Abstract of “A Derivation and Analysis of the N-point Hierarchy For Conservation Laws And Multistable Dynamics with White Noise in the Allen-Cahn Equation”by Carey Caginalp, Ph.D., Brown University, May 2017. The …rst part of this thesis concerns conservation laws subject to random initial conditions. Using a discrete set of values for the solution, one can derive a hierarchy of equations in terms of the states. This hierarchy involves the n-point function, the probability that the solution takes on various values at di¤erent positions, in terms the n+1-point function. A second approach involves obtaining statistics on the n-point function with shocks. This approach yields a hierarchy of equations in which the time derivative of the n-point function is expressed in terms of the spatial derivatives of the n+1 and lower functions. The hierarchy of equations remains valid through collisions of shocks. In the second part of this thesis, we study the in‡uence of multistable dynamics under the in‡uence of white noise in a model of an ODE with a cubic nonlinearity. A prototypical bistable dynamic is replaced by white noise to obtain a stochastic di¤erential equation. A key question is whether there is a steady state, or invariant measure, and if so, what form it takes. We show that under the limit of a vanishing amount of this white noise, there is indeed an invariant measure concentrated as a Dirac type measure at the most stable equilibria if ‡uctations are uniform, or at multiple equilibria if they are not. Abstract of “A Derivation and Analysis of the N-point Hierarchy For Conservation Laws And Multistable Dynamics with White Noise in the Allen-Cahn Equation”by Carey Caginalp, Ph.D., Brown University, May 2017. The …rst part of this thesis concerns conservation laws subject to random initial conditions. Using a discrete set of values for the solution, one can derive a hierarchy of equations in terms of the states. This hierarchy involves the n-point function, the probability that the solution takes on various values at di¤erent positions, in terms the n+1-point function. A second approach involves obtaining statistics on the n-point function with shocks. This approach yields a hierarchy of equations in which the time derivative of the n-point function is expressed in terms of the spatial derivatives of the n+1 and lower functions. The hierarchy of equations remains valid through collisions of shocks. In the second part of this thesis, we study the in‡uence of multistable dynamics under the in‡uence of white noise in a model of an ODE with a cubic nonlinearity. A prototypical bistable dynamic is replaced by white noise to obtain a stochastic di¤erential equation. A key question is whether there is a steady state, or invariant measure, and if so, what form it takes. We show that under the limit of a vanishing amount of this white noise, there is indeed an invariant measure concentrated as a Dirac type measure at the most stable equilibria if ‡uctations are uniform, or at multiple equilibria if they are not. A Derivation and Analysis of the N-point Hierarchy For Conservation Laws And Multistable Dynamics with White Noise in the Allen-Cahn Equation by Carey Caginalp B. Phil. Summa Cum Laude, Honors Mathematics, University of Pittsburgh, 2011 Sc. M., Brown University, 2014 A dissertation submitted in partial ful…llment of the requirements for the Degree of Doctor of Philosophy in the Division of Applied Mathematics at Brown University Providence, Rhode Island May 2017 c Copyright 2017 by Carey Caginalp This dissertation by Carey Caginalp is accepted in its present form by the Department of Computer Science as satisfying the dissertation requirement for the degree of Doctor of Philosophy. Date Govind Menon, Director Recommended to the Graduate Council Date Constantine Dafermos, Reader Date David Kaspar, Reader Approved by the Graduate Council Date Andrew G. Campbell Dean of the Graduate School iii Vita EDUCATION UNDERGRADUATE – UNIVERSITY OF PITTSBURGH, B. PHIL. SUMMA CUM LAUDE, HONORS MATHEMATICS, 2009-11 The B. Phil. degree entails a rigorous program of study beyond the B. Sc. and a thesis that is approved by a committee that includes an external member. I also took numerous physics classes and graduate math courses including: Real and Complex Analysis and Linear Algebra. GRADUATE –BROWN UNIVERSITY –M.SC., APPLIED MATH, 2014 & PH.D., APPLIED MATH, 2017 (EXPECTED) Coursework included: Real Analysis, PDE, ODE, Calculus of Variations and Kinetic Theory, Thermodynamics and Graduate Statistical Mechanics AWARDS & HONORS CULVER PRIZE, DEPARTMENT OF MATHEMATICS, UNIVERSITY OF PITTSBURGH - 2010 Award given to juniors or seniors for exceptional achievement in undergraduate study. BRACKENRIDGE FELLOWSHIP FOR RESEARCH Summer support for collaborative research with a professor. NATIONAL SCIENCE FOUNDATION GRADUATE RESEARCH FELLOW - 2011 Three-year fellowship given to seniors and …rst year graduate students –received as a senior. BROWN UNIVERSITY GRADUATE RESEARCH FELLOWSHIP 2015, 2016 TEACHING TEACHING ASSISTANT 2013-14 -Two semesters of undergraduate Ordinary Di¤erential Equa- tions at the level of Boyce and DiPrima. iv INVITED LECTURES Asymptotic Analysis: Escape Through a Narrow Gate – Brackenridge Fellow Presentation - University of Pittsburgh, June 10, 2010. Program Asymptotics and Computation on Brownian Motion - SIAM Annual Meeting, Pittsburgh, PA, July 2010. Abstract Analytical and Numerical Results for Brownian Motion Through a Material, Workshop on Mod- eling of Materials with Fine Structure, Carnegie-Mellon University, May 2011. Abstract Analytical and Numerical Results on Escape of Brownian particles - The 9th AIMS Conference on Dynamical Systems - Orlando, FL, July 2012. Abstract E¤ects of White Noise in Multistable Dynamics in Di¤erential Equations - AMS Sectional Meet- ing Special Session, Georgetown Univ, Washington, DC, March 7-8, 2015. Abstract Propagation of Special Initial Data Through Burgers’ Equation - Joint Dynamics and PDE Seminar –Brown University, April 25, 2016. Abstract E¤ects of White Noise in Multistable Dynamics, Am. Inst. of Mathematics 11th Int. Conf on Dynamical Systems, Orlando, FL, July 1-5, 2016. Abstract PUBLICATIONS Analytical and numerical results for …rst escape time in 2D (with Xinfu Chen) - Comptes Rendus, C. R. Acad. Sci. Paris, Ser. I 349 (2011) 191–194 Analytical and numerical results on escape (B. Phil. Thesis 2011) - University of Pittsburgh (2011) Analytical and numerical results for an escape problem (with Xinfu Chen) - Archive for Rat. Mech. Analysis 203 (2012) 329–342 E¤ects of white noise in multistable dynamics (with Xinfu Chen, Jianghao Hao and Yajing Zhang) - Discrete and Continuous Dynamical Systems B, 18 (2013)1805-1825 v Preface vi Acknowledgements Part II of the thesis is based on work published in [9]. The author thanks Professor Govind Menon and the support of the NSF and Brown research fellowships, as well as Professor Constantine Dafer- mos and Dr. David Kaspar for valuable discussions. vii Contents List of Tables xi List of Figures xii I Part I - A Derivation and Analysis of the N-point Hierarchy For Conservation Laws 1 1 Introduction 2 1.1 Background on Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Evolution of Conservation Laws with Random Data . . . . . . . . . . . . . . . . . . 4 1.3 Conservation Laws with Polygonal Flux . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Kinetic Equations Derived in This Dissertation . . . . . . . . . . . . . . . . . . . . . 8 1.5 Multistable Dynamics in Allen-Cahn Equations . . . . . . . . . . . . . . . . . . . . . 11 2 Background: A Discrete Example And Generalization for Piecewise Linear Flux 13 2.1 Downward Jump Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2 Upward Jump (Rarefaction) Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.3 Generalization - Downward Jump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.4 Generalization for Upward Jump (Rarefaction) Case . . . . . . . . . . . . . . . . . . 22 3 Derivation of n-point Distribution From Entropy-Entropy Flux Pair Form 27 3.1 Derivation of equation for the 1-point distribution . . . . . . . . . . . . . . . . . . . 28 3.2 Equation for the 2-point distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.3 Equation for the n-point distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 viii 3.4 Reduction to Transport Type Equation and Properties . . . . . . . . . . . . . . . . . 34 3.5 Two-point case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.6 N-point case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4 Solution and Illustration of Equations 43 4.1 Solving the Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.2 Examples: purely deterministic case with shocks . . . . . . . . . . . . . . . . . . . . 44 4.3 Selected examples with randomness introduced into the initial conditions . . . . . . 49 4.4 Example 1: Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 4.5 Example 2: Exponential Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.6 Example 3: Exponential Distribution (with o¤set) . . . . . . . . . . . . . . . . . . . 56 4.7 Analytic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 4.8 Calculating two-point functions for Example 3 . . . . . . . . . . . . . . . . . . . . . 61 5 Utilizing Shocks as the Building Blocks 65 5.1 Compatibility Conditions for n-point Equations . . . . . . . . . . . . . . . . . . . . . 66 5.2 Derivation of 1-point Equation for Shocks . . . . . . . . . . . . . . . . . . . . . . . . 67 5.3 Example with two shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 5.4 Derivation of n-point Equation for Shocks . . . . . . . . . . . . . . . . . . . . . . . . 78 5.5 Reduction of n-point equation to 1-point . . . . . . . . . . . . . . . . . . . . . . . . . 80 5.6 Flavor of a more Complex Example (5 shocks) . . . . . . . . . . . . . . . . . . . . . 81 6 Two interesting limits 88 6.1 Three-state limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 II Part II - Multistable Dynamics in the Allen-Cahn Equation with White Noise 91 7 Introduction 92 8 The Invariant Measure 96 9 Well-posedness of Kolmogorov Equation 98 ix 10 Long Time Behavior 103 11 Limits of the Invariant Measure When White Noise Is Vanishing 106 2 12 The Degenerate Case Where Is Not Everywhere Positive 110 12.1 The Model Case With Special Regularization . . . . . . . . . . . . . . . . . . . . . . 111 12.2 General Regularization When Vanishes at Stable States . . . . . . . . . . . . . . . 114 12.3 The Case When Vanishes Only Beyond All Critical Points . . . . . . . . . . . . . . 118 A Hopf-Lax Style Existence and Uniqueness of Solution 122 A.1 Proving existence of a solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 A.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Bibliography 136 x List of Tables xi List of Figures 2.1 In the discrete case, an upward jump in the initial conditions results in the discretized version of a rarefaction wave. The region beteween splits into values between the two slopes, and the interface between regions is the line with slope prescribed by the Rankine-Hugoniot condition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 5.1 Interactions between possible shocks in the three-state system, categorized by growth and di¤erent kinds of decay. Depending on relative velocities of the shocks, one shock may catch up to another and interact to combine a new shock in the form of (left state, right state) above each …gure. . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 10.1 Graphs of = (u) with b(u) = (u )(1 u2 ) and (u) = 1 + (u 1)2 . When = 0:05, the less stable equilibrium u 1 is selected by the ‡uctuation. . . . . . . . 105 xii Part I Part I - A Derivation and Analysis of the N-point Hierarchy For Conservation Laws 1 Chapter 1 Introduction 1.1 Background on Conservation Laws This dissertation consists of two parts. In the …rst part of this thesis, we develop kinetic equations for scalar conservation laws with a polygonal ‡ux function. The classical prototype for nonlinear conservation laws is Burgers’equation, which has been studied in numerous works (e.g. [2], [3], [4], [5], [6], [14], [16], [20]). This is a simple model that produces shocks, given by: ut + uux = 0 u (x; 0) = g (x) : (1.1) This has also been studied extensively in Menon and Srivisan [31] involving random initial conditions, which is directly relevant to our current work. Under conditions of su¢ cient regularity, the solution of a generalized nonlinear conservation law ut + (f (u))x = 0 on R (0; 1) u (x; 0) = g 0 (x) on R f0g (1.2) 2 3 can be related to that of the Hamilton-Jacobi problem wt + (f (wx )) = 0 on R (0; 1) w = g on R f0g (1.3) by the formula Z w (x; t) = u (x; t) dx; (1.4) assuming f 2 C 2 . One can also consider an analogous problem for a discretized ‡ux function, and we analyze this problem under a wide range of techniques, from Hopf-Lax analysis to front tracking, usage of the theorem of total probability, and …nally the derivation of a formal hierarchy. The classical Hopf-Lax formula is used to introduce the notion of an entropy solution to a scalar conservation law of the form (1.2). One method to construct the solution entails the use of curves (or as in this case, with no source term, straight lines) denoted characteristics in the space-time two-dimensional domain to simplify the PDE to a one-dimensional structure. Along these lines, the slope of which depends on the value at the initial condition, the solution is constant. If a point (x; t) can be traced back to a unique characteristic, the solution at this point is then prescribed by that value. However, it is often the case that multiple characteristics can be drawn back from a given point, leading to a multivalued solution if constructed solely from this method of tracing back characteristics. One method of well-posedness that was resolved by Hopf [22], and generalized by Lax [26], [27], involved the addition of a vanishing viscosity term "uxx as a source to (1.1). Using various transformations and analytic techniques, they showed that there was indeed a closed form solution for the value of the solution. This solution identi…ed the unique characteristic that would yield the solution with the desired properties. This solution formula is known as the Hopf-Lax formula and applies to convex, smooth ‡ux functions f: However, this can also be used as a building block for some of our calculations with a piecewise linear ‡ux, together with molli…cation and other arguments. In particular, if we denote the Legendre transform of the ‡ux function f by f (p) = sup fqp f (q)g ; (1.5) q2R the integrated initial condition as Z p G (p) = g (x) dx; (1.6) 0 4 and the Hopf-Lax functional as x s I (p; x; t) = G0 (p) + tf ; (1.7) t then the inverse Lagrangian point is expressed by the variational form a (x; t) = sup I (p; x; t) = inf I (s; x; t) : (1.8) p2R s2R The solution is then expressed as x a (x; t) u (x; t) = (1.9) t for Burgers’equation, or by 1 x a (x; t) u (x; t) = (f 0 ) (1.10) t in the more general case. Of course, the further growth condition on the functional lim I (p; x; t) = 1 (1.11) p! 1 is also needed in order to ensure a (x; t) is …nite and that the in…mum is achieved. This is known as the Hopf-Lax formula for conservation laws. The notable case of Burgers’equation is achieved by setting f (u) = 21 u2 ; upon which (1.8) may be rewritten as ( ) 2 + (x p) x a (x; t) a (x; t) = arg min G0 (p) + ; u (x; t) = : (1.12) p2R 2t t + The means we pick the largest argument at which the minimum is obtained, should it occur at more than one point. 1.2 Evolution of Conservation Laws with Random Data Burgers proposed his equation [4] as a model for turbulence of incompressible ‡uids. Even though it lacks some of the key features, it is a simple model that exhibits some key aspects of conservation laws. In particular it illustrates the development of shocks, even when the initial conditions are smooth. With the introduction of randomness in the initial conditions, a conservation law such as 5 Burgers’equation is useful to study to understand how the features of the randomness in the initial conditions evolve in time. In particular, questions that can be asked include how do the statistics of the solution evolve, whether Markov properties conserved, and whether there are universality classes for Burgers’equation, as in [30]. In Menon [29] and Menon and Srinivasan [31], further analysis on these equations was performed to gain deeper understanding of particle dynamics and the evolution of the system starting with random initial data. The analysis went beyond Burgers’equation to the more general case of a C 1 ; convex ‡ux. Here, the initial conditions were restricted to processes that were spectrally negative, that is, stochastic processes with jumps but only in one direction. In particular, only downward jumps were permitted, as upward jumps lead to an immediate breakdown of the statistics. Furthermore, this initial stochastic process was also assumed to be Markov. Using the Levy-Khinchine representation for the Laplace exponent and other methods, formal calculations were used to establish a number of results. One remarkable assertion proven states that if one starts with strong Markov, spectrally negative initial data, then this Markov property persists in the entropy solution for any positive time t > 0. This closure result can also hold for a non-stationary process and can obtain an equation for a generator in the case of Feller processes. In general, Feller jump di¤usions are described by a generator acting on Cc1 test functions, taking the form A' (u) = a (u) '00 (u) + b (u) '0 (u) + c (u) ' (u) Z + (' (v) ' (u) (u; v) '0 (u)) n (u; dv) (1.13) Rnfug for certain functions a; b; c; jump kernel n (u; dv) ; and describes certain types of interactions. For example, for a Feller process, the generator (1.13) becomes Z u 0 A (t) ' (u) = b (u; t) ' (u) + (' (v) ' (u)) n (du; v; t) : (1.14) 1 Indeed, one main result of Menon and Srinivasan is that @t A = [A; B] (1.15) 6 where brackets denote the Lie bracket, and the operator B is given by Z u B' (u) = f 0 (u) b (u; t) '0 (u) [f ]u;v (' (v) ' (u)) n (u; dv; t) (1.16) 1 where [f ]uv is the slope given by the Rankine-Hugoniot condition f (u) f (v) [f ]uv := : u v If the process is not stationary, (1.15) can be amended by adding in another term: @t A @x B = [A; B] : (1.17) Another surprising result along these lines is that these equations can admit exact solutions in certain special cases. The work of Groeneboom [20] calculates shock statistics for one such case with white noise initial data. A self-similar solution is obtained with explicit formulae for the jump density in terms of Laplace transforms of Airy functions. Although a closure theorem is proved rigorously, many of the calculations in [31] such as the new kinetic equations and derivations of the Lax equation from four di¤erent approaches are formal. In [23], (1.15) is rigorously established. Speci…cally, they consider the Cauchy problem 8 > < t = H ( )x in R (0; 1) (1.18) > : = on R ft = 0g for a stochastic process = (x) : Uniqueness of the solutions to marginal and kinetic equations was proven under the assumptions: (i) this process is a pure-jump process starting at (0) = 0; evolving according to a speci…c rate kernel, (ii) that the Hamiltonian H : [0; P ] ! R is smooth, convex, has downward jumps, nonnegative right derivative at 0; and nonin…nite left-derivative at P . This is then used to prove the main result. In doing so, they consider a random particle system and write down equations for the in…nite-dimensional system, much like in a statistical mechanics approach from physics. 