The Inviscid Limit for the Navier–Stokes Equations with Data Analytic Only Near the Boundary

Abstract

We address the inviscid limit for the Navier–Stokes equations in a half space, with initial datum that is analytic only close to the boundary of the domain, and that has Sobolev regularity in the complement. We prove that for such data the solution of the Navier–Stokes equations converges in the vanishing viscosity limit to the solution of the Euler equation, on a constant time interval.

Appendix A. Proofs of Some Technical Lemmas

Here we list some technical lemmas.

The next lemma converts an $\ell ^1$ norm in $\xi$ to an $\ell ^2$ norm, which is necessary when converting $S_\mu$ norms to an S and hence a Z norm.

Lemma A.1

Let $\mu \in (0,1)$. We have

$$\begin{aligned} \sum _{i+j\leqq 2}&\Vert \partial _x^i (y\partial _{y})^j\omega \Vert _{S_\mu } \lesssim \sum _{i+j\leqq 2} \left\| \partial _x^i(y\partial _{y})^j\omega \right\| _S + \left\| \partial _x^{i+1}(y\partial _{y})^j\omega \right\| _S . \end{aligned}$$

Proof of Lemma A.1

We have

$$\begin{aligned}&\sum _{\xi } |v_{\xi }| \lesssim \biggl ( \sum _{\xi } (1+|\xi |^2)|v_{\xi }|^2 \biggr )^{1/2} \end{aligned}$$

(A.1)

for every v for which the right side is finite. The inequality (A.1) holds since $\sum _{\xi }(1+|\xi |^2)^{-1}<\infty$. $\quad \square$

Lemma A.2

Assume that the parameters $\mu , \mu _0,\gamma , t > 0$ obey $\mu < \mu _0 - \gamma t$. Then, for $\alpha \in (0,\frac{1}{2})$ we have

$$\begin{aligned} \int _0^t \frac{1}{\sqrt{t-s}}\frac{1}{(\mu _0-\mu -\gamma s)^{1+\alpha }}\,\hbox {d}s \leqq \frac{C}{\sqrt{\gamma }(\mu _0-\mu -\gamma t)^{1/2+\alpha }} \end{aligned}$$

(A.2)

and

$$\begin{aligned} \int _0^t \frac{1}{\sqrt{t-s}}\frac{1}{(\mu _0-\mu -\gamma s)^{\alpha }}\,\hbox {d}s \leqq \frac{C}{\sqrt{\gamma }} , \end{aligned}$$

(A.3)

where $C>0$ is a constant depending on $\mu _0$ and $1/2-\alpha$.

Proof of Lemma A.2

Changing variables $s'=\gamma s$, $t'=\gamma t$, and letting $\mu _0 - \mu = \mu ' > 0$, the left side of (A.2) is rewritten and bounded as

$$\begin{aligned} \int _0^{t'} \frac{\sqrt{\gamma }}{\sqrt{t'- s'}}\frac{1}{(\mu '- s')^{1+\alpha }}\, \frac{\hbox {d} s'}{\gamma }&\leqq \frac{1}{\sqrt{\gamma }(\mu '- t')^{\alpha } } \int _0^{t'} \frac{\hbox {d} s'}{\sqrt{t'- s'}(\mu '- s')} \\&= \frac{2 \arctan \left( \sqrt{\frac{t'}{\mu '-t'}}\right) }{\sqrt{\gamma }(\mu '- t')^{1/2+ \alpha } } \lesssim \frac{1}{\sqrt{\gamma }(\mu '- t')^{1/2+ \alpha } } \\&= \frac{1}{\sqrt{\gamma }(\mu _0-\mu -\gamma t)^{1/2+ \alpha } } . \end{aligned}$$

