= Solution
Let $M=M_1+M_2$. The <centre of mass> condition gives $a_1=aM_2/M$, $a_2=aM_1/M$. For <circular motion> the component <angular momenta> are $J_i=M_ia_i^2\Omega$, so
$$
J=(M_1a_1^2+M_2a_2^2)\Omega
=\frac{M_1M_2}{M}a^2\Omega.
$$
Using <Kepler's third law>, $a^3\Omega^2=GM$, and $\Omega=2\pi/P_{\rm orb}$ gives
$$
\boxed{J=\frac{M_1M_2}{M}a^2\Omega
=\frac{G^{2/3}P_{\rm orb}^{1/3}M_1M_2}{(2\pi)^{1/3}M^{1/3}}.}
$$
This is the <circular-binary orbital angular momentum>. <Mass> received by the companion redistributes <angular momentum> within the <binary star>; only the escaping <stellar wind> removes it under the prescribed model.
The <stellar wind> from the <donor star> has <specific angular momentum>
$$
j_w=\frac{J_1}{M_1}=a_1^2\Omega,\qquad\frac{j_w}{J}=\frac{M_2}{M_1M}.
$$
Here $\dot M_1<0$, $\dot M_2=-f\dot M_1$ and $\dot M=(1-f)\dot M_1$. Thus <donor-wind angular-momentum loss> gives
$$
\frac{\dot J}{J}=(1-f)\dot M_1\frac{M_2}{M_1M}.
$$
Taking the <logarithmic derivative> of the period expression for $J$ gives
$$
\frac{\dot P_{\rm orb}}{P_{\rm orb}}
=3\frac{\dot J}{J}-3\frac{\dot M_1}{M_1}-3\frac{\dot M_2}{M_2}+\frac{\dot M}{M}.
$$
Use $M_2/(M_1M)=1/M_1-1/M$ to simplify this to
$$
\frac{\dot P_{\rm orb}}{P_{\rm orb}}
=-3f\frac{\dot M_1}{M_1}-3\frac{\dot M_2}{M_2}-2\frac{\dot M}{M}.
$$
For constant $f$, <integration> therefore gives
$$
\boxed{P_{\rm orb}\propto M_1^{-3f}M_2^{-3}(M_1+M_2)^{-2}.}
$$
This is the <donor-wind period invariant with fixed retention fraction>. If $f$ changes during the evolution, the differential equation remains valid but the integrated invariant is instead $P_{\rm orb}M_2^3M^2\exp(3\int f\,d\log M_1)=\mathrm{constant}$. The fixed-power expression cannot be used with a time-dependent exponent without this modification.
For the <Roche lobe>, write $q=M_1/M_2$. The equivalent separation expression $J=M_1M_2\sqrt{Ga/M}$ gives
$$
\frac{\dot a}{a}=2\frac{\dot J}{J}-2\frac{\dot M_1}{M_1}-2\frac{\dot M_2}{M_2}+\frac{\dot M}{M}
=\frac{\dot M_1}{M_1}\left[2f(q-1)-(1-f)\frac q{1+q}\right].
$$
Taking the <logarithmic derivative> of $R_L=0.46a(M_1/M)^{1/3}$ then adds $\dot M_1/(3M_1)-\dot M/(3M)$, yielding
$$
\boxed{\frac{\dot R_L}{R_L}=\frac{\dot M_1}{M_1}\left[
f\left(2q+\frac{4q}{3(1+q)}-2\right)+\frac13-\frac{4q}{3(1+q)}\right].}
$$
This <donor-wind Roche-lobe response> uses only the instantaneous value of $f$ and thus still holds if it varies.
Define $A(q)=2q+4q/[3(1+q)]-2$ and $B(q)=1/3-4q/[3(1+q)]$. The <Roche-lobe radius response exponent> is $\zeta_L=B+Af$, whereas the <donor star>'s <stellar radius response exponent> is $\zeta_*=-n$. Keeping $R_1=R_L$ requires the two <logarithmic derivatives> of <radius> to agree, so
$$
\boxed{f\left(2q+\frac{4q}{3(1+q)}-2\right)=\frac{4q}{3(1+q)}-n-\frac13.}
$$
For $A\ne0$, this determines
$$
\boxed{f_{\rm req}=\frac{4q/[3(1+q)]-n-1/3}{2q+4q/[3(1+q)]-2}.}
$$
A physical mixture of <Roche-lobe overflow> and <stellar wind> from the <donor star> needs $0<f<1$. Equivalently, $-n$ must lie strictly between the pure-wind response $B$ and the conservative response $B+A=2q-5/3$. The values $f=0$ and $1$ are valid limiting prescriptions, respectively pure <stellar wind> and <conservative mass transfer>, rather than mixtures. If $f_{\rm req}$ lies outside $[0,1]$, no allowed division of the lost <mass> can maintain contact under this <radius> law and angular-momentum-loss prescription. This is <feasibility of donor-wind binary contact>.
The likely direction away from contact follows without assigning a nonphysical fraction. For any actual $f$, let $\Delta=\log(R_1/R_L)$. Then
$$
\dot\Delta=(-n-B-Af)\frac{\dot M_1}{M_1}
=A(f_{\rm req}-f)\frac{\dot M_1}{M_1}.
$$
Since $\dot M_1<0$, a positive value of $A(f_{\rm req}-f)$ makes the <donor star> move inside its lobe and detach; a negative value increases overfill and promotes stronger transfer. Thus if $A>0$, $f_{\rm req}<0$ gives increasing overfill and $f_{\rm req}>1$ gives detachment; if $A<0$, these outcomes are reversed. A runaway or a new equilibrium would require the <donor star>'s response or the angular-momentum-loss model to change, possibly with additional driving. The contact calculation alone does not prove a particular nonlinear endpoint.
There is also a genuine exceptional denominator: $A=0$ at $q=(\sqrt{10}-1)/3$. At this ratio both limiting lobe responses coincide. Contact is possible only if $-n=B$, namely $n=(\sqrt{10}-2)/(\sqrt{10}+2)$; then every $f$ gives the same instantaneous <radius> response. Otherwise no fraction works, and the sign of $(-n-B)\dot M_1/M_1$ determines detachment or increasing overfill.
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