Some Semi-Markov Processes

Transcription

1 Some Semi-Markov Processes and the Navier-Stokes equations NW Probability Seminar, October 22, 2006 Mina Ossiander Department of Mathematics Oregon State University 1

2 Abstract Semi-Markov processes were of interest to Ron Pyke throughout his career. This interest is documented in articles he authored or co-authored appearing from the early 1960 s to the late 1990 s. This talk describes how some semi- Markov processes can be used to give insight into solutions of the deterministic 3-dimensional Navier-Stokes equations governing the velocity of incompressible fluids. 2

3 Ronald Pyke s students: 1960 s: Betty Flehinger, Simeon Berman, Ronald Schaufele, Michael Wichura, David Root 1970 s: Ravi Nathan, Bruce Lindsay 1980 s: Mina Ossiander, Steven Hubert 1990 s: Joong Kwon 3

4 Semi-Markov proccesses as described by Pyke (1961)... a stochastic process which moves from one to another of a countable number of states with the successive states visited forming a Markov chain, and that the process stays in a given state a random length of time... It is thus a Markov Chain for which the time scale has been randomly transformed. 4

5 1999: The Solidarity of Semi-Markov Processes: Bowls: represent excursions 5

6 Today s talk: Sketch the connection between a semi-markov process: and solutions to the Navier-Stokes equations in 3-d: u t + u u = ν u p + g u = 0, u 0(x) = u(x, 0) 6

7 Outline: some background existence theorem sketch proof comments 7

8 Ties between pde s and stochastic processes: Skorohod (1964) Ikeda, Watanabe, Dawson and others, 1968 and on Savits (1969) McKean (1975) LeJan and Sznitman (1997) Bhattacharya et al (2003) Fontbona (2006) 8

9 Heat equation: u t = u x 2 with u(x, 0) = u 0 (x) Time u(x, t) = u 0 (x y)k(y, t)dy K(y, t) = (2πt) d/2 exp y 2 /2t u(x, t) = E x u 0 (W (t)) Space Brownian motion W 9

10 McKean (1975): KPP equation u t = u x 2 + u2 u with u(x, 0) = f(x), f(x) 1 u(x, t) = E x i N(t) f(w (i) (t)) Markov process W: branching Brownian motion 10

11 LeJan and Sznitman (1997): Fourier transformed NS t û(ξ, t) = e ν ξ 2tû 0 (ξ) + { ipξ û(η, t s) ξ û(ξ η, t s)dη } ds s=0 (2π) 3/2 e ν ξ 2 s û(ξ, 0) = û 0 (ξ) with û 0 (ξ) c ν ξ 2 Time û(ξ, t) = c ν ξ 2 E ξ M(û 0, τ(t)) M multiplicative functional Spectral Space Markov branching random walk 11

12 Continued by Bhattacharya et al (2003): û(ξ, 0) = û 0 (ξ) with sup ξ û 0 (ξ) /h(ξ) < û(ξ, t) = c ν h(ξ)e ξ M(û 0, τ(t)) Time for all t T h h satisfies a convolution condition M multiplicative functional Spectral Space Markov branching random walk 12

13 Navier-Stokes equations in 3-d u t + u u = ν u p + g, pressure p and forcing g incompressible fluids: u = 0 initial data: u 0 (x) = u(x, 0) mild solutions: u = e νt u 0 t 0 e ν(t s) P (u u)(s)ds + t 0 eν(t s) Pg(s)ds P: Leray projection onto divergence free vector fields 13

14 Branching diffusion with jumps Theorem: (O. 05) If the pair (u 0, g) is in F h, then there exists T h > 0 such that for all t T h and x R 3 u(x, t) = h(x)e x Υ(t) is a solution to the NS equations. h: scaling; E x : expectation initialized at x; Υ: functional 14

15 sense of solution? mild solutions u = e νt u 0 + u 0 = 0 t t 0 eν(t s) P (u u)(s)ds 0 eν(t s) Pg(s)ds P: Leray projection onto divergence free vector fields function space where solutions exist? F h follows naturally from method F h = {(v, g) : sup v h <, sup g h < } 15

