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For brevity write $b_1=b_{3/2}^{(1)}(\alpha )$, $b_2=b_{3/2}^{(2)}(\alpha )$, and $K=G{M}_p\alpha /(4a_p)$. The quadratic disturbing function is $\mathcal{R}=K[\frac{1}{2}{b}_1{e}^2-b_2{e}_{p}e\cos ({\varpi }_p-\varpi )]$. Differentiation in Lagrange planetary equations givesThe separate longitude of pericentre equation is singular at $e=0$, because a circular orbit has no distinguished periapsis. The complex eccentricity, $z=e{e}^{i\varpi }$, is regular there. Using $z_p=e_p{e}^{i{\varpi }_p}$ and $\dot{z}=e^{i\varpi }(\dot{e}+ie\dot{\varpi })$ givesSince $n{a}^2=\sqrt{G{M}_{\star }a}$ and $n_p=\sqrt{G{M}_{\star }/a_p^3}$,The approximations are planar, small-orbital eccentricity, first order in the planet-to-star mass ratio, and secular averaging away from important mean-motion resonances. The semimajor axis values remain constant at this order.

For a planet with fixed orbit, set $g=A{b}_1$ and $z_f=(b_2/b_1)z_p$. Then $\dot{z}=ig(z-z_f)$, so $z-z_f=(z(0)-z_f)e^{igt}$. With $z(0)=0$,On the Argand diagram, the complex eccentricity traces a circle centred on the forced eccentricity $z_f$, of radius $|z_f|$. For an interior test particle, $0<\alpha <1$ and $g>0$, so the motion is counterclockwise, starting at the origin. The rotating radius $z-z_f$ is the free eccentricity. The orbital eccentricity ranges from zero to $2|z_f|$, attaining its maximum at half a secular period. If $z_p$ is placed on the positive real axis, the initial tangent points downward, which is consistent with counterclockwise motion from the leftmost point of this circle.

Now let the planetary mass grow as $M_p(t)=M_{p}t/t_p$ for $0\le t\le t_p$, while the planet's orbital elements and the test particle's semimajor axis remain fixed. Both the precession and forcing coefficients acquire the same factor $t/t_p$. Their ratio, and hence $z_f$, remain constant. ThusandThis is the secular phase integral for proportional mass growth. The Argand diagram circle and the free eccentricity amplitude are unchanged; only the rate of traversal changes. The angular speed about $z_f$ starts at zero and rises linearly to $g$. The early displacement is quadratic in time rather than linear. Since $t_p=2\pi /g$, $\Phi (t_p)=\pi$, soA planet already at its final mass would instead return the particle to $z=0$ at $t_p$. After growth stops, the circle continues at rate $g$, with $z(t)=z_f[1+e^{ig(t-t_p)}]$ for $t\ge t_p$. Slow mass growth does not damp the free eccentricity here: the fixed centre of the circle and the conservative equation preserve its amplitude.

For several planets with fixed masses, Laplace-Lagrange secular theory gives $\dot{\mathbf{z}}=i\mathbf{Az}$ for the planets andfor a massless particle, with real secular coupling coefficients ${\nu }_j$. Resolve planetary complex eccentricities into secular eigenmodes, $z_j(t)={\sum }_k{E}_{jk}{e}^{i{g}_{k}t}$. Away from secular resonances, the test-particle solution has the formIts Argand diagram trajectory is a superposition of rotations: generally a quasiperiodic curve with beats, rather than a circle with one fixed centre. Even if the planetary orbits are artificially held fixed, their combined forcing shifts the centre to the sum of the individual forced eccentricities. At $g_k=g_0$, the resonant contribution is proportional to $t{e}^{i{g}_{0}t}$ in linear theory; its growing amplitude eventually invalidates the small-orbital eccentricity approximation.

If one planetary mass changes, the Laplace-Lagrange secular matrix, its eigenvalues, its eigenvectors, and the particle's forced response generally change with time. Planetary modes need not remain independent. Frequencies drift and a secular resonance can sweep across the particle, changing or exciting its free eccentricity. Very slow evolution away from resonances permits an approximate adiabatic mode description; rapid evolution or a resonance passage can mix modes. The simple substitution of a phase integral works only in the special proportional-growth case above, where the entire particle equation is multiplied by one scalar and its forced centre stays fixed. It does not apply to a general many-planet system.

Put $B_i={\sum }_j{A}_j{e}_{ji}$. For $A\ne g_i$, the solution of the linear secular forcing of a test particle equation isThe initial condition gives $e_p=|z(0)-z_f(0)|$ and ${\beta }_p=\arg [z(0)-z_f(0)]$. The proper eccentricity $e_p$ is the constant amplitude of the homogeneous response; its proper longitude of periapsis is $At+{\beta }_p$. The forced eccentricity is the driven vector $z_f(t)$, not necessarily a constant magnitude.

On an Argand diagram, $z-z_f$ describes a circle of radius $e_p$. Its center $z_f$ can itself move under the planetary modes, so $z$ need not trace one fixed circle in the inertial complex plane. At a secular resonance $A=g_i$, that mode instead produces $i{B}_{i}t{e}^{i(At+{\beta }_i)}$, and the undamped linear response grows until the approximation fails.

Insert the planets' two secular eigenmodes and define their forcing strengthsSolving the first-order linear ordinary differential equation givesThe first term is the particle's freely precessing eccentricity. The remaining terms are its forced eccentricity, phase-locked to the planets' modes. In the complex plane, their vector sum makes the eccentricity and longitude of periapsis oscillate. A denominator becomes small at a secular resonance $A=g_k$; the nonresonant formula then ceases to be uniform.

The free precession rate $A(a)$ tends to zero far inside the inner planet, diverges on approaching $a_1$, diverges on both sides of $a_2$, and tends to zero far outside the outer planet. Between $a_1$ and $a_2$ it diverges at both ends and has at least one minimum. A horizontal line $A=g_1$ therefore crosses once inside $a_1$ and once outside $a_2$, plus zero, one tangent, or two times between the planets. There are consequentlylocations of the corresponding secular resonance. This argument uses the smooth Laplace-Lagrange secular theory away from the immediate neighborhoods of the planets and from mean-motion resonances.

For $t\gg \tau$, the free term has disappeared. Exactly at $A=g_1$, the resonant contribution tends towith finite eccentricity $|{\nu }_1|\tau$ and a quarter-cycle phase shift. Without damping, exact resonance instead giveswhose secular resonance amplitude grows linearly in the ideal linear theory.

Changing planetary masses or other system parameters move the frequencies of secular eigenmodes. A particle can cross a secular resonance when its free precession frequency meets a forcing frequency. Passage can change its free eccentricity and mix modes; a single scalar phase replacement generally cannot describe the response.