(MITAKE Hiroyoshi) Joint work with Q. Liu (U. Pittsburgh) Oct/5/2012

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1 (MITAKE Hiroyoshi) ( ) Joint work with ( ), Q. Liu (U. Pittsburgh) Oct/5/2012 in, IV 1

2 0 Introduction (Physical Background) Morphological stability ( ) in crystal growth: Burton, Cabrera, Frank 51 (Micro Level), Chernov 74, E, Yip 01 (Macro Level). V = σ(x)m(p), m(p) = p tanh( p s p s p ), p s = d ε. 2x s V : growth speed in the direction normal to a crystal surface Γ t. m: anisotropy of the kinetic energy. σ: supersaturation ( ). d: step height, x s : surface diffusion distance of a molecule. p: local slope of the crystal surface. 2

3 雪の結晶の初期段階とファセット結晶形状の不安定性 北海道大学低温科学研究所古川研究室撮影

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5 We consider V = σ(x)m( p /ε) f(x) on Γ t. (1) Graph Representation. Introduce the function z ε which satisfies Γ t = {(x, z ε (x, t)) x R N }. Then we have p = Dz ε (x) and V = z ε t 1 + Dz ε 2. Thus, the above surface evolution equation can be written by zt ε σ(x)m( Dzε ) 1 + Dz ε 2 = f(x) 1 + Dz ε 2 (2) ε References: Giga, Surface Evolution Equations, Springer. 3

6 Yokoyama-Giga-Rybka 08 Investigate the behavior of u ε in the ε-time scale, i.e., ũ ε (x, τ) = z ε (x, ετ)/ε (microscopic height) and a new independent variable τ = t/ε (microscopic time). Then ũ ε satisfies ũ ε τ + σ(x)m( Dũ ε ) 1 + ε Dũ ε 2 = f(x) 1 + ε Dũ ε 2 Thus, (at least formally) ũ ε converges to a solution of ũ τ + σ(x)m( Dũ ) = f(x) in R N (0, ). A typical example is σ(x) := σ(1 x 2 ) +, m(r) := r tanh(1/r) (r 1) and, f : nucleation ( ) density. 4

7 1 Main Result We consider HJ equations of the form u t + σ(x)m( Du ) = f(x). (3) (A1) m : [0, ) [0, 1) is strictly increasing, Lipschitz continuous with m(0) =: m 0 [0, 1) and m(r) 1 as r. (A2) f : R n R is continuous and satisfies A := {x R n : f(x) = min f, σ(x) = σ} =, (Revisit) where σ := max σ. (A3) u 0 is Lipschitz continuous in R n. (A4) Set c := σm 0 min f. Ω e := {σ( ) f( ) > c} is bounded and σ C 1 (R N ) satisfies Dσ(x) 0 on Ω e. Remark. Ω e is called a maximal stable region. 5

8 Theorem 1 (Giga-Liu-M., JDE 2012, Trans. AMS to appear). Assume that (A1) (A4). Let u be a solution of (3) with u(, 0) = u 0 W 1, (R N ). Then, u(, t) + ct φ loc. uniformly on Ω e, u(, t) + ct + loc. uniformly on R n \ Ω e as t +, where φ is a solution of Dv = m 1( f(x) + c) =: h(x) in Ωe, σ(x) (S) v n = + on Ω e, sup v(x) < +. Ω e 6

9 Interpretation: Theorem 1 now gives a clear view of z ε on the effective domain Ω e. We have z ε (x, t) = εũ ε (x, t ε ) = εũ(x, t ε ) + o(ε) = ε(φ (x) ct ε + m(ε )) + o(ε) t = εφ (x) + ct + o(ε). Therefore, roughly speaking, the growing facet moving according to (1) is flat up to order ε with speed c on the effective domain Ω e. Remember c = σm 0 min f. 7

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11 Asymptotic Profile on the Effective Domain Ω e. We define the functions φ, φ C(Ω e ) by φ (x) := inf (u(x, t) + ct), t 0 φ (x) := min{d(x, y) + φ (y) y A}, d(x, y) := inf { t h(γ(s)) ds t > 0, γ(t) = x, γ(0) = y, γ(s) 1}. 0 Theorem 2 (Asymptotic Profile). We have φ (x) = lim t (u(x, t) + ct) for all x Ω e. Example. Let n = 1, u 0 = f 0, σ(x) = σ(1 x 2 ) +. Then we have c = σm 0, Ω e = ( 1 m 0, 1 m 0 ), A = {0}. Moreover, φ (x) = min y A {d(x, y) + φ (y)} = d(x, 0). Thus we obtain φ (x) = x 0 m 1( m 0 (1 s 2 ) + ) ds for all x Ωe. 8

12 Discussion 1 (Morphological Stability). In the theory of crystal growth, it is known that as long as the non-uniformity in supersaturation on the facet is not too large, the faceted crystal can grow in a stable manner. Question. How much of non-uniformity implies a stable morphology? Answer. f(x) + c < 1 x R n σ(x) f(x) min f + σm 0 < σ(x) σ x R n. In this case, we can expect we have the large-time asymptotic u(x, t) + ct v(x) uniformly for x R N as t 9

13 Discussion 2 (Step Source). Revisit Assumption (A2): A := {x R n : f(x) = min f, σ(x) = σ} =. First Case: f 0. The set A is considered as a step source in the theory of crystal growth. Mathematically, we can see that u + ct is non-increasing as t. The function f gives a nucleation density of crystal. Our assumption (A2) says that there is a step source in the place where no nucleation occurs. Question. If this is not the case? Reconsider c := max x (σ(x)m 0 f(x)), A := {x σ(x)m 0 f(x) = c}. We don t know the uniform continuity of u on A yet. 10

14 Discussion 3 (Mean Curvature effect). Effect of tension. Consider V ε = ( σ(x) div (n(x)) ) m ( p ) f(x) on Γt. ε If we use the microscopic time and height, i.e., ũ ε (x, τ) = z ε (x, ετ)/ε then we cannot see the difference, since ( ũ ε t +... εdũ ε ) m( Dũε )div ε Dũ = 0. ε Dũ

15 Thank you for your kind attention! 12