Neumann Laplacian in a Perturbed Domain

Abstract

We consider a domain with a small compact set of zero Lebesgue measure removed. Our main result concerns the spectrum of the Neumann Laplacian defined on such domain. We prove that the spectrum of the Laplacian converges in the Hausdorff distance sense to the spectrum of the Laplacian defined on the unperturbed domain.

5. Appendix

In this section we give the proofs of Lemmas 4.1,  4.2,  4.3 and 4.4.

1.1 Proof of Lemma 4.1

Let $(x_0, y_0)\in \omega _{\tilde{K}}$. Assume the validity of the first condition of property$^*$. Without loss of generality assume that $l(x_0)\cap \{y\le y_0\}$ has no intersection with $\tilde{K}$. Let $y_1(x_0)< y_0$ be a point of intersection of $l(x_0)$ and the boundary of $\omega$. One can easily check that for almost any $(x_0, y_0)\in {\omega _{\tilde{K}}}$ there exists $y_2(x_0)\in (y_1(x_0), y_0)$ such that

$$\begin{aligned} |u(x_0, y_0)|\le \frac{1}{\sqrt{y_0-y_1(x_0)}}\sqrt{\int _{y_1(x_0)}^{y_0}|u(x_0, z)|^2\,d z}. \end{aligned}$$

Therefore

$$\begin{aligned}&|u(x_0, y_0)|^2= \left| u(x_0, y_2(x_0))+\int _{y_2(x_0)}^{y_0}\frac{\partial u}{\partial z}(x_0, z)\,d z\right| ^2\\&\quad \le 2|u(x_0, y_2(x_0))|^2+ 2(y_0-y_2(x_0))\int _{y_2({x_0})}^{{y_0}}\left| \frac{\partial u}{\partial z}({x_0}, z)\right| ^2\,d {z}\\&\quad \le \frac{2}{y_0-y_1(x_0)}\int _{y_1(x_0)}^{y_0}|u(x_0, z)|^2\,d z + 2(y_0-y_2(x_0))\int _{y_2(x_0)}^{y_0}\left| \frac{\partial u}{\partial z}(x_0, z)\right| ^2\,d z\\&\quad \le \frac{2}{\mathrm {dist}((x_0, y_0), \partial {\omega })}\int _{y_1(x_0)}^{y_0}|u(x_0, z)|^2\,d z + 2 \mathrm {diam}({\omega })\int _{y_1(x_0)}^{y_0}\left| \frac{\partial u}{\partial z}(x_0, z)\right| ^2\,d z\\&\quad \le \frac{2}{\mathrm {dist}((x_0, y_0), \partial {\omega })}\int _{l(x_0)}|u(x_0, z)|^2\,d z + 2 \mathrm {diam}({\omega })\int _{l(x_0)}\left| \frac{\partial u}{\partial z}(x_0, z)\right| ^2\,d z, \end{aligned}$$

where $\mathrm {diam}({\omega })$ is the diameter of ${\omega }$ and $\mathrm {dist}((x_0, y_0), \partial {\omega })$ is the distance between $(x_0, y_0)$ and the boundary of ${\omega }$. This proves (4.5).

The case when $(x_0, y_0)$ satisfies second condition of property$^*$ can be studied similarly. Repeating the same ideas one gets the validity of (4.6). $\square$

1.2 Proof of Lemma 4.2

Let us estimate separately the curvilinear integrals $\int _{\gamma ^i}|v|_{\partial \gamma ^i}^2\,\text {d} \mu ,\,i=1,2,3,4$, where $\gamma ^i$ are given in polar coordinates as follows

$$\begin{aligned} \gamma ^1=\left\{ r=\tau , \frac{\pi }{4}\le \varphi \le \frac{3\pi }{4}\right\} , \,\, \gamma ^2=\left\{ r=\tau , \frac{3\pi }{4}\le \varphi \le \frac{5\pi }{4}\right\} ,\\ \gamma ^3=\left\{ r=\tau , \frac{5\pi }{4}\le \varphi \le \frac{7\pi }{4}\right\} ,\,\, \gamma ^4=\left\{ r=\tau , -\frac{\pi }{4}\le \varphi \le \frac{\pi }{4}\right\} \end{aligned}$$

and $|v|_{\gamma ^i}$ is the trace of v on $\gamma ^i$.

