# weak degeneracy of regular graphs

Yuxuan Yang  Department of Science, Beijing University of Posts and
Telecommunications, Beijing, China<EMAIL_ADDRESS>
###### Abstract.

Motivated by the study of greedy algorithms for graph coloring, Bernshteyn and
Lee introduced a generalization of graph degeneracy, which is called weak
degeneracy. In this paper, we show the lower bound of the weak degeneracy for
$d$-regular graphs is exactly $\lfloor d/2\rfloor+1$, which is tight. This
result refutes the conjecture of Bernshteyn and Lee on this lower bound.

###### Key words and phrases:

weak degeneracy, chromatic number, graph coloring, regular graph

###### 2010 Mathematics Subject Classification:

05C15

##  1\. Introduction

In this paper, all graphs are finite and simple. Let $G$ be a graph. We denote
by $V(G)$ and $E(G)$ the vertex set and the edge set of the graph $G$,
respectively. Let $S$ be a subset of $V(G)$. Denote by $G-S$ the subgraph
obtained from $G$ by deleting the vertices in $S$ together with their incident
edges. Especially, when $S=\\{v\\}$ we may use $G-v$ instead of $G-\\{v\\}$.
Denote by $G[S]$ the subgraph of $G$ induced by the vertex set $S$. Denote by
$\deg_{G}(v)$ the degree of a vertex $v$ in a graph $G$. If $\deg_{G}(v)=d$
for each vertex $v$ in the graph $G$, we call $G$ a $d$-regular graph.

Recall that for a graph $G$, $\chi(G)$ denotes its chromatic number, that is,
the minimum number of colors necessary to color the vertices of $G$ so that
adjacent vertices are colored differently. A well-studied generalization of
graph coloring is list coloring, which was introduced independently by Vizing
[6] and Erdős et al. [3]. There are further generalizations, such as DP-
chromatic number [2], and DP-paint number [5]. As we shall not discuss these
parameters, other than saying that they are bounded by the weak degeneracy
defined below, we omit the definitions and refer the reader to [5] for the
definitions and discussion about these parameters.

For a graph $G$, the greedy coloring algorithm colors vertices one by one in
order $v_{1},v_{2},\dots,v_{n}$, assigning $v_{i}$ the least-indexed color not
used on its colored neighbors. An upper bound for the number of colors used in
such a coloring is captured in the notion of graph degeneracy. This greedy
algorithm works for all variations of chromatic numbers mentioned above, and
it makes graph degeneracy an important graph parameter.

Frequently, graph degeneracy and chromatic numbers have a big gap between
them. For example, the degeneracy of $d$-regular graph is exactly $d$, and its
chromatic number varies from 2 to $d+1$. It is therefore interesting to see if
we can modify the greedy coloring procedure to “save” some of the colors and
get a better bound. For example, when we color a vertex under the greedy
algorithm, we can choose a color that one of its neighbors can not use if
possible. Motivated by this, Bernshteyn and Lee [1] introduced a new graph
parameter of weak degeneracy. Weak degeneracy is a variation of degeneracy
which shares many nice properties of degeneracy. Since then weak degeneracy
has been paid a lot of attention (see for example [4, 7, 8, 9]). Here is the
definition and notation.

######  Definition (Delete Operation).

Let $G$ be a graph and let $f:V(G)\rightarrow\mathbb{N}$ be a function. For a
vertex $u\in V(G)$, the operation

$\operatorname{Delete}(G,f,u)$

outputs the graph $G^{\prime}=G-u$ and the function
$f^{\prime}:V(G^{\prime})\rightarrow\mathbb{Z}$ given by

$f^{\prime}(v)=\begin{cases}f(v)-1,&\mbox{if }uv\in E(G);\\\
f(v),&\mbox{otherwise}.\end{cases}$ (1.1)

An application of the operation $\operatorname{Delete}$ is legal if the
resulting function $f^{\prime}$ is nonnegative.

######  Definition (DeleteSave Operation).

Let $G$ be a graph and let $f:V(G)\rightarrow\mathbb{N}$ be a function. For a
pair of adjacent vertices $u,w\in V(G)$, the operation

$\operatorname{DeleteSave}(G,f,u,w)$

outputs the graph $G^{\prime}=G-u$ and the function
$f^{\prime}:V(G^{\prime})\rightarrow\mathbb{Z}$ given by

$f^{\prime}(v)=\begin{cases}f(v)-1,&\mbox{if }uv\in E(G)\mbox{ and }v\neq
w;\\\ f(v),&\mbox{otherwise}.\end{cases}$ (1.2)

An application of the operation $\operatorname{DeleteSave}$ is legal if
$f(u)>f(w)$ and the resulting function $f^{\prime}$ is nonnegative.

A graph $G$ is f-degenerate if it is possible to remove all vertices from $G$
by a sequence of legal applications of the operations $\operatorname{Delete}$.
A graph $G$ is weakly f-degenerate if it is possible to remove all vertices
from $G$ by a sequence of legal applications of the operations
$\operatorname{DeleteSave}$ and $\operatorname{Delete}$. Given
$d\in\mathbb{N}$, we say that $G$ is d-degenerate if it is f-degenerate with
respect to the constant function of value $d$. Also, we say that $G$ is weakly
d-degenerate if it is weakly f-degenerate with respect to the constant
function of value $d$. The degeneracy of $G$, denoted by
$\operatorname{d}(G)$, is the minimum integer $d$ such that $G$ is
d-degenerate. The weak degeneracy of $G$, denoted by $\operatorname{wd}(G)$,
is the minimum integer $d$ such that $G$ is weakly d-degenerate.

Bernshteyn and Lee [1] gave the following inequalities.

######  Proposition 1.

[1] For any graph $G$, we always have

$\chi(G)\leqslant\chi_{l}(G)\leqslant\chi_{DP}(G)\leqslant\chi_{DPP}(G)\leqslant\operatorname{wd}(G)+1\leqslant\operatorname{d}(G)+1,$

where $\chi_{DP}(G)$ is the DP-chromatic number of $G$, and $\chi_{DPP}(G)$ is
the DP-paint number of $G$.

From Proposition 1, $\operatorname{wd}(G)+1$ is an upper bound for many graph
coloring parameters. This is actually a direct result from the greedy coloring
algorithm. Then, it is interesting to determine the weak degeneracy of a
graph.

As far as lower bounds on weak degeneracy are concerned, by a double counting
argument, Bernshteyn and Lee [1] gave the following result.

######  Proposition 2.

[1] Let G be a $d$-regular graph with $n\geqslant 2$ vertices. Then
$\operatorname{wd}(G)\geqslant d-\sqrt{2n}$.

In particular, if $n=O(d)$, then $\operatorname{wd}(G)\geqslant
d-O(\sqrt{d})$. Note that the upper bound $d$ is trivial and can be reached by
the complete graph. It indicates that the weak degeneracy of a $d$-regular
graph seems to be close to $d$. Then, they gave a conjecture about this.

######  Conjecture 1.

[1] Every $d$-regular graph $G$ satisfies $\operatorname{wd}(G)\geqslant
d-O(\sqrt{d})$.

In this paper, we will refute this conjecture by constructing a $d$-regular
graph with weak degeneracy $\lfloor d/2\rfloor+1$.

######  Theorem 1.

There exists a $d$-regular graph with weak degeneracy $\lfloor d/2\rfloor+1$.

Also, we will show this is best possible by proving $\lfloor d/2\rfloor+1$ is
exactly the lower bound.

######  Theorem 2.

Every $d$-regular graph $G$ satisfies $\operatorname{wd}(G)\geqslant\lfloor
d/2\rfloor+1$.

Since the degeneracy of $d$-regular graph is exactly $d$ and the weak
degeneracy can be significantly smaller, we can say that in some cases the
weak degeneracy of regular graphs is a better bound for those graph coloring
parameters under Proposition 1.

##  2\. lower bound of weak degeneracy

By a counting argument, we prove Theorem 2 first.

###### Proof of Theorem 2.

Suppose $G$ is a $d$-regular graph with $\operatorname{wd}(G)\leqslant\lfloor
d/2\rfloor$. Let $k=\operatorname{wd}(G)$. By the definition of weak
degeneracy, if we start with the constant function $f_{1}$ of value $k$ on
$G$, there exists a sequence of legal applications of the operations
$\operatorname{Delete}$ and $\operatorname{DeleteSave}$, which removes all
vertices from $G$. We label the vertices

$v_{1},v_{2},\dots,v_{n}$

by the removing order, and let

$f_{1},f_{2},\dots,f_{n}\quad\mbox{ and }\quad G_{1},G_{2},\dots G_{n}$

denotes the corresponding functions and graphs during this procedure. To the
specific, $f_{i}$ is the function after removing $v_{1},v_{2},\dots,v_{i-1}$
and it is a function on $G_{i}=G[\\{v_{i},v_{i+1},\dots,v_{n}\\}]$. We are
interested in the value of the total sum of each function $f_{i}$, denoted by

$s_{i}:=\sum_{v\in G_{i}}f_{i}(v)=\sum_{j=i}^{n}f_{i}(v_{j}).$

At the same time, the operation $\operatorname{DeleteSave}$ plays an important
role in this procedure. We need two sequences to capture its effect. Let

$x_{i}=\begin{cases}1,&\mbox{if the operation of removing }v_{i}\mbox{ is
}\operatorname{DeleteSave};\\\ 0,&\mbox{if the operation of removing
}v_{i}\mbox{ is }\operatorname{Delete}.\end{cases}$

Let $y_{j}$ denotes the number of $i$ that the operation
$\operatorname{DeleteSave}(G_{i},f_{i},v_{i},v_{j})$ we used. It is not hard
to find that

$\sum_{i=1}^{n}x_{i}=\sum_{i=1}^{n}y_{i},$

since they both count the total number of $\operatorname{DeleteSave}$ we used.

