2.1. Introduction

In this subsection, we will describe the mathematical model for Stuart cats’ eyes and Mallier–Maslowe vortices. We consider the two-dimensional Navier–Stokes equations and the continuity equation for incompressible Newtonian fluids in the form

(2.1a)$$\begin{gather} \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + \frac{\partial p}{\partial x}={-} \epsilon \frac{\partial \omega}{\partial y}, \end{gather}$$

(2.1b)$$\begin{gather}\frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} + \frac{\partial p}{\partial y}=\epsilon \frac{\partial \omega}{\partial x}, \end{gather}$$

(2.1c)$$\begin{gather}\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0, \end{gather}$$

in which $(x,y)^\textrm {T}$ are Cartesian coordinates, $t$ is time, $(u,v)^\textrm {T}$ is the velocity vector, $p$ is the pressure, $\omega$ is the vorticity given by

(2.2)\begin{equation} \omega = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}, \end{equation}

$\epsilon =1/Re$ is the reciprocal of the Reynolds number $Re$ and $0 < \epsilon \ll 1$ for high-Reynolds-number flows. The periodic boundary conditions in the streamwise direction are

(2.3)\begin{equation} [u,v,p](0, y, t) = [u,v,p](2 {\rm \pi}, y, t). \end{equation}

The far-field boundary conditions in the lateral direction for Stuart cats’ eyes are

(2.4a,b)\begin{equation} u \rightarrow \pm 1\quad \text{as } y \rightarrow \pm \infty, \quad v \rightarrow 0 \quad\text{as } y \rightarrow \pm \infty, \end{equation}

and for Mallier–Maslowe vortices are

(2.5a,b)\begin{equation} u \rightarrow 0 \quad \text{as } y \rightarrow \pm \infty, \quad v \rightarrow 0 \quad \text{as } y \rightarrow \pm \infty. \end{equation}

2.2. The leading-order solution

We will first perform the perturbation procedure on the nonlinear system (2.1)–(2.3) with (2.4a,b) or (2.5a,b). The solutions for Stuart cats’ eyes and Mallier–Maslowe vortices will then be obtained in §§ 2.2.1 and 2.2.2, respectively. We introduce expansions of the form

(2.6a–d)\begin{equation} u \sim u_0 + \epsilon u_1,\quad v \sim v_0 + \epsilon v_1,\quad p \sim p_0 + \epsilon p_1 ,\quad \omega \sim \omega_0 + \epsilon \omega_1 , \end{equation}

as $\epsilon \rightarrow 0$. At leading order, for the quasi-steady state, we obtain

(2.7a–c)\begin{equation} \bar{L} u_0 + \frac{\partial p_0}{\partial x} = 0,\quad \bar{L} v_0 +\frac{\partial p_0}{\partial y} = 0, \quad \frac{\partial u_0}{\partial x} + \frac{\partial v_0}{\partial y} = 0, \end{equation}

with the differential operator

(2.8)\begin{equation} \bar{L}= u_0 \frac{\partial}{\partial x} +v_0 \frac{\partial}{\partial y}. \end{equation}

We take the $x$-derivative of the second equation in (2.7a–c) minus the $y$-derivative of the first equation in (2.7a–c) to find that $\bar {L} \omega _0 = 0$. A streamfunction $\psi$ is defined by the equations

(2.9a,b)\begin{equation} u_0 = \frac{\partial \psi}{\partial y}, \quad v_0 ={-} \frac{\partial \psi}{\partial x}, \end{equation}

so that the third equation in (2.7a–c) is automatically satisfied and the vorticity equation may be rewritten as

(2.10)\begin{equation} \frac{\partial \psi}{\partial y} \frac{\partial \omega_0}{\partial x} -\frac{\partial \psi}{\partial x} \frac{\partial \omega_0}{\partial y}=0. \end{equation}

2.2.1. Stuart cats’ eyes

The Stuart cats’ eyes correspond to the solution of (2.10) given by $\omega _0 = - \exp \{- 2 \psi \}$, so that $\psi$ satisfies the partial differential equation

(2.11)\begin{equation} \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} = \exp \{- 2 \psi \} \end{equation}

with the boundary conditions

(2.12)\begin{equation} \psi(0, y, \tilde{t})=\psi(2 {\rm \pi}, y, \tilde{t}), \quad \psi \rightarrow \pm y \text{ as } y \rightarrow \pm \infty. \end{equation}

The Stuart cats’ eyes are a solution of this boundary value problem in the form

(2.13)\begin{equation} \psi(x, y, \tilde{t}) = \log [ C(\tilde{t}) \cosh(y) + A(\tilde{t}) \cos(x) ], \end{equation}

where $C(\tilde {t})>1$, $A(\tilde {t})=(C(\tilde {t})^2-1)^{1/2}>0$ and $\tilde {t} = \epsilon t$. The streamlines for $C(\tilde {t})=1.1$ are shown in figure 1(a). We deduce that

(2.14a,b)\begin{equation} u_0 = \frac{C(\tilde{t}) \sinh(y)}{C(\tilde{t}) \cosh(y) + A(\tilde{t}) \cos(x)}, \quad v_0 = \frac{A(\tilde{t}) \sin(x)}{C(\tilde{t}) \cosh(y) + A(\tilde{t}) \cos(x)}. \end{equation}

The leading-order kinetic energy is given by $E_0 = (u_0^2+v_0^2)/2$. We note that $\psi$, $v_0$ and $\omega _0$ ($u_0$) are even (odd) in $y$ about $y=0$. At leading order, the Bernoulli function, $H$, may be written in the form

(2.15)\begin{equation} H(\psi) = p_0 + E_0 = \tfrac{1}{2} \{ 1 - \textrm{e}^{{-}2 \psi} \}. \end{equation}

2.2.2. Mallier–Maslowe vortices

The Mallier–Maslowe vortices correspond to the solution of (2.10) given by $\omega _0 = (1-B^2)\sinh \{ 2 \psi \}/2$, so that $\psi$ satisfies the partial differential equation

(2.16)\begin{equation} \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} ={-}\frac{(1-B^2)}{2} \sinh \{ 2 \psi \} \end{equation}

with the boundary conditions

(2.17)\begin{equation} \psi(0, y, \tilde{t})=\psi(2 {\rm \pi}, y, \tilde{t}), \quad \psi \rightarrow 0 \text{ as } y \rightarrow \pm \infty, \end{equation}

where $0< B(\tilde {t})<1$ and $\tilde {t} = \epsilon t$. The Mallier–Maslowe vortices are a solution of this boundary value problem in the form

(2.18)\begin{equation} \psi(x, y, \tilde{t}) = \log \left [ \frac{\cosh(B(\tilde{t}) y) - B(\tilde{t}) \cos(x)}{\cosh(B(\tilde{t}) y) + B(\tilde{t}) \cos(x)} \right ] ={-}2 \text{arctanh} \left [ \frac{B(\tilde{t}) \cos(x)}{\cosh(B(\tilde{t}) y)} \right ]. \end{equation}

The streamlines for $B(\tilde {t})=1/2$ are shown in figure 1(b). We deduce that

(2.19a,b)\begin{equation} u_0 = \frac{2 B(\tilde{t})^2 \cos(x) \sinh(B(\tilde{t}) y)}{\cosh^2(B(\tilde{t}) y) - B(\tilde{t})^2 \cos^2(x)}, \quad v_0 ={-}\frac{2 B(\tilde{t}) \sin(x) \cosh(B(\tilde{t}) y)}{\cosh^2(B(\tilde{t}) y) - B(\tilde{t})^2 \cos^2(x)}. \end{equation}

We note that $\psi$, $v_0$ and $\omega _0$ ($u_0$) are even (odd) in $y$ about $y=0$. Furthermore, these solutions are such that $\psi$, $u_0$ and $\omega _0$ ($v_0$) are odd (even) in $x$ about zeros of $u_0$. At leading order, the Bernoulli function, $H$, may be written in the form

