16.5 Diocotron modes 475
the wall boundary conditionψ(a) =0. Since the solutions of Eq.(16.82) areψ∼r±l, the
inner solution must be
ψ=α
(r
a
)l
for 0 ≤r<s (16.83)
and the outer solution must be
ψ=β
((
r
a
)l
−
(r
a
)−l)
fors<r≤a (16.84)
where the coefficientsαandβare to be determined.
Integration of Eq.(16.80) across the delta function fromr=s−tor=s+gives the
jump condition [
d
dr
ψ(r,s)
]s+
s−
=− 1 (16.85)
and integrating a second time shows thatψmust be continuous atr=s.These jump and
continuity conditions give two coupled equations inαandβ,
βl
a
((
s
a
)l− 1
+
(s
a
)−l− 1 )
−
αl
a
(s
a
)l− 1
= − 1 (16.86)
β
((
s
a
)l
−
(s
a
)−l)
−α
(s
a
)l
= 0. (16.87)
Solving forαandβgives the Green’s function,
ψ(r,s)=
−
a
2 l
(s
a
)l+1((r
a
)l
−
(r
a
)−l)
forr>s
−
a
2 l
(r
a
)l((s
a
)l+1
−
(s
a
)−l+1)
forr<s;
(16.88)
this satisfies Eq.(16.80) and also the boundary conditions atr= 0andr=a.Using the
Green’s function in Eq.(16.81) gives
̃φl(r) = q
ε 0
[∫r
0
dsψ(r,s) ̃nl(s)+
∫a
r
dsψ(r,s) ̃nl(s)
]
= −
a
2 l
q
ε 0
∫r
0 ds
((
s
a
)l+1((r
a
)l
−
(r
a
)−l))
̃nl(s)
+
∫a
rds
((
r
a
)l((s
a
)l+1
−
(s
a
)−l+1))
n ̃l(s)
(16.89).
Finally, using Eq.(16.77) to substitute for ̃nl(s)gives
φ ̃l(r)= q
2 ε 0 B
∫r
0
ds
sl
al
(
rl
al
−
al
rl
) ̃
φl(s)
ω−lω 0 (s)
dn 0
ds
+
∫a
r
ds
rl
al
(
sl
al
−
al
sl
) ̃
φl(s)
ω−lω 0 (s)
dn 0
ds
. (16.90)
This integral equation forφ ̃l(r)not only prescribes the mode dynamics, but also explic-
itly incorporates the spatial boundary conditions. The resonant denominatorsω−lω 0 (s)in