+ All Categories
Home > Documents > Direct methods

Direct methods

Date post: 29-Jun-2015
Category:
Upload: oscar-mendivelso
View: 261 times
Download: 0 times
Share this document with a friend
Popular Tags:
12
OSCAR EDUARDO MENDIVELSO OROZCO I study Petroleum Engineering Numerical Methods
Transcript
Page 1: Direct methods

OSCAR EDUARDO MENDIVELSO OROZCO I study Petroleum Engineering Numerical Methods

Page 2: Direct methods

INTRODUCTION*The problem to be solved in this chapter is that of a system of m linear equations with n unknowns in a matrix defined by:

*Where A ∈ IR(M, n) and b ∈ IRm are data and x ∈ IRn is the vector incognita. Can be explicitly written as:

(1)

*Usually assume that m = n and that the system has only solution, ie det A =0, or, rg A = n.

Page 3: Direct methods

LU FACTORIZATION METHOD

*The easiest way to explain the LU method is illustrating the basic Gauss method through an example, as is the case of the given matrix and then applying the procedure to a system of four equations with four unknowns:

Page 4: Direct methods

*In the first step, we multiply the first equation by 12 / 6 = 2 and subtract the second, then multiply the first equation by 3 / 6 = 1 / 2 and subtract the third and finally multiply the first equation by -6 / 6 =- 1 and subtract the fourth. The numbers 2, ½ and -1 are the multipliers of the first step in the process of elimination. Number 6 is the pivotal element of this first step and the first row, that remains unchanged is called the pivot row. The system now looks like this:

Page 5: Direct methods

*In the next step of the process, the second row is used as pivot row and -4 as the new pivot element we apply the process: multiply the second row by - 12/-4 = 3 and the remainder of the third and then multiply the second row 2 / (-4) = - 1/2 and subtract the fourth. 

*The multipliers are in this case 3 and -1/2  and the system of equations reduces to:

Page 6: Direct methods

*The last step is to multiply the third equation by 4 / 2 = 2 and subtract the fourth. The resulting system turns out to be:

*The resulting system is upper triangular and equivalent to the original system (the solutions of both systems overlap.) However, this system is easily solvable by applying the backward substitution algorithm. The solution of the system of equations turns out to be:

Page 7: Direct methods

*If we put the multiplier used to transform the system into a unit lower triangular matrix (L) occupying each position of helping to produce zero, we obtain the following matrix:

*Moreover, the upper triangular matrix (U) formed by the coefficients resulting after applying the Gauss algorithm (2) is:

Page 8: Direct methods

*These two matrices give us the LU factorization of the initial matrix of coefficients, A, expressed by equation (1):

Page 9: Direct methods
Page 10: Direct methods
Page 11: Direct methods

REFERENCES

http://www.mty.itesm.mx/etie/deptos/m/ma95-843/lecturas/l843-13.pdf

Page 12: Direct methods

Recommended