Let us start from $\int _{\gamma ^1}|v|_{\gamma ^1}^2\,\text {d} \mu$. We have

$$\begin{aligned}&\int _{\gamma ^1} |v|_{\gamma ^1}^2\,\text {d} \mu =\tau \int _{-\tau /\sqrt{2}}^{\tau /\sqrt{2}}\frac{1}{\sqrt{\tau ^2-x^2}}|v(x, \sqrt{\tau ^2-x^2})|^2\,\text {d} x\\&\quad \le \sqrt{2} \int _{-\tau /\sqrt{2}}^{\tau /\sqrt{2}}|v(x, \sqrt{\tau ^2-x^2})|^2\,\text {d} x. \end{aligned}$$

Since the domain obviously satisfies property$^*$ then in view of estimate 4.5 applied for $(x, \sqrt{\tau ^2-x^2}))$ with $|x|\le \tau /\sqrt{2}$, one gets

$$\begin{aligned}&\int _{\gamma ^1} |v|_{\gamma ^1}^2\,d \mu \le \sqrt{2}C^2 \int _{-\tau /\sqrt{2}}^{\tau /\sqrt{2}} \left( \int _{l(x)}\left| \frac{\partial v}{\partial z}(x, z)\right| ^2\,d z+ \int _{l(x)}|v(x, z)|^2\,d z \right) \,d x\nonumber \\&\quad \le \sqrt{2} C^2 \int _{(\Omega \setminus {\mathbb {B}})\cap \{|x|\le \tau /\sqrt{2}\}}|v(x, z)|^2\,\text {d} x\,\text {d} z\nonumber \\&\qquad +\sqrt{2} C^2 \int _{(\Omega \setminus {\mathbb {B}})\cap \{|x|\le \tau /\sqrt{2}\}}\left| \frac{\partial v}{\partial z}(x, z)\right| ^2\,\text {d} x\,\text {d} z\nonumber \\&\quad \le \sqrt{2} C^2 \int _{\Omega \setminus {\mathbb {B}}}|v(x, z)|^2\,\text {d} x\,\text {d} z + \sqrt{2} C^2\int _{\Omega \setminus {\mathbb {B}}}\left| \frac{\partial v}{\partial z}(x, z)\right| ^2\,d x\,d z. \end{aligned}$$

(5.1)

In the same way we are able to estimate the curvilinear integral

$$\begin{aligned} \int _{\gamma ^3}|v|_{\gamma ^3}^2\,d \mu\le & {} \sqrt{2} C^2 \int _{\Omega \setminus {\mathbb {B}}}|v(x, y)|^2\,\text {d} x\,\text {d} y\nonumber \\&+ \sqrt{2} C^2 \int _{\Omega \setminus {\mathbb {B}}}\left| \frac{\partial v}{\partial y}(x, y)\right| ^2\,\text {d} x\,\text {d} y. \end{aligned}$$

(5.2)

In a similar way one deals with the integrals over curves $\gamma ^2, \gamma ^4$. We establish

$$\begin{aligned}&\int _{\gamma ^2}|v|_{\gamma ^2}^2\,d \mu \le \sqrt{2} C^2 \int _{\Omega \setminus {\mathbb {B}}}|v(x, y)|^2\,\text {d} x\,\text {d} y\\&\quad + \sqrt{2} C^2 \int _{\Omega \setminus {\mathbb {B}}}\left| \frac{\partial v}{\partial x}(x, y)\right| ^2\,\text {d} x\,\text {d} y,\\&\int _{\gamma ^4}|v|_{\gamma ^4}^2\,d \mu \le \sqrt{2} C^2\int _{\Omega \setminus {\mathbb {B}}}|v(x, y)|^2\,\text {d} x\,\text {d} y\\&\quad + \sqrt{2} C^2 \int _{\Omega \setminus {\mathbb {B}}}\left| \frac{\partial v}{\partial x}(x, y)\right| ^2\,\text {d} x\,\text {d} y. \end{aligned}$$

Combining the above bounds together with (5.1) and (5.2) we finish the proof of lemma. $\square$

1.3 Proof of Lemma 4.3

To proceed with a proof we need the following auxiliary material [9, Lemma 4.9]:

