$\displaystyle\ddot{\rho}+\Omega^{2}(t)\rho-\frac{k}{\rho^{3}}$
$\displaystyle=$ $\displaystyle 0$ (14.6)
$\displaystyle\ddot{\alpha}+\Omega^{2}(t)\alpha$ $\displaystyle=$
$\displaystyle F_{1}(t)$ (14.7)

where $k$ is an arbitrary constant. In this notation, the associated QFI
(14.4) becomes

$I=\frac{1}{2}\left[\rho(\dot{\rho}-\dot{\alpha})-\dot{\rho}(q-\alpha)\right]^{2}+\frac{k}{2}\left(\frac{q-\alpha}{\rho}\right)^{2}+\widetilde{G}\left(\frac{q-\alpha}{\rho}\right).$
(14.8)

In the following sections, we consider the problem of central motion and
determine the time-dependent LFIs/QFIs using the direct method of [246]. In
order to do this, we reduce the degrees of freedom to one by means of the LFI
of angular momentum. Finally, we collect our results in Theorem 14.2.1 and
consider various applications.

####  14.2 The integrable time-dependent central potentials $V(t,r)$

The characteristic property of Newtonian central motion is that the angular
momentum is conserved and the motion takes place on a plane normal to the
angular momentum. On that plane, we assume polar coordinates $(r,\theta)$ so
that the Lagrangian becomes

$L=\frac{1}{2}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-V(t,r).$ (14.9)

The E-L equations are:

$\displaystyle\ddot{r}$ $\displaystyle=$ $\displaystyle
r\dot{\theta}^{2}-\frac{\partial V}{\partial r}$ (14.10) $\displaystyle L_{3}$
$\displaystyle=$ $\displaystyle r^{2}\dot{\theta}$ (14.11)

where $L_{3}=r^{2}\dot{\theta}$ is the LFI of the angular momentum. Replacing
$\dot{\theta}$ in (14.10), we find the time-dependent second order ODE in the
variable $r(t)$:

$\ddot{r}=\frac{L_{3}^{2}}{r^{3}}-\frac{\partial V(t,r)}{\partial r}.$ (14.12)

Solving equation (14.12), one finds a solution for $r(t)$ which when replaced
in (14.11), by integration, gives the $\theta(t)$.

In what follows, we study the integrability of the time-dependent system

$\ddot{r}=-\frac{\partial U}{\partial r}$ (14.13)

where the time-dependent potential

$U(t,r)=\frac{L_{3}^{2}}{2r^{2}}+V(t,r).$ (14.14)

so that

$V(t,r)=U(t,r)-\frac{L_{3}^{2}}{2r^{2}}.$ (14.15)

The problem of finding all the integrable potentials $U(t,r)$ has been solved
in sections II and III of [246] using the direct method. In the following, we
update the approach of [246]. Using (7.3) and the dynamical equations (14.1)
to replace the term $\ddot{q}^{a}$ whenever it appears, condition
$\frac{dI}{dt}=0$ leads to the following system of PDEs:

$\displaystyle K_{(ab,c)}$ $\displaystyle=$ $\displaystyle 0$ (14.16)
$\displaystyle K_{ab,t}+K_{(a,b)}$ $\displaystyle=$ $\displaystyle 0$ (14.17)
$\displaystyle K_{a,t}+K_{,a}-2K_{ab}U^{,b}$ $\displaystyle=$ $\displaystyle
0$ (14.18) $\displaystyle K_{,t}-K_{a}U^{,a}$ $\displaystyle=$ $\displaystyle
0$ (14.19) $\displaystyle
K_{a,tt}-2\left(K_{ab}U^{,b}\right)_{,t}+\left(K_{b}U^{,b}\right)_{,a}$
$\displaystyle=$ $\displaystyle 0$ (14.20) $\displaystyle
2\left(K_{[a|c|}U^{,c}\right)_{,b]}-K_{[a,b],t}$ $\displaystyle=$
$\displaystyle 0.$ (14.21)

We note that the PDEs (14.16) and (14.17) are purely geometric because they do
not involve the potential $U(t,q)$. The PDE (14.16) implies that $K_{ab}$ is a
KT of order two (possibly zero) of $\gamma_{ab}$. Moreover, the last two PDEs
(14.20) and (14.21) express the integrability conditions for the scalar $K$.

Because the considered dynamical system (14.13) is 1d, the variable $q^{1}=r$
and the KT $K_{ab}$ is of the form $K_{11}=g_{1}(t)$. Therefore, the system of
PDEs (14.16) - (14.21) becomes:

$\displaystyle K_{11}$ $\displaystyle=$ $\displaystyle g_{1}(t)$ (14.22)
$\displaystyle K_{1}(t,r)$ $\displaystyle=$
$\displaystyle-\dot{g}_{1}r+g_{2}(t)$ (14.23) $\displaystyle\frac{\partial
K}{\partial r}$ $\displaystyle=$ $\displaystyle 2g_{1}\frac{\partial
U}{\partial r}+\ddot{g}_{1}r-\dot{g}_{2}$ (14.24) $\displaystyle\frac{\partial
K}{\partial t}$ $\displaystyle=$
$\displaystyle\left(g_{2}-\dot{g}_{1}r\right)\frac{\partial U}{\partial r}$
(14.25) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\left(\dot{g}_{1}r-g_{2}\right)\frac{\partial^{2}U}{\partial
r^{2}}+2g_{1}\frac{\partial^{2}U}{\partial t\partial
r}+3\dot{g}_{1}\frac{\partial U}{\partial r}+\dddot{g}_{1}r-\ddot{g}_{2}$
(14.26)

where $g_{1}(t)$ and $g_{2}(t)$ are arbitrary smooth functions. The
integrability condition (14.26) is ignored because it is satisfied identically
due to the PDEs (14.24) and (14.25). Replacing $K_{1}$ and $K_{11}$ given by
(14.22) and (14.23), respectively, in the expression (7.3) of the associated
QFI, we find

$I=g_{1}\dot{r}^{2}+\left(g_{2}-\dot{g}_{1}r\right)\dot{r}+K(t,r).$ (14.27)

There remains the scalar $K$ and the corresponding time-dependent potential
$U$ which shall be determined from the PDEs (14.24) and (14.25). There are two
cases to consider: a) $g_{1}(t)=0$ which provides the LFIs, and b)
$g_{1}(t)\neq 0$ which provides the QFIs.

#####  14.2.1 Case $g_{1}(t)=0$ (LFIs)

In this case, $K_{11}=0$, $K_{1}=g_{2}(t)\neq 0$, and the PDEs (14.24) and
(14.25) become:

$\displaystyle\frac{\partial K}{\partial r}$ $\displaystyle=$
$\displaystyle-\dot{g}_{2}$ (14.28) $\displaystyle\frac{\partial K}{\partial
t}$ $\displaystyle=$ $\displaystyle g_{2}\frac{\partial U}{\partial r}.$
(14.29)

Integrating the PDE (14.28), we find that $K(t,r)=-\dot{g}_{2}r+g(t)$ where
$g(t)$ is an arbitrary function. Replacing this function in the PDE (14.29),
we find the potential (see eq. (2.9) in [246])

$U(t,r)=-\frac{\ddot{g}_{2}}{2g_{2}}r^{2}+\frac{\dot{g}}{g_{2}}r.$ (14.30)

Substituting $U(t,r)$ in (14.15), we find the integrable time-dependent
central potential

$V(t,r)=-\frac{\ddot{g}_{2}}{2g_{2}}r^{2}+\frac{\dot{g}}{g_{2}}r-\frac{L_{3}^{2}}{2r^{2}}.$
(14.31)

This potential does not belong to any of the three classes found in [12] due
to the additional term $\frac{\dot{g}}{g_{2}}r$. We observe that it is the sum
of:
a. A repulsive time-dependent oscillator (term
$-\frac{\ddot{g}_{2}}{2g_{2}}r^{2})$.
b. An attractive Newton-Cotes potential (term $-\frac{L_{3}^{2}}{2r^{2}})$.
c. A pure time-dependent central force (term $\frac{\dot{g}}{g_{2}}r$).

The associated LFI is

$I=g_{2}\dot{r}-\dot{g}_{2}r+g$ (14.32)

which coincides with the LFI (2.11) of [246].

#####  14.2.2 Case $g_{1}(t)\neq 0$ (QFIs)

In this case, equation (14.24) can be integrated and gives the potential (see
eq. (3.8) in [246])

$U=\frac{K}{2g_{1}}-\frac{\ddot{g}_{1}}{4g_{1}}r^{2}+\frac{\dot{g}_{2}}{2g_{1}}r+g(t)$
(14.33)

where $g(t)$ is an arbitrary function. Replacing this potential in the PDE
(14.25), we find the function (see eq. (3.14) in [246])
$K=F\left(g_{1}^{-1/2}r+\frac{1}{2}\int
g_{1}^{-3/2}g_{2}dt\right)+\frac{1}{4g_{1}}\left(\dot{g}_{1}r-g_{2}\right)^{2}$
where $F$ is an arbitrary smooth function of its argument.

Substituting the function $K$ in (14.33), we find the potential333We choose
the function $g=-\frac{g_{2}^{2}}{8g_{1}^{2}}$ so as not to have an additive
function of $t$ in the potential.

$U=\left[\frac{1}{8}\left(\frac{\dot{g}_{1}}{g_{1}}\right)^{2}-\frac{\ddot{g}_{1}}{4g_{1}}\right]r^{2}+\frac{1}{2g_{1}}\left(\dot{g}_{2}-g_{2}\frac{\dot{g}_{1}}{2g_{1}}\right)r+\frac{1}{2g_{1}}F\left(g_{1}^{-1/2}r+\frac{1}{2}\int
g_{1}^{-3/2}g_{2}dt\right).$ (14.34)

Replacing $U$ in (14.15), we obtain the integrable time-dependent central
potential

$V(t,r)=\left[\frac{1}{8}\left(\frac{\dot{g}_{1}}{g_{1}}\right)^{2}-\frac{\ddot{g}_{1}}{4g_{1}}\right]r^{2}+\frac{1}{2g_{1}}\left(\dot{g}_{2}-g_{2}\frac{\dot{g}_{1}}{2g_{1}}\right)r+\frac{1}{2g_{1}}F\left(g_{1}^{-1/2}r+\frac{1}{2}\int
g_{1}^{-3/2}g_{2}dt\right)-\frac{L_{3}^{2}}{2r^{2}}.$ (14.35)

The associated QFI (14.27) is

$I=g_{1}\dot{r}^{2}+(g_{2}-\dot{g}_{1}r)\dot{r}+F\left(g_{1}^{-1/2}r+\frac{1}{2}\int
g_{1}^{-3/2}g_{2}dt\right)+\frac{1}{4g_{1}}\left(\dot{g}_{1}r-g_{2}\right)^{2}.$
(14.36)

We note that for $g_{1}=\frac{\phi^{2}}{2}$, $g_{2}=0$ and
$F=\bar{F}+\frac{L_{3}^{2}\phi^{2}}{2r^{2}}$, where $F$ and $\bar{F}$ are
functions of the same argument, we recover the Case III potentials of [12] as
a special case of the potential (14.35).

Using the Inverse Noether Theorem, we have that for the QFI (14.36) the
associated gauged Noether symmetry (14.2) is:

$\displaystyle\eta_{1}$ $\displaystyle=$
$\displaystyle-2g_{1}\dot{r}+\dot{g}_{1}r-g_{2}$ (14.37) $\displaystyle f$
$\displaystyle=$ $\displaystyle-
g_{1}\dot{r}^{2}+F\left(g_{1}^{-1/2}r+\frac{1}{2}\int
g_{1}^{-3/2}g_{2}dt\right)+\frac{1}{4g_{1}}\left(\dot{g}_{1}r-g_{2}\right)^{2}.$
(14.38)

These results coincide with the ones of [12] provided one replaces
$\dot{\theta}$ from $L_{3}=r^{2}\dot{\theta}$.

We collect the results in Theorem 14.2.1.

###### Theorem 14.2.1

The integrable time-dependent central Newtonian potentials $V(t,r)$ with
angular momentum $L_{3}$ are the following:
a. The potentials
$V(t,r)=-\frac{\ddot{g}_{2}}{2g_{2}}r^{2}+\frac{\dot{g}}{g_{2}}r-\frac{L_{3}^{2}}{2r^{2}}$
which admit the LFIs $I=g_{2}\dot{r}-\dot{g}_{2}r+g$, where $g_{2}(t)\neq 0$
and $g(t)$ are arbitrary functions.
b. The potentials
$V(t,r)=\left[\frac{1}{8}\left(\frac{\dot{g}_{1}}{g_{1}}\right)^{2}-\frac{\ddot{g}_{1}}{4g_{1}}\right]r^{2}+\frac{1}{2g_{1}}\left(\dot{g}_{2}-g_{2}\frac{\dot{g}_{1}}{2g_{1}}\right)r+\frac{1}{2g_{1}}F\left(g_{1}^{-1/2}r+\frac{1}{2}\int
g_{1}^{-3/2}g_{2}dt\right)-\frac{L_{3}^{2}}{2r^{2}}$ which admit the QFIs
$I=g_{1}\dot{r}^{2}+(g_{2}-\dot{g}_{1}r)\dot{r}+F\left(g_{1}^{-1/2}r+\frac{1}{2}\int
g_{1}^{-3/2}g_{2}dt\right)+\frac{1}{4g_{1}}\left(\dot{g}_{1}r-g_{2}\right)^{2}$,
where $g_{1}(t)\neq 0$ and $g_{2}(t)$ are arbitrary functions.

####  14.3 Applications of Theorem 14.2.1

From Theorem 14.2.1, it follows that it is possible to classify all integrable
time-dependent central Newtonian potentials in just two cases, according to if
they admit a. LFIs or b. QFIs. In this section, we consider various
applications of Theorem 14.2.1.

#####  14.3.1 The time-dependent oscillator

In this case, the potential is of the form $V=-\omega(t)r^{2}$, where
$\omega(t)$ is an arbitrary function. The LFIs and the QFIs are as follows:

a. For $g(t)=0$ and $L_{3}=0\implies\theta(t)=const$, we have the time-
dependent potential $V=-\frac{\ddot{g}_{2}}{2g_{2}}r^{2}$ with the LFI
$I=g_{2}\dot{r}-\dot{g}_{2}r$.

b. For $g_{2}=0$ and
$F=\frac{c_{0}}{2g_{1}}r^{2}+L_{3}^{2}\frac{g_{1}}{r^{2}}$, where $c_{0}$ is
an arbitrary constant, we obtain the time-dependent potential

$V=-\left[\frac{\ddot{g}_{1}}{4g_{1}}-\frac{1}{8}\left(\frac{\dot{g}_{1}}{g_{1}}\right)^{2}-\frac{c_{0}}{4g_{1}^{2}}\right]r^{2}$
(14.39)

with the QFI

$I=g_{1}\left(\dot{r}^{2}+\frac{L_{3}^{2}}{r^{2}}\right)-\dot{g}_{1}r\dot{r}+\frac{\dot{g}_{1}^{2}}{4g_{1}}r^{2}+\frac{c_{0}}{2g_{1}}r^{2}.$
(14.40)

If we replace $L_{3}=r^{2}\dot{\theta}$, the QFI (14.40) is the sum of the
diagonal components of the Jauch-Hill-Fradkin tensor. If, in addition,
$g_{1}=\frac{\phi^{2}}{2}$ and $c_{0}=\frac{K}{2}$, where $\phi(t)$ is an
arbitrary function and $K$ is a constant, we derive the Case I QFI of [12]

$I=\frac{1}{2}\left(\phi\dot{r}-\dot{\phi}r\right)^{2}+\frac{1}{2}r^{2}\phi^{2}\dot{\theta}^{2}+\frac{K}{2\phi^{2}}r^{2}.$
(14.41)

The above results for the time-dependent oscillator coincide with the ones of
Theorem 6.2 of [173].

#####  14.3.2 The time-dependent generalized Kepler potential

In this case, $V=-\frac{\omega(t)}{r^{\nu}}$ where $\nu$ is an arbitrary non-
zero constant.

a. No new integrable potentials.

b. For $g_{1}=\frac{\phi^{2}}{2}$, $g_{2}=0$, and
$F=\bar{F}+\frac{L_{3}^{2}\phi^{2}}{2r^{2}}$, where $F,\bar{F}$ are functions
of the same argument and $\phi(t)$ is an arbitrary smooth function, we find
the Case III potential of [12]

$V=-\frac{\ddot{\phi}}{2\phi}r^{2}+\phi^{-2}\bar{F}\left(\frac{r}{\phi}\right)$
(14.42)

with the QFI

$I=\frac{1}{2}(\phi\dot{r}-r\dot{\phi})^{2}+\frac{L_{3}^{2}\phi^{2}}{2r^{2}}+\bar{F}\left(\frac{r}{\phi}\right)=\frac{1}{2}(\phi\dot{r}-r\dot{\phi})^{2}+\frac{1}{2}\phi^{2}r^{2}\dot{\theta}^{2}+\bar{F}\left(\frac{r}{\phi}\right)$
(14.43)

where we replaced $L_{3}=r^{2}\dot{\theta}$.

We choose
$\bar{F}\left(\frac{r}{\phi}\right)=k_{1}\frac{r^{2}}{\phi^{2}}-\frac{k\phi^{\nu}}{r^{\nu}}$
with $\phi=\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}$ and
$k_{1}=\frac{b_{0}b_{2}}{2}-\frac{b_{1}^{2}}{8}$, where
$\nu,k,b_{0},b_{1},b_{2}$ are arbitrary constants. Then, the potential (14.42)
becomes the integrable time-dependent generalized Kepler potential

$V=-\frac{\omega_{\nu}(t)}{r^{\nu}},\enskip\omega_{\nu}=k\left(b_{0}+b_{1}t+b_{2}t^{2}\right)^{\frac{\nu-2}{2}}$
(14.44)

which admits the QFI

$J_{\nu}=(b_{0}+b_{1}t+b_{2}t^{2})\left[\frac{1}{2}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-\frac{\omega_{\nu}}{r^{\nu}}\right]-\frac{b_{1}+2b_{2}t}{2}r\dot{r}+\frac{b_{2}r^{2}}{2}.$
(14.45)

This result is in accordance with Propositions 12.9.1 and 12.11.1.

We note that in the case of the standard Kepler potential (i.e. $\nu=1$) for
$k_{1}=0\implies b_{1}^{2}-4b_{0}b_{2}=0$, the ‘frequency’
$\omega_{1}=\frac{k}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}$ degenerates into the
well-known result $\omega_{1}=\frac{k}{a_{0}+a_{1}t}$, where $a_{0}$ and
$a_{1}$ are constants (see Theorem 2.1 of [76] and Proposition 12.10.1).

#####  14.3.3 Integrating the equations of motion for the case a. potentials
of Theorem 14.2.1

In Theorem 14.2.1, we found the new class of time-dependent integrable central
potentials (14.31), which admit the LFIs (14.32).

Using the transformation $R=\frac{g_{2}}{r}$, we compute
$\dot{R}=\frac{\dot{g}_{2}r-g_{2}\dot{r}}{r^{2}}$. Replacing this in the LFI
(14.32), we find the solution

$r(t)=g_{2}\int\frac{I-g}{g_{2}^{2}}dt+c$ (14.46)

where $c$ is an integration constant.

Substituting (14.46) in $L_{3}=r^{2}\dot{\theta}$, we find that

$\theta(t)=\int\frac{L_{3}}{r^{2}(t)}dt+\theta_{0}$ (14.47)

where $\theta_{0}$ is another integration constant.

#####  14.3.4 Integrating the Scrödinger equation for an integrated central
potential

We consider a special class of integrable central potentials with fixed
angular momentum $L_{3}$ generated from case b. of Theorem 14.2.1 for
$g_{1}=\frac{\phi^{2}}{2}$, $g_{2}=0$ and $F=-k\frac{\phi}{r}$, where
$\phi(t)$ is an arbitrary non-zero function and $k\neq 0$ is an arbitrary
constant. Replacing these choices in the defining formula, we obtain the
potential

$V(t,r)=-\frac{\ddot{\phi}}{2\phi}r^{2}-\frac{k}{\phi}r^{-1}-\frac{L_{3}^{2}}{2r^{2}}.$
(14.48)

The associated QFI is

$I=\frac{1}{2}\left(\phi\dot{r}-\dot{\phi}r\right)^{2}-k\frac{\phi}{r}.$
(14.49)

By introducing the transformation $R=\frac{\phi}{r}$, we compute
$\dot{R}=\frac{\dot{\phi}r-\phi\dot{r}}{r^{2}}$ and, replacing in the QFI
(14.49), we find that $\frac{dR}{R^{2}\sqrt{2(I+kR)}}=\pm\frac{dt}{\phi^{2}}$.
Using the LFI of the angular momentum, we find the following orbit:

$L_{3}=r^{2}\dot{\theta}\implies
d\theta=L_{3}R^{2}\frac{dt}{\phi^{2}}\implies\int d\theta=\pm
L_{3}\int\frac{dR}{\sqrt{2(I+kR)}}\implies$
$\theta=\pm\frac{L_{3}}{k}\sqrt{2\left(I+\frac{k\phi}{r}\right)}+\theta_{0}$
(14.50)

where $\theta_{0}$ is an integration constant.

