2. The hybrid method

To understand the deficiencies of classical small- $\mathrm{Re}$ approximations in 2-D, it is beneficial to discuss the difference between 3-D and 2-D in some detail. In the 3-D analysis of a sphere in a uniform stream, the asymptotic expansions are essentially carried out in powers of $\mathrm{Re}$ : while logarithmic terms appear, they may be grouped together with the associated powers, following Fraenkel’s warning [Reference Fraenkel10]. This allows for a straightforward use of the Van Dyke matching rule [Reference Van Dyke31]. Since the asymptotic error is algebraically small, the predictions are reasonably accurate for sufficiently small values of $\mathrm{Re}$ . In that scheme, moreover, the leading-order flow in the outer region is simply a uniform stream. This enables a convenient Oseen linearisation of the Navier–Stokes equations in that region.

In 2-D, the leading-order velocity field in the inner region diverges logarithmically with distance, in accordance with the Stokes paradox. As a consequence, a streaming flow would not constitute a solution to the leading-order outer problem when expanding in powers of $\mathrm{Re}$ . With the desire for a linearisation of the differential equation in the outer region, classical 2-D analyses resorted to an asymptotic expansion in inverse powers of $\log \mathrm{Re}$ (referred hereafter as “inverse-log expansion”). This procedure has serious shortcomings, which are all related. The first is that the asymptotic error is only logarithmically small; thus, unless $\mathrm{Re}$ is extremely small, the accuracy of the scheme is rather poor. The second is the mathematical complexity associated with the need to obtain more than one term in the inner region [Reference Kaplun17]. The third is that Van Dyke’s matching rule cannot be applied [Reference Fraenkel10]. Asymptotic matching thus requires the more delicate apparatus of intermediate variables.

Ideally, one would want to “sum” the inverse-log expansion, thereby obtaining an algebraic accuracy. This goal was effectively accomplished in the (relatively) recent work Ward & coworkers [Reference Kropinski, Ward and Keller21, Reference Hormozi and Ward14]. The key principle in this work is the adherence to expansion in powers of $\mathrm{Re}$ , disregarding its logarithms [Reference Fraenkel10]. This allows for the use of Van Dyke’s matching rule and provides an algebraic accuracy. In the inner region, the streamfunction possesses a simple mathematical form and is derived up to an unknown pre-factor $S$ , a function of $\log \mathrm{Re}$ (representing the summation of an inverse-log expansion). The challenge in that methodology, of course, has to do with the leading-order problem in the outer region, which involves the nonlinear Navier–Stokes equation. Ward & coworkers adopted a hybrid approach, where the nonlinear equation is solved numerically, subject to matching with the inner solution, which appears as a point singularity. The key part in that nonlinear calculation involves the computation of the regular part of the singularity structure as a function of the strength of the singularity. This procedure, in turn, provides the missing pre-factor $S$ as an implicit function of $\log \mathrm{Re}$ . Since the outer region is single-scaled, the hybrid approach is superior to a direct solution of the full Navier–Stokes equation governing the exact problem, which would require the simultaneous resolution of two disparate length scales.

In the papers of Ward & coworkers, the quantity of interest is the hydrodynamic drag. Since that quantity is expressed in terms of the Newtonian stresses at the boundary, it may be calculated using the details of the inner solution. In particular, it is proportional to the pre-factor $S$ . Similarly, the present transport problem, focused at the limit of large $\mathrm{Pe}$ , is effectively restricted to a narrow layer about the boundary of the cylinder – a province contained in the inner region of the flow problem. From our perspective, then, the hybrid method transforms the entire complexity of the small- $\mathrm{Re}$ hydrodynamics to the calculation of the pre-factor, which has already been carried out by Ward & coworkers.

3. Problem formulation

A rigid circular cylinder (radius $a$ ) is held fixed within a viscous fluid (kinematic viscosity $\nu$ ) which is streaming perpendicular to the cylinder axis with velocity $\mathcal U$ . We are concerned with the associated forced convection of a scalar property which undergoes both passive advection and diffusion, say with a uniform diffusivity $D$ . We consider the common scenario where that property is uniformly prescribed on the cylinder boundary at a fixed value, say $\Delta$ , relative to the equilibrium value at large distances. For concreteness, we shall refer to heat transport, where said property is the temperature. The description is equally adequate to describe mass transport, where the transported property is the solute concentration.

