Applied Mathematics/Lagrange Equations

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Lagrange Equation[edit]

\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}