Existence and regularity of optimal shapes for elliptic operators with drift

Abstract

This paper is dedicated to the study of shape optimization problems for the first eigenvalue of the elliptic operator with drift $L = -\Delta + V(x) \cdot \nabla$ with Dirichlet boundary conditions, where V is a bounded vector field. In the first instance, we prove the existence of a principal eigenvalue $\lambda _1(\Omega ,V)$ for a bounded quasi-open set $\Omega$ which enjoys similar properties to the case of open sets. Then, given $m>0$ and $\tau \ge 0$, we show that the minimum of the following non-variational problem

$$\begin{aligned} \min \Big \{\lambda _1(\Omega ,V)\ :\ \Omega \subset D\ \text {quasi-open},\ |\Omega |\le m,\ \Vert V\Vert _{L^\infty }\le \tau \Big \}. \end{aligned}$$

is achieved, where the box $D\subset {\mathbb {R}}^d$ is a bounded open set. The existence when V is fixed, as well as when V varies among all the vector fields which are the gradient of a Lipschitz function, are also proved. The second interest and main result of this paper is the regularity of the optimal shape $\Omega ^*$ solving the minimization problem

$$\begin{aligned} \min \Big \{\lambda _1(\Omega ,\nabla \Phi )\ :\ \Omega \subset D\ \text {quasi-open},\ |\Omega |\le m\Big \}, \end{aligned}$$

where $\Phi$ is a given Lipschitz function on D. We prove that the optimal set $\Omega ^*$ is open and that its topological boundary $\partial \Omega ^*$ is composed of a regular part, which is locally the graph of a $C^{1,\alpha }$ function, and a singular part, which is empty if $d<d^*$, discrete if $d=d^*$ and of locally finite ${\mathcal {H}}^{d-d^*}$ Hausdorff measure if $d>d^*$, where $d^*\in \{5,6,7\}$ is the smallest dimension at which there exists a global solution to the one-phase free boundary problem with singularities. Moreover, if D is smooth, we prove that, for each $x\in \partial \Omega ^{*}\cap \partial D$, $\partial \Omega ^*$ is $C^{1,1/2}$ in a neighborhood of x.

Appendix A. Extremality conditions and Lebesgue density

In this section we prove Proposition A.1, which we use in Proposition 5.12 to show that the Lagrange multiplier $\Lambda _u$ is strictly positive, but the result is of independent interest. For instance, it applies to optimal partition problems (see, for example, [20] and [17]). We first show that a function which is critical for the functional

$$\begin{aligned} J(u):=\int _D|\nabla u|^2e^{-\Phi }\,dx-\lambda \int _Du^2e^{-\Phi }\,dx, \end{aligned}$$

(A.1)

with respect to internal variations that is

$$\begin{aligned} \displaystyle \delta J(u)[\xi ]:=\lim _{t\rightarrow 0}J(u(x+t\xi (x)))=0\quad \text {for every vector field}\quad \xi \in C^\infty _c(D;{\mathbb {R}}^d), \end{aligned}$$

satisfies a monotonicity formula for the associated Almgren frequency function N(r). Now, by the argument of Garofalo and Lin (see [26]) the monotonicity of the frequency function implies that u cannot decay too fast around the free boundary points. If, in addition, u is a solution of $-{{\,\mathrm{div}\,}}(e^{-\Phi }\nabla u)=\lambda u e^{-\Phi }$ on the positivity set $\Omega _u=\{u>0\}$, we can use a Caccioppoli inequality to show that if the Lebesgue density of $\Omega _u$ is too small, then the decay of u on the balls of radius r should be very fast. This, in combination with the monotonicity of the Almgren’s frequency function, shows that the Lebesgue density of $\Omega _u$ should be bounded from below everywhere (and not only on the boundary of $\Omega _u$). In particular, there cannot be points of zero Lebesgue density for $\Omega _u$ in D.

