1 Lagrange Multiplier Method

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1 1 Lagrange Multiplier Method Near a maximum the decrements on both sides are in the beginning only imperceptible. J. Kepler When a quantity is greatest or least, at that moment its flow neither increases nor decreases. I. Newton If one is looking for a maximum or a minimum of some function of many variables subject to the condition that these variables are related by a constraint given by one or more equations, then one should add to the function whose extremum is sought the functions that yield the constraint equations each multiplied by undetermined multipliers and seek the maximum or minimum of the resulting sum as if the variables were independent. The resulting equations combined with constrained equations will serve to determine all unknowns. J. Lagrange 1 These epigrams have been reproduced from the beau- 1 Joseph Louis Lagrange was born on 25th Jan. 1736, in Turin, Italy (and was named Giuseppe Lodovico Lagrangia) and lived in Paris most of his life. He is well known for his many mathematical discoveries, the calculus of variations being one of them.

2 tiful book of Tikhomirov [T]. In this section we would like to study the above statement of Lagrange which goes under the name Lagrange Multiplier Method (LMM). Recall that given a real valued function f on an open subset U of R n and a point p U, we say p is a local maximum (resp. local minimum) if there exists a neighbourhood V of p in U such that f(q) f(p) (resp. f(q) f(p)) for all q V. The point p is called a local extremum point if f(p) is a local maximum or a local minimum. For a smooth function f : U R, a necessary condition for a point p U to be a local extremum of f is that all the 1 st order partial derivatives of f vanish at p. This is deduced by merely restricting the function to various coordinate axes through the point p and then applying the corresponding result from one variable calculus. For future reference let us state and label this result here: Lemma 1 Let f : U R be a smooth function defined on an open subset of R n. A necessary condition for f(p ) to be a local extremum of f at P U is that

3 Df(P ) = 0. Lagrange Multiplier Method deals with such a necessary condition for local extremum of the function restricted to a subspace called the constraint space. Let us first make a little more concise statement out of the above quotation from Lagrange. We are given a smooth function f(x 1,..., x n ) of n variables. The problem is to find the extremal values of this function on the constraint space G given by a number of equations g i (x 1,..., x n ) = 0, i = 1,..., k. The LMM says that the points at which extrema of f may occur are contained in the space of solutions of L L = 0, i = 1,..., n; = 0, j = 1,..., k; (1) x i λ j where L is the Lagrange multiplier function defined by L := L(X, Λ) = f(x) + k λ ig i (X). (2) i=1 When we equate the derivatives with respect to λ j, s to zero, what we get back is the constraint space itself. Thus, if (x 1,..., x n, λ 1,..., λ k ) is a solution of (1), it follows that (x 1,..., x n ) is a point on the constraint space. Con-

4 versely, if (x 1,..., x n ) belongs to the constraint space, and for (λ 1,..., λ k ) the point (x 1,..., x n, λ 1,..., λ k ) satisfies the first n equations of (1) then the other k equations are also satisfied automatically. Thus we see that LMM is a clever way of converting a problem of extrema with constraints, to another problem of extrema without constraints, by increasing the number of variables. Example 1 Before we proceed any further, let us examine a simple example where we have f(x, y) = x and the constraint space is the cuspidal curve y 2 = x 3. This means x 3 0 and hence x 0. Indeed the point (0, 0) is on the curve and f(0, 0) = 0. Thus the minimum is attained precisely at (0, 0). Also we see that for any positive value of x, we can always find two solutions of y 2 = x 3 and hence it follows that f is not bounded above on the curve y 2 = x 3. So, it has no maximum. Now according to LMM, we must look for extrema of f amongst solutions of (1) where, L := L(x, Λ) = x + λ(y 2 x 3 ). (3)