7 1.3 Conservation Laws with Polygonal Flux A polygonal ‡ux function f consists of a …nite number of piecewise linear segments. Denoting these n+1 slopes by fmi gi=1 ;we also require that they are in ascending order to force (non-strict) convexity n of the ‡ux function, and have break points at fci gi=1 . That is, we require m1 < m2 < ::: < mn : More precisely, de…ne f 0 (x) = mi for ci 1 < x < ci ; c0 = 1; cn+1 = +1 (1.19) As is readily calculated, the Legendre transform of such a function is another piecewise linear function on a bounded interval with the roles of the slopes and break points switched, and becomes in…nite outside of that interval. Much of the original Hopf-Lax theory can still be implemented, as it involves minimization problems, which continue to make sense with the obvious correction of taking the minimum over the corresponding domain of the function argument rather than the entire real line. For the initial conditions, we take piecewise constant data whose range of values is speci…ed by n+1 the slopes fmi gi=1 of the polygonal ‡ux function. This sort of setup for the ‡ux function was used in Dafermos [11] in an approximation argument, to build up to a smooth ‡ux using increasingly …ne polygonal approximations Another related approach is that of a sticky particle system as considered by Brenier and Grenier [6]. As particles are discrete, this requires them to express the solution for the velocity …eld as a nonnegative measure rather than a function, and impose an entropy condition @t ( U (u)) + @x ( uU (u)) 0 (1.20) for all convex smooth functions U . They consider n particles on the real line, tracked by their weight (mass), position, and velocity. When two particles collide they "stick" together, and in this manner only a …nite number of collisions (or shocks) can occur in this setup before a steady state is reached. A given pair of particles will either collide within a …nite amount of time or never catch up due to their relative velocities. This is similar to the sort of behavior considered in this dissertation through the study of shocks both at the one and n-point levels. 8 1.4 Kinetic Equations Derived in This Dissertation The main goal of this thesis is to derive kinetic equations that describe the evolution of the statistics of the solution to (1.2) when the ‡ux function is polygonal, and the initial data is piecewise constant. This work improves on [31], as the results therein do not apply when the ‡ux is not C 1 . In this work, we pursue two di¤erent tracks for building a hierarchy of equations to describe the movement and interaction of shocks as they propagate through the system. In essence, the velocities at various points (and to a greater extent, shocks) can be treated as sticky particles. When an interaction occurs, they stick together and will not separate. Under the appropriate assumptions in regards to speci…c permitted initial data, there is no separation (i.e., rarefaction fanning), in such discrete cases. The virtue of this description of the problem and polygonal approximation scheme of Dafermos [11] lies in being able to enumerate possible interactions and illustrate results against a variety of examples. Further, in cases with such discrete states and corresponding initial data, the theorem in [31] does not hold, as the ‡ux is no longer C 1 :We derive results to …ll in the gap and explain the results in our case of interest, with a polygonal ‡ux. Similar work with a di¤erent approach is established in [23] with a …nite boundary condition (that is, restriction of the problem to a box 0 x L) and a representation of statistical solutions with bounded, monotone, and piecewise constant initial conditions. In this dissertation, we consider a piecewise linear ‡ux function as outlined in the proceeding subsections, using two di¤erent methods. In both approaches, a key object of importance to establish the hierarchy is the n-point function. This is the basic building block to track the important properties in a particle system in terms of probability distributions. For example, the one-point function of the form p1 (x; t; ui ) (1.21) describes the probabilities (or distribution, in a continuous case) of a particle at location x with velocity state ui ; occurring at some time t in the system. Similarly, the two-point function may be described as p2 (x1 ; x2 ; t; ui ; uj ) (1.22) and can be thought of in several ways. One interpretation is that we take a cross-section of the solution at time t and view it as a process, and (1.22) provides the information as to the probabilities 9 (or distribution) of the solution with the values at the points x1 and x2 pinned to velocity states ui and uj ; respectively. This de…nition can then easily be extended to any (…nite) number of points, and in particular an n-point function of the form pn (x1 ; :::; xn ; t; ui1 ; :::; uin ) : (1.23) This simply corresponds to the probability that, at time t; the solution pro…le takes the value ui1 at the point x1 ; ui2 at the point x2 ; etc., up through uin at xn : The exact information speci…ed can vary in di¤erent approaches that we consider, but the principle remains. At the n-th level, for a speci…c time t, there are n x-coordinates to which the n-point function ascribes information. Generally, the information at the nth level can be expressed in terms of various interactions at the (n+1) level. The goal is to ultimately provide a derivation of a set of equations known as a hierarchy and perform analysis and work various examples on it to check for consistency. It should be noted that later on in our work it becomes convenient to work with the tail cumulative distribution functions, that is n X F1 (x; t; ui ) = p1 (x; t; ul ) ; l=i Xn n X F2 (x1 ; x2 ; t; ui ; uj ) = p2 (x1 ; x2 ; t; ui ; uj ) ; l1 =i l2 =j and so forth. In the …rst of the approaches we consider, these particles are represented simply as the value of the velocity …eld at various points. When carried through, this method accurately describes the dynamics of the system and initial formation of shocks in a similar manner to the method of characteristics and front tracking. After the time at which shocks …rst encounter each other and combination of shocks occurs, it leads to a multivalued solution much like classical solutions. While this leads formally to a set of kinetic equations, they do not satisfy the entropy condition and do not ultimately give an adequate description for kinetic theory. The latter, more cohesive and comprehensive approach comes from considering shocks rather than individual points in the velocity …eld as the basic building block of the particle system. In doing so, we change the de…nition and notation for the n-point function, writing it in the form fn x1 ; :::; xn ; t; u1 ; :::; un ; v 1 ; :::; v n 10 to represent the probability of the solution pro…le taking on values ui at xi (the left-hand limit of the solution at xi ) and v i at xi : We assume right continuity, so one can regard it as the values at xi and xi +; that is both limits, as one would consider it in a continuous case. In this approach it is necessary to make some reasonable assumptions on the initial data to avoid discretized rarefaction waves. In particular, with strictly ordered states fui g ;the assumption that there are no upward jumps between non-neighboring states is required. With this particle structure, a complete (though not closed) hierarchy is obtained and tested rigorously against several examples. In particular, the n-point equation obtained by this method of focusing on shocks is given by n X @t fn x1 ; :::; xn ; t; u1 ; :::; un ; v 1 ; :::; v n + cui vi @xi fn x1 ; :::; xn ; t; u1 ; :::; un ; v 1 ; :::; v n i=1 n X X = 1fvi =ui +1g (cui w cwvi ) @i fn+1 xi ; xi ; :::; t; ui ; w; w; v i :::: i=1 w n X X (cui v cvi w ) @i0 fn+1 xi ; xi ; :::; t; ui ; v i ; v i ; w; ::: i=1 w2W2i n X X (cwui cui vi ) @i fn+1 xi ; xi ; :::; t; w; ui ; ui ; v i ; ::: : (1.24) i=1 w2W3i On the left-hand side of the equation we have a term representing the change in time of the n- point function, and a second free-streaming term keeping track of the relative motions of each pair ui ; v i : The right-hand side contains three di¤erent types of collision terms. The …rst is a growth n term, enumerating the possibilities by which a trajectory described by ui ; v i i=1 speci…ed on the left-hand side can occur. Namely, this happens when a shock of the form ui ; w collides with w; v i and forms ui ; v i : The characteristic function in front represents the fact that this is only possible if the states ui and vi have an appropriate ordering, so that there is no upward jump between non- nearest neighbors formed, as it should be. The second and third terms refer to the possible ways in which this trajectory can be destroyed, and the sums enumerate the possibilities. Again depending on the ordering of ui and vi ; a particular shock (ui ; vi ) can generally be destroyed by a shock from the right (second term) or shock from the left (third term). Only certain types of interactions are possible, as contained in the sets W2i and W3i for the indices. The intricacies and derivation of these exact lists of possibilities will be explained in greater detail later, but for completeness, we state the 11 de…nitions here: 8 > < w < ui if v i = ui + 1 W2i = w j > : w vi + 1 if v i < ui 8 > < i w > ui + 1 if v i = ui + 1 W3 = w j (1.25) > : w ui 1 if v i < ui Unlike the …rst approach, this hierarchy of equations are consistent through collisions of shocks. When applied to examples with shock interactions, it is shown empirically that solving these equa- tions yields a complete characterization of the solution pro…le. It persists even through shock interactions, an achievement that has eluded many classical methods without some sort of resetting or application of entropy conditions. 1.5 Multistable Dynamics in Allen-Cahn Equations The second part of this thesis concerns a problem involving the introduction of white noise into Allen-Cahn equations. In basic ordinary di¤erential equations, many become familiar with innocuous looking equations of the form du 1 =4 u 1 u2 : (1.26) dt 2 It has a stable equilibrium points at 1 and an unstable equilibrium point at 21 ; and analysis is clear- cut. It is somewhat less obvious, however, that under the addition of even an in…nitesimal amount of randomness, the problem is transformed from simple and deterministic to probabilistic. Based on the relative strength of the associated potential calculated from the expression on the right-hand side, we delve into a rich collection of results regarding this and other stochastic ODEs. These results are interesting in many aspects, not the least of which that they initially counterintuitive to rudimentary 1 ODE theory. We investigate the e¤ect of replacing this bistable forcing term 4 u 2 1 u2 with a generic drift b (u) ; and consider the dynamics of the resulting stochastic di¤erential equation dut = b(ut )dt + " (ut )dWt (1.27) 12 where fWt gt>0 is the Wiener process (Brownian motion), " is a positive constant measuring the size of ‡uctuation, and ( ) is a non-negative function modeling the dependence of ‡uctuation on system state. By standard Stochastic Di¤erential Equation (SDE) theory ([38]), with certain conditions on and b, (1.27) admits a unique solution and the probability density of the random variable is the solution to the Fokker-Plank equation 8 > < @t = @u @u 1 "2 2 b on R (0; 1); 2 (1.28) > : ( ; 0) = 0( ) on R f0g We expect the system to eventually achieve stochastic equilibrium in the sense that there is an invariant distribution formed from the limit lim ( ; t) = ( ): (1.29) t!1 In this second part we establish the existence of a unique invariant measure and analyze the asymp- totic limit of vanishing noise, which in fact leads to Dirac measures. We also consider further results when the regularity and nondegeneracy conditions on the various parameters ( ; b) are relaxed. A more detailed introduction to this problem is presented in Part II. Chapter 2 Background: A Discrete Example And Generalization for Piecewise Linear Flux We start by providing a discrete pedagogical example with a piecewise linear ‡ux, and analyze it using the Hopf-Lax formula. Our intention in doing so serves to illustrate the techniques for dealing with conservation laws with polygonal ‡uxes as in [11]. Although the assumptions of convexity and superlinearity are not met in a strict sense or do not hold, respectively, the solution formula can still be applied. The absence of the latter condition leads to in…nite values for the Legendre transform of the ‡ux function on all of R aside from a compact interval. However, this pathology can be mitigated by simply considering the minimization problem over this compact interval where the Legendre transform takes …nite values. As usual, any sort of example can be decomposed into two basic Riemann-style problems. The …rst is a downward jump in the initial conditions, which results in a single shock propitiating with a velocity determined by the Rankine-Hugoniot condition. We show that the Hamilton-Jacobi solution evaluated through the Legendre transform and minimization problem indeed matches the expected velocity of the shock. The second example is the reverse problem, with an upward jump in the initial solution pro…le. When passed through a smooth ‡ux function, this sort of discontinuity results in a rarefaction fan. We argue that in the piecewise linear ‡ux case, the solution pro…le will imitate 13 14 this rarefaction structure to the best degree of …neness possible based on the available velocities contained in the ‡ux function. In other words, a number of shocks appear between the two initial values, each between nearest neighbors, with velocities that match what would be given by the Rankine-Hugoniot condition. These two cases are illustrated below. Note that in our examples, the left and right slopes do not necessarily match with the slopes of the ‡ux function. Thus, it does not truly correspond to the situation of kinetic theory, but it still provides some valuable insight into the behavior of solutions. 2.1 Downward Jump Case We being by considering the Hamilton-Jacobi problem ut + H (ux ) = 0; in R (0; 1) (2.1) u = g; on R f0g ; (2.2) n+1 n where H is a piecewise linear ‡ux function, with slopes given by fmi gi=1 and break points fci gi=1 ; with some initial condition that is piecewise linear with one break point at zero. In particular, choose the parameters c7 = 2; c6 = 1:5; c5 = :7; c4 = :4; c3 = :2; c2 = :1; c1 = 2; and set 8 > < :5y y<0 g (y) = : (2.3) > : :3y y>0 8 The fmi gi=1 are arbitrary. x y We want to minimize the expression I (y; x; t) = tL t + g (y) : The function I evaluated at the minimum is indeed the solution to the Hamilton-Jacobi equation: For the region x < m2 t; the minimum will be attained at the value y = x m5 t; and we observe u (x; t) = I (x m5 t; x; t) = tL (m5 ) + :5 (x m5 t) (2.4) and @x u (x; t) = :5: (2.5) 15 Similarly, for the region x > m5 t, the minimum is attained at y = x m2 t; and u (x; t) = I (x m2 t; x; t) = tL (m2 ) :3 (x m2 t) (2.6) so that @x u (x; t) = :3: (2.7) The more interesting case occurs when m2 t x m5 t: In this case, we will have two local minima. They will be obtained at the y-values y1 = x m5 t and y2 = x m2 t: We check when the values of I are equal at these two points: I (x m5 t; x; t) = tL (m5 ) + :5 (x m5 t) = tL (m2 ) :3 (x m2 t) = I (x m2 t; x; t) (2.8) 5 )x= t fL (m2 ) L (m5 ) + :5m5 + :3m2 g =: M t: 4 Along M t; the value of y will not be unique. Both y1 and y2 give the value for the minimizer of I (y; x; t) : For M t < x < m5 t; we have I (:::) = tL (m5 ) + :5 (x m5 ) t; and for m2 t < x < M t; we have I (:::) = tL (m2 ) :3 (x m2 t). Hence the derivative is given by 8 > < :5 x>M t @x u (x; t) = : (2.9) > : :3 x < :5y y<0 g (y) = : (2.12) > : :3y y>0 x y The set of points which could potentially be minima for the expression I (y; x; t) = tL t + g (y) for a given (x; t) is limited to a …nite set, namely yi := x mi t and f0g : For x m4 t < 0 < x m3 t; i.e., m3 t < x < m4 t; the minimum is clearly attained at y = 0; so that x y x u (x; t) = I (0; x; t) = min tL + g (y) = tL + g (0) ; (2.13) y2R t t and di¤erentiating yields @x u (x; t) = c3 : (2.14) For x m3 t < 0 < x m2 t; we again have x y x u (x; t) = I (0; x; t) = min tL + g (y) = tL + g (0) ; (2.15) y2R t t and di¤erentiating yields @x u (x; t) = c2 : (2.16) Now, for x m2 t < 0; i.e. x < m2 t; the slope of I (y; x; t) is positive for y > 0; so the minimum 17 must be attained for some y < 0; speci…cally at y = x m2 t. The minimum is then given by I (x m2 t; x; t) = tL (m2 ) + g (x m2 t) = tL (m2 ) :5 (x m2 t) ; (2.17) and the derivative @x u (x; t) = :5: (2.18) For x m4 t > 0; i.e. x > m4 t; the slope of I (y; x; t) is negative for y < 0; so the minimum must be achieved for some y > 0. Speci…cally, the minimum is then attained at y = x m4 t: The value at the minimum is then given by I (x m4 t; x; t) = tL (m2 ) + g (x m4 t) = tL (m4 ) + :3 (x m2 t) ; (2.19) and the derivative @x u (x; t) = :3: (2.20) Putting the value of @x u in all the di¤erent regions together, one has 8 > > :5 x < m2 t > > > > > < c2 m2 t < x < m 3 t @x u (x; t) = : (2.21) > > > > c3 m3 t < x < m 4 t > > > : +:3 m4 t < x From plotting this solution graphically, one can see that we obtain a rarefaction "fan" in a certain discretized sense. More precisely, note that from typical analysis using characteristics, one would expect the behavior in the regions x < m2 t and x > m4 t . The region m2 t < x < m4 t is left empty. For a smooth function, we would expect a rarefaction fan to open up in this wedge. Here, we observe that we still have a fan, in the sense of discretized wave speeds in accordance with the velocities of the ‡ux function. In fact, it picks out the velocities ci that are between the two slopes of the initial condition. In this case, this refers to c1 < :5 < c2 < c3 < +:3 < c4 < ::: The solution pro…le and idea of this rarefaction in a discretized manner are illustrated in the …gure below. 18 Figure 2.1: In the discrete case, an upward jump in the initial conditions results in the discretized version of a rarefaction wave. The region beteween splits into values between the two slopes, and the interface between regions is the line with slope prescribed by the Rankine-Hugoniot condition. 