In order to prove (A.3), we proceed similarly and use $\mu ' > t'$ to deduce

$$\begin{aligned} \int _0^{t'} \frac{\sqrt{\gamma }}{\sqrt{t'- s'}}\frac{1}{(\mu '- s')^{\alpha }}\, \frac{\hbox {d} s'}{\gamma } \leqq \frac{1}{\sqrt{\gamma }} \int _0^{t'} \frac{\hbox {d} s'}{(t'- s')^{1/2+\alpha }} \lesssim \frac{1}{\sqrt{\gamma }} , \end{aligned}$$

where the implicit constant may depend on $\mu _0$ and $1/2-\alpha$. $\quad \square$

Lemma A.3

(Analyticity recovery for the X norm). For ${\widetilde{\mu }}>\mu \geqq 0$, we have

$$\begin{aligned} \sum _{i+j=1}\Vert \partial _{x}^i(y\partial _{y})^jf\Vert _{X_\mu } \lesssim \frac{1}{{\widetilde{\mu }}-\mu }\Vert f\Vert _{X_{{{\widetilde{\mu }}}}} . \end{aligned}$$

Proof of Lemma A.3

First, let $(i,j)=(1,0)$. According to the definition of the $X_\mu$ norm, and using that the bound $({\widetilde{\mu }}-\mu ) |\xi | e^{\epsilon _0 |\xi | ( (1+\mu -y)_+ - (1+{\widetilde{\mu }}-y)_+)} \lesssim 1$ holds on $\Omega _\mu$, we have

$$\begin{aligned} \Vert \partial _{x}f \Vert _{X_\mu }&= \sum _\xi \Vert \xi e^{\epsilon _0(1+\mu -y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^\infty _{\mu ,\nu }} \lesssim \frac{1}{{\widetilde{\mu }}-\mu }\sum _\xi \Vert e^{\epsilon _0(1+{\widetilde{\mu }}-y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^\infty _{\mu ,\nu }} \\&\lesssim \frac{1}{{\widetilde{\mu }}-\mu }\sum _\xi \Vert e^{\epsilon _0(1+{\widetilde{\mu }}-y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^\infty _{{\widetilde{\mu }}, \nu }} = \frac{1}{{\widetilde{\mu }}-\mu }\Vert f \Vert _{X_{{\widetilde{\mu }}}} . \end{aligned}$$

Next, consider $(i,j)=(0,1)$. By the Cauchy integral theorem, we have

$$\begin{aligned} \partial _{y}f_\xi (y) = \int _{C(y, R_y)} \frac{f_\xi (z)}{(y-z)^2}\,\hbox {d}z , \end{aligned}$$

(A.4)

where $C(y, R_y)$ is the circle centered at y with radius $R_y$. Hence, we have

$$\begin{aligned} |\partial _{y}f_\xi (y)| \lesssim \frac{1}{R_y}\sup _{z\in C(y, R_y)} |f_\xi (z)| . \end{aligned}$$

By taking $R_y=C^{-1}({\widetilde{\mu }}-\mu ){{\mathbb {R}}}\mathrm {e}\,y$, for a sufficiently large universal constant $C>0$, we obtain

$$\begin{aligned} \Vert y\partial _{y}f \Vert _{X_\mu }&= \sum _\xi \Vert e^{\epsilon _0(1+\mu -y)_+|\xi |}y\partial _{y}f_\xi \Vert _{{\mathcal {L}}^\infty _{\mu ,\nu }} \lesssim \frac{1}{{\widetilde{\mu }}-\mu }\sum _\xi \Vert e^{\epsilon _0(1+\mu -y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^\infty _{{\widetilde{\mu }}, \nu }} \\&\lesssim \frac{1}{{\widetilde{\mu }}-\mu }\sum _\xi \Vert e^{\epsilon _0(1+{\widetilde{\mu }}-y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^\infty _{{\widetilde{\mu }}, \nu }} = \frac{1}{{\widetilde{\mu }}-\mu }\Vert f \Vert _{X_{{\widetilde{\mu }}}} , \end{aligned}$$

concluding the proof. $\quad \square$

Lemma A.4

(Analyticity recovery for the $Y$ norm). Let $\mu _0\geqq {\widetilde{\mu }}>\mu \geqq 0$. Then we have

$$\begin{aligned} \sum _{i+j=1}\Vert \partial _{x}^i(y\partial _{y})^jf\Vert _{Y_\mu } \lesssim \frac{1}{{\widetilde{\mu }}-\mu }\Vert f\Vert _{Y_{{\widetilde{\mu }}}} . \end{aligned}$$

