Innite Dimensional Quadratic Optimization: Interior-Point Methods and Control Applications January,995 Leonid Faybusovich John B. Moore y Department of Mathematics University of Notre Dame Mail Distribution Center, Notre Dame, IN 46556 USA y Department of Systems Engineering and Cooperative Research Centre for Robust and Adaptive Systems Research School of Information Sciences and Engineering Canberra ACT 000 Australia Abstract An innite-dimensional convex optimization problem with the linear-quadratic cost function and linear-quadratic constraints is considered. We generalize the interior-point techniques of Nesterov-Nemirovsky to this innite-dimensional situation. The obtained complexity estimates are similar to nite-dimensional ones. We apply our results to the linearquadratic control problem with quadratic constraints. It is shown that for this problem the Newton step is basically reduced to the standard LQ-problem. AMS classication: 90C0, 9C05, 49N05 Key words: control problems, quadratic constraints, path-following algorithms
Introduction In their fundamental monograph [9], Nesterov and Nemirovsky have shown that a broad class of a nite-dimensional optimization problems can be solved using interior-point technique based on the notion of the self-concordant barrier. Many control problems (like multi-criteria LQ and LQG problems) could be treated with the help of this technique provided an innite-dimensional version were available. In this paper we make a natural step in this direction. Namely, we consider a convex quadratic programming problem with quadratic constraints in a Hilbert space. The generalization of the technique of Nesterov-Nemirovsky is somewhat straight-forward for this situation. We consider both long-step and short-step versions of a path-following algorithm and obtain estimates on the number of Newton steps analogous to the best known for the nite-dimensional situation. In [] J.Renegar also considers an innite-dimensional version of the Nesterov-Nemirovsky scheme. His approach is quite dierent, however, and aims at the reformulation of complexity estimates in terms of certain generalizations of condition numbers. Of course, the performance of the Newton step in our situation requires solving an innite-dimensional linear-quadratic problem (without inequality constraints). We consider Yakubovich-type problems [] and show that the performance of the Newton step is reduced to solving the standard LQ problem of control plus a system of m linear algebraic equations where m is the number of inequality constraints. It is quite natural to expect that the task of nding the Newton direction will at least require solving the standard LQ-problem. Thus the real situation is as good as one could hope. The case of linear inequality constraints admits a direct treatment without using Nesterov- Nemirovsky technique and leads to sharper complexity estimates [4]. Duality, Central Trajectories and Dynamical Systems Let (H; h; i) be a Hilbert space, X its closed vector subspace. Given self-adjoint nonnegative denite bounded operations Q 0 ; : : :; Q m on H, vectors x 0 ; a 0 ; : : :; a m in H and real numbers b 0 = 0; b ; : : :; b m, consider the following optimization problem q 0 (x)! min; () x x 0 + X; ()
