Pierre-Antoine Giorgi

Anne Nouri
email: anne.nouri@univ-amu.fr

On a Vlasov-Poisson system in a bounded set with direct reflection boundary conditions

Existence (resp. uniqueness) of global (resp. local) in time continuous solutions to the Vlasov-Poisson system is proven in a one-dimensional bounded domain, with direct reflection boundary conditions. Generalized characteristics are used. Electroneutrality is obtained in the limit.

Introduction and main results

We consider the Vlasov-Poisson system

∂ t f + v∂ x f + E∂ v f = 0, t > 0, (x, v) ∈ ]0, 1[×R, (1.1)
∂ t g + v∂ x g -E∂ v g = 0, t > 0, (x, v) ∈ ]0, 1[×R, (1.2
)

ε∂ x E = R f dv - R g dv, t > 0, x ∈]0, 1[, (1.3)
f (0, x, v) = f 0 (x, v), (x, v) ∈]0, 1[×R, (1.4)
g(0, x, v) = g 0 (x, v), (x, v) ∈]0, 1[×R. (1.5) This is a well-known model describing the dynamics of particles in a collisionless, electrostatic and nonrelativistic plasma composed of ions and electrons. f and g respectively denote the ionic and electronic distribution functions. The electric field E = -∂ x φ derives from the potential φ. The parameter ε > 0 is equal to the square of the ratio between the Debye and the characteristic observation lengths. The Debye length is a physical length below which charge separation occurs. In many physical situations, ε is small. The distribution functions f and g satisfy direct reflection boundary conditions at the boundary x ∈ {0, 1}, and E is given and constant at x = 0,

f (t, x, v) = f (t, x, -v), t > 0, x ∈ {0, 1}, v ∈ R, (1.6)
g(t, x, v) = g(t, x, -v), t > 0, x ∈ {0, 1}, v ∈ R, (1.7)
E(t, 0) = E 0 , t > 0.

(1.8)

Many works have already been done on the Vlasov-Poisson system. C. Bardos Using optimal transport, G. Loeper [LOE2006] proved the uniqueness of weak solutions with bounded mass density. A key point in their proofs is the boundedness of the mass density ρ(t, x) = f (t, x, v) dv.

In a domain with boundaries, the treatment of the Vlasov-Poisson system is more complex. Existence of weak solutions to the Vlasov-Poisson system in a bounded domain and given indata were proven by N. Ben Abdallah [BEN1994]. For the Vlasov equation, S. Mischler [MIS1999] extended his result by considering other types of boundary conditions such as specular reflection, proving existence and uniqueness of weak solutions, and studying their traces. On a half-line, H.J. Hwang and J. Schaeffer [HWA2008] proved uniqueness of weak solutions to the one-species Vlasov-Poisson system with specular reflection for the distribution function and given constant electric field at the boundary, pointing inward of the domain, using an approach with characteristics. Considering the Vlasov-Poisson system for two species as it is done in this paper, the electric field points inward (resp. outward) of the domain for one (resp. the other) species. It is known from a counterexample by Y. Guo [GUO1995] that there is in general no C 1 solution to the Vlasov-Maxwell system with direct reflection boundary conditions on a half-line. This counterexample can be adapted to the Vlasov-Poisson system with direct reflection boundary conditions in a bounded domain. In this paper, we prove global in time existence and local in time uniqueness of continuous (resp. continuous in time and C 1 in space) ionic and electronic distribution functions f and g (resp. electric field E).

If the spatial domain in (1.1)-(1.8) were R instead of ]0, 1[, classical characteristics for (1.1) would be defined by X = V, V = E(t, X). In the frame of this paper, bounces may occur at the boundary {0, 1} of the spatial domain. When a characteristics intersects x = 0 (resp. x = 1) with a zero velocity at a time s > 0, it can be seen that it stays in the domain without any discontinuity. And so, it is still considered as a classical characteristics. Bounces at the boundary of the domain occur when a characteristics intersects the boundary with a non zero velocity. Generalized characteristics involving possible bounces [GUO1995, HWA2008] and continuous solutions to (1.1) and (1.6) are defined as follows.

Definition 1.

1

Let E ∈ C([0, +∞[; C 1 ([0, 1])). The generalized backward characteristics (X, V ) from (t, x, v) ∈]0, +∞[×]0, 1[×R related to (1.1) and (1.6)
is defined as the union of the classical characteristics which connect

(t, x, v) to (t 1 , x 1 , v 1 ), (t 1 , x 1 , -v 1 ) to (t 2 , x 2 , v 2 ), . . . , (t n , x n , -v n ) to (t n+1 , x n+1 , v n+1 ), . . ., where x n ∈ {0, 1}, |v n | > 0 and t n > t n+1 ≥ 0.
This gives a set P ⊂ N * counting the number of bounces, and a sequence of bouncing times (t n ) n∈P such that,

X(t n ; t, x, v) = 0 and V (t + n ; t, x, v) = -V (t - n ; t, x, v) > 0, or X(t n ; t, x, v) = 1 and V (t + n ; t, x, v) = -V (t - n ; t, x, v) < 0, n ∈ P. Definition 1.2 Let E ∈ C([0, +∞[; C 1 ([0, 1])). A continuous solution f to (1.1), (1.4) and (1.6) is a function f ∈ C([0, +∞[×[0, 1] × R) such that f (t, x, v) = f 0 (X(0; t, x, v), V (0; t, x, v)), t > 0, (x, v) ∈ [0, 1] × R, (1.9)
where (X(•; t, x, v), V (•; t, x, v)) is the generalized characteristics from (t, x, v) as in Definition 1.1.

The main results of this paper are the following.

Theorem 1.1 Let ε > 0. Let f 0 , g 0 ∈ C([0, 1] × R)
be nonnegative even functions w.r.t. the v variable, with finite kinetic energy, and such that for any R > 0, sup

(x,w)∈[0,1]×R;|w-v|<R f 0 (x, w) ∈ L 1 (R v ), sup (x,w)∈[0,1]×R;|w-v|<R g 0 (x, w) ∈ L 1 (R v ).
(1.10)

Let E 0 ∈ R \ {0} such that E 0 = ε -1 (g 0 -f 0 )(x, v)dxdv. (1.11) There exists a solution (f, g, E) ∈ C([0, +∞[×[0, 1] × R) 2 × C([0, +∞[; C 1 ([0, 1])) of (1.1)-(1.8
) in the sense of Definition 1.2. Moreover, f and g are nonnegative.

Remark 1.1 Due to the mass conservation, E(•, 1) is also a constant function. Assumption (1.11) is satisfied if E 0 = 0 and electroneutrality holds at t = 0. It is done for E(•, 1) to be different from zero.

Remark 1.2 As will be seen in Property 3.1, t → E(t, 0) is assumed to be a constant function in order to ensure the conservation of total energy.

Theorem 1.2

Let f 0 , g 0 be nonnegative Lipschitz functions satisfying (1.10), even w.r.t. the v variable, and such that for some c 0 > 0 and V 0 > 0,

f 0 (x, v) = g 0 (x, v) = 0, x ∈ [0, 1], |v| ≤ c 0 or |v| ≥ V 0 . Let E 0 ∈ R \ {0} satisfying (1.11).
There is a time T 0 > 0 such that the continuous solution to

(1.1)-(1.8) is unique on [0, T ].
The use of generalized characteristics enlightens the solution. The main difficulties to use them are to discard infinitely many bounces accumulating at some boundary point, and to get continuity of the distribution functions f and g from (1.9). The paper organizes as follows. In Section 2, generalized characteristics are studied, taking into account possible bounces. Theorem 1.1 (resp. Theorem 1.2) is proven in Section 3 (resp. Section 4). The quasineutrality equation is obtained when ε → 0 at the end of Section 3.

Generalized characteristics

In this section, T > 0 and E ∈ C([0, T ]; C 1 ([0, 1])) are given. We consider the Cauchy problem for the Vlasov equation,

∂ t f + v∂ x f + E∂ v f = 0, t ∈ [0, T ], (x, v) ∈]0, 1[×R, (2.1)
f (0, x, v) = f 0 (x, v), (x, v) ∈]0, 1[×R, (2.2)
f (t, x, v) = f (t, x, -v), t ∈ [0, T ], (x, v) ∈ {0, 1} × R. (2.3)
As recalled above, existence and uniqueness of weak solutions to the problem have been proven by S.

≤ (t -s) E ∞ , s ∈ [t n+1 , t n ].
Given (t, x, v) ∈ [0, T ] × [0, 1] × R, either the backwards in time generalized characteristics from (t, x, v) reaches {0} × [0, 1] × R at (X(0; t, x, v), V (0; t, x, v)) without any bounce. This is the easy case where P = ∅. Or bounces occur at x = 0 or x = 1. We first prove that there is a finite number of them. This is strongly linked with the sign of the electric field at the boundaries. The following analysis distinguishes two cases. We first deal with the case of a negative value of t → E(t, 1). The case of a positive value of t → E(t, 0) can be treated analogously.