It is convenient to employ a dimensionless notation, where length variables are normalised by $a$ and velocities by $\mathcal U$ . The dimensionless flow problem is governed by a single parameter, the Reynolds number $\mathrm{Re}=a\,\mathcal U/\nu$ . Similarly, the transport problem is governed by a single parameter, the Péclet number $\mathrm{Pe} = a\,\mathcal U/D$ . Their ratio,

(3.1) \begin{equation} \sigma = \mathrm{Pe}/\mathrm{Re}=\nu/D, \end{equation}

is the Prandtl number. Since the flow and transport problems are both two-dimensional, it is only the cylinder cross section that enters the geometry. Hereafter, it is referred to as “the circle.” The two-dimensional problem is schematically depicted in Figure 1.

Figure 1. Problem schematic, conceptually indicating the pertinent asymptotic regions in the double limit (3.15).

In describing the 2-D flow and transport problems we employ polar $(r,\theta )$ coordinates with the origin at the circle centre and the “leading edge” at $\theta =0$ , see Figure 1. We begin with formulating the flow problem, which is independent of the temperature. Writing the velocity field as $\textbf{u} = u\hat{\textbf{e}}_r + v\hat{\textbf{e}}_\theta$ , it is convenient to employ the streamfunction $\psi$ , defined by

(3.2) \begin{equation} u = \frac{1}{r}\frac{\partial \psi }{\partial \theta }, \quad v= - \frac{\partial \psi }{\partial r}, \end{equation}

so the continuity equation is trivially satisfied. In terms of $\psi$ , the Navier–Stokes equations read

(3.3) \begin{equation} \nabla ^4\psi = \frac{\mathrm{Re}}{r} \frac{\partial (\psi,\nabla ^2\psi )}{\partial (\theta,r)}, \end{equation}

wherein

(3.4) \begin{equation} \nabla ^2 = \frac{\partial ^2}{\partial r^2}+ \frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{\partial ^2}{\partial \theta ^2} \end{equation}

is the Laplacian operator in polar coordinates. The streamfunction needs to satisfy the impermeability and no-slip conditions,

(3.5) \begin{equation} \psi = \frac{\partial \psi }{\partial r} = 0 \quad \text{at} \quad r=1, \end{equation}

where we exploit the freedom in the definition of $\psi$ , and the far-field approach to a uniform stream,

(3.6) \begin{equation} \psi \sim -r\sin \theta \quad \text{as} \quad r\to \infty. \end{equation}

It is evident from the problem symmetry that $\psi$ satisfies

(3.7) \begin{equation} \psi (r,-\theta ) = -\psi (r,\theta ). \end{equation}

Once $\textbf{u}$ is calculated from the above nonlinear problem, we can address the transport problem. The dependent variable is the excess temperature, relative to the far-field equilibrium value; normalised by $\Delta$ , it is denoted by $T$ . The transport problem is governed by the advection–diffusion equation,

(3.8) \begin{equation} \nabla ^2{T} = \mathrm{Pe} \,\textbf{u}\boldsymbol{\cdot }\boldsymbol{\nabla }{T} \end{equation}

(where the continuity equation has been used) together with the imposed temperature at the circle,

(3.9) \begin{equation}{T}=1 \quad \text{at} \quad r=1, \end{equation}

and the decay condition

(3.10) \begin{equation} \lim _{r\to \infty }{T}=0. \end{equation}

With $\textbf{u}$ regarded as known, the above problem is linear. It is evident from the problem symmetry that the temperature satisfies

(3.11) \begin{equation}{T}(r,-\theta ) ={T}(r,\theta ). \end{equation}

Our interest lies not in the detailed transport process but rather in the net flux from the cylinder. It is represented by the Nusselt number [Reference Yariv33]

