Optimal Control for Chemical Engineers

The goal of this book is to provide a sufficiently detailed treatment of optimal control that will enable readers to formulate optimal control problems and solve them. With this emphasis, the book provides necessary mathematical analyses and derivations of important results. It is assumed that the reader is at the level of a graduate student.


Simant Ranjan Upreti


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Chemical Engineering

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  • Simant Ranjan Upreti   
  • 310 Pages   
  • 05 May 2015
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    Optimal Control for Chemical Engineers read more..

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    CRC Press is an imprint of the Taylor & Francis Group, an informa business Boca Raton London New York Optimal Control for Chemical Engineers Simant Ranjan Upreti read more..

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    CRC Press Taylor & Fr ancis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2 013 by Taylor & Fr ancis Group, LLC CRC Press is an imprint of Taylor & Fr ancis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20121207 International Standard Book Number: read more..

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    This book is dedicated to my parents, Drs. Hema and Murari Lal Upreti, Grand Master Choa Kok Sui, Shaykh Khwaja Shamsuddin Azeemi, and Lord Gan.apati read more..

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    Contents Preface xiii Notation xv 1 Introduction 1 1.1 Definition .. ... .. .. ... .. .. ... .. .. ... .. . 1 1.2 Optimal Control versus Optimization ... .. .. ... .. . 4 1.3 Examples of Optimal Control Problems . . . . . . . . . . . . 5 1.3.1 Batch Distillation . . . . . . . . . . . . . . . . . . . . 5 1.3.2 Plug Flow Reactor ... .. .. ... .. read more..

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    viii 2.5.2 Application to Optimal Control Problems . . . . . . . 44 2.6 Second Variation . .. .. ... .. .. ... .. .. ... .. . 50 2.6.1 Second Degree Homogeneity . . ... .. .. ... .. . 50 2.6.2 Contribution to Functional Change . . . . . . . . . . . 51 2.A Second-Order Taylor Expansion . . . ... .. .. ... .. . 52 Bibliography . . . ... .. .. ... .. read more..

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    ix 5 Pontryagin’s Minimum Principle 123 5.1 Application . ... .. .. ... .. .. ... .. .. ... .. . 123 5.2 Problem Statement . .. ... .. .. ... .. .. ... .. . 126 5.2.1 Class of Controls . ... .. .. ... .. .. ... .. . 126 5.2.2 New State Variable ... .. .. ... .. .. ... .. . 127 5.2.3 Notation . . . . . . ... .. .. ... .. .. ... .. . 127 5.3 read more..

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    x 7 Numerical Solution of Optimal Control Problems 185 7.1 Gradient Method . . . . ... .. .. ... .. .. ... .. . 185 7.1.1 Free Final Time and Free Final State . . . . . . . . . 185 7.1.2 Iterative Procedure ... .. .. ... .. .. ... .. . 186 7.1.3 Improvement Strategy . . . . . ... .. .. ... .. . 187 7.1.4 Algorithm for the Gradient Method . . . . . . read more..

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    xi 9.7 Autonomous Differential Equations . . . . . . . . . . . . . . 270 9.7.1 Non-Autonomous to Autonomous Transformation . . 270 9.8 Differential . ... .. .. ... .. .. ... .. .. ... .. . 271 9.9 Derivative . ... .. .. ... .. .. ... .. .. ... .. . 272 9.9.1 Directional Derivative . .. .. ... .. .. ... .. . 272 9.10 Leibniz Integral Rule . . ... .. .. read more..

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    Preface It is my pleasure to present this book on optimal control geared toward chem- ical engineers. The application of optimal control is a logical step when it comes to pushing the envelopes of unit operations and processes undergoing changes with time and space. This book is essentially a summary of important concepts I have learned in the last 16 years from the classroom, read more..

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    xiv mended to solve optimal control problems — whether one intends to write one’s own programs or use software and programs developed by others. Optimal control is the result of tremendous contributions of wonderful mathematicians, scientists, and engineers. To list their achievements is a formidable task. What I have presented in this book is what I could under- stand and have read more..

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    Notation Vectors We will use lower case bold face letters for vectors. For example, y ≡ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ y1 y2 .. . yn ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ is a column vector. It has n components. The transpose of y is y ≡ y1 y2 ... yn where y is a row vector. Also, y(t) means that each component of y is time dependent. Function Vectors A function vector is a read more..

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    xvi Matrices We will most often use upper case bold face for matrices, e. g., A ≡ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ a11 a12 ... a1n a21 a22 ... a2n .. . .. . ... .. . am1 am2 ... amn ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ which is an m × n matrix. The matrix components can be functions. Derivatives We will use an over dot ˙ to denote the derivative with respect to time t.A prime read more..

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    xvii The derivative of a vector f with respect to a scalar, say, t, is again a vector of partial derivatives. For example, ft ≡ ∂f1 ∂t ∂f2 ∂t ... ∂fn ∂t The derivative of a vector f with respect to another vector u is a Jacobian matrix of partial derivatives. Thus, fu ≡ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ∂f1 ∂u1 ∂f1 ∂u2 ... ∂f1 read more..

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    Chapter 1 Introduction This chapter introduces optimal control with the help of several exam- ples taken from chemical engineering applications. The examples elucidate the use of control functions to achieve what is desired in those applications. The mathematical underpinnings illustrate the formulation of optimal control problems. The examples help build up the notion of objective read more..

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    2 Optimal Control for Chemical Engineers any time, however, the temperature is the same or uniform throughout the reactor because of perfect mixing. Such a system is described by the mass balances of the involved species or the equations of change. They are differ- ential equations, which have time as the independent variable in the present case. An optimal control problem for read more..

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    Introduction 3 with the initial conditions x(0) = x0,y(0) = y0, and z(0) = 0. In the above equations, k0 is the Arrhenius constant, E is the constant acti- vation energy of the reaction, R is the universal gas constant, and T is the absolute temperature dependent on t. The temperature T (t) is a control func- tion, which is undetermined. It can be suitably changed to read more..

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    4 Optimal Control for Chemical Engineers 1.2 Optimal Control versus Optimization It is easy to perceive from the above example that optimal control involves optimization of an objective functional subject to the equations of change in a system and additional constraints, if any. Because of this fact, optimal control is also known as dynamic or trajectory optimization. The salient read more..

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    Introduction 5 1.3 Examples of Optimal Control Problems To gain further understanding of the applications of optimal control, let us study some examples of optimal control problems. In each problem, there is a system changing with time or some other independent variable. The system is mathematically described or modeled with the help of differential equations. At least one such read more..

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    6 Optimal Control for Chemical Engineers binary liquid mixture bottom still condenser distillate recycle overhead vapors steam u(t) product y(t) m(t), x(t) Figure 1.3 Batch distillation process with distillate production rate versus time or u(t) as the control function subject to Equations (1.7)–(1.9) as well as the purity specification y∗ for the distillate product given by y∗ = tf 0 read more..

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    Introduction 7 minimum reactant) concentration by controlling pressure along the reactor length. z = L z y y0 gas with y = y0 product gas out z = 0 I = y(L) A 2B plug flow reactor k1 k2 Figure 1.4 Gas-phase reaction in a plug flow reactor Assuming the ideal gas law to hold, the model for the reactor of uniform cross-section is given by dy dz = τS −k1yP (z) 2y0 − y read more..

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    8 Optimal Control for Chemical Engineers 1.3.3 Heat Exchanger Figure 1.5 shows a single-tube heat exchanger used to heat (or cool) the fluid flowing inside the tube by controlling its wall temperature Tw as a function of time t (Huang et al., 1969). At any time, Tw is uniform along the z-direction, i. e., the length of the heat exchanger. z = L z T w(tf) T s T0 tube read more..

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    Introduction 9 where T0 is the fluid temperature at the inlet of the heat exchanger. The steady state temperature Ts(z) is the temperature defined by Equa- tion (1.13) with the time derivative set to zero and Tw(t) replaced with θ. Thus, dTs dz = h vρCp [θ − T ] (1.15) Subject to the satisfaction of Equations (1.13)–(1.15), the optimal control problem is to find the read more..

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    10 Optimal Control for Chemical Engineers gas liquid layer P time pressure gas mass in liquid diffusion cell c sat(t) z = 0 z = L Figure 1.6 Diffusion of gas into the liquid layer and boundary conditions: c(0,t)= csat(t)for 0 <t ≤ tf (1.19) ∂c ∂z ⏐ ⏐ ⏐ ⏐ z=L =0 for 0 ≤ t ≤ tf (1.20) where csat is the equilibrium saturation concentration of the gas at the read more..

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    Introduction 11 subject to Equations (1.16)–(1.21). 1.3.5 Periodic Reactor This reactor poses a optimal periodic control problem, which involves periodic control functions. Their application can result in better performance relative to steady state operation and help achieve difficult performance criteria such as those involving molecular weight distribution (MWD). Figure 1.7 shows one such read more..

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    12 Optimal Control for Chemical Engineers x(t), y(t), V output stream with polymer product F x F y t monomer: x f, F x(t) polymer: y f, F y(t) feed streams Figure 1.7 A constant volume stirred tank reactor with periodic feed flow rates like those on the right must be satisfied simultaneously. 1.3.6 Nuclear Reactor Consider nuclear fission in a reactor where neutrons react with read more..

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    Introduction 13 control rod inserted rod reactor core neutrons and fissile material base position u(t) = 0 heat flux retracted rod u(t) < 0 u(t) > 0 Figure 1.8 Control rod positions in a nuclear reactor where x and y are the concentrations of neutrons and precursors, t is time, r = r[u(t)] (1.29) is the degree of change in neutron multiplication as a known function of read more..

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    14 Optimal Control for Chemical Engineers being injected from the top, the solvent drastically reduces the viscosity of oil, causing it to drain under gravity and get produced at the bottom. We would like to maximize the oil production in Vapex by considering solvent pressure versus time as a control function to influence solvent concentration at the top. injection well dra ine d read more..

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    Introduction 15 initial and boundary conditions are ω(z, 0) = 0, 0 ≤ z< L0 (1.32) ω[L(t),t]= ωi[P (t)] (1.33) where L is the height of oil in the reservoir with the initial value L0, ωi is the solvent mass fraction in the solvent–oil interface at the top, and P is the pressure of the solvent. The relationship between ωi and P is known a priori from experiments. read more..

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    16 Optimal Control for Chemical Engineers where ˙y2,in is the constant rate of immune cells that enter the organ to fight cancer cells, and u(t) is the rate of drug injection into the organ (de Pillis and Radunskaya, 2003). The risand βis are constants in the growth terms, while αisand λis are the constants in the decay terms arising due to the action of the drug. read more..

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    Introduction 17 u polymer matrix with drug x tissue in contact x = L diffusive flux of drug x = 0 Figure 1.10 Drug release from a polymer matrix to a tissue in contact At any time, the flux of the drug into the tissue at the point of contact is given by J(t)= −D ∂c ∂x x=L (1.44) It is desired to match J(t) with a specified drug release versus time relation, read more..

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    18 Optimal Control for Chemical Engineers r0 R skin surface at Te(r,t) T(r,t) skin tissue with blood vessels ambient at T Figure 1.11 Cross-section of the skin layer surrounding a cylindrical limb The heat transfer model for the skin layer is given by ∂T ∂t = k rρCp ∂ ∂r r ∂T ∂r + F (t)Cb(Tb − T ) Cp + ΔH(t) ρCp (1.45) where T is temperature, t and r are time read more..

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    Introduction 19 minimizes the difference between the model-predicted T (R, t) and its experi- mental counterpart Te(R, t), i. e., the objective functional I = τ 0 1 − T (R, t) Te(R, t) 2 dt 1.4 Structure of Optimal Control Problems The above examples help us identify the structure of optimal control problems. As shown in Figure 1.12, an optimal control problem involves one or read more..

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    20 Optimal Control for Chemical Engineers Open-Loop Control In most optimal control problems, it is not possible to obtain optimal control laws, i. e., optimal controls as explicit functions of system state. Note that system state is the set of system properties such as temperature, pressure, and concentration. They are subject to change with independent variables like time and space. read more..

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    Introduction 21 control of a continuous “living” anionic polymerization. I. Theoretical study. J. Appl. Poly. Sci., 31:1019–1039, 1986. H.-S. Huang, L.T. Fan, and C.L. Hwang. Optimal wall temperature control of a heat exchanger. Technical Report 15, Institute for Systems Design and Optimization, Kansas State University, Manhattan, 1969. S. Lu, W.F. Ramirez, and K.S. Anseth. Modeling and read more..

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    22 Optimal Control for Chemical Engineers 1.4 Revise the drug delivery example in Section 1.3.9 (p. 16) to minimize the consumption of drug, simultaneously ensuring that its flux does not fall below a threshold value over a certain time duration tf. 1.5 What new conditions would be required if the optimal periodic control problem of Section 1.3.5 (p. 11) is changed to a read more..

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    Chapter 2 Fundamental Concepts This chapter introduces the fundamental concepts of optimal control. Be- ginning with a functional and its domain of associated functions, we learn about the need for them to be in linear or vector spaces and be quantified based on size measures or norms. With this background, we establish the dif- ferential of a functional and relax its definition read more..

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    24 Optimal Control for Chemical Engineers Figure 2.1 Evaluation of a function f for x = a and x = b x = b x = a f(a) f(b) for the evaluation of I(f ), the function f assumes not a single value but the set of all values from f (a)to f (b). This fact becomes explicit in the discrete equivalent of Equation (2.1), i. e., I = n →∞ i=1 fi · Δxi lim Δxi→0 read more..

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    Fundamental Concepts 25 on a set of functions, each of which assumes a set of values for functional evaluation. 2.1.1 Functional as a Multivariable Function A functional dependent on a function is analogous to a multivariable function dependent on a vector of variables. For example, consider the functional K(f )= f1 2f2 f3 2 dependent on a function f comprising three components f1, read more..

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    26 Optimal Control for Chemical Engineers 2.2.1 Linear or Vector Spaces In optimal control, we desire to find the minimum or maximum value of a functional defined over a specified domain. The analytical procedure is to continuously change the associated function from some reference form and examine the corresponding change in the functional. The new form of the function is, in read more..

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    Fundamental Concepts 27 For a continuous function f (x)inthe x-interval [a, b], a definition of the norm could be f = b a [f (x)]2dx (2.4) We can easily verify that Equations (2.3) and (2.4) satisfy the properties of a norm. Thus, given a norm, we can compare the sizes of any two vectors. Con- sidering a change in a vector or a function, which is reflected in its norm, read more..

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    28 Optimal Control for Chemical Engineers where y denotes the derivative dy/dx. The functional J is linear and therefore homogeneous. However, the functional K is homogeneous but not linear. If z = αy,then z = αy and K(αy)= K(z)= b a α2y2 αy − αy dx = α b a y2 y − y dx = αK(y) showing that K is a homogeneous functional. But K does not satisfy the linearity read more..

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    Fundamental Concepts 29 vanish with h . Dividing both sides of Equation (2.6) by h and taking its limit to zero, we get the following equivalent definition for the Fr´echet differential: lim h →0 I(y0 + h) − I(y0) − dI(y0; h) h = lim h →0 (h) h = 0 (2.7) Placing no restriction on the form of h, i. e., the shape of the curve h(x), this requirement implies that read more..

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    30 Optimal Control for Chemical Engineers possible ways — from (i) hk,1 to hl,1, (ii) hk,1 to hl,2, (iii) hk,2 to hl,1,and (iv) hk,2 to hl,1. Figure 2.4 illustrates an analogous example with y as a continuous function and the norm defined by Equation (2.4). In this case, hk,1(x)and hk,2(x)are two different forms of h(x)of norm k with different directions (Section 9.20, p. read more..

