3.2.1. Outer solutions

We attempt to obtain the outer solutions ${\boldsymbol {u}}^o$, denoted by the superscript ‘$o$’, of the governing equations (3.1a)–(3.3). The outer domain is defined as the outside or the inside of the cylinder with respect to $\epsilon$, i.e. $r - 1 = O ( 1 )$. The outer solutions cannot satisfy the boundary condition on the solid–fluid boundary, as mentioned below. The inner solution near the boundary is therefore defined as $r - 1 = O ( \epsilon )$ in the subsequent § 3.2.2.

Following our previous study (see Ueda & Kida Reference Ueda and Kida2021), the velocities, $u_r^o$ and $u_\theta ^o$, in the radial ($r$) and tangential ($\theta$) directions are asymptotically represented by

(3.4a,b)\begin{equation} u_{r}^o =\sum_{n = 0}^\infty \epsilon^n u_{r,n}^o, \quad u_{\theta}^o = \sum_{n = 0}^\infty \epsilon^n u_{\theta,n}^o. \end{equation}

Also, the pressure $p^o$ is asymptotically written as

(3.5)\begin{equation} p^o = \sum_{n = 0}^\infty \epsilon^n p_{n}^o. \end{equation}

Then, (3.1a) and (3.1b) reduce to the following for the first two lowest-order solutions in the $r$ and $\theta$ directions, respectively;

(3.6a)$$\begin{gather} \frac{\partial u_{r0}^o}{\partial T} = \lambda_o \chi( - u_{r0}^o ), \end{gather}$$

(3.6b)$$\begin{gather}\frac{\partial u_{r1}^o}{\partial T} + u_{r0}^o \frac{\partial u_{r0}^o}{\partial r} + u_{\theta 0}^o \frac{\partial u_{r 0}^0}{r\partial \theta}- \frac{u_{\theta 0}^{o2}}{r} =- \frac{1}{\rho} \frac{\partial p_0^o}{\partial r} + \lambda_o \chi ( - u_{r1}^o ), \end{gather}$$

(3.6c)$$\begin{gather}\frac{\partial u_{\theta 0}^o}{\partial T} = \lambda_o \chi ( \varOmega r H ( T ) - u_{\theta 0}^o ), \end{gather}$$

(3.6d)$$\begin{gather}\frac{\partial u_{\theta 1}^o}{\partial T} + u_{r0}^o \frac{\partial u_{\theta 0}^o}{\partial r} + u_{\theta 0}^o \frac{\partial u_{\theta 0}^0}{r\partial \theta} + \frac{u_{r0}^o u_{\theta 0}^{o}}{r} =- \frac{1}{\rho} \frac{\partial p_0^o}{r \partial \theta} + \lambda_o \chi ( - u_{\theta 1}^o ). \end{gather}$$

In the fluid domain, we have $\partial u_{r,0}^o / \partial T = \partial u_{\theta,0}^o / \partial T = 0$, because of $\chi = 0$. This gives $u_{r,0}^o = u_{\theta,0}^0 = 0$ because $u_{r,0}^o$ and $u_{\theta,0}^o$ are zero at $T = 0$. We can obtain the outer solutions in the fluid domain by the recursive calculations for the higher-order approximation of (3.1a) and (3.1b) such as

(3.7)\begin{equation} u_{r,i}^o = u_{\theta,i}^o = 0 \quad \text{for} \ i = 0,1, 2, \ldots. \end{equation}

Assuming that the pressure sufficiently far from the cylinder is zero, we have also

(3.8)\begin{equation} p_i^o = 0 \quad \text{for} \ i = 0, 1, 2, \ldots. \end{equation}

In contrast, since $\chi = 1$ in the body domain ${\mathscr {S}}$, we have $u_{r,i}^o = 0$ and $u_{\theta, i}^o = \varOmega r H ( T ) \delta _{i 0}$ for $i = 0, 1, 2, \ldots$. This relation gives, in the body domain,

(3.9a–c)\begin{equation} u_r^o = 0, \quad u_{\theta}^o = \varOmega r H ( T ), \quad p^o = \frac{\rho}{2} \varOmega^2 r^2 H(T) + C, \end{equation}

where $C$ is an integral constant.

3.2.2. Inner solutions

The outer flow described by (3.7) and (3.9a–c) causes the slip velocity on the surface of the cylinder (at $r = 1$) so that a non-uniform domain can exist near the cylinder surface. To remedy this, the radial coordinate $r$ is stretched as $R = ( r - 1 ) / \epsilon$ where the solution in the stretched domain is called the inner solution that is denoted by the superscript ‘$i$’. The governing equations and the continuity equation of (3.1a)–(3.3) are then described as, in the inner domain,

(3.10a)\begin{align} & \frac{\partial u_r^i}{\partial T} + u_r^i \frac{\partial u_r^i}{\partial R} + \frac{\epsilon}{1 + \epsilon R} u_\theta^i \frac{\partial u_r^i}{\partial \theta} - \epsilon \frac{u_\theta^{i2}}{1 + \epsilon R} =- \frac{1}{\rho} \frac{\partial p^i}{\partial R} - \lambda_o \chi u_r^i \nonumber\\ & \quad + \nu_o \Bigg[ \bigg( \frac{\partial^2}{\partial R^2} + \frac{\epsilon}{1 + \epsilon R} \frac{\partial}{\partial R} + \frac{\epsilon^2}{( 1 + \epsilon R )^2} \frac{\partial^2}{\partial \theta^2} \bigg) u_r^i - \frac{\epsilon^2}{( 1 + \epsilon R )^2} \Bigg( u_r^i + 2 \frac{\partial u_\theta^i}{\partial \theta} \Bigg)\Bigg], \end{align}

(3.10b)\begin{align} & \frac{\partial u_\theta^i}{\partial T} + u_r^i \frac{\partial u_\theta^i}{\partial R} + \frac{\epsilon}{1 + \epsilon R} u_\theta^i \frac{\partial u_\theta^i}{\partial \theta} + \epsilon \frac{u_r^i u_\theta^i}{1 + \epsilon R} \nonumber\\ &\quad =- \frac{1}{\rho} \frac{\epsilon}{1 + \epsilon R} \frac{\partial p^i}{\partial \theta} + \lambda_o \chi [ ( 1 + \epsilon R ) \varOmega H ( T ) - u_\theta^i ]\nonumber\\ & \qquad + \nu_o \left[ \left( \frac{\partial^2}{\partial R^2} + \frac{\epsilon}{1 + \epsilon R} \frac{\partial}{\partial R} + \frac{\epsilon^2}{( 1 + \epsilon R )^2} \frac{\partial^2}{\partial \theta^2} \right) u_\theta^i - \frac{\epsilon^2}{( 1 + \epsilon R )^2} \left( u_\theta^i - 2 \frac{\partial u_r^i}{\partial \theta} \right) \right], \end{align}

(3.10c)\begin{align} &\frac{\partial u_r^i}{\partial R} + \frac{\epsilon}{1 + \epsilon R} u_r^i + \frac{\epsilon}{1 + \epsilon R} \frac{\partial u_\theta^i}{\partial \theta} = 0, \end{align}

in which $u_r^i$, $u_\theta ^i$ and $p^i$ are assumed to be asymptotically represented as

(3.11a–c)\begin{equation} u_r^i = \sum_{n = 0}^\infty \epsilon^n u_{r n}^i, \quad u_\theta^i = \sum_{n = 0}^\infty \epsilon^n u_{\theta n}^i, \quad p^i = \sum_{n = 0}^\infty \epsilon^n p_n^i. \end{equation}

From (3.10c), we have $\partial u_{r0}^i / \partial R = 0$ and, therefore, the first-order solution of $u_r^i$ is found to be independent of $R$. The matching procedure to the outer solution inside the body domain (i.e. $u_r^{io} = u_r^{oi} = 0$) yields

(3.12)\begin{equation} u_{r0}^i = 0. \end{equation}

Equation (3.12) is found to fulfil the matching condition to the outer solution of the fluid domain. The first approximations of (3.10a) and (3.10b) are written as

(3.13a)$$\begin{gather} 0 =- \frac{1}{\rho} \frac{\partial p_0^i}{\partial R}, \end{gather}$$

(3.13b)$$\begin{gather}\frac{\partial u_{\theta 0}^i}{\partial T} = \lambda_o \chi ( \varOmega H ( T ) - u_{\theta 0}^i ) + \nu_o \frac{\partial^2u_{\theta0}^i}{\partial R^2}. \end{gather}$$

From (3.13a), $p_0^i$ is found to be independent of $R$. Furthermore, from the matching procedure to $p_i^o$ of (3.8) in the flow domain (i.e. $p^{io} = p_0^i + O ( \epsilon ) = p^{o i} = 0$), we have $p_0^i = 0$ for $R \geq 0$. For $R < 0$, (3.9a–c) becomes $p^{o i} = ( \rho / 2 ) \varOmega ^2 H ( T ) + C$ and we have

(3.14)\begin{equation} p_0^i = \left\{ \begin{array}{@{}ll} 0 & \text{for} \ R \geq 0,\\ \displaystyle\dfrac{\rho}{2} \varOmega^2 H ( T ) + C & \text{for}\ R < 0. \end{array} \right.\end{equation}

For $R \geq 0$, the governing equation of (3.13b) with respect to $u_{\theta 0}^i$ is written as

(3.15)\begin{equation} \frac{\partial u_{\theta 0}^i}{\partial T} = \nu_o \frac{\partial^2 u_{\theta 0}^i}{\partial R^2}. \end{equation}

The matching condition to the fluid domain gives the following boundary condition:

(3.16)\begin{equation} u_{\theta 0}^i \to 0 \quad \text{as} \ R \to \infty. \end{equation}

To solve (3.15) with (3.16), we employ the Laplace transform, $U_{\theta 0} = {\mathcal {L}} ( u_{\theta 0}^i ) = \int _0^\infty {\rm e}^{- s T} u_{\theta 0}^i ( R, \theta ; T) \,{\rm d} T$, similar to our previous study (see Ueda & Kida Reference Ueda and Kida2021). Then, the solution to (3.15) is easily derived as

(3.17)\begin{equation} U_{\theta 0} = A ( s, \theta ) {\rm e}^{- a R} \quad \text{with} \ a = \sqrt{\frac{s}{\nu_o}}, \end{equation}

where $A$ is an integral constant that is a function of $s$ and $\theta$.