In [12], it has been shown that the wavefunction for the potential
$V_{III}(t,r)=-\frac{\ddot{\phi}}{2\phi}r^{2}+\phi^{-2}F\left(\frac{r}{\phi}\right)$
is given by the relation

$\psi(r,\theta,t)=|\phi|^{-1/2}e^{\frac{i}{2\hbar}\frac{\dot{\phi}}{\phi}r^{2}}\bar{\psi}\left(\phi^{-1}r,\theta,T(t)\right)$
(14.51)

where444We note that in eq. (6.14) of [12] the second term into the brackets
should be multiplied by 2. $\bar{\psi}(R,\Theta,T)=e^{-i\lambda
T/\hbar}e^{im\Theta}R^{-1/2}A(R)$, $\hbar$ is the Planck constant, $\lambda$
and $m$ are arbitrary constants, $R=\phi^{-1}r$, $\Theta=\theta$,
$T(t)=\int\phi^{-2}dt\implies\dot{T}=\phi^{-2}$ and $A(R)$ is an arbitrary
smooth function which satisfies the second order ODE (see eq. (6.6) of [12])

$\frac{d^{2}A}{dR^{2}}+\left(\frac{2\lambda}{\hbar^{2}}-\frac{2}{\hbar^{2}}F-\frac{m^{2}-\frac{1}{4}}{R^{2}}\right)A=0.$
(14.52)

We observe that for $F\left(\frac{r}{\phi}\right)=-k\phi
r^{-1}-\frac{L_{3}^{2}\phi^{2}}{2r^{2}}\implies
F(R)=-\frac{k}{R}-\frac{L_{3}^{2}}{2R^{2}}$, the potential $V_{III}(t,r)$
reduces to the potential (14.48) for which the orbit equation has been found.
For this choice of $F$, the ODE (14.52) becomes

$\frac{d^{2}A}{dR^{2}}+\left(\frac{2\lambda}{\hbar^{2}}+\frac{2k}{\hbar^{2}R}-\frac{m^{2}-\frac{1}{4}-\frac{L_{3}^{2}}{\hbar^{2}}}{R^{2}}\right)A=0.$
(14.53)

From Table 22.6 in p. 781 of [247], we find that (14.53) admits the solution
$A(R)=e^{-R/2}R^{\frac{a+1}{2}}L^{(a)}_{b}(R)$, where $L^{(a)}_{b}(R)$ is a
generalized Laguerre polynomial and the constants $\lambda,k,m$ are fixed as
follows: $\lambda=-\frac{\hbar^{2}}{8}$,
$k=\frac{\hbar^{2}}{4}\left(2b+a+1\right)$ and
$m^{2}=\frac{a^{2}}{4}+\frac{L_{3}^{2}}{\hbar^{2}}$.

Substituting the above results in (14.51), we find the wavefunction

$\psi(r,\theta,t)=|\phi|^{-1/2}e^{\frac{i}{2\hbar}\frac{\dot{\phi}}{\phi}r^{2}}e^{i\hbar
T(t)/8}e^{i\sqrt{\frac{a^{2}}{4}+\frac{L_{3}^{2}}{\hbar^{2}}}\theta}\phi^{1/2}r^{-1/2}e^{-\frac{r}{2\phi}}\left(\frac{r}{\phi}\right)^{\frac{a+1}{2}}L^{(a)}_{b}(\phi^{-1}r)$
(14.54)

provided the defining constant of the potential (14.48) is
$k=\frac{\hbar^{2}}{4}\left(2b+a+1\right)$.

#####  14.3.5 The integrable two-body problem with variable mass

In Newtonian Physics, the motion of a point of variable mass $m(t)$ is
described by the dynamical equation [248, 249]

$m(t)\ddot{\mathbf{R}}=\mathbf{F}+\sum_{i}\dot{m}_{i}\mathbf{u}_{i}$ (14.55)

where $\mathbf{R}$ is the position vector of $m(t)$ wrt a fixed frame of
reference, $\mathbf{F}$ denotes the external forces, $\mathbf{u}_{i}$ is the
relative velocity of the escaping mass from the $i$th-point of the surface
surrounding $m(t)$, and $\dot{m}_{i}$ is the rate of loss of mass from the
$i$th-point. If the loss of mass is continuous, the summation symbol may be
replaced by a double integral over the whole surface around $m$.

It is said that the loss of mass is isotropic iff
$\sum_{i}\dot{m}_{i}\mathbf{u}_{i}=0$. For example, this is the case when the
stars lose mass by radiation.

The two-body problem with variable mass [248] (e.g. a binary system of stars)
consists of two point masses $m_{1}(t)$ and $m_{2}(t)$ with only gravitational
attraction. Assuming isotropic loss of mass, the equations of motion wrt a
fixed inertial frame are:

$m_{1}(t)\ddot{\mathbf{r}}_{1}=\frac{Gm_{1}m_{2}}{r^{2}}\hat{\mathbf{r}},\enskip
m_{2}(t)\ddot{\mathbf{r}}_{2}=-\frac{Gm_{1}m_{2}}{r^{2}}\hat{\mathbf{r}}$
(14.56)

where $G$ is the gravitational constant,
$\mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}$ is the relative position of the
star with mass $m_{2}$ wrt the other star with mass $m_{1}$,
$r=\|\mathbf{r}\|$ and the unit vector
$\hat{\mathbf{r}}=\frac{\mathbf{r}}{r}$. From equations (14.56), we find the
dynamical equation

$\ddot{\mathbf{r}}=-\frac{Gm(t)}{r^{2}}\hat{\mathbf{r}}$ (14.57)

where $m(t)\equiv m_{1}(t)+m_{2}(t)$ is the total mass of the binary system.
If $m(t)=const$, the system is called closed and the orbit is a well-known
conic section. The potential driving the system is the time-dependent Kepler
potential

$V(t,r)=-\frac{\omega(t)}{r}$ (14.58)

where the ‘frequency’ $\omega(t)=Gm(t)$. The problem which has been around for
a long time was the determination of the mass $m(t)$ so that the potential
(14.58) is integrable. It has been found [250, 251, 252, 253, 254] that this
potential is integrable for the following $m(t)$:

$m_{I}(t)=\frac{1}{a_{0}+a_{1}t},\enskip
m_{II}(t)=\frac{1}{\sqrt{b_{0}+b_{1}t}},\enskip
m_{III}(t)=\frac{1}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}.$

Using Theorem 14.2.1, we derive these functions easily. They are subcases of
the ‘frequency’ $\omega_{\nu}$ of the integrable time-dependent generalized
Kepler potential (14.44) for $\nu=1$ and $k=G$. Indeed, we have:

$\omega_{1}(t)=Gm(t)\implies\frac{G}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}=Gm(t)\implies
m(t)=\frac{1}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}=m_{III}(t).$

For $b_{2}=0$, we obtain the mass $m_{II}(t)$, and for vanishing discriminant
$b_{1}^{2}-4b_{0}b_{2}=0$ the mass $m_{I}(t)$.

#####  14.3.6 Time-dependent integrable Yukawa and interatomic potentials

In plasma physics, solid-state physics and nuclear physics, the following
types of central potentials are widely used:

1\. The Yukawa type potentials

$V(r)=A\frac{e^{-Br}}{r}$ (14.59)

where $A$ and $B$ are arbitrary constants. This type of potentials describes
the screened Coulomb potential [255] generated around a positive charged
particle into a neutral fluid (e.g. a plasma of electrons in a background of
heavy positive charged ions [256]), and also models successfully the neutron-
proton interaction [257].

2\. The interatomic pair potentials [258, 259]

$V(r)=\frac{A}{r^{m}}-\frac{B}{r^{n}}$ (14.60)

where $A,B,m,n$ are arbitrary positive constants. These central potentials
manifest between the atoms of diatomic molecules. From (14.60), we observe
that they consist of a repulsive term $\frac{A}{r^{m}}$ and an attractive term
$-\frac{B}{r^{n}}$. The most well-known potential of this form is the Lennard-
Jones potential [260] in which $m=12$ and $n=6$.

Using Theorem 14.2.1, we shall answer to the following problem:
_Assume the parameters $A,B$ of the potentials (14.59) and (14.60) to be time-
dependent. Then, find for which functions $A(t)$ and $B(t)$ the resulting
potentials are integrable._

The case b. potentials of Theorem 14.2.1 for $g_{2}=0$ and
$F=-\frac{c_{1}}{4g_{1}}r^{2}+\frac{L_{3}^{2}g_{1}}{r^{2}}+\bar{F}\left(g_{1}^{-1/2}r\right)$,
where $c_{1}$ is an arbitrary constant and $\bar{F}$ a smooth function of its
argument, reduce to the integrable potentials

$V(t,r)=-\left[\frac{\ddot{g}_{1}}{4g_{1}}-\frac{1}{8}\left(\frac{\dot{g}_{1}}{g_{1}}\right)^{2}+\frac{c_{1}}{8g_{1}^{2}}\right]r^{2}+\frac{1}{2g_{1}}\bar{F}\left(g_{1}^{-1/2}r\right)$
(14.61)

which admit the QFIs

$I=\left(\dot{r}^{2}+\frac{L_{3}^{2}}{r^{2}}\right)g_{1}-\dot{g}_{1}r\dot{r}-\frac{c_{1}}{4g_{1}}r^{2}+\frac{\dot{g}_{1}^{2}}{4g_{1}}r^{2}+\bar{F}\left(g_{1}^{-1/2}r\right).$
(14.62)

From (14.61), we see that integrable time-dependent central potentials of the
form $V(t,r)=\frac{1}{2g_{1}}\bar{F}\left(g_{1}^{-1/2}r\right)$ exist only
when $g_{1}(t)=b_{0}+b_{1}t+b_{2}t^{2}$ and $c_{1}=b_{1}^{2}-4b_{2}b_{0}$,
where $b_{0},b_{1},b_{2}$ are arbitrary constants. For special choices of the
function $\bar{F}$, we obtain the required integrable potentials as follows:

1\. $\bar{F}=\frac{2k\exp{\left(-g_{1}^{-1/2}r\right)}}{g_{1}^{-1/2}r}$, where
$k$ is an arbitrary constant.

We find the new integrable time-dependent Yukawa type potentials

$V(t,r)=\frac{k}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}\frac{e^{-\frac{r}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}}}{r}$
(14.63)

where $A(t)=\frac{k}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}$ and
$B(t)=\frac{1}{\sqrt{b_{0}+b_{1}t+b_{2}t^{2}}}$.

2\. $\bar{F}=\frac{2k_{1}g_{1}^{m/2}}{r^{m}}-\frac{2k_{2}g_{1}^{n/2}}{r^{n}}$,
where $k_{1},k_{2}$ are arbitrary constants.

We find the new integrable time-dependent interatomic pair potentials

$V(t,r)=\frac{k_{1}(b_{0}+b_{1}t+b_{2}t^{2})^{\frac{m-2}{2}}}{r^{m}}-\frac{k_{2}(b_{0}+b_{1}t+b_{2}t^{2})^{\frac{n-2}{2}}}{r^{n}}$
(14.64)

where $A(t)=k_{1}(b_{0}+b_{1}t+b_{2}t^{2})^{\frac{m-2}{2}}$ and
$B(t)=k_{2}(b_{0}+b_{1}t+b_{2}t^{2})^{\frac{n-2}{2}}$.

####  14.4 Conclusions

Using the LFI of angular momentum and the direct method, we have managed to
compute the integrable time-dependent central potentials. These potentials are
widely used in all branches of Physics. It is remarkable that so divergent
potentials can be squeezed into two simple classes, i.e. the ones which admit
LFIs and the ones which admit QFIs. One may ask: Why in [12] not all the
integrable time-dependent central Newtonian potentials were found? The reason
is that in [12] the generalized Killing equations (1.64) and (1.65), instead
of the dynamical equations, were used. Indeed, the former result from the
Noether condition, which is written in the form
$A(t,q,\dot{q})+B_{a}(t,q,\dot{q})\ddot{q}^{a}=0$, by requiring $B_{a}=0$ and
without using the dynamical equations to replace the $\ddot{q}^{a}$ (see
section 1.7).

###  Appendix A Properties of $2\times 2$ real matrices with two equal
eigenvalues $\lambda_{+}=\lambda_{-}$

Let $A=\left(\begin{array}[]{cc}a&b\\\ c&d\\\ \end{array}\right)$
$\in\mathbb{R}^{2\times 2}$. We assume that its eigenvalues $\lambda_{\pm}$
computed in (4.17) are equal, that is,

$\Delta=0\implies
tr(A)^{2}=4\det(A)\implies(a-d)^{2}=-4bc,\enskip\lambda_{\pm}\equiv\lambda=\frac{tr(A)}{2}=\frac{a+d}{2}.$
(A.1)

In order to find the corresponding eigenvectors $v_{\pm}$, we solve the linear
system

$Av=\lambda v\implies\begin{cases}av_{1}+bv_{2}=\frac{a+d}{2}v_{1}\\\
cv_{1}+dv_{2}=\frac{a+d}{2}v_{2}.\end{cases}$ (A.2)

We consider the following cases:

1) Case $b=0$. From equations (A.1), we find that $a=d$ and $\lambda=a$.

1.1. Subcase $c=0$. We have $A=\left(\begin{array}[]{cc}a&0\\\ 0&a\\\
\end{array}\right)=aI$ and $v=\left(\begin{array}[]{c}v_{1}\\\ v_{2}\\\
\end{array}\right)$.

1.2. Subcase $c\neq 0$. We have $A=\left(\begin{array}[]{cc}a&0\\\ c&a\\\
\end{array}\right)$ and $v=\left(\begin{array}[]{c}0\\\ v_{2}\\\
\end{array}\right)$.

2) Case $b\neq 0$. We have $c=-\frac{(a-d)^{2}}{4b}$ and
$\lambda=\frac{a+d}{2}$. The matrix $A=\left(\begin{array}[]{cc}a&b\\\
-\frac{(a-d)^{2}}{4b}&d\\\ \end{array}\right)$ and
$v=\left(\begin{array}[]{c}v_{1}\\\ \frac{d-a}{2b}v_{1}\\\
\end{array}\right)$.

We observe that the matrix $A$ of the subcase 1.1 is the only type of such
matrices which admits two distinct eigenvectors.

We collect the above results in the following proposition.

###### Proposition A.0.1

Consider an arbitrary square matrix $A\in\mathbb{R}^{2\times 2}$ which admits
one double eigenvalue $\lambda_{\pm}=\lambda$. Then, $\lambda=\frac{tr(A)}{2}$
and $tr(A)^{2}=4\det(A)$. There are three different types of such matrices
with the following eigenvectors:
i. $A=\left(\begin{array}[]{cc}a&b\\\ -\frac{(a-d)^{2}}{4b}&d\\\
\end{array}\right)$, $\lambda=\frac{a+d}{2}$ and
$v=\left(\begin{array}[]{c}v_{1}\\\ \frac{d-a}{2b}v_{1}\\\
\end{array}\right)$, where $b\neq 0$ and $v_{1}$ are arbitrary real constants.
ii. $A=\left(\begin{array}[]{cc}a&0\\\ 0&a\\\ \end{array}\right)=aI$,
$\lambda=a$ and $v=\left(\begin{array}[]{c}v_{1}\\\ v_{2}\\\
\end{array}\right)$, where $a,v_{1}$ and $v_{2}$ are arbitrary real constants.
iii. $A=\left(\begin{array}[]{cc}a&0\\\ c&a\\\ \end{array}\right)$,
$\lambda=a$ and $v=\left(\begin{array}[]{c}0\\\ v_{2}\\\ \end{array}\right)$,
where $c\neq 0$, $a$ and $v_{2}$ are arbitrary real constants.

In the case that $tr(A)=0\implies\det(A)=0$, $\lambda=0$, $d=-a$, and
$a^{2}=-bc$. Then, proposition A.0.1 reduces to the following proposition.

###### Proposition A.0.2

An arbitrary square matrix $A\in\mathbb{R}^{2\times 2}$ with $tr(A)=\det(A)=0$
has one eigenvalue $\lambda=0$ with multiplicity two. There are three
different types of such matrices:
i. $A=\left(\begin{array}[]{cc}a&b\\\ -\frac{a^{2}}{b}&-a\\\
\end{array}\right)$ with eigenvector $v=\left(\begin{array}[]{c}v_{1}\\\
-\frac{a}{b}v_{1}\\\ \end{array}\right)$, where $b\neq 0$ and $v_{1}$ are
arbitrary real constants.
ii. $A=0$ (zero matrix) with $v=\left(\begin{array}[]{c}v_{1}\\\ v_{2}\\\
\end{array}\right)$, where $v_{1}$ and $v_{2}$ are arbitrary real constants.
iii. $A=\left(\begin{array}[]{cc}0&0\\\ c&0\\\ \end{array}\right)$ and
$v=\left(\begin{array}[]{c}0\\\ v_{2}\\\ \end{array}\right)$, where $c\neq 0$.

From the conditions (A.1), we deduce that _the matrix $B=A-\lambda I$ has
$tr(B)=\det(B)=0$. This implies that $B^{2}=0$; therefore, $B^{k}=0$ for any
positive integer $k\geq 2$._ Indeed, we have
$tr(B)=tr(A)-2\lambda=tr(A)-2\frac{tr(A)}{2}=0$ and $\det(B)=\det(A-\lambda
I)=0$.

Since $\lambda=\frac{tr(A)}{2}$, the matrix
$B=\left(\begin{array}[]{cc}\frac{a-d}{2}&b\\\ c&\frac{d-a}{2}\\\
\end{array}\right)$ where $(a-d)^{2}=-4bc$. Therefore, for an arbitrary vector
$u=\left(\begin{array}[]{c}u_{1}\\\ u_{2}\\\
\end{array}\right)\in\mathbb{R}^{2\times 1}$ we find that

$B^{2}u=0\implies B(Bu)=0\implies(A-\lambda I)(Bu)=0\implies
A(Bu)=\lambda(Bu)\implies Bu=v$ (A.3)

where $v$ is the eigenvector of $A$ corresponding to the double eigenvalue
$\lambda$ (see proposition A.0.1). _We note that $Bu$ is an eigenvector of
$A$._ If $u$ is linearly dependent with $v$, then $Bu=0$. Therefore, $Bu\neq
0$ only when $u,v$ are linearly independent. Replacing with the three
different types of matrices found in proposition A.0.1, we obtain the
following relations between the vectors $u$ and $v$:

i.

$\left(\begin{array}[]{cc}\frac{a-d}{2}&b\\\
-\frac{(a-d)^{2}}{4b}&\frac{d-a}{2}\\\
\end{array}\right)\left(\begin{array}[]{c}u_{1}\\\ u_{2}\\\
\end{array}\right)=\left(\begin{array}[]{c}v_{1}\\\ \frac{d-a}{2b}v_{1}\\\
\end{array}\right)\implies v_{1}=\frac{a-d}{2}u_{1}+bu_{2}.$

ii. In this case, $B=0$.

iii.

$\left(\begin{array}[]{cc}0&0\\\ c&0\\\
\end{array}\right)\left(\begin{array}[]{c}u_{1}\\\ u_{2}\\\
\end{array}\right)=\left(\begin{array}[]{c}0\\\ v_{2}\\\
\end{array}\right)\implies v_{2}=cu_{1}.$

We observe that $A$ admits two distinct eigenvectors only in the case ii. of
proposition A.0.1. In this case, $Bu=0$ for all vectors $u$. Moreover, the
above relations determine which eigenvector $v$ of $A$ is associated with an
arbitrary vector $u$ if its components $u_{1}$ and $u_{2}$ are given.

###### Proposition A.0.3

Let $A$ be a matrix of either type i. or type iii. of proposition A.0.1. Then,
$A$ admits only one distinct eigenvector $v$ and for an arbitrary vector
$u\in\mathbb{R}^{2\times 1}$ linearly independent with $v$, it holds that

$A^{k}u=\lambda^{k}u+k\lambda^{k-1}v,\enskip\forall k\in\mathbb{N}.$ (A.4)

###  Appendix B Complex solutions for the linear system $\dot{q}=Aq$

Consider the linear system $\dot{q}=Aq$, where $q=(q^{1},...,q^{n})$ and
$A\in\mathbb{R}^{n\times n}$, and let

$q(t)=x(t)+iy(t)\iff q^{a}(t)=x^{a}(t)+iy^{a}(t)$ (B.1)

where $x(t),y(t)\in\mathbb{R}^{n\times 1}$, be a complex solution of the
system. Then, $\dot{x}=Ax$ and $\dot{y}=Ay$. Therefore, the real vectors
$x=Re(q)$ and $y=Im(q)$ are solutions of the system as well.

The complex conjugate $\bar{q}(t)=x(t)-iy(t)$ of the solution $q(t)$ is also a
solution of the system. _The solutions $q(t)$ and $\bar{q}(t)$ are linearly
independent_ because $\bar{q}(t)$ is generated by the complex conjugate
eigenvalue $\bar{\lambda}$ of the eigenvalue $\lambda$, which produces the
$q(t)$. We recall that $Av=\lambda v\implies A\bar{v}=\bar{\lambda}\bar{v}$.

Using $q(t)$ and $\bar{q}(t)$, the real solutions $x(t)$ and $y(t)$ are
written as $x(t)=\frac{q(t)+\bar{q}(t)}{2}$ and
$y(t)=\frac{q(t)-\bar{q}(t)}{2i}$. We shall prove now that the vectors $x(t)$
and $y(t)$ are also linearly independent.

Let $c_{1},c_{2}$ be complex constants such that $c_{1}x(t)+c_{2}y(t)=0$.
Then, it is sufficient to show that $c_{1}=c_{2}=0$. Indeed, we have:
$c_{1}\frac{q(t)+\bar{q}(t)}{2}+c_{2}\frac{q(t)-\bar{q}(t)}{2i}=0\implies(c_{2}+ic_{1})q(t)-(c_{2}-ic_{1})\bar{q}(t)=0$.
The last relation implies that $c_{2}=-ic_{1}$ and $c_{2}=ic_{1}$ due to the
fact that $q(t)$ and $\bar{q}(t)$ are linearly independent. Therefore,
$c_{1}=c_{2}=0$ which implies that $x(t)$ and $y(t)$ are linearly independent.

