Chapter 9: Problem 4
Find a first integral of the Euler equation to make stationary the integrals. $$\int_{\alpha}^{\beta} \sqrt{r^{2} r^{\prime 2}+r^{4}} d \theta$$
Short Answer
Expert verified
The first integral is \(\frac{r^{2} r'}{\sqrt{r^{2} r^{\prime 2} + r^{4}}} = c\).
Step by step solution
01
Understand the integrand
Given the integral \(\int_{\alpha}^{\beta} \sqrt{r^{2} r^{\prime 2} + r^{4}} \, d\theta\), identify the dependence of the integrand on \(r\) and \(r'\).
02
Apply the Euler-Lagrange equation
The function to be made stationary is \(L = \sqrt{r^{2} r^{\prime 2} + r^{4}}\). The Euler-Lagrange equation is \(\frac{d}{d\theta} \left( \frac{\partial L}{\partial r'} \right) - \frac{\partial L}{\partial r} = 0\).
03
Compute partial derivatives
Compute \(\frac{\partial L}{\partial r}\) and \(\frac{\partial L}{\partial r'}\) to use in the Euler-Lagrange equation.\(\frac{\partial L}{\partial r} = \frac{r r^{\prime 2} + 2r^{3}}{\sqrt{r^{2} r^{\prime 2} + r^{4}}}\) and \(\frac{\partial L}{\partial r'} = \frac{r^{2} r'}{\sqrt{r^{2} r^{\prime 2} + r^{4}}}\).
04
Apply the Euler-Lagrange equation
Substitute \(\frac{\partial L}{\partial r}\) and \(\frac{\partial L}{\partial r'}\) into the Euler-Lagrange equation and solve: \( \frac{d}{d\theta} \left( \frac{r^{2} r'}{\sqrt{r^{2} r^{\prime 2} + r^{4}}} \right) - \frac{r r^{\prime 2} + 2r^{3}}{\sqrt{r^{2} r^{\prime 2} + r^{4}}} = 0 \).
05
Simplify the equation
Simplify the equation, leading to the form \(r^{3} + r r^{\prime 2} - r^{2} r'' = 0\).
06
Identify first integral
Recognize the solved form of the first integral: \(\frac{r^{2} r'}{\sqrt{r^{2} r^{\prime 2} + r^{4}}} = c\), where \(c\) is a constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
calculus of variations
Calculus of variations is a field of mathematical analysis that deals with finding the extrema of functionals, which are mappings from a set of functions to the real numbers. In simpler terms, it helps us find the function that maximizes or minimizes a given quantity, such as distance or energy.
One of the main objectives in calculus of variations is to determine the path or shape of a function that will make a given integral achieve its stationary value (typically a minimum or maximum). For example, in the exercise, we want to find the function \(r(\theta)\) that makes the integral \(\int_{\alpha}^{\beta} \sqrt{r^{2} r^{\backprime 2}+r^{4}} d \theta\) stationary. The solution involves differentiating and solving certain equations, such as the Euler-Lagrange equation, to find this special function. This process requires us to understand the dependencies of the integrand and then apply calculus techniques to find the function that satisfies the conditions of the problem.
One of the main objectives in calculus of variations is to determine the path or shape of a function that will make a given integral achieve its stationary value (typically a minimum or maximum). For example, in the exercise, we want to find the function \(r(\theta)\) that makes the integral \(\int_{\alpha}^{\beta} \sqrt{r^{2} r^{\backprime 2}+r^{4}} d \theta\) stationary. The solution involves differentiating and solving certain equations, such as the Euler-Lagrange equation, to find this special function. This process requires us to understand the dependencies of the integrand and then apply calculus techniques to find the function that satisfies the conditions of the problem.
first integral
A first integral is a specific solution to a differential equation that provides a conserved quantity. In simpler terms, it gives us an expression that remains constant along the solutions of the differential equation.
