let's consider the equation $y'(t) = 16-y^2 = f(t,y)$
for all $t\in [0,1]$ and $y(0) = 0$.
How small should the step size $h$ be to ensure the following equation converge for all $i \geq 0 $. This can be achieved by doing numerically using fix point iteration of the Crank-Nicolson scheme given by:
$y_{i+1}^{k+1}$ = $y_i$ + $\frac{h}{2}[f(t_i,y_i) + f(t_{i+1},y_{i+1}^{k})]$
for $k\geq 1$ and $y_{i+1}^{0}$ = $y_i$
I applied fix point iteration which is
$x^{k+1}$ = $F(x^{k})$, so
$F(x^{k})$ = $y_i$ + $\frac{h}{2}f(t_i,y_i) + \frac{h}{2}f(t_{i+1},x_{i+1}^{k})$
By Banach fix point theorem, the iteration converges iff F is a contraction. I need help here how to show this.
b) What if the first equation(Instead of Crank -Nicolson) changed to
$y_{i+1}^{k+1}$ = $y_i$ + $\frac{3h}{4}(4-y_i^2) + \frac{h}{4}(4-(y_{i+1}^k)^2)$
for all $k\geq 1$ and $y_{i+1}^{0}$ = $y_i$
c) If the iterative system has a form
$g(x^k)$ = $y_n$ + $\frac{3h}{4}(4-y_n^2) + \frac{h}{4}(4-(x_{n+1}^k)^2)$
for all $k\geq 1$ and $y_{n+1}^{0}$ = $y_n$
How can I perform
$|\frac{\partial g}{\partial x}| $ $<$ 1
