Normal approximation for the net flux through a random conductor

Abstract

We consider solutions of an elliptic partial differential equation in ${\mathbb R}^d$ with a stationary, random conductivity coefficient. The boundary condition on a square domain of width L is chosen so that the solution has a macroscopic unit gradient. We then consider the average flux through the domain. It is known that in the limit $L \rightarrow \infty$, this quantity converges to a deterministic constant, almost surely. Our main result is about normal approximation for this flux when L is large: we give an estimate of the Kantorovich–Wasserstein distance between the law of this random variable and that of a normal random variable. This extends a previous result of the author (Probab Theory Relat Fields, 2013. doi:10.1007/s00440-013-0517-9) to a much larger class of random conductivity coefficients. The analysis relies on elliptic regularity, on bounds for the Green’s function, and on a normal approximation method developed by Chatterjee (Ann Probab 36:1584–1610, 2008) based on Stein’s method.

Appendix: Estimates for the periodic Green’s function

1.1 $d \ge 3$: Proof of Lemma 4.4

Here we follow ideas used to prove a uniform decay estimate for Green’s functions in ${\mathbb R}^d$, as in Theorem 1.1 of [22] and Lemma 2.8 of [18]; the difference here is the periodicity, so we include a proof for completeness. Let $y \in D_L$ and let u(x) satisfy

$$\begin{aligned} - \nabla \cdot (a \nabla u) + \beta u = \delta _y - |D_L|^{-1}, \quad x \in D_L. \end{aligned}$$

(5.1)

in the weak sense. Here $\beta \ge 0$ is a constant, which we allow for the sake of generality; the Green function in Lemma 4.4 is $u(x) = G(x,y)$ with $\beta = 0$. Suppose that u also satisfies

$$\begin{aligned} |\{ x \in D_L\;|\; u(x) > 0 \}| \le \frac{1}{2} |D_L|. \end{aligned}$$

(5.2)

(If this is not the case, then we could apply the same argument to the function $-u$ instead.) Then, for any $k > 0$, the function $u_k(x) = \max (0,\min (u,k))$ satisfies

$$\begin{aligned} |\{ x \in D_L\;|\; u_k(x) \ne 0 \}| = |\{ x \in D_L\;|\; u(x) > 0 \}| \le \frac{1}{2} |D_L|. \end{aligned}$$

(5.3)

Since $||u_k ||_\infty \le k$, we observe that $u_k$ satisfies

$$\begin{aligned} \int _{D_L} \nabla u_k \cdot a \nabla u_k \,dx = \int _{D_L} \nabla u \cdot a \nabla u_k \,dx \le - \beta \int _{D_L} u u_k \,dx + 2k \le 2k. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{D_L} |\nabla u_k|^2 \,dx \le 2 k/a_* . \end{aligned}$$

(5.4)

Considering (5.3), we know there is a constant C, independent of k, L and $\beta$, such that

$$\begin{aligned} \left( \int _{D_L} |u_k|^q \,dx \right) ^{1/q} \le C \left( \int _{D_L} |\nabla u_k|^2 \,dx \right) ^{1/2} \end{aligned}$$

(5.5)

where $q = 2d/(d-2)$ is the critical Sobolev exponent. By scaling, this is a consequence of the Sobolev imbedding theorem and the Poincaré inequality for functions $v \in H^1_{per}(D_1)$ which also satisfy $|\{ x \in D_1\;|\; v(x) = 0 \}| \ge 1/2$ (for example, see Lemma 4.8 of [23]). By applying Chebychev’s inequality, then (5.5) and (5.4), we obtain the estimate

$$\begin{aligned} | \{ x \in D_L \;|\; u(x) \ge k \}| = | \{ x \in D_L \;|\; u_k(x) \ge k \}|\le & {} k^{-q}\int _{D_L} |u_k|^q \,dx\le C k^{-q/2}.\nonumber \\ \end{aligned}$$

(5.6)

This is a weak-$L^{p}(D_L)$ estimate on $u^+ = \max (u,0)$, for $p = q/2 = d/(d-2)$:

$$\begin{aligned} ||u^+ ||_{L^{p}_W(D_L)} = \sup _{t > 0} \;\;t \;| \{ x \in D_L \;|\; |u^+(x)| > t \}|^{1/p} \le C, \end{aligned}$$

(5.7)

where the constant C is independent of L and $\beta \ge 0$.

