2. Calculation of the bound

A sphere of radius $R$ centred at the origin is surrounded by fluid of viscosity $\nu$ and density $\rho$ which moves in the $z$-direction with velocity $U_\infty$ at infinity. The velocity and pressure fields, $\boldsymbol {v}(\boldsymbol {r},t)$ and $p(\boldsymbol {r},t)$, are determined as a function of position $\boldsymbol {r}$ and time $t$ by the Navier–Stokes equation

(2.1a,b)\begin{equation} \partial_t \boldsymbol{v} + (\boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v} ={-} \frac{1}{\rho} \boldsymbol{\nabla} p +\nu {\nabla}^2 \boldsymbol{v} , \quad \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{v} = 0 \end{equation}

with the boundary conditions that $\boldsymbol {v} = 0$ on $r=|\boldsymbol {r}|=R$ and $\boldsymbol {v} \rightarrow U_\infty \hat {\boldsymbol {z}}$ for $r \rightarrow \infty$, where a hat denotes a unit vector. We choose a background field $\boldsymbol {v}_B(\boldsymbol {r})$ as the sum $\boldsymbol {v}_B = \boldsymbol {v}_P + \boldsymbol {v}_b$ of the potential flow around the obstacle, $\boldsymbol {v}_P(\boldsymbol {r})$, and a boundary-layer contribution $\boldsymbol {v}_b(\boldsymbol {r})$ such that $\boldsymbol {v}_P + \boldsymbol {v}_b = 0$ on $r=R$ and $\boldsymbol {v}_b = 0$ for $r \geqslant R+\delta$, where $\delta$ is a boundary-layer thickness whose size is available for later optimisation. These fields have the properties

(2.2)\begin{equation} \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{v}_B = \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{v}_P = \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{v}_b =0 \end{equation}

(2.3)\begin{equation}\boldsymbol{\nabla} \times\boldsymbol{v}_P =0 \end{equation}

and fulfil the boundary conditions $\boldsymbol {v}_B=0$ and $\hat {\boldsymbol {r}} \boldsymbol {\cdot } \boldsymbol {v}_P=0$ on $r=R$, whereas at infinity, $\boldsymbol {v}_B=\boldsymbol {v}_P=U_\infty \hat {\boldsymbol {z}}$.

The total velocity field is written as $\boldsymbol {v}(\boldsymbol {r},t)=\boldsymbol {v}_B(\boldsymbol {r})+ \boldsymbol {u}(\boldsymbol {r},t)$, so that the fluctuation field $\boldsymbol {u}$ obeys $\boldsymbol {\nabla } \boldsymbol {\cdot } \boldsymbol {u} = 0$ and $\boldsymbol {u} = 0$ both on $r=R$ and at infinity. Inserting this decomposition into the Navier–Stokes equation and taking the dot product with $\boldsymbol {u}$ leads to

(2.4)\begin{align} &\frac{1}{2} \partial_t |\boldsymbol{u}|^2 + \frac{1}{2} \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{u} |\boldsymbol{u}|^2) +\boldsymbol{u} \boldsymbol{\cdot} (\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_B + \frac{1}{2} \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{v}_B |\boldsymbol{u}|^2) +\boldsymbol{u} \boldsymbol{\cdot} (\boldsymbol{v}_B \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_B \nonumber\\ &\quad ={-}\frac{1}{\rho} \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{u} p) + \nu \boldsymbol{u} \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{v}. \end{align}

If we integrate this equation over the entire fluid volume, all divergence terms disappear thanks to the boundary conditions, and

