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.

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