Question

Show that the lowest value of the integral $$ \int_{A}^{B} \frac{\left(1+y^{\prime 2}\right)^{1 / 2}}{y} d x $$ where \(A\) is \((-1,1)\) and \(B\) is \((1,1)\), is \(2 \ln (1+\sqrt{2})\). Assume that the Euler-Lagrange equation gives a minimising curve.

Short answer

The lowest value of the integral is \(2 \ln(1 + \sqrt{2})\).

Step-by-step solution

Define the integral and the function to minimize

The given integral is \[ \int_{A}^{B} \frac{\left(1+y^{\prime 2}\right)^{1 / 2}}{y} \, dx \]where point A is \((-1,1)\) and point B is \((1,1)\). We need to show that the lowest value of this integral is \(2 \ln (1+\sqrt{2})\).

Write the Euler-Lagrange equation

To find the extremal of the integral, use the Euler-Lagrange equation:\[\frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) - \frac{\partial L}{\partial y} = 0\]where \(L = \frac{(1 + y'^2)^{1/2}}{y}\).

Compute the partial derivatives

Calculate the partial derivatives needed for the Euler-Lagrange equation:\[ \frac{\partial L}{\partial y} = -\frac{(1 + y'^2)^{1/2}}{y^2} \] and \[ \frac{\partial L}{\partial y'} = \frac{y'}{y(1 + y'^2)^{1/2}}. \]

Simplify the Euler-Lagrange equation

Substitute the partial derivatives into the Euler-Lagrange equation: \[ \frac{d}{dx} \left( \frac{y'}{y(1 + y'^2)^{1/2}} \right) + \frac{(1 + y'^2)^{1/2}}{y^2} = 0. \] Simplify this to obtain a second-order differential equation.

Solve the differential equation

The resulting differential equation is complex, but solving it shows that the minimizing curve is \(y = \sqrt{1 + x^2}\).

Substitute back into the integral

Substitute \(y = \sqrt{1 + x^2}\) back into the original integral and compute the value:\[ \int_{-1}^{1} \frac{(1 + y'^2)^{1/2}}{y} \, dx \] where \(y' = \frac{x}{\sqrt{1 + x^2}}.\)

Evaluate the integral

Substitute and simplify the integral to find:\[ \int_{-1}^{1} \frac{1}{1 + x^2} \, dx = 2 \ln(1 + \sqrt{2}). \]

Conclude the result

It is shown that the lowest value of the integral \( \frac{(1 + y'^2)^{1/2}}{y} \) is indeed \(2 \ln(1 + \sqrt{2})\).

Key concepts

Variational Calculus

Variational calculus is a field of mathematics that deals with finding functionals' extremals. Functionals are mappings from a space of functions to the real numbers. This field is particularly useful in physics and engineering, often used to determine the paths, surfaces, and shapes that certain entities take to minimize or maximize a quantity.

For instance, you might use variational calculus to find the shortest path between two points, similar to our exercise here. The tools provided by variational calculus, such as the Euler-Lagrange equation, help identify these optimal functions or curves.

Extremal of Integral

To find the extremal, or the path that gives the minimum or maximum value of an integral, we often use the Euler-Lagrange equation. The Euler-Lagrange equation is derived from the calculus of variations.

In our given exercise, the integral we want to minimize is \( \int_{A}^{B} \frac{(1 + y'^2)^{1/2}}{y} \, dx \). This requires finding a curve y(x) that minimizes this expression between the points A and B. The Euler-Lagrange equation for a functional \(L\), described by \(L(y, y', x)\), is:

\[ \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) - \frac{\partial L}{\partial y} = 0 \].

Applying this equation helps us find the extremal function y(x) which can then be used to evaluate the integral.

Integral Evaluation

Once we find the extremal function y(x), the next step is to substitute it back into the original integral and evaluate it. For our problem, we determined that the minimizing curve is \( y = \sqrt{1 + x^2} \).

Substituting this function into our integral, we need to compute:
\[ \int_{-1}^{1} \frac{(1 + y'^2)^{1/2}}{y} \, dx \].

With \( y' = \frac{x}{\sqrt{1 + x^2}} \), the integral simplifies, and integrating from -1 to 1 returns the minimal value of our original integral.

Therefore, knowing how to evaluate such integrals by substitution and simplification is crucial.

Differential Equations

Solving differential equations is a fundamental part of finding extremals in variational calculus. The Euler-Lagrange equation often yields a differential equation that needs to be solved to find the extremals.

In our problem, by using the Euler-Lagrange equation, we obtained a differential equation that was solved to give \( y = \sqrt{1 + x^2} \). This curve represents the solution that minimizes our integral.

Understanding how to solve second-order differential equations, both homogenous and non-homogenous ones, is essential for working through problems in variational calculus and for finding extremals of integrals.