Motion of a Body Immersed in a Stokes Fluid

Abstract

In this chapter we study the unsteady motion of a sphere immersed in a Stokes fluid, that is a linear approximation of a fluid governed by the Navier–Stokes equation. The equation of motion for the sphere leads to an integro-differential equation, and we are interested in the asymptotic behavior in time of the solution. We show that the velocity of the sphere slows down in time with an algebraic law, due to the memory effect of the surrounding fluid. We discuss the case of a sphere moving on a straight line, or executing a rotary motion around a fixed axis.

Appendix

4.1.1 Derivation of the Equation for Rectilinear Motion

We start from the drag on a sphere executing translatory oscillations, which is [11],

$$\displaystyle{ -F_{\omega } = 6\pi \mu R\left (1 + \frac{R} {\delta } \right )U + 3\pi R^{2}\sqrt{\frac{2\mu \rho _{f } } {\omega }} \left (1 + \frac{2R} {9\delta } \right )\frac{\mathrm{d}U} {\mathrm{d}t} \;, }$$

(4.46)

where $U = U_{0}\mathrm{e}^{-\mathrm{i}\omega t}$ and $\delta = \sqrt{2\nu /\omega }$, being $\nu =\mu /\rho _{f}$ the kinematic viscosity. Clearly we perform all the computations in complex notation, and at the end we consider the real part. For ω = 0 this becomes Stokes formula.

To obtain the drag on a sphere moving in an arbitrary manner U(t), we represent it as a Fourier integral,

$$\displaystyle{ U(t) =\int _{ -\infty }^{\infty }\!\mathrm{d}\omega \;U_{\omega }\mathrm{e}^{-\mathrm{i}\omega t}\;,\qquad \qquad U_{\omega } = \frac{1} {2\pi }\int _{-\infty }^{\infty }\!\mathrm{d}\tau \;U(\tau )\mathrm{e}^{\mathrm{i}\omega \tau }\;. }$$

(4.47)

Since the equations are linear, the total drag may be written as the integral of the drag forces for velocities which are the separate Fourier components U _ω e^{−iω t}; these forces are given by (4.46), precisely,

$$\displaystyle{ \pi \rho _{f}R^{3}U_{\omega }\mathrm{e}^{-\mathrm{i}\omega t}\left ( \frac{6\nu } {R^{2}} -\frac{2\mathrm{i}\omega } {3} + \frac{3\sqrt{2\nu }} {R} (1 -\mathrm{i})\sqrt{\omega }\right )\;. }$$

(4.48)

Noticing that $(\mathit{dU}/\mathit{dt})_{\omega } = -\mathrm{i}\omega U_{\omega }$, we can rewrite this as

$$\displaystyle{ \pi \rho _{f}R^{3}\mathrm{e}^{-\mathrm{i}\omega t}\left ( \frac{6\nu } {R^{2}}U_{\omega } + \frac{2} {3}(\dot{U})_{\omega } + \frac{3\sqrt{2\nu }} {R} (\dot{U})_{\omega }\frac{1 + \mathrm{i}} {\sqrt{\omega }} \right )\;. }$$

(4.49)

By integration over ω, the first and second terms give respectively U(t) and $\dot{U}(t)$. To integrate the third term, we notice first of all that for negative ω this term must be written in the complex conjugate form, $(1 + \mathrm{i})/\sqrt{\omega }$ being replaced by $(1 -\mathrm{i})/\sqrt{\vert \omega \vert }$; this is because Eq. (4.46) was derived for a velocity $U = U_{0}\mathrm{e}^{-\mathrm{i}\omega t}$ with ω > 0, and for a velocity U ₀e^{iω t} we obtain the complex conjugate. Therefore, the integration with respect to ω over $\mathbb{R}$ equals twice the real part of the integration restricted on $\mathbb{R}_{+}$; moreover