16 Theorem: If both the forcing and the initial data are uniformly bounded in magnitude and for some p > 0 both lim sup x u 0 (x) x p and lim sup x g(x, t) x 2p+1 are finite, then there exists an unique bounded mild solution to the N-S equations on an interval [0, T ]. Furthermore for each t (0, T ] this solution decays at rate x (p (1 ɛ)) as x. 16

17 Sketch of proof: reformulate integral equation: (integration by parts) u = e νt u 0 + t t 0 eν(t s) P (u u)(s)ds 0 eν(t s) Pg(s)ds recognize probability densities using majorizing kernels build corresponding stochastic process express scaled solution as expectation use dominated convergence to show that the expectation is finite 17

18 Reformulation of the integral equation: u(x, t) = + ( R 3 u 0(x y)k(y, 2νt)dy t s=0 R 3 1 4π 2 νs z 4 {[ z 4νs K(z, 2νs)b 1(z; u(x z, t s), u(x z, t s)) {y: y z } y 2 K(y, 2νs)dy ) b 2 (z; u(x z, t s), u(x z, t s)) ] + [ K(z, 2νs)P z 1 4π z 3 {y: y z } K(y, 2νs)dy(I 3e ze t z) ] g(x z, t s) } dzds 18

19 K(y, t) = (2πt) 3/2 e y 2 /2t e y = y/ y, P y = I e y e t y b 1 (y; u, v) = (u e y )P y v + (v e y )P y u b 2 (y; u, v) = (b 1 (y; u, v) + u (I 3e y e t y)v e y )/2 c 1 = P z c 2 = (I 3e z e t z)/2 Lemma: b k (y; u, v) u v and b k (y; u, v) = h 2 b k (y; u h, v h ) 19

20 Define majorizing kernel pair (h, h): (A) (Ã) 1 sup x h(x) 1 sup x h(x) R 3 h2 (x y) y 1 2γ dy < R 3 h(x y) y 2(1 γ) dy < γ (3/2, 2) a fixed parameter h scales u; dominates u scaled solution: χ(x, t) = h scales g; dominates g u(x, t) h(x) initial data: χ 0 (x) = χ(x, 0) scaled forcing: ϕ(x, t) = g(x, t) h(x) 20

21 Examples of majorizing kernel pairs: ( h(x) = x 2γ 4, h(x) = x 4γ 9) γ (3/2, 2) ( h(x) = (1 + x ) 2γ 4, h(x) = (1 + x ) 4γ 9) γ (3/2, 2) ( h(x) = (1 + x 2 ) 2γ 4, h(x) = 0 ) γ (3/2, 2) Construct more kernels: shift, dilate, take mixtures, bound ratios,... 21

22 Return to the integral equation scaled by h: χ(x, t) = + ( χ 0(y) h(y) K(x y, 2νt)dy R 3 h(x) t s=0 R 3 h 2 (x z) 4π 2 νs z 4 h(x) {[ z h 2 (x z) 4νsh(x) K(z, 2νs)b 1(z; χ(x z, t s), χ(x z, t {y: y z } y 2 K(y, 2νs)dy ) b 2 (z; χ(x z, t s), χ(x z, t s)) ] + [ z) K(z, 2νs) h(x P z h(x) h(x z) 4π z 3 h(x) {y: y z } K(y, 2νs)dy(I 3e ze t z) ] ϕ(x z, t s) } d 22

23 Building blocks of stochastic process: transition densities: diffusion: J(y, t x) = the jump to the velocity branch point: h(y)k(x y, 2νt) h(x) f(y, z x) = (5 2γ) y 2(1 γ) z 4 h 2 (x z)1[ z > y ] 4π R 3 z 1 2γ h 2 (x z)dz the jump to input forcing: f(y, z x) = (5 2γ) y 2(1 γ) z 3 h(x z)1[ z > y ] 4π R 3 z 2(1 γ) h(x z)dz the conditional densities for waiting times between branches: f(s z) = c s γ 1 z 2γ e z 2 /4νs f(s z) = c s γ z 2(γ 1) e z 2 /4νs 23