Proposition A.1

Let $D\subset {\mathbb {R}}^d$ be a bounded open set and $\Phi \in W^{1,\infty } (D)$. Suppose that $\lambda \ge 0$ and $u\in H^1(D)$ is a nonnegative (non-identically-zero) function such that

• (a) u is a solution of the equation

  $$\begin{aligned} -{{\,\mathrm{div}\,}}(e^{-\Phi }\nabla u)=\lambda e^{-\Phi }u\quad \text {in}\ \Omega _u=\{u>0\}; \end{aligned}$$

  (A.2)

• (b) u satisfies the extremality condition

  $$\begin{aligned} \delta J(u)[\xi ]=0\quad \text {for every}\ \xi \in C^\infty _c(D;{\mathbb {R}}^d), \end{aligned}$$

  where J is given by (A.1) and its first variation in the direction $\xi$ is given by

  $$\begin{aligned} \delta J(u)[\xi ] := \int _D \Big [ 2 D\xi (\nabla u)\cdot \nabla u + \big (|\nabla u|^2 - \lambda u^2\big ) (\nabla \Phi \cdot \xi - {{\,\mathrm{div}\,}}\xi ) \Big ] e^{-\Phi } dx. \end{aligned}$$

  (A.3)

Then, $|D\backslash \Omega _u|=0$.

1.1 Reduction to the case $\lambda =0$

In this section we will show that it is sufficient to prove Proposition A.1 for $\lambda =0$. The general case will then follow by an elementary substitution argument. In the next lemma we deal with the first variation of the functional J.

Lemma A.2

Suppose that $D\subset {\mathbb {R}}^d$ is a bounded open set, $a:D\rightarrow {\mathbb {R}}$ is a given Lipschitz function such that $0<{\varepsilon }\le a\le {\varepsilon }^{-1}$ on D. Let $\lambda >0$ and let $\varphi \in H^2(D)$ be such that

$$\begin{aligned} -{{\,\mathrm{div}\,}}(a\nabla \varphi )=\lambda a\varphi \quad \text {in}\ D,\qquad {\varphi \ge {\varepsilon }>0}\quad \text {on}\ D. \end{aligned}$$

For any $u\in H^1(D)$, we set ${{\tilde{a}}}(x):=\varphi ^2(x)a(x)$, ${{\tilde{u}}}:=u/\varphi$,

$$\begin{aligned} J(u):=\int _{D}\left( |\nabla u|^2-\lambda u^2\right) a(x)\,dx\quad \text {and}\quad {{\tilde{J}}}(u):=\int _{D}|\nabla u|^2{{\tilde{a}}}(x)\,dx, \\ \delta J(u)[\xi ]:=\int _D \Big [ 2a D\xi (\nabla u)\cdot \nabla u - \left( |\nabla u|^2-\lambda u^2\right) {{\,\mathrm{div}\,}}(a\xi ) \Big ] dx, \\ \delta {{\tilde{J}}}(u)[\xi ]:=\int _D \Big [ 2{{\tilde{a}}} D\xi (\nabla u)\cdot \nabla u - |\nabla u|^2 {{\,\mathrm{div}\,}}({{\tilde{a}}}\xi ) \Big ] dx\quad \text {for any}\quad \xi \in C^\infty _c(D;{\mathbb {R}}^d). \end{aligned}$$

Then, for every $u\in H^1(D)$ and every $\xi \in C^\infty _c(D;{\mathbb {R}}^d)$, we have

$$\begin{aligned} \delta {{\tilde{J}}} ({{\tilde{u}}})[\xi ]=\delta J(u)[\xi ]-2\int _D\nabla \left( u \xi \cdot \nabla (\ln \varphi ) \right) \cdot \nabla u\,a\,dx+2\int _D\left( u \xi \cdot \nabla (\ln \varphi ) \right) \lambda a u\,dx. \end{aligned}$$

(A.4)

Proof

Notice that we may assume $u\in C^\infty (D)$. First we notice that an integration by parts gives

$$\begin{aligned} \delta {{\tilde{J}}}({{\tilde{u}}})[\xi ]&=\int _D 2\,\partial _i\xi _j\,\partial _i {{\tilde{u}}}\, \partial _j{{\tilde{u}}}\,\tilde{a}\,dx-\int _D|\nabla {{\tilde{u}}}|^2{{\,\mathrm{div}\,}}({{\tilde{a}}}\xi )\,dx\\&=-\int _D 2\,\xi _j\,\partial _i({{\tilde{a}}}\,\partial _i \tilde{u})\,\partial _j{{\tilde{u}}}\,dx-\int _D 2\,\xi _j\,\partial _i \tilde{u}\,\partial _{ij}{{\tilde{u}}}\,{{\tilde{a}}}\,dx-\int _D|\nabla \tilde{u}|^2{{\,\mathrm{div}\,}}({{\tilde{a}}}\xi )\,dx\\&=-\int _D 2\,\xi _j\,\partial _i({{\tilde{a}}}\,\partial _i \tilde{u})\,\partial _j{{\tilde{u}}}\,dx-\int _D {{\,\mathrm{div}\,}}({{\tilde{a}}}|\nabla \tilde{u}|^2\xi )\,dx=-\int _D 2\,\xi _j\,\partial _i({{\tilde{a}}}\,\partial _i {{\tilde{u}}})\,\partial _j{{\tilde{u}}}\,dx\\&=-\int _D 2(\xi \cdot \nabla {{\tilde{u}}}){{\,\mathrm{div}\,}}({{\tilde{a}}}\nabla {{\tilde{u}}})\,dx . \end{aligned}$$

