Applied Mathematics/Lagrange Equations

Lagrange Equation

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0$

where $\dot{x}=\frac{dx}{dt}$
The equation above is called Lagrange Equation.

Let the kinetic energy of the point mass be $T$ and the potential energy be $U$.
$T-U$ is called Lagrangian. Then the kinetic energy is expressed by

$T=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\dot{y}}^2$

$=\frac{m}{2}({\dot{x}}^2+{\dot{y}}^2)$

Thus

$T=T(\dot{x},\dot{y})$

$U=U(x,y)$

Hence the Lagrangian $L$ is

$L=T-U$

$=T(\dot{x},\dot{y})-U(x,y)$

$=\frac{m}{2}({\dot{x}}^2+{\dot{y}}^2)-U(x,y)$

Therefore $T$ relies on only $\dot{x}$ and $\dot{y}$. $U$ relies on only $x$ and $y$. Thus

$\frac{\partial L}{\partial \dot{x}}=\frac{\partial T}{\partial \dot{x}}=m\dot{x}$

$\frac{\partial L}{\partial \dot{y}}=\frac{\partial T}{\partial \dot{y}}=m\dot{y}$

In the same way, we have

$\frac{\partial L}{\partial x}=-\frac{\partial U}{\partial x}$

$\frac{\partial L}{\partial y}=-\frac{\partial U}{\partial y}$