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 Template:Nowrap instead of an integration of Template:Nowrap 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

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Hamilton's principle

As in Lagrangian mechanics, we consider a system with $n$ degrees of freedom. Let $𝒒=(q_1,q_2,\dots ,q_n)$ be its generalized coordinates, and let $𝒖=(u_1,u_2,\dots ,u_n)$ be its generalized velocity. Let $L=L(t,𝒒,𝒖)$ be the Lagrangian function of the physical system. Let $S$ be the action.

Lagrangian mechanics is derived using Hamilton's principle. Fix a starting time and configuration $t_0,{𝒒}_0$ and an ending time and configuration Template:Nowrap Hamilton's principle states that physically real trajectories from are the solutions to the problem of variational calculus:$$\{\begin{array}{c}\delta {\int }_{\gamma }L(t,\gamma (t),\dot{\gamma }(t))dt=0\\ \gamma \text{ has end points }{t}_0,{𝒒}_0,t_1,{𝒒}_1\end{array}$$Equivalently, it can be formulated as$$\{\begin{array}{c}\delta S(t_1)=0\\ \dot{S}(t)=L(t,\gamma (t),\dot{\gamma }(t)),& t\in [t_0,t_1]\\ S(t_0)=S_0\\ \gamma \text{ has end points }{t}_0,{𝒒}_0,t_1,{𝒒}_1\end{array}$$

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 Template:Nowrap depending on $2n+2$ variables. $$\{\begin{array}{c}\delta S(t_1)=0\\ \dot{S}(t)=L(t,\gamma (t),\dot{\gamma }(t),S(t)),& t\in [t_0,t_1]\\ S(t_0)=S_0\\ \gamma \text{ has end points }{t}_0,{𝒒}_0,t_1,{𝒒}_1\end{array}$$

Euler–Lagrange–Herglotz equation

Hamilton's variational principle gives the Euler–Langrange equations.$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{𝒒}}\right)-\frac{\partial L}{\partial 𝒒}=0$$ Similarly, Herglotz's variational principle gives the Euler–Lagrange–Herglotz equations $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{𝒒}}\right)-\frac{\partial L}{\partial 𝒒}=\frac{\partial L}{\partial S}\frac{\partial L}{\partial \dot{𝒒}}$$ which involves an extra term $\frac{\partial L}{\partial S}\frac{\partial L}{\partial \dot{𝒒}}$ that can describe the dissipation of the system. The original Euler–Langrange equations are recovered as a special case when Template:Nowrap

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$$p_i:=\frac{\partial L}{\partial u_i},H:=\sum _i{p}_i{u}_i-L$$ Then the equations of motion are$$\{\begin{array}{c}{\dot{q}}_i=\frac{\partial H}{\partial p_i}\\ {\dot{p}}_i=-(\frac{\partial H}{\partial q_i}+p_i\frac{\partial H}{\partial S})\\ \dot{S}=\sum _{i=1}^n{p}_i\frac{\partial H}{\partial p_i}-H\end{array}$$

