APM526 CLASSROOM NOTES PART IB. APM526, Spring 2018 Last update: MAR 12

Transcription

1 APM526 CLASSROOM NOTES PART IB APM526, Spring 2018 Last update: MAR 12 1

2 PARTICLE METHODS 1. Particle based methods: Large systems of ODEs for individual particles. 2. Partial differential equations: Equations for the density of particles. Example: N particles moving with a speed, dependent on their location. ξ n = a(ξ n ), n = 1 : N The corresponding density function: u(x, t) = N n=1 δ(x ξ n (t)) 2

3 DISTRIBUTIONS g(x) distribution, defined by ψ(x)g(x) dx = α(ψ), ψ C 0 ψ : test function. Example: g(x) = 5δ (x) + cos(x). Example: Discretization of δ (x). Integration against a test function corresponds to physically taking a measurement! 3

4 THE CONCEPT OF A WEAK SOLUTION FOR CON- SERVATION LAWS t u(x, t) + x F (u, x, t) = 0 (1) Integrate against a test function ψ ( the microscope ) ψ(x) t u(x, t) dx F (u, x, t) x ψ(x) dx = 0, ψ C 0 (2) Definition: We call the u(x, t) a weak solution of (1) if (2) holds ψ C 0. This allows for finding solutions if u discontinuous and x u does not exist in the classical sense! 4

5 t u(x, t) = N n=1 δ (x ξ n (t))a(ξ n (t)) (3) u(x, t) = N n=1 δ(x ξ n (t)) t u = N n=1 δ (x ξ n (t))a(ξ n (t)), x u = integrated against any test function ψ: ψ(x) t u(x, t) dx = N n=1 δ (x ξ n (t)) ψ (x)a(x)u(x, t), ψ Satisfies the density equation weakly: t u(x, t) + x [a(x)u(x, t)] = 0 5

6 HOW TO COMPUTE DENSITIES FROM PARTICLES Define a grid and grid cells C n = [x n, x n+1 ]. Define indicator functions χ m (x) = ( ) 1 x Cm 0 else Use χ m as test functions χ m u(x, t) dx = N n=1 χ m (ξ n (t)) Counts the number of particles in C m. ( u(x, t) dx = N). 6

7 THE MEAN FIELD PROBLEM IID case: Independent and Identically Distributed particles t ξ n = a(ξ n ) Assume: particles interact t ξ n = a n ( ξ), ξ = (ξ1,.., ξ n ) Density: u(x, t) = N n=1 δ(x ξ n (t)) ψ(x) t u(x, t) dx = ψ (x) N n=1 a n ( ξ(t))δ(x ξ n (t)) dx 7

8 Mean field: N n=1 a n ( ξ(t))δ(x ξ n (t)) N n=1 a mf (u)δ(x ξ n (t)), u = for N. a mf is called the mean field! N n=1 δ(x ξ n (t)) Given a mean field a mf (u)(???) and N, we have (weakly) t u + x [a mf (u)u] = 0 Example: Gravity, Electrodynamics 8

9 PARTICLES AND DIFFUSION (Einstein theory of Brownian motion) Consider the pure transport (hyperbolic) case d ξ(t) = a(ξ(t)), ξ(t + t) = ξ(t) + ta(ξ(t)) (4) dt Alternative (probabilistic) interpretation: Consider only one single particle. u(x, t) dx = dp[ξ(t) = x]. This gives for the density u(x, t) = δ(x ξ(t)) (in weak form) the equation t u(x, t) + x [a(x)u] = 0 (5) Now consider N >> 1 identical particles, all evolving according to (4), and u(x, t) = N 1 n δ(x ξ n (t). Equation (5) still holds. So, for a system of N >> 1 identical particles, (5) describes the evolution of the density u((x, t). 9

10 Brownian motion: Replace a(ξ) with a random variable q(ξ) ξ(t + t) = ξ(t) + tq(ξ(t)), dp[q(ξ) = x] = φ(x, ξ)dx q(ξ) has a mean a(ξ) and variance σ 2 (ξ) φ(x, ξ) dx = 1, xφ(x, ξ) dx = a(ξ), (x a(ξ)) 2 φ(x, ξ) dx = σ 2 (ξ) Normalize φ: φ(x, ξ) = σ 1 φ 0( x a(ξ) σ(ξ), ξ) with φ 0 normalized distribution with mean 0 and variance 1. φ 0 (x, ξ) dx = 1, xφ 0 (x, ξ) dx = 0, x 2 φ 0 (x, ξ) dx = 1, ξ 10

