2.1 Boundary homogenisation

Suppose we approximate $\phi(r,z)$ by a function of z only, $\psi(z)$ ; see (1.1). Laplace’s equation becomes $\psi''(z)=0$ , so that $\psi(z)=Az+B$ , with arbitrary constants A and B. The mixed boundary conditions at $z=0$ are replaced by (1.2), $\psi'(0)=\kappa\psi(0)$ , where $\kappa$ is a constant. Combining this condition with $\psi(\ell)=\phi_0$ , we obtain

(2.5) \begin{equation} \psi(z)=\phi_0\,\frac{1+\kappa z}{1+\kappa\ell} \quad\mbox{and}\quad {\mathcal J}(\ell)\simeq \pi a^2 \psi' =\frac{\pi a^2\kappa\phi_0}{1+\kappa\ell}.\end{equation}

This calculation is very simple but, of course, we do not know $\kappa$ : the problem of calculating $\phi$ has been replaced by the problem of calculating $\kappa$ .

2.2 Rigid disc in an infinite tube

Returning to problem (2.1), suppose we extend $\phi(r,z)$ into $-\ell<z<0$ as an odd function of z, $\phi(r,-z)=-\phi(r,z)$ , $0<z<\ell$ . Then we see that the problem is equivalent to some kind of potential flow past a thin rigid disc of radius b in a tube of cross-sectional radius a. Note that extending $\phi$ in this way means that (2.1c) is satisfied automatically.

The simplest problem of this kind comes by letting $\ell\to\infty$ , with uniform flow along an infinite tube. For this problem, the conditions $\phi(r,\pm\ell)=\pm\phi_0$ (see (2.1a)) are replaced by

(2.6) \begin{equation} \phi(r,z)= U(z\pm C)+o(1)\quad\mbox{as $z\to\pm\infty$,}\end{equation}

where U is the constant speed of the flow far from the disc and C is the blockage coefficient. (The o(1) terms in (2.6) are exponentially small.) We shall see later that $\kappa$ in (1.2) and C in (2.6) are related.

The problem of uniform flow past a rigid disc has been studied previously [Reference Smythe18, Reference Sherwood and Stone17, Reference Martin and Skvortsov14, Reference Martin and Skvortsov15]. There are approximations when $b/a$ is small and when the gap between the disc and the tube is small ( $b/a\simeq 1$ ) [Reference Sherwood and Stone17, Reference Martin and Skvortsov14]. Here, we begin with a standard method (that could be used to compute C numerically) but then we introduce approximations that lead to analytical estimates.

Two exact formulas are worth noting. Hurley’s formula [Reference Hurley9], [Reference Martin and Skvortsov14, equation (31)] gives

(2.7) \begin{equation} C=\frac{2}{a^2U}\int_0^b \phi(r,0+)\,r\,{\rm d} r.\end{equation}

We also have flux conservation through the gap,

(2.8) \begin{equation} 2\pi \int_b^a \frac{\partial {\phi}}{\partial {z}}(r,0+)\,r\,{\rm d} r=\pi a^2 U.\end{equation}

Now, as it is sufficient to consider $z>0$ , we write

(2.9) \begin{equation} \phi(r,z)=U(z+C) +Ua\sum_{n=1}^\infty c_n J_0(\lambda_nr){\rm e}^{-\lambda_nz},\quad 0\leq r<a,\ z>0,\end{equation}

where $\lambda_n$ are positive solutions of $J_1(\lambda_na)=0$ and $J_n$ are Bessel functions. The (dimensionless) coefficients $c_n$ and $C/a$ are to be found. The representation (2.9) ensures that $\nabla^2\phi=0$ and that (2.1d) and (2.6) are satisfied. Applying (2.1b) and (2.1c) gives

(2.10a) \begin{align}& \sum_{n=1}^\infty c_n (\lambda_na) J_0(\lambda_nr)=1, \quad 0\leq r<b, \end{align}

(2.10b) \begin{align} & \frac{C}{a}+\sum_{n=1}^\infty c_n J_0(\lambda_nr)=0, \quad b<r<a.\end{align}

These are dual series equations for $c_n$ and C [Reference Sneddon19], [Reference Duffy6, Example 3.3.1]. For a numerical treatment, see [Reference Sherwood and Stone17]. Also, if we multiply (2.10b) by r and then integrate from $r=b$ to $r=a$ , we obtain

(2.11) \begin{equation} \frac{C}{a}=\frac{2ab}{a^2-b^2}\sum_{n=1}^\infty c_n \frac{J_1(\lambda_nb)}{\lambda_na}\end{equation}

after use of

(2.12) \begin{equation} \int_p^q J_0(\lambda_nr)\,r\,{\rm d} r =\frac{q}{\lambda_n}\,J_1(\lambda_nq) -\frac{p}{\lambda_n}\,J_1(\lambda_np).\end{equation}

The same formula for C, (2.11), is obtained by substituting (2.9) in (2.7).

Instead of trying to solve (2.10), let $V(r)=\partial {\phi}/\partial {z}$ at $z=0$ for $b<r<a$ ; this is the flow speed in the gap. Thus, using (2.1b) and (2.9),

\begin{equation*} U-U\sum_{n=1}^\infty c_n (\lambda_n a)J_0(\lambda_nr)= \left\{\begin{array}{cl} 0, & 0\leq r<b,\\ \\[-7pt] V(r), & b<r<a. \end{array}\right.\end{equation*}

This is a Dini–Bessel series. Orthogonality gives

(2.13) \begin{equation} \pi a^2 U=2\pi\int_b^a V(r)r\,{\rm d} r\end{equation}

(which is (2.8)) and

(2.14) \begin{equation} c_n=-\frac{2}{U\lambda_na^3 J_0^2(\lambda_na)}\int_b^a V(r)J_0(\lambda_nr)r\,{\rm d} r,\end{equation}

using [16, 10.22.4 and 10.22.5]

(2.15) \begin{equation} \int_0^a J_0(\lambda_m r)J_0(\lambda_nr)\,r\,{\rm d} r= \frac{a^2}{2}\,J_0^2(\lambda_na)\,\delta_{mn}.\end{equation}

Substituting for $c_n$ from (2.14) in (2.10a) leads to an integral equation for V(r). This approach will be pursued later when we return to the finite-tube problem in Section 2.3. However, for the moment, we note that our calculations are exact, but we do not know V.

