Critical Sets of Elliptic Equations with Rapidly Oscillating Coefficients in Two Dimensions

This theorem was more or less proved in [2, 3], although it is not stated explicitly. Also see related work in [1, 6, 4] and the references therein. We give an outline of the proof here.

Step 1. Let $u_{i}=x_{i}+\chi_{i}(x)$ for $i=1,2$, and $U=(u_{1},u_{2})$. Use the continuity and boundedness of $\chi_{i}$ to show that $U:\mathbb{R}^{2}\to\mathbb{R}^{2}$ is onto.

Step 2. Show that $U:\mathbb{R}^{2}\to\mathbb{R}^{2}$ is one-to-one and $U^{-1}$ is continuous. As a result, $U:\mathbb{R}^{2}\to\mathbb{R}^{2}$ is a homeomorphism. The smoothness condition (2.4) is not needed. See [2] for details.

Step 3. Let $\xi\in\mathbb{R}^{2}$ with $|\xi|=1$. Consider the function $u_{\xi}=\langle U,\xi\rangle$. Note that div$(A(x)\nabla u_{\xi})=0$ in $\mathbb{R}^{2}$. To prove $\text{\rm det}(I+\nabla\chi)>0$, it suffices to show that

for any $x\in\mathbb{R}^{2}$. To this end, fix $y_{0}\in\mathbb{R}^{2}$ and $r_{0}>0$. Let $\Omega=U^{-1}(B(y_{0},r_{0}))$. Note that $\text{\rm div}(A\nabla u_{\xi})=0$ in $\Omega$, and $g=u|_{\partial\Omega}$ is unimodal. This implies $|\nabla u_{\xi}(x_{0})|>0$, where $U(x_{0})=y_{0}$. See [3] for details.

Step 4. Use $\text{\rm det}(I+\nabla\chi)>0$ and a compactness argument to show that $\text{\rm det}(I+\nabla\chi)\geq\mu$, where $\mu>0$ depends only on $\lambda$, $\Gamma$ and $(\alpha,M_{\alpha})$. ∎

See Theorem 2.7 and Remark 2.8 in [18]. ∎

This is proved in [18, Theorem 3.1]. ∎

This is proved in [18, Theorem 3.4]. ∎

This is proved in [18, Theorem 3.5]. ∎

This follows from the observation that if $u\in\mathcal{F}(L,\varepsilon,r,0)$ and $v(x)=u(\theta x),$ then $v\in\mathcal{F}(L,\theta^{-1}\varepsilon,\theta^{-1}r,0)$ and

∎

Note that by (3.14),

Since $r^{-1}\varepsilon\geq\varepsilon_{0}$ and $A$ satisfies (1.2) and (1.4), the estimate (3.15) follows readily from [21] for the operator $\mathcal{L}_{1}$ (see [16, 11] for the case $d=2$ and [12] for the case of smooth coefficients). Indeed, the coefficient matrix $\widetilde{A}(x)=A(x/(r^{-1}\varepsilon))$ satisfies the Lipschitz condition $\eqref{smoothness}$ with $M\varepsilon_{0}^{-1}$ in the place of $M$. Moreover, the conditions $N^{*}(u,x,1)\leq 2L$ and $N^{*}(u,x,1/2)\leq L$ for $x\in B(0,1/2)$ implies that

where $C$ depends on $L$. The periodicity condition is not needed. ∎

In view of Lemma 3.4 we may assume $r=1$. By the definition of $\mathcal{E}(\ell-1+\delta_{0},\varepsilon,c_{0})$, it suffices to show that if $u\in\mathcal{F}(\ell-\delta_{0},\varepsilon,1,0)$, then

for any $x_{0}\in B(0,1/2)$, provided that $0<\varepsilon<\varepsilon_{0}$. By covering $B(0,1/4)$ with a finite number of balls $\{B(y_{j},c_{0}/4):j=1,2,\dots,k_{0}\}$, where $k_{0}\leq C(d)/(c_{0})^{d}$ and $y_{j}\in B(0,1/4)$, this would imply that $u\in\mathcal{F}(\ell-1+\delta_{0},\varepsilon,c_{0},y_{j})$ for $1\leq j\leq k_{0}$. As a result,

from which the estimate (3.16) with $r=1$ follows.