$$\displaystyle{\begin{array}{ll} 2\,\text{Re}\left \{(1 + \mathrm{i})\int _{0}^{\infty }\!\mathrm{d}\omega \;\frac{(\dot{U})_{\omega }\mathrm{e}^{-\mathrm{i}\omega t}} {\sqrt{\omega }} \right \}& = \frac{1} {\pi } \,\text{Re}\left \{(1 + \mathrm{i})\int _{-\infty }^{\infty }\int _{ 0}^{\infty }\!\mathrm{d}\omega \,\mathrm{d}\tau \;\frac{\dot{U}(\tau )\mathrm{e}^{\mathrm{i}\omega (\tau -t)}} {\sqrt{\omega }} \right \} \\ & = \frac{1} {\pi } \text{Re}\left \{(1 + \mathrm{i})\int _{-\infty }^{t}\int _{ 0}^{\infty }\!\mathrm{d}\omega \,\mathrm{d}\tau \;\frac{\dot{U}(\tau )\mathrm{e}^{-\mathrm{i}\omega (t-\tau )}} {\sqrt{\omega }} \right. \\ &\quad \left.+\,(1 + \mathrm{i})\int _{t}^{\infty }\int _{0}^{\infty }\!\mathrm{d}\omega \,\mathrm{d}\tau \;\frac{\dot{U}(\tau )\mathrm{e}^{\mathrm{i}\omega (\tau -t)}} {\sqrt{\omega }} \right \} \\ & = \sqrt{\frac{2} {\pi }} \text{Re}\left \{\int _{-\infty }^{t}\!\mathrm{d}\tau \; \frac{\dot{U}(\tau )} {\sqrt{t-\tau }} + \mathrm{i}\int _{t}^{\infty }\!\mathrm{d}\tau \; \frac{\dot{U}(\tau )} {\sqrt{\tau -t}}\right \} \\ & = \sqrt{\frac{2} {\pi }} \int _{-\infty }^{t}\!\mathrm{d}\tau \; \frac{\dot{U}(\tau )} {\sqrt{t-\tau }}\;. \end{array} }$$

Thus, we have finally for the drag,

$$\displaystyle{ F_{\mathrm{drag}} = -2\pi \rho _{f}R^{3}\left (\frac{1} {3} \frac{\mathrm{d}U} {\mathrm{d}t} + \frac{3\nu U} {R^{2}} + \frac{3} {R}\sqrt{\frac{\nu } {\pi }}\int _{-\infty }^{t}\!\mathrm{d}\tau \;\frac{\mathrm{d}U} {\mathrm{d}\tau } \frac{1} {\sqrt{t-\tau }}\right )\;, }$$

(4.50)

which is (4.10).

4.1.2 Derivation of the Equation for Rotary Motion

The total moment may be written as the integral of the moments for angular velocities which are the separate Fourier components $\varOmega _{\omega }\mathrm{e}^{-\mathrm{i}\omega t}$, expressed by (4.21). The single moment is therefore

$$\displaystyle{ M_{\omega } = -\frac{8\pi } {3}\mu R^{3}\varOmega _{ \omega }\mathrm{e}^{-\mathrm{i}\omega t}\left [3 + \frac{2\alpha ^{2}\omega \left [\alpha \sqrt{\omega }-\mathrm{i}(\alpha \sqrt{\omega } + 1)\right ]} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega } \right ]\;, }$$

(4.51)

and since $(\dot{\varOmega })_{\omega } = -\mathrm{i}\omega \varOmega _{\omega }$, we can rewrite this as,

$$\displaystyle\begin{array}{rcl} M_{\omega }& =& -\frac{8\pi } {3}\mu R^{3}\mathrm{e}^{-\mathrm{i}\omega t}\left [3\varOmega _{\omega } + \frac{(\dot{\varOmega })_{\omega }} {-\mathrm{i}\omega } \frac{2\alpha ^{2}\omega \left [\alpha \sqrt{\omega }-\mathrm{i}(\alpha \sqrt{\omega } + 1)\right ]} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega } \right ] \\ & =& -\frac{8\pi } {3}\mu R^{3}\mathrm{e}^{-\mathrm{i}\omega t}\left [3\varOmega _{\omega } + (\dot{\varOmega })_{\omega }\frac{2\alpha ^{2}\left [\alpha \sqrt{\omega } + 1 + \mathrm{i}\alpha \sqrt{\omega }\right ]} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega } \right ]\;.{}\end{array}$$

(4.52)

By integration over ω, the first term gives $-(8\pi \mu R^{3})\varOmega (t)$. To integrate the second term, we notice first of all that for negative ω this term must be written in the complex conjugate form (and replacing ω → | ω | ), because (4.21) was derived for an angular velocity $\varOmega _{0}\mathrm{e}^{-\mathrm{i}\omega t}$ with ω > 0, and for an angular velocity $\varOmega _{0}\mathrm{e}^{\mathrm{i}\omega t}$ we obtain the complex conjugate. Therefore, the integration with respect to ω over $\mathbb{R}$ equals twice the real part of the integration restricted on $\mathbb{R}_{+}$. We then compute,