In the context of our exercise, we aimed to find a first integral of the Euler equation. After solving the Euler-Lagrange equation, we deduced a form of the first integral: \( \frac{r^{2} r'}{\sqrt{r^{2} r^{\backprime 2} + r^{4}}} = c \). Here, \(c\) is a constant, indicating that the expression on the left-hand side remains constant for the solution \(r(\theta)\). This first integral can be used to understand the behavior and properties of the function \(r\) without needing to solve the differential equation explicitly every time.
In the context of our exercise, we aimed to find a first integral of the Euler equation. After solving the Euler-Lagrange equation, we deduced a form of the first integral: \( \frac{r^{2} r'}{\sqrt{r^{2} r^{\backprime 2} + r^{4}}} = c \). Here, \(c\) is a constant, indicating that the expression on the left-hand side remains constant for the solution \(r(\theta)\). This first integral can be used to understand the behavior and properties of the function \(r\) without needing to solve the differential equation explicitly every time.
differential equations
Differential equations are mathematical equations that involve functions and their derivatives. They play a crucial role in modeling various physical phenomena where rates of change are involved. Solving a differential equation means finding the function or functions that satisfy the equation.
In the solution provided, the Euler-Lagrange equation was a differential equation that we needed to solve. The equation itself was obtained by setting the derivative of the partial derivative of the Lagrange function with respect to \(r'\) equal to the partial derivative of the Lagrange function with respect to \(r\). This led to the differential equation:
\[ \frac{d}{d\theta} \left( \frac{r^{2} r'}{\sqrt{r^{2} r^{\backprime 2} + r^{4}}} \right) - \frac{r r^{\backprime 2} + 2r^{3}}{\sqrt{r^{2} r^{\backprime 2} + r^{4}}} = 0 \].
Solving this helped us find a first integral which simplifies the problem by giving us an expression that remains constant and is easier to work with.
In the solution provided, the Euler-Lagrange equation was a differential equation that we needed to solve. The equation itself was obtained by setting the derivative of the partial derivative of the Lagrange function with respect to \(r'\) equal to the partial derivative of the Lagrange function with respect to \(r\). This led to the differential equation:
\[ \frac{d}{d\theta} \left( \frac{r^{2} r'}{\sqrt{r^{2} r^{\backprime 2} + r^{4}}} \right) - \frac{r r^{\backprime 2} + 2r^{3}}{\sqrt{r^{2} r^{\backprime 2} + r^{4}}} = 0 \].
Solving this helped us find a first integral which simplifies the problem by giving us an expression that remains constant and is easier to work with.
lagrange function
The Lagrange function (or Lagrangian) is a function that depends on both the coordinates and their derivatives, encapsulating the dynamics of a system for calculus of variations problems. It is denoted by \(L\).
For the integral presented in the exercise, the Lagrange function was identified as \(L = \sqrt{r^{2} r^{\backprime 2} + r^{4}}\). This function involves both the variable \(r\) and its derivative \(r'\). To find the extremum of the integral, we apply the Euler-Lagrange equation to the Lagrange function.
In this context, the Lagrange function encompasses the quantity that we need to make stationary. By finding the partial derivatives of the Lagrange function with respect to \(r\) and \(r'\) and solving the resulting Euler-Lagrange equation, we determine the function \(r\) that will make the integral stationary.
This process essentially translates a problem of finding extrema (maximum or minimum values) into solving differential equations, which is a core idea within the calculus of variations.
For the integral presented in the exercise, the Lagrange function was identified as \(L = \sqrt{r^{2} r^{\backprime 2} + r^{4}}\). This function involves both the variable \(r\) and its derivative \(r'\). To find the extremum of the integral, we apply the Euler-Lagrange equation to the Lagrange function.
In this context, the Lagrange function encompasses the quantity that we need to make stationary. By finding the partial derivatives of the Lagrange function with respect to \(r\) and \(r'\) and solving the resulting Euler-Lagrange equation, we determine the function \(r\) that will make the integral stationary.
This process essentially translates a problem of finding extrema (maximum or minimum values) into solving differential equations, which is a core idea within the calculus of variations.
