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.

Contents

• Mathematical formulation
• Hamilton's principle
• Herglotz's variational principle
• Euler–Lagrange–Herglotz equation
• Hamiltonian form
• Hamilton–Jacobi equation
• Derivation
• Noether's theorem
• Examples
• Damped particle on a line
• A more general particle on a line
• References

Mathematical formulation

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

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 ${\textstyle \mu (t)=\exp \left(\int _{t_{0}}^{t}{\frac {\partial L}{\partial S}}dt\right)}$, 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.

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$${\displaystyle {\bar {t}}=t+\varepsilon T(t,q),\quad {\bar {q}}=q+\varepsilon Q(t,q)}$$ 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]

Examples

References