in the presence of plasma sheaths Department of Mathematical Sciences Ulsan National Institute of Science and Technology (UNIST) joint work with Masahiro Suzuki (Nagoya) and Chang-Yeol Jung (Ulsan) The 1st Meeting of Young Researchers in PDEs Yonsei University, August 18-19, 2016
Euler-Poisson equations The motion of the positive ions in a plasma is governed by ρ t + (ρu) = 0, (ρu) t + (ρu u) + K ρ = ρ φ, ε φ = ρ e φ, ρ(x, t), u(x, t) and φ(x, t) represent the density, velocity of the positive ions and the electrostatic potential, respectively. K = temperature of positive ion ε is the ratio of the Debye length λd to the radius of an inner sphere R, that is, ε = λ 2 D/R 2. (Lab setting λ D 10 4 )
Plasma physics Plasma (in Greek anything formed ) is one of the four fundamental states of matter. Like gas, plasma does not have a definite shape or a definite volume unless enclosed in a container. Unlike gas, under the influence of a magnetic field, it may form structures such as filaments, beams and double layers. The presence of a significant number of charge carriers makes plasma electrically conductive so that it responds strongly to electromagnetic fields
Plasma physics Quasi-neutrality (quasi, from the Latin, as if, resembling ) describes the apparent charge neutrality of a plasma overall, while at smaller scales, the positive and negative charges making up the plasma, may give rise to charged regions and electric fields. Since electrons are very mobile, plasmas are excellent conductors of electricity, and any charges that develop are readily neutralized, and in many cases, plasmas can be treated as being electrically neutral. The distance over which quasi-neutrality may break down, is often described by the Debye length (or Debye sphere), and varies according to the physical characteristics of the plasma. The Debye length is typically less than a millimetre (ie. charged regions will not exceed a millimetre), in plasmas found in fluorescent light tubes, tokamaks (used in fusion research), and the ionosphere.
From the two-fluid model to our model Full two-fluid model reads as tn ± + (n ±v ±) = 0, n ±m ±( tv ± + (v ± )v ±) + T ± n ± = ±en ± φ, ε φ = 4πe(n + n ) After non-dimensionalization together with Boltzmann-Maxwell relation (m m +), one obtains EP for ion dynamics: ρ t + (ρu) = 0, (ρu) t + (ρu u) + K ρ = ρ φ, ε φ = ρ e φ.
Euler equations of compressible fluids: Euler equations of compressible fluids is given by ρ t + (ρu) = 0, ρ(u t + (u )u) + p = 0, S t + (u )S = 0, where P(ρ, S) = Aρ γ e S is the pressure and γ > 1 is the adiabatic index. σ := P ρ( ρ, S) = Aγ ρ γ 1 e S is the speed of sound. Note that the maximum speed of propagation of the wave of a smooth disturbance is governed by σ. Heuristically, the linear approximation of the Euler equations near the constant state (ρ, u, S) = ( ρ, 0, S) is the wave equation: ρ tt σ 2 ρ = 0. (T. Sideris 1985) C 1 solutions may blow up in finite-time. Hyperbolic + Nonlinear blow up
Dispersion relation EP for ion dynamics: ρ t + (ρu) = 0, (ρu) t + (ρu u) + K ρ γ = ρ φ, ε φ = ρ e φ. Linearizing about the constant state (ρ, u, φ) = (1, 0, 0), we have ρ t + u = 0, u t + K ρ = φ, ε φ = ρ + φ. Furthermore we obtain the wave equation with nonlocal term: ρ tt K ρ = (( 1 + ) 1 ρ). Dispersion relation: K + 1 + K ξ 2 λ = i ξ 1 + ξ 2
Plasma sheaths Physical situation: a negatively charged material is immersed in a plasma and the plasma contacts with its surface. The plasma sheath is the transition from a plasma to a solid surface, where the positive ions outnumber the electrons (Langmuir, 1929) Plasma sheath is emerged. The interaction of electron and positive ion space charges in cathode sheaths, Phys. Rev. 33 (1929), pp. 954 989. (Bohm, 1949) Bohm criterion Minimum ionic kinetic energy for a stable sheath, in The characteristics of electrical discharges in magnetic fields, A. Guthrie and R.K.Wakerling eds., McGraw-Hill, New York, (1949), 77 86.
Plasma sheaths The plasma sheath is a layer in a plasma which has a greater density of positive ions, and hence an overall excess positive charge, that balances an opposite negative charge on the surface of a material with which it is in contact.
