Lecture 12: Detailed balance and Eigenfunction methods
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1 Miranda Holmes-Cerfon Applied Stochastic Analysis, Spring 2015 Lecture 12: Detailed balance and Eigenfunction methods Readings Recommended: Pavliotis [2014] (eigenfunction methods and reversibility), (explicit examples of eigenfunction methods) Gardiner [2009] (detailed balance), 6.5 (eigenfunction methods, multi-dimensional), 5.4 (eigenfunction methods, 1-dimensional) Optional: Pavliotis [2014] 4.9 (reduction to a Schrodinger equation) 12.1 Flux in steady-state Consider a diffusion in R d dx t = b(x t )dt + σ(x t )dw t, (1) and let a = 1 2 σσt. The Fokker-Planck equation for the evolution of probability density p on domain D is t p = L p = j, j = bp (ap). (2) The stationary solution p s solves = 0. (3) In some cases, we have the stronger result that the flux vanishes in steady-state: We will be interested in When? Why? What does this mean, physically? Consider a one-dimensional process. (3) implies that = const. j = 0. (4) s If D = [a,b] and the boundaries are reflecting, then = 0 since it s 0 at the boundary. If D = [a,b] and the boundaries are periodic, then we may have 0. Now consider a >1-dimensional process. Suppose the stationary distribution satisfies (4). What conditions does this impose on the coefficients of the SDE? Then a p s = p s (b a)
2 Suppose a has inverse a 1. Then log p s = a 1 (b a) }{{} Q(x) in components: Q i = a 1 ik (2b k k j a k j ) x j Since the RHS is a gradient, the LHS is too. A necessary and sufficient condition for this is that the curl of Q vanish: Q i = Q j. (5) x j x i When this is the case, we have Here Φ(x) plays the role of a potential energy. p s = Z 1 e Φ(x), with Φ(x) = x Q(x )dx. (6) Example. Suppose the diffusivity is constant and diagonal: a(x) = a i δ i j, where a i are constants. Then a = 0, so the condition for the steady-state flux to vanish is that a 1 b be the gradient of something. This is equivalent to asking that b = φ for some scalar function φ. What this tells us is the force (on the trajectory) must be the gradient of something. If the force has any curl component, the flux cannot vanish in steady-state (given constant diffusivity.) Forces that have curl components cannot typically arise in a system that is thermodynamically isolated. Indeed, if 0, then the steady-state flux will move somewhere, on average, and so by conservation of probability it must contain loops. If this were possible in an isolated system, then we could extract energy indefinitely out of these cycles! Therefore, when 0, the system must be subject to external forcing. One situation where this arises is in active matter: particles that can propel themselves (by swimming, motors, etc). The energy input by the individual motors can sustain a non-zero steady-state flux Detailed Balance The previous discussion is closely related to the concept of detailed balance. Recall that a Markov Chain satisfies detailed balance if π i p i j = π j p ji, which implies that the net flux of probability across each edge is 0 in steady-state. For a continuous Markov process, the concept is very similar. The principle of detailed balance requires that P(a b, forward in time) = P(b a, backward in time), in steady-state, i.e. the probability of making some transition forward in time, equals the probability of making the reverse transition, backward in time, when the system is in steady-state. This principle comes from microscopic reversibility: Newton s laws are time-reversible, so if we coarse-grain them, we want the stochastic system to preserve the same property. 2
3 If we let p(x,t y,s) be the transition density and π(x) be the stationary distribution, then we would like to write the principle of detailed balance as π(y)p(x,t y,s) = π(x)p(y,t x,s). Unfortunately, this won t quite work for a continuous physical system, because not all variables behave the same forward and backward in time. For example, consider a gas of particles with positions x and velocities v (these are both vectors.) Suppose the system moves from state (x,v) (x,v ) in some time interval of length τ. This is not the same as moving from (x,v ) (x,v) backwards in time, because if you change the direction of time, then the velocities of each particle must change sign. What it is physically equivalent to is moving from (x, v ) (x, v) in the same time interval τ. Indeed, to go from x x in a small time t requires v (x x)/ t, and to go from x x requires v (x x )/ t = v. Therefore, for this gas of particles, the requirement that the time-reversed transition probabilities equal the forward transition probabilities in steady-state would be written as p s (r,v,t +τ r,v,t) = p s (r, v,t +τ r, v,t) p(r,v,τ r,v,0)p s (r,v) = p(r, v,τ r, v,0)p s (r,v ). In general, each physical variable is classified as either even or odd, depending on how it transforms under time-reversal. We have x i ε i x i, with ε i = +1( 1) for