Calculus of Variations: Mechanics, Control and Other Applications
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Students will appreciate the text's reader-friendly style, which features gradual advancements in difficulty and starts by developing technique rather than focusing on technical details. The examples and exercises offer many citations of engineering-based applications, and the exercises range from elementary to graduate-level projects, including longer projects and those related to classic papers.
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Calculus of Variations - Charles R. MacCluer
Chapter 1
Preliminaries
This chapter reviews basic tools from calculus that are used in the calculus of variations—directional derivatives, gradients, the chain rules, contour surfaces, sublevel sets, Lagrange multipliers, and the basic notion of convexity. All of these concepts form the basic toolset for attacking optimization problems.
1.1 Directional Derivatives and Gradients
n is denoted by x = (x1, x2, …, xn, which is an n denotes transpose). Suppose a function f of x represents the profit of a commercial enterprise, where x is a vector of the parameters of the operation such as labor costs, production output levels, price of the commodity, and so on. The manager naturally desires to know in which direction from the present operating point x⁰ should the company move in order to obtain the maximum increase in profit. The desired direction is found by using a multi-variable notion of a derivative: For each unit vector u n, the directional derivative of f at x⁰ in direction u is given by
provided the limit exists. Geometrically, this limit is the slope of the line tangent to the curve above x⁰ obtained by cutting the hypersurface z = f(x) with the hyperplane determined by u and the z-axis. See Figure 1.1. The directional derivatives in the directions parallel to the coordinate axes are, of course, the familiar partial derivatives
when u = (0, 0,…, 0, 1 (k.
Figure 1.1 The graph of y = f(x) is cut by a plane perpendicular to the coordinate space in the direction u. The slope of the line tangent to the curve thus cut out is given by the directional derivative Duf(x⁰).
The function f is differentiable at x⁰ provided it is linearly approximated by its tangent plane near x⁰. This means there exists a constant (row) vector a = (a1, a2,…, an) such that for all x in some open ball B = {x; |x − x⁰| < r} of radius r about x⁰,
(x) is such that
If f is differentiable at x⁰, the row vector a is called the gradient of f at x⁰, is necessarily unique (Exercise 1.1), and is denoted by ∇f(x⁰).
Directional derivatives may exist in all directions without a function being differentiable (Exercise 1.2). However, there is a simple formula for the directional derivative in terms of the gradient when f is differentiable.
Theorem A. Suppose f is differentiable at x⁰. The directional derivative in the direction u is obtained as
where the gradient is calculated by
Proof. Exercise 1.3.
The krepresents the sensitivity of f to changes in the kth component variable xk.
Corollary A. For u n with |u| = 1,
is the angle between (the unit vector) u and ∇f(x⁰).
Corollary B. The gradient ∇f(x⁰) points in the direction of maximum increase of f at x⁰. This maximal rate of increase is |∇f(x⁰)|.
Example 1. Let f(x, y) = x² + xy + y³. The directional derivative of f at (1, 2) in the direction u = (a, b) is then
.
1.2 Calculus Rules
The product formula (1.3) for the directional derivative is actually an instance of a much more general result, the chain rule.
Theorem B. (The First Chain Rule) Suppose that each component of the vector curve x = x(t) is differentiable at t = t⁰, and that f = f(x) is differentiable at x⁰ = x(t⁰). Then
In more transparent notation,
Proof. Combine x = x⁰ + (t − t(twith equation (1.2) (Exercise 1.5).
Example 2. Let f(x, y) = x² − 2xy + y³ and x(t) = cos t, y(t) = sin t. Then
The First Chain Rule itself is a special case of the following more general rule.
Theorem C. (The Second Chain Rule) Suppose x = x(un is differentiable at u = u⁰ where u = (u1, …,ur(that is, each component of x is differentiable), and suppose also that f = f(x) is differentiable at x⁰ = x(u⁰). Then at u = u⁰,
Proof. Exercise 1.5.
Example 3. Let f(x, y) = x² − 2xy + y³, x = u² − v², y = u² + v². Then
Corollary. Suppose f(x) = (f1(x), f2(x),…, fm(x)), where x = x(u). Then, where differentiable, we have the matrix relation
That is,
Theorem D. (The Mean Value Theorem) Suppose x⁰ and xn and fn → is differentiable at each point on the line segment [x⁰, x¹] = {tx¹ + (1 − t)x⁰; 0 ≤ t ≤ 1}. Then there exists a point xt [x⁰, x¹] so that
Proof. Apply the usual one-dimensional mean value theorem and the chain rule to t f(tx¹ + (1 − t)x⁰) defined on [0, 1].
