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The backwards Euler method (implicit Euler scheme) is a numerical method for the finding the solution of ordinary differential equations, which is defined as follows, $$ y(t_{n+1}) \approx y(t_n) + hf(t_{n+1}, y(t_{n+1}))$$ where $h = t_{n+1} - t_n$.

How could one find and solve the difference equation resulting from using $f(y) =-y \,$ for this method? What does this then say about the unconditional stability of the backward Euler method?

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  • $\begingroup$ Right out the first few steps explicitly and express $y_(t_j)$ as a function of $j$, $h$, and $y(t_0)$ for $j=1,2,3$. A pattern should emerge that you can then formally prove. $\endgroup$
    – Carl Christian
    Commented Apr 26, 2021 at 9:03
  • $\begingroup$ @CarlChristian Thanks for the comment. I'm not too sure what you mean by this. Could you format an answer so I could see? $\endgroup$
    – Nipster
    Commented Apr 28, 2021 at 9:53
  • $\begingroup$ So you get $$y_{n+1}=y_n+h·(-y_{n+1})$$ and want to know how $y_{n+1}$ computes from $y_n$,... $\endgroup$
    – Lutz Lehmann
    Commented May 2, 2021 at 11:19

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We have $$y(t_{n+1})=y(t_n)-hy(t_{n+1})$$

Hence,

$$(1+h)y(t_{n+1})=y(t_n)$$

$$y(t_{n+1}) = \frac1{1+h}y(t_n)$$

$$y(t_{n})=\left(\frac1{1+h} \right)^n y(t_0)$$

Note that we have $0<\frac1{1+h}<1$ for $h>0$, it is unconditionally stable.

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