Hydrodynamic Limit of the Kawasaki Dynamics on the 1d-lattice with Strong, Finite-Range Interaction

Abstract

We derive the hydrodynamic limit of the Kawasaki dynamics for the one-dimensional conservative system of unbounded real-valued spins with arbitrary strong, quadratic and finite-range interactions. This significantly extends prior results for bounded interaction by Rezakhanlou and complements results obtained by H.T. Yau. The result is obtained by adapting two-scale approach of Grunewald, Otto, Villani and Westdickenberg combined with the authors’ recent approach on conservative systems with strong interactions.

Appendices

Appendix A. Basic Properties of the Grand Canonical Ensemble and the Canonical Ensemble

In this section, we provide useful auxiliary results for the GCE and CE, previously obtained in [15,16,17,18]. The results stated in this section hold under the conditions (2), (3), (4) and (5). Recall that a generalized GCE $\mu ^{\sigma }_N$ with external field $\sigma$ is defined by

$$\begin{aligned} \mu ^{\sigma }_N (\textrm{d}x) : = \frac{1}{Z} \exp \left( \sigma \sum _{i=1}^N x_i - H(x) \right) \textrm{d}x. \end{aligned}$$

Remark A.1

Again, the CE $\mu _{N,m}$ can be thought as a conditional probability distribution which emerges from the GCE $\mu _{N}^{\sigma }$ conditioned on the mean spin

$$\begin{aligned} \frac{1}{N} \sum _{i=1}^{N} x_i = m. \end{aligned}$$

More precisely, we have

$$\begin{aligned} \mu _N^{\sigma } \left( \textrm{d}x \mid \frac{1}{N}\sum _{i =1}^{N} x_i =m \right)&= \frac{1}{Z} \mathbbm {1}_{ \left\{ \frac{1}{N} \sum _{i=1}^{N} x_i =m \right\} }(x) \exp \left( \sigma m N - H(x) \right) \mathcal {L}^{N-1}(\textrm{d}x) \\&= \frac{1}{\widetilde{Z}} \mathbbm {1}_{ \left\{ \frac{1}{N} \sum _{i =1}^{N} x_i =m \right\} }\left( x\right) \exp \left( - H(x) \right) \mathcal {L}^{N-1}(\textrm{d}x) \\&= \mu _{N,m} (\textrm{d}x). \end{aligned}$$

The following statement tells that the variance of the mean spin of the modified GCE $\mu ^{\sigma }_N$ is well behaved.

Lemma A.2

(Lemma 1 in [18]). There exists a constant $C>0$, uniform in N and $\sigma$, such that

$$\begin{aligned} \frac{1}{C} \le \frac{1}{N} {{\,\textrm{var}\,}}_{\mu ^{ \sigma }_N} \left( \sum _{k=1}^N X_k \right) \le C. \end{aligned}$$

Define the (normalized) free energy $A_N : \mathbb {R} \rightarrow \mathbb {R}$ by

$$\begin{aligned} A_N(\sigma ) : = \frac{1}{N} \log \int _{\mathbb {R}^N} \exp \left( \sigma \sum _{i=1}^N x_i - H(x) \right) \textrm{d}x. \end{aligned}$$

The following lemma states that the free energy $A_N$ is uniformly strictly convex.

Lemma A.3

(Lemma 2 in [18]). There is a constant $C >0$, uniform in N and $\sigma$, such that

$$\begin{aligned} \frac{1}{C} \le \frac{\textrm{d}^2}{\textrm{d}\sigma ^2} A_N(\sigma ) \le C. \end{aligned}$$

Let us now introduce the definition of local, intensive and extensive functions.