q i (x) 0; i = ; ; : : :m; () where q i (x) = hx; Q ixi + ha i ; xi + b i. We will assume throughout this paper that the set P H dened by constraints (), () is bounded and has a nonempty interior. More precisely, int(p ) = fx P : q i (x) < 0; i = ; ; : : :; mg (4) is not empty. Along with () - () consider the following (dual) problem: (y; ) = q 0 (y) + 5q 0 (y) + i q i (y)! max; (5) i 5 q i (y) X? ; (6) y x 0 + X; i 0; i = ; ; : : :; m: (7) Proposition. For every x satisfying (), () and every (y; ) satisfying (6),(7) we have: q 0 (x) (y; ): (8) Proof: Indeed, q 0 (x)? (y; ) (x; )? (y; ) h5(y; ); x? yi = 0, where in the second inequality we used the fact that (x; ) is convex in x for nonnegative i (see for example, [0]). The last equality follows by (6) and the fact that x? y X. Remark If f is a smooth function on H, then 5f(x); x H is uniquely determined by the relation Df(x) = h5f(x); i; H: (9) Here Df(x) stands for the Frechet derivative of f at point x evaluated on the vector. Consider the family of optimization problems of the form: f (x) = q 0 (x)? ln(?q i (x))! min; (0) q i (x) < 0; i = ; ; : : :; m () x x 0 + X: () Here 0 is a parameter. In this section we will make the following assumption: D f (x)(; ) a (x) h; i ; 0; ()
for some a (x) > 0. Here x int(p ); X and D f (x)(; ) stands for the second Frechet derivative of f at the point x int(p ) evaluated on the pair (; ) X X. Observe that it is sucient to verify the condition () for the case = 0. One can easily compute: D f (x)(; ) = * ; Q 0? Q i q i (x) We conclude from (4) that f is convex on int(p ).! + + h5q i (x); i h5q i (x); i q i (x) (4) Proposition. Under the assumption () the problem (0) - () has a unique solution x() int(p ); 0: Proof: Fix 0: Let ~x int(p ) and f (~x) =. Consider the set P = fx int(p ) : f (x) g. It is clear that P is a convex and bounded subset of int(p ). Suppose that x i ; i = ; ; : : :; is a sequence of points in P such that lim x i = x; i! + in the strong topology (i.e. in the topology induced by the norm). It is clear that x P. If x @P = P? int(p ), then f (x i )! ; i! +. But f (x i ) for all i. Hence x int(p ). We proved that P is convex, bounded and strongly closed. Hence P is weakly compact (see for example, []). Since f is convex, it is weakly semi-continuous from below on int(p ). We conclude that f attains its minimum x() on P. Observe that infff (x) : x int(p )g = infff (x) : x P g: The uniqueness of the minimum easily follows from the assumption (). Remark In the nite-dimensional situation the curve x(); 0 is usually called the central trajectory and the point x(0) is the analytic center (see for example, [],p.50). Our intermediate goal is to prove that x() is a smooth function of and converges to the optimal solution of () - () when! +. Denote by : H! X the orthogonal projection onto X. The point x() int(p ) is uniquely determined by the condition In other words, a 0 + Q 0 x()? For any 0 consider the map : int(p )! X, 5f (x()) X? : (5) (x) = Q 0 x? 4! 5q i (x()) = 0: (6) q i (x())! 5q i (x) : (7) q i (x)
We can rewrite (6) in the form: (x()) =?a 0 : (8) Proposition. For each 0 the map is a smooth isomorphism of int(p ) onto X such that? depends smoothly on. Proof: By Proposition. for any c X the problem? hc; xi + hq 0x; xi? ln(?q i (x))! min; x int(p ) (9) has a unique solution x c int(p ) and by (8) (x c ) = c. Hence is surjective. On the other hand, if (x ) = (x ); x ; x int(p ), then again by (8) both x and x are solutions to (9). Hence by Proposition. x = x, i.e. is injective. It remains to prove that? smooth and depends smoothly on. Observe that for every X; x int(p ), D (x) = Q 0? If D (x) = 0, then by (0): Qi! q i (x)? h5q i(x); i 5 q i (x) q i (x) is (0) 0 = h; D (x)i = D f (x)(; ): Hence, by () = 0: Thus D (x) induces an injective map from X into X. Further, by (0):! Q i in 5q i (x) 5q i (x) D (x) j X = Q in? + j q i (x) q i (x) X ; () where in : X! H is the canonical immersion. The condition () easily implies that D (x) j X is surjective. We can now apply the implicit function theorem to conclude that? and depends smoothly on. is smooth Corollary. The solution x() of the problem (0) - () depends smoothly on. Proof: By (8), x() =? (?a 0). The result follows by Proposition.. We now introduce a family of Riemannian metrics g which are determined by Hessians of functions f, namely, g (x; ; ) = D f (x)(; ); () 5