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    Fundamental Concepts 31 where the ratio of 1(αh) to a scalar multiplier α vanishes with α itself for all h in the domain of I with 0 < αh <δ.Note that 1(αh)/α is the ratio (αh)/ αh with the introduction of 1(αh) ≡ (αh)/ h . Similar to Equation (2.7), the equivalent definition of the Gˆ ateaux differ- ential is lim α →0 I(y0 + h) − I(y0) − read more..

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    32 Optimal Control for Chemical Engineers x = b hk,2 hk,1 hm,2 hm,1 h = 0 h > 0 h < 0 x = a hl,1 hl,2 Figure 2.6 Changes in a continuous function for the Gˆ ateaux differential from β to (β + α), we get dI(y0; h) = lim Δγ →0 {I[y0 +(γ +Δγ)h] − I(y0 + γh) }γ=0 Δγ = d dγ I(y0 + γh)γ=0 Replacing γ by α, the Gˆ ateaux differential can be read more..

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    Fundamental Concepts 33 Let y ≡ (y0 + αh) be a function in the vicinity of y0. Then Equation (2.9), along with the above definition of y and specifications of y0 and h,provides dI(y0; h)= d dα I( y0 + αh y )α=0 = d dα 1 0 y 2 − 2y +1 α=0 dx = 1 0 d dα y2 − 2y +1 α=0 dx = 1 0 d dy (y2 − 2y +1) dy dα α=0 dx = 1 0 2(y0 + αh y ) − 2 α=0 h read more..

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    34 Optimal Control for Chemical Engineers Let γ be the maximum absolute value of (h − k)inthe x-interval [0, 1]. Then ⏐ ⏐ ⏐dI(y0; h) − dI(y0; k) ⏐ ⏐ ⏐ = ⏐ ⏐ ⏐ ⏐2 1 0 (y0 − 1)(h − k)dx ⏐ ⏐ ⏐ ⏐ ≤ ⏐ ⏐ ⏐ ⏐2 1 0 (y0 − 1)γ dx ⏐ ⏐ ⏐ ⏐ With γ related to h − k as γ = κ h − k where κ is some positive real number, we read more..

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    Fundamental Concepts 35 Example 2.3 Find the Gˆ ateaux differential of the functional I(y)= 1 0 |y| dx corresponding to the reference y-function y0 and the variation h given by y0 =0 and h =2x, respectively. Let y ≡ (y0 + αh) be a function in the vicinity of y0. Applying Equa- tion (2.10), dI(y0; h)= 1 0 d dy |y| α=0 h dx =2 1 0 d dy |y| α=0 x dx Now at α =0, we read more..

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    36 Optimal Control for Chemical Engineers 2.4.3 Variation To deal with an even wider class of functionals, the concept of the Gˆ ateaux differential is further relaxed to the Gˆ ateaux variation, or simply the variation. Similar to the Gˆ ateaux differential, the variation δI(y0; h) of a functional I is 1. defined by I(y0 + αh) − I(y0)= δI(y0; αh)+ 1(αh) (2.11) where read more..

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    Fundamental Concepts 37 The above variation is of course homogeneous with respect to h since δI(y0; αh)= 2 1 0 (y0 − 1)αh dx =2α 1 0 (y0 − 1)h dx = αδI(y0; h) Generalization to Several Functions For a functional dependent on a vector y of several functions in general, the above definitions for differentials apply with the vectors y0 and h replacing y0 and h, read more..

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    38 Optimal Control for Chemical Engineers It can be easily verified that the above variation is homogeneous, i. e., δI(y0; αh)= αδI(y0; h) 2.4.4 Summary of Differentials In summary, we have three types of differentials of a functional: 1.Fr´echet differential, 2.Gˆ ateaux differential, and 3.Gˆ ateaux variation, or simply variation in the order of decreasing strictness for their read more..

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    Fundamental Concepts 39 Example 2.7 Consider the functional I(y)= ⎧ ⎨ ⎩ y1y2 y1 + y2 if y =0 0if y =0 where the function (or equivalently the vector) y has two real components, y1 and y2. Find the variation of I at y0 = 0, i. e., for y0,1 =0 and y0,2 =0. In the vicinity of y0,we have ⎡ ⎣ y1 y2 ⎤ ⎦ y = ⎡ ⎣ 0 0 ⎤ ⎦ y0 + α ⎡ ⎣ h1 h2 ⎤ ⎦ h = read more..

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    40 Optimal Control for Chemical Engineers 2.5 Variation of an Integral Objective Functional Let us derive the variation of the integral objective functional I(y)= b a F (y, y )dx where y and y (or dy/dx) are functions of the independent variable x.The variation has to be obtained in the neighborhood of a reference function y0(x) having its derivative y0(x) with respect to x. If read more..

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    Fundamental Concepts 41 • y is the reference function y0(x), and y is the corresponding derivative function y0(x); and • δy and δy are the functions h(x)and h (x), respectively. In the above equation, the coefficient of δy (i. e., Fy) is called the vari- ational derivative of the functional I with respect to y. Similarly, Fy is called the variational derivative of I read more..

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    42 Optimal Control for Chemical Engineers where y is a function of x.If dF/dy exists and is continuous with respect to x, then the variation of I is also the Fr´echet differential as well as the Gˆ ateaux differential. 2.5.1.1 Equivalence to the Fr´ echet Differential The Fr´echet differential dI(y; h) is defined by lim h →0 I(y + h) − I(y) − dI(y; h) h = 0 (2.7) read more..

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    Fundamental Concepts 43 where ⏐ ⏐¯h ⏐ ⏐ is defined in terms of a positive real number β and the norm h . Multiplying the last two inequalities and integrating with respect to x over the interval [a, b], we get b a ⏐ ⏐ ⏐Fy (¯ y, x) − Fy(y, x) h(x) ⏐ ⏐ ⏐ dx< β(b − a) 1 h where we introduce ≡ β(b − a) 1. From the triangle inequality for read more..

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    44 Optimal Control for Chemical Engineers 2.5.2 Application to Optimal Control Problems In an optimal control problem, we arrive at an integral objective functional having the general form J = tf 0 f (y, ˙y, u, λ)dt (2.19) where the integrand depends on 1. the vector of state variables: y = y1(t) y2(t) ... yn(t) , 2. the derivative of y with respect to time t: ˙y = read more..

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    Fundamental Concepts 45 where δy, δ ˙y, δu,and δ λ are the variation vectors corresponding to y, ˙y, u, and λ, respectively. Thus, for example, δy = δy1(t) δy2(t) ... δyn(t) Example 2.10 A simpler version of the batch reactor problem (p. 2) for the reaction aA −→ cC without any inequality constraints is as follows. Find the T (t) that maximizes I = ck0 tf 0 exp − read more..

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    46 Optimal Control for Chemical Engineers the variation of J is given by δJ = tf 0 (fxδx + f ˙ xδ ˙x + fT δT + fλδλ)dt where fx, f ˙ x, fT ,and fλ are partial derivatives of f with respect to x,˙x, T , and λ, respectively. Therefore, δJ = tf 0 k0 exp − E RT (c − aλ)ax a −1 δx − [λ]δ ˙x + k0E RT 2 exp − E RT (c − aλ)xa δT − ˙x read more..

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    Fundamental Concepts 47 functional: J = L 0 T (x, tf) − T ∗(x) 2 F dx + L 0 tf 0 λ − ˙ T − vT + h ρCp (Tw − T ) G dt dx (2.25) subject to T (z, 0) = T (0,t)= T0 where the undetermined multiplier λ is a function of the two independent variables x and t. Note the presence of the double integral corresponding to these variables. Let us now find the read more..

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    48 Optimal Control for Chemical Engineers subject to ˙ m = −u, (2.27) ˙x = u m (x − y), (2.28) m(0) = m0,and x(0) = x0 with μ as a time invariant multiplier used to incorporate the purity specification, Equation (1.10), in Equation (2.26). The problem at hand has only two constraints, namely, Equations (2.27) and (2.28). With the application of the Lagrange Multiplier Rule, read more..

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    Fundamental Concepts 49 Further Simplification The variations of integral objective functionals can be further simplified at their optima by integrating by parts the terms involving derivative functions such as δ ˙x and δ ˙ m in Equation (2.29). As a matter of fact, this simplification is an important step in deriving the optimality conditions. Therefore, it is very important to read more..

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    50 Optimal Control for Chemical Engineers Upon substituting the last two equations into Equation (2.29), we get δJ = tf 0 ˙λ 1 − λ2u m2 (x − y) δm + ˙λ 2 + λ2u m δx + u μ − λ2 m δy + 1+ μ(y − y∗) − λ1 + λ2 m (x − y) δu dt − λ1δm tf 0 − λ2δx tf 0 The variation of J thus obtained is readily amenable to further analysis for the read more..

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    Fundamental Concepts 51 Replacing h by βh in Equation (2.31), we get δ 2I(y0; βh)= d2 dα2 I(y0 + αβh)α=0 Let γ ≡ αβ. Then using the chain rule of differentiation, δ2I(y0; βh)= d dγ d dγ I(y0 + αβh)γ=0 dγ dα dγ dα = β2 d2 dγ2 I(y0 + γh)γ=0 = β2δ2I(y0; h) 2.6.2 Contribution to Functional Change The functional change, I(y0 + h) − I(y0), canbeexpressed interms read more..

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    52 Optimal Control for Chemical Engineers Generalization to Several Functions The above result can be generalized to a functional dependent on several functions. Thus, I(y0 + h) − I(y0)= δI(y0; h)+ 1 2 δ 2I(y0; h)+ 2(h) (2.33) where I depends on the function vector y given by ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ y1(x) y2(x) .. . yn(x) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ read more..

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    Fundamental Concepts 53 we can write I(y0 + αh) − I(y0)= α 0 I (y0 + βh) v × 1 w dβ = α 0 vw dβ = vw β=α β=0 − α 0 v w dβ (upon integration by parts) = αI (y0 + βh)β=0 δI(y0;h) + α 0 [I (y0 + βh)](α − β)dβ R2 (2.34) where I (y0 + βh) is the second derivative of I(y0 + βh) with respect to β. If m and M are the respective minimum and read more..

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    54 Optimal Control for Chemical Engineers Bibliography D.G. Luenberger. Optimization by Vector Space Methods, Chapter 7, pages 169–175. John Wiley & Sons Inc., New York, 1969. M.Z. Nashed. Some remarks on variations and differentials. Am. Math. Mon., 73(4):63–76, 1966. H. Sagan. Introduction to the Calculus of Variations, Chapter 1, pages 9–29. Dover Publications Inc., New York, read more..

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    Fundamental Concepts 55 2.8 The objective functional of an optimal control problem depends on func- tions as well as their derivatives. Under what circumstances would the varia- tion of such a functional become the differential? 2.9 Show that the Gˆ ateaux differential of the functional in Example 2.3 (p. 35) is (i) −1if y0 < 0, and (ii) 1 if y0 > 1 in the range read more..

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    Chapter 3 Optimality in Optimal Control Problems This chapter presents the conditions related to the optimality of a func- tional. We derive the necessary conditions for optimal control to exist and apply them to optimal control problems. We also present sufficient conditions assuring the optimum under certain conditions. Readers are encouraged to review the logic of the conditional read more..

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    58 Optimal Control for Chemical Engineers 1. greater than or equal to zero when α is greater than zero, or 2. less than or equal to zero when α is less than zero Thus, δI = 0 is the only non-contradicting condition that is applicable for the minimum of I. In fact, we obtain the same condition if I happens to be maximum at ˆ y.In this case, we proceed with the read more..

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    Optimality in Optimal Control Problems 59 through the integrand F , the indirect influence stems from y being affected by u through the state equation constraint, i. e., Equation (3.5). Hence, to solve the problem, we need to first obtain an explicit solution y = y(u) and then substitute it in the expression of F . However, such solutions do not exist for most optimal read more..

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    60 Optimal Control for Chemical Engineers where δF = Fyδy + Fuδu (3.9) and δG = Gyδy + G ˙ y δ ˙y + Guδu (3.10) We rewrite Equation (3.8) as δJ = tf 0 (δF + λδG)dt J1(δy,δu) + tf 0 Gδλ dt J2(δλ) =0 where the first integral (J1) depends on arbitrary functions δy and δu,and the second integral (J2) depends on another arbitrary function δλ. The function δ read more..

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    Optimality in Optimal Control Problems 61 Arising from δJ = 0, both Equations (3.11) and (3.12) are the necessary conditions for the optimum of J. On the basis of the Lagrange Multiplier Rule, these two equations are also the necessary conditions for the constrained optimum of I. Both optima are still subject to the given initial condition, i. e., Equation (3.6). The read more..

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    62 Optimal Control for Chemical Engineers Satisfaction of J3 =0 Upon substituting G ˙ y = −1, which is obtained from the definition of G in Equation (3.5), Equation (3.13) becomes J3 = tf 0 (Fy + λGy)δy dt − tf 0 λδ ˙y dt = 0 (3.16) Before simplifying the above equation further, we need to express the second integral in terms of δy. Applying integration by parts read more..

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    Optimality in Optimal Control Problems 63 • or changes with t,inwhich case pδy attains at least once thesamevalue p(tf)δy(tf)at t1 less than tf (Case 2). Suppose this conclusion is not true, so that p(t1)δy(t1) >p(tf)δy(tf), 0 ≤ t1 <tf Then p(tf)δy(tf) is the minimum integrand value. Thus, tf 0 pδy dt> p(tf)δy(tf)tf which contradicts Equation (3.19). The other situation read more..

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    64 Optimal Control for Chemical Engineers Table 3.1 Necessary conditions for the optimum of I in the problem defined by Equations (3.4)–(3.6) 1. The state equation ˙y = g(y, u), 0 ≤ t ≤ tf (3.5) with initial condition y(0) = y0 (3.6) 2. The costate equation ˙λ = −(Fy + λGy), 0 ≤ t ≤ tf (3.22) with final condition λ(tf)= 0 (3.23) 3. The stationarity Fu + λGu read more..

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    Optimality in Optimal Control Problems 65 Example 3.1 Find the necessary conditions for the minimum of I = tf 0 (u2 − yu)dt subject to the satisfaction of the differential equation constraint ˙y = − √ y + u (3.24) where u(t) is the control and y(t) is the state variable with the initial condition y(0) = 0. The augmented functional for this problem is J = tf 0 u2 − yu read more..

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    66 Optimal Control for Chemical Engineers with T (t) as the control and x(0) = x0 as the initial condition. This is the simple batch reactor control problem (see Example 2.10, p. 45) for which the augmented functional is J = tf 0 ckxa F + λ − ˙x − akxa g G dt with the functions F , g,and G as indicated above. Referring to Table 3.1, T is the control u as a read more..

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    Optimality in Optimal Control Problems 67 Example 3.3 Repeat the previous example utilizing the Hamiltonian. The Hamiltonian is given by H = ckxa + λ( −akxa) Therefore, the necessary conditions for the maximum of I are 1. the state equation, Equation (6.10), obtained from ˙x = Hλ and the initial condition x(0) = 0; 2. the costate equation, Equation (3.26), obtained from ˙λ read more..

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    68 Optimal Control for Chemical Engineers where fyi = Fyi + n j=1 λjGjy i ,f ˙ yi = n j=1 λjGj ˙ yi = − n j=1 λj, and fui = Fui + m j=1 λjGju i The Necessary Conditions Extending the optimal control analysis (Section 3.2.2, p. 59) to the generalized problem with multiple states and controls, and defining the Hamiltonian as H = F + n i=1 λigi (3.34) we get the necessary read more..

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    Optimality in Optimal Control Problems 69 subject to the satisfaction of the state equations ˙x = −kaxayb g1 (3.40) ˙y = −kbxayb g2 (3.41) where T is the control and k = k0 exp − E RT (3.42) The initial conditions are x(0) = x0 and y(0) = y0 This is the batch reactor control problem (seep. 2) without the inequality con- straints. Referring to Table 3.2, x and y are, read more..

