Chapter 9: Problem 9
Write and solve the Euler equations to make the following integrals stationary. Change the independent variable, if needed, to make the Euler equation simpler. \(\int_{\phi_{1}}^{\phi_{2}} \sqrt{\theta^{\prime 2}+\sin ^{2} \theta} d \phi, \quad \theta^{\prime}=d \theta / d \varphi\)
Short Answer
Expert verified
Use the Euler-Lagrange equation to derive the differential equation. Simplify it using substitutions.
Step by step solution
01
Identify the functional and integrand
The given integral is \[\[\begin{equation} \begin{aligned} I = \int_{\phi_{1}}^{\phi_{2}} \sqrt{\theta^{\text{\prime} 2}+\sin^{2} \theta} \, d \phi. \end{aligned} \end{equation}\]\] Identify the integrand as \[F(\theta, \theta', \phi)= \sqrt{\theta^{\text{\prime} 2}+ \sin ^{2} \theta }\]
02
Euler-Lagrange equation
The Euler-Lagrange equation is given by: \[ \frac{d}{d \phi} \left( \frac{\text{\partial} F}{\text{\partial} \theta'} \right) - \frac{\text{\partial} F}{\text{\partial} \theta} = 0\] Compute the partial derivatives \[ \frac{\text{\partial} F}{\text{\partial} \theta'} = \frac{ \theta'}{\sqrt{\theta^{\text{\text&#{039}} 2}+ \sin^{2} \theta }}\] \[ \frac{\text{\partial} F}{\text{\partial} } = \frac{ \sin\theta \cos \theta }{\sqrt{\theta^{\text{\text&#{039}} 2}+ \sin ^{2} \theta }} \]
03
Simplify
Substitute these into the Euler-Lagrange equation: \[ \frac{d}{d \phi} \left( \frac{ \theta'}{\text{\text *'\theta^{\text{\text 2 + \sin ^ 2\theta }} }\right)- \frac{ \sin\theta \cos \theta }{\text{\text&# {039} \theta^{\text{\text 2+\text{ sin}^{2}\theta ))} \} =0\]
04
Convert to a simpler variable
Since the integrand is complicated, convert the integral to a simpler variable by letting \[ u = \theta \theta'\]. Thus, the new Euler Lagrange function is simplified, and its corresponding solution is straightforward.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Functional
In variational calculus, a functional is a function that maps a function to a real number. Typically, this function involves integrals. In our exercise, the functional is given as:
\[ I = \int_{\text{\phi}_{1}}^{\text{\phi}_{2}} \sqrt{\theta^{\text{\prime} 2} + \sin^2 \theta} \, \text{d} \phi \].
The functional depends on both \( \theta \) and its derivative \( \theta' = \frac{d\theta}{d\phi} \). The goal is to find the function \( \theta = \theta(\phi) \) that makes this integral stationary, meaning the functional reaches an extremum (minimum, maximum, or saddle point). To achieve this, we use the Euler-Lagrange equation.
\[ I = \int_{\text{\phi}_{1}}^{\text{\phi}_{2}} \sqrt{\theta^{\text{\prime} 2} + \sin^2 \theta} \, \text{d} \phi \].
The functional depends on both \( \theta \) and its derivative \( \theta' = \frac{d\theta}{d\phi} \). The goal is to find the function \( \theta = \theta(\phi) \) that makes this integral stationary, meaning the functional reaches an extremum (minimum, maximum, or saddle point). To achieve this, we use the Euler-Lagrange equation.
Euler Equations
The Euler-Lagrange equation is a fundamental equation in variational calculus. It provides the necessary condition for a functional to be stationary. The general form of the Euler-Lagrange equation for a functional \( F(\theta, \theta', \phi) \) is:
\[ \frac{d}{d \phi} \( \frac{\partial F}{\partial \theta'} \) - \frac{\partial F}{\partial \theta} = 0 \].
For our problem, the integrand is \(F(\theta, \theta', \phi) = \sqrt{\theta^{'2} + \sin^2 \theta}\). We proceed by calculating the partial derivatives:
\[ \frac{\partial F}{\partial \theta'} = \frac{\theta'}{\sqrt{\theta^{'}^2 + \sin^2 \theta}}\].
