# Euler-Lagrange Equation Trick

Just a quick blast of calculus today.

Consider the Euler-Lagrange equation for a Lagrangian which depends on a single function $y$:

$\displaystyle \frac{\partial L(y,y',x)}{\partial y}=\frac{d}{dx}\Big(\frac{\partial L(y,y',x)}{\partial y'}\Big)$

where $y'\equiv\frac{dy}{dx}$.

In general, substitution of the Lagrangian into the Euler-Lagrange equation yields a second-order differential equation in $y$. It turns out that if the Lagrangian does not explicitly depend on the independent variable $x$, that is

$L=L(y,y'),$

the problem can be reduced to a first-order differential equation.

First we multiply the Euler-Lagrange equation through by the derivative of $y$:

$\displaystyle y'\frac{\partial L}{\partial y}=y'\frac{d}{dx}\Big(\frac{\partial L}{\partial y'}\Big)$

We then use a trick similar to the one used in the derivation of the Euler-Lagrange equation itself. Consider the following:

$\displaystyle \frac{d}{dx}\Big(y'\frac{\partial L}{\partial y'}\Big)=y''\frac{\partial L}{\partial y'}+y'\frac{d}{dx}\Big(\frac{\partial L}{\partial y'}\Big)$

Rearranging for the second term on the right-hande side and substituting into the equation above yields

$\displaystyle y'\frac{\partial L}{\partial y}=\frac{d}{dx}\Big(y'\frac{\partial L}{\partial y'}\Big)-y''\frac{\partial L}{\partial y'}$

$\displaystyle y'\frac{\partial L}{\partial y}+y''\frac{\partial L}{\partial y'}=\frac{d}{dx}\Big(y'\frac{\partial L}{\partial y'}\Big)$

Now consider the differential of the Lagrangian:

$\displaystyle dL=\frac{\partial L}{\partial y}dy+\frac{\partial L}{\partial y'}dy'$

Importantly, there is no partial derivative with respect to $x$ because the Lagrangian does not explicitly depend on it. Using this expression,

$\displaystyle \frac{dL}{dx}=\frac{\partial L}{\partial y}\frac{dy}{dx}+\frac{\partial L}{\partial y'}\frac{dy'}{dx}$

This is effectively the chain rule for a multivariable function. Using prime notation, this is the same as

$\displaystyle \frac{dL}{dx}=\frac{\partial L}{\partial y}y'+\frac{\partial L}{\partial y'}y''$

But this is exactly what appears on the left-hand side of the equation in the previous paragraph. Using this result,

$\displaystyle \frac{dL}{dx}=\frac{d}{dx}\Big(y'\frac{\partial L}{\partial y'}\Big)$

$\displaystyle \frac{d}{dx}\Big(L-y'\frac{\partial L}{\partial y'}\Big)=0$

Hence,

$\displaystyle L(y,y')-y'\frac{\partial L(y,y')}{\partial y'}=\text{constant}$

This is a first-order differential equation which will in general be easier to solve than the second-order Euler-Lagrange equation.