Flux Ratios and Channel Structures

Abstract

We investigate Ussing’s unidirectional fluxes and flux ratios of charged tracers motivated particularly by the insightful proposal of Hodgkin and Keynes on a relation between flux ratios and channel structure. Our study is based on analysis of quasi-one-dimensional Poisson–Nernst–Planck type models for ionic flows through membrane channels. This class of models includes the Poisson equation that determines the electrical potential from the charges present and is in that sense consistent. Ussing’s flux ratios generally depend on all physical parameters involved in ionic flows, particularly, on bulk conditions and channel structures. Certain setups of ion channel experiments result in flux ratios that are universal in the sense that their values depend on bulk conditions but not on channel structures; other setups lead to flux ratios that are specific in the sense that their values depend on channel structures too. Universal flux ratios could serve some purposes better than specific flux ratios in some circumstances and worse in other circumstances. We focus on two treatments of tracer flux measurements that serve as estimators of important properties of ion channels. The first estimator determines the flux of the main ion species from measurements of the flux of its tracer. Our analysis suggests a better experimental design so that the flux ratio of the tracer flux and the main ion flux is universal. The second treatment of tracer fluxes concerns ratios of fluxes and experimental setups that try to determine some properties of channel structure. We analyze the two widely used experimental designs of estimating flux ratios and show that the most widely used method depends on the spatial distribution of permanent charge so this flux ratio is specific and thus allows estimation of (some of) the properties of that permanent charge, even with ideal ionic solutions. The work presented in this paper is a first step showing how measurements of fluxes and flux ratios can give important insights into channel structure and function.

Appendix: Derivations of Results in Sect. 5.2

In this section, we provide a derivation of formula (5.14) in Theorem 5.1 and formula (5.18) in Theorem 5.2 from the basic formula (4.3) in Theorem 4.1 for flux ratios from the one-isotope-setup (Setup 1). The derivation for both formulas are similar so we will provide the detailed derivation for formula (5.14) in Sect. 7.2 and comment on differences in the derivation for formula (5.18) in Sect. 7.3.

1.1 A Reformulation of (4.3) for Cases in Sect. 5.2

We will first recall the setup of the case specified in Sect. 5.2 and give a new form of formula (4.3) for the special case, which is convenient for us to apply previously established results in late part.

Recall that, for the one-isotope-setup (Setup 1), one tracer is used for two experimental settings with two different sets of boundary conditions for tracers. For one experiment, the tracer is injected from the left boundary with concentration $L_1^{[t,i]}=\rho$ and zero right boundary concentration $R_1^{[t,i]}=0$ that produces the influx $J_1^{[t,i]}$. In another experiment, the same tracer is injected from the right boundary with concentration $R_1^{[t,o]}=\rho$ and zero left boundary concentration $L_1^{[t,o]}=0$ that produces the efflux $J_1^{[t,o]}$.

Formula (4.3) in Theorem 4.1 for the flux ratio between $J_1^{[t,i]}$ and $J_1^{[t,o]}$ is

$$\begin{aligned} \begin{aligned} \frac{J_1^{[t,i]}}{J_1^{[t,o]}} =-\frac{L_1^{[t,i]}e^{p_1^{[i]}(0;z_1,d_1)}}{R_1^{[t,o]}e^{p_1^{[o]}(1;z_1,d_1)}}\; \int _0^1\frac{e^{p_1^{[o]}(x;z_1,d_1)}}{D_1(x)h(x)}dx\;\Big (\int _0^1\frac{e^{p_1^{[i]}(x;z_1,d_1)}}{D_1(x)h(x)}dx\Big )^{-1}, \end{aligned} \end{aligned}$$

(7.1)

where $p_1^{[i]}(x;z_1,d_1)$ defined in (2.7) is determined by the profiles of concentrations from the solution of BVP associated with the boundary condition (BVi), and $p_1^{[o]}(x;z_1,d_1)$ defined in (2.7) is determined by the profiles of concentrations from the solution of BVP associated with the boundary condition (BVo).

Remark 7.1

Note that, as $\rho \rightarrow 0$, one has $p_1^{[i]}(x;z_1,d_1)=p_1^{[o]}(x;z_1,d_1)$, and hence,

$$\begin{aligned} \lim _{\rho \rightarrow 0}\frac{J_1^{[t,i]}}{J_1^{[t,o]}} =-\frac{L_1^{[t,i]}e^{p_1^{[i]}(0;z_1,d_1)}}{R_1^{[t,o]}e^{p_1^{[o]}(1;z_1,d_1)}}; \end{aligned}$$

in particular, this limit does not contain information on permanent charge Q. An technical implication is that all terms involving $Q_0$ in the approximation of ${J_1^{[t,i]}}/{J_1^{[t,o]}}$ should have a factor of $\rho$. This explains why there is no $dQ_0$-term in the leading order expansions of ${J_1^{[t,i]}}/{J_1^{[t,o]}}$ in formula (5.14) in Theorem 5.1.

For the case studied in Sect. 5.2, we are able to obtain an approximation for the flux ratio, up to leading orders, explicitly in terms of given quantities of the problem.

It follows from (3.1) in Proposition 3.1 that

$$\begin{aligned} \Big (\int _0^1\frac{e^{p_1^{[i]}(x;z_1,d_1)}}{D_1(x)h(x)}dx\Big )^{-1} =\frac{J_1^{[i]}}{(L_1+L_1^{[t,i]})e^{p_1^{[i]}(0;z_1,d_1)}-R_1e^{p_1^{[i]}(1;z_1,d_1)}}, \end{aligned}$$

where $J_1^{[i]}=J_1^{[m,i]}+J_1^{[t,i]}$ is the sum of the fluxes of the main ion species and its tracer. Similarly,

$$\begin{aligned} \int _0^1\frac{e^{p_1^{[o]}(x;z_1,d_1)}}{D_1(x)h(x)}dx=\frac{L_1e^{p_1^{[o]}(0;z_1,d_1)}-(R_1+R_1^{[t,o]})e^{p_1^{[o]}(1;z_1,d_1)}}{J_1^{[o]}}, \end{aligned}$$

where $J_1^{[o]}=J_1^{[m,o]}+J_1^{[t,o]}$ is the sum of the fluxes of the main ion species and its tracer. Therefore,

$$\begin{aligned} \frac{J_1^{[t,i]}}{J_1^{[t,o]}}=-(I)\cdot (II)\cdot \frac{J_1^{[i]}}{J_1^{[o]}} \end{aligned}$$

(7.2)

where the factors (I) and (II) are given by

$$\begin{aligned} (I)&=\frac{L_1^{[t,i]}e^{p_1^{[i]}(0;z_1,d_1)}}{R_1^{[t,o]}e^{p_1^{[o]}(1;z_1,d_1)}},\nonumber \\ (II)&= \frac{L_1e^{p_1^{[o]}(0;z_1,d_1)}-(R_1+R_1^{[t,o]})e^{p_1^{[o]}(1;z_1,d_1)}}{(L_1+L_1^{[t,i]})e^{p_1^{[i]}(0;z_1,d_1)}-R_1e^{p_1^{[i]}(1;z_1,d_1)}}. \end{aligned}$$

(7.3)

The factors (I) and (II) on the right-hand side of (7.2) are determined by boundary conditions and are independent of the permanent charge. We will first determine these two factors and determine the last factor ${J_1^{[i]}}/{J_1^{[o]}}$ afterwards.

