The Magnetized Vlasov–Ampère System and the Bernstein–Landau Paradox

Abstract

We study the Bernstein–Landau paradox in the collisionless motion of an electrostatic plasma in the presence of a constant external magnetic field. The Bernstein–Landau paradox consists in that in the presence of the magnetic field, the electric field and the charge density fluctuation have an oscillatory behavior in time. This is radically different from Landau damping, in the case without magnetic field, where the electric field tends to zero for large times. We consider this problem from a new point of view. Instead of analyzing the linear magnetized Vlasov–Poisson system, as it is usually done, we study the linear magnetized Vlasov–Ampère system. We formulate the magnetized Vlasov–Ampère system as a Schrödinger equation with a selfadjoint magnetized Vlasov–Ampère operator in the Hilbert space of states with finite energy. The magnetized Vlasov–Ampère operator has a complete set of orthonormal eigenfunctions, that include the Bernstein modes. The expansion of the solution of the magnetized Vlasov–Ampère system in the eigenfunctions shows the oscillatory behavior in time. We prove the convergence of the expansion under optimal conditions, assuming only that the initial state has finite energy. This solves a problem that was recently posed in the literature. The Bernstein modes are not complete. To have a complete system it is necessary to add eigenfunctions that are associated with eigenvalues at all the integer multiples of the cyclotron frequency. These special plasma oscillations actually exist on their own, without the excitation of the other modes. In the limit when the magnetic fields goes to zero the spectrum of the magnetized Vlasov–Ampère operator changes drastically from pure point to absolutely continuous in the orthogonal complement to its kernel, due to a sharp change on its domain. This explains the Bernstein–Landau paradox. Furthermore, we present numerical simulations that illustrate the Bernstein–Landau paradox. In Appendix 2 we provide exact formulas for a family of time-independent solutions.

Appendices

Appendix 1: Technical Formulas

In this Appendix we further study the properties of the secular equation (5.49, 5.45, 5.50). For later use we prepare the following result.

Proposition A.1

Let $a_{n,m}, n \in {\mathbb {Z}}^*, m=1,\dots ,$ be the quantity defined in (5.45). Then, there is a constant, C, that depends on n,  such that,

$$\begin{aligned} a_{n,m} \le C\, \frac{1}{\sqrt{m}}\, \left[ \frac{e n^2}{ 2 \omega _{\text { c}}^2 m }\right] ^m,\qquad m=1,\dots , \end{aligned}$$

(A.1)

where e is Euler’s number. In particular, for any $p >0$ there is a constant C,  that depends on n and p,  such that,

$$\begin{aligned} a_{n,m} \le C \frac{1}{m^p}. \end{aligned}$$

(A.2)

Proof

By equation (10.22.67) in page 245 of [22]

$$\begin{aligned} a_{n,m}= e^{\frac{-n^2}{ \omega _{\text { c}}^2} } \, I_m\left( \frac{n^2}{ \omega _{\text { c}}^2 }\right) , \end{aligned}$$

(A.3)

with $I_n(z)$ a modified Bessel function. Furthermore, by equation (10.41.1) in page 256 of [22],

$$\begin{aligned} I_m\left( \frac{n^2}{ \omega _{\text { c}}^2}\right) = \frac{1}{\sqrt{2 \pi m}}\, \left( \frac{e n^2}{2 \omega _{\text { c}}^2 m }\right) ^m (1+ o(1)), \qquad m \rightarrow \infty . \end{aligned}$$

(A.4)

Equation (A.1) follows from (A.3) and (A.4). Finally, (A.2) follows from (A.1). $\square$

We continue the analysis of the secular equation Let $\lambda _{n,m}, m \ge 2$ be the root given in Lemma  5.6. Recall that $\lambda _{n,m} \in (m \omega _{\mathrm{{c}}}, (m+1)\omega _{\mathrm{{c}}}).$ Then, to isolate terms that can be large as $\lambda _{n,m}$ is close to $m \omega _{\mathrm{{c}}}$ or to ($m+1)\omega _{\mathrm{{c}}},$ we decompose $g( \lambda _{n,m})$ as follows,

$$\begin{aligned} g(\lambda _{n,m})= g^{(1)}( \lambda _{n,m})+g^{(2)}( \lambda _{n,m})+g^{(3)}( \lambda _{n,m})+g^{(4)}( \lambda _{n,m}), \end{aligned}$$

(A.5)

where,

$$\begin{aligned}&g^{(1)}( \lambda _{n,m}):= 4 \pi \,\sum _{1\le q \le m-1 }\, \frac{q^2\,\omega ^2_{\text { c}}}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2} a_{n,q}, \end{aligned}$$

(A.6)

$$\begin{aligned}&g^{(2)}( \lambda _{n,m}):= 4\pi \, \frac{m^2\,\omega ^2_{\text { c}}}{m^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2} a_{n,m}, \end{aligned}$$

(A.7)

$$\begin{aligned}&g^{(3)}( \lambda _{n,m}):= 4 \pi \, \frac{(m+1)^2\,\omega ^2_{\text { c}}}{(m+1)^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2} a_{n,m+1}, \end{aligned}$$