Hamilton–Jacobi equation

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

Derivation

In order to solve this minimization problem, we impose a variation $\delta 𝒒$ on Template:Nowrap and suppose $S(t)$ undergoes a variation $\delta S(t)$ correspondingly, then$$\begin{array}{cc}\delta \dot{S}(t)& =L(t,𝒒(t)+\delta 𝒒(t),\dot{𝒒}(t)+\delta \dot{𝒒}(t),S(t)+\delta S(t))-L(t,𝒒(t),\dot{𝒒}(t),S(t))\\ & =\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)+\frac{\partial L}{\partial \dot{𝒒}}\delta \dot{𝒒}(t)+\frac{\partial L}{\partial S}\delta S(t)\end{array}$$ and since the initial condition is not changed, Template:Nowrap The above equation a linear ODE for the function Template:Nowrap and it can be solved by introducing an integrating factor Template:Nowrap which is uniquely determined by the ODE $$\dot{\mu }(t)=-\mu (t)\frac{\partial L}{\partial S},u(t_0)=1.$$By multiplying $\mu (t)$ on both sides of the equation of $\delta \dot{S}$ and moving the term $\mu (t)\frac{\partial L}{\partial S}\delta S(t)$ to the left hand side, we get $$\mu (t)\delta \dot{S}(t)-\mu (t)\frac{\partial L}{\partial S}\delta S(t)=\mu (t)(\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)+\frac{\partial L}{\partial \dot{𝒒}}\delta \dot{𝒒}(t)).$$ Note that, since Template:Nowrap the left hand side equals to $$\mu (t)\delta \dot{S}(t)+\dot{\mu }(t)\delta S(t)=\frac{d(\mu (t)\delta S(t))}{dt}$$and therefore we can do an integration of the equation above from $t=t_0$ to Template:Nowrap yielding $$\mu (t_1)\delta S_1-\mu (t_0)\delta S_0={\int }_{t_0}^{t_1}\mu (t)(\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)+\frac{\partial L}{\partial \dot{𝒒}}\delta \dot{𝒒}(t))dt$$ 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 $\frac{\partial L}{\partial \dot{𝒒}}\delta \dot{𝒒}(t)$ term to remove the time derivative on Template:Nowrap $$\begin{array}{cc}& {\int }_{t_0}^{t_1}\mu (t)(\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)+\frac{\partial L}{\partial \dot{𝒒}}\delta \dot{𝒒}(t))dt\\ =& {\int }_{t_0}^{t_1}\mu (t)\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)dt+{\int }_{t_0}^{t_1}\mu (t)\frac{\partial L}{\partial \dot{𝒒}}\delta \dot{𝒒}(t)dt\\ =& {\int }_{t_0}^{t_1}\mu (t)\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)dt+\mu (t_1)\frac{\partial L}{\partial \dot{𝒒}}\underset{=0}{\underbrace{\delta 𝒒(t_1)}}-\mu (t_0)\frac{\partial L}{\partial \dot{𝒒}}\underset{=0}{\underbrace{\delta 𝒒(t_0)}}-{\int }_{t_0}^{t_1}\frac{d}{dt}(\mu (t)\frac{\partial L}{\partial \dot{𝒒}})\delta 𝒒(t)dt\\ =& {\int }_{t_0}^{t_1}\mu (t)\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)dt-{\int }_{t_0}^{t_1}\frac{d}{dt}(\mu (t)\frac{\partial L}{\partial \dot{𝒒}})\delta 𝒒(t)dt\\ =& {\int }_{t_0}^{t_1}\mu (t)\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)dt-{\int }_{t_0}^{t_1}(\dot{\mu }(t)\frac{\partial L}{\partial \dot{𝒒}}+\mu (t)\frac{d}{dt}\frac{\partial L}{\partial \dot{𝒒}})\delta 𝒒(t)dt\\ =& {\int }_{t_0}^{t_1}\mu (t)\frac{\partial L}{\partial 𝒒}\delta 𝒒(t)dt-{\int }_{t_0}^{t_1}(-\mu (t)\frac{\partial L}{\partial S}\frac{\partial L}{\partial \dot{𝒒}}+\mu (t)\frac{d}{dt}\frac{\partial L}{\partial \dot{𝒒}})\delta 𝒒(t)dt\\ =& {\int }_{t_0}^{t_1}\mu (t)\underset{\_}{(\frac{\partial L}{\partial 𝒒}+\frac{\partial L}{\partial S}\frac{\partial L}{\partial \dot{𝒒}}-\frac{d}{dt}\frac{\partial L}{\partial \dot{𝒒}})}\delta 𝒒(t)dt,\end{array}$$ and when $S_1$ is minimized, $\delta S_1=0$ for all Template:Nowrap which indicates that the underlined term in the last line of the equation above has to be zero on the entire interval Template:Nowrap 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.

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

We say that the infinitesimal transformation is a symmetry of the action iff the change in $S(t_1)-S(t_0)$ under the infinitesimal transformation is order Template:Nowrap Given such an infinitesimal symmetry, the quantity is a constant of motion$$\exp (-{\int }_{t_0}^t\frac{\partial L(t^{\prime })}{\partial S}d{t}^{\prime })[T(L-\dot{q}\frac{\partial L}{\partial \dot{q}})+Q\frac{\partial L}{\partial \dot{q}}]$$ where ${\partial }_{S}L(t^{\prime })$ is more explicitly written as$${\partial }_{S}L(t^{\prime },q(t^{\prime }),\dot{q}(t^{\prime }),S(t^{\prime }))$$ There is also a version for multiple time dimensions.

Examples

Damped particle on a line

The motion of a particle of mass $m$ in a potential field $V$ with damping coefficient $\gamma$ is$$m\ddot{x}=-V^{\prime }(x)-\gamma \dot{x}.$$It can be produced as the Euler–Lagrange–Herglotz forTemplate:Reference page$$L(t,x,\dot{x},S)=\frac{1}{2}m{\dot{x}}^2-V(x)-\gamma S$$

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 Template:Nowrap and another force proportional to Template:Nowrap Write it as $$\ddot{x}+f(x){\dot{x}}^2+g(t)\dot{x}+h(x)=0$$ This equation is the Euler–Lagrange–Herglotz equation for the Lagrangian $$L=\frac{1}{2}{\dot{x}}^2-[2f(x)\dot{x}+g(t)]S-U(x)$$ where $U(x)$ is any solution of the ODE $$\frac{dU(x)}{dx}+2f(x)U(x)=h(x).$$Some important special cases:

• When $f(x)=0$ and $g(t)$ is constant, it is the damped harmonic oscillator given above.
• When $h(x)=x^n,f(x)=0$ and $g(t)=2/t$, it is the Lane–Emden equation$$\ddot{x}+\frac{2}{t}\dot{x}+x^n=0,n\ne -1.$$ with Lagrangian $$L=\frac{1}{2}{\dot{x}}^2-\frac{x^{n+1}}{n+1}-\frac{2}{t}z.$$
• When $h(x)=0$, it is a Rayleigh-type system$$\ddot{x}+f(x){\dot{x}}^2+g(t)\dot{x}=0.$$ with Lagrangian$$L=\frac{1}{2}{\dot{x}}^2-[2f(x)\dot{x}+g(t)]z.$$

References

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