For $R < 0$, the governing equation of (3.13b) is also written as

(3.18)\begin{equation} \frac{\partial u_{\theta0}^i}{\partial T} = \lambda_o \varOmega H ( T ) - \lambda_o u_{\theta 0}^i + \nu_o \frac{\partial^2 u_{\theta 0}^i}{\partial R^2}. \end{equation}

Taking into account the initial condition of $u_{\theta 0}^i = \varOmega$, the Laplace transform to (3.18) is written as

(3.19)\begin{equation} s U_{\theta 0} - \varOmega = \lambda_o \frac{\varOmega}{s} - \lambda_o U_{\theta 0} + \nu_o \frac{\partial^2 U_{\theta0}}{\partial R^2}. \end{equation}

The boundary condition as $R \to - \infty$ is obtained from the matching condition to the outer solution inside the body domain, i.e. $u_{\theta 0}^i \to \varOmega H ( T )$. Then, the solution to (3.19) is written as

(3.20)\begin{equation} U_{\theta 0} = B ( s, \theta ) {\rm e}^{\bar{a} R} + \frac{\varOmega}{s} \quad \text{with} \ \bar{a} = \sqrt{\frac{s + \lambda_o}{\nu_o}}, \end{equation}

where $B$ is an integral constant that is a function of $s$ and $\theta$. The integral constants, $A$ and $B$, can be determined as $A = B + \varOmega / s$ and $- a A = \bar {a} B$ by the enforcements of the continuity of the velocity and its gradient with respect to $R$ at $R = 0$. Therefore, we have, for (3.17) and (3.20),

(3.21)\begin{equation} U_{\theta 0} = \left\{ \begin{array}{@{}ll} \displaystyle \dfrac{\bar{a}}{a + \bar{a}} \dfrac{\varOmega}{s} {\rm e}^{- a R} & \text{for}\ R \geq 0,\\[9pt] \displaystyle - \dfrac{{a}}{a + \bar{a}} \dfrac{\varOmega}{s} {\rm e}^{\bar{a} R} + \dfrac{\varOmega}{s} & \text{for}\ R < 0. \end{array} \right. \end{equation}

Using the relations of (A1) and (A2) in Appendix A, (3.21) becomes, for $R \geq 0$,

(3.22) \begin{align} u_{\theta 0}^i&= \varOmega \Biggl\{ \text{erfc} \left( \frac{R}{2 \sqrt{\nu_o T}} \right) - \frac{R}{4 \sqrt{{\rm \pi} \nu_o}} \int_0^T \frac{\exp\left({- \dfrac{\lambda_o}{2} \xi}\right)}{( T - \xi )^{3 / 2}} \left[ {\rm I}_1 ( \lambda_o \xi / 2 ) + {\rm I}_0 ( \lambda_o \xi / 2 ) \right]\nonumber\\ &\quad \vphantom{\frac{\exp\left({- \dfrac{\lambda_o}{2} \xi}\right)}{( T - \xi )^{3 / 2}}}\times \exp\left({ - \frac{R^2}{4 \nu_o ( T - \xi )}}\right) {\rm d} \xi \Biggr\}, \end{align}

for $R < 0$,

(3.23)\begin{align} u_{\theta 0}^i &= \varOmega \left\{ 1 - \frac{| R |}{4 \sqrt{{\rm \pi} \nu_o}} \int_0^T \frac{\exp({- \lambda_o ( T - \xi / 2 )})}{( T - \xi )^{3 / 2}} \left[ {\rm I}_1 ( \lambda_o \xi / 2 ) + {\rm I}_0 ( \lambda_o \xi / 2 ) \right] \right.\nonumber\\ &\quad\times\exp\left.\left({- \frac{R^2}{4 \nu_o ( T - \xi )}}\right) {\rm d} \xi \right\}. \end{align}

Here, ${\rm I}_0 ( z )$ and ${\rm I}_1 ( z )$ are the zeroth- and the first-order modified Bessel functions of the first kind, respectively.

Let us carry out further analysis to the second approximation for seeking the solutions of $u_{r1}^i$ and $u_{\theta 1}^i$. Taking into account (3.12) and (3.22)–(3.23), (3.10c) is written as

(3.24)\begin{equation} \frac{\partial u_{r 1}^i}{\partial R} =- \frac{\partial u_{\theta 0}^i}{\partial \theta} = 0. \end{equation}

Hence, we find that

(3.25)\begin{equation} u_{r 1}^i = C ( T, \theta ), \end{equation}

where $C$ is an integral constant that is a function of $T$ and $\theta$. The matching to the outer solution and the enforcement of the continuity of $u_{r1}^i$ at $R = 0$ determine the integral constant of (3.25) as

(3.26)\begin{equation} u_{r 1}^i = 0. \end{equation}

The second approximations of the governing equations (3.10a) and (3.10b) are written as, taking into account (3.12) and (3.14),

(3.27a)$$\begin{gather} u_{\theta 0}^{i 2}= \frac{1}{\rho} \frac{\partial p_1^i}{\partial R}, \end{gather}$$

(3.27b)$$\begin{gather}\frac{\partial u_{\theta 1}^i}{\partial T} = \lambda_o \chi ( R \varOmega H ( T ) - u_{\theta 1}^i ) + \nu_o \Bigg( \frac{\partial^2 u_{\theta 1}^i}{\partial R^2} + \frac{\partial u_{\theta 0}^i}{\partial R} \Bigg). \end{gather}$$

The matching conditions to the outer solutions in the fluid and the solid domains give $p_1^i \to 0$ as $R \to \infty$, and $p_1^i \to \rho R \varOmega ^2 H(T)$ and $u_{\theta 0}^i \to \varOmega H(T)$ as $R \to - \infty$. From (3.27a), the pressure $p_1^i$ is then described as

(3.28)\begin{equation} p_1^i = \left\{ \begin{array}{@{}ll} \displaystyle - \rho \int_R^\infty u_{\theta 0}^{i 2} \,{\rm d} R & \text{for} \ R \geq 0,\\[10pt] \displaystyle \rho \int_{- \infty}^R [ u_{\theta 0}^{i 2} - \varOmega^2 H(T) ] \,{\rm d} R + \rho \varOmega^2 R H ( T ) & \text{for}\ R < 0. \end{array}\right. \end{equation}

Here, we introduce $\hat {u}_{\theta 1}^i$ as

(3.29) \begin{equation} u_{\theta 1}^i = \left\{ \begin{array}{@{}ll} - \dfrac{1}{2} R u_{\theta 0}^i + \hat{u}_{\theta 1}^i & \text{for} \ R \geq 0,\\ - \dfrac{1}{2} R \left[ u_{\theta 0}^i - \varOmega H ( T ) \right] + R \varOmega H ( T ) + \hat{u}_{\theta 1}^i & \text{for}\ R < 0. \end{array} \right. \end{equation}

Then, (3.27b) reduces to

(3.30)\begin{equation} \frac{\partial \hat{u}_{\theta 1}^i}{\partial T} = \left\{ \begin{array}{@{}ll} \displaystyle \nu_o \dfrac{\partial^2 \hat{u}_{\theta 1}^i}{\partial R^2} & \text{for} \ R \geq 0,\\[10pt] \displaystyle \nu_o \dfrac{\partial^2 \hat{u}_{\theta 1}^i}{\partial R^2} - \lambda_o \hat{u}_{\theta 1}^i & \text{for} \ R < 0. \end{array} \right.\end{equation}

Employing the Laplace transform, the solution to (3.30) is obtained as

(3.31)\begin{equation} \hat{u}_{\theta 1}^i = \left\{ \begin{array}{@{}ll} {\mathcal{L}}^{- 1} \left( A_1 ( s, \theta ) {\rm e}^{- a R} \right) & \text{for}\ R \geq 0,\\ {\mathcal{L}}^{- 1} \big( B_1 ( s, \theta ) {\rm e}^{\hat{a} R} \big) & \text{for} \ R < 0, \end{array} \right.\end{equation}

where $A_1$ and $B_1$ are respectively integral constants that are functions of $s$ and $\theta$. Here, for obtaining (3.31), the matching condition to the outer solution, i.e. $\hat {u}_{\theta 1}^i \to 0$ as $R \to \pm \infty$, is used. Equation (3.29) is therefore written as

