Lecture 12: Detailed balance and Eigenfunction methods

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1 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 time-homogeneous diffusion in R d dx t = b(x t )dt + σ(x t )dw t, (1) and let a(x) = 1 2 σ(x)σ T (x). 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? First 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. Therefore (4) should hold for reflecting boundary conditions, and for periodic boundary conditions it doesn t necessarily have to hold. Now consider a higher-dimensional process X t R d. Suppose the stationary distribution satisfies (4). What conditions does this impose on the coefficients of the SDE? 1

2 Then a p s = p s (b a) Suppose a has inverse a 1. Then log p s = a 1 (b a) }{{} Q(x) where 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 (5) holds, we have p s = Z 1 e Φ(x), with Φ(x) = x x 0 Q(x )dx. (6) Here Φ(x) : R d R plays the role of a potential energy. The point x 0 is some arbitrary, fixed point; changing it changes Φ(x) by a constant. Notice that it doesn t matter which path is taken in the integral defining Φ(x) since Q(x) is a gradient. Example. Suppose the diffusivity is constant and diagonal: a(x) = a i δ i j, where a i are constants. Then a(x) i j = 0, so the condition for the steady-state flux to vanish is that a 1 (x)b(x) be the gradient of something. This is equivalent to asking that b(x) = φ(x) for some scalar function φ. What this tells us is the force on the trajectory (namely the deterministic component of dx t dt, which is b(x t)), must be the gradient of something. If the force has any curl component, the flux cannot vanish in steady-state (given constant diffusivity.) Non-zero steady-state fluxes typically cannot 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 external forcing occurs and hence 0 is in active matter, such as particles that can propel themselves (by swimming, motors, etc), or in certain proteins like actin, which can propel themselves via internal motors. The energy input by the individual motors can sustain a non-zero steady-state flux Detailed Balance To understand the physical origins of the condition = 0 we need to consider the concept of reversibility or detailed balance. Recall that a Markov Chain satisfies detailed balance if π i p i j = π j p ji. But π i p i j is the flux of probability from edge i j in steady-state, and π j p ji is the flux of probability from edge j i in steady-state, so the net flux of probability across each edge is 0 in steady-state. 2

3 For a continuous Markov process, the concept is very similar. The principle of detailed balance requires that P(x y, forward in time) = P(y x, backward in time), in steady-state, (7) 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. 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 p s (y)p(x,t y,s) = p s (x)p(y,t x,s). (8) In simple settings this is the correct definition of detailed balance. However in general it needs to be modified. The reason is that for a continuous physical system, not all variables behave the same forward and backward in time. For example, consider a gas of particles with positions contained in vector x and velocities contained in vector v, and remember that dx dt = v. Suppose the system moves from state (x,v) (x,v ) in some time interval of length t. 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 t. 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 in terms of the stationary distribution as p s (r,v,t + τ r,v,t) = p s (r, v,t + τ r, v,t) or in terms of the transition probabilities as 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 we have Definition. A Markov process is reversible or satisfies detailed balance if p s (x,t + τ x,t) = p x (εx,t + τ εx,t) (9) where ε is a matrix with ε i on the diagonal. Alternatively we can use the observation that when τ = 0 then (9) is p s (x) = p s (εx) to write (9) as Notes p(x,τ x,0)p s (x) = p(εx,τ εx,0)p s (x ). (10) 3

4 A common situation where detailed balance holds, but not (4), is the Langevin equation, which you studied on a previous homework. This is because the equation 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. Gardiner [2009] (section 6.3.5) derives conditions on the coefficients b(x), a(x) 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, demonstrating that a diffusion process is reversible it has zero net flux in steady-state. We will return to the connection between detailed balance and the steady-state flux later, after we study the spectral properties of the generator and its adjoint 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 real eigenvalues and a complete orthonormal basis of eigenvectors, and then we could expand many quantities of interest in terms of these eigenvectors. We can derive a similar result for SDEs. We will consider systems where all variables are even, and show that detailed balance implies 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 Overdamped Langevin equation First, we consider a reversible diffusion process with even variables and constant diffusivity tensor proportional to the identify matrix I. Up to constants the corresponding SDE has the form This is a version of the overdamped Langevin equation. dx t = U(X t )dt + 2 dw t, X t,w t R d. (11) The generator, its adjoint, and the stationary distribution π(x) are L f = U f + f L f = ( U f + f ) π = Z 1 e U(x) 4

5 You can verify that the steady-state flux vanishes, j s = Uπ π = 0. We will work in the weighted inner product space Lπ, 2 which uses the stationary distribution as a weight for the inner product: f,g π = f (x)g(x)π(x)dx. (12) Rd We can show the following properties (which are independent of the boundary conditions): 1. First, we show that for g L 2 π. L (gπ) = πl g (13) Remark. This is an important property as it relates solutions to the backward equation to solutions to the forward equation, showing they are related by a factor of π(x). That is, if we have a probability density ρ(x, t) = g(x, t)π(x), so that g(x, t) is the density with respect to the stationary distribution, then ρ solves the backward equation and g solves the forward equation. Proof. 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. 2. L is self-adjoint 1 in L 2 π. Proof. Calculate the inner products: L f,g π = (L f )gπ = f L (gπ) = (L g) f π = f,l g π. 3. 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 0. 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. 5

