# Counterexample: convergence of difference quotient in Backward Euler method

I am pretty sure that, for the iterates of the backward Euler's method for $$y'=f(t,y)$$ one cannot obtain $$\displaystyle \frac{y_n-y_{n-1}}{\delta t}\rightarrow y'(t^*)$$, for a suitable choice of $$n$$. The reason is that $$y_n=y(t_n)$$, but only at first order, so that $$y_n-y_{n-1} = \delta t y'(t^*) + C\delta t$$, and this $$C$$ need not to approach $$0$$.

I was only able, though, to construct examples where we get convergence also of the difference quotients (e.g. $$y'=ay$$ ...). Do you have a counterexample, and possibly a descriptions of the class of functions $$f$$ on which backward Euler is of higher order than $$1$$? It seems that $$f(t,y)=y$$, for instance, is in this class.

Edit

Actually, the derivative approximation property might be true by a bootstrapping argument: $$\displaystyle \frac{y_n-y_{n-1}}{\delta t} = f(t_n,y_n)=f(t^*,y(t^*)) + f_t(t^*,y(t^*))(t-t^*)+f_y(t^*,y(t^*))(y_n-y(t^*)) + o((t-t^*, y_n-y(t^*)))$$.

So that $$f\in C^1$$ suffices, thanks to the convergence $$t_n\rightarrow t^*, y_n \rightarrow y(t^*)$$. What do you think?

The situation is indeed somewhat complicated as you have two convergences to consider, the convergence of the difference quotient and the convergence of the numerical solutions as $$δt\to 0$$. It should be without doubt that $$\frac{y(t_n)-y(t_{n-1})}{t_n-t_{n-1}}\to y'(t^*)$$ where $$y(t)$$ is the exact solution of the IVP, $$t_n\ge t^*\ge t_{n-1}$$ for the selection of $$n$$ and $$δt=t_n-t_{n-1}\to 0$$ for the limit.

What remains is to explore the difference of exact solution and numerical solution for a fixed time discretization. Set $$e_n=y_n-y(t_n)$$. Then indeed $$y(t-δt)=y(t)-y'(t)δt+\frac12y''(t)δt^2\mp… =y(t)-f(t,y(t))δt+\frac12(f_t+f_yf)δt^2\mp…$$ Combined with the backward Euler formula this gives \begin{align} e_{n-1}&=e_n-[f(t_n,y_n)-f(t_n,y(t_n))]δt-\frac12y''(t_n)δt^2\pm…\\ &=e_{n-1}+f_y(t_n,y(t_n))e_nδt-\frac12y''(t_n)δt^2\pm…\\ (I-f_y(t_n,y(t_n))δt)e_n&=e_{n-1}-\frac12y''(t_n)δt^2\pm…\\ \end{align} This should show that $$e_n$$ is a combination of $$e_0$$ with a coefficient that accumulates the factors $$(I-f_y(t_n,y(t_n))δt)^{-1}$$, which is zero if $$e_0=0$$, and additional terms that contain the similarly discounted sum of the higher order Taylor terms. The second order term is thus of size $$O(δt)$$.

Returning to the recursion equation for $$e_n$$ this means that $$e_n-e_{n-1}=O(δt^2)$$. So indeed $$C\sim δt$$ in the notation of the question.

Not really sure what you are asking, but concerning

Do you have a counterexample, and possibly a descriptions of the class of functions 𝑓 on which backward Euler is of higher order than 1?

you can examine the Taylor expansion of

$$\frac{y_n - y_{n-1}}{\Delta t}$$ around $$t_n$$:

$$\frac{y_n - y_{n-1}}{\Delta t} = \frac{y_n - \Big(y_n - \Delta t y'(t_n) + 0.5(\Delta t)^2 y''(t_n) + \dots \Big)}{\Delta t} = y'(t_n) + \sum_{i=1}^\infty \frac{y^{(i+1)}(t_n)}{(i+1)!} (\Delta t)^i .$$

So for solutions with vanishing second derivative, e.g. linear functions like $$f(y) = \alpha y, \alpha \in \mathbb R$$ the approximation is exact. For other functions, you can take single steps of higher order, e.g. when the true solution is for instance $$\sin(t)$$ and you evaluate it at some $$t_n = n \pi, n \in \mathbb Z$$.