(2.20)\begin{equation} H(\psi; B) = p_0 + E_0 ={-}\frac{(1-B^2)}{4} \cosh \{ 2 \psi \}. \end{equation}

2.3. The first correction

We now derive the linear problem for the first correction and the conditions (2.30) under which it may have a solution. At the next order, we have

(2.21) \begin{equation} L \boldsymbol{z} = \left(\begin{array}{@{}c@{}} \displaystyle -\dfrac{\partial u_0}{\partial \tilde{t}}-\dfrac{\partial \omega_0}{\partial y}\\ [10pt] \displaystyle -\dfrac{\partial v_0}{\partial \tilde{t}}+\dfrac{\partial \omega_0}{\partial x}\\ 0 \end{array}\right), \end{equation}

in which

(2.22) \begin{equation} L=\left(\begin{array}{@{}ccc@{}} \displaystyle \bar{L} + \dfrac{\partial u_0}{\partial x} & \displaystyle \dfrac{\partial u_0}{\partial y} & \displaystyle \dfrac{\partial}{\partial x}\\[10pt] \displaystyle \dfrac{\partial v_0}{\partial x} & \displaystyle \bar{L} + \dfrac{\partial v_0}{\partial y} & \displaystyle \dfrac{\partial}{\partial y}\\[10pt] \displaystyle \dfrac{\partial}{\partial x} & \displaystyle \dfrac{\partial}{\partial y} & \displaystyle 0 \end{array}\right),\quad \boldsymbol{z}=\left(\begin{array}{@{}c@{}} u_1\\ v_1\\ p_1 \end{array}\right), \end{equation}

with the periodic and far-field boundary conditions

(2.23a)$$\begin{gather} [u_1,v_1,p_1](0, y, \tilde{t}) = [u_1,v_1,p_1](2 {\rm \pi}, y, \tilde{t}), \end{gather}$$

(2.23b)$$\begin{gather}u_1 \rightarrow 0 \quad \text{as } y \rightarrow \pm \infty, \quad v_1 \rightarrow 0 \quad\text{as } y \rightarrow \pm \infty. \end{gather}$$

A general solution to the linear problem for the first correction (2.21)–(2.23) is very difficult to find. Therefore, we investigate the adjoint problem in order to determine the conditions under which the problem for the first correction has a solution. Following the analysis in Smith (Reference Smith2007), we have an equation with the right-hand side in divergence form

(2.24)\begin{align} \boldsymbol{r}^T L \boldsymbol{z} - \boldsymbol{z}^T L^* \boldsymbol{r}&= \frac{\partial}{\partial x} [ u_0 \{ a u_1 + b v_1 \} +a p_1 + c u_1 ] \nonumber\\ &\quad + \frac{\partial}{\partial y} [ v_0 \{ a u_1 + b v_1 \}+ b p_1 + c v_1 ], \end{align}

in which the adjoint operator $L^*$ and vector $\boldsymbol {r}$ are given by

(2.25) \begin{equation} L^{*}=\left(\begin{array}{@{}ccc@{}} \displaystyle -\bar{L} + \dfrac{\partial u_0}{\partial x} & \displaystyle \dfrac{\partial v_0}{\partial x} & \displaystyle - \dfrac{\partial}{\partial x}\\[10pt] \displaystyle \dfrac{\partial u_0}{\partial y} & \displaystyle -\bar{L} + \dfrac{\partial v_0}{\partial y} & \displaystyle -\dfrac{\partial}{\partial y}\\ [10pt] \displaystyle -\dfrac{\partial}{\partial x} & \displaystyle -\dfrac{\partial}{\partial y} & \displaystyle 0 \end{array}\right),\quad \boldsymbol{r}=\left(\begin{array}{@{}c@{}} a\\ b\\ c \end{array}\right). \end{equation}

Equation (2.24) may be integrated to yield

(2.26)\begin{equation} \langle \boldsymbol{r}^T L \boldsymbol{z} - \boldsymbol{z}^T L^* \boldsymbol{r} \rangle = 0, \end{equation}

where

(2.27)\begin{equation} \langle \ {\bullet}\ \rangle = \int^{2 {\rm \pi}}_{x=0} \int^{\infty}_{y={-}\infty} \ {\bullet}\ \textrm{d} y \,\textrm{d}\kern0.7pt x, \end{equation}

provided that the following conditions are satisfied

(2.28a)$$\begin{gather} [a,b,c](0, y, \tilde{t})= [a,b,c](2 {\rm \pi}, y, \tilde{t}), \end{gather}$$

(2.28b)$$\begin{gather}b \rightarrow 0 \quad \text{as } y \rightarrow \pm \infty. \end{gather}$$

It follows from (2.26) that if

(2.29)\begin{equation} L^* \boldsymbol{r} = \boldsymbol{0}, \end{equation}

subject to the conditions (2.28), then our linear problem for the first correction (2.21)–(2.23) can only have a solution if

(2.30)\begin{equation} \left \langle -a \left [ \frac{\partial \omega_0}{\partial y} + \frac{\partial u_0}{\partial \tilde{t}} \right ] + b \left [ \frac{\partial \omega_0}{\partial x} - \frac{\partial v_0}{\partial \tilde{t}} \right ] \right \rangle=0, \end{equation}

for any $\boldsymbol {r}=(a,b,c)^\textrm {T}$ in the null space of the adjoint problem. It remains to find the linearly independent solution vectors $\boldsymbol {r}$ and substitute them into (2.30).

Two linearly independent solutions of the adjoint problem (2.29) and (2.28) are

(2.31a,b)\begin{equation} \boldsymbol{r}_1 = (0,0,1)^\textrm{T}, \quad \boldsymbol{r}_2 = (1,0,u_0)^\textrm{T}. \end{equation}

One infinite family of linearly independent solutions is given by

(2.32)\begin{equation} \boldsymbol{r}_3 = \left(\frac{\partial }{\partial y} (\omega_0^{n-1}), -\frac{\partial }{\partial x} (\omega_0^{n-1}),-\frac{(n-1)}{n} \omega_0^n \right)^\textrm{T}, \end{equation}

for $n>1$ and $n \in \textrm {I R}$. The vorticity is finite for these flows. The function $\omega _0^n$ is thus continuous on $[-\omega _{max}, \omega _{max}]$, where $\omega _{max}$ bounds the values of vorticity. Given this condition, the Weierstrass approximation theorem ensures that a polynomial

(2.33)\begin{equation} p(\omega_0) = a_0 + a_1 \omega_0 + \cdots + a_q \omega_0^q, \end{equation}

exists with $a_i$ real constants and $q$ a non-negative integer, which uniformly approximates $\omega _0^n$ on $[-\omega _{max}, \omega _{max}]$; that is,

(2.34)\begin{equation} p(\omega_0) - \hat{\epsilon} < \omega_0^n < p(\omega_0) + \hat{\epsilon}, \end{equation}

for any $\hat {\epsilon }>0$. We note that $a_0=0$ in this case and consider the vector space of polynomials $p(\omega _0)$ over the field of real numbers. If we construct a basis for this space of polynomials $\{ \omega _0^n \mid n \in \textrm {I N} \}$, then all functions in $\{ \omega _0^m \mid m \in \textrm {I R}, \ m > 1 \}$ are uniformly approximated by our basis. Hence, it is sufficient to construct modulation equations for the basis, taking the powers to be natural numbers. A countably infinite family of linearly independent solutions of the adjoint problem (2.29) and (2.28) have been deduced; that is, $\boldsymbol {r}_3$ for $n$ in the natural numbers.

A further infinite family of linearly independent solutions is given by

(2.35)\begin{equation} \boldsymbol{r}_4 = \left(f(\psi) u_0, f(\psi) v_0, \int^{\psi} f(s) H^{\prime} (s) \, \textrm{d} s \right)^\textrm{T}, \end{equation}

where $f(s)$ is a differentiable function. As $f(s)$ is not defined on a bounded interval, the Weierstrass approximation theorem does not apply in this case.

