= Solution
The loaded-carrier balance gives $J=k(p_e-p_i)$. Substitute the preceding two complex <concentrations> and collect the current terms:
$$
J\left(1+\frac{2k}{k_-}+\frac{s_i}{K}\right)=\frac kK(s_e-s_i)c_e,
\qquad\boxed{J=\frac{kk_+(s_e-s_i)c_e}{2k+k_-+k_+s_i}.}
$$
Next carrier conservation gives
$$
c_0=\frac{s_i+s_e+2K}{K}c_e+\frac Jk\left(1+\frac{s_i}K\right),
\qquad\boxed{c_e=\frac{c_0K-(J/k)(s_i+K)}{s_i+s_e+2K}.}
$$
Eliminate $c_e$ between these relations and set $K_d=k/k_+$. Simplification yields
$$
\boxed{J=\frac{K_dKk_+c_0}{2}\frac{s_e-s_i}{(s_i+K+K_d)(s_e+K+K_d)-K_d^2}.}
$$
Here $c_0$ must be the \b[total <carrier protein>], as in the conservation <equation>. Calling it total <substrate> in the final line of the scan is a typographical inconsistency; total <substrate> is the different invariant $s_0$.
For <carrier-turnover saturation of passive transport> at fixed $s_i$, write $A=s_i+K+K_d$, $D=A(K+K_d)-K_d^2>0$ and $C=K_dKk_+c_0/2>0$. Then $J=C(s_e-s_i)/(As_e+D)$, so
$$
J'=\frac{C(D+As_i)}{(As_e+D)^2}>0,\qquad J''=-\frac{2AC(D+As_i)}{(As_e+D)^3}<0,
$$
$$
\boxed{J(s_i)=0,\qquad \lim_{s_e\to\infty}J=\frac{K_dKk_+c_0}{2(s_i+K+K_d)}.}
$$
The <flux> is outward when the <substrate> gradient is reversed and inward when $s_e>s_i$. Saturation occurs because there is a finite carrier population and finite binding-release/conformational turnover: arbitrarily abundant exterior <substrate> cannot make carriers cycle arbitrarily quickly. This is passive facilitated transport, so no maintained <flux> remains when the two free <concentrations> agree. Varying $s_e$ at fixed $s_i$ compares reservoir-controlled steady states or systems with different total <substrate>; it is not a trajectory preserving one fixed $s_0$.
\Image[/past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2009-carrier-flux.png]
{title=Carrier-mediated transport: reversal and saturation at fixed interior concentration}
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