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Iterative Method Homework 1 1 0 2 , b , u ( 0) using conjugate gradients. 1. Solve Au = b with A 1 2 2 1 1 Note that the true solutions is u * . 1 2u 2u 0 defined over the square 0 x x 2 y 2 6 and 0 y 6. Divide the square into three subdivisions in each direction (so that the x and y increment is 2) and suppose that the boundary data is 0 except for x=6 it is u = 10. Use the 5 point star stencil to write a system of 4 equations in 4 unknowns u1, u2, u3, u4. 2. Consider the boundary value problem 3. For n=2 use calculus to prove that a potential minimum point of F(u) = (1/ 2 ) uTA u a c e , b and bTu is the solution to A u = b if A is symmetric. Hint: let A f c d x u so that y a c x x (e f ) (1 / 2)ax 2 cxy (1 / 2)dy 2 ex fy F ( x, y ). F (u ) (1 / 2)( x y ) c d y y Set the partial derivatives of F with respect to x and y to zero. 4. Watkins exercise 8.1.12 (3rd edition) or 7.1.12 (2nd edition). 5. For n = 100, b = ones(n,1); and for each of the three matrices below apply 20 steps of my cg_step code. One can do this using k = 0; for i=1:20, cg_step, end in Matlab. The cg_step code prints out ||r|| / ||b|| = || b – Au || / ||b|| which is a measure of how well the current guess solves the system A u = b . Compare the final values of ||r||/||b|| for the three runs. Also for each matrix calculate cond(A) and gamma = (sqrt(cond(A)) – 1 ) / ( sqrt(cond(A)) + 1 ). Discuss. (a) A = diag( 1.01 .^ (0 : n – 1 ) ); (b) A = diag( 1.03 .^ (0 : n – 1 ) ); (c) A = diag( 1.1 .^ (0 : n – 1 ) ); 6. For each of the following values of k = cond(A) predict how many steps are required to reduce the error in conjugate gradients by 10-6 : k=4, 81, 9801, 99801. Use the formulas ( cond ( A) 1) /( cond ( A) 1) and || x m x * || A 2 m || x0 x * || A . According to the second formula the error will be reduced by a factor of 10 6 when 10 6 2 m .