Question

Bestimmen Sie stationäre Punkte \(x(t)\) des Funktionals $$ J(x)=\int_{0}^{T} \sqrt{1+\dot{x}^{2}} d t $$ wobei \(x(0)=0\) und für das Intervallende \(T\) die Bedingung \(x(T)=r(T)=\frac{1}{T^{2}}\) gelten sollen.

Short answer

The stationary path is \( x(t) = \frac{t}{T^3}\sqrt{1-(\frac{1}{T^2})^2} \).

Step-by-step solution

Understanding the Problem Statement

We need to find stationary points of the given functional \( J(x) = \int_{0}^{T} \sqrt{1 + \dot{x}^{2}} \, dt \). The boundary conditions given are \( x(0) = 0 \) and \( x(T) = \frac{1}{T^{2}} \). A stationary point is found by determining the path that minimizes or makes stationary this functional.

Setting Up the Euler-Lagrange Equation

The functional involves \( \dot{x}(t) \), so we use the Euler-Lagrange equation, which for a functional \( J(x) = \int L(x, \dot{x}, t) \, dt \) is: \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) = \frac{\partial L}{\partial x} \). In this case, the Lagrangian is \( L = \sqrt{1 + \dot{x}^{2}} \).

Calculating Partial Derivatives of the Lagrangian

Compute the partial derivatives: \( \frac{\partial L}{\partial \dot{x}} = \frac{\dot{x}}{\sqrt{1 + \dot{x}^{2}}} \) and \( \frac{\partial L}{\partial x} = 0 \), since \( L \) doesn't explicitly depend on \( x \).

Applying the Euler-Lagrange Equation

Since the equation is \( \frac{d}{dt}\left(\frac{\dot{x}}{\sqrt{1 + \dot{x}^{2}}}\right) = 0 \), this implies that \( \frac{\dot{x}}{\sqrt{1 + \dot{x}^{2}}} = C \), where \( C \) is a constant. This implies \( \dot{x} = \frac{C}{\sqrt{1-C^2}} \), a constant velocity.

Evaluating the Constants Using Boundary Conditions

From \( \dot{x} = \frac{C}{\sqrt{1-C^2}} \), integrate to get \( x(t) = \frac{C}{\sqrt{1-C^2}} t + C_1 \). Using \( x(0) = 0 \), we find \( C_1 = 0 \). From \( x(T) = \frac{1}{T^2} \), solve \( \frac{CT}{\sqrt{1-C^2}} = \frac{1}{T^2} \) for \( C \).

Solving for Constant C and Final Expression

Solve the equation \( CT = \frac{1}{T^2}\sqrt{1-C^2} \). After algebraic manipulation and squaring both sides, determine the value of \( C \). Conclude with the solution \( x(t) = \frac{t}{T^3}\sqrt{1-(\frac{1}{T^2})^2} \).

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental principle in the calculus of variations used to find the stationary points of functionals. A functional is an expression that depends on a function and its derivatives. Here, our goal is to determine what path minimizes or makes stationary the given functional.
This problem specifically involves the integral \[ J(x) = \int_{0}^{T} \sqrt{1 + \dot{x}^{2}} \, dt \] where \( \dot{x} \) represents the derivative of \( x(t) \) with respect to time.

To apply the Euler-Lagrange equation, we consider a Lagrangian, labeled \( L \), which combines \( x \), \( \dot{x} \), and any explicit dependence on \( t \). For this problem, the Lagrangian is \( L = \sqrt{1 + \dot{x}^{2}} \), noting there is no explicit dependence on \( x \) or \( t \). The Euler-Lagrange equation is written as: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) = \frac{\partial L}{\partial x} \] In our case, the partial derivative \( \frac{\partial L}{\partial x} \) is zero since \( L \) does not contain \( x \). Thus, simplifying the equation, we're left with:

• \( \frac{d}{dt} \left( \frac{\dot{x}}{\sqrt{1 + \dot{x}^{2}}} \right) = 0 \)

This means the expression inside the time derivative is constant, and so the optimal path \( x(t) \) has a constant velocity.

Stationary Points

Finding stationary points involves identifying the paths that either minimize or make a functional stationary, evident in calculus of variations problems. In our exercise:
\[ J(x) = \int_{0}^{T} \sqrt{1 + \dot{x}^{2}} \, dt \] requires determining such paths given boundary conditions \( x(0) = 0 \) and \( x(T) = \frac{1}{T^2} \).

Stationary points aren't necessarily minimums or maximums, but they're the points where the first derivative of the functional's value with respect to functions vanishes.
In our case, the stationarity condition derived from the Euler-Lagrange equation implies that the expression \( \frac{\dot{x}}{\sqrt{1 + \dot{x}^{2}}} \) is constant.

• This leads to a constant velocity solution \( \dot{x} = \frac{C}{\sqrt{1-C^2}} \).
• Integrating this expression gives \( x(t) = \frac{C}{\sqrt{1-C^2}} t + C_1 \).

Using given boundary conditions, the solution is refined to match these constraints, ensuring it accurately represents a stationary path for the functional.

Functional Analysis

Functional analysis comes into play when handling functionals, like \( J(x) = \int_{0}^{T} \sqrt{1 + \dot{x}^{2}} \, dt \), integrating both calculus and real analysis principles. This method provides a comprehensive view for understanding variations of functions over a specified range, using tools from the calculus of variations.

Functionals map functions to real numbers, capturing potential energy paths, which in turn are comparable using techniques from functional analysis.

• This equips mathematicians with the ability to study functions within infinite-dimensional spaces.
• In such scenarios, it's crucial to identify conditions under which the functional is stationary, using boundary conditions effectively as constraints.

In this specific exercise, analysis turned towards satisfying constraints such as \( x(0)=0 \) and \( x(T)=\frac{1}{T^{2}} \). Adjustments are made in the expression derived from the constant found in the Euler-Lagrange application.
Simply put, functional analysis provides insight and methods necessary for addressing complex problems seen in calculus of variations, helping translate these into practical solutions. By thorough analysis, paths are distinguished as viable solutions within given parameters.