Chapter 2: Problem 6
Find the extremal curve of the functional \(J[y]=\int_{x_{0}}^{x_{1}}\left(2 y+x y^{\prime 2}\right) \mathrm{d} x\), the boundary conditions are \(y\left(x_{0}\right)=y_{0}, y\left(x_{1}\right)=y_{1}\).
Short Answer
Expert verified
Use the Euler-Lagrange equation and solve the differential equation to find the extremal curve: \( y = x \pm C \ln|x| + D \). Adjust according to boundary conditions.
Step by step solution
01
Identify the Euler-Lagrange Equation
To find the extremal curve of the given functional, we apply the Euler-Lagrange equation. The functional provided is \( J[y] = \int_{x_{0}}^{x_{1}} (2y + x (y')^2) \, dx \). The Euler-Lagrange equation is \( \frac{d}{dx} \frac{\partial F}{\partial y'} - \frac{\partial F}{\partial y} = 0 \), where \( F(x, y, y') = 2y + x (y')^2 \).
02
Calculate Partial Derivatives
Calculate the necessary partial derivatives. Firstly, \( \frac{\partial F}{\partial y} = 2 \), and \( \frac{\partial F}{\partial y'} = 2xy' \).
03
Apply the Euler-Lagrange Equation
Now, differentiate \( \frac{\partial F}{\partial y'} = 2xy' \) with respect to \( x \). This gives \( \frac{d}{dx}(2xy') = 2y' + 2xy'' \). Substitute the results into the Euler-Lagrange equation: \( 2y' + 2xy'' - 2 = 0 \).
04
Simplify the Differential Equation
The differential equation simplifies to \( 2y' + 2xy'' = 2 \). Dividing the entire equation by \( 2 \), we have \( y' + xy'' = 1 \).
05
Solve the Differential Equation
To solve the differential equation \( y' + xy'' = 1 \), consider the substitution method. Let \( y' = p \), and hence \( y'' = \frac{dp}{dx} \). Substitute these into the equation: \( p + x \frac{dp}{dx} = 1 \).
06
Solve the First-Order Differential Equation
Separate variables to solve for \( p(x) \): \( \frac{dp}{1-p} = -\frac{dx}{x} \). Integrate both sides: \( \int \frac{dp}{1-p} = \int -\frac{dx}{x} \), giving \( -\ln|1-p| = \ln|x|+ C \).
07
Simplify and Solve for \(p\)
Solve the equation from the last step to find \( p \), by exponentiating: \( 1-p = \pm \frac{C}{x} \). Solve for \( p \) to obtain \( p = 1 \mp \frac{C}{x} \).
08
Obtain \(y\) from \(p\)
Recall that \( p = y' \), hence \( y' = 1 \mp \frac{C}{x} \). Integrate this with respect to \( x \) to find \( y \): \( y = \int (1 \mp \frac{C}{x}) \, dx = x \pm C \ln|x| + D \), where \( D \) is the integration constant.
09
Apply Boundary Conditions and Find Constants
Use the boundary conditions \( y(x_0) = y_0 \) and \( y(x_1) = y_1 \) to find \( C \) and \( D \). Substitute these conditions into the equation \( y = x \pm C \ln|x| + D \) to solve for \( C \) and \( D \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Euler-Lagrange Equation
The Euler-Lagrange Equation is a cornerstone in the field of Calculus of Variations, playing a pivotal role in finding extremal curves for functionals. A functional is an expression that assigns a number to a function, much like an ordinary function assigns a number to a variable. When tasked with finding the extremal curve of a functional, such as \( J[y] = \int_{x_0}^{x_1} (2y + x (y')^2) \, dx \), the Euler-Lagrange equation becomes essential. This equation establishes a necessary condition for a function \( y(x) \) to optimize the given functional.
The Euler-Lagrange equation is defined as:
The Euler-Lagrange equation is defined as:
- \( \frac{d}{dx} \frac{\partial F}{\partial y'} - \frac{\partial F}{\partial y} = 0 \), where \( F(x, y, y') \) is the integrand in the functional.
Extremal Curve
An extremal curve represents the solution to an optimization problem posed by a functional. In our given exercise, the task is to find the extremal curve of the functional \( J[y] = \int_{x_0}^{x_1} (2y + x (y')^2) \, dx \). The idea is to identify a function \( y(x) \) that either minimizes or maximizes this integral.
- After identifying the Euler-Lagrange equation, the next step is solving it to find the differential equation.
- The extremal curve is essentially the solution to this differential equation that also satisfies given boundary conditions \( y(x_0) = y_0 \) and \( y(x_1) = y_1 \).
Differential Equation
Differential equations arise naturally in the process of finding extremal curves using the Euler-Lagrange equation. In our exercise, solving the Euler-Lagrange equation resulted in the differential equation \( y' + xy'' = 1 \).
This type of equation involves derivatives, which describe how the function \( y(x) \) changes as \( x \) changes.
This type of equation involves derivatives, which describe how the function \( y(x) \) changes as \( x \) changes.
- Differential equations may vary in complexity from very simple to extremely challenging to solve.
- Simplification steps are often necessary, as seen where after obtaining \( 2y' + 2xy'' = 2 \), dividing by 2 leads to \( y' + xy'' = 1 \).
- Substitution methods are regularly employed to solve these equations, like \( y' = p \) and \( y'' = \frac{dp}{dx} \) in our case.
Functional Analysis
Functional Analysis is a vast area of mathematics that extends the ideas of functions and calculus to the study of spaces. It becomes particularly significant when dealing with functionals like \( J[y] = \int_{x_0}^{x_1} (2y + x (y')^2) \, dx \).
- This branch of mathematics studies spaces of functions and the functionals that act on these spaces, offering tools and theorems for understanding optimization processes.
- In the context of Calculus of Variations, Functional Analysis helps us rigorously define and solve problems concerning extremal functions that optimize functionals.
- It equips us with the required theoretical framework to handle infinite-dimensional spaces, unlike regular calculus which deals with finite dimensions.