2.4. Modulation equations

The first solution to the adjoint problem, $\boldsymbol {r}_1$, corresponds to a trivial modulation equation. The second solution results in a degenerate modulation equation due to the parity in $y$ of both the Stuart cats’ eyes and Mallier–Maslowe vortices. Only the third and fourth solutions correspond to physical modulation equations to be described in the following.

2.4.1. Vorticity to the power $n$ modulation equations

If we substitute the third vector $\boldsymbol {r}_3$ into (2.30), then we obtain our first secularity conditions

(2.36)\begin{equation} \left \langle -\frac{\partial}{\partial y}(\omega_0^{n-1}) \left [ \frac{\partial \omega_0}{\partial y} + \frac{\partial u_0}{\partial \tilde{t}} \right ] -\frac{\partial}{\partial x}(\omega_0^{n-1}) \left [ \frac{\partial \omega_0}{\partial x} - \frac{\partial v_0}{\partial \tilde{t}} \right ] \right \rangle=0. \end{equation}

After integration by parts in both $x$ and $y$, we derive

(2.37)\begin{equation} \left \langle \omega_0^{n-1} \frac{\partial}{\partial y} \left [ \frac{\partial \omega_0}{\partial y} + \frac{\partial u_0}{\partial \tilde{t}} \right ] + \omega_0^{n-1} \frac{\partial}{\partial x} \left [ \frac{\partial \omega_0}{\partial x} - \frac{\partial v_0}{\partial \tilde{t}} \right ] \right \rangle=0. \end{equation}

This may be rewritten as the modulation equations

(2.38)\begin{equation} \frac{\textrm{d} }{\textrm{d} \tilde{t}} \left \langle \omega_0^n \right\rangle = \left \langle n \omega_0^{n-1} \left [ \frac{\partial^2 \omega_0}{\partial x^2} + \frac{\partial^2 \omega_0}{\partial y^2} \right ] \right\rangle, \end{equation}

for $n \in \textrm {I N}$.

2.4.2. Generalized energy modulation equations

If we substitute the fourth vector $\boldsymbol {r}_4$ into (2.30), then we obtain the further secularity conditions

(2.39)\begin{equation} \left \langle f(\psi) u_0 \left [ -\frac{\partial \omega_0}{\partial y} - \frac{\partial u_0}{\partial \tilde{t}} \right ] + f (\psi) v_0 \left [ \frac{\partial \omega_0}{\partial x} - \frac{\partial v_0}{\partial \tilde{t}} \right ] \right \rangle=0. \end{equation}

This may be rewritten as

(2.40)\begin{equation} \left \langle f (\psi) \left [ \frac{\partial E_0}{\partial \tilde{t}} + u_0 \frac{\partial \omega_0}{\partial y} - v_0 \frac{\partial \omega_0}{\partial x} \right ] \right \rangle=0. \end{equation}

Equations (2.38) and (2.40) are termed as the vorticity to the power $n$ modulation equations and the generalized energy modulation equations, respectively. They are necessary conditions for the solutions of the Euler equations to approximate the solutions for the Navier–Stokes equation.

2.5. Inconsistency of the steady state

We consider the energy solvability condition for the steady state by taking $\partial E_0 / \partial \tilde {t} = 0$ and $f (\psi )=1$ in (2.40); that is,

(2.41)\begin{equation} \left \langle v_0 \frac{\partial \omega_0}{\partial x}- u_0 \frac{\partial \omega_0}{\partial y} \right \rangle=0. \end{equation}

We will show that both Stuart cats’ eyes and Mallier–Maslowe vortices do not satisfy this steady solvability condition (2.41) in §§ 2.5.1 and 2.5.2, respectively. Thus, they do not approximate a steady state of the Navier–Stokes equations.

2.5.1. Stuart cats’ eyes

If we substitute for the leading-order solution (2.13)–(2.15), the solvability condition (2.41) becomes

(2.42)\begin{equation} \int^{2 {\rm \pi}}_{x=0} \int^{\infty}_{y={-}\infty} - \frac{2 \{ A^2 \sin^2 (x) + C^2 \sinh^2(y) \}}{[C \cosh(y) + A \cos(x)]^4} \, \textrm{d} y \,\textrm{d}\kern0.7pt x=0. \end{equation}

The left-hand side of this solvability condition is non-zero and the solvability condition is not satisfied. The problem for the first correction (2.21)–(2.23) does not have a solution by the Fredholm alternative. Stuart cats’ eyes are solutions to the steady Euler equations, but are not a leading-order solution to the steady Navier–Stokes equations for high-Reynolds-number flows. As such, Stuart cats’ eyes do not approximate a steady state of the Navier–Stokes equations.

2.5.2. Mallier–Maslowe vortices

If we substitute for the leading-order solution (2.18)–(2.20), the solvability condition (2.41) becomes

(2.43)\begin{equation} \int^{2 {\rm \pi}}_{x=0} \int^{\infty}_{y={-}\infty} - (1-B^2) E_0 \cosh (2 \psi) \, \textrm{d} y \,\textrm{d}\kern0.7pt x=0. \end{equation}

The left-hand side of this solvability condition is non-zero and the solvability condition is not satisfied. The problem for the first correction (2.21)–(2.23) does not have a solution by the Fredholm alternative. Mallier–Maslowe vortices are the solutions to the steady Euler equations, but do not approximate a steady state of the Navier–Stokes equations.

2.6. Inconsistency of the quasi-steady state

We will show that both Stuart cats’ eyes and Mallier–Maslowe vortices do not satisfy the modulation equations (2.38) in §§ 2.6.1 and 2.6.2, respectively. Thus, they do not approximate quasi-steady states of the Navier–Stokes equations.

2.6.1. Stuart cats’ eyes

We substitute the leading-order solution (2.13)–(2.15) into the vorticity modulation equations (2.38) to obtain

(2.44)\begin{equation} \frac{\textrm{d} C}{\textrm{d} \tilde{t}}=\frac{A \displaystyle \left \langle \frac{1 - 2 \left \{ A^2 \sin^2 (x) + C^2 \sinh^2(y) \right \}}{[C \cosh(y) + A \cos(x)]^{2n+2}} \right \rangle} {\displaystyle \left \langle \frac{A \cosh(y) + C \cos(x)}{[C \cosh(y) + A \cos(x)]^{2n+1}} \right \rangle}, \end{equation}

for $n \in \textrm {I N}$. These equations are inconsistent for different values of $n$ because they contain different terms, figure 2(a) representing an example of the solution for an arbitrary initial condition. The problem for the first correction (2.21)–(2.23) does not have a solution by the Fredholm alternative. Stuart cats’ eyes do not approximate a quasi-steady state of the Navier–Stokes equations. This result could have been anticipated because changes in $C$ only correspond to a redistribution of vorticity (Wu et al. Reference Wu, Ma and Zhou2006).

Figure 2.

Solutions of the vorticity modulation equations using different values of integer $n$ for (a) Stuart cats’ eyes (2.44) and (b) Mallier–Maslowe vortices (2.45).

2.6.2. Mallier–Maslowe vortices

We substitute the leading-order solution (2.18)–(2.20) into the vorticity modulation equations (2.38) to yield

(2.45)\begin{equation} \frac{\textrm{d} B}{\textrm{d} \tilde{t}}=\frac{ \langle \omega_0^{n} [8 E_0 - (1-B^2) \cosh(2 \psi)] \rangle} {\displaystyle \left \langle \omega_0^{n-1} \frac{\partial \omega_0}{\partial B} \right \rangle}, \end{equation}

where $n$ is an even positive integer. If $n$ is an odd positive integer, then $\langle \omega _0^n \rangle =0$ because of the parity in $x$ of $\omega _0$. Therefore the vorticity modulation equations (2.45) are degenerate when $n$ is odd with the numerator in (2.45) being zero and the denominator in (2.45) being non-zero. Equations (2.45) are inconsistent for different even values of $n$ because they contain different terms, figure 2(b) representing an example of the solution for an arbitrary initial condition. The problem for the first correction (2.21)–(2.23) does not have a solution by the Fredholm alternative. Mallier–Maslowe vortices do not approximate a quasi-steady state of the Navier–Stokes equations.

