please help me solve the following question.
Qsn: Solve the following system by Fixed Point Iteration. $$ x^2-2x+y^2-2y=3$$ $$x+y=-1$$
Progress: So I know that we have to assume one of the equations as $f(x,y)$ and the other as $g(x,y)$ then -> test for convergence -> iteration. However, I don't understand how to deal with the squares on x an y.
Thanks,
EEer
Hints: I do not want to spoil you experimenting with this problem, so will provide hints that should suffice for you to proceed.
You want to setup the system as:
$$x_{n+1} = f(x_n, y_n),~~ y_{n+1} = g(x_n, y_n)$$
and choose an initial $(x_0, y_0)$ and iterate. This is analogous to the single variable case.
If $(x_0, y_0)$ is sufficiently close to $(x, y)$ and if:
$$\left| \dfrac{\partial f}{\partial x}(x, y)\right| + \left| \dfrac{\partial f}{\partial y}(x, y)\right| \lt 1, ~ \mbox{and}~ \left| \dfrac{\partial g}{\partial x}(x, y)\right| + \left| \dfrac{\partial g}{\partial y}(x, y)\right| \lt 1$$
then fixed point iteration will converge to the fixed point $(x, y)$.
As an example for your problem, you can try:
$$\tag 1 x_{n+1} = \dfrac{3 + 2 y_n - y_n^2}{x_n - 2},~~ y_{n+1} = -1 - x_n$$
You may need to modify $(1)$ to meet the convergence criteria, so make sure to test that! For example, we could have also tried:
$$\tag 2 x_{n+1} = \pm \sqrt{3 + 2x + 2 y_n - y_n^2},~~ y_{n+1} = -1 - x_n$$
Also note that your problem has two fixed points as:
$$(x, y) = (-1, 0),~ (0, -1)$$
Update
For the root $(x, y) = (-1, 0)$, try:
$$x_{n+1} = -\dfrac{y_n(y_n-2)}{x_n-3} -1,~~ y_{n+1} = -1 - x_n, ~~(x_0, y_0) = (-2.0,2.0)$$
For the root $(x, y) = (0, -1)$, try:
$$x_{n+1} = -1 - y_n,~~ y_{n+1} = \dfrac{-x_n^2+2x_n}{y_n-3} - 1, ~~(x_0, y_0) = (-2.0, -2.0)$$