Prandtl–Batchelor Flows on a Disk

Abstract

For steady two-dimensional flows with a single eddy (i.e. nested closed streamlines), Prandtl (Über Flüssigkeitsbewegung bei sehr kleiner Reibung. Verhandlungen des III. Internationalen Mathematiker Kongresses, Heidelberg, 1905) and Batchelor (J Fluid Mech 7(1):177–190, 1956) proposed that in the limit of vanishing viscosity the vorticity is constant in an inner region separated from the boundary layer. In this paper, by constructing higher order approximate solutions of the Navier–Stokes equations and establishing the validity of Prandtl boundary layer expansion, we give a rigorous proof of the existence of Prandtl–Batchelor flows on a disk with the wall velocity slightly different from the rigid-rotation. The leading order term of the flow is the constant vorticity solution (i.e. rigid rotation) satisfying the Batchelor–Wood formula.

Appendix

1.1 Appendix A: Constant coefficient periodic PDE

In this appendix, we give a brief argument to solve the following problem

$$\begin{aligned} \left\{ \begin{array}{ll} (Q_0)_\theta =u_{e}(1)(Q_0)_{\psi \psi },\\ Q_0(\theta ,\psi )=Q_0(\theta +2\pi ,\psi ),\\ Q_0\big |_{\psi =0}=g(\theta ),\ \ Q_0\big |_{\psi \rightarrow -\infty }=0, \end{array} \right. \end{aligned}$$

(6.1)

where

$$\begin{aligned} g(\theta )=\alpha ^2+2\alpha \eta f(\theta )+\eta ^2 f^2(\theta )-u_{e}^2(1)=2\alpha \eta f(\theta )+\eta ^2 f^2(\theta )-\frac{\eta ^2}{2\pi } \int _0^{2\pi }f^2(\theta )d\theta . \end{aligned}$$

Let $Q_0(\theta ,\psi )=\sum \limits _{k\in \mathbb {Z}}e^{ik\theta }Q_{0k}(\psi )$ and substitute it into (6.1), we obtain

$$\begin{aligned} \left\{ \begin{array}{ll} ikQ_{0k}=u_{e}(1)Q_{0k}'',\\ Q_{0k}\big |_{\psi =0}={\widehat{g}}(k)=\frac{1}{2\pi }\int _0^{2\pi }e^{-ik\theta }g(\theta ) d\theta , \\ Q_{0k}\big |_{\psi \rightarrow -\infty }=0. \end{array} \right. \end{aligned}$$

It is easy to get

$$\begin{aligned} Q_{0k}(\psi )={\widehat{g}}(k)e^{\alpha _k\psi } \end{aligned}$$

with $\alpha _k=\sqrt{\frac{|k|}{2u_e(1)}}(1+\text {sgn}k\cdot i)$. Then

$$\begin{aligned} Q_0(\theta ,\psi )=\sum \limits _{k\in \mathbb {Z}}e^{ik\theta }{\widehat{g}}(k)e^{\alpha _k\psi }\in X \end{aligned}$$

and

$$\begin{aligned} \Vert Q_0\Vert _X\le C\eta . \end{aligned}$$

(6.2)

1.2 Appendix B: Construction of corrector $h(\theta ,r)$

In this section, we give a construction of the corrector $h(\theta ,r)$. Firstly, we give a simple lemma.

Lemma 6.1

Assume that $K(\theta ,r)$ is a $2\pi$-periodic smooth function which satisfies

$$\begin{aligned} \int _0^{2\pi }K(\theta ,r)d\theta =0, \ \forall r\in (0,1]; \ K(\theta , 1)=0, \end{aligned}$$

then there exists a $2\pi$-periodic function $h(\theta ,r)$ such that

$$\begin{aligned}&\partial _\theta h(\theta ,r)=K(\theta ,r); \quad h(\theta , 1)=0;\nonumber \\&\int _0^{2\pi }h(\theta ,r)d\theta =0, \quad \Vert \partial _\theta ^j\partial _r^kh\Vert _2\le C\Vert \partial _\theta ^j\partial _r^kK\Vert _2. \end{aligned}$$

