Scaling Limit of Symmetric Random Walk in High-Contrast Periodic Environment

Abstract

The paper deals with the asymptotic properties of a symmetric random walk in a high contrast periodic medium in $\mathbb Z^d$, $d\ge 1$. From the existing homogenization results it follows that under diffusive scaling the limit behaviour of this random walk need not be Markovian. The goal of this work is to show that if in addition to the coordinate of the random walk in $\mathbb Z^d$ we introduce an extra variable that characterizes the position of the random walk inside the period then the limit dynamics of this two-component process is Markov. We describe the limit process and observe that the components of the limit process are coupled. We also prove the convergence in the path space for the said random walk.

Appendix: Proofs of the Propositions

Proof of Proposition 4

From (33) we obtain the following system of uncoupled equations for the functions $q_j \Big (\frac{z}{\varepsilon }\Big ), \; z \in \varepsilon B^\sharp ,$ and for constants $\alpha _{0 j}$:

$$\begin{aligned}&\Big ((T^0_\varepsilon - I) q_j \Big (\frac{z}{\varepsilon }\Big )\Big ) (f_0(z) - f_j(z)) + \sum _{w \in \varepsilon \{ x_j \}^\sharp } v_\varepsilon (z, w) (f_j(z) - f_0(z)) \nonumber \\&\qquad = \alpha _{0j} (f_j(z) - f_0(z)), \quad j = 1, \ldots , M. \end{aligned}$$

Then, for every $j = 1, \ldots , M$, the function $q_j \Big (\frac{z}{\varepsilon }\Big )$ satisfies the equation

$$\begin{aligned} (T^0_\varepsilon - I) q_j \Big (\frac{z}{\varepsilon }\Big ) = \sum _{w \in \varepsilon \{ x_j \}^\sharp } v_\varepsilon (z, w) - \alpha _{0j} \mathbf{1}_B\Big (\frac{z}{\varepsilon }\Big ), \quad z \in \varepsilon B^{\sharp }, \end{aligned}$$

(50)

which is equivalent to the following equation on $B^\sharp$:

$$\begin{aligned} (P^0 - I) q_j (x) = \sum _{y \in \{ x_j \}^\sharp } v (x, y) - \alpha _{0j} \mathbf{1}_B(x), \quad x \in B^\sharp , \end{aligned}$$

(51)

where $\mathbf{1}_B(x)$ is the indicator function of $B^\sharp$, and $q_j(x)$ is Y-periodic. Using the Fredholm theorem we conclude that Eq. (51) has a unique solution if

$$\begin{aligned} \sum _{y \in \{ x_j \}^\sharp } v (x, y) - \alpha _{0j} \mathbf{1}_B \ \bot \text{ Ker } (P^0 - I)^*. \end{aligned}$$

(52)

Due to the irreducibility of $P^0$ on $B^\sharp$ we have $\mathrm {Ker}(P^0 - I)^*= \mathbf{1}_{B}$. Therefore, the orthogonality condition in (52) implies the following unique choice of constants $\alpha _{0j}$:

$$\begin{aligned} \alpha _{0j} \ = \ \frac{1}{|B|} \ \sum _{x \in B} \sum _{y \in \{ x_j \}^\sharp } v (x, y) \ge 0, \end{aligned}$$

(53)

where |B| is the cardinality of the set B. Thus $\alpha _{0j}$ is defined by (53) for every $j =1, \ldots , M$, and Eq. (51) has a unique up to an additive constant solution $q_j(x), \ x \in B^\sharp ,$ that is a bounded periodic function on the set $B^\sharp$. Proposition 4 is proved. $\square$

Proof of Proposition 3

We say that $y \sim x, \; x,y \in \mathbb Z^d$, if $p_0 (x,y) \ne 0$. Let $\Lambda _x$ be a finite set of $\xi \in \mathbb Z^d$ such that $x+\xi \sim x$. From now on we use the notation

$$\begin{aligned} p_0 (x, x+ \xi ) = p_\xi (x) \; \text{ for } \text{ all } \; x,\,\xi \in \mathbb Z^d, \; \text{ such } \text{ that } \; x \sim x+ \xi . \end{aligned}$$

Then

$$\begin{aligned} \sum _{\xi \in \Lambda _x} p_\xi (x)=1, \end{aligned}$$

and

$$\begin{aligned} (T^0_\varepsilon f)(z) \ = \ \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) f(z+ \varepsilon \xi ), \quad z \in \varepsilon B^{\sharp }. \end{aligned}$$

