# Applied Mathematics/Lagrange Equations

## Lagrange Equation

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

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

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

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

Thus

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

Hence the Lagrangian ${\displaystyle L}$  is

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

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

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

In the same way, we have

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