Principle in mathematical physics
In mathematics and physics, Herglotz's variational principle , named after German mathematician and physicist Gustav Herglotz , is an extension of the Hamilton's principle , where the Lagrangian L explicitly involves the action
S
{\displaystyle S}
as an independent variable, and
S
{\displaystyle S}
itself is represented as the solution of an ordinary differential equation (ODE) whose right hand side is the Lagrangian
L
{\displaystyle L}
, instead of an integration of
L
{\displaystyle L}
.[ 1] [ 2] Herglotz's variational principle is known as the variational principle for nonconservative Lagrange equations and Hamilton equations . It was first proposed in the context of contact geometry .
This presentation is from [ 3] : 108–114
Hamilton's principle[ edit ]
As in Lagrangian mechanics, we consider a system with
n
{\displaystyle n}
degrees of freedom. Let
q
=
(
q
1
,
q
2
,
…
,
q
n
)
{\displaystyle {\boldsymbol {q}}=(q_{1},q_{2},\dots ,q_{n})}
be its generalized coordinates , and let
u
=
(
u
1
,
u
2
,
…
,
u
n
)
{\displaystyle {\boldsymbol {u}}=(u_{1},u_{2},\dots ,u_{n})}
be its generalized velocity. Let
L
=
L
(
t
,
q
,
u
)
{\displaystyle L=L(t,{\boldsymbol {q}},{\boldsymbol {u}})}
be the Lagrangian function of the physical system. Let
S
{\displaystyle S}
be the action .
Lagrangian mechanics is derived using Hamilton's principle. Fix a starting time and configuration
t
0
,
q
0
{\displaystyle t_{0},{\boldsymbol {q}}_{0}}
and an ending time and configuration
t
1
,
q
1
{\displaystyle t_{1},{\boldsymbol {q}}_{1}}
. Hamilton's principle states that physically real trajectories from are the solutions to the problem of variational calculus:
{
δ
∫
γ
L
(
t
,
γ
(
t
)
,
γ
˙
(
t
)
)
d
t
=
0
γ
has end points
t
0
,
q
0
,
t
1
,
q
1
{\displaystyle {\begin{cases}\delta \int _{\gamma }L(t,\gamma (t),{\dot {\gamma }}(t))dt=0\\\gamma {\text{ has end points }}t_{0},{\boldsymbol {q}}_{0},t_{1},{\boldsymbol {q}}_{1}\end{cases}}}
Equivalently, it can be formulated as
{
δ
S
(
t
1
)
=
0
S
˙
(
t
)
=
L
(
t
,
γ
(
t
)
,
γ
˙
(
t
)
)
,
t
∈
[
t
0
,
t
1
]
S
(
t
0
)
=
S
0
γ
has end points
t
0
,
q
0
,
t
1
,
q
1
{\displaystyle {\begin{cases}\delta S(t_{1})=0\\{\dot {S}}(t)=L(t,\gamma (t),{\dot {\gamma }}(t)),&t\in [t_{0},t_{1}]\\S(t_{0})=S_{0}\\\gamma {\text{ has end points }}t_{0},{\boldsymbol {q}}_{0},t_{1},{\boldsymbol {q}}_{1}\end{cases}}}
Herglotz's variational principle[ edit ]
Herglotz's variational principle simply generalizes by allowing the Lagrangian to depend on the action as well. It is of form
L
=
L
(
t
,
q
,
u
,
S
)
{\displaystyle L=L(t,{\boldsymbol {q}},{\boldsymbol {u}},S)}
, depending on
2
n
+
2
{\displaystyle 2n+2}
variables.
{
δ
S
(
t
1
)
=
0
S
˙
(
t
)
=
L
(
t
,
γ
(
t
)
,
γ
˙
(
t
)
,
S
(
t
)
)
,
t
∈
[
t
0
,
t
1
]
S
(
t
0
)
=
S
0
γ
has end points
t
0
,
q
0
,
t
1
,
q
1
{\displaystyle {\begin{cases}\delta S(t_{1})=0\\{\dot {S}}(t)=L(t,\gamma (t),{\dot {\gamma }}(t),S(t)),&t\in [t_{0},t_{1}]\\S(t_{0})=S_{0}\\\gamma {\text{ has end points }}t_{0},{\boldsymbol {q}}_{0},t_{1},{\boldsymbol {q}}_{1}\end{cases}}}
Euler–Lagrange–Herglotz equation[ edit ]
Hamilton's variational principle gives the Euler–Langrange equations .
