2.1 An unsteady Stokeslet

The fundamental solution of unsteady Stokes flow represents the solution due to a concentrated point force $\boldsymbol{f}_{ext}$ , which has a singularity at the point $\boldsymbol{x}_{\mathbf{0}}$ and at time $t=0$ . The governing equation is

(2.1) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0,\\ \displaystyle \unicode[STIX]{x1D70C}\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}t}=-\unicode[STIX]{x1D735}p^{S}+\unicode[STIX]{x1D707}\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}+\boldsymbol{f}_{ext}\end{array}\right\}, & & \displaystyle\end{eqnarray}$$

where $\boldsymbol{f}_{ext}(\boldsymbol{x},t)=\unicode[STIX]{x1D736}\unicode[STIX]{x1D6FF}(\boldsymbol{x}-\boldsymbol{x}_{\mathbf{0}})\unicode[STIX]{x1D6FF}(t)$ is the external force, $\boldsymbol{u}=(u,v,w)$ is the fluid velocity, $\unicode[STIX]{x1D70C}$ is the fluid density, $p^{S}$ is the fluid pressure, $\unicode[STIX]{x1D707}$ is the dynamic viscosity of fluid, $\unicode[STIX]{x1D736}=(\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2},\unicode[STIX]{x1D6FC}_{3})$ is a constant vector and stands for the strength of the point force in a three-dimensional Euclidean space respectively, $\unicode[STIX]{x1D6FF}$ is the Dirac delta function and $\boldsymbol{x}$ is the position vector. The solution in three dimensions for the velocity field is given as

(2.2) $$\begin{eqnarray}\displaystyle u_{i}(\boldsymbol{x},t)=\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ij}^{S}(\hat{\boldsymbol{x}},t)\unicode[STIX]{x1D6FC}_{j}, & & \displaystyle\end{eqnarray}$$

(2.3) $$\begin{eqnarray}\displaystyle G_{ij}^{S}(\hat{\boldsymbol{x}},t) & = & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\frac{r^{2}\unicode[STIX]{x1D6FF}_{ij}-\hat{x}_{i}\hat{x}_{j}}{2\unicode[STIX]{x1D708}tr^{2}}-\frac{2}{r^{2}}\unicode[STIX]{x1D6FF}_{ij}+\frac{3(r^{2}\unicode[STIX]{x1D6FF}_{ij}-\hat{x}_{i}\hat{x}_{j})}{r^{4}}\right)\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{(r^{2}\unicode[STIX]{x1D6FF}_{ij}-3\hat{x}_{i}\hat{x}_{j})}{r^{5}}\text{erf}\left(\frac{r}{\sqrt{4\unicode[STIX]{x1D708}t}}\right).\end{eqnarray}$$

Here $r=|\hat{\boldsymbol{x}}|$ , $\hat{\boldsymbol{x}}=\boldsymbol{x}-\boldsymbol{x}_{\mathbf{0}}$ , $\unicode[STIX]{x1D708}=\unicode[STIX]{x1D707}/\unicode[STIX]{x1D70C}$ is the kinematic viscosity of fluid and $G_{ij}^{S}(\hat{\boldsymbol{x}},t)$ is a free-space Green’s function, also called a fundamental solution or an unsteady Stokeslet. The induced pressure is

(2.4) $$\begin{eqnarray}\displaystyle p^{S}(\boldsymbol{x},t)=\frac{\unicode[STIX]{x1D6FF}(t)}{4\unicode[STIX]{x03C0}}\frac{\hat{x}_{i}}{r^{3}}\unicode[STIX]{x1D6FC}_{i}. & & \displaystyle\end{eqnarray}$$

The surrounding fluid exerts a force on the control surface $S_{c}$ of a sphere with radius $a$ centred at the point impulse $\unicode[STIX]{x1D736}$ . The total forces are given by

(2.5) $$\begin{eqnarray}\displaystyle \boldsymbol{f}=-\frac{\unicode[STIX]{x1D736}}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\frac{a^{3}}{6\unicode[STIX]{x1D708}t^{2}}\right)\exp \left(-\frac{a^{2}}{4\unicode[STIX]{x1D708}t}\right)-\frac{1}{3}\unicode[STIX]{x1D736}\unicode[STIX]{x1D6FF}(t). & & \displaystyle\end{eqnarray}$$

Furthermore, in (2.5) it is observed that the first term and the second term on the right-hand side come from viscous force and the pressure, respectively. The aforementioned formulations have been presented by Hsiao & Young (Reference Hsiao and Young2015).

2.2 An unsteady Stokes doublet

By linear differential equations, the derivatives of $u_{i}(\boldsymbol{x},t)$ , $p^{S}(\boldsymbol{x},t)$ in (2.2), (2.4) are also solutions of (2.1). Therefore, we differentiate an unsteady Stokeslet with respect to $\boldsymbol{x}_{\pmb{{\it 0}}}$ . This means that we take the gradient of (2.3) in the chosen direction. Hence, the induced velocity in three dimensions is

(2.6) $$\begin{eqnarray}\displaystyle u_{i}(\boldsymbol{x},t)=\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijk}^{SD}(\hat{\boldsymbol{x}},t)d_{jk}. & & \displaystyle\end{eqnarray}$$

Following Blake (Reference Blake1971) and Chwang & Wu (Reference Chwang and Wu1975), $G_{ijk}^{SD}(\hat{\boldsymbol{x}},t)$ is called an unsteady Stokeslet doublet and $d_{jk}$ is the strength of singularity by a second-order tensor. Next, $G_{ijk}^{SD}(\hat{\boldsymbol{x}},t)$ is expressed in the form

(2.7) $$\begin{eqnarray}\displaystyle G_{ijk}^{SD}(\hat{\boldsymbol{x}},t) & = & \displaystyle \frac{\unicode[STIX]{x2202}G_{ij}^{S}(\hat{\boldsymbol{x}},t)}{\unicode[STIX]{x2202}x_{0,k}}=-\frac{\unicode[STIX]{x2202}G_{ij}^{S}(\hat{\boldsymbol{x}},t)}{\unicode[STIX]{x2202}\hat{x}_{k}}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\begin{array}{@{}l@{}}\displaystyle \frac{(3\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}+3\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk}+3\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik})}{r^{4}}-\frac{15\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{r^{6}}\\ \displaystyle +\,\frac{(\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}+\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk}+\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik})}{2\unicode[STIX]{x1D708}tr^{2}}-\frac{5\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{2\unicode[STIX]{x1D708}tr^{4}}\\ \displaystyle +\,\frac{\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}}{4\unicode[STIX]{x1D708}^{2}t^{2}}-\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{4\unicode[STIX]{x1D708}^{2}t^{2}r^{2}}\end{array}\right)\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{(3\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}+3\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk}+3\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik})}{r^{5}}\text{erf}\left(\frac{r}{\sqrt{4\unicode[STIX]{x1D708}t}}\right)+\frac{15\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{r^{7}}\text{erf}\left(\frac{r}{\sqrt{4\unicode[STIX]{x1D708}t}}\right).\end{eqnarray}$$

Similarly, the induced pressure of an unsteady Stokeslet doublet is

(2.8) $$\begin{eqnarray}\displaystyle p^{SD}(\boldsymbol{x},t)=\frac{\unicode[STIX]{x2202}p^{S}(\hat{\boldsymbol{x}},t)}{\unicode[STIX]{x2202}x_{0,k}}=\frac{\unicode[STIX]{x1D6FF}(t)}{4\unicode[STIX]{x03C0}}\left(-\frac{\unicode[STIX]{x1D6FF}_{jk}}{r^{3}}+3\frac{\hat{x}_{j}\hat{x}_{k}}{r^{5}}\right)d_{jk}. & & \displaystyle\end{eqnarray}$$

Physically, an unsteady Stokeslet doublet means the induced velocity is due to a point force dipole. Now, by integrating equation (2.6) with respect to $t$ from 0 to $\infty$ , it is found that

(2.9) $$\begin{eqnarray}\displaystyle \int _{0}^{\infty }u_{i}(\boldsymbol{x},t)\,\text{d}t & = & \displaystyle \int _{0}^{\infty }\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijk}^{SD}(\hat{\boldsymbol{x}},t)d_{jk}\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{8\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}}\left(\frac{\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}-\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk}-\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik}}{r^{3}}+\frac{3\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{r^{5}}\right)d_{jk}.\end{eqnarray}$$

Equation (2.9) is in agreement with the steady Stokeslet doublet given by Blake (Reference Blake1971), Chwang & Wu (Reference Chwang and Wu1975) and Pozrikidis (Reference Pozrikidis1992).