(3.32)\begin{equation} u_{\theta 1}^i = \left\{ \begin{array}{@{}ll} - \dfrac{1}{2} R u_{\theta 0}^i + {\mathcal{L}}^{-1} ( A_1( s, \theta ) {\rm e}^{- a R} ) & \text{for} \ R \geq 0,\\ - \dfrac{1}{2} R [ u_{\theta 0}^i - \varOmega H ( T ) ] + R \varOmega H ( T ) + {\mathcal{L}}^{-1} ( B_1 ( s, \theta ) {\rm e}^{\bar{a} R} ) & \text{for}\ R < 0. \end{array} \right.\end{equation}

Here, the enforcements of the continuity of the velocity and its radial derivative at $R = 0$ determine the values of the integral constants $A_1$ and $B_1$ as follows:

(3.33)\begin{equation} A_1 = B_1 =- \frac{3}{2} \frac{\varOmega}{s ( a + \bar{a} )}. \end{equation}

Using the relations of

(3.34)\begin{equation} \frac{1}{s ( a + \bar{a} )} = \frac{\sqrt{\nu_o}}{\lambda_o} \left( \frac{1}{\sqrt{s + \lambda_o}} - \frac{1}{\sqrt{s}} \right) + \frac{\sqrt{\nu_o}}{s \sqrt{s + \lambda_o}} \end{equation}

and (A3) and (A4) in Appendix A, we finally obtain

(3.35a)\begin{align} u_{\theta 1}^i & =- \frac{1}{2} R u_{\theta 0}^i - \frac{3 R \varOmega}{4 \lambda_o {\rm \pi}} \int_0^T \frac{\exp({- \lambda_o \xi}) - 1}{\sqrt{\xi ( T - \xi )^3}} \exp\left({- \frac{R^2}{4 \nu_o ( T - \xi )}}\right) {\rm d} \xi \nonumber\\ &\quad - \frac{3 R \varOmega}{4 \sqrt{{\rm \pi} \lambda_o}} \int_0^T \frac{ \exp\left({- \dfrac{R^2}{4 \nu_o ( T - \xi )}}\right)}{( T - \xi )^{3/2}} \text{erf} ( \sqrt{\lambda_o \xi} ) \,{\rm d} \xi \quad\text{for} \ R \geq 0, \end{align}

(3.35b)\begin{align} u_{\theta 1}^i & =- \frac{1}{2} R \left( u_{\theta 0}^i - \varOmega H ( T ) \right) + R \varOmega H ( T ) \nonumber\\ &\quad - \frac{3 | R | \varOmega}{4 \lambda_o {\rm \pi}} \int_0^T \frac{\exp({- \lambda_o \xi}) - 1}{\sqrt{\xi ( T - \xi )^3}} \exp\left({- \lambda_o ( T - \xi ) - \frac{R^2}{4 \nu_o ( T - \xi )}}\right) {\rm d} \xi \nonumber\\ & \quad - \frac{3 | R | \varOmega}{4 \sqrt{{\rm \pi} \lambda_o}} \int_0^T \frac{\exp\left({- \lambda_o ( T - \xi ) - \dfrac{R^2}{4 \nu_o ( T - \xi )}}\right)}{( T - \xi )^{3/2}} \text{erf} ( \sqrt{\lambda_o \xi} ) \,{\rm d} \xi \quad \text{for} \ R < 0. \end{align}

Figure 2 shows the comparison of $u_{\theta 0}^i / \varOmega$ (first-order solution) and $u_{\theta 1}^i / ( R \varOmega )$ (second-order solution) among three values of $\lambda$ with respect to $\eta = R / ( 2 \sqrt {\nu _o T} )$. To plot the data of figure 2, the numerical calculation of the following integral that is shown in (3.36) is needed. To do so, the integral variable $\xi$ is changed by $\xi = T x$ and, then, the variable $x$ is changed to $y$ by $x = 1 - \eta ^2 / [ (\kern0.7pt y + \eta )^2 ]$;

(3.36)\begin{align} &\int_0^T \frac{f ( \xi )}{( T - \xi )^{3 / 2}} \exp\left({- \frac{R^2}{4 \nu_o ( T - \xi )}}\right) {\rm d} \xi \nonumber\\ &\quad = 2 \frac{\exp({- \eta^2})}{\eta} \int_0^\infty \exp({- y^2 - 2 y \eta}) f \left( 1 - \frac{\eta^2}{(\kern0.7pt y + \eta )^2} \right) {\rm d} y. \end{align}

In the numerical calculation of figure 2, the Hermite quadrature formula of $n = 9$ (see table 25.10 in Abramowitz & Stegun Reference Abramowitz and Stegun1954) is used for the quadrature of the integral on the right-hand side of (3.36).

Figure 2.

Comparison of the tangential velocities $u_{\theta 0}^i$ (first-order solution) and $u_{\theta 1}^i$ (second-order solution) among three values of $\lambda$ with respect to $\eta = R / ( 2 \sqrt {\nu _o T} )$.

3.2.3. Moment exerted by a cylinder

In this subsection we attempt to obtain the moment exerted by the circular cylinder, using three kinds of approaches. Each approach is calculated (I) by the time derivative of the tangential component of the momentum of the entire fluid, (II) by the integration of the shear stress on the cylinder surface, and (III) by the integration of the penalization layer based on (2.3).

In approach (I) we consider the control surface that surrounds a circular domain having a radius of $r_\infty \gg 1$. Then, we can define the fluid domain as ${\mathscr {F}} = [ ( r, \theta ) | 1 \leq r \leq r_\infty, 0 \leq \theta < 2 {\rm \pi}]$. The small fluid element ${\rm d}\kern0.06em x\ {\rm d} y$ has the momentum $\rho u_{\theta } \,{\rm d}\kern0.06em x \,{\rm d} y$ in the tangential direction. The moment induced by the fluid element is therefore described as $| {\boldsymbol x} | \rho [ ( {\rm d} / {\rm d} t ) u_{\theta } ] \,{\rm d}\kern0.06em x \,{\rm d} y$. The moment $M_1$ of the flow exerted by the cylinder rotation is found to be calculated by, noting that the pressure on the control surface does not affect the moment,

(3.37)\begin{equation} M_1 =- \rho \frac{{\rm d}}{{\rm d} t} \int_0^{2 {\rm \pi}} \!\! \int_1^{r_\infty} r^2 u_\theta \,{\rm d} r \,{\rm d} \theta. \end{equation}

To obtain $M_1$, we employ the Laplace transform to (3.37) and take the limit of $r_\infty \to \infty$. Then, we have, taking into account that the outer solution is described as $u_{\theta }^o = 0$ and $u_{\theta }^i$ is independent of $\theta$,

(3.38)\begin{equation} {\mathcal{L}} ( M_1 ) =- 2 {\rm \pi}\rho s \int_0^\infty [ U_{\theta 0} + \epsilon ( 2 R U_{\theta 0} + U_{\theta 1} ) + O ( \epsilon^2 )] {\rm d} R, \end{equation}

where $U_{\theta 1} = {\mathcal {L}} ( u_{\theta 1}^i )$. Using (3.21) and the first equation of (3.32), $M_1$ can be obtained as

(3.39)\begin{equation} {\mathcal{L}} ( M_1 ) =- 2 {\rm \pi}\rho \varOmega \left[ \frac{\bar{a}}{a ( a + \bar{a} )} + \frac{3}{2} \epsilon \left( \frac{\bar{a}}{a^2 ( a + \bar{a} )} - \frac{1}{a ( a + \bar{a} )} \right) + O ( \epsilon^2 ) \right]. \end{equation}

In approach (II) the moment $M_2$ on the fluid is calculated by the integration of the shear stress on the cylinder surface

(3.40)\begin{equation} M_2 = \int_0^{2 {\rm \pi}} 1 \times \tau_{r \theta} \,{\rm d} \theta = \rho \nu \int_0^{2 {\rm \pi}} \left( \frac{\partial u_\theta}{\partial r} - u_\theta + \frac{\partial u_r}{\partial \theta} \right)_{r = 1 + 0} \,{\rm d} \theta. \end{equation}

The Laplace transform to (3.40) becomes, using the relations of $U_{r0} = U_{r1} = 0$,

(3.41)\begin{equation} {\mathcal{L}} ( M_2 ) = 2 {\rm \pi}\rho \nu_o \left[ \frac{\partial}{\partial R} ( U_{\theta 0} + \epsilon U_{\theta 1} ) - \epsilon U_{\theta 0} + O ( \epsilon^2 ) \right]_{R =+ 0}. \end{equation}

Substituting (3.21) and the first equation of (3.32) into (3.41), we have

(3.42)\begin{equation} {\mathcal{L}}( M_2 ) = 2 {\rm \pi}\varOmega \left[ - \frac{\bar{a}}{a ( a + \bar{a} )} + \epsilon \left( - \frac{3}{2} \frac{\bar{a}}{a^2 ( a + \bar{a} )} + \frac{3}{2} \frac{1}{a ( a + \bar{a} )} \right) + O ( \epsilon^2 ) \right]. \end{equation}

Comparing the results between (3.39) and (3.42), it is found that $M_1$ is identical with $M_2$ within the order of $\epsilon$, i.e. the moment calculated from the entire fluid domain is the same as that calculated from the tangential force on the cylinder surface. Making use of the inverse Laplace transform, the moment $M_1$ (or $M_2$) is obtained as

