Herglotz's variational principle

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 ${\displaystyle S}$ as an independent variable, and ${\displaystyle S}$ itself is represented as the solution of an ordinary differential equation (ODE) whose right hand side is the Lagrangian ${\displaystyle L}$, instead of an integration of ${\displaystyle L}$.^[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}

As in Lagrangian mechanics, we consider a system with ${\displaystyle n}$ degrees of freedom. Let ${\displaystyle {\boldsymbol {q}}=(q_{1},q_{2},\dots ,q_{n})}$ be its generalized coordinates, and let ${\displaystyle {\boldsymbol {u}}=(u_{1},u_{2},\dots ,u_{n})}$ be its generalized velocity. Let ${\displaystyle L=L(t,{\boldsymbol {q}},{\boldsymbol {u}})}$ be the Lagrangian function of the physical system. Let ${\displaystyle S}$ be the action.

Lagrangian mechanics is derived using Hamilton's principle. Fix a starting time and configuration ${\displaystyle t_{0},{\boldsymbol {q}}_{0}}$ and an ending time and configuration ${\displaystyle t_{1},{\boldsymbol {q}}_{1}}$. Hamilton's principle states that physically real trajectories from are the solutions to the problem of variational calculus:$${\displaystyle {\begin{cases}\delta \int _{\gamma }L(t,\gamma (t),{\dot {\gamma }}(t))dt=0\\\gamma {\text{ has end points }}t_{0},{\boldsymbol {q}}_{0},t_{1},{\boldsymbol {q}}_{1}\end{cases}}}$$Equivalently, it can be formulated as$${\displaystyle {\begin{cases}\delta S(t_{1})=0\\{\dot {S}}(t)=L(t,\gamma (t),{\dot {\gamma }}(t)),&t\in [t_{0},t_{1}]\\S(t_{0})=S_{0}\\\gamma {\text{ has end points }}t_{0},{\boldsymbol {q}}_{0},t_{1},{\boldsymbol {q}}_{1}\end{cases}}}$$

Herglotz's variational principle simply generalizes by allowing the Lagrangian to depend on the action as well. It is of form ${\displaystyle L=L(t,{\boldsymbol {q}},{\boldsymbol {u}},S)}$, depending on ${\displaystyle 2n+2}$ variables. $${\displaystyle {\begin{cases}\delta S(t_{1})=0\\{\dot {S}}(t)=L(t,\gamma (t),{\dot {\gamma }}(t),S(t)),&t\in [t_{0},t_{1}]\\S(t_{0})=S_{0}\\\gamma {\text{ has end points }}t_{0},{\boldsymbol {q}}_{0},t_{1},{\boldsymbol {q}}_{1}\end{cases}}}$$

Hamilton's variational principle gives the Euler–Langrange equations.$${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)-{\frac {\partial L}{\partial {\boldsymbol {q}}}}=0}$$ Similarly, Herglotz's variational principle gives the Euler–Lagrange–Herglotz equations $${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)-{\frac {\partial L}{\partial {\boldsymbol {q}}}}={\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}}$$ which involves an extra term ${\textstyle {\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}}$ that can describe the dissipation of the system. The original Euler–Langrange equations are recovered as a special case when ${\displaystyle \partial _{S}L=0}$.

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$${\displaystyle p_{i}:={\frac {\partial L}{\partial u_{i}}},\quad H:=\sum _{i}p_{i}u_{i}-L}$$ Then the equations of motion are$${\displaystyle {\begin{cases}\displaystyle {\dot {q}}_{i}={\frac {\partial H}{\partial p_{i}}}\\[1ex]\displaystyle {\dot {p}}_{i}=-\left({\frac {\partial H}{\partial q_{i}}}+p_{i}{\frac {\partial H}{\partial S}}\right)\\[1ex]\displaystyle {\dot {S}}=\sum _{i=1}^{n}p_{i}{\frac {\partial H}{\partial p_{i}}}-H\end{cases}}}$$

If ${\displaystyle S(t,q)}$ is written as a function of time and configuration, then it satisfies a Hamilton–Jacobi equation$${\displaystyle dS=-H\,dt+\sum _{i}p_{i}\,dq^{i}}$$

