Homework 6 Solutions

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1 Homeork 6 Solutions Igor Yanovsky (Math 151B TA) Section 114, Problem 1: For the boundary-value problem y (y ) y + log x, 1 x, y(1) 0, y() log, (1) rite the nonlinear system and formulas for Neton s method We divide [1, ] into N + 1 subintervals hose endpoints are x i 1 + ih, for i 0, 1,, N + 1, and consider the discretization of the boundary-value problem in (1): y (x i ) (y(x i ) ) y(x i ) + log x i () Replacing y (x i ) and y (x i ) by appropriate centered difference formulas, equation () becomes: ( ) y(x i+1 ) y(x i ) + y(x i 1 ) h h y(xi+1 ) y(x i 1 ) 1 y(4) (ξ i ) h h 6 y (η i ) y(x i ) + log x i, for some ξ i and η i in the interval (x i 1, x i+1 ) The difference method results hen the error terms are deleted and the boundary conditions are employed: and 0 0, N+1 log, i+1 i + i 1 h ( ) i+1 i 1 i + log x i 0, (3) h for each i 1,,, N Multiplying (3) by h, e obtain ( ) i+1 i 1 i+1 + i i 1 h i + h log x i 0, hich can be ritten as: or ( ) i+1 i 1 i+1 + i i 1 h i + h log x i 0, i 1 + i i+1 1 4( i 1 i 1 i+1 + i+1) h i + h log x i 0 1

2 Thus, the N N nonlinear system is: ( ) h 1 + h log x 1 0, ( ) h + h log x 0, ( 4 + 4) h 3 + h log x 3 0, N + N 1 N 1 4( N N N + N) h N 1 + h log x N 1 0, N 1 + N log 1 4( N 1 N 1 log + (log ) ) h N + h log x N 0, here e designate the left-hand side of the first equation as F 1 ( 1,, N ), the second equation as F ( 1,, N ),, the last equation as F N ( 1,, N ) Also, e designate F (F 1,, F N ) T and ( 1,, N ) T We use Neton s method for nonlinear systems to approximate the solution to the system F ( ) 0 above A sequence of iterates (k) ( (k) 1, (k),, (k) N )T is generated that converges to the solution of this system The Jacobian matrix J for this system is F 1 F 1 F 1 F N J( 1,, N ) F 1 F F 3 F N 1 1 F N 1 F N 1 F N F N 1 3 F N 3 F N F N 1 N F N N h h N + 1 N h N 1 N h We can no use the Neton s method for nonlinear systems (k) (k 1) J 1( (k 1)) F ( (k 1) )

3 Section 71, Problem 1: Find x and x for the folloing vectors: a) x (3, 4, 0, 3 )T ; c) x (sin k, cos k, k ) T for a fixed positive integer k The L and L norms for the vector x (x 1, x,, x n ) T are defined by x max x i, 1in { } 1 x x i a) For x (3, 4, 0, 3 )T : { x max 3, 4, 0, 3 } 4, ( 3 ) x 3 + ( 4) c) For x (sin k, cos k, k ) T, k is a positive integer : x max { sin k, cos k, k } k, x sin k + cos k + ( k ) k Section 71, Problem (a): Verify that the function 1, defined on R n by x 1 x i, is a norm on R n (i) For all x R n, x 1 x i 0 (ii) If x 0, then x 1 x i 0 0 If x 1 0, e have n x i 0, and thus, x 0 (iii) For all α R and x R n, αx 1 αx i α x i α x i α x 1 (iii) For all x, y R n, x + y 1 x i + y i Thus, 1 is a norm on R n ( xi + y i ) 3 x i + y i x 1 + y 1

4 Section 71, Problem (c): Prove that for all x R n, x 1 x Let x (x 1, x,, x n ) T, and note that ( x1 + x + + x n ) x 1 + x + + x n, or or ( x i ) ( x i x i, x i, hich means that x 1 x Section 71, Problem 4(c): Find for the folloing matrix: 1 0 A Since We have A max 1in a ij a 1j a 11 + a 1 + a , a j a 1 + a + a , a 3j a 31 + a 3 + a , e have A max{3, 4, 3} 4 Section 71, Problem 7: Sho by example that, defined by A does not define a matrix norm max a ij, 1i,jn A function [ ] is a matrix [ norm ] only if it satisfies [ definition ] 78 on page Consider A and B Then, AB We have A , B 1, and AB, and thus, AB A B, hich contradicts one of the conditions for being a norm 4

