Treacherous Euler-Lagrange equation(y′)2=2(1−cos(y)) where y is a function of x subjected to boundary conditions y(x)→0...

Patricia Crane

Answered

2022-01-18

Treacherous Euler-Lagrange equation ${\left({y}^{\prime}\right)}^{2}=2(1-\mathrm{cos}\left(y\right))$ where y is a function of x subjected to boundary conditions $y\left(x\right)\to 0\text{}\text{as}\text{}x\to -\mathrm{\infty}\text{}\text{and}\text{}y\left(x\right)\to 2\pi \text{}\text{as}\text{}x\to +\mathrm{\infty}$, how might I find all its solutions?

Answer & Explanation

Karen Robbins

Expert

2022-01-19Added 49 answers

$2y{}^{\u2033}\left(x\right){y}^{\prime}\left(x\right)=2{y}^{\prime}\left(x\right)\mathrm{sin}y\left(x\right)$ which implies $y{}^{\u2033}\left(x\right)=\mathrm{sin}y\left(x\right)$. After substitution $y\left(x\right)=\pi -\theta \left(x\right)$ this translates into $\theta {}^{\u2033}\left(x\right)=-\mathrm{sin}\theta \left(x\right)$ which is the pendulum equation. Your boundary condition require that $\underset{x\to -\mathrm{\infty}}{lim}\theta \left(x\right)=\pi \text{}\text{and}\text{}\underset{x\to +\mathrm{\infty}}{lim}\theta \left(x\right)=-\pi$. Hence the solution is not periodic.

Becky Harrison

Expert

2022-01-20Added 40 answers

Use $1-\mathrm{cos}y=1-({\mathrm{cos}}^{2}\frac{y}{2}-{\mathrm{sin}}^{2}\frac{y}{2})=2{\mathrm{sin}}^{2}\frac{y}{2}$.