Order of accuracy- Euler's method

[evinda]

Hello! (Smile)

Theorem

Let $f \in C([a,b] \times \mathbb{R})$ a function that satisfies the 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 of $[a,b]$ 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 $M=\max_{a \leq t \leq b} |y''(t)|$.
From the above theorem, we conclude that the order of accuracy of Euler's method is at least $1$.We will show that the order of accuracy of Euler's method is exactly $1$.

We consider the following ODE:$\left\{\begin{matrix}
y'=2t &, 0 \leq t \leq 1 \\
y(0)=0 &
\end{matrix}\right.$Its only solution is $y(t)=t^2,\ \ 0 \leq t \leq 1$.

Let $N \in \mathbb{N}, \ h=\frac{1}{N}, \ \ t^n=nh, \ n=0,1, \dots, N$

$$y^{n+1}=y^n+hf(t^n, y^n) \Rightarrow y^{n+1}=y^n+h2t^n=y^n+h2nh=y^n+2nh^2$$$$y^0=y(0)=0$$

$$y^1=y^0+2 \cdot 0 \cdot h^2=0$$

$$y^2=y^1+2h^2=2h^2$$

$$y^3=y^2+2 \cdot 2 \cdot h^2=2h^2+4h^2=2h^2(1+2)$$

$$y^4=y^3+2 \cdot 3 \cdot h^2=2h^2(1+2)+ 2 \cdot 3 \cdot h^2=2h^2(1+2+3)$$

$$\dots \dots$$

$$y^n=2h^2 \sum_{i=1}^{n-1} i=2h^2 \frac{(n-1)n}{2}=n(n-1)h^2$$For $n=N$: $y^N=N(N-1)h^2=(Nh-h) Nh=1-h$$|y(t^N)-y^N|=|1-1+h|=h$Theorefore, we conclude that the order of accuracy is exactly $1$.
According to the theorem: $$\max_{0 \leq n \leq N} |y(t^n)-y^n| \leq \frac{M}{2L} (e^{L(b-a)}-1)h$$
So, why do we conclude that the order of accuracy of Euler's method is at least $1$ and not at most $1$?
Also, we take into consideration that the order of accuracy of the method is at least $1$ and then we take an example of an ODE and see that the order of accuracy is exactly $1$.
Why do we conclude in this way something for the general case?

How do we deduce that for each ODE the same holds? (Thinking)

 

[jvicens]

I would like to clarify that the order of accuracy of a numerical method is a measure of how closely the numerical solution approximates the exact solution of the differential equation. It is not a characteristic of a specific ODE, but rather a property of the numerical method used to solve the ODE.

In the case of Euler's method, we can see from the theorem that the error in the numerical solution is bounded by a term that is proportional to the step size $h$. This means that as we decrease the step size, the error in the numerical solution also decreases. This is why we can conclude that the order of accuracy of Euler's method is at least $1$.

In your example, you have shown that the order of accuracy is exactly $1$, but this does not hold for all ODEs. There may be cases where the order of accuracy is higher, depending on the Lipschitz condition and the smoothness of the exact solution. However, the theorem guarantees that the order of accuracy will be at least $1$ for any ODE that satisfies the conditions stated. This is why we can conclude that the order of accuracy of Euler's method is at least $1$ in the general case.

Furthermore, we cannot deduce that the order of accuracy will always be exactly $1$ for any ODE. As mentioned before, it depends on the specific ODE and the conditions stated in the theorem. The example you have shown is just one case where the order of accuracy happens to be exactly $1$. In other cases, it may be higher.

In conclusion, the order of accuracy of a numerical method is a property of the method itself, not the ODE being solved. The theorem provides a bound on the error in the numerical solution, and we can conclude that the order of accuracy of Euler's method is at least $1$ based on this bound. However, it is not always exactly $1$, as it depends on the specific ODE being solved.