Let us make a plausible approximation for V. The simplest possibility is to take $V(r)=U_0$ , a constant; from (2.13), $U_0=Ua^2/(a^2-b^2)$ . Then (2.14) gives the approximation

\begin{equation*} c_n =\frac{2bU_0 J_1(\lambda_nb)}{U\lambda_n^2a^3 (a^2-b^2)\,J_0^2(\lambda_na)}\end{equation*}

which, when substituted in (2.11) gives the approximation

(2.16) \begin{equation} C\simeq C_{{\rm disc}}^{{\rm app}}=\frac{4a^3b^2}{(a^2-b^2)^2} \,{\mathbb S}_0(b/a)\end{equation}

where

(2.17) \begin{equation} {\mathbb S}_0(X)= \sum_{n=1}^\infty \frac{J_1^2(\lambda_naX)}{(\lambda_na)^3\,J_0^2(\lambda_na)} =\sum_{n=1}^\infty \frac{J_1^2(j_nX)}{j_n^3J_0^2(j_n)}\end{equation}

and $j_n=\lambda_na$ (so that $J_1(j_n)=0$ ). (The notation ‘app’ in (2.16) indicates ‘approximation’.) The terms in the series ${\mathbb S}_0$ decay as $n^{-3}$ ; as far as we know, ${\mathbb S}_0$ cannot be summed in closed form (although it can be written as an integral [Reference Martin13, equation (16)]). It will appear again later; see (2.24). A generalisation, ${\mathbb S}_q$ will also appear later; see (3.14) and Appendix A.

Although the approximation $V(r)\simeq U_0$ may seem crude, it can be refined, taking into account the fact that V(r) is infinite at the sharp edge of the disc ( $r=b$ ). For details, with an application to a disc that almost fills the tube (small gap, $b/a\simeq 1$ ), see [Reference Martin and Skvortsov15].

2.3 Finite-tube problem: Estimating $\kappa$

2.3.1 An integral equation

Let us return to the finite-tube problem (2.1). Look for a solution in the form

(2.18) \begin{equation} \frac{\phi(r,z)}{\phi_0} =1+c_0\left(\frac{z}{\ell}-1\right) +\sum_{n=1}^\infty c_n J_0(\lambda_nr)\sinh{\{\lambda_n(z-\ell)\}}, \quad 0\leq r<a,\ 0<z<\ell.\end{equation}

This representation satisfies (2.1a) and (2.1d). From (2.2), the flux is given by

(2.19) \begin{equation} {\mathcal J}(\ell)=\pi a^2\phi_0 c_0/\ell.\end{equation}

Applying the boundary conditions at $z=0$ , (2.1b) and (2.1c), we obtain dual series equations,

(2.20a) \begin{align}c_0 +\sum_{n=1}^\infty c_n J_0(\lambda_nr) (\lambda_n\ell) \cosh{\lambda_n\ell} &=0,\quad 0\leq r<b, \end{align}

(2.20b) \begin{align}c_0 +\sum_{n=1}^\infty c_n J_0(\lambda_nr) \sinh{\lambda_n\ell} &=1,\quad b<r<a.\end{align}

Proceeding as in Section 2.2, let $V(r)=\partial {\phi}/\partial {z}$ at $z=0$ for $b<r<a$ . Then, using (2.1b) and (2.18), we obtain

\begin{equation*} c_0 +\sum_{n=1}^\infty c_n J_0(\lambda_nr)(\lambda_n\ell)\cosh{\lambda_n\ell} =\left\{\begin{array}{c@{\quad}l} 0, & 0\leq r<b,\\ \\[-7pt] (\ell/\phi_0) V(r), & b<r<a. \end{array}\right.\end{equation*}

Orthogonality gives

(2.21) \begin{equation} c_0=\frac{2\ell}{\phi_0a^2}\int_b^a V(r)r\,{\rm d} r =\frac{\ell}{\pi a^2\phi_0}{\mathcal J}(\ell)\end{equation}

and

\begin{equation*}c_n \lambda_n\ell \cosh{\lambda_n\ell}=\frac{2\ell}{\phi_0a^2J_0^2(\lambda_na) }\int_b^a V(r)J_0(\lambda_nr)r\,{\rm d} r\end{equation*}

using (2.15). Substitution in (2.20b) gives an integral equation for V,

(2.22) \begin{equation} 1= \frac{2\ell}{\phi_0a^2}\int_b^a V(s)\bigg\{ 1 +\frac{a}{\ell}\sum_{n=1}^\infty \frac{J_0(\lambda_nr)J_0(\lambda_ns)} {(\lambda_na) \,J_0^2(\lambda_na)} \tanh{\lambda_n\ell} \bigg\} s\,{\rm d} s,\quad b<r<a.\end{equation}

This reduction to an integral equation is exact. However, we shall approximate because we are only interested in calculating $c_0$ (see (2.21)) and, moreover, we shall assume that $\ell/a\gg 1$ .

2.3.2 Approximation for large $\ell$

For large $\ell$ , $\tanh{\lambda_n\ell}\simeq 1$ , so we can discard the sum term in (2.22) (because of the factor $a/\ell$ ). Thus, at leading order,

(2.23) \begin{equation} 1\simeq \frac{2\ell}{\phi_0a^2}\int_b^a V(s)s\,{\rm d} s=c_0,\end{equation}

which gives ${\mathcal J}(\ell)\simeq \pi a^2\phi_0/\ell$ . This very simple approximation does not depend on b, although it is exact when $b=0$ .

