# Thread: Euler's implisit - convergence of each step

1. ## Euler's implisit - convergence of each step

Hi,

problem:

the trapezoid method of solving differential equations is,

$y(t_{i+1}) = y(t_i) + h\frac{f(y(t_{i+1})) + f(y(t_i))}{2}$.

Since the method is implicit, we must solve for $y(t_{i+1})$ on each step by some iterative method. If we use the fixed point iteration $x=g(x)$, we know that the method will converge if $|g'(x)|<1$ for $x\in[y(t_i),y(t_{i+1})]$.

Give a condition on $h$ such that the fixed point iteration converges at each step.

attempt:
First of all, I need to write down an expression for $g(x)$. Fixed-point iteration needs an initial guess. I use $y(t_i)$ as my initial guess and write $g$ as,

$g(x) = x + h\frac{f(x)+f(x)}{2} = x + hf(x)$.

But now

$g'(x) = 1 + hf'(x) < 1 \Rightarrow h<0$,

which doesn't make much sense.
What am I doing wrong?

Thanks

2. I think that notation might have me fooled.
At each step, $y(t_i)$ and $f(y(t_i))$ are constants and so,

$|g'(t)| = |\frac{h}{2}f'(y(t))| < |1|.$

This means that,

$h < \frac{2}{f'(y(t))},$

at each step in order to be sure that fixed-point iteration will converge.

3. Originally Posted by Mollier
Hi,

problem:

the trapezoid method of solving differential equations is,

$y(t_{i+1}) = y(t_i) + h\frac{f(y(t_{i+1})) + f(y(t_i))}{2}$.

Since the method is implicit, we must solve for $y(t_{i+1})$ on each step by some iterative method. If we use the fixed point iteration $x=g(x)$, we know that the method will converge if $|g'(x)|<1$ for $x\in[y(t_i),y(t_{i+1})]$.

Give a condition on $h$ such that the fixed point iteration converges at each step.

attempt:
First of all, I need to write down an expression for $g(x)$. Fixed-point iteration needs an initial guess. I use $y(t_i)$ as my initial guess and write $g$ as,

$g(x) = x + h\frac{f(x)+f(x)}{2} = x + hf(x)$.

But now

$g'(x) = 1 + hf'(x) < 1 \Rightarrow h<0$,

which doesn't make much sense.
What am I doing wrong?

Thanks
Try the itteration on $u_k$:

$u_{k+1}=y(x_i)+h \dfrac{f(u_k)+f(x_i)}{2}$

and start with $u_0=y(x_i)+hf(x_i)$ (which is a standard explicit Euler step)

This appears to be equivalent to the 1st order itterated predictor-corrector method.

CB