(6.3)

Proof

Let

$$\begin{aligned} K(\theta ,r)=\sum _{n\ne 0}K_n(r)e^{in\theta }, \ \ K_n(1)=0. \end{aligned}$$

Set

$$\begin{aligned} h(\theta ,r)=\sum _{n\ne 0}\frac{K_n(r)}{in}e^{in\theta }. \end{aligned}$$

It is easy to justify that $h(\theta ,r)$ satisfies (6.3) which completes the proof. $\quad \square$

Next, we construct the corrector $h(\theta ,r)$ by the above lemma. Direct computation gives

$$\begin{aligned} u^a_\theta +rv^a_r+v^a=&\varepsilon ^5\partial _\theta h(\theta ,r)+K(\theta ,r), \end{aligned}$$

where

$$\begin{aligned} K(\theta ,r)=&\varepsilon ^5\chi (r)[Y\partial _Yv_p^{(5)}(\theta ,Y)+v_p^{(5)}(\theta ,Y)] +r\chi '(r)\Big (\sum _{i=1}^5\varepsilon ^i v_p^{(i)}(\theta ,Y)\Big ). \end{aligned}$$

Noticing that $\chi '(r)=0,\ r\in [0,\frac{1}{2}]\cup [\frac{3}{4},1]$ and the property of $v_p^{(i)}$, we deduce that $K(\theta ,r)=O(\varepsilon ^5)$ and

$$\begin{aligned} K(\theta ,1)=0. \end{aligned}$$

Moreover, noticing that

$$\begin{aligned} \int _0^{2\pi }v_p^{i}(\theta ,Y)d\theta =0, \ \forall \ Y\le 0,\quad i=1,\cdot \cdot \cdot ,5, \end{aligned}$$

we deduce that

$$\begin{aligned} \int _0^{2\pi }K(\theta ,r)d\theta =0, \ \forall r\in (0,1]. \end{aligned}$$

Thus, we can choose $h(\theta ,r)$ by Lemma 6.1 such that

$$\begin{aligned} \varepsilon ^5\partial _\theta h(\theta ,r)+K(\theta ,r)=0, \ h(\theta , 1)=0, \ \Vert \partial _\theta ^j\partial _r^kh\Vert _2\le C\varepsilon ^{-k}. \end{aligned}$$

1.3 Appendix C: Prandtl-Batchlor theory on disk

For the convenience of readers, we give an introduction to the Prandtl–Batchelor theory. One can see [1, 20, 21, 30, 42] for more details and physical backgrounds.

Theorem 6.2

We consider the steady Navier–Stokes equations in two dimensional simply-connected domain $\Omega$

$$\begin{aligned} \left\{ \begin{array}{ll} \textbf{u}^\varepsilon \cdot \nabla \textbf{u}^\varepsilon +\nabla p^\varepsilon -\varepsilon ^2\Delta \textbf{u}^\varepsilon =0,\\ \nabla \cdot \textbf{u}^\varepsilon =0,\\ \textbf{u}^\varepsilon \cdot \textbf{n}\big |_{\partial \Omega }=0, \ \ \textbf{u}^\varepsilon \cdot \textbf{t}\big |_{\partial \Omega }=g, \end{array} \right. \end{aligned}$$

(6.4)

where $\textbf{n}$ is the unit normal vector to $\partial \Omega$, $\textbf{t}$ is the unit tangential vector to $\partial \Omega$, and g is a smooth function. We assume that (i) the stream function $\psi ^\varepsilon$ of Eq. (6.4) has no hyperbolic critical point(i.e. nested closed streamlines and single eddy); (ii) for any $\Omega _1\subset \subset \Omega$, there exist $\varepsilon _0>0$ such that for any $0<\varepsilon <\varepsilon _0$, $\Omega _1$ is away from the boundary layer of Eq. (6.4) and $\textbf{u}^\varepsilon \rightarrow \textbf{u}^e$ in $C^2(\Omega _1)$, where $\textbf{u}^e$ is a solution of steady Euler equations in $\Omega$. Then the vorticity $w^e=\nabla \times \textbf{u}^e$ is a constant in $\Omega$.