Now, we consider $f_{i}(v_{i})$. After $i-1$ operations, the function value
decreased from $f_{1}(v_{i})=k$ to $f_{i}(v_{i})$. It means that
$k-f_{i}(v_{i})$ neighbors of $v_{i}$ were removed before $v_{i}$, which led
to the decrease of the function value of $v_{i}$. Additionally, there are
$y_{i}$ neighbors that were removed before $v_{i}$ without changing the
function value of $v_{i}$. In total, $k-f_{i}(v_{i})+y_{i}$ neighbors of
$v_{i}$ were removed, then there are $d-(k-f_{i}(v_{i})+y_{i})$ remaining
neighbors of $v_{i}$ in $G_{i}$. From (1.1) and (1.2), we know the removal of
$v_{i}$ makes that the total function value of $v_{i}$’s neighbors in $G_{i}$
decreases by $d-(k-f_{i}(v_{i})+y_{i})-x_{i}$.

Comparing $s_{i}$ and $s_{i+1}$, we have

$s_{i}-s_{i+1}=f_{i}(v_{i})+d-(k-f_{i}(v_{i})+y_{i})-x_{i}.$

Observe that $f_{i}(v_{i})\geqslant x_{i}$. Since $k\leqslant\lfloor
d/2\rfloor\leqslant d/2$, we know

$s_{i}-s_{i+1}\geqslant d/2+x_{i}-y_{i}.$ (2.1)

Note that (2.1) works for $i\in\\{1,2,\dots,n-1\\}$. For $i=n$, it also works
in principle, but we haven’t defined $s_{n+1}$. Actually, it is natural to let
$s_{n+1}=0$ since it is the total sum of a function with an empty domain. To
avoid confusion, we investigate the case $i=n$ in detail. We have a single-
vertex graph $G_{n}$ and we use $\operatorname{Delete}$ operation for $v_{n}$,
then $x_{n}=0$ and $f_{n}(v_{n})\geqslant 0$. Also, $k-f_{n}(v_{n})+y_{n}$ has
to be $d$, since $v_{n}$ has no neighbors in $G_{n}$. From,
$y_{n}=d-k+f_{n}(v_{n})\geqslant d/2$, we have

$s_{n}=f_{n}(v_{n})\geqslant 0\geqslant d/2+x_{n}-y_{n}.$ (2.2)

Combining (2.1) and $\eqref{eq2.2}$, we know

$s_{1}\geqslant nd/2+\sum_{i=1}^{n}x_{i}-\sum_{i=1}^{n}y_{i}=nd/2.$ (2.3)

On the other hand $s_{1}=kn\leqslant nd/2$. It means that all the equal signs
hold in the analysis above. Then, we have $f_{i}(v_{i})=x_{i}$. However,
$f_{1}$ is a constant function, so we can’t use $\operatorname{DeleteSave}$
for $v_{1}$. Then,

$f_{1}(v_{1})=k\neq 0=x_{1}$

gives a contradiction. ∎

##  3\. regular graphs with low weak degeneracy

To prove Theorem 1, we start with the odd case. For each positive integer
$k\geqslant 0$, we construct a graph $G$ such that
$\operatorname{wd}(G)\leqslant k+1$ and $G$ is $(2k+1)$-regular.

Let $s>2k$ be a large integer to be defined and let $V_{1}$ be the disjoint
union of

$\displaystyle A=\\{a_{1},a_{2},\dots,a_{s}\\},$ $\displaystyle
B=\\{b_{1},b_{2},\dots,b_{s}\\},$ $\displaystyle
C=\\{c_{1},c_{2},\dots,c_{s}\\}.$

Here, $V_{1}=A\cup B\cup C$ is a subset of $V(G)$ and we assign the edges on
$G[V_{1}]$ first. For each $i,j\in\\{1,2,\dots,s\\}$,

$\displaystyle a_{i},a_{j}\mbox{ are
adjacent}\quad\Leftrightarrow\quad|i-j|\in[1,k-1],$ (3.1) $\displaystyle
b_{i},b_{j}\mbox{ are adjacent}\quad\Leftrightarrow\quad|i-j|\in[1,k],$
$\displaystyle c_{i},c_{j}\mbox{ are
adjacent}\quad\Leftrightarrow\quad|i-j|\in[1,k],$ $\displaystyle
a_{i},b_{i}\mbox{ are adjacent},\quad a_{i},c_{i}\mbox{ are adjacent}.$

Figure 1.  the main structure of $G(V_{1})$

All the edges of $G[V_{1}]$ are given by (3.1) and we denote this edge set by
$E_{1}$. Figure 1 shows the main structure of $G[V_{1}]$ by illustrating the
neighbors of $a_{i},b_{i},c_{i}$. As you can see, in this structure
$a_{i},b_{i}$ and $c_{i}$ have $2k,2k+1,2k+1$ neighbors, respectively.
Actually, this main structure only works for $i\in[k+1,s-k]$, and the
structure is fragmentary when $i\leqslant k$ or $i\geqslant s-k+1$. We collect
all these abnormal vertices as

$D:=\\{a_{i}|i\notin[k+1,s-k]\\}\cup\\{b_{i}|i\notin[k+1,s-k]\\}\cup\\{c_{i}|i\notin[k+1,s-k]\\},$
(3.2)

and add pendent edges to them such that their degree is $2k+1$. In other
words, we have a new vertex set $V_{2}$ of size

$|V_{2}|=\sum_{v\in D}(2k+1-\deg_{G[V_{1}]}(v)),$ (3.3)

and each $v\in D$ has $2k+1-\deg_{G[V_{1}]}(v)$ distinct neighbors in $V_{2}$.
This edge set between $D$ and $V_{2}$ is denoted by $E_{2}$, which is shown in
Figure 2.

Figure 2.  The edges between $D$ and $V_{2}$

The last step to construct a $(2k+1)$-regular graph is to add edges between
$V_{2}$ and $V_{1}\backslash D$. By checking the current degree of each vertex
in $V_{1}\cup V_{2}$, we can see that each $v\in V_{2}$ needs $2k$ more
neighbors and $a_{i}$ needs one more neighbor for each $i\in[k+1,s-k]$. We
simply choose a proper $s$ such that their demands match. A simple computation
gives

$\displaystyle|V_{2}|=6\cdot(1+2+\dots+k)=3k(k+1),$ (3.4) $\displaystyle
s=2k+2k|V_{2}|=6k^{3}+6k^{2}+2k.$

To be specific, we have a map

$\phi:\\{a_{i}|i\in[k+1,s-k]\\}\rightarrow V_{2}$

such that each vertex in $V_{2}$ has size-$2k$ preimage and we assign edges
according to $\phi$, which means $v$ is adjacent to $\phi(v)$ for each $v$ in
the domain. We denote the set of these edges by $E_{3}$. The construction of
$G$ is done with $V(G)=V_{1}\cup V_{2}$ and $E(G)=E_{1}\cup E_{2}\cup E_{3}$.
It is trivial to check that $G$ is $(2k+1)$-regular.

We need to show that $\operatorname{wd}(G)$ is at most $k+1$, so we start with
a constant function of value $k+1$. We use $\operatorname{Delete}$ or
$\operatorname{DeleteSave}$ in the following order on the vertice set:

$a_{1},b_{1},c_{1},a_{2},b_{2},c_{2},a_{3},b_{3},c_{3},\dots,a_{s},b_{s},c_{s},V_{2}.$

During this deletion process, we use
$\operatorname{DeleteSave}(G,f,a_{i},\phi(a_{i}))$ for $a_{i}\in
V_{1}\backslash D$ when it is legal, and use $\operatorname{Delete}$ for all
the other vertices. Note that $f$ represents the function on the vertex set at
the current step of deletion, and it keeps changing during the process. We
have the following observations.

######  Lemma 1.

Under the deletion order above, for each $v\in V_{1}$, at the step of deleting
$v$, we have $f(v)\geqslant 0$.

###### Proof.

It suffices to show that there are at most $k+1$ neighbors that have been
deleted at the step of deleting $v$. This is true since the deletion only
happens in $V_{1}$ and the construction of $E_{1}$ in (3.1) ensures this fact.
∎

Lemma 1 shows that the deletions of $V_{1}$ are all legal.

######  Lemma 2.

Under the deletion order above, for each $a_{i}\in V_{1}\backslash D$, at the
step of deleting $a_{i}$, we have $f(a_{i})=2$.

###### Proof.

$a_{i}\notin V_{2}$ indicates that we never use $a_{i}$ as the second vertex
of $\operatorname{DeleteSave}$. The neighbors

$a_{i-k+1},a_{i-k+2},\dots,a_{i-1}$

of $a_{i}$ have been deleted and other neighbors have not. It means that

$f(a_{i})=(k+1)-(k-1)=2.$

at the current step. ∎

From Lemma 2, we know that we only use
$\operatorname{DeleteSave}(G,f,a_{i},\phi(a_{i}))$ for $a_{i}\in
V_{1}\backslash D$ if $f(\phi(a_{i}))\leqslant 1$ at the step of deleting
$a_{i}$.

