Question

Use a Lagrange multiplier to find the curve \(x(y)\) of length \(L=\pi\) on the interval \([0,1]\) which maximizes the integral \(I=\int_{0}^{1} y(x) d x\) and pass through the points \((0,0)\) and \((1,0)\).

Short answer

We formulate a function combining both the integral to be maximized and the constraint, then apply the Euler-Lagrange equation to extremize the function. Solving the differential equation and applying boundary conditions will yield the optimal curve.

Step-by-step solution

Problem Formulation

Firstly, let's write down the functional to be optimized which is \(I=\int_{0}^{1} y(x) dx\). The constraint is a fixed curve length, \(L=\pi\), which can be expressed as \(L=\int_{0}^{1} \sqrt{1+y'(x)^{2}} dx = \pi\). Using Lagrange multipliers, we combine those two equations into a function \(F\), which is \(F= y+\lambda(\sqrt{1+y'^2}-1)\), where \(\lambda\) is a yet undetermined Lagrange multiplier.

Deriving the Euler-Lagrange Equation

We are going to use the Euler-Lagrange equation to extremize the above functional. For a functional \(F\), the equation is \(\frac{\partial F}{\partial y} - \frac{d }{dx}(\frac{\partial F}{\partial y'}) = 0\). Apply this to \(F\) gives the equation \(1 -\frac{d }{dx} \frac{\lambda y'}{\sqrt{1+y'^2}} = 0\). After doing the derivatives and simplifying, the resulting differential equation is \(\frac{d }{dx} \frac{\lambda y'}{\sqrt{1+y'^2}} = 1\).

Solving the Differential Equation

Integrating each side with respect to \(x\), we obtain \(\lambda \sqrt{1+y'^2} = x + A\) where \(A\) is an integration constant. This is a first order nonlinear ordinary differential equation for the derivative \(y'(x)\). Solving this for \(y'(x)\) with the boundary condition \(y(0)=0\) will yield the final solution.

Determining the Lagrange Multiplier and Integrating Constant

The Lagrange multiplier \(\lambda\) and the integration constant \(A\) can be determined by using the second boundary condition \(y(1)=0\) and the constraint \(L=\pi\).

Final Step – Evaluating The Integral

Substituting the results of \(\lambda\) and \(A\) into the differential equation and then integrating it will yield the curve \(x(y)\) that maximizes the integral \(I\) under the given condition and through solving this, a conclusion to the exercise is found.

Key concepts

Euler-Lagrange equation

The Euler-Lagrange equation plays a crucial role in functional optimization problems, like the one in the exercise where we maximize the integral. Essentially, it's a tool used to find functions that make a functional stationary, which means neither increasing nor decreasing. Think of it as an analog to finding maximum or minimum points in a regular function using derivatives.
In the case of our exercise, we have combined the integral we're trying to maximize with a constraint on the curve's length using Lagrange multipliers. This results in a new functional, which we denote as \( F \). The Euler-Lagrange equation helps us find the curve \( y(x) \) that will maximize this functional.
To apply the Euler-Lagrange equation, we need to differentiate \( F \) first partially with respect to \( y \), and then with respect to \( y' \) (which is the derivative of \( y \) with respect to \( x \)), and then subtract the derivative with respect to \( x \) of the latter to find the extremum. In our example, it leads us to a differential equation that we must solve. This gives us a path to find the optimum curve satisfying all given conditions.

ordinary differential equations

Ordinary differential equations, or ODEs, are equations involving a function and its derivatives. In the context of this exercise, once the Euler-Lagrange equation has been applied, we obtain an ODE that describes how the function \( y(x) \) behaves in order to maximize the integral under the constraint.
The resulting equation \( \frac{d }{dx} \frac{\lambda y'}{\sqrt{1+y'^2}} = 1 \) is a first-order nonlinear ordinary differential equation. Solving ODEs often requires knowing initial or boundary conditions, which describe values the solution must adhere to at certain points.
In this exercise, boundary conditions are specified by the points the curve must pass through, namely \((0,0)\) and \((1,0)\). These ensure that the solution for \( y(x) \) not only satisfies the equation but also fits the physical constraints of the problem.

functional optimization

Functional optimization deals with finding a function that maximizes or minimizes an integral. This is a bit more involved than traditional optimization because we're not just looking for maxima or minima for variables, but functions over an entire range.
In our exercise, we are tasked with finding a curve \( x(y) \) with a specific length \( L = \pi \). The goal is to maximize the given integral \( I = \int_{0}^{1} y(x) dx \) over the interval. The Lagrange multiplier technique helps us manage the constraint on curve length while optimizing the integral.
By incorporating the constraint directly into the integral through the multiplier, we transform the problem into one where we can use the Euler-Lagrange equation to find the function that optimizes our goal. The multiplier is essential for maintaining balance between maximizing our objective and satisfying constraints simultaneously, thus leading to the desired solution.