In the case of a negative value of E(t, 1) on [0, T ], let δ ∈]0, 1[ be such that

E(t, x) < 0, (t, x) ∈ [0, T ] × [1 -δ, 1]. (2.6) Let ∆(v) := 1 E ∞ (|v| + T E ∞ ) 2 + 2δ E ∞ -(|v| + T E ∞ ) , v ∈ R. (2.7)
A bound of the number of possible bounces on the generalized characteristics from

(t, x, v) ∈ [0, T ] × [0, 1] × R is given in the following proposition. Proposition 2.1 Assume t → E(t, 1), t ∈ [0, T ], constant and negative.
The number of bounces occuring at x = 1 along the backwards in time generalized characteristics from

(t, x, v) ∈ [0, T ] × [0, 1] × R is finite and bounded by T ∆(v)
. Proof of Proposition 2.1. Let (X, V ) be a backwards in time generalized characteristics from (t, x, v) with at least two bounces at x = 1, occuring at times t 1 and t 2 , with t 2 < t 1 < t. In order to prove the result, it is sufficient to bound

t 1 -t 2 from below. It holds that V (t - 1 ; t, x, v) > 0. As V (•; t, x, v) is decreasing when X(•; t, x, v) is in the interval [1 -δ, 1], X leaves the interval [1 -δ, 1] at a time s 1 ∈]t 2 , t 1 [, and X(s; t, x, v) ≥ 1 -δ for s ∈ [s 1 , t 1 ]. The integration between s and t 1 along the classical characteristics (X(•; t, x, v), V (•; t, x, v)) [t + 2 ,t - 1 ] yields X(s) -1 + (t 1 -s)V (t - 1 ) ≥ - E ∞ 2 (t 1 -s) 2 , s ∈ [s 1 , t 1 ],
so that

E ∞ 2 (t 1 -s 1 ) 2 + (t 1 -s 1 )V (t - 1 ) -δ ≥ 0. This implies that t 1 -t 2 ≥ t 1 -s 1 ≥ V (t - 1 ) 2 + 2δ E ∞ -V (t - 1 ) E ∞ .
And so, by Lemma 2.1,

t 1 -t 2 ≥ 1 E ∞ (|v| + T E ∞ ) 2 + 2δ E ∞ -(|v| + T E ∞ ) .
The number of bounces at

x = 1 on [0, T ] is thus smaller than T ∆(v) .
The opposite case where E(t, 1) > 0, t ∈ [0, T ], is more complicated. It corresponds to an electric field pointing outward of the domain ]0, 1[ at the boundary. It cannot be directly expected that the time between two bounces is bounded from below. An infinite number of bounces is a priori not impossible.

The distance to the boundary of the X component of the characteristics between two bounces could be arbitrarily small. It is proven in the following proposition that this does not occur. We first prove a preliminary lemma.

Lemma 2.2 For any (t, x, v) ∈ [0, T ] × [0, 1] × R, the series n∈P |V (t + n ; t, x, v)| converges.
Proof of Lemma 2.2. Assume P = N * , and first prove that the number of consecutive bounces between x = 0 and x = 1 is finite. Integration along the classical characteristics on [t k+1 , t k ], with (k, k + 1) ∈ P 2 , yields

X(t k ) -X(t k+1 ) = (t k -t k+1 )V (t + k+1 ) + t k t k+1 (t k -r)E(r, X(r))dr. It holds that |X(t k ) -X(t k+1 )| = 1. Hence 1 ≤ (t k -t k+1 )|V (t + k+1 )| + E ∞ 2 (t k -t k+1 ) 2 , so that t k -t k+1 ≥ 1 E ∞ V (t + k+1 ) 2 + 2 E ∞ -|V (t + k+1 )| .
It follows from Lemma 2.1 that

t k -t k+1 ≥ 1 E ∞ (|v| + T E ∞ ) 2 + 2 E ∞ -(|v| + T E ∞ ) .
This implies a finite number of consecutive bounces between x = 0 and x = 1 on [0, T ]. Consequently we can assume X(t k ) = 0 for k ≥ n 0 for some n 0 ∈ N * , the other case X(t k ) = 1 for k ≥ n 0 being analogous. It holds that

2 n k=1 |V (t + k )| = 2 n0-1 k=1 |V (t + k )| + V (t + n0 ) + V (t + n ) + n-1 k=n0 (|V (t + k+1 | + |V (t + k )|) =V (t + k+1 )-V (t - k ) = 2 n0-1 k=1 |V (t + k )| + V (t + n0 ) + V (t + n ) - tn 0 tn E(s, X(s))ds, n ≥ n 0 + 1. By (2.4) applied to s = t + n , n k=1 |V (t + k )| ≤ n0-1 k=1 |V (t + k )| + V (t + n0 ) + |v| 2 + T E ∞ , n ≥ n 0 .
Proposition 2.2 When E(t, 1) takes a constant value E 1 > 0 on [0, T ], the number of bounces occuring at x = 1 along the backwards in time generalized characteristics from

(t, x, v) ∈ [0, T ] × [0, 1] × R is finite.
Proof of Proposition 2.2. It is a proof by contradiction. Assume infinitely many bounces at (t n , 1) n∈N along the backwards in time generalized characteristics from

(t, x, v) ∈ [0, T ] × [0, 1] × R. (t n ) n∈N being a decreasing sequence in [0, T ],
converges to a time t * ≥ 0 when n → +∞. Denote by

V n = V (t - n ) > 0, s n ∈]t n+1 , t n [ such that s n = min s∈[tn+1,tn] X(s), y n = 1 -X(s n ).
It holds that lim n→+∞ s n = t * , lim n→+∞ V n = 0 by Lemma 2.2, and lim n→+∞ y n = 0. Indeed, if for a subsequence

(y n k ) k∈N of (y n ), lim k→+∞ y n k = y * > 0, then E ∞ 2 (t n k -s n k ) 2 + (t n k -s n k ) - y * 2 ≥ tn k sn k (t n k -r)E(r, X(r))dr + V (s n k )(t n k -s n k ) - y * 2 = X(t n k ) -X(s n k ) - y * 2 = y n k - y * 2 ≥ 0,
for k large enough. Consequently,

t n k -s n k ≥ 1 + y * E ∞ -1 E ∞ ,
for k large enough. This would contradict the infinite number of bounces at x = 1. By the continuity of E, there is δ 1 > 0 such that

E(t, x) ≥ E 1 2 , (t, x) ∈ [0, T ] × [1 -δ 1 , 1]. It follows from 0 = tn tn+1 V (s)ds = (t n -t n+1 )V n - tn tn+1 (r -t n+1 )E(r, X(r))dr, that t n -t n+1 ≤ 4 E 1 V n , (2.8)
for n large enough. Moreover,

y n = 1 -X(s n ) = tn sn V (s)ds = (t n -s n )V n - tn sn (r -s n )E(r, X(r))dr ≤ (t n -s n )V n ,
for n large enough. And so,

y n ≤ 4 E 1 V 2 n .
(2.9)

In a neighborhood of (t * , 1), E(s, X(s)) expresses as

E(s, X(s)) = E 1 + (X(s) -1)∂ x E(t * , 1) + (X(s) -1)ε(s), (2.10) with lim s→t * ε(s) = 0. (2.11)
Indeed, the function ε introduced in (2.10) satisfies

ε(s) = 1 X(s) -1 X(s) 1 (∂ x E(s, y) -∂ x E(t * , 1))dy,
which tends to zero when s tends to t * , by the uniform continuity of ∂ x E on [0, T ] × [0, 1]. The case ∂ x E(t * , 1) < 0, i.e. ∂ x E(t * , 1) = -α 2 with α > 0, is treated here. α is taken as 1 for the sake of simplicity. By definition of the characteristics (X, V ) and (2.9)-(2.11), s → (X(s; t, x, v), V (s; t, x, v)) satisfies

X (s) + X(s) = E 1 + 1 + g(s), s ∈ [t n+1 , t n ], V (s) = X (s), s ∈]t n+1 , t n [, X(t n ) = 1, X (t - n ) = V n , where g(s) := (X(s) -1)ε(s) = o(V 2 n ), s ∈ [t n+1 , t n ].
Here

, o(V 2 n ) means that lim n→+∞ o(V 2 n ) V 2 n = 0. Hence, X(s) = E 1 (1 -cos(s -t n )) + V n sin(s -t n ) + 1 + o(V 3 n ), s ∈ [t n+1 , t n ], V (s) = E 1 sin(s -t n ) + V n cos(s -t n ) + o(V 2 n ), s ∈]t n+1 , t n [, or
X(s) = E 1 2 (s -t n ) 2 + V n (s -t n ) + 1 + o(V 3 n ), s ∈ [t n+1 , t n ],
(2.12)