(2.5)\begin{align} \int \boldsymbol{u} \boldsymbol{\cdot} (\boldsymbol{v}_B \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_B \,\mathrm{d}V &= \int \boldsymbol{u} \boldsymbol{\cdot} \left[(\boldsymbol{\nabla} \times \boldsymbol{v}_B)\times \boldsymbol{v}_B + \tfrac{1}{2} \boldsymbol{\nabla} |\boldsymbol{v}_B|^2 \right] \,\mathrm{d}V \nonumber\\ &= \int \boldsymbol{u} \boldsymbol{\cdot} [(\boldsymbol{\nabla} \times \boldsymbol{v}_b)\times \boldsymbol{v}_B] \,\mathrm{d}V. \end{align}

In addition, applying a few times the vector identity $\boldsymbol {b} \boldsymbol {\cdot } (\boldsymbol {\nabla } \times \boldsymbol {a}) = \boldsymbol {\nabla } \boldsymbol {\cdot } (\boldsymbol {a} \times \boldsymbol {b}) + \boldsymbol {a} \boldsymbol {\cdot } (\boldsymbol {\nabla } \times \boldsymbol {b})$, one finds

(2.6)\begin{align} \int \boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla}^2 \boldsymbol{v} \,\textrm{d}V & ={-} \int \boldsymbol{v} \boldsymbol{\cdot} \boldsymbol{\nabla} \times \boldsymbol{\nabla} \times \boldsymbol{v} \,\mathrm{d}V ={-} \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V\nonumber\\ & = \int (\boldsymbol{u} \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{u} + \boldsymbol{v}_B \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{v}_B + \boldsymbol{u} \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{v}_B + \boldsymbol{v}_B \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{u}) \,\mathrm{d}V\nonumber\\ & ={-} \int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\textrm{d}V - \int |\boldsymbol{\nabla} \times \boldsymbol{v}_B|^2 \,\mathrm{d}V + 2 \int \boldsymbol{u} \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{v}_B \,\mathrm{d}V \end{align}

so that

(2.7)\begin{align} \int \boldsymbol{u} \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{v} \,\mathrm{d}V & ={-} \int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\mathrm{d}V + \boldsymbol{u} \boldsymbol{\cdot} {\nabla}^2 \boldsymbol{v}_B \,\mathrm{d}V\nonumber\\ & ={-} \tfrac{1}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V - \tfrac{1}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\mathrm{d}V + \tfrac{1}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{v}_B|^2 \,\mathrm{d}V. \end{align}

Let $V_\delta$ denote the volume $R \leqslant r \leqslant R+\delta$. Inserting $\boldsymbol {v}_B = \boldsymbol {v}_P + \boldsymbol {v}_b$, (2.5) and (2.7) into (2.4), and keeping track of integrands which are zero outside $V_\delta$ and which need to be integrated only over $V_\delta$, we obtain

(2.8)\begin{align} &\partial_t \int \frac{1}{2} |\boldsymbol{u}|^2 \,\mathrm{d}V + \int \boldsymbol{u} \boldsymbol{\cdot}(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_P \,\mathrm{d}V + \frac{\nu}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V - \frac{\nu}{2} \int_{V_\delta} |\boldsymbol{\nabla} \times \boldsymbol{v}_b|^2 \,\mathrm{d}V\nonumber\\ &\quad ={-} \frac{\nu}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\mathrm{d}V - \int_{V_\delta} \boldsymbol{u} \boldsymbol{\cdot}(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_b \,\mathrm{d}V - \int_{V_\delta} \boldsymbol{u} \boldsymbol{\cdot} [(\boldsymbol{\nabla} \times \boldsymbol{v}_b) \times \boldsymbol{v}_B] \,\mathrm{d}V\nonumber\\ &\quad \leqslant - \frac{\nu}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2\, \mathrm{d}V + \left|\int_{V_\delta} \boldsymbol{u} \boldsymbol{\cdot}(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_b \,\mathrm{d}V\right|\nonumber\\ & \qquad + \sqrt{ \int_{V_\delta} |\boldsymbol{u}|^2 \,\mathrm{d}V} \sqrt{ \int_{V_\delta} |(\boldsymbol{\nabla} \times \boldsymbol{v}_b) \times \boldsymbol{v}_B|^2 \,\mathrm{d}V}, \end{align}

where the Schwarz inequality was used in the last line.