(54)

Using (29) we get for all $z \in \varepsilon B^{\sharp }$:

$$\begin{aligned} L_\varepsilon ^0 F_\varepsilon ^P (z) = \frac{1}{\varepsilon ^2} (T_\varepsilon ^0 - I) \left( f_0 (z) + \varepsilon \left( \nabla f_0 (z), h\Big (\frac{z}{\varepsilon }\Big ) \right) \right) + (T_\varepsilon ^0 - I) \left( \nabla \nabla f_0 (z), g\Big (\frac{z}{\varepsilon }\Big )\right) . \end{aligned}$$

(55)

Then the vector function $h\Big (\frac{z}{\varepsilon }\Big )$ should satisfy the relation

$$\begin{aligned} \frac{1}{\varepsilon ^2} (T_\varepsilon ^0 - I) \left( f_0 (z) + \varepsilon \left( \nabla f_0 (z), h\Big (\frac{z}{\varepsilon }\Big ) \right) \right) = O(1). \end{aligned}$$

(56)

Using (54) we rearrange the left-hand side of (56) as follows:

$$\begin{aligned}&\frac{1}{\varepsilon ^2} \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( f_0 (z+ \varepsilon \xi ) - f_0(z) \right) \nonumber \\&+ \frac{1}{\varepsilon } \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( \left( \nabla f_0 (z+\varepsilon \xi ), h\Big (\frac{z}{\varepsilon }+\xi \Big ) \right) - \left( \nabla f_0 (z), h\Big (\frac{z}{\varepsilon }\Big ) \right) \right) \nonumber \\&= \frac{1}{\varepsilon } \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( \nabla f_0 (z), \xi \right) + \frac{1}{\varepsilon } \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( \nabla f_0 (z), h\Big (\frac{z}{\varepsilon }+\xi \Big ) - h\left( \frac{z}{\varepsilon }\right) \right) + O(1)\nonumber \\&= \frac{1}{\varepsilon } \left( \nabla f_0 (z), \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( \xi + h\left( \frac{z}{\varepsilon }+\xi \right) - h\Big (\frac{z}{\varepsilon }\Big ) \right) \right) + O(1). \end{aligned}$$

(57)

Thus the periodic vector function h(x) should solve the equation

$$\begin{aligned} (P^0 - I) \left( l (x) + h(x) \right) =0, \quad x \in B^\sharp , \end{aligned}$$

(58)

where $l(x)=x$ is the linear function. The solvability condition for Eq. (58) reads

$$\begin{aligned} ((P^0 - I)l, \text{ Ker } (P^0 - I)^*) = ((P^0 - I) l ,\ \mathbf{1}_B) = \sum _{x \in B} \sum _{\xi \in \Lambda _{x}} p_\xi (x) \xi = 0. \end{aligned}$$

Since $p_{\xi }(x) = p_{-\xi }(x+\xi )$, this condition holds true, which implies the existence of the unique, up to an additive constant, periodic solution h(x) of Eq. (58).

We follow the similar reasoning to find an equation for the periodic matrix function $g(x), \ x \in B^\sharp$. Collecting in (55) all terms of the order O(1) and using relation (58) on the function h(x) we get:

$$\begin{aligned}&\frac{1}{\varepsilon ^2} \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( f_0 (z+ \varepsilon \xi ) - f_0(z) \right) \nonumber \\&\quad + \frac{1}{\varepsilon } \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( \left( \nabla f_0 (z+\varepsilon \xi ), h\Big (\frac{z}{\varepsilon }+\xi \Big ) \right) - \left( \nabla f_0 (z), h\Big (\frac{z}{\varepsilon }\Big ) \right) \right) \nonumber \\&\quad + \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left( \left( \nabla \nabla f_0 (z + \varepsilon \xi ), g\Big (\frac{z}{\varepsilon } + \xi \Big ) \right) - \left( \nabla \nabla f_0 (z), g\Big (\frac{z}{\varepsilon }\Big ) \right) \right) + O(\varepsilon )\nonumber \\&= \frac{1}{\varepsilon } \bigg ( \nabla f_0 (z), \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon } \Big ) \Big [ \xi + h\Big (\frac{z}{\varepsilon }+\xi \Big ) - h\Big (\frac{z}{\varepsilon }\Big ) \Big ] \bigg )\nonumber \\&\quad + \bigg ( \nabla \nabla f_0 (z), \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left[ \frac{1}{2} \xi \otimes \xi + \xi \otimes h\Big (\frac{z}{\varepsilon }+\xi \Big ) + \Big (g\left( \frac{z}{\varepsilon } + \xi \right) - g\Big (\frac{z}{\varepsilon } \Big ) \Big ) \right] \bigg ) + O(\varepsilon )\nonumber \\&= \bigg ( \nabla \nabla f_0 (z), \sum _{\xi \in \Lambda _{\frac{z}{\varepsilon }}} p_\xi \Big (\frac{z}{\varepsilon }\Big ) \left[ \frac{1}{2} \xi \otimes \xi + \xi \otimes h\Big (\frac{z}{\varepsilon }+\xi \Big ) \right] + (P^0 -I) g\Big (\frac{z}{\varepsilon }\Big ) \bigg ) + O(\varepsilon ). \end{aligned}$$