(A.8)

$$\begin{aligned}&g^{(4)}( \lambda _{n,m}):= 4 \pi \,\sum _{ q \ge m+2 }\, \frac{q^2\,\omega ^2_{\text { c}}}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}} a_{n,q}. \end{aligned}$$

(A.9)

Lemma A.2

Let $g^{(1)}( \lambda _{n,m})$ be the quantity defined in (A.6). Then, there is a constant $C_n$ such that,

$$\begin{aligned} \left| g^{(1)}( \lambda _{n,m})\right| \le C_n\, \frac{1}{m^2}, \qquad m\ge 2. \end{aligned}$$

(A.10)

Proof

First suppose that m is even. Then, m/2 is an integer, and we can decompose $g^{(1)}( \lambda _{n,m})$ as follows,

$$\begin{aligned} g^{(1)}(\lambda _{n,m})= g^{(1,1)}( \lambda _{n,m})+ g^{(1,2)}( \lambda _{n,m}), \end{aligned}$$

(A.11)

where,

$$\begin{aligned} g^{(1,1)}( \lambda _{n,m}):= 4 \pi \,\sum _{1\le q \le m/2 }\, \frac{q^2\,\omega ^2_{\text { c}}}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2} a_{n,q}, \end{aligned}$$

(A.12)

and

$$\begin{aligned} g^{(1,2)}( \lambda _{n,m}):= 4 \pi \,\sum _{m/2 < q \le m-1 }\, \frac{q^2\,\omega ^2_{\text { c}}}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2} a_{n,q}. \end{aligned}$$

(A.13)

Note that,

$$\begin{aligned} \left| \frac{1}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2}\right| \le \frac{2}{m^2 \omega ^2_{\text { c}} }, \qquad q=1,\dots , \frac{m}{2}. \end{aligned}$$

(A.14)

Then, by (A.2), (A.12) and, (A.14)

$$\begin{aligned} \left| g^{(1,1)}( \lambda _{n,m}) \right| \le 4 \pi \, \frac{2}{m^2\omega ^2_{\text { c}}} \sum _{1}^{m/2}\, q^2\,\omega ^2_{\text { c}}\, a_{n,q} \le C \frac{1}{m^2}. \end{aligned}$$

(A.15)

Furthermore, we have

$$\begin{aligned} \left| \frac{1}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2}\right| \le \frac{1}{\omega _{\text { c}}} \, \frac{1}{m {\omega _{\text { c}}}}, \qquad q=\frac{m}{2},\dots , m-1. \end{aligned}$$

(A.16)

Then, by (A.2), (A.13) and, (A.16),

$$\begin{aligned} \left| g^{(1,2)}( \lambda _{n,m}) \right| \le 4 \pi \, \frac{1}{\omega _{\text { c}}} \, \frac{1}{m \omega _{\text { c}}}\, \sum _{m/2 < q \le m-1 }\, q^2\,\omega ^2_{\text { c}}a_{n,q} \le C_p \frac{1}{m^p}, \,p =1,\dots . \end{aligned}$$

(A.17)

Equation (A.10) follows from (A.11), (A.15) and, (A.17). In the case where m is odd, $(m-1)/2$ is an integer, and we decompose $g^{(1)}(\lambda _{n,m})$ as in (A.11) with,

$$\begin{aligned} g^{(1,1)}( \lambda _{n,m}):= 4 \pi \,\sum _{1\le q \le (m-1)/2 }\, \frac{q^2\,\omega ^2_{\text { c}}}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}} a_{n,q}, \end{aligned}$$

(A.18)

and

$$\begin{aligned} g^{(1,2)}( \lambda _{n,m}):= 4 \pi \,\sum _{(m-1)/2 < q \le m-1 }\, \frac{q^2\,\omega ^2_{\text { c}}}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2} a_{n,q}, \end{aligned}$$

(A.19)

and we proceed as in the case of m even. $\square$

In the following lemma we estimate $g^{(4)}( \lambda _{n,m}).$

Lemma A.3

Let $g^{(4)}( \lambda _{n,m})$ be the quantity defined in (A.9). Then, for every $p >0$ there is a constant $C_p$ such that,

$$\begin{aligned} \left| g^{(4)}( \lambda _{n,m})\right| \le C_p\, \frac{1}{m^p}, \qquad m\ge 2. \end{aligned}$$

(A.20)

Proof

Note that,

$$\begin{aligned} \left| \frac{1}{q^2 \omega ^2_{\text { c}}- \lambda _{n,m}^2}\right| \le \frac{1}{\omega _{\text { c}}} \, \frac{1}{(m+3){\omega _{\text { c}}}}, \qquad q \ge m+2. \end{aligned}$$

(A.21)

Equation (A.20) follows from (A.2), (A.9) and, (A.21). $\square$

In the following lemma we estimate how $\lambda _{n,m}$ approaches $m \omega _{\text { c}}$ as $m \rightarrow \pm \infty .$

Lemma A.4

We have,

$$\begin{aligned} \lambda _{n,m}= m \omega _{\text { c}}+ 2 \pi m \, \omega _{\text { c}}\, \frac{a_{n,|m|}}{n^2}+ a_{n,|m|}\, O\left( \frac{1}{|m|}\right) , \qquad m \rightarrow \pm \infty . \end{aligned}$$