24 χ(x, t) = t J(y, t x)χ 0 (y)dy s=0 { m(s, x) f(y2, y 1 x) f(s y k )b k (y 1 ; χ(x y 1, t s), χ(x y 1, t s)) k=1,2 + m(s, x) f(y 2, y 1 x) k=1,2 f(s y k )c k (y 1 )ϕ(x y 1, t s) } dy 1 dy 2 ds χ(x, t) = u(x, t) h(x), ϕ(x, t) = g(x, t) h(x) 24

25 Normalization multipliers: m(s, x) = c ν s γ 3/2 y 1 2γ h 2 (x y)dy h(x) m(s, x) = c ν s γ 3/2 y 2(1 γ) h(x y)dy h(x) Since (h, h) is a majorizing kernel pair, the magnitude of the multipliers is controlled by s γ 3/2 C for some finite C. (Remember that 3/2 < γ < 2.) 25

26 V 10 V 11 (X 1,τ φ + τ 1 ) V 1 V 0 Time V φ (X 0, τ φ +τ 0 ) (X φ, τ φ ) Space 26

27 Semi-Markov process: h excessive: V is a h- Brownian motion with transition density J Transition probabilities for branching random walk determined by spatial densities (f, f) Waiting times to branch points determined by temporal densities (f, f) 27

28 Define the functional Υ iteratively on tree; start with leaves and work down branches: m(τ v, X v ) B v (Υ v 0 (t τ v ), Υ v 1 (t τ v )) Υ v (t) = χ 0 (V v (t))+ if σ v = 1, τ v t m(τ v, X v ) c v ϕ(x v, t τ v ). if σ v = 0, τ v t V v s diffusion X v s Markov branching rw τ v s waiting times σ v s 0 or 1 B v randomized b k s c v randomized 28

29 Theorem: If h has parameter γ, (h, h) is a majorizing kernel pair, and u 0 (x) h(x) and g(x, t) h(x) are both uniformly bounded by 1, then u(x, t) = h(x)e x Υ (t) is a solution to the NS equations on a time interval [0, T ] for T = T (γ) proportional to C 1/(γ 3/2). Furthermore this solution is unique among those in the class {v : sup x v(x, t) h(x) C for t T }. 29

30 Finishing off the proof: Easy contraction argument by induction on finite binary tree shows that Υ is uniformly bounded if the magnitudes of both χ 0 and ϕ are uniformly small. It follows that χ(x, t) = E x Υ (t) exists and is uniformly bounded in magnitude for t in the range specified. (Uniqueness takes a little bit more. ) 30

31 Import of γ parameter: 3/2 < γ < 2 Larger gamma allows forcing and initial data that decays more slowly as x Speeds up waiting times Makes jump sizes larger A larger gamma spreads the random process out farther and faster. 31

32 Theorem: If both the forcing and the initial data are uniformly bounded in magnitude and for some p > 0 both lim sup x u 0 (x) x p and lim sup x g(x, t) x 2p+1 are finite, then there exists an unique bounded solution to the N-S equations on an interval [0, T ] for some T > 0. Furthermore for each t (0, T ] this solution decays at rate x (p (1 ɛ)) as x. 32

33 Regularity (no forcing): Ladyzhenskaya-Prodi-Serrin condition; If a solution u to the NS equations satisfies T u(x, t) q dx) r dt < 0 ( for some q > 3, r r(q) > 0, then u is regular on [0, T ]. Take a majorizing kernel h(x) proportional to 1 (1 + x ) 4(2 γ) for any γ (3/2, 13/8) Corollary: If g = 0 and for some γ (3/2, 13/8) sup x (1 + x ) 4(2 γ) u 0 (x) is finite, then there exists a unique regular solution to the NS equations for t T (γ). 33

34 Comments: Applied mathematicians: Incompressibility is partially captured in the transition probabilities Blow-up? Probabilists: Family of branching diffusions: semi-markov with continuous state space Explosion of branching process? 34

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