$$\displaystyle\begin{array}{rcl} K&:=& 2\,\text{Re}\left \{\int _{0}^{\infty }\!\mathrm{d}\omega \;(\dot{\varOmega })_{\omega }\mathrm{e}^{-\mathrm{i}\omega t} \frac{\alpha \sqrt{\omega } + 1 + \mathrm{i}\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right \} {}\\ & =& \frac{1} {\pi } \text{Re}\left \{\int _{-\infty }^{\infty }\int _{ 0}^{\infty }\!\mathrm{d}\omega \,\mathrm{d}\tau \;\dot{\varOmega }(\tau )\mathrm{e}^{\mathrm{i}\omega (\tau -t)} \frac{\alpha \sqrt{\omega } + 1 + \mathrm{i}\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right \} {}\\ & =& \frac{1} {\pi } \text{Re}\left \{\int _{-\infty }^{t}\int _{ 0}^{\infty }\!\mathrm{d}\omega \,\mathrm{d}\tau \;\dot{\varOmega }(\tau )\mathrm{e}^{\mathrm{i}\omega (\tau -t)} \frac{\alpha \sqrt{\omega } + 1 + \mathrm{i}\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right. {}\\ & & +\left.\int _{t}^{\infty }\int _{ 0}^{\infty }\!\mathrm{d}\omega \,\mathrm{d}\tau \;\dot{\varOmega }(\tau )\mathrm{e}^{\mathrm{i}\omega (\tau -t)} \frac{\alpha \sqrt{\omega } + 1 + \mathrm{i}\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right \}\;. {}\\ \end{array}$$

Defining now, for any $t \in \mathbb{R}$,

$$\displaystyle\begin{array}{rcl} F(t)& =& \int _{0}^{\infty }\!\mathrm{d}\omega \;\cos \left (\omega t\right ) \frac{\alpha \sqrt{\omega } + 1} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega } \\ & & +\int _{0}^{\infty }\!\mathrm{d}\omega \;\sin \left (\omega t\right ) \frac{\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\;,{}\end{array}$$

(4.53)

the quantity K can be expressed as follows,

$$\displaystyle\begin{array}{rcl} K& =& \frac{1} {\pi } \text{Re}\left \{\int _{-\infty }^{t}\!\mathrm{d}\tau \;\dot{\varOmega }(\tau )\left [\int _{ 0}^{\infty }\!\mathrm{d}\omega \;\cos \big(\omega (t-\tau )\big) \frac{\alpha \sqrt{\omega } + 1} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right.\right. {}\\ & & \left.+\int _{0}^{\infty }\!\mathrm{d}\omega \;\sin \big(\omega (t-\tau )\big) \frac{\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right ] {}\\ & & +\,\mathrm{i}\int _{-\infty }^{t}\!\mathrm{d}\tau \;\dot{\varOmega }(\tau )\left [\int _{ 0}^{\infty }\!\mathrm{d}\omega \;\cos \big(\omega (t-\tau )\big) \frac{\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right. {}\\ & & \left.+\int _{0}^{\infty }\!\mathrm{d}\omega \;\sin \big(\omega (t-\tau )\big) \frac{-(\alpha \sqrt{\omega } + 1)} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right ] {}\\ & & +\int _{t}^{\infty }\!\mathrm{d}\tau \;\dot{\varOmega }(\tau )\left [\int _{ 0}^{\infty }\!\mathrm{d}\omega \;\cos \big(\omega (\tau -t)\big) \frac{\alpha \sqrt{\omega } + 1} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right. {}\\ & & \left.-\int _{0}^{\infty }\!\mathrm{d}\omega \;\sin \big(\omega (\tau -t)\big) \frac{\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right ] {}\\ & & +\,\mathrm{i}\int _{t}^{\infty }\!\mathrm{d}\tau \;\dot{\varOmega }(\tau )\left [\int _{ 0}^{\infty }\!\mathrm{d}\omega \;\cos \big(\omega (\tau -t)\big) \frac{\alpha \sqrt{\omega }} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right. {}\\ & & \left.\left.+\int _{0}^{\infty }\!\mathrm{d}\omega \;\sin \big(\omega (\tau -t)\big) \frac{\alpha \sqrt{\omega } + 1} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\right ]\right \} {}\\ & =& \frac{1} {\pi } \int _{-\infty }^{t}\!\mathrm{d}\tau \;\dot{\varOmega }(\tau )F(t-\tau ) + \frac{1} {\pi } \int _{t}^{\infty }\!\mathrm{d}\tau \;\dot{\varOmega }(\tau )F(t-\tau )\;. {}\\ \end{array}$$

Therefore, the total moment of the forces exerted by the fluid on a sphere rotating with an arbitrary angular velocity, Ω(t), is

$$\displaystyle{ M = -8\pi \mu R^{3}\left (\varOmega (t) + \frac{2\alpha ^{2}} {3\pi } \int _{-\infty }^{t}\!\mathrm{d}\tau \;\dot{\varOmega }(\tau )F(t-\tau )\right )\;. }$$

(4.54)