It remains to investigate the function values on $V_{2}$. Note that $V_{2}$ is
an independent set of $G$.

######  Lemma 3.

Under the assumptions above, after the deletion of $V_{1}$, $f(v)\geqslant 0$
for all $v\in V_{2}$.

###### Proof.

For each vertex $v\in V_{2}$, one neighbor $u$ is in $D$ and the remaining
neighbors form the preimage $\phi^{-1}(v)$ with size $2k$. From Lemma 2, the
deletion of each vertex in $\phi^{-1}(v)$ can never reduce its function $f(v)$
to $0$ because $\operatorname{DeleteSave}$ is used when $f(v)=1$. With
considering the deletion of $u$, the value of $v$ is still non-negative. ∎

This completes the proof of $\operatorname{wd}(G)\leqslant k+1$. From Theorem
2, we know $\operatorname{wd}(G)\geqslant k+1$ since $G$ is $(2k+1)$-regular.
As a result, the construction of $G$ establishes the odd case of Theorem 1.

######  Theorem 3.

There exists a $(2k+1)$-regular graph with weak degeneracy $k+1$.

It is not hard to generalize this construction to the even case, but actually
the even case is a direct consequence of the odd case by the following
argument.

######  Theorem 4.

If $G$ is $d$-regular, then there exists a $(d+1)$-regular graph $G^{\prime}$
such that $\operatorname{wd}(G^{\prime})\leqslant\operatorname{wd}(G)+1$.

###### Proof.

We give the construction of $G^{\prime}$ explicitly. We collect $d+1$ copies
of $G$ and add a common neighbor for $d+1$ copies of each vertex in $G$. See
Figure 3.

Figure 3. The construction of $G^{\prime}$

To be specific, let

$V(G^{\prime}):=V(G)\times\\{0,1,2,\dots,d+1\\}.$

For $i\in\\{1,2,\dots,d+1\\}$, $(u,i)$ and $(v,i)$ are adjacent if $uv\in
E(G)$. Also, $(v,i)$ and $(v,0)$ are adjacent for each
$i\in\\{1,2,\dots,d+1\\}$ and $v\in V(G)$. These are all the edges of
$G^{\prime}$. It is trivial to check $G^{\prime}$ is $(d+1)$-regular.

We need to show
$\operatorname{wd}(G^{\prime})\leqslant\operatorname{wd}(G)+1$, so we start
with a constant function of value $\operatorname{wd}(G)+1$. We can use
$\operatorname{Delete}$ operations for each $(v,0)\in V(G^{\prime})$. These
vertices form an independent set and these deletion operations are legal. The
remaining graph is $d+1$ copies of $G$ and the function value is a constant
$\operatorname{wd}(G)$. By the definition of weak degeneracy, we can legally
remove each copy of $G$ by $\operatorname{Delete}$ and
$\operatorname{DeleteSave}$. As a result, we have
$\operatorname{wd}(G^{\prime})\leqslant\operatorname{wd}(G)+1$.

∎

We end with a proof of Theorem 1.

###### Proof of Theorem 1.

When $d$ is an odd number, the statement holds because of Theorem 3. When $d$
is an even number, from Theorem 3, we have a $(d-1)$-regular graph with weak
degeneracy $d/2$. By Theorem 4, we have a $d$-regular graph with weak
degeneracy at most $d/2+1$. By Theorem 2, its weak degeneracy is at least
$d/2+1$. Therefore, we have a $d$-regular graph with weak degeneracy $d/2+1$.
∎

## Acknowledgements

Research supported by “the Fundamental Research Funds for the Central
University” in China.

## References

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††institutetext:  a Department of Physics, Bogazici University, 34342 Bebek,
Istanbul, Turkey
b Institute of Radiation Problems, Azerbaijan National Academy of Sciences,
B.Vahabzade St. 9, AZ1143, Baku, Azerbaijan
c Department of Mathematics, Khazar University, Mehseti St. 41, AZ1096, Baku,
Azerbaijan

# Variational symmetries of Lagrangian systems with higher-order derivatives

Ege Cobana, Ilmar Gahramanova,b,c, and Dilara Kosvaa<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We discuss an elementary derivation of variational symmetries and
corresponding integrals of motion for the Lagrangian systems depending on
acceleration. Providing several examples, we make the manuscript accessible to
a wide range of readers with interest in higher-order Lagrangians and
symmetries. The discussed technique is also applicable to the Lagrangian
systems with higher-order derivatives.

##  1 Introduction

There are many reasons why physicists and mathematicians are interested in
studying Lagrangian systems with higher-order derivatives. The history of such
theories dates back to the study of instabilities due to Ostrogradsky
ostrogradsky1850memoires . Since then, theories with higher-order derivatives
have been studied by many authors from various view points111It is impossible
to cite all relevant papers, we mention here some more modern works for the
interested reader Woodard:2015zca ; de1995symmetries ; ccaugatay2018reductions
; cruz2016hamiltonian ..

In torres2013variational  (see also del2017variational ;
arutyunov2019liouville ) authors presented an elementary derivation of the
variational symmetries and the integrals of motion associated with them. In
this paper, we will consider an extension of this idea to the Lagrangian
systems with higher-order derivatives, especially to the case when the
Lagrangian depends on generalized acceleration. Essentially, we give a
derivation of the equation that determines variational symmetries and then
obtain integrals of motion for some systems. To our knowledge, this technique
is the simplest way to obtain integrals of motion for the Lagrangian systems
with higher-order derivatives. It should be stressed that all computations
presented here rely heavily on group-theoretical studies of differential
equations ibragimov1995crc ; ibragimov1992group ; bluman2008symmetry ;
olver2000applications .

The paper is organized as follows. The variational symmetries and the relevant
techniques are discussed in Sections II and III. Then we present some examples
in Section IV, where we mainly focus on the Lagrangians with second-order
derivatives. In Section V, we shortly discuss higher orders.

##  2 Variational Symmetries for Second-Order Lagrangians

The Euler-Lagrange equation for the Lagrangian depending on second derivatives
of generalized coordinates
$L=L(\mathbf{x},\dot{\mathbf{x}},\ddot{\mathbf{x}};t)$ has the following form,

$\frac{\partial L}{\partial
x_{i}}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial
L}{\partial\dot{x}_{i}}+\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\frac{\partial
L}{\partial\ddot{x}_{i}}=0\;.$ (1)

This is a fourth-order ordinary differential equation. We are going to
construct variational symmetries for this Lagrangian system by solving some
corresponding differential equations.

Consider an action $S=\int
L(\mathbf{x},\dot{\mathbf{x}},\ddot{\mathbf{x}};t)\;\mathrm{d}t$ which is
invariant under the invertible $s$-parametric transformations
$x^{\prime}_{i}=x^{\prime}_{i}(\mathbf{x},t,s)$ and
$t^{\prime}=t^{\prime}(\mathbf{x},t,s)$. These transformations form a one-
parameter group (with the parameter $s$). The corresponding infinitesimal
transformations are

$\displaystyle x^{\prime}_{i}=x_{i}+\eta_{i}(\mathbf{x},t)s\hskip
14.22636pt;\hskip 14.22636pti=1,2,...,n$ (2) $\displaystyle
t^{\prime}=t+\xi(\mathbf{x},t)s\hskip 14.22636pt;\hskip
14.22636pti=1,2,...,n\;,$ (3)

where the functions $\eta_{i}$ and $\xi$ are given by

$\eta_{i}(\mathbf{x};t)=\frac{\partial
x_{i}^{\prime}(\mathbf{x},t,s)}{\partial s}\Bigr{\rvert}_{s=0}\hskip
14.22636pt;\hskip 14.22636pti=1,2,...,n$ (4) $\xi(\mathbf{x};t)=\frac{\partial
t^{\prime}(\mathbf{x},t,s)}{\partial s}\Bigr{\rvert}_{s=0}\hskip
14.22636pt;\hskip 14.22636pti=1,2,...,n\;.$ (5)

The invariance of the action under these transformations (in order to obtain
the same equations of motion) implies that

$L^{\prime}(\mathbf{x}^{\prime},\frac{\mathrm{d}\mathbf{x}^{\prime}}{\mathrm{d}t^{\prime}},\frac{\mathrm{d}^{2}\mathbf{x}^{\prime}}{\mathrm{d}t^{\prime
2}},t^{\prime})\frac{\mathrm{d}t^{\prime}}{\mathrm{d}t}=L(\mathbf{x},\dot{\mathbf{x}},\ddot{\mathbf{x}},t)+\frac{\mathrm{d}}{\mathrm{d}t}F(\mathbf{x},\dot{\mathbf{x}},t,s)\;,$
(6)

where we added the total derivative term to the Lagrangian. If the Lagrangian
transforms under the transformations (2)-(3) as in expression (6) then the
corresponding symmetry is called a variational symmetry of Lagrangian. By
differentiating the last expression with respect to the parameter $s$ and by
setting $s=0$ one obtains

