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.

Mathematical formulation

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

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.

Examples

One simple one-dimensional (${\displaystyle n=1}$) example^[3] is given by the Lagrangian $${\displaystyle L(t,x,{\dot {x}},S)={\frac {1}{2}}m{\dot {x}}^{2}-V(x)-\gamma S}$$The corresponding Euler-Lagrange-Herglotz equation is given as $${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}(m{\dot {x}})+V'(x)=-\gamma {\dot {x}},}$$which simplifies into $${\displaystyle m{\ddot {x}}=-V'(x)-\gamma {\dot {x}}.}$$This equation describes the damping motion of a particle in a potential field ${\displaystyle V}$, where ${\displaystyle \gamma }$ is the damping coefficient.