(3.12) \begin{equation} \mathrm{Nu} = -\frac{1}{2\pi }\int _{-\pi }^{\pi } \left .\frac{\partial{T}}{\partial r}\right |_{r=1}\, d \theta, \end{equation}

or, making use of the symmetry (3.11),

(3.13) \begin{equation} \mathrm{Nu} = -\frac{1}{\pi }\int _0^{\pi } \left .\frac{\partial{T}}{\partial r}\right |_{r=1}\, d \theta. \end{equation}

The Nusselt number is evidently a function of both $\mathrm{Re}$ and $\mathrm{Pe}$ :

(3.14) \begin{equation} \mathrm{Nu}=\mathrm{Nu}(\mathrm{Re},\mathrm{Pe}). \end{equation}

Our interest is in deriving an asymptotic approximation for $\mathrm{Nu}$ in the limit

(3.15) \begin{equation} \mathrm{Re}\to 0, \quad \mathrm{Pe}\to \infty, \end{equation}

which necessitates large Prandtl numbers [recall (3.1)]

(3.16) \begin{equation} \sigma \to \infty. \end{equation}

Given the expected boundary-layer structure, $\mathrm{Nu}$ is anticipated [Reference Leal23] to exhibit a $\mathrm{Pe}^{1/3}$ scaling at large $\mathrm{Pe}$ ,

(3.17) \begin{equation} \mathrm{Nu} \sim f(\mathrm{Re}) \mathrm{Pe}^{1/3} \quad \text{for} \quad \mathrm{Pe}\to \infty. \end{equation}

That scaling relation holds for all $\mathrm{Re}$ . As $\mathrm{Re}\to 0$ , $f(\mathrm{Re})$ does not attain a limit; rather, it becomes a slowly-varying function of $\mathrm{Re}$ . It is convenient to make that dependence explicit by writing

(3.18) \begin{equation} f(\mathrm{Re}) = F(\chi ), \end{equation}

wherein $F$ is a moderately-varying function. There is a freedom is defining $\chi$ , with some natural choices being $\log \mathrm{Re}$ , $\log (1/\mathrm{Re})$ and $1/\log (1/\mathrm{Re})$ . For compatibility with Hormozi & Ward [Reference Hormozi and Ward14], we here define

(3.19) \begin{equation} \chi = \left (\log \frac{1}{\mathrm{Re}}-\frac{1}{2}\right )^{-1}. \end{equation}

We conclude that

(3.20) \begin{equation} \mathrm{Nu} \sim F(\chi ) \mathrm{Pe}^{1/3} \quad \text{for} \quad \mathrm{Pe}\to \infty \quad \text{with} \quad \mathrm{Re}\to 0. \end{equation}

Our goal is accordingly the calculation of the rescaled Nusselt number $F(\chi )$ .

4. Flow problem at small $\mathrm{Re}$

We need an asymptotic approximation for $\textbf{u}$ in the limit $\mathrm{Re}\to 0$ . As explained in Sec. 2, this requires the use of matched asymptotic expansions. In what follows, we summarise briefly the approach of Ward and coworkers [Reference Kropinski, Ward and Keller21, Reference Hormozi and Ward14]. It involves an asymptotic expansion in powers of the small parameter $\mathrm{Re}$ , grouping together terms that differ by $\log \mathrm{Re}$ . In the inner region, where $r=\text{ord}(1)$ , we pose the asymptotic expansion

(4.1) \begin{equation} \psi (r,\theta ;\;\mathrm{Re}) \sim \psi _0(r,\theta ;\;\chi ) + \text{alg} \end{equation}

wherein “alg” represents an asymptotic error which is algebraically small (i.e. smaller than some positive power of $\mathrm{Re}$ ). The leading term $\psi _0$ satisfies the biharmonic equation [cf. (3.3)],

(4.2) \begin{equation} \nabla ^4\psi _0 = 0, \end{equation}

and the boundary conditions [cf. (3.5)]

(4.3) \begin{equation} \psi _0 = \frac{\partial \psi _0}{\partial r} = 0 \quad \text{at} \quad r=1. \end{equation}

Since the far-field condition (3.6) does not apply in the inner region, the symmetry condition [cf. (3.7)]

(4.4) \begin{equation} \psi _0(r,-\theta ) = -\psi _0(r,\theta ) \end{equation}

must be explicitly enforced.