$\sum_{i=1}^{n}\left(\frac{\partial L}{\partial x_{i}}\eta_{i}+\frac{\partial
L}{\partial\dot{x}_{i}}\Bigr{[}\frac{\mathrm{d}\eta_{i}}{\mathrm{d}t}-\dot{x}_{i}\frac{\mathrm{d}\xi}{\mathrm{d}t}\Bigr{]}+\frac{\partial
L}{\partial\ddot{x}_{i}}\Bigr{[}\frac{\mathrm{d}^{2}\eta_{i}}{\mathrm{d}t^{2}}-2\ddot{x}_{i}\frac{\mathrm{d}\xi}{\mathrm{d}t}-\dot{x}_{i}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}\Bigr{]}\right)+\frac{\partial
L}{\partial
t}\xi+L\frac{\mathrm{d}\xi}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}G(\mathbf{x},\dot{\mathbf{x}};t)\;.$
(7)

For the terms inside the square brackets we used the following expressions

$\displaystyle\frac{\partial\dot{x}_{i}^{\prime}}{\partial
s}\Bigr{\rvert}_{s=0}$ $\displaystyle=$
$\displaystyle\frac{\mathrm{d}\eta_{i}}{\mathrm{d}t}-\dot{x}_{i}\frac{\mathrm{d}\xi}{\mathrm{d}t}$
(8) $\displaystyle\frac{\partial\ddot{x}_{i}^{\prime}}{\partial
s}\Bigr{\rvert}_{s=0}$ $\displaystyle=$
$\displaystyle\frac{\mathrm{d}^{2}\eta_{i}}{\mathrm{d}t^{2}}-2\ddot{x}_{i}\frac{\mathrm{d}\xi}{\mathrm{d}t}-\dot{x}_{i}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}\;.$
(9)

and the gauge function $G(\mathbf{x},\dot{\mathbf{x}};t)$ is defined as

$G(\mathbf{x},\dot{\mathbf{x}};t):=\frac{\partial}{\partial
s}F(\mathbf{x},\dot{\mathbf{x}},t,s)\Bigr{\rvert}_{s=0}\;.$ (10)

Now by using the Euler-Lagrange equation in $\eqref{eq1}$, one ends up with
the following expression

$\frac{\mathrm{d}}{\mathrm{d}t} \left[\xi
L+\sum_{i=1}^{n}\left((\eta_{i}-\dot{x}_{i}\xi)\frac{\partial
L}{\partial\dot{x}_{i}}+\frac{\mathrm{d}(\eta_{i}-\dot{x}_{i}\xi)}{\mathrm{d}t}\frac{\partial
L}{\partial\ddot{x}_{i}}-(\eta_{i}-\dot{x}_{i}\xi)\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial
L}{\partial\ddot{x}_{i}}\right)\right)-G\right]=0\;.$ (11)

The expression in the bracket is conserved, i.e. it is integral of motion
under the time evolution torres2013variational ; deriglazov2016classical .
This expression is an analog of the Rund-Trautman identity222In the
literature, sometimes this expression is called Noether-Bassel-Hagen identity,
see e.g. leone2018wonderfulness . trautman1967noether ; neuenschwander2017emmy
; gourieux2021noether .

Now let us compute variational symmetries for some Lagrangian systems with the
second-order derivatives. Using the expression (7) one can easily derive
integrals of motion for relatively simple systems.

###  2.1 Spinning particle

We begin with the most familiar example of the second order Lagrangian
systems, namely with the Lagrangian of the classical spinning particle (for
simplicity we put $m=\omega=1$)

$L=\frac{1}{2}(\ddot{x}^{2}-\dot{x}^{2})\;.$ (12)

The Euler-Lagrange equation for this system is the following fourth-order
differential equation which describes a particle rotating around its
translating center

$x^{(4)}+\ddot{x}=0 \;.$ (13)

This system was discussed in many works, the integrals of motion and related
symmetries can be found, e.g. in cracsmuareanu2000noetherian . By inserting
the Lagrangian of the spinning particle to the expression (7) yields

$\begin{split}\ddot{x}^{2}\Bigr{[}\frac{\partial\eta}{\partial
x}-\frac{5}{2}\dot{x}\frac{\partial\xi}{\partial
x}-\frac{3}{2}\frac{\partial\xi}{\partial
t}\Bigr{]}+\ddot{x}\Bigr{[}\dot{x}^{2}\frac{\partial^{2}\eta}{\partial
x^{2}}+2\dot{x}\frac{\partial^{2}\eta}{\partial t\partial
x}+\frac{\partial^{2}\eta}{\partial
t^{2}}-\dot{x}^{3}\frac{\partial^{2}\xi}{\partial
x^{2}}-2\dot{x}^{2}\frac{\partial^{2}\xi}{\partial t\partial x}\\\
-\dot{x}\frac{\partial^{2}\xi}{\partial
t^{2}}\Bigr{]}+\dot{x}^{3}\frac{1}{2}\frac{\partial\xi}{\partial
x}+\dot{x}^{2}\Bigr{[}-\frac{\partial\eta}{\partial
x}+\frac{1}{2}\frac{\partial\xi}{\partial
t}\Bigr{]}-\dot{x}\frac{\partial\eta}{\partial t}=\frac{\partial
G}{\partial\dot{x}}\ddot{x}+\frac{\partial G}{\partial
x}\dot{x}+\frac{\partial G}{\partial t}\;.\end{split}$ (14)

Since the gauge function $G(x,\dot{x};t)$ is not a function of $\ddot{x}$
coefficients of $\ddot{x}^{2}$ and $\ddot{x}$ terms should vanish. As a result
we obtain the following differential equations

$\frac{\partial\eta}{\partial x}-\frac{5}{2}\dot{x}\frac{\partial\xi}{\partial
x}-\frac{3}{2}\frac{\partial\xi}{\partial t}=0$ (15)
$\dot{x}^{2}\frac{\partial^{2}\eta}{\partial
x^{2}}+2\dot{x}\frac{\partial^{2}\eta}{\partial t\partial
x}+\frac{\partial^{2}\eta}{\partial
t^{2}}-\dot{x}^{3}\frac{\partial^{2}\xi}{\partial
x^{2}}-2\dot{x}^{2}\frac{\partial^{2}\xi}{\partial t\partial
x}-\dot{x}\frac{\partial^{2}\xi}{\partial t^{2}}=\frac{\partial
G}{\partial\dot{x}}$ (16) $\dot{x}^{3}\frac{1}{2}\frac{\partial\xi}{\partial
x}+\dot{x}^{2}\Bigr{[}-\frac{\partial\eta}{\partial
x}+\frac{1}{2}\frac{\partial\xi}{\partial
t}\Bigr{]}-\dot{x}\frac{\partial\eta}{\partial t}=\frac{\partial G}{\partial
x}\dot{x}+\frac{\partial G}{\partial t}\;.$ (17)

In some sense, these equations have the same complexity as the Euler-Lagrange
equations, here the idea is to find symmetries and corresponding integrals of
motion. The equation (15) implies that $\xi(x;t)$ is not a function of $x$ and
moreover $\eta(x;t)$ can be found as

$\eta=\frac{3}{2}x\frac{\mathrm{d}\xi}{\mathrm{d}t}+A(t)\;,$ (18)

where $A(t)$ is a function of time only. If we insert the expression for
$\eta$ in the expressions (16)-(17), we obtain the following equations for the
partial differentials of the gauge function $G(x,\dot{x};t)$

$-2\dot{x}\frac{\mathrm{d}\xi}{\mathrm{d}t}-\frac{3}{2}x\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}-\frac{\mathrm{d}A}{\mathrm{d}t}-\frac{7}{2}\dot{x}\frac{\mathrm{d}^{3}\xi}{\mathrm{d}t^{3}}-\frac{3}{2}x\frac{\mathrm{d}^{4}\xi}{\mathrm{d}t^{4}}-\frac{\mathrm{d}^{3}A}{\mathrm{d}t^{3}}=\frac{\partial
G}{\partial x}$ (19)
$2\dot{x}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}+\frac{3}{2}x\frac{\mathrm{d}^{3}\xi}{\mathrm{d}t^{3}}+\frac{\mathrm{d}^{2}A}{\mathrm{d}t^{2}}=\frac{\partial
G}{\partial\dot{x}}$ (20)
$\dot{x}^{2}\frac{\mathrm{d}\xi}{\mathrm{d}t}+\frac{7}{2}\dot{x}^{2}\frac{\mathrm{d}^{3}\xi}{\mathrm{d}t^{3}}+\frac{3}{2}x\dot{x}\frac{\mathrm{d}^{4}\xi}{\mathrm{d}t^{4}}+\dot{x}\frac{\mathrm{d}^{3}A}{\mathrm{d}t^{3}}=\frac{\partial
G}{\partial t}\;.$ (21)

By comparing them one obtains the following four ordinary differential
equations

$2\frac{\mathrm{d}\xi}{\mathrm{d}t}+5\frac{\mathrm{d}^{3}\xi}{\mathrm{d}t^{3}}=0$
(22)
$2\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}+5\frac{\mathrm{d}^{4}\xi}{\mathrm{d}t^{4}}=0$
(23)
$\frac{\mathrm{d}^{3}\xi}{\mathrm{d}t^{3}}+\frac{\mathrm{d}^{5}\xi}{\mathrm{d}t^{5}}=0$
(24)
$\frac{\mathrm{d}^{2}A}{\mathrm{d}t^{2}}+\frac{\mathrm{d}^{4}A}{\mathrm{d}t^{4}}=0\;.$
(25)