5 Section 71, Problem 9(a): The Frobenius norm (hich is not a natural norm) is defined for an n n matrix A by ( A F a ij Sho that F is a matrix norm For all n n matrices A and B and all real numbers α, e have: (i) ( A F a ij 0 (ii) (iii) ( A F αa F a ij 0 if and only if A is a 0 matrix α a ij αa F α A F (iv) Here, e ill use Cauchy-Scharz Inequality: A + B F ( ( a ij + b ij ( a ij + b ij ) α a ij α ( a ij + a ij b ij + b ij ) a ij + ( a ij + a ij b ij + a ij ( + ( A F + B F ) ( x i y i a ij ( A + B F A F + B F b ij ) a ij α A F x i ( y i b ij (Cauchy-Scharz) b ij + (v) Note that n k1 a n 1kb k1 k1 a n 1kb k k1 a n kb k1 k1 a kb k AB n k1 a n nkb k1 k1 a nkb k n k1 a nkb k,n 1 5 b ij n k1 a 1kb kn n k1 a kb kn n k1 a nkb kn

6 AB ( F a ik b kj a ik b kj ) (Cauchy-Scharz) ( k1 k1 ( a ik b kj ) k1 k1 a ij )( AB F A F B F b ij ) A F B F CONTINUE TO THE NEXT PAGE 6

7 Section 71, Problem 9(c): For any matrix A, sho that A A F n 1/ A The definitions of F and norms are: ( A F a ij A max x 1 Ax Note, that Ax is a vector: n a 1jx j n Ax a jx j n a njx j Thus, e have ( ( ) Ax a ij x j ❶ We first sho that A A F For vector x, such that x 1, e have ( ) Ax a ij x j (Cauchy-Scharz) ( ( ( ( ( ( A F a ij a ij a ij a ij a ij ( )( ) x ) 1 x j x j ) We shoed that, Ax A F for all x, such that x 1 Thus, max Ax A F, or A A F x 1 ❷ We no sho that A F n 1/ A Let x i 1 n for all 1 i n Then, ) ) A max x 1 Ax ( ) a ij x j a ij 1 n 1 n a ij 1 n A F Thus, A F n 1/ A 7

8 Section 73, Problem (c): Find the first to iterations of the Jacobi method for the folloing linear system, using x (0) 0: 4x 1 + x x 3 + x 4, x 1 + 4x x 3 x 4 1, x 1 x + 5x 3 + x 4 0, x 1 x + x 3 + 3x 4 1 The linear system Ax b given by E 1 : 4x 1 + x x 3 + x 4, E : x 1 + 4x x 3 x 4 1, E 3 : x 1 x + 5x 3 + x 4 0, E 4 : x 1 x + x 3 + 3x 4 1 has the unique solution x ( 07534, , 0808, ) To convert Ax b to the form x T x + c, solve equation E 1 for x 1, E for x, E 3 for x 3, E 4 for x 4, to obtain x x x x 4 1, x 1 4 x x x 4 1 4, x x x 1 5 x 4, x x x 1 3 x Then Ax b can be ritten in the form x T x + c, ith T and c For initial approximation, e let x (0) (0, 0, 0, 0) T Then is given by x(0) x(0) x(0) , 1 4 x( x(0) x(0) , x( x(0) 1 5 x(0) 4 0, x( x(0) 1 3 x(0) /3 The next iterate, x (), is given by x () x(1) x(1) x(1) , x () 1 4 x( x(1) x(1) , x () x( x(1) 1 5 x(1) , x () x( x(1) 1 3 x(1)

9 Section 73, Problem 4(c): Find the first to iterations of the Gauss-Seidel method for the folloing linear system, using x (0) 0: 4x 1 + x x 3 + x 4, x 1 + 4x x 3 x 4 1, x 1 x + 5x 3 + x 4 0, x 1 x + x 3 + 3x 4 1 In section 73, Problem (c), e used Jacobi method to solve the linear system above The folloing equations ere used: x(k 1) x(k 1) x(k 1) 4 1, 1 4 x(k x(k 1) x(k 1) 4 1 4, x(k x(k 1) 1 5 x(k 1) 4, x(k x(k 1) 1 3 x(k 1) Hoever, since for i > 1, 1,, x(k) i 1 have already been computed, these are probably better approximations to the actual solutions x 1,, x i 1 than x (k 1) 1,, x (k 1) i 1 Hence, Gauss-Seidel uses the most recently available approximations to x 1,, x i 1 in a calculation of the next iterate: x(k 1) x(k 1) x(k 1) 4 1, 1 4 x(k x(k 1) x(k 1) 4 1 4, x(k x(k) 1 5 x(k 1) 4, x(k x(k) 1 3 x(k) For initial approximation, e let x (0) (0, 0, 0, 0) T Then is given by x(0) x(0) x(0) , 1 4 x( x(0) x(0) , x( x(1) 1 5 x(0) 4 015, x( x(1) 1 3 x(1) The next iterate, x (), is given by x () x(1) x(1) x(1) , x () 1 4 x( x(1) x(1) , x () x( x() 1 5 x(1) 4 05, x () x( x() 1 3 x() Comparing x () to the exact solution x ( 07534, , 0808, ), e see that Gauss-Seidel method gave more accurate results than Jacobi method 9