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Zero-Viscosity Limit of the Navier–Stokes Equations in a Simply-Connected Bounded Domain Under the Analytic Setting

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Abstract

In this paper, we consider the zero-viscosity limit of the 2-D Navier–Stokes equations in a simply-connected bounded domain with non-slip boundary condition. Based on the energy method in Wang et al. (Arch Ration Mech Anal 224(2):555–595, 2017), we justify the zero-viscosity limit under the analytic setting.

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Acknowledgements

C. Wang is supported by NSF of China under Grant 11701016. Y. Wang is supported by China Postdoctoral Science Foundation 8206200009.

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Appendix: Well-Posedness of the Euler Equations and the Prandtl Equation in the Analytic Space

Appendix: Well-Posedness of the Euler Equations and the Prandtl Equation in the Analytic Space

In this appendix, we prove the well-posedness of the Euler equations and the Prandtl equation in the analytic space. The proof of well-poseness of the linearized Euler equations and Prandtl equation is similar, thus they are omitted.

1.1 Well-Posedness of the Euler Equations

Let us first introduce analytic norms

$$\begin{aligned}&\Vert U\Vert _{Y^s_b}^2\buildrel \hbox {def}\over =\sum _{|\alpha |=0}^{s-k}\Vert \nabla _{\theta ,r}^\alpha U\Vert _{X^k_b}^2,\quad \Vert U\Vert _{Y^{s,\frac{1}{2}}_b}^2\buildrel \hbox {def}\over =\sum _{|\alpha |=0}^{s-k}\Vert \nabla _{\theta ,r}^\alpha U\Vert _{X^{k,\frac{1}{2}}_b}^2, \end{aligned}$$

where \(\rho _E(t)=\delta (2-\frac{\lambda _E}{\delta } t)\ge \delta \) with \(\lambda _E\) defined later. What’ s more, operator \(\partial ^\gamma =\partial _\theta ^{\gamma _1}Z^{\gamma _2},~\nabla _{\theta ,r}^\alpha =(\partial _\theta ^{\alpha _1},\partial _r^{\alpha _2})\) . The main difference between definition \(\Vert \cdot \Vert _{X_b^k}\) and \(\Vert \cdot \Vert _{X_b^{k,\frac{1}{2}}}\) is that we add finite derivatives \(\nabla _{\theta ,r}^\alpha .\)

Proposition 9.1

Let initial data \((u_0, v_0)\) (and \((u^0, v^0)\) in polar coordinates) satisfies compatible condition (1.8), (1.9) and bound (1.10). Then there exists \(T_E>0\) so that the Euler equations (2.1) has a unique solution \(U_e^0=(u_e^0,v_e^0)\) in \([0,T_E]\), which satisfies

$$\begin{aligned} \Vert f_e^0(t)\Vert ^2_{X^{20}_i}+\Vert U_e^0(t)\Vert ^2_{Y^{20}_b}\le C,\quad \Vert \partial _t^kf_e^0(t)\Vert _{X^{20-k}_i}+ \Vert \partial _t^kU_e^0(t)\Vert _{Y^{20-k}_b}\le C,\quad k=1,2 \end{aligned}$$

for any \(t\in [0,T_E]\), where \(f_0^e\) is given in (2.16) and \(U^e_0\) is given in (2.1) with relation

$$\begin{aligned} (v_e^0+iu_e^0)(t,\theta ,r)\buildrel \hbox {def}\over =\frac{(\mathfrak {R}f_0^e+i \mathfrak {I}f_0^e)(t, r\cos \theta ,r\sin \theta )}{ e^{i\theta }}. \end{aligned}$$

Proof

Here we just give a priori estimates of the solution. Set \(\omega _{i,e}^0\buildrel \hbox {def}\over =\partial _y \mathfrak {R}f_e^0-\partial _x\mathfrak {I}f_e^0\) be the vorticity of velocity \(f_e^0\) under Eulerian coordinates and \(\omega _{b,e}^0\buildrel \hbox {def}\over =-\frac{\partial _r}{r}(r u^0_e)+\frac{\partial _\theta }{r}v_e^0\) be the vorticity of velocity \((u_e^0,v_e^0)\) under polar coordinates. It is easy to see \(\omega _{i,e}^0\) and \(\omega _{b,e}^0\) satisfy

$$\begin{aligned} \partial _t \omega _{i,e}^0+f_e^0\cdot \nabla \omega _{i,e}^0=0, \end{aligned}$$
(9.1)

and

$$\begin{aligned} \partial _t \omega _{b,e}^0+u_e^0\frac{\partial _\theta }{r}\omega _{b,e}^0+v_e^0\partial _r \omega _{b,e}^0=0, \end{aligned}$$
(9.2)

respectively.