We collect the above results in the following proposition.

###### Proposition B.0.1

If $q(t)=x(t)+iy(t)$ is a complex solution of the linear system
$\dot{q}=Aq(t)$, where $q(t)\in\mathbb{C}^{n\times 1}$ and
$A\in\mathbb{R}^{n\times n}$, then the real vectors $x(t)$ and $y(t)$ are
linearly independent solutions of the system.

Let now $\lambda=\sigma+i\omega$, where $\sigma,\omega\in\mathbb{R}$, be a
complex eigenvalue of $A$ with corresponding complex eigenvector (i.e.
$Av=\lambda v$) $v=w+iz$, where $w,z\in\mathbb{R}^{n\times 1}$. Then, we have
the special complex solution $q(t)=e^{\lambda t}v$ which is written as
follows:

$\displaystyle q(t)$ $\displaystyle=$ $\displaystyle
e^{(\sigma+i\omega)t}(w+iz)=e^{\sigma t}\left[\cos(\omega t)+i\sin(\omega
t)\right](w+iz)$ $\displaystyle=$ $\displaystyle e^{\sigma
t}\left\\{\left[w\cos(\omega t)-z\sin(\omega t)\right]+i\left[w\sin(\omega
t)+z\cos(\omega t)\right]\right\\}$ $\displaystyle=$
$\displaystyle\phi(t)+i\psi(t)$

where we have introduced the real vectors

$\displaystyle\phi(t)$ $\displaystyle=$ $\displaystyle Re(q(t))=e^{\sigma
t}\left[w\cos(\omega t)-z\sin(\omega t)\right]$ (B.2) $\displaystyle\psi(t)$
$\displaystyle=$ $\displaystyle Im(q(t))=e^{\sigma t}\left[w\sin(\omega
t)+z\cos(\omega t)\right].$ (B.3)

From proposition B.0.1, the vector fields $\phi(t)$ and $\psi(t)$ are linearly
independent real solutions of the system. The same result is obtained if we
work with the complex conjugate eigenvalue $\bar{\lambda}$.

Using the above results, we end up with the following proposition.

###### Proposition B.0.2

The general solution of the real linear system (for $n=2$) $\dot{q}=Aq$, where
$q(t)\in\mathbb{R}^{2\times 1}$ and $A\in\mathbb{R}^{2\times 2}$, is of the
form111Since $\phi(t)$ and $\psi(t)$ are linearly independent, the vector
$q(0)=c_{1}\phi(0)+c_{2}\psi(0)$ represents any possible initial condition in
the plane $q^{1},q^{2}$.

$q(t)=c_{1}\phi(t)+c_{2}\psi(t)$ (B.4)

iff $A$ admits the complex eigenvalues $\lambda_{\pm}=\sigma\pm i\omega$ (see
eq. (4.17) ) with corresponding eigenvectors $v_{\pm}=w\pm iz$. The parameters
$c_{1},c_{2}$ are arbitrary real constants computed by the initial conditions
and $\phi(t),\psi(t)$ are given by (B.2) and (B.3), respectively.

Replacing (B.2) and (B.3) in the general solution (B.4), we find

$\displaystyle q(t)$ $\displaystyle=$ $\displaystyle e^{\sigma
t}\left\\{\left[c_{1}\cos(\omega t)+c_{2}\sin(\omega
t)\right]w-\left[c_{1}\sin(\omega t)-c_{2}\cos(\omega t)\right]z\right\\}$
(B.5) $\displaystyle=$ $\displaystyle Re^{\sigma t}\left[\cos(\omega
t+\theta)w-\sin(\omega t+\theta)z\right]$

where $R=\sqrt{c_{1}^{2}+c_{2}^{2}}$, $\cos(-\theta)=\frac{c_{1}}{R}$ and
$\sin(-\theta)=\frac{c_{2}}{R}$. We note that the vectors
$w,z\in\mathbb{R}^{2\times 1}$ are linearly independent due to the fact that
the eigenvectors $v_{\pm}=w\pm iz$ produce the plane $q^{1},q^{2}$. It can be
easily checked that the vector (B.5) does satisfy equation $\dot{q}=Aq$ and,
also, it is compatible with any initial condition $q(0)$. Indeed, we have
$q(0)=R\cos\theta~{}w-R\sin\theta~{}z=c_{1}w+c_{2}z$ and from the eigenvalue
problem of $A$ we get:

$Av_{\pm}=\lambda_{\pm}v_{\pm}\implies A(w\pm iz)=(\sigma\pm i\omega)(w\pm
iz)\implies Aw\pm iAz=(\sigma w-\omega z)\pm i(\omega w+\sigma z)\implies$
$Aw=\sigma w-\omega z,\enskip Az=\omega w+\sigma z.$ (B.6)

Using the last relations, it is straightforward that (B.5) is the general
solution of the 2d linear system.

###  Appendix C Proof of Theorem 7.2.1

We look for solutions in which $g(t)$ and $f(t)$ are analytic functions so
that they can be represented by polynomial functions of $t$:

$\displaystyle g(t)$ $\displaystyle=$
$\displaystyle\sum_{k=0}^{n}c_{k}t^{k}=c_{0}+c_{1}t+...+c_{n}t^{n}$
$\displaystyle f(t)$ $\displaystyle=$
$\displaystyle\sum_{k=0}^{m}d_{k}t^{k}=d_{0}+d_{1}t+...+d_{m}t^{m}$

where $n,m\in\mathbb{N}$ (or infinite) and $c_{k},d_{k}\in\mathbb{R}$.

We recall that the quantities $C_{ab}(q)$ are KTs of order two. In what
follows, we consider various cases111 Equation (7.22) is not necessary because
the integrability condition $K_{,[ab]}=0$ does not intervene in the
calculations. However, it has been checked that equation (7.22) is always
satisfied identically from the solutions of the other equations of the system.
.

I. For both $\mathbf{n}$ and $\mathbf{m}$ finite.

I.1. Case $\mathbf{n=m}$:

Subcase $\mathbf{(n=0,m=0)}$. $g=c_{0}$ and $f=d_{0}$.

$\begin{cases}(\ref{FL.1.a})\implies d_{0}L_{(a;b)}+B_{(a;b)}=0\\\
(\ref{FL.1.b})\implies-2c_{0}C_{ab}V^{,b}+K_{,a}=0\\\ (\ref{FL.1.c})\implies
K_{,t}-d_{0}L_{b}V^{,b}-B_{b}V^{,b}=0\\\ (\ref{FL.1.d})\implies
d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}=0\end{cases}$

We define the vector field $\tilde{L}_{a}\equiv d_{0}L_{a}+B_{a}$. Then,
equation (7.18) implies that $\tilde{L}_{(a;b)}=0$, i.e. $\tilde{L}_{a}$ is a
KV, and (7.21) gives $\tilde{L}_{a}V^{,a}=s_{0}=const$.

Solving equation (7.20), we get $K=s_{0}t+G(q)$, which when replaced into
(7.19) gives $G_{,a}=2c_{0}C_{ab}V^{,b}$.

The QFI is

$I_{00}=c_{0}C_{ab}\dot{q}^{a}\dot{q}^{b}+\tilde{L}_{a}\dot{q}^{a}+s_{0}t+G(q)$

where $c_{0}C_{ab}$ is a KT, $\tilde{L}_{a}$ is a KV such that
$\tilde{L}_{a}V^{,a}=s_{0}$, and $G(q)=2c_{0}\int C_{ab}V^{,b}dq^{a}$.

The QFI $I_{00}$ consists of the independent FIs:

$Q_{1}=C_{ab}\dot{q}^{a}\dot{q}^{b}+G(q),\enskip
Q_{2}=L_{a}\dot{q}^{a}+s_{1}t$

where $C_{ab}$ is a KT, $L_{a}$ is a KV such that $L_{a}V^{,a}=s_{1}=const$,
and $G_{,a}=2C_{ab}V^{,b}$.

Subcase $\mathbf{(n=1,m=1)}$. $g=c_{0}+c_{1}t$ and $f=d_{0}+d_{1}t$ with
$c_{1}d_{1}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies
c_{1}C_{ab}+(d_{0}+d_{1}t)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2c_{1}C_{ab}V^{,b}t-2c_{0}C_{ab}V^{,b}+d_{1}L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left(d_{0}+d_{1}t\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0.\end{cases}$

From (7.18), $L_{a}$ is a KV and $c_{1}C_{ab}+B_{(a;b)}=0$.

From (7.21), we find that $L_{a}V^{,a}=s_{1}$ and
$\left(B_{b}V^{,b}\right)_{;a}=2c_{1}C_{ab}V^{,b}$. Then, (7.20) gives
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)+B_{a}V^{,a}t+G(q)$, which
when substituted into (7.19) yields $G_{,a}=2c_{0}C_{ab}V^{,b}-d_{1}L_{a}$.
Using the relation $\left(B_{b}V^{,b}\right)_{;a}=2c_{1}C_{ab}V^{,b}$, we find
that

$G_{,a}=\frac{c_{0}}{c_{1}}\left(B_{b}V^{,b}\right)_{;a}-d_{1}L_{a}\implies
L_{a}=\frac{c_{0}}{c_{1}d_{1}}\left(B_{b}V^{,b}\right)_{,a}-\frac{1}{d_{1}}G_{,a}.$

The QFI is

$I_{11}=-\frac{1}{c_{1}}\left(c_{0}+c_{1}t\right)B_{(a;b)}\dot{q}^{a}\dot{q}^{b}+\left(d_{0}+d_{1}t\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)+B_{a}V^{,a}t+G(q)$

where $B_{(a;b)}$ is a KT,
$L_{a}=\frac{c_{0}}{c_{1}d_{1}}\left(B_{b}V^{,b}\right)_{,a}-\frac{1}{d_{1}}G_{,a}\equiv\Phi_{,a}$
is a gradient KV such that $L_{a}V^{,a}=s_{1}$ and the vector $B_{a}$
satisfies the condition $\left(B_{b}V^{,b}\right)_{;a}=-2B_{(a;b)}V^{,b}$.

We observe that

$L_{a}=\Phi_{,a}=\frac{c_{0}}{c_{1}d_{1}}\left(B_{b}V^{,b}\right)_{,a}-\frac{1}{d_{1}}G_{,a}\implies
G(q)=\frac{c_{0}}{c_{1}}B_{a}V^{,a}-d_{1}\Phi(q).$

Therefore,

$\displaystyle I_{11}$ $\displaystyle=$
$\displaystyle\frac{c_{0}}{c_{1}}Q_{3}+Q_{4}+d_{0}Q_{2}+d_{1}Q_{5}$

where

$Q_{3}=-B_{(a;b)}\dot{q}^{a}\dot{q}^{b}+B_{a}V^{,a},\enskip
Q_{4}=-tB_{(a;b)}\dot{q}^{a}\dot{q}^{b}+B_{a}\dot{q}^{a}+tB_{a}V^{,a},\enskip
Q_{5}=tL_{a}\dot{q}^{a}+\frac{s_{1}}{2}t^{2}-\Phi(q)$

are independent FIs.

Subcase $\mathbf{(n=2,m=2)}$. $g=c_{0}+c_{1}t+c_{2}t^{2}$ and
$f=d_{0}+d_{1}t+d_{2}t^{2}$ with $c_{2}d_{2}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies(c_{1}+2c_{2}t)C_{ab}+(d_{0}+d_{1}t+d_{2}t^{2})L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2(c_{0}+c_{1}t+c_{2}t^{2})C_{ab}V^{,b}+(d_{1}+2d_{2}t)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies
2d_{2}L_{a}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2(c_{1}+2c_{2}t)C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we get $C_{ab}=0$ and the vectors $L_{a}$, $B_{a}$ are KVs.

From (7.21), we find that $L_{a}V^{,a}=s_{1}$ and
$L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}$, that is, $L_{a}$ is a
gradient KV.

The solution of (7.20) is
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+G(q)$
which when substituted into (7.19) gives

$G_{,a}+d_{1}L_{a}\underbrace{+2d_{2}L_{a}t+\left(B_{b}V^{,b}\right)_{,a}t}_{=0}=0\implies
G_{,a}=-d_{1}L_{a}=\frac{d_{1}}{2d_{2}}\left(B_{b}V^{,b}\right)_{,a}\implies
G(q)=\frac{d_{1}}{2d_{2}}B_{a}V^{,a}.$

The QFI is

$I_{22}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+\frac{d_{1}}{2d_{2}}B_{a}V^{,a}$

where $L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}$ is a gradient KV
such that $L_{a}V^{,a}=s_{1}$, and $B_{a}$ is a KV.

The QFI $I_{22}=d_{0}Q_{2}+d_{1}Q_{5}+F_{1}$. This expression contains the new
independent LFI
$F_{1}=t^{2}X_{a}\dot{q}^{a}+\frac{s}{3}t^{3}+B_{a}\dot{q}^{a}+B_{a}V^{,a}t$,
where $X_{a}\equiv d_{2}L_{a}$ and $s\equiv d_{2}s_{1}$

Subcase $\mathbf{(n=m>2)}$. $c_{n}d_{n}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies(c_{1}+2c_{2}t+...+nc_{n}t^{n-1})C_{ab}+(d_{0}+d_{1}t+..+d_{n}t^{n})L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2(c_{0}+c_{1}t+...+c_{n}t^{n})C_{ab}V^{,b}+(d_{1}+2d_{2}t+...+nd_{n}t^{n-1})L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+...+d_{n}t^{n}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left[2d_{2}+3\cdot
2d_{3}t+...+n(n-1)d_{n}t^{n-2}\right]L_{a}+\left(d_{0}+d_{1}t+...+d_{n}t^{n}\right)\left(L_{b}V^{,b}\right)_{;a}+\\\
\qquad\qquad\quad+\left(B_{b}V^{,b}\right)_{;a}-2(c_{1}+2c_{2}t+...+nc_{n}t^{n-1})C_{ab}V^{,b}=0.\end{cases}$

Equation (7.18) implies that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we find that $L_{a}=0$ and $B_{a}V^{,a}=s_{2}$.

The solution of (7.20) is $K=s_{2}t+G(q)$ which when substituted into (7.19)
gives $G=const$. Such a constant is ignored because any constant can be added
to $I$ without changing the condition $\frac{dI}{dt}=0$.

The FI is (of the form $Q_{2}$) $I_{nn}(n>2)=B_{a}\dot{q}^{a}+s_{2}t$, where
$B_{a}$ is a KV such that $B_{a}V^{,a}=s_{2}$.

We continue with the case $n>m$. This case is broken down equivalently into
the cases $n=m+1$ and $n>m+1$. Both cases are analyzed below222It is much more
convenient to follow this line of analysis because for $n>m+1$ equation (7.18)
implies directly that $C_{ab}=0$ and the derived FIs are linear. For $n=m+1$
all the derived FIs are quadratic..

I.2. Case $\mathbf{n=m+1}$.

Subcase $\mathbf{(n=1,m=0)}$. $g=c_{0}+c_{1}t$ and $f=d_{0}$ with $c_{1}\neq
0$.

$\begin{cases}\eqref{FL.1.a}\implies c_{1}C_{ab}+\tilde{L}_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2c_{1}C_{ab}V^{,b}t-2c_{0}C_{ab}V^{,b}+K_{,a}=0\\\
\eqref{FL.1.c}\implies K_{,t}-\tilde{L}_{a}V^{,a}=0\\\
\eqref{FL.1.d}\implies\left(\tilde{L}_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0.\end{cases}$

Solving (7.20), we get $K=\bar{L}_{a}V^{,a}t+G(q)$ which when substituted into
(7.19) gives $G_{,a}=2c_{0}C_{ab}V^{,b}$. However,
$2C_{ab}V^{,b}=\frac{1}{c_{1}}\left(\tilde{L}_{b}V^{,b}\right)_{;a}$;
therefore,
$G_{,a}=\frac{c_{0}}{c_{1}}\left(\tilde{L}_{b}V^{,b}\right)_{,a}\implies
G(q)=\frac{c_{0}}{c_{1}}\tilde{L}_{a}V^{,a}$.

The QFI is

$I_{10}=-\frac{1}{c_{1}}\left(c_{0}+c_{1}t\right)\tilde{L}_{(a;b)}\dot{q}^{a}\dot{q}^{b}+\tilde{L}_{a}\dot{q}^{a}+\left(t+\frac{c_{0}}{c_{1}}\right)\tilde{L}_{a}V^{,a}$

where $\tilde{L}_{a}$ is a vector such that $\tilde{L}_{(a;b)}$ is a reducible
KT and $\left(\tilde{L}_{b}V^{,b}\right)_{;a}=-2\tilde{L}_{(a;b)}V^{,b}$.

We note that
$I_{10}=\frac{c_{0}}{c_{1}}Q_{3}(\tilde{L}_{a})+Q_{4}(\tilde{L}_{a})$.

Subcase $\mathbf{(n=2,m=1)}$. $g=c_{0}+c_{1}t+c_{2}t^{2}$ and $f=d_{0}+d_{1}t$
with $c_{2}d_{1}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left(c_{1}+2c_{2}t\right)C_{ab}+\left(d_{0}+d_{1}t\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+c_{2}t^{2}\right)C_{ab}V^{,b}+d_{1}L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left(d_{0}+d_{1}t\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2\left(c_{1}+2c_{2}t\right)C_{ab}V^{,b}=0.\end{cases}$

Equation (7.18) gives $2c_{2}C_{ab}+d_{1}L_{(a;b)}=0$ and
$c_{1}C_{ab}+d_{0}L_{(a;b)}+B_{(a;b)}=0$.

From (7.21), we have that
$d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0$
and $d_{1}\left(L_{b}V^{,b}\right)_{;a}=4c_{2}C_{ab}V^{,b}$.

Solving (7.20), we find that
$K=\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)L_{a}V^{,a}+B_{a}V^{,a}t+G(q)$
which when substituted into (7.19) gives

$G_{,a}=2c_{0}C_{ab}V^{,b}-d_{1}L_{a}\implies
G_{,a}=\frac{c_{0}d_{1}}{2c_{2}}\left(L_{b}V^{,b}\right)_{,a}-d_{1}L_{a}\implies
G(q)=\frac{c_{0}d_{1}}{2c_{2}}L_{a}V^{,a}-d_{1}\int L_{a}dq^{a}.$

We note also that: $\begin{cases}2c_{2}C_{ab}+d_{1}L_{(a;b)}=0\\\
c_{1}C_{ab}+d_{0}L_{(a;b)}+B_{(a;b)}=0\end{cases}\implies
B_{(a;b)}=\left(\frac{2c_{2}d_{0}}{d_{1}}-c_{1}\right)C_{ab}$,

$\begin{cases}d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0\\\
d_{1}\left(L_{b}V^{,b}\right)_{;a}=4c_{2}C_{ab}V^{,b}\end{cases}\implies\left(B_{b}V^{,b}\right)_{;a}=2c_{1}C_{ab}V^{,b}-\frac{4c_{2}d_{0}}{d_{1}}C_{ab}V^{,b}$

and
$\frac{c_{1}d_{1}}{2c_{2}}\left(L_{b}V^{,b}\right)_{;a}=d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}$.
Therefore

$\begin{cases}\left(B_{b}V^{,b}\right)_{;a}=-2\left(\frac{2c_{2}d_{0}}{d_{1}}-c_{1}\right)C_{ab}V^{,b}\\\
B_{(a;b)}=\left(\frac{2c_{2}d_{0}}{d_{1}}-c_{1}\right)C_{ab}\end{cases}\implies\left(B_{b}V^{,b}\right)_{;a}=-2B_{(a;b)}V^{,b}\implies[B^{a},V^{,a}]\equiv[\mathbf{B},\mathbf{\nabla}V]^{a}\neq
0.$

The QFI is

$I_{21}=-\frac{d_{1}}{2c_{2}}\left(c_{0}+c_{1}t+c_{2}t^{2}\right)L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+\left(d_{0}+d_{1}t\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)L_{a}V^{,a}+B_{a}V^{,a}t+G(q)$

where $L_{a}$ is a vector such that $L_{(a;b)}$ is a KT and
$\left(L_{b}V^{,b}\right)_{;a}=-2L_{(a;b)}V^{,b}$, $B_{a}$ is a vector
satisfying the relations $\left(B_{b}V^{,b}\right)_{;a}=-2B_{(a;b)}V^{,b}$ and
$B_{(a;b)}=\frac{2c_{2}d_{0}-c_{1}d_{1}}{d_{1}}C_{ab}$, and
$G(q)=\frac{c_{0}d_{1}}{2c_{2}}L_{a}V^{,a}-d_{1}\int L_{a}dq^{a}$.

We observe that
$G_{,a}=\frac{c_{0}d_{1}}{2c_{2}}\left(L_{b}V^{,b}\right)_{,a}-d_{1}L_{a}$.
Therefore, $L_{a}=\Phi_{,a}$, i.e. $L_{a}$ is a gradient, and the function
$G(q)=\frac{c_{0}d_{1}}{2c_{2}}L_{a}V^{,a}-d_{1}\Phi(q)$.

Moreover, the condition
$B_{(a;b)}=\left(\frac{c_{1}d_{1}}{2c_{2}}-d_{0}\right)L_{(a;b)}$ implies that
$-\frac{c_{1}d_{1}}{2c_{2}}L_{(a;b)}=-B_{(a;b)}-d_{0}L_{(a;b)}$ and hence
$B_{(a;b)}$ is a KT.

Substituting the above results in the QFI $I_{21}$, we find

$I_{21}=\frac{c_{0}d_{1}}{2c_{2}}Q_{3}(L_{a})+Q_{4}+d_{0}Q_{4}(L_{a})+d_{1}Q_{6}$

where

$Q_{6}=-\frac{t^{2}}{2}L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+tL_{a}\dot{q}^{a}+\frac{t^{2}}{2}L_{a}V^{,a}-\Phi(q)$

is a new independent QFI.

