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 $S$ as an independent variable, and $S$ itself is represented as the solution of an ordinary differential equation (ODE) whose right hand side is the Lagrangian $L$, instead of an integration of $L$. Herglotz's variational principle is known as the variational principle for nonconservative Lagrange equations and Hamilton equations.

Mathematical formulation

Suppose there is a Lagrangian $L=L(t,{\boldsymbol {q}},{\boldsymbol {u}},S)$ of $2n+2$ variables, where ${\boldsymbol {q}}=(q_{1},q_{2},\dots ,q_{n})$ and ${\boldsymbol {u}}=(u_{1},u_{2},\dots ,u_{n})$ are $n$ dimensional vectors, and $t,S$ are scalar values. A time interval $[t_{0},t_{1}]$ is fixed. Given a time-parameterized curve ${\boldsymbol {q}}={\boldsymbol {q}}(t)$, consider the ODE $${\begin{cases}{\dot {S}}(t)=L(t,{\boldsymbol {q}}(t),{\boldsymbol {\dot {q}}}(t),S(t)),&t\in [t_{0},t_{1}]\\S(t_{0})=S_{0}\end{cases}}$$When $L(t,{\boldsymbol {q}},{\boldsymbol {u}},S),{\boldsymbol {q}}(t),{\boldsymbol {u}}(t)$ are all well-behaved functions, this equation allows a unique solution, and thus $S_{1}:=S(t_{1})$ is a well defined number which is determined by the curve ${\boldsymbol {q}}(t)$. Herglotz's variation problem aims to minimize $S_{1}$ over the family of curves ${\boldsymbol {q}}(t)$ with fixed value ${\boldsymbol {q}}_{0}$ at $t=t_{0}$ and fixed value ${\boldsymbol {q}}_{1}$ at $t=t_{1}$, i.e. the problem $${\underset {{\boldsymbol {q}}:{\boldsymbol {q}}(t_{0})={\boldsymbol {q}}_{0},{\boldsymbol {q}}(t_{1})={\boldsymbol {q}}_{1}}{\arg \min }}S_{1}[{\boldsymbol {q}}]$$Note that, when $L$ does not explicitly depend on $S$, i.e. $L=L(t,{\boldsymbol {q}},{\boldsymbol {u}})$, the above ODE system gives exactly ${\textstyle S(t)=\int _{t_{0}}^{t}L(t,{\boldsymbol {q}}(\tau ),{\boldsymbol {q}}(\tau )){\rm {d}}\tau }$, and thus ${\textstyle S_{1}=S(t_{1})=\int _{t_{0}}^{t_{1}}L(t,{\boldsymbol {q}}(t),{\boldsymbol {q}}(t)){\rm {d}}t}$, which degenerates to the classical Hamiltonian action. The resulting Euler-Lagrange-Herglotz equation is $${\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\right)-{\frac {\partial L}{\partial {\boldsymbol {q}}}}={\frac {\partial L}{\partial S}}{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}$$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.

Derivation

In order to solve this minimization problem, we impose a variation $\delta {\boldsymbol {q}}$ on ${\boldsymbol {q}}$, and suppose $S(t)$ undergoes a variation $\delta S(t)$ correspondingly, then$${\begin{aligned}\delta {\dot {S}}(t)&=L(t,{\boldsymbol {q}}(t)+\delta {\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t)+\delta {\dot {\boldsymbol {q}}}(t),S(t)+\delta S(t))-L(t,{\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t),S(t))\\&={\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)+{\frac {\partial L}{\partial S}}\delta S(t)\end{aligned}}$$and since the initial condition is not changed, $\delta S_{0}=0$. The above equation a linear ODE for the function $\delta S(t)$, and it can be solved by introducing an integrating factor $\mu (t)=\mathrm {e} ^{\int _{t_{0}}^{t}{\frac {\partial L}{\partial S}}\mathrm {d} t}$, which is uniquely determined by the ODE $${\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}},\quad u(t_{0})=1.$$By multiplying $\mu (t)$ on both sides of the equation of $\delta {\dot {S}}$ and moving the term ${\textstyle \mu (t){\frac {\partial L}{\partial S}}\delta S(t)}$ to the left hand side, we get $$\mu (t)\delta {\dot {S}}(t)-\mu (t){\frac {\partial L}{\partial S}}\delta S(t)=\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right).$$Note that, since ${\textstyle {\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}}}$, the left hand side equals to $$\mu (t)\delta {\dot {S}}(t)+{\dot {\mu }}(t)\delta S(t)={\frac {\mathrm {d} (\mu (t)\delta S(t))}{\mathrm {d} t}}$$and therefore we can do an integration of the equation above from $t=t_{0}$ to $t=t_{1}$, yielding $$\mu (t_{1})\delta S_{1}-\mu (t_{0})\delta S_{0}=\int _{t_{0}}^{t_{1}}\mu (t)\left({\frac {\partial L}{\partial {\boldsymbol {q}}}}\delta {\boldsymbol {q}}(t)+{\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)\right)\mathrm {d} t$$where the $\delta S_{0}=0$ so the left hand side actually only contains one term $\mu (t_{1})\delta S_{1}$, 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 ${\textstyle \delta {\boldsymbol {q}}}$:$${\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 $S_{1}$ is minimized, $\delta S_{1}=0$ for all $\delta {\boldsymbol {q}}$, which indicates that the underlined term in the last line of the equation above has to be zero on the entire interval $[t_{0},t_{1}]$, this gives rise to the Euler-Lagrange-Herglotz equation.

Examples

One simple one-dimensional ($n=1$) example is given by the Lagrangian $$L(t,x,{\dot {x}},S)={\frac {1}{2}}m{\dot {x}}^{2}-V(x)-\gamma S$$The corresponding Euler-Lagrange-Herglotz equation is given as $${\frac {\mathrm {d} }{\mathrm {d} t}}(m{\dot {x}})+V'(x)=-\gamma {\dot {x}},$$which simplifies into $$m{\ddot {x}}=-V'(x)-\gamma {\dot {x}}.$$This equation describes the damping motion of a particle in a potential field $V$, where $\gamma$ is the damping coefficient.