Taking \(\langle \cdot , \cdot \rangle _{X^{20}_i}\) on both sides of (9.1) with \(\omega _{i,e}^0\), it is easy to get

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \Vert \omega _{i,e}^0\Vert _{X^{20}_i}^2+\lambda _E\Vert \omega _{i,e}^0\Vert _{X^{{20},\frac{1}{2}}_i}^2 \le \,&C(\Vert f_e^0\Vert _{X^{20}_i}^2+\Vert U_e^0\Vert _{Y^{20}_b}^2)\Vert \omega _{i,e}^0\Vert _{X^{{20},\frac{1}{2}}_i}^2\\ \le \,&C(\Vert \omega _{i,e}^0\Vert _{X^{19}_i}^2+\Vert \nabla _{r,\theta }U_e^0\Vert _{Y^{19}_b}^2+\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2)\Vert \omega _{i,e}^0\Vert _{X^{20,\frac{1}{2}}_i}^2. \end{aligned}$$

Similarly, taking \(\langle \cdot , \cdot \rangle _{Y^{20}_b}\) on both sides of (9.2) with \(\omega _{b,e}^0\), it is easy to obtain

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \Vert \omega _{b,e}^0\Vert _{Y^{20}_b}^2+\lambda _E \Vert \omega _{b,e}^0\Vert _{Y^{20,\frac{1}{2}}_b}^2 \le \,&C(\Vert f_e^0\Vert _{X^{20}_i}^2+\Vert U_e^0\Vert _{Y^{20}_b}^2)\Vert \omega _{b,e}^0\Vert _{Y^{20,\frac{1}{2}}_b}^2\\ \le \,&C(\Vert \omega _{i,e}^0\Vert _{X^{19}_i}^2+\Vert \nabla _{r,\theta }U_e^0\Vert _{Y^{19}_b}^2+\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2)\Vert \omega _{b,e}^0\Vert _{Y^{20,\frac{1}{2}}_b}^2. \end{aligned}$$

To close estimates, it remains to estimate \(\Vert \nabla _{r,\theta }U_e^0\Vert _{Y^{19}_b}^2\) and \(\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2.\) By calculation, we get equation for \(v_e^0:\)

$$\begin{aligned} r\left( \partial _r^2+\frac{\partial _r}{r}+\frac{\partial _\theta ^2}{r^2}\right) v_e^0 = \,&\partial _\theta \omega _{b,e}^0-\frac{v_e^0}{r},\quad v_e^0|_{r=1}=0. \end{aligned}$$
(9.3)

We first estimate norm \(\Vert \nabla _{r,\theta }U_e^0\Vert _{X^{19}_b}^2\) due to there is no boundary term by integrating by parts. We do similar argument as Lemma 6.1 to arrive at the following estimates:

$$\begin{aligned} \Vert \nabla _{r,\theta }v_e^0\Vert _{X^{19}_b}^2 \le \,&C\Vert \omega _{b,e}^0\Vert _{X^{19}_b}\Vert \nabla _{r,\theta }v_e^0\Vert _{X^{19}_b}+C_1\rho _E^2(\Vert \nabla _{r,\theta }v_e^0\Vert _{X^{19}_b}^2+\Vert \omega _{i,e}^0\Vert _{X^{19}_i}^2)+C\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2. \end{aligned}$$

Then we take \(\rho _E\) small enough and use Young’s inequality to obtain

$$\begin{aligned} \Vert \nabla _{r,\theta }v_e^0\Vert _{X^{19}_b}^2 \le \,&C(\Vert \omega _{b,e}^0\Vert _{X^{19}_b}^2+\Vert \omega _{i,e}^0\Vert _{X^{19}_i}^2+\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2). \end{aligned}$$
(9.4)