V (s) = E 1 (s -t n ) + V n + o(V 2 n ), s ∈]t n+1 , t n [. (2.13) Since t n+1 is a solution to X(t n+1 ) = 1, t n -t n+1 satisfies E 1 (t n -t n+1 ) 2 -2V n (t n -t n+1 ) + o(V 3 n ) = 0, i.e. E 1 (t n -t n+1 ) = V n ± V 2 n + o(V 3 n ). By definition of V n+1 , V n+1 = -V (t + n+1 ) = E 1 (t n -t n+1 ) -V n + o(V 2 n ).
Given the positive sign of V n and V n+1 , it results

E 1 (t n -t n+1 ) = V n + V 2 n + o(V 3 n ) = 2V n + o(V 2 n ) and V n+1 = V n + o(V 2 n ).
Thus, for some n 0 large enough,

V n+1 ≥ V n (1 -V n ), n ≥ n 0 . Denote by h : x → x -x 2 and h p = h • • • • • h p times , p ∈ N * . Let n 1 ≥ n 0 be such that V n ≤ 1 2 , n ≥ n 1 . It holds that V n1+p ≥ h p (V n1 ), p ∈ N. (2.14)
Moreover, it can easily be proven by induction that

h p (x) ≥ x p + 1 , p ≥ 1, x ∈ 0, 1 2 . ( 2
V p ≥ n-n1 p=1 h p (V n1 ) ≥ V n1 n-n1 p=1 1 p , n ≥ n 1 + 1,
which contradicts the statement of Lemma 2.2. Hence the number of bounces at x = 1 is finite. The case

∂ x E(t * , 1) > 0 (resp. ∂ x E(t * , 1) = 0
) is similar and also leads to (2.12)-(2.13) (see [GIO2019]).

Proposition 2.2 can similarly be extended to the case where t → E(t, 1) is not a constant function.

We now consider the continuity of the function f defined by (1.9). Despite the continuity of f 0 , f may be discontinuous. Actually, issues arise when X(0; t, x, v) is exactly zero (or one). This is illustrated in the following example.

Example 2.1

Let the field E be a positive constant. Let t ∈]0, T ] and x > Et 2 2 . The map v → V (0; t, x, v) is discontinuous at v = x t + E 2 t and continuous elsewhere. Proof of Example 2.1. The backward in time characteristics from (t, x, v) before any bounce at x = 0 is given by

X(s) = x -v(t -s) + E 2 (t -s) 2 , V (s) = v -E(t -s), s ∈ [0, t]. For v = x t + E 2 t
, there is no bounce on ]0, t], X(0; t, x, v) = 0 and V (0; t, x, v) > 0. For ṽ < v, the backward in time characteristics from (t, x, ṽ) has no bounce and V (0; t, x, ṽ) = ṽ -tE.

For ṽ > v, the backward in time characteristics from (t, x, ṽ) encounters a bounce at time

t 1 = t - ṽ - √ ṽ2 -2xE E > 0, and V (0; t, x, ṽ) = ṽ -tE -2 ṽ2 -2xE.
Thus,

lim ṽ→v - V (0; t, x, ṽ) = v -tE, lim ṽ→v + V (0; t, x, ṽ) = -(v -tE). Hence the map v → V (0; t, x, v) is discontinuous at v = x t + E 2 t. Proposition 2.3 The map (t, x, v) → (X(0; t, x, v), |V (0; t, x, v)|) is continuous on [0, T ] × [0, 1] × R.
Proof of Proposition 2.3. Let ( t, x, ṽ) be given. If the backwards characteristics from ( t, x, ṽ) reaches t = 0 at X(0; t, x, ṽ) ∈]0, 1[, then analogous arguments as for classical characteristics imply that lim (t,x,v)→( t,x,ṽ) X(0; t, x, v), V (0; t, x, v) = X(0; t, x, ṽ), V (0; t, x, ṽ) .

What remains to be proven is the continuity of (t, x, v) → (X(0; t, x, v), |V (0; t, x, v)|) at ( t, x, ṽ) such that its backwards characteristics reaches t = 0 at X(0; t, x, ṽ) ∈ {0, 1}. Assume X(0; t, x, ṽ) = 0. Consider (t, x, v) such that the backwards characteristics (X, V )(•; t, x, v) has an earliest bounce at time t 1 (t, x, v) > 0. For (t, x, v) close enough to ( t, x, ṽ), there is no bounce of the backwards characteristics from ( t, x, ṽ) on the interval [0, t 1 (t, x, v)]. Define the extended electric fiel E e on [0, T ] × R by

E e (t, x) = E(t, x), t ≥ 0, x ∈ [0, 1], E e (t, x) = E(t, 0), t ≥ 0, x < 0, E e (t, x) = E(t, 1), t ≥ 0, x > 1,
and the extended classical characteristics (X e , V e ) from (t,

x, v) ∈ [0, T ] × [0, 1] × R by ∂ s X e = V e , X e (t; t, x, v) = x, ∂ s V e = E e (s, X e ), V e (t; t, x, v) = v.
First consider the case where E 0 < 0. There is no restriction to consider ṽ > 0, and (t, x) (resp. ( t, x)) in the strip close to x = 0 where E < 0.

Let ε > 0 be given, and v ∈] ṽ 2 , 3ṽ 2 [. For (t, x, v) in an appropriate neighborhood of ( t, x, ṽ), it holds that

|X e (s; t, x, v) -X e (s; t, x, ṽ)| + |V e (s; t, x, v) -V e (s; t, x, ṽ)| ≤ ε, s ∈ [0, t]. (2.16)
It follows from

V e (t 1 ; t, x, v) ≥ v, X e (t 1 ; t, x, v) = 0, X e (s; t, x, v) = V e (t 1 ; t, x, v)(s -t 1 ) + t1 s (r -s)E e (r, X e (r; t, x, v))dr, that X e (s; t, x, v) ≤ -2ε, s ≤ t 1 -4ε ṽ .

Together with (2.16) and X(0, t, x, ṽ) = 0, this implies that t 1 < 4ε ṽ . Consequently,

|V (t - 1 ; t, x, v) -V (0; t, x, v)| + |V (t 1 ; t, x, ṽ) -V (0; t, x, ṽ)| ≤ 2 E ∞ t 1 ≤ cε, and
|V (0; t, x, v) + V (0; t, x, ṽ)| ≤ |V (0; t, x, v) -V (t - 1 ; t, x, v)| + |V (0; t, x, ṽ) -V (t 1 ; t, x, ṽ)| + |V (t 1 ; t, x, ṽ) -V (t + 1 ; t, x, v)| ≤ cε.

The inequality

|X(0; t, x, v) -X(0; t, x, ṽ)| ≤ cε, can be proven by bounding |X(0; t, x, v) -X(0; t, x, ṽ)| from above by |X(0; t, x, v) -X(t 1 ; t, x, v)| + |X(0; t, x, ṽ) -X(t 1 ; t, x, ṽ)| + |X(t 1 ; t, x, ṽ) -X(t 1 ; t, x, v)|.

Consider the case where E 0 > 0 and (t, x, v) (resp. ( t, x, ṽ)) such that the backwards characteristics (X, V )(•; t, x, v) (resp. (X, V )(•; t, x, ṽ)) has an earliest bounce at time t 1 (t, x, v) > 0 (resp. has no bounce on [0, t 1 (t, x, v)] and is such that X(0; t, x, ṽ) = 0). There is no restriction to consider ṽ > 0, and (t, x) (resp. ( t, x)) in the strip close to x = 0 where E > E0 2 . Let ε > 0 be given. It holds that for (t, x, v) in an appropriate neighborhood of ( t, x, ṽ), |X e (s; t, x, v) -X e (s; t, x, ṽ)| + |V e (s; t, x, v) -V e (s; t, x, ṽ)| ≤ ε, s ∈ [0, t].

Consider the extreme case where V (t 1 (t, x, v); t, x, v) = 0. Then 0 ≤ V (s; t, x, ṽ) ≤ ε, for s ∈ [0, t 1 (t, x, v)]. And so, E 0 2 t 1 ≤ t1 0 E(r, X(r; t, x, ṽ))dr = V (t 1 ; t, x, ṽ) -V (0; t, x, ṽ) ≤ cε.

From here the proof is analogous to the case where E 0 < 0.

Consequently, taking f 0 continuous and even w.r.t. the v variable and defining

f (t, x, v) = f 0 (X(0; t, x, v), V (0; t, x, v)), (t, x, v) ∈ [0, +∞[×[0, 1] × R,
as in (1.9), makes f continuous.