We note that the expression

$Q_{1}+Q_{6}=-\frac{t^{2}}{2}L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+C_{ab}\dot{q}^{a}\dot{q}^{b}+tL_{a}\dot{q}^{a}+\frac{t^{2}}{2}L_{a}V^{,a}-\Phi(q)+G(q)$

where $\Phi_{,a}=L_{a}$ and $G_{,a}=2C_{ab}V^{,b}$, leads to the new
independent QFI

$Q_{16}=-\frac{t^{2}}{2}L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+C_{ab}\dot{q}^{a}\dot{q}^{b}+tL_{a}\dot{q}^{a}+\frac{t^{2}}{2}L_{a}V^{,a}+H(q)$

where $H_{,a}=2C_{ab}V^{,b}-L_{a}$. The FIs $Q_{1}$ and $Q_{6}$ are derived
from $Q_{16}$ as follows: $Q_{16}(C_{ab}=0)=Q_{6}$ and
$Q_{16}(L_{a}=0)=Q_{1}$.

Subcase $\mathbf{(n=3,m=2)}$. $g=c_{0}+c_{1}t+c_{2}t^{2}+c_{3}t^{3}$ and
$f=d_{0}+d_{1}t+d_{2}t^{2}$ with $c_{3}d_{2}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left(c_{1}+2c_{2}t+3c_{3}t^{2}\right)C_{ab}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+c_{2}t^{2}+c_{3}t^{3}\right)C_{ab}V^{,b}+(d_{1}+2d_{2}t)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies
2d_{2}L_{a}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2\left(c_{1}+2c_{2}t+3c_{3}t^{2}\right)C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we have that $3c_{3}C_{ab}+d_{2}L_{(a;b)}=0$,
$2c_{2}C_{ab}+d_{1}L_{(a;b)}=0$ and $c_{1}C_{ab}+d_{0}L_{(a;b)}+B_{(a;b)}=0$.

From (7.21), we find that
$d_{2}\left(L_{b}V^{,b}\right)_{;a}=6c_{3}C_{ab}V^{,b}$,
$d_{1}\left(L_{b}V^{,b}\right)_{;a}=4c_{2}C_{ab}V^{,b}$ and
$2d_{2}L_{a}+d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0$.

The solution of (7.20) is
$K=\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)L_{a}V^{,a}+B_{a}V^{,a}t+G(q)$
which when substituted into (7.19) and using the above derived conditions
gives

$G_{,a}=2c_{0}C_{ab}V^{,b}-d_{1}L_{a}\implies
G_{,a}=\frac{c_{0}d_{2}}{3c_{3}}\left(L_{b}V^{,b}\right)_{,a}-d_{1}L_{a}\implies
G(q)=\frac{c_{0}d_{2}}{3c_{3}}L_{a}V^{,a}-d_{1}\int L_{a}dq^{a}.$

From the first set of conditions, we get:

$\begin{cases}3c_{3}C_{ab}+d_{2}L_{(a;b)}=0\\\
2c_{2}C_{ab}+d_{1}L_{(a;b)}=0\\\
c_{1}C_{ab}+d_{0}L_{(a;b)}+B_{(a;b)}=0\end{cases}\implies\begin{cases}C_{ab}=-\frac{d_{2}}{3c_{3}}L_{(a;b)}\\\
\left(d_{1}-\frac{2c_{2}d_{2}}{3c_{3}}\right)L_{(a;b)}=0\\\
B_{(a;b)}=\left(\frac{3c_{3}d_{0}}{d_{2}}-c_{1}\right)C_{ab}\end{cases}$

and, from the second set, we get:

$\begin{cases}d_{2}\left(L_{b}V^{,b}\right)_{;a}=6c_{3}C_{ab}V^{,b}\\\
d_{1}\left(L_{b}V^{,b}\right)_{;a}=4c_{2}C_{ab}V^{,b}\\\
2d_{2}L_{a}+d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0\end{cases}\implies\begin{cases}\left(L_{b}V^{,b}\right)_{;a}=\frac{6c_{3}}{d_{2}}C_{ab}V^{,b}\\\
\left(\frac{6c_{3}d_{1}}{d_{2}}-4c_{2}\right)C_{ab}V^{,b}=0\\\
L_{a}=\left(\frac{c_{1}}{6c_{3}}-\frac{d_{0}}{2d_{2}}\right)\left(L_{b}V^{,b}\right)_{;a}-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}.\end{cases}$

Therefore, $L_{a}$ is a gradient vector and the function
$G(q)=\left(\frac{c_{0}d_{2}}{3c_{3}}-\frac{c_{1}d_{1}}{6c_{3}}+\frac{d_{0}d_{1}}{2d_{2}}\right)L_{a}V^{,a}+\frac{d_{1}}{2d_{2}}B_{a}V^{,a}$.

The QFI is

$\displaystyle I_{32}$ $\displaystyle=$
$\displaystyle-\frac{d_{2}}{3c_{3}}\left(c_{0}+c_{1}t+c_{2}t^{2}+c_{3}t^{3}\right)L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+$
$\displaystyle+\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)L_{a}V^{,a}+B_{a}V^{,a}t+\left(\frac{2c_{0}d_{2}-c_{1}d_{1}}{6c_{3}}+\frac{d_{0}d_{1}}{2d_{2}}\right)L_{a}V^{,a}+\frac{d_{1}}{2d_{2}}B_{a}V^{,a}.$

where the vector
$L_{a}=\left(\frac{c_{1}}{6c_{3}}-\frac{d_{0}}{2d_{2}}\right)\left(L_{b}V^{,b}\right)_{;a}-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}$
is a gradient such that $L_{(a;b)}$ is a KT,
$\left(\frac{2c_{2}d_{2}}{3c_{3}}-d_{1}\right)L_{(a;b)}=0$ and
$\left(L_{b}V^{,b}\right)_{;a}=-2L_{(a;b)}V^{,b}$, and $B_{a}$ is a vector
satisfying the relation
$B_{(a;b)}=\left(\frac{c_{1}d_{2}}{3c_{3}}-d_{0}\right)L_{(a;b)}$.

The vector $L_{a}=\Psi_{,a}$ where
$\Psi(q)=\left(\frac{c_{1}}{6c_{3}}-\frac{d_{0}}{2d_{2}}\right)L_{a}V^{,a}-\frac{1}{2d_{2}}B_{a}V^{,a}$.

We observe also that
$\frac{c_{2}d_{2}}{3c_{3}}L_{(a;b)}=\frac{d_{1}}{2}L_{(a;b)}$ and
$\frac{c_{1}d_{2}}{3c_{3}}L_{(a;b)}=B_{(a;b)}+d_{0}L_{(a;b)}$. The last
condition implies that $B_{(a;b)}$ is a reducible KT.

Another useful relation is the following333This is the condition which must be
satisfied in order $Q_{7}$ to be a FI.:

$L_{a}=\Psi_{,a}\implies
2d_{2}L_{a}=\left(\frac{c_{1}d_{2}}{3c_{3}}-d_{0}\right)\left(L_{b}V^{,b}\right)_{,a}-\left(B_{b}V^{,b}\right)_{,a}\implies$
$\left(B_{b}V^{,b}\right)_{,a}=-2B_{(a;b)}V^{,b}-2d_{2}L_{a}.$

Substituting the above relations in the QFI $I_{32}$, we find

$I_{32}=\frac{d_{2}c_{0}}{3c_{3}}Q_{3}(L_{a})+Q_{7}+d_{0}Q_{4}(L_{a})+d_{1}Q_{6}(\Psi)$

where

$Q_{7}=-\frac{t^{3}}{3}d_{2}L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+t^{2}d_{2}L_{a}\dot{q}^{a}+\frac{t^{3}}{3}d_{2}L_{a}V^{,a}-tB_{(a;b)}\dot{q}^{a}\dot{q}^{b}+B_{a}\dot{q}^{a}+tB_{a}V^{,a}$

is a new independent QFI.

Subcase $\mathbf{(n=m+1,m>2)}$. $c_{n}\neq 0$ and $d_{m}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left[c_{1}+2c_{2}t+...+(m+1)c_{n}t^{m}\right]C_{ab}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+...+c_{n}t^{m+1}\right)C_{ab}V^{,b}+(d_{1}+2d_{2}t+...+md_{m}t^{m-1})L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left[2d_{2}+3\cdot
2d_{3}t+...+m(m-1)d_{m}t^{m-2}\right]L_{a}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)\left(L_{b}V^{,b}\right)_{;a}+\\\
\qquad\qquad\quad+\left(B_{b}V^{,b}\right)_{;a}-2\left[c_{1}+2c_{2}t+...+(m+1)c_{n}t^{m}\right]C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find the conditions: $(k+1)c_{k+1}C_{ab}+d_{k}L_{(a;b)}=0$,
where $k=1,2,...,m$, and $c_{1}C_{ab}+d_{0}L_{(a;b)}+B_{(a;b)}=0$. For $k=m$,
we get $C_{ab}=-\frac{d_{m}}{(m+1)c_{m+1}}L_{(a;b)}$ and the remaining
conditions become:

$\left[d_{k}-\frac{(k+1)c_{k+1}d_{m}}{(m+1)c_{m+1}}\right]L_{(a;b)}=0,\enskip
k=1,2,...,m-1$

and

$B_{(a;b)}=\left[\frac{c_{1}d_{m}}{(m+1)c_{m+1}}-d_{0}\right]L_{(a;b)}.$

From (7.21), we find that
$2(k+1)c_{k+1}C_{ab}V^{,b}=d_{k}\left(L_{b}V^{,b}\right)_{;a}$, where
$k=m-1,m$,
$(k+2)(k+1)d_{k+2}L_{a}+d_{k}\left(L_{b}V^{,b}\right)_{;a}-2(k+1)c_{k+1}C_{ab}V^{,b}=0$,
where $k=1,...,m-2$, and
$2d_{2}L_{a}+d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0$,
i.e. $L_{a}$ is a gradient vector.

The first set of equations, for $k=m$, gives

$\left(L_{b}V^{,b}\right)_{,a}=\frac{2(m+1)c_{m+1}}{d_{m}}C_{ab}V^{,b}\implies\left(L_{b}V^{,b}\right)_{,a}=-2L_{(a;b)}V^{,b}$

and, for $k=m-1$,
$\left[d_{m-1}-\frac{mc_{m}d_{m}}{(m+1)c_{m+1}}\right]\left(L_{b}V^{,b}\right)_{,a}=0$.

The second set of equations, for $k=m-2$, gives

$L_{a}=\left[\frac{c_{m-1}}{m(m+1)c_{m+1}}-\frac{d_{m-2}}{m(m-1)d_{m}}\right]\left(L_{b}V^{,b}\right)_{,a}$

and for the remaining values of $k=1,2,...,m-3$ (exist only for $m>3$)

$\left\\{d_{k}+(k+2)(k+1)d_{k+2}\left[\frac{c_{m-1}}{m(m+1)c_{m+1}}-\frac{d_{m-2}}{m(m-1)d_{m}}\right]-\frac{(k+1)c_{k+1}d_{m}}{(m+1)c_{m+1}}\right\\}\left(L_{b}V^{,b}\right)_{,a}=0.$

The third set of equations gives

$\left(B_{b}V^{,b}\right)_{,a}=\left[\frac{c_{1}d_{m}}{(m+1)c_{m+1}}-\frac{2d_{2}c_{m-1}}{m(m+1)c_{m+1}}+\frac{2d_{2}d_{m-2}}{m(m-1)d_{m}}-d_{0}\right]\left(L_{b}V^{,b}\right)_{,a}.$

The solution of (7.20) is
$K=\left(d_{0}t+\frac{d_{1}}{2}t^{2}+...+\frac{d_{m}}{m+1}t^{m+1}\right)L_{a}V^{,a}+B_{a}V^{,a}t+G(q)$
which when substituted into (7.19) gives444Use the conditions:
$2d_{2}L_{a}+d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0$,
$2(k+1)c_{k+1}C_{ab}V^{,b}=d_{k}\left(L_{b}V^{,b}\right)_{;a}$, where
$k=m-1,m$, and
$(k+2)(k+1)d_{k+2}L_{a}+d_{k}\left(L_{b}V^{,b}\right)_{;a}-2(k+1)c_{k+1}C_{ab}V^{,b}=0$,
where $k=1,...,m-2$.

$\displaystyle G_{,a}=$
$\displaystyle-\left(d_{0}t+\frac{d_{1}}{2}t^{2}+...+\frac{d_{m-1}}{m}t^{m}+\frac{d_{m}}{m+1}t^{m+1}\right)\left(L_{b}V^{,b}\right)_{,a}+2\left(c_{0}+c_{1}t+...+c_{m}t^{m}+c_{m+1}t^{m+1}\right)C_{ab}V^{,b}-$
$\displaystyle-\left(B_{b}V^{,b}\right)_{,a}t-(d_{1}+2d_{2}t+...+md_{m}t^{m-1})L_{a}$
$\displaystyle=$ $\displaystyle 2c_{0}C_{ab}V^{,b}-d_{1}L_{a}$
$\displaystyle=$
$\displaystyle\left[\frac{c_{0}d_{m}}{(m+1)c_{m+1}}-\frac{c_{m-1}d_{1}}{m(m+1)c_{m+1}}+\frac{d_{1}d_{m-2}}{m(m-1)d_{m}}\right]\left(L_{b}V^{,b}\right)_{,a}\implies$
$G(q)=\left[\frac{c_{0}d_{m}}{(m+1)c_{m+1}}-\frac{c_{m-1}d_{1}}{m(m+1)c_{m+1}}+\frac{d_{1}d_{m-2}}{m(m-1)d_{m}}\right]L_{a}V^{,a}.$

Therefore,
$K=\left(d_{0}t+\frac{d_{1}}{2}t^{2}+...+\frac{d_{m}}{m+1}t^{m+1}\right)L_{a}V^{,a}+B_{a}V^{,a}t+\left[\frac{c_{0}d_{m}}{(m+1)c_{m+1}}-\frac{c_{m-1}d_{1}}{m(m+1)c_{m+1}}+\frac{d_{1}d_{m-2}}{m(m-1)d_{m}}\right]L_{a}V^{,a}$.

The QFI is

$\displaystyle I_{(m+1)m}(m>2)$ $\displaystyle=$
$\displaystyle-\frac{d_{m}}{(m+1)c_{m+1}}\left(c_{0}+c_{1}t+...+c_{m+1}t^{m+1}\right)L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+$
$\displaystyle+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+B_{a}V^{,a}t+$
$\displaystyle+\left(d_{0}t+\frac{d_{1}}{2}t^{2}+...+\frac{d_{m}}{m+1}t^{m+1}\right)L_{a}V^{,a}+G(q)$

where $c_{m+1}d_{m}\neq 0$ for a finite $m>2$. The vector

$L_{a}=\left[\frac{c_{m-1}}{m(m+1)c_{m+1}}-\frac{d_{m-2}}{m(m-1)d_{m}}\right]\left(L_{b}V^{,b}\right)_{,a}$

is a gradient such that $L_{(a;b)}$ is a KT,
$\left(L_{b}V^{,b}\right)_{,a}=-2L_{(a;b)}V^{,b}$,
$\left[d_{k}-\frac{(k+1)c_{k+1}d_{m}}{(m+1)c_{m+1}}\right]L_{(a;b)}$ $=0$ with
$k=1,2,...,m-1$,
$\left[d_{m-1}-\frac{mc_{m}d_{m}}{(m+1)c_{m+1}}\right]\left(L_{b}V^{,b}\right)_{,a}$
$=0$ and

$\left\\{d_{k}+(k+2)(k+1)d_{k+2}\left[\frac{c_{m-1}}{m(m+1)c_{m+1}}-\frac{d_{m-2}}{m(m-1)d_{m}}\right]-\frac{(k+1)c_{k+1}d_{m}}{(m+1)c_{m+1}}\right\\}\left(L_{b}V^{,b}\right)_{,a}=0$

with $k=1,2,...,m-3$. The vector $B_{a}$ satisfies the conditions
$B_{(a;b)}=\left[\frac{c_{1}d_{m}}{(m+1)c_{m+1}}-d_{0}\right]L_{(a;b)}$ and

$\left(B_{b}V^{,b}\right)_{,a}=\left[\frac{c_{1}d_{m}}{(m+1)c_{m+1}}-\right.\left.\frac{2d_{2}c_{m-1}}{m(m+1)c_{m+1}}+\frac{2d_{2}d_{m-2}}{m(m-1)d_{m}}-d_{0}\right]\left(L_{b}V^{,b}\right)_{,a}.$

The function
$G(q)=\left[\frac{c_{0}d_{m}}{(m+1)c_{m+1}}-\frac{c_{m-1}d_{1}}{m(m+1)c_{m+1}}+\frac{d_{1}d_{m-2}}{m(m-1)d_{m}}\right]L_{a}V^{,a}$.

However, the considered case $m>2$ implies that $m-1>1$ and $m-2>0$;
therefore, there always exist $k$-values for $m-1$ and $m-2$.

For $k=m-2$, the condition
$\left[d_{k}-\frac{(k+1)c_{k+1}d_{m}}{(m+1)c_{m+1}}\right]L_{(a;b)}=0$ gives

$\left[d_{m-2}-\frac{(m-1)c_{m-1}d_{m}}{(m+1)c_{m+1}}\right]L_{(a;b)}=0\implies\left[\frac{c_{m-1}}{m(m+1)c_{m+1}}-\frac{d_{m-2}}{m(m-1)d_{m}}\right]\left(L_{b}V^{,b}\right)_{,a}=0\implies
L_{a}=0$

because $\left(L_{b}V^{,b}\right)_{,a}=-2L_{(a;b)}V^{,b}$.

Since $L_{a}=0$, we have $G=0$, $B_{(a;b)}=0$ (i.e. $B_{a}$ is a KV) and
$B_{a}V^{,a}=s=const$.

Therefore, $I_{(m+1)m}(m>2)=B_{a}\dot{q}^{a}+st=Q_{2}(B_{a})$.

I.3. Case $\mathbf{n>m+1}$.

Subcase $\mathbf{(n>1,m=0)}$. $c_{n}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies(c_{1}+2c_{2}t+...+nc_{n}t^{n-1})C_{ab}+\tilde{L}_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2(c_{0}+c_{1}t+...+c_{n}t^{n})C_{ab}V^{,b}+K_{,a}=0\\\
\eqref{FL.1.c}\implies K_{,t}=\tilde{L}_{b}V^{,b}\\\
\eqref{FL.1.d}\implies\left(\tilde{L}_{b}V^{,b}\right)_{;a}-2(c_{1}+2c_{2}t+...+nc_{n}t^{n-1})C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find that $C_{ab}=0$ and $\tilde{L}_{(a;b)}=0$, i.e.
$\tilde{L}_{a}$ is a KV. Then, (7.21) gives $\tilde{L}_{a}V^{,a}=s_{0}$, and
(7.20) yields $K=s_{0}t+G(q)$. The last result when substituted into (7.19)
gives $G_{,a}=0$ $\implies G=const\equiv 0$.

The FI is (of the form $Q_{2}$) $I_{n0}(n>1)=\tilde{L}_{a}\dot{q}^{a}+s_{0}t$,
where $\tilde{L}_{a}\equiv d_{0}L_{a}+B_{a}$ is a KV such that
$\tilde{L}_{a}V^{,a}=s_{0}$.

Subcase $\mathbf{(n>2,m=1)}$. $c_{n}d_{1}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}+\left(d_{0}+d_{1}t\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+...+c_{n}t^{n}\right)C_{ab}V^{,b}+d_{1}L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left(d_{0}+d_{1}t\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we have that $L_{a}V^{,a}=s_{1}$ and $B_{a}V^{,a}=s_{2}$, where
$s_{1}$ and $s_{2}$ are arbitrary constants.

Then, equation (7.20) gives
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)+s_{2}t+G(q)$ which when
substituted into (7.19) yields $G_{,a}=-d_{1}L_{a}$, that is, $L_{a}$ is a
gradient KV.

The FI is (consists of FIs of the form $Q_{2}$ and $Q_{5}$)

$I_{n1}(n>2)=(d_{0}+d_{1}t)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+\frac{s_{1}d_{1}}{2}t^{2}+(s_{1}d_{0}+s_{2})t-d_{1}\int
L_{a}dq^{a}$

where $L_{a}$ and $B_{a}$ are KVs such that $L_{a}V^{,a}=s_{1}$ and
$B_{a}V^{,a}=s_{2}$.

Subcase $\mathbf{(n>3,m=2)}$. $c_{n}d_{2}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+...+c_{n}t^{n}\right)C_{ab}V^{,b}+(d_{1}+2d_{2}t)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies
2d_{2}L_{a}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-\\\
\qquad\qquad-2\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we have that $L_{a}V^{,a}=s_{1}$ and
$L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}$, that is, $L_{a}$ is a
gradient KV.

Then, equation (7.20) gives
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+G(q)$
which when substituted into (7.19) yields

$G_{,a}=-d_{1}L_{a}=\frac{d_{1}}{2d_{2}}\left(B_{b}V^{,b}\right)_{,a}\implies
G(q)=\frac{d_{1}}{2d_{2}}B_{a}V^{,a}.$

The FI is (consists of FIs of the form $Q_{2}$, $Q_{5}$ and $Q_{7}$)

$I_{n2}(n>3)=(d_{0}+d_{1}t+d_{2}t^{2})L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+\frac{d_{1}}{2d_{2}}B_{a}V^{,a}$

where $L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}$ is a gradient KV
such that $L_{a}V^{,a}=s_{1}$ and $B_{a}$ is a KV.

Subcase $\mathbf{(n>m+1,m>2)}$. $c_{n}d_{m}\neq 0$. Note that $n>n-1>m>2$.

