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

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

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