Solution_CP4.pdf - Solution Class Problem 4 1 Solution of system of non-linear equation[2 independent variables(x y and 2 dependent variables(u v u x y

# Solution_CP4.pdf - Solution Class Problem 4 1 Solution of...

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Unformatted text preview: Solution: Class Problem 4 1. Solution of system of non-linear equation [2 independent variables (x, y) and 2 dependent variables (u, v)] u x, y x 2 x y 0.75 0 v( x, y ) x 2 5 xy y 0 Figure 2: Plot of the non-linear equations u = 0 and v = 0. There are three possible solutions to this system of equations, where curves of u and v intersect. Newton-Raphson method on a system of non-linear eqs For a system of non-linear equations () = , where f is a column vector Here, f = [ (, ) 2 − + − 0.75= [ (, ) 2 − 5 − At iteration step k: (() )∆ = −(() ) where, (() ) = [ Thus, 1 ]( () ,() ) [ u 2 x 1; x v 2 x 5 y; x −( () , () ) ∆ [ ]=[ ∆ −( () , () ) ]( (),()) u 1 y v (5 x 1) y u Iteration 1 (k = 0) x Determinat of Jacobian D v x u y 5 y 2 x (2 x 1)(5 x 1) v y x0 1.2 y0 1.2 u0 u ( x0 , y0 ) 0.69 v0 v( x0 , y0 ) 6.96 x1 x0 u0 Solve y1 y0 v0 v0 u 1.4 1 ((0) ) = [ v0 0 −3.6 −7 2.13 y y 1.2 1.54355 D x0 , y0 6.20 u0 v 1.4 1 ∆ −0.69 u0 0 [ 7.26 ] [ ] = [ x x 1.2 ∆ −3.6 −7 0.02903 −6.96 D x0 , y0 6.20 Perform Gauss Elimination to make the coefficient a21 zero. Multiplication factor l21 = -3.6/1.4 = -2.571 (upto four significant digits) Perform: R2 = R2 – l21*R1 ∆ 1.4 1 −0.69 [[ ] = [ ∆ 0 −4.428 −8.734 Solve by back substitution: ∆ = 1.972; ∆ = −1.902 (+1) = () + ∆ x1 = 1.2 -1.902 = -0.7018; y1 = 1.2 + 1.972 =2.972 Stopping criteria: ‖(1) −(0) ‖ ‖(1) ‖ = ‖∆‖ ‖(1) ‖ = 1.972 2.972 ( ) = 0.66 &gt; = 0.01 Perform next iteration 1 (k = 1) Go on until you converge to one of the three possible solutions shown in the plot above. 2 ...
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