d
d
t
(
∂
L
∂
q
˙
)
−
∂
L
∂
q
=
0
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)-{\frac {\partial L}{\partial {\boldsymbol {q}}}}=0}
Similarly, Herglotz's variational principle gives the Euler–Lagrange–Herglotz equations
d
d
t
(
∂
L
∂
q
˙
)
−
∂
L
∂
q
=
∂
L
∂
S
∂
L
∂
q
˙
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)-{\frac {\partial L}{\partial {\boldsymbol {q}}}}={\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}}
which involves an extra term
∂
L
∂
S
∂
L
∂
q
˙
{\textstyle {\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}}
that can describe the dissipation of the system. The original Euler–Langrange equations are recovered as a special case when
∂
S
L
=
0
{\displaystyle \partial _{S}L=0}
.
Similar to how Lagrangian mechanics is equivalent to Hamiltonian mechanics, the Lagrangian form of Herglotz principle is equivalent to a Hamiltonian form.
Define the momentum and Hamiltonian by taking a Legendre transformation
p
i
:=
∂
u
i
L
,
H
:=
∑
i
p
i
u
i
−
L
{\displaystyle p_{i}:=\partial _{u_{i}}L,\quad H:=\sum _{i}p_{i}u_{i}-L}
Then the equations of motion are
{
q
˙
i
=
∂
p
i
H
p
˙
i
=
−
(
∂
q
i
H
+
p
i
∂
S
H
)
S
˙
=
∑
i
=
1
n
p
i
∂
p
i
H
−
H
{\displaystyle {\begin{cases}{\dot {q}}_{i}=\partial _{p_{i}}H\\{\dot {p}}_{i}=-(\partial _{q_{i}}H+p_{i}\partial _{S}H)\\{\dot {S}}=\sum _{i=1}^{n}p_{i}\partial _{p_{i}}H-H\end{cases}}}
Hamilton–Jacobi equation[ edit ]
If
S
(
t
,
q
)
{\displaystyle S(t,q)}
is written as a function of time and configuration, then it satisfies a Hamilton–Jacobi equation
d
S
=
−
H
d
t
+
∑
i
p
i
d
q
i
{\displaystyle dS=-Hdt+\sum _{i}p_{i}dq^{i}}
In order to solve this minimization problem, we impose a variation
δ
q
{\displaystyle \delta {\boldsymbol {q}}}
on
q
{\displaystyle {\boldsymbol {q}}}
, and suppose
S
(
t
)
{\displaystyle S(t)}
undergoes a variation
δ
S
(
t
)
{\displaystyle \delta S(t)}
correspondingly, then
δ
S
˙
(
t
)
=
L
(
t
,
q
(
t
)
+
δ
q
(
t
)
,
q
˙
(
t
)
+
δ
q
˙
(
t
)
,
S
(
t
)
+
δ
S
(
t
)
)
−
L
(
t
,
q
(
t
)
,
q
˙
(
t
)
,
S
(
t
)
)
=
∂
L
∂
q
δ
q
(
t
)
+
∂
L
∂
q
˙
δ
q
˙
(
t
)
+
∂
L
∂
S
δ
S
(
t
)
{\displaystyle {\begin{aligned}\delta {\dot {S}}(t)&=L(t,{\boldsymbol {q}}(t)+\delta {\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t)+\delta {\dot {\boldsymbol {q}}}(t),S(t)+\delta S(t))-L(t,{\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t),S(t))\\&={\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)+{\frac {\partial L}{\partial S}}\delta S(t)\end{aligned}}}
and since the initial condition is not changed,
δ
S
0
=
0
{\displaystyle \delta S_{0}=0}
. The above equation a linear ODE for the function
δ
S
(
t
)
{\displaystyle \delta S(t)}
, and it can be solved by introducing an integrating factor
μ
(
t
)
=
e
∫
t
0
t
∂
L
∂
S
d
t
{\displaystyle \mu (t)=\mathrm {e} ^{\int _{t_{0}}^{t}{\frac {\partial L}{\partial S}}\mathrm {d} t}}
, which is uniquely determined by the ODE
μ
˙
(
t
)
=
−
μ
(
t
)
∂
L
∂
S
,
u
(
t
0
)
=
1.
{\displaystyle {\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}},\quad u(t_{0})=1.}
By multiplying
μ
(
t
)
{\displaystyle \mu (t)}
on both sides of the equation of
δ
S
˙
{\displaystyle \delta {\dot {S}}}
and moving the term
μ
(
t
)
∂
L
∂
S
δ
S
(
t
)
{\textstyle \mu (t){\frac {\partial L}{\partial S}}\delta S(t)}
to the left hand side, we get
μ
(
t
)
δ
S
˙
(
t
)
−
μ
(
t
)
∂
L
∂
S
δ
S
(
t
)
=
μ
(
t
)
(
∂
L
∂
q
δ
q
(
t
)
+
∂
L
∂
q
˙
δ
q
˙
(
t
)
)
.