5 This yields 1 3λx 2 = 2λy = y 2 x 3 = 0, (4) which, alas, has no solutions at all. Thus we see that the statement of LMM needs some technical correction. One simple way to do this is to put an additional condition on the constraint space: Theorem 1 Let G be the subspace of R n given by some smooth equations g i (x) = 0, for i = 1, 2,..., k, and f be a real valued smooth function on G. Suppose that for every P G, the set { g 1 (P ),..., g k (P )} is independent. Then the local extrema, if any, of a given smooth function f defined on G has to be found amongst the solutions of equation (1), where L is as in (2). Proof: Put g = (g 1,..., g k ) : R n R k. Then G = {x R n : g(x) = 0}. To say that { g 1 (P ),..., g k (P )} is independent is the same as saying that (Dg) P : R n R k is surjective. Let now for some P G, f(p ) be a local extremum for f. In view of the discussion above, we have

6 only to prove that there exists constants λ i, 1 i k such that ( f) P + λ 1 ( g 1 ) P + + λ k ( g k ) P = 0. (5) Apply the surjective form of ImFT (??) to the map x g(x) to obtain a neighbourhood V of 0 in R n and a diffeomorphism φ : V φ(v ) to a neighbourhood φ(v ) of P such that g φ(x 1,..., x n ) = (x 1,..., x k ). Let W be the linear span of {v j := (Dφ) 0 (e j ), : k + 1 j n}. By chain rule, it follows that Dg P Dφ 0 (e j ) = 0, k+1 j n. Therefore, ( g i ) P v j = 0 for 1 i k and k + 1 j n. This means that ( g i ) P W, the orthogonal complement of W. Since these vectors are given to be independent, and since dim W = k, they actually form a basis for W. On the other hand, since φ(0) = P, it follows that 0 is a local extremum for f φ restricted to 0 R n k. Therefore, by lemma 1, D(f φ) 0 (e j ) = 0, k + 1 j n. By chain rule, this means that ( f) P is perpendicular to v j for all k + 1 j n, i.e., ( f) P W. Therefore, (5) follows as required.

7 Remark 1 (i) A global extremum is necessarily a local extremum. Thus to solve a global extremum problem, one uses LMM for the interior of the constraint space and gets all possible local solutions. To this set one adds all local extrema on the boundary of the constraint space. Often LMM itself can be used for the local extrema on the boundary of the constraint space as a problem with an increased number of constraints. This way, we keep cutting down the size of the space on which we need to look for the possible extrema. Hopefully this process will terminate in finitely many steps into a set of finitely many points at which we can simply enumerate the value of the function and find out the actual maximum or minimum. (ii) In general, an extremum problem may not have any solutions and LMM does not address itself to the existence aspect at all. A result of Weierstrass which says that a continuous function at-

8 tains its extrema on a closed and bounded subset comes handy in many situations, to ensure the existence. So, it is good if we have the constraint space as a closed and a bounded subset. (iii) In some situations even if the constraint space is unbounded we can guarantee the existence of the extrema. For instance, suppose we know that the function tends to + as x +. Then this function will attain its minimum. (You may recall that this fact has been used in an elementary canonical proof of the Fundamental Theorem of Algebra. See for example [Sh2].) (iv) The condition that g i are independent ensures that the constraint space is a smooth submanifold of the Euclidean space with its tangent space orthogonal to the gradient lines of the constraint function. In general, a constraint space may not be smooth all over and may have corners e.g., it may be a cube or a polyhedron. Even in such cases the method can be employed for local extrema away

9 from such corner points. At the corners however, you will have to use further analysis of the situation. (v) The condition that g i are independent is not always necessary. However, in the absence of this condition, weird things can happen. From a certain advanced point of view, this happens to be the more interesting case as it allows the study of singularities and degeneracies. In order to have the full force of LMM, it is necessary that we formulate the problem in the projective space of multipliers. Take f = g 0 and put L = k i=0 λ i g i in (2). We are now looking at the projective class of multipliers [λ 0,..., λ k ] IP k 1. [ Two multipliers (λ 1,..., λ k ) and (λ 1,..., λ k) are in the same class if and only if there exists a non zero real number s such that (λ 1,..., λ k ) = s(λ 1,..., λ k). The extra condition that we put on multipliers is that they are not identically zero (i.e., at least one λ i 0]. Thus the theorem 1 takes the form:

