The topics for the course on optimisation will this autumn be: Extrema under Constraints for real-valued functions.

Transcription

1 OPTIMISATION 5 September 2011 Overview no. 1 The topics for the course on optimisation will this autumn be: Extrema under Constraints for real-valued functions. Calculus of Variations. Optimal Control Theory. The purpose of the course is to acquaint you with the theories, their methods and the kind problems they can solve (or address). Ambitions are high in the course, inasmuch as the participants are expected to learn to solve problems themselves, by hand. As you will see, the problems go far beyond what you could treat previously by setting gradients equal to zero. Phrased briefly, we shall deal with general, yet powerful methods for rewriting optimisation problems in terms of ordinary differential equations, which can be solved. But more about this later. The course will be based on several sources: [OC] Optimisation under Constraints, lecture notes. [CF] Convex Functions, an Elementary Approach, lecture notes. [SS] Optimal control theory with economic applications, by Atle Seierstad og Knut Sydsæter; North Holland The book [SS] is out of print, but we have it available in the department. It will be a central issue in this course to work through the exercises. In fact, instead of traditional exams, I will during the course hand out 5 sets of mandatory assignments, which you should complete and get approved. 1st gathering, Wednesday September 7. We meet at 8.15 in G5-109 for the lectures. I will begin with an overview of the course, including a primer on Calculus of Variations and on Optimal Control theory. Then we go through the first topic, which is a deeper understanding of extrema of real-valued functions; we use [OC]. This includes necessary conditions for extrema (second order derivatives) and extrema under constraints think for example of finding the maximum of the function x(y 3 z) arctan(x + 2y 3z) defined on the unit sphere in R 3. From we deal with the exercises: Local maximum: Determine whether f(x, y) = x 3 + x 2 + 3x 2 y + 3xy 2 + y 3 has a local extremum at the origin. 1

2 The integral remainder term: Carry out the induction argument for Taylor s formula with integral remainder. o-functions: Show that if two functions f(t) and g(t) both are o(t n ) for t 0, then so is f(t) + g(t). Would it be justified to write: o(t n ) + o(t n ) = o(t n )? Taylor s limit formula: Determine lim x 0 e x 1 x x 2 /2 x 3 /6 x 4. The lecture notes [OC] are preliminary, but available from my web site. 2nd gathering, Friday 9 September. We meet at 8.15 in G In the lectures we complete the notes [OC], beginning with a reminder on the implicit function theorem please review this at home before the lecture! In the exercises we shall look at Local extrema: Prove that f(x, y) = y 4 x 2 does not have a local maximum at (0, 0). Explain why the necessary condition provided by Theorem 1, which states that every eigenvalue λ 0, is not sufficient for a maximum. (What type of stationary point is (0, 0)?) Find all stationary points of f(x, y) = 4xy x 4 y 4 and determine whether they are local extrema or saddle points. o-functions: Show that Taylor polynomials are unique: If f C n fulfils f(t) = p(t) + o((t ) n ) as well as f(t) = q(t) + o((t ) n ) for t, for two polynomials p, q of degree n, then p(t) and q(t) are identical. Taylor s limit formula: Find the third order Taylor polynomial of e x +1/(1 x) a. [Hint: Use the above exercise]. Constraints: Find the minimum point of f(x, y) = x + y on the unit circle, ie on the level set x 2 + y 2 = 1. Same question for g(x, y) = (x + y) 2. (Note the difference!) Find the dimensions of the largest rectangular box that can be inscribed in the ellipsoid given by ( x a )2 + ( y b )2 + ( z c )2 = 1. (You may take it for granted that this box is parallel to the coordinate axes.) Remaining exercises from the first gathering if time permits. 2

