Fundamentals of Plasma Physics

474 Chapter 16. Non-neutral plasmas

the upper limit has been changed torpbecausen 0 (r)anddn 0 /drare by assumption zero
in the regionrp < r≤a.Ifdn 0 /drhas the same sign throughout the radial interval
0 ≤r≤rp, then the integral would have to be non-zero since the integrand always has
the same sign, and soωiwould have to be zero. Thus, a necessary condition forωito be
finite is fordn 0 /drto change signs in the interval 0 ≤r≤rp. This necessary condition
corresponds ton 0 (r)having a maximum in the interval 0 ≤r≤rp.This sort of profile is
commonly called hollow, becausen 0 (r)starts at some finite value atr=0, increases to a
maximum at some finiter<rp, and then decreases to zero atr=rp.
Further progress can be made by expressing the diocotron equations as a pair of coupled
equations for the density and potential perturbations, i.e.,

̃nl = −

l ̃φl

rB(ω−lω 0 (r))

dn 0

dr

(16.77)

1

r

d

dr

(

r

d ̃φl

dr

)

−

l^2

r^2

̃φl = − ̃nlq

ε 0

. (16.78)

Rather than substitute for ̃nlin order to obtain Eq.(16.38), instead Eq.(16.78) is solved
for ̃φlusing a Green’s function method (Schecter, Dubin, Cass, Driscoll, Lansky and
O’Neil 2000). This approach has the virtue of imposing the perfectly conducting wall
boundary condition onφ ̃lat an earlier stage of the analysis before the entire wave equation
is developed. The set of solutions to Eq.(16.78) are then effectively restricted to those satis-
fying the wall boundary condition, and only this restricted set is used when later combining
Eqs.(16.78) and (16.77) to form a wave equation.
The Green’s function solution to Eq.(16.78) is obtained by first recasting Eq.(16.78)
as

1

r

d

dr

(

r

d ̃φl

dr

)

−

l^2

r^2

φ ̃l=−q

ε 0

∫a

0

dsδ(r−s) ̃nl(s). (16.79)

Thus, ifψ(s,r)is the solution of

1

r

d

dr

(

r

d

dr

ψ(r,s)

)

−

l^2

r^2

ψ(r,s)= −δ(r−s) (16.80)

where 0 ≤s≤athen

̃φl(r)= q

ε 0

∫a

0

dsψ(r,s) ̃nl(s) (16.81)

is the solution of Eq.(16.78). The spatial boundary conditions are accounted forwhen
solving Eq.(16.80) for the Green’s functionψ(r,s)and so are independent of the form of
̃nl.
Equation (16.80) is solved by finding separate solutions to its homogeneous counter-
part
1
r

d

dr

(

r

dψ

dr

)

−

l^2

r^2

ψ=0 (16.82)

for the inner interval 0 ≤r<sand for the outer intervals<r≤aand then appropriately
matching these two distinct homogeneous solutions atr=swhere they meet. The inner
solution must satisfy the regularity conditionψ(0)=0and the outer solution must satisfy