2.3 An unsteady rotlet

We now consider an unsteady rotlet (also called a couplet in steady Stokes flow by Batchelor (Reference Batchelor1970)), which regards the fundamental solution due to a point couple which has a singularity at the point $\boldsymbol{x}_{\pmb{{\it 0}}}$ and at time $t=0$ , in a fluid otherwise at rest. The governing equation is

(2.10) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0,\\ \displaystyle \unicode[STIX]{x1D70C}\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}t}=-\unicode[STIX]{x1D735}p+\unicode[STIX]{x1D707}\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}+\unicode[STIX]{x1D735}\times (\unicode[STIX]{x1D734}\unicode[STIX]{x1D6FF}(\boldsymbol{x}-\boldsymbol{x}_{\pmb{{\it 0}}})\unicode[STIX]{x1D6FF}(t))\end{array}\right\}. & & \displaystyle\end{eqnarray}$$

Since we are looking for rotational flow, the pressure is assumed to be constant. On the other hand, $\unicode[STIX]{x1D734}$ is a constant vector, so that $\unicode[STIX]{x1D735}\times \unicode[STIX]{x1D734}=0$ . After using vector identities, equation (2.10) becomes

(2.11) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}t}=\unicode[STIX]{x1D707}\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}+\unicode[STIX]{x1D735}\unicode[STIX]{x1D6FF}(\hat{\boldsymbol{x}})\times \unicode[STIX]{x1D734}\unicode[STIX]{x1D6FF}(t). & & \displaystyle\end{eqnarray}$$

By taking the Laplace transform of (2.11) with respect to $t$ , we obtain

(2.12) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}s\hat{\boldsymbol{u}}(s,\boldsymbol{x})=\unicode[STIX]{x1D707}\unicode[STIX]{x1D6FB}^{2}\hat{\boldsymbol{u}}(s,\boldsymbol{x})+\unicode[STIX]{x1D735}\unicode[STIX]{x1D6FF}(\hat{\boldsymbol{x}})\times \unicode[STIX]{x1D734}. & & \displaystyle\end{eqnarray}$$

The Laplace transform of a function $\boldsymbol{u}(t,\boldsymbol{x})$ , defined for all real numbers $t\geqslant 0$ , is the function $\hat{\boldsymbol{u}}(s,\boldsymbol{x})$ , defined by

(2.13) $$\begin{eqnarray}\displaystyle \mathscr{L}(\boldsymbol{u}(t,\boldsymbol{x}))=\hat{\boldsymbol{u}}(s,\boldsymbol{x})=\int _{0}^{\infty }\boldsymbol{u}(t,\boldsymbol{x})\exp (-st)\,\text{d}t. & & \displaystyle\end{eqnarray}$$

Again taking the Fourier transform of (2.12) we obtain

(2.14) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}s\bar{\hat{\boldsymbol{u}}}(s,\boldsymbol{k})=-\unicode[STIX]{x1D707}(k_{x}^{2}+k_{y}^{2}+k_{z}^{2})\bar{\hat{\boldsymbol{u}}}+\frac{1}{(2\unicode[STIX]{x03C0})^{3/2}}(\text{i}k_{x}\hat{i} +\text{i}k_{y}{\hat{j}}+\text{i}k_{z}\hat{k})\text{e}^{-\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{x}_{\mathbf{0}}}\times \unicode[STIX]{x1D734}. & & \displaystyle\end{eqnarray}$$

The Fourier transform of a function $\hat{\boldsymbol{u}}(s,\boldsymbol{x})$ is defined by

(2.15) $$\begin{eqnarray}\displaystyle \mathscr{F}(\hat{\boldsymbol{u}}(s,\boldsymbol{x}))=\bar{\hat{\boldsymbol{u}}}(s,\boldsymbol{k})=\frac{1}{(2\unicode[STIX]{x03C0})^{3/2}}\int _{\boldsymbol{E}^{3}}\hat{\boldsymbol{u}}(s,\boldsymbol{x})\exp (-\text{i}(\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{x}))\,\text{d}\boldsymbol{x}, & & \displaystyle\end{eqnarray}$$

where $\boldsymbol{E}^{3}$ denotes the three-dimensional Euclidean space and $\boldsymbol{k}=(k_{x},k_{y},k_{z})$ is a constant vector, respectively. After a straightforward algebra manipulation, equation (2.14) now becomes

(2.16) $$\begin{eqnarray}\displaystyle \bar{\hat{\boldsymbol{u}}}(s,\boldsymbol{k})=\frac{\text{i}}{(2\unicode[STIX]{x03C0})^{3/2}\unicode[STIX]{x1D70C}}\left(\frac{k_{x}}{s+\unicode[STIX]{x1D708}\Vert \boldsymbol{k}\Vert ^{2}}\hat{i} +\frac{k_{y}}{s+\unicode[STIX]{x1D708}\Vert \boldsymbol{k}\Vert ^{2}}{\hat{j}}+\frac{k_{z}}{s+\unicode[STIX]{x1D708}\Vert \boldsymbol{k}\Vert ^{2}}\hat{k}\right)\text{e}^{-\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{x}_{\mathbf{0}}}\times \unicode[STIX]{x1D734}. & & \displaystyle\end{eqnarray}$$

Taking the Laplace and Fourier inversions of (2.16), we obtain the fundamental solution of an unsteady rotlet in the vector form

(2.17) $$\begin{eqnarray}\displaystyle \boldsymbol{u}(\boldsymbol{x},t) & = & \displaystyle \frac{1}{(2\unicode[STIX]{x03C0})^{3/2}\unicode[STIX]{x1D70C}}\frac{1}{(2\unicode[STIX]{x1D708}t)^{3/2}}\unicode[STIX]{x1D735}\left(\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right)\right)\times \unicode[STIX]{x1D734}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}\frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\frac{\unicode[STIX]{x1D734}\times \boldsymbol{r}}{4\unicode[STIX]{x1D708}^{2}t^{2}}\right)\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right),\end{eqnarray}$$

where $r=|\boldsymbol{x}-\boldsymbol{x}_{\mathbf{0}}|$ . Equation (2.17) can also be expressed in a tensor form as follows:

(2.18) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\displaystyle u_{i}(\boldsymbol{x},t)=\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ij}^{R}(\hat{\boldsymbol{x}},t)\unicode[STIX]{x1D6FA}_{j},\\ \displaystyle \text{in which }G_{ij}^{R}(\hat{\boldsymbol{x}},t)=\frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\frac{\unicode[STIX]{x1D700}_{ijk}\hat{x}_{k}}{4\unicode[STIX]{x1D708}^{2}t^{2}}\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right).\end{array}\right\} & & \displaystyle\end{eqnarray}$$