1.2 Derivation of Formula (5.14)

For this formula, the electroneutrality boundary conditions (5.13) are assumed to hold among all ion species including the tracer. It follows from (5.6) that, for the boundary condition associated to (BVi) in (5.13),

$$\begin{aligned} p_1^{[i]}(0;z_1,d_1)&=z_1\zeta _0V_0+ d\Big (2(L_1+\rho )+(1+\lambda ) (L_2-z_1\rho /{z_2})\Big ) +O(d^2),\\&=z_1\zeta _0V_0+ d\Big (2L_1+(1+\lambda ) L_2\Big ) +O(d^2,d\rho ),\\ p_1^{[i]}(1;z_1,d_1)&= d\Big (2R_1+(1+\lambda ) R_2\Big ) +O(d^2); \end{aligned}$$

and, for the boundary condition associated to (BVo) in (5.13),

$$\begin{aligned} p_1^{[o]}(0;z_1,d_1) =&\,z_1\zeta _0V_0+ d(2L_1+(1+\lambda ) L_2) +O(d^2),\\ p_1^{[o]}(1;z_1,d_1) =&\, d(2(R_1+\rho )+(1+\lambda ) (R_2-z_1\rho /{z_2})) +O(d^2)\\ =&\, d(2R_1+(1+\lambda ) R_2) +O(d^2,d\rho ). \end{aligned}$$

Thus, the factor (I) on the right-hand side of (7.2) is

$$\begin{aligned} \begin{aligned} (I)=&\,e^{p_1^{[i]}(0;z_1,d_1)-p_1^{[o]}(1,z_1,d_1)}\\ =&\,e^{z_1\zeta _0V_0} \Big (1+\big (2 (L_1-R_1) +(1+\lambda ) (L_2- R_2) \big )d\Big ) +O(d^2,d\rho ), \end{aligned} \end{aligned}$$

(7.4)

and the factor (II) on the right-hand side of (7.2) is, with $s=L_1/{R_1}$,

$$\begin{aligned} (II)&=\frac{L_1e^{z_1\zeta _0V_0}-(R_1+\rho )+\left[ (2L_1^2+(1+\lambda )L_1L_2) e^{z_1\zeta _0V_0}-(2R_1^2+(1+\lambda ) R_1R_2) \right] d}{(L_1+\rho )e^{z_1\zeta _0V_0}-R_1 +\left[ (2L_1^2+(1+\lambda )L_1L_2)e^{z_1\zeta _0V_0} -(2R_1^2+(1+\lambda )R_1R_2)\right] d}\nonumber \\&\quad +O(\rho ^2, \rho d, d^2)\nonumber \\&= 1 -\frac{ e^{z_1\zeta _0V_0}+1}{se^{z_1\zeta _0V_0}-1}\frac{\rho }{R_1} +O(\rho ^2, \rho d, d^2). \end{aligned}$$

(7.5)

It remains to approximate the factor $J_1^{[i]}/{J_1^{[o]}}$ in (7.2) that depends on the permanent charge Q and hence the full profile of the electric potential and the ion concentrations. Its approximation is complicated even for the simple case we considered in Sect. 5. We start with a general expression

$$\begin{aligned} \frac{J_1^{[i]}}{J_1^{[o]}}=S_0(\rho )+S_1(\rho ) d+S_2(\rho ) Q_0+O(d^2, \rho d Q_0, \rho Q_0^2), \end{aligned}$$

and determine the quantities $S_k(\rho )$’s for $k=0,1,2$. In particular, we are interested in $S_0(\rho )$ up to $O(\rho )$ order, $S_1(\rho )=S_1(0)+O(\rho )$ due to the factor d in $S_1(\rho )d$, and $S_2(\rho )$ up to $O(\rho )$. The reason for the latter is, from Remark 7.1, any term involving $Q_0$ should have a factor $\rho$. The advantage of this expression is as follows. One can assume the ionic solution is ideal to get the term $S_0(\rho ) +S_2(\rho ) Q_0$ (Sect. 7.2.1) and then assume the ionic solution is nonideal but the permanent charge is zero to get $S_0(\rho )+S_1(\rho ) d$ (Sect. 7.2.2). The superposition of them then provides the approximation for $J_1^{[i]}/{J_1^{[o]}}$.

1.2.1 Case with Ideal Ionic Solution for $S_0(\rho ) +S_2(\rho ) Q_0$

We will apply results from [51] for the classical PNP for this case.

It follows from results in [51] (displays (4.4) and (4.5) in [51]) that the influx of the tracer for the boundary concentrations in (BVi) is

$$\begin{aligned} J_1^{[i]}=&J_{10}^{[i]}+J_{11}^{[i]}Q_0+O(Q_0^2), \end{aligned}$$

(7.6)

where

$$\begin{aligned} \begin{aligned} J_{10}^{[i]}&= \frac{D_1( L_1+\rho -R_1)}{H(1)(\ln (L_1+\rho )-\ln R_1)}\frac{ \mu _1^{\delta ,i}}{k_BT},\\ J_{11}^{[i]}&=\frac{D_1A^{[i]}(z_2(1-B^{[i]})\zeta _0 V_0 +\ln (L_1+\rho )-\ln R_1 )}{(z_1-z_2)H(1)\big (\ln (L_1+\rho )-\ln R_1\big )^2}\frac{ \mu _1^{\delta ,i}}{k_BT}, \end{aligned} \end{aligned}$$