3.1. Introduction

In this subsection, we will describe the mathematical model for Hill's spherical vortex. In order to avoid an excessive number of suffices, the same notation as in § 2 will be used for a number of the quantities; these are defined anew in this section. We adopt a spherical polar coordinate system in which $r \leq R$ is the radial coordinate, $\theta$ is the polar angle and $R$ is the radius of a sphere. The radial and polar coordinates of velocity are given by $u$ and $v$, respectively. We denote the pressure by $p$ and the vorticity $\omega$ by

(3.1)\begin{equation} \omega = \frac{1}{r} \frac{\partial}{\partial r} (r v) - \frac{1}{r} \frac{\partial u}{\partial \theta}. \end{equation}

The axisymmetric continuity and Navier–Stokes equations for incompressible Newtonian fluids in spherical polar coordinates become

(3.2)\begin{gather} \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 u) + \frac{1}{r \sin(\theta)} \frac{\partial}{\partial \theta}(v \sin(\theta)) = 0, \end{gather}

(3.3a)\begin{gather} u \frac{\partial u}{\partial r} + \frac{v}{r} \frac{\partial u}{\partial \theta} -\frac{v^2}{r} +\frac{\partial p}{\partial r} ={-}\frac{\epsilon}{r \sin (\theta)} \frac{\partial}{\partial \theta} (\sin(\theta) \omega) , \end{gather}

(3.3b)\begin{gather} u \frac{\partial v}{\partial r} + \frac{v}{r} \frac{\partial v}{\partial \theta} +\frac{uv}{r} +\frac{1}{r} \frac{\partial p}{\partial \theta} = \frac{\epsilon}{r} \frac{\partial}{\partial r} (r \omega) , \end{gather}

where $\epsilon =1/ Re$ is the reciprocal of the Reynolds number and $0 < \epsilon \ll 1$. The boundary conditions are

(3.4a,b)\begin{equation} u(R,\theta) = 0, \quad v(R, \theta) = C \sin ( \theta ), \end{equation}

in which $C$ is a negative constant. Hill's spherical vortex corresponds to $C = - 3 U / 2$, where $U$ is the far-field velocity of the surrounding flow.

3.2. The leading-order solution

We will first perform the perturbation procedure on the nonlinear system (3.3)–(3.4a,b) and obtain the solution for Hill's spherical vortex. We introduce expansions of the form

(3.5a–d)\begin{equation} u \sim u_0 + \epsilon u_1, \quad v \sim v_0 + \epsilon v_1, \quad p \sim p_0 + \epsilon p_1 , \quad \omega \sim \omega_0 + \epsilon \omega_1 , \end{equation}

as $\epsilon \rightarrow 0$. At leading order, we obtain

(3.6a)$$\begin{gather} \bar{L} u_0 - \frac{v_0^2}{r} +\frac{\partial p_0}{\partial r} = 0, \end{gather}$$

(3.6b)$$\begin{gather}\bar{L} v_0 + \frac{u_0 v_0}{r} +\frac{1}{r} \frac{\partial p_0}{\partial \theta} = 0, \end{gather}$$

(3.6c)$$\begin{gather}\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 u_0) + \frac{1}{r \sin(\theta)} \frac{\partial}{\partial \theta}(v_0 \sin(\theta)) = 0, \end{gather}$$

with the differential operator

(3.7)\begin{equation} \bar{L}= u_0 \frac{\partial}{\partial r} + \frac{v_0}{r} \frac{\partial}{\partial \theta} \end{equation}

and the boundary conditions

(3.8a,b)\begin{equation} u_0(R,\theta) = 0, \quad v_0(R, \theta) = C \sin ( \theta ). \end{equation}

Using (3.6a)–(3.6b), the vorticity equation becomes $\bar {L} (\omega _0 / r \sin (\theta ) ) = 0$. Henceforth, we assume a solution of this equation of the form $\omega _0 = \alpha r \sin (\theta )$, where $\alpha$ is a constant to be determined shortly. The Stokes streamfunction is defined by

(3.9a,b)\begin{equation} u_0 = \frac{1}{r^2 \sin (\theta)} \frac{\partial \psi}{\partial \theta}, \quad v_0 ={-} \frac{1}{r \sin (\theta)} \frac{\partial \psi}{\partial r}, \end{equation}

so that (3.6c) is automatically satisfied. The streamfunction $\psi$ satisfies the following partial differential equation

(3.10)\begin{equation} \frac{\partial^2 \psi}{\partial r^2} + \frac{\sin(\theta)}{r^2} \frac{\partial}{\partial \theta} \left [ \frac{1}{\sin(\theta)} \frac{\partial \psi}{\partial \theta} \right ] ={-} \alpha r^2 \sin^2 (\theta) \end{equation}

for $r < R$ with the boundary conditions

(3.11a,b)\begin{equation} \psi (R,\theta) = 0, \quad \frac{\partial \psi}{\partial r} (R,\theta) ={-}C R \sin^2 ( \theta). \end{equation}

Hill's spherical vortex corresponds to the solution of this problem in the form

(3.12)\begin{equation} \psi (r, \theta) = \frac{C r^2}{2} \left ( 1 - \frac{r^2}{R^2} \right ) \sin^2 (\theta) \end{equation}

for $r \leq R$ and provided that $\alpha = 5 C / R^2$. The streamlines for $C=-1$ and $R=1$ are shown in figure 3. We deduce that the velocities are

(3.13a,b)\begin{equation} u_0 = C \left ( 1 - \frac{r^2}{R^2} \right ) \cos (\theta), \quad v_0 ={-}C \left ( 1 - \frac{2 r^2}{R^2} \right ) \sin (\theta). \end{equation}

3.3. The first correction

We now derive the linear problem for the first correction and the conditions (3.23) under which it may have a solution. At next order, we obtain

(3.14) \begin{equation} L \boldsymbol{z} = \left(\begin{array}{@{}c@{}} \displaystyle-\dfrac{1}{r \sin (\theta)} \dfrac{\partial}{\partial \theta} (\sin(\theta) \omega_0)\\[10pt] \displaystyle\dfrac{1}{r} \dfrac{\partial}{\partial r} (r \omega_0)\\ 0 \end{array}\right), \end{equation}

in which

(3.15a,b) \begin{equation} L=\left(\begin{array}{@{}ccc@{}} \displaystyle \bar{L} + \dfrac{\partial u_0}{\partial r} & \displaystyle \dfrac{1}{r} \dfrac{\partial u_0}{\partial \theta} - \dfrac{2 v_0}{r} & \displaystyle \dfrac{\partial}{\partial r}\\[10pt] \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial r} (r v_0) & \displaystyle \bar{L} + \dfrac{1}{r} \dfrac{\partial v_0}{\partial \theta} + \dfrac{u_0}{r} & \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial \theta}\\[10pt] \displaystyle \dfrac{\partial}{\partial r} + \dfrac{2}{r} & \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial \theta}+ \dfrac{\cot(\theta)}{r} & \displaystyle 0 \end{array}\right), \quad \boldsymbol{z}=\left(\begin{array}{@{}c@{}} u_1\\ v_1\\ p_1 \end{array}\right), \end{equation}

with the boundary conditions

(3.16a,b)\begin{equation} u_1 (R,\theta) = 0, \quad v_1 (R, \theta) = 0. \end{equation}

Saffman (Reference Saffman1992) stated one solution to the linear problem for the first correction (3.14)–(3.16a,b). By the Fredholm alternative, there is either a unique solution or an infinite number of solutions. We study the adjoint problem to check our approach and to establish how many solutions exist to this important problem. Following a modified version of the analysis in Smith (Reference Smith2010), we have an equation with the right-hand side in divergence form