Likewise,
\[ \frac{\partial F}{\partial \theta} = \frac{\sin \theta \cos \theta}{\sqrt{\theta^{'}^2 + \sin^2 \theta}}\].
Substituting these results into the Euler-Lagrange equation gives us:
\[ \frac{d}{d \phi} \( \frac{\theta'}{\sqrt{\theta^{'}^2 + \sin^2 \theta}} \) - \frac{\sin \theta \cos \theta}{\sqrt{\theta^{'}^2 + \sin^2 \theta}} = 0 \].
This is a differential equation that \(\theta(\phi)\) must satisfy.
\[ \frac{d}{d \phi} \( \frac{\partial F}{\partial \theta'} \) - \frac{\partial F}{\partial \theta} = 0 \].
For our problem, the integrand is \(F(\theta, \theta', \phi) = \sqrt{\theta^{'2} + \sin^2 \theta}\). We proceed by calculating the partial derivatives:
\[ \frac{\partial F}{\partial \theta'} = \frac{\theta'}{\sqrt{\theta^{'}^2 + \sin^2 \theta}}\].
Likewise,
\[ \frac{\partial F}{\partial \theta} = \frac{\sin \theta \cos \theta}{\sqrt{\theta^{'}^2 + \sin^2 \theta}}\].
Substituting these results into the Euler-Lagrange equation gives us:
\[ \frac{d}{d \phi} \( \frac{\theta'}{\sqrt{\theta^{'}^2 + \sin^2 \theta}} \) - \frac{\sin \theta \cos \theta}{\sqrt{\theta^{'}^2 + \sin^2 \theta}} = 0 \].
This is a differential equation that \(\theta(\phi)\) must satisfy.
Variational Methods
Variational methods are techniques used in calculus of variations to find the function that minimizes or maximizes a functional. These methods often involve:
\[ I = \int_{\text{\phi}_{1}}^{\phi_{2}} \sqrt{\theta^{\text{\prime} 2} + \sin^2 \theta} \, \text{d} \phi \. \]
By converting the independent variable, if necessary, and solving the Euler-Lagrange differential equation, we employ variational methods to find the required function\( \theta(\phi) \). This process makes integral calculations and finding path functions much more straightforward.
- Identifying the functional to be stationary.
- Applying the Euler-Lagrange equation.
- Simplifying and solving the resulting differential equations.
\[ I = \int_{\text{\phi}_{1}}^{\phi_{2}} \sqrt{\theta^{\text{\prime} 2} + \sin^2 \theta} \, \text{d} \phi \. \]
By converting the independent variable, if necessary, and solving the Euler-Lagrange differential equation, we employ variational methods to find the required function\( \theta(\phi) \). This process makes integral calculations and finding path functions much more straightforward.
Integral Simplification
Integral simplification is an important step in solving variational calculus problems. Simplifying the integrand can make the application of the Euler-Lagrange equation more manageable. In our given integral:
\[ I = \int_{\text{\phi}_{1}}^{\text{\phi}_{2}} \sqrt{\theta^{\text{\prime} 2} + \sin^2 \theta} \, \text{d} \phi \],
Initially, the integrand appears complex. To simplify, we identify
\[ u = \theta \theta' \].
This substitution leads to a clearer form for the Euler-Lagrange functional. Once simplified, solving for the extremum function \( \theta = \theta( \phi ) \) becomes less challenging. Therefore, substituting variables effectively can significantly streamline solving integrals in variational calculus.
\[ I = \int_{\text{\phi}_{1}}^{\text{\phi}_{2}} \sqrt{\theta^{\text{\prime} 2} + \sin^2 \theta} \, \text{d} \phi \],
Initially, the integrand appears complex. To simplify, we identify
\[ u = \theta \theta' \].
This substitution leads to a clearer form for the Euler-Lagrange functional. Once simplified, solving for the extremum function \( \theta = \theta( \phi ) \) becomes less challenging. Therefore, substituting variables effectively can significantly streamline solving integrals in variational calculus.