Definition A.4

(Local, intensive and extensive functions/ observables). For a function $f : \mathbb {R}^{\mathbb {Z}} \rightarrow \mathbb {C}$, denote ${{\,\textrm{supp}\,}}f$ by the minimal subset of $\mathbb {Z}$ with $f(x) = f\left( x^{{{\,\textrm{supp}\,}}f} \right)$. We call f a local function if it has a finite support independent of N. A function f is called intensive if there is a positive constant $\varepsilon$ such that $|{{\,\textrm{supp}\,}}f | \lesssim N^{1-\varepsilon }$. A function f is called extensive if it is not intensive.

The following proposition claims the exponential decay of correlations for the GCE.

Proposition A.5

(Lemma 6 in [16]). There exists a constant $C>0$ such that for any intensive functions $f, g : \mathbb {R}^{N} \rightarrow \mathbb {R}$,

$$\begin{aligned}&\left| {{\,\textrm{cov}\,}}_{\mu _N ^{\sigma }} \left( f, g \right) \right| \lesssim \Vert \nabla f\Vert _{L^2 (\mu _N ^{\sigma })}\Vert \nabla g\Vert _{L^2 (\mu _N ^{ \sigma })} \exp \left( -C \text {dist} \left( {{\,\textrm{supp}\,}}f , {{\,\textrm{supp}\,}}g \right) \right) . \end{aligned}$$

The following moment estimate is a consequence of Proposition A.5.

Lemma A.6

(Lemma 3.2 in [17]). For each $k \ge 1$, there exists a constant $C = C(k)>0$ such that for any smooth function $f : \mathbb {R}^{\Lambda } \rightarrow \mathbb {R}$,

$$\begin{aligned} \mathbb {E}_{\mu _N^{ \sigma }} \left[ \left| f(X) - \mathbb {E}_{\mu _N^{ \sigma }} \left[ f(X) \right] \right| ^k \right] \le C(k) \Vert \nabla f \Vert _{\infty }^k. \end{aligned}$$

We now introduce the principle of equivalence of observables. We first relate the external field $\sigma$ of $\mu _N ^{\sigma }$ and the mean spin m of $\mu _{N,m}$ as follows:

Definition A.7

For each $m \in \mathbb {R}$, we choose $\sigma = \sigma _N(m) \in \mathbb {R}$ such that

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}\sigma } A_N( \sigma ) = m. \end{aligned}$$

(85)

Denoting $m_i : = \int x_i \mu _N ^{\sigma } (\textrm{d}x)$ for each $i \in [N]$, we equivalently get

$$\begin{aligned} m = \frac{\textrm{d}}{\textrm{d}\sigma } A_N ( \sigma ) = \frac{1}{N} \frac{ \int _{\mathbb {R}^N} \sum _{i=1}^{N} x_i \exp \left( \sigma \sum _{i=1}^{N} x_i - H(x) \right) \textrm{d}x }{ \int _{\mathbb {R}^N} \exp \left( \sigma \sum _{i=1}^{N} x_i - H(x) \right) \textrm{d}x } = \frac{1}{N}\sum _{i=1}^N m_i . \end{aligned}$$

The following proposition states the equivalence of ensembles results for GCE $\mu _N ^{\sigma }$ and CE $\mu _{N,m}$ with $\sigma$ and m related by (85).

Proposition A.8

(Theorem 2.7 in [15]). There exist constants $C,N_0>0$ such that for any intensive function $f : \mathbb {R}^{N} \rightarrow \mathbb {R}$ and $N \ge N_0$,

$$\begin{aligned} \left| \mathbb {E}_{\mu _N^{\sigma }} \left[ f \right] - \mathbb {E}_{\mu _{N,m}} \left[ f \right] \right| \le C \frac{|{{\,\textrm{supp}\,}}f|}{N} \Vert \nabla f \Vert _{\infty }. \end{aligned}$$

Finally, we state the exponential decay of correlations for the CE.

