Noether’s theorem for variational problems of Herglotz type with real and complex order fractional derivatives

Abstract

A variational principle of Herglotz type with a Lagrangian that depends on fractional derivatives of both real and complex orders is formulated, and the invariance of this principle under the action of a local group of symmetries is determined. By the Noether theorem the conservation law for the corresponding fractional Euler–Lagrange equation is obtained. A sequence of approximations of a fractional Euler–Lagrange equation by systems of integer order equations is used for the construction of a sequence of conservation laws which, with certain assumptions, weakly converge to the one for the basic Herglotz variational principle. Results are illustrated by two examples.

Appendix

We derive the infinitesimal criteria IC$_N$ for the approximate problem (29) given by (40).

Proposition 7.1

Let ${L_N}\in C^{1}([a,T]\times {\mathbb {R}}^{N+2})$ . The necessary and sufficient condition that the vector field $\tau _N\frac{\partial }{\partial t}+\xi _N\frac{\partial }{\partial u}$ generates a local one-parameter symmetry group of ( 29 ) if and only if ( 40 ) holds.

Proof

From (29) one obtains

$$\begin{aligned} z(t) = \gamma +\int _a^t L_N(s,u(s),u^{\prime }(s),\ldots ,u^{N}(s),z(s))\mathrm{d}s,\ t \in [a,T]. \end{aligned}$$

(53)

Recall that $\Delta z(t) = \int _a^t \delta L_N \mathrm{d}s + \left. L_N\tau _N\right| _a^t$ ($\delta \gamma = 0$). Moreover, we have

$$\begin{aligned}&\delta L_N = \frac{\partial L_N}{\partial u}\delta u + \frac{\partial L_N }{\partial u^{\prime }}\delta \left( u^{\prime }\right) + \cdots + \frac{ \partial L_N}{\partial u^{N}}\delta \left( u^{N}\right) + \frac{\partial L_N}{ \partial z}\delta z \end{aligned}$$

(54.1)

$$\begin{aligned}&\quad \qquad \,\, = \frac{\partial L_N}{\partial u}\delta u + \frac{\partial L_N}{\partial u^{\prime }} (\delta u)^{\prime } + \cdots + \frac{\partial L_N}{\partial u^{N} }\delta (\delta u)^{(N)} + \frac{\partial L_N}{\partial z}\delta z \end{aligned}$$

(54.2)

(recall, $\frac{\mathrm{d}}{\mathrm{d}t}\delta (\cdot ) = \delta \frac{\mathrm{d}}{\mathrm{d}t}(\cdot )$.) Using $\Delta u = \delta u + {\dot{u}}\tau$, $\Delta z = \delta z + {\dot{z}}\tau = \delta z + L_N\tau$, from (53) and (54.2) we obtain

$$\begin{aligned} \Delta z(t)= & {} \int _{a}^{t}\left[ \frac{\partial L_{N}}{\partial u}(\Delta u-{\dot{u}} \tau )+\frac{\partial L_{N}}{\partial u^{\prime }}(\Delta u-{\dot{u}}\tau )^{\cdot }+\cdots +\frac{\partial L_{N}}{\partial u^{(N)}}(\Delta u-{\dot{u}} \tau )^{(N)}\right. \\&\left. +\,\frac{\partial L_{N}}{\partial z}(\Delta z-L_{N}\tau )+{\dot{L}}_{N}\tau +L_{N}{\dot{\tau }}\right] \mathrm{d}s, \nonumber \end{aligned}$$

(55)

where we use $L_{N}\tau |_{a}^{t}=\int _{a}^{t}({\dot{L}}_{N}\tau +L_{N}\dot{ \tau })\mathrm{d}s$. Since

$$\begin{aligned} {\dot{L}}_N= & {} \frac{\mathrm{d}}{\mathrm{d}t}L_N(t,u(t),u^{\prime }(t),\ldots ,u^{(N)},z(t)) \nonumber \\= & {} \frac{\partial L_N}{\partial t} + \frac{\partial L_N}{\partial u}{\dot{u}} + \frac{\partial L_N}{\partial {\dot{u}}}\ddot{u} + \cdots + \frac{\partial L_N }{\partial u^{(N)}}u^{(N+1)} + \frac{\partial L_N}{\partial z}L_N \end{aligned}$$