and, analogously,

$$\begin{aligned} \delta J(u)[\xi ] =-\int _D 2(\xi \cdot \nabla u){{\,\mathrm{div}\,}}( a\nabla u)\,dx+\lambda \int _Du^2{{\,\mathrm{div}\,}}(a\xi )\,dx. \end{aligned}$$

Now, since

$$\begin{aligned} {{\,\mathrm{div}\,}}({{\tilde{a}}}\nabla {{\tilde{u}}})={{\,\mathrm{div}\,}}(a(\varphi \nabla u-u\nabla \varphi ))=\varphi {{\,\mathrm{div}\,}}(a\nabla u)-u{{\,\mathrm{div}\,}}(a\nabla \varphi )=\varphi ({{\,\mathrm{div}\,}}(a\nabla u)+\lambda a u), \end{aligned}$$

we get

$$\begin{aligned} \delta {{\tilde{J}}}({{\tilde{u}}})[\xi ]&=-2\int _D\xi \cdot (\nabla u-\frac{u}{\varphi }\nabla \varphi )\big ({{\,\mathrm{div}\,}}(a\nabla u)+\lambda a u\big )\,dx\\&=2\int _D\xi \cdot \nabla \varphi \frac{u}{\varphi }\big ({{\,\mathrm{div}\,}}(a\nabla u)+\lambda a u\big )\,dx-2\int _D(\xi \cdot \nabla u)\big ({{\,\mathrm{div}\,}}(a\nabla u)+\lambda a u\big )\,dx\\&=-2\int _D\nabla \left( \frac{\xi \cdot \nabla \varphi }{\varphi }u\right) \cdot \nabla u\,a\,dx+2\int _D\left( \frac{\xi \cdot \nabla \varphi }{\varphi }u\right) \lambda a u\,dx+\delta J(u)[\xi ], \end{aligned}$$

which is precisely (A.4). $\square$

Let now $D\subset {\mathbb {R}}^d$ and $u\in H^1(D)$ be as in Proposition A.1 for some $\lambda >0$. In order to prove that $|D\backslash \Omega _u|=0$, it is sufficient to prove that $|(D\cap B)\backslash \Omega _u|=0$ for any (small) ball $B\subset D$. Let now $x_0\in D$ and let $R>0$ be such that $\lambda _1(B_R(x_0),\nabla \Phi )=\lambda$. Such a radius exists, since the map $f(r):= \lambda _1(B_r(x_0),\nabla \Phi )$ is continuous, $f(0)=+\infty$ and $f(+\infty )=0$. Notice also that we may assume $\Phi$ to be defined on the entire space ${\mathbb {R}}^d$. Let $\varphi$ be the first eigenfunction on $B_R(x_0)$ and let $r=R/2$. Then, we can apply Lemma A.2 in the set $D\cap B_r(x_0)$ with $a=e^{-\Phi }$. Moreover, since u satisfies (A.2), we get that

$$\begin{aligned} \delta {{\tilde{J}}}({{\tilde{u}}})[\xi ]=\delta J(u)[\xi ]=0,\quad \text {for every}\ \xi \in C^\infty _c(D\cap B_r(x_0);{\mathbb {R}}^d), \end{aligned}$$

which proves that ${{\tilde{u}}}=u/\varphi$ satisfies hypothesis (b) for $\lambda =0$. Finally, in order to prove that ${{\tilde{u}}}$ satisfies hypothesis (a), we notice that on $\Omega _u=\Omega _{\tilde{u}}$ we have (in a weak sense)

$$\begin{aligned} {{\,\mathrm{div}\,}}({{\tilde{a}}}\nabla {{\tilde{u}}})=\varphi {{\,\mathrm{div}\,}}(a\nabla u)-u{{\,\mathrm{div}\,}}(a\nabla \varphi )=\varphi \left( {{\,\mathrm{div}\,}}(a\nabla u)+\lambda a u\right) =0. \end{aligned}$$