2.3 Generalization - Downward Jump We now proceed to generalize the previous example in the form of a Proposition, from [11] as follows. Proposition. Let H ( ) be a piecewise linear ‡ux function with n + 1 di¤ erent slopes, denoted n+1 n by fmi gi=1 , mi < mi+1 ; and n corresponding break points fci gi=1 ; ci < ci+1 , and g be an initial condition of the form 8 > < ca y y<0 g (y) = : (2.22) > : cb y y>0 where cb < ca : Consider the solution u (x; t) to the corresponding Hamilton-Jacobi problem ut + H (ux ) = 0 (2.23) u = g: Then we have 8 > < ca x < mi+1 t @x u (x; t) = : (2.24) > : cb x > mi+1 t Proof. We consider two cases: (i) if there exists i such that ci < cb < ca < ci+1 ; that is, none of the ci are between ca and cb ; and (ii) otherwise, there exists some j 0 such that ci < ca < ci+1 < 19 ::: < ci+j < cb < ci+j+1 : To this end, consider an arbitrary initial condition of the form 8 > < ca y y<0 g (y) = (2.25) > : cb y y>0 with ca > cb constant. We also take a ‡ux function H ( ) that has n + 1 di¤erent slopes denoted n+1 n by fmi gi=1 ; mi < mi+1 , with n corresponding break points fci gi=1 ; ci < ci+1 : Take the Laplace transform in the usual way: L (p) = supq2R fpq H (q)g : Then L (y) has slopes which will be n n+1 given by " 1"; fci gi=1 ; " + 1"; ci < ci+1 ; and break points fmi gi=1 : Now consider f (y; x; t) := x y tL t ; as a function of y: The function will have slopes cn ; cn 1 ; :::; c1 ; with break points at yn+1 ; yn ; :::; y1 ; where yi := yi (x; t) = x mi t; 1 i n + 1: Now, we …nd i; j such that ci < ca < ci+j cb < ci+j+1 ; noting that j 0: If in fact j = 0; it means that there are no such i so that ca < ci < cb ; and if not, then we have ci in this range from i + 1 to i : We split this into cases. Case j = 0: In this case we have ci < cb < ca < ci+1 ; leading to the set of inequalities ca ci > 0; ca ci+1 < 0; cb ci > 0; cb ci+1 < 0: (2.26) When yi+1 > 0; the slope of f near 0 is ci+1 on both sides. Since we have ca ci+1 < 0; cb ci+1 < 0; and cb ci > 0; the minimum will be attained at yi+1 ; and hence x yi+1 t u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) t @x u (x; t) = ca : (2.27) When yi+1 < 0 < yi ; the slope of f near 0 is ci on both sides. Noting that ca ci > 0; ca ci+1 < 0; and cb ci > 0; we note that the minimum is attained at yi+1 (the vertex of the slopes ci+1 and ci of f ), and x yi+1 t u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) t @x u (x; t) = ca (mi t < x < mi+1 t) (2.28) 20 Finally, if yi < 0; the slope of f near 0 is ci 1: Noting that ca ci 1 > 0; cb ci 1 > 0; ca ci > 0; and ca ci+1 < 0; the minimum is attained at yi+1 (the vertex of the slopes ci+1 and ci ), and x yi+1 u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) t @x u (x; t) = ca (since x mi+1 t < x mi t < 0). (2.29) Putting the pieces together, the solution is then given by 8 > < ca x < mi+1 t @x u (x; t) = : (2.30) > : cb x > mi+1 t To check with the Rankine-Hugoniot condition, one observes that [[H (u)]] H (cb ) H (ca ) (cb ca ) mi+1 = = = = mi+1 (2.31) [[u]] cb ca cb ca so it indeed matches. Case j > 0: Then we have the string of inequalities ci < ca < ci+1 < ::: < ci+j < cb < ci+j+1 : This leads to ca ci > 0; ca ci+1 < 0; :::; ca ci+j < 0; ca ci+j+1 < 0; cb ci > 0; cb ci+1 > 0; :::; cb ci+j > 0; cb ci+j+1 < 0: (2.32) For yi+j+1 > 0; the slope of f near 0 is ci+j+1 : Noting that ca ci+j+1 < 0 and cb ci+j+1 > 0; cb ci+j > 0 we see that yi+j+1 is the minimum, hence x yi+j+1 u (x; t) = tL + g (yi+j+1 ) = tL (mi+j+1 ) + g (x mi+j+1 t) t @x u (x; t) = cb : (2.33) For yk+1 < 0 < yk for i < k < i + j; the slope of f near 0 is ck : Noting that ca ck < 0 and 21 cb ck > 0; we have that the minimum is at 0; hence x u (x; t) = tL + g (0) t @x u (x; t) = cb (since x > mk t > mi t). (2.34) For yi+1 < 0 < yi ; the slope of f near 0 is ci : Since ca ci > 0; cb ci > 0; ca ci+1 < 0; the minimum is attained at yi+1 (at the vertex of the slopes ci and ci+1 of f ), so that x yi+1 u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) t @x u (x; t) = ca (2.35) Finally, for yi < 0; the slope of f near 0 is ci 1: We have ca ci 1 > 0; cb ci 1 > 0; ca ci > 0; ca ci+1 , so that the minimum is attained at yi+1 (at the vertex of the slopes ci and ci+1 of f ), so that x yi+1 u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) t @x u (x; t) = ca (since x mi+1 t < x mi t < 0). (2.36) Hence, the solution in this case is still given by 8 > < ca x < mi+1 t @x u (x; t) = : (2.37) > : cb x > mi+1 t although the computations are slightly more involved, as we have seen. Indeed, the Rankine- Hugoniot holds: [[H (u)]] = (2.38) [[u]] mi+1 (ci+1 ca ) + mi+2 (ci+2 ci ) + ::: + mi+j (ci+j ci+j ) + mi+j+1 (cb ca ) = cb ca 22 2.4 Generalization for Upward Jump (Rarefaction) Case Now consider the case with the opposite initial condition, i.e. 8 > < ca y y<0 g (y) = ; (2.39) > : cb y y>0 now with constants ca < cb . We argue that one will obtain a rarefaction fan of discretized slopes if there exist ci in a suitable range, but not otherwise. To this end, we will consider cases. Case (i). First suppose there does not exist i such that we have ca < ci < cb : Then there exists i such that ci < ca < cb < ci+1 : We then have the following inequalities: ca ci > 0; ca ci+1 < 0 cb ci > 0; cb ci+1 < 0: (2.40) Consider …rst the range bounded by yi+1 < 0 < yi : Under the above inequalities, we consider two x y subcases based on the vertex of the function f (y; x; t) := tL t where the slopes ci+1 and ci meet. More precisely, this occurs at the point yi+1 = x mi+1 t: First suppose yi+1 < 0: In this case the slope of f near zero is ci on both sides. Since the sum of the slopes cb ci > 0; the function I (y; x; t) is increasing as one moves in the positive x direction. Further, in the other direction, we note that cb ci < 0; so as one moves in the negative x direction, the function I (y; x; t) is also increasing. This in conjunction with the convexity of the function outside the interval implies that the minimum indeed occurs at yi+1 in this case. Now suppose that in fact yi+1 > 0: Then the slope of f near zero is ci+1 on both sides. Since the sum of the slopes ca ci+1 > 0; the function I (y; x; t) is again increasing in the positive x direction. Furthermore, in the other direction, since cb ci+1 < 0; I (y; x; t) is again increasing. Hence, by the same token, the minimum is attained at yi+1 again in this case. Evaluating I (y; x; t) at the minimum yields x yi+1 t u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) : (2.41) t 23 Di¤erentiating yields 8 > < ca mi t < x < mi+1 t @x u (x; t) = (2.42) > : cb x > mi+1 t which in fact reduces to ca inside the wedge mi t < x < mi+1 t we were considering. One can then show that @x u (x; t) = ca for x < mi t and cb for x > mi+1 t; so that for the …nal result we have 8 > < ca x < mi+1 t @x u (x; t) = : (2.43) > : cb x > mi+1 t The boundary between ca and cb occurs at the line x = mi+1 t; which has slope mi+1 : From the Rankine-Hugoniot condition, we would expect the slope [[H (u)]] (cb ca ) mi+1 = = = mi+1 ; (2.44) [[u]] cb ca which indeed agrees with what we have obtained above. Case (ii). Now suppose there exists exactly one of the ci that is between ca and cb ; that is, ci < ca < ci+1 < cb < ci+2 : This yields ca ci > 0; ca ci+1 < 0; ca ci+2 < 0; cb ci > 0; cb ci+1 > 0; cb ci+2 < 0: (2.45) First suppose that yi+2 > 0: Then near zero, the slope of f (y; x; t) is ci+2 on both sides. Since we have ca ci+2 < 0, cb ci+2 < 0; and cb ci+1 > 0; the minimum is attained at yi+2 (the vertex between the slopes ci+2 and ci+1 ). This means that x yi+2 t u (x; t) = tL + g (yi+2 ) = tL (mi+2 ) + g (x mi+2 t) : (2.46) t Di¤erentiating yields @x u (x; t) = cb ; x > mi+2 t: (2.47) Next consider the case yi+2 < 0 < yi+1 : Then near zero, the slope of f (y; x; t) is ci+1 on both sides. Noting that ca ci+1 < 0 and cb ci+1 > 0; we observe that the minimum of I (y; x; t) occurs 24 at 0; hence x u (x; t) = tL + g (0) t @x u (x; t) = ci+1 ; mi+1 t < x < mi+2 t: (2.48) Now suppose yi+1 < 0 < yi : Then near zero, the slope of f (y; x; t) is ci on both sides. Noting that ca ci > 0; cb ci > 0; we observe that the minimum of I (y; x; t) occurs in this case at yi+1 ; hence x yi+1 t u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) (2.49) t Di¤erentiating then yields @x u (x; t) = ca ; mi t < x < mi+1 t: (2.50) Finally, when yi < 0; we have a minimum at yi by a similar argument, so that x yi t u (x; t) = tL + g (yi ) = tL (mi ) + g (x mi t) t @x u (x; t) = ca ; x < mi t: (2.51) Putting the di¤erent cases together, we have 8 > > ca x < mi t > > < @x u (x; t) = ci mi t < x < mi+1 t : (2.52) > > > > : cb mi+1 t < x This leaves us with two shocks, one on the line x = mi t and the other on x = mi+1 t: We verify that these shocks satisfy the R-H condition: [[H (u)]] (ci ca ) mi i = = = mi [[u]] ci ca [[H (u)]] (cb ci ) mi+1 i+1 = = = mi : (2.53) [[u]] cb ci Case (iii). Suppose there are multiple ci between ca and cb ; i.e., ci < ca < ci+1 < ci+2 < ::: < 25 ci+j < cb < ci+j+1 : We then have the inequalities ca ci > 0; ca ci+1 < 0; :::; ca ci+j < 0; ca ci+j+1 < 0 cb ci > 0; cb ci+1 > 0; :::; cb ci+j > 0; cb ci+j+1 < 0: (2.54) First suppose yi+j+1 > 0: Then near 0; the slope of f (y; x; t) is ci+j+1 on both sides. Since ca ci+j+1 < 0, cb ci+j+1 < 0; and cb ci+j > 0 we have that yi+j+1 (the vertex between slopes ci+j+1 and ci+j ) is the minimum and x yi+j+1 t u (x; t) = tL + g (yi+j+1 ) = tL (mi+j+1 ) + g (x mi+j+1 ) (2.55) t @x u (x; t) = cb : Now suppose that yk+1 < 0 < yk for i < k < i + j: Then near 0; the slope of f (y; x; t) will be ck on both sides. Since ca ck < 0 and cb ck > 0; clearly 0 will be the minimum, and x x u (x; t) = tL + g (0) = tL t t @x u (x; t) = ck : (2.56) Then the value of @x u (x; t) takes on the di¤erent ci bounded between ca and cb in the wedge yi+j+1 > x > yi : Finally we suppose yi < 0: In this case, near 0; the slope of f (y; x; t) is ci 1. We note that ca ci 1 > 0; cb ci 1 > 0; ca ci > 0 and ca ci+1 < 0; so that the minimum is attained at yi+1 (at the vertex between the slopes ci+1 and ci ), and we have x yi+1 u (x; t) = tL + g (yi+1 ) = tL (mi+1 ) + g (x mi+1 t) t @x u (x; t) = ca : (2.57) 26 Now combining all of these cases, we have 8 > > ca x < mi t > > > > > > ci mi t < x < mi+1 t > > < @x u (x; t) = ::: ::: : (2.58) > > > > > > ci+j mi+j t < x < mi+j+1 t > > > > : cb mi+j+1 t < x The last step is to check the Rankine-Hugoniot conditions along the j + 2 shocks. Indeed, we have H (ci ) H (ca ) mi (ci ca ) 0 = = = mi ; ci ca ci ca H (ci+k+1 ) H (ci+k ) mi+k+1 (ci+k+1 ci+k ) k = = = mi+k+1 ; 1 k j; ci+k+1 ci+k ci+k+1 ci+k H (cb ) H (ci+j ) mi+j+1 (cb ci+j ) j = = = mi+j+1 ; (2.59) cb ci+j cb ci+j which indeed matches the expression above. Chapter 3 Derivation of n-point Distribution From Entropy-Entropy Flux Pair Form In attempting to derive equations to represent the statistics of the system, a natural approach entails considering the velocities at certain points (x coordinates) as the basic building block. One might then start from an entropy-entropy ‡ux pair form and derive algebraic identities to express the kinetic behavior at one level in terms of expressions at the next level. As outlined in the introduction, this behavior is expressed in terms of n-point functions. Recall that these expressions take the form, for example, of p1 (x; t; ui ) p2 (x; x+; t; ui ; uj ) p2 (x1 ; x2 ; t; ui ; uj ) pn (x1 ; :::; xn ; t; ui1 ; :::; uin ) : (3.1) The …rst equation in (3.1) represents the probability that the solution at the point x takes on the velocity ui : The second expresses the probability that the solution takes the value ui at x and the value uj is the limiting value from the right - that is, there may be a jump at that point on the right. 27 28 That is, one hopes to obtain and solve an accurate collection of equations for the n-point distributions in terms of various other combinations of the n-point and n+1-point distribution functions. The basis of our approach is the use of a piecewise linear, convex ‡ux f and given test functions. The following sections highlight the steps in obtaining these equations, both for the simplest one- point distribution, the next level of the two-point distribution, and then …nally for the generalized equation for the n-point distribution. 3.1 Derivation of equation for the 1-point distribution We begin our study of the conservation law ut + (f (u))x = 0; (3.2) where f is convex and polygonal (with …nitely many break points) subject to piecewise constant initial conditions. By Corollary 2.8, p.67 of [21], there exists a unique solution that is a piecewise constant function of x for each t; and u (x; t) takes values in the …nite set comprising the break points of f and the values of the initial conditions. Also, there are only …nitely many interactions between the fronts of u. The function u satis…es the Kruzkov entropy condition. For each …xed t as a function of x, the total variation of u is bounded by the initial total variation. In addition, the ku ( ; t) u ( ; s)kL1 kf kLip T V (u0 ) jt sj Using this inequality in conjunction with basic BV results, one sees that u (x; t) is of bounded variation in either x or t with the other variable …xed ([11], p.117, [40]). Thus, u (x; t) is di¤erentiable almost everywhere in x and t, and the derivatives are in L1 ([35], p. 116) Thus, we can calculate @t ' (u) = '0 (u) @t u = '0 (u) (f (u))x = '0 (u) f 0 (u) @x u = 0 (u) @x u: (3.3) In other words, we have @t ' (u (x; t)) = @x (u (x; t)) : (3.4) 29 almost everywhere. To be precise, this is a piecewise linear, cadlag, convex ‡ux f de…ned by its values at points ui together with piecewise linear interpolation; that is f (ui ) = fi ; 1 i M and slopes then given by fk+1 fk ck = : (3.5) uk+1 uk We now take expectations of both sides of equation (3.4), assuming for the moment that we can pass the expectation on the left-hand side inside the time derivative, and have E f@t ' (u (x; t))g = @t E f' (u (x; t))g = E f@x (u (x; t))g (3.6) M We take a test function ' with discrete values on x 2 fxi gi=1 : Then the LHS of (3.6) becomes M X @t ' (ul ) p1 (x; ul ; t) ; (3.7) l=1 and the RHS is M X X M ( (um ) (ul )) p2 (x; x+; ul ; um ; t) : (3.8) l=1 m=1 Therefore, in the general case (3.6) yields M X M X X M @t ' (ul ) p1 (x; ul ; t) = ( (um ) (ul )) p2 (x; x+; ul ; um ; t) : (3.9) l=1 l=1 m=1 Now, we wish to choose the derivative of ' as a discretized version of a -function at uk , i.e. de…ne 1[uk; ;uk+1 ) (u) '0k (u) = : (3.10) uk+1 uk Consequentially, one has (integrating ' up from 1 and holding the left limit at zero) 0 ck 1[uk ;uk+1 ) (u) 'k (u) = 1[uk+1 ;1) (u) ; k (u) = ; (u) = ck 1[uk+1 ;1) (u) (3.11) uk+1 uk 30 Substituting the expressions (3.11) into (3.9), we have M X M X LHS = @t 1[uk+1 ;1) (ul ) p1 (x; ul ; t) = @t p1 (x; ul ; t) l=1 l=k k X M X M X k X RHS = ck p2 (x; x+; ul ; um ; t) + ck p2 (x; x+; ul ; um ; t) (3.12) l=1 m=k+1 l=k+1 m=1 Hence, we have an equation for the 1-point equation in terms of the 2-point equation: M X k X M X M X k X @t p1 (x; ul ; t) = ck p2 (x; x+; ul ; um ; t) + ck p2 (x; x+; ul ; um +; t) (3.13) l=k+1 l=1 m=k+1 l=k+1 m=1 The signi…cance of this expression is that it presents an equation for the change in time of probabilities of a particle having certain velocities in terms of the behavior at the right-hand limit. In a way the terms on the right hand side of (3.13) can be interpreted as a discrete derivative or …nite di¤erence. We show in the following sections that this process can be generalized, for the 2-point and then n-point equations, establishing the pattern of deriving various combinations of the n-point distribution in terms of other expressions involving the n+1 point distributions. Thus, we will see that the change in time at one level of the system is related to changes at the limit points xi + at the next. We outline the procedure …rst for the 2-point distribution and then for the n-point to derive a complete hierarchy of the equations. Once the hierarchy is obtained, our aim is to use the theorem of total probability on the equations to show how they can be reduced to a set of transport equations in n dimensions. One can then solve these systems of equations and then analyze the resulting solutions, comparing them against classically obtained results in di¤erent examples. 3.2 Equation for the 2-point distribution We now take '1 ; '2 that satisfy @t 'i (u (x; t)) = @x i (u (x; t)) ; i = 1; 2 (3.14) 0 with i := f 0 '0i and note that then @t ('1 (u (x1 ; t)) '2 (u (x2 ; t))) = @x 1 (u (x1 ; t)) '2 (u (x2 ; t)) @x 2 (u (x2 ; t)) '1 (u (x1 ; t)) (3.15) 31 Taking expectations and passing to the limit as before, we have @t E f'1 (u (x1 ; t)) '2 (u (x2 ; t))g = E f@x 2 (u (x; t)) '1 (u (x2 ; t))g : (3.16) Now we consider test functions of the form '0i (u) = 1[uki ;uki +1 ) (u) ; 'i (u) = 1[uki +1 ;1) (u) ; i = 1; 2 We then have 8 Z u > < 0 u < uki +1 i (u) = f 0 (s) '0i (s) ds = ; i = 1; 2 (3.17) 1 > : cki u uki +1 We now evaluate both sides of the equation (3.16) and have X M M X M X M X LHS = @t '1 (ul ) '2 (um ) p2 (x1 ; x2 ; ul ; um ; t) = @t p2 (x1 ; x2 ; ul ; um ; t) l=1 m=1 l=k1 +1 m=k2 +1 M X X M M X RHS = ( 1 (u (xm )) 1 (u (xl ))) '2 (u (xp )) l=1 m=1 p=1 M X X M X M ( 2 (u (xm )) 2 (u (xl ))) '1 (u (xp )) l=1 m=1 p=1 k1 X M X M X = ck1 p3 (x1 ; x1 +; x2 ; ul ; um ; up ; t) l=1 m=k1 +1 p=k2 +1 M X k1 X M X + ck1 p3 (x1 ; x1 +; x2 ; ul ; um ; up ; t) l=k1 +1 m=1 p=k2 +1 k2 X M X M X ck2 p3 (x2 ; x2 +; x1 ; ul ; um ; up ; t) l=1 m=k2 +1 p=k1 +1 M X k2 X M X + ck2 p3 (x2 ; x2 +; x1 ; ul ; um ; up ; t) (3.18) l=k2 +1 m=1 p=k1 +1 32 Hence, we have M X M X @t @t p2 (x1 ; x2 ; ul ; um ; t) l=k1 +1 m=k2 +1 k1 X M X M X = ck1 p3 (x1 ; x1 +; x2 ; ul ; um ; up ; t) l=1 m=k1 +1 p=k2 +1 M X k1 X M X + ck1 p3 (x1 ; x1 +; x2 ; ul ; um ; up ; t) l=k1 +1 m=1 p=k2 +1 k2 X M X M X ck2 p3 (x2 ; x2 +; x1 ; ul ; um ; up ; t) l=1 m=k2 +1 p=k1 +1 M X k2 X M X + ck2 p3 (x2 ; x2 +; x1 ; ul ; um ; up ; t) (3.19) l=k2 +1 m=1 p=k1 +1 3.3 Equation for the n-point distribution 0 0 In a similar fashion, we choose test functions 'i with i = fi '0i such that @t 'i (u (x; t)) = @x i (u (x; t)) ; 1 i n (3.20) and note that then ( !) 8 9 n Y < 0 u < uki +1 i (u) = f 0 (s) '0i (s) ds = ; 1 i n: (3.23) 1 > : cki u uki +1 33 We then evaluate both sides of equation (3.21) and have M X M Y X n LHS = @t ::: 'i (ui ) pn (x1 ; :::; xn ; u1 ; :::; un ; t) l1 =1 ln =1 i=1 M X M X = @t ::: pn (x1 ; :::; xn ; uk1 ; :::; ukn ; t) l1 =k1 +1 ln =kn +1 M X M X X n Y RHS = E ::: @x i (u (xli )) 'j u xlj ; t l=1 ln =1 i=1 j6=i 8 9 Pk1 PM PM > Xn >< li =1 li0 =k1 +1 lj =kj +1 cki pn+1 x1 ; x2 ; :::; xi ; xi +; :::; xn ; ul1 ; ul2 ; :::; uli ; uli0 ; :::uln ; t > = j6=i = PM Pk1 PM > > i=1 > : li =ki +1 li0 =1 lj =kj +1 cki pn+1 x1 ; x2 ; :::; xi ; xi +; :::; xn ; ul1 ; ul2 ; :::; uli ; uli0 ; :::uln ; t > ; j6=i (3.24) Hence, M X M X ::: @t pn (x1 ; :::; xn ; ul1 ; :::; uln ; t) l1 =k1 +1 ln =kn +1 8 9 n > > Pk1 PM PM > > X < li =1 lj =kj +1 cki pn+1 li0 =k1 +1 x1 ; x2 ; :::; xi ; xi +; :::; xn ; ul1 ; ul2 ; :::; uli ; uli0 ; :::uln ; t = j6=i = PM Pk1 PM > > > > i=1 : li =ki +1 li0 =1 lj =kj +1 cki pn+1 x1 ; x2 ; :::; xi ; xi +; :::; xn ; ul1 ; ul2 ; :::; uli ; uli0 ; :::uln ; t ; j6=i (3.25) 34 3.4 Reduction to Transport Type Equation and Properties Under the assumption that the theorem of total probability holds without issue at a limit point x+; i.e. an equality of the form X p2 (x; x+; t; ul1 ; ul2 ) = p1 (x; t; ul1 ) ; (3.26) l2 we show that we can reduce and close the n-point hierarchy to a system of transport equations at each level. As expected, the complexity and number of equations increases with n, but the equations all express the time derivative of the n-point function in therms of the other spacial derivatives, as will be shown in the following section. 3.5 Two-point case We begin our calculations with the long form of the two-point equation, which is given by M X M X @t p2 (x1 ; x2 ; t; ul1 ; ul2 ) l1 =k1 +1 l2 =k2 +1 k1 X M X M X = ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =1 l10 =k1 +1 l2 =k2 +1 M X k1 X M X + ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =k1 +1 l10 =1 l2 =k2 +1 k2 X M X M X ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 l2 =1 l20 =k2 +1 l1 =k1 +1 M X k2 X M X + ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 (3.27) l2 =k2 +1 l20 =1 l1 =k1 +1 35 We want to rewrite the terms and rearrange them using the theorem of total probability. We label the terms on the right-hand side as (1) through (4): k1 X M X M X (1) := ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =1 l10 =k1 +1 l2 =k2 +1 M X k1 X M X (2) := ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =k1 +1 l10 =1 l2 =k2 +1 k2 X M X M X (3) := ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 l2 =1 l20 =k2 +1 l1 =k1 +1 M X k2 X M X (4) := ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 (3.28) l2 =k2 +1 l20 =1 l1 =k1 +1 and add or subtract terms to as follows: M X M X M X ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 to (1), l1 =k1 +1 l10 =k1 +1 l2 =k2 +1 M X M X M X + ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 to (2), l1 =k1 +1 l10 =k1 +1 l2 =k2 +1 M X M X M X ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 to (3), l1 =k1 +1 l2 =k2 +1 l20 =k2 +1 M X M X M X + ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 to (4). (3.29) l1 =k1 +1 l2 =k2 +1 l20 =k2 +1 Observe that the term we subtract from (1) is the same as the term we are adding to (2), and that the term we subtract from (3) is the same as the term we subtract add to (4). Therefore, de…ning the new quantities as (1’), (2’), (3’), and (4’), we have M X M X @t p2 (x1 ; x2 ; t; ul1 ; ul2 ) = (10 ) + (20 ) + (30 ) + (40 ) (3.30) l1 =k1 +1 l2 =k2 +1 36 where M X M X M X (10 ) = (1) ck1 p3 x1 ; x1 +; x2 ; ul1 ; ul10 ; ul2 ; t l1 =k1 +1 l10 =k1 +1 l2 =k2 +1 M X M X M X 0 (2 ) = (2) + ck1 p3 x1 ; x1 +; x2 ; ul1 ; ul10 ; ul2 ; t l1 =k1 +1 l10 =k1 +1 l2 =k2 +1 M X M X M X (30 ) = (3) ck2 p3 x2 ; x2 +; x1 ; ul2 ; ul20 ; ul1 ; t l1 =k1 +1 l2 =k2 +1 l20 =k2 +1 M X M X M X (40 ) = (4) + ck2 p3 x2 ; x2 +; x1 ; ul2 ; ul20 ; ul1 ; t (3.31) l1 =k1 +1 l2 =k2 +1 l20 =k2 +1 Then we have k1 X M X M X (10 ) = ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =1 l10 =k1 +1 l2 =k2 +1 M X M X M X ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =k1 +1 l10 =k1 +1 l2 =k2 +1 M X M X M X = ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =1 l10 =k1 +1 l2 =k2 +1 M X M X = ck1 p2 x1 +; x2 ; t; ul10 ; ul2 (3.32) l10 =k1 +1 l2 =k2 +1 whereby in the last step we have used the theorem of total probability. For the other terms, one has M X k1 X M X (20 ) = ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =k1 +1 l10 =1 l2 =k2 +1 M X M X M X + ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =k1 +1 l10 =k1 +1 l2 =k2 +1 M X M X M X = ck1 p3 x1 ; x1 +; x2 ; t; ul1 ; ul10 ; ul2 l1 =k1 +1 l10 =1 l2 =k2 +1 M X M X = ck1 p2 (x1 ; x2 ; t; ul1 ; ul2 ) (3.33) l1 =k1 +1 l2 =k2 +1 37 k2 X M X M X (30 ) = ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; u0l2 ; ul1 l2 =1 l20 =k2 +1 l1 =k1 +1 M X M X M X ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 l1 =k1 +1 l2 =k2 +1 l20 =k2 +1 M X M X M X = ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 l1 =k1 +1 l2 =1 l20 =k2 +1 M X M X = ck2 p2 x2 +; x1 ; t; ul20 ; u11 (3.34) l1 =k1 +1 l20 =k2 +1 M X k2 X M X (40 ) = ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 l2 =k2 +1 l20 =1 l1 =k1 +1 M X M X M X + ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 l1 =k1 +1 l2 =k2 +1 l20 =k2 +1 M X M X M X = ck2 p3 x2 ; x2 +; x1 ; t; ul2 ; ul20 ; ul1 l1 =k1 +1 l2 =k2 +1 l20 =1 M X M X = ck2 p2 (x1 ; x2 ; t; ul1 ; ul2 ) (3.35) l1 =k1 +1 l2 =k2 +1 Relabelling dummy indices, we have M X M X (10 ) = ck1 p2 (x1 +; x2 ; t; ul1 ; ul2 ) l1 =k1 +1 l2 =k2 +1 M X M X (20 ) = ck1 p2 (x1 ; x2 ; t; ul1 ; ul2 ) l1 =k1 +1 l2 =k2 +1 M X M X (30 ) = ck2 p2 (x1 ; x2 +; t; ul1 ; ul2 ) l1 =k1 +1 l2 =k2 +1 M X M X (40 ) = ck2 p2 (x1 ; x2 ; t; ul1 ; ul2 ) : (3.36) l1 =k1 +1 l2 =k2 +1 38 Recalling equation (3.30), we can now put the right-hand side of the original equation back together, yielding M X M X @t p2 (x1 ; x2 ; t; ul1 ; ul2 ) l1 =k1 +1 l2 =k2 +1 8 9 > < ck (p2 (x1 +; x2 ; t; ul ; ul ) > M X M X 1 1 2 p2 (x1 ; x2 ; t; ul1 ; ul2 )) = = : (3.37) > > l1 =k1 +1 l2 =k2 +1 : +ck2 (p2 (x1 ; x2 +; t; ul1 ; ul2 ) p2 (x1 ; x2 ; t; ul1 ; ul2 )) ; Finally, we rewrite this as a derivative of its cumulative distribution function, i.e. @1 F2 (x1; x2 ; t; k1 ; k2 ) M X M X := ck1 p2 (x1 +; x2 ; t; ul1 ; ul2 ) p2 x1 ; x2 ; t; ul10 ; ul2 l1 =k1 +1 l2 =k2 +1 M X M X ck2 (p2 (x1 ; x2 +; t; ul1 ; ul2 ) p2 (x1 ; x2 ; t; ul1 ; ul2 )) (3.38) l1 =k1 +1 l2 =k2 +1 The right-hand side can also be written in terms of a cdf, and so the two-point equation simpli…es to @t F2 (x1 ; x2 ; t; k1 ; k2 ) + c rF2 = 0 where we de…ne the constant vector c as 8 9 < ck > > = 1 c= > : ck2 > ; At this point we have a set of transport equations for F (x1 ; x2 ; ; ; t) : We also want to have consis- tency conditions, i.e. that the functions F2 satisfy 0 F2 (x1 ; x2 ; t; ; ) 1; all t > 0 1 = F2 (x1 ; x2 ; t; 1; 1) ; all t > 0: (3.39) and that they are increasing in the variables k1 and k2 : 39 3.6 N-point case In the previous section we reduced the two-point equation to a set of transport equations for the cumulative distribution function of the two-point equation. Now we will show that by using similar techniques, we can do the same for the n-point equation. Recall the equation for the n-point equation: M X M X ::: @t pn (x1 ; :::; xn ; t; ul1 ; :::; uln ) l1 =k1 +1 ln =kn +1 8 9 > Pki PM PM > n > X < li =1 li0 =ki +1 lj =kj +1 cki pn+1 x1 ; x2 ; :::; xi ; xi +; :::; xn ; t; ul1 ; ul2 ; :::; uli ; uli0 ; :::; uln > = j6=i = PM Pki PM > > i=1 > : li =ki +1 li0 =1 lj =kj +1 cki pn+1 x1 ; x2 ; :::; xi ; xi +; :::; xn ; t; ul1 ; ul2 ; :::; uli ; uli0 ; :::; uln > ; j6=i (3.40) Indeed, for the left-hand side, by de…nition we have M X M X LHS = ::: @t pn (x1 ; :::; xn ; ul1 ; :::; uln ; t) (3.41) l1 =k1 +1 ln =kn +1 =: @t Fn (x1 ; :::; xn ; k1 ; :::; kn ; t) : On the right-hand side, we once again add and subtract o¤ terms and use the theorem of total probability on each term. For brevity we consider ith pair of terms, labelling them as Ai and Bi : These terms are given by ki X M X M X Ai = cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t li =1 li0 =ki +1 lj =kj +1 j6=i M X ki X M X Bi = cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t (3.42) li =ki +1 li0 =1 lj =kj +1 j6=i To the Ai term we add on a sum, de…ning M X M X M X A~i = Ai + cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t (3.43) li =ki +1 li0 =ki +1 lj =kj +1 j6=i 40 To Bi we subtract this same term, writing M X M X M X ~ i = Bi + B cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t (3.44) li =ki +1 li0 =ki +1 lj =kj +1 j6=i These terms cancel, so we have M X M X n X ::: @t pn (x1 ; :::; xn ; ul1 ; :::; uln ; t) = fAi + Bi g (3.45) l1 =k1 +1 ln =kn +1 i=1 Xn n o = A~i + B ~i i=1 Further, we have ki X M X M X A~i = cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t li =1 li0 =ki +1 lj =kj +1 j6=i M X M X M X + cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t li =ki +1 li0 =ki +1 lj =kj +1 j6=i M X M X M X = cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t li =1 li0 =ki +1 lj =kj +1 j6=i M X M X = cki pn xi +; :::; uli0 ; :::; t (3.46) li0 =ki +1 lj =kj +1 j6=i 41 by the theorem of total probability on the index li : For the Bi term, we have M X ki X M X ~i = B cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t li =ki +1 li0 =1 lj =kj +1 j6=i M X M X M X cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t li =ki +1 li0 =ki +1 lj =kj +1 j6=i M X M X M X = cki pn+1 xi ; xi +; :::; uli ; uli0 ; :::; t li =ki +1 li0 =1 lj =kj +1 j6=i M X M X = cki pn (xi ; :::; uli ; :::; t) (3.47) li =ki +1 lj =kj +1 j6=i using the theorem of total probability on the index li0 . We relabel the li0 index in the ith term, so we have M X M X A~i = cki pn (xi +; :::; uli ; :::; t) (3.48) li =ki +1 lj =kj +1 j6=i and hence 8 9 M > < pn (x1 ; :::; xi +; :::; xn ; ul ; :::; ul :::; ul ; t) > = X A~i + B ~i = 1 i n cki (3.49) > : > lj =kj +1 pn (x1 ; :::xi ; :::; xn ; ul1 ; :::; uli ; :::; uln ; t) ; 1 j n Using (3.41), (3.45), and (3.49), we may write the n-point equation as n X @t Fn (x1 ; :::; xn ; k1 ; :::; kn ; t) = cki A~i + B ~i i=1 0 1 n X B Fn (x1 ; :::xi +; :::; xn ; k1 ; :::; kn ; t) C cki @ A (3.50) i=1 Fn (x1 ; :::; xi ; :::; xn ; k1 ; :::; kn ; t) or, more simply as @t Fn (x1 ; :::; xn ; k1 ; ::; kn ; t) + c rFn = 0; (3.51) 42 again a series of transport equations, where the vector c is given by 8 9 > > ck > > > > > > < 1 = c= ::: (3.52) > > > > > > > > : ck ; n In this way, under the assumption that the theorem of total probability can be applied at the right- hand limit of some point, denoted by x+; the n-point equations can be reduced to a system of transport equations. As can be see in a number of examples following the solution of the equations, these algebraic identities yield a solution that persists until shock interactions form. Because the transport equation in both the two and higher dimensional cases is time-reversible, we know that the solution will break down once any of the initial shocks collide. This is explored further in several examples that follow. However, using a method similar to front tracking, one can use a bootstrapping method to continue the solution at each intersection of shocks. In essence, when shocks collide one state is eliminated and a recalibration of the set of discrete states is necessitated. The process continues until the next collision, etc. Chapter 4 Solution and Illustration of Equations 4.1 Solving the Equations The natural next step after working out the form for these equations is to solve them and apply the solutions to some simple cases to evaluate the results. For the one-point case, we drop the indices on x and k and have the set of one-dimensional transport equations given by @t F1 (x; k; t) + ck @1 F1 (x; k; t) = 0 F1 (x; k; 0) = gk (x) (4.1) where gk (x) are prescribed initial conditions satisfying the conditions gM (x) = 1; 0 gk (x) gl (x) 1; x 2 R; l k 2 N: (4.2) The solution to the kth equation is given by F1 (x; k; t) = gk (x ck t) (4.3) 43 44 In n dimensions, we have @t Fn (x1 ; :::; xn ; k1 ; ::; kn ; t) + c rFn = 0 Fn (x1 ; :::; xn ; k1 ; :::; kn ; 0) = gk1 ;:::;kn (x) : (4.4) with an associated initial and consistency conditions given by gk1 ;:::;kn (x1 ; :::; xn ) jki =M = gk1 ;:::;ki 1 ;ki+1 ;:::;kn (x1 ; :::; xk 1 ; xk+1 ; :::; xn ) gM;:::;M (x1 ; :::; xn ) = 1; 0 gk (x) 1; x 2 R (4.5) The solution to this transport equation is given by Fn (x1 ; :::; xn ; k1 ; ::; kn ; t) = gk1 ;:::;kn (x ct) (4.6) 4.2 Examples: purely deterministic case with shocks Now we want to consider some examples with purely deterministic cases. In these examples only the one-point functions are relevant, so we drop the subscript 1 on F . We consider several cases and observe behavior consistent with classical results. Example. First consider the case of one shock with only two possible states. We have an initial condition of the form 8 > < 1 x 0 g (x) = (4.7) > : 0 x>0 which leads to the cdfs given by 8 > < 1 x 0 g1 (x) = 1; g2 (x) = (4.8) > : 0 x>0 Note that we are labelling the states in this and following examples in increasing order, i.e. 0 is the 45 …rst and 1 is the second. Then the solution is given by 8 > < 1 x c1 t F (x; 2; t) = g2 (x c2 t) = > : 0 x > c1 t F (x; 1; t) = g1 (x c1 t) = 1: (4.9) The values can then be calculated by subtraction: 8 > < 1 x c1 t p1 (x; 2; t) = F (x; 2; t) = > : 0 x > c1 t 8 > < 0 x c1 t p1 (x; 1; t) = F (x; 1; t) F (x; 2; t) = (4.10) > : 1 x > c1 t Hence, the solution is 8 > < 1 x c1 t u (x; t) = : (4.11) > : 0 x > c1 t Example. Now consider a rarefaction type case, where we have three possible states (u = 0; 1) and an initial condition of the form 8 > < 1 x 0 g (x) = ; (4.12) > : 1 x>0 leading to the cdfs given by 8 > < 0 x 0 g1 (x) = 1; g2 (x) = g3 (x) = : (4.13) > : 1 x>0 46 We then have 8 > < 0 x c3 t F (x; 3; t) = g3 (x c3 t) = > : 1 x > c3 t 8 > < 0 x c2 t F (x; 2; t) = g2 (x c2 t) = > : 1 x > c2 t F (x; 1; t) = g1 (x c1 t) = 1 (4.14) and probabilities are computed by 8 > < 0 x c3 t p1 (x; 3; t) = F (x; 3; t) = > : 1 x > c3 t 8 > > 0 x c2 t > > < p1 (x; 2; t) = F (x; 2; t) F (x; 3; t) = 1 c2 t < x c3 t > > > > : 0 c3 t < x 8 > < 1 x c2 t p1 (x; 1; t) = F (x; 1; t) F (x; 2; t) = (4.15) > : 0 x > c2 t Therefore, the solution in this case is then given by 8 > > 1 x c1 t > > < u (x; t) = 0 c1 t < x c2 t : (4.16) > > > > : 1 c2 t < x Example. Now consider the evolution of the equations under a more complicated, 3-state case given by 8 > > 3 x 1 > > < g (x) = 2 1 > > > : 1 2 < 1 x 2 g2 (x) = > : 0 x>2 8 > < 1 x 1 g3 (x) = (4.18) > : 0 x>1 The resulting 1-point equation is then a series of simple transport equations that are given by @t F1 (x; k; t) + ck @1 F1 (x; k; t) = 0 (4.19) F1 (x; k; 0) = gk (x) and the solution to each of these equations is F1 (x; k; t) = gk (x ck t) (4.20) The individual probabilities can be computed directly from the cumulative distribution functions, as before: 8 > < 1 x 1 + c3 t F (x; 3; t) = g3 (x c3 t) = > : 0 x 1 + c3 t 8 > < 1 x > 2 + c2 t F (x; 2; t) = g2 (x c2 t) = > : 0 x 2 + c2 t F (x; 1; t) = g1 (x c1 t) = 1; all x (4.21) 48 so that 8 > < 1 x 1 + c3 t p1 (x; 3; t) = F1 (x; 3; t) = > : 0 x 1 + c3 t p1 (x; 2; t) = F1 (x; 2; t) F1 (x; 3; t) 8 > > 1 2 + c2 t < x 1 + c3 t > > < = 0 x min (1 + c3 t; 2 + c2 t) or x > max (1 + c3 t; 2 + c2 t) > > > > : 1 2 + c2 t < x 1 + c3 t 8 > < 0 x > 2 + c2 t p1 (x; 1; t) = F1 (x; 1; t) F1 (x; 2; t) = (4.22) > : 1 x 2 + c2 t This leads to the solution 8 > > 3 x 1 + 5t > > < f (x) = 2 = 1 + 5t < x 2 + t (4.23) > > > > : 1 2+t > 3 x 1 > > < g (x) = 2 1 > > > : 1 2 (4.27) 4 We now consider several examples where we induce randomness by prescribing it in the initial condition g for the location of the shocks to blur out the sharp discontinuity. We consider a uniform distribution, and an exponential distribution with and without an o¤set. As the uniform distribution 50 is absolutely continuous with respect to the exponential, we expect qualitative similarities to arise in the results, as indeed becomes apparent through the calculations. More precisely, the prescription is to rewrite (4.26) as 8 > > 3 x 0 > > < g~ (x) = 2 0 > > > : 1 x < 1 1 : 0 otherwise As given by (4.28), our new initial condition is prescribed by 8 > > 3 x 0 > > < g~ (x) = 2 0 > > > : 1 x > 1 x 1 > > < g2 (x) = 1 x 1 > > > : 0 2 < 1 x 0 g3 (x) = (4.32) > : 0 0 > 1 x 1 + c2 t > > < F (x; 2; t) = g2 (x c2 t) = 1 + c2 t x 1 + c2 t < x 2 + c2 t > > > > : 0 2 + c2 t < x 8 > < 1 x c3 t F (x; 3; t) = g3 (x c3 t) = (4.33) > : 0 c3 t < x The expressions in (4.33) can now be used to compute the probability of each state at any point with a few simple computations. One has 8 > < 1 x c3 t p1 (x; 3; t) = F (x; 3; t) = ; (4.34) > : 0 c3 t < x 52 p1 (x; 2; t) = F (x; 2; t) F (x; 3; t) 8 > > 1 x 1 + c2 t 8 > > > < < 1 x c3 t = 1 + c 2 t x 1 + c2 t < x 2 + c2 t > > > : 0 c3 t < x > > : 0 2 + c2 t < x 8 > > 1 c3 t < x 1 + c2 t > > > > > > 1 + c2 t x max f1 + c2 t; c3 tg < x 2 + c2 t > > < = 0 max f2 + c2 t; c3 tg < x or x min f1 + c2 t; c3 tg ; (4.35) > > > > > > c2 t x 1 + c2 t < x min f2 + c2 t; c3 tg > > > > : 1 2 + c2 t < x c3 t p1 (x; 1; t) = F (x; 1; t) F (x; 2; t) 8 > > 1 x 1 + c2 t > > < =1 1 + c2 t x 1 + c2 t < x 2 + c2 t > > > > : 0 2 + c2 t < x 8 > > 0 x 1 + c2 t > > < = x c2 t 1 + c2 t < x 2 + c2 t (4.36) > > > > : 1 2 + c2 t < x Making a sketch of the xt-plane, one has six separate regions bounded by the x-axis and the lines x = 1 + c2 t; x = 2 + c2 t; x = c3 t: Labelling them counterclockwise and using the shorthand p (i) := p (x; ui ; t) ; we can describe the probabilities of the solution to take on certain values and summarize the information obtained in equations (4.34)-(4.36). Speci…cally, let R1 := f(x; t) j x min f1 + c2 t; c3 tgg R2 := f(x; t) j c3t < x 1 + c2 tg R3 := f(x; t) j max f1 + c2 t; c3 tg < x 2 + c2 tg R4 := f(x; t) j max f2 + c2 t; c3 tg < xg R5 := f(x; t) j 2 + c2 t < x c3 tg R6 := f(x; t) j 1 + c2 t < x < min f2 + c2 t; c3 tgg : (4.37) 53 For shorthand, denote p ( ) = p1 (x; ; t) : Then for u (x; t) ; we have p (1) = 0; p (2) = 0; p (3) = 1 in R1 p (1) = 0; p (2) = 1; p (3) = 0 in R2 p (1) = x c2 t; p (2) = 1 + c2 t x; p (3) = 0 in R3 p (1) = 1; p (2) = 0; p (3) = 0 in R4 p (1) = 1; p (2) = 1; p (3) = 1 in R5 p (1) = x c2 t; p (2) = c2 t x; p (3) = 1 in R6 (4.38) In the regions R1 ; R2 ; and R4 ; the solution proceeds as one might expect. However, in regions R3 ; R5 ; and R6 ; we obtain probabilities outside the range [0; 1]. For example, in R3 by de…nition x c2 t > 1; implying that p (1) > 1 and p (2) < 0; both of which are physical impossibilities. 4.5 Example 2: Exponential Distribution Now consider a second example with a di¤erent kind of randomness. This time we choose an exponential distribution for the placement of the second shock. Recall the initial condition 8 > > 3 x 0 > > < g~ (x) = 2 0 > > > : 1 x < e x 0 : 0 otherwise 54 As before, we write out the expressions for gi (x) and have g1 (x) = 1; all x 8 > < 1 x 0 g2 (x) = > : x e 0 < 1 x 0 g3 (x) = (4.41) > : 0 0 < 1 x c2 t F (x; 2; t) = g2 (x c2 t) = > : ec2 t x c2 t < x 8 > < 1 x c3 t F (x; 3; t) = g3 (x c3 t) = : (4.42) > : 0 c3 t < x The probabilities of given states can then be calculated as follows: 8 > < 1 x c3 t p1 (x; 3; t) = F (x; 3; t) = (4.43) > : 0 c3 t < x 55 p1 (x; 2; t) = F (x; 2; t) F (x; 3; t) 8 8 > < > < 1 1 x c2 t x c3 t = > : ec2 t x > : 0 c2 t < x c3 t < x 8 > > 1 c3 t < x c2 t > > > > > < ec2 t x 1 c2 t < x c3 t = > > > ec2 t > x c3 t < x > > > : 0 x c2 t 8 > > 0 x c2 t > > < = ec2 t x 1 c2 t < x c3 t (4.44) > > > > : ec2 t x c3 t < x 8 > < 1 x c2 t p1 (x; 1; t) = F (x; 1; t) F (x; 2; t) = 1 > : ec2 t x c2 t < x 8 > < 0 x c2 t = (4.45) > : 1 ec2 t x c2 t < x Labelling regions from left to right as R1 ; R2 ; and R3 ; where R1 = f(x; t) j x c2 tg R2 = f(x; t) j c2 t < x c3 tg R3 = f(x; t) j c3 t < xg (4.46) we then have p (1) = 0; p (2) = 0; p (3) = 1 in R1 ; p (1) = 1 ec2 t x ; p (2) = ec2 t x 1; p (3) = 1 in R2 ; p (1) = 1 ec2 t x ; p (2) = ec2 t x ; p (3) = 0 in R3 : (4.47) 56 Again, one observes that physical impossibilities arise in the cone between the two shocks, wher- ever they would collide (somewhere in the region R2 ). For the other two regions, the computed probabilities make intuitive sense. 4.6 Example 3: Exponential Distribution (with o¤set) We consider the exponential distribution again, but this time put an o¤set to introduce a set of …xed measure where the initial condition takes the intermediate value with probability 1: That is, modify the initial condition to the following: 8 > > 3 x 1 > > < g~ (x) = 2 1 > > > : 1 x < e x 0 : 0 otherwise We then have g1 (x) = 1; all x 8 > < 1 x 0 g2 (x) = > : x e 0 < 1 x 1 g3 (x) = (4.50) > : 0 1 < 1 x c2 t F (x; 2; t) = g2 (x c2 t) = > : ec2 t x c2 t < x 8 > < 1 x c3 t 1 F (x; 3; t) = g3 (x c3 t) = : (4.51) > : 0 c3 t 1 < 1 x c3 t 1 p (x; 3; t) = F (x; 3; t) = (4.52) > : 0 c3 t 1 < > < 1 x c3 t 1 1 x c2 t = > : ec2 t x c2 t < x > : 0 c3 t 1 < x 8 > > 1 c3 t 1 < x c2 t > > > > > < 1 ec2 t x c2 t < x c3 t 1 = (4.53) > > > ec2 t x > max fc2 t; c3 t 1g < x > > > : 0 x min fc2 t; c3 t 1g 8 > < 1 x c2 t p1 (x; 1; t) = F (x; 1; t) F (x; 2; t) = 1 > : ec2 t x c2 t < x 8 > < 0 x c2 t = (4.54) > : 1 ec2 t x c2 t < x 58 The space is now separated into several regions bounded by c2 t and c3 t 1; label them going counterclockwise as follows: R1 = f(x; t) j x min fc2 t; c3 t 1gg R2 = f(x; t) j c3 t 1 > 3 x 1 > > < g~ (x) = 2 1 > > > : 1 x > 3 x c3 t 1 > > < u (x; t) = 2 c3 t 1 > > > : 1 x + c2 t < x 8 > < 3 x x0 + c~t = for t0 < t: (4.61) > : 1 x0 + c~t < x 60 Of course, we want the result in terms of probabilities rather than x ; so we use the distribution to obtain this result. Recall the de…nition of the four regions used in the third example: R1 = f(x; t) j x min fc2 t; c3 t 1gg R2 = f(x; t) j c3 t 1 < 3 x +1 x c3 c2 + c~t u (x; t) = : (4.64) > : 1 x +1 c3 c2 + c~t < x Hence, x +1 p (1) = P + c~t < x = P fx < (c3 c2 ) (x c~t) 1g c3 c2 Z (c3 c2 )(x c~t) 1 y = 1(0;1) (y) e dy = 1 exp (max f0; 1 (c3 c2 ) (x c~t)g) : (4.65) 0 61 We can therefore represent the full solution probabilistically as p (1) = 0; p (2) = 0; p (3) = 1 in R1 p (1) = 0; p (2) = 1; p (3) = 0 in R2 ; p (1) = 1 ec2 t x ; p (2) = ec2 t x 1; p (3) = 0 in R3 8 > > p (1) = 1 exp (min f0; 1 (c3 c2 ) (x c~t)g) > > < p (2) = 0 in R4 (4.66) > > > > : p (3) = exp (min f0; 1 (c3 c2 ) (x c~t)g) It is not quite clear how to reconcile (4.56) with (4.66). 4.8 Calculating two-point functions for Example 3 To remedy the nonphysical probabilities we obtained in the examples, we now calculate out expres- sions for the two-point functions in the pursuit of a better understanding of the missing decay term on the right-hand side of (4.24). We perform our calculations in terms of x and the point of …rst collision, which we label as (x0 ; t0 ) ; and modify the notation to de…ne several regions of xt-space: R1 = f(x; t) j x min fx + c2 t; c3 t 1g ; t < t0 g R2 = f(x; t) j c3 t 1 > 3 in R1 > > > > > > 2 in R2 > > < u (x; t) = 1 in R3 (4.69) > > > > > > 1 in R4 > > > > : 3 in R5 For brevity, in this section we adopt the notation F2 (k1 ; k2 ) := F2 (x1 ; x2 ; k1 ; k2 ; t) = P fu (x1 ; t) uk1 ; u (x2 ; t) u k2 g (4.70) and where further uk = k; within the scope of this example. Clearly, with this setup in terms of x ; one can write the two-point function in terms of the discretized heaviside function: F2 (k1 ; k2 ) = Hu(x1 ;t);k1 Hu(x2 ;t);k2 ; 8 > < 1 i j Hi;j = (4.71) > : 0 i u + 1; (5.8) assuming that the states are labelled by integers and in increasing order. Indeed, by simple convexity arguments, one can conclude further that no larger upward jumps (that is, an upward jump between two states that are not nearest neighbors) can ever form. 5.2 Derivation of 1-point Equation for Shocks We want to write down an equation for the one-point density in (5.1) as a function of the two-point density. On the left-hand side we clearly have a term of the form @t f1 (x ; x; t; u; v) (5.9) for the creation/destruction of the species. There will also be a free-streaming term, as the shocks will be moving with speed cuv given from the Rankine-Hugoniot condition based on the values taken on by the solution on either side of the shock. This term will be given by cuv @x f1 (x ; x; t; u; v) : (5.10) On the right-hand side of the equation, we must now account for creation or destruction of shocks of the ‡avor with u on the left side and v on the right side. Indeed, there is a birth term of the following sort: X f2 (x; x + x; t; u; w; w; v) : (5.11) w 68 That is, when we have one shock between u and an intermediate value w; and a second shock between that intermediate value w and the original terminal value v; these shocks will combine to produce a new member of the species uv. This can only occur when the shock between u and w can catch the shock between w and v, which is possible when f (u) f (w) f (w) f (v) = cuw > cwv = : (5.12) u w w v We can also have a destruction of a shock with speed cuv ; upon collision with another shock. This leads to a term of the following X f2 (x; x + x; t; u; v; v; w) : (5.13) w There is a negative sign as this term leads to a destruction of the shock rather than creation as in (5.11). Again, this can only happen if the left shock can overtake the right one, i.e. if f (u) f (v) f (v) f (w) = cuv > cvw = (5.14) u v v w Further, there can also be a loss if we have the shock between u and v collides with a di¤erent shock between w and u from the left, leading to a term of the form X f2 (x x; x; t; w; u; u; v) : (5.15) w which can only occur if f (w) f (u) f (u) f (v) = cwu > cuv = (5.16) w u u v We now want to express these terms in a Taylor