(56)

where we use ${\dot{z}} = L_N$, it follows

$$\begin{aligned} \Delta z(t)= & {} \int _{a}^{t}\left\{ \frac{\partial L_{N}}{\partial u}(\Delta u-{\dot{u}} \tau )+\frac{\partial L_{N}}{\partial u}(\Delta u-{\dot{u}}\tau )^{\cdot }+\cdots +\frac{\partial L_{N}}{\partial u^{(N)}}(\Delta u-{\dot{u}}\tau )^{(N)} \right. \nonumber \\&+\,\frac{\partial L_{N}}{\partial z}\Delta z-\frac{\partial L_{N}}{ \partial z}L_{N}\tau +\frac{\partial L_{N}}{\partial t}\tau +\frac{\partial L_{N}}{\partial u} {\dot{u}}\tau +\frac{\partial L_{N}}{\partial {\dot{u}}}\ddot{u}\tau +\cdots + \frac{\partial L_{N}}{\partial u^{(N)}}u^{(N+1)}\tau \nonumber \\&\left. +\,\frac{\partial L_{N}}{ \partial z}L_{N}\tau +L_{N}{\dot{\tau }} \right\} \mathrm{d}s, \end{aligned}$$

(57)

and after cancellation of terms ${\mp } \frac{\partial L_{N}}{\partial u}{\dot{u}} \tau$ ${\mp } \frac{\partial L_{N}}{\partial z}L_{N}\tau$ and differentiation we obtain

$$\begin{aligned} ({\Delta z})^{\cdot } - \frac{\partial L_N}{\partial z}\Delta z = \frac{ \partial L_N}{\partial u}\xi + \sum _{i=1}^{N}(\xi - {\dot{u}}\tau )^{(i)}\frac{ \partial L_N}{\partial u^{(i)}} + \frac{\partial L_N}{\partial t}\tau + \left( \sum _{i=1}^{N}\frac{\partial L_N}{\partial u^{(i+1)}} \right) \tau + L_N{\dot{\tau }} \end{aligned}$$

(58)

since $\Delta u = \xi$. Next, by multiplying (58) with $\lambda _N$ and using $\frac{\mathrm{d}}{\mathrm{d}t}\lambda _N(t) = -\frac{\partial L_N}{\partial z} \lambda _N(t)$, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left( \lambda _N\Delta z \right) = \lambda _N\left( \frac{\partial L_N}{ \partial u}\xi + \sum _{i=1}^{N}(\xi - {\dot{u}}\tau )^{(i)}\frac{\partial L_N}{ \partial u^{(i)}} + \frac{\partial L_N}{\partial t}\tau + \left( \sum _{i=1}^{N}\frac{\partial L_N}{\partial u^{(i+1)}} \right) \tau + L_N\dot{ \tau }\right) , \end{aligned}$$

(59)

and then after integrating and using $\Delta z(a)=0$, we have

$$\begin{aligned} \lambda _N(t)\Delta z(t)= & {} \int _a^t \lambda _N(s)\left( \frac{\partial L_N}{\partial u}\xi + \sum _{i=1}^{N}(\xi - {\dot{u}}\tau )^{(i)}\frac{\partial L_N}{\partial u^{(i)}} + \frac{\partial L_N}{\partial t}\tau \right. \nonumber \\&+ \,\left. \left( \sum _{i=1}^{N}\frac{ \partial L_N}{\partial u^{(i+1)}} \right) \tau + L_N{\dot{\tau }}\right) \mathrm{d}s. \end{aligned}$$

(60)

Since $\lambda (t) > 0$, $t\in [a,T]$ and $\Delta z(t)=0$, the invariance criterion is equivalent to

$$\begin{aligned} \frac{\partial L_N}{\partial u}\xi + \sum _{i=1}^{N}(\xi - {\dot{u}}\tau )^{(i)} \frac{\partial L_N}{\partial u^{(i)}} + \frac{\partial L_N}{\partial t}\tau + \left( \sum _{i=1}^{N}\frac{\partial L_N}{\partial u^{(i+1)}} \right) \tau + L_N{\dot{\tau }} = 0,\; t\in [a,T]. \end{aligned}$$

This is equivalent to (40). $\square$