One can straightforwardly solve these differential equations, and find that

$\xi=c_{1}\hskip 14.22636pt;\hskip
14.22636ptA=c_{2}\sin{t}+c_{3}\cos{t}+c_{4}t+c_{5}\;.$ (26)

Therefore $\eta(x;t)$ has the form

$\eta=c_{2}\sin{t}+c_{3}\cos{t}+c_{4}t+c_{5}\;,$ (27)

where $c_{2}$, $c_{3}$, $c_{4}$ and $c_{5}$ are arbitrary constants. It is
easy to compute the gauge function $G(x,\dot{x};t)$

$G=-c_{2}\dot{x}\sin{t}-c_{3}\dot{x}\cos{t}-c_{4}x\;.$ (28)

Now by inserting $\eta$, $\xi$ and $G$ into the (11) one ends up with the
following expression

$\begin{split}\frac{\mathrm{d}}{\mathrm{d}t}\Bigr{[}c_{1}(-\frac{\ddot{x}^{2}}{2}+\frac{\dot{x}^{2}}{2}+\dot{x}\dddot{x})+c_{2}(\ddot{x}\cos{t}-\dddot{x}\sin{t})+c_{3}(-\dddot{x}\cos{t}-\ddot{x}\sin{t})\\\
+c_{4}(-\dot{x}t-\dddot{x}t+\ddot{x}+x)+c_{5}(-\dot{x}-\dddot{x})\Bigr{]}=0\;.\end{split}$
(29)

Since $c_{1}$, $c_{2}$, $c_{3}$, $c_{4}$ and $c_{5}$ are arbitrary constants,
terms within the parenthesis are integrals of motion with the following
symmetry generators

$\displaystyle\Gamma_{1}$ $\displaystyle=\frac{\partial}{\partial t}\hskip
14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{1}$
$\displaystyle=\ddot{x}^{2}-2\dot{x}\dddot{x}-\dot{x}^{2}$ (30)
$\displaystyle\Gamma_{2}$ $\displaystyle=\sin{t}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{2}$
$\displaystyle=\ddot{x}\cos{t}-\dddot{x}\sin{t}$ (31)
$\displaystyle\Gamma_{3}$ $\displaystyle=\cos{t}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{3}$
$\displaystyle=\ddot{x}\sin{t}+\dddot{x}\cos{t}$ (32)
$\displaystyle\Gamma_{4}$ $\displaystyle=t\frac{\partial}{\partial x}\hskip
14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{4}$
$\displaystyle=x+\ddot{x}-t(\dot{x}+\dddot{x})$ (33) $\displaystyle\Gamma_{5}$
$\displaystyle=\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{5}$ $\displaystyle=\dot{x}+\dddot{x}\;.$
(34)

Note that in the two-dimensional case, i.e. if the Lagrangian of the spinning
particle has a form

$L=\frac{1}{2}(\ddot{x}^{2}-\dot{x}^{2}+\ddot{y}^{2}-\dot{y}^{2})\;,$ (35)

we will get one more integral of motion (the third component of the angular
momentum)

$\Gamma_{6}=x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\hskip
14.22636pt;\hskip
14.22636ptI_{6}=y(\dot{x}+\dddot{x})-x(\dot{y}+\dddot{y})+\ddot{y}\dot{x}-\ddot{x}\dot{y}\;.$
(36)

As one can see, we constructed the infinitesimal transformations by solving
corresponding differential equations333Of course, if one knows the symmetries
in advance, then it is easier to use the Noether theorem to obtain the
integrals of motion..

##  3 Cyclic coordinates

If we have a one-parameter symmetry group of the action, then we can discard
one of the coordinates and consider the parameter $s$ as a new coordinate.
Then the coordinate $s$ turns out to be cyclic (ignorable) and the
corresponding momentum is conserved. Actually, this is an idea of Noether’s
theorem.

We would like to show that for
$L(\mathbf{x},\mathbf{\dot{x}},\mathbf{\ddot{x}};t)$, there exists a
coordinate system $({\mathbf{x}}^{\prime},t^{\prime})$ such that for a fixed
$k$ the momentum conjugate corresponding to ${x_{k}}^{\prime}$ is conserved.

$\frac{\mathrm{d}p^{\prime}_{k}}{\mathrm{d}t^{\prime}}=0\;.$ (37)

Under coordinate transformations
$x_{i}=x_{i}({\mathbf{x}}^{\prime},t^{\prime})$ and
$t=t({\mathbf{x}}^{\prime},t^{\prime})$, Lagrangian transforms as;

$L(\mathbf{x},\mathbf{\dot{x}},\mathbf{\ddot{x}};t)\frac{\mathrm{d}t}{\mathrm{d}t^{\prime}}=L^{\prime}(\mathbf{x}^{\prime},\mathbf{\dot{x}}^{\prime},\mathbf{\ddot{x}}^{\prime};t)$
(38) $\frac{\partial
L^{\prime}}{\partial{x_{k}}^{\prime}}=L\frac{\partial}{\partial
x_{k}^{\prime}}\left(\frac{\mathrm{d}t}{\mathrm{d}t^{\prime}}\right)+\frac{\mathrm{d}t}{\mathrm{d}t^{\prime}}\left(\sum_{i=1}^{n}\left(\frac{\partial
L}{\partial x_{i}}\frac{\partial x_{i}}{\partial
x_{k}^{\prime}}+\frac{\partial
L}{\partial\dot{x}_{i}}\frac{\partial\dot{x}_{i}}{\partial
x_{k}^{\prime}}+\frac{\partial
L}{\partial\ddot{x_{i}}}\frac{\partial\ddot{x_{i}}}{\partial{x_{k}}^{\prime}}\right)+\frac{\partial
L}{\partial t}\frac{\partial t}{\partial x_{k}^{\prime}} \right)=0\;.$ (39)

In order to simplify this expression, we make the following definitions and
follow the same method used in Section II,

$\xi:=\frac{\partial t}{\partial x_{k}^{\prime}}\;\; \eta_{i}:=\frac{\partial
x_{i}}{\partial x_{k}^{\prime}}$ (40)
$L\frac{\mathrm{d}\xi}{\mathrm{d}t}+\sum_{i=1}^{n}\left[\frac{\partial
L}{\partial x_{i}}\eta_{i}+\frac{\partial
L}{\partial\dot{x}_{i}}\left(\frac{\mathrm{d}\eta_{i}}{\mathrm{d}t}-\dot{x}_{i}\frac{\mathrm{d}\xi}{\mathrm{d}t}\right)+\frac{\partial
L}{\partial\ddot{x_{i}}}\left(\frac{\mathrm{d}^{2}\eta_{i}}{\mathrm{d}t^{2}}-2\ddot{x_{i}}\frac{\mathrm{d}\xi}{\mathrm{d}t}-\dot{x_{i}}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}\right)\right]+\frac{\partial
L}{\partial t}\xi=0 \;,$ (41)

Then the corresponding momentum takes the following form

$p^{\prime}_{k}=\xi L+\sum_{i=1}^{n}(\eta_{i}-\dot{x}_{i}\xi)\frac{\partial
L}{\partial\dot{x}_{i}}+\frac{\mathrm{d}(\eta_{i}-\dot{x}_{i}\xi)}{\mathrm{d}t}\frac{\partial
L}{\partial\ddot{x}_{i}}-(\eta_{i}-\dot{x}_{i}\xi)\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial
L}{\partial\ddot{x}_{i}}\right)\;.$ (42)

One can also consider the case where $x^{\prime}_{k}$ is not necessarily
cyclic. If,

$\frac{\partial L}{\partial x_{k}}=\frac{\mathrm{d}G}{\mathrm{d}t}\;,$ (43)

Then from the Euler-Lagrange equations,

$\frac{\mathrm{d}(p_{k}-G)}{\mathrm{d}t}=0 \;.$ (44)

In the extended configuration space the expression (43) corresponds to

$L\frac{\mathrm{d}\xi}{\mathrm{d}t}+\sum_{i=1}^{n}\left[\frac{\partial
L}{\partial x_{i}}\eta_{i}+\frac{\partial
L}{\partial\dot{x}_{i}}\left(\frac{\mathrm{d}\eta_{i}}{\mathrm{d}t}-\dot{x}_{i}\frac{\mathrm{d}\xi}{\mathrm{d}t}\right)+\frac{\partial
L}{\partial\ddot{x_{i}}}\left(\frac{\mathrm{d}^{2}\eta_{i}}{\mathrm{d}t^{2}}-2\ddot{x_{i}}\frac{\mathrm{d}\xi}{\mathrm{d}t}-\dot{x}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}\right)\right]+\frac{\partial
L}{\partial t}\xi=\frac{\mathrm{d}G}{\mathrm{d}t}\;.$ (45)

Hence the conserved quantity takes the form given in the expression (11).