In order to improve our approximations, we use an iterative approach. Thus, suppose that $V(r)=V_0/\ell+V_1/\ell^2+\cdots$ , where $V_0$ is a constant defined by $\phi_0a^2=2V_0\int_b^ar\,{\rm d} r$ , which gives $V_0=\phi_0a^2/(a^2-b^2)$ . Substitution in (2.22) then gives

\begin{equation*} 1\simeq \frac{2\ell}{\phi_0a^2}\int_b^a \left(\frac{V_0}{\ell}+\frac{V_1}{\ell^2}\right)s\,{\rm d} s +\frac{2\ell}{\phi_0a^2}\int_b^a \frac{V_0a}{\ell^2} \sum_{n=1}^\infty \frac{J_0(\lambda_nr)J_0(\lambda_ns)} {(\lambda_na) \,J_0^2(\lambda_na)} s\,{\rm d} s,\quad b<r<a.\end{equation*}

Rearranging

\begin{align*} c_0&=1 -\frac{2V_0}{\phi_0a\ell} \sum_{n=1}^\infty \frac{J_0(\lambda_nr)} {(\lambda_na) \,J_0^2(\lambda_na)} \int_b^a J_0(\lambda_ns)s\,{\rm d} s\\[3pt] &=1 +\frac{2ba^2}{\ell(a^2-b^2)} \sum_{n=1}^\infty \frac{J_0(\lambda_nr)J_1(\lambda_nb)} {(\lambda_na)^2 \,J_0^2(\lambda_na)}.\end{align*}

But the right-hand side is a function of r, so average over the annulus: multiply by $2r/(a^2-b^2)$ and integrate between $r=b$ and $r=a$ :

(2.24) \begin{align}c_0&=1 +\frac{4ba^2}{\ell(a^2-b^2)^2} \sum_{n=1}^\infty\frac{J_1(\lambda_nb)} {(\lambda_na)^2 \,J_0^2(\lambda_na)}\int_b^aJ_0(\lambda_nr)r\,{\rm d} r\nonumber\\[3pt] &=1 -\frac{4b^2a^3}{\ell(a^2-b^2)^2} \sum_{n=1}^\infty\frac{J_1^2(\lambda_nb)} {(\lambda_na)^3 \,J_0^2(\lambda_na)}.\end{align}

The series seen here is ${\mathbb S}_0(b/a)$ , defined by (2.17). It occurred in our approximation to the blockage coefficient C for potential flow past a thin rigid disc in a tube of infinite length, $C_{{\rm disc}}^{{\rm app}}$ , given by (2.16). If we accept that approximation, we obtain the relation

(2.25) \begin{equation} c_0=1-C_{{\rm disc}}^{{\rm app}}/\ell\end{equation}

and then (2.19) gives an approximation for the flux ${\mathcal J}(\ell)$ .

2.3.3 Estimating $\kappa$

We have two estimates for ${\mathcal J}$ , namely

\begin{equation*} {\mathcal J}(\ell)=\frac{\pi a^2\phi_0}{\ell}\left(1-\frac{C_{{\rm disc}}^{{\rm app}}}{\ell}\right) \quad\mbox{and}\quad {\mathcal J}(\ell) =\frac{\pi a^2\phi_0\kappa}{1+\kappa\ell};\end{equation*}

see (2.5) for the second. Equating these and solving for $\kappa$ , we obtain

(2.26) \begin{equation} \kappa(\ell)=\frac{1}{C_{{\rm disc}}^{{\rm app}}} -\frac{1}{\ell}.\end{equation}

Recall that $C_{{\rm disc}}^{{\rm app}}$ pertains to an infinite tube whereas $\kappa(\ell)$ is for a finite tube of length $\ell$ . We see that the difference decays slowly as $\ell\to\infty$ .

The observation that $\kappa(\infty)=C^{-1}$ is to be expected. If we approximate $\phi(r,z)$ for an infinite pipe (with far field given by (2.6)) by a one-dimensional function $\psi(z)$ , we must have $\psi(z)=U(z+C)$ . Suppose the rigid object in the tube is confined to the region $|z|<h$ for some h, and then apply the condition $\psi'(z_0)=\kappa_\infty^0\psi(z_0)$ (compare with (1.2)), where $z_0\geq h$ ; we have written $\kappa_\infty^0$ to indicate that we are representing the solution in an infinite pipe with a boundary condition applied at $z_0$ . Doing this gives $\kappa_\infty^0=1/(C+z_0)$ . (For a thin disc, we can take $z_0=h=0$ .) This simple calculation shows that we should expect $\kappa$ to depend on the location at which the boundary condition $\psi'=\kappa\psi$ is imposed. Indeed, if we use a different location, using $\psi'(z_1)=\kappa_\infty^1\psi(z_1)$ , we obtain $\kappa_\infty^1=1/(C+z_1)$ and then, eliminating C, we obtain

(2.27) \begin{equation} \frac{1}{\kappa_\infty^0}-\frac{1}{\kappa_\infty^1}=z_0-z_1.\end{equation}

We conclude that, unlike the blockage coefficient C, $\kappa$ is not dependent solely on the shape of the rigid object in the tube.

2.4 Finite-tube problem: The one-dimensional approximation $\psi$

Let us construct $\psi$ from $\phi$ , using (1.1), which reduces to

\begin{equation*} \psi(z)=\frac{2}{a^2}\int_0^a \phi(r,z)r\,{\rm d} r=\phi_0\left\{1+c_0\left(\frac{z}{\ell}-1\right) \right\},\end{equation*}

an exact formula, after using (2.18). Evidently, $\psi(\ell)=\phi_0$ whereas imposition of $\psi'(0)=\kappa\psi(0)$ , (1.2), gives a relation between $\kappa$ and $c_0$ ,

(2.28) \begin{equation} \kappa=\frac{c_0}{\ell(1-c_0)}.\end{equation}

Now, the leading-order estimate for $c_0$ , (2.23), is $c_0\simeq 1$ , and so the exact formula (2.28) does not provide a useful estimate of $\kappa$ . However, if we use the refined estimate for $c_0$ , (2.25), in (2.28), we recover the formula (2.26).

