Let $\phi:\mathbb R^n \to \mathbb R$ be a twice continuously differentiable function and let
$$L_{\phi}(t,x,\dot x) =\nabla\phi(x)^T\dot x = \sum_i\frac{\partial \phi}{\partial x_i}(x_1,..,x_n)\dot x_i.$$
Let $L:[a,b] \times \mathbb R^n \times \mathbb R^n \to \mathbb R$ another arbitrary $C^1$-Lagrange function. How can one show that the Euler-Lagrange-Equations $L$ and $L+L_{\phi}$ have the same solutions?
Best Answer
First, notice$$\begin{align}\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L_\phi&=\sum_i\left(\frac{\partial^2\phi}{\partial x_i\partial x_j}\dot{x_i}-\frac{d}{dt}\left(\frac{\partial\phi}{\partial x_i}\delta_{ij}\right)\right)\\&=\sum_i\frac{\partial^2\phi}{\partial x_i\partial x_j}\dot{x_i}-\frac{d}{dt}\left(\frac{\partial\phi}{\partial x_j}\right)\\&=0.\end{align}$$So$$\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)(L+L_\phi)-\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L=\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L_\phi=0.$$So$$\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L=0\iff\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)(L+L_\phi)=0,$$i.e. the two systems of ELEs are equivalent.