Proposition A.9

(Theorem 2.10 in [15]). There exist constants $C,N_0>0$ such that for any intensive functions $f, g : \mathbb {R}^{N} \rightarrow \mathbb {R}$ and $N \ge N_0$,

$$\begin{aligned}&\left| {{\,\textrm{cov}\,}}_{\mu _{N,m} } \left( f, g \right) \right| \\&\quad \le C \ \Vert \nabla f \Vert _{L^{\infty }(\mu _N^{\sigma })}\Vert \nabla g \Vert _{L^{\infty }(\mu _N^{\sigma })} \left( \frac{ |{{\,\textrm{supp}\,}}f | + |{{\,\textrm{supp}\,}}g | }{N} + \exp \left( -C\text {dist}\left( {{\,\textrm{supp}\,}}f, {{\,\textrm{supp}\,}}g \right) \right) \right) . \end{aligned}$$

Here, $\sigma$ and m are related by (85).

Appendix B. Derivatives of Coarse-Grained Hamiltonian

In this section, we provide the first and second derivative formula for the coarse-grained Hamiltonian. First, we state an explicit formula for $\bar{H}$, obtained in [25].

Lemma B.1

(Lemma 1 in [25]). For $z \in X$ with $Pz=0$ and $y \in Y$, define $H_{M}(z,y)$ by

$$\begin{aligned} H_{M} (z,y) : = \frac{1}{2}\langle z, (Id +M)z \rangle + \langle z, MNP^* y\rangle + \sum _{i=1}^N \psi _b (z_i + ( NP^* y )_i ), \end{aligned}$$

(86)

where $M=(M_{ij})_{1\le i,j\le n}$ is an interaction matrix in the Hamiltonian (cf. (1)). Then

$$\begin{aligned} \bar{H}(y) = \frac{1}{2} \langle y, (Id + PM NP^*) y \rangle _Y - \frac{1}{N}\log \int _{Px=0} \exp \left( -H_{M} (x,y) \right) \mathcal {L}(\textrm{d}x). \end{aligned}$$

Proof of Lemma B.1

For $x \in X$ with $Px=y$ and $y \in Y$, let $z = x - NP^*y$. Recalling the identity $PNP^* = Id_Y$, we have $Pz=0$. We then write the Hamiltonian H as

$$\begin{aligned} H(x)&= \sum _{i =1 }^{N} \left( \psi (x_i) +\frac{1}{2}\sum _{j : \ 1 \le |j-i| \le R } M_{ij}x_i x_j \right) \nonumber \\&= \frac{1}{2} \langle x , (\text {Id} + Mx \rangle + \sum _{i=1}^{N} \psi _b ( x_i) \nonumber \\&= \frac{1}{2} \langle z + NP^* y , (\text {Id} + M ) ( z + NP^* y) \rangle + \sum _{i=1}^{N} \psi _b (z_i + (NP^* y)_i ) \nonumber \\&= \frac{1}{2} \langle z , (\text {Id} + M) z \rangle + \langle z, NP^* y \rangle + \langle z , M NP^* y \rangle + \frac{1}{2} \langle NP^* y, (\text {Id} + M ) NP^* y \rangle \nonumber \\&\quad + \sum _{i=1}^{N} \psi _b (z_i + (NP^* y)_i ). \end{aligned}$$

(87)

Because $Pz = 0$, we have

$$\begin{aligned} \langle z , NP^* y \rangle = N \langle Pz, y \rangle _Y = 0. \end{aligned}$$

(88)

It also holds by $PNP^* = \text {Id}_Y$ that

$$\begin{aligned} \frac{1}{2} \langle NP^* y , (\text {Id} +M)NP^* y \rangle&= \frac{N}{2} \langle y , PNP^* y + PM NP^ * y \rangle _Y \nonumber \\&= \frac{N}{2} \langle y , (\text {Id} + PMNP^*) y \rangle _Y. \end{aligned}$$

(89)

Plugging (88) and (89) into (87) yields

$$\begin{aligned} H(x) = \frac{N}{2} \langle y, (\text {Id} + PM NP^*) y \rangle _Y + H_{M} (z, y) , \end{aligned}$$

and hence,

$$\begin{aligned} \bar{H}(y)&= - \frac{1}{N} \log \int _{Px = y} \exp \left( - H(x) \right) \mathcal {L} (\textrm{d}x) \\&= \frac{1}{2} \langle y, (\text {Id} + PM NP^*) y \rangle _Y - \frac{1}{N}\log \int _{Pz=0} \exp \left( -H_{M} (z,y) \right) \mathcal {L}(\textrm{d}z). \end{aligned}$$

$\square$

Next, we compute the first derivative of the coarse-grained Hamiltonian $\bar{H}$ using Lemma B.1.

