《结构工程中的数学方法》课程教学课件（讲稿）example_Jacobi_iteration

2AlgebraicproblemsinmatrixformSystemsofLinearEquationsJacobiiterativemethod-example (fromslide33a)systemofequationswith:a=2.50m,b=1.25m,g=32.0kN/m072572.5 1.25(E.)X1251.251.25-1.0(E2)X2A.x=b0 -1.0-40(E:)0.6X3.checksystemproperties:diagonallydominant,i.e.,a>zak,i=1,.,n》 (E1):[1.25] *[1.25| +-1.0]10.6|*0+-1.0[2.5|>1.25 +0》(E2):》 (E3):notdiagonallydominantconvergenceproblemsre-arrangement of thesystemtomake it diagonally dominant[2.51.25025(E.)X1》rowswitch(E)(E)0(E2)-1.00.6-40X2》add up (E)-0.5·(E)→(E)012.5(E,)0.625-1.0X3》(E1):2.5>|1.25+|0[-1.0| [0| +[0.6]》 (E3):-1.0 >[0| +|0.625》 (E2):70aMichael Beer,Engineering Mathematics

Michael Beer, Engineering Mathematics system of equations with: Systems of Linear Equations Jacobi iterative method − example (from slide 33a) 2 Algebraic problems in matrix form 70a ● a 2.50 m, b 1.25 m, q 32.0 kN/m = = = ● check system properties: diagonally dominant, i.e., ? ( ) ( ) ( )           −⋅ =       − −          1 1 2 2 3 3 2.5 1.25 0 x 25 E 1.25 1.25 1.0 x 25 E 0 1.0 0.6 x 40 E Ax b ⋅= ⇔ » (E1): 2.5 1.25 0 > + » (E2): » (E3): ≠ ii > ∑ ik k i a a , i=1,.,n 1.25 1.25 1.0 > +− / 0.6 0 1.0 > +− / ● re-arrangement of the system to make it diagonally dominant not diagonally dominant convergence problems (E 0.5 E E 3 13 ) −⋅ → ( ) ( ) » row switch (E E 2 3 ) ↔ ( ) » add up » (E1): 2.5 1.25 0 > + » (E2): » (E3 − >+ 1.0 0 0.6 ): − >+ 1.0 0 0.625 ( ) ( ) ( )           − ⋅ =−       −          1 1 2 2 3 3 2.5 1.25 0 x 25 E 0 1.0 0.6 x 40 E 0 0.625 1.0 x 12.5 E

2AlgebraicproblemsinmatrixformSystemsof Linear EquationsJacobi iterative method- example (cont'd)re-structuringofthesystem[k]K-1xXra[2.51.25025X=-0.5.X,+10X0-40-1.00.6X,X2=0.6.X,+4000.62512.5-1.0X,=0.625X,-12.5X[k][k-1]00-0.510.0XiX1[k][k-1]0040.00.6》matrixform:X=T.x+C←X2X,十000.625-12.5X3X70bMichaelBeer,EngineeringMathematics

Systems of Linear Equations Jacobi iterative method − example (cont'd) 2 Algebraic problems in matrix form 70b ● re-structuring of the system − =⋅ + ⇔ [k] [k 1] x Tx c − = ≠     =− + =   ∑         [k] n [k 1] i ij j i j 1 ii j i 1 x a x b , i 1,.,n a [ ] [ − ]   −           = ⋅ +     −       k k 1 1 1 2 2 3 3 x 0 0.5 0 x 10.0 x 0 0 0.6 x 40.0 x 0 0.625 0 x 12.5 x 0.5 x 10 1 2 =− ⋅ + x 0.6 x 40 2 3 = ⋅+ x 0.625 x 12.5 3 2 = ⋅− » matrix form:           − ⋅ =−       −          1 2 3 2.5 1.25 0 x 25 0 1.0 0.6 x 40 0 0.625 1.0 x 12.5 Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformSystemsof Linear EquationsJacobi iterative method- example (cont'd)initial assumption for solution vector[0][0][0]X1= 0X3=0》k=0:X2=0iterativesolution》k=1:[0][1][1][0][1][0]=-0.512.5X2+10.0=0.6.XX3+40.0=0.625X3X2X2X110.040.0-12.5》k=2:[1][2] [1][2][1][2]= -0.5=0.6X3+40.012.5X2+10.0X2=0.625.X3X1X21=-10.0= 32.5=12.5?o.?L6》 k = 15:[15][14][15][14][15][14]X1=-0.5.X2+10.0=0.6.X3+40.0=0.625.x2-12.5X2X3=51.987(52)=19.966(20)=-15.973(-16)70cMichaelBeer,EngineeringMathematics

initial assumption for solution vector Systems of Linear Equations Jacobi iterative method − example (cont'd) 2 Algebraic problems in matrix form 70c ● » k = 0: = ⋅+ [1] [0] x 0.6 x 40.0 2 3 ● iterative solution » k = 1: =− ⋅ + [2] [1] x 0.5 x 10.0 1 2 = ⋅− [2] [1] x 0.625 x 12.5 3 2 » k = 2: = = = [0] [0] [0] x 0, x 0, x 0 123 = ⋅+ [2] [1] x 0.6 x 40.0 2 3 = ⋅− [1] [0] =− ⋅ + x 0.625 x 12.5 3 2 [1] [0] x 0.5 x 10.0 1 2 = 10.0 = 40.0 = −12.5 = −10.0 = 32.5 = 12.5 ● ● ● ● ● ● ● ● ● » k = 15: =− ⋅ + [15] [14] x 0.5 x 10.0 1 2 = ⋅− [15] [14] = ⋅+ x 0.625 x 12.5 3 2 [15] [14] x 0.6 x 40.0 2 3 = −15.973 (−16) = 51.987 (52) = 19.966 (20) Michael Beer, Engineering Mathematics