$\begin{cases}\eqref{FL.1.a}\implies\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+...+c_{n}t^{n}\right)C_{ab}V^{,b}+(d_{1}+2d_{2}t+...+md_{m}t^{m-1})L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left[2d_{2}+3\cdot
2d_{3}t+...+m(m-1)d_{m}t^{m-2}\right]L_{a}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)\left(L_{b}V^{,b}\right)_{;a}+\\\
\qquad\qquad\quad+\left(B_{b}V^{,b}\right)_{;a}-2\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we have that $L_{a}=0$ and $B_{a}V^{,a}=s_{2}$.

Then, the solution of (7.20) is $K=s_{2}t+G(q)$ which when substituted into
(7.19) gives $G=const\equiv 0$.

The FI is (again of the form $Q_{2}$)
$I_{nm}(n>m+1,m>2)=B_{a}\dot{q}^{a}+s_{2}t$, where $B_{a}$ is a KV such that
$B_{a}V^{,a}=s_{2}$.

I.4. Case $\mathbf{n<m}$.

Subcase $\mathbf{(n=0,m=1)}$. $g=c_{0}$ and $f=d_{0}+d_{1}t$ with $d_{1}\neq
0$.

$\begin{cases}\eqref{FL.1.a}\implies(d_{0}+d_{1}t)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2c_{0}C_{ab}V^{,b}+d_{1}L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies K_{,t}=(d_{0}+d_{1}t)L_{b}V^{,b}+B_{b}V^{,b}\\\
\eqref{FL.1.d}\implies(d_{0}+d_{1}t)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{;b}\right)_{;a}=0.\end{cases}$

Equation (7.18) implies that $L_{a}$ and $B_{a}$ are KVs.

Equation (7.21) gives $L_{b}V^{,b}=s_{1}=const$ and $B_{b}V^{,b}=s_{2}=const$.

Then, (7.20) gives
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)+s_{2}t+G(q)$ which when
substituted into (7.19) yields $G_{,a}=2c_{0}C_{ab}V^{,b}-d_{1}L_{a}$.

The QFI is ($c_{0}$ is absorbed by $C_{ab}$)

$I_{01}=C_{ab}\dot{q}^{a}\dot{q}^{b}+\left(d_{0}+d_{1}t\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)+s_{2}t+G\left(q\right)$

where $d_{1}\neq 0$, $L_{a}$ and $B_{a}$ are KVs such that $L_{a}V^{,a}=s_{1}$
and $B_{a}V^{,a}=s_{2}$, and $C_{ab}$ is a KT such that
$G_{,a}=2C_{ab}V^{,b}-d_{1}L_{a}$.

We see that $I_{01}=Q_{2}(B_{a})+d_{0}Q_{2}+Q_{8}$, where

$Q_{8}=C_{ab}\dot{q}^{a}\dot{q}^{b}+d_{1}tL_{a}\dot{q}^{a}+d_{1}\frac{s_{1}}{2}t^{2}+G(q)$

is just a subcase of $Q_{16}$ because $d_{1}L_{a}$ is a KV.

Subcase $\mathbf{(n=0,m=2)}$. $g=c_{0}$ and $f=d_{0}+d_{1}t+d_{2}t^{2}$ with
$d_{2}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2c_{0}C_{ab}V^{,b}+\left(d_{1}+2d_{2}t\right)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{b}V^{,b}+B_{b}V^{,b}\\\
\eqref{FL.1.d}\implies
2d_{2}L_{a}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}=0.\end{cases}$

From (7.18), we have that $L_{a}$ and $B_{a}$ are KVs.

From (7.21), we get $L_{b}V^{,b}=s_{1}$ and
$L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{,a}$, that is, $L_{a}$ is a
gradient KV.

Equation (7.20) yields
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{b}V^{,b}t+G(q)$
and (7.19) gives $G_{,a}=2c_{0}C_{ab}V^{,b}-d_{1}L_{a}$. Using the relation
$L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{,a}$, we find that
$G(q)=\frac{d_{1}}{2d_{2}}B_{a}V^{,a}+2c_{0}\int C_{ab}V^{,b}dq^{a}$.

The QFI is ($c_{0}$ is absorbed by $C_{ab}$)

$I_{02}=C_{ab}\dot{q}^{a}\dot{q}^{b}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+G\left(q\right)$

where $d_{2}\neq 0$, $B_{a}$ is a KV,
$L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{,a}$ is a gradient KV such
that $L_{a}V^{,a}=s_{1}$, and $C_{ab}$ is a KT satisfying the relation
$G_{,a}-2C_{ab}V^{,b}+d_{1}L_{a}=0$.

We note that $I_{02}=Q_{8}+d_{0}Q_{2}+Q_{7}(L_{a}=KV,B_{a}=KV)$.

Subcase $\mathbf{(n=0,m>2)}$. $d_{m}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2c_{0}C_{ab}V^{,b}+\left(d_{1}+2d_{2}t+...+md_{m}t^{m-1}\right)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{b}V^{;b}+B_{b}V^{;b}=0\\\
\eqref{FL.1.d}\implies\left[2d_{2}+3\cdot
2d_{3}t+...+m(m-1)d_{m}t^{m-2}\right]L_{a}+\\\
\qquad\qquad\quad+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)\left(L_{b}V^{;b}\right)_{;a}+\left(B_{b}V^{;b}\right)_{;a}=0.\end{cases}$

From (7.18), $L_{a}$ and $B_{a}$ are KVs.

From (7.21), $L_{a}=0$ and $B_{a}V^{,b}=s_{2}$.

Solving (7.20), we find $K=s_{2}t+G(q)$ which when substituted into (7.19)
gives $G_{,a}=2c_{0}C_{ab}V^{,b}$.

The QFI is (of the form $Q_{8}$)

$I_{0m}(m>2)=c_{0}C_{ab}\dot{q}^{a}\dot{q}^{b}+B_{a}\dot{q}^{a}+s_{2}t+2c_{0}\int
C_{ab}V^{,b}dq^{a}$

where $C_{ab}$ is a KT and $B_{a}$ is a KV such that $B_{a}V^{,a}=s_{2}$.

Subcase $\mathbf{(n=1,m=2)}$. $g=c_{0}+c_{1}t$ and $f=d_{0}+d_{1}t+d_{2}t^{2}$
with $c_{1}d_{2}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies
c_{1}C_{ab}+(d_{0}+d_{1}t+d_{2}t^{2})L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t\right)C_{ab}V^{,b}+\left(d_{1}+2d_{2}t\right)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies
2d_{2}L_{a}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0.\end{cases}$

Equation (7.18) implies that $L_{a}$ is a KV and $c_{1}C_{ab}=-B_{(a;b)}$.

From (7.21), we have $L_{a}V^{,a}=s_{1}$ and
$2d_{2}L_{a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0$.

The solution of (7.20) is
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+G(q)$
which when replaced into (7.19) gives

$-2\left(c_{0}+c_{1}t\right)C_{ab}V^{,b}+\left(d_{1}+2d_{2}t\right)L_{a}+G_{,a}+\left(B_{b}V^{,b}\right)_{,a}t=0\implies$
$G_{,a}-2c_{0}C_{ab}V^{,b}+d_{1}L_{a}+\underbrace{\left[-2c_{1}C_{ab}V^{,b}+2d_{2}L_{a}+\left(B_{b}V^{,b}\right)_{,a}\right]}_{=0}t=0\implies$
$G_{,a}=\underbrace{2c_{0}C_{ab}V^{,b}}-d_{1}L_{a}=c_{0}\underbrace{\frac{2d_{2}c_{0}}{c_{1}}L_{a}+\frac{c_{0}}{c_{1}}\left(B_{b}V^{,b}\right)_{;a}}-d_{1}L_{a}.$

The QFI is

$\displaystyle I_{12}$ $\displaystyle=$
$\displaystyle-\frac{1}{c_{1}}\left(c_{0}+c_{1}t\right)B_{(a;b)}\dot{q}^{a}\dot{q}^{b}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+$
$\displaystyle+B_{a}V^{,a}t+G\left(q\right)$

where $c_{1}d_{2}\neq 0$, $B_{(a;b)}$ is a KT and
$L_{a}=-\frac{1}{2d_{2}}\left[\left(B_{b}V^{,b}\right)_{,a}+2B_{(a;b)}V^{,b}\right]$
is a KV such that $L_{a}V^{,a}=s_{1}$. The function $G\left(q\right)$ is
defined by the condition
$G_{,a}-\frac{c_{0}}{c_{1}}\left(B_{b}V^{,b}\right)_{,a}+\left(d_{1}-\frac{2d_{2}c_{0}}{c_{1}}\right)L_{a}$
$=0$ which is analysed as follows:

1) For $d_{1}\neq\frac{2d_{2}c_{0}}{c_{1}}$.

Then, $L_{a}$ is a gradient KV, that is, $L_{a}=\Phi_{,a}\implies
G(q)=\frac{c_{0}}{c_{1}}B_{a}V^{,a}-d_{1}\Phi(q)+\frac{2d_{2}c_{0}}{c_{1}}\Phi(q)$.

The QFI becomes

$I_{12(1)}=\frac{c_{0}}{c_{1}}Q_{9}+d_{0}Q_{2}+d_{1}Q_{6}(L_{a}=KV)+Q_{7}(L_{a}=KV)$

where $Q_{9}=-B_{(a;b)}\dot{q}^{a}\dot{q}^{b}+B_{a}V^{,a}+2d_{2}\Phi(q)$. We
observe that $Q_{3}=Q_{9}(\Phi=0)$ and
$\left(B_{b}V^{,b}\right)_{,a}=-2B_{(a;b)}V^{,b}-2d_{2}\Phi_{,a}$.

2) For $d_{1}=\frac{2d_{2}c_{0}}{c_{1}}$.

We have $G(q)=\frac{c_{0}}{c_{1}}B_{a}V^{,a}$ and the QFI becomes
$I_{12(2)}=\frac{c_{0}}{c_{1}}Q_{3}+d_{0}Q_{2}+Q_{8}(C_{ab}=-B_{(a;b)})+Q_{7}(L_{a}=KV)$.

Subcase $\mathbf{(n=1,m>2)}$. $g=c_{0}+c_{1}t$ and $c_{1}d_{m}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies
c_{1}C_{ab}+(d_{0}+d_{1}t+...+d_{m}t^{m})L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t\right)C_{ab}V^{,b}+\left(d_{1}+2d_{2}t+...+md_{m}t^{m-1}\right)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left[2d_{2}+2\cdot
3d_{3}t+...+(m-1)md_{m}t^{m-2}\right]L_{a}+\\\
\qquad\qquad+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0.\end{cases}$

Equation (7.18) implies that $L_{a}$ is a KV and $c_{1}C_{ab}=-B_{(a;b)}$.

From (7.21), we have $L_{a}=0$ and
$\left(B_{b}V^{,b}\right)_{;a}=2c_{1}C_{ab}V^{,b}$.

The solution of (7.20) is $K=B_{a}V^{,a}t+G(q)$ which when substituted into
(7.19) gives

$-2c_{0}C_{ab}V^{,b}\underbrace{-2c_{1}C_{ab}V^{,b}t+\left(B_{b}V^{,b}\right)_{;a}t}_{=0}+G_{,a}=0\implies
G_{,a}=2c_{0}C_{ab}V^{,b}.$

But $\left(B_{b}V^{,b}\right)_{;a}=2c_{1}C_{ab}V^{,b}$; therefore,
$G_{,a}=\frac{c_{0}}{c_{1}}\left(B_{b}V^{,b}\right)_{;a}\implies
G(q)=\frac{c_{0}}{c_{1}}B_{a}V^{,a}$.

The QFI is (consists of FIs of the form $Q_{1}$ and $Q_{4}$)

$I_{1m}(m>2)=\left(c_{0}+c_{1}t\right)C_{ab}\dot{q}^{a}\dot{q}^{b}+B_{a}\dot{q}^{a}+B_{a}V^{,a}t+\frac{c_{0}}{c_{1}}B_{a}V^{,a}$

where $C_{ab}=-\frac{1}{c_{1}}B_{(a;b)}$ is a KT and $B_{a}$ is a vector such
that $\left(B_{b}V^{,b}\right)_{;a}+2B_{(a;b)}V^{,b}=0$.

Subcase $\mathbf{(n>1,m>n)}$. $c_{n}d_{m}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+...+c_{n}t^{n}\right)C_{ab}V^{,b}+(d_{1}+2d_{2}t+...+md_{m}t^{m-1})L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left[2d_{2}+3\cdot
2d_{3}t+...+m(m-1)d_{m}t^{m-2}\right]L_{a}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)\left(L_{b}V^{,b}\right)_{;a}+\\\
\qquad\qquad\quad+\left(B_{b}V^{,b}\right)_{;a}-2\left(c_{1}+2c_{2}t+...+nc_{n}t^{n-1}\right)C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we have that $L_{a}=0$ and $B_{a}V^{,a}=s_{2}$.

Then, the solution of (7.20) is $K=s_{2}t+G(q)$ which when substituted into
(7.19) gives $G=const\equiv 0$.

The FI is (of the form $Q_{2}$) $I_{nm}(n>1,m>n)=B_{a}\dot{q}^{a}+s_{2}t$,
where $B_{a}$ is a KV such that $B_{a}V^{,a}=s_{2}$.

II. For $\mathbf{n\to\infty}$ and $\mathbf{m}$ finite.

We find the equivalences: $(n=\infty,m=0)\equiv(n>1,m=0)\equiv(g=e^{\lambda
t},m=0)$, $(n=\infty,m=1)\equiv(n>2,m=1)\equiv(g=e^{\lambda t},m=1)$,
$(n=\infty,m=2)\equiv(n>3,m=2)\equiv(g=e^{\lambda t},m=2)$,
$(n=\infty,m>2)\equiv(n>m+1,m>2)\equiv(g=e^{\lambda t},m>2)$.

Then, for each case, we have the following.

II.1. Case $\mathbf{(g=e^{\lambda t}}$, $\mathbf{f=d_{0})}$. $\lambda\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\lambda e^{\lambda
t}C_{ab}+d_{0}L_{(a;b)}+B_{(a;b)}=0\\\ \eqref{FL.1.b}\implies-2e^{\lambda
t}C_{ab}V^{,b}+K_{,a}=0\\\ \eqref{FL.1.c}\implies
K_{,t}=d_{0}L_{a}V^{,a}+B_{a}V^{,a}\\\ \eqref{FL.1.d}\implies
d_{0}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2\lambda
e^{\lambda t}C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we get $C_{ab}=0$ and $\tilde{L}_{a}\equiv d_{0}L_{a}+B_{a}$ is a
KV.

From (7.21), we have that $\tilde{L}_{a}V^{,a}=s_{0}$.

Equation (7.20) gives $K=s_{0}t+G(q)$ which when substituted into (7.19)
yields $G=const\equiv 0$.

The FI is (of the form $Q_{2}$) $I_{e0}=\tilde{L}_{a}\dot{q}^{a}+s_{0}t$,
where $\tilde{L}_{a}$ is a KV such that $\tilde{L}_{a}V^{,a}=s_{0}$.

II.2. Case $\mathbf{(g=e^{\lambda t}}$, $\mathbf{f=d_{0}+d_{1}t)}$. $\lambda
d_{1}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\lambda e^{\lambda
t}C_{ab}+\left(d_{0}+d_{1}t\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2e^{\lambda t}C_{ab}V^{,b}+d_{1}L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left(d_{0}+d_{1}t\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2\lambda
e^{\lambda t}C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we have that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we get that $L_{a}V^{,a}=s_{1}$ and $B_{a}V^{,a}=s_{2}$.

Then, equation (7.20) gives
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}\right)+s_{2}t+G(q)$ which when
substituted into (7.19) gives $G_{,a}=-d_{1}L_{a}$.

The FI is (consists of $Q_{2}$, $Q_{5}$)

$I_{e1}=\left(d_{0}+d_{1}t\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+(s_{1}d_{0}+s_{2})t+\frac{s_{1}d_{1}}{2}t^{2}-d_{1}\int
L_{a}dq^{a}$

where $L_{a}=-\frac{1}{d_{1}}G_{,a}$ is a gradient KV such that
$L_{a}V^{,a}=s_{1}$ and $B_{a}$ is a KV such that $B_{a}V^{,a}=s_{2}$.

II.3. Case $\mathbf{(g=e^{\lambda t}}$, $\mathbf{f=d_{0}+d_{1}t+d_{2}t^{2})}$.
$\lambda d_{2}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies\lambda e^{\lambda
t}C_{ab}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2e^{\lambda
t}C_{ab}V^{,b}+(d_{1}+2d_{2}t)L_{a}+K_{,a}=0\\\ \eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies
2d_{2}L_{a}+\left(d_{0}+d_{1}t+d_{2}t^{2}\right)\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2\lambda
e^{\lambda t}C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we have that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we get that $L_{a}V^{,a}=s_{1}$ and
$\left(B_{b}V^{,b}\right)_{;a}=-2d_{2}L_{a}$, that is, $L_{a}$ is a gradient
KV.

Then, equation (7.20) gives
$K=s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+G(q)$
which when substituted into (7.19) yields $G_{,a}=-d_{1}L_{a}$.

We observe that

$\begin{cases}G_{,a}=-d_{1}L_{a}\\\
L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}\end{cases}\implies
G_{,a}=\frac{d_{1}}{2d_{2}}\left(B_{b}V^{,b}\right)_{,a}\implies
G=\frac{d_{1}}{2d_{2}}B_{b}V^{,b}.$

The FI is (consists of $Q_{2}$, $Q_{5}$ and $Q_{7}$)

$I_{e2}=\left(d_{0}+d_{1}t+d_{2}t^{2}\right)L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+s_{1}\left(d_{0}t+\frac{d_{1}}{2}t^{2}+\frac{d_{2}}{3}t^{3}\right)+B_{a}V^{,a}t+\frac{d_{1}}{2d_{2}}B_{b}V^{,b}$

where $L_{a}=-\frac{1}{2d_{2}}\left(B_{b}V^{,b}\right)_{;a}$ is a gradient KV
such that $L_{a}V^{,a}=s_{1}$ and $B_{a}$ is a KV.

II.4. Case $\mathbf{(g=e^{\lambda t}}$, $\mathbf{m>2)}$. $\lambda d_{m}\neq
0$.

$\begin{cases}\eqref{FL.1.a}\implies\lambda e^{\lambda
t}C_{ab}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2e^{\lambda
t}C_{ab}V^{,b}+\left(d_{1}+2d_{2}t+...+md_{m}t^{m-1}\right)L_{a}+K_{,a}=0\\\
\eqref{FL.1.c}\implies
K_{,t}=\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)L_{a}V^{,a}+B_{a}V^{,a}\\\
\eqref{FL.1.d}\implies\left[2d_{2}+3\cdot
2t+...+m(m-1)d_{m}t^{m-2}\right]L_{a}+\left(d_{0}+d_{1}t+...+d_{m}t^{m}\right)\left(L_{b}V^{,b}\right)_{;a}+\\\
\qquad\qquad+\left(B_{b}V^{,b}\right)_{;a}-2\lambda e^{\lambda
t}C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we have that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we get that $L_{a}=0$ and $B_{a}V^{,a}=s_{2}$.

Then, equation (7.20) gives $K=s_{2}t+G(q)$ which when substituted into (7.19)
gives $G=const\equiv 0$.

Therefore, the FI is (of the form $Q_{2}$)
$I_{em}(m>2)=B_{a}\dot{q}^{a}+s_{2}t$, where $B_{a}$ is a KV such that
$B_{a}V^{,a}=s_{2}$. This is a time-dependent LFI.

III. For $\mathbf{n}$ finite and $\mathbf{m\to\infty}$.

We distinguish between two cases because in condition (7.21) we have to
compare polynomial coefficients of the infinite sums $f_{,tt}$ and $f$.

III.1. Case with $\mathbf{f_{,tt}\neq\lambda^{2}f}$.

$(n=0,m=\infty)\equiv(n=0,m>2),\enskip(n=1,m=\infty)\equiv(n=1,m>2),\enskip(n>1,m=\infty)\equiv(n>1,m>n).$

III.2. Case with $\mathbf{f_{,tt}=\lambda^{2}f}$.

$(n=0,m=\infty)\equiv(n=0,f=e^{\lambda
t}),\enskip(n=1,m=\infty)\equiv(n=1,f=e^{\lambda
t}),\enskip(n>1,m=\infty)\equiv(n>1,f=e^{\lambda t}).$

For each subcase, we have the following.

Subcase $\mathbf{(g=c_{0}}$, $\mathbf{f=e^{\lambda t})}$. $\lambda\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies e^{\lambda t}L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2c_{0}C_{ab}V^{,b}+\lambda e^{\lambda
t}L_{a}+K_{,a}=0\\\ \eqref{FL.1.c}\implies K_{,t}=e^{\lambda
t}L_{b}V^{,b}+B_{b}V^{,b}\\\ \eqref{FL.1.d}\implies\lambda^{2}e^{\lambda
t}L_{a}+e^{\lambda
t}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}=0.\end{cases}$

Equation (7.18) implies that $L_{a}$ and $B_{a}$ are KVs.

From (7.21), we find that $B_{a}V^{,b}=s_{2}$ and
$L_{a}=-\frac{1}{\lambda^{2}}\left(L_{b}V^{,b}\right)_{,a}$, that is, $L_{a}$
is a gradient KV.

From (7.20), we find $K=\frac{1}{\lambda}e^{\lambda t}L_{b}V^{,b}+s_{2}t+G(q)$
which when substituted into (7.19) gives $G_{,a}=2c_{0}C_{ab}V^{,b}$.