; H and q i (x) < 0; i = ; ; : : :; m. We clearly have: (x) = Q 0? g (x)(; ) = h (x); i; () Q i q i (x) + m X 5q i (x) 5q i (x) q i (x) ; (4) 0: We can think of int(p ) (see (4)) as a Riemannian submanifold of the open subset in H dened by the conditions: q i (x) < 0; i = ; ; : : :; m: Given a smooth function ' dened on an open neighborhood of int(p ) in H, denoted by 5 '(x) X the gradient of ' relative to the metric g, the gradient 5 '(x) X is uniquely determined by conditions: D'(x) = h5'(x); i = g (x; ; 5 '(x)) = D f (x)(; 5 '(x)) (5) X; x int(p ): The existence and uniqueness of the 5 '(x) X satisfying (5) follows easily by (). Proposition.4 We have (x) 5 '(x)? 5'(x) X? : (6) Proof: For X: h; (x) 5 '(x)? 5'(x)i = g (x; ; 5 '(x))? h; 5'(x)i = h; 5'(x)i? h; 5'(x)i = 0 The result follows. Corollary. We have: D (x) 5 '(x) = (5'(x)): (7) Proof: By (0),(4) we have: D (x) 5 '(x) = ( (x) 5 '(x)): The result follows by (6). The next Proposition describes a dierential equation which has the central trajectory as its solution. 6
Proposition.5 Let '(x) =?q 0 (x). Then Proof: Set () = (x()). We have (see (7)): On the other hand, by (8) Hence, dx() d =? 5 '(x); 0: (8) d d () = D (x()) dx() d + (Q 0x()) : d d () =?a 0: D (x()) dx() d =? (5q 0(x())) : Comparing this with (7) and using Proposition., we arrive at (8). Corollary. The function q 0 (x()) is a monotonically decreasing function of for 0. Proof: Indeed, d d q 0(x()) = 5q 0 (x()); dx() d = g x(); 5 q 0 (x()); dx() d = g (x(); 5 q 0 (x()); 5 q 0 (x())) 0; where in the second equality we used (5) and in the third equality we used (8). Consider now the dual of the problem (0) - (): (y; ) = (y; ) + 5q 0 (y) + ln i + m( + ln)! max; (9) i 5 q i (y) X? ; (0) i > 0; i = ; ; : : :; m; () y x 0 + X Here (y; ) was dened in (5). 7
Proposition.6 Let x() be a solution to (0) - (). Set y() = x(), i () =? ; i = ; ; : : :; m: q i (x()) Then f (x()) = (y(); ()): () Proof: A direct computation. Proposition.7 For every x satisfying (), () and every (y; ) satisfying (0), () we have: f (x) (y; ): Proof: The proof is quite similar to the one of Lemma.9 in []. Corollary.4 The pair of (x(); ()) is an optimal solution to (9) - (). Corollary.5 Let x be the optimal solution to the problem () - (). Then q 0 (x()) q 0 (x ) q 0 (x())? m ; () lim q 0 (x()) = q 0 (x );! +: (4) Proof: Observe that (x(); ()) is a feasible solution to (5) - (7). By Proposition. q 0 (x ) (x(); ()) = q 0 (x())? m. Since by Proposition.5 q 0 (x()) is a monotonically decreasing function of, (4) follows from (). Corollary.6 The function (y(); ()) is a monotonically increasing function of for 0. Proof: The proof is quite similar to one of Theorem.8 in []. Corollary.7 If > > 0; then q 0 (x( ))? q 0 (x( )) m? m : (5) 8
Proof: By the previous Corollary (x( ); ( )) (x( ); ( )). We also know that q(x())? (x(); ()) = m : (6) Hence q 0 (x())? q 0 (x()) [q 0 (x( ))? (x( ); ( ))]? [q 0 (x()? (x( )); ( ))] = m? m : Complexity Estimates for Path-Following Algorithms The main idea behind path-following algorithms is to produce a nite-step approximation to the central trajectory. Using the interpretation of points on the central trajectory as optimal solutions to the family of optimization problems (0) - (), one can reduce the construction of path-following algorithms to the performance of Newton steps for functions (0) and updating the parameter. The choice of logarithmic barriers enables us to use the powerful technique of self-concordant functions [9]. In this section we show that an innite-dimensional generalization of the Nesterov-Nemirovsky scheme