By the relation:

$$\begin{aligned} \partial _\theta u_e^0=-\,r\partial _r v_e^0-v_e^0, \quad \partial _r u_e^0=\frac{\partial _\theta }{r}v_e^0-\omega _{b,e}^0-\frac{u_e^0}{r}, \end{aligned}$$
(9.5)

and along with estimate (9.4), we have

$$\begin{aligned} \Vert \nabla _{r,\theta }u_e^0\Vert _{X^{19}_b}^2 \le \,&C\left( \Vert \omega _{b,e}^0\Vert _{X^{19}_b}^2+\Vert \omega _{i,e}^0\Vert _{X^{19}_i}^2+\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2\right) +C_1\rho _E^2\Vert \nabla _{r,\theta }u_e^0\Vert _{X^{18}_b}^2. \end{aligned}$$
(9.6)

Taking \(\rho _E\) small enough, we have

$$\begin{aligned} \Vert \nabla _{r,\theta }U_e^0\Vert _{X^{19}_b}^2 \le \,&C(\Vert \omega _{b,e}^0\Vert _{X^{19}_b}^2+\Vert \omega _{i,e}^0\Vert _{X^{19}_i}^2+\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2). \end{aligned}$$
(9.7)

Next, we estimate norm \(\Vert \nabla _{r,\theta }U_e^0\Vert _{Y^{19}_b}^2.\) Here we heavily use (9.5) and the equation of \(v_e^0\) [ see (9.3)] to obtain

$$\begin{aligned} \Vert \nabla _{r,\theta }U_e^0\Vert _{Y^{19}_b}^2\le C(\Vert \omega _{b,e}^0\Vert _{Y^{19}_b}^2+\Vert \omega _{i,e}^0\Vert _{X^{19}_i}^2+\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2). \end{aligned}$$

On the other hand, we use

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t f^0_e +f^0_e\cdot \nabla _{x,y} f^0_e +\bar{G^0_e}=0, \\&\mathfrak {R}(\partial _xf_e^0)+\mathfrak {I}(\partial _y f_e^0)=0. \end{aligned} \right. \end{aligned}$$
(9.8)

and an easy argument which is same as Lemma 5.13 that

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2 \le \,&C(\Vert (f_e^0)_{\rho _E}\Vert _{H^2_{x,y}}^3. \end{aligned}$$

In all, we get that

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}&(\Vert \omega _{i,e}^0\Vert _{X^{20}_i}^2+\Vert \omega _{b,e}^0\Vert _{Y^{20}_b}^2+\Vert (f_e^0)_{\rho _E}\Vert _{L^2}^2)+\lambda _E\left( \Vert \omega _{i,e}^0\Vert _{X^{20,\frac{1}{2}}_i}^2+\Vert \omega _{i,e}^0\Vert _{X^{20,\frac{1}{2}}_i}^2\right) \\ \le \,&C\left( \Vert \omega _{i,e}^0\Vert _{X^{20}_i}^2+\Vert \omega _{b,e}^0\Vert _{Y^{20}_b}^2\right) \Vert \omega _{i,e}^0\Vert _{X^{20,\frac{1}{2}}_i}^2+C(\Vert \omega _{i,e}^0\Vert _{X^{20}_i}+\Vert \omega _{b,e}^0\Vert _{Y^{20}_b})^3. \end{aligned}$$

Here, we take \(\lambda _E=4CM\) and \(T_E>0\) so that

$$\begin{aligned} \rho _E(t)\ge \delta \quad \text {for}\quad t\in [0,T_E],\quad 4CT_EM\le \frac{\delta }{2}. \end{aligned}$$

Then continuous argument ensures that

$$\begin{aligned} \Vert f_e^0\Vert _{X^{20}_i}^2+ \Vert U_e^0\Vert _{Y^{20}_b}^2\le C. \end{aligned}$$

This gives the first estimate. The second estimate can be deduced by using the equation. \(\square \)