(3.43) \begin{align} M_1= M_2 & =- 2{\rm \pi} \rho \varOmega \left\{ \sqrt{\frac{\nu_o}{{\rm \pi} T}} + \frac{1}{2} \sqrt{\frac{\nu_o}{{\rm \pi} T^3}} \frac{1 - {\rm e}^{- \lambda_o T}}{\lambda_o}\right.\nonumber\\ &\quad\left.+\, \frac{3}{2} \epsilon \nu_o [ H ( T ) - {\rm e}^{- \lambda_o T / 2} \left( {\rm I}_0 ( \lambda_o T / 2 ) + {\rm I}_1 ( \lambda_o T / 2 ) \right)] + O ( \epsilon^2 ) \right\}. \end{align}

In approach (III) the moment $M_3$ on the fluid is calculated by the integration of the penalization layer, based on (2.3),

(3.44) \begin{align} M_3& =- \rho \lambda \int_0^{2 {\rm \pi}} \!\! \int_0^1 r ( r \varOmega - u_\theta ) r \, {\rm d} r\, {\rm d} \theta\nonumber\\ & =- \rho \lambda \epsilon \int_0^{2 {\rm \pi}} \!\! \int_{- \infty}^0 [ ( 1 + \epsilon R ) \varOmega - u_\theta^i ] ( 1 + \epsilon R )^2 \,{\rm d} R\, {\rm d} \theta. \end{align}

Similar to the calculation of $M_1$ or $M_2$, the Laplace transform to $M_3$ is written as, using (3.21) and the second equation of (3.32),

(3.45)\begin{equation} {\mathcal{L}} ( M_3 )=- 2 {\rm \pi}\rho \lambda_o \varOmega \left[ \frac{a}{s \bar{a} ( a + \bar{a} )} + \frac{3}{2} \epsilon \left( - \frac{a}{s \bar{a}^2 ( a + \bar{a} )} + \frac{1}{s \bar{a} ( a + \bar{a} )} \right) \right]. \end{equation}

The inverse Laplace transform to (3.45) yields

(3.46) \begin{align} M_3 =- 2 {\rm \pi}\rho \varOmega \sqrt{\nu_o} \left\{ \frac{1 - {\rm e}^{- \lambda_o T}}{\sqrt{{\rm \pi} T}} + \epsilon \frac{3 \nu_o}{2 \lambda_o} [ H ( T ) + {\rm e}^{- \lambda_o T} - 2 {\rm e}^{- \lambda_o T / 2} {\rm I}_0 ( \lambda_o T / 2 ) ] \right\}. \end{align}

From the results of (3.43) and (3.46), each moment is found to behave like, as $\lambda _o \to \infty$,

(3.47a–c)\begin{equation} M_1, \quad M_2 \to - 2 {\rm \pi}\rho \varOmega \sqrt{\frac{\nu_o}{{\rm \pi} T}}, \quad M_3 \to - 2 {\rm \pi}\rho \varOmega \sqrt{\frac{\nu_o}{{\rm \pi} T}}. \end{equation}

It can therefore be found that the moment $M_3$ obtained by the integration of the penalization layer is identical to $M_1$ and $M_2$ as $\lambda _o \to \infty$. Because the pressure is independent of the moment in this pure rotation problem, the pressure is found to play an important role to the variance of the drag force, which is demonstrated in our previous study (see Ueda & Kida Reference Ueda and Kida2021). Note that (3.47a–c) is the same as the result of Badr & Dennis (Reference Badr and Dennis1985). Figure 3 shows the comparisons of the values between $M_1$ and $M_3$ with respect to $t$ among the three values of $\lambda$ at $Re = 100$ (i.e. $\nu = 1 / 100$). It seems that the variance between $M_1$ (or $M_2$) and $M_3$ decreases exponentially with an increase in the value of $\lambda$.

Figure 3.

Initial behaviours of the moments $M_1$ and $M_3$ with respect to $t$ among the different values of $\lambda$ at $\nu = 1 / 100$.

3.3.1. Outer solutions

The outer solutions in the fluid domain are described by (3.6a)–(3.6d) in the same manner as § 3.2.1, and the first approximation reduces to

(3.48)\begin{equation} \frac{\partial u_{r 0}^o}{\partial T} = \frac{\partial u_{\theta 0}^o}{\partial T} = 0. \end{equation}

The velocities $u_{r0}^0$ and $u_{\theta 0}^o$ are then found to be independent of $T$. The first-order outer solution in the fluid domain is known to exhibit a potential flow (i.e. inviscid solution). This fact can be confirmed by performing the recursive calculation. Based on this fact and the initial condition of the fluid velocity that $u_{r}^o = \cos {\theta } H ( T )$ and $u_{\theta }^0 = - \sin {\theta } H ( T )$ for $r \geq 1$, the outer streamfunction $\psi ^o$ can be described as

(3.49)\begin{equation} \psi^o = r \sin{\theta} H ( T ) + \sum_{n = 1}^\infty \frac{1}{r^n} ( a_n^c \cos{n \theta} + a_n^s \sin{n \theta} ), \end{equation}

where $a_n^c$ and $a_n^s$ are functions of $T$ and $\epsilon$, respectively.

The outer solutions inside the body domain are also described by (3.6a)–(3.6d). Taking into account that the initial condition is given by $u_r^o = 0$ and $u_{\theta }^o = r \varOmega H ( T )$, we readily have

(3.50a–c)\begin{equation} u_r^o = 0, \quad u_{\theta}^o = r \varOmega H ( T ), \quad \frac{p^o}{\rho} = \frac{1}{2} \varOmega^2 r^2 H ( T ) + C, \end{equation}

where $C$ is an integral constant that is a function of $T$.

3.3.2. Inner solutions

The inner solutions are described by the same governing equations as (3.10a)–(3.10c). The continuity equation (3.10c) gives

(3.51)\begin{equation} \frac{\partial u_{r 0}^i}{\partial R} = 0, \end{equation}

in which $u_{r 0}^i$ is found to be independent of $R$ (i.e. a function of $T$ and $\theta$). Because the outer solutions inside the body domain were $u_r^o = 0$ and $u_\theta ^o = r \varOmega H ( T )$ as obtained in (3.50a–c), we have the following first-order inner solution, for $R \geq 0$ and $R < 0$, taking into account the matching condition to the outer solutions:

(3.52)\begin{equation} u_{r 0}^i = 0. \end{equation}

The first-order outer solution in the fluid domain described by (3.49) can be determined from the matching procedure to the inner solution of (3.52) and, therefore, we have $u_{r 0}^{o i} = u_{r 0}^{i o} = 0$. Here, the inner (or outer) expansion of the outer (or inner) solution is denoted by the superscript ‘$oi$’ (or ‘$io$’). Taking into account that the outer solution of $\psi ^o$ is written by (3.49), the above-mentioned matching condition (i.e. $u_{r 0}^{o i} = u_{r 0}^{i o} = 0$) yields the first-order streamfunction in the outer fluid domain as

(3.53)\begin{equation} \psi_0^o = \left( r - \frac{1}{r} \right) H ( T ) \sin{\theta}, \end{equation}

which is known to be the solution of an inviscid uniform flow past a circular cylinder. The governing equations that describe the first-order solutions in the inner fluid domain are the same as (3.13a) and (3.13b), i.e.

(3.54a)$$\begin{gather} 0 =- \frac{1}{\rho} \frac{\partial p_0^i}{\partial R}, \end{gather}$$

(3.54b)$$\begin{gather}\frac{\partial u_{\theta 0}^i}{\partial T} = \lambda_o \chi ( \varOmega H ( T ) - u_{\theta 0}^i ) + \nu_o \frac{\partial^2 u_{\theta0}^i}{\partial R^2}. \end{gather}$$

Here, let us consider $u_{\theta 0}^i$ of (3.54b). From (3.53), the inner expansion of $u_{\theta 0}^o$ in the fluid domain is written as $u_{\theta 0}^{o i} = - 2 \sin {\theta } H ( T )$ and, therefore, we can write

(3.55)\begin{equation} u_{\theta 0}^i \to - 2\sin{\theta} H ( T ) \quad \text{as} \ R \to \infty. \end{equation}

In contrast, since $u_{\theta 0}^o = r \varOmega H ( T )$ inside the body domain, we can write

(3.56)\begin{equation} u_{\theta 0}^i \to \varOmega H ( T ) \quad \text{as}\ R \to - \infty. \end{equation}

The Laplace transform to (3.54b) for $R \geq 0$ is written as, taking into account the initial condition of $u_{\theta 0}^i = - 2 \sin {\theta }$,

(3.57)\begin{equation} s U_{\theta 0} + 2 \sin{\theta} = \nu_o \frac{\partial^2 U_{\theta 0}}{\partial R^2}, \end{equation}

where $U_{\theta 0} = {\mathcal {L}} ( u_{\theta 0}^i )$. The solution to (3.55) can be then obtained as (see Appendix B for the detailed derivation)

(3.58)\begin{equation} U_{\theta 0} =- \frac{2}{s} \sin{\theta} + A_0 ( s, \theta ) {\rm e}^{- a R} \quad \text{with}\ a = \sqrt{\frac{s}{\nu_o}}, \end{equation}

where $A_0$ is an integral constant that is a function of $s$ and $\theta$. For $R < 0$, the governing equation is given in (3.54b) as $\chi = 1$, and the initial condition is written as $u_{\theta 0}^i = \varOmega$. Therefore, the Laplace transform to the governing equation yields, for $R < 0$,

(3.59)\begin{equation} s U_{\theta 0} - \varOmega = \lambda_o \left( \frac{\varOmega}{s} - U_{\theta 0} \right) + \nu_o \frac{\partial^2 U_{\theta 0}}{\partial R^2}. \end{equation}