Lemma 5.1

Let $\Pi \subset {\mathbb {R}}^n$ be a convex set and let G and Q be arbitrary measurable sets in $\Pi$ with $\mu \, (G)\ne 0$. Then, for all $v\in {\mathcal {H}}^1(\Pi )$, the following inequality holds:

$$\begin{aligned}&\int _Q |v|^2\,\text {d} x\,\text {d} y\nonumber \\&\quad \le \frac{2\mu \, (Q)}{\mu \,(G)} \int _G |v|^2\,\text {d} x\,\text {d} y +\frac{C(n) (d(\Pi ))^{n+1} (\mu \, (Q))^{1/n}}{\mu \,(G)} \int _\Pi |\nabla v|^2\, \text {d} x\,\text {d} y,\qquad \quad \end{aligned}$$

(5.3)

where $\text {d}(\Pi )$ is the parameter of $\Pi$, $\mu$ is the Lebesque measure on ${\mathbb {R}}^n$, and the constant C(n) depends only on the dimension of ${\mathbb {R}}^n$.

Let $G_\epsilon$ be a convex subset of $\Omega$ to be chosen later which contains ${{\mathbb {B}}}_\epsilon$. Since $g\in {\mathcal {H}}^2_{\mathrm {loc}}(\Omega )$ in view of the interior regularity theory then applying (5.3) for $Q={{\mathbb {B}}}_\epsilon$ and $G=\Pi = G_\epsilon$ one infers that

$$\begin{aligned}&\int _{{{\mathbb {B}}}_\epsilon } \left| \frac{\partial g}{\partial x}\right| ^2 \,\text {d} x\, \text {d} y\le \frac{2 \mu \,({{\mathbb {B}}}_\epsilon )}{\mu \, (G_\epsilon )} \int _{G_\epsilon } \left| \frac{\partial g}{\partial x}\right| ^2\,\text {d} x\,\text {d} y\\&\qquad +\frac{C(2) (d\,(G_\epsilon ))^3 (\mu \, ({{\mathbb {B}}}_\epsilon ))^{1/2}}{\mu \,(G_\epsilon )} \int _{G_\epsilon }\left| \nabla \left( \frac{\partial g}{\partial x}\right) \right| ^2\,\text {d} x\,\text {d} y \\&\quad =\frac{2 \mu \,({{\mathbb {B}}}_\epsilon )}{\mu \, (G_\epsilon )} \int _{G_\epsilon } \left| \frac{\partial g}{\partial x}\right| ^2\,\text {d} x\,\text {d} y+ \frac{C(2) (d\,(G_\epsilon ))^3 (\mu \, ({{\mathbb {B}}}_\epsilon ))^{1/2}}{\mu \,(G_\epsilon )}\\&\qquad \times \int _{G_\epsilon }\left( \left| \frac{\partial ^2 g}{\partial x^2}\right| ^2+ \left| \frac{\partial ^2 g}{\partial x \partial y}\right| ^2\right) \,\text {d} x\,\text {d} y \end{aligned}$$

and

$$\begin{aligned}&\int _{{{\mathbb {B}}}_\epsilon } \left| \frac{\partial g}{\partial y}\right| ^2 \,\text {d} x\, \text {d} y \le \frac{2 \mu \,({{\mathbb {B}}}_\epsilon )}{\mu \, (G_\epsilon )} \int _{G_\epsilon } \left| \frac{\partial g}{\partial y}\right| ^2\,\text {d} x\,\text {d} y\\&\qquad + \frac{C(2) (d\,(G_\epsilon ))^3 (\mu \, ({{\mathbb {B}}}_\epsilon ))^{1/2}}{\mu \,(G_\epsilon )} \int _{G_\epsilon }\left| \nabla \left( \frac{\partial g}{\partial y}\right) \right| ^2\,\text {d} x\,\text {d} y \\&\quad =\frac{2 \mu \,({{\mathbb {B}}}_\epsilon )}{\mu \, (G_\epsilon )} \int _{G_\epsilon } \left| \frac{\partial g}{\partial y}\right| ^2\,\text {d} x\,\text {d} y+ \frac{C(2) (d\,(G_\epsilon ))^3 (\mu \, ({{\mathbb {B}}}_\epsilon ))^{1/2}}{\mu \,(G_\epsilon )}\\&\qquad \times \int _{G_\epsilon }\left( \left| \frac{\partial ^2 g}{\partial y \partial x}\right| ^2+ \left| \frac{\partial ^2 g}{\partial y^2}\right| ^2\right) \,\text {d} x\,\text {d} y. \end{aligned}$$