4.1.3 Computation of F(t)

We want to show that the function F(t) expressed in (4.53) is equal, for t > 0, to

$$\displaystyle{ F(t) = \frac{\sqrt{\pi }} {\sqrt{2t}} - \frac{\pi } {2}\mathrm{e}^{t/2}\,\text{Erfc}\sqrt{ \frac{t} {2}}\;, }$$

(4.55)

and it is zero for t < 0. Let us compute first

$$\displaystyle{ \int _{0}^{\infty }\!\mathrm{d}\omega \;\cos \left (\omega t\right ) \frac{\alpha \sqrt{\omega } + 1} {1 + 2\alpha \sqrt{\omega } + 2\alpha ^{2}\omega }\;, }$$

(4.56)

which is clearly an even function of t, so in the sequel we consider t > 0. By the change of variable ω = x ², and putting equal 1 the irrelevant parameter α, (4.56) becomes

$$\displaystyle{ \int _{0}^{\infty }\!\mathrm{d}x\;\cos \left (x^{2}t\right ) \frac{2x^{2} + 2x} {1 + 2x + 2x^{2}}\;. }$$

(4.57)

In order to evaluate this integral, we consider the contour $\mathcal{C}$ of the complex plane defined as follows:

$$\displaystyle\begin{array}{rcl} \mathcal{C}& =& \left \{z \in \mathbb{C}: 0 \leq \text{Re}(z) \leq R,\,\,\text{Im}(z) = 0\right \} \cup \left \{z \in \mathbb{C}: z = R\mathrm{e}^{i\theta },\,\,0 \leq \theta \leq \frac{\pi } {4}\right \} \\ & & \cup \left \{z \in \mathbb{C}: z = r\mathrm{e}^{\mathrm{i}t/4},\,\,0 \leq r \leq R\right \} = \mathcal{C}_{ 1} \cup \mathcal{C}_{2} \cup \mathcal{C}_{3}\;, {}\end{array}$$

(4.58)

where R is a fixed positive real number. By the Cauchy theorem we have

$$\displaystyle{ \oint _{\mathcal{C}}\!\mathrm{d}z\;\mathrm{e}^{\mathrm{i}tz^{2} } \frac{2z^{2} + 2z} {1 + 2z + 2z^{2}} = 0\;, }$$

(4.59)

which splits into the integrals over $\mathcal{C}_{1}$, $\mathcal{C}_{2}$, and $\mathcal{C}_{3}$. It is not difficult to see that the integral over $\mathcal{C}_{2}$ goes to zero when R → ∞, so we have

$$\displaystyle{ \int _{0}^{\infty }\!\mathrm{d}x\;\mathrm{e}^{\mathrm{i}tx^{2} } \frac{2x^{2} + 2x} {1 + 2x + 2x^{2}} =\int _{ 0}^{\infty }\!\mathrm{d}r\;\mathrm{e}^{\mathrm{i}tr^{2}\mathrm{e}^{\mathrm{i}t/2} } \frac{2\left (r\mathrm{e}^{\mathrm{i}t/4}\right )^{2} + 2r\mathrm{e}^{\mathrm{i}t/4}} {1 + 2r\mathrm{e}^{\mathrm{i}t/4} + 2\left (r\mathrm{e}^{\mathrm{i}t/4}\right )^{2}}\mathrm{e}^{\mathrm{i}t/4}\;, }$$

(4.60)

and the integral we are interested in, (4.57), coincides with the real part of (4.60). After some elementary algebra and taking the real part of (4.60), we arrive at

$$\displaystyle\begin{array}{rcl} \int _{0}^{\infty }\!\mathrm{d}x\;\cos \left (x^{2}t\right ) \frac{2x^{2} + 2x} {1 + 2x + 2x^{2}}& =& \int _{0}^{\infty }\!\mathrm{d}r\;\mathrm{e}^{-tr^{2} } \frac{\sqrt{2}r^{2}} {1 + 2r^{2}} \\ & =& \frac{\sqrt{2\pi }} {4\sqrt{t}} - \frac{\pi } {4}\mathrm{e}^{t/2}\text{Erfc}\sqrt{ \frac{t} {2}}.{}\end{array}$$

(4.61)

For the other integral constituting F(t) in (4.53) we have, by the same technique, for t > 0,

$$\displaystyle{ \int _{0}^{\infty }\!\mathrm{d}\omega \;\sin \left (\omega t\right ) \frac{\sqrt{\omega }} {1 + 2\sqrt{\omega } + 2\omega } = \frac{\sqrt{2\pi }} {4\sqrt{t}} - \frac{\pi } {4}\mathrm{e}^{t/2}\text{Erfc}\sqrt{ \frac{t} {2}}\;, }$$

(4.62)

hence exactly the same expression obtained before, while for t < 0, since sin(ω t) is odd, we get a cancellation and so F(t) = 0.