1.3 Contour Surfaces and Sublevel Sets
The locus of all points x satisfying f(x) = f(x⁰) is called the contour surface (or a contour curve if n = 2) of f through the point x = x⁰.
Under mild assumptions on the differentiability of f near x = x⁰, we may generically, in theory, solve for one of the components xj of x, say for xn, in terms of the remaining xk [i.e., xn = xn(x1, …, xn−1) so that the portion of the contour surface f(x) = f(x⁰) near x = x⁰ is the graph of z = f(x1, x2,…, xn −1, xn(x1, …,xn.
Example 4. Consider the contour curve x² − y² = 1 of f(x, y) = x² − y² through the point (1, 0).
The locus has two disjoint branches—one in the first and fourth quadrant, the other in the second and and third quadrant. But only one branch passes through (1, 0), where we may solve for x in terms of y:
valid for all y. See Figure 1.2.
The technical result that validates the preceding intuition in the general case is a workhorse of mathematics, the so-called Implicit Function Theorem. This theorem is easily deducible from another workhorse, the so-called inverse function theorem. See Chapter 4 and Appendix A. See also Exercises 1.15 and 5.13.
Figure 1.2 The contour curve of x² − y² = 1.
The value f(x(t)) along any curve C : x = x(t) passing through x⁰ = x(0) that lies within the contour surface f(x) = f(x⁰) must, of course, be constantly f(x⁰). But then by the first chain rule,
where of course v is tangent to the curve C given by x = x(t) (i.e., the gradient is normal to the curve C). But since C was an arbitrary curve in the contour surface f(x) = f(x⁰),
The gradient is normal to the contour surface.
that is tangent to the contour surface
at x = x⁰ has normal vector ∇f(xare those x that satisfy the equation
Example 6. The plane tangent to the unit sphere f(x, y, z) = x² + y² + z² = 1 at
has equation
that is,
Example 7. The plane tangent to the surface z = f(x, y) at (x0, y0) is the plane tangent to the contour surface z − f(x, y) = 0 at (x0, y0, z0), which must be perpendicular to the gradient (−fx, − fy, 1) and so has equation
Compare with (1.10).
A sublevel set of f is of the form {x : f(x) ≤ c} and can be thought of as the side of a contour surface that contains the points where f is smaller than the values on the surface. The preceding considerations (specifically Corollary B) show that the gradient direction at x⁰ points outward from the sublevel set formed by setting c = f(x⁰).
1.4 Lagrange Multipliers
The following result is a recurring motif of the calculus of variations.
Theorem E. Suppose both f and g are realvalued functions of the real vector variable x = (x1, …, xn, n > 1, with continuous first partial derivatives on an open neighborhood of x = x⁰. Suppose also that
(a) x = x⁰ is a local minimum or maximum of f on the contour surface g(x) = g(x⁰) and
(b) ∇g(x⁰) ≠ 0.
,
Geometric Intuition. Unless ∇f(x⁰) is normal to the contour surface g(x) = g(x⁰), by (1.5) there are curves through x⁰ lying within the contour surface along which f has both larger and smaller values than f(x⁰). See Figure 1.3.
Formal Proof. If ∇f(x= 0. So assume the result false, that ∇f and ∇g are not multiples of one another at x = xare not multiples at x = x⁰ (Exercise 1.17). Consider the mapping
continuously differentiable in a neighborhood of x = x⁰. The Jacobian
of this mapping is nonzero, and so by the inverse function theorem of §5.3 and Appendix A, there are neighborhoods of x⁰ and its image on which the mapping (1.12) is bijective with a continuous inverse. But this means we can choose points x near x⁰ where g(x) = g(x⁰) but where f(x) is larger and smaller than f(x⁰).
Figure 1.3 Unless the gradients are parallel, by (1.5) there will be nearby points on the contour where f is both larger and smaller than its value at x⁰.
Example 8. What point on the unit sphere x² + y² + z² = 1 is closest to the point (2, 1, 1)?
. Let us obtain this via Lagrange multipliers.
