g}}{\partial {h}}\right| _{{x}}\right)\right\rangle _{{h}}\nonumber\\[5pt]& =4\pi {k}\frac{{e}^{2}{\phi }_{0}^{2}}{{m}}{\int }_{\kern-3.4pt 0}^{{\infty }}{\mathrm{d}v}_{\bot }{v}_{\bot }\varDelta {\int }_{\kern-3.4pt 0}^{{\infty }}{\mathrm{d}h}\left.\frac{\partial }{\partial {h}}\right| _{{x}}\left\langle {j\sec}{h}^{2}{x}\right\rangle _{{h}}\frac{\partial {g}_{1}}{\partial {h}}.\end{align}

Then using

(4.4) \begin{align}{\int }_{\kern-3.4pt 0}^{{\infty }}{{\rm d}h}\frac{\partial }{\partial {h}}\bigg| _{{x}}(\langle {j\sec}{h}^{2}{x}\rangle _{{h}}\frac{\partial {g}_{1}}{\partial {h}}=\langle {j\sec}{h}^{2}{x}\rangle _{{h}}{\frac{\partial {g}_{1}}{\partial {h}}\bigg| }_{{h}=0}^{{h}\gg 1}\end{align}

and integrals no. 2.261 and no. 2.264 from Gradshteyn & Ryzhik (Reference Gradshteyn and Ryzhik2000) to find

(4.5) \begin{align}\left\langle {j\sec}{h}^{2}{x}\right\rangle _{{h}}=\begin{cases} 2\sqrt{2{h}}\\[4pt] \pi /\sqrt{2} \end{cases} \begin{array}{c} {h}\gg 1,\\ {h}=0, \end{array}\end{align}

then

(4.6) \begin{align}\left\langle {j\sec}{h}^{2}{x}\right\rangle _{{h}}{\left.\frac{\partial {g}}{\partial {h}}\right| }_{{h}=0}^{{h}\gg 1}=\left({X}-2\right)\left(\frac{{e}\phi _{0}}{{m}}\right)^{1/2}\frac{{mV}}{{T}}\bar{{f}}_{0}.\end{align}

The length 2X appears because the hole, and therefore the dissipation region, extends to infinity. Normalising using $\langle {P}\rangle _{{h}}=2{X\kern-2pt P}$ gives the power dissipated to be

(4.7) \begin{align}{P}=\left(1-\frac{2}{{X}}\right)\frac{{mn}_{0}{kV}\Delta _{0}}{\pi ^{1/2}{{v}}_{{e}}^{2}}\left(\frac{{e}\phi _{0}}{{T}}\right)^{5/2}{e}^{-{{V^{2}}}/{{v}}_{{e}}^{2}}{\int }_{\kern-3.4pt 0}^{{\infty }}{{\rm d}v}_{\bot }{v}_{\bot }{{v}}_{\bot {z}}^{2}{e}^{-{m}{{v}}_{\bot }^{2}/2{T}},\end{align}

where $\varDelta =2^{3/2}(\nu {{v}}_{\bot {z}}^{2}/{k}{{v}}_{{e}}^{3})({T}/{e}\phi _{0})^{3/2}\equiv {2^{3/2}}\Delta _{0}{{v}}_{\bot {z}}^{2}/{{v}}_{{e}}^{2}$ and $\nu \approx \nu _{{e}}{{v}}_{{e}}^{3}/2{V}^{3}$ . Integrating gives

(4.8) \begin{align}{P}=\left(1-\frac{2}{{X}}\right)\frac{({Z}+2)\nu _{{e}}{n}_{0}{T}{{v}}_{{e}}^{2}}{2\pi ^{1/2}({Z}+1){V}^{2}}\left(\frac{{e}\phi _{0}}{{T}}\right){e}^{-{{V^{2}}}/{{v}}_{{e}}^{2}}.\end{align}

To see this result implies the dissipation or damping rate $\gamma _{{hole}}$ is very small within a finite hole region of extent $2X\gg 1$ , the island power dissipated is balanced by the slow decay of the hole potential to obtain

(4.9) \begin{align}{P}=\gamma _{{hole}}\left\langle {{E}}_{| | }^{2}\right\rangle _{{h}}/4\pi =\gamma _{{hole}}{n}_{0}{Tk}^{2}{\lambda }_{{D}}^{2}{{e}^{2}}\left\langle (\partial \phi /\partial {x})^{2}\right\rangle _{{h}}/{T}^{2}.\end{align}