We now have to specify expressions for $\boldsymbol {v}_P$ and $\boldsymbol {v}_b$ in order to evaluate the integrals in the last equation. The potential flow around a sphere is given by

(2.9)\begin{equation} \boldsymbol{v}_P = \hat{\boldsymbol{r}} U_\infty \cos \theta \left[ 1-\left( \frac{R}{r} \right)^3 \right] - \hat{\boldsymbol{\theta}} U_\infty \sin \theta \left[ 1 + \frac{1}{2}\left( \frac{R}{r} \right)^3 \right] \end{equation}

in spherical polar coordinates $(r, \theta , \varphi )$. Let us choose $\boldsymbol {v}_b$ as

(2.10)\begin{equation} \boldsymbol{v}_b = \left\{\begin{array}{@{}ll} \boldsymbol{\nabla} \times \boldsymbol{\nabla} \times \left[ -\dfrac{3}{2} U_\infty \cos \theta R(r-R) \left( 1 - \dfrac{r-R}{\delta} \right)^2\hat{\boldsymbol{r}} \right], & R \leqslant r \leqslant R+\delta \\ 0, & r \geqslant R+\delta \end{array} \right. . \end{equation}

With this choice, $\boldsymbol {v}_b$ is continuous everywhere, and all the integrations in the following can be carried out in closed form. However, we will finally be interested in the limit of high Reynolds numbers and the limit $\delta /R \ll 1$. The results of all algebraic manipulations will therefore be given only to the leading order in $\delta /R$ because this leads to considerably more transparent expressions. This procedure is sufficient for the ultimate goal of the calculation and alleviates the at any rate tedious algebra. For instance,

(2.11)\begin{equation} \int_{V_\delta} |\boldsymbol{\nabla} \times \boldsymbol{v}_b|^2 \,\mathrm{d}V = 24 {\rm \pi}U_\infty^2 \frac{R^2}{\delta} \left[1+{O}\left( \frac{\delta^2}{R^2} \right) \right] \end{equation}

and

(2.12)\begin{equation} \int_{V_\delta} |(\boldsymbol{\nabla} \times \boldsymbol{v}_b) \times \boldsymbol{v}_B|^2 \,\mathrm{d}V = \frac{3816}{175} {\rm \pi}U_\infty^4 \frac{R^2}{\delta} \left[1+{O}\left( \frac{\delta}{R} \right) \right]. \end{equation}

Let us call $\boldsymbol{\mathsf{e}}_b$ the deformation-rate tensor of the boundary-layer flow $\boldsymbol {v}_b$ so that ${e}_{b,ij} = (\partial _j v_{b,i} + \partial _i v_{b,j})/{2}$ and $\boldsymbol {u} \boldsymbol {\cdot }(\boldsymbol {u} \boldsymbol {\cdot } \boldsymbol {\nabla }) \boldsymbol {v}_b = \boldsymbol {u} \boldsymbol {\cdot } \boldsymbol{\mathsf{e}}_b \boldsymbol {u}$. If we write down $\boldsymbol{\mathsf{e}}_b$ in spherical coordinates and order the components within $\boldsymbol {u}$ in the sequence $(u_r, u_\theta , u_\varphi )$, we find

(2.13)\begin{equation} \boldsymbol{\mathsf{e}}_b = f_1(r) \boldsymbol{\mathsf{M}}_1 + f_2(r) \boldsymbol{\mathsf{M}}_2 \end{equation}

with matrices $\boldsymbol{\mathsf{M}}_1$ and $\boldsymbol{\mathsf{M}}_2$ defined by

