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 {\displaystyle S} as an independent variable, and S {\displaystyle S} itself is represented as the solution of an ordinary differential equation (ODE) whose right hand side is the Lagrangian L {\displaystyle L} , instead of an integration of L {\displaystyle L} . 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
Hamilton's principle As in Lagrangian mechanics, we consider a system with n {\displaystyle n} degrees of freedom. Let q = ( q 1 , q 2 , … , q n ) {\displaystyle {\boldsymbol {q}}=(q_{1},q_{2},\dots ,q_{n})} be its generalized coordinates, and let u = ( u 1 , u 2 , … , u n ) {\displaystyle {\boldsymbol {u}}=(u_{1},u_{2},\dots ,u_{n})} be its generalized velocity. Let L = L ( t , q , u ) {\displaystyle L=L(t,{\boldsymbol {q}},{\boldsymbol {u}})} be the Lagrangian function of the physical system. Let S {\displaystyle S} be the action. Lagrangian mechanics is derived using Hamilton's principle. Fix a starting time and configuration t 0 , q 0 {\displaystyle t_{0},{\boldsymbol {q}}_{0}} and an ending time and configuration t 1 , q 1 {\displaystyle t_{1},{\boldsymbol {q}}_{1}} . Hamilton's principle states that physically real trajectories from are the solutions to the problem of variational calculus: { δ ∫ γ L ( t , γ ( t ) , γ ˙ ( t ) ) d t = 0 γ has end points t 0 , q 0 , t 1 , q 1 {\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 { δ S ( t 1 ) = 0 S ˙ ( t ) = L ( t , γ ( t ) , γ ˙ ( t ) ) , t ∈ [ t 0 , t 1 ] S ( t 0 ) = S 0 γ has end points t 0 , q 0 , t 1 , q 1 {\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 L = L ( t , q , u , S ) {\displaystyle L=L(t,{\boldsymbol {q}},{\boldsymbol {u}},S)} , depending on 2 n + 2 {\displaystyle 2n+2} variables.
{ δ S ( t 1 ) = 0 S ˙ ( t ) = L ( t , γ ( t ) , γ ˙ ( t ) , S ( t ) ) , t ∈ [ t 0 , t 1 ] S ( t 0 ) = S 0 γ has end points t 0 , q 0 , t 1 , q 1 {\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. d d t ( ∂ L ∂ q ˙ ) − ∂ L ∂ q = 0 {\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
d d t ( ∂ L ∂ q ˙ ) − ∂ L ∂ q = ∂ L ∂ S ∂ L ∂ q ˙ {\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 ∂ L ∂ S ∂ L ∂ q ˙ {\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 ∂ S L = 0 {\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 p i := ∂ L ∂ u i , H := ∑ i p i u i − L {\displaystyle p_{i}:={\frac {\partial L}{\partial u_{i}}},\quad H:=\sum _{i}p_{i}u_{i}-L} Then the equations of motion are { q ˙ i = ∂ H ∂ p i p ˙ i = − ( ∂ H ∂ q i + p i ∂ H ∂ S ) S ˙ = ∑ i = 1 n p i ∂ H ∂ p i − H {\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}}}
Hamilton–Jacobi equation If S ( t , q ) {\displaystyle S(t,q)} is written as a function of time and configuration, then it satisfies a Hamilton–Jacobi equation d S = − H d t + ∑ i p i d q i {\displaystyle dS=-H\,dt+\sum _{i}p_{i}\,dq^{i}}
Derivation In order to solve this minimization problem, we impose a variation δ q {\displaystyle \delta {\boldsymbol {q}}} on q {\displaystyle {\boldsymbol {q}}} , and suppose S ( t ) {\displaystyle S(t)} undergoes a variation δ S ( t ) {\displaystyle \delta S(t)} correspondingly, then δ S ˙ ( t ) = L ( t , q ( t ) + δ q ( t ) , q ˙ ( t ) + δ q ˙ ( t ) , S ( t ) + δ S ( t ) ) − L ( t , q ( t ) , q ˙ ( t ) , S ( t ) ) = ∂ L ∂ q δ q ( t ) + ∂ L ∂ q ˙ δ q ˙ ( t ) + ∂ L ∂ S δ S ( t ) {\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, δ S 0 = 0 {\displaystyle \delta S_{0}=0} . The above equation a linear ODE for the function δ S ( t ) {\displaystyle \delta S(t)} , and it can be solved by introducing an integrating factor μ ( t ) = exp ( ∫ t 0 t ∂ L ∂ S d t ) {\textstyle \mu (t)=\exp \left(\int _{t_{0}}^{t}{\frac {\partial L}{\partial S}}dt\right)} , which is uniquely determined by the ODE μ ˙ ( t ) = − μ ( t ) ∂ L ∂ S , u ( t 0 ) = 1. {\displaystyle {\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}},\quad u(t_{0})=1.} By multiplying μ ( t ) {\displaystyle \mu (t)} on both sides of the equation of δ S ˙ {\displaystyle \delta {\dot {S}}} and moving the term μ ( t ) ∂ L ∂ S δ S ( t ) {\textstyle \mu (t){\frac {\partial L}{\partial S}}\delta S(t)} to the left hand side, we get
μ ( t ) δ S ˙ ( t ) − μ ( t ) ∂ L ∂ S δ S ( t ) = μ ( t ) ( ∂ L ∂ q δ q ( t ) + ∂ L ∂ q ˙ δ q ˙ ( t ) ) . {\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 μ ˙ ( t ) = − μ ( t ) ∂ L ∂ S {\textstyle {\dot {\mu }}(t)=-\mu (t){\frac {\partial L}{\partial S}}} , the left hand side equals to μ ( t ) δ S ˙ ( t ) + μ ˙ ( t ) δ S ( t ) = d ( μ ( t ) δ S ( t ) ) d t {\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 t = t 0 {\displaystyle t=t_{0}} to t = t 1 {\displaystyle t=t_{1}} , yielding μ ( t 1 ) δ S 1 − μ ( t 0 ) δ S 0 = ∫ t 0 t 1 μ ( t ) ( ∂ L ∂ q δ q ( t ) + ∂ L ∂ q ˙ δ q ˙ ( t ) ) d t {\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 δ S 0 = 0 {\displaystyle \delta S_{0}=0} so the left hand side actually only contains one term μ ( t 1 ) δ S 1 {\displaystyle \mu (t_{1})\delta S_{1}} , and for the right hand side, we can perform the integration-by-part on the ∂ L ∂ q ˙ δ q ˙ ( t ) {\textstyle {\frac {\partial L}{\partial {\dot {\boldsymbol {q}}}}}\delta {\dot {\boldsymbol {q}}}(t)} term to remove the time derivative on δ q {\textstyle \delta {\boldsymbol {q}}} :
∫ t 0 t 1 μ ( t ) ( ∂ L ∂ q δ q ( t ) + ∂ L ∂ q ˙ δ q ˙ ( t ) ) d t = ∫ t 0 t 1 μ ( t ) ∂ L ∂ q δ q ( t ) d t + ∫ t 0 t 1 μ ( t ) ∂ L ∂ q ˙ δ q ˙ ( t ) d t = ∫ t 0 t 1 μ ( t ) ∂ L ∂ q δ