In order to verify (2.18), we integrate the equation with respect to $t$ from 0 to $\infty$ , and get

(2.19) $$\begin{eqnarray}\displaystyle \int _{0}^{\infty }\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ij}^{R}(\hat{\boldsymbol{x}},t)\unicode[STIX]{x1D6FA}_{j}\,\text{d}t=\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}}\frac{\unicode[STIX]{x1D734}\times \boldsymbol{r}}{r^{3}}. & & \displaystyle\end{eqnarray}$$

Equation (2.19) agrees with the steady rotlet equation obtained by Blake (Reference Blake1971), Chwang & Wu (Reference Chwang and Wu1975) and Pozrikidis (Reference Pozrikidis1992). The surrounding fluid exerts a force on the control surface $S_{c}$ of a sphere with radius $a$ centred at a point impulse with strength $\unicode[STIX]{x1D734}$ . The force is given by

(2.20) $$\begin{eqnarray}\displaystyle \boldsymbol{f}=\int _{S_{c}}\unicode[STIX]{x1D748}\boldsymbol{\cdot }\boldsymbol{n}\,\text{d}A, & & \displaystyle\end{eqnarray}$$

where $\unicode[STIX]{x1D748}$ is the stress per unit area of the spherical surface and $\boldsymbol{n}$ is the outward unit vector normal to the surface. Due to the symmetry, the hydrodynamic force in (2.20) is $\boldsymbol{f}=\mathbf{0}$ . Next, the hydrodynamic torque on the sphere is given by

(2.21) $$\begin{eqnarray}\displaystyle \boldsymbol{c}=\int _{S_{c}}\boldsymbol{r}\times (\unicode[STIX]{x1D748}\boldsymbol{\cdot }\boldsymbol{n})\,\text{d}A. & & \displaystyle\end{eqnarray}$$

Making use of the constitutive relation for stress in a Newtonian fluid,

(2.22) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70E}_{ij}=-p\unicode[STIX]{x1D6FF}_{ij}+\unicode[STIX]{x1D707}\left(\frac{\unicode[STIX]{x2202}u_{j}}{\unicode[STIX]{x2202}x_{i}}+\frac{\unicode[STIX]{x2202}u_{i}}{\unicode[STIX]{x2202}x_{j}}\right), & & \displaystyle\end{eqnarray}$$

then (2.21), after some reduction we obtain the hydrodynamic torque as

(2.23) $$\begin{eqnarray}\displaystyle \boldsymbol{c}=-\frac{\unicode[STIX]{x1D734}}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\frac{a^{5}}{12\unicode[STIX]{x1D708}^{2}t^{3}}\right)\exp \left(-\frac{a^{2}}{4\unicode[STIX]{x1D708}t}\right). & & \displaystyle\end{eqnarray}$$

In a similar way, we integrate equation (2.23) with respect to $t$ from 0 to $\infty$ , and the torque due to a steady rotlet is given as

(2.24) $$\begin{eqnarray}\displaystyle \boldsymbol{C}=-\int _{0}^{\infty }\frac{\unicode[STIX]{x1D734}}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\frac{a^{5}}{12\unicode[STIX]{x1D708}^{2}t^{3}}\right)\exp \left(-\frac{a^{2}}{4\unicode[STIX]{x1D708}t}\right)\,\text{d}t=-2\unicode[STIX]{x1D734}. & & \displaystyle\end{eqnarray}$$

The result also agrees with the steady rotlet equation obtained by Chwang & Wu (Reference Chwang and Wu1975).

2.4 An unsteady stresslet

Equation (2.7) demonstrates that an unsteady Stokeslet doublet can be divided into a symmetrical part and an antisymmetrical part with respect to the indices $j$ and $k$ , thus

(2.25) $$\begin{eqnarray}\displaystyle G_{ijk}^{SD}(\hat{\boldsymbol{x}},t)=G_{ijk}^{SD-S}(\hat{\boldsymbol{x}},t)+G_{ijk}^{SD-AS}(\hat{\boldsymbol{x}},t), & & \displaystyle\end{eqnarray}$$

where $G_{ijk}^{SD-S}(\hat{\boldsymbol{x}},t)$ and $G_{ijk}^{SD-AS}(\hat{\boldsymbol{x}},t)$ denote the symmetrical part and the antisymmetrical part, respectively. The symmetrical part is

(2.26) $$\begin{eqnarray}\displaystyle G_{ijk}^{SD-S}(\hat{\boldsymbol{x}},t) & = & \displaystyle \frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\begin{array}{@{}c@{}}\displaystyle \frac{(3\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}+3\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik}+3\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk})}{r^{4}}-\frac{15\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{r^{6}}\\ \displaystyle +\,\frac{(\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}+\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik}+\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk})}{2\unicode[STIX]{x1D708}tr^{2}}-\frac{5\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{2\unicode[STIX]{x1D708}tr^{4}}\\ \displaystyle +\,\frac{\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}}{8\unicode[STIX]{x1D708}^{2}t^{2}}+\frac{\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik}}{8\unicode[STIX]{x1D708}^{2}t^{2}}-\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{4\unicode[STIX]{x1D708}^{2}t^{2}r^{2}}\end{array}\right)\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{(3\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk}+3\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}+3\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik})}{r^{5}}\text{erf}\left(\frac{r}{\sqrt{4\unicode[STIX]{x1D708}t}}\right)+\frac{15\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{r^{7}}\text{erf}\left(\frac{r}{\sqrt{4\unicode[STIX]{x1D708}t}}\right)\end{eqnarray}$$

and the antisymmetrical part is

(2.27) $$\begin{eqnarray}\displaystyle G_{ijk}^{SD-AS}(\hat{\boldsymbol{x}},t)=\frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\frac{\hat{x}_{k}\unicode[STIX]{x1D6FF}_{ij}}{8\unicode[STIX]{x1D708}^{2}t^{2}}-\frac{\hat{x}_{j}\unicode[STIX]{x1D6FF}_{ik}}{8\unicode[STIX]{x1D708}^{2}t^{2}}\right)\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right). & & \displaystyle\end{eqnarray}$$

Substituting (2.26) and (2.27) into (2.6) yields

(2.28) $$\begin{eqnarray}\displaystyle u_{i}(\boldsymbol{x},t)=\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijk}^{SD-S}(\hat{\boldsymbol{x}},t)d_{jk}+\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijk}^{SD-AS}(\hat{\boldsymbol{x}},t)d_{jk}. & & \displaystyle\end{eqnarray}$$

The first term on the right-hand side in (2.28) is called an unsteady stresslet, which stands for the velocity field corresponding to the straining motion. By integrating with respect to $t$ from 0 to $\infty$ , we get

(2.29) $$\begin{eqnarray}\displaystyle \int _{0}^{\infty }\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijk}^{SD-S}(\hat{\boldsymbol{x}},t)d_{jk}\,\text{d}t=\frac{1}{8\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}}\left(-\frac{\hat{x}_{i}\unicode[STIX]{x1D6FF}_{jk}}{r^{3}}+\frac{3\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{r^{5}}\right)d_{jk}. & & \displaystyle\end{eqnarray}$$

The result also agrees with the steady stresslet equation obtained by Blake (Reference Blake1971) and Chwang & Wu (Reference Chwang and Wu1975). If we define $\unicode[STIX]{x1D734}=-1/2\unicode[STIX]{x1D700}_{ijk}d_{jk}$ , then the second term in (2.28) is given by

