View source for Applied Mathematics/Lagrange Equations

== Lagrange Equation ==
:<math>\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) -\frac{\partial L}{\partial x}=0</math>
where <math>\dot{x} = \frac{dx}{dt}</math><br />
The equation above is called '''Lagrange Equation'''.

Let the kinetic energy of the point mass be <math>T</math> and the potential energy be <math>U</math>.<br />
<math>T-U</math> is called '''Lagrangian'''. Then the kinetic energy is expressed by
:<math>T =\frac{1}{2}m \dot{x}^2 + \frac{1}{2}m \dot{y}^2</math>
:<math>=\frac{m}{2}(\dot{x}^2+\dot{y}^2)</math>

Thus
:<math>T=T(\dot{x},\dot{y})</math>
:<math>U=U(x,y)</math>

Hence the Lagrangian <math>L</math> is 
:<math>L=T-U</math>
:<math>=T(\dot{x},\dot{y})-U(x,y)</math>
:<math>=\frac{m}{2}(\dot{x}^2+\dot{y}^2)-U(x,y)</math>

Therefore <math>T</math> relies on only <math>\dot{x}</math> and <math>\dot{y}</math>. <math>U</math> relies on only <math>x</math> and <math>y</math>. Thus
:<math>\frac{\partial L}{\partial \dot{x}} = \frac{\partial T}{\partial \dot{x}} = m\dot{x}</math>
:<math>\frac{\partial L}{\partial \dot{y}} = \frac{\partial T}{\partial \dot{y}} = m\dot{y}</math>

In the same way, we have
:<math>\frac{\partial L}{\partial x} = - \frac{\partial U}{\partial x}</math>
:<math>\frac{\partial L}{\partial y} = - \frac{\partial U}{\partial y}</math>
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