is quite straight-forward. We use some simplications of this scheme proposed by F. Jarre [6] and D. den Hertog []. Interior-point methods for nitedimensional convex quadratic programming problems with quadratic constraints have been also considered in [7], [8] and [5]. Let q(x) = hx; Qxi + ha; xi + b be a convex quadratic form on H. Given x; h H, consider the function We obviously have (t) = q(x + th); t IR: (7) h i t (t) = q(x) + (Dq(x) h)t + D q(x)(h; h) : (8) Suppose that D q(x)(h; h) > 0 and (0) = q(x) < 0. Since is convex we conclude that (t) = `(t? t )(t? t ); (9) where l = D q(x)(h; h)= and t < 0 < t. In particular, (t) < 0, t < t < t : (40) 9
Furthermore,?ln(?(t)) =?lnl? ln(t? t )? ln(t? t) (4) Using (4) we can easily compute the Taylor expansion of the function (x) =?ln(?q(x)): Indeed, by (4): (x + th) = X n=0 t n n! Dn (x)(h; : : :h) =?ln(` t (?t )) + X n= t n n t n + t n (4) Hence D n (x)(h; : : :h) = (n? )! t n + t n ; n (4) Here D n (x) denotes the nth Frechet derivative of at the point x H. If D q(x)(h; h) = 0 but Dq(x) h 6= 0, then (t) is a linear function with a single root t 6= 0. The corresponding expansion for (t) will have the form: (t) =?ln(?q(x)) + X n t n n t n : (44) Return now to the problem (0) - (). Given x int(p ), introduce a measure of proximity of x to the point x() on the central trajectory: (x) = D f (x) (p (x); p (x)) ; (45) where (see (5)). p (x) =? 5 f (x); (46) Proposition. Suppose that q i (x) < 0 for i = ; ; : : :; m and h H is such that D f 0 (x)(h; h) < ; (47) then q i (x + h) < 0 for i = ; ; : : :; m. Proof: Consider the functions i (t) = q i (x + th) and i (t) =?ln(? i (t)); i = ; ; : : :; m: Fix i [; m]. Consider rst, the case D q i (x)(h; h) > 0. We have by (4) D i (x)(h; h) = i + i ; (48) ht (i) ht (i) 0
where t (i) ; t(i) are roots of the quadratic polynomial i (t): Since we conclude by (48): or min n t (i) ; t (i) D f 0 (x)(h; h) = ht (i) o < : Hence, ]? ; []t (i) D i (x)(h; h); i + i < ht (i) ; t(i) [. But i(t) < 0 for t ]t (i) ; t(i) [. This proves q i (x + h) < 0. If D q i (x)(h; h) = 0 but Dq i (x) h 6= 0, then i (t) is a linear (nonconstant) function with the root t (i) such that The case of the constant i is trivial. t (i) >. Since i (0) < 0, we necessarily have i () < 0. Lemma. Given A > 0; : : :A m > 0, we have for 0 < y < x. (A x + A x + : : : + A x m) x (A y + Ay + : : : + Ay m) y Proof: It is sucient to prove that the function is monotonically decreasing for x > 0. But X 0 (x) = X (x) = x ln (Ax + : : : + A x m) x (A x + : : : + Ax m) [(Ax ln(a x ) + : : : + A x mln(a x m))? (A x + : : : + A x m) ln (A x + A x + : : : + A x m) ] : The function S(p ; : : :p m ) = P m p i lnp i is convex on the positive orthant p i 0. Hence, maxfs(p) : p i 0; i = ; ; : : :; m; p + p + : : : + p m = cg is attained at one of the extreme points of the simplex fp IR m : p i 0; i = ; ; : : :; m; p + : : :+ p m = cg. We conclude S(p ; : : :p m ) (p + : : : + p m )ln(p + : : : + p m ): Hence X 0 (x) 0 for any positive x. Proposition. For any x int(p ); h X; 0 we have: D j f (x)(h; : : :h) (j? )!! j Dk f (x)(h; : : :h) (k? )!! k (49) for any k; j such that k is even and k j. In particular D f (x)(h; h; h) D f (x)(h; h) (50)