(7.7)

and

$$\begin{aligned} \frac{1}{k_BT} \mu _1^{\delta ,i} :=\frac{1}{k_BT}\left( \mu ^{[i]}(0)-\mu ^{[i]}(1)\right) = z_1 \zeta _0V_0+ \ln (L_1+\rho )-\ln R_1, \end{aligned}$$

(7.8)

is the difference between the boundary electrochemical potentials for (BVi), and

$$\begin{aligned} \begin{aligned} A^{[i]}&=-\frac{(\beta -\alpha )(L_1+\rho -R_1)^2}{((1-\alpha )(L_1+\rho )+\alpha R_1)((1-\beta )(L_1+\rho )+\beta R_1)\ln \frac{L_1+\rho }{ R_1}}, \\ B^{[i]}&=\frac{1}{A^{[i]}}\ln \frac{ (1-\beta )(L_1+\rho )+\beta R_1}{(1-\alpha )(L_1+\rho )+\alpha R_1}. \end{aligned} \end{aligned}$$

(7.9)

Similarly, the efflux of the tracer for the boundary concentrations in (BVo) is

$$\begin{aligned} J_1^{[o]}= J_{10}^{[o]}+J_{11}^{[o]}Q_0+O(Q_0^2), \end{aligned}$$

(7.10)

where

$$\begin{aligned} \begin{aligned} J_{10}^{[o]}&= \frac{D_1( L_1-(R_1+\rho ))}{H(1)(\ln L_1-\ln (R_1+\rho ))}\frac{ \mu _1^{\delta ,o}}{k_BT}, \\ J_{11}^{[o]}&=\frac{D_1A^{[o]}(z_2(1-B^{[o]})\zeta _0 V_0 +\ln L_1-\ln (R_1+\rho ) ) }{(z_1-z_2)H(1)\big (\ln L_1-\ln (R_1+\rho )\big )^2}\frac{ \mu _1^{\delta ,o}}{k_BT}, \end{aligned} \end{aligned}$$

(7.11)

and

$$\begin{aligned} \frac{1}{k_BT} \mu _1^{\delta ,o}:=\frac{1}{k_BT}\left( \mu ^{[o]}(0)-\mu ^{[o]}(1)\right) = z_1 \zeta _0V_0+ \ln L_1-\ln (R_1+\rho ) \end{aligned}$$

(7.12)

is the differences between the boundary electrochemical potentials (BVo), and

$$\begin{aligned} \begin{aligned} A^{[o]}&=-\frac{(\beta -\alpha )(L_1-R_1-\rho )^2}{((1-\alpha )L_1+\alpha (R_1+\rho ))((1-\beta )L_1+\beta (R_1+\rho ))\ln \frac{L_1}{R_1+\rho }}, \\ B^{[o]}&=\frac{1}{A^{[o]}}\ln \frac{ (1-\beta )L_1+\beta (R_1+\rho )}{(1-\alpha )L_1+\alpha (R_1+\rho )}. \end{aligned} \end{aligned}$$

(7.13)

Therefore,

$$\begin{aligned} \begin{aligned} \frac{J_1^{[i]}}{J_1^{[o]}}&= \frac{J_{10}^{[i]}+J_{11}^{[i]}Q_0}{J_{10}^{[o]}+J_{11}^{[o]}Q_0}+O(Q_0^2)\\&= F_0\left( 1+\frac{F_1}{F_0}Q_0\right) +O(Q_0^2), \end{aligned} \end{aligned}$$

(7.14)

where

$$\begin{aligned} \begin{aligned} F_0&=\frac{J_{10}^{[i]}}{J_{10}^{[o]}} = \frac{ L_1+\rho -R_1}{ L_1-(R_1+\rho )} \frac{\ln L_1-\ln (R_1+\rho )}{\ln (L_1+\rho )-\ln R_1}\frac{ \mu _1^{\delta ,i}}{ \mu _1^{\delta ,o}},\\ \frac{F_1}{F_0}&=\frac{J_{11}^{[i]}}{J_{10}^{[i]}}-\frac{J_{11}^{[o]}}{J_{10}^{[o]}} =\frac{ A^{[i]}\Big (z_2(1-B^{[i]}) \zeta _0V_0 +\ln (L_1+\rho )-\ln R_1\Big )}{(z_1-z_2)(L_1+\rho -R_1) \big (\ln (L_1+\rho )-\ln R_1\big )}\\&\qquad -\frac{ A^{[o]}\Big (z_2(1-B^{[o]}) \zeta _0V_0 +\ln L_1-\ln (R_1+\rho )\Big )}{(z_1-z_2)(L_1-(R_1+\rho )) \big (\ln L_1-\ln (R_1+\rho )\big )}. \end{aligned} \end{aligned}$$

(7.15)

Lemma 7.2

With $s=L_1/{R_1}$, one has

$$\begin{aligned} F_0 = 1+ \left( f_1(s)+\frac{s+1}{s(z_1\zeta _0V_0+\ln s)}\right) \frac{\rho }{R_1}+O(\rho ^2), \end{aligned}$$

where $f_1(s)$ is defined in (5.8), and

$$\begin{aligned} \frac{F_1}{F_0}=g_1(V_0,s,\alpha ,\beta )\frac{\rho }{R_1^2}+O(\rho ^2), \end{aligned}$$

where $g_1(V_0,s,\alpha ,\beta )$ is given in (5.9). As a consequence,

$$\begin{aligned} S_0(\rho )=1+ \Big (f_1(s)+\frac{s+1}{s(z_1\zeta _0V_0+\ln s)}\Big )\frac{\rho }{R_1}\; \text{ and } \; S_2(\rho ) =g_1(V_0,s,\alpha ,\beta )\frac{\rho }{R_1^2}. \end{aligned}$$

(7.16)

Proof

For the term $F_0$, it follows (7.15), (7.8) and (7.12) that

$$\begin{aligned} F_0&= \frac{s-1+\frac{\rho }{R_1}}{s-1- \frac{\rho }{R_1}}\cdot \frac{\ln s-\ln (1+\frac{\rho }{R_1})}{\ln s+\ln (1+\frac{\rho }{sR_1})}\frac{z_1\zeta _0V_0+\ln (s+\frac{\rho }{R_1})}{z_1\zeta _0V_0+\ln s-\ln (1+\frac{\rho }{R_1})},\\&=1+ \left( \frac{2s\ln s-s^2+1}{s(s-1)\ln s}+\frac{s+1}{s(z_1\zeta _0V_0+\ln s)}\right) \frac{\rho }{R_1}+O(\rho ^2), \end{aligned}$$

which is the claimed formula for $F_0$.

