164 Chapter 5. Streaming instabilities and the Landau problem
contour may be deformed so that the vertical portion is pushed to the far left, except
where there is a polepj;the contour “snags” around each polepjas shown in Fig.5.2(c).
ForRep→−∞,the numeratorN(p)→ 0 ,while the denominatorD(p)→ 1 .Since
exp(pt)→ 0 for Rep→−∞and positivet,the left hand vertical line does not contribute
to the integral and Eq.(5.52) simply consists of the sum of the residues of all the poles,
i.e.
φ 1 (t)=∑
jlim
p→pj[
(p−pj)N(p)
D(p)
ept]
. (5.62)
Where do the polespjcome from? Upon examining Eq.(5.62), it is clear that poles could
come either from (i)N(p)having an explicit pole, i.e.N(p)contains a term∼ 1 /(p−pj),
or (ii) fromD(p)containing a factor∼(p−pj), i.e.,pj is a root of the equationD(p)=
0 .The integrand in Eq. (5.60) has a pole in thev‖plane;this pole is “used up” as a
residue upon performing thev‖integration, and so does not contribute a pole toN(p).
The only other possibility is that the initial valueFσ 1 (v‖,0)somehow provides a pole, but
Fσ 1 (v‖,0)is a physical quantity with a bounded integral [i.e.,
∫
Fσ 1 (v‖,0)dv‖is finite] and
so cannot contribute a pole inN(p).It is therefore concluded that all poles inN(p)/D(p)
must come from the roots (also called zeros) ofD(p).
The problem can be simplified by deciding to be content with alessthan complete so-
lution. Instead of attempting to calculateφ 1 (t)for all positive times (i.e., all the poles
pjcontribute to the solution), we restrict ourselves to the less burdensome problem of find-
ing the long time asymptotic behavior ofφ 1 (t).Because each term in Eq.(5.62) has a factor
exp(ipjt), the least damped term [i.e., the term with pole furthest to the right in Fig.5.2(c)],
will dominate all the other terms at larget. Hence, in order to find the long-term asymptotic
behavior, all that is required is to find the rootpjhaving the largest real part.
The problem is thus reduced to finding the roots ofD(p);this requires performing the
v‖integration sketched in Fig.5.4. Before doing this, it is convenient to integrate out the
perpendicular velocity dependence fromD(p)so that
D(p) = 1−1
k^2∑
σq^2 σ
ε 0 mσ∫
d^3 vik·∂fσ 0
∂v
(p+ik·v)= 1−
1
k^2∑
σq^2 σ
ε 0 mσ∫∞
−∞dv‖∂Fσ 0
∂v‖
(v‖−ip/k). (5.63)
Thus, the relationD(p)=0can be written in terms of susceptibilities as
D(p)=1+χi+χe=0 (5.64)since the quantities being summed in Eq.(5.63) are essentially the electron and ion pertur-
bations associated with the oscillation, andD(p) is the Laplace transform analog of the
the Fourier transform of Poisson’s equation. In the special case where the equilibrium dis-
tribution function is Maxwellian, the susceptibilities can be written in a standardized form