The solution to (3.59) is obtained as, taking into account (3.56) (see Appendix B),

(3.60)\begin{equation} U_{\theta 0} = B_0 ( s, \theta ) {\rm e}^{\bar{a} R} + \frac{\varOmega}{s} \quad \text{with}\ \bar{a} = \sqrt{\frac{s + \lambda_o}{s}}, \end{equation}

where $B_0$ is an integral constant that is a function of $s$ and $\theta$. The indeterminate functions, $A_0$ and $B_0$, are determined by the enforcements of the continuity of the velocity and its derivative with respect to $R$ at $R = 0$. Therefore, we can finally obtain

(3.61)\begin{equation} U_{\theta 0} = \left\{ \begin{array}{@{}ll} \displaystyle \dfrac{\bar{a}}{a + \bar{a}} \dfrac{\varOmega + 2 \sin{\theta}}{s} {\rm e}^{- a R} - \dfrac{2}{s} \sin{\theta} & \text{for} \ R \geq 0,\\[10pt] \displaystyle - \dfrac{{a}}{a + \bar{a}} \dfrac{\varOmega + 2 \sin{\theta}}{s} {\rm e}^{\bar{a} R} + \dfrac{\varOmega}{s} & \text{for} \ R < 0. \end{array} \right. \end{equation}

By virtue of the form of (3.61) in the Laplace space, the solution $u_{\theta 0}^i$ in the real space is found to be written as

(3.62)\begin{equation} u_{\theta 0}^i = \hat{u}_{\theta 0}^0 + \hat{u}_{\theta 0}^s \sin{\theta}. \end{equation}

Here, $\hat {u}_{\theta 0}^0$ is the same as (3.22) and (3.23) that are the solutions for the pure rotation problem analysed in § 3.2. In addition, $\hat {u}_{\theta 0}^s$ is the same as the solution for the pure translating motion that was derived in our previous study (see Ueda & Kida Reference Ueda and Kida2021).

For the second approximation, $u_{r1}^i$ is described by (3.10c), and it reduces to

(3.63)\begin{equation} \frac{\partial u_{r 1}^i}{\partial R} =- \frac{\partial u_{\theta 0}^i}{\partial \theta}. \end{equation}

Using the matching condition to the outer solution inside the body domain ($u_{r 1}^i \to 0$ as $R \to - \infty$), (3.63) becomes

(3.64)\begin{equation} u_{r 1}^i =- \frac{\partial}{\partial \theta} \int_{- \infty}^R u_{\theta 0}^i \,{\rm d} R. \end{equation}

Making use of the Laplace transform and the first-order solution of (3.61) in the Laplace space, (3.64) can be written as, for $R < 0$,

(3.65)\begin{equation} U_{r 1} =- \frac{\partial}{\partial \theta} \int_{- \infty}^R U_{\theta 0}\,{\rm d} R = 2 \frac{a}{\bar{a} ( a + \bar{a} )} \frac{1}{s} {\rm e}^{\bar{a} R} \cos{\theta}, \end{equation}

where $U_{r 1} = {\mathcal {L}} ( u_{r 1}^i )$. For $R \geq 0$, the function $U_{r 1}$ is continuous at $R = 0$ and, therefore, (3.64) can be written as

(3.66) \begin{align} U_{r 1}& =- \frac{\partial}{\partial \theta} \Bigg( \int_{- \infty}^0 U_{\theta 0} \,{\rm d} R + \int_0^R U_{\theta 0} \,{\rm d} R \Bigg)\nonumber\\ & = 2 \frac{a}{\bar{a} ( a + \bar{a} )} \frac{1}{s} \cos{\theta} - 2 \frac{\bar{a}}{a ( a + \bar{a} )} \frac{1 - {\rm e}^{- a R}}{s} \cos{\theta} + \frac{2}{s} R \cos{\theta}. \end{align}

By virtue of (3.65) and (3.66), the function $u_{r 1}^i$ is found to be expressed as the form of

(3.67)\begin{equation} u_{r 1}^i = \hat{u}_{r 1} \cos{\theta}. \end{equation}

Here, we consider the matching to the outer solution. Equation (3.66) tends to, as $R \to \infty$,

(3.68)\begin{equation} U_{r 1} \to \frac{2}{s} R \cos{\theta} + 2 \frac{a - \bar{a}}{a \bar{a} s} \cos{\theta}. \end{equation}

Taking into account that the outer streamfunction is asymptotically represented as $\psi ^o = \psi _0^o + \epsilon \psi _1^o + O ( \epsilon ^2 )$, i.e.

(3.69)\begin{equation} \psi^o = \left( r - \frac{1}{r} \right) H ( T ) \sin{\theta} + \epsilon \sum_{n = 1}^\infty \frac{1}{r^n} ( a_n^c \cos{n \theta} + a_n^s \sin{n \theta} ) + O ( \epsilon^2 ), \end{equation}

the matching procedure between (3.68) and (3.69) determines the values of the coefficients as

(3.70a,b)\begin{equation} a_n^c = 0, \quad {\mathcal{L}} ( a_n^s ) = 2 \frac{a - \bar{a}}{a \bar{a} s} \delta_{n 1}. \end{equation}

The streamfunction in the outer fluid domain is therefore written as

(3.71)\begin{equation} \psi^o = \left( r - \frac{1}{r} \right) H ( T ) \sin{\theta} + 2 \epsilon \frac{A}{r} \sin{\theta} + O ( \epsilon^2 ), \end{equation}

where $A$ is defined by

(3.72)\begin{equation} A = {\mathcal{L}}^{-1} \left( \frac{a - \bar{a}}{a \bar{a} s} \right). \end{equation}

The inner expansion of the outer solution (3.71) is calculated as, in the Laplace space,

(3.73)\begin{equation} {\mathcal{L}} ( u_\theta^{o i} ) =- 2 \frac{1}{s} \sin{\theta} + 2 \epsilon \frac{R}{s} \sin{\theta} + 2 \epsilon \frac{a - \bar{a}}{a \bar{a} s} \sin{\theta}. \end{equation}

The matching to (3.73) (i.e. ${\mathcal {L}} ( u_\theta ^{o i} ) = {\mathcal {L}} ( u_\theta ^{i o} )$) gives the following boundary condition of $u_{\theta 1}^i$ as $R \to \infty$:

(3.74)\begin{equation} {\mathcal{L}} ( u_{\theta 1}^i ) \to 2 \frac{R}{s} \sin{\theta} + 2 \frac{a - \bar{a}}{a \bar{a} s} \sin{\theta} \quad \text{as} \ R \to \infty. \end{equation}

Similarly, since $u_\theta ^o = r \varOmega H ( T )$ inside the body domain, we have

(3.75)\begin{equation} {\mathcal{L}} ( u_{\theta 1}^i ) \to \frac{R}{s} \varOmega \quad \text{as} \ R \to - \infty. \end{equation}

The second approximation of (3.10a) and (3.10b) are written as

(3.76a)$$\begin{gather} \frac{\partial u_{r 1}^i}{\partial T}-u_{\theta 0}^{i 2} =- \frac{1}{\rho} \frac{\partial p_1^i}{\partial R} - \lambda_o \chi u_{r 1}^i + \nu_o \frac{\partial^2 u_{r 1}^i}{\partial R^2}, \end{gather}$$

(3.76b)$$\begin{gather}\frac{\partial u_{\theta 1}^i}{\partial T} + u_{r 1}^i \frac{\partial u_{\theta 0}^i}{\partial R} + u_{\theta 0}^i \frac{\partial u_{\theta 0}^i}{\partial \theta} =- \frac{1}{\rho} \frac{\partial p_0^i}{\partial \theta} + \lambda_o \chi ( R \varOmega H ( T ) - u_{\theta 1}^i) + \nu_o \Bigg( \frac{\partial^2 u_{\theta 1}^i}{\partial R^2} + \frac{\partial u_{\theta 0}^i}{\partial R} \Bigg). \end{gather}$$

The pressure $p_0^i$ is found to be independent of $R$ from (3.54a), and $p^o$ is given by (3.50a–c) in the outer domain inside the body. In the fluid domain, since the outer solution is described by a potential flow (inviscid solution), the following unsteady Bernoulli's equation is valid:

(3.77)\begin{equation} \frac{p^o}{\rho} =- \frac{\partial \phi^o}{\epsilon \partial T} - \frac{1}{2} ( | {\boldsymbol{u}} |^2 - 1 ). \end{equation}

Here, $\phi ^o$ is the velocity potential in the outer fluid domain, and it is given from (3.71), as

(3.78)\begin{equation} \phi^o = \left( r + \frac{1}{r} \right) H ( T ) \cos{\theta} - 2 \epsilon A \frac{1}{r} \cos{\theta} + O ( \epsilon^2 ). \end{equation}

Substituting (3.78) into (3.77) yields

(3.79)\begin{equation} \frac{p^o}{\rho} = \frac{2}{r} \frac{{\rm d} A}{{\rm d} T} \cos{\theta} + \left( \frac{1}{r^2} - \frac{1}{2 r^4} - \frac{2}{r^2} \sin^2{\theta} \right) H ( T ) + O ( \epsilon ). \end{equation}

Setting $r = 1 + \epsilon R$, we have

(3.80)\begin{equation} \frac{p^{o i}}{\rho} = 2 \frac{{\rm d} A}{{\rm d} T} \cos{\theta} - 2 H(T) \sin^2{\theta} + \frac{1}{2} H ( T ) + O ( \epsilon ). \end{equation}

