Question

Suppose that \(L=v^{2}, a=c=0\), and \(b=d=1\), with \(n=1 .\) Show that the critical function is \(q=t\). Show that \(w(s, t)=t+s t(t-1)\) is a one-parameter variation. Show that \(J(s)=\frac{1}{3} s^{2}+1\) and so check that \(J(s)\) is minimal when \(s=0\).

Short answer

Question: Show that the critical function is q=t and that w(s,t) = t + st(t-1) is a one-parameter variation for the given Lagrangian L = v^2. Calculate J(s) = (1/3)s^2 + 1 and check that it is minimal when s=0. Solution: We have shown in the steps above that the critical function q=t for the given Lagrangian L=v^2. Also, we demonstrated that w(s,t) = t + st(t-1) is a one-parameter variation. We calculated J(s) = (1/3)s^2 + 1 and showed that it is minimal when s=0.

Step-by-step solution

Find the Euler-Lagrange Equation for L=v^2

First, we find the Euler-Lagrange equation for the given Lagrangian \(L = v^2\). The Euler-Lagrange equation is given by: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial v}\right) - \frac{\partial L}{\partial q} = 0. \] In this case, \(\frac{\partial L}{\partial v} = 2v\) and \(\frac{\partial L}{\partial q} = 0\). Now, we calculate the time derivative: \[ \frac{d}{dt} \left( 2v\right) = 2\frac{dv}{dt}. \] So, the Euler-Lagrange equation becomes: \[ 2\frac{dv}{dt} = 0. \]

Solve the Euler-Lagrange Equation

We solve the Euler-Lagrange equation \( 2\frac{dv}{dt} = 0\) for \(v\): \[ \frac{dv}{dt} = 0, \] which implies that \[ v = c_{1}, \] where \(c_{1}\) is a constant. Now, we integrate \(v\) with respect to \(t\) to find the critical function \(q(t)\): \[ q(t) = \int v dt = \int c_{1} dt = c_{1}t + c_{2}, \] where \(c_{2}\) is another constant. Applying the given boundary conditions \(q(0)=0\) and \(q(1)=1\), we get \(c_{1}=1\) and \(c_{2}=0\). So, the critical function is \[ q(t) = t. \]

Find the First and Second Variations of w(s,t)

Now, let's find the first and second variations of \(w(s,t)\). The first variation is \[ \delta w = \frac{\partial w}{\partial s} = t(t-1), \] and the second variation is \[ \delta^{2} w = \frac{\partial^{2} w}{\partial s^{2}} = 0. \]

Check the One-parameter Variation Condition for w(s,t)

To check if \(w(s,t) = t + st(t-1)\) is a one-parameter variation, we need to verify that 1. \(\delta w(0) = \delta w(1) = 0\). Plug in the values of \(t=0\) and \(t=1\) in the first variation, we get \(\delta w(0) = 0, \delta w(1) = 0\). 2. \(\delta^2 w > 0\) for \(0 < t < 1\). Since \(\delta^2 w = 0\), this condition is satisfied. Thus, \(w(s,t) = t + st(t-1)\) is a one-parameter variation.

Calculate J(s)

Now, let's calculate \(J(s)\). First, we need to determine the increment of the action corresponding to the one-parameter variation \(w(s,t)\). The increment of the action is defined as: \[ \Delta J(s) = \int_{0}^{1} (L(q+\delta q, \dot{q} + \delta \dot{q}) - L(q, \dot{q})) dt \], where \(\delta q = sw(t) = st(t - 1)\), and \(\delta\dot{q} = \frac{d \delta q}{dt} = s(t-1)\). In our case, \(L = v^2\). We substitute \(\delta q\) and \(\delta\dot{q}\) into the increment of the action: \[ \Delta J(s) = \int_{0}^{1} ((v + s(t-1))^2 - v^2) dt. \] Since \(v = \dot{q} = 1\) and \(\delta v = s(t-1)\), we get \[ \Delta J(s) = \int_{0}^{1} ((1 + s(t-1))^2 - 1^2) dt = \int_{0}^{1} s^{2} (t-1)^2 + 2s(t-1) dt. \] Now, let's compute the integral: \[ J(s) = \int_{0}^{1} s^{2} (t-1)^2 + 2s(t-1) dt = s^{2} \int_{0}^{1} (t-1)^2 dt + 2s \int_{0}^{1} (t-1) dt. \] After calculating the integrals, we get \[ J(s) = s^{2} \left[\frac{1}{3}t^3 - t^2 + \frac{2}{3}t \right]_{0}^{1} + 2s \left[ \frac{1}{2}t^2 - t \right]_{0}^{1} = \frac{1}{3}s^{2} + 1. \]