expansion and rewrite the quantity x using the relative speeds of the collisions. Note that, starting with term under the summand in (5.11) we observe f2 (x; x + x; t; u; w; w; v) =f ~ 2 (x; x; t; u; w; w; v) + x@2 f2 (x; x; t; u; w; w; v) (5.17) We want to set x = (cuw cwv ) t (5.18) 69 yielding f2 (x; x + x; t; u; w; w; v) = f2 (x; x; t; u; w; w; v) + (cwv cuw ) t@2 f2 (x; x; t; u; w; w; v) (5.19) so the term we want is X (cuw cwv ) @2 f2 (x; x; u; w; w; v) : (5.20) w The …rst term vanishes since whenever u; v; w are not distinct, there are no shocks. For the time being, we drop the …rst term in the right-hand side of (5.19) as it does not make a physical contribution. If u; v; w are not distinct, there is by de…nition no shock, and if they are equal, there are by de…nition no shocks. Similarly to this, we will obtain terms X (cuv cvw ) @2 f2 (x; x; t; u; v; v; w) (5.21) w and X (cwu cuv ) @1 f2 (x; x; t; w; u; u; v) (5.22) w There is an additional matter of which values the summand can take; that is, for which values of w do we indeed have growth and decay terms, and which simply give no contribution due to the left shock not being able to catch the right shock. These di¤erent cases can be separated into u < v and v < u; and further bifurcated by the relative position of w with respect to the two values. Using the notion of the (piecewise linear) convexity of f; one can perform the bookkeeping to see which cases are possible and which are not. For example, for the term (5.20), we have shocks between the values (u; w) and (w; v). With u < v we have f (u) f (w) f (w) f (v) w cwv = u w w v f (u) f (w) f (w) f (v) v cwv = u w w v f (u) f (w) f (w) f (v) v cwv = (5.24) u w w v meaning that the term can occur for any value of w as long as v < u: Next, for the decay term with collisions from the right (5.21), the relevant shocks are (u; v) and (v; w) : We compute, for u < v : w < u < v ) cuv > cwv u < w < v ) cuv < cwv u < v < w ) cuv < cwv (5.25) and for v < u : w < v < u ) cuv > cwv v < w < u ) cuv > cwv v < u < w ) cuv < cwv (5.26) so this term will have a nonzero contribution if u < v and w < u or if v < u and w < u. Finally, for the decay term with collisions from the left (5.22), the relevant shocks are (w; u) and (u; v) : When u < v we have w < u < v ) cwu < cuv u < w < v ) cwu < cuv u < v < w ) cwu > cuv (5.27) 71 and for v < u we have w < v < u ) cwu < cuv v < w < u ) cwu > cuv v < u < w ) cwu > cuv (5.28) so that this term can occur for u < v if w > v or if v < u and w > v: Putting together the terms (5.9), (5.10), and (5.20)-(5.22), we have @t f1 (x; t; u; v) + cuv @x f1 (x; t; u; v) X = (cuv cvw ) @2 f2 (x; x; t; u; v; v; w) wu+1 for u < v, wherein the only meaningful case is when u and v are neighbors. We use the shorthand v = u + 1 to mean that for u = ui ; we set v = ui+1 ; that is, the next state in increasing order. We use the same notation (where applicable) with the quantity u 1 below in the equation for v < u 72 and have @t f1 (x; t; u; v) + cuv @x f1 (x; t; u; v) X X = (cuw cwv ) @2 f2 (x; x; t; u; w; w; v) (cuv cvw ) @2 f2 (x; x; t; u; v; v; w) w u+1 w u2 > u1 : The various possible interactions are pictured in the …gure on the next page. By convexity, one has f (u3 ) f (u2 ) f (u2 ) f (u1 ) = c32 > c21 = ; (5.34) u3 u2 u2 u1 73 ensuring that the left shock will catch up to the right one in …nite time. More precisely, write an initial condition of the form 8 > > u3 x < x1 > > < g (x) = u2 x1 x < x2 (5.35) > > > > : u1 x2 x We will apply the one-point equation to all six species of shocks, denoting them by the notation (u; v) ; where u is the value of the "left" state and v the value of the "right" state. First we do this in the general framework of a three-state system, then apply it speci…cally to the initial condition in (5.35). We begin with the case u = 3; v = 1 (i.e., what is the equation describing the creation and destruction of shocks between the values u = u3 and v = u1 ), one has X @t f1 (x; 3; 1) + c31 @x f1 (x; 3; 1) = (c3w cw1 ) @2 f2 (x; x; t; 3; w; w; 1) w 4 X (c31 c1w ) @2 f2 (x; x; t; u; v; v; w) w 2 X (cw3 c31 ) @1 f2 (x; x; w; 3; 3; 1) : (5.36) w 2 Labelling the terms on the right-hand side of equation (4.3) as (I) and (II), and (III), we observe that each sum contributes only one term (using the restriction that u; v; w must all be distinct): (I) = (c32 c21 ) @2 f2 (x; x; 3; 2; 2; 1) (II) = (c31 c12 ) @2 f2 (x; x; 3; 1; 1; 2) (III) = (c23 c31 ) @1 f2 (x; x; 2; 3; 3; 1) (5.37) Hence, (5.36) yields @t f1 (x; 3; 1) + c31 @x f1 (x; 3; 1) = (c32 c21 ) @2 f2 (x; x; 3; 2; 2; 1) (c31 c12 ) @2 f2 (x; x; 3; 1; 1; 2) (c23 c31 ) @1 f2 (x; x; 2; 3; 3; 1) (5.38) which conforms to our expectations of the right equation; on the left hand side we have a creation and translation term between the states u3 = 3 and u1 = 1; and on the right we have a growth term 74 Figure 5.1: Interactions between possible shocks in the three-state system, categorized by growth and di¤erent kinds of decay. Depending on relative velocities of the shocks, one shock may catch up to another and interact to combine a new shock in the form of (left state, right state) above each …gure. 75 when the pairs (3; 2) ; (2; 1) collide, and decay when (3; 1) hits (1; 2) or (2; 3) hits (3; 1) : In contrast, for the case u = 1; v = 3; we are considering a hypothetical upward jump between non-neighboring states, so we end up with a trivial equation of the form @t f1 (x; 1; 3) + c13 @x f1 (x; 1; 3) = 0; (5.39) so that since the shock does not exist in the initial conditions, the @x term is zero, hence the @t term is also zero and such a shock can never be created, which is consistent with our assumptions and intuition. If we try for example u = 1; v = 2; we have @t f1 (x; 1; 2) + c12 @x f1 (x; 1; 2) X X = (c12 c2w ) @2 f2 (x; x; 1; 2; 2; w) (cw1 c12 ) @1 f2 (x; x; w; 1; 1; 2) (5.40) w<1 w>2 and the right-hand side reduces to (c31 c12 ) @1 f2 (x; x; 3; 1; 1; 2) (5.41) leading to the equation @t f1 (x; 1; 2) + c12 @x f1 (x; 1; 2) = (c31 c12 ) @1 f2 (x; x; 3; 1; 1; 2) : (5.42) That is, shocks between u1 = 1 and u2 = 2 can be destroyed by a collision between (3; 1) from the left with (1; 2) ; but not created. For the case u = 2; v = 3; we have @t f1 (x; 2; 3) + c23 @x f1 (x; 2; 3) X = (c23 c3w ) @2 f2 (x; x; 2; 3; 3; w) w<2 X (cw2 c23 ) @1 f2 (x; x; w; 2; 2; 3) : (5.43) w>3 76 or @t f1 (x; 2; 3; ) + c23 @x f1 (x; 2; 3) = (c23 c31 ) @2 f2 (x; x; 2; 3; 3; 1) (5.44) so shocks of this form can only be destroyed by an interaction from the right between (2; 3) and (3; 1) : The next case is u = 3; v = 2; for which we have @t f1 (x; 3; 2) + c32 @x f1 (x; 3; 2) X X = (c3w cw2 ) @2 f2 (x; x; 3; w; w; 2) (c32 c2w ) @2 f2 (x; x; 3; 2; 2; w) w 4 w<3 X (cw3 c32 ) @1 f2 (x; x; w; 3; 3; 2) : (5.45) w 2 or @t f1 (x; 3; 2) + c32 @x f1 (x; 3; 2) = (c31 c12 ) @2 f2 (x; x; 3; 1; 1; 2) (c32 c21 ) @2 f2 (x; x; 3; 2; 2; 1) (5.46) so this shock can be created by a collision between (3; 1) and (1; 2) and destroyed by a collision from the right with (2; 1) : The …nal case is u = 2; v = 1; and we have @t f1 (x; 2; 1) + c21 @x f1 (x; 2; 1) X X = (c2w cw1 ) @2 f2 (x; x; 2; w; w; 1) (c21 c1w ) @2 f2 (x; x; 2; 1; 1; w) w 3 w 2 X (cw2 c21 ) @1 f2 (x; x; w; 2; 2; 1) : (5.47) w 1 or @t f1 (x; 2; 1) + c21 @x f1 (x; 2; 1) = (c23 c31 ) @2 f2 (x; x; 2; 3; 3; 1) (c32 c21 ) @1 f2 (x; x; ; 3; 2; 2; 1) (5.48) so this shock can be created by a collision between (2; 3) and (3; 1) or decay via a collision from the left by (3; 2) : 77 In the case of (5.35), the densities of f1 and f2 for a speci…c time t are expressed as measures concentrated at discrete points, and can be written formally as -functions. It is known that the solution pro…le is given as 8 > > u3 x < min fx1 + c23 t; x + c13 (t t )g > > < u (x; t) = u2 x1 + c23 t < x < x2 + c12 t (5.49) > > > > : u1 x > max fx2 + c12 t; x + c13 (t t )g where x2 x1 t = c23 c12 x2 x1 (c23 c12 ) x1 + c23 x2 c23 x1 x = x1 + c23 = c23 c12 c23 c12 c23 x2 c12 x1 = (5.50) c23 c12 Speci…cally, for the one- and two-point functions, one then has f1 (x; t; 3; 1) = (x x c13 (t t )) 1ft>t g f1 (x; t; 3; 2) = (x x1 c23 t) 1ft t g f1 (x; t; 2; 1) = (x x2 c12 t) 1 ft t g (5.51) and f2 (x; y; t; 3; 2; 2; 1) = (x x1 c23 t) (y x2 c12 t) 1ft t g (5.52) Evaluating these functions for any other combination of values in this problem yields identically zero. Testing this against (5.38), (5.46), and (5.48) yields @t f1 (x; t; 3; 1) + c13 @x f1 (x:t; 3; 1) = (c23 c21 ) @2 f2 (x; x; t; 3; 2; 2; 1) @t f1 (x; t; 3; 2) + c23 @x f1 (x:t; 3; 2) = (c23 c21 ) @2 f2 (x; x; t; 3; 2; 2; 1) @t f1 (x; t; 2; 1) + c21 @x f1 (x; t; 2; 1) = (c23 c21 ) @1 f2 (x; x; t; 3; 2; 2; 1) (5.53) These equations formally give the correct conclusions. For t > t , the …rst equation describes the 78 steady-state of the shock (3; 1) persisting inde…nitely and moving to the left at speed c13 : @t f1 (x; 3; 1) + c13 @x f1 (x; 3; 1) = 0 (5.54) and for t < t yields identically zero. The second and third equations yield identically zero for t > t and the steady-state of the shocks (3; 2) and (2; 1) persisting for t < t : @t f1 (x; t; 3; 2) + c23 @x f1 (x; t; 3; 2) = 0 @t f1 (x; t; 2; 1) + c12 @x f1 (x; t; 2; 1) = 0: (5.55) At the point (x ; t ) where the two shocks (3; 2) and (2; 1) collide and form (3; 1), 5.4 Derivation of n-point Equation for Shocks In the same way as for the section on the derivation for the one-point equation, we now want to generalize the procedure and derive the equations for the n -point equation. For the left-hand side of the equation, the analog of the time-derivative term should clearly be represented by @t fn x1 ; :::xn ; t; u1 ; :::; un ; v 1 ; :::; v n : (5.56) We use superscripts to distinguish the di¤erent states fui g from the 2n arguments in this term giving the left and right hand values of the solution on either side of the shocks located at the positions fxi g : That is, ui ; v i 2 fu1 ; u2 ; :::g for each 1 i n: For the free-streaming term, as a replacement for the single term dealing with the interaction between two states in the one-point term, we now need to account for every possible combination of pairs of states, leading to the term cui vi @xi fn x1 ; :::; xn ; t; u1 ; :::; un ; v 1 ; :::; v n summed over all combinations: n X cui vi @xi fn x1 ; :::; xn ; t; u1 ; :::; un ; v 1 ; :::; v n : (5.57) i=1 We now want to consider the terms that will form on the right-hand side of the equation. In the one-point equation, these consisted of birth or decay terms for the two-point function arising from interactions of those shocks, modi…ed by the appropriate rate constants cuv . Our hypothesis here is 79 that we have similar terms with a possible interaction involving any of the n shocks, with the same appropriate rate constants. For the birth term, we stipulate 0 1 n X X B xi ; xi x1 ; :::; xi 1 ; xi+1 ; :::; xn ; t; C (cui w cwvi ) @i0 fn+1 @ A (5.58) i=1 w ui ; w; w; vi ; u1 ; :::; ui 1 ; ui+1 ; :::un ; v 1 ; :::; v i 1 ; v i+1 ; :::; v n We have rearranged the order of arguments to put the ith component and intermediate shock …rst. The notation @i0 refers to the derivative with respect to the index that appears second under f in (5.3), and @i to the one appearing …rst (the ith component). For brevity, we will use abbreviated notation henceforth and suppress the remainder of the arguments, i.e. rewrite (5.58) as n X X (cui w cwvi ) @i fn+1 xi ; xi ; :::; t; ui ; w; w; v i :::: (5.59) i=1 w The range for w is precisely the same as for the one-point equation case, that is for w ui + 1 as long as v i < ui : We can also have a decay term formed by collisions from the right, given by n X X (cui v cvi w ) @i0 fn+1 xi ; xi ; :::; t; ui ; v i ; v i ; w; ::: (5.60) i=1 w and for collisions from the left, given by n X X (cwui cui vi ) @i fn+1 xi ; xi ; :::; t; w; ui ; ui ; v i ; ::: (5.61) i=1 w The appropriate ranges for the terms (5.60) and (5.61) are given by w < ui and w > ui + 1 for ui < v i (i.e. when v i = ui + 1) and w v i + 1 and w ui 1 for v i < ui : For ease of notation, de…ne the sets 8 > < w < ui if v i = ui + 1 W2i = w j > : w vi + 1 if v i < ui 8 > < i w > ui + 1 if v i = ui + 1 W3 = w j (5.62) > : w ui 1 if v i < ui 80 Then the n-point equation is given by n X @t fn x1 ; :::; xn ; t; u1 ; :::; un ; v 1 ; :::; v n + cui vi @xi fn x1 ; :::; xn ; t; u1 ; :::; un ; v 1 ; :::; v n i=1 n X X = 1fvi =ui +1g (cui w cwvi ) @i fn+1 xi ; xi ; :::; t; ui ; w; w; v i :::: i=1 w n X X (cui v cvi w ) @i0 fn+1 xi ; xi ; :::; t; ui ; v i ; v i ; w; ::: i=1 w2W2i n X X (cwui cui vi ) @i fn+1 xi ; xi ; :::; t; w; ui ; ui ; v i ; ::: : (5.63) i=1 w2W3i In this fully written out n-point hierarchy, one can see all the expected interactions and dynamics between the modelled particle system of shocks. On the left-hand side we have two terms: the …rst of which the change in time of the distribution of a particular n-point state, a collection of n shocks between di¤erent state values at prescribed points. The other term is a free-streaming term representing the motion of each of these n shocks with relative velocities fcui vi g along the x direction as time advances. On the right-hand side, we have three terms relating to the creation or destruction of this object by various types of interactions. The summand of the …rst term involves a growth by a shock of states ui ; w combining with w; v i ; forming the shock ui ; v i ; hence the positive term. The latter two summands involve destruction of the shock ui ; v i from the right by a shock v i ; w or from the left with the shock w; ui , respectively. The sums then cover interactions with any of the n shocks involved. In this derivation, it should also be noted we have also implicitly ruled out the possibility of three or more shocks intersecting at the same point, which is a reasonable assumption for kinetic theory. 5.5 Reduction of n-point equation to 1-point One important check that can be performed via a straightforward calculation is the reduction of the n-point equation down to the one-point. Substituting in n = 1 into equation (5.63) yields for the left-hand side @t f1 x1 ; t; u1 ; v 1 + cu1 v1 f1 x1 ; t; u1 ; v 1 (5.64) 81 and for the right-hand side X 1fu1 =v1 g (cu1 w cwv1 ) @10 f2 x1 ; x1 ; t; u1 ; w; w; v 1 w X (cu1 v1 cwv1 ) @10 f2 x1 ; x1 ; t; u1 ; v 1 ; w w2W21 X (cwu1 cu1 v1 ) @1 f2 x1 ; x1 ; t; w; u1 ; u1 ; v 1 (5.65) w2W31 Dropping the subscripts and breaking into cases, we have for u < v (v = u + 1) that @t f1 (x; t; u; v) + cuv f (x; t; u; v) X X = (cuw cwv ) @2 f2 (x; x; t; u; w; w; v) (cuv cwv ) @2 f2 (x; x; t; u; v; w) w wu+1 and for u > v; we have @t f1 (x; t; u; v) + cuv f (x; t; u; v) X = (cuv cwv ) @2 f2 (x; x; t; u; v; w) w v+1 X (cwv cuv ) @1 f2 (x; x; t; w; u; u; v) ; (5.67) w u 1 matching the one-point equation. 5.6 Flavor of a more Complex Example (5 shocks) One can consider more complex examples and in theory, perform more computations on them to further verify the n-point hierarchy. As we will see in the following section, however, such examples become increasingly more complicated to work out. We limit ourselves to writing out the probability densities and showing how the equations match up intuitively with the expected terms. To this end 82 we consider initial condition of the form 8 > > u5 x x1 > > > > > > u4 x1 < x x2 > > < g (x) = u1 x2 < x x3 (5.68) > > > > > > u2 x3 < x x4 > > > > : u3 x4 < x Initially, there are four shocks with speeds c45 ; c14 ; c12 ; and c23 (in order from left to right). The …rst interaction expected is c15 with c14 ; forming a shock of speed c15 : This shock then collides subsequently with c12 ; then c23 ; to form a shock c35 which then persists inde…nitely. Denoting the collision points by (xi ; ti ) ; the expected solution pro…le is given by the following: 8 > > u5 x x1 + c15 t > > > > > > u4 x1 + c15 t < x x2 + c14 t > > < u (x; t) = u1 x2 + c14 t < x x3 + c12 t for t t1 (5.69) > > > > > > u2 x3 + c12 t < x x4 + c23 t > > > > : u3 x4 + c13 t < x 8 > > u5 x x1 + (t t1 ) c15 > > > > > < u1 x1 + (t t1 ) c15 < x x3 + c12 t u (x; t) = for t1 t t2 (5.70) > > > > u2 x3 + c12 t < x x4 + c23 t > > > : u 3 x4 + c13 t < x 8 > > u5 x x2 + (t t2 ) c25 > > < u (x; t) = u2 x2 + (t t2 ) c25 < x x4 + c23 t for t2 t t3 (5.71) > > > > : u3 x4 + c13 t < x 8 > < u5 x x3 + (t t3 ) c35 u (x; t) = for t3 t (5.72) > : u3 x3 + (t t3 ) c35 < x 83 Given the solution pro…le (5.69)-(5.72), we can now write out the n-point probability distributions. Starting with the one-point distribution, we have f1 (x; t; 5; 4) = (x x1 c15 t) 1ft t1 g f1 (x; t; 4; 1) = (x x2 c14 t) 1ft t1 g f1 (x; t; 1; 2) = (x x3 c12 t) 1ft t2 g f1 (x; t; 2; 3) = (x x4 c23 t) 1ft t3 g f1 (x; t; 5; 1) = (x x1 (t t1 ) c15 ) 1ft t t2 g 1 f1 (x; t; 5; 2) = (x x2 (t t2 ) c25 ) 1ft t3 g 2 f1 (x; t; 5; 3) = (x x3 (t t3 ) c35 ) 1ft tg (5.73) 3 and zero for any other arguments. For the two-point distributions, we compute f2 (x; y; t; 5; 4; 4; 1) = (x x1 c15 t) (y x2 c14 t) 1ft t1 g f2 (x; y; t; 5; 4; 1; 2) = (x x1 c15 t) (y x3 c12 t) 1ft t2 g f2 (x; y; t; 5; 4; 2; 3) = (x x1 c15 t) (y x4 c23 t) 1ft t2 g f2 (x; y; t; 4; 1; 1; 2) = (x x2 c14 t) (y x3 c12 t) 1ft t2 g f2 (x; y; t; 4; 1; 2; 3) = (x x2 c14 t) (y x4 c23 t) 1ft t2 g f2 (x; y; t; 1; 2; 2; 3) = (x x3 c12 t) (y x4 c23 t) 1ht t2 i f2 (x; y; t; 5; 1; 1; 2) = (x x1 (t t1 ) c15 ) (y x3 c12 t) 1ft t t2 g 1 f2 (x; y; t; 5; 1; 2; 3) = (x x1 (t t1 ) c15 ) (y x4 c23 t) 1ft t t2 g 1 f2 (x; y; t; 5; 2; 2; 3) = (x x2 (t t2 ) c25 ) (y x4 c23 t) 1ft t t3 g (5.74) 2 Aside from simple permutation of the arguments (i.e. f2 (y; x; t; 4; 1; 5; 4) = f2 (x; y; t; 5; 4; 4; 1)), the two-point distributions will be zero for other combinations of ui in their arguments. Next, for the 84 three-point distributions, we have f3 (x; y; z; t; 5; 4; 4; 1; 1; 2) = (x x1 c15 t) (y x2 c14 t) (z x3 c12 t) 1ft t1 g f3 (x; y; z; t; 5; 4; 4; 1; 2; 3) = (x x1 c15 t) (y x2 c14 t) (z x4 c23 t) 1ft t1 g f3 (x; y; z; t; 5; 4; 1; 2; 2; 3) = (x x1 c15 t) (y x3 c12 t) (z x4 c23 t) 1ft t1 g f3 (x; y; z; t; 4; 1; 1; 2; 2; 3) = (x x2 c14 t) (y x3 c12 t) (z x4 c23 t) 1ft t1 g f3 (x; y; z; t; 5; 1; 1; 2; 2; 3) = (x x1 (t t1 ) c15 ) (y x3 c12 t) (z x4 c23 t) 1ft t t2 g 1 (5.75) and zero otherwise, except for permutations of the arguments. Finally, for the four-point distribution, we have simply f4 (x; y; z; w; t; 5; 4; 4; 1; 1; 2; 2; 3) = (x x1 c15 t) (y x2 c14 t) (z x3 c12 t) (w x4 c23 t) 1ft t1 g (5.76) Now we want to substitute the expressions (5.73)-(5.76) into the n-point equations. Recall the one-point equation is expressed by @t f1 (x; t; u; u + 1) + cu;u+1 @x f1 (x; t; u; u + 1) X = (cu;u+1 cu+1;w ) @2 f2 (x; x; t; u; u + 1; u + 1; w) wu+1 for v = u + 1 and @t f1 (x; t; u; v) + cuv @x f1 (x; t; u; v) X X = (cuw cwv ) @2 f2 (x; x; t; u; w; w; v) (cuv cvw ) @2 f2 (x; x; t; u; v; v; w) w u+1 w v+1 X (cwu cuv ) @1 f2 (x; x; t; w; u; u; v) : (5.77) w u 1 85 for v < u: We start with equation (5.77) for u = 1; 2; :::, and have @t f1 (x; t; 1; 2) + c12 @x f1 (x; t; 1; 2) X X = (c12 c2w ) @2 f2 (x; x; t; 1; 2; 2; w) + (cw1 c12 ) @1 f2 (x; x; t; w; 1; 1; 2) w<1 w>2 (c31 c12 ) f2 (x; x; t; 3; 1; 1; 2) (c41 c12 ) f2 (x; x; t; 4; 1; 1; 2) (c51 c12 ) f2 (x; x; t; 5; 1; 1; 2) (5.78) @t f1 (x; t; 2; 3) + c23 @x f1 (x; t; 1; 2) X X = (c23 c3w ) @2 f2 (x; x; t; 2; 3; 3; w) (cw2 c23 ) @1 f2 (x; x; t; w; 2; 2; 3) w<2 w>3 = (c23 c31 ) @2 f2 (x; x; t; 2; 3; 3; 1) (c42 c23 ) @1 f2 (x; x; t; 4; 2; 2; 3) (c52 c23 ) @1 f (x; x; t; 5; 2; 2; 3) (5.79) @t f1 (x; t; 3; 4) + c34 @x f1 (x; t; 3; 4) X X = (c34 c4w ) @2 f2 (x; x; t; 3; 4; 4; w) (cw3 c34 ) @2 f2 (x; x; t; 3; 4; 4; w) w<3 w>4 (c34 c41 ) @2 (x; x; t; 3; 4; 4; 2) (c34 c42 ) @2 f2 (x; x; t; 3; 4; 4; 2) (c53 c34 ) @2 f2 (x; x; t; 3; 4; 4; 5) (5.80) @t f1 (x; t; 4; 5) + c12 @x f1 (x; t; 4; 5) X X = (c45 c5w ) @2 f2 (x; x; t; ; 4; 5; 5; w) (cw4 c45 ) @1 f2 (x; x; t; w; 4; 4; 5) w<4 w>5 = (c45 c51 ) @2 f2 (x; x; t; 4; 5; 5; 1) (c45 c52 ) @s f2 (x; x; t; 4; 5; 5; 2) (c45 c53 ) @2 f2 (x; x; t; 4; 5; 5; 3) (5.81) 86 Now we consider equation (5.77), the case v < u; …rst considering v = 1: This yields @t f1 (x; t; u; 1) + cu1 @x f1 (x; t; u; 1) X X = (cuw cw1 ) @2 f2 (x; x; t; u; w; w; 1) (cu1 c1w ) @2 f2 (x; x; t; u; 1; 1; w) w u+1 w 2 X (cwu cu1 ) @1 f2 (x; x; t; w; u; u; 1) : (5.82) w u 1 and substituting in for u = 2; 3; ::: will yield @t f1 (x; t; 2; 1) + c21 @x f1 (x; t; 2; 1; ) X X = (c2w cw1 ) @2 f2 (x; x; t; 2; w; w; 1) (c21 c1w ) @2 f2 (x; x; t; 2; 1; 1; w) w 3 w 2 X (cw2 c21 ) @1 f2 (x; x; t; w; 2; 2; 1) w 1 = (c23 c31 ) @2 f2 (x; x; t; 2; 3; 3; 1) (c32 c21 ) @1 f2 (x; x; t; 