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    70 Optimal Control for Chemical Engineers is given by (Ray and Soliman, 1970) ˙x = Q V (xf − x) − kxy (3.43) ˙y = m − Qy V (3.44) ˙ T = Q V (Tf − T ) − ΔH ρCp kxy − hA + K(T − Ts) VρCp (T − Tc) (3.45) where V ,ΔH, ρ,and Cp are, respectively, the reactor volume, heat of reaction, liquid density, and specific heat capacity. The subscript f denotes the read more..

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    Optimality in Optimal Control Problems 71 2. the costate equations ˙λ 1 = −Hx = −2(x − xs)+ λ1 Q V + ky + λ3 ΔH ρCp ky ˙λ 2 = −Hy = −2(y − ys)+ λ1kx + λ2 Q V + λ3 ΔH ρCp kx ˙λ 3 = −HT = −2(T − Ts)+ λ1 kE RT 2 xy + λ3 Q V + ΔH ρCp kE RT 2 xy + hA + K(2T − Ts − Tc) VρCp with the final conditions λi(tf)= 0, i =1, 2, 3. read more..

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    72 Optimal Control for Chemical Engineers 3. improvement of each control function utilizing the gradient information from Hui — the variational derivative of the objective functional with respect to the control function ui. The improvement in control functions causes the objective functional value to get closer to the optimum. Therefore, iterative application of the above steps leads to read more..

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    Optimality in Optimal Control Problems 73 3.3.1 Presence of Several Local Optima When solving an optimal control problem, it has to be kept in mind that several local optima may exist. Consider for example a problem with a single control function. The objective functional value may be locally optimal, i. e., optimal only in a vicinity of the obtained optimal control function. In read more..

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    74 Optimal Control for Chemical Engineers conditions minimum (I ≥ ˆ I) or maximum (I ≤ ˆ I) minimum maximum or ˆ I an optimum: ˆ I a global optimum: necessary sufficient conditions Figure 3.3 Necessary and sufficient conditions for the optimum of the objective functional I Naturally, we would like to determine the optimum that is global, i. e., valid over the entire space of read more..

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    Optimality in Optimal Control Problems 75 1. the controls satisfy the necessary conditions for the minimum, 2. the costates at the minimum are non-positive if g [i. e., the vector of gis in Equation (3.32) on p. 67] is nonlinear with respect to the states or controls, and 3. the functions F [Equation (3.31), p. 67] and g are convex jointly with respect to the states and read more..

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    76 Optimal Control for Chemical Engineers The Condition If we place a condition that the error ratio 2/ δu 2 vanishes with δu ,and δ2I ≥ α δu 2 where α is some positive number, then I(y, u) − I(ˆ y, ˆ u) ≥ α 2 δu 2 + 2 ≥ δu 2 α 2 + 2 δu 2 If we can find a sufficiently small neighborhood of ˆ u such that all δu therein have norms sufficiently read more..

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    Optimality in Optimal Control Problems 77 discontinuity jump discontinuity y t u3 u1 corner tf 0 u2 t1 t2 ut 1,t2 jump Figure 3.4 A piecewise continuous control ut 1,t2 and the accompanying state y(t) Let us use ut 1,t2 to denote a piecewise continuous control with jump dis- continuities at times t1 and t2.Thus, ut 1,t2 is made up of three continuous controls — u1, u2,and u3, as read more..

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    78 Optimal Control for Chemical Engineers discontinuities. Therefore, to find the control ut 1,t2 that minimizes I, we apply the developed optimal control analysis to the following three subproblems: 1.Find the u1 that minimizes I1 subject to ˙y = g(y, u),y(0) = y0, 0 ≤ t ≤ t1 2.Find the u2 that minimizes I2 subject to ˙y = g(y, u),t1 ≤ t ≤ t2 where y(t1) is read more..

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    Optimality in Optimal Control Problems 79 Introducing φ(t) ≡ t 0 (Fy + λGy)dt and considering that δy is zero at t = 0 due to the initial condition [Equa- tion (3.6), p. 58], we can write Equation (3.47) as δy(tf)φ(tf) − tf 0 [φ(t)+ λ]δ ˙y dt = 0 (3.48) Note that the above equation is valid for the entire class of continuous δy functions that are zero at t = read more..

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    80 Optimal Control for Chemical Engineers conforms to the current conditions of δy being zero at t =0 and tf.For this choice, δ ˙y is given by δ ˙y = c(t) − c0 (3.50) and is admissible. We will use it in the following result tf 0 [c(t) − c0]δ ˙y dt = tf 0 c(t)δ ˙y dt − c0 δy tf 0 = 0 (3.51) which is due to Equation (3.49) and the conditions δy(0) = read more..

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    Optimality in Optimal Control Problems 81 3.B Vanishing of (Fy + λGy + ˙λ) at t =0 We need to show that the coefficient (Fy + λGy + ˙λ)of δy in the following equation on p. 63: tf 0 Fy + λGy + ˙λ δy dt = 0 (3.20) has to be zero at t = 0 even though δy there is zero. Suppose that the coefficient (Fy + λGy + ˙λ) is greater than zero at t read more..

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    82 Optimal Control for Chemical Engineers (Fy + λGy + ˙λ)δy t1 δy (Fy + λGy + ˙λ) 0 tf 0 Figure 3.5 The functions Fy + λGy + ˙λ , δy, and their product, i. e., the inte- grand of Equation (3.20) 2. all non-positive (non-negative) elements of the costate vector λ in the interval [0,tf] at the minimum (maximum) if g = g1 g2 ... gn is nonlinear with respect to read more..

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    Optimality in Optimal Control Problems 83 Since ˆ u satisfies the necessary optimality conditions (see Table 3.2, p. 68), λ = − ˆ Hy ⇒ ˆ Fy = − ˙λ − ˆ gy λ ⇒ ˆ Fy = − ˙λ − λ ˆ gy (3.54) ˆ Hu =0 ⇒ ˆ Fu = −ˆgu λ ⇒ ˆ Fu = −λ ˆ gu (3.55) Using Equations (3.54) and (3.55) in Inequality (3.52), and integrating both sides provides I − ˆ read more..

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    84 Optimal Control for Chemical Engineers W.H. Ray and M.A. Soliman. The optimal control of processes containing pure time delays — I. Necessary conditions for an optimum. Chem. Eng. Sci., 25:1911–1925, 1970. H. Sagan. Introduction to the Calculus of Variations, Chapter 1, pages 32–43. Dover Publications Inc., New York, 1969. D.R. Smith. Variational Methods in Optimization, Chapter read more..

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    Optimality in Optimal Control Problems 85 by (Jensen, 1964) ˙y1 = u4 − (u1 + u2 + u4)y1 − a1y1y2 − a2y1y6u3 ˙y2 = u1 − (u1 + u2 + u4)y2 − a1y1y2 − 2a3y2y3 ˙y3 = u2 − (u1 + u2 + u4)y3 − a3y2y3 ˙y4 = −(u1 + u2 + u4)y4 +2a1y1y2 − a4y4y5 ˙y5 = −(u1 + u2 + u4)y5 + a5y2y3 − a4y4y5 ˙y6 = −(u1 + u2 + u4)y6 +2a4y4y5 − a2y1y6u3 ˙y7 = read more..

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    Chapter 4 Lagrange Multipliers In this chapter, we introduce the concept of Lagrange multipliers. We show how the Lagrange Multiplier Rule and the John Multiplier Theorem help us handle the equality and inequality constraints in optimal control problems. 4.1 Motivation Consider the simplest optimal control problem, in which we wish to find the control function u that optimizes I = tf read more..

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    88 Optimal Control for Chemical Engineers y = y(u) into Equation (3.4) and obtain I in terms of u alone. However, this approach fails in most problems where analytical solutions of the involved constraints are simply not possible. 4.2 Role of Lagrange Multipliers The above difficulty is surmounted by introducing an undetermined function, λ(t), called Lagrange multiplier, in the read more..

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    Lagrange Multipliers 89 then the following conditions are necessary: 1. The constraint is satisfied, i. e., K(ˆ y)= k0 (4.2) 2. There exists a constant multiplier λ such that δI(ˆ y; δy)+ λδK(ˆ y; δy) = 0 (4.3) for all variations δy. Remarks 1. The provision that δK(ˆ y; δz) = 0 along at least one variation δz is called the normality condition or the constraint read more..

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    90 Optimal Control for Chemical Engineers Note that most of the optimal control problems have continuous partial derivatives of the involved integrands and, therefore, satisfy the require- ment of weak continuity. Outline of the Proof As shown in Figure 4.1, together with the givens and an assumption of a non- zero Jacobian of I and K at the optimum, we will invoke the Inverse read more..

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    Lagrange Multipliers 91 be, respectively, i(μ, ν) ≡ I(ˆ y + μδy + νδz) (4.4) and k(μ, ν) ≡ K(ˆ y + μδy + νδz) (4.5) for all μ and ν within a region R that is sufficiently small to allow for the weak continuity of δI as well as δK at y =(ˆ y + μy + νz). Figure 4.2 shows the (μ, ν) coordinates in R mapped to the (i, k) coordinates by read more..

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    92 Optimal Control for Chemical Engineers The last equation follows from the definition of the variation [see Equa- tion (2.12), p. 36]. Similar results are obtained for the remaining partial derivatives. Using them in the definition of D,we obtain D(μ, ν)= δI(ˆ y + μδy + νδz; δy) δI(ˆ y + μδy + νδz; δz) δK(ˆ y + μδy + νδz; δy) δK(ˆ y + μδy + read more..

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    Lagrange Multipliers 93 which leads to the second necessary condition for the minimum of I(y)at ˆ y subject to the constraint K(y)= k0. The same condition is obtained for the constrained maximum of I by considering i1 greater than I(ˆ y). Expanding Equation (4.9), we get δI(ˆ y; δy)δK(ˆ y; δz) − δI(ˆ y; δz)δK(ˆ y; δy) = 0 (4.10) for all functions δy and δz. With read more..

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    94 Optimal Control for Chemical Engineers 1. The variations of both I and K are weakly continuous near ˆ y. 2. The constraint qualification that the variation of K at y =ˆ y is non-zero for at least one variation of y. Observe that Equations (4.2) and (4.3), which arise from δJ(ˆ y; δy) = 0, are also the necessary conditions for the optimum of J. This fact gives read more..

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    Lagrange Multipliers 95 weak continuity of δI, δK1,and δK2 near ˆ y, then from Equation (4.6) it follows that D(μ, ν1,ν2) ≡ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ∂i ∂μ ∂i ∂ν1 ∂i ∂ν2 ∂k1 ∂μ ∂k1 ∂ν1 ∂k1 ∂ν2 ∂k2 ∂μ ∂k2 ∂ν1 ∂k2 ∂ν2 ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ = ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ δI(y; δy) δI(y; read more..

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    96 Optimal Control for Chemical Engineers Lagrange Multiplier Theorem for Several Equality Constraints Generalizing the above results for m constraints, Ki(y)= ki, i =1, 2,... ,m, the necessary conditions for the optimum of I(y)at ˆ y Ki(ˆ y)= ki,i =1, 2,... ,m δI(ˆ y; δy)+ m i=1 λiδKi(ˆ y; δy)= 0 subject to the following preconditions: 1. The variations of I and Ki, i =1, read more..

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    Lagrange Multipliers 97 Let I be a functional dependent on the vector of n functions y = y1 y2 ... yn and subject to m constraints Ki(y)= ki,i =1, 2,... ,m If satisfying the above constraints I is optimum at y = ˆ y such that 1. the functionals δI and all δKis are weakly continuous near ˆ y,and 2. there exists a set of m variations (δz1,δz2, ..., δzm) where δzi read more..

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    98 Optimal Control for Chemical Engineers where Kiˆy is the vector of partial derivatives of Ki with respect to y and is evaluated at y = ˆ y,i. e., Kiˆy = ∂Ki ∂y1 ∂Ki ∂y2 ... ∂Ki ∂yn ˆ y Thus, the constraint qualification can be written as η0 = ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ K1ˆyδz1 K1ˆyδz2 ... K1ˆyδzm K2ˆyδz1 K2ˆyδz2 ... K2ˆyδzm .. . .. . read more..

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    Lagrange Multipliers 99 4.3.3 Application to Optimal Control Problems We will apply the Lagrange Multiplier Rule to obtain the set of necessary conditions for the optimum in an optimal control problem constrained by a differential equation. In Section 3.2, we asserted the rule and obtained the following necessary conditions (see p. 60): G = 0 (3.11) and tf 0 (δF + λδG)dt = 0 read more..

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    100 Optimal Control for Chemical Engineers equation constrains the corresponding values of y to y0 + t1 t0=0 g dt y1 ,y1 + t2 t1 g dt y2 , ... , yn−2 + tn−1 tn−2 g dt yn−1 , and yn−1 + tn=tf tn−1 g dt yn where for each i-th subinterval [ti,ti+1], the length Δti ≡ (ti+1 − ti) tends to zero. The i-thsuchconstraintis ti+1 ti g(y, u)dt = yi+1 − yi which can be read more..

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    Lagrange Multipliers 101 Thus, the optimum of J2 is equivalent to the optimum of I subject to the first two constraints. In this way, we sequentially arrive at J ≡ Jn = I + n i=1 λiKi, lim n →∞ whose optimum is equivalent to the optimum of I subject to all constraints Ki(zi)= 0,i =1, 2,... ,n, lim n →∞ (4.11) Consequently, the necessary conditions for the optimum read more..

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    102 Optimal Control for Chemical Engineers where λ is an undetermined function of the independent variable t.The bracketed term multiplied by λ is δG obtained from the definition of G. Hence, the above equation can be expressed as tf 0 (δF + λδG)dt = 0 (3.12) To summarize, the above equation and G = 0 (3.11) along with y(0) = y0 are necessary for the constrained optimum read more..

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    Lagrange Multipliers 103 1. The partial derivatives of F and g with respect to y and u are continuous in the vicinity of the minimum. 2. The following constraint qualification is satisfied for G ≡− ˙y + g.There exists at least one set of variations (δ ˙y, δy, δu)at each t in (0,tf]for which the variation of G is not zero at the minimum. Example 4.3 Similarly in read more..

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    104 Optimal Control for Chemical Engineers subject to the autonomous ordinary differential equations ˙y = g(y, u) with the initial conditions y(0) = y0 where g is the function vector g1(y, u) g2(y, u) ... gn(y, u) The above problem is equivalent to optimizing the augmented functional J = tf 0 F + λ ( − ˙y + g) G dt subject to the initial conditions where λ is the vector read more..

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    Lagrange Multipliers 105 where the controls u1 and u2 are the volumetric flow rate of the CSTR and the catalyst mass flow rate, respectively. The aim is to minimize the deviation of the state of the CSTR with minimum control action, i. e., to minimize I = tf 0 2 i=1 (xi − xsi) 2 +(ui − usi) 2 dt where the superscript s denotes the steady state value. The above read more..

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    106 Optimal Control for Chemical Engineers subject to ˙y = g(y, u), y(0) = y0 To the base problem we add the following algebraic equality constraints: hi(y, u)=0,i =1, 2,... ,l or h(y, u)= 0;0 ≤ t ≤ tf where l is less than m, the dimension of u. From Section 4.3.3.2 (p. 103), the base problem is equivalent to the optimization of J = tf 0 F + λ ( − ˙y + g) read more..

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    Lagrange Multipliers 107 Example 4.5 In the last example, let us say we want u2 to be increasingly higher for a smaller amount of the catalyst-to-reactant ratio. For this purpose we enforce the following equality constraint at all times: x2 x1 = a 1 bu2−c − 1 where a, b,and c are some suitable parameters. The objective is to minimize I subject to state equations and the read more..

