CSCI3656: Numerical Computation Homework 6 solution

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Consider the following nonlinear system of equations with two equations and two unknowns. The math
problem can be stated as follows. Given f1(x1, x2) and f2(x1, x2) defined as
f1(x1, x2) = x
3
1 − x
3
2 + x1,
f2(x1, x2) = x
2
1 + x
2
2 − 1.
(1)
Find r1 and r2 such that f1(r1, r2) = 0 and f2(r1, r2) = 0.
1. Note that all the points such that f2 = 0 define a circle of radius 1 centered at the origin. Make
a plot that shows (i) all the points that satisfy f1 = 0 and (ii) all the points that satisfy f2 = 0.
Identify the points on the plot that satisfy both f1 = 0 and f2 = 0.
2. By hand, calculate the 2 × 2 Jacobian matrix of the system (f1, f2).
3. Use Newton’s method for systems to find the two solutions to the system of equations (f1 =
0, f2 = 0). Try several (10 or so) different initial guesses. Make a table of the answer that
Newton’s method gives—something like:
Initial guess (x
(0)
1
, x
(0)
2
) Newton’s answer (r1, r2)
####, #### ####, ####
The superscript in the column heading indicates the iteration number, i.e., 0 means the initial
guess. Check the plot you made in problem 1 to see whether the answers you’re getting make
sense.
4. Find a starting point where Newton’s method fails. Why did it fail?
5. BONUS (10%): Plot the iterates from Newton’s method in (x1, x2) space. Interpret what you
see (related to the solution that Newton’s method gives).
6. BONUS (10%): Set up a grid on the (x1, x2) space; you can use meshgrid in Matlab/Numpy.
For each point in the grid, compute the Newton direction
p = −J(x1, x2)
−1
f(x1, x2), (2)
where J is the Jacobian matrix and f = (f1, f2). Make a quiver plot of the Newton directions
on the grid. How does this relate to the solution that Newton’s method gives?
7. BONUS (20%): Derive a fixed point iteration for nonlinear systems. (That is: show how to
transform the nonlinear system problem to a fixed point problem and then write the iteration).
Implement your method. Solve the same two-equation system with your fixed point method.
Anything interesting?
1