Fundamentals of Plasma Physics

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