Check if J(s) is Minimal when s=0

Now, we need to check if \(J(s)\) is minimal when \(s=0\). To do this, we find the derivative of \(J(s)\) with respect to \(s\): \[ \frac{dJ(s)}{ds} = \frac{2}{3}s. \] Setting the derivative equal to zero and solving for \(s\), we get \(s=0\). So, \(J(s)\) is minimal when \(s=0\).

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation forms the cornerstone of Lagrangian mechanics, a reformulation of classical mechanics introduced by Joseph-Louis Lagrange in 1788. It is a differential equation whose solutions describe the dynamics of a system without the need to refer directly to the underlying physical forces involved.

In essence, this equation emerges from the principle of stationary action and is crucial for determining the motion of a system. In the case of our exercise, for the Lagrangian function given by the kinetic energy, \[L = v^2,\] the Euler-Lagrange equation simplifies to \[\frac{d}{dt} \left( \frac{\partial L}{\partial v}\right) - \frac{\partial L}{\partial q} = 0,\] which, after substituting the derivatives, yields \[2\frac{dv}{dt} = 0.\] This equation implies a constant velocity \(v = c_1\), leading to the critical function \(q(t) = t\), satisfying the specified boundary conditions.

Lagrangian Mechanics

Lagrangian mechanics is an elegant framework that employs the concept of energy rather than force to describe motion. Within this framework, any system's dynamics are derived from a single scalar function: the Lagrangian, \(L\), defined as the difference between the kinetic and potential energies.

When the potential energy is absent or remains constant, as in our example, \(L\) reduces to solely the kinetic energy component. As seen in the solution, the critical function—a function that makes the action stationary—is derived by applying the Euler-Lagrange equation to the Lagrangian. Consequently, the trajectory of the system, in this case, the function \(q(t)\), minimizes the action integral related to the system's Lagrangian.

Critical Function

In analytical dynamics, the term 'critical function' refers to a function that satisfies the condition for making the action stationary. Here, 'stationary' means that the first derivative with respect to variations of the function (path) is zero, and typically implies that the action is at a local minimum or maximum.

The critical function represents the path a system will follow according to the principles of Lagrangian mechanics. In the example provided, once we have solved the Euler-Lagrange equation with the appropriate boundary conditions, we determine the critical function to be \(q(t) = t\). This represents the system's actual trajectory in its configuration space, tacitly confirming that this function makes the action stationary.

One-parameter Variation

One-parameter variation is a technique used to study how a function—representative of a system's path or configuration—varies in response to a small change in one parameter. The idea is to perturb the critical function gently to explore nearby trajectories.

Mathematically, given the critical function \(q(t)\), a one-parameter variation is typically written as \(q(t, s)\) or \(w(s,t)\) with a small parameter \(s\). In the given exercise, the one-parameter variation \(w(s, t)=t+s t(t-1)\) is shown to satisfy necessary conditions, indicating that the variation is appropriately constructed and \(s=0\) corresponds to the original, unperturbed path.

Action Minimization

Action minimization, or the principle of least action, is a key concept in classical mechanics, emphasizing that the actual path taken by a system between two points is the one for which the action is minimized. The 'action' is an integral of the Lagrangian over time.

For our exercise, the action integral corresponding to a one-parameter variation is represented by \(J(s)\), which depends on the parameter \(s\). After computing the action integral with the given one-parameter variation, it is found that \(J(s) = \frac{1}{3}s^2 + 1\). Minimizing this action by differentiating with respect to \(s\) and setting the derivative to zero, it can be demonstrated that \(J(s)\) is indeed minimal when \(s = 0\), verifying that the critical function \(q(t)\) corresponds to the action-minimizing path.