3 Proof of the existence Theorem 1.1

Up to Property 3.1 in this section, the parameter ε in (1.3) is taken as one for the sake of simplicity. Let T > 0 be given. Theorem 1.1 is proven with a fixed point argument for the map S defined on

K := a ∈ C([0, T ] × [0, 1]) ; 1 0 a(t, x)dx = f 0 L 1 -g 0 L 1 , t ∈ [0, T ] , by S = S 3 • S 2 • S 1 . Here, S 1 (a)(t, x) = E 0 + x 0 a(t, y)dy, (t, x) ∈ [0, T ] × [0, 1], S 2 (E) := (f, g),
where f (resp. g) is the solution to the linear Vlasov equation with force field E (resp. -E), initial datum f 0 (resp. g 0 ) and direct reflection boundary conditions, and

S 3 (f, g) := (f -g)(•, •, v)dv.
In Lemma 3.1 we prove that S maps K into K. Lemma 3.2 proves the compactness of S in C([0, T ]×[0, 1]). Lemma 3.3 proves its continuity. We then conclude that there is a fixed point a of S, such that (f, g) = S 2 • S 1 (a) is a solution to the Cauchy problem (1.1)-(1.8).

Lemma 3.1 The map S is well defined and maps K into K.

Proof of Lemma 3.1. S 1 (a) := E is continuous on [0, T ] × [0, 1] like a, and globally Lipschitz with respect to x since

|E(t, x ) -E(t, x)| ≤ a ∞ |x -x|, t ∈ [0, T ], (x, x ) ∈ [0, 1] 2 . (3.1)
Moreover, E(t, 0) = E 0 and E(t, 1) = E 0 + f 0 L 1 -g 0 L 1 are constants different from zero by (1.11). The analysis from Section 2 and the evenness of (f 0 , g 0 ) with respect to v allow to define S 2 (E) = (f, g), where f (resp. g) is the solution to the linear Vlasov equation with force field E (resp. -E), initial datum f 0 (resp. g 0 ) and direct reflection boundary conditions. Recall that

f (t, x, v) = f 0 (X(0; t, x, v), |V (0; t, x, v)|), resp. g(t, x, v) = g 0 (Y (0; t, x, v), |W (0; , t, x, v)|) , (t, x, v) ∈ [0, T ] × [0, 1] × R,
where (X, V ) (resp. (Y, W )) are the generalized characteristics associated to E (resp. -E), such that

(t, x, v) → (X(0; t, x, v), |V (0; t, x, v)|, Y (0; t, x, v), |W (0; t, x, v)|) (3.2)
is continuous. Consequently, f (resp. g) is continuous, and nonnegative like f 0 (resp. g 0 ).

S(a) = (f -g)(•, •, v)dv belongs to C([0, T ] × [0, 1]). Indeed, let R = (|E 0 |+ a ∞ ) T.
By (1.10), there is U > 0 such that 2) and (f 0 , g 0 ), that the map

(t, x) → |v|<U (f -g)(t, x, v)dv = |v|<U f 0 (X(0; t, x, v), |V (0; t, x, v)|) -g 0 (Y (0; t, x, v), |W (0; t, x, v)|) dv is continuous on [0, T ] × [0, 1].
Finally, the mass conservation of f (resp. g) implies that

1 0 (f -g)(t, x, v)dvdx = f 0 L 1 -g 0 L 1 , t ∈ [0, T ].
Consequently S(a) belongs to K.

Lemma 3.2 S is compact in C([0, T ] × [0, 1]). Proof of Lemma 3.2. Let (a n ) n∈N with a n ∈ C([0, T ] × [0, 1]
) bounded by M . Denote by Prove its uniform equicontinuity. Let η > 0 be given. By (1.10), there is U > 0 such that,

E n = S 1 (a n ) and (f n , g n ) = (S 2 • S 1 )(a n ).
|v|>U sup x∈[0,1],|w-v|<(|E0|+M )T f 0 (x, w)dv + |v|>U sup x∈[0,1],|w-v|<(|E0|+M )T g 0 (x, w)dv < η 2 . Let (X n , V n ) (resp. (Y n , W n ))
be the generalized characteristics associated to the field S 1 (a n ) (resp.

-S 1 (a n )). The existence of h 0 > 0 such that

sup (t,x)∈[0,T ]×[0,1] |v|<U f 0 (X n (0; t + h, x + k, v), |V n (0; t + h, x + k, v)|) -f 0 (X n (0; t, x, v), |V n (0; t, x, v)|) dv < η 4 , (resp. sup (t,x)∈[0,T ]×[0,1] |v|<U g 0 (Y n (0; t + h, x + k, v), |W n (0; t + h, x + k, v)|) -g 0 (Y n (0; t, x, v), |W n (0; t, x, v)|) dv < η 4 , n ∈ N, |h| + |k| < h 0 , follows from the uniform continuity on [0, T ] × [0, 1] × [-U, U ] of the map (t, x, v) → X n (0; t, x, v), |V n (0; t, x, v)|, Y n (0; t, x, v), |W n (0; t, x, v)| ,
and its continuous dependence with respect to the fields. The Ascoli theorem applies, which ends the proof of the lemma.

Lemma 3.3 S is continuous in C([0, T ] × [0, 1]). Proof of Lemma 3.3. Let (a n ) n∈N converging to a in C([0, T ] × [0, 1]).
Denote by

E n = S 1 (a n ), (f n , g n ) = (S 2 • S 1 )(a n ), S(a n ) = R (f n -g n )dv, E = S 1 (a), (f, g) = (S 2 • S 1 )(a), S(a) = R (f -g)dv.
The sequence

(E n ) n∈N converges to E in C([0, T ] × [0, 1]), because max [0,T ]×[0,1] |E n -E| ≤ max [0,T ]×[0,1]
|a n -a|.

Let η > 0 be given. By (1.10), there is U > 0 such that,

|v|>U sup x∈[0,1],|w-v|<(|E0|+M )T f 0 (x, w)dv + |v|>U sup x∈[0,1],|w-v|<(|E0|+M )T g 0 (x, w)dv < η.
And so,

|v|>U f 0 (X n (0; t, x, v), |V n (0; t, x, v)|) + g 0 (Y n (0; t, x, v), |W n (0; t, x, v)|) dv + |v|>U f 0 (X(0; t, x, v), |V (0; t, x, v)|) + g 0 (Y (0; t, x, v), |W (0; t, x, v)|) dv ≤ 4η, n ∈ N * . The convergence of |v|<U f 0 (X n (0; t, x, v), |V n (0; t, x, v)|) -g 0 (Y n (0; t, x, v), |W n (0; t, x, v)|) dv to |v|<U f 0 (X(0; t, x, v), |V (0; t, x, v)|) -g 0 (Y (0; t, x, v), |W (0; t, x, v)|) dv in C([0, T ] × [0, 1]) when n → +∞ follows from the continuous dependence of [0, T ] × [0, 1] × [-U, U ] (t, x, v) → (X(0; t, x, v), |V (0; t, x, v)|, Y (0; t, x, v), |W (0; t, x, v)|)
with respect to the fields.

End of the proof of Theorem 1.1. The existence of a solution (f, g, E) to the Cauchy problem (1.1)-(1.8) follows from Lemmas 3.2-3.3 and the Schauder fixed-point theorem for the map S in K.

The obtention of quasineutrality from the Vlasov-Poisson system, i.e. the passage to the limit when ε → 0 in (1.1)-(1.8) is a difficult problem. The formal limit does not hold for unstable profiles, as proven by D.

Han-Kwan and M. Hauray in [HAN2015]. In [HAN2016], D. Han-Kwan and F. Rousset justified the quasineutral limit of a Vlasov-Poisson system with adiabatic electrons for small times in Sobolev spaces, and for initial data satisfying a Penrose stability condition.

In the following property, we prove that the electroneutraliy equation (3.6) holds at the limit ε → 0.

Property 3.1

For every ε > 0 let (f ε , g ε , E ε ) be a solution to the Vlasov-Poisson system (1.1)-(1.8) with initial datum (f 0 ε , g 0 ε , E ε,0 ) satisfying the assumptions of Theorem 1.1, and

[0,1]×R v 2 (f 0 ε + g 0 ε )dxdv ≤ C, ε > 0, (finite initial kinetic energy), (3.3)
√ εE ε,0 ≤ C, ε > 0, (3.4)
R f 0 ε (x, v)dv = R g 0 ε (x, v)dv, a.a. (t, x) ∈ [0, T ] × [0, 1], ε > 0, (initial electroneutrality), (3.5)
f 0 ε L 2 + g 0 ε L 2 ≤ C, ε > 0,
for some C > 0. There exist a subsequence (f εn , g εn ) of (f ε , g ε ), a subsequence (E εn ) of (E ε ), and functions (f, g) such that (f εn , g εn ), (resp.

