Question

Find and describe the path \(y=y(x)\) for which the integral \(\int_{x_{1}}^{x_{2}} \sqrt{x} \sqrt{1+y^{\prime 2}} d x\) is stationary.

Short answer

The path is a linear function, representing a straight line.

Step-by-step solution

- Understand the Problem

The task is to find the function that makes the given integral stationary. This typically involves calculus of variations, where we need to find a function that will make an integral take on an extremum value.

- Identify the Functional

The integral is given by \( \int_{x_{1}}^{x_{2}} \sqrt{x} \sqrt{1+y^{\prime 2}} \, dx \). In this problem, the functional to minimize or maximize is \( F(x, y, y') = \sqrt{x} \sqrt{1+y^{\prime 2}} \).

- Apply Euler-Lagrange Equation

The Euler-Lagrange equation is \( \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) - \frac{\partial F}{\partial y} = 0 \). We will apply this to our functional to find the desired path.

- Compute Partial Derivatives

Calculate the partial derivatives. Since \( F \) does not depend on \( y \), \( \frac{\partial F}{\partial y} = 0 \). For \( y' \), we have \( F = \sqrt{x} \sqrt{1 + (y')^2} \). The partial derivative is \( \frac{\partial F}{\partial y'} = \frac{y' \sqrt{x}}{\sqrt{1 + (y')^2}} \).

- Simplify Euler-Lagrange Equation

Substitute the partial derivatives into the Euler-Lagrange equation. Since \( \frac{\partial F}{\partial y} = 0 \), it simplifies to \( \frac{d}{dx} \left( \frac{y' \sqrt{x}}{\sqrt{1 + (y')^2}} \right) = 0 \).

- Solve the Differential Equation

Since the derivative \( \frac{d}{dx}(...) = 0 \), \( \frac{y' \sqrt{x}}{\sqrt{1 + (y')^2}} \) must be a constant, say \( c \). Simplify this equation to find \( y' = c \sqrt{1 + (y')^2} / \sqrt{x} \).

- Integrate to Find y(x)

Solve for \( y' \) to integrate and find \( y(x) \). This can result in a differential equation that can be separated and integrated to find the form of \( y \), generally a linear function since \( c \) is a constant, implying \( y' \) is constant.

Key concepts

Euler-Lagrange equation

In calculus of variations, the Euler-Lagrange equation is a fundamental tool. It helps us find the function that makes a given integral stationary, meaning that function either minimizes or maximizes the integral. The process involves a functional, which is a function of other functions. The Euler-Lagrange equation is given as:\[\frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) - \frac{\partial F}{\partial y} = 0,\]where \( F(x, y, y') \) is the integrand, or the function inside the integral. A functional becomes stationary when it satisfies this condition. The essential part is identifying \( F \), which is derived from the integrand. Then, the partial derivatives \( \frac{\partial F}{\partial y} \) and \( \frac{\partial F}{\partial y'} \) are computed for substitution into the Euler-Lagrange equation.Understanding this equation allows us to transition from the integral to a differential equation. This step is crucial because solving this differential equation reveals the function \( y = y(x) \) that gives the stationary integral.

Stationary Integral

The idea of a stationary integral is central to problems solved by calculus of variations. But what exactly is a stationary integral? It means the value of the integral remains constant for small variations in the function, often sought at maximal or minimal values.In our exercise, the integral \(\int_{x_{1}}^{x_{2}} \sqrt{x} \sqrt{1+y'^2} \, dx\) must be stationary. To find this, we explore the functional \( F(x, y, y') \), which simplifies searches in function space. Achieving stationarity involves setting up the Euler-Lagrange equation and solves for \( y(x) \). The functional serves as a lens through which we view the behavior of many functions under the constraint imposed by the Euler-Lagrange equation.Thus, a stationary integral highlights the path independent of small variations, focusing our search on extremal paths, be they minima or maxima.

Differential Equation Solving

Differential equation solving is the final piece in finding the function that makes an integral stationary. Once we have the differential equation from the Euler-Lagrange process, the next goal is to solve it.The differential equation derived from our problem was \(\frac{d}{dx} \left( \frac{y' \sqrt{x}}{\sqrt{1 + (y')^2}} \right) = 0\). When a function's derivative equals zero, this suggests the function itself is constant. Therefore, \(\frac{y' \sqrt{x}}{\sqrt{1 + (y')^2}} = c\), where \( c \) is a constant, results in a solvable form of \( y' = c \sqrt{1 + (y')^2} / \sqrt{x} \).Solving differential equations often involves simple integration, factoring, or even separating variables. In our case, the solution indicated a linear relationship as \( y' \) itself turned out to be constant, presenting \( y(x) \) as a simple straight line. Thus, differential equation solving translates the abstract possibilities into concrete functions fulfilling the desired criteria of stationarity.