(2.14)\begin{equation} \boldsymbol{\mathsf{M}}_1 = \begin{bmatrix} 0 & \sin \theta & 0 \\ \sin \theta & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{equation}

and

(2.15)\begin{equation} \boldsymbol{\mathsf{M}}_2 = \begin{bmatrix} 2 \cos \theta & \sin \theta & 0 \\ \sin \theta & - \cos \theta & 0 \\ 0 & 0 & - \cos \theta \end{bmatrix} . \end{equation}

The functions $f_1(r)$ and $f_2(r)$ are of course zero for $r \geqslant R+\delta$ and, using the shorthand notation $s=r-R$, they are given within $V_\delta$ by

(2.16)\begin{equation} f_1(r)={-} \frac{3 R U_\infty}{2 \delta^{2} \left(R + s\right)^{3}} \left[R^{2} \left(2 \delta - 3 s\right) + R \left(4 \delta s - 6 s^{2}\right) + \delta^{2} s - 2 s^{3}\right] \end{equation}

and

(2.17)\begin{equation} f_2(r)= \frac{3 R U_\infty}{\delta^{2} \left(R + s\right)^{3}} \left[R \left(- \delta^{2} + 4 \delta s - 3 s^{2}\right) + \delta^{2} s - s^{3}\right] . \end{equation}

The eigenvalues $\lambda _1$ of $\boldsymbol{\mathsf{M}}_1$ obey $|\lambda _1| \leqslant 1$, whereas for the eigenvalues $\lambda _2$ of $\boldsymbol{\mathsf{M}}_2$, the inequality $|\lambda _2| \leqslant 2$ holds, so that

(2.18)\begin{equation} |\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\mathsf{e}}_b \boldsymbol{u} | \leqslant |\boldsymbol{u}|^2 \left\{ |f_1(r)| + 2 |f_2(r)| \right\} \leqslant |\boldsymbol{u}|^2 3 \frac{U_\infty}{\delta} \left[1+\textit{O}\left( \frac{\delta}{R} \right) \right] \end{equation}

and

(2.19)\begin{equation} \left|\int_{V_\delta} \boldsymbol{u} \boldsymbol{\cdot}(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_b \,\mathrm{d}V\right| \leqslant 3 \frac{U_\infty}{\delta} \int_{V_\delta} |\boldsymbol{u}|^2 \,\mathrm{d}V \left[1+{O}\left( \frac{\delta}{R} \right) \right] . \end{equation}

We now want to find the maximum of the last two lines in the sequence (2.8). To this end, we first relate $\int _{V_\delta } |\boldsymbol {u}|^2 \,\mathrm {d}V$ to $\int |\boldsymbol {\nabla } \times \boldsymbol {u}|^2 \,\mathrm {d}V = \int |\boldsymbol {\nabla } \boldsymbol {u}|^2 \,\mathrm {d}V$ with the notation $|\boldsymbol {\nabla } \boldsymbol {u}|^2 = \sum _{i,j} |\partial _j u_i|^2$. A simple approach to achieve this goal is to first note that

(2.20)\begin{equation} |\boldsymbol{u}(\boldsymbol{r},t)|^2 = \left|\int_0^s \partial_r \boldsymbol{u}(\boldsymbol{r}',t) \,\mathrm{d}s' \right|^2, \end{equation}

where $\boldsymbol {r} = (R+s, \theta , \varphi )$ and $\boldsymbol {r}' = (R+s', \theta , \varphi )$. Inserting a factor 1 in the integral in (2.20) and applying Schwarz's inequality yields

(2.21)\begin{equation} |\boldsymbol{u}(\boldsymbol{r},t)|^2 \leqslant \left( \sqrt{s} \sqrt{\int_0^s |\partial_r \boldsymbol{u}(\boldsymbol{r}',t)|^2 \,\mathrm{d}s'} \right)^2 \leqslant s \int_0^s |\boldsymbol{\nabla} \boldsymbol{u}|^2 \,\mathrm{d}s' . \end{equation}

The integral of $|\boldsymbol {u}|^2$ over $V_\delta$ can therefore be estimated as follows:

