Case 2 - Application to ion-acoustic solitons and shocks:
Take:
v eo = v io = u o
And: Electron, ion streaming at same speed, i.e.
Ti= 0, Te ≠ 0
Electron thermal speed:
v eo 2 = 2Te /me
Ion sound speed: cs 2 = Te /mi
We can then write:
n i = n o /Ö 1 - 2e j /mi uo2
With electron distribution function:
f s ~ n s = A n so- v 2 - 2e j /me v oe2 )
Making assumption:
uo2 < < v oe2
We get for the number density:
n e = n o exp [e j /Te ]
Note: Boltzmann distribution, given:
exp ( 2e j /me v oe2 ) = exp [e j /Te ]
Therefore:
d 2j/ d x2 = 4 p e (n e – n i) =
4 p n o e (exp [e j /Te ] - 1 /Ö 1 - 2e j /mi uo2 )
We now seek to reframe this equation non-dimensionally using:
F = e j/ Te
Proceeding:
{2 e j/ Te }/mi uo2 / Te) = 2F / uo2 / cs 2 = 2 F/M s 2
Where M s is the sonic shock number.
Then:
4 p n o e (exp [e j /Te ] - 1 /Ö 1 - 2e j /mi uo2 ) =
4 p n o e (exp [F] - 1 /Ö 1 - 2 F/M s 2 )
Whence:
d 2/ d x2 {exp [e F /Te } = 4 p n o e 2/Te [exp [F] - 1 /Ö 1 - 2 F/M s 2 ]
Or:
d 2j/ d x2 = 1/l e 2 [exp [F] - 1 /Ö 1 - 2 F/M s 2 ]
We have in summary:
F = e j/ Te , x = x/ le And: M s = uo / cs
Then: d 2 F/ d x 2 = exp (F) - 1Ö(1 – (2 e F)/ (M s )2 = - ¶ y/ ¶ j
Which is the dimensionless Poisson equation
Where: Y = exp (F) - 1/ Ö(1 – (2 e F)/ M s 2) + C
We require for self-consistent solutions:
1) y (F =0) = 0 Þ C = 1 + M s 2
2) d 2 Y / d F 2 ‖f =0 < 0
3) y (F c ) > 0
Which leads to: 1 - exp [M s 2 /2] + M s 2 > 0
Hence:
M s < 1.6
Thus for a soliton (shock) solution associated with case 2 to exist:
1 < M s < 1.6
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