The approximate formula for $F_1/F_0$ is lengthy but otherwise straightforward.

With $s=L_1/{R_1}$ and $L_1^{[t,i]}=R_1^{[t,o]}=\rho$, it follows from (7.15) that

$$\begin{aligned} \frac{F_1}{F_0}&=\frac{ z_2(A^{[i]}-A^{[i]}B^{[i]}) \zeta _0V_0 }{(z_1-z_2)R_1(s-1+\rho /{ R_1})\big (\ln s+ \rho / (sR_1)\big )}\\&\quad +\frac{A^{[i]}}{(z_1-z_2)R_1(s-1+\rho /{ R_1})}\\&\quad -\frac{ z_2(A^{[o]}-A^{[o]}B^{[o]}) \zeta _0V_0 }{(z_1-z_2)R_1(s-1-\rho /{ R_1})\big (\ln s- \rho /{ R_1}\big )}\\&\quad -\frac{A^{[o]}}{(z_1-z_2)R_1(s-1-\rho /{ R_1})} +O(\rho ^2). \end{aligned}$$

Next, we fix $A^{[i]}, B^{[i]}, A^{[o]}$ and $B^{[o]}$, and expand the above in $\rho$ to get

$$\begin{aligned} \frac{F_1}{F_0}&=\frac{ z_2 \zeta _0V_0 (A^{[i]}-A^{[i]}B^{[i]}-A^{[o]}+A^{[o]}B^{[o]}) }{(z_1-z_2)R_1(s-1) \ln s }\\&\quad +\frac{(s-1)(A^{[i]}-A^{[o]})}{(z_1-z_2)R_1(s-1)^2 }\\&\quad -\frac{ z_2 \zeta _0V_0[(A^{[i]}-A^{[i]}B^{[i]})(\ln s+1-1/s)+(A^{[o]}-A^{[o]}B^{[o]})(\ln s+s-1)]}{(z_1-z_2) (s-1 )^2(\ln s)^2} \frac{\rho }{ R_1^2}\\&\quad -\frac{A^{[i]}+A^{[o]}}{(z_1-z_2) (s-1)^2 }\frac{\rho }{ R_1^2} +O(\rho ^2). \end{aligned}$$

Now, we expand $A^{[i]}, A^{[o]}, B^{[i]}$ and $B^{[o]}$ defined in (7.9) and (7.13) in $\rho$ as

$$\begin{aligned} A^{[i]}&=A_0^{[i]}+A_1^{[i]}\frac{\rho }{R_1}+O(\rho ^2),\\ A^{[o]}&=A_0^{[o]}+A_1^{[o]}\frac{\rho }{R_1}+O(\rho ^2),\\ A^{[i]}B^{[i]}&=(A^{[i]}B^{[i]})_0+(A^{[i]}B^{[i]})_1\frac{\rho }{R_1}+O(\rho ^2),\\ A^{[o]}B^{[o]}&=(A^{[o]}B^{[o]})_0+(A^{[o]}B^{[o]})_1\frac{\rho }{R_1}+O(\rho ^2 ). \end{aligned}$$

It follows directly that

$$\begin{aligned} \begin{aligned} A_0^{[i]}&=A_0^{[o]}=-(\beta -\alpha )\frac{(s-1)^2}{[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]\ln s},\\ A_1^{[i]}&=-(\beta -\alpha )\frac{2(s-1) }{[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]\ln s}\\&\quad +(\beta -\alpha )\frac{(s-1)^2[2(1-\alpha )(1-\beta )s+\alpha +\beta -2\alpha \beta ] }{[(1-\alpha )s+\alpha ]^2[(1-\beta )s+\beta ]^2\ln s}\\&\quad +(\beta -\alpha )\frac{(s-1)^2 }{[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]s(\ln s)^2},\\ A_1^{[o]}&=(\beta -\alpha )\frac{2(s-1) }{[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]\ln s}\\&\quad +(\beta -\alpha )\frac{(s-1)^2[(\alpha +\beta -2\alpha \beta )s+ 2\alpha \beta ] }{[(1-\alpha )s+\alpha ]^2[(1-\beta )s+\beta ]^2\ln s}\\&\quad -(\beta -\alpha )\frac{(s-1)^2 }{[(1-\alpha )s+\alpha ][(1-\beta )t+\beta ] (\ln s)^2},\\ (A^{[i]}B^{[i]})_0&=(A^{[o]}B^{[o]})_0=\ln \frac{(1-\beta )s+\beta }{(1-\alpha )s+\alpha },\\ (A^{[i]}B^{[i]})_1&= \frac{1-\beta }{(1-\beta ) s+\beta } - \frac{1-\alpha }{(1-\alpha ) s+\alpha },\\ (A^{[o]}B^{[o]})_1&=\frac{\beta }{(1-\beta ) s+\beta } - \frac{\alpha }{(1-\alpha ) s+\alpha }. \end{aligned} \end{aligned}$$

(7.17)

In particular,

$$\begin{aligned} A_0^{[i]}-A_0^{[o]}&=0,\quad (A^{[i]}B^{[i]})_0-(A^{[o]}B^{[o]})_0=0,\\ A_0^{[i]}-(A^{[i]}B^{[i]})_0&= -\frac{(\beta -\alpha )(s-1)^2}{[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]\ln s}-\ln \frac{(1-\beta )s+\beta }{(1-\alpha )s+\alpha },\\ A_1^{[i]}-A_1^{[o]}&=-\frac{4(\beta -\alpha )(s-1) }{[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]\ln s}\\&\quad +\frac{(\beta -\alpha )(s-1)^2[(2-3\alpha -3\beta +4\alpha \beta )s+\alpha +\beta -4\alpha \beta ] }{[(1-\alpha )s+\alpha ]^2[(1-\beta )s+\beta ]^2\ln s}\\&\quad +\frac{ (\beta -\alpha )(s+1)(s-1)^2}{[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]s(\ln s)^2},\\ (A^{[i]}B^{[i]})_1-(A^{[o]}B^{[o]})_1&=\frac{1-2\beta }{(1-\beta ) s+\beta } - \frac{1-2\alpha }{(1-\alpha )s+\alpha }. \end{aligned}$$