It is still possible to write an equivalent Lagrangian $\tilde{L}^{\prime}$
where $x_{k}^{\prime}$ is a cyclic coordinate. For that reason we define a
function $F(\mathbf{x}^{\prime},\dot{\mathbf{x}}^{\prime};t^{\prime})$ as
follows

$G=-\frac{\partial F}{\partial x^{\prime}_{k}}\;.$ (46)

Now, it is easy to see that defining a new Lagrangian as
$\tilde{L^{\prime}}=L^{\prime}+\dfrac{\mathrm{d}F}{\mathrm{d}t^{\prime}}$ will
result in an equivalent Lagrangian with a cyclic coordinate $x^{\prime}_{k}$
since

$\frac{\partial}{\partial
x_{k}^{\prime}}\left(L^{\prime}+\frac{\mathrm{d}F}{\mathrm{d}t^{\prime}}\right)=0\;.$
(47)

##  4 Examples

###  4.1 Higher-derivative harmonic oscillator

As an elementary example, we will consider the following second-order
Lagrangian

$L=\frac{1}{2}({\ddot{x}^{2}-x^{2}})\;.$ (48)

This system was studied as a toy model in Smilga:2005gb ; Boulanger:2018tue .
We need to find variational symmetries satisfying the condition (41).
Substituting the above Lagrangian into the equation (41), we obtain

$-x\eta+\ddot{x}\left(\frac{\mathrm{d}^{2}\eta_{i}}{\mathrm{d}t^{2}}-2\ddot{q_{i}}\frac{\mathrm{d}\xi}{\mathrm{d}t}-\dot{q}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}\right)+\frac{1}{2}\left(\ddot{x}^{2}-x^{2}\right)\frac{\mathrm{d}\xi}{\mathrm{d}t}=\frac{\mathrm{d}G}{\mathrm{d}t}\;,$
(49)

where the gauge function $G(x,\dot{x};t)$ is a function of $t,x$ and
$\dot{x}$. Following the same method that we have explained in our previous
example, we can express our functions as follows

$\displaystyle\xi$ $\displaystyle=$ $\displaystyle c_{1}$ (50)
$\displaystyle\eta$ $\displaystyle=$ $\displaystyle
c_{2}e^{t}+c_{3}e^{-t}+c_{4}\cos{t}+c_{5}\sin{t}$ (51) $\displaystyle G$
$\displaystyle=$ $\displaystyle
c_{2}e^{t}(\dot{x}-x)+c_{3}e^{-t}(\dot{x}+x)+c_{4}(-\dot{x}\cos{t}-x\sin{t})+c_{5}(-\dot{x}\sin{t}+x\cos{t})\;.$
(52)

Corresponding symmetry generators and integrals of motion are

$\displaystyle\Gamma_{1}$ $\displaystyle=\frac{\partial}{\partial t}\hskip
14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{1}$
$\displaystyle=\dot{x}\dddot{x}-\frac{x^{2}}{2}-\frac{\ddot{x}^{2}}{2}$ (53)
$\displaystyle\Gamma_{2}$ $\displaystyle=e^{t}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{2}$
$\displaystyle=(x-\dot{x}+\ddot{x}-\dddot{x})e^{t}$ (54)
$\displaystyle\Gamma_{3}$ $\displaystyle=e^{-t}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{3}$
$\displaystyle=(x+\dot{x}+\ddot{x}+\dddot{x})e^{-t}$ (55)
$\displaystyle\Gamma_{4}$ $\displaystyle=\cos{t}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{4}$
$\displaystyle=(\dot{x}-\dddot{x})\cos{t}+(x-\ddot{x})\sin{t}$ (56)
$\displaystyle\Gamma_{5}$ $\displaystyle=\sin{t}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{5}$
$\displaystyle=(\dot{x}-\dddot{x})\sin{t}+(-x+\ddot{x})\cos{t} \;.$ (57)

In order to simplify the problem of finding a Lagrangian with a cyclic
coordinate, we will consider the case where $c_{i}$’s are all equal to zero
except $c_{2}$ which equals to $1$. Then

$\xi=0\;,\;\eta=e^{t}\;,\;G=\dot{x}e^{t}-xe^{t}\;.$ (58)

The coordinate transformations satisfying these equations are

$t=t^{\prime} \;,\;x=x^{\prime}e^{t^{\prime}}\;.$ (59)

Under these transformations, the Lagrangian (48) transforms as

$L^{\prime}=\frac{1}{2}\left((\ddot{x}^{\prime}e^{t^{\prime}}+2\dot{x}^{\prime}e^{t^{\prime}}+x^{\prime}e^{t^{\prime}})^{2}-{x^{\prime}}^{2}e^{2t^{\prime}}\right)\;.$
(60)

The Lagrangian $\tilde{L}^{\prime}$ where $x^{\prime}$ is a cyclic coordinate
is given by

$\tilde{L}^{\prime}=L^{\prime}+\frac{\mathrm{d}F}{\mathrm{d}t^{\prime}}\;,$
(61)

where $F(x,\dot{x},t)$ is defined as in (46). After expressing
$G(x,\dot{x},t)$ and $F(x,\dot{x},t)$ in terms of primed coordinates, we
simply follow the definition and $\tilde{L}^{\prime}$ takes the following
form,

$\tilde{L}^{\prime}=\frac{e^{2t^{\prime}}}{2}\left(\ddot{x}^{\prime
2}+2\dot{x}^{\prime 2}+4\ddot{x}^{\prime}\dot{x}^{\prime}\right)\;,$ (62)

where $x$ is clearly cyclic. The momentum conjugate corresponding to the
coordinate $x$ is

$p^{\prime}=-e^{2t^{\prime}}\left(\dddot{x}^{\prime}+
2\ddot{x}^{\prime}+2\dot{x}^{\prime}\right)\;.$ (63)

Returning to our initial coordinate system, the conserved quantity corresponds
to,

$I=(x-\dot{x}+\ddot{x}-\dddot{x})e^{t}\;,$ (64)

which is also given by (11).

###  4.2 An illustrative example

Now, we would like to move on to a slightly more complicated case. As a next
example, consider a system with the following Lagrangian

$L=\dot{x}^{4}+3x^{2}\ddot{x}^{2}\;.$ (65)

The Euler-Lagrange equation for this system is the following ordinary
differential equation

$3\ddot{x}^{2}+xx^{(4)}+4\dot{x}\dddot{x}=0\;.$ (66)

By applying (7) we obtain the following partial differential equation

$x\eta+x^{2}\frac{\partial\eta}{\partial
x}-\frac{3}{2}x^{2}\frac{\mathrm{d}\xi}{\mathrm{d}t}=0\;,$ (67)

from which one can easily find that $\xi=\xi(t)$. Then

$\frac{\partial(x\eta)}{\partial
x}=\frac{3}{2}x\frac{\mathrm{d}\xi}{\mathrm{d}t}\;.$ (68)

Therefore $\eta(x;t)$ has the form
$\eta=\dfrac{3}{4}x\dfrac{\mathrm{d}\xi}{\mathrm{d}t}+\dfrac{A(t)}{x}$.

The partial differential equations involving derivatives of the gauge
functions are

$3x^{2}\dot{x}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}+\frac{9}{2}x^{3}\frac{\mathrm{d}^{3}\xi}{\mathrm{d}t^{3}}+6x\frac{\mathrm{d}^{2}A}{\mathrm{d}t^{2}}-12\dot{x}\frac{\mathrm{d}A}{\mathrm{d}t}+12\frac{\dot{x}^{2}}{x}A=\frac{\partial
G}{\partial\dot{x}}$ (69)
$3x\dot{x}^{3}\frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}}+4\frac{\dot{x}^{3}}{x^{2}}\frac{\mathrm{d}A}{\mathrm{d}t}-4\frac{\dot{x}^{4}}{x^{2}}A=\frac{\partial
G}{\partial x}\dot{x}+\frac{\partial G}{\partial t}\;.$ (70)

By solving these equations one finds $\xi$ and $A$ as

$\xi=c_{1}t^{2}+c_{2}t+c_{3}\hskip 14.22636pt;\hskip
14.22636ptA=c_{4}t^{3}+c_{5}t^{2}+c_{6}t+c_{7}\;,$ (71)

and the gauge function $G(x,\dot{x};t)$ as

$\begin{split}G=c_{1}3x^{2}\dot{x}^{2}+&c_{4}\left(36x\dot{x}-18\dot{x}^{2}t^{2}+4\frac{\dot{x}^{3}t^{3}}{x}-18x^{2}\right)\\\
+&c_{5}\left(12x\dot{x}-12\dot{x}^{2}t+4\frac{\dot{x}^{3}t^{2}}{x}\right)+c_{6}\left(-6\dot{x}^{2}+4\frac{\dot{x}^{3}t}{x}\right)+c_{7}4\frac{\dot{x}^{3}}{x}\;.\end{split}$
(72)

