Question

Write and solve the Euler equations to make the following integrals stationary. In solving the Euler equations, the integrals in Chapter 5, Section 1, may be useful.

1.${\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\sqrt{\mathbf{x}}\sqrt{\mathbf{1}\mathbf{+}\mathbf{y}{\mathbf{\text{'}}}^{\mathbf{2}}}\mathit{d}\mathit{x}$

Short answer

The Euler equation for the given integral${\int }_{x_1}^{x_2}\sqrt{x}\sqrt{1+y{\text{'}}^2}dx$ is ${\left(y-B\right)}^2=4C^2\left(x-C^2\right)$.

Step-by-step solution

Given Information.

Thegiven integral is ${\int }_{x_1}^{x_2}\sqrt{x}\sqrt{1+y{\text{'}}^2}dx$.

Definition ofEuler equations

The solutions of the Euler-Lagrange equations, which are stationary points of the defined action functional in the calculus of variations and classical mechanics, are a set of second-order ordinary differential equations.

Write and solve Euler equation.

Let $F=\sqrt{x}\sqrt{1+y{\text{'}}^2}$

First, write the Euler equation as $\frac{d}{dx}\frac{\partial F}{\partial y\text{'}}-\frac{\partial F}{\partial y}=0$.

Now calculate the required derivatives.

$$\begin{gathered} \frac{\partial F}{\partial y\text{'}}=\frac{\sqrt{x}y\text{'}}{\sqrt{1+y{\text{'}}^2}} \\ \frac{\partial F}{\partial y}=0 \end{gathered}$$

Further, there is no need to calculate the derivative with respect to $x$because it is zero in the context of the Euler equation and therefore the whole expression is constant.

$$\begin{gathered} \frac{d}{dx}\left[\frac{\sqrt{x}y\text{'}}{\sqrt{1+y{\text{'}}^2}}\right]=0 \\ \frac{\sqrt{x}y\text{'}}{\sqrt{1+y{\text{'}}^2}}=C \end{gathered}$$

Solve for $y\text{'}$. Square both sides of the equation and multiply by denominator to obtain:

$$\begin{gathered} xy{\text{'}}^2=C^2\left(1+y{\text{'}}^2\right) \\ xy{\text{'}}^2-C^{2}y{\text{'}}^2=C^2 \\ \left(x-C^2\right)y{\text{'}}^2=C^2 \\ y{\text{'}}^2=\frac{C^2}{x-C^2} \end{gathered}$$

Therefore,

$y\text{'}=\pm \frac{C^2}{\sqrt{x-C^2}}$

Integrate the expression to obtain $B$.

$$\begin{gathered} y=\int \frac{C^2}{\sqrt{x-C^2}}dx \\ y=2C\sqrt{x-C^2}+B \end{gathered}$$

Let’s rewrite the expression in a simple way by moving $B$to the left side and then square both the sides.

$$\begin{gathered} y-B=2C\sqrt{x-C^2} \\ {\left(y-B\right)}^2=4C^2\left(x-C^2\right) \end{gathered}$$which corresponds to a parabola.

Therefore, the Euler equation is ${\left(y-B\right)}^2=4C^2\left(x-C^2\right)$.