Since $p_0^{o i} = C + ( \rho / 2 ) \varOmega ^2 H ( T ) + O ( \epsilon )$ for $R < 0$ from (3.50a–c), the function $p_0^i$ is written as, taking into account (3.54a),

(3.81)\begin{equation} \frac{p_0^i}{\rho} = \left\{ \begin{array}{@{}ll} \displaystyle 2 \dfrac{{\rm d} A}{{\rm d} T} \cos{\theta} - 2 H ( T ) \sin^2{\theta} + \dfrac{1}{2} H ( T ) & \text{for} \ R \geq 0,\\[10pt] \displaystyle \dfrac{C}{\rho} + \dfrac{1}{2} \varOmega^2 H ( T ) & \text{for} \ R < 0. \end{array} \right.\end{equation}

Equation (3.76b) can then be rewritten as

(3.82)\begin{align} \frac{\partial u_{\theta 1}^i}{\partial T} - \nu_o \frac{\partial^2 u_{\theta 1}^i}{\partial R^2} + \lambda_o \chi u_{\theta 1}^i & = \left\{\begin{array}{ll} \displaystyle 2\dfrac{{\rm d} A}{{\rm d} T} \sin{\theta} + 4 H ( T ) \sin{\theta} \cos{\theta} & \text{for}\ R \geq 0 \\[10pt] \displaystyle 0 & \text{for}\ R < 0 \end{array}\right\} - u_{r 1}^i \frac{\partial u_{\theta 0}^i}{\partial R} \nonumber\\ &\quad - u_{\theta 0}^i \frac{\partial u_{\theta 0}^i}{\partial \theta} + \nu_o \frac{\partial u_{\theta 0}^i}{\partial R} + \lambda_o \chi ( \varOmega R H ( T ) ). \end{align}

The second-order solution $u_{\theta 1}^i$ is therefore found to be expressed as, with respect to $\theta$,

(3.83)\begin{equation} u_{\theta 1}^i = \hat{u}_{\theta 1}^0 + \hat{u}_{\theta 1}^c \cos{\theta} + \hat{u}_{\theta 1}^s \sin{\theta} + \hat{u}_{\theta 1}^{c s} \sin{\theta} \cos{\theta}. \end{equation}

Substituting (3.83) into (3.82), we can write (3.82) separately:

(3.84a)$$\begin{gather} \frac{\partial \hat{u}_{\theta 1}^0}{\partial T} - \nu_o \frac{\partial^2 \hat{u}_{\theta 1}^0}{\partial R^2} + \lambda_o \chi \hat{u}_{\theta 1}^0 = \nu_o \frac{\partial \hat{u}_{\theta 0}^0}{\partial R} + \lambda_o \chi R \varOmega H ( T ), \end{gather}$$

(3.84b)$$\begin{gather}\frac{\partial \hat{u}_{\theta 1}^c}{\partial T} - \nu_o \frac{\partial^2 \hat{u}_{\theta 1}^c}{\partial R^2} + \lambda_o \chi \hat{u}_{\theta 1}^c =- \hat{u}_{r 1} \frac{\partial \hat{u}_{\theta 0}^0}{\partial R} - \hat{u}_{\theta 0}^0 \hat{u}_{\theta 0}^s, \end{gather}$$

(3.84c)$$\begin{gather}\frac{\partial \hat{u}_{\theta 1}^s}{\partial T} - \nu_o \frac{\partial^2 \hat{u}_{\theta 1}^s}{\partial R^2} + \lambda_o \chi \hat{u}_{\theta 1}^s = \nu_o \frac{\partial \hat{u}_{\theta 0}^s}{\partial R} + \left\{\begin{array}{@{}ll@{}} \displaystyle2 \dfrac{{\rm d} A}{{\rm d} T} & \text{for} \ R \geq 0\\[9pt] \displaystyle 0 & \text{for}\ R < 0 \end{array} \right\}, \end{gather}$$

(3.84d)$$\begin{gather}\frac{\partial \hat{u}_{\theta 1}^{c s}}{\partial T} - \nu_o \frac{\partial^2 \hat{u}_{\theta 1}^{c s}}{\partial R^2} + \lambda_o \chi \hat{u}_{\theta 1}^{c s} =- \hat{u}_{r 1} \frac{\partial \hat{u}_{\theta 0}^s}{\partial R} - \hat{u}_{\theta 0}^{s 2} + \left\{ \begin{array}{@{}ll}\displaystyle 4 H ( T ) & \text{for}\ R \geq 0\\ \displaystyle 0 & \text{for} \ R < 0 \end{array} \right\}. \end{gather}$$

The boundary condition are written as, $R \to \infty$,

(3.85a–d)\begin{equation} \hat{u}_{\theta 1}^0 \to 0, \quad \hat{u}_{\theta 1}^c \to 0, \quad \hat{u}_{\theta 1}^s \to 2 R H ( T ) + 2 A, \quad \hat{u}_{\theta 1}^{c s} \to 0. \end{equation}

As $R \to -\infty$, the boundary conditions are also written as

(3.86a–d)\begin{equation} \hat{u}_{\theta 1}^0 \to R \varOmega H ( T ), \quad \hat{u}_{\theta 1}^c \to 0, \quad \hat{u}_{\theta 1}^s \to 0, \quad \hat{u}_{\theta 1}^{c s} \to 0. \end{equation}

Using the Laplace transformation and the similar procedure for the derivation of the first approximations, we finally obtain the following solutions (the detailed derivation is described in Appendix C):

(3.87)\begin{equation} U_{\theta 1}^0 = \left\{ \begin{array}{@{}ll} \displaystyle - \dfrac{1}{2} \dfrac{\varOmega}{s ( a + \bar{a} )} \left( \bar{a} R - 3 \right) {\rm e}^{- a R} & \text{for} \ R \geq 0,\\[10pt] \displaystyle \dfrac{1}{2} \dfrac{\varOmega}{s ( a + \bar{a} )} \left( a R - 3 \right) {\rm e}^{\bar{a} R} + \dfrac{\varOmega}{s} R & \text{for} \ R < 0. \end{array} \right.\end{equation}

The velocity, $\hat {u}_{\theta 1}^0$, is then found to be the same as the solution to the pure rotation problem obtained in (3.32).

For $u_{\theta 1}^s$, we have

(3.88) \begin{equation} U_{\theta 1}^s = \left\{ \begin{array}{@{}ll} \displaystyle 2 \dfrac{R}{s} + 2 \dfrac{a - \bar{a}}{s a \bar{a}} - \dfrac{\bar{a}}{s ( a + \bar{a} )} R {\rm e}^{- a R} + D_s {\rm e}^{- a R} & \text{for} \ R \geq 0,\\[10pt] \displaystyle \dfrac{a}{s ( a + \bar{a} )} R {\rm e}^{\bar{a} R} + E_s {\rm e}^{\bar{a} R} & \text{for} \ R < 0, \end{array} \right. \end{equation}

where

(3.89a,b) \begin{equation} D_s = \frac{2}{s a} - \frac{3}{s ( a + \bar{a} )}, \quad E_s = \frac{2}{s \bar{a}} - \frac{3}{s ( a + \bar{a} )}. \end{equation}

It is then found that the velocity $\hat {u}_{\theta 1}^s$ is the same as the solution to the pure translation problem analysed in our previous paper (see Ueda & Kida Reference Ueda and Kida2021).

We define the following relations that are written as

(3.90a,b)\begin{equation} U_{\theta 1}^c = {\mathcal{L}} ( \hat{u}_{\theta 1}^c ) \quad \text{and} \quad F^c = {\mathcal{L}} \Bigg( \hat{u}_{r 1} \frac{\partial \hat{u}_{\theta 0}^0}{\partial R} + \hat{u}_{\theta 0}^0 \hat{u}_{\theta 0}^s \Bigg). \end{equation}

Then, we can have the following relations:

(3.91)\begin{equation} U_{\theta 1}^c = \left\{\begin{array}{@{}ll} - \displaystyle\dfrac{1}{2 \nu_o a} \left( {\rm e}^{a R} \int_R^\infty F^c {\rm e}^{- a R} \,{\rm d} R + {\rm e}^{- a R} \int_0^R F^c {\rm e}^{a R} \,{\rm d} R \right) + D_c {\rm e}^{- a R}, & R\geq 0, \\[10pt] \displaystyle- \dfrac{1}{2 \nu_o \bar{a}} \left( {\rm e}^{- \bar{a} R} \int_{- \infty}^R F^c {\rm e}^{\bar{a} R} \,{\rm d} R - {\rm e}^{\bar{a} R} \int_0^R F^c {\rm e}^{- \bar{a} R} \,{\rm d} R \right) + E_c {\rm e}^{\bar{a} R}, & R<0, \end{array} \right. \end{equation}

where $D_c$ and $E_c$ are respectively given by

(3.92a)$$\begin{gather} D_c = \frac{\bar{a} - a}{2 \nu_o a ( a + \bar{a} )} \int_0^\infty F^c {\rm e}^{- a R} \,{\rm d} R - \frac{1}{\nu_o ( a + \bar{a} )} \int_{- \infty}^0 F^c {\rm e}^{\bar{a} R} \,{\rm d} R, \end{gather}$$

(3.92b)$$\begin{gather}E_c =- \frac{1}{\nu_o ( a + \bar{a} )} \int_0^\infty F^c {\rm e}^{- a R} {\rm d} R + \frac{a - \bar{a}}{2 \nu_o \bar{a} ( a + \bar{a} )} \int_{- \infty}^0 F^c {\rm e}^{\bar{a} R}\,{\rm d} R. \end{gather}$$