With $\psi _0$ governed by a homogeneous problem, it is determined up to a multiplicative constant. The solution of (4.2)–(4.4) that is least singular as $r\to \infty$ is of the form

(4.5) \begin{equation} \psi _0(r,\theta ;\;\chi )={S}(\chi )\psi _c(r,\theta ), \end{equation}

where the canonical solution,

(4.6) \begin{equation} \psi _c =\left (\frac{r}{2}-r\log r-\frac{1}{2r} \right )\sin \theta, \end{equation}

is $O(r\log r)$ at large $r$ ,

(4.7) \begin{equation} \psi _c \sim r\left (\frac{1}{2}-\log r\right )\sin \theta + \text{alg} \quad \text{as} \quad r \to \infty. \end{equation}

The pre-factor $S$ can only be determined via asymptotic matching. Consider now the outer region, at distances $r=\text{ord}(1/\mathrm{Re})$ , where inertial and viscous forces are comparable (see Figure 1). Using the stretched radial coordinate $\rho = \mathrm{Re} \,r$ , the outer streamfunction $\Psi = \mathrm{Re}^{-1}\psi$ is governed by the differential equation,

(4.8) \begin{equation} \nabla ^4_\rho \Psi = \frac{1}{\rho } \frac{\partial (\Psi,\nabla ^2_\rho \Psi )}{\partial (\theta,\rho )}, \end{equation}

wherein

(4.9) \begin{equation} \nabla ^2_\rho = \frac{\partial ^2}{\partial \rho ^2}+ \frac{1}{\rho }\frac{\partial }{\partial \rho }+\frac{1}{\rho ^2}\frac{\partial ^2}{\partial \theta ^2}. \end{equation}

is the Laplacian with respect to the stretched coordinates [cf. (3.4)]. In addition, it satisfies the approach to a uniform stream

(4.10) \begin{equation} \Psi \sim -\rho \sin \theta \quad \text{as} \quad \rho \to \infty \end{equation}

and the requirement of asymptotic matching with the inner solution.

Similarly to (4.1), we introduce the outer expansion,

(4.11) \begin{equation} \Psi (\rho,\theta ;\;\mathrm{Re}) \sim \Psi _0(\rho,\theta ;\;\chi ) + \text{alg}. \end{equation}

The leading-order field $\Psi _0$ satisfies the original equation (4.8),

(4.12) \begin{equation} \nabla ^4_\rho \Psi _0 = \frac{1}{\rho } \frac{\partial (\Psi _0,\nabla ^2_\rho \Psi _0)}{\partial (\theta,\rho )}, \end{equation}

and the remote condition (4.10),

(4.13) \begin{equation} \Psi _0 \sim -\rho \sin \theta \quad \text{as} \quad \rho \to \infty. \end{equation}

The separation by powers of $\mathrm{Re}$ allows for the use of Van-Dyke matching. Making use of (4.5) and (4.7) in conjunction with (3.19), it follows that

(4.14) \begin{equation} \Psi _0 \sim -{S}(\chi ) \rho \left (\log \rho +{1}/{\chi }\right )\sin \theta \quad \text{as} \quad \rho \to 0. \end{equation}

The nonlinear problem (4.12)–(4.14) was solved numerically in [Reference Hormozi and Ward14]. That solution provides $\Psi _0$ , and, more importantly, ${S}(\chi )$ . The function ${S}(\chi )$ is tabulated in Table 1, using values obtained in [Reference Hormozi and Ward14]. It is considered hereafter as known. Accordingly, so is $\psi _0$ .