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    108 Optimal Control for Chemical Engineers Step 1 Let J be optimal at ˆ u, which depends on k0, the value of the con- straint. Then at any t in the t-interval, Taylor’s first order expansion gives ˆ u(k0 +Δk0)= ˆ u(k0)+ ˆ uk 0 Δk0 + where /Δk0 vanishes with Δk0. We can rewrite the above equation as ˆ u(k0 +Δk0)= ˆ u(k0)+Δk0 (ˆ uk 0 + /Δk0) δu Step 2 read more..

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    Lagrange Multipliers 109 Step 3 The augmented objective functional is M ≡ J + μK where μ is a Lagrange multiplier. From the Lagrange Multiplier Theorem, assuming that δK[ˆ u(k0); ˆ uk 0 ] =0, we have δM [ˆ u(k0); ˆ uk 0 ]= δJ[ˆ u(k0); ˆ uk 0 ]+ μδK[ˆ u(k0); ˆ uk 0 ]= 0 The last two equations yield Jk 0 (ˆ u)= −μδK[ˆu(k0); ˆ uk 0 ] (4.18) Repeating Step read more..

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    110 Optimal Control for Chemical Engineers J and K are Gˆ ateaux differentiable since we will need the continuity of the differentials. Let J be minimum at ˆ u among all u satisfying the inequality constraint. Then K(ˆ u) could be either k0 or less. We will consider these two cases as follows: Case 1 Here K(ˆ u)= k0 and the inequality constraint is said to be active. read more..

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    Lagrange Multipliers 111 The necessary condition for the minimum at ˆ u(t)is δM (ˆ u)= δJ(ˆ u)+ μδK(ˆ u)= 0 K(ˆ u) ≤ k0 μ ≥ 0,μ[K(ˆ u) − k0]= 0 complementary slackness condition The following preconditions must be satisfied: 1. The Gˆ ateaux differentials of both J and K are weakly continuous near ˆ u(t). 2. The constraint qualification — There exists a δu for read more..

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    112 Optimal Control for Chemical Engineers 1. The Gˆ ateaux differential dK(ˆ y, u; δu) is weakly continuous near ˆ u for t = tf. This means that the partial derivative Ku at ˆ y is continuous. 2. The constraint qualification — If the constraint is active, there exists a control variation δu for which δK(ˆ y, ˆ u; δu) =0 for t = tf.It means that Ku =0 at read more..

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    Lagrange Multipliers 113 4.5.1 Generalized John Multiplier Theorem The approach in the preceding section may be followed to arrive at the fol- lowing the John Multiplier Theorem for several inequality constraints: The necessary conditions for the minimum of J(y, u) subject to K(y, u) ≤ k at ˆ u(t)are δM (ˆ y, ˆ u)= δJ(ˆ y, ˆ u)+ μ δK(ˆ y, ˆ u)= 0 K(ˆ y, ˆ u) ≤ k read more..

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    114 Optimal Control for Chemical Engineers and the following algebraic equality and inequality constraints: x2 x1 = a 1 bu2−c − 1 −u2 + umin ≤ 0 u2 − umax ≤ 0 The equivalent problem is to minimize J given by Equation (4.15) on p. 105 subject to the initial conditions and the algebraic constraints. It is assumed that the preconditions for the minimum of J (see p. read more..

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    Lagrange Multipliers 115 4.5.2 Remark on Numerical Solutions When solving an inequality-constrained optimal control problem numerically, it is impossible to determine which constraints are active. The reason is one cannot obtain a μ exactly equal to zero. This difficulty is surmounted by considering a constraint to be active if the corresponding μ ≤ α where α is a small positive read more..

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    116 Optimal Control for Chemical Engineers Outline of Proof Based on the given function f , we will develop an auxiliary function g.We will show it to be a contraction, which is associated to a unique fixed point. This property will then lead to the existence of the inverse function f inv.We start with the description of a contraction and its fixed point. Contraction A read more..

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    Lagrange Multipliers 117 Step 3 Using Inequality (4.24) for each right-hand side term above, we get xi − xj ≤ (c i + c i+1 + ··· + c j −2 + c j −1) P x1 − x0 Because c is a positive fraction, the coefficient on the right-hand side P ≤ P + cj + cj+1 + ··· = ci(1 + c + c2 + ... )= ci 1 − c because of which xi − xj ≤ ci 1 − c x1 − read more..

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    118 Optimal Control for Chemical Engineers Step 1 The continuity of f at x0 implies that there exists an open set X in which x − x0 <δ, δ > 0 (4.26) such that f (x) − D < 1, 1 > 0 (4.27) Differentiating g(x) with respect to x,we get g (x)= I − D− 1f (x)= D− 1[D − f (x)] where I is the identity matrix. Taking the norm on both sides of the read more..

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    Lagrange Multipliers 119 Step 3 Let h(s) ≡ [g(1) − g(0)] g(s). Then for 0 ≤ s ≤ 1, the Mean Value Theorem for derivatives (Section 9.14.1, p. 276) yields h(1) − h(0) = dh ds s (1 − 0) = [g(1) − g(0)] dg ds s Also, from the definition of h(s), h(1) − h(0) = [g(1) − g(0)] [g(1) − g(0)] = g(1) − g(0) 2 From the last two equations, g(1) − g(0) 2 read more..

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    120 Optimal Control for Chemical Engineers Injection of f (x) Consider g(x) defined by Equation (4.25) in the open set described by In- equality (4.26). Since g(x) is a contraction, it is associated with a unique fixed point given by x = g(x)or x = x − f (x) − y D Since the determinant of D is not zero, further simplification yields f (x)= y which holds for exactly read more..

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    Lagrange Multipliers 121 A. Takayama. Mathematical Economics, Chapter 8, pages 646–667. Cam- bridge University Press, Cambridge, UK, 1985. Exercises 4.1 Show that the costate variables in the optimal control problem a. are continuous with respect to time, and b. have piecewise continuous time derivatives 4.2 Show that the augmented functional formed by adjoining even the initial condition read more..

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    read more..

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    Chapter 5 Pontryagin’s Minimum Principle One of the most profound results of applied mathematics, Pontryagin’s minimum principle provides the necessary conditions for the minimum of an optimal control problem. The elegance of the principle lies in the simplicity of its application to a vast variety of optimal control problems. Boltyanskii et al. (1956) developed the principle originally read more..

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    124 Optimal Control for Chemical Engineers with the initial condition y(0) = y0 (3.6) This problem is the same as the one in Section 3.2 (p. 58). The Hamiltonian for this problem is defined as H(y, λ, u)= F (y, u)+ λg(y, u) According to Pontryagin’s minimum principle, if ˆ u is optimal, then the cor- responding Hamiltonian H(ˆ y, ˆ λ, ˆ u) at each time instant is read more..

  • Page - 144

    Pontryagin’s Minimum Principle 125 for sufficiently small variation δu in ˆ u. Now, for sufficiently δu at any time, we can express the Hamiltonian at (ˆ y, ˆ λ, ˆ u + δu) using the first order Taylor expansion as H(ˆ y, ˆ λ, ˆ u + δu u )= H(ˆ y, ˆ λ, ˆ u)+ Hu(ˆ y, ˆ λ, ˆ u)δu where u is the perturbed control (ˆ u + δu). From the last two read more..

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    126 Optimal Control for Chemical Engineers Case 5 The integrand F in Equation (3.4) is a function of |u|.Now, when u = 0 the partial derivative of Fu does not exist, and we cannot apply the stationarity condition, Hu = 0. However, Pontryagin’s minimum principle does not require the partial derivatives Fu and gu to exist. According to the principle, H(ˆ y, ˆ λ, u) ≥ read more..

  • Page - 146

    Pontryagin’s Minimum Principle 127 1.The controls uisare piecewise continuous with respect to t. Figure 3.4 (p. 77) showed one such control, which is made of three continuous curves. When two such curves meet, there is a jump discontinuity,e. g., at time t1, as shown in the figure. On either side of t1,the control is provided by the curve on that side. At t ≥ t1 we read more..

  • Page - 147

    128 Optimal Control for Chemical Engineers 3.The i-th component of g is the function gi(y, u) ≡ gi(y1,y2,... ,yn,u1,u2,...,um) It is implicitly understood that any gi does not depend on y0. 4. The costate at the minimum is denoted by λ without the hat ˆ. Using the above notation, the state equations can be written as ˙y ≡ dy dt = g(y, u) (5.3) and the Hamiltonian can read more..

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    Pontryagin’s Minimum Principle 129 5.3.2 Statement If ˆ u is the optimal control resulting in the optimal state ˆ y over the time interval [0, ˆ tf]inwhich I is minimum, then there exists a non-zero vector λ such that the following conclusions hold: 1. The corresponding Hamiltonian ˆ H ≡ H(ˆ y, λ, ˆ u) is minimum over the set of all admissible controls at each point in read more..

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    130 Optimal Control for Chemical Engineers given a pulse perturbation given a temporal show that λ0 is a constant conclude that the Hamiltonian is zero at the final time corresponding to ˆ u,i. e., H(ˆy, λ, ˆ u), a. constant in a subinterval where ˆ u is continuous b. the infimum of H(ˆy, λ, u) w.r.t. the value of u 2. the infimum is continuous throughout the entire read more..

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    Pontryagin’s Minimum Principle 131 perturbed ˆ ui ˆ ui Δt1 Δtf ˆ ui(ˆtf) perturbation t1 ˆ tf perturbation temporal at t1 t0 vi ˆ ui(t) t ˆ ui is continuous pulse Figure 5.2 Two types of perturbations in ˆ ui,the i-th component of the optimal control ˆ u temporal perturbation in the final time. These perturbations change the optimal state generated by the optimal control. The time read more..

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    132 Optimal Control for Chemical Engineers Since there is no perturbation prior to (t1 − Δt1), y at this time is the same as ˆ y. Hence, the difference between the above two equations gives y(t1) − ˆ y(t1)= t1 t1−Δt1 [g(y, u) − g(ˆ y, ˆ u)] dt (5.6) Step 2 Being a continuous function of (y0,y1,...,yn) as per Assumption 1 (Section 5.3.1, p. 128), g(y, v) read more..

  • Page - 152

    Pontryagin’s Minimum Principle 133 y(t1)= ˆ y(t1)+ t1 t1−Δt1 g[ˆ y(t1), v] − g[ˆ y(t1), ˆ u(t1)] + O(Δt1) dt = ˆ y(t1)+ g[ˆ y(t1), v] − g[ˆ y(t1), ˆ u(t1)] Δt1 first order term + O(Δt 2 1) second order term (5.7) We consider Δt1 to be sufficiently small so that the second order term vanishes. With this provision, the change of state at t1 is Δy(t1) ≡ y(t1) − read more..

  • Page - 153

    134 Optimal Control for Chemical Engineers which is first order accurate with respect to Δt1, the interval of pulse per- turbation. We take Δt1 to be small enough for the above equation to be true. 5.4.2 Temporal Perturbation of Optimal Control As shown in Figure 5.2, we now consider a truncation or extension of the optimal control ˆ u at the final time ˆ tf.If Δtf is read more..

  • Page - 154

    Pontryagin’s Minimum Principle 135 x θ ≥ π/2 (yf − ˆ yf) ˜ y z on the line y1 yf ˆ yf boundary point is any point y2 convex set S hyperplane P normal p = x − ˆ yf Figure 5.3 Aconvexset S of final states with a supporting hyperplane P .A hyperplane is a tangent line in two dimensions Supporting Hyperplane Now for a convex set, there exists a supporting hyperplane read more..

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    136 Optimal Control for Chemical Engineers 5.4.4 Choice of Final Costate Let us choose λ(ˆtf)= −p. Then the above inequality becomes λ(ˆtf) (yf − ˆ yf) ≥ 0 (5.13) Simply put, we choose the final costate to be a vector opposite to the normal p and pointing toward the convex set of final states (see Figure 5.3).∗ The first component of the costate, λ0, is a read more..

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    Pontryagin’s Minimum Principle 137 relevant to an engineering optimal control problem, we implicitly assume that λ0 is not zero. It could be zero in a rare problem where the solution is independent of the objective functional, i. e., the latter is inconsequential. See Kamien and Schwartz (1991) for an example. In practice, therefore, we expect to get solutions with any non-zero read more..

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    138 Optimal Control for Chemical Engineers where v is any admissible control value at t1, including ˆ u(t1). Hence, the above inequality shows that at t1, the Hamiltonian achieves the minimum with the optimal control function. Note that t1 is any arbitrary time instant when the optimal control is contin- uous. Hence the optimal control, whenever continuous, minimizes the Hamil- read more..

  • Page - 158

    Pontryagin’s Minimum Principle 139 3. Due to the above two reasons, y and λ, being respective solutions of Equation (5.3) on p. 128 and Equation (5.14) on p. 136, are continuous functions of t. As a consequence, H is a continuous function of t in the subinterval [ti,tj]. Next, we examine a function defined as ˆ H(t) ≡ H[ˆ y(t), λ(t), ˆ u(t)] and ˆ H(t, tc) ≡ H[ˆ read more..

  • Page - 159

    140 Optimal Control for Chemical Engineers Minimality of the Hamiltonian Being continuous in the subinterval [ti,tj], ˆ u minimizes the Hamiltonian so that H[ˆ y(t), λ(t), ˆ u(t)] ˆ H(t) ≤ H[ˆ y(t), λ(t), ˆ u(tc)] ˆ H(t,tc) Subtracting ˆ H(tc) from both the sides of the above inequality, ˆ H(t) − ˆ H(tc) ≤ ˆ H(t, tc) − ˆ H(tc) (5.19) Depending on t we have the read more..

  • Page - 160

    Pontryagin’s Minimum Principle 141 in the subinterval. Expanding the above inequality, we get ˆ H(tj) − ˆ H(ti) tj − ti ≤ ˆ H(t) − ˆ H(tmin) t − tmin (5.22) Now Inequality (5.21), which is also satisfied for tc = tmin,provides ˆ H(t) − ˆ H(tmin) t − tmin < (5.23) From Inequalities (5.22) and (5.23), ˆ H(tj) − ˆ H(ti) tj − ti < or ˆ H(tj) − ˆ H(ti) < read more..

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    142 Optimal Control for Chemical Engineers Constancy of ˆ H in [ti,tj] Since t is arbitrary within the closed subinterval [ti,tj], the intersection of Inequalities (5.24) and (5.28) yields ˆ H(ti)= ˆ H(tj). Hence if tk is an arbitrary time in [ti,tj], then ˆ H(ti)= ˆ H(tk)= ˆ H(tj) By necessity therefore, ˆ H is constant throughout the subinterval where the optimal control is read more..

  • Page - 162

    Pontryagin’s Minimum Principle 143 Step 2 We need to show that h(t) is continuous in the time interval [0, ˆ tf]. Thus, for each > 0 there exists a δ> 0 such that |s − t | <δ implies |h(s) − h(t) | < In terms of the function denoted by H[t, u(t)] ≡ H[ˆ y(t), λ(t), u(t)] (5.32) we can write h(t)= H[ˆ y(t), λ(t), w(t)] = H[t, w(t)] (5.33) Because of read more..

  • Page - 163

    144 Optimal Control for Chemical Engineers The Bolzano–Weierstrass Theorem (Section 9.17, p. 277) assures that this se- quence contains a subsequence that is convergent on some value, say, w0.∗ As wn approaches w0, sn approaches t, thereby implying H[t, w0] ≤ h(t) − which contradicts h(t) being the infimum. Hence, Inequality (5.35) is true. Combining it with Inequality (5.34) read more..

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    Pontryagin’s Minimum Principle 145 Step 3b Now ˆ H(t) is a constant, not necessarily the same, in the left as well as the right part of the neighborhood corresponding to t< ti and t> ti. Let the respective constants be ˆ Hi−1 and ˆ Hi, as shown in Figure 5.4. Then the above set of equations becomes ˆ Hi−1 = h(ti)= ˆ Hi which shows that ˆ H(t) is constant read more..

