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evinda
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Hello! (Smile)
Theorem: Let $f \in C([a,b] \times \mathbb{R})$ function that satifies the total Lipschitz condition and let $y \in C^2([a,b])$ the solution of the ODE $\left\{\begin{matrix}
y=f(t,y(t)) &, a \leq t \leq b \\
y(a)=y_0 &
\end{matrix}\right.$
If $y^0, y^1, \dots, y^N$ are the approximations of Euler's method for uniform partition with step $h=\frac{b-a}{N}$ then
$$\max_{0 \leq n \leq N} |y(t^n)-y^n| \leq \frac{M}{2L}(e^{L(b-a)}-1)h$$
where $M=\max_{a \leq t \leq b} |y''(t)|$.
Then there is the following remark:
$$y'=1, 0 \leq t \leq 1 \\ y(0)=0 \\ y(t)=t $$
When the function is linear, there is no error.Can we justify as follows?Suppose that we have a linear function $y(t)=at+b$.
Then $y'(t)=a$ and $y''(t)=0$
So $M=0$ and thus $\max_{0 \leq n \leq N} |y(t^n)-y^n| \leq 0 \Rightarrow \max_{0 \leq n \leq N} |y(t^n)-y^n|=0$.
Also don't we always have a uniform partition at Euler's method? (Thinking)
Theorem: Let $f \in C([a,b] \times \mathbb{R})$ function that satifies the total Lipschitz condition and let $y \in C^2([a,b])$ the solution of the ODE $\left\{\begin{matrix}
y=f(t,y(t)) &, a \leq t \leq b \\
y(a)=y_0 &
\end{matrix}\right.$
If $y^0, y^1, \dots, y^N$ are the approximations of Euler's method for uniform partition with step $h=\frac{b-a}{N}$ then
$$\max_{0 \leq n \leq N} |y(t^n)-y^n| \leq \frac{M}{2L}(e^{L(b-a)}-1)h$$
where $M=\max_{a \leq t \leq b} |y''(t)|$.
Then there is the following remark:
$$y'=1, 0 \leq t \leq 1 \\ y(0)=0 \\ y(t)=t $$
When the function is linear, there is no error.Can we justify as follows?Suppose that we have a linear function $y(t)=at+b$.
Then $y'(t)=a$ and $y''(t)=0$
So $M=0$ and thus $\max_{0 \leq n \leq N} |y(t^n)-y^n| \leq 0 \Rightarrow \max_{0 \leq n \leq N} |y(t^n)-y^n|=0$.
Also don't we always have a uniform partition at Euler's method? (Thinking)