Question

Write and solve the Euler equations to make the following integrals stationary. Change the independent variable, if needed, to make the Euler equation simpler.

${\mathbf{\int }}_{{\mathbf{t}}_{\mathbf{1}}}^{{\mathbf{t}}_{\mathbf{2}}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\sqrt{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{s}{\mathbf{\text{'}}}^{\mathbf{2}}}\mathbf{dt}\mathbf{,}\mathbf{}\mathbf{s}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{ds}}{\mathbf{dt}}$

Short answer

$t=Cln\left(s^2+\sqrt{s^2-C^2}\right)+B$, where$B$ is the integration constant.

Step-by-step solution

Given Information.

The given integral is${\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{s}}^{-1}\sqrt{{\mathrm{s}}^2+\mathrm{s}{\text{'}}^2}\mathrm{dt},\mathrm{s}\text{'}=\frac{\mathrm{ds}}{\mathrm{dt}}$.The given integral is to be made stationary by using Euler equations.

Definition of Euler equation

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.

Use Euler equation

The given integral is ${\int }_{t_1}^{t_2}{s}^{-1}\sqrt{s^2+s{\text{'}}^2}dt,s\text{'}=\frac{ds}{dt}$

Euler equation for polar coordinates $\left(s,t\right)$is $\frac{d}{ds}\left(\frac{\partial F}{\partial t\text{'}}\right)-\frac{\partial F}{\partial t\text{'}}=0$where $s\text{'}=\frac{ds}{dt}$

$$\begin{gathered} t\text{'}=\frac{dt}{ds} \\ \Rightarrow s\text{'}=\frac{1}{t\text{'}} \end{gathered}$$

Use the two equations above to change the given integral

$$\begin{gathered} \int s^{-1}\sqrt{s^2+s{\text{'}}^2}dt=\int s^{-1}\sqrt{s^2+s{\text{'}}^2}t\text{'}ds \\ =\int s^{-1}\sqrt{s^2+t{\text{'}}^{-2}}t\text{'}ds \\ =\int \frac{\sqrt{s^{2}t{\text{'}}^2+1}}{s}ds \end{gathered}$$

Now, let$F=s^{-1}\sqrt{s^{2}t{\text{'}}^2+1}$.

Use the Euler equation , $\frac{d}{ds}\left(\frac{\partial F}{\partial t\text{'}}\right)-\frac{\partial F}{\partial t}=0$

Calculate the required derivatives

$$\begin{gathered} \frac{\partial F}{\partial t\text{'}}=\frac{s^{2}t\text{'}}{s\sqrt{s^{2}t{\text{'}}^2+1}} \\ =\frac{st\text{'}}{\sqrt{s^{2}t{\text{'}}^2+1}} \\ \frac{\partial F}{\partial t}=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{ds}\left[\frac{st\text{'}}{\sqrt{s^{2}t{\text{'}}^2+1}}\right]=0 \\ \Rightarrow \frac{st\text{'}}{\sqrt{s^{2}t{\text{'}}^2+1}}=C \\ \Rightarrow t{\text{'}}^2=\frac{C^2}{s^2-C^2} \end{gathered}$$

Where $C$is constant.

Integrate $t\text{'}=\frac{C}{\sqrt{s^2-C^2}}$to get the desired result

$$\begin{gathered} t=\int \frac{C}{\sqrt{s^2-C^2}}ds \\ =Cln\left(s^2+\sqrt{s^2-C^2}\right)+B \end{gathered}$$

Where $B$is integration constant.

Therefore, $t=Cln\left(s^2+\sqrt{s^2-C^2}\right)+B$, where$B$ is the integration constant.