Program: hydrodynamic description of the formation of plasma sheaths via the Euler-Poisson system Plasma sheaths (sharp transition layers near the boundary, stationary boundary layer solutions) Bohm criterion (physical criterion) Existence of the stationary solution (plasma sheath) Euler-Poisson system its quasi-neutral limiting equations (Euler system) Stability (time-asymptotic) vs blow-up Local existence (hyperbolic initial-boundary value problem) Global existence (uniform estimates) or Spectral analysis towards blow-up Quasi-neutral limit Stationary and time-dependent problem Uniform decay estimate (in terms of ε = Debye length) Correctors (information on boundary layers)
Euler-Poisson system and its quasi-neutral limiting eqns E-P system: ρ ε t + (ρ ε u ε ) = 0, (ρ ε u ε ) t + (ρ ε u ε u ε ) + K ρ ε = ρ ε φ ε, ε φ ε = ρ ε e φε. The associated quasi-neutral limiting equations (ε = 0) ρ 0 t + (ρ 0 u 0 ) = 0, u 0 t + u 0 u 0 + (K + 1) log ρ 0 = 0.
Euler-Poisson system in an annular domain Annulus domain Ω: Ω := {(x 1, x 2, x 3) R 3 1 < where L > 0 is the width of the annulus domain. We consider E-P in Ω: wehre x Ω and t 0. Initial conditions: Boundary conditions: where 1, u + and φ b are constants. ρ t + (ρu) = 0, 3 j=1 x 2 i < 1 + L}, (ρu) t + (ρu u) + K ρ = ρ φ, ε φ = ρ e φ, (ρ, u)(0, x) = (ρ 0, u 0)(x), inf x Ω ( u0 2 Kρ 0 )(x) > 0, inf ρ0(x) > 0, x Ω sup u 0 ν(x) < 0, x Ω (ρ, u)(t, 1 + L, ω) = (1, u +ω), ω := x x, φ(t, 1, ω) = φ b, φ(t, 1 + L, ω) = 0.
Existence of stationary radial solutions Condition (Bohm s Criterion for the annulus domain) u 2 + > a (1 + K), where a > 1 is a larger root of 1 + 4 log(1 + L) x + log x = 0. Theorem (Existence for the stationary Q-N-L) Let u 2 + > a (1 + K) hold. Then the SLP has a unique solution ( ρ 0, ũ 0 ) C (I ) C(I ). Moreover, ρ 0 is a monotone decreasing function with the bound 1 = ρ 0 (1 + L) ρ 0 (r) ρ 0 (1). Theorem (Existence for the stationary E-P eqns.) There are constants ε 2 > 0 and δ 1 > 0 such that if ε ε 2 and φ b log ρ 0 (1) + δ 1, the SEP has a unique solution φ C 2 ([1, 1 + L]) satisfying φ(r) G(r, ρ 0 (r)) r [1, 1 + L] φ(r) log ρ 0 (r) + δ 1 r [1, 1 + L]
Quasi-neutral limit in the presence of plasma sheaths Theorem (Quasi-neutral limit) Under the same assumptions, the difference ( ρ ρ 0 θ 0 φ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) satisfies the decay estimates ( ρ ρ 0 θ 0 ρ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) L 2 Cε3/4, ( ρ ρ 0 θ 0 ρ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) H 1 Cε1/4, ( ρ ρ 0 θ 0 ρ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) L Cε1/2, where C > 0 is a constant independent of ε. Proposition (pointwise estimates for the correctors) Under the same assumptions, for any r I, there hold φ b φ ( 0 (1) e C 0 /ε(r 1) e ) C 0 /εl θ 0 φ (r) φ b φ 0 (1) e c 0 /ε(r 1), where c, C, c 0 and C 0 are some positive constants independent of ε. This implies that the thickness of the boundary layer is of order O( ε) = O(λ D ). By a direct calculation, it seems that the best L 2 and H 1 estimates for φ φ 0 are φ φ 0 L 2 Cε 1/4 and φ φ 0 H 1 Cε 1/4, respectively.
Numerical experiments in radial symmetry setting (Jung-K-Suzuki 2016)
Existence for the stationary Euler-Poisson system Condition Recall the physical assumptions: (Bohm criterion) u+ 2 > a (1 + K) φ b G(1, ρ 0 (1)) Theorem There are constants ε 2 > 0 and δ 1 > 0 such that if ε ε 2 and φ b log ρ 0 (1) + δ 1, the SEP has a unique solution φ C 2 ([1, 1 + L]) satisfying φ(r) G(r, ρ 0 (r)) r [1, 1 + L] φ(r) log ρ 0 (r) + δ 1 r [1, 1 + L]
Quasi-neutral limit problem Study the quasi-neutral limit problem (ε 0): For (ρ = ρ ε, u = u ε, φ = φ ε ), (P ε ) (ρu) r = 2 r ρu, (ρu 2 + Kρ) r = ρφ r 2 r ρu2, εr 2 (r 2 φ r ) r = (ρ e φ ), φ = φ b at r = 1, (ρ, u, φ) = (1, u +, 0) at r = 1 + L. (0.1) For (ρ = ρ 0, u = u 0, φ = φ 0 ), (P 0 ) (ρu) r = 2 r ρu, (ρu 2 + Kρ) r = ρφ r 2 r ρu2, 0 = ρ e φ, (ρ, u) = (1, u +) at r = 1 + L. (0.2) discrepancy bet n the boundary data φ b and log ρ(1) causes a sharp transition layer
Quasi-neutral limit in the presence of plasma sheaths Theorem (Quasi-neutral limit) Under the same assumptions, the difference ( ρ ρ 0 θ 0 φ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) satisfies the decay estimates ( ρ ρ 0 θ 0 ρ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) L 2 Cε3/4, ( ρ ρ 0 θ 0 ρ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) H 1 Cε1/4, ( ρ ρ 0 θ 0 ρ, ũ ũ0 θ 0 u, φ φ 0 θ 0 φ ) L Cε1/2, where C > 0 is a constant independent of ε, and θ 0 is a corrector which is a solution to the dominating equations: εθ 0 rr = g(r, φ0 + θ 0 ) e φ0 θ 0, θ 0 (1) = φ b + log ρ 0 (1), θ 0 (1 + L) = 0. RMK. By a direct calculation, it seems that the best L 2 and H 1 estimates for φ φ 0 are φ φ 0 L 2 Cε 1/4 and φ φ 0 H 1 Cε 1/4, respectively.
