Treacherous Euler-Lagrange equation(y′)2=2(1−cos⁡(y)) where y is a function of x subjected to boundary conditions y(x)→0 as x→−∞ and y(x)→2π as x→+∞, how might I find all its solutions?

Name: Treacherous Euler-Lagrange equation(y′)2=2(1−cos⁡(y)) where y is a function of x subjected to boundary conditions y(x)→0 as x→−∞ and y(x)→2π as x→+∞, how might I find all its solutions?
Uploaded: 2022-02-23T17:22:57+00:00
Description: Treacherous Euler-Lagrange equation(y′)2=2(1−cos⁡(y)) where y is a function of x subjected to boundary conditions y(x)→0 as x→−∞ and y(x)→2π as x→+∞, how might I find all its solutions?

Question

Treacherous Euler-Lagrange equation(y′)2=2(1−cos⁡(y)) where y is a function of x subjected to boundary conditions y(x)→0 as x→−∞ and y(x)→2π as x→+∞, how might I find all its solutions?

Karen Robbins · Accepted Answer

2y″(x)y′(x)=2y′(x)sin⁡y(x)which implies y″(x)=sin⁡y(x). After substitution y(x)=π−θ(x) this translates into θ″(x)=−sin⁡θ(x) which is the pendulum equation. Your boundary condition require that limx→−∞θ(x)=π and limx→+∞θ(x)=−π. Hence the solution is not periodic.

Becky Harrison · Accepted Answer

Use 1−cos⁡y=1−(cos2y2−sin2y2)=2sin2y2.

alenahelenash · Accepted Answer

1−cos⁡(y)=2sin2⁡(y2). Hence, y′2=4sin2⁡(y2)⇒y′=±2sin⁡(y2)