3.1 Boundary homogenisation

As in Section 2.1, suppose we approximate $\phi(r,z)$ by a function $\psi(z)$ , with $\psi''(z)=0$ . Imposing $\psi(\ell_{{n}})=\phi_0$ and $\psi(-\ell_{{w}})=0$ gives

\begin{equation*} \psi(z)=A(z+\ell_{{w}}),\quad -\ell_{{w}}<z<0 \quad\mbox{and}\quad \psi(z)=\phi_0 +B(z-\ell_{{n}}),\quad 0<z<\ell_{{n}},\end{equation*}

where A and B are arbitrary constants. The flux is $\pi a^2 A$ for $-\ell_{{w}}<z<0$ and $\pi b^2 B$ for $0<z<\ell_{{n}}$ . For these to match, $a^2A=b^2B$ . Thus, in terms of A,

(3.4) \begin{equation} \psi'(0-)=A, \quad \psi'(0+)=(a/b)^2A,\quad \psi(0-)=A\ell_{{w}},\quad \psi(0+)=\phi_0-(a/b)^2A\ell_{{n}}.\end{equation}

We need one more condition to determine A.

In [Reference Berezhkovskii, Barzykin and Zitserman1, equation (2.3)], the following conditions are imposed at $z=0$ :

(3.5) \begin{equation} G_{{w}}^{\prime}(0-)=G_{{n}}^{\prime}(0+)=\kappa_{{n}} G_{{n}}(0+)-\kappa_{{w}} G_{{w}}(0-).\end{equation}

These two relations involve the functions $G_{{w}}$ and $G_{{n}}$ , and the parameters $\kappa_{{w}}$ and $\kappa_{{n}}$ .

The constants $\kappa_{{w}}$ and $\kappa_{{n}}$ are known as ‘boundary trapping rates’ [Reference Berezhkovskii, Barzykin and Zitserman1]. Thus $\kappa_{{w}}$ is the trapping rate for diffusing particles approaching $z=0$ from the wide part of the structure (that is, from $z<0$ ) and $\kappa_{{n}}$ is the rate for particles arriving from the narrow part. These two rates are related, as we shall see.

The first of (3.5) is identified in [Reference Berezhkovskii, Barzykin and Zitserman1] as flux conservation, so we put $G_{{w}}(z)=a^2\psi(z)$ for $-\ell_{{w}}<z<0$ and $G_{{n}}(z)=b^2\psi(z)$ for $0<z<\ell_{{n}}$ . The second of (3.5) then becomes

(3.6) \begin{equation} a^2\psi'(0-)=\kappa_{{n}} b^2\psi(0+)-\kappa_{{w}} a^2 \psi(0-).\end{equation}

In [Reference Berezhkovskii, Barzykin and Zitserman1, equation (2.5)], it is argued that the boundary trapping rates satisfy

(3.7) \begin{equation} \kappa_{{n}} = (a/b)^2\kappa_{{w}} =\kappa,\end{equation}

say. (The authors appeal to ‘the condition of detailed balance’ but an elementary argument suffices. If we replace the boundary condition (3.1e) by $\phi(r,-\ell_{{w}})=\phi_0$ , the exact solution is $\phi(r,z)=\phi_0$ with zero flux. The corresponding one-dimensional solution is $\psi(z)=\phi_0$ , but this solution will only satisfy (3.6) if (3.7) holds.) Using (3.7) reduces (3.6) to

(3.8) \begin{equation} a^2\psi'(0-)=b^2\kappa \left\{ \psi(0+)-\psi(0-)\right\}.\end{equation}

We notice that the right-hand side of (3.8) contains the discontinuity in $\psi$ across the junction at $z=0$ whereas $\phi(r,0+)=\phi(r,0-)$ for $0\leq r<b$ .

Substituting from (3.4) in (3.8), we find that $A=\kappa b^2\phi_0/(a^2+\kappa\Omega)$ and

(3.9) \begin{equation} {\mathcal J}=\pi a^2 A=\frac{\pi a^2b^2\phi_0}{a^2\kappa^{-1}+\Omega} \quad\mbox{with}\quad \Omega=a^2\ell_{{n}}+b^2\ell_{{w}}.\end{equation}

It is known [Reference Berezhkovskii, Barzykin and Zitserman1] that $\kappa\to\infty$ as $b\to a$ , and so the correct result is obtained in this limit.

3.2 Two semi-infinite tubes

Returning to problem (3.1), suppose we let $\ell_{{w}}\to\infty$ and $\ell_{{n}}\to\infty$ , and consider uniform flow along two coaxial semi-infinite tubes, joined together with a rigid annulus at $z=0$ . For this problem, the boundary conditions (3.1a) and (3.1e) are replaced by far-field conditions,

\begin{equation*} \phi(r,z)=\left\{\begin{array}{l@{\quad}l@{\quad}l} U_{{n}}(z+C)+o(1)\quad &\mbox{as $z\to\infty$,} &0\leq r<b,\\ \\[-6pt] Uz+o(1)\quad& \mbox{as $z\to-\infty$,} & 0\leq r<a. \end{array}\right.\end{equation*}

Flux conservation shows that the constants U and $U_{{n}}$ are related by $\pi a^2U=\pi b^2 U_{{n}}$ . The constant C is unknown; it is the blockage coefficient. As all boundary conditions are Neumann conditions, we are free to exclude an additive constant in the behaviour of $\phi(r,z)$ as $z\to -\infty$ .

