Question

The purpose of this problem is to derive the shortest path \(y(x)\) between the points \(\left(x_{0}, y_{0}\right)\) and \(\left(x_{1}, y_{1}\right) .\) Consider an arbitrary path between these points as shown in the figure. We can break the path into differential elements \(d \vec{l}=d x \hat{a}_{x}+d y \hat{a}_{y} .\) The magnitude of each differential element is $$|d \vec{l}|=\sqrt{(d x)^{2}+(d y)^{2}}=d x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}$$ The distance between the points can be described by the action $$\mathbb{S}=\int_{x_{0}}^{x_{1}} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x .$$ To find the path \(y(x)\) that minimizes the action, we can solve the Euler- Lagrange equation, with \(\mathcal{L}=\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\) as the Lagrangian, for this shortest path \(y(x)\). This approach can be used because we want to minimize the integral of some functional \(\mathcal{L}\) even though this functional does not represent an energy difference \([163,\) p. 33\(]\). Set up the Euler-Lagrange equation, and solve it for the shortest path, \(y(x)\).

Short answer

The shortest path between two points is a straight line.

Step-by-step solution

Set Up the Problem

To find the shortest path between two points using the Euler-Lagrange equation, we need to first identify the Lagrangian, which is given as \(\mathcal{L} = \sqrt{1 + \left( \frac{dy}{dx} \right)^2 }\). We are tasked with finding the path \(y(x)\) that minimizes the action \(\mathbb{S} = \int_{x_0}^{x_1} \mathcal{L} \, dx\).

Formulate the Euler-Lagrange Equation

The general form of the Euler-Lagrange equation is given by \( \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial \left( \frac{dy}{dx} \right)} \right) = \frac{\partial \mathcal{L}}{\partial y} \). For \(\mathcal{L} = \sqrt{1 + \left( \frac{dy}{dx} \right)^2 }\), we calculate its partial derivatives.

Compute the Partial Derivatives

Compute \(\frac{\partial \mathcal{L}}{\partial y} = 0\) since \(\mathcal{L}\) does not explicitly depend on \(y\). Next, compute \(\frac{\partial \mathcal{L}}{\partial \left( \frac{dy}{dx} \right)} = \frac{\frac{dy}{dx}}{\sqrt{1 + \left( \frac{dy}{dx} \right)^2 }}\).

Solve the Euler-Lagrange Equation

Substitute the derivatives into the Euler-Lagrange equation: \(\frac{d}{dx} \left( \frac{\frac{dy}{dx}}{\sqrt{1 + \left( \frac{dy}{dx} \right)^2 }} \right) = 0\). This implies \( \frac{\frac{dy}{dx}}{\sqrt{1 + \left( \frac{dy}{dx} \right)^2 }} = C \), where \(C\) is a constant.

Simplify and Integrate

The equation \( \frac{\frac{dy}{dx}}{\sqrt{1 + \left( \frac{dy}{dx} \right)^2 }} = C \) simplifies to \( \left( \frac{dy}{dx} \right)^2 = \frac{C^2}{1 - C^2} \). Solving, we have \( \frac{dy}{dx} = \pm \frac{C}{\sqrt{1 - C^2}} \). Integrating both sides, we get \( y = \pm \frac{C}{\sqrt{1 - C^2}} x + D \) where \(D\) is the integration constant.

Determine the Path Equation

The solution \( y = \pm \frac{C}{\sqrt{1 - C^2}} x + D \) represents a straight line equation: \( y = mx + b \) where slope \(m = \pm \frac{C}{\sqrt{1 - C^2}}\) and intercept \(b = D\), confirming that the shortest path is a straight line between the two points.

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental equation in calculus of variations. It provides a necessary condition for a solution to be an extremum (minimum or maximum) of a functional. Imagine a scenario where you have a functional \[ \mathbb{S}[y] = \int_{x_0}^{x_1} \mathcal{L}(y, y', x) \, dx \]which you are aiming to optimize. Here, \(\mathcal{L}\) is known as the Lagrangian, and it depends on the variable \(y\), its derivative \(y' = \frac{dy}{dx}\), and \(x\).

To find the function \( y(x) \) that minimizes or maximizes \(\mathbb{S}\), you apply the Euler-Lagrange equation, given by:\[ \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial y'} \right) = \frac{\partial \mathcal{L}}{\partial y} \]It involves calculating the partial derivatives of the Lagrangian \(\mathcal{L}\) with respect to \(y\) and its derivative \(y'\).

In the context of the shortest path problem, the Euler-Lagrange equation was used to minimize the integral representing the path length between two points. The Lagrangian in this problem was \(\mathcal{L} = \sqrt{1 + \left( \frac{dy}{dx} \right)^2 }\), making calculations straightforward due to the absence of \(y\) in \(\mathcal{L}\). Apply the equation correctly, and you are guided to the solution that the shortest path is indeed a straight line.

Shortest Path Problem

The shortest path problem revolves around determining the most efficient path between two points. Given two points in a plane, such as \((x_0, y_0)\) and \((x_1, y_1)\), the task is to find the path \(y(x)\) that minimizes the total travel distance.

To solve this, we express the path in terms of its differential elements. Each segment of the path can be described with a differential arc length \(d \vec{l} = dx \hat{a}_x + dy \hat{a}_y\), whose magnitude can be written as:\[|d \vec{l}| = \sqrt{(dx)^2 + (dy)^2} = dx \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\]

The shortest path is then obtained by minimizing the total length, represented by the integral:\[\mathbb{S} = \int_{x_0}^{x_1} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]Using the Euler-Lagrange equation, the functional form describes minimizing the integral by solving for \(y(x)\). In the simplest case, it reveals that the shortest path between two points in a plane is a straight line, described by the equation \(y = mx + b\), confirming our intuitive understanding of geometry.

Lagrangian Mechanics

Lagrangian mechanics is a powerful reformulation of classical mechanics. It uses the principle of least action to provide insights into physical systems. The central component of this formalism is the Lagrangian, \(\mathcal{L}\), which is a function that summarizes the dynamics of a system.

In the shortest path problem, the Lagrangian was used to capture the essence of minimizing path length. While in physics, the Lagrangian typically represents the difference between kinetic and potential energy, in this mathematical problem, it encapsulates the quantity of interest: the arc length.

• Lagrangian Definition: In mechanical terms, \(\mathcal{L} = T - V\), where \(T\) is kinetic energy and \(V\) is potential energy.
• Principle of Least Action: The path between two configurations of a system is the one for which the action integral is stationary.

Calculus of variations extends these ideas, allowing the application of Lagrangian principles to a variety of fields beyond physics. It shines in problems involving optimizing functionals for any aspect of interest. This demonstrates the inherent flexibility and the wide range of applicability of Lagrangian techniques, such as in path optimization.