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    146 Optimal Control for Chemical Engineers This function has pulse perturbations v1 and v2 in the two disjunct subinter- vals I1 and I2, respectively. Elsewhere in [0, ˆ tf], the function is the same as the optimal control. Derived earlier in Section 5.4.1 (p. 131), the state change at time t1 is Δy(t1)= g[ˆ y(t1), v] − g[ˆ y(t1), ˆ u(t1)] Δt1 (5.8) This state change, read more..

  • Page - 166

    Pontryagin’s Minimum Principle 147 Hence, for sufficiently small Δt2, the above equation when substituted in the result of Step 1, yields [compare with Equation (5.8), p. 133] Δy(t2) ≡ y(t2) − ˆ y(t2)= g[ˆ y(t2), v2] − g[ˆ y(t2), ˆ u(t2)] Δt2 δy(t2) +Δy(t2 − Δt2) In the above equation, δy(t2) indicates the state change at t2 if there were no previous perturbation. read more..

  • Page - 167

    148 Optimal Control for Chemical Engineers 1. arbitrarily fixed but admissible pulse perturbations v1, v2,..., vn at a finite number of time instants, t1,t2,... ,tn;∗ 2. corresponding intervals Δt1, Δt2,... Δtn — each of variable size, includ- ing zero; and 3. intervals of temporal perturbations at ˆ tf, which also have variable sizes including zero. In terms of ai introduced read more..

  • Page - 168

    Pontryagin’s Minimum Principle 149 and z(ˆ tf)= ˆ y(ˆ tf)+ αy(ˆ tf)+ (1 − α)˜ y(ˆ tf) = ˆ y(ˆ tf)+ n i=1 ai [αΔti +(1 − α)Δ˜ ti] Δ˘ ti + g[ˆ y(ˆ tf), ˆ u(ˆ tf)] [αΔtf +(1 − α)Δ˜ tf] Δ˘ tf Thus, z(ˆ tf) is the result of a control ˘ u(t) from the collection of admissible controls with non-negative time intervals Δ˘ tisand Δ˘ tf.∗ Therefore, the read more..

  • Page - 169

    150 Optimal Control for Chemical Engineers 1. In a closed convex set S in Rn, there exists a unique point ˆ y,which is closest to a given point x outside the set S. 2.If ˆ y in the set S is closest to x, then for all y in S the dot product (x − ˆ y) (y − ˆ y) ≤ 0 and vice versa. 3. Given any boundary point ˆ y of the set S, there exists a read more..

  • Page - 170

    Pontryagin’s Minimum Principle 151 If γ = −1, then x =(ˆ y + ya)/2, which belongs to the convex set S.This is contradictory since x lies outside S.Thus, γ = −1. Hence, γ =1, thereby implying that ˆ y = ya.In other words, ˆ y is a unique point of S for which the distance x − ˆ y is minimum. Step 2 Next, we show that if ˆ y is closest to x, then read more..

  • Page - 171

    152 Optimal Control for Chemical Engineers Consequently, p y < p x. In other words, given a point xi outside the convex set S, there is a unique point ˆ yi in S, corresponding to the unit vector pi =(xi − ˆ yi)/ xi − ˆ yi , such that pi y < pi xi for each y in S. Now consider an infinite sequence of xi tending to a point ˆ y on the boundary of read more..

  • Page - 172

    Chapter 6 Different Types of Optimal Control Problems In this chapter, we engage with different types of optimal control problems and derive the necessary conditions for the minimum. The constraints in the problems are handled using the Lagrange Multiplier Rule and the John Multiplier Theorem. Since derivatives are available in most of the engineering problems, we assume the functions read more..

  • Page - 173

    154 Optimal Control for Chemical Engineers Based on the Lagrange Multiplier Rule (see Section 4.3.3.2, p. 103), the above problem is equivalent to minimizing the augmented functional J = tf 0 F + λ ( − ˙y + g) dt (6.4) subject to the initial conditions. The vectors y, ˙y, g,and λ are each of dimension n with the first element indexed 1. For example, λ = λ1 λ2 ... read more..

  • Page - 174

    Different Types of Optimal Control Problems 155 the above equation in the expression for δJ,weobtain δJ = tf 0 (Hy + ˙λ) δy +(Hλ − ˙y) δ λ + Hu δu dt − λ (tf) δy(tf) + H − λ ˙y tf δtf (6.5) Note that δy(tf) is the variation in the optimal state ˆ y at time tf,i. e., δy(tf)= y(tf) − ˆ y(tf) (6.6) This variation is different from the variation read more..

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    156 Optimal Control for Chemical Engineers After combining the last two equations and discarding the second-order term δ ˙y(tf)δtf,we get δyf = δy(tf)+ ˙ˆy(t f)δtf or δy(tf)= δyf − ˙ˆy(t f)δtf (6.8) Finally, substituting the above equation in Equation (6.5), we obtain at the minimum of J [where δJ =0, u = ˆ u, y = ˆ y, and, in particular, ˙y(tf)= ˙ˆy(t f)] tf 0 read more..

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    Different Types of Optimal Control Problems 157 Example 6.1 For the batch reactor problem in Example 2.10 (p. 45), find the necessary conditions for the minimum of I = − tf 0 ckxa dt, k = ko exp − E RT subject to the satisfaction of the state equation ˙x = −akxa (6.10) with T (t) as the control, x(0) = x0 as the initial condition, and tf as the unspecified final read more..

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    158 Optimal Control for Chemical Engineers Example 6.2 Let the final state in Example 6.1 be specified as x(tf)= xf.In this case, the necessary conditions for the minimum are the same except λ(tf) = 0, which is replaced with x(tf)= xf. 6.1.3 Final State on Hypersurfaces In this case, the objective is to minimize the functional I = tf 0 F [y(t), u(t)] dt subject to ˙y = read more..

  • Page - 178

    Different Types of Optimal Control Problems 159 at the minimum. In the above equation, qy(t f) is the matrix of partial deriva- tives of q with respect to y(tf), and the right-hand side is the result of sub- stituting δy(tf) given by Equation (6.8). Substituting Equations (6.9) and (6.12) in the expression for δM ,we obtain δM = tf 0 (Hy + ˙λ) δy +(Hλ − ˙y) δ read more..

  • Page - 179

    160 Optimal Control for Chemical Engineers The necessary conditions for the minimum of I are ⎡ ⎢ ⎢ ⎢ ⎣ ˙y1 ˙y2 ˙y3 ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ −a1y2 1 eE1/T a1y21e E1/T − a2y2eE2/T a2y2eE2/T − a3y3eE3/T ⎤ ⎥ ⎥ ⎥ ⎦ Hλ , ⎡ ⎢ ⎢ ⎢ ⎣ y1 y2 y3 ⎤ ⎥ ⎥ ⎥ ⎦ t=0 = ⎡ ⎢ ⎢ ⎢ ⎣ 1 0 0 ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎣ q1 q2 ⎤ ⎦ = ⎡ ⎣ read more..

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    Different Types of Optimal Control Problems 161 6.2 Fixed Final Time In this section, we take up problems in which the final time is specified or fixed. The control functions are the only optimization parameters. 6.2.1 Free Final State Consider again the optimal control problem of Section 6.1.1 (p. 153). If the final time is fixed, then its variation δtf must be zero in read more..

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    162 Optimal Control for Chemical Engineers along with the respective initial and final conditions y(0) = y0 and y(tf)= yf Example 6.5 Let both final time and final state be fixed in Example 6.1 (p. 157). Thus, for fixed tf and x(tf)= xf, the necessary conditions for the minimum are as follows: ˙λ = akxa−1(c + λ)˙x = −akxa − kE RT 2 xa(c + aλ)= 0 x(tf)= xf read more..

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    Different Types of Optimal Control Problems 163 6.3 Algebraic Constraints This section deals with optimal control problems constrained by algebraic equalities and inequalities. 6.3.1 Algebraic Equality Constraints Consider the objective to minimize the functional I = tf 0 F [y(t), u(t)] dt subject to the state equations ˙y = g[y(t), u(t)], y(0) = y0 and the equality constraints fi(y, u)= 0,i read more..

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    164 Optimal Control for Chemical Engineers Considering that both final state and time are free, i. e., unspecified, the vari- ation of M upon simplification is given by [compare with Equation (6.9) on p. 156] δM = tf 0 (Ly + ˙λ) δy +(Lλ − ˙y) δ λ + Lμδμ + Lu δu dt − λ (tf) δyf + L(tf)δtf (6.17) Since δM should be zero at the minimum, the read more..

  • Page - 184

    Different Types of Optimal Control Problems 165 Based on Equation (6.16), the Lagrangian for this problem is L = 2 i=1 (yi − y s i ) 2 +(ui − u s i ) 2 + λ1[u1(yf − y1) − ky1y2] + λ2[u2 − u1y2]+ μ y2 y1 − a 1 bu2−c − 1 f Then the necessary conditions for the minimum of I are as follows: ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ˙y1 ˙y2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ read more..

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    166 Optimal Control for Chemical Engineers 6.3.2 Algebraic Inequality Constraints Consider the optimal control problem in which the objective is to minimize the functional I = tf 0 F [y(t), u(t)] dt subject to ˙y = g[y(t), u(t)], y(0) = y0 and the inequality constraints fi(y, u,t) ≤ 0,i =1, 2,... ,l or f (y, u,t) ≤ 0 where each fi and its derivative with respect to y and u read more..

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    Different Types of Optimal Control Problems 167 • the ones in the previous example except Lμ = 0, which is replaced with Lμ = y2 y1 − a 1 bu2−c − 1 ≤ 0 and • μ ≥ 0, and the complimentary slackness condition, i. e., μ y2 y1 − a 1 bu2−c − 1 =0 Example 6.9 Let us consider Example 6.3 (p. 159) without the selectivity constraints and impose the constraints read more..

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    168 Optimal Control for Chemical Engineers LT = − 1 T 2 ( −λ1 + λ2)a1y 2 1 E1e E1/T +( −1 − λ2 + λ3)a2y2E2e E2/T +(1 − λ3)a3y3E3e E3/T − μ1 + μ2 =0 ⎡ ⎣ Lμ 1 Lμ 2 ⎤ ⎦ = ⎡ ⎣− T + Tmin T − Tmax ⎤ ⎦ ≤ ⎡ ⎣ 0 0 ⎤ ⎦ L(tf)= ( −λ1 + λ2)a1y 2 1 eE1/T +( −1 − λ2 + λ3)a2y2e E2/T +(1 − λ3)a3y3e E3/T + μ1( −T + Tmin)+ read more..

  • Page - 188

    Different Types of Optimal Control Problems 169 and the integral equality constraints tf 0 Fi(y, u,t)dt = ki,i =1, 2,... ,l where each Fi and its partial derivative with respect to y and u are continuous. According to the Lagrange Multiplier Rule, the above problem is equivalent to the minimization of the augmented functional M = tf 0 F + λ ( − ˙y + g) dt + l i=1 μi read more..

  • Page - 189

    170 Optimal Control for Chemical Engineers Remarks The integral constraints are equivalent to having additional state equations dyn+i dt = Fi,yn+i(0) = 0,yn+i(tf)= ki; i =1, 2,... ,l with state variables yn+i, i =1, 2,... ,l. The reader can easily verify this equivalence by taking the integral on both sides of the above equations and applying the boundary conditions. Hence, an read more..

  • Page - 190

    Different Types of Optimal Control Problems 171 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ˙λ 1 ˙λ 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ −λ2u 1 − ∂y ∂z1 + μu ∂y ∂z1 λ2u ∂y ∂z2 + μu ∂y ∂z2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ −Lz , ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ λ1 λ2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ tf = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 0 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ Lu = −1 − λ1 + λ2(z1 − y)+ read more..

  • Page - 191

    172 Optimal Control for Chemical Engineers Example 6.11 Let us consider Example 6.10 with its equality constraint replaced with tf 0 (y∗ − y)u dt ≤ 0 In other words, we want distillate purity greater than or equal to y∗.Then the necessary conditions for the minimum are • those of Example 6.10 except tf 0 Lμ dt = 0, which is replaced with tf 0 Lμ dt = tf 0 (y∗ read more..

  • Page - 192

    Different Types of Optimal Control Problems 173 is the vector of functions depending on y(t1)and t1, which is an unspecified or free time in the interval [0,tf]. With the help of Lagrange multipliers, the above problem is equivalent to that of minimizing the augmented functional M = tf 0 F + λ ( − ˙y + g) dt + μ q = t1 0 (H − λ ˙y)dt + tf t1 (H − λ ˙y)dt read more..

  • Page - 193

    174 Optimal Control for Chemical Engineers the integrals having the same integrand, we get δM = tf 0 (a + ˙λ δy − ˙y δ λ)dt − λ δy t1− 0 − λ δy tf t1+ + H − λ ˙y t1− δt1 − H − λ ˙y t1+ δt1 + H − λ ˙y tf δtf + q δ μ + μ qy(t 1 )δy1 + qt 1 δt1 Expanding a and using Equation (6.8) on p. 156 as well as its equivalents read more..

  • Page - 194

    Different Types of Optimal Control Problems 175 equations dy1 dt = u − γ6y1 dy2 dt =˙y2,in + r2 y2y4 β2 + y4 − γ3y2y4 − γ4y2 − α2y2 1 − e−y1λ2 dy3 dt = r3y3(1 − β3y3) − γ5y3y4 − α3y3 1 − e−y1λ3 dy4 dt = r1y4(1 − β1y4) − γ1y3y4 − γ2y2y4 − α1y4 1 − e−y1λ1 along with the initial conditions yi(0) = yi,0,i =1, 2, 3, 4 and the interior read more..

  • Page - 195

    176 Optimal Control for Chemical Engineers The necessary conditions for the minimum are as follows: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ˙y1 ˙y2 ˙y3 ˙y4 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ u − γ6y1 ˙y2,in + r2 y2y4 β2 + y4 − γ3y2y4 − γ2y4 − α2y2 1 − read more..

  • Page - 196

    Different Types of Optimal Control Problems 177 ⎡ ⎣ q1 q2 ⎤ ⎦ = ⎡ ⎣ y2 − ay4 y3 − by4 ⎤ ⎦ t1 = ⎡ ⎣ 0 0 ⎤ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ λ1 λ2 λ3 λ4 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ t1− = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ λ1 λ2 λ3 λ4 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ t1+ + ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ∂q1 ∂y1 ∂q1 ∂y2 ∂q1 ∂y3 ∂q1 ∂y4 ∂q2 ∂y1 ∂q2 read more..

  • Page - 197

    178 Optimal Control for Chemical Engineers which are also known as Weierstrass–Erdmann corner conditions.They imply the continuity of H and λ on the corners at the minimum. 6.7 Multiple Integral Problems Optimal control problems involving multiple integrals are constrained by par- tial differential equations. A general theory similar to the Pontryagin’s mini- mum principle is not read more..

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    Different Types of Optimal Control Problems 179 and the boundary conditions c(0,t)= csat(t) cz(L)= 0 for 0 <t ≤ tf The state equation, G = 0, constitutes a partial differential equation con- straint. Applying the Lagrange Multiplier Rule, the equivalent problem is to find the control function D(c) that minimizes the augmented objective functional J = I + tf 0 L 0 λG dz dt read more..

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    180 Optimal Control for Chemical Engineers Substituting the above expressions for δI and δG in Equation (6.22), we obtain δJ = tf 0 L 0 2A(mc − me)+ λgc δc + λ − δ ˙c + gczδcz + gczzδczz + gDδD dz dt = 0 (6.23) Next, we simplify the terms containing δ ˙c, δcz,and δczz. Integrating these terms by parts, we get − tf 0 L 0 λδ ˙c dz dt = − L 0 tf 0 read more..