Now let $\alpha \in (1, p)$, $x_0 \in D_L$, $R < dist(x_0,y)$. The weak bound (5.7) implies that $u^+ \in L^\alpha (B_R(x_0))$. By using the identity

$$\begin{aligned} \int _{B_R} |u^+|^{\alpha }\,dx= & {} \alpha \int _0^\infty t^{\alpha -1} |\{ x \in B_R \;|\; u^+(x) \ge t \}| \,dt \le |B_R| s^\alpha \\&+\,\alpha \int _s^\infty t^{\alpha -1} |\{ x \in B_R \;|\; u^+(x) \ge t \}| \,dt \end{aligned}$$

and optimizing in s, we see that

$$\begin{aligned} ||u^+ ||_{L^{\alpha }(B_R)} \le C \left( \frac{p}{p-\alpha }\right) ^{1/\alpha }|B_R|^{\frac{p-\alpha }{p\alpha }}, \end{aligned}$$

(5.8)

where the constant C depends on $\alpha$ and p, but not on L or R or $\beta \ge 0$. Since $-\nabla \cdot (a \nabla u) + \beta u = -|D_L|^{-1}$ in $B_R$, the estimates of De Giorgi and Moser give us a bound on $u^+(x)$ in terms of $||u^+ ||_{L^\alpha (B_R(x_0))}$. Specifically, Theorem 4.1 of [23] (or Theorem 8.17 of [13]) implies that u is locally bounded and satisfies:

$$\begin{aligned} \sup _{x \in B_{R/2}(x_0)} u^+(x)\le & {} C R^{-d/\alpha } \left( \int _{B_R} (u^+(y))^{\alpha } \,dy \right) ^{1/\alpha } + C R^{2}|D_L|^{-1}, \end{aligned}$$

(5.9)

with a constant C that depends only on d, $a_*$, $a^*$, and $\alpha$. Note that in Theorem 4.1 of [23], the constant depends on $|\beta |R^2$. However, it is easy to see from the proof (method 1) that if $\beta$ is known to be non-negative, then the bound is independent of $\beta$, so the same bound holds under rescaling (as in Theorem 4.14 of [23]).

By combining (5.8) and (5.9) we have

$$\begin{aligned} \sup _{x \in B_{R/2}(x_0)} u^+(x) \!\le \! C R^{-d/\alpha } |B_R|^{\frac{p-\alpha }{p\alpha }} \!+\! C R^{2}|D_L|^{-1} \!\le \! C R^{-d/p} \!+\! C R^{2}L^{-d} \le C R^{2-d}, \end{aligned}$$

where the constant C depends on the dimension, but not on L, $\beta \ge 0$, R. In particular,

$$\begin{aligned} u^+(x) \le C \left( dist(x,y)\right) ^{2-d}. \end{aligned}$$

(5.10)

Now, assuming (5.2) holds for u (otherwise, replace u by $-u$), let us choose $r \le 0$ such that both

$$\begin{aligned} |\{ x \in D_L\;|\; u(x) > r \}| \le \frac{1}{2}|D_L| \quad \text {and} \quad |\{ x \in D_L\;|\; u(x) < r \}| \le \frac{1}{2}|D_L| \end{aligned}$$

hold. Consider the function $\bar{u} = r - u$ which satisfies

$$\begin{aligned} - \nabla \cdot (a \nabla \bar{u}) +\beta \bar{u} = - \delta _y + |D_L|^{-1} - \beta |r| \end{aligned}$$

and $|\{ x \in D_L\;|\; \bar{u}(x) > 0 \}| \le \frac{1}{2}|D_L|$. To the functions $\bar{u}_k = \max (0,\min (\bar{u},k))$ and $\bar{u}^+ = \max (0,\bar{u})$ we apply the same argument used to obtain (5.10). The result is:

$$\begin{aligned} \bar{u}^+(x) \le C \left( dist(x,y)\right) ^{2-d}. \end{aligned}$$

(5.11)

In deriving (5.9) for $\bar{u}^+$, we must use the fact that $\bar{u}$ is a subsolution of $- \nabla \cdot (a \nabla \bar{u}) +\beta \bar{u} = |D_L|^{-1}$ away from y, since $-\beta |r| \le 0$. That is,

$$\begin{aligned} \int _{B_R} \nabla \varphi \cdot a \nabla \bar{u}+ \beta \bar{u} \varphi \,dx \le |D_L|^{-1} \int _{B_R} \varphi \,dx \end{aligned}$$

holds for all $\varphi \in H^1_{0}(B_R)$ which satisfy $\varphi \ge 0$. Thus, Theorem 4.1 of [23] (or Theorem 8.17 of [13]) still applies. Apart from this detail, the argument is identical. By combining (5.10) and (5.11) we obtain

$$\begin{aligned} r \!-\! C \left( dist(x,y)\right) ^{2-d} \le r - \bar{u}^+(x) \!\le \! u(x) \!\le \! u^+(x) \le C \left( dist(x,y)\right) ^{2-d}\qquad \end{aligned}$$

