The Variational Problem

Prerequisites: Variational Calculus. Full path (3 pages)

• Intuitive
• Concrete
• Abstract

The core idea of this problem is to find a function that is a stationary point for a certain process, expressed as an integral functional, this means that we need to set the functional derivative of the integral to zero.

$$0 = \delta S[q|\phi] = \left.\frac{\partial}{\partial \varepsilon} \int_a^b F\circ \Gamma(q+\varepsilon \phi)\, dt\right|_{\varepsilon=0} = \int_a^b \left.\frac{\partial}{\partial \varepsilon} F(q+\varepsilon \phi, \dot q+\varepsilon \dot\phi)\right|_{\varepsilon=0} dt =$$ $$\int_a^b \frac{\partial F}{\partial q}\phi + \frac{\partial F}{\partial \dot q}\dot\phi dt = \left[\frac{\partial F}{\partial \dot q}\phi\right]^b_a + \int_a^b \left(\frac{\partial F}{\partial q} - \frac{d}{dt}\frac{\partial F}{\partial \dot q}\right)\phi dt$$

We haven't made any assumption of the form of $q$ and the space $\Omega$ where it resides. On the standard variational problem $\Omega$ is the space of the functions that $q(a)=q_a$ and $q(b)=q_b$, i.e, fixed boundaries, and are sufficiently smooth. Also, we assumed a standard lift: $\Gamma(q) = (q, \dot q)$.

Saying that, now the fundamental lemma of variational calculus enters. The variation, $\phi$, is arbitrary but with $\phi(a)=\phi(b)=0$. The later assertion, makes the term outside the integral zero. On the other hand, by the fundamental lemma, the interior parenthesis is zero, giving:

$$\frac{\partial F}{\partial q} - \frac{d}{dt}\frac{\partial F}{\partial \dot q} = 0$$

that is, the Euler-Lagrange equations.

Some variations worth studying

\1. Change $\Omega$: the functions do not need to be continous at all points, this gives collisions on a variational setting.

\2. Variable end points, i.e. $a$ and $b$ vary with $\varepsilon$

\3. Change the interdependencies of the lift $\Gamma$. This will lead to Vakonomic Mechanics.

Intuitive

The variational problem asks: among all curves that start at a given point and end at a given point, which one makes some quantity (a length, a travel time, an energy) as small or as large as possible, or at least stationary?

Think of standing in a landscape and looking for the bottom of a valley. At the bottom the ground is flat, so taking a tiny step in any direction does not change your height to first order. The variational problem uses the same test on curves. Take a candidate curve and push it slightly out of shape somewhere in the middle, keeping its two ends pinned down. If the quantity does not change to first order, for every such small push, the candidate is a solution.

Because the push can be placed anywhere along the curve, the condition has to hold at every point separately. That turns one condition on the whole curve into a local rule, a differential equation that the curve must obey at each point. This equation is the Euler-Lagrange equation. The fixed ends are part of the problem. If an end is allowed to move, the test has to include pushes that move it, and this gives an extra condition at that end.

Concrete

Statement

Given a function $F(q, \dot q, t)$, find among all sufficiently smooth $q: [a,b] \to \mathbb{R}$ with $q(a) = q_a$ and $q(b) = q_b$ those that make $S[q] = \int_a^b F(q, \dot q, t)\, dt$ stationary. As derived above, they satisfy

$$\frac{\partial F}{\partial q} - \frac{d}{dt}\frac{\partial F}{\partial \dot q} = 0 .$$

For several functions $q_1, \dots, q_n$ there is one such equation for each $q_i$, since each can be varied independently.

A first integral

If $F$ does not depend explicitly on $t$, then along solutions

$$\frac{d}{dt}\left( F - \dot q \frac{\partial F}{\partial \dot q} \right) = \dot q \left( \frac{\partial F}{\partial q} - \frac{d}{dt}\frac{\partial F}{\partial \dot q} \right) = 0 ,$$

so $F - \dot q\, \partial F/\partial \dot q$ is constant (the Beltrami identity). For $F = L$ a mechanical Lagrangian this is minus the energy.