( √ ε n E εn )) weakly converges to (f, g) in L 2 ([0, T ] × [0, 1] × R) (resp. L 2 ([0, T ] × [0, 1])). Moreover, ( R f εn dv) (resp. ( R g εn dv)) weakly converges in L 2 ([0, T ] × [0, 1]) to R f dv (resp. R gdv), and R f (t, x, v)dv = R g(t, x, v)dv, a.a. (t, x) ∈ [0, T ] × [0, 1]. (3.6)
Proof of Property 3.1. The energy associated to the Vlasov-Poisson system (1.1)-(1.8) is the sum of the kinetic and potential energies,

E ε (t) = 1 2 [0,1]×R v 2 (f ε + g ε )(t, x, v)dxdv + ε 2 [0,1] E ε (t, x) 2 dx. (3.7)
The boundary conditions have been chosen in order to ensure the conservation of the energy. This classically follows from the multiplication by v 2 of the Vlasov equations (1.1)-(1.2), their integration with respect to (x, v), and the use of the continuity equations,

∂ t R (f ε -g ε )dv + ∂ x R v(f ε -g ε )dv = 0.
More details can be found in [GIO2019]. And so,

E ε (t) = 1 2 [0,1]×R v 2 (f 0 ε + g 0 ε )(x, v)dxdv + ε 2 E 2 ε,0 , t ∈ [0, T ]. The family ( √ εE ε ) ε>0 being uniformly bounded in L 2 ([0, T ] × [0, 1]), there is a sequence (ε n ) tending to zero when n → +∞ such that ( √ ε n E εn ) weakly converges in L 2 ([0, T ] × [0, 1]). Hence (ε n E εn ) weakly converges to zero in L 2 ([0, T ] × [0, 1]) when n → +∞. The family (f ε ) and (g ε ) being uniformly bounded in L 2 ([0, T ] × [0, 1] × R),
there is a subsequence of (ε n ), still denoted by (ε n ) for the sake of simplicity, and functions f and

g in L 2 ([0, T ] × [0, 1] × R), such that (f εn ) (resp. (g εn )) weakly converges in L 2 ([0, T ] × [0, 1] × R) to f (resp. g), and
v 2 (f + g)(t, x, v)dxdv < +∞.
Moreover, ( f εn (t, x, v)dv) n∈N (resp. ( g εn (t, x, v)dv) n∈N ) weakly converges in L 2 ([0, T ] × [0, 1]) to f (t, x, v)dv (resp. g(t, x, v)dv). Indeed, for any function

α ∈ L 2 ([0, T ] × [0, 1]), α(t, x) (f εn -f )(t, x, v)dvdxdt ≤ α ∞ K 2 v 2 (f εn + f )(t, x, v)dvdxdt + α(t, x)χ K (v)(f εn -f )(t, x, v)dvdxdt , (3.8)
where χ K denotes the characteristic function of ] -K, K[. The first term in the r.h.s. of (3.8) tends to zero when K → +∞ uniformly with respect to n. The second term in the r.h.s. of (3.8) tends to zero for any fixed K, given that the map α(t, x)χ K (v) belongs to L 2 ([0, T ] × [0, 1] × R). The passage to the limit when n → +∞ in (1.3) leads to the electroneutrality equation (3.6).

4 Proof of the uniqueness Theorem 1.2

This section splits into two lemmas. Under the assumptions of Theorem 1. Lemma 4.1 Assume that for some c 0 > 0 and V 0 > c 0 ,

f 0 (x, v) = 0, x ∈ [0, 1], |v| ≤ c 0 or |v| ≥ V 0 . (4.1)
Let T ∈ 0, c0 2ci , where

c i = |E 0 | + 1 ε 1 0 (f 0 + g 0 )(x, v)dvdx. (4.2)
Then the number of possible bounces of the generalized characteristics of a solution to (1.1)-(1.8) at the boundaries x = 0 and x = 1 of the domain is bounded by

T c i V 2 0 + 2c i -2V 0 .
Proof of Lemma 4.1. By the mass conservations,

f (t, x, v)dxdv = f 0 (x, v)dxdv, g(t, x, v)dxdv = g 0 (x, v)dxdv, t ∈ [0, T ].
Hence, c i is a bound from above of E ∞ . Let x 0 ∈ [0, 1] and c 0 < |v 0 | < V 0 . By (2.5) and (4.2), the velocities V (s; 0, x 0 , v 0 ) along the generalized characteristics starting at (0, x 0 , v 0 ) satisfy

c 0 2 ≤ |V (s; 0, x 0 , v 0 )| ≤ 2V 0 , s ∈ [0, T ]. (4.3)
Assume a first bounce on the generalized characteristics starting at (0, x 0 , v 0 ) occurs at s 1 on the x = 1 boundary. It follows from the positive bound from below of |V (s; 0, x 0 , v 0 )| in (4.3) that the next possible bounce will occur at x = 0. Denote by s 2 the time of such a bounce. Since X(s 2 ; 0, x 0 , v 0 ) = X(s 2 ; s 1 , 1, V (s + 1 ; 0, x 0 , v 0 )), it holds that

1 + V (s + 1 ; 0, x 0 , v 0 )(s 2 -s 1 ) + s2 s1
(s 2 -r)E(r, X(r; 0, x 0 , v 0 ))dr = 0.

Taking into account that |V (s + 1 ; 0, x 0 , v 0 )| (resp. E(r, X(r; 0, x 0 , v 0 ))) is bounded from above (resp. from below) by 2V 0 (resp. -c i ) implies that

c i (s 2 -s 1 ) 2 + 2V 0 (s 2 -s 1 ) -2 ≥ 0.
The result of the lemma follows. Lemma 4.2 Assume f 0 Lipschitz with respect to the (x, v) variable, and even with respect to the variable v. For T > 0 small enough, there is a unique solution to the Cauchy problem (1.1)-(1.8).

Proof of Lemma 4.2. Let (f, g) and ( f , g) be two solutions to the problem. Denote by (X(

•; t, x, v), V (•; t, x, v)) (resp. (Y (•; t, x, v), W (•; t, x, v)), resp. ( X(•; t, x, v), Ṽ (•; t, x, v)), resp. ( Ỹ (•; t, x, v), W (•; t, x, v))) the generalized characteristics associated to f (resp. g, resp. f , resp. g) starting at (t, x, v). Let us prove that (X(0; t, x, v), |V (0; t, x, v)|, Y (0; t, x, v), |W (0; t, x, v)|) = ( X(0; t, x, v), | Ṽ (0; t, x, v)|, Ỹ (0; t, x, v), | W (0; t, x, v)|), (t, x, v) ∈ [0, T ] × [0, 1] × R.
By Lemma 4.1, the number of bounces of the generalized characteristics starting at (0, x 0 , v 0 ) is uniformly bounded. Let us first consider the case where no bounce (resp. at most one bounce) occurs at x = 0 (resp. at x = 1). Denote by A 0 (resp. A 2 ) the set of (r, y, u) ∈ [0, T ] × [0, 1] × R such that the (X, V ) and ( X, Ṽ ) characteristics respectively associated to f and f , passing at (r, y, u), both have no bounce (resp. one bounce) at x = 1 on [0, T ]. Denote by A 1 the set of (r, y, u) ∈ [0, T ] × [0, 1] × R such that the (X, V ) (resp. the ( X, Ṽ )) characteristics passing at (r, y, u) has one bounce (resp. no bounce) at Using the Lipschitz assumption on f 0 and g 0 , and their evenness when considering A 1 and B 1 , notice that

|E -Ẽ|(r, z) = z 0
f 0 (X(0; r, y, u), V (0; r, y, u)) -f 0 ( X(0; r, y, u), Ṽ (0; r, y, u)) dudy z 0 g 0 (Y (0; r, y, u), W (0; r, y, u)) -g 0 ( Ỹ (0; r, y, u), W (0; r, y, u)) dudy

≤ cα, (r, z) ∈ [0, T ] × [0, 1].
Let (t, x, v) ∈ A 0 . Both generalized characteristics (X, V ) and ( X, Ṽ ) without backward bounce, starting at (t, x, v), are given by Consequently,

X(s; t, x, v) = x + v(s -t) + t s (r -s)E(r, X(r; t, x, v))dr, V (s; t, x, v) = v - t s E(r, X(r; t, x, v))dr, X(s; t, x, v) = x + v(s -t) + t s (r -s) Ẽ(r, X(r; t, x, v))dr, Ṽ (s; t, x, v) = v - t s Ẽ(
|X -X| + |V -Ṽ | (s; t, x, v) ≤ cT α, s ∈ [0, T ], (t, x, v) ∈ A 0 .
In particular,

|X -X| + |V -Ṽ | (0; t, x, v) ≤ cT α, (t, x, v) ∈ A 0 . (4.4) Let (t, x, v) ∈ A 2 .
Both generalized characteristics with a backward bounce at (t 1 (t, x, v), 1), starting at (t, x, v), are given by

X(s; t, x, v) = x + v(s -t) + t s (r -s)E(r, X(r; t, x, v))dr, V (s; t, x, v) = v - t s E(r, X(r; t, x, v))dr, s ∈ [t 1 (t, x, v), t], X(s; t, x, v) = 1 + (t 1 -s) v - t t1 E(r, X(r; t, x, v))dr + t1 s (r -s)E(r, X(r; t, x, v))dr, V (s; t, x, v) = -v + t t1 E(r, X(r; t, x, v))dr - t1 s E(r, X(r; t, x, v))dr, s ∈ [0, t 1 (t, x, v)[,
and similar equations for ( X, Ṽ ) with a backward bounce at ( t1 (t, x, v), 1). Assume t 1 (t, x, v) ≤ t1 (t, x, v).