Remark The condition (50) means that the function f is -self-concordant. It is crucial in the Nesterov-Nemirovsky scheme [], [9]. Proof: Observe that f (x) = q 0 (x) + P m i(x) with i (x) =?ln(?q i (x)). Inequalities (49) easily follows by (4) and Lemma.. Proposition. Suppose that x int(p ); t = + (x). Then x + tp (x) int(p ) and f (x + tp (x)) f (x) + ln( + (x))? (x): (5) Remark Recall (46) that p (x) =? 5 f (x) is the Newton's direction for the f at the point x; (x) was dened in (45). Proof: Set p (x) = p; (x) = in this proof. We have f (x + tp) = f (x) + tdf (x) p + t D f (x)(p; p) + X j Let i (t) = q i (x + tp) = `i t? t (i) (t? t (i) D j f (x)(p; : : :p) = (j? )! ) with t(i) < 0 < t (i). By (4): 0 B @ i j + ht (i) ht (i) D j f (x)(p; : : :p) tj j! : (5) i C j A ; (5) for j and (x) = D f (x)(p; p) = D q 0 (x)(p; p) + 0 B @ i + ht (i) ht (i) i C A : (54) (If i (t) is linear, the formulas (5), (54) hold true if we set t (i) = +. See (44)). Observe, further, that by (5): and by (5), (54) and Lemma.: X j Df (x) p =?D f (x)(p; p) =? (x) (55) D j f (x)(p : : :p) tj j! Combining (5), (55), (56), we obtain X j jtj j j! (j? )! (x) j (56) f (x + tp)? f (x)?t(? )? ln(? ); (57)
t 0. For t = t we obtain (5). Observe now that D f 0 (x)(tp; tp) D f (x)(tp; tp) = Hence, x + tp int(p ) by Proposition.. (x) ( + (x)) < : Lemma. Suppose that F is a symmetric trilinear form, G is a symmetric bilinear form on a Hilbert space H such that: F (h; h; h) G(h; h) (58) for some > 0 and every h H. Then F (h ; h ; h ) G(h ; h )G(h ; h )G(h ; h ) (59) for any h ; h ; h H: Proof: Given h ; h ; h H consider the vector subspace V H spanned by vectors h ; h ; h. It is clear that V is at most -dimensional. Consider the restrictions of F; G to V. It is sucient to establish (59) for any three vectors in V. For the proof of this see for example, [6]. Proposition.4 Let x int(p ); (x) <. Then x + = x + p (x) int(p ) and (x + ) Proof: Given H, consider the function (x) (? (x)) : (60) ' (t) = Df (x + tp)? (? t)df (x) ; (6) where p = p (x): It is clear that ' () = Df (x + p); ' (0) = 0. We have ' 0 (t) = D f (x + tp)(; p) + Df (x) ; (6) Using inequality (50) and Lemma., we obtain: ' 00 ' 00 (t) = D f (x + tp)(; p; p): (6) (t) D f (x + tp)(p; p) h i D f (x + tp)(; ) : (64)
Observe now that D f (x)(; ) h i? t D f (x)(p; p) D f (x + tp)(; ) D f (x)(; )? t [D f (x)(p; p)] (65) for 0 t <. Indeed, let (t) = D f (x + tp)(; ). Using inequality (50) and Lemma., we have : 0 (t) (t) (t) ; (66) where (t) = D f (x + tp)(p; p). Now (again using (50) and Lemma.) 0 (t) (t) : (67) By (67) (see [], p. 54) (t) (0)? t (0) ; 0 t < : By (66) we will have [], p.54: 0 (t) (t) (0) ; 0 t < : (68)? t (0) By (68), [], p.55: (0)(? t (0) ) (t) (0)? t (0) ; (69) which coincides with (65). Observe now that ' 0 (0) = D f (x)(p; ) + Df (x) =?Df (x) + Df (x) = 0, where we used (5), (46). Hence from (64), (65) we obtain: ' 0 (t) = Z t ' 00 0 ()d hd f (x)(; ) (? t)? Z t ' 00 0 i Z t 0 () d d h (? ) = D f (x)(; ) ; (70) where = (x). Further, by taking into account ' (0) = 0, we obtain h i Z t j' (t)j D f (x)(; ) (? )? d 0 = t D f (x)(; )? t i : (7) 4
Take t = ; =?p (x + ). We have ' () =?Df (x + ) p (x + ) = D f (x + )(p (x + ); p (x + )) = (x + ) : Hence, by (7) (x + ) hd? i f (x) p (x + ); p (x + )?? (x + )? ; which coincides with (60). In the last inequality we used (65) again. Remark The main idea of the proof of Proposition.4 is due to Y. Nesterov and A. Nemirovsky [9]. We used simplications suggested by [6] and []. Given IR, denote by P, the set fx int(p ) : f (x) g. We know (see the proof of Proposition.) that P is a convex, closed and bounded set. Suppose that for every IR there exist positive constants m and M such that M jjjj D f (x)(; ) m jjjj ; X; x P (7) The next proposition states the convergence of the Newton method for the function f. Proposition.5 Given 0; x int(p ); (x) <, consider the sequence x 0 = x; x = x 0 + p (x 0 ); : : :; x i = x i? + p (x i? ); i : Then x i int(p ) for all i and moreover under the assumption (7) lim x i = x(); i! + (7) Proof: By (60) (x ) (x 0 ) (? (x 0 )) 9 4 (x 0 ) : Using induction on i, we can easily conclude from (60): (x i ) 9 4 i? (x 0 ) i 4 i? (74) In particular, (x i )! 0; i! +: (75) 5
