= Solution
For <Roche-lobe overflow>, $R_2=R_L$. Cubing the <Roche lobe> approximation and using the donor's <mass-radius relation> gives
$$
a^3=\frac{R_2^3}{0.46^3}\frac M{M_2}=\frac{R_\odot^3}{0.46^3}\frac{M M_2^2}{M_\odot^3}.
$$
<Kepler's third law> therefore yields the <Roche-lobe period-mass relation for a linear donor radius law>:
$$
\boxed{\frac P{P_0}=\frac{M_2}{M_\odot}},\qquad P_0=2\pi\sqrt{\frac{R_\odot^3}{0.46^3GM_\odot}}.
$$
The component distances from the <centre of mass> are $a_1=aM_2/M$ and $a_2=aM_1/M$. Summing their <angular momenta> gives the <circular-binary orbital angular momentum>
$$
\boxed{J=(M_1a_1^2+M_2a_2^2)\Omega=\frac{M_1M_2}{M}a^2\Omega=M_1M_2\sqrt{\frac{Ga}{M}}}.
$$
During <conservative binary mass transfer>, $\dot M=0$ and $\dot M_1=-\dot M_2$. Logarithmic differentiation at constant <angular momentum> gives $\dot a/a=-2(1-q)\dot M_2/M_2$. Thus the <Roche-lobe radius response exponent> gives
$$
\boxed{\frac{\dot R_L}{R_L}=\left(2q-\frac53\right)\frac{\dot M_2}{M_2}},\qquad \frac{\dot R_2}{R_2}=\frac{\dot M_2}{M_2}.
$$
Their difference is $d\log(R_L/R_2)/dt=(2q-8/3)\dot M_2/M_2$. Since the donor loses <mass>, this is positive for $q<4/3$: \b[the donor shrinks relative to its Roche lobe, so transfer shuts off without an external driver.] The <Roche lobe> itself expands only for $q<5/6$; for $5/6<q<4/3$ it shrinks more slowly than the donor.
<Gravitational-wave emission from a binary system> provides such a driver: the orbit emits <gravitational waves> carrying <energy> and <angular momentum>, and the shrinking <Roche lobe> can maintain contact. Allowing $\dot J<0$, the <binary mass-transfer contact equation> becomes
$$
\frac{\dot R_L}{R_L}=2\frac{\dot J}{J}+\left(2q-\frac53\right)\frac{\dot M_2}{M_2}=\frac{\dot M_2}{M_2},\qquad \boxed{\frac{\dot M_2}{M_2}=\frac{3}{4-3q}\frac{\dot J}{J}}.
$$
It gives a negative transfer rate for $q<4/3$.
For the <classical nova>, assume <isotropic re-emission from a binary star>: ejecta carry the accretor's <specific angular momentum> $j_1=a_1^2\Omega$. The eruption lasts many <orbital periods>, so the orbit responds approximately adiabatically and remains nearly circular; an instantaneous asymmetric kick would be a different model. To first order,
$$
\frac{\delta J}{J}=-\frac{j_1\delta m}{J}=-q\frac{\delta m}{M},\qquad \delta M_1=-\delta m,\qquad\delta M_2=0,\qquad\delta M=-\delta m.
$$
Differentiating the <circular-binary orbital angular momentum> formula during the ejection gives
$$
-q\frac{\delta m}{M}=-\frac{\delta m}{M_1}+\frac12\frac{\delta a}{a}+\frac12\frac{\delta m}{M},\qquad \boxed{\frac{\delta a}{a}=\frac{\delta m}{M}}.
$$
The accompanying <Roche lobe> change is therefore
$$
\boxed{\frac{\delta R_L}{R_L}=\frac{\delta a}{a}+\frac13\frac{\delta m}{M}=\frac{4\delta m}{3M}}.
$$
The donor does not change <mass> in the eruption and moves inside its expanded <Roche lobe>: this is <nova-induced binary detachment>.
Put $\Gamma=-\dot J>0$ for the constant external loss. During detachment the component masses are fixed, so the <Roche lobe> shrinks at fractional rate $2\Gamma/J$. Closing the fractional gap $4\delta m/(3M)$ takes $t_d=2J\delta m/(3M\Gamma)$. During the subsequent <semidetached binary> phase, the <binary mass-transfer contact equation> gives $|\dot M_2|=3M_2\Gamma/[J(4-3q)]$, so accumulating the next eruption's <mass> takes $t_s=J(4-3q)\delta m/(3M_2\Gamma)$. Consequently
$$
\boxed{\frac{t_d}{t_s}=\frac{2M_2}{M(4-3q)}=\frac{2q}{(4-3q)(1+q)}}.
$$
Both times are evaluated to first order in $\delta m/M$, with the same equilibrium donor <mass-radius relation>; changes of $q,J$ and the transfer rate within one cycle affect higher orders.
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