The QFI is ($c_{0}$ is absorbed by $C_{ab}$)

$I_{0e}=C_{ab}\dot{q}^{a}\dot{q}^{b}+e^{\lambda
t}L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+\frac{1}{\lambda}e^{\lambda
t}L_{a}V^{,a}+s_{2}t+G\left(q\right)$

where $\lambda\neq 0$,
$L_{a}=-\frac{1}{\lambda^{2}}\left(L_{b}V^{,b}\right)_{,a}$ is a gradient KV,
$B_{a}$ is a KV such that $B_{a}V^{;a}=s_{2}$, and $C_{ab}$ is a KT such that
$G_{,a}-2C_{ab}V^{,b}=0$.

We note that $I_{0e}=Q_{1}+Q_{2}(B_{a})+Q_{10}$ where

$Q_{10}=e^{\lambda
t}\left(L_{a}\dot{q}^{a}+\frac{1}{\lambda}L_{a}V^{,a}\right)$

is a new independent LFI.

Subcase $\mathbf{(g=c_{0}+c_{1}t}$, $\mathbf{f=e^{\lambda t})}$. $\lambda
c_{1}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies c_{1}C_{ab}+e^{\lambda
t}L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t\right)C_{ab}V^{,b}+\lambda
e^{\lambda t}L_{a}+K_{,a}=0\\\ \eqref{FL.1.c}\implies K_{,t}=e^{\lambda
t}L_{b}V^{,b}+B_{b}V^{,b}\\\ \eqref{FL.1.d}\implies\lambda^{2}e^{\lambda
t}L_{a}+e^{\lambda
t}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2c_{1}C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find that $L_{a}$ is a KV and $c_{1}C_{ab}=-B_{(a;b)}$.

From (7.21), we get the conditions
$L_{a}=-\frac{1}{\lambda^{2}}\left(L_{b}V^{,b}\right)_{;a}$ and
$\left(B_{b}V^{,b}\right)_{;a}=2c_{1}C_{ab}V^{,b}$.

Equation (7.20) gives $K=\frac{1}{\lambda}e^{\lambda
t}L_{b}V^{,b}+B_{b}V^{,b}t+G(q)$. Substituting $K$ into (7.19), we obtain

$G_{,a}=2c_{0}C_{ab}V^{,b}=\frac{c_{0}}{c_{1}}\left(B_{b}V^{,b}\right)_{;a}\implies
G(q)=\frac{c_{0}}{c_{1}}B_{b}V^{,b}.$

The QFI is

$I_{1e}=-\frac{1}{c_{1}}\left(c_{0}+c_{1}t\right)B_{(a;b)}\dot{q}^{a}\dot{q}^{b}+e^{\lambda
t}L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+\frac{1}{\lambda}e^{\lambda
t}L_{a}V^{,a}+B_{a}V^{,a}t+\frac{c_{0}}{c_{1}}B_{a}V^{,a}$

where $\lambda c_{1}\neq 0$,
$L_{a}=-\frac{1}{\lambda^{2}}\left(L_{b}V^{,b}\right)_{,a}$ is a gradient KV,
and $B_{a}$ is such that $B_{(a;b)}$ is a KT and
$\left(B_{b}V^{,b}\right)_{,a}=-2B_{(a;b)}V^{,b}$.

We note that $I_{1e}=\frac{c_{0}}{c_{1}}Q_{3}+Q_{4}+Q_{10}$.

Subcase $\mathbf{(n>1}$, $\mathbf{f=e^{\lambda t})}$. $\lambda c_{n}\neq 0$.

$\begin{cases}\eqref{FL.1.a}\implies(c_{1}+2c_{2}t+...+nc_{n}t^{n-1})C_{ab}+e^{\lambda
t}L_{(a;b)}+B_{(a;b)}=0\\\
\eqref{FL.1.b}\implies-2\left(c_{0}+c_{1}t+...+c_{n}t^{n}\right)C_{ab}V^{,b}+\lambda
e^{\lambda t}L_{a}+K_{,a}=0\\\ \eqref{FL.1.c}\implies K_{,t}=e^{\lambda
t}L_{b}V^{,b}+B_{b}V^{,b}\\\ \eqref{FL.1.d}\implies\lambda^{2}e^{\lambda
t}L_{a}+e^{\lambda
t}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2(c_{1}+2c_{2}t+...+nc_{n}t^{n-1})C_{ab}V^{,b}=0.\end{cases}$

From (7.18), we find that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs. Then,
(7.21) implies that $B_{a}V^{,a}=s_{2}$ and
$L_{a}=-\frac{1}{\lambda^{2}}\left(L_{b}V^{,b}\right)_{;a}$, i.e. $L_{a}$ is a
gradient KV.

The solution of (7.20) is $K=\frac{1}{\lambda}e^{\lambda
t}L_{b}V^{,b}+s_{2}t+G(q)$ which when substituted into (7.19) gives
$G=const\equiv 0$.

The FI is (consists of $Q_{2}$ and $Q_{10}$)

$I_{ne}(n>1)=e^{\lambda
t}L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+\frac{1}{\lambda}e^{\lambda
t}L_{a}V^{,a}+s_{2}t$

where $L_{a}=-\frac{1}{\lambda^{2}}\left(L_{b}V^{,b}\right)_{;a}$ is a
gradient KV and $B_{a}$ is a KV such that $B_{a}V^{,a}=s_{2}$.

IV. Both $\mathbf{n}$ and $\mathbf{m}$ are infinite.

We consider three cases.

IV.1. Case where $\mathbf{f_{,tt}=\lambda^{2}f}$ and
$\mathbf{g_{,t}\neq\lambda f}$.

$(n=\infty,m=\infty)\equiv(g=e^{\mu t},f=e^{\lambda
t},\lambda\neq\mu)\equiv(n>1,f=e^{\lambda t}).$

Subcase $\mathbf{(g=e^{\lambda t}}$, $\mathbf{f=e^{\mu t})}$. $\lambda\mu\neq
0$.

$\begin{cases}\eqref{FL.1.a}\implies\lambda e^{\lambda t}C_{ab}+e^{\mu
t}L_{(a;b)}+B_{(a;b)}=0\\\ \eqref{FL.1.b}\implies-2e^{\lambda
t}C_{ab}V^{,b}+\mu e^{\mu t}L_{a}+K_{,a}=0\\\ \eqref{FL.1.c}\implies
K_{,t}=e^{\mu t}L_{a}V^{,a}+B_{a}V^{,a}\\\ \eqref{FL.1.d}\implies\mu^{2}e^{\mu
t}L_{a}+e^{\mu
t}\left(L_{b}V^{,b}\right)_{;a}+\left(B_{b}V^{,b}\right)_{;a}-2\lambda
e^{\lambda t}C_{ab}V^{,b}=0.\end{cases}$

a) For $\lambda\neq\mu$:

From (7.18), we have that $C_{ab}=0$ and $L_{a}$, $B_{a}$ are KVs.

From (7.21), we find that $\mu^{2}L_{a}+\left(L_{b}V^{,b}\right)_{;a}=0$ and
$B_{b}V^{,b}=s_{2}$.

The solution of (7.20) is $K=\frac{1}{\mu}e^{\mu t}L_{a}V^{,a}+s_{2}t+G(q)$
which when replaced into (7.19) and using the relation
$\mu^{2}L_{a}+\left(L_{b}V^{,b}\right)_{;a}=0$ gives $G=const\equiv 0$.

The FI is (consists of $Q_{2}$ and $Q_{10}$)

$I_{ee}(\lambda\neq\mu)=e^{\mu
t}L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+\frac{1}{\mu}e^{\mu t}L_{a}V^{a,}+s_{2}t$

where $L_{a}=-\frac{1}{\mu^{2}}\left(L_{b}V^{,b}\right)_{;a}$ is a gradient KV
and $B_{a}$ is a KV such that $B_{b}V^{,b}=s_{2}$. This is a time-dependent
LFI.

b) For $\lambda=\mu$:

From (7.18), we have that $\lambda C_{ab}+L_{(a;b)}=0$ and $B_{a}$ is a KV.

From (7.21), we find that
$\lambda^{2}L_{a}+\left(L_{b}V^{,b}\right)_{;a}-2\lambda C_{ab}V^{,b}=0$ and
$B_{b}V^{,b}=s_{2}$.

The solution of (7.20) is $K=\frac{1}{\lambda}e^{\lambda
t}L_{a}V^{,a}+s_{2}t+G(q)$ which when substituted into (7.19) and using the
relation $\lambda^{2}L_{a}+\left(L_{b}V^{,b}\right)_{;a}-2\lambda
C_{ab}V^{,b}=0$ gives $G=const\equiv 0$.

The QFI is

$I_{ee}(\lambda=\mu)=-\frac{1}{\lambda}e^{\lambda
t}L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+e^{\lambda
t}L_{a}\dot{q}^{a}+B_{a}\dot{q}^{a}+\frac{1}{\lambda}e^{\lambda
t}L_{a}V^{,a}+s_{2}t$

where $\lambda\neq 0$, $L_{a}$ is such that $L_{(a;b)}$ is a KT,
$\lambda^{2}L_{a}+\left(L_{b}V^{,b}\right)_{,a}+2L_{(a;b)}V^{,b}=0$, and
$B_{a}$ is a KV such that $B_{a}V^{,a}=s_{2}$.

We note that $I_{ee}(\lambda=\mu)=Q_{2}(B_{a})+Q_{11}$, where

$Q_{11}=e^{\lambda
t}\left(-\frac{1}{\lambda}L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+L_{a}\dot{q}^{a}+\frac{1}{\lambda}L_{a}V^{,a}\right)$

is a new independent QFI. Then, $Q_{10}=Q_{11}(L_{a}=KV)$.

IV.2. Case where $\mathbf{f_{,tt}=\lambda^{2}f}$ and $\mathbf{g_{,t}=\lambda
f}$.

$(n=\infty,m=\infty)\equiv(g=e^{\lambda t},f=e^{\lambda t}).$

IV.3. Case where $\mathbf{f_{,tt}\neq\lambda^{2}f}$ and
$\mathbf{g_{,t}\neq\lambda f}$ or $\mathbf{g_{,t}=\lambda f}$.

$(n=\infty,m=\infty)\equiv(n>1,m>n)\equiv(n>m+1,m>2)\equiv(g=e^{\lambda
t},m>2)\equiv(n=m,m>2).$

By collecting all the above LFIs/QFIs $Q_{A}$, the derivation of Theorem 7.2.1
is straightforward. Specifically, we cover all the FIs mentioned in Theorem
7.2.1 as follows:

$I_{1}=Q_{16},\enskip I_{2}=Q_{7},\enskip I_{3}=Q_{11}.$

The FIs $Q_{1}$, $Q_{3}$, $Q_{5}$, $Q_{6}$, $Q_{8}$, $Q_{9}$ are subcases of
$I_{1}$; the $Q_{2}$, $Q_{4}$ are subcases of $I_{2}$; and, finally, $Q_{10}$
is a subcase of $I_{3}$.

After a careful and extensive study, we have shown that the LFIs/QFIs of an
autonomous conservative dynamical system can be produced by the three
parameterized FIs listed in Theorem 7.2.1.

###  Appendix D Proof of Theorem 9.4.1

Recall that

$K_{ab}(t,q)=C_{(0)ab}(q)+C_{(1)ab}(q)t+C_{(2)ab}(q)\frac{t^{2}}{2}+...+C_{(n)ab}(q)\frac{t^{n}}{n}$

and

$K_{a}(t,q)=L_{(0)a}(q)+L_{(1)a}(q)t+L_{(2)a}(q)t^{2}+...+L_{(m)a}(q)t^{m}.$

We consider various cases111Equation (9.24) is not necessary because the
integrability condition $K_{,[ab]}=0$ does not intervene in the calculations.
However, it has been checked that equation (9.24) is always satisfied
identically from the solutions of the other equations of the system..

I. Case $\mathbf{n=m}$ (both $n$ and $m$ are finite)

Equation (9.20) implies that: $C_{(1)ab}=-L_{(0)(a;b)}-2C_{(0)c(a}A_{b)}^{c}$,
$C_{(k)ab}=-L_{(k-1)(a;b)}-\frac{2}{k-1}C_{(k-1)c(a}A_{b)}^{c}$ with
$k=2,...,n$, and $L_{(n)(a;b)}=-\frac{2}{n}C_{(n)c(a}A_{b)}^{c}$.

Equation (9.23) gives: $L_{(n)a}Q^{a}=s$,
$\left(L_{(n-1)b}Q^{b}\right)_{,a}=2C_{(n)ab}Q^{b}-nL_{(n)b}A^{b}_{a}$, and
$\left(L_{(k-2)b}Q^{b}\right)_{,a}=2C_{(k-1)ab}Q^{b}-k(k-1)L_{(k)a}-(k-1)L_{(k-1)b}A^{b}_{a}$
with $k=2,...,n$.

The solution of (9.22) is
$K=L_{(0)a}Q^{a}t+L_{(1)a}Q^{a}\frac{t^{2}}{2}+...+L_{(n)a}Q^{a}\frac{t^{n+1}}{n+1}+G(q)$
which when replaced into (9.21) gives
$G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}-L_{(0)b}A^{b}_{a}$.

The QFI is

$\displaystyle I_{n}$ $\displaystyle=$
$\displaystyle\left(\frac{t^{n}}{n}C_{(n)ab}+...+\frac{t^{2}}{2}C_{(2)ab}+tC_{(1)ab}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{n}L_{(n)a}\dot{q}^{a}+...+t^{2}L_{(2)a}\dot{q}^{a}+$
$\displaystyle+tL_{(1)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+\frac{t^{n+1}}{n+1}L_{(n)a}Q^{a}+...+\frac{t^{2}}{2}L_{(1)a}Q^{a}+tL_{(0)a}Q^{a}+G(q)$

where $C_{(N)ab}$ are KTs, $C_{(1)ab}=-L_{(0)(a;b)}-2C_{(0)c(a}A_{b)}^{c}$,
$C_{(k+1)ab}=-L_{(k)(a;b)}-\frac{2}{k}C_{(k)c(a}A_{b)}^{c}$ for $k=1,...,n-1$,
$L_{(n)(a;b)}=-\frac{2}{n}C_{(n)c(a}A_{b)}^{c}$, $L_{(n)a}Q^{a}=s$,
$\left(L_{(n-1)b}Q^{b}\right)_{,a}=2C_{(n)ab}Q^{b}-nL_{(n)b}A^{b}_{a}$,
$\left(L_{(k-1)b}Q^{b}\right)_{,a}=2C_{(k)ab}Q^{b}-k(k+1)L_{(k+1)a}-kL_{(k)b}A^{b}_{a}$
for $k=1,...,n-1$, and $G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}-L_{(0)b}A^{b}_{a}$.

We note that $I_{0}<I_{1}<I_{2}<I_{3}<I_{4}<...$, that is, each QFI $I_{k}$ is
a subcase of the next QFI $I_{k+1}$ for all $k\in\mathbb{N}$. Therefore, we
have only one independent QFI the $I_{n}$. The value of $n$ is determined by
the symmetries of the kinetic metric and the dynamics of each specific system.

Observe that for $A^{a}_{b}=0$, the QFI $I_{n}$ reduces to

$\displaystyle I_{ns}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{n}}{n}L_{(n-1)(a;b)}-...-\frac{t^{2}}{2}L_{(1)(a;b)}-tL_{(0)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{n}L_{(n)a}\dot{q}^{a}+...+t^{2}L_{(2)a}\dot{q}^{a}+$
$\displaystyle+tL_{(1)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+\frac{t^{n+1}}{n+1}L_{(n)a}Q^{a}+...+\frac{t^{2}}{2}L_{(1)a}Q^{a}+tL_{(0)a}Q^{a}+G(q)$

where $C_{(0)ab}$ and $L_{(N)(a;b)}$ are KTs, $L_{(n)a}$ is a KV,
$L_{(n)a}Q^{a}=s$, $\left(L_{(n-1)b}Q^{b}\right)_{,a}=-2L_{(n-1)(a;b)}Q^{b}$,
$\left(L_{(k-1)b}Q^{b}\right)_{,a}=-2L_{(k-1)(a;b)}Q^{b}-k(k+1)L_{(k+1)a}$ for
$k=1,...,n-1$, and $G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}$.

We shall prove that $I_{n}(A^{a}_{b}=0)$ consists of two independent FIs.

In the case that $A^{a}_{b}=0$, we have the following:

\- For $n=0$.

$I_{0}=C_{(0)ab}\dot{q}^{a}\dot{q}^{b}+L_{(0)a}\dot{q}^{a}+st+G(q)$

where $C_{(0)ab}$ is a KT, $L_{(0)a}$ is a KV, $L_{(0)a}Q^{a}=s$, and
$G_{,a}=2C_{(0)ab}Q^{b}$.

This QFI consists of the independent FIs:
$I_{01}=C_{(0)ab}\dot{q}^{a}\dot{q}^{b}+G(q)$ and
$I_{02}=L_{(0)a}\dot{q}^{a}+st$.

\- For $n=1$.

$\displaystyle I_{1}$ $\displaystyle=$
$\displaystyle\left(-tL_{(0)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+tL_{(1)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+\frac{t^{2}}{2}s+tL_{(0)a}Q^{a}+G(q)$

where $C_{(0)ab}$ and $L_{(0)(a;b)}$ are KTs, $L_{(1)a}$ is a KV,
$L_{(1)a}Q^{a}=s$, $\left(L_{(0)b}Q^{b}\right)_{,a}=-2L_{(0)(a;b)}Q^{b}$, and
$G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}$.

This QFI consists of the independent FIs:

$\displaystyle I_{11}$ $\displaystyle=$ $\displaystyle
C_{(0)ab}\dot{q}^{a}\dot{q}^{b}+tL_{(1)a}\dot{q}^{a}+\frac{t^{2}}{2}s+G(q)$
$\displaystyle I_{12}$ $\displaystyle=$ $\displaystyle-
tL_{(0)(a;b)}\dot{q}^{a}\dot{q}^{b}+L_{(0)a}\dot{q}^{a}+tL_{(0)a}Q^{a}.$

\- For $n=2$.

$\displaystyle I_{2}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{2}}{2}L_{(1)(a;b)}-tL_{(0)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{2}L_{(2)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+$
$\displaystyle+\frac{t^{3}}{3}s+\frac{t^{2}}{2}L_{(1)a}Q^{a}+tL_{(0)a}Q^{a}+G(q)$

where $C_{(0)ab}$ and $L_{(M)(a;b)}$ for $M=0,1$ are KTs, $L_{(2)a}$ is a KV,
$L_{(2)a}Q^{a}=s$, $\left(L_{(1)b}Q^{b}\right)_{,a}=-2L_{(1)(a;b)}Q^{b}$,
$\left(L_{(0)b}Q^{b}\right)_{,a}=-2L_{(0)(a;b)}Q^{b}-2L_{(2)a}$, and
$G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}$.

This QFI consists of the independent FIs:

$\displaystyle I_{21}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{2}}{2}L_{(1)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+tL_{(1)a}\dot{q}^{a}+\frac{t^{2}}{2}L_{(1)a}Q^{a}+G(q)$
$\displaystyle I_{22}$ $\displaystyle=$ $\displaystyle-
tL_{(0)(a;b)}\dot{q}^{a}\dot{q}^{b}+t^{2}L_{(2)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+\frac{t^{3}}{3}s++tL_{(0)a}Q^{a}.$

\- For $n=3$.

$\displaystyle I_{3}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{3}}{3}L_{(2)(a;b)}-\frac{t^{2}}{2}L_{(1)(a;b)}-tL_{(0)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{3}L_{(3)a}\dot{q}^{a}+t^{2}L_{(2)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+$
$\displaystyle+L_{(0)a}\dot{q}^{a}+\frac{t^{4}}{4}s+\frac{t^{3}}{3}L_{(2)a}Q^{a}+\frac{t^{2}}{2}L_{(1)a}Q^{a}+tL_{(0)a}Q^{a}+G(q)$

where $C_{(0)ab}$ and $L_{(M)(a;b)}$ for $M=0,1,2$ are KTs, $L_{(3)a}$ is a
KV, $L_{(3)a}Q^{a}=s$, $\left(L_{(2)b}Q^{b}\right)_{,a}=-2L_{(2)(a;b)}Q^{b}$,
$\left(L_{(1)b}Q^{b}\right)_{,a}=-2L_{(1)(a;b)}Q^{b}-6L_{(3)a}$,
$\left(L_{(0)b}Q^{b}\right)_{,a}=-2L_{(0)(a;b)}Q^{b}-2L_{(2)a}$, and
$G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}$.

This QFI consists of the independent FIs:

$\displaystyle I_{31}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{2}}{2}L_{(1)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{3}L_{(3)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+\frac{t^{4}}{4}s+\frac{t^{2}}{2}L_{(1)a}Q^{a}+G(q)$
$\displaystyle I_{32}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{3}}{3}L_{(2)(a;b)}-tL_{(0)(a;b)}\right)\dot{q}^{a}\dot{q}^{b}+t^{2}L_{(2)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+\frac{t^{3}}{3}L_{(2)a}Q^{a}+tL_{(0)a}Q^{a}.$

\- For $n=4$.