If we apply Green’s theorem to $\phi(r,z)$ and z in the narrow tube, we obtain

\begin{equation*} U_{{n}} C\pi b^2=2\pi\int_0^b \phi(r,0+)r\,{\rm d} r.\end{equation*}

If we do the same in the wide tube, we obtain

\begin{equation*} 2\pi\int_0^a \phi(r,0-)r\,{\rm d} r =0.\end{equation*}

As $U_{{n}} b^2=Ua^2$ and $\phi(r,0+)=\phi(r,0-)$ for $0\leq r<b$ , we obtain

(3.10) \begin{equation} C= -\frac{2}{Ua^2}\int_b^a \phi(r,0-)r\,{\rm d} r.\end{equation}

Our goal is to obtain an approximation to C; (3.10) shows that we are mainly interested in $\phi$ in the wide tube. (Of course, a full solution would require matching to $\phi$ in the narrow tube through the aperture at $z=0$ ; see [Reference Yeh20] for some details.)

As the flow is axisymmetric, write $\phi$ in the wide tube as

(3.11) \begin{equation} \phi(r,z)=Uz +Ua\sum_{n=1}^\infty c_n J_0(\lambda_nr)\,{\rm e}^{\lambda_nz}, \quad z<0,\quad 0\leq r<a,\end{equation}

where $J_1(\lambda_na)=0$ , as before. Differentiating,

\begin{equation*} U+U\sum_{n=1}^\infty c_n(\lambda_na) J_0(\lambda_nr)=\left\{ \begin{array}{c@{\quad}l} V(r), & 0\leq r<b\\ \\[-7pt] 0, & b<r<a, \end{array}\right.\end{equation*}

where V(r) is the unknown speed in the aperture, $V(r) =\partial {\phi}/\partial {z}$ at $z=0$ , $0\leq r<b$ and we have used (3.12c). Orthogonality gives

(3.12) \begin{equation} \pi a^2U=2\pi\int_0^bV(r)r\,{\rm d} r\end{equation}

(which is flux conservation again) and

\begin{equation*} c_n =\frac{2}{U\lambda_na^3\,J_0^2(\lambda_na)}\int_0^b V(r)J_0(\lambda_nr)r\,{\rm d} r.\end{equation*}

Also, substituting (3.11) in (3.10) gives

(3.13) \begin{align}C&= -\frac{2}{a}\sum_{n=1}^\infty c_n \int_b^aJ_0(\lambda_nr)r\,{\rm d} r= \frac{2b}{a}\sum_{n=1}^\infty \frac{c_n}{\lambda_n} J_1(\lambda_nb) \nonumber\\[3pt] &=\frac{4b}{Ua^2}\sum_{n=1}^\infty \frac{J_1(\lambda_nb)}{(\lambda_na)^2 J_0^2(\lambda_na)}\int_0^b V(r)J_0(\lambda_nr)r\,{\rm d} r, \end{align}

using (2.12). The formula (3.13) is exact, but we do not know V(r).

To proceed, we approximate and try $V(r)=U_q(b^2-r^2)^{-q}$ , with $0\leq q<1$ ; conservation gives $U_q=(1-q)a^2b^{2q-2}U$ . A local analysis near the corner at $(r,z)=(b,0)$ gives $q=\frac{1}{3}$ [Reference Lamb12, Section 63]. Note that $q=0$ would mean that V(r) has been approximated by a constant whereas $q=\frac{1}{2}$ would be appropriate when two identical semi-infinite tubes are joined with a rigid iris (a thin screen with a hole of radius b). Then (3.13) gives

(3.14) \begin{align}C \simeq C_{{\rm tube}}^{{\rm app}}&= \frac{4bU_q}{Ua^2}\sum_{n=1}^\infty \frac{J_1(\lambda_nb)}{(\lambda_na)^2 J_0^2(\lambda_na)}\int_0^b (b^2-r^2)^{-q} J_0(\lambda_nr)r\,{\rm d} r \nonumber\\&=2^{2-q} b \,\Gamma(2-q)\,(b/a)^{q-1}\,{\mathbb S}_q(b/a),\end{align}

where

(3.15) \begin{equation} {\mathbb S}_q(X)= \sum_{n=1}^\infty \frac{J_1(\lambda_naX)\,J_{1-q}(\lambda_naX)}{(\lambda_na)^{3-q}\,J_0^2(\lambda_na)} =\sum_{n=1}^\infty \frac{J_1(j_nX)\,J_{1-q}(j_nX)}{j_n^{3-q}J_0^2(j_n)},\end{equation}

$j_n=\lambda_na$ and we have used [Reference Gradshteyn and Ryzhik7, 6.567.1]

(3.16) \begin{equation} \int_0^b (b^2-r^2)^{-q} J_0(\lambda_nr)r\,{\rm d} r =\frac{b^{2-2q}\,\Gamma(1-q)}{2^q (\lambda_nb)^{1-q}}\,J_{1-q}(\lambda_nb).\end{equation}

Note that the series ${\mathbb S}_0$ appeared earlier; see (2.17). Asymptotic approximations to C for $b/a\ll 1$ are developed in Appendix A.

3.3 Junction problem: Estimating $\kappa$

Let us return to the junction problem (3.1).