(2.30) $$\begin{eqnarray}\displaystyle \frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijk}^{SD-AS}(\hat{\boldsymbol{x}},t)d_{jk}=\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}\frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\frac{\unicode[STIX]{x1D734}\times \boldsymbol{r}}{4\unicode[STIX]{x1D708}^{2}t^{2}}\right)\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right). & & \displaystyle\end{eqnarray}$$

Equation (2.30) corresponds to the fundamental solution of an unsteady rotlet, which is the same as (2.17). The induced pressure of an unsteady stresslet is equivalent to an unsteady Stokeslet doublet as shown in (2.8), since the pressure of an unsteady rotlet is constant. Furthermore, a steady Stokeslet doublet has been already separated into the symmetrical part (namely, a stresslet) and the antisymmetrical part (namely, a rotlet) by Batchelor (Reference Batchelor1970) and Blake (Reference Blake1971). This property also exists in unsteady Stokes flow. Due to the symmetry, a steady Stokeslet doublet results in no net force and a steady stresslet results in no net force and no net torque to the fluid, as presented by Blake (Reference Blake1971) and Chwang & Wu (Reference Chwang and Wu1975). Similarly, an unsteady Stokeslet doublet and an unsteady stresslet also show the identical properties as expected.

2.5 An unsteady Stokeslet quadrupole

Next, we differentiate again an unsteady Stokeslet doublet with respect to $\boldsymbol{x}_{\mathbf{0}}$ . The induced velocity in three dimensions is

(2.31) $$\begin{eqnarray}\displaystyle u_{i}(\boldsymbol{x},t)=\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijkl}^{SQ}(\hat{\boldsymbol{x}},t)Q_{jkl}, & & \displaystyle\end{eqnarray}$$

where $G_{ijkl}^{SQ}(\hat{\boldsymbol{x}},t)$ is called an unsteady Stokeslet quadrupole and $Q_{jkl}$ is the strength of singularity by a third-order tensor. $G_{ijkl}^{SQ}(\hat{\boldsymbol{x}},t)$ is defined by

(2.32) $$\begin{eqnarray}\displaystyle & & \displaystyle G_{ijkl}^{SQ}(\hat{\boldsymbol{x}},t)=\frac{\unicode[STIX]{x2202}G_{ijk}^{SD}(\hat{\boldsymbol{x}},t)}{\unicode[STIX]{x2202}x_{0,l}}=-\frac{\unicode[STIX]{x2202}G_{ijk}^{SD}(\hat{\boldsymbol{x}},t)}{\unicode[STIX]{x2202}\hat{x}_{l}}\nonumber\\ \displaystyle & & \displaystyle =\frac{1}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}t}}\left(\begin{array}{@{}l@{}}\displaystyle -3\frac{\unicode[STIX]{x1D6FF}_{kl}\unicode[STIX]{x1D6FF}_{ij}+\unicode[STIX]{x1D6FF}_{il}\unicode[STIX]{x1D6FF}_{jk}+\unicode[STIX]{x1D6FF}_{jl}\unicode[STIX]{x1D6FF}_{ik}}{r^{4}}+15\frac{\hat{x}_{k}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ij}+\hat{x}_{i}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{jk}+\hat{x}_{j}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ik}}{r^{6}}\\ \displaystyle +\,15\frac{\hat{x}_{j}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{il}+\hat{x}_{i}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{jl}+\hat{x}_{i}\hat{x}_{j}\unicode[STIX]{x1D6FF}_{kl}}{r^{6}}-105\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{8}}\\ \displaystyle +\,\frac{1}{2\unicode[STIX]{x1D708}t}\left(\begin{array}{@{}l@{}}\displaystyle -\,\frac{\unicode[STIX]{x1D6FF}_{kl}\unicode[STIX]{x1D6FF}_{ij}+\unicode[STIX]{x1D6FF}_{il}\unicode[STIX]{x1D6FF}_{jk}+\unicode[STIX]{x1D6FF}_{jl}\unicode[STIX]{x1D6FF}_{ik}}{r^{2}}+5\frac{\hat{x}_{k}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ij}+\hat{x}_{i}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{jk}+\hat{x}_{j}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ik}}{r^{4}}\\ \displaystyle +\,5\frac{\hat{x}_{j}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{il}+\hat{x}_{i}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{jl}+\hat{x}_{i}\hat{x}_{j}\unicode[STIX]{x1D6FF}_{kl}}{r^{4}}-35\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{6}}\end{array}\right)\\ \displaystyle +\,\frac{1}{(2\unicode[STIX]{x1D708}t)^{2}}\left(\begin{array}{@{}l@{}}\displaystyle -\unicode[STIX]{x1D6FF}_{kl}\unicode[STIX]{x1D6FF}_{ij}+\frac{\hat{x}_{k}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ij}+\hat{x}_{i}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{jk}+\hat{x}_{j}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ik}}{r^{2}}\\ \displaystyle +\,\frac{\hat{x}_{j}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{il}+\hat{x}_{i}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{jl}+\hat{x}_{i}\hat{x}_{j}\unicode[STIX]{x1D6FF}_{kl}}{r^{2}}-7\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{4}}\end{array}\right)\\ \displaystyle +\,\frac{1}{\left(2\unicode[STIX]{x1D708}t\right)^{3}}\left(\hat{x}_{k}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ij}-\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{2}}\right)\end{array}\!\right)\exp \left(-\frac{r^{2}}{4\unicode[STIX]{x1D708}t}\right)\nonumber\\ \displaystyle & & \displaystyle +\,\text{erf}\left(\frac{r}{\sqrt{4\unicode[STIX]{x1D708}t}}\right)\left(\begin{array}{@{}l@{}}\displaystyle 3\frac{\unicode[STIX]{x1D6FF}_{kl}\unicode[STIX]{x1D6FF}_{ij}+\unicode[STIX]{x1D6FF}_{il}\unicode[STIX]{x1D6FF}_{jk}+\unicode[STIX]{x1D6FF}_{jl}\unicode[STIX]{x1D6FF}_{ik}}{r^{5}}+105\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{9}}\\ \displaystyle -\,15\frac{\hat{x}_{k}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ij}+\hat{x}_{i}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{jk}+\hat{x}_{j}\hat{x}_{l}\unicode[STIX]{x1D6FF}_{ik}}{r^{7}}-15\frac{\hat{x}_{j}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{il}+\hat{x}_{i}\hat{x}_{k}\unicode[STIX]{x1D6FF}_{jl}+\hat{x}_{i}\hat{x}_{j}\unicode[STIX]{x1D6FF}_{kl}}{r^{7}}\end{array}\right).\end{eqnarray}$$

By integrating equation (2.32) with respect to $t$ from 0 to $\infty$ , we obtain

(2.33) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }u_{i}(\boldsymbol{x},t)\,\text{d}t=\int _{0}^{\infty }\frac{1}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}}G_{ijkl}^{SQ}(\hat{\boldsymbol{x}},t)Q_{jkl}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{8\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}}\left(\begin{array}{@{}l@{}}\displaystyle \frac{\unicode[STIX]{x1D6FF}_{ik}\unicode[STIX]{x1D6FF}_{jl}+\unicode[STIX]{x1D6FF}_{il}\unicode[STIX]{x1D6FF}_{jk}-\unicode[STIX]{x1D6FF}_{ij}\unicode[STIX]{x1D6FF}_{kl}}{r^{3}}\\ \displaystyle -\,\frac{3}{r^{5}}(\unicode[STIX]{x1D6FF}_{kl}\hat{x}_{i}\hat{x}_{j}+\unicode[STIX]{x1D6FF}_{jl}\hat{x}_{i}\hat{x}_{k}+\unicode[STIX]{x1D6FF}_{jk}\hat{x}_{i}\hat{x}_{l}+\unicode[STIX]{x1D6FF}_{il}\hat{x}_{j}\hat{x}_{k}+\unicode[STIX]{x1D6FF}_{ik}\hat{x}_{j}\hat{x}_{l}-\unicode[STIX]{x1D6FF}_{ij}\hat{x}_{k}\hat{x}_{l})\\ \displaystyle +\,15\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{7}}\end{array}\right)Q_{jkl}.\end{eqnarray}$$