3 OPTIMISATION 12 September 2011 Overview no. 2 Last time we formulated the problem of optimising under constraints. In particular we saw that this may involve functions defined on closed sets (like the unit sphere S n 1 in R n, with empty interior so that differentiation is dubious). Or the gradient need not vanish anyhere on the surface defining the constrains. Consquently some altogether new methods will be needed for such problems. And we formulated the main result and gave an example. 3rd gathering, Wednesday 14 September. Here we meet in C3-204 at In the lectures we complete the material in [OC] on constrained optimisation, giving a proof based on the Implicit Function Theorem. We shall also have a first look at the calculus of variations, pp in [SS]. These pages in [SS] are also covered in my small lecture note Notes on calculus of variations, which is available from my web site. Exercises: Local extrema: Show that f(x, y) = 4x 2 e y 2x 4 e 4y has exactly 2 critical points; find them. Show that they are both local maxima ( Two mountains without a valley ). Surprised? could this happen in dimension 1? Degenerate critical point: Look up monkey saddle on Wikipedia (english) and study the graph (it is f(x, y) = x 3 3xy 2 ) to learn why it is called so. Can you prove that the origin is not a saddle point for this function? Constraints: Find the minimum point of f(x, y) = x + y on the unit circle, ie on the level set x 2 + y 2 = 1. Same question for g(x, y) = (x + y) 2. (Note the difference!) Find the dimensions of the largest rectangular box that can be inscribed in the ellipsoid given by ( x a )2 + ( y b )2 + ( z c )2 = 1. (You may take it for granted that this box is parallel to the coordinate axes.) Consider an electric current I that branches into three, so that I = I 1 + I 2 + I 3, where each I j runs through a resistor R j, where the power dissipation is R j I 2 j. Supposing that the actual sizes of I 1, I 2, I 3 will minimise the total loss of energy, find the ratios I 1 /I 2, I 2 /I 3 and I 1 /I 3. 3

4 4th gathering, Friday 16 September. In the lecture we shall complete pp in [SS], including the remainder of the proof for the Euler equation, which is the (lengthy) argument for differentiation under the integral sign (that is, formula (8) in [SS]). If time permits, we shall continue with pp in [SS]. Here we shall meet other, less strict terminal conditions, as well as the transversality conditions they give rise to. In the exercises you can work with Constraints: Find the points on the surface xy z 2 = 1 that has the least distance to the origin. (Hint: Which function should you minimise?) Consider the production of a buoy from steel plates; it should have circular cross section of radius r and height H, while its ends should be conical og height h (much like a pencil sharpened at both ends!). Minimise the surface area A (and hence the production costs) under the constraint that the volume V is a given constant V 0. Hint: Write both A and V as functions of r, H, h. (This is exercise in the calculus book of Edwards and Penney.) Basics: Do in [SS] (preferably at home along with your reading!). Euler equation: Solve from [SS] in the groups. 4

5 OPTIMISATION 26 September 2011 Overview no. 3 Last time we covered all essential parts of p In particular we covered the deduction of the Euler equation as a necessary condition. The special cases in Section 1.4 is left for you to read. 5th gathering, Wednesday 28 September. We continue in the lectures with Section 1.3 and 1.5. Later we proceed with Section 1.6 and some material on convex/concave functions; for the latter we can, at least for now, refer to Appendix B in [SS]. In the exercises we look at the following: Euler equation: Solve Exercise Legendre s necessary condition: Do Exercise Special cases: Solve Exercise (easy if you have read Section 1.4!) 6th gathering, Friday 30 September. The lectures will continue with Section In particular we shall give a proof for the transversality condition (iii) by means of the Implicit Function Theorem. The exercises will deal with: Special cases: Complete from last time or do Exercise Transversality conditions: Begin with Exercise Continue with Applications: Confirm your conception of the world by doing Exercise 1.4.5! Continue with (which should have general appeal... ). 5