11 Let u(x, t)dx = dp[ξ(t) = x] Sum up over all possibilities: u(x, t + t) = δ(y + tq x)φ(q, y)u(y, t) dydq = δ(y + tq x) 1 σ(y) φ 0( q a(y), y)u(y, t) dydq σ(y) weak formulation: ψ(x)u(x, t+ t) dx = for all test functions ψ C 0. ψ(x)u(x, t+ t) dx = Taylor expand the test function: ψ(x)u(x, t + t) dx = ψ(y+ tq) 1 σ(y) φ 0( q a(y), y)u(y, t) dydq σ(y) ψ(y+ ta+ tσz)φ 0 (z, y)u(y, t) dzdy [ψ(y)+( ta+ tσz)ψ (y)+ 1 2 ( ta+ tσz)2 ψ (y)]φ 0 (z, y)u(y, t) dzdy+h 11

12 [ψ(y)+ ta(y)ψ (y)+ 1 2 ( t2 a(y) 2 + t 2 σ(y) 2 )ψ (y)]u(y, t) dy+h.o.t. strong form: u(x, t+ t) = u(x, t) t x [au(x, t)] x [( t2 a(x) 2 + t 2 σ(x) 2 )u(x, t)] set σ 2 t σ2. so q(ξ) has a mean a(ξ) and variance σ2 (ξ) t Limiting equation for t 0: t u(x, t) = x [au(x, t)] x[σ(x) 2 )u(x, t)] 12

13 Theorem: A stochastic process of the form ξ(t + t) = ξ(t) + tq(ξ(t)) where q(ξ) is a random variable with dp[q(ξ) = x] = φ(x, ξ)dx and E(q(ξ)) = a(ξ), var(q(ξ)) = σ(ξ)2 t i.e. φ(x, ξ) dx = 1, xφ(x, ξ) dx = a(ξ), results in the limit t 0 in the equation t u(x, t) = x [au(x, t)] x [σ(x)2 u(x, t)] for the probability density u(x, t)dx = dp[ξ(t) = x] (x a(ξ)) 2 φ(x, ξ) dx = σ(ξ)2 t 13

14 THINK IN REVERSE: Given a PDE t u(x, t) = x [au(x, t)] x[σ(x) 2 )u(x, t)] Define a Monte Carlo particle method in the following way: Use N particles ξ n (t), n = 1 : N solve ξ n (t + t) = ξ n (t) + tq n (ξ n (t)) with dp[q n (ξ) = y] = φ(y, ξ)dx (1, x, x 2 )φ(x, ξ) dx = (1, a(ξ), a(ξ) 2 + σ(ξ)2 t ) 14

15 JOB: Given a probability distribution φ in [a, b] (a, b can be ±!) φ(x), φ(x) 0, b a φ dx = 1 and a random number generator, producing numbers uniformly distributed in (0, 1), compute random numbers q with dp[q = x] = φ(x)dx SOLUTION: set q = g(y), y (0, 1). g monotone (0, 1) (a, b) P[q < x] = P[g(y) < x] = P[y < g 1 (x)] = g 1 (x) Φ (x) = φ(x), Φ(x) = P[q < x] = g 1 (x) g(y) = Φ 1 (y) Algorithm: Take φ(x), compute the anti - derivative Φ(x) = x a φ(y) dy, Φ(a) = 0, Φ(b) = 1. set g(x) = Φ 1 (x), g(x) (a, b), compute y (0, 1) (with c.f. Matlab rand) set q = g(y) 15

16 EXAMPLE 1: exponential distribution a = 0, b =, φ(x) = e x anti - derivative: Φ(x) = H(x)[1 e x ] computing the inverse: g(y) = ln(1 y). compute y with rand and set q = ln(1 y) EXAMPLE 2: Gaussian: a =, b =, φ(x) = 1 2π exp( x2 2 ) anti - derivative: Φ(x) = 1 x exp( y2 ) dy = erf(x). 2π compute y with rand and set q = erf 1 (y) 2 16

