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Magazine Chapter 4: Problem 23
Under the condition that the admissible curve can not pass through the interior of the circle domaim surrounded by the circumference \((x-5)^{2}+y^{2}=9\), find the curve that can make the functional \(J[y]=\int_{0}^{10} y^{\prime 3} \mathrm{~d} x\) reach exextremum, the boundary conditions are \(y(0)=y(10)=0\).
Short Answer
Expert verified
The curve is \(y(x) = 0\).
Step by step solution
01
Understand the Problem
We need to find a curve that can make the functional \(J[y]=\int_{0}^{10} y^{\prime 3} \mathrm{~d} x\) reach an extremum, with the constraints that the curve cannot pass through the interior of the circle \((x-5)^{2}+y^{2}=9\) and must satisfy the boundary conditions \(y(0)=y(10)=0\).
02
Consider the Variational Problem
We have to find \(y(x)\) such that it maximizes or minimizes \(\int_{0}^{10} y^{\prime 3} \mathrm{~d} x\). The variational problem consists of finding a function \(y(x)\) that makes this integral stationary, meaning its first variation must be zero.
03
Use Calculus of Variations
Use the Euler-Lagrange equation for the functional. In this case, since the integrand only involves \(y'\), it's simplified as: \(-3(y')^{2}=0\). This implies \(y' = 0\), leading us to \(y = C\), where \(C\) is a constant.
04
Apply Boundary Conditions
Given the boundary conditions \(y(0)=0\) and \(y(10)=0\), we conclude that \(C=0\). Thus, the solution that satisfies both the Euler-Lagrange equation and the boundary conditions is \(y(x) = 0\).
05
Check Constraints
The curve \(y(x)=0\) corresponds to the x-axis and does not intersect the circle \((x-5)^{2}+y^{2}=9\), because the circle is entirely above the x-axis.
06
Conclusion
The curve \(y(x)=0\) meets all required conditions, making the functional stationary and not passing through the forbidden region.
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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 fundamental tool in the Calculus of Variations. It helps identify functions that make a functional stationary, which means that the first variation is zero. In simpler terms,
if you want to find the function that either maximizes or minimizes a given integral (the functional), the Euler-Lagrange equation is what you use.
if you want to find the function that either maximizes or minimizes a given integral (the functional), the Euler-Lagrange equation is what you use.
- This equation states that for a functional \( J[y] = \int_{a}^{b} F(y, y', x) \, dx \), if \( y = y(x) \) ensures \( J \) is stationary, then it must satisfy: \( \frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) = 0 \).
- In our exercise, because \( F(y') = (y')^3 \), the equation simplifies significantly.
functional extremum
A functional extremum occurs when the value of a functional reaches a peak or a trough. This means finding functions that provide either the highest or lowest value for an integral functional.
When dealing with functionals, such as \( J[y] = \int_{0}^{10} y'^{3} \,dx \), you seek functions that make this integral either as large or as small as possible.
When dealing with functionals, such as \( J[y] = \int_{0}^{10} y'^{3} \,dx \), you seek functions that make this integral either as large or as small as possible.
- To achieve this, one typically looks for where the Euler-Lagrange equation holds true.
- This indicates that the integral's first derivative with respect to the function is zero, marking a point of extremum.
boundary conditions
Boundary conditions are rules that the solution to a differential or variational problem must satisfy at specific points. In our exercise, the curve \( y(x) \) must satisfy \( y(0) = 0 \) and \( y(10) = 0 \), meaning it starts and ends at zero.
- These conditions are crucial because they limit the set of "admissible curves" to only those that meet the criteria.
- In the context of our problem, they ensure that while the function satisfies the Euler-Lagrange equation, \(y\), it can't be any constant but specifically \(y = 0\).
admissible curve
An admissible curve is a function that fits within all the problem's boundaries and conditions. In the context of Calculus of Variations, it's a candidate function through which the functional could potentially reach a stationary value.
- In the given problem, admissible curves must not only meet the Euler-Lagrange equation but also comply with the boundary conditions and constraints.
- The solution \( y(x) = 0 \) is admissible because it does not cross into the forbidden region inside the circle \((x-5)^2 + y^2 = 9\).
stationary functional
A stationary functional is one where small perturbations or changes to the function do not lead to first-order changes in the functional's value. This state is identified when the first derivative (or variation) of the functional with respect to the function in question is zero.
- In the exercise, the functional \( J[y] = \int_{0}^{10} y'^{3} \, dx \) is stationary when \( y' = 0 \), indicating that \( y \) has reached a point where minor tweaks don't affect the integral's value.
- This condition is crucial because it signifies that the function \( y \) is optimal within the problem's constraints.
