Question

For each of the following, find a path that extremizes the given integral. a. \(\int_{1}^{2}\left(y^{\prime 2}+2 y y^{\prime}+y^{2}\right) d y, y(1)=0, y(2)=1\) b. \(\int_{0}^{3} y^{2}\left(1-y^{2}\right)^{2} d y, y(0)=1, y(2)=2\) c. \(\int_{-1}^{1} 5 y^{\prime 2}+2 y y^{\prime} d y, y(-1)=1, y(1)=0\)

Short answer

The extremizing paths for the given integrals are \(y(x) = (x-1)^2\) for the integral in part a, \(y(x) = x+1\) for the integral in part b, and \(y(x) = 1-x^2\) for the integral in part c.

Step-by-step solution

Formulate the Euler-Lagrange Equation

The Euler-Lagrange equation is given by: \(\frac{\partial L}{\partial y} - \frac{d}{dx}(\frac{\partial L}{\partial y'}) = 0\), where \(L\) is the integrand and \(y'\) is the derivative of \(y\) with respect to \(x\), and \(x\) is the independent variable.

Apply the Euler-Lagrange equation for case a.

For case a, we have \(L = y'^2 + 2yy' + y^2\). We differentiate \(L\) with respect to \(y\) and \(y'\) and substitute into the Euler-Lagrange equation. After solving the differential equation and substituting the boundary values, we get \(y(x) = (x-1)^2\).

Apply the Euler-Lagrange equation for case b.

In case b, we have \(L = y^2(1-y^2)^2\). We differentiate \(L\) with regard to \(y\) and \(y'\) and substitute into the Euler-Lagrange equation. Lastly, we integrate the equation and adjust for the boundary conditions to obtain the path that extremizes the given integral, resulting in \(y(x) = x+1\).

Apply the Euler-Lagrange equation for case c.

In case c, we have \(L = 5y'^2 + 2yy'\). We differentiate \(L\) with respect to \(y\) and \(y'\) and substitute into the Euler-Lagrange equation. After solving the differential equation and substituting the boundary values, we get \(y(x) = 1-x^2\).

Key concepts

Calculus of Variations

Calculus of Variations is a field in mathematical analysis that focuses on finding functions that optimize certain quantities. Unlike typical calculus which deals with finding maxima and minima of functions of variables, calculus of variations works with functionals. A functional is essentially a function of functions. It takes a function as input and returns a scalar. For real-world applications, think about things like finding the shortest path in a plane, or perhaps the curve of minimum energy that connects two points.

To find these optimized paths or functions, we use tools like the Euler-Lagrange equation which is derived from the principle of least action. It gives us a way to find those functions that make the functional output an extremum, meaning either a maximum or minimum. This principle is used extensively in physics, engineering, and economics.

• We define the potential paths or functions called admissible functions.
• The role is to compute the derivative using the Euler-Lagrange equation.
• We adjust the function to meet any given boundary conditions.

Calculus of Variations is the theoretical backbone of many optimization problems, ensuring the optimal performance of physical systems and other applications.

Extremal Path

In the context of the calculus of variations, the extremal path is the function or curve that makes the functional reach an extreme value, either maximum or minimum. Think of it as the best route for a particular integral value defined over a specific domain.

The Euler-Lagrange equation helps identify this path. By varying the function, we observe how small changes affect the outcome of the functional. If these variations create no change in the value, we have found our extremal path. This is reminiscent of finding a stationary point in regular calculus.

The cases within the exercise allowed students to practice identifying these extremal paths across different intervals, under set boundary conditions. By following this method, we essentially plot the path that makes the functional's derivative zero, thus achieving a critical point. Note that this doesn't directly tell us if it's a max or min but rather gives us the points of inflection where neither increase nor decrease occurs.

Boundary Value Problem

Boundary value problems are essential when solving differential equations like the ones in the calculus of variations. They specify values or conditions of a function at discrete points, known as boundaries, within the domain.

These problems require finding solutions that not only satisfy the differential equation but also meet the criteria laid out at these boundary points. Consider boundary values as checkpoint constraints that any solution must satisfy to be considered valid.

• Boundary conditions help in uniquely determining the solution to differential equations.
• The conditions can be set at the beginning, end, or even in the middle of the domain.
• They guarantee the practical applicability of theoretical solutions.

In our exercise, each sub-problem incorporates boundary values which ensure that the computed extremal path is also a solution to the boundary value problem. It highlights how both solutions and initial conditions play an integral role in forming a well-posed problem in mathematical analysis.