{\displaystyle \mu (t)\delta {\dot {S}}(t)-\mu (t){\frac {\partial L}{\partial S}}\delta S(t)=\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right).}
Note that, since
μ
˙
(
t
)
=
−
μ
(
t
)
∂
L
∂
S
{\textstyle {\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}}}
, the left hand side equals to
μ
(
t
)
δ
S
˙
(
t
)
+
μ
˙
(
t
)
δ
S
(
t
)
=
d
(
μ
(
t
)
δ
S
(
t
)
)
d
t
{\displaystyle \mu (t)\delta {\dot {S}}(t)+{\dot {\mu }}(t)\delta S(t)={\frac {\mathrm {d} (\mu (t)\delta S(t))}{\mathrm {d} t}}}
and therefore we can do an integration of the equation above from
t
=
t
0
{\displaystyle t=t_{0}}
to
t
=
t
1
{\displaystyle t=t_{1}}
, yielding
μ
(
t
1
)
δ
S
1
−
μ
(
t
0
)
δ
S
0
=
∫
t
0
t
1
μ
(
t
)
(
∂
L
∂
q
δ
q
(
t
)
+
∂
L
∂
q
˙
δ
q
˙
(
t
)
)
d
t
{\displaystyle \mu (t_{1})\delta S_{1}-\mu (t_{0})\delta S_{0}=\int _{t_{0}}^{t_{1}}\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right)\mathrm {d} t}
where the
δ
S
0
=
0
{\displaystyle \delta S_{0}=0}
so the left hand side actually only contains one term
μ
(
t
1
)
δ
S
1
{\displaystyle \mu (t_{1})\delta S_{1}}
, and for the right hand side, we can perform the integration-by-part on the
∂
L
∂
q
˙
δ
q
˙
(
t
)
{\textstyle {\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)}
term to remove the time derivative on
δ
q
{\textstyle \delta {\boldsymbol {q}}}
:
∫
t
0
t
1
μ
(
t
)
(
∂
L
∂
q
δ
q
(
t
)
+
∂
L
∂
q
˙
δ
q
˙
(
t
)
)
d
t
=
∫
t
0
t
1
μ
(
t
)
∂
L
∂
q
δ
q
(
t
)
d
t
+
∫
t
0
t
1
μ
(
t
)
∂
L
∂
q
˙
δ
q
˙
(
t
)
d
t
=
∫
t
0
t
1
μ
(
t
)
∂
L
∂
q
δ
q
(
t
)
d
t
+
μ
(
t
1
)
∂
L
∂
q
˙
δ
q
(
t
1
)
⏟
=
0
−
μ
(
t
0
)
∂
L
∂
q
˙
δ
q
(
t
0
)
⏟
=
0
−
∫
t
0
t
1
d
d
t
(
μ
(
t
)
∂
L
∂
q
˙
)
δ
q
(
t
)
d
t
=
∫
t
0
t
1
μ
(
t
)
∂
L
∂
q
δ
q
(
t
)
d
t
−
∫
t
0
t
1
d
d
t
(
μ
(
t
)
∂
L
∂
q
˙
)
δ
q
(
t
)
d
t
=
∫
t
0
t
1
μ
(
t
)
∂
L
∂
q
δ
q
(
t
)
d
t
−
∫
t
0
t
1
(
μ
˙
(
t
)
∂
L
∂
q
˙
+
μ
(
t
)
d
d
t
∂
L
∂
q
˙
)
δ
q
(
t
)
d
t
=
∫
t
0
t
1
μ
(
t
)
∂
L
∂
q
δ
q
(
t
)
d
t
−
∫
t
0
t
1
(
−
μ
(
t
)
∂
L
∂
S
∂
L
∂
q
˙
+
μ
(
t
)
d
d
t
∂
L
∂
q
˙
)
δ
q
(
t
)
d
t
=
∫
t
0
t
1
μ
(
t
)
(
∂
L
∂
q
+
∂
L
∂
S
∂
L
∂
q
˙
−
d
d
t
∂
L
∂
q
˙
)
_
δ
q
(
t
)
d
t
,
{\displaystyle {\begin{aligned}&\int _{t_{0}}^{t_{1}}\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right)\mathrm {d} t\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)\mathrm {d} t+\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\mathrm {d} t\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)\mathrm {d} t+\mu (t_{1}){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\underbrace {\delta {\boldsymbol {q}}(t_{1})} _{=0}-\mu (t_{0}){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\underbrace {\delta {\boldsymbol {q}}(t_{0})} _{=0}-\int _{t_{0}}^{t_{1}}{\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mu (t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)\mathrm {d} t\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)\mathrm {d} t-\int _{t_{0}}^{t_{1}}{\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mu (t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)\mathrm {d} t\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)\mathrm {d} t-\int _{t_{0}}^{t_{1}}\left({\dot {\mu }}(t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}+\mu (t){\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)\mathrm {d} t\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)\mathrm {d} t-\int _{t_{0}}^{t_{1}}\left(-\mu (t){\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}+\mu (t){\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)\mathrm {d} t\\=&\int _{t_{0}}^{t_{1}}\mu (t){\underline {\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}+{\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)}}\delta {\boldsymbol {q}}(t)\mathrm {d} t,\end{aligned}}}
and when
S
1
{\displaystyle S_{1}}
is minimized,
δ
S
1
=
0
{\displaystyle \delta S_{1}=0}
for all
δ
q
{\displaystyle \delta {\boldsymbol {q}}}
, which indicates that the underlined term in the last line of the equation above has to be zero on the entire interval
[
t
0
,
t
1
]
{\displaystyle [t_{0},t_{1}]}
, this gives rise to the Euler–Lagrange–Herglotz equation.
Generalizations of Noether's theorem and Noether's second theorem apply to Herglotz's variational principle.[ 4] [ 5] [ 6]
An infinitesimal transformation is
t
¯
=
t
+
ϵ
T
(
t
,
q
)
,
q
¯
=
q
+
ϵ
Q
(
t
,
q
)
{\displaystyle {\bar {t}}=t+\epsilon T(t,q),\quad {\bar {q}}=q+\epsilon Q(t,q)}
where
T
,
Q
{\displaystyle T,Q}
are smooth functions of time and configuration, and
ϵ
{\displaystyle \epsilon }
is an infinitesimal. The transformation deforms a trajectory
(
t
,
q
(
t
)
)
{\displaystyle (t,q(t))}
to
(
t
+
ϵ
T
(
t
,
q
)
,
q
(
t
)
+
ϵ
Q
(
t
,
q
)
)
{\displaystyle (t+\epsilon T(t,q),q(t)+\epsilon Q(t,q))}
, and accordingly deforms the action integral as well.
We say that the infinitesimal transformation is a symmetry of the action iff the change in
S
(
t
1
)
−
S
(
t
0
)
{\displaystyle S(t_{1})-S(t_{0})}
under the infinitesimal transformation is order
O
(
ϵ
2
)
{\displaystyle O(\epsilon ^{2})}
. Given such an infinitesimal symmetry, the quantity is a constant of motion
e
−
∫
t
0
t
∂
S
L
(
t
′
)
d
t
′
(
T
(
L
−
q
˙
∂
q
˙
L
)
+
Q
∂
q
˙
L
)
{\displaystyle e^{-\int _{t_{0}}^{t}\partial _{S}L(t')dt'}(T(L-{\dot {q}}\partial _{\dot {q}}L)+Q\partial _{\dot {q}}L)}
where
∂
S
L
(
t
′
)
{\displaystyle \partial _{S}L(t')}
is more explicitly written as
∂
S
L
(
t
′
,
q
(
t
′
)
,
q
˙
(
t
′
)
,
S
(
t
′
)
)
{\displaystyle \partial _{S}L(t',q(t'),{\dot {q}}(t'),S(t'))}
There is also a version for multiple time dimensions .[ 7]
Damped particle on a line [ edit ]
The motion of a particle of mass
m
{\displaystyle m}
in a potential field
V
{\displaystyle V}
with damping coefficient
γ
{\displaystyle \gamma }
is
m
x
¨
=
−
V
′
(
x
)
−
γ
x
˙
.