Combining the above inequalities we arrive

$$\begin{aligned}&\int _{{{\mathbb {B}}}_\epsilon }|\nabla g|^2 \,\text {d} x\, \text {d} y\le \frac{2 \mu \,({{\mathbb {B}}}_\epsilon )}{\mu \, (G_\epsilon )} \int _{G_\epsilon } |\nabla g|^2\,\text {d} x\,\text {d} y \nonumber \\&\qquad + \frac{C(2) (d\,(G_\epsilon ))^3 (\mu \, ({{\mathbb {B}}}_\epsilon ))^{1/2}}{\mu \,(G_\epsilon )} \int _{G_\epsilon }\left( \left| \frac{\partial ^2 g}{\partial x^2}\right| ^2+ 2\left| \frac{\partial ^2 g}{\partial x \partial y}\right| ^2+ \left| \frac{\partial ^2 f}{\partial y^2}\right| ^2\right) \,\text {d} x\,\text {d} y\nonumber \\&\quad \le \frac{(\mu \,({{\mathbb {B}}}_\epsilon ))^{1/2}}{\mu \, (G_\epsilon )}\left( 2(\mu \,({{\mathbb {B}}}_\epsilon ))^{1/2}+ C(2) (d\,(G_\epsilon ))^3 \right) \Vert g\Vert _{{\mathcal {H}}^2(G_\epsilon )}^2 \nonumber \\&\quad = \frac{\sqrt{\pi } \epsilon }{\mu \, (G_\epsilon )}\left( 2 \sqrt{\pi } \epsilon + C(2) (d\,(G_\epsilon ))^3 \right) \Vert g\Vert _{{\mathcal {H}}^2(G_\epsilon )}^2. \end{aligned}$$

(5.4)

Let us choose $G_\epsilon =\{r:\, 0\le r\le \epsilon ^{1-\alpha }\}$ with some $\alpha \in \left( \frac{1}{2}, 1\right)$ to be chosen later. One can easily notice that ${{\mathbb {B}}}_\epsilon \subset G_\epsilon$. The inequality (5.4) performs to

$$\begin{aligned} \int _{{{\mathbb {B}}}_\epsilon }|\nabla g|^2 \,\text {d} x\, \text {d} y\le 2\epsilon \left( \epsilon ^{2\alpha -1}+ \frac{4 C(2)}{\sqrt{\pi }} \epsilon ^{1-\alpha }\right) \, \Vert g\Vert ^2_{{\mathcal {H}}^2(G_\epsilon )}. \end{aligned}$$

(5.5)

To make the further estimates we need the interior regularity theorem [1] and the following lemma:

Theorem 5.2

(Interior Regularity Theorem). Suppose that $h\in {\mathcal {H}}^1(\Omega )$ is a weak solution to $-\Delta h=w$. Then $h\in {\mathcal {H}}^2_{\mathrm {loc}}(\Omega )$ and for each $\Omega _0\subset \Omega$ there is a constant $c= c(\Omega _0)$ independent of h and w such that:

$$\begin{aligned} \Vert h\Vert _{{\mathcal {H}}^2(\Omega _0)}\le c \left( \Vert h\Vert _{L^2(\Omega )}+\Vert w\Vert _{L^2(\Omega )}\right) . \end{aligned}$$

(5.6)

Lemma 5.3

(The proof will be given in the next subsection). For any $z\in \mathrm {Dom}(-\Delta _N^\Omega )$ the following estimate is valid

$$\begin{aligned} \int _\Omega |-\Delta z + z|^2\,\text {d} x\,\text {d} y \ge \frac{1}{16} \int _\Omega (|\Delta z|^2+ |z|^2)\,\text {d} x\,\text {d} y. \end{aligned}$$