We are to minimize f(x, y, z) = (x − 2)² + (y − 1)² + (z − 1)² subject to the constraint g(x, y, z) = x² + y² + z² = 1. By the theorem,
Thus
But then by the constraint,
.
1.5 Convexity
For x⁰, xn and 0 ≤ t ≤ 1, let
A set C n is convex provided that the line segment connecting any two points in C is itself in C (i.e., whenever x⁰, xC, then xt C for all t [0, 1]). A function f n → is convex provided that
for all x⁰, x¹, and t [0, 1].
An important aspect of convexity is that the tangent plane not only approximates the function locally but in fact sits below the function globally.
Theorem F. Suppose f is convex and is differentiable at x⁰. Then for all x n,
Proof. Fix x = xn and let xt be as in (1.16). Then by (1.17),
which can be rearranged to
Letting t 0 implies (1.18).
Corollary A. If ∇f(x⁰) = 0 and f is convex, then f has a global minimum at x = x⁰.
Thus convexity plays a special role in minimization—the first-order necessary condition that the gradient vanishes is sufficient under convexity. This will be a recurring theme of this book. For example, Theorem E has a convexity converse.
Corollary B. Suppose f and g are both convex and differentiable at x⁰. Consider the problem of minimizing f on the contour g(x) = g(x≥ 0 so that
then x⁰ is a solution to the problem—that is, f(x) ≥ f(x⁰) for all x satisfying g(x) = g(x⁰).
Geometrical interpretation. The minimum value in the optimization problem is the infimum over c that satisfy
Since sublevel sets of convex functions are convex (Exercise 1.29), if the gradients of f and g at x⁰ are pointing in opposite directions, these level sets must be tangential, and so the sets in (1.19) will become disjoint if c < f(x⁰).
Proof of Corollary B. ≥ 0, the function h(x) = f(xg(xn (Exercise 1.28). By Corollary A, f(xg(x) ≥ f(xg(x⁰) for all x n. In particular, then, f(x) ≥ f(x⁰) for all x satisfying g(x) = g(x⁰).
The final result of this chapter is a set analogue to the function result in Theorem F and is fundamental in optimization. It in effect says that closed convex sets can be separated by a hyperplane from any points not in it.
Theorem G. (The Separation Theorem) Suppose C is closed and convex and xCn and r so that
Thus a closed convex set C can be separated from a point x⁰ not in it by a hyperplane {x, x〉 = r}.
Proof.¹ Suppose xC, and let cC be such that
Such a c⁰ exists since C = x⁰ − c⁰, and fix r , c⁰〉 < r , x, x⁰ − c²| > 0. The theorem will be easily derived from the following claim:
To prove (1.20), let cC, ct and c¹ − cand ct − c⁰. Since c⁰ is the closest point in C to x⁰, we have by the law of cosines
Rearranging terms and dividing by 0 < t ≤ 1 gives
Letting t ≤ 0 and consequently that the claim (1.20) is valid.
To finish the proof, we simply note from (1.20) that for each c C,
which is (1.19).
EXERCISES
1.1 Prove that the row vector a in (1.2a) is unique.
1.2 Show that f(x) = |x| is not differentiable at x⁰ = 0 yet its directional derivative exists in every direction from x⁰ = 0.
1.3 Prove the formula (1.3) for computing the directional derivative by combining (1.1) with (1.2).
1.4 Compute the directional derivative of f = x³ + 4xy² + 3y² + z.
1.5 Prove Theorems B and C.
1.6 For polar coordinates x = r , y = r , compute the Jacobians
and
What is their matrix product AB?
1.7 in three ways for the spherical coordinates x , y , z , as follows:
(a) Make the substitutions and then take the partial derivative.
(b) Use the second chain rule.
(c) Without computation via a geometric interpretation.
1.8 Find the plane tangent to the surface z = x² − 2xy + y³ above (x, y) = (1, −1).
Answer: z = 4x + y − 1.
1.9 Parametrize the contour curve y² − 2xy − x² = −2 near (1, 1) by solving for one variable in terms of the other. Why is it possible to solve for one in terms of the other but not vice versa?
Hint: Look up the statement of the implicit function theorem in Appendix A. Or, graph the equation.
1.10 Show that u(x, t) = f(x − ct) + g(x + ct) satisfies the wave equation
for any sufficiently smooth f and g.
1.11 Conversely, suppose u satisfies the wave equation