3.3.1 The wide tube

In the wide tube, we can write

(3.17) \begin{equation} \frac{\phi(r,z)}{\phi_0}=c_0\left(\frac{z}{\ell_{{w}}}+1\right) +\sum_{m=1}^\infty c_m J_0(\lambda_mr)\sinh{\{\lambda_m(z+\ell_{{w}})\}}\end{equation}

for $-\ell_{{w}}<z<0$ and $0\leq r<a$ , where $J_1(\lambda_ma)=0$ . This representation satisfies (3.1d) and (3.1e). At $z=0$ , we have

(3.18) \begin{equation} c_0+\sum_{m=1}^\infty c_m J_0(\lambda_mr)\sinh{\lambda_m\ell_{{w}}}=\frac{\varphi(r)}{\phi_0}, \quad 0\leq r<b,\end{equation}

where $\varphi(r)=\phi(r,0)$ , $0\leq r<b$ ; $\varphi(r)$ is the unknown potential in the aperture. Differentiating (3.17),

(3.19) \begin{equation} c_0+\sum_{m=1}^\infty c_m J_0(\lambda_mr)(\lambda_m\ell_{{w}})\cosh{\lambda_m\ell_{{w}}} =\left\{\begin{array}{cl} (\ell_{{w}}/\phi_0) V(r), &0\leq r<b, \\ \\[-7pt] 0, & b<r<a, \end{array}\right.\end{equation}

where $V(r)=\partial {\phi}/\partial {z}$ at $z=0$ , $0\leq r<b$ and we have used (3.1c). Orthogonality gives

(3.20) \begin{equation} c_0=\frac{2\ell_{{w}}}{\phi_0a^2}\int_0^b V(r)r\,{\rm d} r =\frac{\ell_{{w}}}{\phi_0\pi a^2}\,{\mathcal J}\end{equation}

using (3.2) and, using (2.15),

\begin{equation*}c_m \lambda_m\ell_{{w}}\cosh{\lambda_m\ell_{{w}}}=\frac{2\ell_{{w}}}{\phi_0a^2 J_0^2(\lambda_ma) } \int_0^b V(r) J_0(\lambda_mr) r\,{\rm d} r.\end{equation*}

If we eliminate $c_m$ from (3.17) and assume that V(r) is constant, we obtain [Reference Keller and Stein11, equation (3)]. Instead of doing that, we eliminate $c_m$ from (3.18), giving an equation relating two unknown functions, V and $\varphi$ ,

(3.21) \begin{equation} \frac{2\ell_{{w}}}{a^2}\int_0^b V(s) \bigg\{ 1 + \frac{a}{\ell_{{w}}}\sum_{m=1}^\infty \frac{J_0(\lambda_mr) J_0(\lambda_ms)}{ (\lambda_ma) J_0^2(\lambda_ma)}\tanh{\lambda_m\ell_{{w}}} \bigg\} s\,{\rm d} s =\varphi(r), \quad 0\leq r<b.\end{equation}

3.3.2 The narrow tube

Let us make a similar calculation for the narrow tube, starting with

(3.22) \begin{equation} \frac{\phi(r,z)}{\phi_0}=1+d_0\left(\frac{z}{\ell_{{n}}}-1\right) +\sum_{m=1}^\infty d_m J_0(\mu_m r)\sinh{\{\mu_m(z-\ell_{{n}})\}},\end{equation}

for $0<z<\ell_{{n}}$ and $0\leq r<b$ , where $\mu_m$ are positive solutions of $J_1(\mu_mb)=0$ and the coefficients $d_m$ are to be determined. This representation for $\phi$ ensures that (3.1a) and (3.1b) are satisfied. At $z=0$ ,

(3.23) \begin{equation} 1-d_0-\sum_{m=1}^\infty d_m J_0(\mu_mr)\sinh{\mu_m \ell_{{n}}}=\frac{\varphi(r)}{\phi_0}, \quad 0\leq r<b.\end{equation}

Differentiating (3.22),

(3.24) \begin{equation} d_0 +\sum_{m=1}^\infty d_m J_0(\mu_mr) (\mu_m\ell_{{n}})\cosh{\mu_m\ell_{{n}}} =(\ell_{{n}}/\phi_0) V(r), \quad 0\leq r<b.\end{equation}

From this equation, we obtain

(3.25) \begin{equation} d_0=\frac{2 \ell_{{n}}}{\phi_0b^2} \int_0^b V(r)r\,{\rm d} r =\frac{\ell_{{n}}}{\phi_0\pi b^2}\,{\mathcal J}\end{equation}

and, using (2.15),

\begin{equation*}d_m \mu_m\ell_{{n}}\cosh{\mu_m\ell_{{n}}}=\frac{2\ell_{{n}}}{\phi_0b^2 J_0^2(\mu_mb)}\int_0^b V(r) J_0(\mu_mr)r\,{\rm d} r.\end{equation*}

Eliminating $d_m$ from (3.23),

(3.26) \begin{equation} -\frac{2 \ell_{{n}}}{b^2} \int_0^b V(s) \bigg\{1 + \frac{b}{\ell_{{n}}} \sum_{m=1}^\infty\frac{ J_0(\mu_mr) J_0(\mu_ms) }{(\mu_mb) J_0^2(\mu_mb)} \tanh{\mu_m\ell_{{n}}} \bigg\}s\,{\rm d} s =\varphi(r)-\phi_0 \quad 0\leq r<b.\end{equation}

Chapman and Parker [Reference Chapman and Parker3] eliminate $\varphi$ between (3.18) and (3.23) and V between (3.19) and (3.24), giving a set of dual series equations for $c_m$ and $d_m$ . They solve these equations numerically, with a focus on computing $c_0$ , which is related to the flux by (3.20). See also [Reference Yeh20, Section III.B].

3.3.3 An integral equation for V

Eliminating $\varphi$ between (3.21) and (3.26), we obtain

(3.27) \begin{align}& \frac{2\ell_{{w}}}{a^2}\int_0^b V(s) \bigg\{ 1 + \frac{a}{\ell_{{w}}}\sum_{m=1}^\infty \frac{J_0(\lambda_mr) J_0(\lambda_ms)}{ (\lambda_ma) J_0^2(\lambda_ma)}\tanh{\lambda_m\ell_{{w}}} \bigg\} s\,{\rm d} s \nonumber\\[3pt] & \quad\mbox{} +\frac{2 \ell_{{n}}}{b^2} \int_0^b V(s) \bigg\{1 + \frac{b}{\ell_{{n}}} \sum_{m=1}^\infty\frac{ J_0(\mu_mr) J_0(\mu_ms) }{(\mu_mb) J_0^2(\mu_mb)} \tanh{\mu_m\ell_{{n}}} \bigg\}s\,{\rm d} s =\phi_0,\end{align}

which is an exact integral equation for V(r), $0\leq r<b$ . It is unlikely that analytical solutions can be found, so we seek approximations.