$\displaystyle I_{4}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{4}}{4}L_{(3)(a;b)}-\frac{t^{3}}{3}L_{(2)(a;b)}-\frac{t^{2}}{2}L_{(1)(a;b)}-tL_{(0)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{4}L_{(4)a}\dot{q}^{a}+$
$\displaystyle+t^{3}L_{(3)a}\dot{q}^{a}+t^{2}L_{(2)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+\frac{t^{5}}{5}s+\frac{t^{4}}{4}L_{(3)a}Q^{a}+\frac{t^{3}}{3}L_{(2)a}Q^{a}+$
$\displaystyle+\frac{t^{2}}{2}L_{(1)a}Q^{a}+tL_{(0)a}Q^{a}+G(q)$

where $C_{(0)ab}$ and $L_{(M)(a;b)}$ for $M=0,...n-1$ are KTs, $L_{(4)a}$ ia a
KV, $L_{(4)a}Q^{a}=s$, $\left(L_{(3)b}Q^{b}\right)_{,a}=-2L_{(3)(a;b)}Q^{b}$,
$\left(L_{(2)b}Q^{b}\right)_{,a}=-2L_{(2)(a;b)}Q^{b}-12L_{(4)a}$,
$\left(L_{(1)b}Q^{b}\right)_{,a}=-2L_{(1)(a;b)}Q^{b}-6L_{(3)a}$,
$\left(L_{(0)b}Q^{b}\right)_{,a}=-2L_{(0)(a;b)}Q^{b}-2L_{(2)a}$, and
$G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}$.

This QFI consists of the independent FIs:

$\displaystyle I_{41}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{4}}{4}L_{(3)(a;b)}-\frac{t^{2}}{2}L_{(1)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{3}L_{(3)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+$
$\displaystyle+\frac{t^{4}}{4}L_{(3)a}Q^{a}+\frac{t^{2}}{2}L_{(1)a}Q^{a}+G(q)$
$\displaystyle I_{42}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{3}}{3}L_{(2)(a;b)}-tL_{(0)(a;b)}\right)\dot{q}^{a}\dot{q}^{b}+t^{4}L_{(4)a}\dot{q}^{a}+t^{2}L_{(2)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+$
$\displaystyle+\frac{t^{5}}{5}s+\frac{t^{3}}{3}L_{(2)a}Q^{a}+tL_{(0)a}Q^{a}.$

-For $n=5$.

$\displaystyle I_{5}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{5}}{5}L_{(4)(a;b)}-\frac{t^{4}}{4}L_{(3)(a;b)}-\frac{t^{3}}{3}L_{(2)(a;b)}-\frac{t^{2}}{2}L_{(1)(a;b)}-tL_{(0)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+$
$\displaystyle+t^{5}L_{(5)a}\dot{q}^{a}+t^{4}L_{(4)a}\dot{q}^{a}+t^{3}L_{(3)a}\dot{q}^{a}+t^{2}L_{(2)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+\frac{t^{6}}{6}s+$
$\displaystyle+\frac{t^{5}}{5}L_{(4)a}Q^{a}+\frac{t^{4}}{4}L_{(3)a}Q^{a}+\frac{t^{3}}{3}L_{(2)a}Q^{a}+\frac{t^{2}}{2}L_{(1)a}Q^{a}+tL_{(0)a}Q^{a}+G(q)$

where $C_{(0)ab}$ and $L_{(M)(a;b)}$ for $M=0,...n-1$ are KTs, $L_{(5)a}$ is a
KV, $L_{(5)a}Q^{a}=s$, $\left(L_{(4)b}Q^{b}\right)_{,a}=-2L_{(4)(a;b)}Q^{b}$,
$\left(L_{(3)b}Q^{b}\right)_{,a}=-2L_{(3)(a;b)}Q^{b}-20L_{(5)a}$,
$\left(L_{(2)b}Q^{b}\right)_{,a}=-2L_{(2)(a;b)}Q^{b}-12L_{(4)a}$,
$\left(L_{(1)b}Q^{b}\right)_{,a}=-2L_{(1)(a;b)}Q^{b}-6L_{(3)a}$,
$\left(L_{(0)b}Q^{b}\right)_{,a}=-2L_{(0)(a;b)}Q^{b}-2L_{(2)a}$, and
$G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}$.

The QFI consists of the independent FIs:

$\displaystyle I_{51}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{4}}{4}L_{(3)(a;b)}-\frac{t^{2}}{2}L_{(1)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{5}L_{(5)a}\dot{q}^{a}+t^{3}L_{(3)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+$
$\displaystyle+\frac{t^{6}}{6}s+\frac{t^{4}}{4}L_{(3)a}Q^{a}+\frac{t^{2}}{2}L_{(1)a}Q^{a}+G(q)$
$\displaystyle I_{52}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{5}}{5}L_{(4)(a;b)}-\frac{t^{3}}{3}L_{(2)(a;b)}-tL_{(0)(a;b)}\right)\dot{q}^{a}\dot{q}^{b}+t^{4}L_{(4)a}\dot{q}^{a}+t^{2}L_{(2)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+$
$\displaystyle+\frac{t^{5}}{5}L_{(4)a}Q^{a}+\frac{t^{3}}{3}L_{(2)a}Q^{a}+tL_{(0)a}Q^{a}.$

If we continue in the same way, we prove that for $A^{a}_{b}=0$ the QFI
$I_{n}$ consists of the independent FIs:

$\displaystyle I_{\ell 1}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{2\ell}}{2\ell}L_{(2\ell-1)(a;b)}-...-\frac{t^{4}}{4}L_{(3)(a;b)}-\frac{t^{2}}{2}L_{(1)(a;b)}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{2\ell-1}L_{(2\ell-1)a}\dot{q}^{a}+...+$
$\displaystyle+t^{3}L_{(3)a}\dot{q}^{a}+tL_{(1)a}\dot{q}^{a}+\frac{t^{2\ell}}{2\ell}L_{(2\ell-1)a}Q^{a}+...+\frac{t^{4}}{4}L_{(3)a}Q^{a}+\frac{t^{2}}{2}L_{(1)a}Q^{a}+G(q)$

where $C_{(0)ab}$ and $L_{(M)(a;b)}$ for $M=1,3,...,2\ell-1$ are KTs,
$\left(L_{(2\ell-1)b}Q^{b}\right)_{,a}=-2L_{(2\ell-1)(a;b)}Q^{b}$,
$\left(L_{(k-1)b}Q^{b}\right)_{,a}=-2L_{(k-1)(a;b)}Q^{b}-k(k+1)L_{(k+1)a}$ for
$k=2,4,...,2\ell-2$, and $G_{,a}=2C_{(0)ab}Q^{b}-L_{(1)a}$; and

$\displaystyle I_{\ell 2}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{2\ell+1}}{2\ell+1}L_{(2\ell)(a;b)}-...-\frac{t^{3}}{3}L_{(2)(a;b)}-tL_{(0)(a;b)}\right)\dot{q}^{a}\dot{q}^{b}+t^{2\ell}L_{(2\ell)a}\dot{q}^{a}+...+t^{2}L_{(2)a}\dot{q}^{a}+$
$\displaystyle+L_{(0)a}\dot{q}^{a}+\frac{t^{2\ell+1}}{2\ell+1}L_{(2\ell)a}Q^{a}+...+\frac{t^{3}}{3}L_{(2)a}Q^{a}+tL_{(0)a}Q^{a}$

where $L_{(M)(a;b)}$ for $M=0,2,...,2\ell$ are KTs,
$\left(L_{(2\ell)b}Q^{b}\right)_{,a}=-2L_{(2\ell)(a;b)}Q^{b}$, and
$\left(L_{(k-1)b}Q^{b}\right)_{,a}=-2L_{(k-1)(a;b)}Q^{b}-k(k+1)L_{(k+1)a}$ for
$k=1,3,...,2\ell-1$.

We note that the set of the constraints of the QFI $I_{n}(A^{a}_{b}=0)$ is
divided into: a) One set involving the odd vectors $L_{(2k+1)a}$, the KT
$C_{0ab}$ and the function $G(q)$; and b) A second set involving only the even
vectors $L_{(2k)a}$. This explains why the QFI $I_{n}(A^{a}_{b}=0)$ breaks
into two independent FIs.

II. Case $\mathbf{n\neq m}$. ($n$ or $m$ may be infinite)

We find QFIs that are subcases of those found in Case I and Case III below.

III. Both $\mathbf{n}$ and $\mathbf{m}$ are infinite.

In this case, we consider the solution to have the form222To find a solution,
we consider $C_{(0)ab}=c_{0}C_{ab}$, $C_{(1)ab}=c_{1}C_{ab}$, …,
$C_{(n)ab}=nc_{n}C_{ab}$, and $L_{(0)a}=d_{0}L_{a}$, $L_{(1)a}=d_{1}L_{a}$,
…., $L_{(m)a}=d_{m}L_{a}$.: $K_{ab}(t,q)=g(t)C_{ab}(q)$ and
$K_{a}(t,q)=f(t)L_{a}(q)$, where the functions $g(t)$ and $f(t)$ are analytic
so that they may be represented by polynomial functions as follows:

$g(t)=\sum^{n}_{k=0}c_{k}t^{k}=c_{0}+c_{1}t+...+c_{n}t^{n}\enskip\text{and}\enskip
f(t)=\sum^{m}_{k=0}d_{k}t^{k}=d_{0}+d_{1}t+...+d_{m}t^{m}.$

Only the following subcase give a new (non-trivial) independent QFI333This is
the QFI $J_{2}$ of Theorem 9.4.1..

Subcase $\mathbf{(g=e^{\lambda t}}$, $\mathbf{f=e^{\mu t})}$. $\lambda\mu\neq
0$.

$\begin{cases}\eqref{eq.veldep6}\implies\lambda e^{\lambda t}C_{ab}+e^{\mu
t}L_{(a;b)}+2e^{\lambda t}C_{c(a}A^{c}_{b)}=0\\\
\eqref{eq.veldep7}\implies-2e^{\lambda t}C_{ab}Q^{b}+\mu e^{\mu
t}L_{a}+K_{,a}+e^{\mu t}L_{b}A^{b}_{a}=0\\\ \eqref{eq.veldep8}\implies
K_{,t}=e^{\mu t}L_{a}Q^{a}\\\ \eqref{eq.veldep9}\implies\mu^{2}e^{\mu
t}L_{a}+\mu e^{\mu t}L_{b}A^{b}_{a}+e^{\mu
t}\left(L_{b}Q^{b}\right)_{;a}-2\lambda e^{\lambda
t}C_{ab}Q^{b}=0.\end{cases}$

We consider the following subcases:

a) For $\lambda\neq\mu$:

From (9.20), we have that $C_{ab}=-\frac{2}{\lambda}C_{c(a}A^{c}_{b)}$ and
$L_{a}$ is a KV.

From (9.23), we find that $C_{ab}Q^{b}=0$ and $\mu^{2}L_{a}+\mu
L_{b}A^{b}_{a}+\left(L_{b}Q^{b}\right)_{,a}=0$.

The solution of (9.22) is $K=\frac{1}{\mu}e^{\mu t}L_{a}Q^{a}+G(q)$ which when
replaced in (9.21) gives $G(q)=const\equiv 0$.

The QFI is

$I_{e}(\lambda\neq\mu)=e^{\lambda t}C_{ab}\dot{q}^{a}\dot{q}^{b}+e^{\mu
t}L_{a}\dot{q}^{a}+\frac{1}{\mu}e^{\mu t}L_{a}Q^{a}$

where $C_{ab}=-\frac{2}{\lambda}C_{c(a}A_{b)}^{c}$ is a KT such that
$C_{ab}Q^{b}=0$, and
$L_{a}=-\frac{1}{\mu^{2}}\left(L_{b}Q^{b}\right)_{,a}-\frac{1}{\mu}L_{b}A_{a}^{b}$
is a KV.

We note that the QFI $I_{e}(\lambda\neq\mu)$ consists of the two independent
FIs:

$J_{2a}=e^{\lambda t}C_{ab}\dot{q}^{a}\dot{q}^{b},\enskip J_{2b}=e^{\mu
t}L_{a}\dot{q}^{a}+\frac{1}{\mu}e^{\mu t}L_{a}Q^{a}$

b) For $\lambda=\mu$:

From (9.20), we have that
$C_{ab}=-\frac{1}{\lambda}L_{(a;b)}-\frac{2}{\lambda}C_{c(a}A^{c}_{b)}$.

From (9.23), we find that $\lambda^{2}L_{a}+\lambda
L_{b}A^{b}_{a}+\left(L_{b}Q^{b}\right)_{,a}-2\lambda C_{ab}Q^{b}=0$.

The solution of (9.22) is $K=\frac{1}{\lambda}e^{\lambda t}L_{a}Q^{a}+G(q)$
which when replaced in (9.21) gives $G(q)=const\equiv 0$.

The QFI is

$I_{e}(\lambda=\mu)=e^{\lambda t}C_{ab}\dot{q}^{a}\dot{q}^{b}+e^{\lambda
t}L_{a}\dot{q}^{a}+\frac{1}{\lambda}e^{\lambda t}L_{a}Q^{a}\equiv J_{2}$

where $C_{ab}=-\frac{1}{\lambda}L_{(a;b)}-\frac{2}{\lambda}C_{c(a}A_{b)}^{c}$
is a KT, and the vector
$L_{a}=-\frac{1}{\lambda^{2}}\left(L_{b}Q^{b}\right)_{,a}-\frac{1}{\lambda}L_{b}A_{a}^{b}+\frac{2}{\lambda}C_{ab}Q^{b}$.

We observe that the FIs $J_{2a}$ and $J_{2b}$ found previously are subcases of
the new QFI $J_{2}$; indeed,
$J_{2a}=$ $J_{2}(L_{a}=0)$ and $J_{2b}=$ $J_{2}(C_{ab}=0)$. Therefore, the
Case III leads to only one independent QFI the $J_{2}$.

The above complete the proof of Theorem 9.4.1.

###  Appendix E Proof of Theorem 11.4.1

Substituting (11.9) and (11.10) in the system of PDEs111Equation (11.3) is
identically satisfied since the quantities $C_{(N)i_{1}...i_{m}}$ are assumed
to be $m$th-order KTs. (11.4) - (11.8), we obtain the following system of
equations222Equation (11.8) is not necessary, because the integrability
condition $M_{,[i_{1}i_{2}]}=0$ does not intervene in the calculations. In any
case, it has been checked that equation (11.8) is always satisfied (as an
identity $0=0$) from the solutions of the other equations of the system.:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-2L_{(1)i_{1}i_{2}}Q^{i_{2}}-4L_{(2)i_{1}i_{2}}Q^{i_{2}}t-...-2nL_{(n)i_{1}i_{2}}Q^{i_{2}}t^{n-1}+2L_{(2)i_{1}}+6L_{(3)i_{1}}t+...+$
(E.1)
$\displaystyle+n(n-1)L_{(n)i_{1}}t^{n-2}+\left(L_{(0)c}Q^{c}\right)_{,i_{1}}+\left(L_{(1)c}Q^{c}\right)_{,i_{1}}t+...+\left(L_{(n)c}Q^{c}\right)_{,i_{1}}t^{n}$
$\displaystyle 0$ $\displaystyle=$ $\displaystyle
M_{,t}-L_{(0)i_{1}}Q^{i_{1}}-L_{(1)i_{1}}Q^{i_{1}}t-...-L_{(n)i_{1}}Q^{i_{1}}t^{n}$
(E.2) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-2L_{(0)i_{1}i_{2}}Q^{i_{2}}-2L_{(1)i_{1}i_{2}}Q^{i_{2}}t-...-2L_{(n)i_{1}i_{2}}Q^{i_{2}}t^{n}+L_{(1)i_{1}}+2L_{(2)i_{1}}t+...+$
(E.3) $\displaystyle+nL_{(n)i_{1}}t^{n-1}+M_{,i_{1}}$ $\displaystyle 0$
$\displaystyle=$
$\displaystyle-(r+1)L_{(0)i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}-(r+1)L_{(1)i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}t-...-$
(E.4)
$\displaystyle-(r+1)L_{(n)i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}t^{n}+L_{(1)i_{1}...i_{r}}+2L_{(2)i_{1}...i_{r}}t+...+nL_{(n)i_{1}...i_{r}}t^{n-1}+$
$\displaystyle+L_{(0)(i_{1}...i_{r-1};i_{r})}+L_{(1)(i_{1}...i_{r-1};i_{r})}t+...+L_{(n)(i_{1}...i_{r-1};i_{r})}t^{n},\enskip
r=2,3,...,m-2$ $\displaystyle 0$ $\displaystyle=$ $\displaystyle-
mC_{(0)i_{1}...i_{m-1}i_{m}}Q^{i_{m}}-mC_{(1)i_{1}...i_{m-1}i_{m}}Q^{i_{m}}t-mC_{(2)i_{1}...i_{m-1}i_{m}}Q^{i_{m}}\frac{t^{2}}{2}-...-$
(E.5) $\displaystyle-
mC_{(n)i_{1}...i_{m-1}i_{m}}Q^{i_{m}}\frac{t^{n}}{n}+L_{(1)i_{1}...i_{m-1}}+2L_{(2)i_{1}...i_{m-1}}t+...+nL_{(n)i_{1}...i_{m-1}}t^{n-1}+$
$\displaystyle+L_{(0)(i_{1}...i_{m-2};i_{m-1})}+L_{(1)(i_{1}...i_{m-2};i_{m-1})}t+...+L_{(n)(i_{1}...i_{m-2};i_{m-1})}t^{n}$
$\displaystyle 0$ $\displaystyle=$ $\displaystyle
C_{(1)i_{1}...i_{m}}+C_{(2)i_{1}...i_{m}}t+...+C_{(n)i_{1}...i_{m}}t^{n-1}+L_{(0)(i_{1}...i_{m-1};i_{m})}+L_{(1)(i_{1}...i_{m-1};i_{m})}t+$
(E.6) $\displaystyle+...+L_{(n)(i_{1}...i_{m-1};i_{m})}t^{n}$

where –without loss of generality– the polynomial expressions (11.9) and
(11.10) of $t$ are assumed to be of the same degree, that is, $n=n_{r}$ for
all values of $r$. All the results with $n\neq n_{r}$ are derived as subcases
from the case $n=n_{r}$.

We consider the following cases.

I. Case with $n$ finite.

From (E.1), we obtain $L_{(n)i_{1}}Q^{i_{1}}=s=const$,
$\left(L_{(n-1)c}Q^{c}\right)_{,i_{1}}=$ $2nL_{(n)i_{1}i_{2}}Q^{i_{2}}$ and

$\left(L_{(k-2)c}Q^{c}\right)_{,i_{1}}=2(k-1)L_{(k-1)i_{1}i_{2}}Q^{i_{2}}-k(k-1)L_{(k)i_{1}},\enskip
k=2,3,...,n.$

The solution of (E.2) is
$M=L_{(0)c}Q^{c}t+L_{(1)c}Q^{c}\frac{t^{2}}{2}+...+L_{(n-1)c}Q^{c}\frac{t^{n}}{n}+s\frac{t^{n+1}}{n+1}+G(q)$,
where $G(q)$ is an arbitrary function. Substituting the function $M$ in (E.3),
we find that $G_{,i_{1}}=2L_{(0)i_{1}i_{2}}Q^{i_{2}}-L_{(1)i_{1}}$.

Equation (E.4) implies that
$L_{(n)(i_{1}...i_{r-1};i_{r})}=(r+1)L_{(n)i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}$
and

$L_{(k-1)(i_{1}...i_{r-1};i_{r})}=(r+1)L_{(k-1)i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}-kL_{(k)i_{1}...i_{r}},\enskip
k=1,2,...,n$

where $r=2,3,...,m-2$.

Equation (E.5) gives
$L_{(n)(i_{1}...i_{m-2};i_{m-1})}=\frac{m}{n}C_{(n)i_{1}...i_{m-1}i_{m}}Q^{i_{m}}$
and

$L_{(k-1)(i_{1}...i_{m-2};i_{m-1})}=\frac{m}{k-1}C_{(k-1)i_{1}...i_{m-1}i_{m}}Q^{i_{m}}-kL_{(k)i_{1}...i_{m-1}},\enskip
k=1,2,...,n.$

We note that _when the undefined term $\frac{C_{(0)i_{1}...i_{m}}}{0}$ appears
in the calculations, it must be replaced by $C_{(0)i_{1}...i_{m}}$ in order to
have a consistent result._

Finally, equation (E.6) implies that $L_{(n)(i_{1}...i_{m-1};i_{m})}=0$ and
$L_{(k-1)(i_{1}...i_{m-1};i_{m})}=-C_{(k)i_{1}...i_{m}}$, where $k=1,2,...,n$.