It holds that The distance from t 1 to t1 can be controlled in the following way. The definition of (t 1 (t, x, v), V (t + 1 )),

(X -X)(s) = t s (r -s) E(r, X(r)) -Ẽ(r, X(r)) dr, (V -Ṽ )(s) = t s E(r, X(r)) -Ẽ(r, X(r)) dr, s ∈ [ t1 (t, x, v), t].
x -1 + v(t 1 -t) + since, using that X is non-increasing on [0, t] because the sign of Ṽ cannot change, 0 ≤ X(0) -X(t 1 ) ≤ 1 -X(t 1 ) = (X -X)(t 1 ) ≤ cT α.

Here, the last inequality results from (4.12). Moreover, X(0) -X(t The case where more than zero (resp. one) bounce occurs at x = 0 (resp. at x = 1) can be analogously treated by splitting the characteristics between those bouncing an even (resp. odd) number of times. And so, α ≤ cT α. Hence α = 0 for T small enough. Consequently, (X, Y )(0; t, x, v) = ( X, Ỹ )(0; t, x, v), (V, W )(0; t, x, v) = ±( Ṽ , W )(0; t, x, v), (t, x) ∈ [0, T ] × [0, 1], c 0 2 ≤ |v| ≤ 2V 0 , so that (f, g) = ( f , g).

  and P. Degond [BAR1985] proved global in time existence and uniqueness of classical solutions to the Cauchy problem related to the Vlasov-Poisson system in the whole three-dimensional space for small initial data. Using characteristics, K. Pfaffelmoser [PFA1992] proved existence of classical solutions to the Vlasov-Poisson system for general initial data. Together with the study of propagation of moments |v| m f (t, x, v) dxdv with m > 3, P.-L. Lions and B. Perthame [LIO1991] proved existence and uniqueness of an L 1 ∩ L ∞ weak solution to the Vlasov-Poisson system in the whole space. C. Pallard [PAL2012] proved an analogous result for m > 2.

  1],|w-v|<R f 0 (x, w)dv and |v|>U sup x∈[0,1],|w-v|<R g 0 (x, w)dv are arbitrarily small. It follows from the continuity of (3.

By ( 1 .

 1 10), (S(a n )) n∈N is bounded in C([0, T ] × [0, 1]) by sup x∈[0,1],|w-v|<(|E0|+M )T f 0 (x, w)dv + sup x∈[0,1],|w-v|<(|E0|+M )T g 0 (x, w)dv.

  2 and locally in time, Lemma 4.1 provides a bound on the number of possible bounces at the boundary of the domain of the generalized characteristics associated to a solution to (1.1)-(1.8). Lemma 4.2 proves the local in time uniqueness result of Theorem 1.2.

x = 1

 1 on [0, T ]. Denote by (B i ) 0≤i≤2 analogous sets relative to the (Y, W ) and ( Ỹ , W ) characteristics respectively associated to g and g. Denote by α = sup (r,y,u)∈A0∪A2 |(X -X)(0; r, y, u)| + |(V -Ṽ )(0; r, y, u)| + sup (r,y,u)∈A1 |(X -X)(0; r, y, u)| + |(V + Ṽ )(0; r, y, u)| + sup (r,y,u)∈B0∪B2 |(Y -Ỹ )(0; r, y, u)| + |(W -W )(0; r, y, u)| + sup (r,y,u)∈B1 |(Y -Ỹ )(0; r, y, u)| + |(W + W )(0; r, y, u)| .

  + |V -Ṽ | (s; t, x, v) ≤ cT α, s ∈ [ t1 (t, x, v), t], (t, x, v) ∈ A 2 , (4.5)as in the A 0 case. Moreover,(X -X)(s) = 2( t1 -s) -v + t t1 Ẽ(r, X(r))dr + t1 s (r -s)(E(r, X(r)) -Ẽ(r, X(r))dr + t t1 (r -s)(E(r, X(r)) -Ẽ(r, X(r))dr, s ∈ [t 1 (t, x, v), t1 (t, x, v)[. (4.6)

  t 1 )E(r, X(r))dr = 0, v -t t1 E(r, X(r))dr = V (t + 1 ), implies thatV (t - 1 )( t1 -t 1 ) = t1 t1 (r -t1 )E(r, X(r))dr + t t1 (r -t1 ) Ẽ(r, X(r)) -E(r, X(r)) dr.Hence, for T small enough,0 ≤ t1 -t 1 ≤ cT sup r∈[ t1,t] |X -X|(r) + α ≤ cT α,(4.7)by(4.5). Consequently,|X -X|(s) ≤ c| t1 -t 1 | ≤ cT α, s ∈ [t 1 (t, x, v), t1 (t, x, v)[.(4.8)Finally,(X -X)(s) = v(t 1 -t1 ) -(t 1 -s) t t1 E(r, X(r))dr + ( t1 -s) s)E(r, X(r))dr -t1 s (r -s) Ẽ(r, X(r))dr = ( t1 -t 1 ) -v + t t1 E(r, X(r))dr -( t1 -s) t t1 E(r, X(r)) -Ẽ(r, X(r)) dr -( t1 -s) t1 t1 E(r, X(r))dr + t1 s (r -s) E(r, X(r)) -Ẽ(r, X(r)) dr + t1 t1 (r -s)E(r, X(r))dr, (V -Ṽ )(s) = 2 t1 t1 E(r, X(r))dr + t t1 E(r, X(r)) -Ẽ(r, X(r) dr -t1 s E(r, X(r)) -Ẽ(r, X(r) dr, s ∈ [0, t 1 [.Hence, |X -X|(s) ≤ c| t1 -t 1 | + cT α + cT sup r∈[t1,t] |X -X|(r) + c t1 s |X -X|(r)dr ≤ cT α + c t1 s |X -X|(r)dr, by (4.7), (4.5) and (4.8), ≤ cT α, s ∈ [0, t 1 [, (4.9) |V -Ṽ |(s) ≤ c| t1 -t 1 | + cT α + cT sup r∈[0,t] |X -X|(r) ≤ cT α, s ∈ [0, t 1 [, by (4.5), (4.8) and (4.9). (4.10) It follows from (4.9)-(4.10) taken at s = 0 that|X -X| + |V -Ṽ | (0; t, x, v) ≤ cT α, (t, x, v) ∈ A 2 .(4.11)Let us consider the last case, where (t, x, v) ∈ A 1 . Again,|X -X| + |V -Ṽ | (s; t, x, v) ≤ cT α, s ∈ [t 1 (t, x, v), t]. (4.12)Moreover,X(s) = X(t 1 ) + (t 1 -s)V (t + 1 ) + t1 s (r -s)E(r, X(r))dr, V (s) = -V (t + 1 ) -t1 s E(r, X(r))dr, X(s) = X(t 1 ) -(t 1 -s) Ṽ (t 1 ) + t1 s (r -s) Ẽ(r, X(r))dr, Ṽ (s) = Ṽ (t 1 ) -t1 s Ẽ(r, X(r))dr, s ∈ [0,t 1 (t, x, v)[. Consequently, |X -X| + |V + Ṽ | (0; t, x, v) ≤ |X -X| + |V -Ṽ | (t + 1 (t, x, v); t, x, v) + ct 1 ≤ cT α, (t, x, v) ∈ A 1 , (4.13)

  if P is finite (resp. s ∈ [t 1 , t] ∪ n∈P [t n+1 , t n ] if P is not finite).

	Mischler in [MIS1999], using a variety of test functions. Our approach differs from his by considering
	generalized characteristics and continuous distribution functions. Excluding the case where infinitely
	many bounces would accumulate at a boundary point, we prove in Propositions 2.1 and 2.2 that the
	backwards in time generalized characteristics from any (t, x, v) ∈ [0, T ] × [0, 1] × R,
	s → (X(s; t, x, v), V (s; t, x, v)),
	has a finite number of bounces at the boundary, hence reaches time zero. Example 2.1 exhibits a case
	where the map (t, x, v) → V (0; t, x, v) is discontinuous. In Proposition 2.3, the continuity of the map
	(x, v) → (X(0; t, x, v), | V (0; t, x, v) |) is proven, for any t > 0.
	Lemma 2.1 For t > 0 and (x, v) ∈ [0, 1] × R, the second component V of the generalized characteristics
	from (t, x, v) satisfies
	|V (s; t, x, v)| ≤ |v| + T E ∞ ,	(2.4)
	for s ∈ [0, t] Proof of Lemma 2.1. The case where P is finite is classical. Assume P = N * . Denote by t 0 = t and prove
	by induction on n ∈ N that
	|v| -|V (s; t, x, v)| ≤ (t -s) E ∞ , s ∈ [t n+1 , t n ].	(2.5)
	For n = 0 and s ∈ [t 1 , t], the equation ∂ s V (s; t, x, v) = E(s, X(s; t, x, v)) yields
	t
	|v| -|V (s; t, x, v)| ≤	|E(r, X(r; t, x, v))|dr ≤ (t -s) E ∞ .
	s
	Assuming that |v| -|V (s; t, x, v)| ≤ (t -s) E ∞ for s ∈ [t n , t n-1 ] and n ≥ 1, it holds
	|v| -|V (s; t, x, v)| ≤ |v| -|V (t + n ; t, x, v)| + |V (t -n ; t, x, v)| -|V (s; t, x, v)|
		tn
	≤ (t -t n ) E ∞ +	|E(r, X(r; t, x, v))|dr
		s

Acknowledgements

The project leading to this publication has received funding from the Excellence Initiative of Aix-Marseille University -A*MIDEX, a French 'Investissements d'Avenir' program.