10 Theorem 2 The extrema of the function g 0 on the constraint space G are contained in the set of solutions of (2), where L = k i=0 λ i g i. Equation (5) says that the vector ( f) P = ( g 0 ) P belongs to the linear span of {( g 1 ) P,..., ( g k ) P }. This condition is now replaced by saying that the set of vectors, {( g 0 ) P,..., ( g k ) P } is linearly dependent. Then the case that we have discussed in theorem 1 corresponds to the restricted classes in which λ 0 0. The advantage in the new formulation is that we need not put the additional hypothesis that {( g 1 ) P,..., ( g k ) P } are linearly independent. Geometrically, this amounts to allowing the constraint space to have singularities. We shall not go into more details here and be content with just discussing how this helps us to resolve the difficulty that we faced in example 1. Setting L = λ 0 x + λ 1 (y 2 x 3 ) with the only restriction that (λ 0, λ 1 ) (0, 0), we then obtain λ 0 3λ 1 x 2 = 2λ 1 y = y 2 x 3 = 0. This then allows the solution λ 0 = 0 together with (x 1, x 2 ) = (0, 0) as a probable solution,

11 which we know happens to be the actual solution! Example 2 Physical interpretation of grad Consider any linear function φ : R n R, φ(x) = i α i x i, and the problem of finding its maxima on the unit sphere S n 1. The Lagrangian in this case is n α ix i λ i=1 i=1 x2 i 1 which leads to the solution that n x i = ± α i i αi 2, i = 1, 2,..., n. (6) Let us now apply this to a specific case where φ = Df 0, and f : U R is a smooth function in a neighbourhood U of 0 R n. To each v S n 1 we can consider the path ( ɛ, ɛ) U given by t tv and look at the function t f(tv). The derivative of this map at 0 is nothing but Df 0 (v). Now Df 0 (x) = i f x i x i. Therefore, from (6), it follows that the extrema of the function v Df 0 (v) occur at ± f f. Thus f is the direction in which the increment in f is the maximum.

12 Example 3 Let x, y, z be non negative real numbers subject to the condition x + y + z = 1. Find the maximum value of f(x, y, z) = xy + yz + zx 2xyz. Observe that part of the constraint (viz., non negativity) is given by a number of inequalities. We shall ignore this part in the beginning and consider them only at a later stage. The other part of constraint is given by a linear equation. So, it seems easier to eliminate one of the variables and treat the problem with two variables only and without any constraint. I leave it to you to work out this problem in this method and see which one is actually easier. We shall carry out the LMM below. Thus, here L(x, y, z, λ) = xy + yz + zx 2xyz λ(x + y + z 1). Putting L = 0, we obtain x + y + z 1 = 0 and y + z 2yz = x + z 2xz = x + y 2xy = λ.

13 This means (y x)(1 2z) = 0 etc. and yields the following: y = x or z = 1/2 z = y or x = 1/2 x = z or y = 1/2. Observe that if none of x, y, z is equal to 1/2 then x = y = z = 1/3 and f(x, y, z) = 7/27. So, consider the case when z = 1/2. Then x = y and f(x, y, 1/2) = (x + y)/2 = 1/4 < 7/27. By symmetry, this is case with x = 1/2 as well as with y = 1/2. One should remember that the above set of solutions gives all possible local extrema, in the interior of the region under consideration. Since a global extrema which is in the interior of the constraint domain has to be amongst these, but not those which are on the boundary of the constraint domain, we should keep in mind to look at the boundary also separately. So, let us look at one of the boundary pieces say given by z = 0. This is the same as adding another

14 constraint to the above situation. Therefore, we may simply consider L(x, y, z, Λ) = f(x, y, z) λ 1 (x + y + z 1) λ 2 z. But, in this case, we see that plugging simply z = 0 in the given problem makes it easier to handle. We get L(x, y, Λ) = xy λ(x + y 1). Putting L = 0 we get y = x = λ = 1/2. Hence f(1/2, 1/2, 0) = 1/4 < 7/27. Other constraints such as x = 0, or y = 0 yield the same result because of the symmetry. The problem is not over yet. We have to consider the boundary constraints two at a time, say y = z = 0. But then f(x, 0, 0) = 0 < 7/27. Therefore the maximum value is 7/27. Incidentally we have also found out that minimum value is 0. Thus we have proved an inequality: 0 xy + yz + zx 2xyz 7 27 whenever x, y, z 0 and x + y + z = 1.