an even (odd) variable respectively. For example, position is an even variable, and velocity is an odd variable. Then detailed balance requires p s (x,t + τ x,t) = p x (εx,t + τ εx,t) (7) where ε is a matrix with ε i on the diagonal. Using the fact that when τ = 0 the condition is p s (x) = p s (εx), (7) can also be written as p(x,τ x,0)p s (x) = p(εx,τ εx,0)p s (x ). (8) Either of (7), (8) can be taken as the condition for detailed balance, for an arbitrary Markov process. Notes Gardiner [2009] (section 6.3.5) derives conditions on the coefficients b, a of a time-homogeneous SDE, in order for detailed balance to be satisfied. He shows that necessary and sufficient conditions are: (i) ε i b i (εx)p s (x) = b i (x)p s (x) + j j (2a i j (x)p s (x)) (ii) ε i ε j a i j (εx) = a i j (x). Notice that these are formulated in terms of p s (x), so you need to know this first in order to check the conditions. When all variables are even, these conditions reduce to = 0. In this case, (4) is equivalent to detailed balance (8). A common situation where detailed balance holds, but not (4), is the Langevin equation (see HW10). This is because it contains velocities, which are odd. Therefore this system will contain non-zero fluxes in steady-state, even though it can be shown to satisfy detailed balance. 3
4 12.3 Eigenfunction methods Recall that for Markov chains that satisfy detailed balance, we were able to symmetrize the transition matrix by the similarity transformation D 1/2 PD 1/2, where D i j = π i δ i j is the matrix with the stationary distribution on its diagonal. We used this to argue that P has a complete orthonormal basis of eigenvectors, with real eigenvalues, and then we could expand many quantities of interest in terms of these eigenvectors. We can derive a similar result for SDEs. We will show that for systems that satisfy detailed balance (at least when all variables are even), the generator is self-adjoint and negative semi-definite when working in the appropriate function space. Then, we will be able to expand many quantities of interest in terms of the eigenfunctions, and obtain exponential bounds for the decay to stationary values Smoluchowski equation First, we consider a system with even variables and constant diffusivity tensor proportional to I, that satisfies detailed balance. This has the form dx t = U(X t )dt + 2 dw t, X t,w t R d. (9) This is sometimes called the Smoluchowski equation. The generator, its adjoint, and the stationary distribution are L f = U f + f L f = ( U f + f ) π = Z 1 e U(x) Let s work in the weighted inner product space Lπ. 2 This uses the stationary distribution as a weight for the inner product: f,g π = f (x)g(x)π(x)dx. R d Then we can show the following properties: 1. L is self-adjoint 1 in L 2 π Proof. (a) First, we show that L (gπ) = πl g. This is just algebra. L (gπ) = ( Ugπ + (gπ)) = Ugπ + U gπ + Ug π + gπ + 2 g π + πg = π ( ug + U g U Ug + g 2 g U Ug + U Ug) = π( U g + g) = πl g. 1 What we will actually show is that it is symmetric. While it is also true that it is self-adjoint, this is harder to show. A symmetric operator is one such that L f,g = f,l g for all f,g in the domain of L, and it is self-adjoint if the operator is defined everywhere on the Hilbert space. Sometimes one can extend a symmetric operator to a self-adjoint one; when this extension is possible and unique the operator is said to be essentially self-adjoint. 4
5 (b) Next, calculate: L f,g π = (L f )gπ = f L (gπ) = (L g) f π = f,l g π. 2. L is positive semi-definite. Proof. L f, f π = U f f π + f f π = U f f π + ( f f π) f ( f π) = U f f π f f π f π f }{{} U f f π = f 2 π L is ergodic (provided it comes with reflecting boundary conditions.) Proof. We will show its null space consists of constants. Suppose L f = 0. Substituting into the expression above, we find f πdx = 0. Therefore f is constant. From 1, 2 we know that the eigenvalues of L are real and non-positive, and the eigenfunctions are orthonormal in L 2 π. Similarly, the eigenfunctions of L are orthonormal in L 2 π 1. In order to know that these eigenfunctions are complete, we need more information about L. For example, being compact would suffice, as then we could use the spectral theorem for compact self-adjoint operators. Another sufficient condition is that L have a spectral gap: Var( f ) L f, f = D L ( f ). (10) for all f in the domain of definition of D L ( f ), and for some fixed > 0. The quantity on the right-hand side is called the Dirichlet form of the generator L. Condition (10) is equivalent to the Poincaré inequality f 2 π f 2 π, f : f π = 0. For the Smoluchowski equation, this holds for potentials U(x) such that ( ) U(x) 2 lim U(x) =, (11) x 2 and functions f C 1 (R d ) L 2 π with f π = 0. This is the content of Theorem 4.3 (p.112) in Pavliotis [2014], where it is discussed (but not proved.) The spectral gap estimate (10) can be used to show the probability density converges exponentially fast to equilibrium in L 2 π 1. See Pavliotis [2014], Theorem 4.4, p
