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As part of a derivation in Bernstein '58 [1] a linear first-order (eqn. (9) in the image) appears:

Bernstein '58

But the general solution I would usually take (as appears in Gradshteyn and Ryzhik and checked in Mathematica) here would be

$$F=c_1G+\frac1G\int^\phi_1\frac{G(\phi^\prime)\Phi(\phi^\prime)}\Omega$$ where $\Phi$ is the RHS of eqn. (9). I suppose it depends on the requirement of periodicity and that $s$ has positive complex part (it comes from a Laplace transform) but I can't see how Bernstein's solution is obtained or equivalent, i.e. where the factor of $1/G$ goes. I'm sure I'm missing something silly, but I've wasted a bit of time on this already and can't get it to work out. The same derivation appears in slightly neater form in Montgomery and Tidman[2] but without further insight. Any help justifying this result would be greatly appreciated.

[1] Bernstein, I. (1958). Waves in a plasma in a magnetic field. Physical Review. Retrieved from http://journals.aps.org/pr/abstract/10.1103/PhysRev.109.10

[2] Montgomery D.C., Tidman, D.A. (1964). Plasma Kinetic Theory, McGraw-Hill advanced physics monograph series

1 Answers1

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In general, one can write the linearized and Fourier transformed Vlasov equation, assuming $F_{s}(\mathbf{v})$ $\rightarrow$ $F_{so}(\mathbf{v})$ + $\delta F_{s}(\mathbf{v})$ , as:
$$ \partial_{\phi} F_{s} - i \ (\alpha_{s} + \beta_{s} \ cos \ \phi) \ F_{s} = \frac{q_{s}}{m_{s} \Omega_{cs}} \left[ \mathbf{E} + \mathbf{v} \ \times \ \left( \frac{\mathbf{k} \ \times \ \mathbf{E}}{\omega} \right) \right] \cdot \nabla_{\mathbf{v}} \ F_{so} $$ where subscript $s$ defines the particle species, $\mathbf{k}$ came from the linearized assumption that $\nabla$ $\rightarrow$ $i \ \mathbf{k}$, and $\alpha_{s}$ and $\beta_{s}$ are given by:
$$ \alpha_{s} = \frac{k_{\parallel} \ v_{\parallel} - \omega}{\Omega_{cs}} \\ \beta_{s} = \frac{k_{\perp} \ v_{\ perp}}{\Omega_{cs}} $$ If we limit ourselves to electrostatic waves, then $\mathbf{k} \ \times \ \mathbf{E}$ $\rightarrow$ 0 and $\mathbf{E}$ can be approximated by $-i \ \mathbf{k} \ \Phi$. Then the Vlasov equation above goes to:
$$ \partial_{\phi} F_{s} - i \ (\alpha_{s} + \beta_{s} \ cos \ \phi) \ F_{s} = \frac{-i \ q_{s} \ \Phi}{m_{s} \ \Omega_{cs}} \left( \mathbf{k} \ \cdot \ \nabla_{\mathbf{v}} \right) \ F_{so} $$ which can be rewritten in a general form as:
$$ \frac{d F}{dx} + P(x) \ F = Q(x) $$ This has a general solution given by:
$$ F = e^{-\int \ dx' \ P(x')} \left[ \int \ dx' \ Q(x') \ e^{ \int dx'' \ P(x'') } \right] $$ where the term $\int \ dx \ P(x)$ goes to:
$$ \int \ d\phi \ P(\phi) = -i \ \int \ d\phi \ \left( \alpha_{s} + \beta_{s} \ cos \ \phi \right) \\ = -i \ \left( \alpha_{s} \phi + \beta_{s} \ sin \ \phi \right) $$ which lets us rewrite the solution for F as:
$$ F_{s} = \frac{-i \ q_{s} \ n_{s} \ \Phi}{m_{s} \ \Omega_{cs}} \ e^{ i \ \left( \alpha_{s} \phi + \beta_{s} \ sin \ \phi \right) } \ \int \ d\phi' \ \left( \mathbf{k} \ \cdot \ \nabla_{\mathbf{v}} \right) \ F_{so} \ e^{ -i \ \left( \alpha_{s} \phi' + \beta_{s} \ sin \ \phi' \right) } $$

This is the same form as found in Bernstein's work.

Edit
After discussing with a colleague, I realized that the principle difference between the above derivation and Bernstein's work is that I did not use a Laplace transform. Physically, the Laplace transform gives you causality in the equation, it will give you a formal initial condition. The derivation I showed is only using a Fourier transform and imposing a priori the zero on the right-hand side of the Vlasov equation.

In many cases, the method used above is okay because $\Im [\omega]$ $\ll$ $\Re [\omega]$, in which case one can get away with assuming zero initial amplitudes.

The isolated $F^{*}(\mathbf{k}, \mathbf{v}, 0)$ on the right-hand side comes from a Laplace transform and integrating by parts. The the only surviving term is the Laplace transform at t = 0.

This is the reason why my result differs from Bernstein's work.