(A.22)

Proof

Note that since $\lambda _{n,-m}= - \lambda _{n,m}$ it is enough to prove Eq. (A.22) when $m \rightarrow \infty .$

Using (A.10) and (A.20) we write (5.50) as follows

$$\begin{aligned} 4 \pi \, \frac{m^2\,\omega ^2_{\text { c}}}{ \lambda _{n,m}^2 - m^2 \omega ^2_{\text { c}}} a_{n,m} =n^2 + g^{(3)}( \lambda _{n,m}) +O\left( \frac{1}{m^2}\right) , \qquad m \rightarrow \infty . \end{aligned}$$

(A.23)

Moreover, as $g^{(3)}(\lambda _{n,m}) \ge 0,$ we get,

$$\begin{aligned} 4 \pi \, \frac{m^2\,\omega ^2_{\text { c}}}{ \lambda _{n,m}^2 - m^2 \omega ^2_{\text { c}}} a_{n,m} \ge n^2+O\left( \frac{1}{m^2}\right) , \qquad m \rightarrow \infty . \end{aligned}$$

Then, there is an $m_0$ such that $4 \pi \, \frac{m^2\,\omega ^2_{\text { c}}}{ \lambda _{n,m}^2 - m^2 \omega ^2_{\text { c}}} a_{n,m} \ge \frac{\pi }{4}$, $m \ge m_0$, and then, $\lambda _{n,m}^2 \le m^2 \omega ^2_{\text { c}} + 16 m^2\,\omega ^2_{\text { c}} \, a_{n,m}$, $m \ge m_0$, and taking the square root we obtain

$$\begin{aligned} m \omega _{\text { c}} \le \lambda _{n,m} \le m \omega _{\text { c}} \sqrt{ 1 + 16 \, a_{n,m}}, \qquad m \ge m_0. \end{aligned}$$

(A.24)

This already shows that $\lambda _{n,m}$ is asymptotic to $\omega _{\text { c}}$ for large m. However, we can improve this estimate to obtain (A.22). By (A.2) and (A.24) for every $p >0,$

$$\begin{aligned} \left( (m+1) \omega _{\text { c}} - \lambda _{n,m} \right) ^{-1}= \frac{1}{ \omega _{\text { c}}}\left( 1+ O\left( \frac{1}{m^p}\right) \right) , \qquad m \rightarrow \infty . \end{aligned}$$

(A.25)

Further, introducing (A.8) and (A.25) into (A.23), and using (A.2) we obtain,

$$\begin{aligned} 4 \pi \, \frac{m^2\,\omega ^2_{\text { c}}}{ \lambda _{n,m}^2 - m^2 \omega ^2_{\text { c}}} a_{n,m} =n^2 +O\left( \frac{1}{m^2}\right) , \qquad m \rightarrow \infty . \end{aligned}$$

(A.26)

We rearrange (A.26) as follows,

$$\begin{aligned} \lambda _{n,m} - m \omega _{\text { c}}= \frac{4\pi }{n^2}\, \frac{m^2\,\omega ^2_{\text { c}}}{ \lambda _{n,m}+ m \omega _{\text { c}}} a_{n,m}+ \frac{1}{n^2}\, ( \lambda _{n,m} - m \omega _{\text { c}})\, O\left( \frac{1}{m^2}\right) , \qquad m \rightarrow \infty . \end{aligned}$$

(A.27)

By (A.24)

$$\begin{aligned} \lambda _{n,m}- m \omega _{\text { c}} \le m \omega _{\text { c}} \, O\left( a_{n,m} \right) , \qquad m \rightarrow \infty . \end{aligned}$$

(A.28)

Further,

$$\begin{aligned} \left( \lambda _{n,m}+ m \omega _{\text { c}}\right) ^{-1}= \left( 2 m \omega _{\text { c}} + \lambda _{n,m}- m \omega _{\text { c}}\right) ^{-1}= \frac{1}{2m \omega _{\text { c}}}\, (1+ O\left( a_{n,m} \right) ), \qquad m \rightarrow \infty . \end{aligned}$$

(A.29)

Expansion (A.22) follows from (A.28) and, (A.29). $\square$

Appendix 2: A Family of Stationary Solutions

In this appendix we construct explicitly a family of time-independent solutions to the linearized magnetized Vlasov–Poisson system. We first construct the family in dimension 1+2 (one dimension in space, two dimensions in velocity), that is the situation that we consider in our work. Then, we generalize our family of solutions to the case dimension 3+3 (three dimensions in space, three dimensions in velocity) that is the case considered by Bedrossian and Wang [5].

