2.4: An Important First Integral of the Euler-Lagrange Equation
- Page ID
- 29526
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(y^{\prime}=d y / d x\)
\[\dfrac{\partial f\left(y, y^{\prime}\right)}{\partial y} \dfrac{d y}{d x}-\dfrac{d}{d x}\left(\dfrac{\partial f\left(y, y^{\prime}\right)}{\partial y^{\prime}}\right) y^{\prime}=0\]
Since \(f\) doesn’t depend explicitly on \(x\), we have
\[\dfrac{d f}{d x}=\dfrac{\partial f}{\partial y} \dfrac{d y}{d x}+\dfrac{\partial f}{\partial y^{\prime}} \dfrac{d y^{\prime}}{d x}\]
and using this to replace \(\dfrac{\partial f\left(y, y^{\prime}\right)}{\partial y} \dfrac{d y}{d x}\) in the preceding equation gives
\[\dfrac{d f}{d x}-\dfrac{\partial f}{\partial y^{\prime}} \dfrac{d y^{\prime}}{d x}-\dfrac{d}{d x}\left(\dfrac{\partial f\left(y, y^{\prime}\right)}{\partial y^{\prime}}\right) y^{\prime}=0\]
then multiplying by − (to match the equation as usually written) we have
\[\dfrac{d}{d x}\left(y^{\prime} \dfrac{\partial f}{\partial y^{\prime}}-f\right)=0\]
giving a first integral
\[y^{\prime} \dfrac{\partial f}{\partial y^{\prime}}-f=\mathrm{constant.}\]
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
\[\sqrt{1+y^{\prime 2}}-\dfrac{d}{d x} \dfrac{y y^{\prime}}{\sqrt{1+y^{\prime 2}}}=0\]
and has first integral
\[y^{\prime} \dfrac{\partial f}{\partial y^{\prime}}-f=\dfrac{y y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}-y \sqrt{1+y^{\prime 2}}=-\dfrac{y}{\sqrt{1+y^{\prime 2}}}=\mathrm{constant.}\]
We’ll write
\[\dfrac{y}{\sqrt{1+y^{\prime 2}}}=a\]
with a the constant of integration, which will depend on the endpoints.
This is a first-order differential equation, and can be solved.
Rearranging,
\[\dfrac{d y}{d x}=\sqrt{\left(\dfrac{y}{a}\right)^{2}-1}\]
or
\[d x=\dfrac{a d y}{\sqrt{y^{2}-a^{2}}}\]
The standard substitution here is \(y=a \cosh \xi\) from which
\[y=a \cosh \left(\dfrac{x-b}{a}\right)\]
Here \(b\) is the second constant of integration, the fixed endpoints determine \(a,b\).