Question

Find a first integral of the Euler equation to make stationary the integrals. $$\int_{a}^{b} \frac{y y^{\prime 2} d x}{\sqrt{1+y^{\prime 2}}}$$

Short answer

\[ \frac{2y y'}{(1 + (y')^2)^{3/2}} = C \]

Step-by-step solution

Formulate the Lagrangian

Identify the integrand in the given integral as the Lagrangian, which is a function of y and y'. Here, the Lagrangian is: \ \[ L(y, y') = \frac{y (y')^2}{\sqrt{1 + (y')^2}} \ \]

Euler-Lagrange Equation

Write down the Euler-Lagrange equation specific to the given Lagrangian: \ \[ \frac{\text{d}}{\text{d}x} \frac{\text{d}L}{\text{d}y'} - \frac{\text{d}L}{\text{d}y} = 0 \ \]

Compute the Partial Derivatives

Compute the partial derivatives: \ \[ \frac{\text{d}L}{\text{d}y} = \frac{(y')^2}{\sqrt{1 + (y')^2}} \ \] \ \[ \frac{\text{d}L}{\text{d}y'} = \frac{2y y' \sqrt{1 + (y')^2} - y (y')^3 \frac{1}{2 (1 + (y')^2)^{-1/2}} }{(1 + (y')^2)^{3/2}} = \frac{2y y'}{(1 + (y')^2)^{3/2}} \ \]

Substitute into Euler-Lagrange Equation

Substitute the partial derivatives into the Euler-Lagrange equation to get: \ \[ \frac{\text{d}}{\text{d}x} \left( \frac{2y y'}{(1 + (y')^2)^{3/2}} \right) - \frac{(y')^2}{\sqrt{1 + (y')^2}} = 0 \ \]

First Integral

The equation simplifies because the term involving \(\frac{\text{d}}{\text{d}x}\) suggests that the expression inside the parenthesis must be a constant: \ \[ \frac{2y y'}{(1 + (y')^2)^{3/2}} = C \ \] where C is a constant.

Key concepts

calculus of variations

The calculus of variations is a field of mathematical analysis that deals with optimizing functionals. Functionals are mappings from a set of functions to the real numbers. In simpler terms, it means finding a function that makes the value of an integral the largest or smallest possible.

In our exercise, we aim to make a specific integral stationary, meaning we want to find the function that either maximizes or minimizes this integral value. This involves finding a function that satisfies the Euler-Lagrange equation, which is a necessary condition for an extremum in calculus of variations. By solving this, we can deduce the first integral that makes our integral stationary.

first integral

A first integral simplifies our original differential equation and helps us solve for the necessary conditions. In Lagrangian mechanics, a first integral can often be related to conservation laws, such as the conservation of energy or momentum.

For our problem, we reduce the complex Euler-Lagrange equation to a simpler form involving a constant. This simpler form is referred to as a first integral. Specifically, we identify: \( \frac{2y y'}{(1 + (y')^2)^{3/2}} = C \), where C is a constant. Recognizing this first integral is crucial because it gives us a direct route to solving our problem.

Lagrangian mechanics

Lagrangian mechanics is a reformulation of classical mechanics introduced by Joseph Louis Lagrange. The main idea is to derive the equations of motion using the Lagrangian, a function that summarizes the dynamics of the system.

The Lagrangian is typically defined as the kinetic energy minus the potential energy. In variational problems, we use the Lagrangian to derive the Euler-Lagrange equation, which provides the conditions necessary for finding the function that optimizes our integral.

For our specific problem, we identified the Lagrangian as: \( L(y, y') = \frac{y (y')^2}{\text{sqrt}(1 + (y')^2)} \). This function is fundamental in setting up the conditions for stationarity.

partial derivatives

Partial derivatives are essential in multivariable calculus. They measure how a function changes as one of its input variables changes, while keeping the other variables constant.

In the context of the Euler-Lagrange equation, we need to take partial derivatives of the Lagrangian with respect to both the function y and its derivative y'. This helps set up the equation used to find the function that makes our integral stationary.

For our problem, the partial derivatives were computed as: \( \frac{\text{d}L}{\text{d}y} = \frac{(y')^2}{\text{sqrt}(1 + (y')^2)} \) and \( \frac{\text{d}L}{\text{d}y'} = \frac{2y y'}{(1 + (y')^2)^{3/2}} \). Calculating these derivatives correctly is crucial for accurately solving the Euler-Lagrange equation.