2.4: An Important First Integral of the Euler-Lagrange Equation

Last updated
Save as PDF

Page ID: 29526

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

It turns out that, since the function \(f \) does not contain x explicitly, there is a simple first integral of this equation. Multiplying throughout by \(y^{\prime}=d y / d x\)

\[\frac{\partial f\left(y, y^{\prime}\right)}{\partial y} \frac{d y}{d x}-\frac{d}{d x}\left(\frac{\partial f\left(y, y^{\prime}\right)}{\partial y^{\prime}}\right) y^{\prime}=0\]

Since \(f\) doesn’t depend explicitly on \(x\), we have

\[\frac{d f}{d x}=\frac{\partial f}{\partial y} \frac{d y}{d x}+\frac{\partial f}{\partial y^{\prime}} \frac{d y^{\prime}}{d x}\]

and using this to replace \(\frac{\partial f\left(y, y^{\prime}\right)}{\partial y} \frac{d y}{d x}\) in the preceding equation gives

\[\frac{d f}{d x}-\frac{\partial f}{\partial y^{\prime}} \frac{d y^{\prime}}{d x}-\frac{d}{d x}\left(\frac{\partial f\left(y, y^{\prime}\right)}{\partial y^{\prime}}\right) y^{\prime}=0\]

then multiplying by − (to match the equation as usually written) we have

\[\frac{d}{d x}\left(y^{\prime} \frac{\partial f}{\partial y^{\prime}}-f\right)=0\]

giving a first integral

\[y^{\prime} \frac{\partial f}{\partial y^{\prime}}-f=\mathrm{constant.}\]

For the soap film between two rings problem,

\[f\left(y, y^{\prime}\right)=y \sqrt{1+y^{\prime 2}}\]

so the Euler-Lagrange equation is

\[\sqrt{1+y^{\prime 2}}-\frac{d}{d x} \frac{y y^{\prime}}{\sqrt{1+y^{\prime 2}}}=0\]

and has first integral

\[y^{\prime} \frac{\partial f}{\partial y^{\prime}}-f=\frac{y y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}-y \sqrt{1+y^{\prime 2}}=-\frac{y}{\sqrt{1+y^{\prime 2}}}=\mathrm{constant.}\]

We’ll write

\[\frac{y}{\sqrt{1+y^{\prime 2}}}=a\]

with a the constant of integration, which will depend on the endpoints.

This is a first-order differential equation, and can be solved.

Rearranging,

\[\frac{d y}{d x}=\sqrt{\left(\frac{y}{a}\right)^{2}-1}\]

\[d x=\frac{a d y}{\sqrt{y^{2}-a^{2}}}\]

The standard substitution here is \(y=a \cosh \xi\) from which

\[y=a \cosh \left(\frac{x-b}{a}\right)\]

Here \(b\) is the second constant of integration, the fixed endpoints determine \(a,b\).