Combining the above estimates, one has

$$\begin{aligned} \frac{F_1}{ F_0}&=-\frac{4(\beta -\alpha )z_2 \zeta _0V_0 }{(z_1-z_2) [(1-\alpha )s+\alpha ][(1-\beta )s+\beta ](\ln s)^2}\frac{\rho }{(R_1)^2}\\&\quad +\frac{(\beta -\alpha )z_2 \zeta _0V_0(s-1) [(2-3\alpha -3\beta +4\alpha \beta )s+\alpha +\beta -4\alpha \beta ] }{(z_1-z_2) [(1-\alpha )s+\alpha ]^2[(1-\beta )s+\beta ]^2(\ln s)^2}\frac{\rho }{(R_1)^2}\\&\quad +\frac{ (\beta -\alpha )z_2 \zeta _0V_0(s+1)(s-1)}{(z_1-z_2) [(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]s(\ln s)^3}\frac{\rho }{(R_1)^2}\\&\quad -\frac{ z_2 \zeta _0V_0 }{(z_1-z_2) (s-1)\ln s}\Big [\frac{1-2\beta }{(1-\beta )s+\beta } - \frac{1-2\alpha }{(1-\alpha )s+\alpha }\Big ]\frac{\rho }{(R_1)^2} \\&\quad +\frac{ (\beta -\alpha )z_2 \zeta _0V_0(2\ln s+s-1/s) }{(z_1-z_2)[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ] (\ln s)^3} \frac{\rho }{(R_1)^2} \\&\quad +\frac{ z_2 \zeta _0V_0(2\ln s+s-1/s)}{(z_1-z_2) (s-1 )^2(\ln s)^2} \ln \frac{(1-\beta )s+\beta }{(1-\alpha )s+\alpha }\frac{\rho }{(R_1)^2} \\&\quad -\frac{(\beta -\alpha ) [(\alpha +\beta -2\alpha \beta )(s-1)^2+2s] }{(z_1-z_2)[(1-\alpha )s+\alpha ]^2[(1-\beta )s+\beta ]^2\ln s} \frac{\rho }{(R_1)^2}\\&\quad +\frac{ (\beta -\alpha )(s+1)(s-1)}{(z_1-z_2)[(1-\alpha )s+\alpha ][(1-\beta )s+\beta ]s(\ln s)^2} \frac{\rho }{(R_1)^2} +O(\rho ^2). \end{aligned}$$

The fifth term above can be split and combine with the first and the third terms to get the formula for $F_1/{F_0}$ claimed in the statement of the lemma. $\square$

1.2.2 A Case with a Hard-Sphere Component for $S_0(\rho )+S_1(\rho )d$

In this section, we examine the flux ratio for PNP models with a hard-sphere component (point-charge with volume exclusion) and assume zero permanent charge.

Recall the local hard-sphere potential (5.5)

$$\begin{aligned} \frac{1}{k_BT}\mu _k^{LHS}(x)&= - \ln \Big (1-\sum _{j=1}^{n}d_jc_j(x)\Big )+ \frac{d_k\sum _{j=1}^nc_j(x)}{1-\sum _{j=1}^{n}d_jc_j(x)}\\&= \sum _{j=1}^{n}(d_j+d_k)c_j(x) +O(d_kd_j) \end{aligned}$$

where $d_j$ is the diameter of the jth ion species.

For two ion species with diameters $d_1$ and $d_2$, set $d=d_1$ and $d_2=\lambda d$, and expand the fluxes as $J_k=J_{k0}+J_{k1}d+O(d^2), k=1,2$. As observed in [29], the Nernst–Planck equations imply that the flux $J_i$ is proportional to the transmembrane electrochemical potential $\mu _i^{\delta }=\mu _i(0)-\mu _i(1)$ where

$$\begin{aligned} \begin{aligned} \frac{1}{k_BT}\mu _1^{\delta }&=z_1\zeta _0V_0 +\ln L_1-\ln R_1 \\&\quad +\frac{2z_2-(1+\lambda )z_1}{z_2} (L_1-R_1)d+O(d^2),\\ \frac{1}{k_BT}\mu _2^{\delta }&=z_2\zeta _0V_0 +\ln L_2-\ln R_2\\&\quad -\frac{(1+\lambda )z_2-2\lambda z_1}{z_1} (L_2-R_2)d+O(d^2). \end{aligned} \end{aligned}$$

(7.18)

It follows from results in [50, 54], for example, from Corollary 3.6 in [54] (with different but equivalent expressions) that the flux $J_1^{[i]}$ is given by

$$\begin{aligned} J_1^{[i]} = \left( 1+ \frac{2(\lambda z_1-z_2)}{z_1z_2} w_1^{[i]} d\right) \frac{D_1w_0^{[i]}}{z_1H(1)}\frac{\mu _1^{\delta ,i}}{k_BT} +O(d^2), \end{aligned}$$

where, with $\rho =L_1^{[t,i]}=R_1^{[t,o]}$,

$$\begin{aligned} \frac{1}{k_BT}\mu _1^{\delta ,i}&=z_1\zeta _0V_0 +\ln (L_1+\rho )-\ln R_1\\&\quad +\frac{2z_2-(1+\lambda )z_1}{z_2} (L_1+\rho - R_1)d+O(d^2) \end{aligned}$$

is the difference between the boundary electrochemical potentials for (BVi), and

$$\begin{aligned} w_0^{[i]}&=\frac{z_1(L_1+\rho - R_1)}{\ln (L_1+\rho )-\ln R_1}\; \text{ and } \\ w_1^{[i]}&= \frac{z_1(L_1+\rho - R_1)}{\ln (L_1+\rho )-\ln R_1}-\frac{z_1(L_1+\rho +R_1)}{2}. \end{aligned}$$

Similarly, the flux $J_1^{[o]}$ is given by

$$\begin{aligned} J_1^{[o]} = \left( 1+ \frac{2(\lambda z_1-z_2)}{z_1z_2} w_1^{[o]} d\right) \frac{D_1w_0^{[o]}}{z_1H(1)}\frac{\mu _1^{\delta ,o}}{k_BT} +O(d^2), \end{aligned}$$

where

$$\begin{aligned} \frac{1}{k_BT}\mu _1^{\delta ,o}= & {} z_1\zeta _0V_0 +\ln L_1-\ln (R_1+\rho )\\&+\frac{2z_2-(1+\lambda )z_1}{z_2} (L_1- R_1-\rho )d+O(d^2) \end{aligned}$$

is the difference between the boundary electrochemical potentials for (BVo), and

$$\begin{aligned} w_0^{[o]}= & {} \frac{z_1(L_1- R_1-\rho )}{\ln L_1-\ln (R_1+\rho )} \; \text{ and } \\ w_1^{[o]}= & {} \frac{z_1(L_1- R_1-\rho )}{\ln L_1-\ln (R_1+\rho )}-\frac{z_1(L_1+R_1+\rho )}{2}. \end{aligned}$$