We also define $\hat {U}_{\theta 1}^{c s}$ and $F^{c s}$ as

(3.93a,b)\begin{align} U_{\theta 1}^{c s} = {\mathcal{L}} ( \hat{u}_{\theta 1}^{c s}) \quad \text{and} \quad F^{c s} = {\mathcal{L}} \left( \hat{u}_{r1} \frac{\partial \hat{u}_{\theta 0}^s}{\partial R} + \left\{ \begin{array}{@{}cc@{}} \displaystyle \hat{u}_{\theta 0}^{s 2} - 4H(T) & \text{for}\ R \geq 0\\ \displaystyle \hat{u}_{\theta 0}^{s 2} & \text{for} \ R < 0 \end{array} \right\} \right). \end{align}

Then, we can obtain

(3.94)\begin{align} U_{\theta 1}^{c s} = \left\{ \begin{array}{@{}ll} \displaystyle - \dfrac{1}{2 \nu_o a} \left( {\rm e}^{a R} \int_R^\infty F^{c s} {\rm e}^{- a R} \,{\rm d} R + {\rm e}^{- a R} \int_0^R F^{c s} {\rm e}^{a R} \,{\rm d} R \right) + D_{c s} {\rm e}^{- a R} & \text{for} \ R \geq 0,\\[10pt] \displaystyle - \dfrac{1}{2 \nu_o \bar{a}} \left( {\rm e}^{- \bar{a} R} \int_{- \infty}^R F^{c s} {\rm e}^{\bar{a} R} \,{\rm d} R - {\rm e}^{\bar{a} R} \int_0^R F^{c s} {\rm e}^{- \bar{a} R} \,{\rm d} R \right) + E_{c s} {\rm e}^{\bar{a} R} & \text{for}\ R < 0, \end{array} \right. \end{align}

in which the constants $D_{c s}$ and $E_{c s}$ are respectively given by

(3.95a)$$\begin{gather} D_{c s} = \frac{\bar{a} - a}{2 \nu_o a ( a + \bar{a} )} \int_0^\infty F^{c s} {\rm e}^{- a R} \,{\rm d} R - \frac{1}{\nu_o ( a + \bar{a} )} \int_{- \infty}^0 F^{c s} {\rm e}^{\bar{a} R} \,{\rm d} R, \end{gather}$$

(3.95b)$$\begin{gather}E_{c s} =- \frac{1}{\nu_o ( a + \bar{a} )} \int_0^\infty F^{c s} {\rm e}^{- a R} \,{\rm d} R + \frac{a - \bar{a}}{2 \nu_o \bar{a} ( a + \bar{a} )} \int_{- \infty}^0 F^{c s} {\rm e}^{\bar{a} R} \,{\rm d} R. \end{gather}$$

Based on the solutions obtained, each velocity component is found to asymptotically behave like, for $\lambda _o \gg 1$,

(3.96a)\begin{gather} \hat{u}_{\theta 0}^0 \sim \left\{ \begin{array}{@{}ll} \displaystyle \varOmega \, \text{erfc} \left( \dfrac{R}{2 \sqrt{\nu_o T}} \right): & R \geq 0,\\[10pt] \displaystyle \varOmega H ( T ) : & R < 0. \end{array}\right.\end{gather}

(3.96b)\begin{gather} \hat{u}_{\theta 0}^s \sim \left\{ \begin{array}{@{}ll} \displaystyle 2 \, \text{erfc} \left( \dfrac{R}{2 \sqrt{\nu_o T}} \right)- 2 H ( T ) : & R \geq 0,\\ \displaystyle 0 : & R < 0. \end{array} \right.\end{gather}

(3.96c)\begin{gather} \hat{u}_{r1} \sim \left\{ \begin{array}{@{}ll} \displaystyle 2 R H ( T ) - 4 \sqrt{\dfrac{\nu_o T}{\rm \pi}} + 4 \sqrt{\dfrac{\nu_o T}{\rm \pi}} \exp\left({- \dfrac{R^2}{4 \nu_o T}}\right) - 2 R \, \text{erfc} \left( \dfrac{R}{2 \sqrt{\nu_o T}} \right) : & R \geq 0,\\ \displaystyle 0 : & R < 0. \end{array} \right.\end{gather}

(3.96d)\begin{gather} \hat{u}_{\theta 1}^0 \sim \left\{ \begin{array}{@{}ll} \displaystyle - \dfrac{1}{2} \varOmega R \, \text{erfc} \left( \dfrac{R}{2 \sqrt{\nu_o T}} \right) : & R \geq 0,\\ \displaystyle \varOmega R H ( T ) : & R < 0. \end{array} \right. \end{gather}

(3.96e)\begin{gather} \hat{u}_{\theta 1}^s \sim \left\{ \begin{array}{@{}ll} \displaystyle 2 R H ( T ) - 4 \sqrt{\dfrac{\nu_o T}{\rm \pi}} + 4 \sqrt{\dfrac{\nu_o T}{\rm \pi}} \exp\left({- \dfrac{R^2}{4 \nu_o T}}\right) - 3 R \, \text{erfc} \left( \dfrac{R}{2 \sqrt{\nu_o T}} \right) : & R \geq 0,\\ \displaystyle 0 : & R < 0. \end{array} \right. \end{gather}

(3.96f)\begin{gather} \hat{u}_{\theta 1}^c \sim \left\{ \begin{array}{@{}ll} \displaystyle - \dfrac{1}{2\sqrt{{\rm \pi} \nu_o}} \int_0^\infty \,{\rm d} R' \,\int_0^T \dfrac{f^c ( \xi, R' )}{\sqrt{T - \xi}}\left[ \exp\left({- \dfrac{( R - R' )^2}{4 \nu_o ( T - \xi )}}\right) \right. & \\[10pt] \quad -\left. \exp\left({- \dfrac{( R + R' )^2}{4 \nu_o ( T - \xi )}}\right) \right] {\rm d} \xi : & R \geq 0,\\[10pt] \displaystyle 0 : & R < 0. \end{array} \right.\end{gather}

(3.96g)\begin{gather} \hat{u}_{\theta 1}^{c s} \sim \left\{ \begin{array}{@{}ll} \displaystyle - \dfrac{1}{2\sqrt{{\rm \pi} \nu_o}} \int_0^\infty \, {\rm d} R' \,\int_0^T \,\dfrac{f^{c s} ( \xi, R' )}{\sqrt{T - \xi}} \left[ \exp\left({- \dfrac{( R - R' )^2}{4 \nu_o ( T - \xi )}}\right)\right. & \\[10pt] \quad - \left.\exp\left({- \dfrac{( R +R' )^2}{4 \nu_o ( T - \xi )}}\right) \right] {\rm d} \xi : & R \geq 0,\\[10pt] \displaystyle 0 : & R < 0. \end{array} \right. \end{gather}

Figure 4(a) shows the velocity distribution of $u_x$ on the $y$ axis as $\lambda _o \to \infty$ in the case of $\nu = 1 / 500$ and $\varOmega = 0.5$. Note here that the definition of the Reynolds number accounts for the fact that the quantity $1 / \nu$ is replaced by $2 / \nu$ for the Reynolds number used in Badr & Dennis (Reference Badr and Dennis1985). Since the velocity $u_x$ on the y axis is equal to $u_\theta$ at $\theta = {\rm \pi}/ 2$, it can be obtained as the composite velocity, $u_{\theta }^i + u_\theta ^o - u_\theta ^{o i}$ at $\theta = {\rm \pi}/2$, i.e.

(3.97)\begin{align} u_x & \approx ( \varOmega + 2 ) \,\text{erfc} ( \eta ) - 1 - \frac{1}{r^2} - \frac{\varOmega}{2} ( r - 1 ) \,\text{erfc} ( \eta ) \end{align}

(3.98)\begin{align} &\quad + 4 \sqrt{\frac{\nu t}{\rm \pi}} \left( {\rm e}^{- \eta^2} - \frac{1}{r^2} \right) - 3 ( r - 1 ) \,\text{erfc} ( \eta ) \quad \text{with}\ \eta = \frac{r - 1}{2 \sqrt{\nu t}}. \end{align}

The vorticity distribution on the cylinder surface is shown on figure 4(b). The solid line denotes the present results and the dotted line is the result of Badr & Dennis (Reference Badr and Dennis1985) that is written as

(3.99)\begin{align} \omega |_{r = 1}& = \frac{\varOmega}{\sqrt{\nu t}} \left( \frac{1}{\sqrt{\rm \pi}} - \frac{\sqrt{\nu t}}{2} \right) + \left( \frac{2}{\sqrt{{\rm \pi} \nu t}} + 1 \right) \sin{\theta} \end{align}

(3.100)\begin{align} &\quad + \varOmega t \left( 2.7844 - \frac{8}{3 \sqrt{{\rm \pi} \nu t}} \right) \cos{\theta} + t \left[ 6.5577 - \frac{2}{\sqrt{{\rm \pi} \nu t}} \left( 1 + \frac{4}{3{\rm \pi}} \right) \right] \sin{2 \theta}. \end{align}

Since the vorticity distribution is then obtained from the inner solutions, i.e.