Therefore the symmetry generators and corresponding integrals of motion are

$\displaystyle\Gamma_{1}$ $\displaystyle=t^{2}\frac{\partial}{\partial
t}+\frac{3}{2}xt\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{1}$
$\displaystyle=3x^{3}\ddot{x}-x^{2}\dot{x}^{2}+(2x\dot{x}^{3}-7x^{2}\dot{x}\ddot{x}-3x^{3}\dddot{x})t$
(73)
$\displaystyle+(-\dot{x}^{4}-x^{2}\ddot{x}^{2}+4x\dot{x}^{2}\ddot{x}+2x^{2}\dot{x}\dddot{x})t^{2}$
$\displaystyle\Gamma_{2}$ $\displaystyle=t\frac{\partial}{\partial
t}+\frac{3}{4}x\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{2}$
$\displaystyle=2x\dot{x}^{3}-7x^{2}\dot{x}\ddot{x}-3x^{3}\dddot{x}$ (74)
$\displaystyle+(-2\dot{x}^{4}-2x^{2}\ddot{x}^{2}+8x\dot{x}^{2}\ddot{x}+4x^{2}\dot{x}\dddot{x})t$
$\displaystyle\Gamma_{3}$ $\displaystyle=\frac{\partial}{\partial t}\hskip
14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{3}$
$\displaystyle=-\dot{x}^{4}-x^{2}\ddot{x}^{2}+4x\dot{x}^{2}\ddot{x}+2x^{2}\dot{x}\dddot{x}$
(75) $\displaystyle\Gamma_{4}$
$\displaystyle=\frac{t^{3}}{x}\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{4}$
$\displaystyle=3x^{2}-6x\dot{x}t+(3\dot{x}^{2}+3x\ddot{x})t^{2}+(-x\dddot{x}-3\dot{x}\ddot{x})t^{3}$
(76) $\displaystyle\Gamma_{5}$
$\displaystyle=\frac{t^{2}}{x}\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{5}$
$\displaystyle=-2x\dot{x}+(2\dot{x}^{2}+2x\ddot{x})t+(-x\dddot{x}-3\dot{x}\ddot{x})t^{2}$
(77) $\displaystyle\Gamma_{6}$
$\displaystyle=\frac{t}{x}\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{6}$
$\displaystyle=\dot{x}^{2}+x\ddot{x}+(-3\dot{x}\ddot{x}-x\dddot{x})t$ (78)
$\displaystyle\Gamma_{7}$ $\displaystyle=\frac{1}{x}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{7}$
$\displaystyle=3\dot{x}\ddot{x}+x\dddot{x}\;.$ (79)

In order to pass into primed coordinates where $x^{\prime}$ is a cyclic
coordinate, for simplicity we set $c_{1}=1$ and other constants to zero. Then
we make a suitable coordinate transformation,

$t=-\frac{1}{x^{\prime}}\hskip 14.22636pt;\hskip 14.22636pt
x=t^{\prime}{x^{\prime}}^{-\frac{3}{2}}\;.$ (80)

The gauge function $G(x,\dot{x};t)$ in old coordinates is given by the
expression

$G=3x^{2}\dot{x}^{2}\;,$ (81)

and

$F=3\frac{{t\prime}^{2}}{{\dot{x}\prime}^{2}{x\prime}}-\frac{9}{2}\frac{{t\prime}^{3}}{{\dot{x}\prime}{x\prime}^{2}}+\frac{9}{4}\frac{{t\prime}^{4}}{{x\prime}^{3}}\;.$
(82)

From the equation (47) the new Lagrangian $\tilde{L}^{\prime}$ gives to the
desired result

$\tilde{L}\prime=3\frac{{\ddot{x}\prime}^{2}{t\prime}^{2}}{{\dot{x}\prime}^{5}}+\frac{1}{{\dot{x}\prime}^{3}}\;.$
(83)

Here $x^{\prime}$ is a cyclic coordinate and the corresponding conserved
momentum gives the expression $\eqref{I1}$.

###  4.3 Free particle with a Chern-Simons like term

Now let us examine a system with two degrees of freedom. We consider the
following Lagrangian system introduced by Lukierski, Stichel, and Zakrzewski
1997

$L=\frac{\lambda}{2}(\dot{y}\ddot{x}-\dot{x}\ddot{y})+\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2})\;,$
(84)

where $m$ stands for the mass, and $\lambda$ is some parameter. Since the
latter parameter labels the central extension 1997  and has interesting
properties, we specially keep the parameters $m$ and $\lambda$. The Euler-
Lagrange equations are

$\displaystyle m\ddot{x}-\lambda\dddot{y}$ $\displaystyle=0\;,$ (85)
$\displaystyle m\ddot{y}+\lambda\dddot{x}$ $\displaystyle=0\;.$ (86)

Since in general $\eta_{1}$, $\eta_{2}$ and $\xi$ depend on $x,y$ and $t$
using our prescription one obtains the following three partial differential
equations

$\frac{\lambda}{2}\dot{y}\frac{\partial\eta_{2}}{\partial
y}+\frac{\lambda}{2}\frac{\partial\eta_{2}}{\partial
t}+\frac{\lambda}{2}\dot{y}\frac{\partial\eta_{1}}{\partial
x}-\lambda\dot{y}\frac{\mathrm{d}\xi}{\mathrm{d}t}=\frac{\partial
G}{\partial\dot{x}}\;,$ (87)
$-\frac{\lambda}{2}\dot{x}\frac{\partial\eta_{1}}{\partial
x}-\frac{\lambda}{2}\frac{\partial\eta_{1}}{\partial
t}+\lambda\dot{x}\frac{\mathrm{d}\xi}{\mathrm{d}t}-\frac{\lambda}{2}\dot{x}\frac{\partial\eta_{2}}{\partial
y}=\frac{\partial G}{\partial\dot{y}}\;,$ (88)
$\begin{split}m\dot{x}\frac{\mathrm{d}\eta_{1}}{\mathrm{d}t}+m\dot{y}\frac{\mathrm{d}\eta_{2}}{\mathrm{d}t}+\frac{\lambda}{2}\dot{y}\Bigr{[}\dot{x}^{2}\frac{\partial^{2}\eta_{1}}{\partial
x^{2}}+2\dot{x}\dot{y}\frac{\partial^{2}\eta_{1}}{\partial x\partial
y}+2\dot{x}\frac{\partial^{2}\eta_{1}}{\partial x\partial
t}+\dot{y}^{2}\frac{\partial^{2}\eta_{1}}{\partial
y^{2}}+2\dot{y}\frac{\partial^{2}\eta_{1}}{\partial y\partial
t}+\frac{\partial^{2}\eta_{1}}{\partial t^{2}}\Bigr{]}\\\
-\frac{\lambda}{2}\dot{x}\Bigr{[}\dot{x}^{2}\frac{\partial^{2}\eta_{2}}{\partial
x^{2}}+2\dot{x}\dot{y}\frac{\partial^{2}\eta_{2}}{\partial x\partial
y}+2\dot{x}\frac{\partial^{2}\eta_{2}}{\partial x\partial
t}+\dot{y}^{2}\frac{\partial^{2}\eta_{2}}{\partial
y^{2}}+2\dot{y}\frac{\partial^{2}\eta_{2}}{\partial y\partial
t}+\frac{\partial^{2}\eta_{2}}{\partial
t^{2}}\Bigr{]}+\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2})\frac{\mathrm{d}\xi}{\mathrm{d}t}\\\
=\frac{\partial G}{\partial x}\dot{x}+\frac{\partial G}{\partial
y}\dot{y}+\frac{\partial G}{\partial t}\;.\end{split}$ (89)

By solving these equations, one finds infinitesimal transformations and the
gauge function as

$\xi=c_{1}\;,$ (90)
$\eta_{1}=c_{2}y+c_{3}\sin{(\frac{m}{\lambda}t)}+c_{4}\cos{(\frac{m}{\lambda}t)}+c_{5}t+c_{6}\;,$
(91)
$\eta_{2}=-c_{2}x-c_{3}\cos{(\frac{m}{\lambda}t)}+c_{4}\sin{(\frac{m}{\lambda}t)}+c_{7}t+c_{8}\;,$
(92)
$\begin{split}G=c_{3}\left(\frac{m}{2}\dot{x}\sin{(\frac{m}{\lambda}t)}-\frac{m}{2}\dot{y}\cos{(\frac{m}{\lambda}t)}\right)+&c_{4}\left(\frac{m}{2}\dot{y}\sin{(\frac{m}{\lambda}t)}+\frac{m}{2}\dot{x}\cos{(\frac{m}{\lambda}t)}\right)\\\
+&c_{5}\left(mx-\frac{\lambda}{2}\dot{y}\right)+c_{7}\left(my+\frac{\lambda}{2}\dot{x}\right)\;.\end{split}$
(93)

Then the integrals of motion with the corresponding symmetry generators are

$\displaystyle\Gamma_{1}$ $\displaystyle=\frac{\partial}{\partial t}\hskip
14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{1}$
$\displaystyle=\lambda\left(\dot{x}\ddot{y}-\dot{y}\ddot{x}\right)-\frac{m}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)$
(94) $\displaystyle\Gamma_{2}$ $\displaystyle=y\frac{\partial}{\partial
x}-x\frac{\partial}{\partial y}\hskip 14.22636pt$ $\displaystyle;\hskip
14.22636ptI_{2}$
$\displaystyle=\lambda\left(\frac{\dot{x}^{2}}{2}+\frac{\dot{y}^{2}}{2}-x\ddot{x}-y\ddot{y}\right)+m\left(\dot{x}y-\dot{y}x\right)$
(95) $\displaystyle\Gamma_{3}$
$\displaystyle=\sin{(\frac{m}{\lambda}t)}\frac{\partial}{\partial
x}-\cos{(\frac{m}{\lambda}t)}\frac{\partial}{\partial y}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{3}$
$\displaystyle=\lambda\left(\ddot{y}\sin{(\frac{m}{\lambda}t)}+\ddot{x}\cos{(\frac{m}{\lambda}t)}\right)$
(96) $\displaystyle\Gamma_{4}$
$\displaystyle=\cos{(\frac{m}{\lambda}t)}\frac{\partial}{\partial
x}+\sin{(\frac{m}{\lambda}t)}\frac{\partial}{\partial y}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{4}$
$\displaystyle=\lambda\left(\ddot{y}\cos{(\frac{m}{\lambda}t)}-\ddot{x}\sin{(\frac{m}{\lambda}t)}\right)$
(97) $\displaystyle\Gamma_{5}$ $\displaystyle=t\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{5}$
$\displaystyle=\lambda\left(\dot{y}-\ddot{y}t\right)+m\left(-x+\dot{x}t\right)$
(98) $\displaystyle\Gamma_{6}$ $\displaystyle=\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{6}$
$\displaystyle=-\lambda\ddot{y}+m\dot{x}$ (99) $\displaystyle\Gamma_{7}$
$\displaystyle=t\frac{\partial}{\partial y}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{7}$
$\displaystyle=\lambda\left(-\dot{x}+\ddot{x}t\right)+m\left(-x+\dot{x}t\right)$
(100) $\displaystyle\Gamma_{8}$ $\displaystyle=\frac{\partial}{\partial
y}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{8}$
$\displaystyle=\lambda\ddot{x}+m\dot{y}$ (101)