6 OPTIMISATION 3 October 2011 Overview no. 4 Last time we covered Chapter 1.5 including details on p.39 and Example 11 in [SS]. To understand the latter one has to know the concept of the elasticity of a function. In case f : I R is differentiable on a interval I R, this is defined as el(f) = xf (x) f(x). (1) This is a dimensionless quantity that measures how sensitive f is to changes in x. The point is to measure changes relatively to the size of f(x), and that larger values of x tend to produce larger fluctuations in x; whence the factor x. 7th gathering, Wednesday 5 October. We will cover Section in [SS] and some material on convex and concave functions. For the exercises we consider: Applications: Confirm your conception of the world by doing Exercise 1.4.5! Transversality conditions: Do Exercise Continue with Sufficient conditions: Try Exercise And previous exercises if time permits. 8th gathering, Friday 7 October. We proceed in the lectures with Sections 1.3 and 5 in the notes on convex functions. Also Section 1.8 in [SS] on variational problems with several unknowns will be mentioned. The exercises will focus on Scrap functions: Do Exercise Continue with Sufficient conditions: Solve Exercise and Convex functions: Do the easy Exercise 1.6 in [CF]. Convex sets: Exercise 1.1 and 1.2 in [CF]. Applications: Continue Exercise by going through

7 OPTIMISATION October 24, 2011 Overview no. 5 Because of the failures of the computer system in week 41, the following 2 programmes were not issued prior to the gatherings: 9th gathering, Wednesday 12 October. Here we covered both Section 4 and the closely related Proposition 2.2 in [CF]. The main theme was tangents and halftangents of convex functions. The exercises 1.1, in [CF] were supplemented by in [SS]. 10th gathering, Friday 14 October. The lectures concerned Section in [SS] on Optimal Control Theory. Emphasis was laid on Example 1 and 2. In the exercises we did Exercise 2.1, 2.3, 2.4 and 4.1 in [CF] together with in [SS]. This week we shall have the 11th gathering, Wednesday 26 October. The lectures will focus on Section 2.4 in [SS], which concerns control problems with several state variables and controls. The main focus will be on Example 3 and the difficulties in exploiting the maximum principle. As the exercises we first do and then 2.3.1, th gathering, Friday 28 October. We proceed with section 2.5, where we revisit the calculus of variations and apply the maximum principle to that sphere of problems. We shall also cover section 2.6 on sufficient conditions for a maximum again the concave functions play a prominent role, so you should recall the notion in preparation for the lecture. Exercises: Do part (i) and (ii) plus verification of (80). Also and

8 OPTIMISATION 31 October 2011 Overview no. 6 Addendum after last week s lectures: Concerning Example 3 page 89 92: As noted the first line on p. 91 seems to be a misinterpretation of the necessary condition for t > 0 that T > 2/a. Another problem is that formally t = T if one takes the written definition of t literally: this is obvious since p 1 and p 2 both equal 0 at T! To overcome these shortcomings of the text one can proceed from line 5 prior to formula (74) as follows: For some small δ > 0 we have that p 1 (t) < p 2 (t) whenever t ]T δ, T [. Therefore we can define t = sup{ t [0, T δ] p 1 (t) p 2 (t) }; (2) this is with the modification that t = 0 in case the set is empty. If t > 0, obviously p 1 < p 2 on ]t, T [, so it follows that u = 0 on ]t, T ]; and moreover by continuity of p 2 p 1 that p 1 (t ) p 2 (t ), and the opposite inequality follows from the definition of t, i.e. p 1 (t ) = p 2 (t ). So by solving the differential equation for p 1 on [t, T ], one has that a(t t ) 2 = 2(T t ), hence t = T 2 a. (3) Therefore t > 0 implies that T > 2/a. Conversely, given that T > 2/a, it necessary that t > 0. For if it is assumed that t = 0, then p 1 < p 2 holds on the entire interval, so that u = 1 and thence a 2 (t T )2 < T t holds for all t in [0, T ]; which gives the contradiction that T 2 a < t for all t [0, T ]. (4) So, knowing that t > 0, we note from the above that t = T 2/a and proceed to find the behaviour of p 1 on [0, t ]. Clearly { ap 1 when p 1 > p 2, ṗ 1 = (5) ap 2 when p 1 p 2, which in both cases leads to the inequality ṗ 1 (t) ap 2 (t) = a(t T ). (6) In particular one has for t < t that ṗ 1 (t) a(t T ) = 2 whereas ṗ 2 (t) = 1. Hence (p 2 p 1 ) 2 1 > 0, so the fundamental theorem gives for such t, p 2 (t) p 1 (t) = p 2 (t ) p 1 (t ) t t 8 (p 2 (s) p 1 (s)) ds < = 0. (7)