The FI is

$\displaystyle I^{(m)}_{n}$ $\displaystyle=$
$\displaystyle\left(-\frac{t^{n}}{n}L_{(n-1)(i_{1}...i_{m-1};i_{m})}-...-\frac{t^{2}}{2}L_{(1)(i_{1}...i_{m-1};i_{m})}-tL_{(0)(i_{1}...i_{m-1};i_{m})}+C_{(0)i_{1}...i_{m}}\right)\dot{q}^{i_{1}}...\dot{q}^{i_{m}}+$
(E.7)
$\displaystyle+\sum_{r=1}^{m-1}\left(t^{n}L_{(n)i_{1}...i_{r}}+...+tL_{(1)i_{1}...i_{r}}+L_{(0)i_{1}...i_{r}}\right)\dot{q}^{i_{1}}...\dot{q}^{i_{r}}+s\frac{t^{n+1}}{n+1}+$
$\displaystyle+L_{(n-1)c}Q^{c}\frac{t^{n}}{n}+...+L_{(1)c}Q^{c}\frac{t^{2}}{2}+L_{(0)c}Q^{c}t+G(q)$

where $C_{(0)i_{1}...i_{m}}$ and $L_{(N)(i_{1}...i_{m-1};i_{m})}$ for
$N=0,...,n-1$ are $m$th-order KTs, $L_{(n)i_{1}...i_{m-1}}$ is an $(m-1)$th-
order KT, $s$ is an arbitrary constant and the following conditions are
satisfied:

$\displaystyle L_{(n)(i_{1}...i_{m-2};i_{m-1})}$ $\displaystyle=$
$\displaystyle-\frac{m}{n}L_{(n-1)(i_{1}...i_{m-1};i_{m})}Q^{i_{m}}$
$\displaystyle L_{(k-1)(i_{1}...i_{m-2};i_{m-1})}$ $\displaystyle=$
$\displaystyle-\frac{m}{k-1}L_{(k-2)(i_{1}...i_{m-1};i_{m})}Q^{i_{m}}-kL_{(k)i_{1}...i_{m-1}},\enskip
k=2,3,...,n$ $\displaystyle L_{(0)(i_{1}...i_{m-2};i_{m-1})}$ $\displaystyle=$
$\displaystyle mC_{(0)i_{1}...i_{m-1}i_{m}}Q^{i_{m}}-L_{(1)i_{1}...i_{m-1}}$
$\displaystyle L_{(n)(i_{1}...i_{r-1};i_{r})}$ $\displaystyle=$
$\displaystyle(r+1)L_{(n)i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}},\enskip
r=2,3,...,m-2$ $\displaystyle L_{(k-1)(i_{1}...i_{r-1};i_{r})}$
$\displaystyle=$
$\displaystyle(r+1)L_{(k-1)i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}-kL_{(k)i_{1}...i_{r}},\enskip
k=1,2,...,n,\enskip r=2,3,...,m-2$ $\displaystyle L_{(n)i_{1}}Q^{i_{1}}$
$\displaystyle=$ $\displaystyle s$
$\displaystyle\left(L_{(n-1)c}Q^{c}\right)_{,i_{1}}$ $\displaystyle=$
$\displaystyle 2nL_{(n)i_{1}i_{2}}Q^{i_{2}}$
$\displaystyle\left(L_{(k-2)c}Q^{c}\right)_{,i_{1}}$ $\displaystyle=$
$\displaystyle 2(k-1)L_{(k-1)i_{1}i_{2}}Q^{i_{2}}-k(k-1)L_{(k)i_{1}},\enskip
k=2,3,...,n$ $\displaystyle G_{,i_{1}}$ $\displaystyle=$ $\displaystyle
2L_{(0)i_{1}i_{2}}Q^{i_{2}}-L_{(1)i_{1}}.$

The notation $I^{(m)}_{n}$ means the $m$th-order FI (upper index) with time-
dependence $n$ (lower index). For example, in this notation the $I^{(2)}_{n}$
is a QFI whose coefficients are expressed as polynomials of $t$ of degree
fixed by $n$.

We note that all totally symmetric tensors $C_{(0)i_{1}...i_{m}}$ and
$L_{(N)(i_{1}...i_{m-1};i_{m})}$ for $N=0,...,n-1$ are $m$th-order KTs;
therefore, in order to find the $m$th-order FIs, we need to calculate _all_
KTs (both reducible and irreducible) of the kinetic metric $\gamma_{ab}$ of
all orders (i.e. $0,1,...,m$).

II. Case with $n$ infinite.

In this case, we consider the $m$th-order KT $M_{i_{1}...i_{m}}(t,q)$ and the
$r$-rank totally symmetric tensors $M_{i_{1}...i_{r}}(t,q)$, where
$r=1,2,...,m-1$, to have the form:

$M_{i_{1}...i_{m}}(t,q)=f_{(m)}(t)C_{i_{1}...i_{m}}(q),\enskip
M_{i_{1}...i_{r}}(t,q)=f_{(r)}(t)L_{i_{1}...i_{r}}(q),\enskip r=1,2,...,m-1$

where $L_{i_{1}...i_{r}}(q)$ are $r$-rank totally symmetric tensors,
$C_{i_{1}...i_{m}}(q)$ is an $m$th-order KT of $\gamma_{ab}$ and the functions
$f_{(s)}(t)$, $s=1,2,...,m$, are analytic so that they may be represented by
polynomial functions of $t$ as follows:

$f_{(s)}(t)=\sum^{n}_{k=0}d_{(s)k}t^{k}=d_{(s)0}+d_{(s)1}t+...+d_{(s)n}t^{n}.$

The coefficients $d_{(s)0},d_{(s)1},...,d_{(s)n}$ are arbitrary constants.

It is found that all subcases give results already found in the previous case
with $n$ finite except for the subcase considered below.

Subcase $f_{(s)}=e^{\lambda_{s}t}$ with $\lambda_{s}\neq 0$.

The system of equations (E.1) - (E.6) becomes:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-2\lambda_{2}e^{\lambda_{2}t}L_{i_{1}i_{2}}Q^{i_{2}}+\lambda_{1}^{2}e^{\lambda_{1}t}L_{i_{1}}+e^{\lambda_{1}t}\left(L_{c}Q^{c}\right)_{,i_{1}}$
(E.8) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
M_{,t}-e^{\lambda_{1}t}L_{i_{1}}Q^{i_{1}}$ (E.9) $\displaystyle 0$
$\displaystyle=$
$\displaystyle-2e^{\lambda_{2}t}L_{i_{1}i_{2}}Q^{i_{2}}+\lambda_{1}e^{\lambda_{1}t}L_{i_{1}}+M_{,i_{1}}$
(E.10) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(r+1)e^{\lambda_{r+1}t}L_{i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}+\lambda_{r}e^{\lambda_{r}t}L_{i_{1}...i_{r}}+e^{\lambda_{r-1}t}L_{(i_{1}...i_{r-1};i_{r})},\enskip
r=2,3,...,m-2$ (E.11) $\displaystyle 0$ $\displaystyle=$ $\displaystyle-
me^{\lambda_{m}t}C_{i_{1}...i_{m-1}i_{m}}Q^{i_{m}}+\lambda_{m-1}e^{\lambda_{m-1}t}L_{i_{1}...i_{m-1}}+e^{\lambda_{m-2}t}L_{(i_{1}...i_{m-2};i_{m-1})}$
(E.12) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\lambda_{m}e^{\lambda_{m}t}C_{i_{1}...i_{m}}+e^{\lambda_{m-1}t}L_{(i_{1}...i_{m-1};i_{m})}.$
(E.13)

We note that in the calculations that follow, without loss of generality, all
the constants $\lambda_{s}$ are fixed to the same non-zero constant $\lambda$.
The FI produced from this assumption contains as subcases all the FIs arising
from constants $\lambda_{s}$ which are not all the same.

Equation (E.8) implies that $\left(L_{c}Q^{c}\right)_{,i_{1}}=2\lambda
L_{i_{1}i_{2}}Q^{i_{2}}-\lambda^{2}L_{i_{1}}$.

The solution of (E.9) is $M=\frac{e^{\lambda
t}}{\lambda}L_{i_{1}}Q^{i_{1}}+G(q)$ which when replaced in (E.10) gives
$G(q)=const\equiv 0$.

From the remaining equations (E.11) - (E.13), we find the following
conditions:

$C_{i_{1}...i_{m}}=-\frac{1}{\lambda}L_{(i_{1}...i_{m-1};i_{m})}$
$L_{(i_{1}...i_{r-1};i_{r})}=(r+1)L_{i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}-\lambda
L_{i_{1}...i_{r}},\enskip r=2,3,...,m-2$

and

$L_{(i_{1}...i_{m-2};i_{m-1})}=mC_{i_{1}...i_{m-1}i_{m}}Q^{i_{m}}-\lambda
L_{i_{1}...i_{m-1}}.$

The FI is

$I^{(m)}_{e}=\frac{e^{\lambda
t}}{\lambda}\left(-L_{(i_{1}...i_{m-1};i_{m})}\dot{q}^{i_{1}}...\dot{q}^{i_{m}}+\lambda\sum_{r=1}^{m-1}L_{i_{1}...i_{r}}\dot{q}^{i_{1}}...\dot{q}^{i_{r}}+L_{i_{1}}Q^{i_{1}}\right)$
(E.14)

where $\lambda\neq 0$, $L_{(i_{1}...i_{m-1};i_{m})}$ is an $m$th-order KT and
the following conditions are satisfied:

$\displaystyle L_{(i_{1}...i_{m-2};i_{m-1})}$ $\displaystyle=$
$\displaystyle-\frac{m}{\lambda}L_{(i_{1}...i_{m-1};i_{m})}Q^{i_{m}}-\lambda
L_{i_{1}...i_{m-1}}$ $\displaystyle L_{(i_{1}...i_{r-1};i_{r})}$
$\displaystyle=$
$\displaystyle(r+1)L_{i_{1}...i_{r}i_{r+1}}Q^{i_{r+1}}-\lambda
L_{i_{1}...i_{r}},\enskip r=2,3,...,m-2$
$\displaystyle\left(L_{c}Q^{c}\right)_{,i_{1}}$ $\displaystyle=$
$\displaystyle 2\lambda L_{i_{1}i_{2}}Q^{i_{2}}-\lambda^{2}L_{i_{1}}.$

We note that the FI $I^{(m)}_{e}$ requires only the reducible KTs of
$\gamma_{ab}$ of all orders (i.e. $0,1,...,m$).

###  Appendix F Proof of Theorem 12.5.1

Substituting the polynomial function (12.35) in the system of PDEs (12.30) -
(12.34), we have the following cases.

I. Case $\mathbf{n=m}$ (both $n$ and $m$ are finite)

From (12.30), we find that $C_{(k)ab}=-L_{(k-1)(a;b)}$, $k=1,...,n$, and
$L_{(n)a}$ is a KV of $\gamma_{ab}$.

Equation (12.33) gives

$\displaystyle 0$ $\displaystyle=-2\left(b_{1}+2b_{2}t+...+\ell
b_{\ell}t^{\ell-1}\right)\left(C_{(0)ab}Q^{b}+C_{(1)ab}Q^{b}t+...+C_{(n)ab}Q^{b}\frac{t^{n}}{n}\right)+2L_{(2)a}+$
$\displaystyle\quad+6L_{(3)a}t+...+n(n-1)L_{(n)a}t^{n-2}-2\left(b_{0}+b_{1}t+...+b_{\ell}t^{\ell}\right)\left(C_{(1)ab}Q^{b}+C_{(2)ab}Q^{b}t+\right.$
$\displaystyle\quad\left.+...+C_{(n)ab}Q^{b}t^{n-1}\right)+\left(b_{0}+b_{1}t+...+b_{\ell}t^{\ell}\right)\left[\left(L_{(0)b}Q^{b}\right)_{,a}+\left(L_{(1)b}Q^{b}\right)_{,a}t+...+\right.$
$\displaystyle\quad\left.+\left(L_{(n-1)b}Q^{b}\right)_{,a}t^{n-1}+\left(L_{(n)b}Q^{b}\right)_{,a}t^{n}\right].$

This is a polynomial equation of the general form
$P_{(0)a}(q)+P_{(1)a}(q)t+...+P_{(n+\ell)a}(q)t^{n+\ell}=0$; therefore, from
the vanishing of the coefficients $P_{(k)a}(q)$, we get the following
conditions:

$L_{(n)a}Q^{a}=c=const$ (F.1)

and

$\displaystyle 0$
$\displaystyle=\sum_{s=0}^{\ell-1}\left[-\frac{2(k+s)b_{(k+s\leq\ell)}}{n-s}C_{(n-s\geq
0)ab}Q^{b}-2b_{(k+s\leq\ell)}C_{(n-s>0)ab}Q^{b}+\right.$
$\displaystyle\quad\left.+b_{(k+s\leq\ell)}\left(L_{(n-s-1\geq
0)b}Q^{b}\right)_{,a}\right],\enskip k=1,2,...,\ell$ (F.2) $\displaystyle 0$
$\displaystyle=-\sum_{s=1}^{\ell}\left[\frac{2sb_{s}}{n-s}C_{(n-s\geq
0)ab}Q^{b}\right]+\sum_{s=0}^{\ell}\left[-2b_{s}C_{(n-s>0)ab}Q^{b}+b_{s}\left(L_{(n-s-1\geq
0)b}Q^{b}\right)_{,a}\right]$ (F.3) $\displaystyle 0$
$\displaystyle=\sum_{s=0}^{\ell}\left[-2b_{s}C_{(k-s-1>0)ab}Q^{b}+b_{s}\left(L_{(k-s-2\geq
0)b}Q^{b}\right)_{,a}\right]-\sum_{s=1}^{\ell}\left[\frac{2sb_{s}}{k-s-1}C_{(k-s-1\geq
0)ab}Q^{b}\right]+$ $\displaystyle\quad+k(k-1)L_{(k)a},\enskip k=2,3,...n.$
(F.4)

We note that in the $n+\ell+1$ equations (F.2) - (F.4), when the undefined
quantity $\frac{C_{(0)ab}}{0}$ appears in the calculations, it must be
replaced by $C_{(0)ab}$ in order to have a consistent result.

We continue with the remaining constraints (12.31) and (12.32) in order to
determine the scalar coefficient $K(t,q)$.

The solution of (12.32) is

$\displaystyle K_{,t}$ $\displaystyle=$ $\displaystyle
L_{(0)a}Q^{a}\left(b_{0}+b_{1}t+...+b_{\ell}t^{\ell}\right)+L_{(1)a}Q^{a}\left(b_{0}t+b_{1}t^{2}+...+b_{\ell}t^{\ell+1}\right)+...+$
$\displaystyle+L_{(n-1)a}Q^{a}\left(b_{0}t^{n-1}+b_{1}t^{n}+...+b_{\ell}t^{n+\ell-1}\right)+s\left(b_{0}t^{n}+b_{1}t^{n+1}+...+b_{\ell}t^{n+\ell}\right)\implies$
$\displaystyle K$ $\displaystyle=$ $\displaystyle
L_{(0)a}Q^{a}\left(b_{0}t+b_{1}\frac{t^{2}}{2}+...+b_{\ell}\frac{t^{\ell+1}}{\ell+1}\right)+L_{(1)a}Q^{a}\left(b_{0}\frac{t^{2}}{2}+b_{1}\frac{t^{3}}{3}+...+b_{\ell}\frac{t^{\ell+2}}{\ell+2}\right)+...+$
$\displaystyle+L_{(n-1)a}Q^{a}\left(b_{0}\frac{t^{n}}{n}+b_{1}\frac{t^{n+1}}{n+1}+...+b_{\ell}\frac{t^{n+\ell}}{n+\ell}\right)+s\left(b_{0}\frac{t^{n+1}}{n+1}+b_{1}\frac{t^{n+2}}{n+2}+...+b_{\ell}\frac{t^{n+\ell+1}}{n+\ell+1}\right)+G(q).$

Replacing $K$ in (12.31) and using the conditions (F.1) - (F.4), we find that
$G_{,a}=2b_{0}C_{(0)ab}Q^{b}-L_{(1)a}$.

Condition (12.34) is satisfied trivially from the above solutions.

The QFI is

$\displaystyle I$ $\displaystyle=$
$\displaystyle\left(\frac{t^{n}}{n}C_{(n)ab}+...+tC_{(1)ab}+C_{(0)ab}\right)\dot{q}^{a}\dot{q}^{b}+t^{n}L_{(n)a}\dot{q}^{a}+...+tL_{(1)a}\dot{q}^{a}+L_{(0)a}\dot{q}^{a}+$
$\displaystyle+L_{(0)a}Q^{a}\left(b_{0}t+b_{1}\frac{t^{2}}{2}+...+b_{\ell}\frac{t^{\ell+1}}{\ell+1}\right)+L_{(1)a}Q^{a}\left(b_{0}\frac{t^{2}}{2}+b_{1}\frac{t^{3}}{3}+...+b_{\ell}\frac{t^{\ell+2}}{\ell+2}\right)+...+$
$\displaystyle+L_{(n-1)a}Q^{a}\left(b_{0}\frac{t^{n}}{n}+b_{1}\frac{t^{n+1}}{n+1}+...+b_{\ell}\frac{t^{n+\ell}}{n+\ell}\right)+s\left(b_{0}\frac{t^{n+1}}{n+1}+b_{1}\frac{t^{n+2}}{n+2}+...+b_{\ell}\frac{t^{n+\ell+1}}{n+\ell+1}\right)+G(q)$

where $C_{(0)ab}$ is a KT, the KTs $C_{(k)ab}=-L_{(k-1)(a;b)}$ for
$k=1,...,n$, $L_{(n)a}$ is a KV such that $L_{(n)a}Q^{a}=s$,
$G_{,a}=2b_{0}C_{(0)ab}Q^{b}-L_{(1)a}$, and the conditions (F.2) - (F.4) are
satisfied.

II. Case $\mathbf{n\neq m}$. (either $n$ or $m$ may be infinite)

We find QFIs that are subcases of those found in Case I and Case III below.

III. Both $\mathbf{n}$ and $\mathbf{m}$ are infinite.

In this case, we consider the solution to have the form
$K_{ab}(t,q)=g(t)C_{ab}(q)$ and $K_{a}(t,q)=f(t)L_{a}(q)$, where the functions
$g(t)$ and $f(t)$ are analytic so that they may be represented by polynomial
functions as follows:

$g(t)=\sum^{n}_{k=0}c_{k}t^{k}=c_{0}+c_{1}t+...+c_{n}t^{n},\enskip
f(t)=\sum^{m}_{k=0}d_{k}t^{k}=d_{0}+d_{1}t+...+d_{m}t^{m}.$

The coefficients $c_{0},c_{1},...,c_{n}$ and $d_{0},d_{1},...,d_{m}$ are
arbitrary constants. We find that only the following subcase gives a new
independent QFI.

Subcase $\mathbf{(g=e^{\lambda t}}$, $\mathbf{f=e^{\mu t})}$,
$\mathbf{\lambda\mu\neq 0}$.

In this case, the system of PDEs (12.30) - (12.33) becomes111Equation (12.34)
is satisfied trivially from the solutions found below.:

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\lambda e^{\lambda
t}C_{ab}+e^{\mu t}L_{(a;b)}$ (F.5) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-2\left(b_{0}+b_{1}t+...+b_{\ell}t^{\ell}\right)e^{\lambda
t}C_{ab}Q^{b}+\mu e^{\mu t}L_{a}+K_{,a}$ (F.6) $\displaystyle 0$
$\displaystyle=$ $\displaystyle
K_{,t}-(b_{0}+b_{1}t+...+b_{\ell}t^{\ell})e^{\mu t}L_{a}Q^{a}$ (F.7)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-2\left(b_{1}+2b_{2}t+...+\ell
b_{\ell}t^{\ell-1}\right)e^{\lambda
t}C_{ab}Q^{b}-2\lambda(b_{0}+b_{1}t+...+b_{\ell}t^{\ell})e^{\lambda
t}C_{ab}Q^{b}+$ (F.8) $\displaystyle+\mu^{2}e^{\mu
t}L_{a}+(b_{0}+b_{1}t+...+b_{\ell}t^{\ell})e^{\mu
t}\left(L_{b}Q^{b}\right)_{,a}.$

We consider the following subcases:

a. For $\lambda\neq\mu$:

From (F.5), we have that $C_{ab}=0$ and $L_{a}$ is a KV. Equation (F.8)
implies that $L_{a}=0$. Therefore, the QFI $I_{e}(\lambda\neq\mu)=const$ which
is trivial.

b. For $\lambda=\mu$:

From (F.5), we have that $C_{ab}=-\frac{1}{\lambda}L_{(a;b)}$; therefore,
$L_{(a;b)}$ is a reducible KT.

We consider two cases according to the degree $\ell$ of the polynomial
$\omega(t)$.

\- Case $\ell=1$.

From (F.8), we get the conditions:

$\displaystyle\left(L_{b}Q^{b}\right)_{,a}$ $\displaystyle=$ $\displaystyle
2\lambda C_{ab}Q^{b}$ (F.9)
$\displaystyle\lambda^{2}L_{a}+b_{0}\left(L_{b}Q^{b}\right)_{,a}-2(b_{1}+\lambda
b_{0})C_{ab}Q^{b}$ $\displaystyle=$ $\displaystyle 0.$ (F.10)

Replacing with $C_{ab}=-\frac{1}{\lambda}L_{(a;b)}$ and by substituting (F.9)
in (F.10), we obtain:

$\displaystyle\left(L_{b}Q^{b}\right)_{,a}$ $\displaystyle=$
$\displaystyle-2L_{(a;b)}Q^{b}$ (F.11)
$\displaystyle\lambda^{3}L_{a}+2b_{1}L_{(a;b)}Q^{b}$ $\displaystyle=$
$\displaystyle 0.$ (F.12)

The solution of (F.7) is
$K=\left(\frac{b_{0}}{\lambda}-\frac{b_{1}}{\lambda^{2}}\right)e^{\lambda
t}L_{a}Q^{a}+\frac{b_{1}}{\lambda}te^{\lambda t}L_{a}Q^{a}+G(q)$ which when
replaced in (F.6) gives $G=const\equiv 0$.

The QFI is

$I_{e}(\ell=1)=-e^{\lambda t}L_{(a;b)}\dot{q}^{a}\dot{q}^{b}+\lambda
e^{\lambda
t}L_{a}\dot{q}^{a}+\left(b_{0}-\frac{b_{1}}{\lambda}\right)e^{\lambda
t}L_{a}Q^{a}+b_{1}te^{\lambda t}L_{a}Q^{a}$ (F.13)

where $L_{(a;b)}$ is a KT,
$\left(L_{b}Q^{b}\right)_{,a}=\frac{\lambda^{3}}{b_{1}}L_{a}$ and
$\lambda^{3}L_{a}=-2b_{1}L_{(a;b)}Q^{b}$.

\- Case $\ell>1$.

From (F.8), we find that $\left(L_{b}Q^{b}\right)_{,a}=2\lambda C_{ab}Q^{b}$,
$C_{ab}Q^{b}=0$ and $\lambda^{2}L_{a}=2b_{1}C_{ab}Q^{b}$. Therefore, $L_{a}=0$
$\implies$ $C_{ab}=-\frac{1}{\lambda}L_{(a;b)}=0$ and we end up with the
trivial FI $I_{e}=const$.

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