1.2 Well-Posedness of the Prandtl Equation

Let us introduce some weighted analytic norms

$$\begin{aligned} \Vert U\Vert _{X^s_{w} }^2\buildrel \hbox {def}\over =&\sum _{m=0}^{\infty } \frac{\rho _P(t)^{2m}(1+m)^{2s} }{(m!)^2}\sum _{|\gamma |=m}\Vert e^{\phi }A^m{\tilde{\partial }}^\gamma U \Vert _{L^2_{\theta ,R}}^2,\\ \Vert U\Vert _{X^{s,\frac{1}{2}}_{w}}^2\buildrel \hbox {def}\over =&\sum _{m=1}^{\infty }\frac{ \rho _P(t)^{2m-1}(1+m)^{2s}m}{(m!)^2}\sum _{|\gamma |=m}\Vert e^{\phi }A^m{\tilde{\partial }}^\gamma U \Vert _{L^2_{\theta ,R}}^2\\&+\, \sum _{m=0}^{\infty } \frac{\rho _P(t)^{2m}(1+m)^{2s} }{(m!)^2}\sum _{|\gamma |=m}\Vert Re^{\phi }A^m{\tilde{\partial }}^\gamma U \Vert _{L^2_{\theta ,R}}^2, \end{aligned}$$

where \(\rho _P(t)=\delta (2-\frac{\lambda _p}{\delta } t)\ge \delta \) and \(\phi (t,R)=\rho _P(t)R^2\) with \(\lambda _P\) defined later.

Proposition 9.2

Let \((u_e^0,v_e^0)\) be given by Proposition 9.1. There exists \(T_P>0\) so that the Prandtl equation (2.6) has a unique solution \((u^p,v^p)\) in \([0,T_P]\), which satisfies

$$\begin{aligned}&\Vert u^p\Vert _{X^{17}_{w}}+\Vert v^p\Vert _{X^{16}_{w}}\le C,\\&\Vert \partial _R^2u^p\Vert _{X^{15}_w}+\Vert \partial _R^2v^p\Vert _{X^{14}_{w}}\le C. \end{aligned}$$

Proof

Again, we just present a priori estimates of the solution. We introduce a new function

$$\begin{aligned} {\overline{u}}^p=u^{p}+e^{-2\phi (t,R)}u_{e}^0(t,\theta ,1)\triangleq u^{p}+g. \end{aligned}$$

It is easy to verify that \({\overline{u}}^p\) satisfies

$$\begin{aligned}&\partial _t {\overline{u}}^p-\partial _R^2 {\overline{u}}^p+F^p=0, \end{aligned}$$
(9.9)

where \(F^p\) is given by

$$\begin{aligned} F^p = \,&{\overline{u}}^p \partial _\theta {\overline{u}}^p+{\overline{u}}^p \partial _\theta u^{e}(t,\theta ,1)+(u_{e}^0(t,\theta ,1)-g)\partial _\theta ({\overline{u}}^p-g)-{\overline{u}}^p\partial _\theta g\\&-\,g\partial _\theta u_{e}^0(t,\theta ,1)+\Big (R\partial _rv_{e}^0 (t,\theta ,1)-\int _{R}^{+\infty } \partial _\theta u^{p} dR'\Big )\partial _R ({\overline{u}}^p -g)-\partial _t g+\partial _R^2 g. \end{aligned}$$

Taking \(\langle \cdot , \cdot \rangle _{X^{17}_w}\) on both sides of (9.9) with \({\overline{u}}^p\), it is easy to get

$$\begin{aligned} \frac{d}{dt}\Vert {\overline{u}}^p\Vert ^2_{X^{17}_{w}}+\lambda _P\Vert {\overline{u}}^p\Vert _{X^{17,\frac{1}{2}}_{w}}^2 -\big \langle \partial _R^2 {\overline{u}}^p, {\overline{u}}^p\big \rangle _{X^{17}_w} =-\,\,\big \langle F^p,{\overline{u}}^p\big \rangle _{X^{17}_w}. \end{aligned}$$

