Let be $$y'= \frac{t}{y^2+1} $$ $y(0)=0 $
I want to determine an approximate solution $ \tilde{y} $ from the starting point $ \phi_0 =0 $ so that the error is : $ \lVert \tilde{y} -y \rVert_{ C[-1,1]} \leq 0,14 $
Well, I want to use Picard- Lindelöf's Theorem:
Let be$ R:= [x_o-a,x_o+a] \times[y_o-b,y_0+b] $ a rectangle $ \in \mathbb{R^2} $ and $f: R \rightarrow \mathbb{R} $ a continous function.
If $f$ is lipschitz-continous in the second argument ($y$) then the initial value problem $$y'=f(x,y) $$, $y(x_0)=y_0$ has a solution $y$ on the intervall $ J=[x_0- \alpha,x_0+ \alpha] $ where $ \alpha:= min( a, \frac{b}{ \lVert f\rVert_{C(R)}} ) $
The solution is the uniformly limes of the series of the iterations $ \phi_n $
$ \phi_n(x)= y_0 + \int_{x_0}^x f(t, \phi_{n-1}(t)) dt $ with the error estimates
$$ | y(x) - \phi_n (x) | \leq \sum_{k=n}^{ \infty} \frac{ ( \alpha L)^k }{k!} \lVert \phi_1 - \phi_o \rVert_{ C(J) } $$ and
$$ |y(x) - \phi_n (x) | \leq \frac{ (\alpha L)^n }{n!} e^{ \alpha L } \lVert \phi_1 - \phi_0 \rVert $$First, I want to use the last equations to see how many iteration steps I need to do, to get $ \leq 0,14 $ but here I do get suck.
$ \tilde{y}_1(x)= \int_0^t f(x, \phi_0 (x)) dt= \int_0^t x dx= \frac{1}{2} t^2 $ so
$ \lVert \phi_1- \phi_0 \rVert_{ C[-1,1] } = max_{t \in [-1,1]} \frac{1}{2} t^2= \frac{1}{2} $
and for the Lipschitz-Constant I do get:
$ max_{t,y \in \mathbb{R} }| \frac{ \partial f }{ \partial y } | = max_{t,y \in \mathbb{R} } | - \frac{2ty}{(y^2+1)^2 } | =2=L$ is this right?
then putting together: $ | y(x)- \phi_n | \leq \sum_{k=n}^{ \infty} \frac{ ( \alpha L)^k }{k!} \lVert \phi_1 - \phi_o \rVert_{ C[-1,1] } = \sum_{k=n}^{ \infty } \frac{2^k}{2k!} $
What is wrong? I want to be able to determine the correct error estimation, without having to determine the actual solution. I appreciate any help and thank you in advance !