(2.22)\begin{align} \int_{V_\delta} |\boldsymbol{u}|^2 \,\mathrm{d}V & = \int_0^\delta \,\mathrm{d}s \int_0^{2{\rm \pi}} \,\mathrm{d} \varphi \int_0^{\rm \pi} \,\mathrm{d} \theta \sin \theta (R+s)^2 |\boldsymbol{u}(\boldsymbol{r},t)|^2\nonumber\\ & \leqslant \int_0^\delta \,\mathrm{d}s s (R+s)^2 \int_0^{2{\rm \pi}} \,\mathrm{d} \varphi \int_0^{\rm \pi} \,\mathrm{d} \theta \sin \theta \int_0^s \,\mathrm{d}s' \frac{(R+s')^2}{(R+s')^2} |\boldsymbol{\nabla} \boldsymbol{u}|^2\nonumber\\ & \leqslant \int_0^\delta \,\mathrm{d}s s (R+s)^2 \frac{1}{R^2} \int_{V_s} |\boldsymbol{\nabla} \boldsymbol{u}|^2 \,\mathrm{d}V, \end{align}

where $V_s$ denotes the volume $R \leqslant r \leqslant R+s$. As the integrand $|\boldsymbol {\nabla } \boldsymbol {u}|^2$ is always positive, we obtain another valid inequality by extending the volume of integration from $V_s$ to the entire fluid volume:

(2.23)\begin{align} \int_{V_\delta} |\boldsymbol{u}|^2 \,\mathrm{d}V& \leqslant \left\{ \int |\boldsymbol{\nabla} \boldsymbol{u}|^2 \,\mathrm{d}V \right\} \left\{ \int_0^s \,\mathrm{d}s s (R+s)^2 \frac{1}{R^2} \right\}\nonumber\\ & \leqslant \left\{ \int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\mathrm{d}V \right\} \left\{ \frac{1}{2} \delta^2 \left[1+{O}\left( \frac{\delta}{R} \right) \right] \right\} . \end{align}

The leading order in $\delta /R$ of the last right-hand side in (2.8) is therefore bounded from above by

(2.24)\begin{equation} \left( -\frac{\nu}{2} + \frac{3}{2} U_\infty \delta \right) \int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\mathrm{d}V + \frac{6}{35} \sqrt{371 {\rm \pi}} R U_\infty^2 \sqrt{\delta} \sqrt{\int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\mathrm{d}V} . \end{equation}

We now regard $\int |\boldsymbol {\nabla } \times \boldsymbol {u}|^2 \,\mathrm {d}V$ as the variable in expression (2.24) and consider what the maximum of this expression is if $\int |\boldsymbol {\nabla } \times \boldsymbol {u}|^2\,\mathrm {d}V$ is varied. In order for a maximum to exist, the condition

(2.25)\begin{equation} 3 U_\infty \delta \leqslant \nu \end{equation}

needs to be satisfied. If this is the case, the maximum of (2.24) is achieved for

(2.26)\begin{equation} \sqrt{\int |\boldsymbol{\nabla} \times \boldsymbol{u}|^2 \,\mathrm{d}V} = \frac{6}{35} \sqrt{371 {\rm \pi}} R U_\infty^2 \sqrt{\delta} \frac{1}{\nu-3 U_\infty \delta} . \end{equation}

Inserting the maximum of (2.24) together with (2.11) into (2.8) yields

(2.27)\begin{align} &\partial_t \int \frac{1}{2} |\boldsymbol{u}|^2 \,\mathrm{d}V + \int \boldsymbol{u} \boldsymbol{\cdot}(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_P \,\mathrm{d}V + \frac{\nu}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V\nonumber\\ &\quad \leqslant 12 {\rm \pi}R^2 U_\infty^3 \left( \frac{\nu}{U_\infty \delta} - \frac{159}{350} \dfrac{1}{3-\dfrac{\nu}{U_\infty \delta}} \right) \left[ 1 + {O}\left( \dfrac{\delta}{R} \right) \right] . \end{align}