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    Different Types of Optimal Control Problems 181 tf 0 λgcz − ∂ ∂z (λgczz) z=L δc(L, t)dt δJ4 + tf 0 λgczzδcz L 0 dt δJ5 + tf 0 L 0 λgDδD dz dt δJ6 =0 We will now find the equations that eliminate the terms δJ1 to J6,as indicated in the above equation, so that the necessary condition δJ =0 is satisfied for the minimum. Elimination of δJ1 This term is eliminated read more..

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    182 Optimal Control for Chemical Engineers Elimination of δJ4 Since c(L, t) is not specified, δc(L, t) is not necessarily zero. Therefore, we eliminate δJ4 by specifying λ(L, t)= 0, 0 ≤ t ≤ tf (6.26) which is the first boundary condition for Equation (6.24). Elimination of δJ5 The above equation along with the enforcement λ(0,t)=0, 0 ≤ t ≤ tf (6.27) eliminates δJ5. This read more..

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    Different Types of Optimal Control Problems 183 N.G. Long and D. L´eonard. Optimal Control Theory and Static Optimization in Economics, Chapters 6 and 7. Cambridge University Press, Cambridge, UK, 1992. A. Takayama. Mathematical Economics, Chapter 1. Cambridge University Press, Cambridge, UK, 1985. Exercises 6.1 Utilize the results of Section 6.1.1 to find the necessary conditions for the read more..

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    read more..

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    Chapter 7 Numerical Solution of Optimal Control Problems The solution of an optimal control problem requires the satisfaction of dif- ferential equations subject to initial as well as final conditions. Except when the equations are linear and the objective functional is simple enough, an an- alytical solution is impossible. This is the reality of most of the problems for which read more..

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    186 Optimal Control for Chemical Engineers subject to ˙y = g[y(t), u(t)], y(0) = y0 The above problem is equivalent to minimizing the augmented functional J = tf 0 F + λ ( − ˙y + g) dt (6.4) subject to y(0) = y0. From Section 6.1.1, the necessary conditions for the minimum of I are ˙y = g, y(0) = y0 state equations and initial conditions , ˙λ = −Hy, λ(tf)= 0 read more..

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    Numerical Solution of Optimal Control Problems 187 7.1.3 Improvement Strategy With the state and costate equations satisfied at the end of Step 3 above, the variation of the augmented objective functional becomes [see Equation (6.9), p. 156] δJ = tf 0 Hu δu dt + H(tf)δtf (7.1) In the above equation, Hu or the coefficient of δu is the variational derivative of J with respect read more..

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    188 Optimal Control for Chemical Engineers of subintervals] to calculate δJ as well as J.Thus, J or I is rendered into a function dependent on the vector of optimization parameters p ≡ u(t0) u(t1) ... u(tN ) tf (7.4) where t0,t1,...,tN form the time-grid of (N + 1) equispaced grid points in thetimeinterval [0,tf]. However, a complication arises with this approach. An improvement read more..

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    Numerical Solution of Optimal Control Problems 189 subject to y(0) = y0. The variation δJ is given by [compare with Equa- tion (6.14) on p. 161] δJ = 1 0 (Hy + d λ dσ ) δy +(Hλ − dy dσ ) δ λ + Hu δu dσ − λ (1) δy(1) + 1 0 H tf dσ Jt f δtf (7.5) where Jt f , as indicated above, is the partial derivative of J with respect to the final time read more..

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    190 Optimal Control for Chemical Engineers δJ = 1 3N Hu (σ0)δu(σ0)+4 N i=1,3,5,... Hu (σi)δu(σi)+2 N i=2,4,6,... Hu (σi)δu(σi)+ Hu (σN )δu(σN ) + 1 3Ntf H(σ0)+ 4 N i=1,3,5,... H(σi)+2 N i=2,4,6,... H(σi)+ H(σN ) Jt f δtf where the vector ∇J ≡ Hu(σ0) Hu(σ1) ... Hu(σN ) Jt f is the gradient of J. Next, we consider the vector of optimization parameters, p, definedinEqua- read more..

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    Numerical Solution of Optimal Control Problems 191 which reduce J. Subsequent reduction of J is achieved as follows. Based on the improved controls and final time (namely, unext and tf,next), Steps 1–3 of Section 7.1.2 (p. 186) are repeated, and ∇J is recalculated and utilized to repeat the im- provements and reduce J. This iterative procedure continued until the reduc- tion in read more..

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    192 Optimal Control for Chemical Engineers equi-spaced grid points. Assume a value of the control function u at each grid point. 3. Integrate state equations forward using the initial conditions and the control function values. Save the values of state variables at each grid point. 4. Evaluate the objective functional using the values of control functions and state variables. Save the read more..

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    Numerical Solution of Optimal Control Problems 193 • u2 (K) is the reactor temperature as a function of time. It is desired to find the optimal controls u1(t)and u2(t) that maximize y4(tf), i. e., the final concentration of C. Thus, the objective is to minimize I = −y4(tf)= − y4(tf) 0 dy4 = − tf 0 dy4 dt dt = tf 0 u1y4 y1 − k0e a/u2 y2y3 dt subject to the state read more..

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    194 Optimal Control for Chemical Engineers Necessary Conditions for the Minimum The necessary conditions for the minimum of I are ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ dy1 dσ dy2 dσ dy3 dσ dy4 dσ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = tf ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ u1 u1(y0 − y2) y1 − κy2y3 − read more..

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    Numerical Solution of Optimal Control Problems 195 Gradient Algorithm 1. Set the iteration counter k = 0. Assume tk f and an even N . Obtain the fixed σ-grid of (N + 1) equi-spaced grid points σ0(= 0),σ1,σ2, ... , σN−1,σN (= 1) We use k in a superscript to denote the iteration number. At each grid point, assume control function values as follows: uki ≡ uk(σi)= read more..

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    196 Optimal Control for Chemical Engineers of costate variables at the grid points as λk i ≡ λk(σi)= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ λk1(σi) λk2(σi) λk3(σi) λk4(σi) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ,i =0, 1,... ,N 6. Evaluate the gradient of Jk by calculating the partial derivatives Hk u,i ≡ Hk u(σi)= ⎡ ⎣ Hu 1 (yki, uki, λk i ) Hu 2 (yki, uki, λk i ) ⎤ ⎦; i read more..

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    Numerical Solution of Optimal Control Problems 197 • The optimal objective functional value is Ik. • The optimal control ˆ u(t) is represented by uki, i =0, 1,... ,N . • The optimal final time is tk f . • The optimal state ˆ y(t) is represented by yki, i =0, 1,... ,N . Results Table 7.1 lists the parameters used to solve the optimal control problem with the read more..

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    198 Optimal Control for Chemical Engineers With the initially guessed final time of 60 min, the initial controls pro- vided I = −4.12, which corresponds to the final product concentration of 4.12 g/cm3. 0 1020 30405060 t (min) 0 10 20 30 40 50 60 ˆ y1 /10 2 ,ˆ y2 ,ˆ y3 ,ˆ y4 ˆ y1 ˆ y2 ˆ y3 ˆ y4 Figure 7.2 The optimal states versus time for Example 7.1. The final time read more..

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    Numerical Solution of Optimal Control Problems 199 The application of the gradient algorithm minimized I, i. e., maximized the product concentration to 7.56 g/cm3. The final time was reduced from 60 to 50.3 min. Figure 7.4 plots I versus the iteration of the gradient algorithm. 0 102030 40 iteration −8 −7 −6 −5 −4 J Figure 7.4 The objective functional versus iteration for read more..

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    200 Optimal Control for Chemical Engineers Results For the same set of parameters and initial controls used in Example 7.1, Fig- ure 7.5 shows the optimal states with the fixed final time. The corresponding controls are shown in Figure 7.6. 0 10203040 5060 t (min) 0 10 20 30 40 50 60 ˆ y1 /10 2 ,ˆ y2 ,ˆ y3 ,ˆ y4 ˆ y1 ˆ y2 ˆ y3 ˆ y4 Figure 7.5 The optimal states versus read more..

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    Numerical Solution of Optimal Control Problems 201 The optimal I is −7.51, corresponding to the final product concentration of 7.51 g/cm3. When the final time was free and available for optimization in the last example, we got better results — more production in less time. The reason is that free final time affords more freedom (than when it is fixed) in the optimal read more..

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    202 Optimal Control for Chemical Engineers where μ is the vector of Lagrange multipliers corresponding to the hypersur- face. From Section 6.1.3 (p. 158), the necessary conditions for the minimum of M are ˙y = gy(0) = y0 ˙λ = −Hy λ (tf)= μ qy(t f ) Hu = 0 H(tf)= 0 q = 0 The gradient algorithm as such cannot work with the above equations. With an assumed control if read more..

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    Numerical Solution of Optimal Control Problems 203 2. Set the weighting matrix W = αr1 where 1 is the l × l identity matrix given by 1 = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 10 ... 0 01 ... 0 00 . . . 0 00 ... 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 3. On the problem transformed to the fixed σ-interval [0, 1], apply the gradient algorithm of Section 7.1.4 (p. 191) with λ (1) = read more..

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    204 Optimal Control for Chemical Engineers Computational Algorithm 1. Set the counter r = 0. Initialize the weighting matrix as W = ⎡ ⎣ αr 0 0 αr ⎤ ⎦ 2. Apply the gradient algorithm of Section 7.1.4 (p. 191) with λ (1) = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ λ1 λ2 λ3 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ σ=1 = ⎡ ⎢ ⎢ ⎣ y4 − b1y2 y4 − b2y3 ⎤ ⎥ ⎥ ⎦ σ=1 q ⎡ ⎢ read more..

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    Numerical Solution of Optimal Control Problems 205 Results We present the results for b1 =5 and b2 = 2 and the same set of parameters and initial controls as used in Example 7.1. Figure 7.7 shows the convergence of the penalty function method. At each outer iteration r, the plotted objective functional value is the optimal with cor- responding constraint violation quantified as read more..

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    206 Optimal Control for Chemical Engineers 0 1020 30405060 t (min) 0 10 20 30 40 50 ˆ y1 /10 2 ,ˆ y2 ,ˆ y3 ,ˆ y4 ˆ y1 ˆ y2 ˆ y3 ˆ y4 Figure 7.8 The optimal states versus time for Example 7.3 0 10 2030405060 t (min) 0 100 200 300 400 500 u 1 , u 2 ,ˆ u 1 ,ˆ u 2 u1 u2 ˆ u1 ˆ u2 Figure 7.9 The initially guessed and optimal controls versus time for Example 7.3 read more..

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    Numerical Solution of Optimal Control Problems 207 7.2.2 Free Final Time but Fixed Final State The objective in this problem is to find the control u that minimizes I = tf 0 F [y(t), u(t)] dt subject to ˙y = g[y(t), u(t)], y(0) = y0, and y(tf)= yf The final state fixed as y(tf)= yf is the vector of hypersurfaces q[y(tf)] ≡ y(tf) − yf = 0 at the final time tf. read more..

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    208 Optimal Control for Chemical Engineers and the final costate vector λ (1) = ⎡ ⎢ ⎢ ⎢ ⎣ λ1 λ2 λ3 ⎤ ⎥ ⎥ ⎥ ⎦ σ=1 = ⎡ ⎢ ⎢ ⎢ ⎣ y1 − y1,f y2 − y2,f y3 − y3,f ⎤ ⎥ ⎥ ⎥ ⎦ σ=1 q ⎡ ⎢ ⎢ ⎢ ⎣ αr 00 0 αr 0 00 αr ⎤ ⎥ ⎥ ⎥ ⎦ W = αr ⎡ ⎢ ⎢ ⎢ ⎣ y1 − y1,f y2 − y2,f y3 − y3,f ⎤ ⎥ ⎥ ⎥ ⎦ σ=1 Results This read more..

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    Numerical Solution of Optimal Control Problems 209 0 10 2030405060 t (min) 0 100 200 300 400 500 u 1 , u 2 ,ˆ u 1 ,ˆ u 2 u1 u2 ˆ u1 ˆ u2 Figure 7.11 The initially guessed and optimal controls corresponding to Fig- ure 7.10 7.2.3 Algebraic Equality Constraints The objective in this problem is to find the control u that minimizes I = tf 0 F [y(t), u(t)] dt subject to ˙y = read more..

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    210 Optimal Control for Chemical Engineers where μ are the time dependent multipliers associated with the algebraic equalities and L = F + λ g + μ f = H + μ f is the Lagrangian. The necessary conditions for the minimum of M ,and equivalently of I, are ˙y = gy(0) = y0 ˙λ = −Ly λ(tf)= 0 f = 0 Lu = 0 L(tf)= 0 Since the algebraic equations f = 0 are not read more..

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    Numerical Solution of Optimal Control Problems 211 where L(σi)= tf F (yi, ui)+ λ i g(yi, ui) +[f (yi, ui)] W μi f (yi, ui); i =0, 1,... ,N Using composite Simpson’s 1/3 Rule, given that the state equations are satisfied, M = 1 3N A0 +4 N 1,3,5,... Ai +2 N i=2,4,6,... Ai + AN where Ai ≡ tfF (yi, ui)+[f (yi, ui)] Wf (yi, ui); i =0, 1,... ,N The improvements in u read more..

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    212 Optimal Control for Chemical Engineers Example 7.5 Consider the batch reactor problem in Example 7.1 (p. 192) subject to the following algebraic equality constraint throughout the time interval [0,tf]: u1 = b1y 2 3 + b2y3 In this case, f ≡ u1 − b1y 2 3 − b2y3 =0 and the weighting matrix is just an element αr at the r-th outer iteration. The corresponding Lagrange read more..

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    Numerical Solution of Optimal Control Problems 213 The optimal states obtained at the convergence are shown in Figure 7.13. Figure 7.14 shows the corresponding optimal controls. 0 1020 3040 t (min) 0 10 20 30 40 50 ˆ y1 /10 2 ,ˆ y2 ,ˆ y3 ,ˆ y4 ˆ y1 ˆ y2 ˆ y3 ˆ y4 Figure 7.13 The optimal states when the algebraic constraints are satisfied upon convergence in Example 7.5 0 10 read more..

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    214 Optimal Control for Chemical Engineers 7.2.4 Integral Equality Constraints The objective in this problem is to find the control u that minimizes I = tf 0 F [y(t), u(t)] dt subject to ˙y = g[y(t), u(t)], y(0) = y0 and l integral equality constraints tf 0 Fj(y, u,t)dt = kj,j =1, 2,... ,l As shown in Section 6.4.1 (p. 168), the equivalent problem is to find the min- imum read more..

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    Numerical Solution of Optimal Control Problems 215 where α is a real number greater than unity. With this specification, the augmented functional becomes M = tf 0 H − λ ˙y dt + l j=1 tf 0 Fj dt − kj 2 α penalty function The summation term above is the penalty function, which is positive whenever any integral equality constraint is violated. Example 7.6 Consider the batch read more..

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    216 Optimal Control for Chemical Engineers 0 10203040 5060 t (min) 0 10 20 30 40 50 ˆ y1 /10 2 ,ˆ y2 ,ˆ y3 ,ˆ y4 ˆ y1 ˆ y2 ˆ y3 ˆ y4 Figure 7.15 The optimal states when the integral equality constraints are satisfied upon convergence in Example 7.6 0 10203040 5060 t (min) 0 100 200 300 400 500 u 1 , u 2 ,ˆ u 1 ,ˆ u 2 u1 u2 ˆ u1 ˆ u2 Figure 7.16 The initially guessed and read more..

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    Numerical Solution of Optimal Control Problems 217 7.2.5 Algebraic Inequality Constraints This problem is similar to that in Section 7.2.3 (p. 209) except that the alge- braic equality constraints are replaced with the inequalities f (y, u) ≤ 0 Recall from Section 6.3.2 (p. 166) that the necessary conditions for the mini- mum are ˙y = gy(0) = y0 ˙λ = −Ly λ(tf)= 0 f ≤ 0 read more..

