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