Question

Bestimmen Sie stationäre Punkte \(x(t)\) mit \(x(0)=1\) und \(x(2)=2\) des Funktionals $$ J(x)=\int_{0}^{2}\left[\frac{1}{2} \dot{x}^{2}+x \dot{x}+\dot{x}\right] d t $$

Short answer

The stationary point is \( x(t) = \frac{1}{2}t + 1 \).

Step-by-step solution

Understand the Problem

We are given a functional \( J(x) \) that we need to find stationary points for, subject to the conditions \( x(0)=1 \) and \( x(2)=2 \). This requires using the calculus of variations to find functions \( x(t) \) that satisfy these boundary conditions and make \( J(x) \) have a stationary value.

Identify the Lagrangian

The integrand of the functional \( J(x) \) is \( L(x, \dot{x}) = \frac{1}{2} \dot{x}^2 + x \dot{x} + \dot{x} \). This is the Lagrangian we will use in the Euler-Lagrange equation.

Apply Euler-Lagrange Equation

The Euler-Lagrange equation for a functional \( J(x) = \int L(x, \dot{x}) \, dt \) is \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) - \frac{\partial L}{\partial x} = 0 \). We need to compute the partial derivatives \( \frac{\partial L}{\partial \dot{x}} \) and \( \frac{\partial L}{\partial x} \).

Compute Partial Derivatives

Calculate \( \frac{\partial L}{\partial \dot{x}} = \dot{x} + x + 1 \) and \( \frac{\partial L}{\partial x} = \dot{x} \).

Solve the Euler-Lagrange Equation

Substitute the partial derivatives into the Euler-Lagrange equation: \( \frac{d}{dt} (\dot{x} + x + 1) - \dot{x} = 0 \). Differentiating \( \dot{x} + x + 1 \) with respect to \( t \) gives \( \ddot{x} + \dot{x} = \dot{x} \).

Simplify and Solve the Second Order ODE

The equation simplifies to \( \ddot{x} = 0 \). This is a second-order differential equation with a general solution \( x(t) = At + B \) for constants \( A \) and \( B \).

Apply Boundary Conditions

Using the boundary condition \( x(0) = 1 \), we have \( B = 1 \). Using \( x(2) = 2 \), we find that \( 2A + 1 = 2 \), giving \( A = \frac{1}{2} \). Thus, \( x(t) = \frac{1}{2}t + 1 \).

Key concepts

Euler-Lagrange equation

To find the stationary points of a functional, the Euler-Lagrange equation is a fundamental tool. It comes from the calculus of variations and helps us determine which functions make a given functional reach a stationary value.
In simple terms, the Euler-Lagrange equation is used to find the best path or shape that minimizes or maximizes an integral.
Specifically, if we have a functional of the form \(J(x) = \int L(x, \dot{x}) \, dt\), the Euler-Lagrange equation is:

• \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) - \frac{\partial L}{\partial x} = 0 \)

Here's how it works:1. **Identify the Lagrangian**: This is the expression under the integral sign, \(L(x, \dot{x})\).2. **Compute partial derivatives**: Calculate \(\frac{\partial L}{\partial \dot{x}}\) and \(\frac{\partial L}{\partial x}\).3. **Solve**: Substitute these into the Euler-Lagrange equation.These steps will help you find a function \(x(t)\) that makes the functional reach a stationary value. It's crucial for solving many physics and engineering problems.

functional analysis

Functional analysis is a branch of mathematical analysis that studies spaces of functions and their properties. In the context of calculus of variations, it helps us understand how different functions behave under various operations.
Functional analysis provides tools to deal with infinite-dimensional spaces. These are spaces where each point is actually a function rather than a number.
The purpose of functional analysis in the calculus of variations includes:

• Understanding functionals: It helps describe and analyze functionals, which are functions of functions.
• Studying continuity and derivatives: It provides a framework for thinking about the smoothness and change of functionals.
• Solving optimization problems: It is crucial for finding functions that optimize (maximize or minimize) a certain criterion, such as energy or length.

In practical terms, functional analysis serves as a language and toolbox for dealing with situations where functions themselves are the unknowns we need to solve for, much like we solve equations for numbers in elementary math.

boundary conditions

Boundary conditions are essential when solving problems involving differential equations or functionals. They set the constraints or limits that the solution must adhere to at specific points in the domain.Consider them as 'rules' the solution must follow:

• They specify values of a function or its derivatives at the boundaries.
• In our exercise, boundary conditions were \(x(0) = 1\) and \(x(2) = 2\).

These conditions ensure that the solutions we find are not just mathematically valid, but also practically meaningful in the context of the problem.For example, in physics, boundary conditions can specify how a string is attached or how a fluid behaves at a boundary. When working with these conditions:

• Apply them after solving the differential equations.
• Use them to determine constants in the solution.

By explicitly setting these conditions, we can better control the problem's outcomes and ensure the solution fits real-world scenarios or criteria.