1.1 2.1 Dimension 1+2

For the purpose of making the comparison with [5] more transparent we consider the Vlasov equation,

$$\begin{aligned} \partial _t f + v_1 \partial _ x f +{\mathbf {F}} \cdot \nabla _v f=0, \end{aligned}$$

(B.1)

with the electromagnetic Lorenz force,

$$\begin{aligned} {\mathbf {F}}(t, x) = \frac{q}{m} \left( {\mathbf {E}} (t, x)+ v \times {\mathbf {B}}_0(t, x)\right) . \end{aligned}$$

(B.2)

Taking ${\mathbf {B}}_0\ne 0$ the plasma is magnetized. The variable x is in the periodic torus $x\in {\mathcal {T}}=[0,2\pi ]_{\mathrm{per}}$. The velocity variables are $(v_1,v_2)\in {\mathbb {R}}^2$. We take the charge $q >0$ (Remark 2.1) the mass $m>0,$ and we assume, as before, that the magnetic field ${\mathbf {B}} (t,{\mathbf {x}})={\mathbf {B}}_0$ is constant in space-time. We suppose again that the two-dimensional velocity v is perpendicular to the constant magnetic field, i.e., ${\mathbf {B}}_0= (0,0, B_0), B_0 >0.$ Moreover, we assume that the electric field is directed along the first coordinate axis, ${\mathbf {E}}(t,x)=(E(t,x), 0, 0 ),$ that it has mean zero,

$$\begin{aligned} \int _{\mathcal {T}}\, E(t,x)\, dx=0, \end{aligned}$$

and that it satisfies the Gauss law,

$$\begin{aligned} \partial _x E(t,x)= \frac{1}{4\pi } \,\left( -1 + q \int _{{\mathbb {R}}^2}\, f dv\right) , \end{aligned}$$

(B.3)

where as in [5] we introduced the factor $\frac{1}{4 \pi }$ in the right-hand side of the Gauss law, and we have taken the charge of the heavy particles equal to minus one. See Remark 2.1.

We linearize the equations around a homogeneous Maxwellian equilibrium state ${\tilde{f}}_0(v),$ where,

$$\begin{aligned} {\tilde{f}}_0(v):= \frac{1}{2\pi } e^{\frac{-v^2}{2}}. \end{aligned}$$

We take as equilibrium state ${\tilde{f}}_0(v):= \frac{1}{2 \pi } e^{\frac{-v^2}{2}}$ to make the comparison with the results of [5] more transparent. This corresponds to the expansion,

$$\begin{aligned} f(t,x,v)={\tilde{f}}_0(v) + \varepsilon h(t,x,v) +O(\varepsilon ^2), \end{aligned}$$

(B.4)

and

$$\begin{aligned} E(t,x)=E_0 + \varepsilon F(t,x)+O(\varepsilon ^2), \end{aligned}$$

(B.5)

with a null reference electric field $E_0=0$. Inserting (B.4) and (B.5) into (B.1)–(B.3), and keeping the terms up to linear in $\varepsilon ,$ we obtain the linearized magnetized Vlasov–Poisson system

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t h+ v_1 \partial _x h - \frac{q}{m} F v_1 {{\tilde{f}}_0} - \frac{q}{m} B_0 \left( -v_2 \partial _{v_1} +v_1 \partial _{v_2} \right) h= 0, \\&\partial _x F= \frac{q}{4\pi } \, \int _{{\mathbb {R}}^2} h \, dv_1 dv_2, \int _{\mathcal {T}}\, F(t,x)\, dx=0. \end{aligned} \right. \end{aligned}$$

(B.6)

As we look for time-independent solutions, we have to solve,

$$\begin{aligned} \left\{ \begin{aligned}&v_1 \partial _x h - \frac{q}{m} F v_1 {{\tilde{f}}_0}- \frac{q}{m} B_0 \left( -v_2 \partial _{v_1} +v_1 \partial _{v_2} \right) h= 0, \\&\partial _x F= \frac{q}{4 \pi } \int _{{\mathbb {R}}^2} h\, dv_1 \,dv_2, \qquad \int _{\mathcal {T}}\, F(t,x)\, dx=0. \end{aligned} \right. \end{aligned}$$

(B.7)

Note that,

$$\begin{aligned} \left( -v_2 \partial _{v_1} +v_1 \partial _{v_2} \right) {\tilde{f}}_0(v_1,v_2) =0. \end{aligned}$$

(B.8)

Our objective hereafter is to construct a family of non trivial smooth solutions to (B.7) that have fast decay in velocity.

Lemma B.1

There exists an explicit family of non trivial smooth solutions (h, F) to the time-independent linearized magnetized Vlasov–Poisson system (B.7), where $F= - \varphi '(x),$ with $\varphi \in C^\infty ({\mathcal {T}}),$ and where the function h can be taken with l continuous derivates with respect to $v, l=1,2,\dots ,$ or infinitely differentiable with respect to v. Moreover, for each fixed $x \in {\mathcal {T}}, h\in L^1({\mathbb {R}}^2).$ Further, the absolute value of h and of all its derivatives can be taken bounded by Gaussian functions of v,  uniformly in $x \in {\mathcal {T}}.$ Moreover, $h+\frac{q}{m} \varphi f_0$ can be taken with compact support in ${\mathbf {v}}$, uniformly in $x \in {\mathcal {T}}.$

