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Part 7 Optimization in Functional Space 7.0 Motivation Example Maximizing Yield of Batch Reaction Consider the following reactions A BC D The rate equation for each species can be expressed as dC A dCB 2 k1C 1.5 k C ; r k1C 1.5 A 2 A B A k3C B ; dt dt dCC dCD 2 rC k3CB ; rD k3C A dt dt 11600 29600 where, k1 exp 17.2 ; k exp 41.9 2 ; RT RT 20000 k3 exp 30.5 ; R 1.987. RT rA Maximizing Yield of Batch Reaction If the above reactions are to be carried out in a batch reactor for 1000 minutes and the initial conentrations of the four components are: C A 0 1.0 mole/liter and CB 0 CC 0 CD 0 0, determine the temperature profile T (t ) for 0 t 1000 that maximizes the yield of reactant B, i.e., CB 1000 max T t 1.0 C 1000 A 0t 1000 Note that it can be assumed that the volume of reaction mixture remains unchanged. Part 7 Optimization in Functional Space 7.1 Calculus of Variation Objective Functions Lagrange Form J1 x t x t , x t , t dt t0 tf Bolza Form J 2 x t x t f , t f x t , x t , t dt t0 tf Mayer Form J 3 x t x t f , t f Equivalence of Lagrange and Bolza Forms Let x t0 , t0 0 J 2 x t f , t f dt dt t0 t0 tf x t , x t , t dt J1 t0 tf tf Equivalence of Bolza and Mayer Forms Let u t x t , x t , t u t0 0 u t f dt If x t f , t f x t f , t f u t f then tf t0 J 2 x t x t f , t f u t f u t f x t f , t f J 3 x t Example /2 min x ( t ),0 t / 2 I 0 dx 2 2 x 2tx dt dt s.t. x 0 x 0 2 Problem Statement Consider min I F t , x, x dt b a s.t. x( a ) A x(b) B Assume that F t , x, x has continuous 2nd-order derivatives wrt its three arguments and x(t ) possesses two derivatives everywhere in (a, b). Intuitive Interpretation Let’s visualize a competition, to which only functions which have 2 derivatives in (a,b) and which take on the prescribed end values are permissible. Let’s further assume that there exists a x*(t) such that I is the smallest. Variation Let's consider a family of admissible functions, i.e., x t x* t t where, t is any arbitrarily chosen twice-differentiable function which vanishes at the end points of the interval (a, b), i.e. a b 0 The increment t , representing the difference between the varied function and the optimal function, is often called the variation of x* (t ). Necessary Condition For a specific t , I I F t , x* , x* dt b dI d b a a b F x I x I x F x dt a x x x x F t , x* , x* F t , x* , x* dt * * x x dI d 0 * * * * F t , x , x F t , x , x b 0 dt * * a x x Integration by Parts - 2nd Term b a F t dt * x u dv t b b d F F * t t dt * x t a a dt x b d F t dt * a dt x Euler-Lagrange Equation d F F * * t dt 0 a dt x x Since this equation is valid for arbitrary t as long as b it is twice-differentable and a b 0. d F F 0 dt x x (Euler-Lagrange equation) It is the necessary condition for the extremal solution x* t . Example min I xt /2 0 F t , x, x dt dx F x 2 2tx; x 0 x 0 dt 2 d F F d 2 x 2t 2 x dt x x dt 2 x 2 x 2t 0 x x t 0 2 x c1 cos t c2 sin t t x 0 0 c1 ; x t 2 x 0 c2 2 2 sin t t I min 2 1 2 12 Transversality Conditions In the previous development, the values of the dependent variable, x(t ), are fixed at end points. If this