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    218 Optimal Control for Chemical Engineers 3. Apply the gradient algorithm of Section 7.1.4 (p. 191). Use the following initial guesses if r> 0: uri; i =0, 1,... ,N and trf Note In this case, the augmented objective functional is M whose gra- dient in the transformed σ-interval [0, 1] is ∇M ≡ Lu(σ0) Lu(σ1) ... Lu(σN ) Mt f where L(σi)= tf F (yi, ui)+ λ i g(yi, ui) read more..

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    Numerical Solution of Optimal Control Problems 219 6. Given a positive real number ε3 close to zero, if the error E = N i=0 [eri] f (yri, uri) >ε3 (7.6) then some constraints are violated. Therefore, go to Step 3. Otherwise, the constraints f ≤ 0 are satisfied, and the values y r i , u r i ,i =0, 1,... ,N ;and t r f correspond to the minimum. Example 7.7 Consider read more..

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    220 Optimal Control for Chemical Engineers 0 1020 30405060 t (min) 0 10 20 30 40 50 60 ˆ y1 /10 2 ,ˆ y2 ,ˆ y3 ,ˆ y4 ˆ y1 ˆ y2 ˆ y3 ˆ y4 Figure 7.17 The optimal states when the inequality constraints are satisfied upon convergence in Example 7.7 0 10 2030405060 t (min) 0 100 200 300 400 u 1 , u 2 ,ˆ u 1 ,ˆ u 2 u1 u2 ˆ u1 ˆ u2 Figure 7.18 The initially guessed and optimal read more..

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    Numerical Solution of Optimal Control Problems 221 7.2.6 Integral Inequality Constraints This problem is similar to that in Section 7.2.4 (p. 214) except that the integral equality constraints are replaced with the inequalities Ij ≡ tf 0 Fj(y, u,t)dt ≤ kj,j =1, 2,... ,l From Section 6.4.2 (p. 171), the necessary conditions for the minimum are ˙y = gy(0) = y0 ˙λ = −Ly λ(tf)= read more..

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    222 Optimal Control for Chemical Engineers 0 10203040 5060 t (min) 0 10 20 30 40 50 60 ˆ y1 /10 2 ,ˆ y2 ,ˆ y3 ,ˆ y4 ˆ y1 ˆ y2 ˆ y3 ˆ y4 Figure 7.19 The optimal states when the integral inequality constraints are satis- fieduponconvergence in Example 7.8 0 10203040 5060 t (min) 0 100 200 300 400 u 1 , u 2 ,ˆ u 1 ,ˆ u 2 u1 u2 ˆ u1 ˆ u2 Figure 7.20 The initially guessed and read more..

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    Numerical Solution of Optimal Control Problems 223 gence, the optimal objective functional was −6.9, which corresponds to the final product concentration of 6.9 g/cm3. The optimal final time in this case increased from 60 to 62.9 min. Problems with Fixed Final Time The final time is not an optimization parameter in these problems. Thus, these problems are straightaway solvable by read more..

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    224 Optimal Control for Chemical Engineers subject to y = g[y(t),u(t)],y(0) = y0 The necessary conditions for the minimum are dy dt = g, y(0) = y0 state equation and initial condition , dλ dt = −Hy,λ(tf)= 0 costate equation and final condition , and Hu =0 stationarity condition Two Point Boundary Value Problem Suppose that it is possible to solve Hu = 0 and obtain an explicit read more..

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    Numerical Solution of Optimal Control Problems 225 Equations Governing ∂λ/∂λ0 Differentiating the state and costate equations with respect to λ0,we get ∂ ∂λ0 dy dt = ∂g ∂λ0 ∂ ∂λ0 dλ dt = ∂h ∂λ0 or d dt ∂y ∂λ0 = ∂g ∂y ∂y ∂λ0 + ∂g ∂λ ∂λ ∂λ0 d dt ∂λ ∂λ0 = ∂h ∂y ∂y ∂λ0 + ∂h ∂λ ∂λ ∂λ0 where ∂y ∂λ0 (0) = 0 ∂λ ∂λ0 (0) = read more..

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    226 Optimal Control for Chemical Engineers The Hamiltonian for this problem is given by H =(y − ys) 2 +(u − us) 2 + λ u(yf − y) − ky 2 The necessary conditions to be satisfied at the minimum are the process model or the state equation with the initial condition, the costate equation with the final condition, i. e., dλ dt = −2(y − ys)+ λ(u +2ky),λ(tf)= 0 and read more..

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    Numerical Solution of Optimal Control Problems 227 to obtain the following derivative equations: dyλ 0 dt = − λ(yf − y) 2 + us yλ 0 + (y − yf)λλ 0 + λyλ 0 (yf − y) 2 − 2kyλ 0 dλλ 0 dt = −2yλ 0 + λ(yf − y) 2 + us +2ky λλ 0 + (y − yf)λλ 0 +(λ +4k)yλ 0 λ 2 along with the respective initial conditions yλ 0 (0) = 0 and λλ 0 (0) = 1 read more..

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    228 Optimal Control for Chemical Engineers 00.20.40.60.81 t (min) 3 5 7 9 y, u,ˆ y,ˆ u y u ˆ y ˆ u Figure 7.21 The initial and optimal states and controls in Example 7.9 00.20.40.60.81 t (min) −0.2 0 0.5 1 1.5 λ λ initial λ/10 Figure 7.22 The initial and optimal costates in Example 7.9. The initial costate is in gray read more..

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    Numerical Solution of Optimal Control Problems 229 7.A Derivation of Steepest Descent Direction Consider a continuously differentiable function f = f (x)where x is the vari- able vector x1 x2 ... xn For a small change dx = dx1 dx2 ... dxn the change in f is given by df = n i=1 ∂f ∂xi dxi Let the magnitude of dx be denoted by ds.Thus, ds = n i=1 dx2 i Then the change in read more..

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    230 Optimal Control for Chemical Engineers Applying the Lagrange Multiplier Rule, the equivalent problem is to find the w that minimizes J = n i=1 viwi + λ n i=1 w2i − 1 where λ is a Lagrange multiplier. Now, at the minimum of J dJ = n i=1 (vi +2λwi)dwi +dλ n i=1 w2i − 1 =0 Hence, the necessary conditions for the minimum are vi +2λwi =0; i =1, 2,... ,n and n i=1 read more..

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    Numerical Solution of Optimal Control Problems 231 Thus, J( ˆ w,λ), i. e., df/ds corresponding to ˆ w, is indeed the minimum. The above result shows that df/ds is minimized by ˆ w = − v1 v v2 v ... vn v which is a unit vector in the direction opposite to the gradient of the func- tion f (x). Simply put, the rate of function decrease is maximum along the steepest read more..

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    232 Optimal Control for Chemical Engineers minimum of I with respect to tf: dI dt = F (y, u)=0 d2I dt2 = n i=1 ∂F ∂yi dyi dt gi + m i=1 ∂F ∂ui dui dt > 0 7.3 A CSTR carrying out a first order exothermic reaction is described by (Luus and Lapidus, 1967) dy1 dt = −2y1 − a1 +(y1 + a1)exp a2y1 y1 +4a1 − (y1 +0.5a1)u, y1(0) = y1,0 dy2 dt = a1 − y2 − (y2 + read more..

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    Numerical Solution of Optimal Control Problems 233 Using a suitable computational algorithm, solve the above problem for a1 = a2 = yf = tf =1 Compare the optimal u with the analytical solution (Kamien and Schwartz, 1991) given by u = a2(2t − tf) 4a1 + yf tf ,u ≥ 0 7.6 The integral equality constraints of the optimal control problem in Sec- tion 7.2.4 (p. 214) may be read more..

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    Chapter 8 Optimal Periodic Control Optimal periodic control involves a periodic process, which is characterized by a repetition of its state over a fixed time period. Examples from nature include the circadian rhythm of the core body temperature of mammals and the cycle of seasons. Man-made processes are run periodically by enforcing periodic control inputs such as periodic feed rate read more..

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    236 Optimal Control for Chemical Engineers 8.1.1 Necessary Conditions Similar to Section 6.1.1 (p. 153), the variation of J is δJ = tf 0 (Hy + ˙λ) δy +(Hλ − ˙y) δ λ + Hu δu dt + λ (0) δy(0) − λ (tf) δyf + H(tf)δtf Note that the above equation has one additional term, λ (0) δy(0), when compared to Equation (6.9) on p. 156. The reason is that δy(0) read more..

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    Optimal Periodic Control 237 Note that if τ is not fixed, then we have the additional necessary condition H(τ )= − ckx a + λakx a τ =0 Example 8.2 Consider the CSTR described in Example 6.7 (p. 164) but under periodic operation in which the objective is to minimize I = τ 0 2 i=1 (yi − ysi) 2 +(ui − usi) 2 dt where τ is a fixed time period. The state is read more..

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    238 Optimal Control for Chemical Engineers I are as follows: ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ˙y1 ˙y2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ u1(yf − y1) − ky1y2 u2 − u1y2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ Lλ , ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ y1 y2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ t=0 = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ y1 y2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ τ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ˙λ 1 ˙λ 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = read more..

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    Optimal Periodic Control 239 • μ ≥ 0, and the complimentary slackness condition μ τ 0 u1 dt − aτ =0 8.2 Solution Methods The solution of a optimal periodic control problem requires the integration of state and costate equations, both subject to periodicity conditions. Other than this integration aspect, the solution methods for optimal periodic control problems are similar to read more..

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    240 Optimal Control for Chemical Engineers whereweuse y0 to denote the initial state vector y(0). Starting with some guess, y(0) is improved iteratively using the Newton–Raphson method, i. e., y0,next = y0 − [fy 0 ] −1 τ f (y0) (8.1) where [fy 0 ] τ is the Jacobian ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ∂y1 ∂y1,0 − 1 ∂y1 ∂y2,0 ... ∂y1 ∂yn,0 read more..

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    Optimal Periodic Control 241 4. Repeat the above steps for costate equations. Example 8.3 Consider a CSTR under periodic operation governed by the state equations dy1 dt = u1(yf − y1) −R,y1(0) = y1(τ ) dy2 dt = u2 − u1y2,y2(0) = y2(τ ) where y1 and y2 are concentrations of the reactant and catalyst, R = k0y21y2 is the rate of production, yf is the reactant concentration read more..

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    242 Optimal Control for Chemical Engineers The necessary conditions for the minimum are 1. the state equations in the σ-interval 2. the costate equations dλ1 dσ =2k0y1y2 + τλ1(u1 +2k0y1y2),λ1(0) = λ1(1) dλ2 dσ = k0y 2 1 + τ (λ1k0y 2 1 + λ2u1),λ2(0) = λ2(1) 3. the stationarity conditions with respect to the controls u1 and u2,i. e., ⎡ ⎣ Hu 1 Hu 2 ⎤ ⎦ = τ ⎡ read more..

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    Optimal Periodic Control 243 Derivative Costate Equations Similar differentiation of the costate equations yields the following derivative costate equations: d dσ ∂λ1 ∂λ1,0 = τ (u1 +2k0y1y2) ∂λ1 ∂λ1,0 , ∂λ1 ∂λ1,0 (0) = 1 d dσ ∂λ1 ∂λ2,0 = τ (u1 +2k0y1y2) ∂λ1 ∂λ2,0 , ∂λ1 ∂λ2,0 (0) = 0 d dσ ∂λ2 ∂λ1,0 = τ k0y 2 1 ∂λ1 ∂λ1,0 + u1 ∂λ2 ∂λ1,0 , ∂λ2 read more..

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    244 Optimal Control for Chemical Engineers 2.b Integrate forward from σ = 0 to 1, the state equations along with derivative state equations using the control functions ukis. 2.c Improve the initial state by applying Equations (8.1) and (8.2), i. e., ⎡ ⎣ y s+1 1,0 y s+1 2,0 ⎤ ⎦ ys+1 0 = ⎡ ⎣ ys1,0 ys2,0 ⎤ ⎦ ys 0 − ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ∂y1 ∂y1,0 − 1 ∂y1 read more..

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    Optimal Periodic Control 245 6.a Set the counter s = 0. Assume the initial costate λs 0 = λk1,0 λk2,0 6.b Integrate forward from σ = 0 to 1, the costate equations along with the derivative costate equations using the control functions ukis and the state variables ykis. 6.c Improve the initial costate by applying Equations (8.1) and (8.2), i. e., ⎡ ⎣ λ s+1 1,0 λ s+1 2,0 read more..

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    246 Optimal Control for Chemical Engineers Check the magnitude of the gradient. Given a small positive real number ε2, if the norm of the gradient N i=0 2 j=1 Huj (yk i , uk i , λk i ) 2 + Jk τ 2 <ε2 then go to Step 12. 9. Improve control functions by calculating u k+1 i = u k i − H k u,i,i =0, 1,... ,N where is a positive real number causing maximum reduction read more..

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    Optimal Periodic Control 247 01234567 t (min) 0 1 2 3 4 5 y1 , y2 /10, ˆ y1 ,ˆ y2 y1 y2 ˆ y1 ˆ y2 Figure 8.1 The initial and optimal states for Example 8.3 01 2345 67 t (min) 0 5 10 15 ˆ u 1 × 10, ˆ u 2 , u 1 × 10, u 2 u1 u2 ˆ u1 ˆ u2 Figure 8.2 The initially guessed and optimal controls for Example 8.3 With the initially guessed time period of 6 min, the initial read more..

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    248 Optimal Control for Chemical Engineers concentration to 0.12 g/min. The optimal time period increased to 6.8 min. Figure 8.3 shows the convergence of I to the minimum. 0 1020304050 60 iteration −0.13 −0.12 −0.11 −0.1 −0.09 −0.08 −0.07 I Figure 8.3 The objective functional versus iteration for Example 8.3 8.3 Pi Criterion The pi criterion is a sufficient condition for the read more..

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    Optimal Periodic Control 249 subject to the following constraints: ˙y = g(y, u), y(0) = y(τ ) (8.5) 1 τ τ 0 v(y, u)dt = 0 (8.6) 1 τ τ 0 w(y, u)dt ≤ 0 (8.7) Integrating Equation (8.5) after dividing it by τ yields 1 τ τ 0 g(y, u)dt = 0 Using the following notation for the integrals in the problem I(y, u) ≡ 1 τ τ 0 F (y, u)dt Ii(y, u) ≡ 1 τ τ 0 gi(y, u)dti read more..

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    250 Optimal Control for Chemical Engineers J at Optimal Steady State Under the steady state, both u and y are time independent. Then the above problem reduces to the minimization of I = F (y, u) subject to ˙y = g(y, u)= 0 1 τ τ 0 v(y, u)dt = v(y, u)= 0 1 τ τ 0 w(y, u)dt = w(y, u) ≤ 0 Let the pair (¯ y, ¯ u) denote the optimal steady state solution with read more..

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    Optimal Periodic Control 251 The first assumption implies that the constraint qualification or normality condition (see Section 4.3.2, p. 96) is satisfied. It ensures that an infinite number of solution pairs exist near the optimal steady state solution pair (¯ y, ¯ u). This pair, as well as ¯ u, is called normal when the normality condition is satisfied. Next, we retain the read more..

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    252 Optimal Control for Chemical Engineers We shall return to the above equation after coming up with an admissible pair (y, u), or equivalently, an admissible variational pair (δy,δu). Admissibility Criteria The pair (y, u) is admissible if and only if it satisfies the constraints — Equa- tions (8.5) and (8.6) and Inequality (8.7). If (y, u) is admissible, then the related read more..

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    Optimal Periodic Control 253 We now apply the admissibility criteria of the previous section. For the variation pair (δy,δu) to be admissible, δu(t) should generate a state variation δy governed by Equation (8.15). Moreover, the variation pair must satisfy the remaining constraints, i. e., Equation (8.16) and Inequality (8.17). In the following treatment, we use the Fourier transform, read more..