3; 2; 2; 1) (c42 c21 ) @1 f2 (x; x; t; 4; 2; 2; 1) (c52 c21 ) @1 f2 (x; x; t; 5; 2; 2; 1) (5.83) @t f1 (x; t; 3; 1) + c31 @x f1 (x; t; 3; 1) X X = (c4w cw1 ) @2 f2 (x; x; t; 3; w; w; 1) (c31 c1w ) @2 f2 (x; x; t; 3; 1; 1; w) w 4 w 2 X (cw3 c31 ) @1 f2 (x; x; t; w; 3; 3; 1) w 2 = (c42 c21 ) @2 f2 (x; x; t; 3; 2; 2; 1) + (c43 c41 ) @2 f2 (x; x; t; 3; 4; 4; 1) (c31 c12 ) @2 f2 (x; x; t; 3; 1; 1; 2) (c23 c31 ) @1 f2 (x; x; t; 2; 3; 3; 1) (c43 c31 ) @1 f2 (x; x; t; 4; 3; 3; 1) (c53 c31 ) @1 f2 (x; x; t; 5; 3; 3; 1) (5.84) 87 @t f1 (x; t; 4; 1) + c41 @x f1 (x; t; 4; 1) X X = (c4w cw1 ) @2 f2 (x; x; t; 4; w; w; 1) (c41 c1w ) @2 f2 (x; x; t; 4; 1; 1; w) w 5 w 2 X (cw4 c41 ) @1 f2 (x; x; t; w; 4; 4; 1) w 3 = (c42 c21 ) @2 f2 (x; x; t; 4; 2; 2; 1) + (c43 c31 ) @2 f2 (x; x; t; 4; 3; 3; 1) + (c45 c51 ) @2 f2 (x; x; t; 4; 5; 5; 1) (c41 c12 ) @2 f2 (x; x; t; 4; 1; 1; 2) (c34 c41 ) @1 f2 (x; x; t; 3; 4; 4; 1) (c54 c41 ) @1 f2 (x; x; t; 5; 4; 4; 1) (5.85) @t f1 (x; t; 5; 1) + c51 @x (x; t; 5; 1) X X = (c5w cw1 ) @2 f2 (x; x; t; 5; w; w; 1) (c51 c1w ) @2 f2 (x; x; t; 5; 1; 1; w) w 6 w 2 X (cw5 c51 ) @1 f2 (x; x; t; w; 5; 5; 1) w 4 = (c54 c41 ) @2 f2 (x; x; t; 5; 4; 4; 1) (c53 c31 ) @2 f2 (x; x; t; 5; 3; 3; 1) (c52 c21 ) @2 f2 (x; x; t; 5; 2; 2; 1) (c51 c12 ) @2 f2 (x; x; t; 5; 1; 1; 2) (c45 c51 ) @1 f2 (x; x; t; 4; 5; 5; 1) (5.86) These calculations help to give a ‡avor of future direction and serve to illustrate the various inter- actions by which perturbations at the one-point level can occur in a more complex system. Based on whether the velocity of one shock is su¢ cient to catch up to that of another, shocks of certain species can be created or destroyed, with rate constants speci…ed by the relative velocities between the two. There is a free-streaming term on the right-hand side that tracks the movement of any existing shocks in that species. For future study, this lays the groundwork for either numerical computation or further analysis under special cases, such as more stringent initial conditions. Chapter 6 Two interesting limits In the previous section, we derived a set of exact equations for the n-point equations. Of particular interest is the application of this framework to two speci…c limits. The …rst such limit is the simplest nontrivial case, where we have three possible velocity states and hence, three di¤erent shock speeds in the system. The second limit of interest is obtained as the number of states tends to in…nity, creating a …ner approximation by piecewise linear functions to a smooth ‡ux function, and passing to continuum limits in all the expressions, using theory of semigroups and generators. One would expect that in this latter limit and under appropriate assumptions, the discrete model in fact recovers the correct continuum dynamics of Menon and Srinivisan [31], etc. 6.1 Three-state limit First consider the limit of the three velocity state system. Speci…cally, this means that the indices ui ; v i all take values from the set fu1 ; u2 ; u3 g : We may assume without loss of generality that u1 < u2 < u3 : (6.1) The three possible species of shock are between u1 and u2 ; between u2 and u3 ; and …nally between u1 and u3 : We allow upward jumps only between nearest neighbors, so a shock between u1 and u3 is allowed only with u3 as the left state and u1 as the right state, and the other two are allowed with 88 89 either orientation. The shocks then have speeds given by f (u2 ) f (u1 ) c12 = u2 u1 f (u3 ) f (u2 ) c23 = u3 u2 f (u3 ) f (u1 ) c13 = (6.2) u3 u1 and by the usual convexity assumption on f; we have c12 < c13 < c23 : (6.3) We start with the sets W2i ; W3i de…ned in (5.62). Because of the restrictions on ui ; v i ; we may rewrite them based on cases. We have ui = u1 ; v i = u2 ) W2i = fw j w < u1 g = ; W3i = fw j w > u2 g = fu3 g ui = u2 ; v i = u1 ) W2i = fw j w u2 g = fu1 ; u2 g ; W3i = fw j w u0 g = fu1 ; u2 ; u3 g ui = u2 ; v i = u3 ) W2i = fw j w < u2 g = fu1 g ; W3i = fw j w > u3 g = ui = u3 ; v i = u1 ) W2i = fw j w u2 g = fu1 ; u2 g ; W3i = fw j w u2 g = fu2 ; u3 g ui = u3 ; v i = u2 ) W2i = fw j w u3 g = fu1 ; u2 ; u3 g ; W3i = fw j w u2 g = fu2 ; u3 g (6.4) We can write out the exact form for the one-point equation, in the case of three states, as follows: @t p1 x1 ; t; u1 ; v 1 + cu1 v1 @x1 f1 x1 ; t; u1 ; v 1 X = 1hv1 u1 +1i (cu1 w cwv1 ) @2 f2 x1 ; x1 ; t; u1 ; w; w; v 1 w u1 +1 X (cu1 v1 cv1 w ) @2 f2 x1 ; x1 ; t; u1 ; v 1 ; v 1 ; w w2W21 X (cwu1 cu1 v1 ) @1 f2 x1 ; x1 ; t; w; u1 ; u1 ; v 1 (6.5) w2W31 90 and the two-point equation as 2 X @t p2 x1 ; x2 ; t; u1 ; u2 ; v 1 ; v2 + cui vi @xi f2 x1 ; x2 ; t; u1 ; u2 ; v 1 ; v 2 i=1 2 X X = 1fvi ui +1g (cui w cwvi ) @i0 f3 xi ; xi ; :::; t; ui ; w; w; v i ; ::: i=1 w ui +1 2 X X (cui vi cvi w ) @i0 f3 xi ; xi ; :::; t; ui ; v i ; v i ; w; ::: i=1 w2W2i 2 X X (cwui cui vi ) @i f3 xi ; xi ; :::; t; w; ui ; ui ; v i ; ::: : (6.6) i=1 w2W3i As one can see, in the simplest case the equations do simplify somewhat, as the range of indices in and number of Wji are reduced. However, in terms of notation, the two-point equation cannot be simpli…ed much further, as there are certain combinations of possibilities allowed at the values of the given x-coordinates. For example, in the one-point equation with u1 and v 1 taking on the values u1 and u2 , respectively, one would have @t p1 (x1 ; t; u1 ; u2 ) + cu1 v1 @x1 f1 (x1 ; t; u1 ; v1 ) X = (c1w cw2 ) @2 f2 (x1 ; x1 ; t; u1 ; w; w; u2 ) w 2 X 0 (cw1 c12 ) @1 f2 (x1 ; x1 ; t; w; 1; 1; 2) w2fu3 g = (c31 c12 ) @1 f2 (x1 ; x1 ; t; 3; 1; 1; 2) : (6.7) In particular (6.7) indicates that, within the three-state system, the a shock between (1; 2) can decay via an interaction with a shock of the species (3; 1) from the left. It cannot decay within this limited state system from an interaction from the right or form from a collision of two other shocks. Part II Part II - Multistable Dynamics in the Allen-Cahn Equation with White Noise 91 Chapter 7 Introduction The Allen-Cahn equation ut = u+u u3 can also be studied without the presence of the di¤usion term to see how solutions evolve toward the equilibrium points. In the theory of ordinary di¤erential equations, the concept of stable equilibrium points plays a key role. The initial condition determines uniquely the evolution to the particular stable equilibrium point. An important question involves the introduction of white noise. In fact, as we will see, even a very small amount of noise will result in a small perturbation that in‡uences which particular stable equilibrium point to which the trajectory evolves. As a practical exam- ple, multistable dynamics are well-known in the description of patterns in physics, chemistry, and biology; see for example a recent experimental work of Chang, Oh, Ingber, and Huang [10] analyz- ing a multistep dynamics of mammalian cell di¤erentiation with multistability. Mathematically a prototype bistable dynamics is du = 4(u )(1 u2 ) or du = 4(u )(1 u2 )dt (7.1) dt where 2 [0; 1) is a constant and u := 1 and u := 1 are two stable equilibria. In the context of phase transition, the stability of an equilibrium is described by the smallness of the potential R P (u) = 4 (u )(u2 1)du: In particular, when 2 (0; 1), the equilibrium u 1 is regarded as more stable than u 1 since P ( 1) < P (1): 92 93 It is commonplace that ‡uctuations gives rise to many interesting phenomena; see for example, molecular collisions in gasses and liquids [19] and electronic ‡uctuations in solids [25]. It has been long observed that external noise can induce a phase transition; see for example, Bulsara, Schieve, and Gragg [7], Porrá, Masoliver and Lindenberg [32, 33], Schimansky-Geier, Hesse, and Zülick [36, 37], Xiao, Yan, and Zhang [43], Zhang, Cao, and Wu [44], de Rueda, Izús and Borzi [39], Weidlich and Grabert [42], and references therein. It is well-known that white noise best models the ‡uctuations generated by microscopic e¤ects in a homogeneous physical system. In the case of Allen–Cahn [1] bistable dynamics describing phase transition in binary alloys, interesting phenomena induced by white noises have been studied by Brassesco, De Masi, and Presutti [8], Erbar [13], Funika [15], Da Prato and J. Zabczyk [12], Fatkullin and Vanden-Eijnden [17], Katsoulakis, Kossioris, and Lakkis [24], Liu [28], Rezniko¤ & G. Vanden-Eijnden [34], Weber [41], etc. In this part we investigate the e¤ect of white noise on multistable dynamics. Replacing the bistable forcing 4(u )(1 u2 ) by a generic multistable forcing (or drift) b(u), we consider the dynamics with white noise modeled by the stochastic di¤erential equation (SDE) dut = b(ut )dt + " (ut )dWt (7.2) where fWt gt>0 is the Wiener process (Brownian motion), " is a positive constant measuring the size of ‡uctuation, and ( ) is a non-negative function modeling the dependence of ‡uctuation on system state. By a standard theory of std (e.g. [38]), under certain technical conditions on and b, (7.2) subject to an initial condition admits a unique solution and the probability density = (u; t) of the random variable ut is the solution of the following Fokker-Plank or forward Kolmogorov equation 8 > < @t = @u @u 1 "2 2 b on R (0; 1); 2 (7.3) > : ( ; 0) = 0( ) on R f0g where @t and @u are partial derivatives and 0 is the probability density function (pdf) of u0 . We expect that the system eventually achieve a stochastic equilibrium in the sense that lim ( ; t) = ( ): (7.4) t!1 When is non-trivial, it must be a pdf and is well-known as the invariant measure which describes 94 the observed distribution density after a certain amount of initiation time that is used to get rid of the initial e¤ect. If a non-trivial exists then it must be a solution of the following linear second order ordinary di¤erential equation (ode) subject to the pdf conditions: 8 n o0 > 1 0 < 2 "2 2 b = 0 on R; (7.5) > : R > 0 on R; R (u)du = 1: Suppose there exists a unique solution and that (7.4) holds. Then we conclude that ‡uctuation eventually drives a multistable system into an invariant state that does not depend on any initial setup. This is totally di¤erent from the ode dynamics (7.1) whose long time behavior of the solution crucially depends on the initial condition. This is indeed one of the main reason that multistable systems are studied with ‡uctuations. In this part, we establish the existence of a unique invariant measure and analyze its asymptotic limit as the white noise vanishes (i.e. as " & 0). The limit is shown to be Dirac measures. When is a constant, the Dirac measure is concentrated on the most stable equilibria of the potential Z P (u) := b(u)du: When is biased, the measure may support on those lesser stable equilibria but with smaller . The overall selection is the most stable equilibria of the e¤ective potential Q de…ned by Z b(u) Q(u) = 2 (u) du: Observe that Q shares the same set of equilibria and set of stable equilibria as P , but with di¤erent stability. The smaller the , the more stable the stable equilibrium of Q. We shall also address a few fundamental mathematical issues related to the degenerate case when is not everywhere positive and to the well-posedness of (7.3). The rest of Part II is organized as follows. In the Chapter 8 we establish the existence of a unique solution of (7.5) under the following regularity, non-degeneracy, and stability conditions: ; b 2 C(R); " 2 R; " > 0; ( ) > 0; lim b(u) < 0 < lim b(u): (7.6) u!1 u! 1 95 In Chapter 9 we convert (7.3) to its classically equivalent version for the cumulative distribution function (cdf) to establish its well-posedness under the conditions only listed in (7.6). In Chapter 10 we show that every solution of (7.3) with 0 being a pdf approaches as t ! 1. In Chapter 11 we study the asymptotic limit of the invariant measure as the size " of the white noise approaches zero. Finally, in Chapter 12 we consider the case when is not everywhere positive. We regard the invariant measure as the limit of invariant measures with replaced by an approximation sequence of positive functions. We show that if vanishes at a set S of (non-degenerate) stable equilibria, then for any probability measure supported on S, there exists a sequence of positive functions approximating uniformly such that the corresponding invariant measures approach . On the other hand, if vanishes only outside of all stable equilibria, the limit, although highly non-trivial, does not depend the approximation in general, so an invariant measure can be uniquely de…ned. Chapter 8 The Invariant Measure In this section we prove the following: Theorem 8.1. Assume that the regularity, non-degeneracy, and stability conditions listed in (7.6) hold. Then (7.5) admits a unique solution and the solution is given by Z u Z 1 eq(u) 2b(x) eq(x) (u) = ; q(u) := dx; Z := dx: (8.1) Z 2 (u) 0 "2 2 (x) R 2 (x) Proof. Existence. We show that de…ned in (8.1) is a solution. By the positivity assumption "; > 0 and the regularity assumption ; b 2 C(R), the function q is well-de…ned. To see that is a pdf, we need only show that Z is …nite. For this we use the stability assumption limu!1 b(u) < 0 and limu! 1 b(u) > 0. There exist positive constants and M such that b(u) 6 for every u > M and b(u) > for every u 6 M . It then follows that Z 1 Z 1 Z 1 eq(x) 2b(x) eq(x) "2 q 0 (x)eq(x) "2 eq(M ) 2 (x) dx 6 2 (x) dx = dx 6 : (8.2) M M 2 M 2 2 R M 2 Similarly, 1 (x)eq(x) dx is also …nite. Thus, de…ned in (8.1) is a pdf. By di¤erentiation, one can verify that satis…es (7.5) and therefore is a solution of (7.5). Uniqueness. Let be a generic solution of (7.5). The ode in (7.5) can be solved by an q integration followed by another integration with integrating factor e , giving the general solution Z u eq(u) q(x) (u) = 2 (u) C1 e dx + C2 (8.3) 0 96 97 with integration constants C1 and C2 . For such to be non-negative, C1 must be 0. Indeed, Z u Z M Z u q(x) q(x) q(x) lim e dx = e dx + lim e dx = 1 (8.4) u! 1 0 0 u! 1 M since q(x) 6 q(M ) for x > M (as b < 0 in [M; 1)) and q(x) 6 q( M ) for x 6 M . Thus, C1 = 0. R The condition R (x)dx = 1 then gives C2 = 1=Z. This completes the proof. Chapter 9 Well-posedness of Kolmogorov Equation Under the conditions listed in (7.6), the problem (7.3) may not admit a classical solution, i.e., a solution such that all derivatives appeared in (7.3) exist in the classical sense and the di¤erential equation in (7.3) is satis…ed. Hence, a certain notion of weak solutions is needed. From the proba- bilistic point of view and also to treat measure type initial values, it is natural to convert (7.3) to the cumulative distribution functions (cdf) de…ned by Z u Z u f (u; t) := (x; t)dx; f0 (u) := P(fu0 < ug); f (u) := (x)dx: (9.1) 1 1 Here P stands for the probability. Setting a = "2 2 =2, one can derive formally that f satis…es 8 > > Lf := @t f @u (a@u f ) + b @u f = 0 in R (0; 1); > > > > < lim f (u; t) = f0 (u) a.e. u 2 R; (9.2) > > t&0 > > > > : lim f (u; t) = 0; lim f (u; t) = 1 locally uniformly in t 2 [0; 1): u! 1 u!1 We de…ne a solution of (9.2) as follows: f is called a classical solution of (9.2) if f; @u f; @t f; @u (a@u f ) 2 C(R (0; 1)); @u f > 0 (9.3) 98 99 and each equation in (9.2) is satis…ed. Also, when f0 is continuous, we require f 2 C(R [0; 1)). We say that is a weak solution of (7.3) if = @u f and f is a classical solution of (9.2). Theorem 9.1. Assume that "; ; and b satisfy (7.6) and a = "2 2 =2. Then for each given cumulative distribution function f0 , (9.2) admits a unique classical solution. The classical theory of linear parabolic pde of divergence form (e.g. [18]) involves weak solutions in Sobolev spaces and does not seem to be enough to provide the stated assertion of the theorem. There are two concerns: (i) the regularity (9.3) is often lacking in high space dimensions, and (ii) the asymptotic behavior of the solution as u ! 1 depends on behavior of b; that is, the stability property of b is crucial in the analysis since without this property, drifted Brownian particles could escape from 1 in …nite time, so the solution may not be a cdf. Proof . We divide the proof into four steps. In the …rst step we establish the existence of a weak solution of Lf = 0 in R (0; 1) with initial condition f = f0 on R f0g. In the second step we show that the weak solution satis…es the boundary condition at u = 1. In the third step we establish the regularity (9.3). Finally, we prove uniqueness. 1. The existence of a weak solution is standard so we keep it brief. Let f(an ; bn )g1 n=1 , where an > 0, be smooth functions that approach (a; b) locally uniformly in R as n ! 1. Let ff0n g1 n=1 be smooth increasing functions such that f0n ( n) = 0, f0n (n) = 1, and f n ! f0 a.e. in R as n ! 1. Let f n be the smooth solution of the initial boundary value problem 8 > > Ln f n := @t f n @u (an @u f n ) + bn @u f n = 0 in ( n; n) (0; 1); > > > > < > f n ( n; t) = 0; f n (n; t) = 1 8 t > 0; (9.4) > > > > > : f n (u; 0) = f n (u) 0 8 u 2 [ n; n]: The existence of a unique smooth solution f n follows from the classical theory of parabolic pde [18]. By the maximum principle, 0 6 f n 6 1 on [ n; n] [0; 1), so f n ( n; t) = 0 is the global minimum, f n (n; t) = 1 is the global maximum, and @u f n ( n; t) > 0 for every t > 0. By the maximum principle again, @u f n > 0 in [ n; n] (0; 1). Standard local regularity estimates (c.f. Step 3 below) allows us to pass to a limit to obtain a weak (in the sense of distribution) solution of Lf = 0 in R (0; 1) 100 with initial condition f ( ; 0) = f0 . The solution has the property @u f 2 L2loc (R [0; 1)); @u f > 0; 0 6 f 6 1: (9.5) 2. We show that the solution obtained from Step 1 has the desired limit as u ! 1. It is in this step that the stability assumption on b is needed, by means of using the invariant measure. Let be any small positive constant. Since f0 is a cdf, there exist x < 0 and y > 0 such that f0 (u) 6 8u 6 x ; f0 (u) > 1 8u > y : (9.6) Also, since 0 6 f0 6 1, for every u 2 R we have fn (u) 1 fn (u) + =: f1 (u) > f0 (u) > f2 (u) := 1 (9.7) fn (x ) 1 fn (y ) where fn is the analog of f , the cdf of the invariant measure de…ned in the previous section with (a; b) replaced by (an ; bn ). Note that Ln f1 = 0 and Ln f2 = 0 on R [0; 1). Also, when n > maxfjx j; jy jg, f1 > f n > f2 on the parabolic boundary of ( n; n) [0; 1). Hence, applying the comparison principle on [ n; n] [0; 1) we derive that f2 (u) 6 f n (u; t) 6 f1 (u) 8 u 2 [ n; n]; t > 0: (9.8) Passing to the limit, we see that 1 f (u) f (u) 1 6 f (u; t) 6 + 8 u 2 R; t 2 [0; 1): (9.9) 1 f (y ) f (x ) Since 0 6 f 6 1, …rst sending u ! 1 and then & 0 we conclude that lim sup jf (u; t)j = 0; lim sup jf (u; t) 1j = 0: (9.10) u! 1 t2[0;1) u!1 t2[0;1) 3. Next we establish the needed regularity (9.3) for the solution obtained in Step 1. We use energy estimates. For notational simplicity, we omit the superscript n so in the following energy estimates, (f; a; b) is indeed meant to be (f n ; an ; bn ) with n > R + 2. 101 Let R > 1 be any constant and let be a cut-o¤ function: 2 C 1 (R); 0 6 ; j 0 j 6 1 on R; = 1 on [ R; R]; (u) = 0 if juj > R + 2: (9.11) 4k+2 For any non-negative integer k, integrating (u)[@tk f ][@tk Ln f ] = 0 over R ftg and using inte- gration by parts we obtain Z Z Zh i 1 d 4k+2 j@tk f j2 + a 4k+2 j@tk @u f j2= (4k+2)a 4k+1 0 +b 4k+2 (@tk f )(@tk @u f ) 2 dt Z Zh 1 b2 i 4k k 2 6 a 4k+2 j@tk @u f j2 + (4k+2)2 a+ j@t f j (9.12) 2 a by Cauchy inequality. Multiplying the resulting inequality by 2t2k we obtain Z Z Zh i d 2b2 t t 2k 4k+2 j@tk f j2 + at 2k 4k+2 j@tk @u f j2 6 2(4k+2)2 at+ + 2k t2k 1 4k j@tk f j2 : (9.13) dt a Similarly, integrating 2t2k+1 4k+4 (@tk+1 f )(@tk Ln f ) = 0 over R ftg we obtain Z Z Zh i d 2b2 t t2k+1 4k+4 j@tk+1 f j2+ at2k+1 4k+4 j@tk @u f j2 6 2(4k+4)2 at+ +2k+1 at2k 4k+2 j@tk @u f j2 : (9.14) dt a Alternately integrating these two inequalities with k = 0; 1 over [0; T ] we obtain Z Z TZ Z TZ R+2 Z 2b2 2 2 2 f (u; T )du + a 2 j@u f j2 dudt 6 8a + f dudt + 2 2 f0 du; 0 0 R 2 a Z Z TZ Z TZ 2b2 t 4 2 T a j@u f j du + t j@t f j dudt 6 4 2 32at + + 1 a 2 j@u f j2 dudt; 0 0 a Z Z TZ Z TZ 2b2 t T 2 6 2 j@t f j du + at j@t @u f j dudt 6 2 6 2 72at + + 2 t 4 j@t f j2 dudt; 0 0 a Z Z TZ Z TZ 2b2 t T 3 a 8 j@t @u f j2 du + t3 8 j@t2 f j2 6 128a2 t + + 3 at2 6 j@t @u f j2 dudt: (9.15) 0 0 a These inequalities imply that Z TZ R Z R t3 j@t (@t f )j2 dudt + sup t3 j@u (@t f (u; t))j2 du 6 C(R; T ) (9.16) 0 R t2[0;T ] R 102 where C(R; T ) depends only on T and on ka + b2 =a + 1=akL1 ([ R 2;R+2]) : Thus, by Sobolev imbed- ding, changing notation f back to f n which is what f meant to be here, we obtain k@t f n kC 1=4 ([ R;R] [T;1=T ]) 6 C(R; ^ T ) 8 R > 1; T > 1; n > R + 2: (9.17) Passing this estimate to the limit, we see that @t f 2 C 1=4 (R (0; 1)). Regarding @t f as a known q q q function, writing Lf = 0 as @u (ae @u f ) = e @t f , and using @t (ae @u f ) 2 L2loc (R (0; 1)) we q derive that ae @u f 2 C 1=4 (R (0; 1)). Then, @u (a@u f ) = @t f + b@u f 2 C(R (0; 1)). Thus (9.2) admits a classical solution. 