In order to solve this minimization problem, we impose a variation ${\displaystyle \delta {\boldsymbol {q}}}$ on ${\displaystyle {\boldsymbol {q}}}$, and suppose ${\displaystyle S(t)}$ undergoes a variation ${\displaystyle \delta S(t)}$ correspondingly, then$${\displaystyle {\begin{aligned}\delta {\dot {S}}(t)&=L(t,{\boldsymbol {q}}(t)+\delta {\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t)+\delta {\dot {\boldsymbol {q}}}(t),S(t)+\delta S(t))-L(t,{\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t),S(t))\\&={\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)+{\frac {\partial L}{\partial S}}\delta S(t)\end{aligned}}}$$ and since the initial condition is not changed, ${\displaystyle \delta S_{0}=0}$. The above equation a linear ODE for the function ${\displaystyle \delta S(t)}$, and it can be solved by introducing an integrating factor ${\textstyle \mu (t)=\exp \left(\int _{t_{0}}^{t}{\frac {\partial L}{\partial S}}dt\right)}$, which is uniquely determined by the ODE $${\displaystyle {\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}},\quad u(t_{0})=1.}$$By multiplying ${\displaystyle \mu (t)}$ on both sides of the equation of ${\displaystyle \delta {\dot {S}}}$ and moving the term ${\textstyle \mu (t){\frac {\partial L}{\partial S}}\delta S(t)}$ to the left hand side, we get $${\displaystyle \mu (t)\delta {\dot {S}}(t)-\mu (t){\frac {\partial L}{\partial S}}\delta S(t)=\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right).}$$ Note that, since ${\textstyle {\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}}}$, the left hand side equals to $${\displaystyle \mu (t)\delta {\dot {S}}(t)+{\dot {\mu }}(t)\delta S(t)={\frac {d(\mu (t)\delta S(t))}{dt}}}$$and therefore we can do an integration of the equation above from ${\displaystyle t=t_{0}}$ to ${\displaystyle t=t_{1}}$, yielding $${\displaystyle \mu (t_{1})\delta S_{1}-\mu (t_{0})\delta S_{0}=\int _{t_{0}}^{t_{1}}\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right)dt}$$ where the ${\displaystyle \delta S_{0}=0}$ so the left hand side actually only contains one term ${\displaystyle \mu (t_{1})\delta S_{1}}$, and for the right hand side, we can perform the integration-by-part on the ${\textstyle {\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)}$ term to remove the time derivative on ${\textstyle \delta {\boldsymbol {q}}}$: $${\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)dt\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)dt+\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)dt\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)dt+\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 {d}{dt}}\left(\mu (t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)dt\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)dt-\int _{t_{0}}^{t_{1}}{\frac {d}{dt}}\left(\mu (t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)dt\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)dt-\int _{t_{0}}^{t_{1}}\left({\dot {\mu }}(t){\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}+\mu (t){\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)dt\\=&\int _{t_{0}}^{t_{1}}\mu (t){\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)dt-\int _{t_{0}}^{t_{1}}\left(-\mu (t){\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}+\mu (t){\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)\delta {\boldsymbol {q}}(t)dt\\=&\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 {d}{dt}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)}}\delta {\boldsymbol {q}}(t)dt,\end{aligned}}}$$ and when ${\displaystyle S_{1}}$ is minimized, ${\displaystyle \delta S_{1}=0}$ for all ${\displaystyle \delta {\boldsymbol {q}}}$, which indicates that the underlined term in the last line of the equation above has to be zero on the entire interval ${\displaystyle [t_{0},t_{1}]}$, 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$${\displaystyle {\bar {t}}=t+\varepsilon T(t,q),\quad {\bar {q}}=q+\varepsilon Q(t,q)}$$ where ${\displaystyle T,Q}$ are smooth functions of time and configuration, and ${\displaystyle \varepsilon }$ is an infinitesimal. The transformation deforms a trajectory ${\displaystyle (t,q(t))}$ to ${\displaystyle (t+\varepsilon T(t,q),q(t)+\varepsilon Q(t,q))}$, and accordingly deforms the action integral as well.

We say that the infinitesimal transformation is a symmetry of the action iff the change in ${\displaystyle S(t_{1})-S(t_{0})}$ under the infinitesimal transformation is order ${\displaystyle O(\varepsilon ^{2})}$. Given such an infinitesimal symmetry, the quantity is a constant of motion$${\displaystyle \exp \left(-\int _{t_{0}}^{t}{\frac {\partial L(t')}{\partial S}}\,dt'\right)\left[T{\left(L-{\dot {q}}{\frac {\partial L}{\partial {\dot {q}}}}\right)}+Q{\frac {\partial L}{\partial {\dot {q}}}}\right]}$$ where ${\displaystyle \partial _{S}L(t')}$ is more explicitly written as$${\displaystyle \partial _{S}L(t',q(t'),{\dot {q}}(t'),S(t'))}$$ There is also a version for multiple time dimensions.^[7]

The motion of a particle of mass ${\displaystyle m}$ in a potential field ${\displaystyle V}$ with damping coefficient ${\displaystyle \gamma }$ is$${\displaystyle m{\ddot {x}}=-V'(x)-\gamma {\dot {x}}.}$$It can be produced as the Euler–Lagrange–Herglotz for^[3]: 114$${\displaystyle L(t,x,{\dot {x}},S)={\tfrac {1}{2}}m{\dot {x}}^{2}-V(x)-\gamma S}$$

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 ${\displaystyle {\dot {x}}}$, and another force proportional to ${\displaystyle {\dot {x}}^{2}}$. Write it as^[2] $${\displaystyle {\ddot {x}}+f(x){\dot {x}}^{2}+g(t){\dot {x}}+h(x)=0}$$ This equation is the Euler–Lagrange–Herglotz equation for the Lagrangian $${\displaystyle L={\tfrac {1}{2}}{\dot {x}}^{2}-\left[2f(x){\dot {x}}+g(t)\right]S-U(x)}$$ where ${\textstyle U(x)}$ is any solution of the ODE $${\displaystyle {\frac {dU(x)}{dx}}+2f(x)U(x)=h(x).}$$Some important special cases:

• When ${\displaystyle f(x)=0}$ and ${\displaystyle g(t)}$ is constant, it is the damped harmonic oscillator given above.
• When ${\textstyle h(x)=x^{n},f(x)=0}$ and ${\textstyle g(t)=2/t}$, it is the Lane–Emden equation$${\displaystyle {\ddot {x}}+{\frac {2}{t}}{\dot {x}}+x^{n}=0,\quad n\neq -1.}$$ with Lagrangian $${\displaystyle L={\frac {1}{2}}{\dot {x}}^{2}-{\frac {x^{n+1}}{n+1}}-{\frac {2}{t}}z.}$$
• When ${\textstyle h(x)=0}$, it is a Rayleigh-type system$${\displaystyle {\ddot {x}}+f(x){\dot {x}}^{2}+g(t){\dot {x}}=0.}$$ with Lagrangian$${\displaystyle L={\tfrac {1}{2}}{\dot {x}}^{2}-\left[2f(x){\dot {x}}+g(t)\right]z.}$$