Worked example: the brachistochrone

A bead slides without friction from the origin to a lower point, along a curve $y(x)$ with $y$ measured downward. Energy conservation gives the speed $v = \sqrt{2gy}$, so the travel time is

$$T[y] = \int \frac{\sqrt{1 + y'^2}}{\sqrt{2 g y}}\, dx ,$$

where $x$ now plays the role of $t$. The integrand does not depend on $x$, so the Beltrami identity gives

$$\frac{\sqrt{1+y'^2}}{\sqrt{2gy}} - \frac{y'^2}{\sqrt{1+y'^2}\sqrt{2gy}} = \frac{1}{\sqrt{1+y'^2}\sqrt{2gy}} = \text{const} ,$$

that is $y\,(1 + y'^2) = 2R$ for a constant $R$. One checks by substitution that the cycloid

$$x = R(\varphi - \sin\varphi) , \qquad y = R(1 - \cos\varphi)$$

solves this: $y' = \sin\varphi/(1 - \cos\varphi)$ and $1 + y'^2 = 2/(1-\cos\varphi)$. The constant $R$ is fixed by the endpoint.

Changing the boundary conditions

If the value $q(b)$ is not prescribed, the variation $\phi(b)$ is arbitrary and the boundary term $\left[\frac{\partial F}{\partial \dot q}\phi\right]_a^b$ in the derivation above no longer vanishes automatically. Stationarity then requires, in addition to the Euler-Lagrange equation, the natural boundary condition

$$\left.\frac{\partial F}{\partial \dot q}\right|_{t = b} = 0 .$$

If the admissible functions must also satisfy a constraint of the form $\int_a^b G(q, \dot q, t)\, dt = \text{const}$, as for the hanging chain of fixed length, one makes $\int (F - \lambda G)\, dt$ stationary with a Lagrange multiplier $\lambda$.

Abstract

Let $Q$ be a manifold and consider the space of paths $\mathcal{P} = \{ q: [a,b] \to Q \mid q(a) = q_a, \, q(b) = q_b \}$ of a suitable regularity class (for example $H^1$ or $C^2$). This is an infinite-dimensional manifold, and its tangent space at $q$ consists of the vector fields $\phi(t) \in T_{q(t)}Q$ along $q$ with $\phi(a) = 0 = \phi(b)$. The variations $\phi$ of the derivation above are exactly these tangent vectors, and the computation shows

$$dS_q(\phi) = \int_a^b \langle \mathcal{E}(q)(t), \phi(t) \rangle \, dt ,$$

where in local coordinates $\mathcal{E}(q)_i = \partial F/\partial q^i - \frac{d}{dt}\partial F/\partial \dot q^i$. The Euler-Lagrange expression $\mathcal{E}(q)(t)$ is a covector in $T^*_{q(t)}Q$, which is why the equations take the same form in every chart. The variational problem is the search for critical points of the function $S$ on $\mathcal{P}$.

Geodesics

On a Riemannian manifold $(Q, g)$ the critical points of the energy $E[q] = \frac{1}{2}\int_a^b g(\dot q, \dot q)\, dt$ are the geodesics with constant speed. The length functional $\int \sqrt{g(\dot q, \dot q)}\, dt$ has the same critical curves, but because it is invariant under reparametrization its critical points come in infinite-dimensional families of reparametrizations, which is why the energy is used in practice.

Stationary versus minimal

The variational problem asks for critical points. In mechanics, with $F = L = T - V$ and $T$ positive definite, the physical trajectories are local minima of the action as long as the time interval contains no point conjugate to the initial point, which holds for sufficiently short intervals. Beyond the first conjugate point they are no longer minima. The second variation and the Morse index theorem, which counts the negative directions of the second variation by conjugate points, make this precise.

Variants

The modifications listed above change the space $\mathcal{P}$ or the form of $S$. Free or moving endpoints lead to natural boundary and transversality conditions. For nonholonomic constraints (constraints on $\dot q$ that cannot be integrated to constraints on $q$), imposing the constraint on the space of paths before varying gives vakonomic mechanics, which in general yields different equations of motion than the Lagrange-d'Alembert principle used for mechanical systems with rolling constraints.

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1. Calculus
2. Functional
3. Variational Calculus