#### You've reached the end of your free preview.

Want to read both pages?

**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 > = 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 ...

View
Full Document

- Fall '19