Table 1. The function $S(\chi)$ , tabulated using the calculation of [Reference Hormozi and Ward14]. The range of $\chi$ -values is restricted to $\mathrm{Re}\lt 1/2$

5. The limit $\mathrm{Pe}\to \infty$

In analysing the transport problem (3.8)–(3.10) in the limit $\mathrm{Pe}\to \infty$ , we find from (3.8)

(5.1) \begin{equation} \textbf{u}\boldsymbol{\cdot }\boldsymbol{\nabla }{T}=0. \end{equation}

It follows that $T$ is constant along streamlines. Since the streamlines are open, originating at infinity, condition (3.10) gives

(5.2) \begin{equation}{T}\equiv 0. \end{equation}

In fact, with (5.2) being an exact solution of (3.8) and (3.10), the asymptotic correction to (5.2) is exponentially small.

Since the Dirichlet condition (3.9) is incompatible with (5.2), a boundary layer is formed about the boundary $r=1$ . In that layer, whose width $\delta (\ll 1)$ is yet undetermined, the diffusive and advective terms in (3.8) are comparable. To determine these terms, we need to approximate the velocity components in the layer, where $r-1\ll 1$ . We readily find from (4.5)–(4.6)

(5.3) \begin{equation} u_0 \sim -{S}(\chi )(r-1)^2\cos \theta, \quad v_0\sim 2{S}(\chi )(r-1)\sin \theta. \end{equation}

In what follows we denote the leading-order boundary-layer temperature by $\Theta$ . Defining the stretched radial coordinate,

(5.4) \begin{equation} Y = \frac{r-1}{\delta (\mathrm{Pe})}, \end{equation}

we find using (5.3) that the advective term (i.e. the right-hand side) of (3.8) is

(5.5) \begin{equation} \sim{S}(\chi ) \mathrm{Pe}\, \delta \left (2Y\frac{\partial \Theta }{\partial \theta }\sin \theta - Y^2\frac{\partial \Theta }{\partial Y}\cos \theta \right ) \end{equation}

within the boundary layer, while the diffusive term (i.e. the left-hand side) of (3.8) is $\sim \delta ^{-2} \partial ^2\Theta/\partial Y^2$ there, dominated by transverse variations. Balancing the terms provides the anticipated scaling

(5.6) \begin{equation} \delta = \mathrm{Pe}^{-1/3} \end{equation}

of the boundary-layer thickness, see Figure 1.

We conclude that, within the boundary layer, $\Theta$ is governed by the parabolic equation

(5.7) \begin{equation} \frac{\partial ^2\Theta }{\partial Y^2} ={S}(\chi ) \left (2Y\frac{\partial \Theta }{\partial \theta }\sin \theta - Y^2\frac{\partial \Theta }{\partial Y}\cos \theta \right ). \end{equation}

This equation is subject to the Dirichlet condition [cf. (3.9)],

(5.8) \begin{equation} \Theta =1 \quad \text{at} \quad Y=0, \end{equation}

and matching with (5.2),

(5.9) \begin{equation} \lim _{Y\to \infty }\Theta =0. \end{equation}

The associated approximation for the Nusselt number (3.13) is

(5.10) \begin{equation} \mathrm{Nu}\sim -\frac{\mathrm{Pe}^{1/3}}{\pi }\int _0^{\pi } \left .\frac{\partial \Theta }{\partial Y}\right |_{Y=0}\, d \theta. \end{equation}

6. Similarity solution

It is convenient to employ the independent variable $\eta =\cos \theta$ instead of $\theta$ . Thus, (5.7) becomes

(6.1) \begin{equation} \frac{\partial ^2\Theta }{\partial Y^2} = -{S}(\chi ) \left [2Y(1-\eta ^2)\frac{\partial \Theta }{\partial \eta } +Y^2\eta \frac{\partial \Theta }{\partial Y}\right ] \end{equation}

while (5.10) gives

(6.2) \begin{equation} \mathrm{Nu} \sim -\frac{\mathrm{Pe}^{1/3}}{\pi }\int _{-1}^1 \left .\frac{\partial \Theta }{\partial Y}\right |_{Y=0}\, \frac{d \eta }{(1-\eta ^2)^{1/2}}. \end{equation}

It is well known that problems governed by equations of the form (6.1) adopt a similarity solution [Reference Acrivos1]. We define the similarity variable,

(6.3) \begin{equation} \zeta = \frac{Y}{g(\eta )}, \end{equation}

where the function $g$ remains to be determined. The temperature becomes a function of the single variable $\zeta$ , say $\Omega (\zeta )$ . In terms of that function, (6.2) becomes

(6.4) \begin{equation} \mathrm{Nu}\sim -\frac{\mathrm{Pe}^{1/3}\Omega '(0)}{\pi }\int _{-1}^1 \frac{d \eta }{g(\eta )(1-\eta ^2)^{1/2}}, \end{equation}

where the prime denotes differentiation.