For the dissipation term, we have

$$\begin{aligned} -\,\big \langle \partial _R^2 {\overline{u}}^p, {\overline{u}}^p\big \rangle _{X^{17}_w} \ge \big (1-\delta \big )\Vert \partial _R{\overline{u}}^p\Vert ^2_{X^{17}_{w}}-C\Vert {\overline{u}}^p\Vert ^2_{X^{17}_{w}}. \end{aligned}$$

For the other term, \((u_{e}^0,v_{e}^0)\) is estimated in Proposition 9.1 and is bounded by constant in norm \(\Vert \cdot \Vert _{Y_b^k}\) and \(\Vert \cdot \Vert _{Y_b^{k,\frac{1}{2}}}\). Similar argument in Lemma 4.8, we have

$$\begin{aligned}&\langle F^p,{\overline{u}}^p\big \rangle _{X^{17}_w}\le C\big (1+\Vert {\overline{u}}^p\Vert ^2_{X^{17}_{w}}\big )\big (1+\Vert {\overline{u}}^p\Vert ^2_{X^{17,\frac{1}{2}}_{w}}\big )+\frac{1}{2}\Vert \partial _R{\overline{u}}^p\Vert ^2_{X^{17}_{w}}. \end{aligned}$$

In all, we deduce that

$$\begin{aligned}&\frac{d}{dt}\Vert {\overline{u}}^p\Vert ^2_{X^{17}_{w}}+\big (\lambda _P-C-C\Vert {\overline{u}}^p\Vert _{X^{17}_{w}}\big )\Vert {\overline{u}}^p\Vert _{X^{17,\frac{1}{2}}_{w}}^2 +\big (\frac{1}{2}-\delta \big )\Vert \partial _R{\overline{u}}^p\Vert ^2_{X^{17}_{w}}\le C. \end{aligned}$$

With this, a continuous argument ensures that there exists \(T_p>0\) so that

$$\begin{aligned} \rho _P(t)\ge \delta ,\quad \Vert {\overline{u}}^p(t)\Vert ^2_{X^{17}_{w}}\le C \end{aligned}$$

for any \(t\in [0,T_p]\). This implies the first estimate. For the second estimate, one can first prove that

$$\begin{aligned} \Vert \partial _t{\overline{u}}^p(t)\Vert ^2_{X^{15}_{w}}\le C. \end{aligned}$$

Then the desired estimate can be deduced by using the equation of \(u^p\). \(\square \)

1.3 Proof of Lemmas 5.1, 5.3 and 5.4

First of all, Proposition 9.1 gives the existence of the solution \(U_e^0\) of (2.1) and the solution \(f_0^e\) of (2.16) with the bound

$$\begin{aligned} \Vert f_e^0(t)\Vert ^2_{X^{20}_i}+\Vert U_e^0(t)\Vert ^2_{Y^{20}_b}\le C. \end{aligned}$$

With \(U_e^0\) in hand, Proposition 9.2 gives the existence of the solution \((u_p^{0},v^{1}_p)\) of (2.6), (2.7) with the bound

$$\begin{aligned} \Vert u_p^{0}\Vert _{X^{17}_w}+\Vert v_p^{1}\Vert _{X^{16}_w}\le C. \end{aligned}$$

Next, we can solve the linearized Euler equation (2.2), (2.3) of \(U_e^1\) in \(Y^{15}_b\) and \(f_e^1\) in \(X^{15}_i\). Finally, we solve the linearized Prandtl equation (2.10) of \((u_p^1, v_p^2)\) in \(X^{12}_w\). Then Lemma 5.1 follows easily.

While, recalling the definition of \((R_1,R_2,R_3)\) in (3.3)–(3.5) and \((f^a,\widetilde{{\widetilde{R}}})\) in (3.20) and (3.23), we know \((R_1,R_2,R_3)\) and \((f^a,\widetilde{{\widetilde{R}}})\) are composed of approximate solutions which are estimated in Lemma 5.1. As a result, Lemmas 5.3 and 5.4 can be deduced by using Lemmas 4.11 and 5.1. Here we omit the details.

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Wang, C., Wang, Y. Zero-Viscosity Limit of the Navier–Stokes Equations in a Simply-Connected Bounded Domain Under the Analytic Setting. J. Math. Fluid Mech. 22, 8 (2020). https://doi.org/10.1007/s00021-019-0471-0

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