1.2 Proof of Proposition A.1 in the case $\lambda =0$

Let $\lambda =0$. Then we have

$$\begin{aligned}&J(u):=\int _D|\nabla u|^2e^{-\Phi }\,dx, \end{aligned}$$

(A.5)

$$\begin{aligned}&\delta J(u)[\xi ] := \int _D \Big [ 2 D\xi (\nabla u)\cdot \nabla u + |\nabla u|^2 (\nabla \Phi \cdot \xi - {{\,\mathrm{div}\,}}\xi ) \Big ] e^{-\Phi } dx. \end{aligned}$$

(A.6)

Let $x_0=0\in D$ and $\tau =\Vert \nabla \Phi \Vert _{L^\infty (D)}$. We set

$$\begin{aligned} H(r):=\int _{\partial B_r}u^2e^{-\Phi }d{\mathcal {H}}^{d-1},\qquad D(r):=\int _{B_r}|\nabla u|^2e^{-\Phi }dx\quad \text {and}\quad N(r):=\frac{rD(r)}{H(r)}. \end{aligned}$$

Step 1 Derivative of H. We calculate

$$\begin{aligned} H'(r)&=\frac{d-1}{r}H(r)+r^{d-1}\frac{d}{dr}\int _{\partial B_1}u^2(rx)e^{-\Phi (rx)}d{\mathcal {H}}^{d-1}(x)\\&=\frac{d-1}{r}H(r)+2\int _{\partial B_r}u\frac{\partial u}{\partial n}e^{-\Phi }d{\mathcal {H}}^{d-1}-\int _{\partial B_r}u^2(n\cdot \nabla \Phi )e^{-\Phi }d{\mathcal {H}}^{d-1}\\&=\frac{d-1}{r}H(r)+2\int _{B_r}|\nabla u|^2e^{-\Phi }dx-\int _{\partial B_r}u^2(n\cdot \nabla \Phi )e^{-\Phi }d{\mathcal {H}}^{d-1}, \end{aligned}$$

which we rewrite as

$$\begin{aligned} H'(r)=\frac{d-1}{r}H(r)+2 D(r)-H_\Phi (r). \end{aligned}$$

(A.7)

where we have set

$$\begin{aligned} H_\Phi (r):=\int _{\partial B_r}u^2(n\cdot \nabla \Phi )e^{-\Phi }d{\mathcal {H}}^{d-1}\quad \text {and}\quad |H_\Phi (r)|\le \tau H(r). \end{aligned}$$

Step 2 Equidistribution of the energy. Let $\phi _{\varepsilon }$ be a radially decreasing function such that $0\le \phi _{\varepsilon }\le 1$ on $B_r$, $\phi _{\varepsilon }=1$ on $B_{r(1-{\varepsilon })}$, $\phi _{\varepsilon }=0$ on $\partial B_r$ and $|\nabla \phi _{\varepsilon }|\le C(r{\varepsilon })^{-1}$. The vector field $\xi (x):=x\phi _{\varepsilon }(x)$ satisfies ${{\,\mathrm{div}\,}}\xi (x)=d\phi _{\varepsilon }(x)+x\cdot \nabla \phi _{\varepsilon }$ and $\partial _i\xi _j=\delta _{ij}\phi _{\varepsilon }(x)+x_j\partial _i\phi _{\varepsilon }(x)$. Since $\lambda =0$ we have

$$\begin{aligned} \delta J(u)[\xi ]&= \int _D \Big [ 2 D\xi (\nabla u)\cdot \nabla u + |\nabla u|^2 (\nabla \Phi \cdot \xi - {{\,\mathrm{div}\,}}\xi ) \Big ] e^{-\Phi } dx\\&= \int _D \Big [ 2 |\nabla u|^2\phi _{\varepsilon }+2(x\cdot \nabla u)(\nabla \phi _{\varepsilon }\cdot \nabla u) - |\nabla u|^2 (d\phi _{\varepsilon }(x)+x\cdot \nabla \phi _{\varepsilon }) \Big ] e^{-\Phi } dx\\&\qquad +\int _D|\nabla u|^2(\nabla \Phi \cdot x)\phi _{\varepsilon }e^{-\Phi } dx, \end{aligned}$$