Substituting of (6.3) into conditions (5.8)–(5.9) gives

(6.5) \begin{equation} \Omega (0)=1, \quad \Omega (\infty )=0. \end{equation}

Also, substitution of (6.3) into (6.1) yields

(6.6) \begin{equation} \Omega ''(\zeta ) ={S}(\chi ) \zeta ^2 \Omega '(\zeta ) \left [\frac{2}{3}(1-\eta ^2)h'-\eta h\right ] \end{equation}

wherein

(6.7) \begin{equation} h=g^3. \end{equation}

For consistency with the similarity ansatz we require,

(6.8) \begin{equation} \frac{2}{3}(1-\eta ^2)h'-\eta h = \text{const}. \end{equation}

The constant cannot be zero, as the resulting equation $\Omega ''=0$ is incompatible with (6.5). For the same reason it cannot be positive. There is a freedom in choosing the negative constant: since any choice amounts to a rescaling of $g$ , it is merely equivalent to a redefinition of $\zeta$ . Here we find it convenient to set the constant to $-3$ , whereby (6.6) reduces to the ordinary differential equation,

(6.9) \begin{equation} \Omega ''(\zeta ) + 3{S}\zeta ^2\Omega '(\zeta ) = 0. \end{equation}

The solution of that equation subject to conditions (6.5) is

(6.10) \begin{equation} \Omega = 1-\frac{1}{ \Gamma (4/3)}\int _0^{{S}^{1/3}\zeta }e^{-t^3}\,d t, \end{equation}

wherein $\Gamma$ is the Gamma function. In particular, $\Omega '(0)= -{{S}^{1/3}}/{\Gamma (4/3)}$ . Substitution into (6.4) and comparing with (3.18) yields

(6.11) \begin{equation} F(\chi )= \frac{{S}^{1/3}(\chi )}{\pi \Gamma (4/3)}\int _{-1}^1 \frac{d \eta }{g(\eta )(1-\eta ^2)^{1/2}}. \end{equation}

All that remains is to determine the function $g$ , or, equivalently, $h$ . The latter is governed by the first-order differential equation (6.8), with the constant set to $-3$ . Integration gives

(6.12) \begin{equation} h = \frac{K}{(1-\eta ^2)^{3/4}} - \frac{9\eta }{2(1-\eta ^2)^{3/4}} F_2^1\left (\frac{1}{4},\frac{1}{2},\frac{3}{2},\eta ^2\right ), \end{equation}

wherein $F_2^1$ is the hypergeometric function. The integration constant $K$ is determined by the requirement that the solution does not blow up at the “leading edge” $\eta =1$ . Using

(6.13) \begin{equation} \lim _{\eta \to 1} F_2^1\left (\frac{1}{4},\frac{1}{2},\frac{3}{2},\eta ^2\right ) = \frac{\pi ^{3/2}}{2^{1/2} \Gamma (1/4) \Gamma (5/4)} \end{equation}

we obtain

(6.14) \begin{equation} K = \frac{9\pi ^{3/2}}{8^{1/2} \Gamma (1/4) \Gamma (5/4)}. \end{equation}

Note that $h$ still blows up at the “trailing edge” $\eta =-1$ . The associated divergence there of $g$ , like $(1-\eta ^2)^{-1/4}$ , is weaker than that found in the 3-D problem [Reference Leal23].