(59)

Let $\frac{z}{\varepsilon } = y \in B$, and denote by $\Phi (h)$ the following matrix function

$$\begin{aligned} \Phi (h)(y) = \frac{1}{2} \sum _{\xi \in \Lambda _y} p_\xi (y)\, \xi \otimes \xi + \sum _{\xi \in \Lambda _y} p_\xi (y)\, \xi \otimes h(y+ \xi ), \quad y \in B. \end{aligned}$$

(60)

In order to ensure the convergence in (32) we should find a matrix $\Theta$ and a periodic matrix function g(y) such that

$$\begin{aligned} \Phi (h)_{km}(y) + (P^0 - I) g_{km}(y) = \Theta _{km}, \end{aligned}$$

(61)

The solvability condition for (61) reads

$$\begin{aligned} (-\Phi (h)_{km} + \Theta _{km}, \ Ker (P^0 - I)^*) = (-\Phi (h)_{km} + \Theta _{km} ,\ \mathbf{1}_B) = 0, \end{aligned}$$

thus $\Theta$ is uniquely defined as follows:

$$\begin{aligned} \Theta _{k m} = \frac{1}{|B|} \sum _{y \in B} \Phi _{k m}(h) (y), \end{aligned}$$

and g(y) is a solution of Eq. (61). This solution is uniquely defined up to a constant matrix.

Proposition 7

The matrix $\Theta$ defined by

$$\begin{aligned} \Theta = \frac{1}{|B|} \sum _{y \in B} \Phi (h) (y) \quad \text{ with } \quad \Phi (h)(y) = \sum _{\xi \in \Lambda _y} p_\xi (y)\, \xi \otimes \left( \frac{1}{2} \xi + h(y + \xi ) \right) \end{aligned}$$

(62)

is positive definite, i.e. $(\Theta \eta , \eta )>0 \; \forall \eta \ne 0$.

Proof

Step 1. Here we show that

$$\begin{aligned} \sum _{\xi \in \Lambda _y} \partial _{-\xi } \big (a(y) \partial _{\xi } g(y)\big ) = -2(P^0 - I)g(y), \end{aligned}$$

(63)

where $\partial _{\xi } g(y) = g(y+\xi ) - g(y)$ for every $\xi$, and $a(y) = a_{\xi \xi }(y) = p_{\xi } (y)$ is the diagonal matrix. Using

$$\begin{aligned} a(y) \partial _{\xi } g(y) = p_\xi (y) (g (y+\xi ) - g(y)) \quad \text{ and } \; p_{\xi } (y- \xi ) = p_{-\xi } (y), \end{aligned}$$

we obtain (63):

$$\begin{aligned} \sum _{\xi \in \Lambda _y} \partial _{-\xi } a(y) \partial _\xi g (y)= & {} \sum _{\xi \in \Lambda _y} \left( p_\xi (y-\xi ) (g (y) - g(y- \xi )) - p_\xi (y) (g (y+\xi ) - g(y)) \right) \\= & {} \sum _{\xi \in \Lambda _y} \left( p_{-\xi } (y) (g (y) - g(y- \xi )) - p_{\xi } (y) (g (y+\xi ) - g(y)) \right) \\= & {} -2(P^0 - I) g(y). \end{aligned}$$