{\displaystyle m{\ddot {x}}=-V'(x)-\gamma {\dot {x}}.}
It can be produced as the Euler–Lagrange–Herglotz for[ 3] : 114
L
(
t
,
x
,
x
˙
,
S
)
=
1
2
m
x
˙
2
−
V
(
x
)
−
γ
S
{\displaystyle L(t,x,{\dot {x}},S)={\frac {1}{2}}m{\dot {x}}^{2}-V(x)-\gamma S}
A more general particle on a line [ edit ]
More generally, consider a particle on a line under the influence of 3 forces: a conservative force due to a potential field, a dissipative force proportional to
x
˙
{\displaystyle {\dot {x}}}
, and another force proportional to
x
˙
2
{\displaystyle {\dot {x}}^{2}}
. Write it as[ 2]
x
¨
+
f
(
x
)
x
˙
2
+
g
(
t
)
x
˙
+
h
(
x
)
=
0
{\displaystyle {\ddot {x}}+f(x){\dot {x}}^{2}+g(t){\dot {x}}+h(x)=0}
This equation is the Euler–Lagrange–Herglotz equation for the Lagrangian
L
=
1
2
x
˙
2
−
(
2
f
(
x
)
x
˙
+
g
(
t
)
)
S
−
U
(
x
)
{\displaystyle L={\frac {1}{2}}{\dot {x}}^{2}-(2f(x){\dot {x}}+g(t))S-U(x)}
where
U
(
x
)
{\textstyle U(x)}
is any solution of the ODE
d
U
(
x
)
d
x
+
2
f
(
x
)
U
(
x
)
=
h
(
x
)
.
{\displaystyle {\frac {\mathrm {d} U(x)}{\mathrm {d} x}}+2f(x)U(x)=h(x).}
Some important special cases:
When
f
(
x
)
=
0
{\displaystyle f(x)=0}
and
g
(
t
)
{\displaystyle g(t)}
is constant, it is the damped harmonic oscillator given above.
When
h
(
x
)
=
x
n
,
f
(
x
)
=
0
{\textstyle h(x)=x^{n},f(x)=0}
and
g
(
t
)
=
2
/
t
{\textstyle g(t)=2/t}
, it is the Lane–Emden equation
x
¨
+
2
t
x
˙
+
x
n
=
0
,
n
≠
−
1.
{\displaystyle {\ddot {x}}+{\frac {2}{t}}{\dot {x}}+x^{n}=0,\quad n\neq -1.}
with Lagrangian
L
=
1
2
x
˙
2
−
x
n
+
1
n
+
1
−
2
t
z
.
{\displaystyle L={\frac {1}{2}}{\dot {x}}^{2}-{\frac {x^{n+1}}{n+1}}-{\frac {2}{t}}z.}
When
h
(
x
)
=
0
{\textstyle h(x)=0}
, it is a Rayleigh-type system
x
¨
+
f
(
x
)
x
˙
2
+
g
(
t
)
x
˙
=
0.
{\displaystyle {\ddot {x}}+f(x){\dot {x}}^{2}+g(t){\dot {x}}=0.}
with Lagrangian
L
=
1
2
x
˙
2
−
(
2
f
(
x
)
x
˙
+
g
(
t
)
)
z
.
{\displaystyle L={\frac {1}{2}}{\dot {x}}^{2}-(2f(x){\dot {x}}+g(t))z.}
^ Gaset, Jordi; Lainz, Manuel; Mas, Arnau; Rivas, Xavier (2022-11-30), "The Herglotz variational principle for dissipative field theories" , Geometric Mechanics , 01 (2): 153– 178, arXiv :2211.17058 , doi :10.1142/S2972458924500060 , retrieved 2025-05-06
^ a b Georgieva, Bogdana (2012). The Variational Principle of Hergloz and Related Results (Report). GIQ. doi :10.7546/giq-12-2011-214-225 .
^ a b Guenther, R. B.; Gottsch, J. A.; Guenther, C. M. (1996). The Herglotz Lectures on Contact Transformations and Hamiltonian Systems (PDF) . Toruń, Poland: Juliusz Center for Nonlinear Studies. Archived (PDF) from the original on 2 Aug 2023.
^ Georgieva, Bogdana; Guenther, Ronald (2002). "First Noether-type theorem for the generalized variational principle of Herglotz" . Topological Methods in Nonlinear Analysis . 20 (2): 261– 273. MR 1962221 . Zbl 1032.58007 .
^ Georgieva, Bogdana A. (2001). Noether-type theorems for the generalized variational principle of Herglotz (PhD thesis). Oregon State University.
^ Georgieva, Bogdana; Guenther, Ronald B. (2005). "Second Noether-type theorem for the generalized variational principle of Herglotz" . Topological Methods in Nonlinear Analysis . 26 (2). Juliusz Schauder Center for Nonlinear Studies: 307– 314. MR 2197762 . Zbl 1191.70007 .
^ Georgieva, Bogdana; Guenther, Ronald; Bodurov, Theodore (2003-09-01). "Generalized variational principle of Herglotz for several independent variables. First Noether-type theorem" . Journal of Mathematical Physics . 44 (9): 3911– 3927. doi :10.1063/1.1597419 . ISSN 0022-2488 .