(3.17)\begin{align} \boldsymbol{r}^T L \boldsymbol{z} - \boldsymbol{z}^T L^* \boldsymbol{r}&= \frac{1}{r^2} \frac{\partial}{\partial r} [ r^2 u_0 \{ a u_1 + b v_1 \} +r^2 a p_1 +r^2 c u_1 ] \nonumber\\ &\quad +\frac{1}{r \sin(\theta)} \frac{\partial}{\partial \theta} [ \sin(\theta) ( v_0 \{ a u_1 + b v_1 \} + b p_1 + c v_1 )], \end{align}

in which the adjoint operator $L^*$ and vector $\boldsymbol {r}$ are given by

(3.18a,b)\begin{equation} L^{*}=\left(\begin{array}{ccc} \displaystyle -\bar{L} + \dfrac{\partial u_0}{\partial r} & \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial r} (r v_0) & \displaystyle -\dfrac{\partial}{\partial r}\\[8pt] \displaystyle \dfrac{1}{r} \dfrac{\partial u_0}{\partial \theta} - \dfrac{2 v_0}{r} & \displaystyle -\bar{L} + \dfrac{1}{r} \dfrac{\partial v_0}{\partial \theta} + \dfrac{u_0}{r} & \displaystyle -\dfrac{1}{r} \dfrac{\partial}{\partial \theta}\\[8pt] \displaystyle -\dfrac{\partial}{\partial r} - \dfrac{2}{r} & \displaystyle - \dfrac{1}{r} \dfrac{\partial}{\partial \theta} - \dfrac{\cot (\theta)}{r} & \displaystyle 0 \end{array}\right),\quad \boldsymbol{r}=\left(\begin{array}{c} a\\ b\\ c \end{array}\right). \end{equation}

Equation (3.17) may be integrated to yield

(3.19)\begin{equation} \langle \boldsymbol{r}^T L \boldsymbol{z} - \boldsymbol{z}^T L^* \boldsymbol{r} \rangle = 0, \end{equation}

where

(3.20)\begin{equation} \langle \ {\bullet}\ \rangle = \int^{\rm \pi}_{\theta=0} \int^{R}_{r=0} \ {\bullet}\ r^2 \sin(\theta) \,\textrm{d} r \,\textrm{d} \theta, \end{equation}

provided that the following condition is satisfied on the boundary of the sphere:

(3.21)\begin{equation} a(R, \theta) = 0. \end{equation}

It follows from this that if

(3.22)\begin{equation} L^* \boldsymbol{r} = \boldsymbol{0}, \end{equation}

subject to the condition (3.21), then our linear problem for the first correction (3.14)–(3.16a,b) can only have a solution if

(3.23)\begin{equation} \left \langle - \frac{a}{r \sin(\theta)} \frac{\partial}{\partial \theta} (\sin(\theta) \omega_0) + \frac{b}{r} \frac{\partial}{\partial r} (r \omega_0) \right \rangle=0, \end{equation}

for any $\boldsymbol {r}=(a,b,c)^\textrm {T}$ in the null space of the adjoint problem. It remains to find the linearly independent solution vectors $\boldsymbol {r}$ and substitute them into (3.23).

A linearly independent solution of the adjoint problem (3.22) and (3.21) is

(3.24)\begin{equation} \boldsymbol{r}_1 = (0,0,1)^\textrm{T}. \end{equation}

An infinite family of linearly independent solutions is given by

(3.25)\begin{equation} \boldsymbol{r}_2 = \left(f(\psi) u_0, f(\psi) v_0, -\alpha \int^{\psi} f(s) \, \textrm{d} s \right)^\textrm{T}. \end{equation}

The function $f(\psi )$ is differentiable on the domain $[C R^2 / 8, 0]$. Given this condition, the Weierstrass approximation theorem ensures that a polynomial,

(3.26)\begin{equation} p(\psi) = a_0 + a_1 \psi + \cdots + a_q \psi^q, \end{equation}

exists with $a_i$ real constants and $q$ non-negative integer which uniformly approximates $f(\psi )$ on the domain $[C R^2 / 8,0]$. We consider the vector space of polynomials $p(\psi )$ over the field of real numbers. If we construct a basis for this space of polynomials $\{ \psi ^n \mid n \in \textrm {I N} \cup \{ 0 \} \}$, then all functions $f(\psi )$ are uniformly approximated by our basis. Hence, it is sufficient to construct solvability conditions for the basis. A countably infinite family of linearly independent solutions of the adjoint problem (2.29) and (2.28) have been deduced; that is, $f(\psi )=\psi ^n$ for $n$ in the non-negative integers.

A family of vorticity equations are required for Stuart cats’ eyes, Mallier–Maslowe vortices, travelling waves in two-dimensional plane Poiseuille flow (Smith & Wissink Reference Smith and Wissink2015) and two-dimensional Kolmogorov flow (Smith & Wissink Reference Smith and Wissink2018). However, in the case of Hill's spherical vortex, there are no solutions corresponding to vorticity.

3.4. Solvability conditions

Only the second solution corresponds to physical solvability conditions. The corresponding solvability conditions are

(3.27)\begin{equation} \left \langle - \frac{\psi^{n} u_0}{r \sin(\theta)} \frac{\partial}{\partial \theta} (\sin(\theta) \omega_0) + \frac{\psi^{n} v_0}{r} \frac{\partial}{\partial r} (r \omega_0) \right \rangle=0, \end{equation}

where $n$ is a non-negative integer. In the remainder of this subsection, the leading-order solution will be shown to satisfy this countably infinite set of solvability conditions.

We substitute the leading-order solution (3.12)–(3.13a,b) into the left-hand side of the solvability conditions (3.27) to yield

(3.28)\begin{align} &\left \langle - \frac{\psi^{n} u_0}{r \sin(\theta)} \frac{\partial}{\partial \theta} (\sin(\theta) \omega_0) + \frac{\psi^{n} v_0}{r} \frac{\partial}{\partial r} (r \omega_0)\right \rangle \nonumber\\ &\quad ={-}\frac{40}{R^2} \left ( \frac{C}{2} \right )^{n+2} \int^{R}_{r=0} r^{2n+2} \left ( 1 - \frac{r^2}{R^2} \right )^{n+1} \textrm{d} r \int^{\rm \pi}_{\theta=0} \sin^{2 n + 1}(\theta) \cos^2 (\theta) \,\textrm{d} \theta \nonumber\\ &\qquad -\frac{40}{R^2} \left ( \frac{C}{2} \right )^{n+2} \int^{R}_{r=0} r^{2n+2} \left ( 1 - \frac{r^2}{R^2} \right )^{n} \left ( 1 - \frac{2 r^2}{R^2} \right ) \textrm{d} r \int^{\rm \pi}_{\theta=0} \sin^{2 n + 3}(\theta) \,\textrm{d} \theta. \end{align}

The second term on the right-hand side may be rewritten

(3.29)\begin{align} &\left \langle - \frac{\psi^{n} u_0}{r \sin(\theta)} \frac{\partial}{\partial \theta} (\sin(\theta) \omega_0) + \frac{\psi^{n} v_0}{r} \frac{\partial}{\partial r} (r \omega_0) \right \rangle \nonumber\\ &\quad ={-}\frac{40}{R^2} \left ( \frac{C}{2} \right )^{n+2} \int^{R}_{r=0} r^{2n+2} \left ( 1 - \frac{r^2}{R^2} \right )^{n+1} \textrm{d} r \int^{\rm \pi}_{\theta=0} \sin^{2 n + 1}(\theta) \cos^2 (\theta) \,\textrm{d} \theta \nonumber\\ &\qquad -\frac{40}{R^2} \left ( \frac{C}{2} \right )^{n+2} \int^{R}_{r=0} r^{2n + 2} \left ( 1 - \frac{r^2}{R^2} \right )^{n+1} \textrm{d} r \int^{\rm \pi}_{\theta=0} \sin^{2 n + 3}(\theta) \,\textrm{d} \theta\nonumber\\ &\qquad +\frac{40}{R^4} \left ( \frac{C}{2} \right )^{n+2} \int^{R}_{r=0} r^{2(n+2)} \left ( 1 - \frac{r^2}{R^2} \right )^{n} \textrm{d} r \int^{\rm \pi}_{\theta=0} \sin^{2 n + 3}(\theta) \,\textrm{d} \theta. \end{align}