It is also clear from (74) that (x i ) < for all i. Hence, by Proposition. x i int(p ) for all i. Moreover, by (74) (x i ) < for all i. Since the function + + ln(? ) is positive for 0 < <, we conclude by (57) that f (x i ) < f (x i? ); i. Hence, all x i belong to the set P for an appropriately chosen. Since by (7) (x i ) m jjp (x i )jj; i ; we have by (75) p (x i )! 0; i! + (76) Recall that by (6) (x i )p (x i ) = 5 f (x i ): The condition (7) implies that jj (x)jj M for all x in P. Hence, (76) implies 5 f (x i )! 0; i! +:) (77) Observe now that (Df (y)? Df (x)) (y? x) m jjx? yjj for all x; y P (see for example, [0]). Take x = x(); y = x i. We have: (Df (y)? Df (x)) (y? x) = h5f (x i )? 5f (x()); x i? x()i = h5f (x i )? 5f (x(); (x i? x()))i = h 5 f (x i )? 5 f (x()); x i? x()i = h 5 f (x i ); x i? x()i m jjx i? x()jj ; (78) where in the second equality we used x i?x() X, and 5f (x()) = 0 in the fourth equality. Now (77), (78) and boundedness of P imply jjx i? x()jj! 0; i! +. Proposition.6 Given x int(p ); (x) < ; we have: f (x)? f (x()) (x) h i 9? : (79) 4 (x) Proof: Set p = p (x) in this proof. Since f is convex, we have: f (x + p) f (x) + Df (x) p = f (x)? (x) ; (80) 6
where we used (55). Let x 0 = x and x ; x ; : : : be the sequence of points obtained by repeating Newton's steps. We have by (74): (x i ) and x i! x(); i! +; by Proposition.5. 9 4 i? (x 0 ) i (8) Hence, f (x)? f (x()) = X i=0 (x i ) X i=0 X (x 0 ) h i 9? ; 4 (x 0 ) i=0 9 4 [f (x i )? f (x i+ )] i+? (x 0 ) i+ where we used (80) in the rst inequality and (8) in the second inequality. Proposition.7 If x int(p ); (x) <, then jq 0 (x)? q 0 (x())j? (x) +? 9 4 (x) (x) p m? (x) (8) Proof: The proof is quite similar to one of Lemma. in []. We are now in the position to describe a version of the large-step path-following algorithm for solving () - (). Suppose we are given 0 > 0 and x 0 int(p ) such that (x 0 ). Given # > 0, we perform what is called outer iterations = ( + #) 0 ; : : : i = ( + #) i 0. Suppose we are able to construct a sequence of points x ; x ; : : : in int(p ) such that i (x i ). Theorem. Suppose " > 0 is given and i ln 4m " 0 ln( + #) (8) Then q 0 (x i )? q 0 (x ) ", where x is an optimal solution to the problem () - (). Proof: Observe that (8) is equivalent to i 4m " : (84) By () q 0 (x( i ))? q 0 (x ) m i : 7
By (8) with (x) we obtain: q 0 (x i )? q 0 (x( i ))? 9 4 +? p m p m : i i Thus q 0 (x i )? q 0 (x ) = q 0 (x i )? q 0 (x( i )) + q 0 (x( i ))? q 0 (x ) p m + m i " where in the last inequality we used (84). It remains to describe a procedure for updating x i. Suppose we are given x i int(p ) such that i (x i ). Set i+ = ( + #) i. We have to nd x i+ int(p ) such that i+ (x i+ ). To do that we perform several Newton steps for the function f i+ Each such a Newton step is called an inner iteration. using x i as a starting point. Theorem. Each outer iteration requires at most + # 5 p! m + # + # m (85) inner iterations. point Proof: We start with a point x i int(p ) such that i (x i ). By Proposition. the belongs to int(p ) and moreover x = x i + tp i+ (x i ); t = + i+ (x i ) f i+ (x i )? f i+ (x) i+ (x i )? ln( + i+ (x i ))? ln + > : (86) We continue to perform iterations described above until i+ (x i+ ). By (86) the value of the function f i+ is decreased by at least such iterations, we obviously have: after such an iteration. If N is the number of N f i+ (x i )? f i+ (x( i+ )) : (87) Let (x; ) = f (x)? f (x()): We wish to estimate (x i ; i+ ). By the mean value theorem: (x i ; i+ ) = (x i ; i ) + @ @ (x; ^)# i (88) 8
