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.

Mathematical formulation

This presentation is from ^{[3]: 108–114}

Hamilton's principle

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

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.$${\displaystyle {\begin{cases}\delta S(t_{1})=0\\{\dot {S}}(t)=L(t,\gamma (t),{\dot {\gamma }}(t),S(t)),&t\in [t_{0},t_{1}]\\S(t_{0})=S_{0}\\\gamma {\text{ has end points }}t_{0},{\boldsymbol {q}}_{0},t_{1},{\boldsymbol {q}}_{1}\end{cases}}}$$

Euler–Lagrange–Herglotz equation

Hamilton's variational principle gives the Euler–Langrange equations.$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)-{\frac {\partial L}{\partial {\boldsymbol {q}}}}=0}$$ Similarly, Herglotz's variational principle gives the Euler–Lagrange–Herglotz equations$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)-{\frac {\partial L}{\partial {\boldsymbol {q}}}}={\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}}$$ which involves an extra term ${\textstyle {\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}}$ that can describe the dissipation of the system. The original Euler–Langrange equations are recovered as a special case when ${\displaystyle \partial _{S}L=0}$.

Hamiltonian form

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

Hamilton–Jacobi equation

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

Derivation

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 ${\displaystyle \mu (t)=\mathrm {e} ^{\int _{t_{0}}^{t}{\frac {\partial L}{\partial S}}\mathrm {d} t}}$, 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 $${\displaystyle \mu (t)\delta {\dot {S}}(t)+{\dot {\mu }}(t)\delta S(t)={\frac {\mathrm {d} (\mu (t)\delta S(t))}{\mathrm {d} t}}}$$and therefore we can do an integration of the equation above from ${\displaystyle t=t_{0}}$ to ${\displaystyle t=t_{1}}$, yielding $${\displaystyle \mu (t_{1})\delta S_{1}-\mu (t_{0})\delta S_{0}=\int _{t_{0}}^{t_{1}}\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right)\mathrm {d} t}$$ 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)\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 ${\displaystyle S_{1}}$ 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.

Noether's theorem

Generalizations of Noether's theorem and Noether's second theorem apply to Herglotz's variational principle.^[4][5][6]

An infinitesimal transformation is$${\displaystyle {\bar {t}}=t+\epsilon T(t,q),\quad {\bar {q}}=q+\epsilon Q(t,q)}$$ where ${\displaystyle T,Q}$ are smooth functions of time and configuration, and ${\displaystyle \epsilon }$ is an infinitesimal. The transformation deforms a trajectory ${\displaystyle (t,q(t))}$ to ${\displaystyle (t+\epsilon T(t,q),q(t)+\epsilon Q(t,q))}$, and accordingly deforms the action integral as well.

We say that the infinitesimal transformation is a symmetry of the action iff the change in ${\displaystyle S(t_{1})-S(t_{0})}$ under the infinitesimal transformation is order ${\displaystyle O(\epsilon ^{2})}$. Given such an infinitesimal symmetry, the quantity is a constant of motion$${\displaystyle e^{-\int _{t_{0}}^{t}\partial _{S}L(t')dt'}(T(L-{\dot {q}}\partial _{\dot {q}}L)+Q\partial _{\dot {q}}L)}$$ where ${\displaystyle \partial _{S}L(t')}$ 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]

Examples

Damped particle on a line

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$${\displaystyle L(t,x,{\dot {x}},S)={\frac {1}{2}}m{\dot {x}}^{2}-V(x)-\gamma S}$$

A more general particle on a line

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$${\displaystyle L={\frac {1}{2}}{\dot {x}}^{2}-(2f(x){\dot {x}}+g(t))S-U(x)}$$where ${\textstyle U(x)}$ is any solution of the ODE$${\displaystyle {\frac {\mathrm {d} U(x)}{\mathrm {d} x}}+2f(x)U(x)=h(x).}$$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$${\displaystyle L={\frac {1}{2}}{\dot {x}}^{2}-(2f(x){\dot {x}}+g(t))z.}$$