3.3.4 Approximation for long tubes

When $\ell_{{w}}/a$ and $\ell_{{n}}/b$ are large, we can replace both hyperbolic tangents in (3.27) by 1. Moreover, at leading order, we can discard the infinite series because they are multiplied by $a/\ell_{{w}}$ or $b/\ell_{{n}}$ . Doing this leaves the simple approximation

(3.28) \begin{equation}\left(\frac{2\ell_{{w}}}{a^2}+\frac{2\ell_{{n}}}{b^2}\right)\int_0^bV(s)s\,{\rm d} s=\phi_0\end{equation}

which gives

(3.29) \begin{equation} {\mathcal J}\simeq \pi a^2b^2\phi_0/\Omega \quad\mbox{where}\quad \Omega=a^2\ell_{{n}}+b^2\ell_{{w}}\end{equation}

appeared in (3.9). Also, comparing (3.29) with (3.20) gives

(3.30) \begin{equation} c_0\simeq b^2\ell_{{w}}/\Omega.\end{equation}

To improve our approximation, we proceed as in Section 2.3.2 and use an iterative approach. Suppose that $V=V_0\ell_{{w}}^{-1}+V_1\ell_{{w}}^{-2}+\cdots$ , where $V_0$ is chosen so that $V(s)=V_0(s)\ell_{{w}}^{-1}$ satisfies (3.28). If we choose $V_0(r)=W_q(b^2-r^2)^{-q}$ , we obtain $W_q=\phi_0\ell_{{w}} a^2b^{2q}(1-q)/\Omega$ . Hence, we find that $V_1$ satisfies

\begin{align*}& \frac{2\ell_{{w}}}{a^2}\int_0^b \bigg\{ \frac{V_1(s)}{\ell_{{w}}^2} + \frac{aV_0(s)}{\ell_{{w}}^2}\sum_{m=1}^\infty \frac{J_0(\lambda_mr) J_0(\lambda_ms)}{ (\lambda_ma) J_0^2(\lambda_ma)} \bigg\} s\,{\rm d} s \\[3pt] &\quad\mbox{}+ \frac{2 \ell_{{n}}}{b^2} \int_0^b \bigg\{\frac{V_1(s)}{\ell_{{w}}^2} + \frac{bV_0(s)}{\ell_{{n}}\ell_{{w}}} \sum_{m=1}^\infty\frac{ J_0(\mu_mr) J_0(\mu_ms) }{(\mu_mb) J_0^2(\mu_mb)} \bigg\}s\,{\rm d} s =0,\quad 0\leq r<b.\end{align*}

Using (3.16), we have

\begin{equation*} \int_0^b V_0(s)\,J_0( \lambda s)s\,{\rm d} s=\frac{\phi_0\ell_{{w}} a^2b^2 \,\Gamma(2-q)}{2^q\,\Omega\,(\lambda b)^{1-q}}\,J_{1-q}(\lambda b)\end{equation*}

whence

\begin{align*} \frac{2}{\ell_{{w}}^2}\left(\frac{\ell_{{w}}}{a^2}+\frac{\ell_{{n}}}{b^2}\right)\int_0^b V_1(s)s\,{\rm d} s&=-\frac{\phi_0a^2b (b/a)^q\Gamma(2-q)}{2^{q-1}\,\Omega} \sum_{m=1}^\infty \frac{J_0(\lambda_mr)\,J_{1-q}(\lambda_mb)}{(\lambda_ma)^{2-q}J_0^2(\lambda_ma)}\\[3pt] &\quad\mbox{} - \frac{\phi_0a^2b \,\Gamma(2-q)}{2^{q-1}\,\Omega} \sum_{m=1}^\infty \frac{J_0(\mu_mr)\,J_{1-q}(\mu_mb)}{(\mu_ma)^{2-q}J_0^2(\mu_ma)}.\end{align*}

But, as the right-hand side depends on r, we average over the disc: multiply by $2r/b^2$ and integrate between $r=0$ and $r=b$ , using

\begin{equation*} \int_0^b J_0(\lambda_mr)r\,{\rm d} r=\frac{b}{\lambda_m}J_1(\lambda_mb) \quad\mbox{and}\quad \int_0^b J_0(\mu_mr)r\,{\rm d} r =0.\end{equation*}

Doing this gives

\begin{equation*} \frac{2}{\ell_{{w}}^2}\left(\frac{\ell_{{w}}}{a^2}+\frac{\ell_{{n}}}{b^2}\right)\int_0^b V_1(s)s\,{\rm d} s =-\frac{\phi_0a^3 (b/a)^q\Gamma(2-q)}{2^{q-2}\,\Omega} \,{\mathbb S}_q(b/a),\end{equation*}

where ${\mathbb S}_q(X)$ is defined by (3.15). For the flux, we obtain

\begin{align*} {\mathcal J}&=2\pi\int_0^bV(s)s\,{\rm d} s=2\pi\int_0^b \left(\frac{V_0}{\ell_{{w}}}+\frac{V_1}{\ell_{{w}}^2}\right)s\,{\rm d} s\\[3pt] & =\frac{\pi a^2 b^2\phi_0}{\Omega}\left(1 -\frac{a^3(b/a)^q}{2^{q-2}\Omega}\,\Gamma(2-q)\,{\mathbb S}_q(b/a) \right),\end{align*}

where $\Omega$ is defined by (3.29). The sum appearing here also appears in a comparable approximation for the blockage coefficient C for uniform flow along two coaxial tubes joined together, $C_{{\rm tube}}^{{\rm app}}$ , defined by (3.14). Using this, we obtain