Recall that $s=L_1/{R_1}$. One gets

$$\begin{aligned} \frac{1}{k_BT}\mu _1^{\delta ,i}&=z_1\zeta _0V_0 +\ln s +\frac{\rho }{sR_1} +\frac{2z_2-(1+\lambda )z_1}{z_2} (s-1) R_1d +O(\rho d, d^2),\\ w_0^{[i]}&= \frac{s-1}{\ln s}z_1R_1+\frac{s\ln s-(s-1)}{s(\ln s)^2} z_1\rho +O(\rho ^2),\\ w_1^{[i]}&=\frac{s-1}{\ln s}{z_1R_1}-\frac{s+1}{2}{z_1R_1}+\left( \frac{s\ln s-(s-1)}{s(\ln s)^2}-\frac{1}{2}\right) {z_1\rho }+O(\rho ^2);\\ \frac{1}{k_BT}\mu _1^{\delta ,o}&=z_1\zeta _0V_0 +\ln s -\frac{\rho }{R_1} +\frac{2z_2-(1+\lambda )z_1}{z_2} (s-1) R_1d +O(\rho d, d^2),\\ w_0^{[o]}&= \frac{s-1}{\ln s}{z_1R_1}+\frac{s-1-\ln s}{(\ln s)^2}z_1 \rho +O(\rho ^2),\\ w_1^{[o]}&=\frac{s-1}{\ln s}{z_1R_1}-\frac{s+1}{2}{z_1R_1}+\left( \frac{s-1-\ln s}{(\ln s)^2}-\frac{1}{2}\right) {z_1\rho }+O(\rho ^2). \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\mu _1^{\delta ,i}}{\mu _1^{\delta ,o}}&=1+\frac{s+1}{s(z_1\zeta _0V_0 +\ln s)}\frac{\rho }{R_1}+O(\rho ^2, \rho d, d^2),\\ \frac{w_0^{[i]}}{w_0^{[o]}}&= 1+\frac{2s\ln s-s^2+1}{s(s-1)\ln s}\frac{\rho }{R_1}+O(\rho ^2),\\ w_1^{[i]}-w_1^{[o]}&=\frac{2s\ln s-s^2+1}{s(\ln s)^2}z_1\rho +O(\rho ^2). \end{aligned}$$

Therefore, the ratio between the fluxes $J_1^{[i]}$ and $J_1^{[o]}$ is given by

$$\begin{aligned} \frac{J_1^{[i]}}{J_1^{[o]}}&=\frac{z_1z_2+2(\lambda z_1-z_2)w_1^{[i]}d}{z_1z_2+2(\lambda z_1-z_2)w_1^{[o]} d}\cdot \frac{w_0^{[i]} }{w_0^{[o]}}\frac{\mu _1^{\delta ,i}}{\mu _1^{\delta ,o}}+O( d^2)\\&=\left( 1+\frac{2(\lambda z_1-z_2)}{z_1z_2}(w_1^{[i]}-w_1^{[o]})d\right) \frac{w_0^{[i]}}{w_0^{[o]}}\frac{\mu _1^{\delta ,i}}{\mu _1^{\delta ,o}} +O(d^2) \\&=1+\left( \frac{2s\ln s-s^2+1}{s(s-1)\ln s}+\frac{s+1}{t(z_1\zeta _0V_0 +\ln s)}\right) \frac{\rho }{R_1}+O(\rho ^2, \rho d,d^2). \end{aligned}$$

As a consequence,

$$\begin{aligned} \begin{aligned} S_0(\rho ) =1+\left( \frac{2s\ln s-s^2+1}{s(s-1)\ln s}+\frac{s+1}{s(z_1\zeta _0V_0 +\ln s)}\right) \frac{\rho }{R_1}\; \text{ and } \; S_1(0)=0. \end{aligned} \end{aligned}$$

(7.19)

Note that $S_0(\rho )$ obtained here agrees with that in (7.16), as expected. The conclusion $S_1(0)=0$ implies that there is not linear term in d in the approximation for $J_1^{[i]}/{J_1^{[o]}}$.

1.2.3 Approximation of $J_1^{[i]}/{J_1^{[o]}}$

The superposition of (7.16) and (7.19) then gives

$$\begin{aligned} \begin{aligned} \frac{ J_1^{[i]}}{J_1^{[o]}}&=1+\left( \frac{2s\ln s-s^2+1}{s(s-1)\ln s}+\frac{s+1}{s(z_1\zeta _0V_0 +\ln s)}\right) \frac{\rho }{R_1}\\&\quad +g_1(V_0,s,\alpha ,\beta )\frac{\rho Q_0}{R_1^2}+O(\rho ^2, d^2, \rho d Q_0). \end{aligned} \end{aligned}$$

(7.20)

Finally, formula (5.14) in Theorem 5.1 can be obtained from the formula (7.2) by multiplying the three factors estimated in (7.4), (7.5) and (7.20).

1.3 Derivation of Formula (5.18)

We now derive formula (5.18) for the case where approximate electroneutrality boundary conditions are assumed; that is, we assume the boundary concentrations $L_1$ and $R_1$ for the main ion species and those $L_2$ and $R_2$ for the counterion species stay the same when tracer is added. Therefore, the electroneutrality boundary conditions are only approximate with the tracer concentration $\rho$ being small. In this approximate electroneutrality boundary conditions, there will be boundary layers to compensate the imperfect electroneutrality.

More precisely, for (BVi) with tracer concentrations $c_1^{[t,i]}(0)=L_1^{[t,i]}=\rho >0$ at $x=0$ and $c_1^{[t,i]}(1)=R_1^{[t,i]}=0$ at $x=1$, one has

$$\begin{aligned} \text{(BVi): }\quad z_1(L_1+\rho )+z_2L_2=z_1\rho \approx 0\; \text{ and } \; z_1R_1+z_2R_2=0,\end{aligned}$$

and there will be one boundary layer at $x=0$.

For (BVo) with tracer concentrations $c_1^{[t,o]}(0)=L_1^{[t,o]}=0$ at $x=0$ and $c_1^{[t,o]}(1)=R_1^{[t,o]}=\rho >0$ at $x=1$, one has

$$\begin{aligned} \text{(BVo): }\quad z_1L_1+z_2L_2= 0\; \text{ and } \; z_1(R_1+\rho )+z_2R_2=z_1\rho \approx 0, \end{aligned}$$

and there will be one boundary layer at $x=1$.

