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Find the extremal of the functional $$ J[y]=\int_{0}^{\pi}\left(2 x \sin t-\dot{x}^{2}\right) d t $$ that satisfies \(x(0)=x(\pi)=0\). Show that this extremal provides the global maximum of \(J\)

Short Answer

Expert verified
The extremal of the functional is \(x(t) = 0\), and this extremal provides the global maximum of \(J\).

Step by step solution

01

Set up the Euler-Lagrange equation

The Euler-Lagrange equation for this problem involves the Lagrangian of the functional. The Lagrange's equation is derived from \(\frac{\partial L}{\partial x}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=0\). The given functional's Lagrangian \(L=2 x \sin t-\dot{x}^2\).
02

Substitute Lagrange's equation and solve for \(x(t)\)

When we substitute Lagrange's equation, we obtain \(\sin t - 2\ddot{x} = 0\). The solution to this differential equation is \(x(t) = A \sin t + B \cos t\). Using the condition \(x(0) = x(\pi) = 0\), we get \(A = 0\) and \(B = 0\). Thus, the extremal is \(x(t) = 0\).
03

Show Extremal Provides Global Maximum of J

With \(x(t) = 0\), substitute this back to the functional \(J[y] = \int_{0}^{\pi}(2x \sin t - \dot{x}^2)dt = \int_{0}^{\pi}0 dt = 0\). For any other \(x(t) \neq 0\), the integral of \(-\dot{x}^2\) will always be less than 0 because it's the integral of a square term with a negative sign. Therefore, \(J[y] \leq 0\) for any y. And we know the maximum of \(J[y]\) is 0 when \(x(t) = 0\). Hence, this extremal provides the global maximum of \(J\).

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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 concept in the calculus of variations. It is used to find the function, often called the extremal, that will maximize or minimize a given functional. Essentially, it transforms the problem of finding an extremal into solving a differential equation.
When dealing with a functional, the Euler-Lagrange equation is formed as \( \frac{\partial L}{\partial x} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) = 0 \), where \(L\) is the Lagrangian, a function of \(x\), \(\dot{x}\), and \(t\). Solving this equation yields the conditions the extremal must satisfy.
Here's a helpful way to think about it:
  • The Euler-Lagrange equation takes into account the derivative of the Lagrangian with respect to the function \(x(t)\), and the derivative with respect to its rate of change \(\dot{x}(t)\).
  • By solving the equation, you find a function \(x(t)\) that extremizes the given functional.
Functional Extremals
In the context of the calculus of variations, finding a functional extremal is like finding the best possible path or configuration that minimizes or maximizes a given quantity. These "best paths" are functions which the Euler-Lagrange equation helps us to find.
For a functional \( J[y] \), which depends on a function \( y \), the extremal \( x(t) \) is a critical solution that either maximizes or minimizes \( J \). In our given problem, we seek to maximize the integral of \(2x \sin t - \dot{x}^2\) between 0 and \(\pi\).
Key considerations when determining a functional extremal:
  • **Initial conditions**: These are values that the solution must satisfy at the endpoints of the domain.
  • **Boundary conditions**: Specific requirements that help shape the solution, such as \( x(t) = 0 \) at \(t = 0\) and \(t = \pi\) in our problem.
Together, these conditions help ensure you find the correct extremal function like \( x(t) = 0 \) in our solution.
Boundary Conditions
Boundary conditions are crucial in solving problems involving the calculus of variations, particularly through the Euler-Lagrange equation. They act as constraints that the extremal function must satisfy at the endpoints of the interval considered.
In the provided exercise, the functional considered must satisfy the conditions \(x(0)=0\) and \(x(\pi)=0\). These are known as Dirichlet boundary conditions.
  • Boundary conditions help ensure the extremal function is valid within the context of the problem.
  • They provide necessary information to determine constants that appear in the general solution of the differential equation derived from the Euler-Lagrange equation.
Without the inclusion of appropriate boundary conditions, it would be impossible to uniquely ascertain the behavior of the extremal across the interval, which could result in an incorrect or incomplete solution.

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Most popular questions from this chapter

An aircraft flies in the \((x, z)\)-plane from the point \((-a, 0)\) to the point \((a, 0) .(z=0\) is ground level and the \(z\)-axis points vertically upwards.) The cost of flying the aircraft at height \(z\) is \(\exp (-k z)\) per unit distance of flight, where \(k\) is a positive constant. Find the extremal for the problem of minimising the total cost of the journey. [Assume that \(k a<\pi / 2\).]

Solve the problem of finding the path of quickest descent for a skier from the point \(P\left(x_{0}, y_{0}, z_{0}\right)\) to the point \(Q\left(x_{1}, y_{1}, z_{1}\right)\) on a snow covered mountain whose profile is given by \(z=G(x, y)\), where \(G\) is a known function. [Assume that the skier starts from rest and that the total energy of the skier is conserved in the descent.] Let \(\mathcal{C}\) be a path connecting \(P\) and \(Q\). Show that the time taken to descend by this route is given by $$ T[y]=(2 g)^{-1 / 2} \int_{x_{0}}^{x_{1}}\left(\frac{1+\dot{y}^{2}+\left(G,_{x}+G,{ }_{y} \dot{y}\right)^{2}}{G\left(x_{0}, y_{0}\right)-G(x, y)}\right)^{1 / 2} d x $$ where \(y=y(x)\) is the projection of \(\mathcal{C}\) on to the plane \(z=0, \dot{y}\) means \(d y / d x\), and \(G,_{x}, G, y\) are the partial derivatives of \(G\) with respect to \(x, y\). Obtain the E-L equation with computer assistance and solve it numerically with the initial conditions \(y\left(x_{1}\right)=y_{1}, y^{\prime}\left(x_{1}\right)=\lambda\) and choose \(\lambda\) so that the path passes through the \(P\). [The numerical ODE integrator finds it easier to integrate the equation starting from the bottom. Why?] Plot the quickest route using 3D graphics. The author used the profile of the Cosine Valley resort, for which \(G(x, y)=\cos ^{2}(\pi x / 2) \cos ^{2}(\pi y / 4)\). The skier had to descend from \(P(1 / 3,0,3 / 4)\) to \(Q(2,2,0)\). The computued quickest route down the valley is shown in Figure 13.7. Those who lose their nerve at the summit can walk down by the shortest route (shown dashed). You may make up your own mountain profile, but keep it simple.

By using Hamilton's principle, show that, if the Lagrangian \(L(\boldsymbol{q}, \dot{\boldsymbol{q}}, t)\) is modified to \(L^{\prime}\) by any transformation of the form $$ L^{\prime}=L+\frac{d}{d t} g(\boldsymbol{q}, t) $$ then the equations of motion are unchanged.

Find the extremal of the path length functional $$ L[y]=\int_{0}^{1}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{1 / 2} d x $$ that satisfies \(y(0)=y(1)=0\) and show that it does provide the global minimum for \(L\).

A certain oscillator with generalised coordinate \(q\) has Lagrangian $$ L=\dot{q}^{2}-4 q^{2} $$ Verify that \(q^{*}=\sin 2 t\) is a motion of the oscillator, and show directly that it makes the action functional \(S[q]\) satationary in any time interval \([0, \tau]\). For the time interval \(0 \leq t \leq \pi\), find the variation in the action functional corresponding to the variations (i) \(h=\epsilon \sin 4 t\), (ii) \(h=\epsilon \sin t\), where \(\epsilon\) is a small parameter. Deduce that the motion \(q^{*}=\sin 2 t\) does not make \(S\) a minimum or a maximum.

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