Equation (2.33) is in agreement with the fundamental solution of the steady Stokeslet quadrupole as given by Pozrikidis (Reference Pozrikidis1992). Similarly, the induced pressure of an unsteady Stokeslet quadrupole is obtained by

(2.34) $$\begin{eqnarray}\displaystyle p^{SQ}(\hat{\boldsymbol{x}},t)=\frac{\unicode[STIX]{x2202}p^{SD}(\hat{\boldsymbol{x}},t)}{\unicode[STIX]{x2202}x_{0,l}}=\frac{\unicode[STIX]{x1D6FF}(t)}{4\unicode[STIX]{x03C0}}\left(-\frac{3\unicode[STIX]{x1D6FF}_{jk}\hat{x}_{l}+3\unicode[STIX]{x1D6FF}_{jl}\hat{x}_{k}+3\unicode[STIX]{x1D6FF}_{kl}\hat{x}_{j}}{r^{^{5}}}+\frac{15\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{7}}\right)Q_{jkl}. & & \displaystyle\end{eqnarray}$$

2.6 The potential singularity of Stokes flow

We differentiate successively a point source and can obtain higher-order singularities. Thus a potential dipole (appendix A, equation (A 5)) is given by

(2.35) $$\begin{eqnarray}\displaystyle G_{ij}^{PD}(\hat{\boldsymbol{x}})=-\frac{\unicode[STIX]{x1D6FF}_{ij}}{r^{3}}+3\frac{\hat{x}_{i}\hat{x}_{j}}{r^{5}}. & & \displaystyle\end{eqnarray}$$

A potential quadrupole $G_{ijk}^{PQ}(\hat{\boldsymbol{x}})$ obtained by differentiating a potential dipole is given as

(2.36) $$\begin{eqnarray}\displaystyle G_{ijk}^{PQ}(\hat{\boldsymbol{x}})=\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x_{0,k}}\left(-\frac{\unicode[STIX]{x1D6FF}_{ij}}{r^{3}}+3\frac{\hat{x}_{i}\hat{x}_{j}}{r^{5}}\right)=-\frac{3\unicode[STIX]{x1D6FF}_{ij}\hat{x}_{k}+3\unicode[STIX]{x1D6FF}_{ik}\hat{x}_{j}+3\unicode[STIX]{x1D6FF}_{jk}\hat{x}_{i}}{r^{^{5}}}+\frac{15\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}}{r^{7}}. & & \displaystyle\end{eqnarray}$$

In addition, a potential octupole, that is, the second derivative of a potential dipole, is given by

(2.37) $$\begin{eqnarray}\displaystyle G_{ijkl}^{PO}(\hat{\boldsymbol{x}}) & = & \displaystyle \frac{\unicode[STIX]{x2202}^{2}G_{ij}^{PD}(\hat{\boldsymbol{x}})}{\unicode[STIX]{x2202}x_{0,k}\unicode[STIX]{x2202}x_{0,l}}=\frac{\unicode[STIX]{x2202}^{2}G_{ij}^{PD}(\hat{\boldsymbol{x}})}{\unicode[STIX]{x2202}\hat{x}_{k}\unicode[STIX]{x2202}\hat{x}_{l}}\nonumber\\ \displaystyle & = & \displaystyle \left(\begin{array}{@{}c@{}}\displaystyle 3\frac{\unicode[STIX]{x1D6FF}_{ij}\unicode[STIX]{x1D6FF}_{kl}+\unicode[STIX]{x1D6FF}_{ik}\unicode[STIX]{x1D6FF}_{jl}+\unicode[STIX]{x1D6FF}_{jk}\unicode[STIX]{x1D6FF}_{il}}{r^{^{5}}}-15\frac{\unicode[STIX]{x1D6FF}_{ij}\hat{x}_{k}\hat{x}_{l}+\unicode[STIX]{x1D6FF}_{ik}\hat{x}_{j}\hat{x}_{l}+\unicode[STIX]{x1D6FF}_{jk}\hat{x}_{i}\hat{x}_{l}}{r^{^{7}}}\\ \displaystyle -\,15\frac{\unicode[STIX]{x1D6FF}_{il}\hat{x}_{j}\hat{x}_{k}+\unicode[STIX]{x1D6FF}_{jl}\hat{x}_{i}\hat{x}_{k}+\unicode[STIX]{x1D6FF}_{kl}\hat{x}_{i}\hat{x}_{j}}{r^{^{7}}}+105\frac{\hat{x}_{i}\hat{x}_{j}\hat{x}_{k}\hat{x}_{l}}{r^{9}}\end{array}\right).\end{eqnarray}$$

These singularities are standard and have been addressed in the literature (for example, Kim & Karrila Reference Kim and Karrila1991 and Pozrikidis Reference Pozrikidis1992).

Figure 1.

Schematic representation of a sphere which rotates with an angular velocity $\unicode[STIX]{x1D74E}(t)$ .

6.1 Contribution of the unsteady Stokeslet

The force contribution from the unsteady Stokeslet can be decomposed as viscous and pressure parts $\boldsymbol{F}_{\boldsymbol{s}}(t)=\boldsymbol{F}_{\boldsymbol{s},\boldsymbol{v}}(t)+\boldsymbol{F}_{\boldsymbol{s},\boldsymbol{p}}(t)$ . In order to obtain the viscous force $\boldsymbol{F}_{\boldsymbol{s},\boldsymbol{v}}(t)$ , we can use the first term on the right-hand side in (2.5) which shows that the viscous force exerted by an unsteady Stokeslet on the sphere in time $t$ , and the convolution integral will yield

(6.9) $$\begin{eqnarray}\displaystyle \boldsymbol{F}_{\boldsymbol{s},\boldsymbol{v}}(t)=-\int _{0}^{t}\frac{\unicode[STIX]{x1D736}(\unicode[STIX]{x1D70F})}{\sqrt{\unicode[STIX]{x03C0}\unicode[STIX]{x1D708}}}\left(\frac{a^{3}}{6\unicode[STIX]{x1D708}}\right)\frac{\exp \left(-\displaystyle \frac{a^{2}}{4\unicode[STIX]{x1D708}(t-\unicode[STIX]{x1D70F})}\right)}{(t-\unicode[STIX]{x1D70F})^{5/2}}\,\text{d}\unicode[STIX]{x1D70F}. & & \displaystyle\end{eqnarray}$$

Taking the Laplace transform of both sides, we have

(6.10) $$\begin{eqnarray}\displaystyle \boldsymbol{F}_{\boldsymbol{s},\boldsymbol{v}}(s)=-\frac{2}{3}\hat{\unicode[STIX]{x1D736}}(s)\left(1+\frac{a}{\sqrt{\unicode[STIX]{x1D708}}}\sqrt{s}\right)\exp (-\sqrt{a^{2}s/\unicode[STIX]{x1D708}}). & & \displaystyle\end{eqnarray}$$

By substituting $\hat{\unicode[STIX]{x1D6FC}}_{1}(s)$ of (6.3) into (6.10), we get