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    254 Optimal Control for Chemical Engineers where ¯ a is, respectively, ¯ vi or ¯ wi. Substituting for δu and δy from Equa- tions (8.19) and (8.21), we get δ ¯ Ii = 1 τ τ 0 (¯ ay GUe−iωt + GUeiωt +¯ au U e−iωt + eiωt )dt =0 after integrating and applying the limits. Basically, we have applied the result τ 0 cekiωt dt = c kiω eki(2π/τ)t τ 0 = 0 (8.23) read more..

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    Optimal Periodic Control 255 1. u1, the time dependent concentration of A fed to the reactor and 2. u2, the volumetric flow rate in and out of the reactor both per unit reactor volume. For the periodic operation of time period τ , the objective is to maximize the average concentration of the product B,i. e., to find the minimum of I = − 1 τ τ 0 x2 dt subject to read more..

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    256 Optimal Control for Chemical Engineers where U = U1 U1 is the amplitude of δu = δu1 δu2 ,which is the periodic control variation given by Equation (8.18) for some frequency ω =2π/τ .Both U and ω need to be specified. The matrices in the above inequality are as follows. The matrices G and G are given by G = G(iω)and G = G( −iω) where for k = ±1, we read more..

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    Optimal Periodic Control 257 definiteness of the pi matrix is the sufficient condition for the neighboring controls to be optimally better. This condition does not hold if an optimal steady state control ¯ ui lies at its boundary ui,max, as shown in Figure 8.4. In this case, δui given by Equation (8.19) becomes invalid since ui =¯ ui + δui trespasses the boundary for any read more..

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    258 Optimal Control for Chemical Engineers Satisfaction of Equation (8.15) Substituting the above Fourier coefficients of δu in Equation (8.20), we obtain δy = GUe−iωt + GV + GUeiωt (8.27) where G ≡ G(0) = − ¯ g−1 y ¯ gu Observe that δy so obtained satisfies Equation (8.15) as well as the periodicity condition δy(0) = δy(τ ). Satisfaction of Remaining Constraints To read more..

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    Optimal Periodic Control 259 Properness of (¯ y, ¯ u) For the pair (¯ y, ¯ u) to be proper, I − ¯ I ≤ 0. From Equation (8.14) I − ¯ I = 1 τ τ 0 δ ¯ F dt δ ¯ I + 1 2τ τ 0 δ 2 ¯ L dt δ2 ¯ I Then (I − ¯ I) is less than zero if • either δ ¯ I =0 and δ2 ¯ I is less than zero, or • both δ ¯ I and δ2 ¯ I are less than read more..

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    260 Optimal Control for Chemical Engineers then an appropriate control variation is given by Equation (8.26) in which the vector V = V1 V2 ensures that u = ¯ u + δu does not exceed umax. Hence, Vi is • zero if the corresponding ¯ ui <ui,max,or • some suitable value to ensure that the neighboring perturbed control ui <ui,max if ¯ ui = ui,max. Thus, provided that read more..

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    Optimal Periodic Control 261 ¯ νi ≥ 0when wi(¯ y, ¯ u)= 0 and ¯ νi =0 when wi(¯ y, ¯ u) < 0 for i = p +1,p +2,... ,q Ly(¯ y, ¯ u, ¯ λ, ¯ μ, ¯ ν)= Hy(¯ y, ¯ u, ¯ λ, ¯ μ, ¯ ν)+ ¯ μ vy(¯ y, ¯ u)+ ¯ ν wy(¯ y, ¯ u)= 0 Lu(¯ y, ¯ u, ¯ λ, ¯ μ, ¯ ν)= Hu(¯ y, ¯ u, ¯ λ, ¯ μ, ¯ ν)+ ¯ μ vu(¯ y, ¯ u)+ ¯ ν wu(¯ y, read more..

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    262 Optimal Control for Chemical Engineers for some real numbers and β1, β2, ... , βk. Dependent on these numbers there are infinite such controls. From the Embedding Theorem (Hestenes, 1966) there exists for each u a unique solution y of the differential equations ˙y = g(y, u); y(0) = ¯ y(0) + k j=1 βjδyj(0) + δz(0) (8.30) Thus, there are infinitely many admissible read more..

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    Optimal Periodic Control 263 As a result, the Jacobian of F ≡ F1 F2 ... Fk with respect to β ≡ β1 β2 ... βk at α = 0 is the non-zero determinant D in Equation (8.29). From the Implicit Function Theorem (Section 9.16, p. 277), Equations (8.32) are valid in an open region around β = 0 and have solutions βj = bj( ); | | <δ1 > 0; j =1, 2,... ,k (8.34) read more..

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    264 Optimal Control for Chemical Engineers In terms of the variations δy0 and δu0 as indicated, we obtain Ij(y, u)= δIj(¯ y, ¯ u; δy0,δu0); j =1, 2,... ,k Since δy0 and δu0 are, after all, respective variations in y and u,we can rewrite the above equations as Ij(y, u)= δIj(¯ y, ¯ u; δy,δu); j =1, 2,... ,k (8.12) Step 5 Finally, we need to show that the family read more..

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    Optimal Periodic Control 265 8.D Derivation of Equation (8.25) We need to show that each right-hand side term of Equation (8.24) on p. 254 is equivalent to the corresponding term in Equation (8.25). We will do that for the first term, i. e., derive 1 2τ τ 0 δy ¯ Lyyδy dt = U G ¯ LyyGU The rest of the equivalences may be similarly obtained by the reader. In the read more..

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    266 Optimal Control for Chemical Engineers F.J.M. Horn and R.C. Lin. Periodic processes: A variational approach. Ind. Eng. Chem. Des. Dev., 6(1):21–30, 1967. D. Sinˇci´c and J.E. Bailey. Analytical optimization and sensitivity analysis of forced periodic chemical processes. Chem. Eng. Sci., 35:1153–1161, 1980. Exercises 8.1 Find the necessary conditions for the minimum of I = 1 τ τ 0 read more..

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    Chapter 9 Mathematical Review 9.1 Limit of a Function Consider a function f (x) defined in the vicinity of x = x0. Then the limit of f (x)at x = x0 is a real number L approached by f (x)as x approaches x0. Symbolically, lim x →x0 f (x)= L During the approach process, x and f (x) may be greater or less than the respective targets x0 and L. The limit L may not read more..

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    268 Optimal Control for Chemical Engineers With the help of absolute differences, |f(x) − f (x0) | and |x − x0 |,the above concepts are expressed more precisely as follows. A function f (x) is defined to be continuous at x = x0 when given any > 0 for which |f(x) − f (x0) | < there exists a δ> 0 such that |x − x0 | <δ. 9.2.1 Lower and Upper read more..

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    Mathematical Review 269 9.4 Bounds A lower bound of a set is a number with the following property: A member of a set is either less than or equal to the lower bound. Thus, 1 and 2 are lower bounds of [2, 3] as well as (2, 3). There may be multiple lower bounds. Moreover, a lower bound may not be a member of the set. The greatest of the lower bounds of a set read more..

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    270 Optimal Control for Chemical Engineers 9.6 Taylor Series and Remainder Theestimateof a function f (x)at a distance h from the reference point x = x0 is given by the n-th order Taylor series or expansion f (x0 + h)= f (x0)+ h df dx x0 + h2 2! d2f dx2 x0 + h3 3! d3f dx3 x0 + ··· + hn n! dnf dxn x0 + hn+1 (n +1)! dn+1f dxn+1 ζ remainder (9.1) where ζ lies in the read more..

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    Mathematical Review 271 the last non-autonomous differential equation is transformed into the following set of autonomous differential equations: dy1 dt = y1y2 − uy22 and dy2 dt =1 9.8 Differential Let f be a function of the independent variable x. Then the differential of f at x = x0 is defined as the change df in f corresponding to the change h in x from x0 meeting the read more..

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    272 Optimal Control for Chemical Engineers 9.9 Derivative The definition of the derivative follows from the differential. We multiply and divide df (x0; h)by h in the last equation to obtain f (x0 + h)= f (x0)+ df (x0; h) h derivative h + (h) where the derivative is the coefficient of h and is defined as lim h →0 df (x0; h) h ≡ df dx ⏐ ⏐ ⏐ ⏐ x0 = lim h →0 read more..

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    Mathematical Review 273 9.10 Leibniz Integral Rule Consider the following definite integral I = b a f dx Then the derivative of I with respect to its upper limit is given by dI db = lim Δb →0 I(b +Δb) − I(b) Δb = lim Δb →0 b+Δb a f dx − b a f dx Δb = lim Δb →0 b+Δb b f dx Δb = lim Δb →0 f (b)Δb Δb = f (b) Similarly, the derivative of I with respect to read more..

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    274 Optimal Control for Chemical Engineers Quadratic Convergence Sufficiently close to the root xr, the second order Taylor expansion gives f (xr) =0= f (xi)+ f (xi)(xr − xi)+ (xr − xi)2 2 f (ζ) where ζ lies in the interval (xi,xr). From the above equation, we subtract Equation (9.2) to get xr − xi+1 new error = − f (ζ) 2f (xi) (xr − xi old error )2 Thus, the new read more..

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    Mathematical Review 275 9.12 Composite Simpson’s 1/3 Rule The basic Simpson’s 1/3 Rule rule provides the value of the integral I = b a f (x)dx = b − a 6 f (a)+4f a + b 2 + f (b) by approximating the integrand f (x) with the quadratic function that inter- polates the function values f (a), f [(a + b)/2], and f (b). The value of I is improved by dividing the read more..

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    276 Optimal Control for Chemical Engineers Moreover, if in the closed interval [a, b] the function f is real, but not necessarily continuous, and has the antiderivative F ,then F (b) − F (a)= b a f dx = b a dF dx dx 9.14 Mean Value Theorem If f (x) is a continuous function in the x-interval [a, b], then there is a real number c in the interval such that (b − a)f read more..

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    Mathematical Review 277 9.16 Implicit Function Theorem Consider the set of functions f1(x1,x2,... ,xn)= c1 f2(x1,x2,... ,xn)= c2 .. . fm(x1,x2,... ,xn)= cm where m<n. Then according to this theorem, provided that • the vector of functions f (x) ≡ f1(x) f2(x) ... fm(x) is differentiable near a, • the derivatives of f with respect to x are continuous at a,and • the Jacobian read more..

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    278 Optimal Control for Chemical Engineers Point of Accumulation Not necessarily in a set, its point of accumulation has in its each neighborhood, at least one non-identical point from the set. A closed set contains all of its accumulation points. Subsequence Given a sequence x1,x2,x3,x4,x5, ... its subsequence is, e. g., x1,x3,x4,x6,x8, ... which is contained in the original sequence. read more..

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    Mathematical Review 279 ii. α(x + y)= αx + αy for any two vectors x and y 3. For any two real numbers α and β and any vector x i. (α + β)x = αx + βx ii. α(βx)= (αβ)x 4. The space contains i. the zero vector, 0, such that x +0 = x ii. the unit vector, 1, such that 1x = x iii. the negative of x, denoted by −x, such that x +( −x)=0 for read more..

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    280 Optimal Control for Chemical Engineers is characterized by the set of n direction ratios {d1,d2, ..., di, ... , dn } where di = vi v Two vectors along the same direction have an identical set of direction ratios and vice versa. 9.21 Parallelogram Identity For any two vectors a and b and a scalar α a + αb 2 = n i=1 (ai + αbi) 2 = n i=1 (a2i + α2b2i +2αaibi) = a read more..

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    Mathematical Review 281 9.23 Cauchy–Schwarz Inequality The inequality is a b ≤ a b where a and b are two n-dimensional vectors. For n =2, let a ≡ ⎡ ⎣ a1 a2 ⎤ ⎦ and b ≡ ⎡ ⎣ b1 b2 ⎤ ⎦ Then (a21 + a22) a 2 (b21 + b22) b 2 − (a1b1 + a2b2) 2 (a b)2 =(a1b2 − a2b1) 2 ≥0 Therefore, a b ≤ a b , which can be easily generalized for higher dimen- sions. read more..

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    282 Optimal Control for Chemical Engineers Similarly, t2 ≤ (a 2 21 + a 2 22)(b 2 1 + b 2 2) Adding the last two inequalities, we get t1 + t2 Ab 2 ≤ (a211 + a212 + a221 + a222) A 2 (b21 + b22) b 2 The above result in terms of A and b is Ab ≤ A b which can be easily generalized for higher dimensions. 9.25 Conditional Statement A conditional statement is a read more..

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    Mathematical Review 283 9.26 Fundamental Matrix Consider the homogeneous linear differential equation dx dt = Ax in a time interval where x = x(t)is an n-dimensional vector, and A is an n ×n matrix of constants. Then n-linearly independent solutions of the differential equation exist. Let us denote the solutions by column vectors: x1(t), x2(t), ..., xn(t) Then the fundamental matrix read more..

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    284 Optimal Control for Chemical Engineers E. Kreyszig. Introductory Functional Analysis with Applications. John Wiley & Sons, US, 1978. W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, Inc., New York, 1976. B.S.W. Schr¨ oder. Fundamentals of Mathematics — An Introduction to Proofs, Logic, Sets, and Numbers. Wiley, New Jersey, 2010. A.E. Taylor and W.R. Mann. Advanced Calculus. read more..

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    Index In this Index, a page number in bold face corresponds to the main informa- tion of the index entry. An underlined page number refers to a numerically solved example. adjoint equation, 63 admissibility criteria, 252 admissible pair, 252 augmented functional, xvii, 45, 54, 59, 65, 66, 67, 93, 94, 96, 98, 102–104, 111, 121, 124, 153, 154, 162, 163, 169, 173, 186, 188, 202, read more..

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    286 Optimal Control for Chemical Engineers continuity, 33 linearity, 33 differential equations autonomous, 270, 271 non-autonomous, 270 directional derivative, 32, 36, 272, 272 discontinuous controls, 177 Dubois–Reymond Lemma, 79 dynamic optimization, 4 Embedding Theorem, 262 Euler’s formula, 253 Euler–Lagrange equation, 63 Extreme Value Theorem, 278 fixed point, 117 Fourier transform, 253, 257, 264, read more..

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    Index 287 linear transformation, 188 inequality constraints active, 110, 111–115, 166 inactive, 110, 111, 250, 251 infimum, 138, 142, 144, 269 integration by parts, 49 Intermediate Value Theorem, 53, 276 interval closed, 268 open, 268 inverse Fourier transform, 253, 264 Inverse Function Theorem, 90, 92, 95, 115, 117 Jacobian, xvii, 191, 240, 243, 263, 274 determinant, 90–92, 277 John read more..

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    288 Optimal Control for Chemical Engineers properties, 26 normal control, 251 pair, 251 normality condition, 89, 251 objective functional, 3 autonomous, 126 operator inequality, 118, 281 optimal control, 1 advantage, 4 laws, 20 subject, 1,4 optimal control problems an outline of solution, 71 batch distillation, 5 with integral equality constraint, 170 with integral inequality constraint, 172 batch read more..

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    Index 289 parallelogram identity, 280 penalty function, 202, 210, 215, 217, 219, 221 penalty function method, 185, 201, 201, 202, 205, 207, 214, 217, 219, 233 algorithm, 203, 266 weighting matrix, 202–204, 210, 212, 217 perturbation effect of final state, 134 pulse, 130, 131 temporal, 131, 134 pi criterion, 248 assumptions, 250 periodic CSTR, 254 with control constraints, 256 periodic CSTR, read more..

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    290 Optimal Control for Chemical Engineers variation of functional, 23, 36, 45, 47, 48, 50, 52, 57, 62 of integral objective, 40, 49 variation of integral equivalence to Fr´echet differential, 42 equivalence to Gˆ ateaux differential, 43 variational derivatives, 41, 44, 72, 76, 182, 186, 187 pair, 252, 253, 258 vector spaces, 23, 25, 26, 278 Weierstrass Theorem, 140, 150, 278 read more..

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