15 Example 4 The Inequality of Arithmetic and Geometric Means: Given non negative real numbers a 1,..., a n show that a 1 + a a n a 1 a 2 a n n. n We set S = i a i and x j = a j /S. Then i x i = 1, x i 0 and we must prove that x 1 x 2 x n (1/n) n. Thus we have converted the given problem into an extremal problem: L = x 1 x 2 x n λ(x 1 + x x n 1). Putting L = 0 gives x 1 λ = x 2 λ = = x n λ = x 1 x 2 x n. Since we are interested in the maximum of x 1 x n clearly we can assume that none of x i is 0. Then λ 0 and x 1 = x 2 = = x n = 1/n. Therefore the maximum value x 1 x n is at this point which is equal to (1/n) n as required. Example 5 Inequality of Geometric and Quadratic Means:

16 Given non negative real numbers a 1,..., a n that (a 1 a 2 a n ) 1/n a a 2 n n 1/2. show We set S = (a a 2 n) 1/2 and x j = a j /S. Then i x 2 j = 1. And we have to prove that (x 1 x 2 x n ) 1/n 1/n. So we take f(x 1,..., x n ) = x 1 x 2 x n and maximize it subject to the constraint i x 2 i = 1. As before the maximum is seen to have attained at x 1 = x 2 = = x n = 1/n and is equal to 1/n. This has a nice geometric interpretation: Find the box of maximum size inscribed in a sphere. The above solution tells us that this box is actually a cube and its volume is r n n/2 1, where r is the radius of the n sphere. Example 6 Cauchy s Inequality: For arbitrary real numbers

17 a 1,..., a n, b 1,..., b n show that i a ib i i a2 i 1/2 Put A = i a 2 i. If all the a i inequality is obvious. i b2 i 1/2. = 0 then the given So, we assume A 0. Put x i = a i /A so that i x 2 i = 1. We now fix b 1,..., b n and put B = i b 2 i. Observe that we may assume that B 0. We have to prove that i b i x i B subject to the constraint i x 2 i = 1. Thus: L = i b ix i λ Equating L = 0 we get i x2 i 1. 2λx i = b i, i = 1, 2,..., n. Since some b i 0, we have λ 0. Therefore 1 = i x 2 i = B 2 /4λ 2. This means λ = ±B/2. Correspondingly, we have x i = ±b i /B. The negative sign corresponds to the value B whereas the positive sign gives B. Therefore, we get B i b i x i B. Example 7 Hölder s Inequality: Given real numbers p, q > 1 such that 1 p + 1 q = 1. Let a i, b i be any non

18 negative real numbers. Then i a ib i i ap i 1/p i bq i 1/q. The proof is similar to the above after we set x j = a j /A, where A = ( i a p i ) 1/p etc.. Exercises 1. Examine the map f(x, y, z) = xyz for extremum values on the unit sphere x 2 + y 2 + z 2 = Examine the function x 2 +y 2 +z 2 for extremum values on the surface z = xy Find the maximum value of 8x 2 + 4yz 16z + 10 on the ellipsoid 4x 2 + y 2 + 4z 2 = 16. a b 4. Consider the space SL(2, R) of 2 2 matrices c d over the real numbers and with determinant ad bc = 1. Show that the Euclidean distance of SL(2, R) from

19 the origin viz., minimum of (a 2 + b 2 + c 2 + d 2 ) 1/2 is 2. What about the same problem if R is replaced by C?

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