is not given, then F t 0 x for t a and b. Transversality Conditions This equation must be examined in conjunction with the given boundary conditions: (1) x a given x(b) free (2) x a free x(b) given (3) x a free x(b) free a 0 F x 0 t b F x 0 t a b 0 F x F x 0 t a 0 t b Example 1 Find the curve with minimum arc length between the point x(0) 1 and the line t 2. min I x (t ) 0t 2 2 0 s.t. x(0) 1 x(2) free 1 x 2 dt Solution of Example 1 1 d F F 2 2 0 and F 1 x , From Euler-Lagrange equation dt x x x d x 2 const x const x const 0 2 2 dt 1 x 1 x x c1t c2 From transversality conditions: x 0 1 c2 1 F x t 2 x 1 x 2 0 x 2 c1 0 t 2 x t 1 for 0 t 2 Example 2 1 2 min I x xx x x dt 0 2 x(0) free 2 x(2) free Solution of Example 2 1 2 F x xx x x 2 d F F d x x 1 x 1 x 1 0 dt x x dt t2 x t c1t c2 2 F Transversality conditions: x x 1 0 at t 0, 2 x t2 t c1 c1t c2 1 0 at t 0, 2 2 c1 2, c2 1 t2 x t 2t 1 2 for 0 t 2 Dependent Boundary Conditions If G x a , x b 0 G x* a a , x* b b 0 then dG 0 d G x* a a , x* b b G 0 and dG d x a a * a G x* a a , x* b b G x a , x b G x a , x b dG a b d 0 x a x b a b G x a , x b x b G x a , x b x a r x b b * b Dependent Boundary Conditions From the previous derivation, it is required that F F t x x a b F b x t b a 0 t a F a F b 0 x t b b x t a Since b can be chosen arbitrarily, 0 x(a ) t a x(b) G x(a ), x(b) 0 F x F r x t b Unspecified Terminal Time We now consider a generalized problem where the final time is defined as the first time after the initial time t0 that the state trajectory is a member of a target set or terminal manifold. Problem Definition min I x, x, t dt tf t0 t0 given x t0 may or may not be given t f unspecified x t f C t f Variations of Optimal Trajectory and Terminal Time Consider a family of curves x t x* t t Since the terminal time is unspecified, it must be treated as a variable. Thus, t f t *f t *f Necessary Conditions I t *f t *f t0 x* t t , x* t t , t dt t *f t *f dI * * * t f x t f t f , x t t f , t f t t dt t 0 d x x dI d t x t * f 0 t x t * f * * f * * f , x t , t * , x t , t * * f * f * f * f t *f t0 t * x t *f t0 t t dt x* x* d t * dt * t0 x dt x t *f t *f * * d * * * * * * t f x t f , x t f , t f t f * t f t0 * t0 t * dt * t 0 x x x dt x 0 0 0 0 Terminal Constraint x t f x* t f t f C t f x* t t t t C t * f * f * f * f LHS * f t *f RHS d LHS t f LHS LHS LHS t *f t *f t *f d t f t f d LHS t *f x* t *f t *f d 0 d dC dt f RHS C t f t *f d dt f d d RHS C t *f t *f d 0 t *f t *f C t f x* t *f Necessary Conditions d (1) 0 * * x dt x (2) t0 * t0 0 x (3) t x t * f * * f t0 t t f , x t , t * * f * f * t * t f x * f * * * * * * * * t x t f , x t f , t f C t f x t f * t f x 0 * f Example I tf 0 1 2 2 1 x dt x 0 1 x t f 2 t f Example From Euler-Lagrange equation x at b x 0 1 b 1 1 x x t tf C x 0 1 x2 0 x 1 x2 x 1 a 1 x t 1 x t f t f 1 2 t f 1 t f 2 Vector Formulation d 0 x j dt x j j 1, 2, ,n j t0 t0 0 x j j 1 n j t f tf 0 x j j 1 n Example J x, x, t dt f tf t0 where x x1 , x2 T x t0 x 0 x t f x t f 1 2 1 2 2 when t t f x * 1 1 x 2 1 x1* t f 1 t f x2* t f 2 t f 0 2 * 2 2 x2 x2 1 2 2 0 2 2 x1 x2 x1 x1 x2 x1 x1 x2 x2 0 x1 x1 x2 x1 t f , x2 t f x12 x22 1