(5.12)

On the other hand, (5.10) implies that

$$\begin{aligned} \int _{D_L} u^+(x) \,dx \le C L^2. \end{aligned}$$

We combine this with the fact that $\int _{D_L} u \,dx = 0$ to conclude that

$$\begin{aligned} 0= & {} \int _{D_L} u^+(x) \,dx + \int _{\{ r < u \le 0\}} u(x) \,dx + \int _{\{ u \le r \}} u(x) \,dx \nonumber \\\le & {} CL^2 + r |\{ x \in D_L \;|\; u(x) \le r \}| \nonumber \\= & {} CL^2 + r |\{ x \in D_L \;|\; \bar{u}(x) \ge 0 \}| \le C L^2 + r L^{d}/2. \nonumber \end{aligned}$$

Hence $|r| \le 2CL^{2 - d}$. Combining this with (5.12) we obtain $|u(x)| \le C dist(x,y)^{2-d}$, as desired.

1.2 $d=2$: Proof of Lemma 4.5

Lemma 4.5 relies on the following oscillation estimate, which is a version of Lemma 2.8(i) of [18]:

Lemma 5.1

Let $d =2$. For any $q \ge 1$, there is a constant $C > 0$ such that

$$\begin{aligned} R^{-2} \int _{B_R(x_0)} |G(x,y)- \bar{G}_R(y)|^q \,dx \le C \end{aligned}$$

holds for all $x_0 \in D_L$, $y \in D_L \setminus B_{2R}(x_0)$, $R > 0$, $L > 1$, where $\bar{G}_R(y)$ is the average of $G(\cdot ,y)$ over the ball $B_R(x_0)$.

Proof of Lemma 5.1

This is proved as in Lemma 2.8 of [18] (see part (i), Step 2) for the free-space Green’s function (see Step 2 in the proof therein); here we include the proof in the contiuum, periodic setting just for completeness. Fix $y \in D_L$. Let u(x) satisfy

$$\begin{aligned} - \nabla \cdot (a \nabla u ) + \beta u = \delta _y - |D_L|^{-1}, \end{aligned}$$

where $\beta \ge 0$ is a constant (as in the proof of Lemma 4.4, we have $G(x,y) = u(x)$ with $\beta = 0$. Let $\overline{u_R}$ be the average of u over the ball $B_R$. Without loss of generality, suppose $\overline{u_R} \ge 0$. For $k \ge 0$, define

$$\begin{aligned} u_k = \max ( \min (u, \overline{u_R} + k), \overline{u_R} - k). \end{aligned}$$

We claim that

$$\begin{aligned} \int _{D_L} |\nabla u_k|^2 \,dx \le \frac{2k}{a_*}. \end{aligned}$$

(5.13)

To see this, observe that for any constant $c \in {\mathbb R}$,

$$\begin{aligned} \int _{D_L} \nabla u_k \cdot a \nabla u_k \,dx= & {} \int _{D_L} \nabla (u_k + c) \cdot a \nabla u \,dx \nonumber \\ \!= & {} \! u_k(y) \!-\! |D_L|^{-1}\int _{D_L} u_k(x) \,dx - \beta \int _{D_L} u (u_k \!+\! c) \,dx.\qquad \end{aligned}$$

(5.14)

If $\overline{u_R} \in [0,k]$, let $c = 0$. Then $u(x)(u_k(x) + c) \ge 0$ at every point $x \in D_L$. Hence

$$\begin{aligned} \beta \int _{D_L} u (u_k + c) \,dx \ge 0. \end{aligned}$$

(5.15)

Therefore, (5.13) follows from (5.14). If $\overline{u_R} > k$, let $c = k - \overline{u_R}$. Then $u_k + c \ge 0$. Also, $u(x) > \overline{u_R} - k > 0$ must hold wherever $(u_k(x) + c) > 0$. Hence (5.15) still holds. Moreover, $0 \le u_k(x) + c \le 2k$, so again (5.13) follows from (5.14).