6 4. L is ergodic if has reflecting boundary conditions. Proof. We will show its null space consists of constants. Suppose L f = 0. Substituting into the expression above from item 3, we find f πdx = 0. Therefore f is constant. Remark. These properties hold for a more general diffusion process such that j s = 0. You can try to show this as an exercise. From items 2, 3 we know 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. To finish our analogy with a Markov chain, we need to know the eigenfunctions are complete, i.e. they span L 2 π. This is a question in functional analysis so we won t delve into it here, but rather briefly mention some sufficient conditions on L that would imply completeness. L has a spectral gap: Var( f ) L f, f = D L ( f ). (14) for all f in the domain of definition of D L ( f ), and for some fixed > 0. The quantity on the righthand side is called the Dirichlet form of the generator L. Condition (14) implies the spectrum of L is discrete, so the eigenfunctions span the space L 2 π. The Poincaré inequality holds: f 2 π f 2 π, f : f π = 0. The Poincaré inequality can be shown to be equivalent to condition (14). For (11), the Poincaré inequality holds for potentials U(x) such that ( ) U(x) 2 lim U(x) =, (15) 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.) Basically, (15) forces U(x) to grow quickly enough at so that the stationary distribution e U(x) decays quickly enough. The spectral gap estimate (14) 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 Henceforth we will assume that L has a complete set of eigenfunctions which span the space L 2 π. Let us use the following notation: {} 0 eigenvalues {φ } eigenfunctions of L {ψ } eigenfunctions of L These satisfy the following properties. 6

7 By (13), we have that πl φ = L (πφ ). But we also have πl φ = πφ. Therefore if φ is an eigenfunction of L with eigenvalue, then πφ is an eigenfunction of L with eigenvalue. Therefore eigenfunction dual pairs satisfy ψ = πφ. (16) If you work carefully through the derivation, you will notice that (16) is true, even when the boundary conditions for the eigenfunctions are not reflecting, e.g. when they are absorbing or a mixture of absorbing and reflecting. Note however that π remains the unique stationary distribution found by solving L π = 0 with reflecting boundary conditions. φ,ψ form a bi-orthogonal set: φ,ψ L 2 = δ, (17) 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 y,0) solves the forward equation p t = L p. We can solve this using separation of variables. Writing p(x,t y,0) = a ψ (x)g (t) and substituting into the forward equation shows that g (t) = e t. The coefficients a are found from the initial condition p(x, 0 y, 0) = δ(x y), so we must have From (17) we have δ(x y),φ (x) L 2 = a ψ (x),φ (x) L 2. a = p(x,0 y,0),φ (x) L 2 = δ(x y),φ (x) L 2 = φ (y). Putting this all together shows the transition probability can be written as 2. (Covariance matrix in steady-state) p(x,t y,0) = φ (y)ψ (x)e t. (18) 7

8 Suppose E π X t = 0 (if not, subtract the mean). Then the covariance matrix is C(t) = EX t X0 T = xy T p(x,t y,0)π(y)dxdy = xy T π(y) ψ (x)φ (y)e t dxdy ( )( ) T = xψ (x)dx xψ (x)dx e t since πφ = ψ = C e t, (19) where C is a matrix with components (C ) i j = x x = x x T, with x i = x i ψ (x)dx = Notice that the covariance function is a sum of decaying exponentials. x i φ (x)π(x)dx. Exercise Consider Brownian motion on an interval [0, 1] with reflecting boundary conditions. Calculate the eigenvalues and eigenfunctions. Suppose you start the process with initial condition ρ(x,0) = 2 1 x [0, 1 2 ] s. Write down an expression for the probability density ρ(x, t) at time t. Estimate how quickly this decays to the stationary distribution. Example (Ornstein-Uhlenbeck process, 1D). The generator and its adjoint are The eigenfunctions of the generator solve L f = αx f x σ 2 f xx, L p = x (αxp) σ 2 xx p. 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 φ (x) = (n!) 1/2 H n (x 2α/σ 2 ), = nα, n = 0,1,... ρ(x,t) = 1 x2 α 2πσ 2 /2α e σ 2 (n!) 1/2 A n H n (x 2α/σ 2 )e nαt n 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 probabilist s definition by a rescaling: Hn p (x) = 2 n/2 H n ( 2x). 8

9 where the coefficients are calculated as A n = 1 ρ 0 (x),h n (x 2α/σ 2 ) L 2. n! The probability density 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, it is given by C(t) = n=0 c 2 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 of the form (11) will be a sum of exponentials More general reversible diffusions So far we have considered only the overdamped Langevin equation. One can extend all of these results to more general reversible diffusions with even variables. Pavliotis [2014] shows (formally) that for a general diffusion process X t with even variables, X t is reversible L is self-adjoint in L 2 π j s = 0. The first implication is shown (formally) in [Pavliotis, 2014, Section 4.6, Theorem 4.5]. A similar calculation is done in Gardiner [2009], section 6.5.2, p.159. The second implication is shown in [Pavliotis, 2014, Section 4.6, Proposition 4.5]. A lot of the calculations are similar. For example, you can show that (13) still holds, i.e. we have From this relation you can show L is symmetric in L 2 π. Exercise Show the statements above. L (gπ) = πl g. (20) What about eigenfunctions? Recall from sections 12.1,12.2 that if X t satisfies detailed balance, then we can write π(x) = e Φ(x) where Φ(x) is defined in (6). Using this explicit representation of the stationary distribution one can manipulate the generator and show it can be written as L = π 1 (a(x) π(x) ). (21) See Pavliotis [2014], section 4.6, p The corresponding Dirichlet form is D L ( f ) = a f, f π. 9

10 If the diffusion matrix a(x) is uniformly positive definite, and the generalized potential Φ satisfies condition (15), then π satisfies a Poincaré inequality, so L has a spectral gap (14). 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 overdamped Langevin 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 Gardiner, C. (2009). Stochastic methods: A handbook for the natural sciences. Springer, 4th edition. Pavliotis, G. A. (2014). Stochastic Processes and Applications. Springer. 10

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