Let us go back to inequality (5.5). Using (5.6) and Lemma 5.3 one establishes that

$$\begin{aligned}&\int _{{{\mathbb {B}}}_\epsilon }|\nabla g|^2 \,\text {d} x\, \text {d} y\le 4\, c^2 \epsilon \left( \epsilon ^{2\alpha -1}+ \frac{4 C(2)}{\sqrt{\pi }} \epsilon ^{1-\alpha }\right) \, (\Vert \Delta g\Vert _{L^2(\Omega )}^2+ \Vert g\Vert _{L^2(\Omega )}^2) \nonumber \\&\quad \le 64\, c^2 \epsilon \left( \epsilon ^{2\alpha -1}+ \frac{4 C(2)}{\sqrt{\pi }} \epsilon ^{1-\alpha }\right) \, \Vert -\Delta g+ g\Vert _{L^2(\Omega )}^2\nonumber \\&\quad =64\, c^2 \epsilon \left( \epsilon ^{2\alpha -1}+ \frac{4 C(2)}{\sqrt{\pi }} \epsilon ^{1-\alpha }\right) \, \Vert g\Vert _2^2\le \tilde{C} \left( \epsilon ^{2\alpha }+ \epsilon ^{2-\alpha }\right) \, \Vert g\Vert _2^2, \end{aligned}$$

(5.7)

where $\tilde{C}:= 64 c^2 \mathrm {max}\left\{ 1, \frac{4 C(2)}{\sqrt{\pi }}\right\}$ and c depends on the distance of ${\mathbb {B}}_\varepsilon$ from the boundary of $\Omega$.

Let us investigate the function $F(\alpha ):= \epsilon ^{2\alpha }+ \epsilon ^{2-\alpha }$ in (5.7) on interval $\left( \frac{1}{2}, 1\right)$. It attains its minimum at $\alpha _0= \frac{2}{3}- \frac{1}{3}\frac{\ln 2}{\ln \epsilon }$ and takes the value $F(\alpha _0)=\epsilon ^{4/3}\left( \frac{1}{4^{1/3}}+ 2^{1/3}\right)$.

Let us now choose $\alpha = \alpha _0$ in inequality (5.7). Then

$$\begin{aligned} \int _{{{\mathbb {B}}}_\epsilon }|\nabla g|^2 \,\text {d} x\, \text {d} y\le \epsilon ^{4/3} \left( \frac{1}{4^{1/3}}+ 2^{1/3}\right) \tilde{C}\,\Vert g\Vert _2^2, \end{aligned}$$

which concludes the proof of Lemma 4.3 with $C'=\left( \frac{1}{4^{1/3}}+ 2^{1/3}\right) \tilde{C}$. $\square$

1.4 Proof of Lemma 4.4

We first prove the following auxiliary statement: there exists $\tau \in (\varepsilon , 2\varepsilon )$ such that

$$\begin{aligned} \int _0^{2\pi } \left| \nabla g(\tau \cos \varphi , \tau \sin \varphi )\right| ^2\,d \varphi \le \frac{1}{\varepsilon ^2} \Vert g\Vert _1^2. \end{aligned}$$

(5.8)

Let us assume the opposite: for any $r\in (\varepsilon , 2\varepsilon )$ we have

$$\begin{aligned} \int _0^{2\pi } \left| \nabla g(r \cos \varphi , r \sin \varphi )\right| ^2\,d \varphi >\frac{1}{\varepsilon ^2} \Vert g\Vert _1^2. \end{aligned}$$

Passing to polar coordinates in integral $\int _{{{\mathbb {B}}}_{2\varepsilon }\backslash {{\mathbb {B}}}_\varepsilon } |\nabla g|^2\,\text {d} x\,\text {d} y$ and using the above bound we get

$$\begin{aligned} \int _{{{\mathbb {B}}}_{2\varepsilon }\backslash {{\mathbb {B}}}_\varepsilon } |\nabla g|^2\,\text {d} x\,\text {d} y= \int _\varepsilon ^{2\varepsilon } \,\int _0^{2\pi } r \left| \nabla g(r \cos \varphi , r \sin \varphi )\right| ^2\,d \varphi \, d r>\Vert g\Vert _1^2. \end{aligned}$$

