= Solution
For <Roche-lobe overflow> in the stipulated <stellar thermal equilibrium>, $R_2=R_L$. Using the <mass-radius relation> and <Kepler's third law>,
$$
a^3=\frac{R_\odot^3}{0.46^3}\frac{M M_2^2}{M_\odot^3},\qquad P^2=\frac{4\pi^2a^3}{GM}.
$$
The total <mass> cancels, giving
$$
\boxed{\frac{P}{P_0}=\frac{M_2}{M_\odot},\qquad P_0=2\pi\sqrt{\frac{R_\odot^3}{0.46^3GM_\odot}}\simeq8.91\ \mathrm{h}.}
$$
Here $R_\odot$ is the <solar radius>. This is the <Roche-lobe-filling period-density relation> specialized to $R_2\propto M_2$.
Neglecting spin, the <angular momentum> of a <circular orbit> is $J=M_1M_2\sqrt{Ga/M}$. Between <classical novae> the transfer conserves total <mass>, so $\dot M=0$ and $\dot M_1=-\dot M_2$. Its logarithmic derivative gives
$$
\frac{\dot J}{J}=(1-q)\frac{\dot M_2}{M_2}+\frac12\frac{\dot a}{a},\qquad
\frac{\dot R_L}{R_L}=2\frac{\dot J}{J}+\left(2q-\frac53\right)\frac{\dot M_2}{M_2}.
$$
In particular, for <conservation of angular momentum>,
$$
\boxed{\frac{\dot R_L}{R_L}=\left(2q-\frac53\right)\frac{\dot M_2}{M_2},\qquad \frac{\dot R_2}{R_2}=\frac{\dot M_2}{M_2}.}
$$
The source PDF has $\dot J=0$ here; the TeX's $J=0$ is a transcription defect. For $q>4/3$, the <Roche-lobe radius response exponent> exceeds the equilibrium <stellar radius response exponent> $1$. Since $\dot M_2<0$, the <Roche lobe> then shrinks faster than the <donor star>; the overfill increases and drives more transfer, giving positive feedback. Equality at $q=4/3$ is marginal in this linear response test.
Two qualifications matter. First, $q>4/3$ is a formal extrapolation outside the stated $q<1$ range, and the supplied <Roche lobe> approximation need not remain accurate there. Second, true <dynamical stability of binary mass transfer> compares $\zeta_L$ with the adiabatic <stellar radius response exponent> $\zeta_{\rm ad}$, not $\zeta_{\rm eq}=1$. The <stellar thermal equilibrium> argument supplies the displayed feedback threshold under the imposed radius law, rather than a necessary and sufficient physical dynamical threshold. For example, the <polytropic mass-radius relation> for a fully convective <adiabatic stellar polytrope> with index $3/2$ gives $\zeta_{\rm ad}=-1/3$; within this same <Roche lobe> approximation its dynamical threshold is $q=2/3$. Thus $q<4/3$ by itself does not guarantee dynamical stability.
<Gravitational-wave emission from a binary system> removes orbital <energy> and <angular momentum>. <Magnetic braking of a binary star> provides another important loss: a magnetized <stellar wind> carries away donor spin, while <tidal locking> couples that spin to the orbit. Tides alone redistribute <angular momentum> and are not an external sink. Matter expelled from the system can also carry orbital <angular momentum>, although appreciable continuous mass loss would require nonconservative modifications of the preceding equations. Setting $\dot R_L/R_L=\dot R_2/R_2$ gives the <binary mass-transfer contact equation>
$$
\boxed{\frac{\dot M_2}{M_2}=\frac{3\dot J}{J(4-3q)}.}
$$
For $q<4/3$, a negative external $\dot J$ therefore sustains negative $\dot M_2$ along the stipulated <stellar thermal equilibrium> contact sequence, provided the system is otherwise stable and can remain thermally relaxed.
During the <classical nova>, the ejecta carry the <white dwarf>'s <specific angular momentum>, not zero <angular momentum>. With $\Omega^2=GM/a^3$ and the accretor's distance from the <centre of mass> $a_1=aM_2/M$, isotropic escape gives
$$
j_1=a_1^2\Omega,\qquad \frac{j_1}{J}=\frac{M_2}{M_1M}=\frac qM,\qquad \frac{\delta J}{J}=-\frac{q\delta m}{M}.
$$
The slow ejection relative to the <orbital period> permits the adiabatic <circular orbit> approximation; the <donor star> does not transfer appreciable additional mass during it. Take $\delta M_1=-\delta m$, $\delta M_2=0$ and $\delta M=-\delta m$. Differentiating $J=M_1M_2\sqrt{Ga/M}$ gives
$$
-\frac{q\delta m}{M}=-\frac{\delta m}{M_1}+\frac12\frac{\delta a}{a}+\frac{\delta m}{2M}.
$$
Since $M/M_1=1+q$, this yields
$$
\boxed{\frac{\delta a}{a}=\frac{\delta m}{M},\qquad \frac{\delta R_L}{R_L}=\frac{\delta a}{a}+\frac13\left(\frac{\delta M_2}{M_2}-\frac{\delta M}{M}\right)=\frac{4\delta m}{3M}.}
$$
These are first-order formulae with errors of order $(\delta m/M)^2$. The orbital widening agrees with <Jeans-mode mass loss>. The <donor star>'s radius is unchanged to this order by the assumed ejection, so its <Roche lobe> expands away from it, causing <nova-induced binary detachment>. This idealization neglects changes in the donor from irradiation or interaction with ejecta; those are not supplied in the model.
Let $\Gamma=-\dot J>0$ be the fixed external loss rate. During the <detached binary> phase both component masses are fixed, and $\dot R_L/R_L=2\dot J/J=-2\Gamma/J$. Closing the fractional gap $4\delta m/(3M)$ takes
$$
t_d=\frac{2J\delta m}{3M\Gamma}.
$$
In the subsequent <semidetached binary>, the <binary mass-transfer contact equation> gives $-\dot M_2=3M_2\Gamma/[J(4-3q)]$. Accumulating a fresh layer of <mass> $\delta m$ therefore takes
$$
t_s=\frac{J(4-3q)\delta m}{3M_2\Gamma},\qquad
\boxed{\frac{t_d}{t_s}=\frac{2M_2}{M(4-3q)}=\frac{2q}{(1+q)(4-3q)}.}
$$
Throughout these first-order estimates $J,M,q$ can be evaluated at the start of the cycle: their fractional changes during it are of order $\delta m/M$. The limit $\Gamma=0$ is excluded, since no finite reconnection time follows without an external shrinkage mechanism.
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