Proof

We introduce an electric potential $\varphi \in C^1({\mathcal {T}})$ as

$$\begin{aligned} F(x)=-\varphi '(x), \end{aligned}$$

with $\varphi (2\pi )=\varphi (0)$ and $\varphi '(2 \pi )= \varphi '(0).$ Plugging in the first equation in (B.7), we obtain

$$\begin{aligned} v_1\partial _x \left[ h(x,v_1,v_2) + \frac{q}{m} \varphi (x) {\tilde{f}}_0(v_1,v_2) \right] - \frac{q}{m} B_0 \left( -v_2 \partial _{v_1} +v_1 \partial _{v_2} \right) h= 0. \end{aligned}$$

Let us define

$$\begin{aligned} G(x,v_1,v_2)=h(x,v_1,v_2) + \frac{q}{m} \varphi (x) {\tilde{f}}_0(v_1,v_2) . \end{aligned}$$

(B.9)

Since we have (B.8), G satisfies the equation

$$\begin{aligned} v_1\partial _x G(x,v_1,v_2) - \frac{q}{m} B_0 \left( -v_2 \partial _{v_1} +v_1 \partial _{v_2} \right) G(x,v_1,v_2)=0. \end{aligned}$$

(B.10)

Let us make another change of function which is valid since $B_0\ne 0,$

$$\begin{aligned} H(x,v_1,v_2)= G(x- \frac{m \,v_2}{\, qB_0 },v_1,v_2) \Longleftrightarrow G(x,v_1,v_2)= H(x+ \frac{m \, v_2}{q \,B_0},v_1,v_2). \end{aligned}$$

(B.11)

Equation (B.10) is rewritten as

$$\begin{aligned} \frac{q}{m} \omega _c \left( -v_2 \partial _{v_1} +v_1 \partial _{v_2} \right) H(x,v_1,v_2)=0. \end{aligned}$$

(B.12)

Under this form, it is easy to find a general solution which writes

$$\begin{aligned} H(x,v_1,v_2)= K(x, v_1^2+v_2^2) \end{aligned}$$

where K is an arbitrary smooth function which decreases sufficiently fast at infinity with respect to its second variable. For example, we can take,

$$\begin{aligned} K(x, v_1^2+v_2^2)= e^{inx} g( v_1^2+v_2^2), \qquad n \in {\mathcal {Z}}\setminus \{0\}, \end{aligned}$$

(B.13)

where $g(v_1^2+v_2^2) \in C^l({\mathbb {R}}^2), l=1,2,\dots ,$ or $g(v_1^2+v_2^2) \in C^\infty ({\mathbb {R}}^2),$ and $g(v_1^2+v_2^2) \in L^1({\mathbb {R}}^2).$ For example, g can be taken with compact support, or a Gaussian. Going back to the perturbation h, one obtains the representation formula

$$\begin{aligned} h(x,v_1,v_2) = -\frac{q}{m} \varphi (x) {\tilde{f}}_0(v_1,v_2) + K(x+ \frac{m \,v_2}{q\, B_0}, v_1^2+v_2^2), \end{aligned}$$

where the electric potential $\varphi$ remains to be determined. All functions h of this form satisfy the first equation of (B.7).

It remains to verify the Gauss law, that is the second equation of (B.7). The right-hand side of the Gauss law is

$$\begin{aligned} \frac{q}{4\pi } \int _{{\mathbb {R}}^2} h \,dv_1\,dv_2= \frac{q}{4\pi } \left( -\frac{q}{m}\, \varphi (x) +\int _{{\mathbb {R}}^2}\, K\left( x+ \frac{m\, v_2}{q\, B_0}, v_1^2+v_2^2\right) \,dv_1 \, dv_2\right) . \end{aligned}$$

So the Gauss law is rewritten as

$$\begin{aligned} -\varphi ''(x) + \frac{q^2}{4\pi m} \, \varphi (x) = \frac{q}{4\pi } \, \int _{{\mathbb {R}}^2} K\left( x+ \frac{m \,v_2}{q\, B_0}, v_1^2+v_2^2\right) \,dv_1 \,dv_2. \end{aligned}$$

Using (B.13), one gets

$$\begin{aligned} -\varphi ''(x)+ \frac{q^2}{4\pi m } \varphi (x) = \frac{q}{4\pi } e^{inx} \int _{{\mathbb {R}}^2} \displaystyle e^{ in \frac{m v_2}{q B_0}} g( v_1^2+v_2^2) \, dv_1 \, dv_2. \end{aligned}$$

This is an equation for the electric potential. The periodic solution is explicit,

$$\begin{aligned} \varphi (x) = \frac{1}{n^2+\frac{q^2}{4\pi m}}\, \frac{q}{4\pi } \,e^{inx} \, \int _{{\mathbb {R}}^2} \displaystyle e^{in \frac{m v_2}{q B_0} } g( v_1^2+v_2^2) dv_1dv_2. \end{aligned}$$

(B.14)

The remaining properties of the solution (h, F) follow immediately from the explicit representation of (h, F). $\square$

Remark that the solutions given by Lemma B.1 satisfy,

$$\begin{aligned} \int _{{\mathcal {T}}} h(x,v_1,v_2)\, dx=0, \end{aligned}$$

and in particular,

$$\begin{aligned} \int _{{\mathcal {T}} \times {\mathbb {R}}^2}\, h(x,v_1,v_2)\, dx\, dv_1\, dv_2=0. \end{aligned}$$

The solutions given by Lemma B.1 are in agreement with (6.1), (6.2), (6.4), (6.7), and also with (6.14), (6.15), (6.16).