9 Consequently p 1 (t) > p 2 (t) whenever 0 t t. So by the above, the differential equation for p 1 reduces on [0, T 2/a] to ṗ 1 (t) = ap 1 (t), which is solved by p 1 (t) = Ce at. Since p 1 (t ) = p 2 (t ) = T t = 2/a, this yields p 1 (t) = 2 a e a(t T +2/a) for 0 t T 2/a. Moreover, we have that u (t) = 1 on the interval [0, T 2/a]. Altogether the Maximum Principle yields only one candidate for T > 2/a, namely { u 1 for t [0, T 2/a], (t) = (8) 0 for t ]T 2/a, T ]. (Cf. the convention that u(t) is continuous from the left at each point.) It is now an easy task to find the associated state variables, that is, the functions x 1(t) and x 2(t) etc. (see the book). In case T 2/a, then it necessarily holds that t = 0. Or, in other words, that p 1 < p 2 on the entire interval [0, T ]. And in this case, it was seen that u (t) = 1 must hold for all t; i.e. there is only the single candidate u 1. 13th gathering, Wednesday 2 November. We continue in Section 2.5 of [SS] with the sufficiency theorem of Arrow, which applies to cases without the crucial concavity of H(x, u) assumed in the theorem of Mangasarian. For this subject we need to invoke the existence of supergradients of concave functions, which closely related to the subgradients of convex functions. The discussion of this will be based on Section 2 in [CF]. The exercises will concern Hamiltonians vs. variations: Do Exercise (Straightforward!) Maxima of concave functions: Solve (hint: use the Mean Value Theorem for (i)). Abnormal problems: Work through Exercise (or the rest of it). Note to Excercise 2.6.5: In the Mangasarian Theorem it is implicitly assumed that the control region U is a subset of an open set, say A so that f j (t, x, u) is of class C 1 (R n A) for each fixed t and j {0, 1,..., n}. (Reason: partial derivatives are only defined for functions defined on open sets.) With this proviso, the exercise accounts for the statement in (119). 14th gathering, Friday 4 November. We complete the review of convex funtions by going through the remaining parts of [CF]. This comprises the enveloping halfspaces, subdifferentiability and continuity in Section 1 3. Exercises will be made in Sufficient conditions: Solve Exercise 2.6.1, Concavity: Do Subdifferentials: Solve Exercise 2.3 and 2.5 in [CF]. 9

10 OPTIMISATION 7 November 2011 Overview no. 7 15th gathering, Wednesday 9 November. We continue with Section 2.8 on conditions for existence of an optimal control (pp ). For good reasons (to be explained) this will involve so-called measurable functions a non-trivial class of functions. In the exercises we look at the following: Convex sets: Solve Exercise 1.4 in [CF] (easy!). Proceed with Exercise 1.2. Mangasarian/Arrow: Show explicitly that the Hamiltonian function in Example 5 p in [SS] is not concave with respect to (x, u). Old exercises in the remaining time. 16th gathering, Friday 11 november. In the lectures we proceed with section 2.9 (ca pp ) on problems in which the final time t 1 is a variable, so that solutions of the control problem have the form (u (t), t 1). As time permits, we also take up a discussion of the results on linear problems, which are briefly summarised in section For the exercises we consider Convex sets: Try your hand with Exercise 1.3 there. (Hint: Make a drawing first.) Existence of solutions: Solve Exercise and Previous exercises if time permits. 10