The choice of $\delta$ that minimises the right-hand side of this inequality, yet respects condition (2.25), is

(2.28)\begin{equation} \frac{U_\infty \delta}{\nu} = \frac{5}{2991}(210-\sqrt{2226}) \end{equation}

so that (2.8) now turns into

(2.29)\begin{align} &\partial_t \int \frac{1}{2} |\boldsymbol{u}|^2 \,\mathrm{d}V + \frac{\nu}{2} \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V \nonumber\\ &\quad \leqslant \left|\int \boldsymbol{u} \boldsymbol{\cdot}(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_P \,\mathrm{d}V\right| + 12 {\rm \pi}\left(3+\frac{\sqrt{2226}}{35} \right) R^2 U_\infty^3 \left[ 1 + {O}\left( \frac{\delta}{R} \right) \right] . \end{align}

Finally, if we compute the deformation-rate tensor for the potential flow $\boldsymbol {v}_P$, we may conclude that

(2.30)\begin{align} \left|\int \boldsymbol{u} \boldsymbol{\cdot}(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}) \boldsymbol{v}_P \,\mathrm{d}V\right| & = \left|\int \frac{3}{2} \frac{U_\infty}{r} \left( \frac{R}{r} \right)^3 \boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\mathsf{M}}_2 \boldsymbol{u} \,\mathrm{d}V\right|\nonumber\\ & \leqslant 3 \frac{U_\infty}{R} \int \left( \frac{R}{r} \right)^4 |\boldsymbol{u}|^2 \,\mathrm{d}V . \end{align}

Let us use angle brackets $\langle ~ \rangle$ to denote the time average of any function $f(t)$:

(2.31)\begin{equation} \langle f(t) \rangle = \lim_{T \rightarrow \infty} \frac{1}{T} \int_0^T f(t) \,\mathrm{d}t. \end{equation}

Assuming a statistically stationary state in which the time average of a time derivative disappears, we deduce from (2.29) that

(2.32)\begin{align} \nu \left\langle \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V \right\rangle &\leqslant 6 \frac{U_\infty}{R} \int \left( \frac{R}{r} \right)^4 \langle |\boldsymbol{u}|^2 \rangle\, \mathrm{d}V \nonumber\\ &\quad + 24 {\rm \pi}\left(3+\frac{\sqrt{2226}}{35} \right) R^2 U_\infty^3 \left[ 1 + {O}\left( \frac{\delta}{R} \right) \right] . \end{align}

The viscosity does not appear explicitly on the right-hand side of this inequality, but there still is a viscosity dependence hidden in the definition of $\boldsymbol {u}$. To make this dependence apparent, we note that

(2.33)\begin{equation} |\boldsymbol{u}|^2 =|(\boldsymbol{v} - \boldsymbol{v}_P) - \boldsymbol{v}_b|^2 \leqslant (|\boldsymbol{v} - \boldsymbol{v}_P| + |\boldsymbol{v}_b|)^2 \leqslant 2 |\boldsymbol{v} - \boldsymbol{v}_P|^2 + 2 |\boldsymbol{v}_b|^2 \end{equation}

and

(2.34)\begin{equation} \int \frac{|\boldsymbol{u}|^2}{r^4} \,\mathrm{d}V \leqslant 2 \int \frac{|\boldsymbol{v} - \boldsymbol{v}_P|^2}{r^4} \,\mathrm{d}V + \frac{2}{R^4} \int |\boldsymbol{v}_b|^2 \,\mathrm{d}V \end{equation}

together with

(2.35)\begin{equation} \int |\boldsymbol{v}_b|^2 \,\mathrm{d}V = \frac{4{\rm \pi}}{5} R^2 U_\infty^2 \delta \left[ 1 + {O}\left( \frac{\delta^2}{R^2} \right) \right]. \end{equation}