This contradicts with the fact that the left-hand side of the above inequality does not exceed $\Vert g\Vert _1^2$. Hence there exists at least one number $\tau \in (\varepsilon , 2\varepsilon )$ such that

$$\begin{aligned} \int _0^{2\pi } |\nabla g(\tau \cos \varphi , \tau \sin \varphi )|^2\,d \varphi \le \frac{1}{\varepsilon ^2} \Vert g\Vert _1^2. \end{aligned}$$

(5.9)

in view of the representation of the derivative of function $\tilde{g}(r, \varphi )=g(r\cos \varphi , r\sin \varphi )$

$$\begin{aligned} \frac{\partial \widetilde{g}}{\partial \varphi }(\tau , \varphi )= -\tau \frac{\partial g}{\partial x}(\tau \cos \varphi , \tau \sin \varphi ) \sin \varphi +\tau \frac{\partial g}{\partial y}(\tau \cos \varphi , \tau \sin \varphi ) \cos \varphi \end{aligned}$$

and the Cauchy inequality we conclude that

$$\begin{aligned} \left| \frac{\partial \widetilde{g}}{\partial \varphi }(\tau , \varphi )\right| \le 2\varepsilon \left| \nabla g(\tau \cos \varphi , \tau \sin \varphi )\right| . \end{aligned}$$

Hence, employing (5.9) we establish that

$$\begin{aligned} \int _0^{2\pi }\left| \frac{\partial \tilde{g}}{\partial \varphi }(\tau , \varphi )\right| ^2\,d \varphi \le 4\varepsilon ^2 \int _0^{2\pi }\left| \nabla g(\tau \cos \varphi , \tau \sin \varphi )\right| ^2\,d \varphi \le 4\Vert g\Vert _1^2, \end{aligned}$$

which concludes the proof. $\square$

1.5 Proof of Lemma 5.3

It is straightforward to check that

$$\begin{aligned} \int _\Omega |-\Delta u+ u|^2\,\text {d} x\,\text {d} y= & {} \int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y+2 \int _\Omega |\nabla u|^2\,\text {d} x\,\text {d} y+\int _\Omega |u|^2\,\text {d} x\,\text {d} y\nonumber \\\ge & {} \int _\Omega |u|^2\,\text {d} x\,\text {d} y.\end{aligned}$$

(5.10)

Let us consider two cases:

$$\begin{aligned}&\int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y \ge 4\int _\Omega |u|^2\,\text {d} x\,\text {d} y, \end{aligned}$$

(5.11)

$$\begin{aligned}&\int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y< 4\int _\Omega |u|^2\,\text {d} x\,\text {d} y. \end{aligned}$$

(5.12)

Starting from the first one and employing (5.10) we have

$$\begin{aligned}&\sqrt{\int _\Omega |-\Delta u+ u|^2\,\text {d} x\,\text {d} y}\ge \sqrt{\int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y}- \sqrt{\int _\Omega |u|^2\,\text {d} x\,\text {d} y} \\&\quad \ge \frac{1}{2} \sqrt{\int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y}\ge \frac{1}{4}\sqrt{\int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y}+\frac{1}{2}\sqrt{\int _\Omega |u|^2\,\text {d} x\,\text {d} y}\\&\quad \ge \frac{1}{4}\left( \sqrt{\int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y}+\sqrt{\int _\Omega |u|^2\,\text {d} x\,\text {d} y}\right) . \end{aligned}$$

Hence we arrive at the bound

$$\begin{aligned} \int _\Omega |-\Delta u+ u|^2\,\text {d} x\,\text {d} y\ge \frac{1}{16}\left( \int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y+\int _\Omega |u|^2\,\text {d} x\,\text {d} y\right) . \end{aligned}$$

(5.13)

Now let us consider the case (5.12). In view of the inequality (5.10) we conclude

$$\begin{aligned} \int _\Omega |-\Delta u+ u|^2\,\text {d} x\,\text {d} y\ge & {} \int _\Omega |u|^2\,\text {d} x\,\text {d} y\ge \frac{1}{2}\int _\Omega |u|^2\,\text {d} x\,\text {d} y+ \frac{1}{8} \int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y\\\ge & {} \frac{1}{8}\left( \int _\Omega |\Delta u|^2\,\text {d} x\,\text {d} y+ \int _\Omega |u|^2\,\text {d} x\,\text {d} y\right) . \end{aligned}$$

Combining the above estimate together with (5.13) we complete the proof of the lemma. $\square$