Lemma B.2

For each $l \in [M]$, it holds that

$$\begin{aligned} \frac{\partial }{\partial y_l } \bar{H} (y) = \frac{1}{M} y_l + \frac{1}{N}\mathbb {E}_{\mu _{N,m}(\textrm{d}x|y)}\left[ \sum _{i=1}^N \sum _{j \in B(l)} M_{ij}X_i + \sum _{i \in B(l)} \psi _b ' (X_i)\right] . \end{aligned}$$

Proof of Lemma B.2

Recall that the inner product $\langle \cdot , \cdot \rangle _Y$ is given by

$$\begin{aligned} \langle x, y \rangle _Y = \frac{1}{M} \sum _{l =1}^M x_l y_l. \end{aligned}$$

First of all, noting that

$$\begin{aligned} \langle y, PMNP^* y \rangle _Y =\frac{1}{N} \sum _{l, n = 1}^{M}\sum _{i \in B(l), j \in B(n)} M_{ij} y_l y_n, \end{aligned}$$

we have

$$\begin{aligned} \frac{\partial }{\partial y_l} \left( \frac{1}{2} \langle y , (\text {Id}+ PMNP^* )y \rangle _Y \right) = \frac{1}{M} y_l + \frac{1}{N} \sum _{n=1}^M \sum _{i \in B(l), j \in B(n)} M_{ij} y_n. \end{aligned}$$

(90)

In addition, differentiating (86) yields

$$\begin{aligned}&\frac{\partial }{\partial y_l}\left( H_{M} (x,y) \right) \\&\quad = \frac{\partial }{\partial y_l} \left( \langle x, MNP^* y \rangle \right) + \frac{\partial }{\partial y_l} \left( \sum _{i=1}^N \psi _b ( x_i + (NP^* y)_i ) \right) \\&\quad = \sum _{i=1}^N \sum _{j \in B(l)} M_{ij} x_i + \sum _{i \in B(l)} \psi _b ' (x_i + (NP^* y)_i ) \\&\quad = \sum _{i=1}^N \sum _{j \in B(l)} M_{ij} (x_i + (NP^* y) _i )\\&\qquad - \sum _{i=1}^N \sum _{j \in B(l)} M_{ij} (NP^* y) _i + \sum _{i \in B(l)} \psi _b ' (x_i + (NP^* y)_i ) \\&\quad = \sum _{i=1}^N \sum _{j \in B(l)} M_{ij} (x_i + (NP^* y) _i ) + \sum _{i \in B(l)} \psi _b ' (x_i + (NP^* y)_i )\\&\qquad - \sum _{n=1}^M \sum _{ i \in B(n), j \in B(l)}M_{ij} y_n. \end{aligned}$$

As a consequence, we obtain

$$\begin{aligned}&\frac{\partial }{\partial y_l} \left( - \frac{1}{N}\log \int _{Px=0} \exp \left( -H_{M} (x,y) \right) \mathcal {L}(\textrm{d}x) \right) \nonumber \\&\qquad = \frac{1}{N} \frac{\int _{Px=0} \frac{\partial }{\partial y_l} \left( H_{M} (x,y) \right) \exp \left( -H_{M} (x,y) \right) \mathcal {L}(\textrm{d}x)}{\int _{Px=0} \exp \left( -H_{M} (x,y) \right) \mathcal {L}(\textrm{d}x)} \nonumber \\&\qquad = \frac{1}{N} \mathbb {E}_{\mu _{N,m}(\textrm{d}x|y)}\left[ \sum _{i=1}^N \sum _{j \in B(l)} M_{ij}X_i + \sum _{i \in B(l)} \psi _b ' (X_i) - \frac{1}{N} \sum _{n=1}^M \sum _{i \in B(n), j \in B(l)} M_{ij} y_n \right] . \end{aligned}$$