Substitution into (6.11) of (6.7), (6.12) and (6.14) yields the rescaled Nusselt number,

(6.15) \begin{equation} F(\chi )= \frac{1}{\Gamma (4/3)} \left [\frac{3{S}(\chi )}{2\Gamma ^2(1/4)\Gamma ^2(5/4)}\right ]^{1/3}. \end{equation}

Numerical evaluation of the Gamma function gives

(6.16) \begin{equation} F(\chi )= 0.5799 [{S}(\chi )]^{1/3}, \end{equation}

which corresponds to the expression quoted in the abstract. The error incurred by the use of (3.20) with (6.15) is algebraically small in both $\mathrm{Re}$ and $\mathrm{Pe}$ .

7. Comparison with numerical solution

It is illustrative to compare our approximation with the numerical calculations of Dennis et al. [Reference Dennis, Hudson and Smith9], who solved the full Navier–Stokes equations and the advection–diffusion equation governing the companion transport problem. (Note that their Reynolds and Nusselt numbers are based upon the circle diameter, rather than radius.) The only data in [Reference Dennis, Hudson and Smith9] relevant to the double limit (3.15) considered herein are presented in their table III, which provides $\mathrm{Nu}/\sigma ^{1/3}$ as a function of $\mathrm{Re}$ and $\sigma$ . The smallest Reynolds number in that table is (in the present notation) 0.25; at that $\mathrm{Re}$ , it gives (in the present notation) the value $0.256$ for $\mathrm{Nu}/\sigma ^{1/3}$ at the largest $\sigma$ -value employed there [corresponding to numerical convergence to the asymptotic limit (3.15)]. Comparing with (3.1) and (3.20), this corresponds to

(7.1) \begin{equation} F=0.4064. \end{equation}

Now, for $\mathrm{Re}=0.25$ , we find from (3.19) $\chi =1.1283$ . This value is close to $\chi =1.1469$ in the present Table 1, for which $S=0.332$ . Substituting that $S$ -value into (6.15) gives $F(\chi )=0.4016$ .

Since the Reynolds number is rather modest, the above agreement with (7.1) is quite remarkable. Given the singular nature of the small- $\mathrm{Re}$ limit, it is hardly surprising that [Reference Dennis, Hudson and Smith9] did not report numerical data for smaller values of $\mathrm{Re}$ : indeed, a numerical solution at small $\mathrm{Re}$ needs to resolve two disparate length scales, namely $1$ (cylinder scale) and $1/\mathrm{Re}$ (Oseen scale). The smaller is $\mathrm{Re}$ , the more difficult is the numerical challenge. On the other hand, the present approximation (6.16), where the asymptotic error is algebraically small, only improves as $\mathrm{Re}$ diminishes.

It is interesting to compare our asymptotic prediction with that provided by an inverse-log expansion [Reference Kaplun17, Reference Proudman and Pearson30]. The leading-order approximation resulting from that expansion is equivalent to replacing $S$ by $a_1(\log \mathrm{Re})^{-1}$ with $a_1=-1$ . Substitution into (6.16) and (3.20) gives

(7.2) \begin{equation} \mathrm{Nu} \sim 0.5799 \left (\frac{-\mathrm{Pe}}{\log{\mathrm{Re}}}\right )^{1/3}. \end{equation}

This is a very simple approximation that does not require the use of the function $S(\chi )$ . It is, however, rather crude. For $\mathrm{Re}=0.25$ , it gives the value $0.5201$ for $\mathrm{Nu}/\mathrm{Pe}^{1/3}$ , about $25\%$ larger than the numerical prediction (7.1).

An attempt to improve the approximation by incorporating more terms only makes matters worse. (Hardly surprising, given the modest value of $\mathrm{Re}$ .) Thus, using a 2-term expansion amounts to replacing $S$ by

(7.3) \begin{equation}{a_1}({\log \mathrm{Re}})^{-1} +{a_2}({\log \mathrm{Re}})^{-2}, \end{equation}

where $a_2=\gamma _E-1/2-\log 4$ , $\gamma _{\text{E}}=0.57721\ldots$ being the Euler–Mascheroni constant. That “improved” approximation gives the value $0.1986$ for $\mathrm{Nu}/\mathrm{Pe}^{1/3}$ .