(3.31) \begin{equation} {\mathcal J} =\frac{\pi a^2\phi_0}{\ell_{{w}}}\,c_0\simeq \frac{\pi a^2 b^2\phi_0}{\Omega}\left(1 -\frac{a^2C_{{\rm tube}}^{{\rm app}}}{\Omega} \right).\end{equation}

3.3.5 Estimating $\kappa$

We have two estimates for ${\mathcal J}$ , namely (3.9) and (3.31). Equating these gives

\begin{equation*}\frac{1}{a^2\kappa^{-1}+\Omega} =\frac{1}{\Omega}\left(1 -\frac{a^2C_{{\rm tube}}^{{\rm app}}}{\Omega} \right).\end{equation*}

Solving for $\kappa$ , we obtain

(3.32) \begin{equation} \kappa=\frac{1}{C_{{\rm tube}}^{{\rm app}}}-\frac{a^2}{a^2\ell_{{n}}+b^2\ell_{{w}}}.\end{equation}

Again, we see the slow decay of $\kappa$ with increasing tube lengths.

If we combine the leading-order estimate for $\kappa$ with (3.7), we obtain

\begin{equation*} \kappa_{{w}} = \frac{b^2}{a^2}\kappa \simeq \frac{b^2}{a^2C_{{\rm tube}}^{{\rm app}}}. \end{equation*}

For small $b/a$ , we can use the asymptotic approximation (A3), giving

(3.33) \begin{equation} \kappa_{{w}}\simeq \frac{bA}{a^2}\quad\mbox{with}\quad A=\left(\frac{3}{2}-q\right)\!\left(\frac{\Gamma(\frac{3}{2}-q)}{\Gamma(2-q)}\right)^2,\end{equation}

where it was assumed that V(r) is proportional to $(b^2-r^2)^{-q}$ with $0\leq q<1$ . The proper choice for q is $q=\frac{1}{3}$ ; this gives $A\simeq 1.23$ .

If we take $q=0$ (so that V(r) is approximated by a constant, the average flux into the narrow pipe), we obtain $A=\frac{3}{8}\pi\simeq 1.18$ .

If we take $q=\frac{1}{2}$ , we obtain $A=4/\pi\simeq 1.27$ . This is the limiting value in [Reference Berezhkovskii, Barzykin and Zitserman1, equation (2.4a)]. The inverse square-root behaviour at $r=b$ implies that the underlying flow problem has been replaced by the problem of flow through a hole in a thin rigid screen, leading to an error that can be estimated using (3.33). (For details and references on this ‘iris problem’, see [Reference Martin and Skvortsov14, Reference Martin and Skvortsov15].) Qualitatively, this is a mistake because it ignores the presence of the narrow tube; however, quantitatively, the values of A for $q=\frac{1}{3}$ ( $A\simeq 1.23$ ) and $q=\frac{1}{2}$ ( $A\simeq 1.27$ ) are close.

Of course, the assumption that V(r) is proportional to $(b^2-r^2)^{-q}$ is itself an approximation: in principle, the exact solution for V could be found by solving the integral equation (3.27), but this seems to be difficult analytically without further approximations. Numerical solutions could be sought, but doing this would be expensive (and it would be outside the scope of this paper).

3.4 Junction problem: The one-dimensional approximation $\boldsymbol\psi$

Let us construct $\psi$ from $\phi$ , using (1.1), which reduces to

(3.34a) \begin{align} \psi(z)&=\frac{2}{a^2}\int_0^a \phi(r,z)r\,{\rm d} r=\phi_0c_0\left(\frac{z}{\ell_{{w}}}+1\right), \qquad -\ell_{{w}}<z<0, \end{align}

(3.34b) \begin{align} \psi(z)&=\frac{2}{b^2}\int_0^b \phi(r,z)r\,{\rm d} r=\phi_0\left\{1+d_0\left(\frac{z}{\ell_{{n}}}-1\right) \right\}, \quad 0<z<\ell_{{n}},\end{align}

after use of (3.17) and (3.22). These satisfy $\psi(-\ell_{{w}})=0$ and $\psi(\ell_{{n}})=\phi_0$ . Flux continuity, $a^2\psi'(0-)=b^2\psi'(0+)$ gives $a^2c_0/\ell_{{w}}=b^2d_0/\ell_{{n}}$ , which is seen to hold; compare (3.20) and (3.25). Then, applying the condition (3.8) at $z=0$ gives

\begin{equation*} a^2c_0=\kappa b^2\ell_{{w}}(1-c_0-d_0)=\kappa(b^2\ell_{{w}}-\Omega c_0),\end{equation*}

which gives an exact formula for $\kappa$ ,

(3.35) \begin{equation} \kappa=\frac{a^2c_0}{b^2\ell_{{w}}-\Omega c_0}.\end{equation}

Now, the leading-order estimate for $c_0$ , (3.30), is $c_0\simeq b^2\ell_{{w}}/\Omega$ , and so (3.35) does not give a useful estimate of $\kappa$ . However, if we use the refined estimate for $c_0$ , (3.31), we recover (3.32).

As ${\mathcal J}=\pi a^2\phi_0c_0/\ell_{{w}}=\pi b^2\phi_0d_0/\ell_{{n}}$ , we can write (3.34) as

\begin{align*}& \pi a^2\psi(z)={\mathcal J}(z+\ell_{{w}}),\qquad -\ell_{{w}}<z<0,\\[3pt] & \pi b^2\psi(z)={\mathcal J}(z-\ell_{{n}} +\ell_{{n}}/d_0),\quad 0<z<\ell_{{n}}.\end{align*}

Kalinay and Percus [Reference Kalinay and Percus10, equation (39)] obtained similar formulas. However, their analysis of (3) does not take account of finite tube lengths and does not gives estimates of $\kappa$ .