for some ^ ( i ; i+ ). We have But 5f (x()) X? ; dx() d X. Hence, By (88), (89) of. @ @ (x; ) = q 0(x) + 5f (x()); dx() d @ @ (x; ) = q 0(x)? q 0 (x()): (89) : (x i ; i+ ) (x i ; i ) + ( i+? i ) q 0 (x i )? q 0 (x( ^)) (x i ; i ) + # i (jq 0 (x()))? q 0 (x i )j + (q 0 (x())? q 0 (x( i+ )) : (90) In the last inequality we used the fact that q 0 (x()) is a monotonically decreasing function By (8) Further, by (79) Finally, by (5): + jq 0 (x i )? q 0 (x( i ))j? 9? 4 (x i ; i )? 9 4 q 0 (x())? q 0 (x( i+ )) m i? Substituting (9)-(9) into (90), we obtain: (x i ; i+ ) + # Combining (94), (87) we arrive at (85). From Theorems.,. we immediately obtain: 5p m p m 5 p m : (9) < : (9) m = m# (9) i+ i+! + m# : (94) + # Theorem. An upper bound for the total number of Newton iterations is given by: 4m " 0 ln ln( + #) + # 5 p m + m# + # We now briey consider the situation where the number of inner iterations (per one outer iteration) does not exceed. 9!! :
Proposition.8 Given x int(p ); > 0; # > 0. If + = ( + #), then +(x) ( + #) (x) + # p m: (95) Proof: Given X, D f +(x)(; ) = + D q 0 (x)(; ) + Introduce a vector ~p X by the condition: [Dq i (x) ] D f q i (x) (x)(; ): (96) D f (x)(; ~p) =?Df +(x) ; X (97) We have Hence, by (97): Df +(x) = ( + #)Df (x) + # D f (x)(; ~p) = ( + #)D f (x)(; p (x))? # h5q i (x); i : q i (x) h5q i (x); i : (98) q i (x) Taking = ~p, we obtain from (98): h i D f (x)(~p; ~p) ( + #) D f (x)(~p; ~p) + p m h5q i (x); ~pi q i (x)! h D f (x)(~p; ~p) hd i f (x)(p (x); p (x)) i? (x)( + #) + p m : (99) Here we used the Cauchy-Schwarz inequality twice. It remains to prove that D f +(x)? p +(x); p +(x) D f (x)(~p; ~p): (00) Indeed, by (97):?Df +(x) = D f (x)(; ~p) = D f +(x)? ; p +(x) : In particular, for = p +(x):? D f +(x) p +(x); p +(x) = D f (x) (~p; p (x)) hd? i f (x) p +(x); p +(x) h D f (x)(~p; ~p) i hd f +(x)? p +(x); p +(x) i hd f (x)(~p; ~p) i where in the last inequality we used (96). The result follows by (99), (00). 0
Theorem.4 Let x int(p ); (x) and + = ( + #), where # = p m, for suciently small > 0. Then x + = x + p (x) is such that +(x + ). Proof: By Propositions.4,.8: +(x + ) +(x)? +(x) 9 4 +(x) 9 ( + #) (x) + # p m 4 + p (x) + = () (0) m = 9 4 Observe that ()! 9 4 (x) 4 for! 0. Hence +(x+ ) for suciently small. 4 Example Consider the linear-quadratic time-dependent control problem with quadratic inequality constraints. Namely, let q i (y; u) = Z T 0 h i y T (t)q i (t)y(t) + u(t) T R i (t)u(t) dt + b i ; i = ; ; : : :; m (0) where (u; y) L p [0; T ]Ln [0; T ]; y is absolutely continuous function with values in IR n and such that _y L n[0; T ]. We use the standard notation Ln [0; T ] for the Hilbert space of measurable square integrable functions on [0; T ] with values in IR n. Further, Q i (t) (respectively, R i (t)) is a symmetric positive denite n by n (respectively, p by p) matrix which depends continuously on t [0; T ]; A(t) (respectively, B(t)) is a n by n (respectively, n by p) matrix which depends continuously on t [0; T ]; b i ; i = ; ; : : :; m are real numbers and b 0 = 0; T is a xed positive number. Consider the problem q 0 (y; u)! min (0) q i (y; u) 0; i = ; ; : : :; m (y; u) x 0 + X where X = f(y; u) L n[0; T ] Lp[0; T ] : y is absolutely continuous on [0; T ]; _y Ln [0; T ]; y(0) = 0 and _y(t) = A(t)y(t) + B(t)u(t); t [0; T ]g. Theorem. gives a rather sharp estimate for the number of Newton steps for the path following algorithm. From the computational viewpoint it is crucial to see that the performance