For (BVi), let $(\phi _{\infty }^{[i]}, c_{1,\infty }^{[i]}, c_{1,\infty }^{[t,i]},c_{2,\infty }^{[i]})$ be the corresponding values of $(\phi ^{[i]}, c_1^{[i]},c_1^{[t,i]},c_2^{[i]})$ at the limiting point of the boundary layer at $x=0$. Then, it follows from [56] (Lemma 3.2 and Proposition 3.3 in [56]) and (5.17) that

$$\begin{aligned} \begin{aligned} \phi _{\infty }^{[i]}&=V_0-\frac{1}{\zeta _0(z_1-z_2)}\ln \frac{-z_2L_2}{z_1(L_1+\rho )}=V_0+\frac{1}{\zeta _0(z_1-z_2)}\frac{\rho }{L_1}+O(\rho ^2),\\ c_{1,\infty }^{[i]}&=L_1\left( \frac{L_1}{L_1+\rho } \right) ^{\frac{z_1}{z_1-z_2}} =L_1-\frac{z_1}{z_1-z_2} \rho +O(\rho )^2, \\ c_{1,\infty }^{[t,i]}&=\rho \left( \frac{L_1}{L_1+\rho } \right) ^{\frac{z_1}{z_1-z_2}}=\rho -\frac{z_1}{z_1-z_2} \frac{\rho ^2}{L_1} +O(\rho )^3\\ c_{2,\infty }^{[i]}&=L_2 \left( \frac{L_1}{L_1+\rho } \right) ^{\frac{z_2}{z_1-z_2}}=L_2-\frac{z_1L_2}{(z_1-z_2)L_1}\rho +O(\rho )^2. \end{aligned} \end{aligned}$$

(7.21)

It is easy to check that $z_1(c_{1,\infty }^{[i]}+ c_{1,\infty }^{[t,i]})+z_2c_{2,\infty }^{[i]}=0$, which is known to be true for the limiting point [56].

Similarly, for (BVo), let $(\phi _{\infty }^{[o]}, c_{1,\infty }^{[o]}, c_{1,\infty }^{[t,o]},c_{2,\infty }^{[o]})$ be the corresponding values of $(\phi ^{[o]}, c_1^{[o]},c_1^{[t,o]},c_2^{[o]})$ at the limiting point of the boundary layer at $x=1$. Then,

$$\begin{aligned} \begin{aligned} \phi _{\infty }^{[o]}&= -\frac{1}{\zeta _0(z_1-z_2)}\ln \frac{-z_2R_2}{z_1(R_1+\rho )}=\frac{1}{\zeta _0(z_1-z_2)}\frac{\rho }{R_1}+O(\rho ^2),\\ c_{1,\infty }^{[o]}&=R_1\left( \frac{R_1}{R_1+\rho } \right) ^{\frac{z_1}{z_1-z_2}}=R_1-\frac{z_1}{z_1-z_2} \rho +O(\rho )^2, \\ c_{1,\infty }^{[t,o]}&=\rho \left( \frac{R_1}{R_1+\rho } \right) ^{\frac{z_1}{z_1-z_2}}=\rho -\frac{z_1}{z_1-z_2}\frac{\rho ^2}{R_1} +O(\rho )^3,\\ c_{2,\infty }^{[o]}&=R_2\left( \frac{R_1 }{R_1+\rho } \right) ^{\frac{z_2}{z_1-z_2}}=R_2-\frac{z_1R_2}{(z_1-z_2)R_1} \rho +O(\rho )^2. \end{aligned} \end{aligned}$$

(7.22)

One can derive the formula (5.18) following exactly the same procedure as in Sect. 7.2. Another approach is to make use of the derivation in Sect. 7.2 with necessary modifications. We have used both approaches and, not surprisingly, got the same approximate formula (5.18). The latter is simpler and is presented below.

More precisely, we need some modifications in the derivation in Sect. 7.2.

For (BVi), the boundary conditions should be replaced by

$$\begin{aligned} (\phi _{\infty }^{[i]}, c_{1,\infty }^{[i]}, c_{1,\infty }^{[t,i]}, c_{2,\infty }^{[i]})\; \text{ at } \; x=0;\quad (0, R_1, 0, R_2)\; \text{ at } \; x=1.\end{aligned}$$

For (BVo), the boundary conditions should be replaced by

$$\begin{aligned} (V_0, L_1, 0, L_2)\; \text{ at } \; x=0;\quad (\phi _{\infty }^{[o]}, c_{1,\infty }^{[o]}, c_{1,\infty }^{[t,o]}, c_{2,\infty }^{[o]})\; \text{ at } \; x=1.\end{aligned}$$

With these modifications, it is not hard to get, from (5.6) that, for the boundary condition associated to (BVi) in (5.17),

$$\begin{aligned} p_1^{[i]}(0^+;z_1,d_1)&=z_1\zeta _0\phi _{\infty }^{[i]}+ d\Big (2(c_{1,\infty }^{[i]}+c_{1,\infty }^{[t,i]}) +(1+\lambda ) c_{2,\infty }^{[i]}\Big ) +O(d^2)\\&=z_1\zeta _0V_0+\frac{z_1}{z_1-z_2}\frac{\rho }{L_1}+ d\Big (2L_1 +(1+\lambda ) L_2\Big ) +O(d^2,d\rho ),\\ p_1^{[i]}(1;z_1,d_1)&= d\Big (2R_1+(1+\lambda ) R_2\Big ) +O(d^2); \end{aligned}$$

and, for the boundary condition associated to (BVo) in (5.17),

$$\begin{aligned} p_1^{[o]}(0;z_1,d_1)&=z_1\zeta _0V_0+ d(2L_1+(1+\lambda ) L_2) +O(d^2),\\ p_1^{[o]}(1^{-};z_1,d_1)&=z_1\zeta _0\phi _{\infty }^{[o]}+ d(2(c_{1,\infty }^{[o]}+c_{1,\infty }^{[t,o]})+(1+\lambda ) c_{2,\infty }^{[o]}) +O(d^2)\\&= \frac{z_1}{z_1-z_2}\frac{\rho }{R_1}+ d(2R_1+(1+\lambda ) R_2) +O(d^2,d\rho ). \end{aligned}$$