(91)

Combining (90) and (91) with symmetry of $M_{ij}$, i.e., $M_{ij} = M_{ji}$, we have the desired equation

$$\begin{aligned} \frac{\partial }{\partial y_l } \bar{H} (y) = \frac{1}{M} y_l + \frac{1}{N}\mathbb {E}_{\mu _{N,m}(\textrm{d}x|y)}\left[ \sum _{i=1}^N \sum _{j \in B(l)} M_{ij}X_i + \sum _{i \in B(l)} \psi _b ' (X_i)\right] . \end{aligned}$$

$\square$

The second derivatives of the coarse-grained Hamiltonian $\bar{H}$ follow from a similar calculations.

Lemma B.3

(Lemma 2 in [25]). For $l, n \in [M]$, we have

$$\begin{aligned} \left( {{\,\textrm{Hess}\,}}_Y \bar{H} (y) \right) _{ln}&= \delta _{ln} + \delta _{ln} \frac{1}{K} \int \sum _{i \in B(l)}\psi _b '' (x_i) \mu _{N, m} (\textrm{d}x | y) + \frac{1}{K} \sum _{i \in B(l), j \in B(n)}M_{ij} \\&\quad - \frac{1}{K} {{\,\textrm{cov}\,}}_{\mu _{N, m}(\textrm{d}x|y)} \left( \sum _{j \in B(l)} \left( \sum _{i=1}^{N} M_{ij} X_i + \psi _b ' (X_j) \right) , \right. \\&\quad \left. \sum _{j \in B(n)} \left( \sum _{i=1}^{N} M_{ij} X_i + \psi _b ' (X_j) \right) \right) . \end{aligned}$$

Appendix C. Criteria for the Logarithmic Sobolev Inequality

In this section, we state several standard criteria for deducing a LSI. For proofs we refer to the literature. For a general introduction and more comments on the LSI, we refer the reader to [2, 20, 21, 30].

Theorem C.1

(Tensorization Principle [11]). Let $\mu _1$ and $\mu _2$ be probability measures on Euclidean spaces $X_1$ and $X_2$, respectively. Suppose that $\mu _1$ and $\mu _2$ satisfy LSI$(\rho _1)$ and LSI$(\rho _2)$, respectively. Then the product measure $\mu _1 \bigotimes \mu _2$ satisfies LSI$(\rho )$, where $\rho = \min \{\rho _1, \rho _2\}$.

Theorem C.2

(Holley–Stroock Perturbation Principle [13]). Let $\mu _1$ be a probability measure on Euclidean space X and $\psi _b : X \rightarrow \mathbb {R}$ be a bounded function. Define a probability measure $\mu _2$ on X by

$$\begin{aligned} \mu _2 (\textrm{d}x) : = \frac{1}{Z} \exp \left( - \psi _b (x) \right) \mu _1 (\textrm{d}x). \end{aligned}$$

Suppose that $\mu _1$ satisfies LSI$(\rho _1)$. Then $\mu _2$ also satisfies LSI with constant

$$\begin{aligned} \rho _2 = \rho _1 \exp \left( - osc \psi _b \right) , \end{aligned}$$

where $osc \psi _b : = \sup \psi _b - \inf \psi _b$.