17 2 dimensional distributions: f(x, y) JOB: compute a random pair (q 1, q 2 ) distributed according to f(x, y) conditional probabilities: f(x, y) = φ(y x)φ 1 (x), φ 1 (x) = f(x, y) dy φ 1 (x) is the probability density for x. φ(y x) is the probability density for y when x is already known. 1. compute a random number q 1 according to φ given q 1, compute q according to φ(y q 1 ). 17

18 % compute an exponential distribution with N particles clear format N=1000 % number of particles K=10 % number of grid points X=3% length of interval dx=x/k xi=-log(rand(1,n)) x=x*(0:k)/k % hh(k,n)=(x(k) xi(n)) 18

19 h=(x *ones(1,n) ones(k+1,1)*xi) hh=sum(h,2) u=diff(hh) u=u/n/dx figure(1) plot(x(1:k),u)

20 19

21 RELATION TO DIFFERENCE SCHEMES Given a discretization of t u = x (au)+ x (b x u) of the form u n (t+ t) = u n + ta 2 x (u n 1 u n+1 )+ tb x 2(u n+1 2u n +u n 1 )(t) u n (t + t) = αu n 1 + βu n + γu n 1 (t) α = ta 2 x + tb x2, β = 1 2 tb x 2, γ = ta 2 x + tb x 2, α + β + γ = 1 rewrite as a stochastic process: u n (t + t) = u n 1 (t) u n (t) u n+1 (t) P = α P = β P = γ CFL condition: α, β, γ 0 Modeling option: Add interaction between cells α = α( u(t))... Works for first order in time schemes (like LaxFriedrichs) and not for any higher order in time schemes (like LaxWendroff) 20

22 CONVERGENCE ISSUES Q: What are good approximations to t, x and good values for t, x for difference schemes as t, x 0? Q: What are good choices for φ(x, ξ) and the random variable q(ξ) for the random motion of the particle as N? 21

23 CONVERGENCE OF PARTICLE METHODS: Start with a single particle at position ξ ξ(t + t) = ξ(t) + tq(ξ), dp[q(ξ) = x] = φ(x, ξ)dx Probability: u(x, t)dx = dp[ξ(t) = x] Observable: Mesh: x j, indicator function χ j (x) = 1 if x {x j, x j+1 }, χ j (x) = 0, else Take a Counter: g j (t) = 1 x χ j(ξ(t)) Expectation: E[g j ] = 1 x x j+1 x j χ j (x)u(x, t) dx u(x j, t) 22

24 Compute N realizations of the process: Blackbox: given ξ = (ξ 1,.., ξ N ), compute N identically distributed random variables q n with dp[q = x] = φ(x, ξ(t)) dx Compute averages: G j (t) = 1 N n g n j (t) = 1 N 1 x G(x, t) should approx- Since all realizations are independent, imate E[g(x, t)] u(x, t) n χ j (ξ n (t)) 23

25 LAW OF LARGE NUMBERS AND CENTRAL LIMIT THE- OREM: Let ξ n, n = 1 : N be identically distributed random variables with mean = 0 and variance = σ 2 dp[ξ n = x] = f(x)dx, n, Compute the average q = 1 N Nn=1 ξ n. (1, x, x 2 )f(x) dx = (1, 0, σ 2 ) Then F (y)dy = dp[q = y] is given for N approximately by the Gaussian by N F (y) σ exp( Ny2 2σ 2 ) independent of the structure of f(x)! 24

26 COROLLARY: Let ξ n, n = 1 : N be identically distributed random variables with mean = a and variance = σ 2 dp[ξ n = x] = f(x)dx, n, Compute the average q = 1 N Nn=1 ξ n. (1, x, x 2 )f(x) dx = (1, a, a 2 + σ 2 ) Then F (y)dy = dp[q = y] is given for N approximately by the Gaussian by N a)2 F (y) exp( N(y σ 2σ 2 ) independent of the structure of f(x)! 25

27 THE 1 N PROBLEM: Compute N realizations of a stochastic process. Compute an observable of the process by averaging. The observable will converge to the correct mean a with a rate 1. N 26