Thus, it follows from the definitions of the factors in (7.2) that

$$\begin{aligned} \begin{aligned} (I)&=\frac{c_{1,\infty }^{[t,i]}}{c_{1,\infty }^{[t,o]}}\left( e^{p_1^{[i]}(0;z_1,d_1)-p_1^{[o]}(1^-;z_1,d_1)}\right) =\left( 1+\frac{z_1(s-1)}{(z_1-z_2)s}\frac{\rho }{R_1}\right) e^{z_1\zeta _0V_0} \\&\quad \times \Big (1+\frac{z_1(1-s)}{(z_1-z_2)s}\frac{\rho }{R_1}+\big (2 (L_1-R_1) +(1+\lambda ) (L_2- R_2) \big )d\Big )\\&\quad +O(d^2,d\rho , \rho ^2)\\&=e^{z_1\zeta _0V_0} \Big (1+\big (2 (L_1-R_1) +(1+\lambda ) (L_2- R_2) \big )d\Big )+O(d^2,d\rho , \rho ^2), \end{aligned} \end{aligned}$$

(7.23)

and

$$\begin{aligned} \begin{aligned} (II)&=\frac{L_1e^{p_1^{[o]}(0;z_1,d_1)}-(c_{1,\infty }^{[o]}+c_{1,\infty }^{[t,o]})e^{p_1^{[o]}(1^-;z_1,d_1)}}{(c_{1,\infty }^{[i]}+c_{1,\infty }^{[t,i]})e^{p_1^{[i]}(0^+;z_1,d_1)}-R_1e^{p_1^{[i]}(1;z_1,d_1)}}+O(\rho ^2, \rho d, d^2)\\&=1 -\frac{ e^{z_1\zeta _0V_0}+1}{se^{z_1\zeta _0V_0}-1}\frac{\rho }{R_1} +O(\rho ^2, \rho d, d^2). \end{aligned} \end{aligned}$$

(7.24)

Also, $F_0$ in (7.15) should be replaced by

$$\begin{aligned} F_0&=\frac{c_{1,\infty }^{[i]}+c_{1,\infty }^{[t,i]}-R_1}{L_1-(c_{1,\infty }^{[o]}+c_{1,\infty }^{[t,o]})} \frac{\ln L_1-\ln (c_{1,\infty }^{[o]}+c_{1,\infty }^{[t,o]})}{\ln (c_{1,\infty }^{[i]}+c_{1,\infty }^{[t,i]})-\ln R_1} \\&\quad \times \frac{z_1\zeta _0 \phi _{\infty }^{[i]}+\ln (c_{1,\infty }^{[i]}+c_{1,\infty }^{[t,i]})-\ln R_1}{z_1\zeta _0 \phi _{\infty }^{[o]}+\ln L_1-\ln (c_{1,\infty }^{[o]}+c_{1,\infty }^{[t,o]})}. \end{aligned}$$

A straightforward calculations gives

$$\begin{aligned} F_0 =1+ \left( \frac{-z_2}{z_1-z_2}f_1(s)+\frac{s+1}{s(z_1\zeta _0V_0+\ln s)}\right) \frac{\rho }{R_1}+O(\rho ^2), \end{aligned}$$

where $f_1(s)$ is defined in (5.8).

The most complicate term is $F_0/{F_1}$ in (7.15). By a close examination of the terms involved in the expression, the following modifications are needed. One is to replace $\rho$ with $-z_2\rho /{(z_1-z_2)}$. The other is, in the last expression of $F_0/{F_1}$ in (7.15), to replace $V_0$ by $\phi _{\infty }^{[i]}$ in the first term and to replace $V_0$ by $V_0-\phi _{\infty }^{[o]}$ in the second term. The combined effect of the latter is the extra term

$$\begin{aligned} (E)=&\frac{A^{[i]}(1-B^{[i]})}{(z_1-z_2)\left( L_1-\frac{z_2}{z_1-z_2}\rho -R_1\right) \left( \ln (L_1-\frac{z_2}{z_1-z_2}\rho )-\ln R_1\right) }\frac{z_2}{z_1-z_2}\frac{\rho }{sR_1}\\&\quad +\frac{A^{[o]}(1-B^{[o]})}{(z_1-z_2)\left( L_1-R_1+\frac{z_2}{z_1-z_2}\rho \right) \left( \ln L_1-\ln (R_1-\frac{z_2}{z_1-z_2}\rho )\right) }\frac{z_2}{z_1-z_2} \frac{\rho }{R_1}\\&=\frac{z_2}{(z_1-z_2)^2}\frac{A_0^{[i]}+A_0^{[0]}s}{s(s-1)\ln s}\frac{\rho }{R_1^2}-\frac{z_2}{(z_1-z_2)^2}\frac{A_0^{[i]}B_0^{[i]}+A_0^{[0]}B_0^{[0]}s}{s(s-1)\ln s}\frac{\rho }{R_1^2}. \end{aligned}$$

Using the expressions of $A_0^{[i]}, A_0^{[0]}, A_0^{[i]}B_0^{[i]}$ and $A_0^{[0]}B_0^{[0]}$ in (7.17), one gets

$$\begin{aligned} (E)=g_2(s,\alpha ,\beta )\frac{-z_2}{z_1-z_2}\frac{\rho }{R_1^2}, \end{aligned}$$

where $g_2(s,\alpha ,\beta )$ is given in (5.9). Therefore, we have

$$\begin{aligned} \frac{F_1}{F_0}&=\frac{-z_2}{z_1-z_2}g_1(V_0,s,\alpha ,\beta )\frac{\rho }{R_1^2}+(E)\\&=\frac{-z_2}{z_1-z_2}\Big (g_1(V_0,s,\alpha ,\beta )+g_2(s,\alpha ,\beta )\Big )\frac{\rho }{R_1^2}+O(\rho ^2), \end{aligned}$$

and hence,

$$\begin{aligned} \begin{aligned} \frac{ J_1^{[i]}}{J_1^{[o]}}&=F_0\left( 1+\frac{F_1}{F_0}Q_0\right) =1\\&\quad +\left( \frac{-z_2}{z_1-z_2}f_1(s)+\frac{s+1}{s(z_1\zeta _0V_0 +\ln s)}\right) \frac{\rho }{R_1}\\&\quad +\Big (g_1(V_0,s,\alpha ,\beta )+g_2(s,\alpha ,\beta )\Big )\frac{-z_2}{z_1-z_2}\frac{\rho Q_0}{R_1^2}+O(\rho ^2, d^2, \rho d Q_0). \end{aligned} \end{aligned}$$

(7.25)

The formula (5.18) in Theorem 5.2 then follows from (7.2), and the estimates (7.23), (7.24) and (7.25).