and passing to the limit as ${\varepsilon }\rightarrow 0$, rearranging the terms and using the property (b), we get

$$\begin{aligned} 0&=-(d-2)\int _{B_r}|\nabla u|^2e^{-\Phi }dx+r\int _{\partial B_r}|\nabla u|^2e^{-\Phi } d{\mathcal {H}}^{d-1}\\&\qquad -2r\int _{\partial B_r}\left( \frac{\partial u}{\partial n}\right) ^2e^{-\Phi } d{\mathcal {H}}^{d-1}+\int _{B_r}|\nabla u|^2 (\nabla \Phi \cdot x) e^{-\Phi } dx, \end{aligned}$$

which we rewrite as

$$\begin{aligned} -(d-2)D(r)+rD'(r)= 2r\int _{\partial B_r}\left( \frac{\partial u}{\partial n}\right) ^2e^{-\Phi } d{\mathcal {H}}^{d-1}-rD_\Phi (r), \end{aligned}$$

where

$$\begin{aligned} D_\Phi (r):=\frac{1}{r}\int _{B_r}|\nabla u|^2 (\nabla \Phi \cdot x) e^{-\Phi } dx\qquad \text {and}\qquad |D_\Phi (r)|\le \tau D(r). \end{aligned}$$

Step 3 The derivative of N. We notice that N(r) is only defined for r such that $H(r)>0$. In what follows we fix $r_0>0$ such that $B_{r_0}(x_0)\subset D$ and $H(r_0)>0$. Since $u\in H^1(D)$, there is an interval $(a,b)\ni r_0$, on which $H>0$.

$$\begin{aligned} N'(r)&=\frac{D(r)H(r)+rD'(r)H(r)-rD(r)H'(r)}{H^2(r)}\nonumber \\&= \frac{D(r)H(r)+rD'(r)H(r)-r D(r)\left( \frac{d-1}{r}H(r)+2D(r)- H_\Phi (r)\right) }{H^2(r)}\nonumber \\&=\frac{-(d-2)D(r)H(r)+rD'(r)H(r)-2rD^2(r)+r D(r)H_\Phi (r)}{H^2(r)}\nonumber \\&=\frac{2r}{H^2(r)}\left( H(r)\int _{\partial B_r}\left( \frac{\partial u}{\partial n}\right) ^2e^{-\Phi } d{\mathcal {H}}^{d-1}-D^2(r)\right) +\frac{r \left( D(r)H_\Phi (r)-D_\Phi (r)H(r)\right) }{H^2(r)} \end{aligned}$$

(A.8)

Now we notice that, since u solves (A.2) on $\Omega _u$, we have

$$\begin{aligned} D(r)&=\int _{B_r}|\nabla u|^2e^{-\Phi }dx=\int _{\partial B_r}u\frac{\partial u}{\partial n}e^{-\Phi } d{\mathcal {H}}^{d-1}, \end{aligned}$$

and so, by the Cauchy-Schwarz inequality and (A.8) we obtain

$$\begin{aligned} N'(r)\ge \frac{r \left( D(r)H_\Phi (r)-D_\Phi (r)H(r)\right) }{H^2(r)}\ge -2\tau N(r). \end{aligned}$$

(A.9)

Step 4 A bound on N(r). Using the estimate (A.9) from the previous step we get that the function $\,r\mapsto e^{2\tau r}N(r)\,$ is non-decreasing in r and so

$$\begin{aligned} N(r)\le e^{2\tau (r_0-r)}N(r_0)\le e^{2\tau r_0}N(r_0)\quad \text {for every}\ a<r\le r_0. \end{aligned}$$

Step 5 Strict positivity and doubling inequality for H(r). By the step 4 we have

$$\begin{aligned} \frac{d}{dr}\left[ \log \left( \frac{H(r)}{r^{d-1}}\right) \right] =2\frac{N(r)}{r}-\frac{H_\Phi (r)}{H(r)}\le \frac{2e^{2\tau r_0}N(r_0)}{r}+\tau , \end{aligned}$$

(A.10)

and integrating we get

$$\begin{aligned} \log \left( \frac{H(r_0)}{r_0^{d-1}}\right) -\log \left( \frac{H(r)}{r^{d-1}}\right) \le \log \Big (\frac{r_0}{r}\Big )\,2e^{2\tau r_0}N(r_0)+\tau r_0,\quad \text {for every}\ a<r\le r_0. \end{aligned}$$