(3.101)\begin{equation} \omega = \frac{1}{\epsilon} \frac{\partial u_{\theta 0}^i}{\partial R} + \frac{\partial u_{\theta 1}^i}{\partial R} + u_{\theta 0}^i + O ( \epsilon ), \end{equation}

we have

(3.102)\begin{equation} \omega |_{r = 1} \approx- \frac{\varOmega + 2 \sin{\theta}}{\sqrt{{\rm \pi} \nu t}} - \frac{1}{2} \varOmega - \sin{\theta} - 8 \varOmega \sqrt{\frac{t}{{\rm \pi} \nu}} c_1 \cos{\theta} - 2 \sqrt{\frac{t}{{\rm \pi} \nu}} c_2 \sin{2 \theta}, \end{equation}

in which the coefficients $c_1$ and $c_2$ are given as $c_1 = 0.0452900$ and $c_2 = - 1.34657$. We find that the leading term of the vorticity distribution is equal to the results of Badr & Dennis (Reference Badr and Dennis1985).

Figure 4.

Velocity distribution of $u_x$ on the $y$ axis (a) and the vorticity distribution on the cylinder surface (b) in the case of $\nu = 1 / 500$ and $\varOmega = 0.5$.

We obtain the second-order solution of the pressure $p_1^i$ that is governed by

(3.103)\begin{equation} \frac{\partial u_{r 1}^i}{\partial T} - u_{\theta 0}^{i 2} =- \frac{1}{\rho} \frac{\partial p_1^i}{\partial R} - \lambda_o \chi u_{r 1}^i + \nu_o \frac{\partial^2 u_{r 1}^i}{\partial R^2}. \end{equation}

Inside the body domain (i.e. $R < 0$), the inner expansion of the outer solution of the pressure is readily found to be written as

(3.104)\begin{equation} \frac{p_1^{i o}}{\rho} = \varOmega^2 R H ( T ). \end{equation}

Furthermore, because of $u_{\theta 0}^{i o} = \varOmega H(T)$ and $u_{r 1}^{i o} = 0$, (3.103) is rewritten as

(3.105)\begin{align} \frac{p_1^i}{\rho}& =- \frac{\partial}{\partial T} \int_{- \infty}^R u_{r 1}^i \,{\rm d} R - \lambda_o \int_{- \infty}^R u_{r 1}^i \,{\rm d} R + \nu_o \frac{\partial u_{r 1}^i}{\partial R} \nonumber\\ &\quad + \int_{- \infty}^R ( u_{\theta 0}^{i 2} - \varOmega^2 H(T) ) \,{\rm d} R + \varOmega^2 R H ( T ). \end{align}

For $R < 0$, the functions $u_{r 1}$ and $u_{\theta 0}$ are given by (3.96a)–(3.96c) for $\lambda _o \gg 1$. Therefore, (3.105) can be calculated as

(3.106)\begin{equation} \frac{p_1^i}{\rho} \sim \varOmega^2 R H ( T ) + O ( 1 / \sqrt{\lambda_o} ) \quad \text{for}\ \lambda_o \gg 1. \end{equation}

For $R > 0$, we find that $u_{r 1}^i \sim 2 ( A + RH(T) ) \cos {\theta }$ and $u_{\theta 0}^i \sim - 2 \sin {\theta }H(T)$ as $R \to \infty$ from (3.61)–(3.68). Therefore, (3.103) is rewritten as

(3.107)\begin{align} \frac{1}{\rho} \frac{\partial p_1^i}{\partial R} & =- \frac{\partial}{\partial T} [ u_{r 1}^i - ( 2 A +2 R H ( T ) ) \cos{\theta}] + ( u_{\theta 0}^{i 2} - 4 H ( T ) \sin^2{\theta} ) \nonumber\\ &\quad + \nu_o \frac{\partial^2 u_{r 1}^i}{\partial R^2} - 2 \frac{{\rm d} A}{{\rm d} T} \cos{\theta} + 4 H ( T ) \sin^2{\theta}. \end{align}

Integrating (3.107) with respect to $R$, we have

(3.108)\begin{align} \frac{p_1^i}{\rho} & = \frac{\partial}{\partial T} \int_R^\infty [ u_{r 1}^i - ( 2 A + 2R H ( T ) ) \cos{\theta}] {\rm d} R - \int_R^\infty [ u_{\theta 0}^{i 2} - 4 H ( T ) \sin^2{\theta}] {\rm d} R\nonumber\\ &\quad + \nu_o \frac{\partial u_{r 1}^i}{\partial R} - 2 \frac{{\rm d} A}{{\rm d} T} R \cos{\theta} + 4 R H ( T ) \sin^2{\theta} + C_1 ( \theta, T ), \end{align}

where $C_1$ is an integral constant that can be determined from the matching procedure to the inner expansion of the outer pressure $p^{oi}$. The derivations of the pressure $p_1^o$ and $p^{oi}$ are described in Appendix D. Therefore, we have

(3.109)\begin{equation} C_1=- 2 \nu_o H ( T ) \cos{\theta} + 4 A \sin^2{\theta} - \frac{{\rm d}}{{\rm d} T} ( a_1^c \cos{\theta} + a_1^s \sin{\theta} ) - \frac{{\rm d}}{{\rm d} T} ( a_2^c \cos{2 \theta} ), \end{equation}

where $a_1^{c}$, $a_1^{s}$ and $a_2^c$ are given in Appendix D. Using (3.105) and (3.108), we have the pressure jump of $\Delta p_1 = p_1^i |_{r = 1 + 0} - p_1^i |_{r = 1 - 0}$ as

(3.110)\begin{equation} \frac{\Delta p_1}{\rho} = \frac{\partial}{\partial T} \text{pf} \int_{- \infty}^\infty u_{r 1}^i \,{\rm d} R + \lambda_o \int_{- \infty}^0 u_{r 1}^i \,{\rm d} R - \text{pf} \int_{- \infty}^\infty u_{\theta 0}^{i 2} \,{\rm d} R + C_1, \end{equation}

where the symbol $\text {pf}$ denotes the integration in the finite part sense of Hadamard (see Hadamard Reference Hadamard1932).

Figure 5 shows the pressure distribution on the cylinder surface. The present result is given by

(3.111) \begin{align} \frac{p - p_\infty}{\rho} & \approx- 2 \sqrt{\frac{\nu}{{\rm \pi} t}} \cos{\theta} + \frac{1}{2} - 2 \sin^2{\theta}\nonumber\\ &\quad - 16 \sqrt{\frac{\nu t}{\rm \pi}} ( \varOmega + \sin{\theta}) \left[ \varOmega \left( 1 - \frac{1}{\sqrt{2}} \right) - \frac{1}{\sqrt{2}} \sin{\theta} \right]\nonumber\\ &\quad - \nu \cos{\theta} - 8 \sqrt{\frac{\nu t}{\rm \pi}} \sin^2{\theta} + \Bigg[ 4 \varOmega \sqrt{\frac{\nu t}{\rm \pi}} f_m^c - 4 \varOmega \sqrt{\frac{\nu t}{\rm \pi}} ( 3 - 2 \sqrt{2} ) \Bigg] \sin{\theta}\nonumber\\ &\quad+ \Bigg[ 2 \sqrt{\frac{\nu t}{\rm \pi}} f_m^{c s} - 8 \sqrt{\frac{\nu t}{\rm \pi}} ( 3 - 2 \sqrt{2}) \Bigg] \cos{2 \theta}. \end{align}

Thus, we have the alternative expression to compare with the result of Badr & Dennis (Reference Badr and Dennis1985):

(3.112) \begin{align} \Delta C_p & = \frac{p ( \theta )- p ( {\rm \pi})}{\rho} \approx- 2 \sqrt{\frac{\nu}{{\rm \pi} t}} ( 1 + \cos{\theta} ) - 2 \sin^2{\theta} - 16 \varOmega \sqrt{\frac{\nu t}{\rm \pi}} ( 1 - \sqrt{2} ) \sin{\theta}\nonumber\\ &\quad - 8 \sqrt{\frac{\nu t}{\rm \pi}} \sin^2{\theta} + \nu ( \cos{\theta} - 1 ) + \sqrt{\frac{\nu t}{\rm \pi}} [ 4 \varOmega f_m^c - 4 \varOmega ( 3 - 2 \sqrt{2} )] \sin{\theta}\nonumber\\ &\quad + \sqrt{\frac{\nu t}{\rm \pi}} [ 2 f_m^{c s} - 8 ( 1 - 2 \sqrt{2})] ( \cos{2 \theta} - 1 ). \end{align}

Here $f_m^c = 0.21483$ and $f_m^{c s} = -2.69143$. In contrast, the result of Badr & Dennis (Reference Badr and Dennis1985) is given by

(3.113)\begin{equation} \Delta C_p \approx- 2 \sqrt{\frac{\nu}{{\rm \pi} t}} ( 1 + \cos{\theta} ) - \left( 1 - \frac{\sqrt{\nu t}}{4} q_2 \right) ( 1 - \cos{2 \theta} ) + \frac{1}{2} \sqrt{\nu t} \varOmega q_1 \sin{\theta}, \end{equation}

in which the coefficients $q_1$ and $q_2$ are written as $q_1 = 5.79901$ and $q_2 = 14.3122$. Note here that the leading term is found to be the same between both results.

Figure 5.

Comparison of the pressure distribution $\Delta C_p = (\kern0.7pt p( \theta ) - p ( {\rm \pi}) ) / \rho$ on the cylinder surface between the present results and Badr & Dennis (Reference Badr and Dennis1985) in the case of $\nu = 1 / 500$ and $\varOmega = 0.5$.