28 MARKOV PROCESSES: ω: frequency (time scale). t: time step ξ(t + t) = (1 κ)ξ(t) + κ[ξ(t) + q], κ {0, 1}, P[κ = 1] = ω t, P[q = y] = φ(y, ξ(t)) Define u(x, t) dx = dp[ξ(t) = x] t u(x, t) = ωφ(x, t) 27

29 STABILITY IN THE - NORM U = max n U n Matrix norms: A 1 = max n A = max x 0 m Ax x A mn, A = max m Ax A x n A mn, A 2 = λ max (A T A) 28

30 Difference scheme (linear): u(x, t + t) = n α(x)t n u(x, t) A(m, m + n) = α(x m ) u(x m, t) := U m (t) U m (t + t) = n α m U m+n (t) U(t + t) (max x α(x) ) U(t) Consistency: m α(x m ) = 1 + O( t) If α(x m ) 0 we have α(x) = α(x) and stability U(t + t) (1 + O( t)) U(t). α m 0 gives the CFL condition! 29

31 MONOTONE OPERATORS U(t + t) = Φ(U(t)), U = (u 1,.., u N ), Φ = (φ 1,.., φ N ) Definition: Φ is monotone : consistency: φ j u k 0, j, k, t φ j (C 1) = C, C max j min j u j (t+ t) = max j u j (t+ t) = min j φ j (U(t)) max j φ j (U(t)) min j φ j ( 1 max k φ j ( 1 min k U(t + t) U(t) {u k(t)}) = max {u k(t)} {u k (t)}) = min{u k (t)} Example: Viscous Burger s equation t u+ x ( 1 2 u2 s x u) = 0 k k Remark: Explicit monotone schemes retain positivity! 30

32 Remark: Explicit monotone schemes can be interpreted as the solution of a particle method where particles jump only with multiples of x! Remark: Monotonicity works also for nonlinear scalar equations but not in general for systems!

33 INVERSE MONOTONE OPERATORS AND IMPLICIT SCHEMES Definition: B is inverse monotone (an M-matrix) if ( B jk 0 k B jk K > 0 j for j k (strictly diagonally dominant) ) Theorem: B is an M matrix B

34 Generalization: B is inverse monotone (an M-matrix) if B jk 0 k B jk 0 j α : k B αk > 0 for j k diagonally dominant) Remark: α usually given by the boundary conditions. 32

35 IMPLICIT SCHEMES (LINEAR) BU(t + t) = AU(t) B M-matrix and A > 0. B jk 0 k B jk 1 + tk B, K B > 0 j A jk 0 k A jk 1 tk A, K A 0, j for j k (strictly diagonally dominant) B 1 A 0 and U(t + t 1 tk A 1 + tk B U(t + t 33

36 Example: (damped heat equation) t u = 2 xu cu u(x, t+ t) = u(x, t)+ t x 2(T 2+T 1 )u(x, t + t) c tu(x, t + t) involves solving a system of equations at each time step write this as k k B jk U k (t + t) = k A jk U k (t) B jk = 1 + c t, B jj 0, B jk 0 for j k, A = I, U(t + t) c t U(t) This scheme requires the solution of equations at each time step, but is unconditionally stable (no CFL condition) 34

37 IMPLICIT SCHEMES (NONLINEAR) φ U φ(u(t + t)) = ψ(u(t)) M-matrix and ψ U > 0. φ j U 0 for j k k φ(c 1) (1 + tk φ )c, K φ > 0 c ψ j U 0 k ψ(c 1) c(1 tk ψ ), K ψ 0, c U(t + t 1 tk φ 1 + tk ψ U(t + t Example: Cranck - Nicholson (trapezoidal rule) t u = xu 2 u(t+ t) t 2 x 2(T 2+T 1 )u(t+ t) = u(t)+ t 2 x 2(T 2+T 1 )u(t) 35

38 ARTIFICIAL DISPERSION Consider t u + a x u = 0. Plane wave ansatz: u(x, t) = exp[iξ(x vt) rt] Exact: v = a, r = 0 Difference scheme: v = v(ξ) (incorrect wave speeds for high frequency components!) Not a problem in the presence of artificial diffusion (r > 0)! Example: Cranck Nicholson for t u + a x u = 0 will give the wrong wave speeds due the lack of artificial diffusion. 36