In particular, $H>0$ on every interval $[{\varepsilon }r_0,r_0]$ and so, $H>0$ on $(0,r_0]$ and we might take $a=0$. Moreover, integrating once again the inequality (A.10) from $r<r_0/2$ to 2r, we get

$$\begin{aligned} \log \left( \frac{H(2r)}{H(r)}\right) \le ((d-1)\log 2+\tau r_0)+2\log 2\,e^{2\tau r_0}N(r_0)\quad \text {for every}\ 0<r\le \frac{r_0}{2}. \end{aligned}$$

Taking $r_0\le 1$, there is a constant C, depending only on d and $\tau$, such that

$$\begin{aligned} H(2r)\le C\exp (CN(r_0))H(r)\quad \text {for every}\ 0<r\le \frac{r_0}{2}. \end{aligned}$$

(A.11)

Integrating once more in r we get

$$\begin{aligned} \int _{B_{2r}}u^2e^{-\Phi }\,dx\le C\exp (CN(r_0))\int _{B_{r}}u^2e^{-\Phi }\,dx\qquad \text {for every}\qquad 0<r\le \frac{r_0}{2}. \end{aligned}$$

(A.12)

Step 6. Caccioppoli inequality and conclusion. Let $r\in (0,r_0/2]$ and let $\phi \in C^\infty _0(B_{2r})$ be such that $\phi =1$ in $B_r$, $\phi =0$ on $\partial B_{2r}$, $0\le \phi \le 1$ and $|\nabla \phi |\le 2/r$ on $B_{2r}\backslash B_r$. Using the fact that u is a solution of $-{{\,\mathrm{div}\,}}(e^{-\Phi }\nabla u)=0$ in $\Omega _u$, we get the following Caccioppoli inequality:

$$\begin{aligned} \int _{B_r}|\nabla u|^2e^{-\Phi }\,dx&\le \int _{B_{2r}}|\nabla (u\phi )|^2e^{-\Phi }\,dx=\int _{B_{2r}}\left( u^2|\nabla \phi |^2+\nabla u\cdot \nabla (u\phi ^2)\right) e^{-\Phi }\,dx\nonumber \\&= \int _{B_{2r}}u^2|\nabla \phi |^2e^{-\Phi }\,dx-\int _{B_{2r}} u\phi ^2{{\,\mathrm{div}\,}}\left( e^{-\Phi }\nabla u)\right) \,dx\nonumber \\&= \int _{B_{2r}}u^2|\nabla \phi |^2e^{-\Phi }\,dx.\nonumber \\&\le \frac{4}{r^2}\int _{B_{2r}}u^2e^{-\Phi }\,dx. \end{aligned}$$

(A.13)

On the other hand, there are dimensional constants $C_d$ and ${\varepsilon }_d>0$ such that, if $|\Omega _u\cap B_r|\le {\varepsilon }_d|B_r|$, then the following inequality does hold (see [15, Lemma 4.4])

$$\begin{aligned} \int _{B_r}u^2\,dx\le C_d r^2\bigg (\frac{|\Omega _u\cap B_r|}{|B_r|}\bigg )^{2/d}\int _{B_r}|\nabla u|^2\,dx, \end{aligned}$$

which, taking $C:=C_d\exp (\max \Phi -\min \Phi )$, implies

$$\begin{aligned} \int _{B_r}u^2e^{-\Phi }\,dx\le C r^2\bigg (\frac{|\Omega _u\cap B_r|}{|B_r|}\bigg )^{2/d}\int _{B_r}|\nabla u|^2e^{-\Phi }\,dx. \end{aligned}$$

This, together with (A.13) and the doubling inequality (A.12), gives that there are constants $C_1$ and $C_2$, depending only on d and $\tau$ such that

$$\begin{aligned} \min \left\{ {\varepsilon }_d,C_1\exp (-C_2N(r_0))\right\} \le \frac{|\Omega _u\cap B_r|}{|B_r|}\quad \text {for every}\ 0<r\le \frac{r_0}{2}, \end{aligned}$$

where to be precise we recall that we assumed $r_0\le 1$. In particular, we have a lower density bound for $\Omega _u$ at every point of D, which implies that $|D\backslash \Omega _u|=0$ and concludes the proof.

$\square$