Theorem C.3

(Bakry–Émery criterion [1]) Let X be a N-dimensional Euclidean space and $H \in C^2 (X)$. Define a probability measure $\mu$ on X by

$$\begin{aligned} \mu (\textrm{d}x): = \frac{1}{Z} \exp \left( - H(x)\right) \textrm{d}x. \end{aligned}$$

Suppose that there is a constant $\rho >0$ such that ${{\,\textrm{Hess}\,}}H \ge \rho$, in other words

$$\begin{aligned} \langle v, {{\,\textrm{Hess}\,}}H(u) v \rangle \ge \rho |v|^2,\quad \forall v\in T_x X. \end{aligned}$$

Then $\mu$ satisfies LSI$(\rho )$.

Theorem C.4

(Otto-Reznikoff Criterion [28]) Let $X = X_1 \times \cdots \times X_N$ be a direct product of Euclidean spaces and $H \in C^2 (X)$. Define a probability measure $\mu$ on X by

$$\begin{aligned} \mu (\textrm{d}x) : = \frac{1}{Z} \exp \left( - H(x) \right) \textrm{d}x. \end{aligned}$$

Assume that

• For each $i \in [N]$, the conditional measures $\mu (\textrm{d}x_i | \bar{x}_i )$ satisfy LSI$(\rho _i)$ for any $\bar{x}_i := (x_1,\ldots ,x_{i-1},x_{i+1},\ldots ,x_N)\in X_1\times \cdots \times X_{i-1} \times X_{i+1} \times \cdots \times X_N$.

• For each $i \ne j \in [N]$ there is a constant $\kappa _{ij}>0$ such that

  $$\begin{aligned} \left| \nabla _i \nabla _j H(x) \right| \le \kappa _{ij}, \qquad \forall x \in X. \end{aligned}$$

  Here, $| \cdot |$ denotes the operator norm of a bilinear form.

• Define a symmetric matrix $A= (A_{ij})_{1 \le i, j \le N}$ by

  $$\begin{aligned} A_{ij} = {\left\{ \begin{array}{ll} \rho _i, \qquad &{} i=j, \\ - \kappa _{ij}, \qquad &{} i \ne j. \end{array}\right. } \end{aligned}$$

  Assume that there is a constant $\rho >0$ with

  $$\begin{aligned} A \ge \rho {{\,\textrm{Id}\,}}, \end{aligned}$$

  in the sense of quadratic forms.

Then $\mu$ satisfies LSI$(\rho )$.

Theorem C.5

(Two-Scale Criterion for LSI [9]). Let X and Y be Euclidean spaces. Consider a probability measure $\mu$ on X defined by

$$\begin{aligned} \mu (\textrm{d}x) : = \frac{1}{Z} \exp \left( -H(x)\right) \textrm{d}x. \end{aligned}$$

Let $P : X \rightarrow Y$ be a linear operator such that for some $N \in \mathbb {N}$,

$$\begin{aligned} PNP^* = {{\,\textrm{Id}\,}}_Y. \end{aligned}$$

Define

$$\begin{aligned}&\kappa : = \max \left\{ \langle {{\,\textrm{Hess}\,}}H(x) \cdot u, v \rangle : \right. \\ {}&\left. u \in Ran (NP^* P), v \in Ran ({{\,\textrm{Id}\,}}_X - NP^* P), |u|=|v|=1 \right\} . \end{aligned}$$

Assume that

• $\kappa < \infty$.

• There is $\rho _1>0$ such that the conditional measure $\mu (\textrm{d}x | Px=y)$ satisfies LSI$(\rho _1)$ for all $y \in Y$.

• There is $\rho _2>0$ such that the marginal measure $\bar{\mu }= P_{\#}\mu$ satisfies LSI$(\rho _2 N)$.

Then $\mu$ satisfies LSI$(\rho )$, where

$$\begin{aligned} \rho : = \frac{1}{2} \left( \rho _1 +\rho _2 + \frac{ \kappa ^2}{\rho _1} - \sqrt{ \left( \rho _1 + \rho _2 + \frac{\kappa ^2 }{\rho _1} \right) ^2 - 4 \rho _1 \rho _2 } \right) >0. \end{aligned}$$