To see (3.17), we note that $N^{*}(u,x_{0},1)\leq 2(\ell-\delta_{0})$ and $N^{*}(u,x_{0},1/2)\leq\ell-\delta_{0}$ for any $x_{0}\in B(0,1/2)$. By Theorem 3.2 we have $N^{*}(u,x_{0},c_{0})\leq\ell-1+\delta_{0}$, where $c_{0}=\delta_{0}/(8\ell)$. Observe that if $\varepsilon$ is sufficiently small, we may use Theorem 3.1 to obtain $N^{*}(u,x_{0},2c_{0})\leq C(L)$. Applying Theorem 3.1 again gives $N^{*}(u,x_{0},c_{0}/2)\leq\ell-1+\delta_{0}$. ∎

By Lemma 3.4 we may assume $r=1$. Let $u\in\mathcal{F}(3/2,\varepsilon,1,0)$. Then $N^{*}(u,x_{0},1)\leq 3$ and $N^{*}(u,x_{0},1/2)\leq 3/2$ for any $x_{0}\in B(0,1/2)$. By Theorem 3.3 we obtain $|\nabla u(x_{0})|\neq 0$, if $0<\varepsilon<\varepsilon_{0}$. Thus $\mathcal{C}(u)\cap B(0,1/4)=\emptyset$ and consequently, $\mathcal{E}(3/2,\varepsilon,1)=0$. ∎

By translation and dilation it suffices to consider the case $x_{0}=0$ and $r=1$. To prove (4.2), we argue by contradiction. Suppose there exist sequences $\{\varepsilon_{j}\}\subset\mathbb{R}_{+}$ and $\{u_{j}\}\subset H^{1}(B(0,1))$ such that $\varepsilon_{j}\to 0$, $\text{\rm div}(A^{j}(x/\varepsilon_{j})\nabla u_{j})=0$ in $B(0,1)$ for some $A^{j}$ satisfying (1.2), (1.3), (1.4) and (2.6), $u_{j}$ is not constant, $\nabla u_{j}(y_{j})=0$ for some $y_{j}\in B(0,3/4)\setminus B(0,1/128)$,

and that

We may assume that $u_{j}(0)=0$ and

Since $N^{*}(u_{j},0,1)\leq 2\ell+2$, this implies that $\{u_{j}\}$ is bounded in $L^{2}(\partial B(0,1))$. It follows that $\{u_{j}\}$ is bounded in $L^{2}(B(0,1))$. Thus, in view of Theorem 2.3, by passing to a subsequence, we may assume that $u_{j}\to u_{0}$ weakly in $L^{2}(B(0,1))$ and strongly in $L^{2}(B(0,r))$ for any $0<r<1$, where $u_{0}$ is harmonic in $B(0,1)$. Moreover,

and

where $\chi^{j}$ denotes the first-order correctors for the matrix $A^{j}$.

Next, by letting $j\to\infty$, we obtain $u_{0}(0)=0$ and

Hence, $u_{0}$ is not constant. Moreover,

By the monotonicity of $N^{*}(u_{0},0,r)$ for harmonic functions, we obtain

It follows that $u_{0}$ is a homogeneous harmonic polynomial of degree $\ell$. Since $d=2$, this implies that $|\nabla u_{0}(x)|\neq 0$ for any $x\neq 0$. However, since $|\nabla u_{j}(y_{j})|=0$ and

in view of (4.4), we conclude that $|\nabla u_{0}(y_{j})|\to 0$ as $j\to\infty$. Since $1\geq|y_{j}|\geq(1/128)$, we obtain a contradiction. ∎

By Lemma 3.4 we may assume $r=1$. Let $u\in\mathcal{F}(\ell+\delta_{0},\varepsilon,1,0)$, where $\delta_{0}\in(0,1/4)$ is given by Lemma 4.1. Consider the cover

Let $\{B(y_{j},1/40):j=1,2,\dots,k_{0}\}$ be a Vitali subcover; i.e., $y_{j}\in\mathcal{C}(u)\cap B(0,1/4)$,