1.2 2.2: Dimension 3+3

We now consider solutions to the magnetized Vlasov–Poisson system in ${\mathcal {T}}^3 \times {\mathbb {R}}^3,$ where ${\mathcal {T}}^3$ is the three-dimensional torus, ${\mathcal {T}}^3:= [0,2\pi ]_{\mathrm{per}}^3.$ We denote ${\mathbf {x}}:=(x,y,z)\in {\mathcal {T}}^3,$ and ${\mathbf {v}}=(v_1,v_2,v_3)\in {\mathbb {R}}^3.$ The Vlasov equation is given by,

$$\begin{aligned} \partial _t f(t,{\mathbf {x}},{\mathbf {v}}) + {\mathbf {v}} \cdot \nabla _ {{\mathbf {x}}} f(t, {\mathbf {x}},{\mathbf {v}}) +{\mathbf {F}}(t, {\mathbf {x}}) \cdot \nabla _{{\mathbf {v}}} f(t,{\mathbf {x}},{\mathbf {v}})=0, \end{aligned}$$

(B.15)

with the electromagnetic Lorenz force,

$$\begin{aligned} {\mathbf {F}}(t, {\mathbf {x}}) = \frac{q}{m} \left( {\mathbf {E}} (t, {\mathbf {x}})+ {\mathbf {v}} \times {\mathbf {B}}_0(t,{\mathbf {x}})\right) , \end{aligned}$$

(B.16)

where $q>0, m >0,$ and we assume, as before, that the magnetic field ${\mathbf {B}} (t,{\mathbf {x}})={\mathbf {B}}_0$ is constant in space-time, and that it is directed along the third coordinate, i.e., ${\mathbf {B}}_0= (0,0, B_0), B_0 >0.$ Moreover, we assume that the electric field satisfies the Gauss law,

$$\begin{aligned} \nabla _{{\mathbf {x}}}\cdot {\mathbf {E}}(t, {\mathbf {x}})= \frac{1}{4\pi } \rho (t,{\mathbf {x}}), \qquad \rho (t,{\mathbf {x}}):= \,[-1 + q \int _{{\mathbb {R}}^3}\, f (t,{\mathbf {x}}, {\mathbf {v}})\, d^3 {\mathbf {v}}]. \end{aligned}$$

(B.17)

Further, we assume that the electric field has mean zero,

$$\begin{aligned} \int _{{\mathcal {T}}^3}\, {\mathbf {E}}(t,{\mathbf {x}})\, d^3{\mathbf {x}}=0. \end{aligned}$$

(B.18)

We linearize equations (B.15)–(B.18) around a homogeneous Maxwellian equilibrium state $f^0({\mathbf {v}}),$ where,

$$\begin{aligned} f^0({\mathbf {v}}):= {\tilde{f}}_0(v_1,v_2)\ \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }}, \end{aligned}$$

where $T_\parallel >0$ is the temperature along the magnetic field. This amounts to take ${{\tilde{f}}}=0$ in (1.2). This corresponds to the expansion,

$$\begin{aligned} f(t,{\mathbf {x}},{\mathbf {v}})=f^0({\mathbf {v}}) + \varepsilon {\mathcal {G}}(t,{\mathbf {x}},{\mathbf {v}}) +O(\varepsilon ^2), \end{aligned}$$

(B.19)

and

$$\begin{aligned} {\mathbf {E}}(t,{\mathbf {x}})={\mathbf {E}}_0 + \varepsilon {\mathcal {F}}(t, {\mathbf {x}})+O(\varepsilon ^2), \end{aligned}$$

(B.20)

with a null reference electric field ${\mathbf {E}}_0=0$. Inserting (B.19) and (B.20) into (B.15)–(B.17), and keeping the terms up to linear in $\varepsilon ,$ we obtain the linearized magnetized Vlasov–Poisson system,

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t {\mathcal {G}}(t,{\mathbf {x}},{\mathbf {v}})+ {\mathbf {v}} \cdot \nabla _{{\mathbf {x}}} {\mathcal {G}}(t,{\mathbf {x}}, {\mathbf {v}}) + \frac{q}{m} {\mathcal {F}}(t,{\mathbf {x}})\cdot \nabla _{{\mathbf {v}}} {f^0({\mathbf {v}})}+ \frac{q}{m} {\mathbf {v}} \times {\mathbf {B}}_0\cdot \nabla _{{\mathbf {v}}} {\mathcal {G}}(t,{\mathbf {x}},{\mathbf {v}})= 0, \\&\nabla _{{\mathbf {x}}} \cdot {\mathcal {F}}(t,{\mathbf {x}})= \,\frac{q}{4\pi } \, \rho (t,{\mathbf {x}}), \qquad \rho (t,{\mathbf {x}}):= \int _{{\mathbb {R}}^3}\, {\mathcal {G}}(t,{\mathbf {x}},{\mathbf {v}})\, d^3{\mathbf {v}}, \qquad \int _{{\mathcal {T}}^3}\, {\mathcal {F}}(t,{\mathbf {x}})\, d^3{\mathbf {x}}=0. \end{aligned} \right. \end{aligned}$$

