Solving a System of Equations
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1 Solving a System of Eqations Objectives Understand how to solve a system of eqations with: - Gass Elimination Method - LU Decomposition Method - Gass-Seidel Method - Jacobi Method A system of linear algebraic eqations Linear algebraic eqations are of the general form, a x a x a x b n n 1 a x a x a x b n n 2 a x a x a x b n1 1 n2 2 nn n n (1) Where the a s are constant coefficients, the b s are constants, the x s are nknowns, and n is the nmber of eqation. The algebraic eqations can be briefly rewritten in the matrix form as follows: A x b (2)
2 Gass Elimination Method There are many techniqes for solving a system of linear algebraic eqations. One of basic techniqes that can be applied to large sets of eqation and can be formalized and programmed for the compter is Gass Elimination Method. The procedre consists of two major steps which are 1. Forward Elimination the eqations are maniplated to eliminate the nknowns from the eqations ntil we have one eqation with one nknown. 2. Back sbstittion The eqation with one nknown can be solved directly. The reslt is back-sbstitted into one of the original eqations to solve for the remaining nknown. Fig. 1 The steps of Gass Elimination
3 Example 1 Use Gass Elimination to solve 2x x 5x 21 x 2x 2x 15 x 4x x
4
5 Example 2 Use Gass Elimination to solve 2x x x 6x 7x 3 2x 3x 6x
6
7 Example 3 Use Gass Elimination to solve x x x x Exact Soltions: x x 1 2 1/ / If we rearrange the pivoting eqation by setting the largest element as the pivot element x x x x
8 LU Decomposition Method The forward elimination of Gass elimination comprises the blk of the comptational effort, especially the large system of eqations. LU decomposition method separates the time consming elimination of the matrix [A] by the following steps. 1. LU decomposition step The matrix [A] is decomposed into lower trianglar matrix [L], and pper trianglar matrix [U]. 2. Sbstittion step [L] and [U] are sed to determine a soltion {x} for a righthand side {b}. This step consists of the forward and back sbstittions. 2.1 The forward sbstittion is condcted to determine the intermediate vector {d} from [L]{d}={b}. 2.2 Then, the reslt of {d} is sbstitted into [U]{x}={d}, and solve for {x} throgh the back sbstittion.
9 Fig.2 The steps in LU decomposition. 1. The LU decomposition/ Factorization step [ A] [ L][ U] l l a 11 a a l l l a a 22 a 32 l 31 l l a a a 32 [L] and [U] can be obtained by solving the above 9 eqations.
10 2. The sbstittion step We start from [ A]{ x} { b} (1) After LU decomposition step, we have Now, let [ L][ U]{ x} { b} (2) [ U]{ x} { d} (3) Eq. (2) can be rewritten as Or [ L]{ d} { b} (4) d b 1 1 l d2 b2 l31 l32 1 d 3 b 3 (5) Then, determine {d} throgh forward sbstittion. After we get {d}, we sbstitte {d} into eq. (3). We have x d x2 d x 3 d 3 (6) Use back sbstittion to determine {x}.
11 Example 4 Use LU decomposition to solve 10x 2x x 27 3x 6x 2x 61.5 x x 5x 21.5 LU decomposition/factorization [ A] Sbstittion step for determine {d} and {x}
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14 Iterative Methods for Systems of Eqations - Jacobi Method - Gass-seidel Method Jacobi Method Assme that we have a 3x3 set of eqations. Or a a a x b a21 a22 a 23x2 b2 a31 a32 a33 x 3 b 3 a x a x a x b a x a x a x b a x a x a x b (1) If the diagonal elements are all nonzero, the first eqation can be solved for x 1, the second for x 2, the third for x 3. Then, we have j1 j j b a x a x x1 a (2a) 11 j1 j j b a x a x x2 a (2b) 22 j1 j j b a x a x x3 a (2c) 33
15 where j and j-1 are the present and previos iterations. To solve the soltions, we j j j need the initial gesses for x1, x2, x 3 for the first iteration. Then, the iteration is contined ntil or soltions converge closely enogh to the tre vales or the error of the approximation less than or eqal to tolerance. j j1 xi j i xi εa, i 100 ε x s (3) where i = 1, 2, 3 for the x s. Example 5 Use Jacobi method to solve 2x x 5x 21 x 2x 2x 15 x 4x x 18
16
17 Gass-seidel Method This method is the most commonly sed iterative method for solving linear algebraic eqations. Assme that we have a 3x3 set of eqations. a x a x a x b a x a x a x b a x a x a x b (1) If the diagonal elements are all nonzero, the first eqation can be solved for x 1, the second for x 2, the third for x 3. Then, we have j1 j j b a x a x x1 a (2a) 11 j j j b a x a x x2 a (2b) 22 j j j b a x a x x3 a (2c) 33 where j and j-1 are the present and previos iterations. To solve the soltions, we j j j need the initial gesses for x1, x2, x 3 for the first iteration. Then, the iteration is contined ntil or soltions converge closely enogh to the tre vales or the error of the approximation less than or eqal to tolerance. j j1 xi j i xi εa, i 100 ε x s (3) where i = 1, 2, 3 for the x s.
18 Example 6 Use Gass-Seidel method to solve 10x 2x x 27 3x 6x 2x 61.5 x x 5x 21.5
19 Fig. 3 The difference procedre between (a) the Gass-Seidel method and (b) the Jacobi method. Exercise x 7x x 7x 30 0 x 7x 40 3x 5x Solve the above system of eqations sing Gass elimination, Jacobi, and Gass-Seidel methods. Use three iterations for the Jacobi, and Gass-Seidel iterations.
20 2. The position of three masses sspended vertically by series of identical springs can be modeled by the following steady-state force balances: 0 k( x x ) m g kx k( x x ) m g k( x x ) m g k( x x ) If g = 9.81 m/s2, m1 = 2 kg, m2 = 3 kg, m3 = 2.5 kg, and the k s = 10 N/m. Use the Gass elimination, Jacobi, and Gass-Seidel methods to solve for position (the x s) of masses. Note that three iterations are performed for the Jacobi, and Gass-Seidel iterations. 3. Use the Jacobi, and Gass-Seidel methods to solve the following system ntil the percent relative error falls below εs 5% x y z 105
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