Using the two identities involving definite integrals in Appendix A, we find that

(3.30)\begin{align} &\left \langle - \frac{\psi^{n} u_0}{r \sin(\theta)} \frac{\partial}{\partial \theta} (\sin(\theta) \omega_0) + \frac{\psi^{n} v_0}{r} \frac{\partial}{\partial r} (r \omega_0)\right \rangle \nonumber\\ &\quad ={-}\frac{40}{R^2} \left ( \frac{C}{2} \right )^{n+2} \int^{\rm \pi}_{\theta=0} \sin^{2 n + 1}(\theta) \,\textrm{d} \theta \nonumber\\ &\qquad \times\left \{ \int^{R}_{r=0} r^{2n + 2} \left ( 1 - \frac{r^2}{R^2} \right )^{n+1} \textrm{d} r - \frac{2 n + 2}{(2 n + 3)} \frac{1}{R^2} \int^{R}_{r=0} r^{2(n+2)} \left ( 1 - \frac{r^2}{R^2} \right )^{n} \textrm{d} r \right \}. \end{align}

If we apply integration by parts to the first of the two integrals in the curly brackets, then the first integral is shown to be equal and opposite to the second integral. Therefore, all of the solvability conditions are satisfied. Provided that there are no further solutions to the adjoint problem (3.21)–(3.22), the problem for the first correction (3.14)–(3.16a,b) does have an infinite family of solutions by the Fredholm alternative. In fact, one member of this infinite family of solutions of (3.14)–(3.16a,b) is given by $u_1=v_1=0$ and $p_1=-2 \alpha r \cos (\theta )$, which was described by Saffman (Reference Saffman1992). Hill's spherical vortex has been confirmed to be consistent with the Navier–Stokes equations. We now consider a graphical interpretation in the case of Hill's spherical vortex. Equations (3.27) represent a manifold in infinite-dimensional phase space for steady states of the Navier–Stokes equations, which is denoted by an (orange) surface in figure 4. The leading-order solution (3.12) is represented by a (green) dot in figure 4, which resides on the manifold. In the case of Stuart cats’ eyes or the Mallier–Maslowe vortices, the manifold is given by the steady version of (2.38)–(2.40) and the leading-order solutions are (2.13) and (2.18), respectively. However, these leading-order solutions are now represented by the (red) cross in figure 4, which do not reside on the appropriate manifold.

Figure 4.

A schematic of the structure of infinite-dimensional phase space for steady states of the Navier–Stokes equations. The (orange) surface denotes the viscous manifold for asymptotic solutions of the Navier–Stokes equations in the high-Reynolds-number limit. The (green) dot represents the leading-order solution in the case of Hill's spherical vortex and the (red) cross represents the leading-order solution for either the Stuart cats’ eyes or the Mallier–Maslowe vortices.

4.1. Introduction

In this subsection, we will describe the mathematical model for the Lamb–Chaplygin dipole. In order to avoid an excessive number of suffices, the same notation as in §§ 2 and 3 will be adopted for a number of the quantities; these are defined anew in this section. The system of dimensionless equations to be studied are now introduced. We consider the two-dimensional Navier–Stokes equations for incompressible Newtonian fluids in plane polar coordinates

(4.1a)$$\begin{gather} \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial r} + \frac{v}{r} \frac{\partial u}{\partial \theta} -\frac{v^2}{r} +\frac{\partial p}{\partial r} ={-} \frac{\epsilon}{r} \frac{\partial \omega}{\partial \theta}, \end{gather}$$

(4.1b)$$\begin{gather}\frac{\partial v}{\partial t} + u \frac{\partial v}{\partial r} + \frac{v}{r} \frac{\partial v}{\partial \theta} +\frac{uv}{r} +\frac{1}{r} \frac{\partial p}{\partial \theta} = \epsilon \frac{\partial \omega}{\partial r}, \end{gather}$$

and the continuity equation of the form

(4.2)\begin{equation} \frac{1}{r} \frac{\partial}{\partial r}(ru) + \frac{1}{r} \frac{\partial v}{\partial \theta} = 0, \end{equation}

in which $r < R$ is the radial coordinate, $R$ is the radius of the circle, $\theta$ is the azimuthal coordinate, $t$ is time, $u$ is the radial velocity component, $v$ is the azimuthal velocity component, $p$ is the pressure, $\omega$ is the vorticity given by

(4.3)\begin{equation} {\omega} = \frac{1}{r} \frac{\partial}{\partial r}(r v)- \frac{1}{r} \frac{\partial u}{\partial \theta}, \end{equation}

$\epsilon = 1/ Re$ is the reciprocal of the Reynolds number and $0 < \epsilon \ll 1$. The boundary conditions are

(4.4) \begin{equation} \left.\begin{array}{c@{}} u(R,\theta,t) = 0, \quad v(R, \theta,t) = C (\epsilon t) \sin ( \theta ), \\[4pt] {[}u,v,p](r,0,t)=[u,v,p](r,2 {\rm \pi}, t), \end{array}\right\} \end{equation}

in which $C (\epsilon t)$ is a non-zero slowly varying function. The Lamb–Chaplygin dipole corresponds to $C = - 2 U$, where $U$ is the far-field velocity of the surrounding flow.

4.2. The leading-order solution

We will first perform the perturbation procedure on the nonlinear system (4.1)–(4.4) and obtain the solution for Lamb–Chaplygin dipole. We introduce expansions of the form

(4.5a–d)\begin{equation} u \sim u_0 + \epsilon u_1, \quad v \sim v_0 + \epsilon v_1, \quad p \sim p_0 + \epsilon p_1 , \quad \omega \sim \omega_0 + \epsilon \omega_1 , \end{equation}

as $\epsilon \rightarrow 0$. At leading order, for the quasi-steady state, we obtain

(4.6a)$$\begin{gather} \bar{L} u_0 - \frac{v_0^2}{r} + \frac{\partial p_0}{\partial r} = 0, \end{gather}$$

(4.6b)$$\begin{gather}\bar{L} v_0 + \frac{u_0 v_0}{r} + \frac{1}{r} \frac{\partial p_0}{\partial \theta} = 0, \end{gather}$$

(4.6c)$$\begin{gather}\frac{1}{r} \frac{\partial}{\partial r}(ru_0) + \frac{1}{r} \frac{\partial v_0}{\partial \theta} = 0, \end{gather}$$

where the differential operator

(4.7)\begin{equation} \bar{L}=u_0 \frac{\partial}{\partial r} + \frac{1}{r} v_0 \frac{\partial}{\partial \theta}. \end{equation}

We take the radial partial derivative of (4.6b) in the form

(4.8)\begin{equation} u_0 \frac{\partial}{\partial r} (r v_0) + v_0 \frac{\partial v_0}{\partial \theta} + \frac{\partial p_0}{\partial \theta} = 0 \end{equation}

minus the azimuthal partial derivative of (4.6a) to obtain the vorticity equation $\bar {L} \omega _0 =0$. A streamfunction $\psi$ is defined by the equations

(4.9a,b)\begin{equation} u_0 = \frac{1}{r} \frac{\partial \psi}{\partial \theta}, \quad v_0 ={-} \frac{\partial \psi}{\partial r}, \end{equation}

so that (4.6c) is automatically satisfied and the vorticity equation may be rewritten as

(4.10)\begin{equation} \frac{\partial \psi}{\partial \theta} \frac{\partial \omega_0}{\partial r} -\frac{\partial \psi}{\partial r} \frac{\partial \omega_0}{\partial \theta}=0. \end{equation}

The Lamb–Chaplygin dipole corresponds to the solution of (4.10) given by $\omega _0 = k^2 \psi$ with k being an arbitrary constant, so that $\psi$ satisfies the partial differential equation

