at time t, in dimension d. The index i varies in a countable set I. We call configuration the family, denoted generically by Φ: U (x i (t) x j (t))


 Elmer Harmon
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1 Notations In this chapter we investigate infinite systems of interacting particles subject to Newtonian dynamics Each particle is characterized by its position an velocity x i t, v i t R d R d at time t, in dimension d The index i varies in a countable set I We call configuration the family, denoted generically by Φ: Φ = x i, v i Particles, as said above, are subject to the classical dynamics x i t = U x i t x j t where U : R d R is the so called potential Unless differently stated, we shall simplify some aspect of our investigation to concentrate on the diffi culties coming from the infinite number of particles and assume U Cc R d namely twice continuously differentiable with compact support Denote by R > the radius of a ball B, R such that U = outside B, R As usual, we reformulate the system of secondorder equations as a system of firstorder ones { x i t = v i t v i t = U x i t x j t and we consider the Cauchy problem given by these equations and the initial conditions Intuitions about blowup x i t= = x i, v i t= = v i, i I The interaction potential U is smooth and compact support, hence we do not have troubles similar to those of point vortices However, since the number of particles in infinite, we need that the sum U x i t x j t has only a finite number of terms, for every i and t otherwise the sum of vectors could even have no meaning At time t = we impose this condition; are we sure that this local finiteness is maintained during the evolution? If not, we speak of blowup
2 In both the following intuitive examples consider the case when I = Z d and x i = i, namely partices start uniformly distributed on the lattice Consider initial velocities which increase when i is larger Assume U = It is not diffi cult to define values v i such that all particles are in B, at a given time t > Similarly, we may create infinite sectors S n Z d that reach B, at times t n, with lim n t n = We thus see not only that blowup is possible, without any restriction on U and velocities, but it may also happen immediately, namely a solution may not exist By the lattice Z d, the space R d is divided in hypercubes call them cubes of side one; color every second cube by red Define the velocities v i at the corners of each red cube in such a way that they are of intensity bounded, say v i and point in the direction of the interior of the red cube, and lead to a collision such that one particle gather almost all the kinetic energy of all the others This way, after a short time, we have much faster particles moving from the red cubes, one for every cube Repeating this construction in a selfsimilar way, we may guess it is possible to produce faster and faster particles in shorter and shorter times and thus concentrate in finite time infinitely many particles in B, Locally finite and uniform configuration spaces The object Φ = x i, v i belongs to the product space R d R d I However, the sum U x i x j should contain only a finite number of terms, otherwise the equations may have no meaning Thus it is necessary to restrict configurations Φ to the smaller set of locally finite configurations L f R d R d I defined as { L f = Φ = x i, v i : } {xi B,R} < for every R > Remark With a suitable metric, L f is a complete separable metric space and all functionals F : L f R of the form F Φ = f x i, v i Φ = x i, v i are continuous, where f is continuous bounded and f x, v = for x outside a bounded set We shall not use this remark below The space of configurations where we look for solutions will be a subspace of L f X L f R d R d I
