Hi good people, I'm doing some old exam questions and here's one I don't quite get.

We have the ordinary diff. eq,

$\displaystyle y'(t) = e^{y(t)}$

where $\displaystyle y(0)=1$ and $\displaystyle t\in [0,1]$.

If we want to solve this by using Euler's implicit formula, we need solve an equation by say fixed-point iteration in each step. How small must the step size $\displaystyle h$ be in order for the fixed-point iteration to converge on each step.

First of all, the problem does not say what kind of implicit formula I need to use. I get to the formula in the following manner,

$\displaystyle \int^{t_{i+1}}_{t_i} y'(t)dt = \int^{t_{i+1}}_{t_i}f(t,y(t))dt$,

and so,

$\displaystyle y(t_{i+1}) = y(t_i) + \int^{t_{i+1}}_{t_i}f(t,y(t))dt$.

To get an implicit formula from this I could simply approximate the integral by,

$\displaystyle f(t_{i+1},y(t_{i+1}))(t_{i+1}-t_i) = e^{y(t_{i+1})}h$,

or I could approximate it by the Trapezoidal rule.. Let's assume I do the former such that,

$\displaystyle y(t_{i+1}) = y(t_i) + e^{y(t_{i+1})}h.$

Now I need to use fixed-point iteration to solve this equation. If I let my first guess at $\displaystyle y(t_{i+1})$ be $\displaystyle y(t_i)$ which I believe is a reasonable guess, then I could write this thing in the familiar way;

$\displaystyle x_{n+1} = x_n + e^{x_n}h = g(x_n)$.

The sequence defined by the expression above converges if $\displaystyle 0\leq|g'(x)|<1$ for all $\displaystyle x$ in the interval. But $\displaystyle |g'(x)|=|1+e^xh|<1$ and so $\displaystyle h<0$ but that doesn't make much sense does it? I don't think the step size must be zero Hope someone can help me out here.

Thanks.