4. Finally we prove the uniqueness of the solution. The argument relies on the required behavior of the solution near u = 1. Suppose f1 and f2 are two solutions of (9.2). Let T > 0 be any positive constant. Fixing > 0, we derive from the boundary condition limu! 1 fi (u; t) = (1 1)=2 uniformly on [0; T ] that jf1 (u; t) f2 (u; t)j 6 for all juj > y ; t 2 [0; T ], for some large enough y . On the domain [ y ; y ] [0; T ] we can apply the maximum principle for weak solutions [18] to conclude that jf1 f2 j 6 in [ y ; y ] [0; T ]. Sending ! 0 we obtain f1 = f2 in R (0; T ]. As T is arbitrary, we see that f1 f2 , and therefore obtain the uniqueness of the solution. This completes the proof of Theorem 9. Remark. One can further show that @t (a@u f ) 2 C(R (0; 1)). Then t = @t @u f = @u j with j = @u [a ] b is also continuous, so is indeed a classical solution too, though it may not be even di¤erentiable in u. The delicacy here is that the combinations a and j are di¤erentiable in u. These properties in general hold only in one space dimension. Chapter 10 Long Time Behavior In this section we prove (7.4). Theorem 10.1. Assume that "; ; b satisfy (7.6) and that 0 is a probability density function. Let be de…ned as in (8.1) and be the weak solution of (7.3) de…ned in the previous section. Then lim ( ; t) = ( ) locally uniformly in R and in L1 (R): (10.1) t!1 Consequently, for any Borel set A in R, the solution fut gt>0 of the SDE (7.2) satis…es Z Z lim P(fut 2 Ag) = lim (x; t)dx = (x)dx: (10.2) t!1 t!1 A A Note that de…ned in (8.1) depends only on "; ; and b; in particular it does not depend on the initial distribution of u0 . Theorem 10 proclaims that regardless of initial setup, any solution of the std (7.2) approaches the same statistical equilibrium distribution as in the long run. Proof. Let f ( ; t) and f ( ) be the cdfs of ( ; t) and ( ) respectively. Set g = f f . Then Lg = 0; g( ; 0) = f0 ( ) f ( ); lim sup jg(u; t)j = 0 (10.3) juj!1 t2[0;1) where the last equation follows from (9.10) (as the solution is unique). Let > 0 be any small positive constant. We shall show that kg( ; t)kL1 (R) 6 2 when t is large 103 104 enough, by using comparison principle. The above asymptotic behavior of g as juj ! 1 implies that there exists y > 0 such that jg(u; t)j 6 8 u 2 ( 1; y ] [ [y ; 1); t 2 [0; 1): (10.4) Next we focus the solution on the spatially bounded domain [ y ; y ] [0; 1). Let n o t 2 '(u; t) := e sin( f (u)); := min a(u) (u)2 : (10.5) u2[ y ;y ] For u 2 [ y ; y ] and t > 0, direct computation gives L' = [ +a 2 (@u f )2 ]' + e t cos( f )Lf = [ + 2 a 2 ]' > 0: (10.6) On the set [ y ; y ] (0; 1) let t e sin( f (u)) (u; t) := + : (10.7) minfsin( f ( y )); sin( f (y ))g It is easy to see that L > 0 on [ y ; y ] [0; 1), (u; 0) > 1 > jg(u; 0)j for all u 2 [ y ; y ] and ( y ; t) > > jg( y ; t)j for all t > 0. Hence, comparing with g on [ y ; y ] [0; 1) we see that g < on [ y ; y ] [0; 1). Since this estimate is already known to be true when u 62 [ y ; y ], we hence conclude that jg(u; t)j 6 (u; t) for every u 2 R; t 2 [0; 1): Therefore, there exists T > 0 such that jgj 6 2 on R [T ; 1). Since is arbitrary, we conclude that lim kg( ; t)kL1 (R) = 0: (10.8) t!1 Finally, we apply the energy estimates in Step 3 of the regularity proof in the previous section to the function g( ; h + ) in the domain R [0; 2] for any h > 0 to conclude that for every R > 1, lim k ( ; t) ( )kL1 (( R;R)) = lim k@u g( ; t)kL1 (( R;R)) = 0: (10.9) t!1 t!1 This completes the proof. 105 Figure 10.1: Graphs of = (u) with b(u) = (u )(1 u2 ) and (u) = 1 + (u 1)2 . When = 0:05, the less stable equilibrium u 1 is selected by the ‡uctuation. Chapter 11 Limits of the Invariant Measure When White Noise Is Vanishing In this section, we consider the asymptotic behavior of as " & 0. We write it as 2 Z u (u) eq(u) 2Q(u) b(x) (u) = R ; q(u) := ; Q(u) = dx: (11.1) 2 (y)eq(y) dy "2 0 2 (x) Figure 10.1 displays the probability density function for the case b(u) = (u )(1 u2 ) and (u) = 1 + (u 1)2 with ( ; "; ) = (0:01; 0:1; 0:05) and (0:01; 0:1; 0) respectively. Basic observation indicates that as " & 0, approaches a Dirac type mass concentrated on the most stable equilibria of the e¤ective potential Q. We con…rm this observation into two Theorems, beginning with the model problem with constant . Theorem 11.1. Let (u) 1, b(u) = (u )(1 u2 ), 0 < " 1, and fut gt>0 be any solution of (7.2). R 1 1=" 1. If = 0, then p ju 1j6 " (u)du = 2 + O(e ): Consequently, for all t su¢ ciently large, p P(fjut 1j 6 "g) = 1 2 + O(e 1=" ): 106 107 R 1=" 2. If 2 (0; 1), then p ju+1j> " (u)du = O(e ) and for each su¢ ciently large t, p P(fjut + 1j > "g) = O(e 1=" ): The proof follows by a straightforward evaluation of the potential and therefore is omitted. Notice that without the white noise, i.e., for the ode solution of (7.1), we have 8 > > 1 if u(0) > ; > > < lim u(t) = if u(0) = ; t!1 > > > > : 1 if u(0) < : Thus, adding an arbitrary small white noise totally eliminates the initial e¤ect in the long run. Statistically, it is important to …nd the mean time needed for the system to switch from one stable state to another by ‡uctuations; see [32, 33, 36, 37, 39, 43, 44] and references therein. Next we consider the general case. We denote by ( x) the Dirac mass concentrated at x. Ru Theorem 11.2. Let and b be in C 1 (R) and satisfy (7.6), Q(u) = 0 b(x) 2 (x)dx, and S be the set of global minimizers of Q. Assume that b0 (x) 6= 0 for each x 2 S. Then in the sense of measure 1 X 1 X 1 lim ()= p ( x); where A = p : (11.2) "&0 A 0 jb (x)j (x) 0 jb (x)j (x) x2S x2S Remark. Note that S is a subset of the stable steady states fu j P 0 (u) = 0 6 P 00 (u)g since Q0 = b= 2 = P 0= 2 and on S, Q00 = b0 = 2 = P 00 = 2 . Proof. Assume, without loss of generality, that 0 2 S. Then minu2R Q(u) = Q(0) = 0: Denote by M and the positive constants such that b(u) > in ( 1; M + ] and b(u) < in [M ; 1). Then Q0 (u) = b(u)= 2 (u) > 0 in [M ; 1) and Q0 (u) < 0 in ( 1; M + ], so that S ( M + ;M ). As Q00 (x) = b0 (x)= 2 (x) is non-zero and therefore must be positive for every x 2 S, S is non-empty and has …nitely many points. By taking smaller if necessary we 108 can assume that Q00 (y) > 12 Q00 (x) if jy xj < and x 2 S. We introduce d(y; S) = min jx yj; c1 = min Q(y); c2 = min Q00 (x): (11.3) x2S d(y;S)> x2S We now calculate the contributions of the integrand towards the integral Z 1 eq(y) 2Q(y) Z := 2 (y) dy with q(u) := : (11.4) " "2 p We shall show that the contributions from the set fy j d(y; S) > 2 "g is negligible, whereas the p p contribution from each ball fy j d(x; y) < 2 "g is proportional to 1= jb0 (x)j 2 (x), if x 2 S. (i) For the set ( 1; M ] [ [M; 1), we use the estimate in the proof of Theorem 8 to derive Z 2c1 ="2 1 eq(y) "[eq( M) + eq(M ) ] "e 2 (y) dy 6 6 : (11.5) " jyj>M 2 2 (ii) In the bounded set fy 2 [ M; M ] j d(y; S) > g we have Q > c1 so that Z 2 1 eq(y) 2M e 2c1 =" 2 (y) dy 6 : (11.6) " jyj " minf 2 (u) j u 2 [ M; M ]g p (iii) Suppose 2 " < jy xj < for some x 2 S. Then Q(x) = 0 and Q0 (x) = 0 < Q00 (x). Using y ) > 21 Q00 (x) for every y~ 2 [x Q00 (~ ; x + ] and Taylor expansion, we derive that y )jy xj2 > Q00 (x)" > c2 "; Q(y) = 12 Q00 (~ Z 1 eq(y) 2M e 2c2 =" p 2 dy 6 2 : (11.7) " 2 "2 " "&0 Z " d(y;S)>2 " 2 (y) p The assertion of Theorem 11.2 thus follows. Chapter 12 The Degenerate Case Where 2 Is Not Everywhere Positive Here we consider the degenerate case where is not everywhere positive. Our prototype is b(u) = (u )(1 u2 ) ( 2 [0; 1) ); (12.1) (u) = (maxf`2 u2 ; 0g) ( > 0; ` > 0 ): For the invariant measure in (8.1) to be well-de…ned, we go through a regularization process de…ning as the & 0 limit of those regular obtained by replacing by positive : Z x Z 1 eq (x) b(s) eq (u) (x) := 2 (x) ; q (x) := 2 (s) ds; Z := 2 (u) du: (12.2) Z 0 p Here by absorbing " into , we assume for simplicity that " = 2: It turns out that when ` 2 (0; 1], the limit depends sensitively on the regularization process, whereas when ` > 1, the limit does not depend on the regularization process and therefore is well-de…ned. We shall consider a special regularization for the case ` = 1 in the …rst subsection. Then consider a general regularization covering the case ` 2 (0; 1] in second subsection, showing that lim () = ( + 1) + [1 ] ( 1) (12.3) &0 110 111 where 2 [0; 1] is a constant that depends on the regularization. Indeed, for any constant 2 [0; 1], there is a regularization such that (12.3) is true. The conclusion is then generalized to a much broader situation where vanishes at certain stable equilibria of a quite general potential. In the last subsection we consider a general regularization covering the case ` > 1, proving that m1 ( + `) + m2 ( `) + ^( ) ( `;`) ( ) lim ()= (12.4) &0 m1 + m2 + m where A is the characteristic function of the set A, ^ = eq = 2 , m is a positive constant, and m1 ; m2 are non-negative constants depending only on ; `; and ; m1 = m2 = 0 if and only if > 1=2. 12.1 The Model Case With Special Regularization We begin with a special regularization for a special case where vanishes at the stable equilibria. Theorem 12.1. Let 2 [0; 1) and > 0 be constants and b(u) = (u )(1 u2 ) and = (maxf1 u2 ; 0g) be functions of u 2 R. For 2 (0; 1] de…ne by (12.2) with = maxf ; g. Then (12.3) holds in the sense of measure with 8 > > 1 if = 0; > > > > 1 < q R1 b0 (1) b(u) = ; m= exp 2 (u) ds if 2 (0; 1); 2 (0; 1); 1+m > > b0 ( 1) 1 > > > > : 0 if 2 (0; 1); > 1: To prove the theorem, we …rst prove a lemma. For this we introduce + eq (x) q (1) eq (x) q ( 1) (x) := 2 (x) e [ ;1) (x); (x) := 2 (x) e ( 1; ) (x): (12.5) Lemma 12.1. As & 0, () !M ( 1) where, with k = jb0 ( 1)j=2, 8 > R1 2 > > e kz dz if > 1; > > 1 > < R 1 kz 2 R 1=2 e k=4 dz M = e dz + if = 1; > > 1=2 1 j2zjk=2+2 > > > R1 > : 2 e kz dz if 2 (0; 1): 0 112 R Proof of Theorem 12.1. Setting M = (x)dx and m = M + eq (1) =[M eq ( 1) ]. Then + m (x) 1 (x) (x) = + : (12.6) 1+m M+ 1+m M The assertion of Theorem 12.1 thus follows from Lemma 12.1 and the computation Z M+ q (1) q ( 1) M+ 1 b(u) lim m = lim e = lim exp 2 (u) du = m: (12.7) &0 &0 M M &0 1 + Proof of Lemma 12.1. First consider . We shall show that the major contribution toward the total mass M + of + comes from the interval (1 j ln j; 1 + j ln j) for the case > 1, and from the interval [1; 1 + j ln j) for the case 2 (0; 1). p Set = 1 1 1= . Then (u) = when juj > 1 and (u) = (1 u2 ) when juj 6 1 . Denote k = b0 (1)=2. Then for x > 1 , Z x b(u) x 1 2 b00 (^ x) q (x) q (1) = 2 du = k 1+ (x 1) (12.8) 1 3b0 (1) ^ lies between 1 and x. Since b0 (1) < 0 and b00 < 0 in [1; 1), by Taylor expansion, where x Z 1 Z 1 Z 1 eq (x) q (1) kz 2 lim + (x)dx = lim dx 6 lim e dz = 0 (12.9) &0 1+ j ln j &0 1+ j ln j &0 j ln j by the change of variable x = 1 + z. Also, by Lebesgue’s Dominated Convergence Theorem, 8 > R1 2 > > e kz dz if > 1; > > 1 Z 1+ j ln j Z j ln j > < R + kz 2 (1+O(z )) 1 2 lim (x)dx = lim e dz = e kz dz if = 1; &0 1 &0 = > > 1=2 > > > > R 1 kz2 : e dz if 2 (0; 1): 0 (12.10) 1 1 1 Finally consider the integral on ( ; 1 ]. We shall use = 2 [1 + O( )] and 2 b00 (^ x) hk i 2= 2 q (1 ) q (1) = k 1 0 = + O( ) : (12.11) 3b (1) 4 113 (a) When > 1, using q (x) 6 q (1 ) and (x) > (1 + ) (1 x) for x 2 ( ; 1 ], we obtain Z 1 Z 1 2= 2 eq (1 ) q (1) e [k=4+O( )] + (x)dx 6 dx 6 2 1 ! 0 as & 0: (1 + )2 (1 x)2 (2 1)(1 + )2 (12.12) (b) When 2 (0; 1), considering separately cases 2 (0; 1=2); = 1=2; and 2 (1=2; 1) we derive that Z 1 Z 1 dx + (x)dx 6 6 O(1) [1 + j ln j+ 1 2 ] ! 0 as & 0: (12.13) (1 x2 )2 (c) When = 1, we have = 2 + O( 2 ) and for x 2 ( ; 1 ], Rx (u ) du 1+ + eq (1 ) q (1) e1 (1 u2 ) ek=4+O( ) k 2 +2 2 2 +2 (x) = = : (12.14) (1 x2 )2 1 x 1+x Using the change of variable x = 1 + z we obtain Z 1 Z = h i Z 1=2 + k 2+2 k=4 e k=4 dz lim (x)dx = lim e +O( ln ) dz = ; &0 1 j ln j &0 j ln j z 1 j2zjk=2+2 Z 1 j ln j Z j ln j O(1) lim + (x)dx 6 lim dz = 0: (12.15) &0 &0 1 jzjk=2+2 Rx + In conclusion, de…ning f (x) = 1 (u)du, we …nd that lim &0 f (x) = 0 if x < 1 and = M + + if x > 1. Thus, ( ) ! M+ ( 1). Similarly, by considering ~ (x) = ( x) with k = b0 ( 1)=2, we can show that ()!M ( + 1). This completes the proof of Lemma 12.1. Remark. From the proof, one sees that has the following limiting pro…le: 8 > > e kz 2 if > 1; > > > < lim ( 1 z )= kz 2 (12.16) > e [0;1) if 2 (0; 1); &0 > > > > : e kz 2 e k=4 [ 1 2 ;1) + j2zjk=2+2 ( 1; 1 2) if = 1: 114 12.2 General Regularization When Vanishes at Stable States Here we extend Theorem 12.1 to a general regularization for the general case when vanishes at stable equilibria. We assume that vanishes near 1 where 1 is the smallest and 1 is the largest zeros of b. Also is positive in [x1 ; x2 ] ( 1; 1) where (x1 ; x2 ) is an interval contains all zeros of b except 1. Theorem 12.2. Let b be a continuous function that satis…es, for some x2 2 (0; 1) and x1 2 ( 1; 0), b < 0 in ( 1; x1 ] [ (1; 1); b(u) > 0 in ( 1; 1) [ [x2 ; 1); lim jb(u)j > 0: (12.17) juj!1 Let be a non-negative continuous function such that, for some > 0, (u) > 0 in [x1 ; x2 ]; (u) = 0 in [ 1 ; 1] [ [1; 1 + ]: (12.18) For to be speci…ed below, let be de…ned as in (12.2). The following holds: 1. Suppose f g 2(0;1] is a family of continuous positive functions satisfying lim &0 ()= () uniformly on [ 1 ; 1 + ]. Then for some 2 [0; 1], (12.3) holds along a sequence & 0; 2. For every 2 [0; 1], there exists a family f g 2(0;1] of continuous positive functions satisfying lim &0 = uniformly on R such that (12.3) holds. Proof. Similar to the proof of Theorem 12.1, we introduce Z eq (x) q ( 1) (x) = 2 (x) [0;1) ( x); M = (x)dx: (12.19) 1. We …rst show that lim &0 M + = 1. For each x > 1 setting b (x) = maxu2[1;x] jb(u)j and 115 using q 0 = b= 2 , b < 0 in (1; x), lim &0 = 0 on [1; 1 + ], and b(1) = 0, we derive that Z Rx b(u) x jb(u)j 1 exp( 1 2 (u) du) M + > + (u)du = ; 1 b (x) b (x) Rx b(u) 1 exp( 1 2 (u) du) 1 lim M + > lim lim = lim = 1: (12.20) &0 x&1 &0 b (x) x&1 b (x) 2. Next we estimate the mass on [x; 1) for each x > 1. Setting b (x) = inf u>x jb(u)j we have Z 1 Z 1 jb(u)j lim + (u)du 6 lim + (u)du &0 x &0 x b (x) q (x) q (1) Z x e 1 b(u) 6 lim = lim exp 2 du = 0: (12.21) &0 b (x) &0 b (x) 1 3. For any x 2 (x2 ; 1), setting B (x) = minu2[x+ ;x] b(u) we derive that Z x Z x Z x b(u) e q (1) 1 + (u)du6 + (u)du = q 0 (u)eq (u) du 6 ; x+ x+ B (x) B (x) x+ B (x) Z x Z x2 1 + 1 eq (u) q (x2 ) 1 lim (u)du6lim + du + = 0: (12.22) &0 M + 1 &0 M 0 2 (u) B (x) + 4. The above three estimates shows that as & 0, (x)=M + ! (x 1). Similarly, as & 0, (x)=M ! (x + 1): Finally, note that + (x) (x) M eq ( 1) (x) = + [1 ] ; := : (12.23) M M +( ) M + eq (1) + M eq ( 1) As f g 2(0;1] is a family in (0; 1), there exists a sequence f i g1 i=1 in (0; 1) such that as i ! 1, i & 0 and i ! for some 2 [0; 1]. This implies that i () ! ( + 1) + [1 ] ( 1) as i ! 1. The …rst assertion of the theorem thus follows. As the second assertion is a special case of the theorem to be presented next, this completes the proof. Remark. Following the same proof, one sees that the …rst assertion of Theorem 12.2 remains true if condition (12.17) is replaced by the following more applicable conditions: Z x2 Z x2 Z 1 Z 1+" du du du du 2 (u) 6 lim 2 (u) <1= 2 (u) = 2 (u) du 8 " > 0: (12.24) x1 &0 x1 1 " 1 p For example, (u) = juj j ln u2 j. 116 R 2 Since the e¤ective potential is Q(u) = b(u)= (u)du, the second assertion of Theorem 12.2 can be extended to a much more general case where vanishes at multiple stable equilibria. Theorem 12.3. Let be a non-negative continuous function and b be a C 1 (R) function satis- fying lim b(u) < 0 < lim b(u): (12.25) u!1 u! 1 Assume that the set S := fu 2 R j b(u) = 0; b0 (0) < 0; (u) = 0g is non-empty. Then for every probability measure supported on S, these exists a family of continuous positive functions f g0< <1 such that lim &0 = uniformly on R and the function de…ned in (12.2) satis…es lim = in the sense of measure: (12.26) &0 Proof. 1. Let 2 (0; 1] be any constant. Observe that S is bounded and has only …nitely many points. As = 0 and b0 < 0 on S, there exists > 0 such that (y) 6 and jb0 (y)=b0 (x) 1j 6 1=2 when y 2 [x ; x + ] and x 2 S. The distance between di¤erent points in S exceeds 2 since b(x ) > 0 > b(x + ) for each x 2 S. Denote A = [x2S [x ; x + ]; Ac := R n A; ^ (u) = maxf (u); g; Z u Z Z b(y) ^= eq^(y) eq^(y) q^(u) = 2 dy; Z 2 dy; Z^c = 2 dy: (12.27) 0 ^ (y) R ^ (y) Ac ^ (y) Here without loss of generality, we assume that 0 2 Ac . Recall that the stability assumption on b implies that Z^ is …nite. For non-negative functions s(x) and L(x; u) to be chosen later, we de…ne X ^ (u) (u) = ^ (u) Ac (u) + p [x ";x+"] (u) (12.28) 2 1 + ^ (u)s(x)L(x; u) x2S and the corresponding ; q ; Z as in (12.2). The central idea here is to choose an appropriate L such that q = q^ on Ac . Then the integral of on Ac is still Z^c , whereas in each interval [x ; x+ ] with x 2 S, we still have the freedom to choose s(x). For this we notice the following 1 1 X 2 (u) = + s(x)L(x; u) [x ;x+ ] (u); (12.29) ^ 2 (u) x2S X Z u q (u) = q^(u) + s(x) b(y)L(x; y) [x ;x+ ] (y)dy: (12.30) x2S 0 117 R x+ Hence, for q q^ on Ac , we need only x b(y)L(x; y)dy = 0 for each x 2 S. More details will be given later. Here we point out that if L(x; u) is continuous in u and L(x; x ") = 0 for each x 2 S, then is positive and continuous. In addition, since 6 on A, k kL1 (R) 6 : (12.31) 2. Now we de…ne L(x; ) for each x 2 S as follows: L(x; ) is zero in ( 1; x =2], linear with slope 2 in [x =2; x], linear with slop k in [x; x + =k] and zero in [x + =k; 1); more precisely, L(x; u) = [2(u x) + ] [x =2;x] (u) + [ + (x u)k] (x;x+ =k] (u): (12.32) Here k = k(x) > 1 is the root of the equation j(k) = 0 where Z Z x Z x+ =k j(k) := L(x; u)b(u)du = [2(u x) + ]b(u)du + b(u)[ + (x u)k]du x =2 x Z h b(x 2 1 s=2) b(x + s=k) i = (1 s) + ds 0 2 k Z 1 Z 1 h b0 (x + 3 s=k) b0 (x s=2) = (1 s)s ]d ds: (12.33) 0 0 k2 4 1j 6 1=2 for y 2 [x ; x + ]. There exists a (unique) k 2 [1; 4] such that j(k) = 0 since jb0 (y)=b0 (x) Ru 6 A, we see from (12.29) that q = q^ on Ac . In addition, since x " b(y)L(x; y)dy is Since 0 2 increasing in ( 1; x] and decreasing in [x; 1), it is non-negative in R. 3. Now we de…ne s(x) = s for x 2 S as the root of Z Z x+ 1 u Z^ [ (fxg) + ] + sL(x; u) exp q^(y) + s b(y)L(x; y)dy du = : (12.34) x ^ 2 (u) x There exists a unique solution s > 0 since the left-hand side is an increasing function of s 2 [0; 1) Ru (as x " b(y)L(x; y) > 0 for each u 2 [x ; x + ]), approaches 1 as s ! 1, and is less than the right-hand side when s = 0. 4. Now we are ready to complete the proof. Since q = q^ on Ac , we have Z Z eq (u) eq^(u) 2 (u) du = du = Z^c 6 Z: ^ (12.35) Ac Ac ^ 2 (u) 118 Also, for each x 2 S using (12.28), (12.29) and the de…nition of s = s(x) in (12.34) we …nd that Z x+ eq (y) Z^ [ (fxg) + ] 2 (x) dx = : (12.36) x Consequently, Z XZ X Z^ [ (fxg) + ] eq (y) x+ eq (y) Z^ Z = 2 (y) dy + 2 (y) dy = Z^c + = Z^c + jSjZ^ + Ac x2S x x2S P since x2S (fxg) = (R) = 1, where jSj is number of points of S. Hence, Z Z 1 eq Z^c (y)dy = dy = ; Ac Z Ac 2 (1 + jSj)Z^ + Z^c Z x+ Zx+2 1 eq fxg + (y)dy = dy = 8 x 2 S: (12.37) x Z x 2 2 1 + jSj + Z^c =Z^ Sending & 0 we then conclude that ! . This completes the proof. 