(B.21)

We look for solutions to (B.21) that satisfy,

$$\begin{aligned} \int _{{\mathcal {T}}^3 \times {\mathbb {R}}^3}\, {\mathcal {G}}(t,{\mathbf {x}},{\mathbf {v}})\, d^3{\mathbf {x}}\,d^3 {\mathbf {v}}=0. \end{aligned}$$

(B.22)

Under the condition (B.22) the Gauss law, that is the second equation in (B.21), is equivalent to the following equation,

$$\begin{aligned} {\mathcal {F}}(t,{\mathbf {x}})=- \nabla _{{\mathbf {x}}} \, \int _{{\mathcal {T}}^3}\, W({\mathbf {x}}-{\mathbf {y}})\, \rho (t,{\mathbf {y}})\, d^3{\mathbf {y}}, \end{aligned}$$

(B.23)

where,

$$\begin{aligned} W({\mathbf {x}}):= \frac{q}{4\pi }\, \frac{1}{(2\pi )^3}\, \sum _{ {\mathbf {k}} \in {\mathbb {Z}}^3\setminus \{0\}}\, \frac{1}{| {\mathbf {k}}|^2}\, e^{ik\cdot {\mathbf {x}}}. \end{aligned}$$

Since we are looking for time-independent solutions we have to solve,

$$\begin{aligned} \left\{ \begin{aligned}&{\mathbf {v}} \cdot \nabla _{{\mathbf {x}}} {\mathcal {G}}(t,{\mathbf {x}}, {\mathbf {v}}) + \frac{q}{m} {\mathcal {F}}(t,{\mathbf {x}})\cdot \nabla _{{\mathbf {v}}} {f^0({\mathbf {v}})}+ \frac{q}{m} {\mathbf {v}} \times {\mathbf {B}}_0\cdot \nabla _{{\mathbf {v}}} {\mathcal {G}}(t,{\mathbf {x}},{\mathbf {v}})= 0, \\&\nabla _{{\mathbf {x}}}\cdot {\mathcal {F}}(t,{\mathbf {x}})= \frac{q}{4\pi } \, \rho (t,{\mathbf {x}}), \qquad \rho (t,{\mathbf {x}}):= \int _{{\mathbb {R}}^3}\, {\mathcal {G}}(t,{\mathbf {x}}, {\mathbf {v}})\, d^3{\mathbf {v}}, \qquad \int _{{\mathcal {T}}^3}\, {\mathcal {F}}(t,{\mathbf {x}})\, d^3{\mathbf {x}}=0. \end{aligned} \right. \end{aligned}$$

(B.24)

Let (h, F) be one of the solutions to (B.7) given by Lemma B.1. We define,

$$\begin{aligned} {\mathcal {G}} ({\mathbf {x}},{\mathbf {v}})= h(x,v_1,v_2)\, \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }}, \text{ and } {\mathcal {F}}({\mathbf {x}})=\left( \begin{array}{ccc} F(x) \\ 0 \\ 0 \end{array} \right) . \end{aligned}$$

(B.25)

Lemma B.2

Let $(h(x,v_1,v_2),F(x))$ be one of the solutions to the time-independent linearized magnetized Vlasov–Poisson system (B.7) given by Lemma B.1. Then, the pair $({\mathcal {G}}({\mathbf {x}},{\mathbf {v}}),{\mathcal {F}}({\mathbf {x}}))$ defined in (B.25) is a solution to the time-independent linearized magnetized Vlasov–Poisson system (B.24) in ${\mathcal {T}}^3\times {\mathbb {R}}^3,$ with ${\mathcal {F}} \in C^\infty ({\mathcal {T}}^3),$ and where the function h can taken with l continuous derivates with respect to ${\mathbf {v}}, l=1,2,\dots ,$ or infinitely differentiable with respect to ${\mathbf {v}}.$ Moreover, for each fixed $x \in {\mathcal {T}}^3, h\in L^1({\mathbb {R}}^3).$ Further, the absolute value of h and of all its derivatives can be taken bounded by Gaussian functions of ${\mathbf {v}},$ uniformly in ${\mathbf {x}} \in {\mathcal {T}}^3.$ Further, the solution $({\mathcal {G}}({\mathbf {x}},{\mathbf {v}}),{\mathcal {F}}({\mathbf {x}}))$ satisfies (B.22).