Now let $v(x) = u(x) - \overline{u_R}$. Let $v_k(x) = \max (\min (v(x),k),-k) = u_k(x) - \overline{u_R}$. Let $\overline{v_R}$ and $\overline{v_{k,R}}$ be the average of v and $v_k$ over $B_R$, respectively. Hence $\overline{v_R} = 0$. Then the goal is to bound

$$\begin{aligned} \left( R^{-2} \int _{B_R} |v|^q \,dx \right) ^{1/q}= & {} \left( R^{-2} \int _{B_R \cap \{ |v| \le k\}} |v_k|^q \,dx + R^{-2} \int _{B_R \cap \{ |v| > k\}} |v|^q \,dx \right) ^{1/q} \nonumber \\\le & {} C \left( R^{-2} \int _{B_R \cap \{ |v| \le k\}} |v_k - \overline{v_{k,R}}|^q \,dx\right) ^{1/q} + C |\overline{v_{k,R}}| \nonumber \\&+\, C \left( R^{-2} \int _{B_R \cap \{ |v| > k\}} |v|^q \,dx \right) ^{1/q}. \nonumber \end{aligned}$$

Since $\overline{v_R} = 0$, we have

$$\begin{aligned} |\overline{v_{k,R}}| \le 2 \left( R^{-2} \int _{B_R \cap \{|v| \ge k\}} |v|^q \,dx \right) ^{1/q}. \end{aligned}$$

Therefore,

$$\begin{aligned} \left( R^{-2} \int _{B_R} |v|^q \,dx \right) ^{1/q}\le & {} C \left( R^{-2} \int _{B_R} |v_k - \overline{v_{k,R}}|^q \,dx\right) ^{1/q} \nonumber \\&+\, C \left( R^{-2} \int _{B_R \cap \{ |v| > k\}} |v|^q \,dx \right) ^{1/q}. \end{aligned}$$

(5.16)

By the Sobolev inequality and then (5.13), we know that for any $s \in [1, \infty )$ there is a constant $C_s$ (depending only on s) such that

$$\begin{aligned}&\left( R^{-2} \int _{B_R} |v_k - \overline{v_{k,R}}|^s \,dx \right) ^{1/s} \le C_s \left( \int _{B_R} |\nabla v_k|^2 \,dx \right) ^{1/2}\\&\quad = C_s \left( \int _{B_R} |\nabla u_k|^2 \,dx \right) ^{1/2} \le C k^{1/2}. \end{aligned}$$

To estimate the last integral appearing in (5.16) we use

$$\begin{aligned} \int _{B_R \cap \{ |v| > k\}} |v|^q \,dx= & {} \int _0^\infty q t^{q - 1}|\{ |v| \ge \max (t,k) \}| \,dt\\\le & {} |\{ |v| \ge k \}| k^{q} + q \int _k^\infty t^{q - 1} |\{ |v| \ge t \}| \,dt, \end{aligned}$$

and

$$\begin{aligned} |\{ |v| \ge k \}| \le |\{ |v_k| \ge k \}| \le k^{-s} \int _{B_R} |v_k|^s \,dx. \end{aligned}$$

Let $s > 2q$. Then

$$\begin{aligned} \int _{B_R} |v_k|^s \,dx\le & {} C \int _{B_R} |v_k - \overline{v_{k,R}}|^s \,dx + C R^2 (\overline{v_{k,R}})^s \nonumber \\\le & {} C \int _{B_R} |v_k - \overline{v_{k,R}}|^s \,dx + C R^2\left( R^{-2} \int _{B_R} |v|^q \,dx \right) ^{s/q} \nonumber \\\le & {} C R^2 k^{s/2} + C R^2\left( R^{-2} \int _{B_R} |v|^q \,dx \right) ^{s/q}. \nonumber \end{aligned}$$

So, if $I_q = (R^{-2} \int _{B_R} |v|^q \,dx)^{1/q}$, we have

$$\begin{aligned} \int _{B_R} |v_k|^s \,dx \le C R^2 k^{s/2} + C R^2 I_q^s \end{aligned}$$

and $|\{ |v| \ge k \}| \le R^2 k^{-s/2} + C k^{-s} R^2 I_q^s$.

