Preliminary analysis

According to Equation (G.6), we have

$$\frac{dX}{dt} = \frac{\partial f_1}{\partial t} + \sum_{k=1,6}\frac{\partial f_1}{\partial c_k}\frac{d c_k}{dt}.$$ (G.13)

If this expression, and the analogous expressions for $dY/dt$ and $dZ/dt$, were differentiated with respect to time, and the results substituted into Equations (G.3)–(G.5), then we would obtain three time evolution equations for the six variables $c_1,\cdots,c_6$. In order to make the problem definite, three additional conditions must be introduced into the problem. It is convenient to choose

$$\sum_{k=1,6}\frac{\partial f_l}{\partial c_k}\frac{dc_k}{dt} = 0,$$ (G.14)

for $l=1,2,3$. Hence, it follows from Equations (G.12) and (G.13) that

$$\frac{dX}{dt} = \frac{\partial f_1}{\partial t} = g_1,$$ (G.15)

$$\frac{dY}{dt} = \frac{\partial f_2}{\partial t} = g_2,$$ (G.16)

$$\frac{dZ}{dt} = \frac{\partial f_3}{\partial t} = g_3.$$ (G.17)

Differentiation of these equations with respect to time yields

$$\frac{d^{\,2}X}{dt^{\,2}} = \frac{\partial^{\,2} f_1}{\partial t^{\,2}} +\sum_{k=1,6} \frac{\partial g_1}{\partial c_k}\,\frac{d c_k}{dt},$$ (G.18)

$$\frac{d^{\,2}Y}{dt^{\,2}} = \frac{\partial^{\,2} f_2}{\partial t^{\,2}}+\sum_{k=1,6} \frac{\partial g_2}{\partial c_k}\,\frac{d c_k}{dt},$$ (G.19)

$$\frac{d^{\,2}Z}{dt^{\,2}} = \frac{\partial^{\,2} f_3}{\partial t^{\,2}}+\sum_{k=1,6} \frac{\partial g_3}{\partial c_k}\,\frac{d c_k}{dt}.$$ (G.20)

Substitution into Equations (G.3)–(G.5) gives

$$\frac{\partial^{\,2} f_1}{\partial t^{\,2}} +\mu\,\frac{f_1}{r^{\,3}} +\sum_{k=1,6} \frac{\partial g_1}{\partial c_k}\,\frac{d c_k}{dt} = \frac{\partial {\cal R}}{\partial X},$$ (G.21)

$$\frac{\partial^{\,2} f_2}{\partial t^{\,2}}+\mu\,\frac{f_2}{r^{\,3}}+\sum_{k=1,6} \frac{\partial g_2}{\partial c_k}\,\frac{d c_k}{dt} = \frac{\partial {\cal R}}{\partial Y},$$ (G.22)

$$\frac{\partial^{\,2} f_3}{\partial t^{\,2}}+\mu\,\frac{f_3}{r^{\,3}}+\sum_{k=1,6} \frac{\partial g_3}{\partial c_k}\,\frac{d c_k}{dt} = \frac{\partial {\cal R}}{\partial Z},$$ (G.23)

where $r=(f_1^{\,2}+f_2^{\,2}+f_3^{\,2})^{1/2}$. Because $f_1$, $f_2$, and $f_3$ are the respective solutions to Equation (G.3)–(G.5) when the right-hand sides are zero, and the orbital elements are thus constants, it follows that the first two terms in each of the preceding three equations cancel one another. Hence, writing $f_1$ as $X$, and $g_1$ as $\dot{X}$, and so on, Equations (G.14) and (G.21)–(G.23) yield

$$\sum_{k=1,6}\frac{\partial X}{\partial c_k} \,\frac{dc_k}{dt} = 0,$$ (G.24)

$$\sum_{k=1,6}\frac{\partial Y}{\partial c_k} \,\frac{dc_k}{dt} = 0,$$ (G.25)

$$\sum_{k=1,6}\frac{\partial Z}{\partial c_k} \,\frac{dc_k}{dt} = 0,$$ (G.26)

$$\sum_{k=1,6}\frac{\partial \dot{X}}{\partial c_k} \,\frac{dc_k}{dt} = \frac{\partial {\cal R}}{\partial X},$$ (G.27)

$$\sum_{k=1,6}\frac{\partial \dot{Y}}{\partial c_k} \,\frac{dc_k}{dt} = \frac{\partial {\cal R}}{\partial Y},$$ (G.28)

$$\sum_{k=1,6}\frac{\partial \dot{Z}}{\partial c_k} \,\frac{dc_k}{dt} = \frac{\partial {\cal R}}{\partial Z}.$$ (G.29)

These six equations are equivalent to the three original equations of motion [(G.3)–(G.5)].