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THE PONTRYAGIN MAXIMUM PRINCIPLE-REVIEW OF LAGRANGE MULTIPLIERS.
المؤلف:
Lawrence C. Evans
المصدر:
An Introduction to Mathematical Optimal Control Theory
الجزء والصفحة:
45-47
9-10-2016
356
CONSTRAINTS AND LAGRANGEMULTIPLIERS.What first strikes us about general optimal control problems is the occurence of many constraints, most notably that the dynamics be governed by the differential equation
This is in contrast to standard calculus of variations problems, where we could take any curve x(.) as a candidate for a minimizer.
Now it is a general principle of variational and optimization theory that “constraints create Lagrange multipliers” and furthermore that these Lagrange multipliers often “contain valuable information”. This section provides a quick review of the standard method of Lagrange multipliers in solving multivariable constrained optimization problems.
UNCONSTRAINED OPTIMIZATION. Suppose first that we wish to find a maximum point for a given smooth function f : Rn → R. In this case there is no constraint, and therefore if f(x∗) = maxx∈Rn f(x), then x∗ is a critical point of f:
∇f(x∗) = 0.
CONSTRAINED OPTIMIZATION. We modify the problem above by introducing the region
R := {x ∈ Rn | g(x) ≤ 0},
determined by some given function g : Rn → R. Suppose x∗ ∈ R and f(x∗) = maxx∈R f(x). We would like a characterization of x∗ in terms of the gradients of f and g.
Case 1: x∗ lies in the interior of R. Then the constraint is inactive, and so
(1.1) ∇f(x∗ ) = 0.
Case 2: x∗ lies on ∂R. We look at the direction of the vector ∇f(x∗). A geometric picture like Figure 1 is impossible; for if it were so, then f(y∗) would be greater that f(x∗) for some other point y∗ ∈ ∂R. So it must be ∇f(x∗) is perpendicular to ∂R at x∗, as shown in Figure 2.
Since ∇g is perpendicular to ∂R = {g = 0}, it follows that ∇f(x∗) is parallel to ∇g(x∗). Therefore
(1.2) ∇f(x∗) = λ∇g(x∗)
for some real number λ, called a Lagrange multiplier.
CRITIQUE. The foregoing argument is in fact incomplete, since we implicitly assumed that ∇g(x∗) = 0, in which case the Implicit Function Theorem implies that the set {g = 0} is an (n − 1)-dimensional surface near x∗ (as illustrated).
If instead ∇g(x∗) = 0, the set {g = 0} need not have this simple form near x∗; and the reasoning discussed as Case 2 above is not complete.
The correct statement is this:
(1.3)
If μ= 0, we can divide by μ and convert to the formulation (1.2). And if ∇g(x∗) = 0, we can take λ = 1, μ = 0, making assertion (1.3) correct (if not particularly useful).
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