(6.11) $$\begin{eqnarray}\displaystyle \boldsymbol{F}_{\boldsymbol{s},\boldsymbol{v}}(s)=\left\{\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a^{3}K(s)+\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\unicode[STIX]{x1D708}}\sqrt{s}K(s)+\frac{14}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\frac{s\left(1+\displaystyle \frac{a}{\sqrt{\unicode[STIX]{x1D708}}}\sqrt{s}\right)}{1+\displaystyle \frac{a}{\sqrt{\unicode[STIX]{x1D708}}}\sqrt{s}+\frac{a^{2}}{3\unicode[STIX]{x1D708}}s}K(s)\right\}\boldsymbol{e}_{\boldsymbol{x}}. & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Using the convolution theorem and inverse Laplace transform (solved in appendix D), equation (6.11) including the initial condition becomes

(6.12) $$\begin{eqnarray}\displaystyle \boldsymbol{F}_{\boldsymbol{s},\boldsymbol{v}}(t) & = & \displaystyle \left\{\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a^{3}K(t)+\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}K}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}\right.\nonumber\\ \displaystyle & & \displaystyle +\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}K}{\text{d}\unicode[STIX]{x1D70F}}S_{F1}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}+\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\int _{0}^{t}\frac{\text{d}^{2}K}{\text{d}\unicode[STIX]{x1D70F}^{2}}S_{F2}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}K(0)\frac{1}{\sqrt{t}}+\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}K(0)S_{F1}(t)\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\left(K(0)\frac{\text{d}}{\text{d}t}(S_{F2}(t))+\left.\frac{\text{d}K}{\text{d}t}\right|_{t=0}S_{F2}(t)\right)\right\}\boldsymbol{e}_{\boldsymbol{x}},\end{eqnarray}$$

where

(6.13) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\displaystyle S_{F1}(t)=\text{Im}\left(\sqrt{\unicode[STIX]{x03C0}}\left(\frac{\sqrt{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D6FD}}{a}\right)\text{e}^{(\unicode[STIX]{x1D708}/a^{2})\unicode[STIX]{x1D6FD}^{2}t}\text{erfc}\left(\frac{\sqrt{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D6FD}}{a}\sqrt{t}\right)\right),\\ \displaystyle S_{F2}(t)=-\text{Im}\left(\text{e}^{(\unicode[STIX]{x1D707}/a^{2})\unicode[STIX]{x1D6FD}^{2}t}\text{erfc}\left(\frac{\sqrt{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D6FD}}{a}\sqrt{t}\right)\right),\quad \text{and}\quad \unicode[STIX]{x1D6FD}=\frac{3}{2}+\frac{\sqrt{3}}{2}\text{i}.\end{array}\right\} & & \displaystyle\end{eqnarray}$$

To determine the pressure force $\boldsymbol{F}_{\boldsymbol{s},\boldsymbol{p}}(t)$ , we note that the second term on the right-hand side in (2.5) shows the pressure exerted by an unsteady Stokeslet. We substitute $\hat{\unicode[STIX]{x1D6FC}}_{1}(s)$ of (6.3) into the second term in (2.5) and use the convolution integral to obtain

(6.14) $$\begin{eqnarray}\displaystyle \boldsymbol{F}_{\boldsymbol{s},\boldsymbol{p}}(t)=-\frac{1}{3}\int _{0}^{t}\unicode[STIX]{x1D6FC}_{1}(\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6FF}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}\boldsymbol{e}_{\boldsymbol{x}}=-\frac{1}{3}\unicode[STIX]{x1D6FC}_{1}(t)\boldsymbol{e}_{\boldsymbol{x}}. & & \displaystyle\end{eqnarray}$$

Therefore, the total force acting on the sphere due to an unsteady Stokeslet is the summation of (6.12) and (6.14) as follows:

(6.15) $$\begin{eqnarray}\displaystyle \boldsymbol{F}_{\boldsymbol{s}}(t) & = & \displaystyle \boldsymbol{F}_{\boldsymbol{s},\boldsymbol{v}}(t)+\boldsymbol{F}_{\boldsymbol{s},\boldsymbol{p}}(t)\nonumber\\ \displaystyle & = & \displaystyle \left\{\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a^{3}K(t)+\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}K}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}\right.\nonumber\\ \displaystyle & & \displaystyle +\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}K}{\text{d}\unicode[STIX]{x1D70F}}S_{F1}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}+\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\int _{0}^{t}\frac{\text{d}^{2}K}{\text{d}\unicode[STIX]{x1D70F}^{2}}S_{F2}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{8}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}K(0)\frac{1}{\sqrt{t}}+\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}K(0)S_{F1}(t)\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\left(K(0)\frac{d}{dt}(S_{F2}(t))+\left.\frac{\text{d}K}{\text{d}t}\right|_{t=0}S_{F2}(t)\right)-\frac{1}{3}\unicode[STIX]{x1D6FC}_{1}(t)\right\}\boldsymbol{e}_{\boldsymbol{x}}.\end{eqnarray}$$

6.2 Contribution of the unsteady potential dipole

The surrounding fluid exerts a force on the sphere of radius $a$ centred at an unsteady potential dipole as (appendix A, equation (A 7))

(6.16) $$\begin{eqnarray}\displaystyle F_{i}(t)=-\frac{\unicode[STIX]{x1D70C}}{3}\frac{\text{d}}{\text{d}t}(d_{i}(t)). & & \displaystyle\end{eqnarray}$$

By taking the Laplace transform of (6.16), the surface force due to an unsteady potential dipole in the Laplace transform domain is

(6.17) $$\begin{eqnarray}\displaystyle \hat{F}_{i}(s)=-\frac{\unicode[STIX]{x1D70C}}{3}s\hat{d}_{i}(s). & & \displaystyle\end{eqnarray}$$

Substituting $\hat{d}_{1}(s)$ of (6.5) into (6.17) and making use of (6.3), we get

(6.18) $$\begin{eqnarray}\displaystyle \hat{\boldsymbol{F}}_{\boldsymbol{d}}(s)=\left\{\frac{4}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a^{3}K(s)+\frac{4}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}sK(s)+\frac{4}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\unicode[STIX]{x1D708}}\sqrt{s}K(s)+\frac{1}{3}\hat{\unicode[STIX]{x1D6FC}}_{1}(s)\right\}\boldsymbol{e}_{\boldsymbol{x}}. & & \displaystyle\end{eqnarray}$$

Using the inverse Laplace transform, we obtain

(6.19) $$\begin{eqnarray}\displaystyle \boldsymbol{F}_{\boldsymbol{d}}(t) & = & \displaystyle \left\{\frac{4}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a^{3}K(t)+\frac{4}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\frac{\text{d}K}{\text{d}t}+\frac{4}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}K}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}+\frac{1}{3}\unicode[STIX]{x1D6FC}_{1}(t)\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{4}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}K(0)+\frac{4}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}K(0)\frac{1}{\sqrt{t}}\right\}\boldsymbol{e}_{\boldsymbol{x}}.\end{eqnarray}$$

It should be mentioned that an unsteady Stokeslet quadrupole and an unsteady potential octupole result in no net force and no net torque to the fluid due to the symmetry property.

