91Ó°ÊÓ

The given integral is${∫}_{{\mathrm{Ï•}}_1}^{{\mathrm{Ï•}}_2}\sqrt{\mathrm{θ}{\text{'}}^2+{\sin }^2\mathrm{θ}}d\mathrm{Ï•},\mathrm{θ}\text{'}=\frac{\mathrm{»åθ}}{\mathrm{»åÏ•}}$.The given integral is to be made stationary by using Euler equations.

In the calculus of variations andclassical mechanics, the Euler equations is a system of second-orderordinary differential equations whose solutions arestationary points of the givenaction functional.

The given integral is${∫}_{{\mathrm{Ï•}}_1}^{{\mathrm{Ï•}}_2}\sqrt{\mathrm{θ}{\text{'}}^2+{\sin }^2\mathrm{θ}}d\mathrm{Ï•},\mathrm{θ}\text{'}=\frac{\mathrm{»åθ}}{\mathrm{»åÏ•}}$

Euler equation for polar coordinates$\left(\mathrm{θ},\mathrm{ϕ}\right)$is$\frac{d}{d\mathrm{θ}}\left(\frac{∂F}{∂\mathrm{ϕ}\text{'}}\right)-\frac{∂F}{∂\mathrm{ϕ}}=0$ where $\mathrm{θ}\text{'}=\frac{d\mathrm{θ}}{d\mathrm{ϕ}}$

$$\begin{gathered} \mathrm{θ}\text{'}=\frac{d\mathrm{θ}}{d\mathrm{ϕ}} \\ ⇒\mathrm{θ}\text{'}=\frac{1}{\mathrm{ϕ}\text{'}} \end{gathered}$$

Use the two equations above to change the given integral

$$\begin{gathered} ∫\sqrt{\mathrm{θ}{\text{'}}^2+{\sin }^2\mathrm{θ}}d\mathrm{ϕ}=∫\sqrt{\mathrm{θ}{\text{'}}^2+{\sin }^2\mathrm{θ}}\mathrm{ϕ}\text{'}d\mathrm{θ} \\ =∫\sqrt{\mathrm{ϕ}{\text{'}}^{-2}+{\sin }^2\mathrm{θ}}\mathrm{ϕ}\text{'}d\mathrm{θ} \\ =∫\sqrt{1+\mathrm{ϕ}{\text{'}}^2+{\sin }^2\mathrm{θ}}d\mathrm{θ} \end{gathered}$$

Now, let$F=\sqrt{1+\mathrm{ϕ}{\text{'}}^2+{\sin }^2\mathrm{θ}}$.

Use the Euler equation , $\frac{d}{d\mathrm{θ}}\left(\frac{∂F}{∂\mathrm{ϕ}\text{'}}\right)-\frac{∂F}{∂\mathrm{ϕ}}=0$

Calculate the required derivatives

$$\begin{gathered} \frac{∂F}{∂\mathrm{ϕ}\text{'}}=\frac{\mathrm{ϕ}\text{'}{\sin }^2\mathrm{θ}}{\sqrt{1+\mathrm{ϕ}{\text{'}}^2{\sin }^2\mathrm{θ}}} \\ \frac{∂F}{∂\mathrm{ϕ}}=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{d\mathrm{θ}}\left[\frac{\mathrm{ϕ}\text{'}{\sin }^2\mathrm{θ}}{\sqrt{1+\mathrm{ϕ}{\text{'}}^2{\sin }^2\mathrm{θ}}}\right]=0 \\ ⇒\frac{\mathrm{ϕ}\text{'}{\sin }^2\mathrm{θ}}{\sqrt{1+\mathrm{ϕ}{\text{'}}^2{\sin }^2\mathrm{θ}}}=C \\ ⇒\mathrm{ϕ}\text{'}=\frac{C}{\sin \mathrm{θ}\sqrt{{\sin }^2\mathrm{θ}-C^2}} \end{gathered}$$

Where C is constant.

Integratelocalid="1665053818462" $\mathrm{ϕ}\text{'}=\frac{C}{\sin \mathrm{θ}\sqrt{{\sin }^2\mathrm{θ}-C^2}}$to get the desired result

$$\begin{gathered} \mathrm{ϕ}\text{'}=∫\frac{C}{sin\mathrm{θ}\sqrt{si{n}^2\mathrm{θ}-C^2}}d\mathrm{θ} \\ =-\frac{CArcTanh\left({\displaystyle \frac{\sqrt{2}\sqrt{-C^2}cos\mathrm{θ}}{\sqrt{1-2C^2-cos2\mathrm{θ}}}}\right)}{\sqrt{-C^2}}+B \end{gathered}$$

Wherelocalid="1665053113180" $B$is integration constant.

The solution is real if

$1-2C^2-\cos 2\mathrm{θ}≤0$

This means that spectrum of $\mathrm{θ}$is bound and it can be moved by value of $C$

Therefore, $\mathrm{ϕ}=∫\frac{C}{\sin \mathrm{θ}\sqrt{{\sin }^2\mathrm{θ}-C^2}}d\mathrm{θ}=-\frac{CArcTanh\left({\displaystyle \frac{\sqrt{2}\sqrt{-C^2}\cos \mathrm{θ}}{\sqrt{1-2C^2-\cos 2\mathrm{θ}}}}\right)}{\sqrt{-C^2}}+B$, where$B$is the integration constant.