Proof

We detail the calculations for the convenience of the reader. One has

$$\begin{aligned} \begin{array}{lll} {\mathbf {v}} \cdot \nabla _{{\mathbf {x}}} {\mathcal {G}} (x,y,z,v_1,v_2,v_3)= v_1 \partial _x h (x,v_1,v_2) \ \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }}, \\ \\ \nabla _{{\mathbf {v}}} {\mathcal {G}} (x,y,z,v_1,v_2,v_3)= \left( \begin{array}{ccc} \partial _{v_1} h(x,v_1,v_2) \ \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }} \\ \\ \partial _{v_2} h (x,v_1,v_2) \ \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }} \\ \\ h(x,v_1,v_2) \ \partial _{v_3} \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }} \\ \end{array} \right) , \\ \\ {\mathcal {F}}(x,y,z)\cdot \nabla _{{\mathbf {v}}} f^0 = -F(x) \,v_1 \,f^0, \\ {\mathbf {v}} \times {\mathbf {B}}_0= \left( \begin{array}{ccc} B_0 v_2 \\ \\ - B_0 v_1 \\ \\ 0 \end{array} \right) , \\ \\ ({\mathbf {v}} \times {\mathbf {B}}_0)\cdot \nabla _{{\mathbf {v}}} {\mathcal {G}} (x,y,z,v_1,v_2,v_3)= B_0 \left( v_2\partial _{v_1} - v_1\partial _{v_2} \right) h(x, v_1, v_2) \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }} . \end{array} \end{aligned}$$

(B.26)

Therefore, by (B.26) one gets the first equation in the linearized magnetized Vlasov–Poisson system (B.24),

$$\begin{aligned} {\mathbf {v}} \cdot \nabla _{{\mathbf {x}}} {\mathcal {G}}(t,{\mathbf {x}}, {\mathbf {v}}) + \frac{q}{m} {\mathcal {F}}(t,{\mathbf {x}})\cdot \nabla _ {{\mathbf {v}}} {f^0(v)}+ \frac{q}{m} {\mathbf {v}} \times {\mathbf {B}}_0\cdot \nabla _{{\mathbf {v}}} {\mathcal {G}}(t,{\mathbf {x}},{\mathbf {v}})= 0. \end{aligned}$$

(B.27)

Moreover, one has $\nabla _{{\mathbf {x}}} \cdot {\mathcal {F}}({\mathbf {x}})= \partial _x F(x),$ and

$$\begin{aligned} \begin{array}{lll} \int _{{\mathbb {R}}^3} {\mathcal {G}} ({\mathbf {x}},{\mathbf {v}})\, d^3{\mathbf {v}}&{} =&{} \int _{{\mathbb {R}}^2} h(x,v_1,v_2)dv_1dv_2\times \int _{{\mathbb {R}}} \, \frac{1}{\sqrt{2\pi T_\parallel }}\, e^{\frac{-v^2_3}{2 T_\parallel }} \, dv_3 \\ &{} =&{} \int _{{\mathbb {R}}^2} h(x,v_1,v_2)dv_1dv_2 . \end{array} \end{aligned}$$

So one obtains immediately the Gauss law

$$\begin{aligned} \nabla _{{\mathbf {x}}} \cdot {\mathcal {F}}({\mathbf {x}} )=\frac{q}{4\pi }\, \rho (t,{\mathbf {x}}). \end{aligned}$$

(B.28)

The fact that (B.22) holds, and the properties of the solution $({\mathcal {G}}, {\mathcal {F}})$ stated in the lemma hold, follow immediate from the definition of the pair (h, F). $\square$

The solutions $({\mathcal {G}}, {\mathcal {F}}),$ given by Lemma B.2, to the time-independent linearized magnetized Vlasov–Poisson system (B.24), and that fulfill (B.22), satisfy the assumption of Theorem 1 of Bedrossian and Wang, [5] (see Theorem 1.1 above). For example, we can take $g(v_1^2+v_2^2)= e^{-v_1^2+v_2^2}.$ Moreover, the charge density fluctuation is independent of y and z,  and is given by,

$$\begin{aligned} \rho (x)= e^{inx} \left[ \frac{1}{n^2+\frac{q^2}{4\pi m}}\, \frac{-q^2}{4\pi m} + 1 \right] \int _{{\mathbb {R}}^2} \, e^{in \frac{m \,v_2}{q\, B_0}}\, g( v_1^2+v_2^2) \, dv_1 dv_2. \end{aligned}$$

(B.29)

The Fourier coefficient ${\hat{\rho }}(n,0,0)$ is, in general, not zero, and it is given by,

$$\begin{aligned} {\hat{\rho }}(n,0,0)= (2 \pi )^3\, \left[ \frac{1}{n^2+\frac{q^2}{4\pi m}}\, \frac{-q^2}{4\pi m} + 1 \right] \int _{{\mathbb {R}}^2} \, e^{in \frac{m \,v_2}{q\, B_0}}\, g( v_1^2+v_2^2) \, dv_1 dv_2. \end{aligned}$$

(B.30)

1.3 2.3 Limit $B_0 \rightarrow 0$

An interesting question is passing to the limit $B_0 \rightarrow 0$ in the right-hand side of (B.14). One has weak convergence to zero of the right-hand side under standard integrability conditions on g since

$$\begin{aligned} \underset{B_0 \rightarrow 0}{\lim }\int _{{\mathbb {R}}^2} e^{i n\frac{ mv_2}{q B_0}} g( v_1^2+v_2^2) dv_1dv_2 =0 \end{aligned}$$

because of the oscillating term $e^{i n\frac{ mv_2}{q B_0}}.$ Therefore, by (B.14) the solutions of (B.7) given by Lemma B.1 satisfy

$$\begin{aligned} \underset{ B_0 \rightarrow 0}{\varphi }= 0. \end{aligned}$$

The weak limit recovers the classical results in the non magnetized case.