(A.5)

Proof of Lemma A.4

By the same argument which yielded the X norm estimate in Lemma A.3, we obtain

$$\begin{aligned} \Vert \partial _{x}f \Vert _{Y_\mu }&= \sum _\xi \Vert \xi e^{\epsilon _0(1+\mu -y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^1_\mu } \lesssim \frac{1}{{\widetilde{\mu }}-\mu }\sum _\xi \Vert e^{\epsilon _0(1+{\widetilde{\mu }}-y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^1_\mu } \\&\lesssim \frac{1}{{\widetilde{\mu }}-\mu }\sum _\xi \Vert e^{\epsilon _0(1+{\widetilde{\mu }}-y)_+|\xi |} f_\xi \Vert _{{\mathcal {L}}^1_{{\widetilde{\mu }}}} = \frac{1}{{\widetilde{\mu }}-\mu }\Vert f \Vert _{Y_{{\widetilde{\mu }}}} . \end{aligned}$$

In order to prove the estimate (A.5) for $(i,j)=(0,1)$, we use (A.4) to bound

$$\begin{aligned} \Vert y\partial _{y}f_\xi \Vert _{L^1(\partial \Omega _{\theta })} = \int _{\partial \Omega _{\theta }} \left| \int _{C(y, R_y)} \frac{yf_\xi (z)}{(y-z)^2}\,\hbox {d}z\right| \hbox {d}y \lesssim \int _{\partial \Omega _{\theta }} \int _{C(y, R_y)} \frac{|yf_\xi (z)|}{R_y^2}\,\hbox {d}z\hbox {d}y \end{aligned}$$

for any $0\leqq \theta <\mu$. By taking $R_y=C^{-1} ({\widetilde{\mu }}-\mu ){{\mathbb {R}}}\mathrm {e}\,y$ for a sufficiently large universal constant $C>0$, using that |y| is comparable to ${{\mathbb {R}}}\mathrm {e}\,y$ in this region, and applying Fubini’s theorem, we obtain

$$\begin{aligned} \Vert y\partial _{y}f_\xi \Vert _{L^1(\partial \Omega _{\theta })}&\lesssim \frac{1}{{\widetilde{\mu }}-\mu }\int _{\partial \Omega _{\theta }} \int _{C(y, R_y)} \frac{|f_\xi (z)|}{R_y}\,\hbox {d}z\hbox {d}y \\&\lesssim \frac{1}{{\widetilde{\mu }}-\mu }\int _{\partial \Omega _{\theta }} \int _{0}^{2\pi } |f_\xi (y + R_y e^{i\phi } )| \,\hbox {d}\phi \hbox {d}y \\&\lesssim \frac{1}{{\widetilde{\mu }}-\mu } \sup _{{\bar{\theta }} \in \left( \theta - \frac{2({\widetilde{\mu }}-\mu )}{C}, \theta + \frac{2({\widetilde{\mu }}-\mu )}{C}\right) } \left\| f_\xi \right\| _{L^1(\partial \Omega _{{\bar{\theta }}})} \\&\lesssim \frac{1}{{\widetilde{\mu }}-\mu } \Vert f_\xi \Vert _{{\mathcal {L}}^1_{{\widetilde{\mu }}}} , \end{aligned}$$

which proves the claim. Since $e^{\epsilon _0(1+\mu -y)_{+}|\xi |} \leqq e^{\epsilon _0(1+{\widetilde{\mu }}-y)_{+}|\xi |}$, for every $y \in \Omega _{\mu }$, the lemma follows. $\quad \square$