If we insert these expressions into (2.32), we find that the term in $\int |\boldsymbol {v}_b|^2 \,\mathrm {d}V$ is smaller than the last term in (2.32) by a factor $\delta /R$ and is of the same order as previously neglected terms, so that this term has to be omitted for consistency as well. Let us simplify a numerical prefactor according to $327 < 24 {\rm \pi}(3+{\sqrt {2226}}/{35} ) < 328$ to finally arrive at

(2.36)\begin{equation} \nu \left\langle \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V \right\rangle \leqslant 12 \frac{U_\infty}{R} \int \left( \frac{R}{r} \right)^4 \langle |\boldsymbol{v} - \boldsymbol{v}_P|^2 \rangle \,\mathrm{d}V + 328 R^2 U_\infty^3 + {O}\left( \frac{\nu}{U_\infty R} \right) . \end{equation}

Relation (2.36) now has the characteristic features that Kolmogorov scaling demands. For any given velocity field, the right-hand side of (2.36) becomes asymptotically independent of viscosity as the Reynolds number tends to infinity, and it depends only on the velocity field itself, not on any of its derivatives. Deviations from the potential flow, such as the wake, contribute to the first term in the bound. This term weights the deviations such that the region close to the sphere contributes more than the far-away regions. The second term in the bound depends solely on $U_\infty$, which sets the amplitude of the irrotational flow. A purely irrotational flow does not cause any drag, a fact known as d'Alembert paradox. However, a purely irrotational flow is not compatible with the no-slip boundary conditions assumed here. Any non-zero $U_\infty$ thus implies a boundary layer or other rotational component, which explains the presence of the second term.

The Kolmogorov phenomenology originates from the study of homogeneous and isotropic turbulence. In this context, the dissipation per volume plays a central role and is proportional to $U^3/l$, where $U$ and $l$ are a characteristic velocity and length of the flow. The flow around a sphere is neither homogeneous nor isotropic, and its dissipation is finite and occurs in an infinite volume, so that the ratio of dissipation over flow volume is zero. Nonetheless, for the purpose of comparison with the Kolmogorov scaling, we may divide (2.36) by the reference volume $R^3$ to find that in the limit of $\nu /(U_\infty R)$ tending to zero,

(2.37)\begin{equation} \frac{\nu}{R^3} \left\langle \int |\boldsymbol{\nabla} \times \boldsymbol{v}|^2 \,\mathrm{d}V \right\rangle \leqslant 328 \frac{U^3}{R} \end{equation}

with

(2.38)\begin{equation} U = \left[ U_\infty^2 + \frac{3}{82}\frac{1}{R^3} \int \left( \frac{R}{r} \right)^4 \langle |\boldsymbol{v} - \boldsymbol{v}_P|^2 \rangle \,\mathrm{d}V \right]^{1/2} \end{equation}

as a possible choice for the characteristic velocity.

The empirical argument that the drag force should be proportional to $\rho U_\infty ^2 R$ requires, in addition, that $U$ is less than a bound proportional to $U_\infty$. The calculation presented here does not allow such a conclusion.

There are different ways in which the background field could be altered to produce relations similar to (2.36). The boundary-layer contribution $\boldsymbol {v}_b$ can obviously be optimised. However, there are also different possible choices for the irrotational part of the background field. For example, the potential flow around any obstacle enclosing the sphere is a viable option. Such a field could emulate the presence of a wake. If one aims for a short derivation, one might select the unperturbed velocity $U_\infty \hat {\boldsymbol {z}}$ as the irrotational part of the background. All the terms in (2.8) still allow for estimates of the required form if $\delta$ is fixed at a viscosity-independent value as for example $R$, so that $\boldsymbol {v}_b$ and $\boldsymbol {\nabla } \times \boldsymbol {v}_b$ have viscosity-independent supremum norms.