Combining these bounds and returning to (5.16), we obtain

$$\begin{aligned} I_q\le & {} C k^{1/2} + C R^{-2/q} \left( |\{ |v| \ge k \}| k^{q } + q \int _k^\infty t^{q - 1} |\{ |v| \ge t \}| \,dt \right) ^{1/q} \nonumber \\\le & {} C k^{1/2} + C R^{-2/q} \left( R^2 k^{-s/2}k^q + C k^{-s} R^2 I_q^s k^{q}\right) ^{1/q} \nonumber \\&+\,C R^{-2/q} \left( q \int _k^\infty t^{q - 1} |\{ |v| \ge t \}| \,dt \right) ^{1/q} \nonumber \\\le & {} C k^{1/2} + C R^{-2/q} \left( R^2 k^{-s/2}k^q + C k^{-s} R^2 I_q^s k^{q}\right) ^{1/q} \nonumber \\&+\, C R^{-2/q} \left( q \int _k^\infty t^{q - 1} (R^2 t^{-s/2} + C t^{-s} R^2 I_q^s) \,dt \right) ^{1/q} \nonumber \\\le & {} C k^{1/2} + C k^{1 - s/(2q)} + C I_q^{s/q} k^{1 - s/q}. \end{aligned}$$

(5.17)

By choosing $k = \alpha I_q$ with $\alpha > 0$ sufficiently large, we see that this implies $I_q \le C$. $\square$

Now we continue with the proof of Lemma 4.5. By assumption, $dist(x_0,y) > 2R$. Let $\varphi$ be a smooth function supported in $B_{2R}(x_0)$ and satisfying: $0 \le \varphi (x) \le 1$ for all x, $\varphi (x) = 1$ for $x \in B_{R}(x_0)$, and $|\nabla \varphi | \le C/R$. Applying Lemma 3.2 to $u(x) = G(x,y)$ with this choice of $\varphi$, we conclude

$$\begin{aligned} \int _{B_{R}} |\nabla u |^2 \, dx\le & {} K_1 |D_L|^{-1} \int _{B_{2R}} (u - b) \varphi ^2 \,dx -K_1 \beta \int _{B_{2R}} u(u-b) \varphi ^2 \,dx \nonumber \\&+\, K_2 \int _{B_{2R}} |\nabla \varphi |^2 (u- b)^2\,dx. \end{aligned}$$

(5.18)

If we choose $b = \left( \int _{B_{2R}} \varphi ^2 \,dx \right) ^{-1} \int _{B_{2R}} u \varphi ^2 \,dx$ then Jensen’s inequality implies $\int _{B_{2R}} u(u - b) \varphi ^2 \,dx \ge 0$. Therefore, since $\beta \ge 0$,

$$\begin{aligned} \int _{B_{R}} |\nabla u |^2 \, dx\le & {} K_1 |D_L|^{-1} \int _{B_{2R}} (u - b) \varphi ^2 \,dx + K_2 \int _{B_{2R}} |\nabla \varphi |^2 (u- b)^2\,dx \nonumber \\\le & {} K_1 R^{-2} \int _{B_{2R}} |u - b| \,dx + C R^{-2} \int _{B_{2R}} (u- b)^2\,dx. \end{aligned}$$

(5.19)

On the other hand, if $\bar{u}$ denotes the average of u(x) over $B_{2R}(x_0)$, we know from Lemma 5.1 that

$$\begin{aligned} R^{-2} \int _{B_{2R}} |u - \bar{u}| \,dx \le C. \end{aligned}$$

Hence

$$\begin{aligned} |\bar{u} - b| \!\le \!\left( \int _{B_{2R}} \varphi ^2 \,dx \right) ^{-1} \int _{B_{2R}} |\bar{u} \!-\! u(x)| \varphi ^2 \,dx \le C R^{-2} \int _{B_{2R}} |\bar{u} \!-\! u(x)| \varphi ^2 \,dx \le C. \end{aligned}$$

Applying Lemma 5.1 again, we obtain

$$\begin{aligned} R^{-2} \int _{B_{2R}} (u- b)^2\,dx \le C R^{-2} \int _{B_{2R}} (u- \bar{u})^2\,dx + C R^{-2} \int _{B_{2R}} (\bar{u}- b)^2\,dx \le C. \end{aligned}$$

Similarly,

$$\begin{aligned} R^{-2} \int _{B_{2R}} |u- b| \,dx \le C R^{-2} \int _{B_{2R}} |u- \bar{u}| \,dx + C R^{-2} \int _{B_{2R}} (\bar{u}- b)\,dx \le C. \end{aligned}$$

In view of (5.19) and the fact that C is independent of R, L and $\beta \ge 0$, we have proved the desired result. $\square$