(4.11)\begin{equation} \frac{\partial^2 \psi}{\partial r^2} + \frac{1}{r} \frac{\partial \psi}{\partial r} + \frac{1}{r^2} \frac{\partial^2 \psi}{\partial \theta^2} ={-}k^2 \psi \end{equation}

for $r < R$ with the boundary conditions

(4.12a,b)\begin{equation} \psi (R,\theta, \tilde{t}) = 0, \quad \frac{\partial \psi}{\partial r} (R,\theta, \tilde{t}) ={-}C (\tilde{t}) \sin ( \theta), \quad \psi(r,0, \tilde{t})=\psi(r,2 {\rm \pi}, \tilde{t}), \end{equation}

in which $\tilde {t} =\epsilon t$ is the slowly varying time scale. The Lamb–Chaplygin dipole corresponds to the solution of this boundary value problem in the form

(4.13)\begin{equation} \psi (r, \theta, \tilde{t}) ={-}\frac{C (\tilde{t}) J_1 (k r) \sin (\theta)}{k J_0(k R)}, \end{equation}

for $r \leq R$. The first zero of the Bessel function $J_1 (k R)=0$ fixes $k R \approx 3.8317$. The streamlines for $C (\tilde {t})=1$ and $R=1$ are shown in figure 5. We deduce that the velocities are given by

(4.14a,b)\begin{equation} u_0 (r, \theta, \tilde{t}) ={-}\frac{C (\tilde{t}) J_1 (k r) \cos (\theta)}{k r J_0(k R)}, \quad v_0 (r, \theta, \tilde{t}) = \frac{C (\tilde{t}) J_1^{\prime} (k r) \sin (\theta)}{J_0(k R)}. \end{equation}

4.3. The first correction

We now derive the linear problem for the first correction and the conditions (4.24) under which it may have a solution. At next order, we obtain

(4.15) \begin{equation} L \boldsymbol{z} = \left(\begin{array}{@{}c@{}} \displaystyle -\dfrac{\partial u_0}{\partial \tilde{t}} - \dfrac{1}{r} \dfrac{\partial \omega_0}{\partial \theta}\\[10pt] \displaystyle -\dfrac{\partial v_0}{\partial \tilde{t}} + \dfrac{\partial \omega_0}{\partial r}\\ 0 \end{array}\right)=\left(\begin{array}{@{}c@{}} \displaystyle -\dfrac{\partial u_0}{\partial \tilde{t}} - k^2 u_0\\[10pt] \displaystyle -\dfrac{\partial v_0}{\partial \tilde{t}} - k^2 v_0\\ 0 \end{array} \right), \end{equation}

in which

(4.16) \begin{equation} L=\left(\begin{array}{@{}ccc@{}} \displaystyle \bar{L} + \dfrac{\partial u_0}{\partial r} & \displaystyle \dfrac{1}{r} \dfrac{\partial u_0}{\partial \theta} - \dfrac{2 v_0}{r} & \displaystyle \dfrac{\partial}{\partial r}\\[10pt] \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial r} (r v_0) & \displaystyle \bar{L} + \dfrac{1}{r} \dfrac{\partial v_0}{\partial \theta} + \dfrac{u_0}{r} & \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial \theta}\\[10pt] \displaystyle \dfrac{\partial}{\partial r} + \dfrac{1}{r} & \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial \theta} & \displaystyle 0 \end{array}\right),\quad \boldsymbol{z}=\left(\begin{array}{@{}c@{}} u_1\\ v_1\\ p_1 \end{array}\right), \end{equation}

with the boundary conditions

(4.17) \begin{equation} \left.\begin{array}{c@{}} u_1 (R,\theta, \tilde{t}) = 0, \quad v_1 (R, \theta, \tilde{t}) = 0, \\[4pt] {[}u_1,v_1,p_1](r,0, \tilde{t})=[u_1,v_1,p_1](r,2 {\rm \pi}, \tilde{t}). \end{array}\right\} \end{equation}

A general solution to the linear problem for the first correction (4.15)–(4.17) is very difficult to find. Therefore, we study the adjoint problem in order to determine the conditions under which the problem for the first correction has a solution. Following a modified version of the analysis in Smith (Reference Smith2007), we have the following divergence formulation:

(4.18)\begin{align} \boldsymbol{r}^T L \boldsymbol{z} - \boldsymbol{z}^T L^* \boldsymbol{r}&= \frac{1}{r} \frac{\partial}{\partial r} \left[ r u_0 \{ a u_1 + b v_1 \} +r a p_1 +r c u_1 \right] \nonumber\\ &\quad +\frac{1}{r} \frac{\partial}{\partial \theta} \left[ v_0 \{ a u_1 + b v_1 \} + b p_1 + c v_1 \right], \end{align}

in which the adjoint operator $L^*$ and vector $\boldsymbol {r}$ are given by

(4.19) \begin{equation} L^{*}=\left(\begin{array}{@{}ccc@{}} \displaystyle -\bar{L} + \dfrac{\partial u_0}{\partial r} & \displaystyle \dfrac{1}{r} \dfrac{\partial}{\partial r} (r v_0) & \displaystyle -\dfrac{\partial}{\partial r}\\[10pt] \displaystyle \dfrac{1}{r} \dfrac{\partial u_0}{\partial \theta} - \dfrac{2 v_0}{r} & \displaystyle -\bar{L} + \dfrac{1}{r} \dfrac{\partial v_0}{\partial \theta} + \dfrac{u_0}{r} & \displaystyle -\dfrac{1}{r} \dfrac{\partial}{\partial \theta}\\[10pt] \displaystyle -\dfrac{\partial}{\partial r} - \dfrac{1}{r} & \displaystyle - \dfrac{1}{r} \dfrac{\partial}{\partial \theta} & \displaystyle 0 \end{array}\right), \quad \boldsymbol{r}=\left(\begin{array}{@{}c@{}} a\\ b\\ c \end{array}\right). \end{equation}

Equation (4.18) may be integrated to yield

(4.20)\begin{equation} \langle \boldsymbol{r}^T L \boldsymbol{z} - \boldsymbol{z}^T L^* \boldsymbol{r} \rangle = 0, \end{equation}

where

(4.21)\begin{equation} \langle \ {\bullet} \ \rangle = \int^{2 {\rm \pi}}_{\theta=0} \int^{R}_{r=0}\ {\bullet} \ r \,\textrm{d} r \,\textrm{d} \theta, \end{equation}

provided that the following periodicity conditions and boundary condition on the radius of the circle are satisfied:

(4.22a)$$\begin{gather} [a,b,c](r,0, \tilde{t}) = [a,b,c](r,2 {\rm \pi}, \tilde{t}), \end{gather}$$

(4.22b)$$\begin{gather}a(R, \theta, \tilde{t}) = 0. \end{gather}$$

It follows from this that if

(4.23)\begin{equation} L^* \boldsymbol{r} = \boldsymbol{0}, \end{equation}

subject to the conditions (4.22), then our linear problem for the first correction (4.15)–(4.17) can only have a solution if

(4.24)\begin{equation} \left \langle a \left [ -\frac{\partial u_0}{\partial \tilde{t}} - k^2 u_0 \right ] + b \left [ -\frac{\partial v_0}{\partial \tilde{t}} - k^2 v_0 \right ] \right \rangle=0, \end{equation}

for any $\boldsymbol {r}=(a,b,c)^\textrm {T}$ in the null space of the adjoint problem. It remains to find the linearly independent solution vectors $\boldsymbol {r}$ and substitute them into (4.24).

