Fundamentals of Plasma Physics

14.3 Echoes 419

ping terms higher than second-order gives the second-order equation

∂f 2

∂t

+v

∂f 2

∂x

−

eE 2

m

∂f 0

∂v

−

e

m

E 1

∂f 1

∂v

=0. (14.113)

The special case whereE 1 f 1 has a zero frequency component was discussed in the previous
section on quasilinear diffusion and so here only the situation whereE 1 f 1 has a finite
frequency will be considered. Since the first-order system has been solved, the last term in
Eq.(14.113) can be considered as a source term and so we re-arrange the equationto be

∂f 2

∂t

+v

∂f 2

∂x

−

eE 2

m

∂f 0

∂v

=

e

m

∂

∂v

(E 1 f 1 ) (14.114)

whereE 1 andf 1 are known andf 2 is to be determined. The specification thatf=f 0 +f 1
att= 0provides the boundary conditionf 2 = 0at timet= 0. Because Eq. (14.114)
has both a self-consistent electric field contribution (E 2 term on the left hand side) and a
prescribed electric field term (E 1 term on the right hand side), it has aspects of both the
self-consistent problem and of the problem where the electric field is prescribed.

The linear problem Suppose periodic grids are inserted into a plasma and the grids are
transiently pulsed thereby creating the potentialφext(x,t).The charged particles will move
in response to this applied potential and the resulting displacement produces aperturbation
of the plasma charge density. Poisson’s equation must therefore take intoaccount both the
charge density on the grid and the resulting plasma charge density and so has the form

∇^2 φ 1 =−

1

ε 0

(grid charge density+plasma charge density). (14.115)

Since the grid charge density is related to the grid potential by

∇^2 φext=−

1

ε 0

(grid charge density) (14.116)

Poisson’s equation can be recast as

∇^2 φ 1 =∇^2 φext+

e

ε 0

∫

f 1 dv. (14.117)

A Fourier-Laplace transform operation is then applied to Eq.(14.117) to obtain

−k^2 ̃φ 1 =−k^2 φ ̃ext+

e

ε 0

∫

f ̃ 1 dv. (14.118)

However, usingE 1 =−∇φ 1 the Fourier Laplace transform of Eq.(14.112) gives

f ̃ 1 =−ie

m

k ̃φ 1

p+ikv

∂f 0

∂v

(14.119)

so that Eq.(14.118) becomes

φ ̃ 1 (p,k)=

̃φext(p,k)

D(p,k)

(14.120)

where

D(p,k)=1−i

e^2

k^2 mε 0

∫

k

p+ikv

∂f 0

∂v

dv (14.121)