HW5A_98F - Solutions of Homework#5 1998 Fall MEAM 501...

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Solutions of Homework #5, 1998 Fall MEAM 501 Analytical Methods in Mechanics and Mechanical Engineering 1. Consider a constrained minimization problem ( 29 ( 29 n n n n T T K F F R b R A R v b v Av v v v v - = × , , , 2 1 , min { } m n m n K R g R B 0 g Bv R v - = × , , : where a matrix A is symmetric, that is, A A = T , and the constrained set K is non- empty. (1) Find the necessary condition that an element K u is a minimizer of the functional F on the constrained set K . ( 29( 29 ( 29 ( 29 ( 29 ( 29 K F F T T 2200 - - - = - + = - v b u v Av u v u v u u v u , 0 lim 0 a a d a (2) Obtain the Lagrangian L to this constrained minimization problem, and set up the "equivalent" unconstrained problem on the primal variable v by considering the necessary condition of the problem obtained by the Lagrange multiplier method. Lagrangian L is defined by ( 29 ( 29 ( 29 ( 29 0 R g Bv b v Av v g Bv v v - - - = - - = m T T T T F L , 2 1 , and we shall consider the problem ( 29 v 0 v , max min L Suppose that ( 29 0 R R u × , , m n is a solution of the min-max problem. Then we have ( 1 ) ( 29 ( 29 n L L R v v u 2200 , , , ( 2 ) ( 29 ( 29 0 R u u 2200 m L L , , , From (1), we have the following necessary condition ( 29( 29 ( 29 ( 29 ( 29 ( 29 n T T L L R v b B Au u v u v u u v u 2200 - - - = - + = - , 0 , lim , 0 a a d a Since v is arbitrary, and since we can take u u v d ± = , we have
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n T T R u b B Au u 2200 - - ± d d , 0 that is n T in R b B Au = - Similarly, from the second inequality relation (2), we have ( 29( 29 ( 29 ( 29 ( 29 ( 29 0 R g Bu u u 2200 - - - = + = - m T L L , 0 - , lim , 0 a a d a that is ( 29 ( 29 0 R g Bu 2200 - - m T , 0 Combining these, the necessary condition of the Lagrange multiplier formulation becomes ( a ) n T in R b B Au = - ( b ) ( 29 ( 29 0 R g Bu 2200 - - m T , 0 The second inequality yields also the KKT condition ( 29 0 , , = - - g Bu 0 0 g Bu T (3) Solve the problem for A , B , and g obtained by the following MATLAB program. = - - - - - - - - - - - - - - - - - - = 1 1 1 1 1 1 1 1 1 1 , 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 b A = = 8
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HW5A_98F - Solutions of Homework#5 1998 Fall MEAM 501...

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