6.3 Net hydrodynamic force on the sphere

By combining (6.15) and (6.19), we finally have the hydrodynamic force on the sphere as

(6.20) $$\begin{eqnarray}\displaystyle \boldsymbol{F} & = & \displaystyle \left\{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a^{3}K(t)+\frac{4}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\frac{\text{d}K}{\text{d}t}+4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}K}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}\right.\nonumber\\ \displaystyle & & \displaystyle +\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}K}{\text{d}\unicode[STIX]{x1D70F}}S_{F1}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}+\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\int _{0}^{t}\frac{\text{d}^{2}K}{\text{d}\unicode[STIX]{x1D70F}^{2}}S_{F2}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{4}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}K(0)+4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}K(0)\frac{1}{\sqrt{t}}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{4}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}K(0)S_{F1}(t)\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{5}\left(K(0)\frac{\text{d}}{\text{d}t}(S_{F2}(t))+\left.\frac{\text{d}K}{\text{d}t}\right|_{t=0}S_{F2}(t)\right)\right\}\boldsymbol{e}_{\boldsymbol{x}}.\end{eqnarray}$$

For steady Stokes flow, the drag on the sphere when the ambient field has a paraboloidal profile with the axis passing through the centre of the sphere is the same as that on the same sphere moves with an ‘equivalent velocity’ in a uniform flow (Chwang & Wu Reference Chwang and Wu1975). The role is based on the calculation of drag force. In (6.20), we follow the equivalent velocity in steady Stokes flow given by Chwang & Wu (Reference Chwang and Wu1975) to put it in the following form:

(6.21) $$\begin{eqnarray}\displaystyle \boldsymbol{U}_{\boldsymbol{e}}=\frac{2}{3}K(t)a^{2}\boldsymbol{e}_{\boldsymbol{ x}}. & & \displaystyle\end{eqnarray}$$

Therefore, equation (6.20) can be rewritten in an equivalent velocity form:

(6.22) $$\begin{eqnarray}\displaystyle \boldsymbol{F} & = & \displaystyle 6\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a\boldsymbol{U}_{\boldsymbol{e}}(t)+\frac{2}{5}\unicode[STIX]{x03C0}a^{3}\unicode[STIX]{x1D70C}\frac{\text{d}\boldsymbol{U}_{\boldsymbol{e}}}{\text{d}t}+6\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}\boldsymbol{U}_{\boldsymbol{e}}}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{14\sqrt{3}}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}\boldsymbol{U}_{\boldsymbol{e}}}{\text{d}\unicode[STIX]{x1D70F}}S_{F1}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}+\frac{14\sqrt{3}}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{3}\int _{0}^{t}\frac{\text{d}^{2}\boldsymbol{U}_{\boldsymbol{ e}}}{\text{d}\unicode[STIX]{x1D70F}^{2}}S_{F2}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{2}{5}\unicode[STIX]{x03C0}a^{3}\unicode[STIX]{x1D70C}\boldsymbol{U}_{\boldsymbol{ e}}(0)+6\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\boldsymbol{U}_{\boldsymbol{ e}}(0)\frac{1}{\sqrt{t}}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{14\sqrt{3}}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\boldsymbol{U}_{\boldsymbol{ e}}(0)S_{F1}(t)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{14\sqrt{3}}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{3}\left(\boldsymbol{U}_{\boldsymbol{ e}}(0)\frac{\text{d}}{\text{d}t}(S_{F2}(t))+\left.\frac{\text{d}\boldsymbol{U}_{\boldsymbol{e}}}{\text{d}t}\right|_{t=0}S_{F2}(t)\right).\end{eqnarray}$$

Equation (6.22) is compared with (1.1), when $\boldsymbol{U}_{\boldsymbol{e}}(0)=\mathbf{0}$ and $\text{d}\boldsymbol{U}_{\boldsymbol{e}}/\text{d}t_{t=0}=\mathbf{0}$ is taken. It shows that the rule is not hydrodynamically equivalent in unsteady Stokes flow. The first three terms of (6.22) are the standard forms for the quasi-steady Stokes drag force, added mass force and Basset force term, respectively. But results also show that the coefficient of added mass force of (6.22) does not correspond to (1.1). It should be noted that a very important feature of the result is to produce two additional memory force terms in addition to the Basset memory force. The fourth term, like the Basset force, is a memory force integral. More specifically, comparing (1.3) with (6.12) and (6.13), it is surprising that the kernel function $S_{F1}(t)$ is also found in the hydrodynamic force on a slightly deformed translating spheroid given by Lawrence & Weinbaum (Reference Lawrence and Weinbaum1986), and where the constants $\sqrt{\unicode[STIX]{x1D702}}=\unicode[STIX]{x1D6FD}$ . In addition, when the flow starts from rest, the initial condition at $t=0$ , $K=\text{d}K(t)/\text{d}t=0$ , is applied and (6.22) is simply reduced to

(6.23) $$\begin{eqnarray}\displaystyle \boldsymbol{F} & = & \displaystyle 6\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a\boldsymbol{U}_{\boldsymbol{e}}(t)+\frac{2}{5}\unicode[STIX]{x03C0}a^{3}\unicode[STIX]{x1D70C}\frac{\text{d}\boldsymbol{U}_{\boldsymbol{e}}}{\text{d}t}+6\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}\boldsymbol{U}_{\boldsymbol{e}}}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{14\sqrt{3}}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}\boldsymbol{U}_{\boldsymbol{e}}}{\text{d}\unicode[STIX]{x1D70F}}S_{F1}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}+\frac{14\sqrt{3}}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{3}\int _{0}^{t}\frac{\text{d}^{2}\boldsymbol{U}_{\boldsymbol{ e}}}{\text{d}\unicode[STIX]{x1D70F}^{2}}S_{F2}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

By making the simple change of variable, we use the dimensionless time variable $T^{\ast }=(\unicode[STIX]{x1D708}/a^{2})t$ which is normalized by the viscous time scale $a^{2}/\unicode[STIX]{x1D708}$ , to rewrite the two new memory kernels $S_{F1}(t)$ , $S_{F2}(t)$ in the following form:

(6.24) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\displaystyle S_{F1}(T^{\ast })=\frac{\sqrt{\unicode[STIX]{x1D708}}}{a}\text{Im}\left(\sqrt{\unicode[STIX]{x03C0}}\unicode[STIX]{x1D6FD}\text{e}^{\unicode[STIX]{x1D6FD}^{2}T^{\ast }}\text{erfc}(\unicode[STIX]{x1D6FD}\sqrt{T^{\ast }})\right),\\ S_{F2}(T^{\ast })=-\text{Im}\left(\text{e}^{\unicode[STIX]{x1D6FD}^{2}T^{\ast }}\text{erfc}(\unicode[STIX]{x1D6FD}\sqrt{T^{\ast }})\right),\\ \displaystyle 1/\sqrt{t}=\frac{\sqrt{\unicode[STIX]{x1D708}}}{a}\left(1/\sqrt{T^{\ast }}\right).\end{array}\right\} & & \displaystyle\end{eqnarray}$$

It is should be noticed in particular that $S_{F1}(T^{\ast })$ is dimensional while $S_{F2}(T^{\ast })$ is dimensionless. The Imaginary parts of both equations are all non-dimensional though. The behaviour of $S_{F1}(x)$ as $x\rightarrow \infty$ is well approximated by using the asymptotic expansions of $\text{erfc}(x)$ :

(6.25) $$\begin{eqnarray}\displaystyle \text{erfc}(x)\sim \frac{\text{e}^{-x^{2}}}{\sqrt{\unicode[STIX]{x03C0}}x}\mathop{\sum }_{n=0}^{\infty }(-1)^{n}\frac{(2n-1)!!}{(2x^{2})^{n}}\quad \text{as}~x\rightarrow \infty , & & \displaystyle\end{eqnarray}$$

where $(2n-1)!!=1\times 3\times 5\times 7\times (2n-1)$ . Thus $S_{F1}(T^{\ast })$ in (6.24) exhibits the behaviour that this term ultimately decays as $(\sqrt{3}/12)(\sqrt{\unicode[STIX]{x1D708}}/a)T^{\ast -3/2}$ . Moreover, we use the asymptotic expansions of $\text{erfc}(x)$ as $x\rightarrow 0^{+}$ ,

