Question

Find the equation of the path joining the origin \(O\) to the point \(P(1,1)\) in the \(x y\) plane that makes the integral \(\int_{o}^{P}\left(y^{2}+y y^{\prime}+y^{2}\right) d x\) stationary.

Short answer

The steps lead to a contradiction, requiring re-evaluation or problem clarification.

Step-by-step solution

Identify the objective functional

The functional that should be minimized or maximized is given by \[J[y] = \int_{0}^{1} (y^2 + y y' + y^2) \, dx.\]Simplify the integrand by combining like terms:\[J[y] = \int_{0}^{1} (2y^2 + y y') \, dx.\]

Apply the Euler-Lagrange equation

We use the Euler-Lagrange equation for functional derivatives, given by:\[\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) - \frac{\partial L}{\partial y} = 0,\]where \(L = 2y^2 + y y'\). First, find \(\frac{\partial L}{\partial y'}\) and \(\frac{\partial L}{\partial y}\):\[\frac{\partial L}{\partial y'} = y\]\[\frac{\partial L}{\partial y} = 4y + y'.\]

Differentiate \(\frac{\partial L}{\partial y'}\)

Calculate the total derivative of \(\frac{\partial L}{\partial y'} = y\) with respect to \(x\):\[\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) = y'.\]

Substitute into Euler-Lagrange equation

Now plug the derivatives back into the Euler-Lagrange equation:\[y' - (4y + y') = 0.\]Simplify the equation:\[y' - 4y = 0.\]

Solve the differential equation

Solve the first-order linear ordinary differential equation:\[y' = 4y.\]The solution is a standard exponential function given by:\[y = Ce^{4x}.\]

Apply boundary conditions

Given the path goes from \((0,0)\) to \((1,1)\), apply these boundary conditions. Specifically, at \(x = 0\), \(y = 0\):\[0 = Ce^{0} \rightarrow C = 0.\]This is a contradiction because it implies \(y = 0\) along the entire path, conflicting with the final point \((1,1)\). Consider re-checking the problem statement or given information. Another interpretation may require finding both constant and variable solutions.

Key concepts

Functional Derivatives

Functional derivatives are a critical concept in solving problems involving functionals. A functional is a mapping from a space of functions to the real numbers. In the problem you're working on, the functional is \[ J[y] = \int_{0}^{1} (2y^2 + yy') \, dx. \]Functional derivatives help us determine how this functional changes with slight variations in the function\( y(x) \).

To find the functional derivative, we use the Euler-Lagrange equation, which is central in calculus of variations. This equation helps identify functions that make a given functional stationary (usually a minimum or maximum). The Euler-Lagrange equation is \[\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) - \frac{\partial L}{\partial y} = 0,\] where \( L \) is the integrand of our functional. This equation essentially provides us with a road map to find optimal paths or surfaces that are crucial in physics and engineering.

Understanding how to take functional derivatives and apply the Euler-Lagrange equation allows one to solve a variety of optimization problems, including ones in physics like finding the shortest path or least action.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve functions and their derivatives. They are pivotal in expressing and solving problems related to rates of change. In the context of the Euler-Lagrange equation, solving ODEs is a step towards finding functions that make a given functional stationary.

In our exercise, after applying the Euler-Lagrange equation to the functional, we end up with the differential equation \( y' - 4y = 0 \). This is a first-order linear ODE, which is usually straightforward to solve.

• The solution to this type of ODE is generally an exponential function.
• Such problems often hinge on applying boundary conditions to refine the constants in the solution.

This means that despite finding a general solution, you need additional information to find the specific solution relevant to the problem at hand. In our case, to match the path from \((0,0)\) to \((1,1)\), we had a contradiction which means we need to re-evaluate our approach or conditions.

Path Optimization

Path optimization refers to finding the best possible path between two points, given certain criteria or constraints. This is a key aspect of many engineering and physics problems and comes into play when dealing with the calculus of variations.

The simplified problem is finding a path where a functional (often representing energy, time, or another quantity) is minimized or maximized. For our exercise, optimizing the path between the origin and point \((1,1)\) involves ensuring that the integral \[ \int_{0}^{1} (2y^2 + y y') \, dx \] is stationary.

• This involves applying boundary conditions and verifying the viability of solutions over the course of optimizing.
• Often this requires creative thinking or reconsidering constants if initial assumptions are incorrect.

The eventual outcome is to not only calculate minimized paths but also check for practical consistency with given or real-world conditions.