11 OPTIMISATION 9 November 2011 Overview no. 8 We have today substantiated Note 16 on page 133 in [SS], first of all by adding the requirements that a(t0 0 and b(t) 0 there (these were most likely forgotten by the authors). Secondly we showed that the claim in note 16 was a simple consequence of the well-known lemma below (it has been further adapted to the piecewise continuous functions that appear in control theory): Lemma (Grönwall s inequality). For two non-negative, piecewise continuous functions, say f(t) 0 and k(t) 0, on an interval [, T [ (possibly with T = ) the first of the following inequalities implies the second: f(t) c(t) + t k(s)f(s) ds c(t) exp( t k(s) ds); (9) hereby c(t) 0 can be any monotone increasing function on [, T [. Proof: In case f, k are continuous, the proof may be conducted by showing (9) for the right endpoint of an interval of the form [, t 1 ] for arbitrary t 1 >. Then c(t) may be replaced by the constant c(t 1 ). Moreover, define F (t) = c(t 1 )+ t k(s)f(s) ds for t t 1 ; then F (t) is differentiable with F (t) = k(t)f(t). One may therefore use k(t) exp( t k(s) ds) as an integration factor: the obvious inequality f(t) F (t) implies that f(t)k(t) exp( t This is by the Leibniz rule equivalent to k(s) ds) F (t)k(t) exp( t d (F (t) exp( dt t k(s) ds)) 0. k(s) ds) 0. By integration over [, t 1 ] it follows that F (t 1 ) F ( ) exp( t 1 k(s) ds), hence that f(t 1 ) F (t 1 ) c(t 1 ) exp( t 1 k(s) ds), as was to be shown. In general, if t 1 > is the first point of discontinuity of f or k, then for every t in the next interval of continuity, say for t [t 1, t 2 ], one has by splitting the integral over [, t] and using the above that f(t) = ( c(t 1 ) + c(t 1 ) exp( c(t) exp( t1 k(s)f(s) ds ) + ( c(t) c(t 1 ) + t1 t 0 t k(s) ds) + (c(t) c(t 1 )) exp( k(s) ds). 11 t t t 1 k(s)f(s) ds ) t 1 k(s) ds)

12 Indeed, the last inequality is seen by using the positivity of k to estimate both integrals by the integral over the full interval [, t]. With finitely many similar steps, the validity of the second inequality of the lemma is extended beyond any point of discontinuity of f and k; hence to the whole interval [, T [. 12

13 OPTIMISATION 11 November 2011 Overview no. 9 17th gathering, Monday 14 November, at We continue with section 2.10 on linear problems. Observe, however, that this section is rather sketchy. We therefore follow notes supplied by . The exercises comprise Convexity: Do Exercise in [SS]. Non-existence: Elucidate this possibility by solving Exercise Time minimisation: Solve Exercise th gathering, Wednesday 16 November. Here we complete the theory of linear control systems in Section 2 of the lecture notes [NOCP] handed out: these are called Some notes on Optimal Control Prolems. Exercises: Solve the problems in the notes [NOCP]. 13

14 OPTIMISATION 21 November 2011 Overview no th gathering, Wednesday 23 November. Here we focus on the time minimisation problem for linear control systems; cf. Section 3 in [NOCP]. The exercises will be 3.1, 3.2 and 3.3 in [NOCP]. 20th gathering, Friday 25 November. The main theme will be Section 3.7 in [SS] on problems with infinite time horizons. In particular the 4 different criteria for optimality: OT, CU, SCU and PW. The exercise will be to give a proof that OT = CU = SCU = PW. (10) Syllabus. In Optimal control theory with economic applications by Seierstad og Sydsæter we have read Chapter 1 (except 1.10) and Chapter 2 (except 2.11) and Chapter 3.7. However, Chapter 2.10 of [SS] was replaced by Section 1 3 in Some Notes on Optimal Control Problmes (that were handed out), Notions and results for convex sets and functions were covered by reading Notes on Convex Functions an elementary approach (handed out). Finally some standard results for optimisation of real-valued functions, with or without constraints, were convered as well from the notes Optimisation under Constraints. 14