From the form of the Lagrangian (84), it can be seen that $x$ and $y$ are
cyclic coordinates. We will pursue the same method to find the Lagrangian
$\tilde{L}^{\prime}$, where the momentum conjugate corresponding to
$x^{\prime}$ is given by (11). Consider the case in which $c_{2}=1$ is the
only constant that does not equal to zero. Then

$\frac{\partial t}{\partial x^{\prime}}=0\hskip 14.22636pt;\hskip 14.22636pt
\frac{\partial x}{\partial x^{\prime}}=y\hskip 14.22636pt;\hskip 14.22636pt
\frac{\partial y}{\partial x^{\prime}}=-x\hskip 14.22636pt;\hskip
14.22636ptG=0\;.$ (102)

It is clear that these symmetries correspond to the transformation to polar
coordinates

$t=t^{\prime}\hskip 14.22636pt;\hskip 14.22636pt
x=y^{\prime}\sin{x^{\prime}}\hskip 14.22636pt;\hskip 14.22636pt
y=y^{\prime}\cos{x^{\prime}}\;.$ (103)

After calculating the first and second derivatives of the primed coordinates,
we can rewrite our Lagrangian as

$L\prime=\frac{\lambda}{2}\left({\dot{x}\prime}^{3}{y\prime}^{2}+2\dot{x}\prime{\dot{y}\prime}^{2}+\ddot{x}\prime\dot{y}\prime
y\prime-\dot{x}\prime\ddot{y}\prime
y\prime\right)+\frac{m}{2}\left({\dot{y}\prime}^{2}+{\dot{x}\prime}^{2}{y\prime}^{2}\right)\;.$
(104)

Obviously, $x^{\prime}$ is a cyclic coordinate and the corresponding conserved
quantity to the symmetry is as follows

$I_{2}=\lambda\left(\frac{\dot{x}^{2}}{2}+\frac{\dot{y}^{2}}{2}-x\ddot{x}-y\ddot{y}\right)+m\left(\dot{x}y-\dot{y}x\right)
\;.$ (105)

which is also given by (95).

##  5 Variational Symmetries for High-Order Lagrangians

Finally, let us generalize what we have done in the previous sections for
Lagrangian systems with $N$th order derivatives

$L=L(\mathbf{x},\dot{\mathbf{x}},...,\mathbf{x}^{(N)};t)\;.$ (106)

Using the same procedure one find the generalized version of the equation
$\eqref{eq1}$

$\sum_{i=1}^{n}\sum_{k=0}^{N}\frac{\partial L}{\partial
x_{i}^{(k)}}\Bigr{[}\frac{\mathrm{d}^{k}}{\mathrm{d}t^{k}}\Bigr{(}\eta_{i}-\dot{x}_{i}\xi\Bigr{)}+x_{i}^{(k+1)}\xi\Bigr{]}+\frac{\partial
L}{\partial
t}\xi+L\frac{\mathrm{d}\xi}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}G(\mathbf{x},\dot{\mathbf{x}},...,\mathbf{x}^{(N-1)};t)\;.$
(107)

The generalized form of the equation $\eqref{Q}$ can be written as follows
Woodard:2015zca

$\frac{\mathrm{d}}{\mathrm{d}t}\left(\xi
L-\sum_{i=1}^{n}\sum_{k=1}^{N}\sum_{j=1}^{k}(-1)^{j}\frac{\mathrm{d}^{k-j}(\eta_{i}-\dot{x}_{i}\xi)}{\mathrm{d}t^{k-j}}\frac{\mathrm{d}^{j-1}}{\mathrm{d}t^{j-1}}\left(\frac{\partial
L}{\partial x_{i}^{(k)}}\right)-G\right)=0\;.$ (108)

As an example let us consider the system with the following Lagrangian

$L=\frac{\dddot{x}^{2}}{2}\;.$ (109)

The Euler-Lagrange equation of the system is the sixth order ordinary
differential equation

$x^{(6)}=0\;.$ (110)

Substituting the Lagrangian into the expression (107) we obtain the following
result

$\xi(t)=c_{1}t^{2}+c_{2}t+c_{3}$ (111)
$\eta(x,t)=c_{1}5xt+c_{2}\frac{5}{2}x+c_{4}t^{5}+c_{5}t^{4}+c_{6}t^{3}+c_{7}t^{2}+c_{8}t+c_{9}$
(112)
$G(x,\dot{x},\ddot{x},t)=c_{1}\frac{9}{2}\ddot{x}^{2}+c_{4}\left(60\ddot{x}t^{2}-120\dot{x}t+120x\right)+c_{5}\left(24\ddot{x}t-24\dot{x}\right)+c_{6}6\ddot{x}\;.$
(113)

Using these expressions we find symmetry generators and integrals of motion

$\displaystyle\Gamma_{1}$ $\displaystyle=t^{2}\frac{\partial}{\partial
t}+5xt\frac{\partial}{\partial x}\hskip 14.22636pt$ $\displaystyle;\hskip
14.22636ptI_{1}$
$\displaystyle=-\frac{9}{2}\ddot{x}^{2}+8\dot{x}\dddot{x}-5xx^{(4)}$ (114)
$\displaystyle+(\ddot{x}\dddot{x}-3\dot{x}x^{(4)}+5xx^{(5)})t+(-\frac{\dddot{x}^{2}}{2}+\ddot{x}x^{(4)}-\dot{x}x^{(5)})t^{2}$
(115) $\displaystyle\Gamma_{2}$ $\displaystyle=t\frac{\partial}{\partial
t}+\frac{5}{2}x\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{2}$
$\displaystyle=\ddot{x}\dddot{x}-3\dot{x}x^{(4)}+5xx^{(5)}+(2\ddot{x}x^{(4)}-2\dot{x}x^{(5)}-\dddot{x}^{2})t$
(116) $\displaystyle\Gamma_{3}$ $\displaystyle=\frac{\partial}{\partial
t}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{3}$
$\displaystyle=-\dddot{x}^{2}+2\ddot{x}x^{(4)}-2\dot{x}x^{(5)}$ (117)
$\displaystyle\Gamma_{4}$ $\displaystyle=t^{5}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{4}$
$\displaystyle=-120x+120\dot{x}t-60\ddot{x}t^{2}+20\dddot{x}t^{2}-5x^{(4)}t^{4}+x^{(5)}t^{5}$
(118) $\displaystyle\Gamma_{5}$ $\displaystyle=t^{4}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{5}$
$\displaystyle=24\dot{x}-24\ddot{x}t+12\dddot{x}t^{2}-4x^{(4)}t^{3}+x^{(5)}t^{4}$
(119) $\displaystyle\Gamma_{6}$ $\displaystyle=t^{3}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{6}$
$\displaystyle=-6\ddot{x}+6\dddot{x}t-3x^{(4)}t^{2}+x^{(5)}t^{3}$ (120)
$\displaystyle\Gamma_{7}$ $\displaystyle=t^{2}\frac{\partial}{\partial
x}\hskip 14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{7}$
$\displaystyle=2\dddot{x}-2x^{(4)}t+x^{(5)}t^{2}$ (121)
$\displaystyle\Gamma_{8}$ $\displaystyle=t\frac{\partial}{\partial x}\hskip
14.22636pt$ $\displaystyle;\hskip 14.22636ptI_{8}$
$\displaystyle=-x^{(4)}+x^{(5)}t$ (122) $\displaystyle\Gamma_{9}$
$\displaystyle=\frac{\partial}{\partial x}\hskip 14.22636pt$
$\displaystyle;\hskip 14.22636ptI_{9}$ $\displaystyle=x^{(5)} \;.$ (123)

##  6 Conclusions

This paper dealt with an elementary derivation of integrals of motion for
higher-order Lagrangian systems based on variational symmetries technique. The
methods used here to compute variational symmetries and related integrals of
motion are an extension of those used in torres2013variational  and
del2017variational . We provided variational symmetries of higher-order
Lagrangian systems which are well-known in the literature and studied from
different viewpoints.

The symmetries discussed here can be useful also in the context of the
quantization of the higher derivative Lagrangian systems. Another possible
direction is the derivation of symmetries for field theories with higher-
derivative terms.

## Acknowledgements

We would like to thank all the participants of the seminar series on “Higher-
derivative systems” held at Bogazici University in summer 2020. The work of
Ilmar Gahramanov is partially supported by the Bogazici University Research
Fund under grant number 20B03SUP3. Ege Çoban and Dilara Kosva are supported by
the 2209-A TUBITAK National/International Research Projects Fellowship
Programme for Undergraduate Students under grant number 1919B012000987.

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