Pointwise estimates for the correctors Proposition (pointwise estimates) Under the same assumptions, for any r I, there hold ( φ b φ 0 (1) e C 0 /ε(r 1) e ) C 0 /εl θ 0 φ (r) φ b φ 0 (1) e c 0 /ε(r 1), where c, C, c 0 and C 0 are some positive constants independent of ε. This implies that the thickness of the boundary layer is of order O( ε) = O(λ D ).
Numerical Experiments for Quasi-neutral limit problem
Sketch of proof To find the boundary layer structures, we decompose φ = φ 0 + θ. ε d 2 dr ( φ 0 + θ) + 2 2 r ε d dr ( φ 0 + θ) = g(r, φ 0 + θ) e φ 0 θ. Since φ 0 is independent of ε, the terms εd 2 φ0 /dr 2 and εd φ 0 /dr are small and we drop them. In addition, to zoom the behavior of θ near r = 1, using a stretched variable r = (r 1)/ ε, we obtain d 2 d r θ + 2 d ε 2 ε r + 1 d r θ = g(r, φ 0 + θ) e φ 0 θ. (0.3) Then we look for the leading asymptotic term θ 0 φ with θ θ0 φ + εθ 1 φ. Substituting this expansion for θ in (0.3) and using the Taylor expansion for g(r, φ) and e φ about φ = φ 0 + θ 0 φ, we find that d 2 d r 2 θ0 φ = g(r, φ 0 + θ 0 φ ) φ 0 θ 0 e φ + O( ε). At the leading order, i.e. O(1), we obtain the equation for the boundary layers.
Sketch of proof Define a corrector function by θ(r) := (φ b φ 0 (1)) exp ( r 1 ε ) χ(r). Then θ 0 φ θ satisfies ε(θ 0 φ θ) = g(r, φ 0 + θ 0 φ ) φ 0 e θ φ 0 θ R 1, (0.4a) (θ 0 φ θ)(1) = (θ 0 φ θ)(1 + L) = 0, (0.4b) where R 1 := ( 2 εχ + εχ ) (φ b φ 0 (1))e (r 1)/ ε. On the other hand, by a straightforward calculation, one can estimate θ and R 1 as where C > 0 is a constant independent of ε. θ L 2 + R 1 L 2 Cε 1/4, (0.5)
By L 2 estimate in a straightforward fashion, we have 1+L 1+L ( ) ε (θ 0 φ θ) 2 dr + θ 0 φ g(r, φ 0 + θ 0 φ ) φ 0 e θ φ 0 dr 1 1 1+L ( ) = θ g(r, φ 0 + θ 0 φ ) φ 0 e θ φ 0 g(r, φ 0 ) + e φ 0 1+L dr + ( θ + R 1)(θ 0 φ θ)dr. 1 1 This gives ε (θ 0 φ θ) 2 L 2 + θ 0 φ 2 L 2 Cε 1/4. It is easy to check that θ L 2 Cε 1/4 Combining these, we have ε (θ 0 φ ) 2 L 2 + θ 0 φ 2 L 2 Cε 1/4.
Moreover, define Then it satisfies w := φ φ 0 θ 0 φ. ε(r 2 w ) = r 2 {K(r, φ, φ 0 + θ 0 φ ) + J(r, φ, φ 0 + θ 0 φ )}w + R2, where R 2 = ε{2r( φ 0 + θ 0 φ ) + r 2 ( φ 0 ) }. w(1) = w(1 + L) = 0, Note that R 2 L 2 Cε 3/4. Again, by energy estimate, we arrive at ε w 2 L 2 + w 2 L 2 C R 2 2 L 2 Cε 3/2.
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