3 In this section we make the simplest choice, which has strong limitations for applications but will lead to a first set of simple results Given Φ = x j, j I L f, set { } = Φ L f : sup x j x j + v j < j I Elements of are not too far from Φ and have not too large velocities, uniformly in j I Over this set we define the distance d x j, v j j I, x j, v j = sup x j I j x j, v j v j j I which makes, d a complete separable metric space The set is not a vector space and, d is not a Banach space, in spite of the shape of d which looks like a norm Following [], we shall change variables and reduce the problem to a Banach space 3 Change of variable In order to have a Banach space, it is suffi cient to change variable We set ξ j t = x j t x j ξ i = x j x j y ij = x i x j and consider the new system { ξ i t = v i t v i t = U ξ i t ξ j t + y ij and we consider the Cauchy problem given by these equations and the initial conditions ξ i t= = ξ i, v i t= = v i, i I Let us repeat the notations, adding the symbol Ψ for ξ j, v j j I : Ψ = Φ Φ ξ j,v j x j,v j x j, We call Y the set L f read in the new variables: { Y = Ψ = ξ i, v i : {ξi +x i B,R} < for every R > } 3
4 and, more important, we introduce the Banach space Y : { } Y = Ψ = ξ i, v i : Ψ := sup ξ i v i < With the norm, Y is a separabe Banach space We have Then we introduce the operator ξj B, v j j I i Y + Φ = B : Y R d R d I = v i, U ξ i ξ j + y ij and we recognize that our original Cauchy problem is equivalent to { Ψ t = B Ψ t Ψ = Ψ 4 About the number of particles Counting the number of particles in a set is fundamental, in this topic For the intuition, it is much better to use the orginal variables Φ = x j, v j for this purpose Definition For a R d, Φ = x j, v j j I L f and r, we set N a, Φ, r = j I {xj Ba,r} N Φ, r = sup a R d N a, Φ, r In the particular case of Φ = x j, j I L f, we simply write N r = N Φ, r Definition 3 We say that Φ has bounded density if for every r N Φ, r < 4
5 Example 4 If I = Z d and x i = i, then Φ = x j, j I has bounded density The following lemma is very useful Lemma 5 Given Φ, Φ = Y + Φ, and r Φ Φ for every i I we have N x i, Φ, r N x i, Φ, r + Φ Φ If Φ has bounded density and Ψ Y, then also Φ = Φ + Ψ has bounded density and N Φ, r N r + Ψ Proof One has x i x j x i x j + x i x i + x j x j x i x j + Φ Φ If j I\ {i} satisfies x i x j r, then it also satisfies x i x j r + Φ Φ This implies the inequality between the cardinality of points The final claim comes simply from taking Φ = Φ and the inequality for every s and i I 5 Properties of B in Y N x i, Φ, s N s A priori, we have defined B as an operator from Y or more generally from Y to R d R d I In fact, it maps Y into Y and it is also locally Lipschitz continuous Lemma 6 Assume that Φ has bounded density Then, for every Ψ, Ψ Y we have B Ψ Ψ U N R + Ψ B Ψ B Ψ Ψ Ψ D U N R + Ψ + N R + Ψ Proof B Ψ Ψ sup Ψ U sup = Ψ U sup U ξ i ξ j + y ij { ξi ξ j +y ij R } { xi x j R } Ψ U sup N x i, Φ, R Ψ U N R + Ψ 5
6 B Ψ B Ψ Ψ Ψ [ sup U ξi ξ j + y ij U ξ i ξ ] j + y ij Ψ Ψ D U sup ξi ξ i + ξj ξ j { ξi ξ j +y ij R } + { ξ i ξ j Ψ Ψ D U Ψ Ψ sup { ξi ξ j +y ij R } + { ξ i ξ j +y ij R } = Ψ Ψ D U Ψ Ψ sup Ψ Ψ D U Ψ Ψ sup { xi x j R } + { x i x j R } N xi, Φ, R + N x i, Φ, R Ψ Ψ D U Ψ Ψ N R + Ψ + N R + Ψ 6 Local well posedness and global results in In the space of bounded displacements ξ i and bounded velocities v i we may prove local well posedness The differential system, under the change of variables, in integral form is Ψ t = Ψ + that we investigate in the space C [, T ] ; Y t B Ψ s ds Theorem 7 Assume