MHB Bound of Euler method- nonuniform partition

[evinda]

Hello! (Wave)

Consider a nonuniform partition $a=t_0< t_1< \dots < t_{\nu}=b$ and assume that if $h_n=t^{n+1}-t^n, 0 \leq n \leq N-1 $ is the changeable step, then $\min_{n} h_n > \lambda \max_{n} h_n, \lambda>0$ independent of $n$.
Show a bound of the error of Euler method analogous to

$$||y^n-y(t^n)|| \leq h c_1 \frac{e^{L(t^n-a)}}{2L} \max_{t \in [a,t^n]} ||y^{(2)} (t)||$$

where $h=\max_{n} h_n$.

According to my lecture notes:We will use the following lemma:

Let $\delta$ be a positive number and $K, d_0, d_1, \dots$ non-negative numbers such that

$$d_{i+1} \leq (1+ \delta) d_i+K, i=0,1, \dots$$

Then it holds

$$d_n \leq e^{n \delta} d_0+ K \frac{e^{n \delta}-1}{\delta}, n=0,1,2, \dots$$

$$\left\{\begin{matrix}
y^{n+1}=y^n+h_n f(t^n,y^n) &, n=0,1, \dots, N-1 \\
y^0=y(0) &
\end{matrix}\right.$$

Expanding $y$ with Taylor, we have:

$$y(t^{n+1})=y(t^n)+ h_n y'(t^n)+ \frac{h^2}{2} y''(\xi_n), \xi_n \in (t^n, t^{n+1})$$

From the differential equation we have:

$$y(t^{n+1})=y(t^n)+ h_n f(t^n, y(t^n))+ \frac{h^2}{2} y''(\xi_n), \xi_n \in (t^n, t^{n+1})$$

$$y(t^{n+1})-y^{n+1}=y(t^n)-y^n+h_n [f(t^n, y(t^n))-f(t^n,y^n)]+ \frac{h h_n}{2} y''( \xi_n)$$

$$\epsilon^{n+1}= \epsilon^n+ h_n [f(t^n, y(t^n))-f(t^n,y^n)]+ \frac{h h_n}{2} y''(\xi_n)$$

and so on...

Why is it as follows?

$$y(t^{n+1})=y(t^n)+ h_n y'(t^n)+ \frac{h^2}{2} y''(\xi_n), \xi_n \in (t^n, t^{n+1})$$Shouldn't we have $h_n$ at each point where we have the step, i.e. shouldn't it be as follows?

$$y(t^{n+1})=y(t^n)+ h_n y'(t^n)+ \frac{h_n^2}{2} y''(\xi_n), \xi_n \in (t^n, t^{n+1})$$

Or am I wrong? (Worried)

 

[I like Serena]

Hey! (Mmm)

Shouldn't we have $h_n$ at each point where we have the step, i.e. shouldn't it be as follows?

$$y(t^{n+1})=y(t^n)+ h_n y'(t^n)+ \frac{h_n^2}{2} y''(\xi_n), \xi_n \in (t^n, t^{n+1})$$

Or am I wrong? (Worried)

You're right. Looks like a bit of sloppiness. (Nod)