6 Condition (10) implies the spectrum of L is discrete, so the eigenfunctions span the space L 2 π. Let 0 denote the eigenvalues, let {φ } be the eigenfunctions of L, and let {ψ } be the eigenfunctions of L. We have the following properties. By 1(a), we have ψ = πφ. (12) φ,ψ form a bi-orthogonal set: φ,ψ L 2 = δ, where δ is the Kronecker delta. This is because {φ } is orthonormal with respect to, π. When L has reflecting boundary conditions, = 0 is an eigenvalue. The corresponding eigenfunctions are φ 0 = 1, ψ 0 = π. We expect ψ (x) to decay at x = ±, but not φ (x). Note that all of the above is independent of the boundary conditions associated with L, apart from the statement about = 0 being an eigenvalue. Many quantities can be expressed using eigenfunctions. Examples 1. (Transition probability) p(x,t x 0,0) solves p t = L p, so using separation of variables it can be expressed using the eigenfunctions of L as p(x,t x 0,0) = a ψ (x)e t. The coefficients are found by projecting the initial condition onto the eigenfunctions of L : a = p(x,0 x 0,0),φ (x) L 2 = δ(x x 0 ),φ (x) L 2 = φ (x 0 ). Therefore p(x,t x 0,0) = φ (x 0 )ψ (x)e t. 2. (Covariance matrix in steady-state) This calculation assumes that E π X t = 0. C(t) = EX t X0 T = xx0 T p(x,t x 0,0)π(x 0 )dxdx 0 = xx0 T π(x 0 ) ψ (x)φ (x 0 )e t dxdx 0 ( )( ) T = xψ (x)dx xψ (x)dx e t since πφ = ψ = C e t, where (C ) i j = x x = xx T, with x i = x i ψ (x)dx = x i φ (x)π(x)dx. Note in particular that the covariance function is a sum of decaying exponentials. 6
7 Example (Ornstein-Uhlenbeck process, 1D). The eigenfunctions of the generator solve L p = x (αxp) σ 2 xx p, L f = αx f x σ 2 f xx. Let y = x 2α σ 2. The eigenfunction equation becomes φ 2αx σ 2 φ + 2 σ 2 φ = 0. φ yφ + α φ = 0. This is exactly the eigenvalue equation for the Hermite polynomials 2 (with eigenvalues scaled by α): Therefore H n (x) = ( 1) n e x2 /2 dn dx n e x2 /2, H n = {1, x, x 2 1, x 3 3x, x 4 6x 2 + 3,...} The probability density at time t is ρ(x,t) = 1 x2 α 2πσ 2 /2α e σ 2 φ (x) = (n!) 1/2 H n (x 2α/σ 2 ), = nα, n = 0,1,... (n!) 1/2 A n H n (x n 2α/σ 2 )e nαt, A n = 1 ρ 0 (x),h n (x 2α/σ 2 ) L 2. n! This relaxes to the stationary probability exponentially quickly, with a rate governed by the smallest non-zero eigenvalue = α. Let s calculate the covariance function. From item 2 above, this is given by n=0 c2 ne nαt, where c n = x,ψ n L 2. Note that x = σ 2 /2αφ 1 (x), and φ 1,ψ j L 2 = 0 if j 1. Therefore c 1 = σ 2 /2α, and c n = 0 for n 1. Putting this together gives σ 2 C(t) = 2α e αt. The covariance function is just a single exponential, because x itself is an eigenfunction of the generator! In general, the covariance function for a process satisfying the Smoluchowski equation will be a sum of exponentials More general reversible diffusions One can extend all of these results to more general reversible diffusions with even variables. Recall from sections 12.1,12.2 that a diffusion (1) is reversible (satisfies detailed balance) if = 0, i.e. the flux is zero in steady-state. A necessary and sufficient condition for this is the curl of Q(x) = a 1 (b a) vanish. Then, we can write π = e Φ, where Φ(x) = x Q(x )dx. This is in fact related to the symmetry properties of the generator of the process. 2 This is the probabilist s definition of Hermite polynomials. You may also encounter the physicist s definition: Hn p (x) = ( 1) n e x2 d n dx n e x2, which differs from the math one by a rescaling: Hn p (x) = 2 n/2 H n ( 2x). 7
8 Theorem (Theorem 4.5 in Pavliotis [2014]). A stationary diffusion process X t in R d with generator L and invariant measure π is reversible if and only if its generator is self-adjoint in L 2 π. Proof. See Pavliotis [2014], section 4.6. The calculation is also done in Gardiner [2009], section 6.5.2, p.159. Using the explicit representation of the invariant measure as π = e Φ, one can manipulate the generator and show it can be written as L = π 1 (a π ). (13) See Pavliotis [2014], section 4.6, p The corresponding Dirichlet form is D L ( f ) = a f, f π. If the diffusion matrix a(x) is uniformly positive definite, and the generalized potential Φ satisfies condition (11), then π satisfies a Poincaré inequality, so L has a spectral gap (10). Then L is a negative semi-definite self-adjoint operator in L 2 π with a discrete spectrum and a complete basis of eigenfunctions. Therefore we can perform all the same calculations as for the Smoluchowski equation, and expand the transition probability in terms of these eigenfunctions. Again, you find that the covariance function for this reversible diffusion is a sum of decaying exponentials. See Pavliotis [2014], section 4.7. References C. Gardiner. Stochastic methods: A handbook for the natural sciences. Springer, 4th edition, G. A. Pavliotis. Stochastic Processes and Applications. Springer,
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