that Φ has bounded density Given Ψ Y or equivalently given Φ, there exists T > and a unique solution Ψ C [, T ] ; Y equivalently Φ C [, T ] ; Given r >, one can choose T > depending on r such that the previous result is true for all Ψ Y with Ψ r Proof Step For every T >, denote by Ψ the supremum norm sup t [,T ] Ψ t, whch makes C [, T ] ; Y a Banach space; we hope there is no danger of confusion due to the use of the same symbol for different objects Consider the map Γ : C [, T ] ; Y C [, T ] ; Y defined as Γ Ψ t = Ψ + t B Ψ s ds It is easy to check, based on Lemma 6 that Γ really maps C [, T ] ; Y into itself, for every choice of T > 6
7 Given r >, denote by Y T,r the closed ball or center zero and radius r in C [, T ] ; Y : the set of all Ψ C [, T ] ; Y such that sup Ψ t r t [,T ] Let us check that, for T small enough depending only on r, this ball is invariant by Γ We have, from Lemma 6, Γ Ψ t Ψ + Ψ + t t B Ψ s ds Ψ s U N R + Ψ s ds r + [ r U N R + 4r ] T hence r for every T T with T satisfying [ r U N R + 4r ] T = r Step For every T T, for every Ψ, Ψ Y T,r, from Lemma 6 we have Γ Ψ t Γ Ψ t t B Ψ s B Ψ s ds t Ψ s Ψ s D U N R + Ψ s + N R + Ψ s ds 4 D U N R + 4r t Ψ s Ψ s ds 4 D U N R + 4r T Ψ Ψ Hence for any T T such that 4 D U N R + 4r T < the map Γ is a contraction, in the complete metric space Y T,r Hence it has a fixed point, that is the unique solution claimed by the theorem 6 Criteria for global solutions Proposition 8 Assume that Φ has bounded density Given Φ, assume there exists C > such that, for every continuousin solution Φ on some interval [, T ] we have Ψ C, namely sup d Φ s, Φ C s [,T ] 7
8 Then the unique solution provided by Theorem 7 is global following conditions is siffi cient: Moreover, each one of the sup sup N x i, Φ s, R C s [,T ] sup sup x i s x i C s [,T ] sup sup v i s C s [,T ] Proof A unique local solution exists on an interval T satisfying to preserve Y T,C [ C U N R + 4C ] T C and to be a contraction 4 D U N R + 4C T < The value Φ T, however, satisfies the same condition d Φ T, Φ C, by assumption, hence we may solve the equation on the interval [T, T ] with the same value of T found above In a finite number of steps we cover any predefined interval of time We leave the reader to check that the other conditions are suffi cient 6 Global solution in d = Lemma 9 There exists C d > with the following property If Φ has bounded density, then N Φ, r C d N Φ, + r d Proof It is suffi cient to prove it for r Cover B a, r by C d r d balls of the form B a,, with suitable centers a Then N a, Φ, r a N a, Φ, N Φ, C d r d Changing C d if becessary, we get the result Theorem In d =, assume Φ has bounded density The the local solutions of Theorem 7 are global Proof By the previous lemma we have N r C + r 8
9 for a suitable constant C > If Φ is a solution on any time interval [, T ], then, using the bound of Lemma 6, we have Ψ t Ψ + Ψ + Ψ + t t t and thus, by Gronwall lemma, Ψ s U N R + Ψ s ds Ψ s + U C + R + Ψ s ds C U + R ds + Ψ t e +C U t Ψ + t t + C U Ψ s ds C U + R ds On any time interval [, T ] the assumption of Proposition 8 is satisfied, hence the solution is global in [, T ], hence on [, Remark In generic dimension d, let us say that Φ has strongly decaying density if N r C + r for a suitable constant C > In this case the proof of the previous theorem works and global existence holds Of course the initial conditions Φ allowd have also strongly decaying density, namely N Φ, r has at most linear growth, and this is very restrictive, in dimension d > 7 Translation invariant measures Except in dimension d =, we are not able to prove a globalintime result in the class Y However, similarly