of a Newton step is a reasonable problem. Set x = (y; u). We will frequently omit the time variable t in what follows. We have: (" _p + A T p (y; u) f (x) = q 0 (y; u)? 5f (y; u) = # " # = d i = ln(?q i (y; u)); " P m Q0 y? P Q i y q i (y;u) m R R 0 y? i u q i (y;u) 6 P 4 Q m 0? Z T o # Q i + d iq i x q i (y;u) qi (y;u) R 0? P m R i q i (y;u) + d ir i y q i (y;u) y T Q i + u T R i p (x) = " # ; ; 7 5 ; dt; i = ; ; : : :; m; (04) X? = B T : p is absolutely continuous on [0; T ] and _p L p n [0; T ]; p(t ) = 0g. The equation (6) will take form: where We further have _p + A T p? y = Q (x) + B T p? u = R (x) + Q (x) = Q 0? R (x) = R 0? Q i q i (x) ; R i q i (x) ; (Q i y)d i ; (05) qi (x) (R i u)d i ; (06) qi (x) _ = A + B; (0) = 0 (07) " # y p(t ) = 0; 5f (x) = : u We have to solve (05) - (07) with respect to ;. The problem (05) - (07) is quite similar to one arising in connection with the standard linear-quadratic control problem (see for example, [], []). The only dierence is that the constants d i are unknown. We proceed, nevertheless as in the standard linear-quadratic case. We are looking for the solution (05) - (07) in the form: p(t) = K(t)(t) + (t): (08)
By (06) (t) = R (x)? B T p? u? Substituting (08), (09) in (00), (06), respectively, we obtain: d i R (x)? R i u q i (x) : (09) _K + KA + A T K + KL (x)k? Q (x) = 0 (0) K(T ) = 0 where L (x) = B(t)R (x)? B T (t), _p =?A T p + Q (x) + (t); p(t ) = 0; () _ = A + L (x) p + (t); (0) = 0; () (Q i y)d i (t) = y + ; () q i (x)?! R (x)? R (t) =?B R (x)? i u d i u + q i (x) ; (4)! = R (x)? B T (R i u)d i p? u? q i (x) : (5) _ =? A T + KL + (t) + (t); (T ) = 0: (6) From (0) we nd K(t); 0 t T. Let (t; ) be the fundamental solution to _ =?(A T + KL ): (7) Then (t; ) = (t; )?T is the fundamental solution to _ = (A + L K): By (6): (t) = Z t (See () for the known functions v; v i.) By () T (t; )(() + ())d = v(t) + (t) = Z t 0 = (t) + for known functions ; i (see (4)). (t; ) (L () + ()) d d i v i (t) (8) d i i (t) (9)
By (08), (8), (9) p(t) = (t) + for the known functions i ; and by (5): (t) = (t) + for the known functions ; i. Substituting (9), () into (04), we obtain: d i = j= d i i (t) (0) d i i (t) () c ij d j + " i ; i = ; ; : : :; m; () where c ij = " i = Z T 0 Z T 0 y T Q i j + u T R i j dt y T Q i + u T R i dt: Finally, we nd d i solving the system of linear algebraic equations (). We see that the performance of the Newton's step involves three essential steps: (a) The numerical integration of the Riccati equation (0) (b) Finding the fundamental matrix for the linear system for the linear time-dependent system (7) (c) Solving a system of m by m linear algebraic equations (). The rst two steps are the same as for the solving the standard time-dependent LQ-problem. We see that the performance of the Newton step in our situation is essentially equivalent to solving the standard LQ-problem plus the system of linear algebraic equations. The standard numerical treatment of the problem (0) involves introducing the Lagrange multipliers and considering the dual problem. From the computational viewpoint nding the value of the cost function of the dual problem at a point is more or less equivalent to the performing the Newton step of the primal problem. 4
5 Concluding Remarks In the present paper we have considered both long-step and short-step versions of a pathfollowing algorithm for the innite-dimensional quadratic programming problem with quadratic constraints. We found an estimate for the number of Newton steps which coincides with the best known [] in the nite-dimensional case. We then considered an example of the linear-quadratic timedependent control problem with quadratic constraints. We have shown that in this situation the performance of the Newton step is essentially equivalent to solving the standard LQ-problem without inequality constraints. Thus the interior-point methodology has been extended to an innite-dimensional situation useful for control applications. Acknowledgment. This research was supported in part by the Cooperative Research Centre for Robust and Adaptive Systems while the rst author visited the Australian National University and by the NSF grant DMS 94-79. 5
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