content_image/339126.jpg content_image/339128.jpg content_image/339129.jpg
these singularities read, 

<img src='content_image/339126.jpg'>

For K = 2 in Fig. 3a, we have Ω 1 < Ω 2 < Ω 3 < Ω 4 . The system is here stable because ϵ (Ω) > 0, which gives 4 real roots to the 4th degree dispersion equation. For K = 3, the ordering of the singularities is still the same, but ϵ (0) < 0, rendering the system unstable. Finally, for K = 4, the ordering is Ω 1 < Ω 3 < Ω 2 < Ω 4 , namely, Ω 3 and Ω 2 switched their positions. The stability in this case is necessary from the behavior of the dispersion function near its singularities. When their order is the one of Fig. 3c, i.e. “1,3,2,4”, the dispersion equation necessarily crosses the Ω axis 4 times and the 4th degree dispersion equation has therefore only purely real roots. If the ordering is “1,2,3,4”, like the one found for K = 2 or 3, stability relies on the sign of F (0). From Figs. 3a to 3b, the system is therefore unstable until Ω 2 = Ω 3 which gives, 

<img src='content_image/339128.jpg'>

defining thus the largest unstable K in a given direction. For θ = 0, the quantity 2 /H is exactly the upper limit of the large K instability two-stream branch [6]. For an arbitrary angle, the equation above describes a circle of diameter 2 /H and center ( K$_{x}$, K$_{z}$ ) = (0 , 1 /H ) which fits perfectly as is observed on Figs. 1. Finally, the expression of K ∗ breaks down for θ ∼ π/ 2; according to Eq. (12), K $^{∗}$( π/ 2) = 0 while the fluid quantum theory of the filamentation instability gives a finite value for this quantity [10]. 

Turning now the sign of ϵ (0), a little algebra gives ϵ (0) = 0 for, 

From θ $_{= 0 to arccos}$√$_{H}$$_{,}$ K + ( θ ) is monotonically decreasing, and systems with K < K + ( θ ) are stable. In the same interval, K − ( θ ) is monotonically increasing, and systems with K < 

<img src='content_image/339129.jpg'>