to the case of point vortices, it could be true that singularities are avoided for ae initial condition, with respect to a suitable measure Let us start the investigation of this topic The most obvious idea would be to use a potentially invariant measure However, the natural ones for this purpose, the Gibbs measures, having a Maxwell distribution of velocities, is not supported on Y ; more precisely Y has measure zero This is a main reason to investigate larger spaces than Y ; however, let us insist on Y becuase of its simplicity and replace the concept of invariant measure with the concept of translationinvariant measure These new objects are not as powerful as the previous ones but may lead to interesting results Consider the Polish space, d defined above, with the Borel σalgebra B A probability measure µ on, B is called translation invariant if τ a µ = µ 9
10 for every a R d, where τ a µ is the pushforward of µ under τ a and τ a will be now defined On R d, τ a is the map defined as τ a x = x + a It induces a map, still denoted by τ a, on R d R d I : τ a x i, v i = x i + a, v i Restrictred to, we see that τ a XΦ = XΦ This map is measurable, hence the pushforward of a measure on, d is well defined Equivalent is to ask that F τ a Φ µ dφ = F Φ µ dφ for every point a R d and bounded measurable functions F : R As a particular case of F let us consider F Φ = f x i, v i with f bounded measurable on R d R d, equal to zero for x > R for some R > We get f x i + a, v i µ dφ = f x i, v i µ dφ If µ is a Borel measure on, a Borel measure on Y = Φ is naturally defined as the push forward under the map Φ Ψ := Φ Φ Let us call µ Y such measure We may reformulate the theory of translation invariant measures µ on by means of measures on Y with a suitable property: Lemma The measure µ on is translation invariant if an only if the measure µ Y on Y satisfies G Ψ µ Y dψ = G τ a Ψ + τ a Φ Φ µ Y dψ Y or equivalently G τ a Ψ µ Y dψ = Y Y Y G Ψ + Ψ a µy dψ where Ψ a = Φ τ a Φ, for every bounded measurable functions G : Y R
11 Proof Denoting the function Φ G Φ Φ by F Φ, G Ψ µ Y dψ = G Φ Φ µ dφ = F Φ µ dφ Y X Φ = F τ a Φ µ dφ = G τ a Φ Φ µ dφ X Φ = G τ a Φ Φ + τ a Φ Φ µ dφ X Φ = G τ a Ψ + τ a Φ Φ µ Y dψ Y Remark 3 However, the intuition is better on, hence we leave the translation for investigations which profit from the Banach space property Remark 4 The vcomponent plays an auxiliary role in this framework A general way to construct translation invariant measure on, d is to construct them with such a property on the purely spatial component, and then take product measure on the vcomponent, for instance independent identically distributed velocities to all particles; with bounded distribution, to be included in If we denote the spatial component of by X x, that of Φ Φ by Φ x and the projection of µ on the spatial component by µ x, we have f x i + a µ x dφ x = f x i µ x dφ x X x Φ X x Φ for every bounded measurable compact support function f on R d Remark 5 Let I = Z d, Φ be given by x i = i, for every i Z d An example of translation invariant measure on the spatial component of, d is given by the convex combination µ dx = [, ] δ d x+q dx dq where I = Z d, x = x i, x = x i, x + q = x i + q, q R d For this measure we have f x i + a µ x dφ x = f x i + a δ x+q dx dq X x Φ [, ] d X x Φ = f i + q + a dq = f i + q + a dq [, ] d [, d ] = f x + a dx = f x dx R d R d independently of a R d