(6.26) $$\begin{eqnarray}\displaystyle \text{erfc}(x)\sim 1-\frac{2}{\sqrt{\unicode[STIX]{x03C0}}}\mathop{\sum }_{n=1}^{\infty }(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)\times (n-1)!}\quad \text{as }x\rightarrow 0^{+} & & \displaystyle\end{eqnarray}$$

and the limiting value is $S_{F1}(0)=\sqrt{3\unicode[STIX]{x03C0}}/2(\sqrt{\unicode[STIX]{x1D708}}/a)$ . Similarly, another important memory term $S_{F2}(T^{\ast })$ exhibits the behaviour that this term ultimately decays as $(1/2\sqrt{3\unicode[STIX]{x03C0}})T^{\ast -1/2}-(1/6\sqrt{3\unicode[STIX]{x03C0}})T^{\ast -3/2}$ and the limiting value is $S_{F2}(0)=0$ . The fourth term (new memory force with kernel $S_{F1}(T^{\ast })$ ) decays as $T^{\ast -3/2}$ , which is much faster than the third term (the Basset force kernel) by decaying as $T^{\ast -1/2}$ , as the motions tend to become steady, as shown in figure 4. It is noticed that the vertical coordinate of figure 4 has a dimension of $\sqrt{\unicode[STIX]{x1D708}}/a$ . We compare the fifth term on the right-hand side in (6.23) (another new memory force with kernel $S_{F2}(T^{\ast })$ ) with the second term (added mass force). This new memory force is found the first time and the result is illustrated by figure 5. The behaviour of $S_{F2}(T^{\ast })$ is bounded at short times and has a maximum value of 0.2414 at $T^{\ast }\approx 0.229$ .

Figure 4.

Comparison of the fourth term (new memory force with kernel $S_{F1}(T^{\ast })$ ), with the third term (the Basset force), in (6.23).

Figure 5.

Comparison of the fifth term (new memory force with kernel $S_{F2}(T^{\ast })$ , with the second term (added mass force), in (6.23).

Figure 6.

The hydrodynamic force that a rigid spherical particle moving axially without spinning in a circular cylinder tube with the boundary condition of unsteady Poiseuille flow $\boldsymbol{u}=u_{max}(t)(1-(y^{2}+z^{2})/R_{0}^{2})\boldsymbol{e}_{\boldsymbol{x}},u_{max}(t)=K(t)R_{0}^{2}$ , at $x=\pm \infty$ and $R_{0}\gg a$ .

Now we are going to extend the results from the parabolic flow to obtain similar results for Poiseuille flow in a cylindrical tube. For steady Stokes flow, the steady hydrodynamic force from a single spherical particle, fixed and kept from rotating in a circular cylinder tube with the boundary condition of Poiseuille flows, $\boldsymbol{u}=U_{0}(1-(y^{2}+z^{2})/R_{0}^{2})\boldsymbol{e}_{\boldsymbol{x}}$ at $x=\pm \infty$ , is well known and was obtained by Simha (Reference Simha1936) and Brenner & Happel (Reference Brenner and Happel1958). By the method of reflections given by Brenner & Happel (Reference Brenner and Happel1958), the steady force exerted on the sphere is

(6.27) $$\begin{eqnarray}\displaystyle \boldsymbol{F}=-6\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}a\left(-U_{0}+\frac{2}{3}U_{0}\left(\frac{a}{R_{0}}\right)^{2}\right)\boldsymbol{e}_{\boldsymbol{x}}, & & \displaystyle\end{eqnarray}$$

where $R_{0}$ is the tube radius, $a$ is the sphere radius and $U_{0}$ is the maximum axial velocity. The centre of the sphere is situated at the cylinder axis. The force in (6.27) is asymptotically correct only in the limit of small $a/R_{0}\ll 1$ . The method of reflections is an approximation method in the context of steady Stokes flow or featured in reference textbooks (e.g. Happel & Brenner Reference Happel and Brenner1965; Kim & Karrila Reference Kim and Karrila1991). It is an iterative procedure for solving linear boundary value problems based on the Stokes equations. Thus the velocity fields can be approximated by a truncated expansion and each partial velocity satisfying the physical boundary conditions. However, this concept is seldom extended to the unsteady Stokes flows. Following the work of Happel & Brenner (Reference Happel and Brenner1965, p. 300), the incident field is unsteady, and the velocity profile in the tube is given as

(6.28) $$\begin{eqnarray}\displaystyle \boldsymbol{u}=u_{max}(t)(1-(y^{2}+z^{2})/R_{0}^{2})\boldsymbol{e}_{\boldsymbol{x}}, & & \displaystyle\end{eqnarray}$$

where $u_{max}(t)$ is the maximum flow velocity as shown in figure 6. If we set

(6.29) $$\begin{eqnarray}\displaystyle u_{max}(t)=K(t)R_{0}^{2}, & & \displaystyle\end{eqnarray}$$

then the incident velocity becomes $\boldsymbol{u}=\{u_{max}(t)-K(t)(y^{2}+z^{2})\}\boldsymbol{e}_{\boldsymbol{x}}$ and can be decomposed as the sum of a uniform flow and a paraboloidal flow with a negative direction. By making use the initial condition at $t=0$ , $K=\text{d}K(t)/\text{d}t=0$ into (6.20), and also (1.1), the transient hydrodynamic force for the spherical particle is obtained as

(6.30) $$\begin{eqnarray}\displaystyle \boldsymbol{F} & = & \displaystyle \left\{6\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}au_{max}+\frac{2}{3}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{3}\frac{\text{d}u_{max}}{\text{d}t}+6\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\int _{0}^{t}\frac{\text{d}u_{max}}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}\right.\nonumber\\ \displaystyle & & \displaystyle -\,4\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}au_{max}\left(\frac{a}{R_{0}}\right)^{2}-\frac{4}{15}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{3}\left(\frac{a}{R_{0}}\right)^{2}\frac{\text{d}u_{max}}{\text{d}t}\nonumber\\ \displaystyle & & \displaystyle -\,4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\left(\frac{a}{R_{0}}\right)^{2}\int _{0}^{t}\frac{\text{d}u_{max}}{\text{d}\unicode[STIX]{x1D70F}}\frac{1}{\sqrt{t-\unicode[STIX]{x1D70F}}}\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle -\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}\sqrt{\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x03C0}}}\left(\frac{a}{R_{0}}\right)^{2}\int _{0}^{t}\frac{\text{d}u_{max}}{\text{d}\unicode[STIX]{x1D70F}}S_{F1}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle \left.-\,\frac{28\sqrt{3}}{45}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{3}\left(\frac{a}{R_{0}}\right)^{2}\int _{0}^{t}\frac{\text{d}^{2}u_{max}}{\text{d}\unicode[STIX]{x1D70F}^{2}}S_{F2}(t-\unicode[STIX]{x1D70F})\,\text{d}\unicode[STIX]{x1D70F}\right\}\boldsymbol{e}_{\boldsymbol{x}}.\end{eqnarray}$$

It should be noted that the force in (6.30) is also an approximated solution, and the corresponding contribution is due to the first reflected flow field. Similarly, it is also asymptotically correct only in the limit of small $a/R_{0}\ll 1$ . When the boundary condition of the Poiseuille flow is steady and $u_{max}(t)=\text{constant}=U_{0}$ , the force in (6.30) agrees with (6.27), as expected.