12.3 The Case When Vanishes Only Beyond All Critical Points Here we consider a case based on the prototype b(u) = (u )(1 u2 ) and (u) = (maxf`2 u2 ; 0g) with ` > 1. Let [ 1; 1] be an interval that contains all critical points of the potential, i.e., the set fx 2 R j b(x) = 0g. We consider a degenerate case where is positive in an open interval (x1 ; x2 ) that contains [ 1; 1] and vanishes outside (x1 ; x2 ). It turns out that the limit of the regularized invariant measure does not depend on the modi…cation. Therefore, the limit can be de…ned as the invariant measure. Theorem 12.4. Let b be a continuous function on R that satis…es b > 0 in ( 1; 1); b < 0 in (1; 1); lim jb(u)j > 0: (12.38) juj!1 Let be a continuous function on R such that, for some x1 < 1, x2 > 1 and > 0, > 0 in (x1 ; x2 ); = 0 in [x1 ; x1 ] [ [x2 ; x2 + ]: (12.39) 119 Let ( ) be de…ned as in (12.2) where f g 2(0;1] is any family of continuous positive functions in R that satis…es lim &0 ( ) = ( ) uniformly on [x1 ; x2 + ]. Then in measure 1 eq(x) lim (x) = m1 (x x1 ) + m2 (x x2 ) + 2 (x) (x1 ;x2 ) (x) (12.40) &0 m1 + m2 + m where Z x Z x2 b(s) eq(xi ) eq(x) q(x) := 2 (x) ds 8 x 2 [x1 ; x2 ]; mi := ; m := 2 (x) dx: (12.41) 0 jb(xi )j x1 Before the proof, we remark on the sizes of m; m1 and m2 . Rx Remark. Clearly, m1 ; m2 , and m are …nite positive constants if x12 2 (x)dx < 1. Rx Suppose 0 2 2 (x)dx = 1. Then q(x2 ) = 1 since b(x2 ) < 0. This implies that m2 = 0. In addition, setting 1 1 := minfx2 1; 1 x1 g; b := inf jb(x)j; 2 x2( 1;x1 + 1 ][[x2 1 ;1) we have b > 0 and Z x2 Z y eq(x) 1 eq(x2 1) 2 (x) dx 6 q 0 (x)eq(x) dx = < 1: x2 1 b x2 " b R0 2 R x1 +"1 2 q(x) Similarly, if x1 (x)dx = 1, then q(x1 ) = 1; m1 = 0; and x1 e dx < 1. Thus, m is a …nite positive constant whereas m1 and m2 are non-negative …nite constants. Remark. The limit does not depend on regularization, so the right-hand side of (12.40) can well be de…ned as the invariant measure . Here the contribution of the point masses at x1 and x2 is highly non-trivial and has to be obtained through a regularization process. Proof of Theorem 12.4. 1. Introduce Z x Z x eq (u) eq(u) F (x) := 2 (u) du 8 x 2 R; F (x) := 2 (u) du 8 x 2 [x1 ; x2 ]: (12.42) 0 0 120 By Remark 12.3, F is continuous on [x1 ; x2 ]. Since > 0 in (x1 ; x2 ) and lim &0 = uniformly, lim q (x) = q(x); lim F (x) = F (x) 8 x 2 (x1 ; x2 ): (12.43) &0 !0 2. Next we consider F (1) F (x) for x 2 (x2 ; 1). Let b be as above. For > 0, Z 1 Z 1 Z 1 eq (u) jb(u)j q 1 eq (x2 + ) 2 (u) du 6 e (x) du = q 0 (u)eq (u) du 6 : (12.44) x2 + x2 + b 2 (u) b x2 + b In addition, since lim &0 = 0 in [x2 ; x2 + ] and b < 0 in (1; 1), Z x b(s) lim q (x2 + ) = q(1) + lim 2 (s) ds = 1: (12.45) &0 &0 1 It then follows that Z 1 eq (u) lim 2 (u) du = 0 8 > 0: (12.46) &0 x2 + 3. Now we consider the jump of F across x2 . For every 2 (0; "1 ), by the mean value theorem, there exists x ; 2 [x2 ; x2 + ] such that Z x2 + Z x1 + q (x2 ) q (x2 + ) 0 q (x) b(x) q (x) e e = q (x)e dx = 2 (x) e dx x2 x2 Z x2 + eq (x) = b(x ; ) 2 (x) dx = jb(x ; )j[F (x2 + ) F (x2 )]: (12.47) x2 Thus, for any …xed x > x2 , in view of (12.46), (12.45) and (12.43), we have lim F (x) = lim lim F (x2 + ) &0 &0 &0 eq (x2 ) eq (x2 + ) = lim lim F (x2 )+ &0 &0 jb(x ; )j ! eq(x2 ) eq(x2 ) = lim F (x2 )+ = F (x2 ) + = F (x2 ) + m2 : (12.48) &0 lim &0 jb(x ; )j jb(x2 )j Similarly, we derive lim &0 F (x) = F (x2 )+m2 . Hence, lim &0 F (x) = F (x2 )+m2 for each x > x2 : 121 In summary, we have 8 > < F (x) if x 2 (x1 ; x2 ); lim F (x) = (12.49) &0 > : m2 + F (x2 ) if x > x2 : After a similar analysis for F (x) for x 6 x1 , we then obtain the assertion of the Theorem. Remark. Following the same proof, one sees that Theorem 12.3 remains true if the assumptions are replaced by the following more applicable conditions: Z x2 Z x2 Z x1 Z x2 +" du du du du lim 2 (u) = 2 (u) <1= 2 (u) = 2 (u) du 8 2 (0; x2 2 x1 ): &0 x1 + x1 + x1 x2 (12.50) p For example, (u) = j(u x1 )(x2 u)j j sin(2 u)j1=4 with x1 < 1 and x2 > 1. ... Appendix A Hopf-Lax Style Existence and Uniqueness of Solution By relating the conservation law to the corresponding Hamilton-Jacobi problem, it is possible to prove both uniqueness and existence of such a solution. The argument involves using molli…cation techniques and analysis, as well as a solution candidate from the conservation law. We have the following. Theorem. Let H be a piecewise linear function with a …nite number of break points, and g 0 be a step function with a …nite number of steps, and …x T > 0. Consider the PDE given by wt + (H (w))x = 0 on R (0; 1) w = g 0 on R f0g : Then there exists an integral solution w 2 L1 (R [0; T )) which satis…es Z 1 Z 1 Z 1 w t + H (w) x dxdt + w dxjt=0 = 0 0 1 1 for each test function where : R [0; T ) ! R is smooth with compact support. Furthermore, this solution is unique. 122 123 A.1 Proving existence of a solution I. Clari…cation of su¢ cient conditions for existence in Evans text We go over Lemma 1 in Section 3.3 without the assumption of superlinearity on the ‡ux function on H. This means that some quantities in L will be in…nite, but this will not matter since we are then taking the minimum which is not a¤ected by the in…nite values. Also, the Lemma 1 in Evans depends on part of Lemma 2, and we write this in a more coherent manner using Lemma A, B etc. The …rst thing to do is to make sure that we have the key theorem that H and L are Legendre transforms of one another. We do not need to use any of the theorems in Evans that rely on superlinearity. We only assume that H is Lipschitz continuous, which is obvious from the de…nition. We also assume that g (which will be the initial condition for the Hamilton-Jacobi equation) is Lipschitz on R. In all of the theorems we start with the assumption that H (q) is polygonal convex with the last segments having slopes m1 at the left and mN +1 at the right, with c1 < c2 < ::: < cN at the break points with c1 < 0 < cN . Assume m1 < 0 < mN +1 . De…nition. We de…ne the usual Legendre transform, and let L (p) denote it: L (p) := H (q) := sup fpq H (q)g (A.1) q2R A computation shows that this is a convex polygonal shape such that L (p) < 1 i¤ p 2 (m1 ; mN + 1) : It has break points at m1 < m2 < ::: < mN +1 and slopes c1 ; c2 ; ::; cN . The last break point is at mN +1 where the slope and L (mN +1 ) become in…nite. Note that L is Lipschitz on (m1 ; mN +1 ) : Prop. P1. Let L (p) be as de…ned above (with L (p) < 1 i¤ p 2 (m1 ; mN +1 ) ). Then the Legendre transform (de…ned above) of L (p) is L (q) := sup fpq L (q)g = H (q) : (A.2) p2R Remark. In other words, if we de…ne L (p) as polygonal, convex between p 2 (m1 ; mN +1 ) as in the …gure, with L (p) = 1 for p 62 (m1 ; mN +1 ) then the operation supp2R fpq L (q)g yields the function H de…ned above. This is all that one needs in order to prove (below and in Evans, Theorem x y 5 or 6) that if u is de…ned by the Hopf-Lax formula u (x; t) := miny2R tL t + g (y) ; then it 124 satis…es the Hamilton-Jacobi equation, ut + H (Du) = 0 at points where u is di¤erentiable. Lemma A. Suppose L and g are Lipschitz continuous and L ( ) < 1 only on (m1 ; mN +1 ) : Then u de…ned by the Hopf-Lax formula is Lipschitz continuous in x: Proof. Fix t > 0; x; x ^ 2 R . Choose y 2 R (depending on x; t) such that x z x y u (x; t) = min tL + g (z) = tL + g (y) : (A.3) z t t The minimum is attained since both L and g are continuous. Note that while there may be values of (x z) =t such that L ((x z) =t) = 1; these are irrelevant, as there are some …nite values, and L ((x y) =t) will be one of those. Now use (1) to write x ^ z~ x y u (^ x; t) u (x; t) = inf tL + g (~ z) tL g (y) : (A.4) z~ t t We de…ne z := x ^ x + y such that x y x ^ z = (A.5) t t and substitute this z in place of z~ in (A.4) which can only increase the RHS. This yields the inequality x ^ z x y u (^ x; t) u (x; t) tL + g (z) tL g (y) t t x y x y = tL + g (^ x x + y) tL g (y) t t = g (^ x x + y) g (y) : Using the assumption that g is Lipschitz, one has u (^ x; t) u (x; t) Lip (g) j^ x xj : (A.6) Note that in obtaining this inequality, x and x ^ were arbitrary (without any assumption on order). Hence, we can interchange them. I.e., we start by de…ning y such that, instead of (1), it satis…es x ^ y u (^ x; t) = tL + g (y) : t 125 Thus, we obtain the same inequality as (A.6) with the x and x ^ interchanged, yielding ju (^ x; t) u (x; t)j Lip (g) j^ x xj : (5) Lemma 1. Suppose L and g are Lipschitz continuous and L ( ) < 1 only on (m1 ; mN ) : For each x 2 R and 0 s < t we have x y u (x; t) = min (t s) L + u (y; s) : y2R t s Proof of Lemma 1. Part 1. Fix y 2 R, 0 < s < t: Since u and L are continuous, the minimum in the de…nition of u (x; t) is attained on the interval (m1 ; mN +1 ) where L is …nite. Thus we can …nd z 2 R such that y z u (y; s) = sL + g (z) : (A.7) s Note that since (y z) =s is the minimizer, we know that L ((y z) =s) is …nite. By convexity of L we can write (as in Evans) using x z s x y s y z = 1 + (A.8) t t t s t s x z s x y s y z L 1 L + L : (A.9) t t t s t s Note that L ((y z) =s) is …nite, and, from the geometry of convexity it is clear that for any …xed y 2 R, 0 < s < t we have x z x y L = 1 if f L = 1: (A.10) t t s Next, we have from our basic assumption that u (x; t) is de…ned by the Hopf-Lax formula, the identity x z^ u (x; t) = min tL + g (^ z) (A.11) z^ t so substituting the z de…ned above in (A.7) yields the inequality x z u (x; t) tL + g (z) (A.12) t 126 and now using (A.9) yields x y y z u (x; t) (t s) L + sL + g (z) : (A.13) t s s Now note that the last two terms, by (A.7) are u (y; s) : This yields x y u (x; t) (t s) L + u (y; s) : (A.14) t s Now we take the minimum over all y 2 R. Note that y has been arbitrary. As we take the minimum over y we note that there are values of y for which ?? is in…nite, but there are also values for which it is …nite. Thus in taking the minimum, the values for which it is in…nite are irrelevant. Hence, we can write x y u (x; t) min (t s) L + u (y; s) : (A.15) y2R t s Part 2. Next, we know again that there exists w 2 R (depending on x and t that we regard as …xed) such that x w u (x; t) = tL + g (w) : (A.16) t We choose y := st x + 1 s t w; which implies x y x w y w = = : (A.17) t s t s We know that w is the minimizer (and of course, g is …nite in the domain (m1 ; mN )) so that L ((x w) =t) is …nite. Thus, by the identity (A.17) above, so are L ((x w) =t) and L ((y w) =s) : Thus we can write, using the basic de…nition of u (y; s), the equality x y x y y z^ (t s) L + u (y; s) = (t s) L + min sL + g (^ z) t s t s z^ x x y y w (t s) L + sL + g (w) (A.18) t s x where the inequality is obtained simply by substituting a particular value for z^; namely the w that we de…ned above in (A.16): We can use (A.17) in order to re-write the arguments of L in equivalent forms. By the equality and the fact that L ((x w) =t) is …nite, so are so are L ((x w) =t) and L ((y w) =s) : Hence, 127 replacing the two expressions involving L on the RHS of (A.18) yields x y x w x w (t s) L + u (y; s) (t s) L + sL + g (w) t s t t x w = tL + g (w) = u (x; t) (8) t where the last identity follows from the expression (A.16) that de…nes w: Thus (A.18) gives us an identity for a particular y that we de…ned, namely x y (t s) L + u (y; s) u (x; t) : (A.19) t s If we replace y be a minimum over all y^ we obtain the inequality x y min (t s) L + u (y; s) u (x; t) : (A.20) y^ t s Combining (A.15) and (A.20) proves Lemma 1. === Lemma B (This is the analog of Lemma 2 - proof in part 2 of Evans. Part 2 is essentially the same; one needs only pay attention to …niteness of the terms) Under the same conditions (L and g Lipschitz continuous), one has u (x; 0) = g (x) : Proof. Since 0 2 (m1 ; mN +1 ), the interval on which L is …nite, one has x y u (x; t) = min tL + g (y) tL (0) + g (x) ; (A.21) y t upon choosing y = x: Also, x y u (x; t) = min tL + g (y) y t x y min tL + g (x) Lip (g) jx yj y t x y = g (x) + min tL Lip (g) jx yj y t jx yj x y = g (x) t max Lip (g) L y t t = g (x) t max fjzj Lip (g) L (z)g : (A.22) z 128 Note that maxz fjzj Lip (g) L (z)g is a …nite number since L (z) is bounded above and is 1 outside of the range (m1 ; mN ). Thus we de…ne n o C := max jL (0)j ; max fjzj Lip (g) L (z)g (A.23) z and combine (A.21) and (A.22) to write ju (x; t) g (x)j Ct for all x 2 R and t > 0: === (A.24) Lemma C (This is the analog of Lemma 3–proof in part 3 of Evans. This is just sketched out, even in the case he considers). If L and g are Lipschitz, one has for any x 2 R and 0 < t^ < t; the inequalities, u (x; t) u x; t^ C t t^ n o C := max jL (0)j ; max fjzj Lip (g) L (z)g : z Proof. Let x 2 R and 0 < t^ < t: By Lemma A one has ju (x; t) u (^ x; t)j Lip (g) jx x ^j i.e., Lip (u ( ; t)) = Lip (g) : From Lemma 1, we have for 0 s = t^ < t; the inequality x y u (x; t) = min (t s) L + u (y; s) y t s x y min (t s) L + u (x; s) Lip (u) jx yj y t s x y = u (x; s) + min (t s) L Lip (g) jx yj y t s x y jx yj = u (x; s) + (t s) min L Lip (g) y t s t s = u (x; s) + (t s) min fL (z) Lip (g) jzjg (A.25) z where we have just de…ned z := jx yj = (t s) : We can also write this as u (x; t) u (x; s) (t s) max fLip (g) jzj L (z)g : (A.26) z 129 Using C1 := maxz fLip (g) jzj L (z)g which, as discussed above, is clearly …nite, one has then u (x; t) u (x; s) C1 (t s) : (A.27) The other direction in the inequality is given by using Lemma 1 directly. Substituting x in place of the y in the minimizer, we have x y u (x; t) = min (t s) L + u (y; s) y t s x x (t s) L + u (x; s) t s = (t s) L (0) + u (x; s) (A.28) yielding the inequality u (x; t) u (x; s) (t s) L (0) : (A.29) Combining (A.28) and (A.29) yields the Lipschitz inequality in t; namely, ju (x; t) u (x; s)j C jt sj : (A.30) Lemma D. Under the conditions of Lemmas A and C one has ^; t^ u x u (x; t) C ^; t^ x (x; t) 2 (A.31) where k k2 is the usual Euclidean norm. Proof. We use the triangle inequality and Lemmas A and C, one has ^; t^ u x u (x; t) ^; t^ u x u x; t^ + u x; t^ u (x; t) ^j + C t t^ C jx x n 2 o 2 2C jx x ^j + t t^ = 2C ^; t^ x (x; t) 2 : (A.32) Lemma E. If L and g are Lipschitz continuous then u : R2 ! R is di¤erentiable almost everywhere on R (0; 1) : Proof. Follows from Rademacher’s theorem and Lemma D. 130 Theorem 5 (Modi…ed Evans’Sec. 3.3) Let x 2 R and t > 0. Let u be de…ned by the Hopf-Lax formula and di¤erentiable at (x; t) 2 R (0; 1) : Then ut (x; t) + H (Du (x; t)) = 0: (A.33) Proof. Part 1. Fix q 2 (m1 ; mN +1 ) and h > 0: By Lemma 1, we have x + hq y u (x + hq; t + h) = min hL + u (y; t) : (A.34) y2R h (Once again since there are some …nite values over which we are taking the minimum, the expression is well-de…ned). Upon setting y as x; we can only obtain a larger quantity on the RHS, yielding u (x + hq; t + h) hL (q) + u (x; t) : (A.35) Hence, we have that for q 2 (m1 ; mN +1 ) the inequality u (x + hq; t + h) u (x; t) L (q) : (A.36) h Since we are assuming that u is di¤erentiable at (x; t) we have the existence of the limit of the LHS of (A.36) thereby yielding @t u (x; t) + qDu (x; t) L (q) ; i:e:; @t u (x; t) + qDu (x; t) L (q) 0: (A.37) We now use the de…nition, supq2R fqw L (q)g =: H (w) that we made earlier (note that we are taking the sup over q and not p now), and write H (Du (x; t)) = sup fqDu L (q)g : (A.38) q2R Note that the values of q for which L (q) = 1 are clearly not candidates for the supp since L (q) = 1. Hence, we can take the sup over all q in (2) ; (which is equivalent to taking the sup 131 over q 2 (m1 ; mN +1 )) to obtain @t u (x; t) + sup f@t u (x; t) + qDu (x; t) L (q)g 0; or; q2R @t u (x; t) + H (u (x; t)) 0: (A.39) Part 2. Now use the de…nition x y u (x; t) = min tL + g (y) : y2R t Since L and g are continuous, the minimizer exists and for some z 2 R depending on (x; t) we have x z u (x; t) = tL + g (z) : (A.40) t s s x z y z De…ne s := t h; y := tx + 1 t z so t = s : Then we can write, using the de…nition of u (y; s), x z y y^ u (x; t) u (y; s) = tL + g (z) min sL + g (^ y) : (A.41) t y^ s By substituting z (de…ned by (A.40)) in place of y^ in this expression, we subtract out at least as much and obtain the inequality x z y z u (x; t) u (y; s) tL + g (z) sL + g (z) t s x z = (t s) L (A.42) t x z y z x z y z by virtue of the equality t = s : Note that by de…nition, L t =L s < 1; so there is no divergence problem there. Now replace y with its de…nition above, and use t s = h to write (5) as u (x; t) u x + ht (z x) ; t h x z L : (A.43) h t Since we are assuming that the derivative exists at (x; t) we can take the limit as h ! 0+ and obtain z x x z Du (x; t) + @t u (x; t) L (A.44) t t 132 We use the de…nition of H again, and write ut (x; t) + H (Du (x; t)) = ut (x; t) + max fqDu L (q)g q2R z x x z ut (x; t) + Du (x; t) L (A.45) t t where we have chosen one value of q; namely (x z) =t to obtain the inequality. Note, again, that x z from the original de…nition in (4) ; L t must be …nite. Hence, the RHS of (8) is well-de…ned. Combining (7) and (8) yields the inequality @t u (x; t) + H (Du (x; t)) 0: (A.46) Combining (A.46) with the result from Part 1 of the proof, namely (A.40); we obtain the result that u satis…es the Hamilton-Jacobi equation at (x; t). === Thus, Theorem 5 in Evans (Section 3.3) is …ne as stated for the polynomial H: Theorem 6. The function u (x; t) de…ned by the Hopf-Lax formula is Lipschitz continuous, di¤erentiable a.e. in R (0; 1) and solves the initial value problem ut + H (Du) = 0 a:e: in R (0; 1) u (x; 0) = g (x) f or all x 2 R . (A.47) De…nition. We say that w 2 L1 (R (0; 1)) is an integral solution of wt + H (w)x = 0 in R (0; 1) w (x; 0) = h (x) f or all x 2 R (A.48) if for all test functions : R [0; 1) ! R (i.e., that are smooth and have compact support) one has the identity Z 1 Z 1 Z 1 Z 1 Z 1 w t dxdt + w dxjt=0 + H (w) x dxdt = 0: (A.49) 0 1 1 0 1 Theorem 2 (Section 3.4 of Evans, but statement of the theorem is somewhat di¤erent) Under 133 the assumptions that g is Lipschitz and that H is polygonal and convex (as described above), the function w (x; t) := @x u (x; t) where u is the Hopf-Lax function is an integral solution of the initial value problem for the conservation law above. Proof. From Theorem 6 above we know that u is Lipschitz continuous, di¤erentiable a.e. in R (0; 1) and solves the Hamilton-Jacobi initial value problem subject to initial condition h (x) Rx where g (x) = 0 h (z) dz. We multiply ut + H (ux ) = 0 with a test function x and integrate over R1R1 0 1 :::dxdt. Upon integrating by parts one obtains the relation above in the de…nition of integral solution. The integration by parts operations are justi…ed by the fact that u (x; t) is Lipschitz in both x and t. Also, w (x; 0) = @x u (x; 0) = g 0 (x) = h (x) : (A.50) A.2 Uniqueness Suppose that we have two integral solutions, w and w, ~ to the conservation law IVP: wt + H (w)x = 0; w (x; 0) = g 0 (x) : (A.51) I.e., these satisfy the integral identity that we obtain by multiplying by the derivative of a test function, x: We also assume that they satisfy the condition 1 w (x + z; t) w (x; t) C 1+ z: (A.52) t Rx Rx We de…ne u (x; t) := 0 w (~ x; t) d~ x and u ~ (x; t) := 0 w ~ (~ x; t) d~ x: We have seen that u and u ~ will then be solutions to the Hamilton-Jacobi equations a.e. in (x; t) : We then take the equations and mollify them in R2 , i.e., (ut )" + (H (ux ))" = 0; (~ ut )" + (H (~ ux ))" = 0: (A.53) Note that (ut )" = (u" )t and similarly for x. 134 Let v" (x; t) := u ~" (x; t) u" (x; t) so that @t v" = (H (ux ))" (H (~ ux ))" ; v" (x; 0) = 0: (A.54) We add and subtract terms to write @t v" = H " (u";x ) + f(H (ux ))" H " (u";x )g (A.55) H " (~ u";x ) f(H (~ ux ))" H " (~ u";x )g =: H " (u";x ) H " (~ u";x ) + R1 (") + R2 (") (A.56) By the usual theorems on molli…cation, we note that R1 and R2 converge (uniformly on compact sets in (x; t) ) to zero as " ! 0: We de…ne Z" (x; t) by leaving out the R1 and R2 terms: @t Z" = H " (u";x ) H " (~ u";x ) ; Z" (x; 0) = 0: (A.57) We then use the arguments of Evans (Section 3.3 Theorem 7). Note that these use convexity, but not uniform convexity or superlinearity. Also, (1) implies, for small z > 0; 1 u (x + z; t) 2u (x; t) + u (x z; t) C 1+ z2; (A.58) t which is the key condition needed for uniqueness. The inequality is inherited by the molli…ed functions u" and u ~" : Note that all we need for the proof is for (A.58) to hold for small, positive z; since we use it to bound smoothed out derivatives. To establish that (A.52) implies (A.58) for small, positive z; note the following. By de…nition of w one has from (A.52); 1 @x u (x + z) @x u (x) C 1+ z: (A.59) t For su¢ ciently small h > 0; one must have for some C 0 u (x + z) u (x + z h) u (x) u (x h) 1 C0 1 + z: (A.60) h h t 135 Setting z at h; and dividing by h; one has u (x + h) 2u (x) + u (x h) 1 C0 1 + h: (A.61) h2 t Hence, when (A.52) applies, we can use (A.58) at least for small h > 0: Thus, we can use the proof of Evans identically on (A.57), noting that H is convex and u" and u ~" satisfy (5) : This allows us to conclude that Z" (x; t) = 0 and so @t Z" (x; t) = 0 both on R [0; 1): From (4) this means that H " (u";x ) H " (~ u";x ) = 0: We can then use this in (A.56) to obtain vt (x; t) = R1 (") + R2 (") ; v (x; 0) = 0: (A.62) From the uniform convergence properties of Ri we conclude that jv (x; t)j CT "; on R [0; T ] : (A.63) Since this di¤erence can be made arbitrarily small, we have uniqueness. Another approach to proving these results is to follow [21]. ... Bibliography [1] S.M. Allen & J.W. 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