Step 2. From (58) and (63) it follows that the vector function $l+h$ satisfies the equation

$$\begin{aligned} \sum _{\xi \in \Lambda _y} \partial _{-\xi } a(y) \partial _\xi (l+h)(y) = 0, \quad y \in B. \end{aligned}$$

Consequently, for all $\eta \in \mathbb R^d$ we get

$$\begin{aligned} 0 = \bigg (\Big ( \sum _{y \in B} h(y) \otimes \sum _{\xi \in \Lambda _y} \partial _{-\xi } a(y) \partial _\xi (l+h)(y)\Big ) \eta , \eta \bigg ) = \bigg ( \sum _{y \in B} \sum _{\xi \in \Lambda _y} \partial _{\xi } h(y)\otimes a(y) \partial _\xi (l+h)(y) \eta , \eta \bigg ), \end{aligned}$$

(64)

where we have denoted

$$\begin{aligned} \Big \{\sum _{y \in B} \sum _{\xi \in \Lambda _y} \partial _{\xi } h(y)\otimes a(y) \partial _\xi (l+h)(y)\Big \}_{km} =\sum _{y \in B} \sum _{\xi \in \Lambda _y} \partial _{\xi } h_k(y) a(y) \partial _\xi (l+h)_m(y). \end{aligned}$$

Step 3. Let us check that the following quadratic form is positive definite:

$$\begin{aligned} \bigg ( \sum _{y \in B} \sum _{\xi \in \Lambda _y} \partial _{\xi } (l+h)(y)\otimes a(y) \partial _\xi (l+h)(y) \eta , \eta \bigg )>0 \quad \forall \eta \ne 0. \end{aligned}$$

(65)

To this end, taking into account the fact that $a(y) = \{p_\xi (y)\}$ is the diagonal matrix, we rearrange the left-hand side of (65) as follows

$$\begin{aligned} \bigg ( \sum _{y \in B} \sum _{\xi \in \Lambda _y} \partial _{\xi } (l+h)(y)\otimes a(y) \partial _\xi (l+h)(y) \eta , \eta \bigg ) = \sum _{y \in B} \sum _{\xi \in \Lambda _y}p_\xi (y) \bigg (\sum \limits _{k=1}^d \partial _{\xi } (l+h)_k(y)\eta _k\bigg )^2. \end{aligned}$$

Since $l(x)=x$, and h is a periodic function, the expression on the right-hand side is strictly positive for any $\eta \not =0$.

Subtracting (64) from (65) and using the relation $\partial _\xi l(y)=\xi$ we have

$$\begin{aligned} \bigg ( \sum _{y \in B} \sum _{\xi \in \Lambda _y} \partial _{\xi } l(y)\otimes a(y) \partial _\xi (l+h)(y) \eta , \eta \bigg ) =\bigg (\sum _{y \in B} \sum _{\xi \in \Lambda _y} p_\xi (y)\, \xi \otimes \left( \xi + h(y+ \xi ) - h(y) \right) \eta , \eta \bigg ) >0. \end{aligned}$$

(66)

Observe that

$$\begin{aligned} - \sum _{ y \in B} \sum _{\xi \in \Lambda _y} p_\xi (y)\, \xi \otimes h(y)= & {} \sum _{y \in B} \sum _{\xi \in \Lambda _y} p_{-\xi } (y+\xi ) (-\xi )\otimes h(y)\\= & {} \sum _{u \in B} \sum _{\xi \in \Lambda _u} p_{\xi } (u)\, \xi \otimes h(u+ \xi ); \end{aligned}$$

here we have set $u = y + \xi$ and uses the identity $p_\xi (y) = p_{-\xi }(u)$. Finally, the expression in (66) can be written as

$$\begin{aligned}&\bigg (\sum _{y \in B}\sum _{\xi \in \Lambda _y} p_\xi (y)\, \xi \otimes \left( \xi + 2 h(y+ \xi ) \right) \eta , \eta \bigg ) \nonumber \\&\quad = 2 \bigg (\sum _{y \in B} \sum _{\xi \in \Lambda _y} p_\xi (y)\, \xi \otimes \Big ( \frac{1}{2} \xi + h(y+ \xi ) \Big ) \eta , \eta \bigg ) \nonumber \\&\quad = 2 \bigg (\sum _{y \in B} \Phi (h)(y) \eta , \eta \bigg ) >0, \end{aligned}$$

and the positive definiteness of the matrix $\Theta$ follows. $\square$

This complete the proof of Proposition 3.