Marie Ladier-Fouladi

MOBILIZATION AND PARTICIPATION IN HYBRID REGIMES COMPARATIVE

Over the last recent decades, the adoption of the electoral system by many states, particularly in developing countries, has raised debates about the nature of these new political regimes.

Although most of these states fail to meet the most elementary rules of democracy, namely

freedom of expression, these elections have been taken into consideration by observers and scholars, given that the verdict from the ballot box is universally regarded as the very essence of a democratic regime. They then try to define these regimes using various terms like "competitive authoritarian system", "hegemonic-party system" or "hybrid regimes".

Whatever the definition, the common point shared by these regimes is the organization of political elections, more or less free. This led to invent terms like electoral authoritarianism or pseudo-democratic, in order to somehow define the new feature of these regimes. Without going into details and discuss what democracy is and is not, because it might lead me astray, but I will answer your questions during the discussion-time if you wish to discuss the topic, I will only say that, in my opinion, this effort to name or to distinguish various types of political regimes among the "non-democratic" ones, seems to serve, eventually, to grant a certain legitimacy, and thus a place in the concert of nations to these "elected" leaders.

Having said that, the case of Iran seems to me eloquent in this respect. Thus, for my presentation, I suggest investigating the electoral system set up since the establishment of the Islamic Republic in Iran in 1979, and then I'll present and analyze the last presidential election held in 2013 as a concrete example.

MARIE LADIER-FOULADI

18/11/15 Institutional arrangements between religion and politics, Hybrid political system: The case of Iran 2 Beforehand though, it is important to remember that the Islamic Republic of Iran is built on a dual sovereignty: divine and popular. While the former is the basis and essence of the government of jurisconsult (velayat-e faghih), it is the latter which gives its political legitimacy to the Islamic State. Indeed, referring to the people as a legitimate source of power is inherent to the Islamic State as far as it was born in the wake of and thanks to a popular revolution. Khomeini himself became the Guide following a revolutionary process, and he did not hesitate to exploit this popular support, bypassing the classic institutional procedures. In return, the Iranians, driven by the strength and the enthusiasm the revolution generated, remained active on the political scene. In particular, during the first two years following the overthrow of the monarchy, they always responded positively to the calls of the new leaders and expressed their will to participate in the political life effectively to design the country's destiny. To meet the Iranians' demand for democracy the new leaders were thus forced to adopt the elective principle. Since the referendum of March 1979 that gave birth to the Islamic Republic, the leaders resorted systematically to universal suffrage. Between March 1979 and June 2013, 29 elections and 3 referendums were held. This relatively high frequency might suggest that, paradoxically, democracy is exercised without difficulty in the Islamic Republic of Iran, inasmuch as democracy is often equated with the electoral competition of political parties. But don't be mistaken: in Iran, legitimization through the ballot box actually raises questions on the effective exercise of popular sovereignty, as well as on the representativeness of the elected, given the system of the Islamic Republic in which political space is locked and the Supreme Guide (or the Vali-faghih) has absolute power to act and intervene in all the areas of public life.

To better understand this contradictory situation and what really the elections represent in Iran, it is necessary to examine the electoral procedures first.

I. The main characteristics of the Islamic Republic's electoral procedure

First, let's turn to the authorities organizing the elections, and their prerogatives:

A. The organizing authorities of the elections and their prerogatives

According to electoral law (precisely the principle number 99 of the Constitution), the Council of the Guardians of the Constitution and the Ministry of the Interior are in charge of the organization and the supervision of the elections.

The Council of the Guardians of the Constitution is composed of 12 jurists, including six clerics appointed by the Supreme Guide, and six jurists elected by Parliament among the Muslim jurists nominated by the Head of the Judicial System: that is a non-elective authority.

As its name suggests, the primary mission of the Council of the Guardians is to interpret the Constitution and it may reject bills voted by Parliament because they are deemed incompatible with the constitution or sharia (Islamic law).

Its second mission is to supervise the elections. Between 1980 and 1990, the Council of the Guardians supervised the conformity of the electoral vote, according to that principle. But in 1991 the Council suggested a new interpretation of the 99 th principle, imposing the notion of nezarat-e estesvabi (fair supervising): it endorsed the right to ban the candidates from the election if their Islamic credentials, in the Council's view, were not adequate. That way, all those who belonged to a group that was "un-Islamic" to the Council's view could be excluded without any further justification. The Council has become since then a formidable machine for the exclusion of those candidates deemed unreliable to the Hardliners. The Islamic Parliament, made up mainly of the conservatives, voted in favor of this interpretation in 1995 by modifying article 3 of the legislative electoral law thus allowing the Council of the Guardians to exert discretionary supervision over the entire electoral process, which implies the endorsement or the exclusion of the candidates at will. 1 As for the Ministry of the Interior, in every election it sets up a "National Bureau of Elections" (setad entekhabat keshvar) directed by someone named by the Minister. In close cooperation with the Ministry of the Interior and the Council of the Guardians, the governors, prefects and other local authorities constitute the "executive councils"(hey'athayeejra'ee) that assess the candidates' judiciary past and credentials, and validate or reject their candidacy.

Two major conditions are mentioned for the acceptance of a candidate:

-to have faith in Islam and a real engagement towards it; -to have faith in the foundations of the Islamic Republic and its Constitution.

These conditions are not defined through objective criteria and therefore, the members of the "executive councils" decide, according to their own views and interpretations, whether a candidate fulfills those two conditions or not. The members of the executive councils closely follow the views of the Council of the Guardians. For instance, in the 2008 parliamentary elections2 , the executive councils invalidated some 3,200 over 7,168 candidates, mostly among the reformist candidates, many of them being well-known and being even incumbent MPs for some of them. The hemorrhage was so important that Hachemi Rafsanjani and Mohammad Khatami, the two previous Presidents, and Mehdi Karrubi, the previous President of Parliament urged the Supreme Guide to step forward, and he succeeded in convincing the Council of the Guardians to approve of 4,954 candidates, that is 69% of the registered candidates. The established laws are of little avail and the arbitrariness of their interpretation allows the Council of the Guardians to reject or accept a candidate according to the wishes of the ruling oligarchy, at the top of which is the Supreme Guide. This is one of the major features of the electoral procedure in Iran: many candidates are excluded before the voting, due to the clout of the Council of the Guardians that makes a fair election impossible. Before any vote is cast, the pre-selection of the candidates tips the balance in favor of those singled out by the oligarchy, according to the conjuncture.

As for the candidates' representatives in the polling stations, up to the year 2000, in accordance with articles 80 to 93 of the electoral law, the Council of the Guardians had the prerogative to monitor any electoral process through the Surveillance Committees, from the local up to the central polling stations. In 2000, the Sixth legislature (where the majority lay with the Reformists) voted a single-article bill stipulating that the candidates be granted the right to have their representatives in the polling stations in order to watch the electoral process and in particular, supervise the counting of the ballots. According to this law the candidates were required to introduce their representatives to the prefects and governors in the districts at least 5 days before the elections, and the latter should deliver them a permit at least 48 hours before the polling day. But it is not required that they should be present and the voting process can begin even in their absence. That is why in Statement number 22, where the electoral process is recorded, there is no box for the signature of the candidates' representatives. This procedure, which existed before Ahmadinejad, highlights the fact that the Islamic Republic has never provided the necessary devices to ensure the transparency of elections.

The Iranian electoral procedure is indeed particularly opaque and ambiguous, and it offers a wide scope of opportunities to falsify the elections and rig the vote. It is up to the authorities whether to use these "options" or not. For instance, on the occasion of the 1997 presidential election, the balance of power between the two opposite factions were fairly narrow so that Khamenei and Rafsanjani both delivered speeches urging the ballot boxes to speak freely3 . It was also the case in the 2000 legislative and in the 2001 presidential elections: indeed, the Reformists were still popular and the balance of power was in their favor, the Hardliners then were not able to use these "options" to rig the ballots. From the 2003 municipal election onwards, the Hardliners, who had managed to achieve unity eventually, began to regain the control over the electoral procedure and succeeded in using its various "options," as the ballots held since 2004 have confirmed. In October 2008, with the approval of the Supreme Guide, he incorporated 4,000 new members into the body of the "Political Guides" (hadiyan siyasi) 4 . From then on, that body comprised 12,000 Pasdaran and Bassijis whose formal task was to encourage people to take part in the vote. In reality, their task is to talk people into voting for the Supreme Guide's candidate, as the daily newspaper Keyhan confirmed clearly. In its editorial of August 19, 2009, it is explained that there were "manufacturers of the vote" (ra'ysazan) who played a key role in the election of Ahmadinejad in 2005: These Bassijis "spontaneously rallied and traveled across the country up to the remotest villages, in order to bring the message of change to the people."