Proof

Let $\textbf{u}^\varepsilon =(u^\varepsilon ,v^\varepsilon )$ and $w^\varepsilon =\partial _yu^\varepsilon -\partial _xv^\varepsilon$ be the vorticity, then it is easy to obtain that

$$\begin{aligned} \textbf{u}^\varepsilon \cdot \nabla w^\varepsilon -\varepsilon \Delta w^\varepsilon =0. \end{aligned}$$

The boundary is taken to be defined by $\psi ^\varepsilon =0$ and $0<\psi ^\varepsilon <c_1$ throughout the interior of the eddy. For any $0<c<c_1$, integrating the Navier–Stokes equations over the domain which is surrounded by the closed streamline $\{(x,y)|\psi ^\varepsilon (x,y)=c\}$ and using the divergence theorem, we obtain

$$\begin{aligned} \int _{\{\psi ^\varepsilon =c\}}\frac{\partial w_\varepsilon }{\partial n}dl=0, \ \ \forall c\in (0,c_1). \end{aligned}$$

(6.5)

Moreover, due to $\textbf{u}^\varepsilon \rightarrow \textbf{u}^e$ in $C^2$, then there holds $w^\varepsilon \rightarrow w^e$ in $C^1$.

Let $\psi ^e$ be the associated stream function of Euler equations, then $w^e=F(\psi ^e).$ In fact, we introduce the action-angle transform $(x,y)\rightarrow (r^{\varepsilon },\theta ^{\varepsilon })$, where

$$\begin{aligned} r^{\varepsilon }=\psi ^{\varepsilon }(x,y) ,\quad \ \frac{2\pi }{v^{\varepsilon }(r)}=\oint _{\left\{ \psi ^{\varepsilon }=r^{\varepsilon }\right\} }\frac{1}{\left| \nabla \psi ^{\varepsilon }\right| },\quad \theta ^{\varepsilon }=v^{\varepsilon }(r) \int _{0}^{l}\frac{dl^{\prime }}{\left| \nabla \psi ^{\varepsilon }\right| }, \end{aligned}$$

and l is the arc-length variable on the curve $\left\{ \psi ^{\varepsilon }=r^{\varepsilon }\right\}$. Then the transformation $(x,y) \rightarrow (r^{\varepsilon },\theta ^{\varepsilon })$ has Jacobian 1 and the operator $u^{\varepsilon }\cdot \nabla$ becomes $v^{\varepsilon }(r)\partial _{\theta ^{\varepsilon }}$. In $\Omega _{1}$, $( r^{\varepsilon },\theta ^{\varepsilon }) \rightarrow ( r^{0} ,\theta ^{0})$ in $C^{1}$ and $r^{0}=\psi ^{e}$. Then $(\psi ^{e},\theta ^{0})$ is a coordinate system and the steady vorticity $\omega ^{e}$ is a single-valued function $F(\psi ^{e})$.

Taking $\varepsilon \rightarrow 0$ in (6.5), we obtain

$$\begin{aligned} F'(c)=0, \ \forall c\in (0,c_1). \end{aligned}$$

Thus, $w_e$ is a constant in $\Omega$. $\quad \square$

Remark 6.3

On the disk $B_1(0)$: due to $\vec {u}_e=u_e(\theta ,r)\vec {e}_\theta +v_e(\theta ,r)\vec {e}_r$, we deduce

$$\begin{aligned} w_e=\frac{1}{r}(\partial _r(ru_e)-\partial _\theta v_e)=a, \quad v_e|_{\partial \Omega }=0. \end{aligned}$$

Solving this equation, we obtain

$$\begin{aligned} (u_e,v_e)=\Big (\frac{a}{2}r, 0\Big ), \end{aligned}$$

where a is any constant.