Travelling bubbles in a moving boundary problem of Hele-Shaw type

Transcription

1 Travelling bubbles in a moving boundary problem of Hele-Shaw type G. Prokert, TU Eindhoven Center for Analysis, Scientific Computing, and Applications (CASA) g.prokert@tue.nl Joint work with M. Günther, Universität Leipzig 1/21

2 Modeling background I: Hele-Shaw bubble with kinetic undercooling ψ =0 ψ=o(1) in R 2 \ Ω (t) as x large Ω (t) n V n = ψ.n+n 1 ψ ψ. n = x +ln l 1 1 ψ := φ x 1, l 0 small: kinetic undercooling parameter 2/21

3 Modeling Background II: Electric streamers PDE Model: (EBERT, ARRAYAS e.a. 02, 04,... ) σ t div (σe) = σf( E ), ρ t = σf( E ), E = φ, div E = ρ σ in R 2 E e 1 as x 1 σ: electron density ρ: ion density E: electric field f: ionization rate 3/21

4 Front solutions Minimal model: (EBERT, ARRAYAS, MEULENBROEK e.a.) Ω (t) (ionized, conductive) R 2 \ Ω (t) (nonionized: ρ=σ=0 ) φ=0 φ=0 n φ~ e 1 as x 1large V φ. n = n φ l φ. n =0 l 0 small interface thickness. Boundary conditions from an analysis of planar fronts in the PDE model. But: The PDE streamer model only admits fronts that advance into the nonionized domain. 4/21

5 Known results: Area is conserved: Vol Ω(t) = const. l = 0: problem is strongly linearly ill-posed l > 0: short-time existence of solutions for sufficiently regular initial boundaries (P. 99) 5/21

6 Travelling bubbles (TB) Ansatz: Ω(t) = Ω + νt, comoving frame: Ω ν ψ = 0 in R 2 \ Ω, ψ l n ψ = x 1 + ln 1 on Ω, ψ = O(1) as x, n ψ + ( e 1 ν)n = 0 on Ω. stationary FBP for (Ω, ν). 6/21

7 Trivial TBs: Circles Ω 0 = B(0, 1), ν 0 = 2 l + 1 e 1, ψ = l 1 x 1 l + 1 x 2. Translation and scaling invariance three-parameter family Ω = B(a, R), ν = 2R l + R e 1. 7/21

8 Nontrivial TBs: l = 0: Ellipses Ω := {(x 1, x 2 ) x 2 1/a 2 + a 2 x 2 2 < 1}, ν = (1 + a 2 ) e 1. l = 1: ν = e 1 : purely geometric problem: n 1 = x 1 on Ω various families of nonsmooth domains bounded by circular arcs 8/21

9 The problem Find all TB solutions (Ω, ν) near the unit circle for fixed l. Small perturbation approach: Unknown: u : S R + S Ω u Ω u := {x R 2 0 < x < u(x/ x )} {0} Transformation via x(u)(θ) := u(θ)(cos θ, sin θ) T. Note: x(u) Diff(S, Ω u ) 9/21

10 Operator equation TB problem F (u, ν) := l ν n(u) x 1 (u) + A l (u) [ x 1 (u) ln 1 (u) ] = 0 on S. n(u): transformed outer unit normal (nonlinear DO of order 1) A l (u): transformed exterior Robin-Dirichlet operator (nonlinear ΨDO of order 1) Normalization conditions: u 2 (θ) dθ = 2π, S S u 3 (θ)(cos θ, sin θ) T dθ = 0. 10/21

11 The nonlocal operator A l Properties: A l C (H s +(S), L(H t 1 (S), H t (S))), s > 3/2, t [1/2, s] + corresponding locally uniform estimates for Fréchet derivatives. Linearization: A l (u) = ψ x(u) A l(u)[h] = ψ x(u) + ( r ψ x(u))h, where ψ = ψ (u)[h] solves ψ = 0 in R 2 \ Ω u, ψ l n ψ = r ψh + l ( r n ψh + ψ n (u)[h] ) on Ω u, ψ = O(1) as x. r : radial directional derivative. 11/21

12 The linearized TB problem Formally: F (u( ν), ν) = 0 solved by u = u( ν) = 1 + h + o( ν ν 0 ) where where ˆL := D u F (1, ν 0 ). Solvable in terms of Fourier series: ˆLh = l( ν ν 0 ) (cos θ, sin θ) T, h, 1 = h, cos = h, sin = 0, h = ( ν ν 0 ) φ, φ = with c k (l) k l+1 l, so s := 1 l c k (l)(cos(2kθ), sin(2kθ)) T k=1 φ H s (S) for s < s, φ / H s (S) for s > s. 12/21

13 Degeneration Observe: ˆL 2l l + 1 sin θ θ. The leading order coefficient has zeros on S. ˆL is Fredholm in the scale {H s (S)}, but its kernel depends on s. Even for s < s, ˆLh H s 1 does not imply h H s. The nonlinear problem is not straightforwardly tractable by the Implicit Function theorem. 13/21

14 The nonlinear problem: Quasilinearization Trick: Differentiate the problem to get a quasilinear structure: where F (1 + v, ν) = 0 θ F (1 + v, ν) = 0 L(v, ν)v = l( ν ν 0 ) ( sin θ, cos θ) T, L(v, ν)w = θ ˆLw + a(v, v, ν)w + R(v, ν)w, } a(v, v, ν) H s 1 (S) R(v, ν) L(H s (S), H s 1 small, vanish for v = 0. (S)) Caution: dom L(v, ν) depends on v, perturbation does not work straightforwardly here. 14/21

15 The core: Degenerate first order DOs Localized case: Consider with γ(0) = 0, γ < 0. Assume L 0 (γ, b)u := γu + bu = f in ( 1, 1) (D) b + (σ 1/2)γ µ > 0. Then, for f H σ ( 1, 1), supp f ( δ, δ) there is precisely one u H σ ( 1, 1) satisfying (D) and supp u ( δ, δ). Estimate: u σ C f σ. Proof: Extend to R, show degenerate estimate [L 0 u, u] H σ c u 2 H σ use Galerkin approximations. Remark: σ θ L 0 (γ, b) L 0 (γ, b+σγ ) σ θ 15/21

16 Globalize to S, include normalization conditions and perturb (carefully): Estimates for linearized operator Assume v 1 s, ν ν 0, w 1 1 0, w 2 cos 0, w 3 sin 0 small. Then uniquely solvable, L(v, ν)u = f + c, w i, u = g i u t + c C( f t 1 + g ), t [2, s]. 16/21

17 The nonlinear problem: Fixed point iteration Define (v n ) by v 0 = 0, L(v n, ν)v n+1 = l( ν ν 0 ) ( sin θ, cos θ) T + c n+1, w i (v n ), v n+1 = 0. (v n ) remains small in H s, converges in H s 1. c n 0. 17/21

18 Result Fix l (0, 2/7), s [4, 1/l + 1/2) N. For ν ν 0 small, the TB problem has a unique normalized solution u = u( ν) H s with u( ν) = 1 + l( ν ν 0 ) φ + o( ν ν 0 ) in H s 1. 18/21

19 Numerics (in fact, just Mathematica c...) Solutions to the linearized problem: ν ν 0 = ±0.1 e 1 ν ν 0 = ±0.1 e l = 1/ l = 4/ /21

20 Recall TB problem is degenerate elliptic. Upper smoothness bound: s (l) as l 0. Kinetic undercooling regularizes the evolution but deregularizes the travelling bubble profiles. 20/21

21 Open questions Stability / long time behavior of the evolution problem actual smoothness of the TBs 3D case 21/21