B. Additional means that Ahmadinejad put at his disposal

Ahmadinejad invigorated the machinery to secure the vote of the people in his favor through the Pasdaran Army.

Now, I suggest presenting the conditions to be an eligible voter and the voter turnout.

MARIE LADIER-FOULADI

18/11/15 Institutional arrangements between religion and politics, Hybrid political system: The case of Iran

6

C. The electors and voter turnout

In order to be a voter in Iran, one has to be an Iranian citizen, and of sound mind too. The voting age is generally set at 16, but the government decides to lower or increase it at the last minute, according to what is at stake. For instance in the 2000 legislative poll, this age was set at 17, then in the 2005 presidential elections, 15 and in the 2008 parliamentary elections, at 18.

It is very important to know that, in Iran, there is neither an electoral roll nor voter registration card. The voters can vote wherever they like, provided they show their identity card, their identity booklet that will be stamped by the authorities, their forefinger being tainted with indelible ink. This allows the government to register the votes, without any possible supervision at the local or regional level (many towns registered more than 100% voters; the government subsequently argued that voters from other regions had cast their votes in those districts). Furthermore, the pre-printed ballots with the name of the candidates do not exist in Iran. Voters have to write the names and codes corresponding to the candidate of their choice on the blank ballot specifically designed and printed by the Ministry of Interior on the occasion of each vote. It goes without saying that this process leads to many errors of various kinds (inaccurate, illegible or incomplete names of the candidates, etc.) and, consequently, provides great opportunities for manipulation of the ballots.

Since there is no electoral roll, the Interior Ministry's National Office of Elections has to use the population census data (every decade, the last one being 2011) in order to calculate the number of Iranian citizens of legal voting age. The turnout rate is then the ratio between the number of ballots cast and the number of estimated Iranian citizens of legal voting age.

Up to 2005 the Interior Ministry's estimates were based on the statistics provided by the Iranian Statistical Center. But since then, it resorts particularly to the data provided by the Office of Registration and Records (sazeman sabt ahval), whose direction is ideologically very close to the Hardliners, to make an estimate of the number of Iranian citizens of legal voting age. Now, this Office used its own statistics which are not published, cannot be verified, and are subject to manipulation. According to my own estimates, 5 the Office's estimates show a gap of 3 to 6 million with those of the Iranian Statistical Center. By underestimating the number of potential electors, the government artificially increases the turnout 5 Our estimates are based on an extrapolation of the two last Iranian censuses of 1996 and 2006 by means of the demographic growth rate between these two censuses. The Afghan immigrants, who have no right to vote, were naturally taken into account in this calculation.

rate. Here are a few examples: in the 2009 presidential elections, the Interior Ministry declared 46.2 million the number of Iranian citizens of legal voting age, whereas, according to my calculation, their number was 51.3 million, that is 5 million more (10% of the electorate).

After the election, the Ministry of Interior declared that the number of votes cast amounted 39.3 million6 ; therefore, from its perspective the turnout rate was 85%. Assuming that 39.3 million people actually voted, according to my estimate, the turnout rate would rather be around 76%. On top of that, those 5 million "potential electors" are liable to have been used by the government in a "phantom vote", cast in its favor, without supervision by independent observers.

All in all, the lack of an electoral roll, the fallacious number of Iranian citizens of legal voting age, the ever changing legal voting age, and the lack of serious supervision by nongovernmental bodies cast indeed serious doubts over the conformity/legality of the entire electoral system. In the absence of representatives of the opposition candidates in the polling stations, it is obvious that members of the executive councils may have influenced the voters and particularly were free to rig the votes cast during the counting.

II. The 2013 presidential election: a minimal consensus

In my opinion, based on an in-depth analysis of the process of the elections, the results of the 2013 presidential election was the result of a minimal consensus between Khamenei and Rafsanjani. The two men shared a common goal: to remove Ahmadinejad and his team from the political scene. In short, it is likely that negotiations and bargaining have taken place between Khamenei and Rafsanjani. As far as it is possible to know, the dice were cast after the third television debate and during the second week of the election campaign. Khamenei and Rafsanjani agreed that Rohani should succeed to Ahmadinejad. Then, they managed to stage his election. A real master stroke! By a sleight of hand, they were able to deceive everyone, especially the Iranian voters, and made them believe that this time, unlike in 2009, their votes had been taken into consideration. Everybody was satisfied with the results, both inside and outside the country.

Shall I remind that the presidential election of June 2013, since it took place in an uncommon explosive sociopolitical and economic context, represented crucial stakes for the regime? First, on the domestic level, the Islamic Republic was experiencing the worst political crisis of its history characterized, on the one hand, by the weakening of the legitimacy of its power and, on the other hand, by the struggle for power between the Supreme Guide, Ali Khamenei, and Mahmoud Ahmadinejad. Both first and second characteristics were linked to the presidential election of 2009 and its aftermath. Very briefly, Mahmoud Ahmadinejad had begun to avail himself of a robust political legitimacy against the Guide, because, allegedly, 24 million ballots had been cast in his favor. He tried to provide himself with the means to remodel the Islamic Republic so that the Republic remains Islamic for ever but without the active participation of the clergy, and thus without the system of the velâyât-e faqih, the governance of jurisconsult. This project, visible through the positions, speeches and political practices of Ahmadinejad and his team, is at the origin of struggle for power between the two men.

On the international level, then, the discussions on the nuclear program were stalled while international sanctions had tightened since 2009, and had increased the endemic economic crisis, which shook he country for more than three decades.

The Guide was thus in an extremely awkward situation and his choice for a successor to Ahmadinejad seemed critical as far as the survival of the regime was concerned and, consequently, regarding his own power Choosing Hassan Rohani was wise. He has been very close to the political power since the institution of the Islamic Republic in Iran. He was the first to name Khomeini Imam Khomeini, a former nuclear negotiator, and the Guide's representative to the Iranian Security Council. In short, Rohani is a reliable person to both Khamenei and Rafsanjani, and he does not represent the opposition whatsoever.

Rohani hardly exceeded the threshold of 50% of the votes. Yet, his opponents who conceded their defeat recognized his narrow victory. That way, a very risky second round, for it may have divided the electorate deeply, was avoided eventually Undoubtedly, it is the supreme Guide, Ali Khamenei, who was strengthened from the very long political battles of the 2000s, particularly from the test of the 2013 presidential election, so that he emerged as the real master of the country again.

It is indeed the intervention of Rafsanjani and Khatami in the last weeks of the election campaign that lent a helping hand to Khomeini to mobilize the voters, and that's why the MARIE LADIER-FOULADI 18/11/15 Institutional arrangements between religion and politics, Hybrid political system: The case of Iran 9 turnout could reach a decent threshold (73% according to the Ministry of Interior) for a regime that had just experienced the fiercest disputes in the previous presidential election.

It goes without saying that Rafsanjani and Khatami didn't help out the Supreme Guide without compensation. The latter had to support their candidate, Hassan Rohani.

But this choice, with regard to internal and external stakes, benefited mainly to the Guide. This helped him recover his legitimacy and that of the Islamic regime with the national and international public opinion. The strength of the Islamic regime was no longer an issue.

  Despite this powerful lever, Mahmoud Ahmadinejad felt compelled to set up additional devices in order to control and orientate the vote according to his will. After the 2008 parliamentary elections, he replaced the Interior Minister by someone closer to him, and in March 2009, for the third time since 2006, he changed the director of the National Bureau of Elections, appointing Kamran DANESHJOO, one of his closest right-hand men.

For instance, in

2004, it invalidated the candidacy of half of the incumbent members of the parliament who asked for the renewal of their election. Or, in the elections, the Council invalidates sometime hundreds of thousands of votes, mainly in the large districts like Tehran, when the results of the election do not fit its views. See Ladier-Fouladi M., 2009, Iran Un monde de paradoxes, Nantes, l'Atalante, coll. Comme un accordéon,

See Marie-Ladier Fouladi, "La face cachée des législatives iraniennes de 2008", in La Vie des idées, July

30th 2008, http://www.laviedesidees.fr/La-face-cachee-des-legislatives.html.

See the Rafsadjnai's speech at Friday prayer on May 16, 1997 and Khamaeni's speech on May 21, 1997 on the eve of the close of the electoral campaign period, quoted in Ladier-Fouladi M., 2009, Iran Un monde de paradoxes, op. cit.

A paramilitary's organization created in 2003 within the Pasdaran Army.

It goes without saying that there is no way to check the statistics on votes cast, declared by the Interior Ministry.