Using $\langle \phi ^{2}\rangle _{{h}}=4{\phi }_{0}^{2}/3$ , $\langle \phi ^{3}\rangle _{{h}}=16{\phi }_{0}^{3}/15$ , (3.10) and (3.11) gives

(4.10) \begin{align}{{k}^{2}}{\lambda }_{{D}}^{2}\left\langle \left(\frac{\partial \phi }{\partial {x}}\right)^{2}\right\rangle _{{h}} & =-{{k}^{2}\lambda }_{{D}}^{2}\left\langle \phi \frac{\partial ^{2}\phi }{\partial {x}^{2}}\right\rangle =\left[\frac{{{v}}_{{e}}^{2}}{{V}^{2}}\left(1+\frac{3{{v}}_{{e}}^{2}}{2{V}^{2}}\right)-\frac{{ZT}}{{T}_{{i}}}\right]\left\langle \phi ^{2}\right\rangle _{{h}}\nonumber\\[4pt]& \quad +\frac{{e}}{2{T}}\left(\frac{{Z}^{2}{T}^{2}}{{{T}}_{{i}}^{2}}+\frac{{{v}}_{{e}}^{2}}{2{V}^{2}}\right)\left\langle \phi ^{3}\right\rangle _{{h}}=\frac{4}{3}\left[\frac{{{v}}_{{e}}^{2}}{{V}^{2}}\left(1+\frac{3{{v}}_{{e}}^{2}}{2{V}^{2}}\right)-\frac{{ZT}}{{T}_{{i}}}\right]{\phi }_{0}^{2}\nonumber\\[4pt]& \quad +\frac{8{e}}{15{T}}\left(\frac{{Z}^{2}{T}^{2}}{{{T}}_{{i}}^{2}}+\frac{{{v}}_{{e}}^{2}}{2{V}^{2}}\right){\phi }_{0}^{3}=\frac{16}{15}{{k}^{2}\lambda }_{{D}}^{2}{\phi }_{0}^{2}\end{align}

and thereby yields

(4.11) \begin{align}{P}=\frac{16}{15}\gamma _{{hole}}{n}_{0}{T}{{k}^{2}\lambda }_{{D}}^{2}\left(\frac{{e}\phi _{0}}{{T}}\right)^{2}.\end{align}

As a result, the island or hole decay or damping rate is

(4.12) \begin{align}\gamma _{{hole}} & =\frac{15({Z}+2)\nu _{{e}}{{v}}_{{e}}^{2}{e}^{-{{V^{2}}}/{{v}}_{{e}}^{2}}}{32\pi ^{1/2}({Z}+1){k}^{2}{\lambda }_{{D}}^{2}{V}^{2}({{e}\phi _{0}}/{{T}})}\nonumber\\[5pt]& =\frac{15\sqrt{3}({Z}+2)\nu _{{e}}{{v}}_{{e}}^{2}{e}^{-{{V^{2}}}/{{v}}_{{e}}^{2}}}{16\pi ^{1/2}({Z}+1){k}\lambda _{{D}}{V}^{2}({Z}^{2}{T}^{2}/{{T}}_{{i}}^{2}+{{v}}_{{e}}^{2}/2{V}^{2})^{1/2}({{e}\phi _{0}}/{{T}})^{3/2}},\end{align}

with the last form more suggestive than the first. The limit considered here assumes $\Delta _{0}=(\nu _{{e}}/{kv}_{{e}})({{v}}_{{e}}^{3}/2{V}^{3})({T}/{e}\phi _{0})^{3/2}\ll 1$ , giving $\gamma _{{hole}}\sim ({V}\Delta _{0}/\lambda _{{D}}){e}^{-{{V^{2}}}/{{v}}_{{e}}^{2}}$ if ${V}^{2}/{{v}}_{{e}}^{2}\approx 3$ and ${ZT}/{T}_{{i}}\gt 1/2$ . As a result, this decay rate can be very small – much smaller than the sinusoidal Landau (Reference Landau1946) damping rate for a Maxwellian

(4.13) \begin{align}\gamma _{{L}}=\frac{\pi ^{1/2}{kV}^{2}}{2{k}^{2}{\lambda }_{{D}}^{2}{v}_{{e}}}{e}^{-3/2-1/2{{k^{2}}}{\lambda }_{{D}}^{2}}.\end{align}

Consequently, a hole is allowed to be very long lived and be many Debye lengths long.

The rate $\gamma _{{hole}}$ differs from the BGK damping rate used by Valentini (Reference Valentini2008) because he employed the sinusoidal results of Zakharov & Karpman (Reference Zakharov and Karpman1963) rather than a solitary hole as considered here and for which ${\phi }_{0}^{2}$ terms are required.