Two linearly independent solutions of the adjoint problem (4.23) and (4.22) are

(4.25)\begin{equation} \boldsymbol{r}_1 = (0, 0, 1)^\textrm{T}, \quad \boldsymbol{r}_2 = (0,r,r v_0)^\textrm{T}. \end{equation}

An infinite family of linearly independent solutions is given by

(4.26)\begin{equation} \boldsymbol{r}_3 = \left(f(\psi) u_0, f(\psi) v_0, -k^2 \int^{\psi} sf(s) \, \textrm{d} s \right)^\textrm{T}. \end{equation}

The function $f(\psi )$ is differentiable on $[-\psi _{max},\psi _{max}]$, where $\psi _{max}$ is the bound on the streamfunction. Given this condition, the Weierstrass approximation theorem ensures that a polynomial,

(4.27)\begin{equation} p(\psi) = a_0 + a_1 \psi + \cdots + a_q \psi^q, \end{equation}

exists with $a_i$ real constants and $q$ non-negative integer which uniformly approximates $f(\psi )$ on $[-\psi _{max},\psi _{max}]$. We consider the vector space of polynomials $p(\psi )$ over the field of real numbers. If we construct a basis for this space of polynomials $\{ \psi ^n \mid n \in \textrm {I N} \cup \{ 0 \} \}$, then all functions $f(\psi )$ are uniformly approximated by our basis. Hence, it is sufficient to construct solvability conditions for the basis. A countably infinite family of linearly independent solutions of the adjoint problem (4.23) and (4.22) have been deduced; that is, $f(\psi )=\psi ^n$ for $n$ in the non-negative integers. We note that there are no linearly independent solutions corresponding to vorticity.

4.4. Modulation equations

The first solution to the adjoint problem, $\boldsymbol {r}_1$, corresponds to a trivial modulation equation. Only the second and third solutions correspond to physical modulation equations, which are described in the following subsections.

4.4.1. Angular momentum modulation equation

If we substitute the second vector $\boldsymbol {r}_2$ into (4.24), then we obtain our first secularity condition

(4.28)\begin{equation} \left \langle -\frac{\partial}{\partial \tilde{t}} (r v_0) - k^2 r v_0 \right \rangle=0. \end{equation}

This may be rewritten as the modulation equation

(4.29)\begin{equation} \frac{\textrm{d} }{\textrm{d} \tilde{t}} \langle r v_0 \rangle ={-}k^2 \langle r v_0 \rangle. \end{equation}

4.4.2. Generalized energy modulation equations

If we substitute the third vector, $\boldsymbol {r}_3$, into (4.24) with $f(\psi )=\psi ^n$, then we obtain the further secularity condition

(4.30)\begin{equation} \left \langle \psi^n u_0 \left [ -\frac{\partial u_0}{\partial \tilde{t}} - k^2 u_0 \right ] + \psi^n v_0 \left [ -\frac{\partial v_0}{\partial \tilde{t}} - k^2 v_0 \right ] \right \rangle=0. \end{equation}

This may be rewritten as

(4.31)\begin{equation} \left \langle \psi^n \left [ \frac{\partial E_0}{\partial \tilde{t}} + 2 k^2 E_0 \right ] \right \rangle=0, \end{equation}

for $n$ in the non-negative integers.

4.5. Inconsistency of the steady state

Using (4.31) when $n=0$, we have the solvability condition

(4.32)\begin{equation} \langle E_0 \rangle =0. \end{equation}

The left-hand side of this solvability condition is positive and the solvability condition is not satisfied. The problem for the first correction (4.15)–(4.17) does not have a solution by the Fredholm alternative. The Lamb–Chaplygin dipole is a solution to the steady Euler equations, but it is not a leading-order solution to the steady Navier–Stokes equation for high-Reynolds-number flows. As such, the Lamb–Chaplygin dipole does not approximate a steady state of the Navier–Stokes equations.

4.6. Consistency of the quasi-steady state

If we substitute the leading-order solution (4.13)–(4.14a,b) into (4.29), then we have a trivial identity. Substitution of the leading-order solution (4.13)–(4.14a,b) into (4.31) produces the equation

(4.33)\begin{equation} \left \langle \psi^n E_0 \left [ \frac{\textrm{d} }{\textrm{d} \tilde{t}}(C^2) + 2 k^2 C^2 \right ] \right \rangle=0, \end{equation}

for $n$ in the non-negative integers. This may be rewritten as

(4.34)\begin{equation} \langle \psi^n E_0 \rangle \left [ \frac{\textrm{d} C}{\textrm{d} \tilde{t}} + k^2 C \right ]=0. \end{equation}

These modulation equations are all satisfied providing

(4.35)\begin{equation} C(\tilde{t}) = C_0 \,\textrm{e}^{- k^2 \tilde{t}}, \end{equation}

where $C_0$ is a constant of integration. Provided that there are no further solutions to the adjoint problem (4.22)–(4.23), the problem for the first correction (4.15)–(4.17) does have an infinite family of solutions by the Fredholm alternative. In fact, one member of this infinite family of solutions of (4.15)–(4.17) is given by $u_1=v_1=p_1=0$. Furthermore, our leading-order solution turns out to be an exact solution of the Navier–Stokes equations as described below.

4.7. An exact solution

If we rewrite the leading-order solution (4.13)–(4.14a,b) in terms of the original dimensionless variables, we obtain

(4.36a,b)$$\begin{gather} u (r, \theta, t) ={-}\frac{C_0 \,\textrm{e}^{- k^2 \epsilon t} J_1 (k r) \cos (\theta)}{k r J_0(k R)}, \quad v (r, \theta, t) = \frac{C_0 \,\textrm{e}^{- k^2 \epsilon t} J_1^{\prime} (k r) \sin (\theta)}{J_0(k R)}, \end{gather}$$

(4.37a,b)$$\begin{gather}\omega (r, \theta, t) ={-}\frac{k^2 C_0 \,\textrm{e}^{- k^2 \epsilon t} J_1 (k r) \sin (\theta)}{k J_0(k R)}, \quad p (r, \theta, t) ={-}\frac{\omega^2}{2 k^2}- \frac{1}{2}(u^2+v^2), \end{gather}$$

for $r \leq R$. This is an original exact solution of the Navier–Stokes equations (4.1)–(4.4) provided that $C (\epsilon t) = C_0 \,\textrm {e}^{- k^2 \epsilon t}$.

This exact solution can be seen to be one of the most important because it is time dependent; it has non-zero nonlinear convective terms; it is restricted to a finite domain; and it has a decay rate dependent on the dipole radius. We do not know of another exact solution with this combination of properties (Drazin & Riley Reference Drazin and Riley2006). Numerical and experimental evidence also exists which indicate that it is linearly stable (see Couder & Basdevant Reference Couder and Basdevant1986).

The decay rate may be written in the form

(4.38)\begin{equation} \left ( \frac{3.8317}{R} \right )^2 \epsilon. \end{equation}

The decay rate varies with the reciprocal of the Reynolds number, $\epsilon = 1/ Re$, and with the reciprocal of the square of the dipole radius $R$ (see figure 6). The decay rate increases rapidly as the dipole radius decreases. If the dipole radius $R \ll \epsilon ^{1/2} = Re^{-1/2}$, then the dissipation is exponentially fast. If the dipole radius $R \sim \epsilon ^{1/2} = Re^{-1/2}$, then the exponential decay is order one. Larger values of the dipole radius, $R \gg \epsilon ^{1/2} = Re^{-1/2}$, result in an exponentially slow dissipation rate and the solution may easily be mistaken for a steady state.

Figure 6.

Decay rate of the exact solution of the Navier–Stokes equations (4.36a,b)–(4.37a,b) as a function of the dipole radius $R$ for three values of the Reynolds number.

The long time scale viscous evolution of symmetric two-dimensional dipoles has been studied elsewhere in which the domain of rotational flow was not restricted to a circle. Kizner, Khvoles & Kessler (Reference Kizner, Khvoles and Kessler2010) have shown that viscosity first takes the dipole to an intermediate asymptotic state and then slowly moves away. Kizner et al. (Reference Kizner, Khvoles and Kessler2010) claim that, in the limit of vanishing viscosity (large Reynolds number), a unique elliptical dipole solution is selected with a separatrix aspect ratio of $1.037$. Our exact solution of the full Navier–Stokes equations has shown that there is an alternative evolution when the domain of rotational flow is restricted to a circle.