12 8 Evolution of translation invariant measures Recall from the proof of Theorem 7 that T can be taken the same for all Φ having the same distance from Φ Thus, if a translation invariant measure µ is supported on a ball B XΦ Φ, r in XΦ of center Φ and radius r, its time evolution is well defined for t [, T ], where T is a value defined by Theorem 7 with respect to r If we denote by Φ t; Φ, for t [, T ], the unique solution starting from Φ B XΦ Φ, r, call µt the pushforward of µ under the map Φ Φ t; Φ Lemma 6 µ t is translation invariant Proof Using the property τ a Φ t; Φ = Φ t; τ a Φ easy to check and the invariance of µ : F τ a Φ µ t dφ = F τ a Φ t; Φ µ dφ X Φ = F Φ t; τ a Φ µ dφ X Φ = F Φ t; Φ µ dφ X Φ = F Φ µ t dφ 9 Specific energy By energy of particle i in configuration Φ we mean e i, Φ = v i + U x i x j By energy of configuration Φ in the Borel set B of R d we mean e Φ, B = e i, Φ {xi B} Given a translation invariant measure µ, by specific energy we mean e µ = e Φ, W µ dφ
13 where W = [, ] d More generally we may define the average energy in B R d as e µ, B = e Φ, B µ dφ The setfunction B e µ, B is additive Lemma 7 e µ, B = e µ, B + a for every a R d Proof e Φ, B + a µ dφ = v i + U x i x j {xi B+a}µ dφ = v i + U x i a x j a {xi a B}µ dφ = v i + U x i x j {xi B}µ dφ = e Φ, B µ dφ The name specific energy can now be understood: Lemma 8 where W N = [ N, N + ] d e µ = lim N N + d e µ, W N Proof Decompose W N in N + d disjoint hypercubes and apply additivity of B e µ, B along with the previous lemma Now, let µ be a translation invariant measure supported on the ball B XΦ Φ, r and let µ t be its timeevolution, defined for t [, T ], also translation invariant by the lemma above Define the specific energy at time t as e t := e µ t Lemma 9 Assume that µ is translation invariant and supported on the ball B XΦ Φ, r Then the function t e t is constant 3
14 Proof Step By the previous lemma e t = lim N = lim N = lim N N + d e µ t, W N N + d e Φ, W N µ t dφ X Φ N + d e Φ t; Φ, W N µ dφ Heuristically, e Φ t; Φ, R d = e Φ, R d but both are infinite!, namely e Φ t; Φ, W N e Φ, W N for large N, hence N + d e Φ t; Φ, W N µ dφ N + d e Φ, W N µ dφ e The rigorous proof requires a control of the error We have e Φ t; Φ, W N = v i t + U x i t x j t {xi t W N } Since where g i t : = v i t we have = d v i t + dt e Φ t; Φ, W N = U x i t x j t + U x i t x j t = g i t U x i t x j t v i t + v j t v i t + v i + U x i t x j t v i t v j t U x i t x j t {xi t W N } {x i W N} U x i x j = e Φ, W N + AN Φ + B N Φ 4 {x i W N} + g i t {x i W N}
15 where A N Φ = v i t + B N Φ = g i t {x i W N} U x i t x j t {xi t W N } {x i W N} Thus A N Φ + B N Φ e t = e + lim N N + d µ dφ We have to prove that this limit is equal to zero Step We give only the idea of the computation We have AN Φ r + U N i, Φ t; Φ, R {xi t W N } {x i W N} Since d Φ, Φ r, for t [, T ] we have d Φ t; Φ, Φ R for a certain chosen R > r, hence N i, Φ t; Φ, R N R + R, and thus, denoting by C > a constant independent of N and Φ, we get AN Φ C {xi t W N } {x i W N} Since v i s R, xi t x i RT, hence only indexes i such that either x i W N\W N RT or x i t W N \W N RT may contribute to the sum; the volume of W N \W N RT is of order N + d and thanks to the assumption on N r the numebr of i s with the previous properties is also of order equal or less than N + d It follows that A N Φ N + d C N + Step 3 One has B N Φ = = i,j I;i j U x i t x j t v i t + v j t {x i W N} U x i t x j t v i t {x j W N} {x i W N} hence B N Φ U r i,j I;i j {x j W N} {x i W N} { xi t x j t R } 5
16 With arguments similar to those of Step we deduce that B N Φ is at most of order N + d and that B N Φ N + d C N + The results of Steps and 3 prove the final claim of Step References [] O E Lanford III, The classical mechanics of onedimensional systems of infinitely many particles, Comm Math Phys 9 968,
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