# Implicit Euler scheme and stability

Gold Member
Find the fixed points of the implicit Euler scheme
\begin{equation} y_{n+1}-y_{n}= hf(t_{n+1},y_{n+1})
\end{equation}
when applied to the differential equation $y'=y(1-y)$ and investigate their stability?

=>
implicit Euler scheme
\begin{equation} y_{n+1}-y_{n}= hf(t_{n+1},y_{n+1})
\end{equation}

$y'=y(1-y)$
\begin{equation} y_{n+1}=y_{n}+hy_{n+1}(1-y_{n+1})
\end{equation}
\begin{equation} y_{n+1}=y_{n}+hy_{n+1}-hy^2_{n+1}
\end{equation}

For fixed points
$y_{n+1}=y_{n}$
\begin{equation} y_{n}=y_{n}+hy_{n}-hy^2_{n}
\end{equation}
$y_{n}=0$ or $1$

I got problem with stability but this is what I have done

$y_{n}= \alpha +\epsilon^n$, $y_{n+1}= \alpha +\epsilon^{n+1}$,
\begin{equation} \alpha +\epsilon^{n+1}= \alpha +\epsilon^n + h (\alpha +\epsilon^{n+1})(1-\alpha -\epsilon^{n+1}) \end{equation}
\begin{equation} \epsilon^{n+1}= \epsilon^n + h (\alpha +\epsilon^{n+1})(1-\alpha -\epsilon^{n+1}) \end{equation}
When $y_{n}=0=\alpha$
\begin{equation} \epsilon^{n+1}= \epsilon^n + h \epsilon^{n+1}(1-\epsilon^{n+1}) \end{equation}

I don't what to say or do after that to determine the stability.

When $y_{n}=1=\alpha$
\begin{equation} \epsilon^{n+1}= \epsilon^n - h \epsilon^{n+1}(1+\epsilon^{n+1}) \end{equation}

same again what can say about with my answer to investigate the stability.

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pasmith
Homework Helper
Find the fixed points of the implicit Euler scheme
\begin{equation} y_{n+1}-y_{n}= hf(t_{n+1},y_{n+1})
\end{equation}
when applied to the differential equation $y'=y(1-y)$ and investigate their stability?

=>
implicit Euler scheme
\begin{equation} y_{n+1}-y_{n}= hf(t_{n+1},y_{n+1})
\end{equation}

$y'=y(1-y)$
\begin{equation} y_{n+1}=y_{n}+hy_{n+1}(1-y_{n+1})
\end{equation}
\begin{equation} y_{n+1}=y_{n}+hy_{n+1}-hy^2_{n+1}
\end{equation}

For fixed points
$y_{n+1}=y_{n}$
\begin{equation} y_{n}=y_{n}+hy_{n}-hy^2_{n}
\end{equation}
$y_{n}=0$ or $1$

I got problem with stability but this is what I have done

$y_{n}= \alpha +\epsilon^n$, $y_{n+1}= \alpha +\epsilon^{n+1}$,
I think you mean $y_n = \alpha + \epsilon_n$, etc.

\begin{equation} \alpha +\epsilon^{n+1}= \alpha +\epsilon^n + h (\alpha +\epsilon^{n+1})(1-\alpha -\epsilon^{n+1}) \end{equation}
\begin{equation} \epsilon^{n+1}= \epsilon^n + h (\alpha +\epsilon^{n+1})(1-\alpha -\epsilon^{n+1}) \end{equation}
When $y_{n}=0=\alpha$
\begin{equation} \epsilon^{n+1}= \epsilon^n + h \epsilon^{n+1}(1-\epsilon^{n+1}) \end{equation}

I don't what to say or do after that to determine the stability.
If $$y_{n+1} = g(y_n)$$ and $y^{*} = g(y^{*})$ is a fixed point, then one can set $y_n = y^{*} + \epsilon_n$ and expand $g$ in a taylor series to obtain $$\epsilon_{n+1} = g'(y^{*})\epsilon_n + O(\epsilon_n^2).$$
Thus if $g'(y^{*}) \neq 0$ then for $|\epsilon_0| \ll 1$ one has initially
$$\epsilon_n = (g'(y^{*}))^n \epsilon_0$$ and the fixed point will then be stable if $0 < |g'(y^{*})| < 1$ and unstable if $|g'(y^{*})| > 1$. If $|g'(y^{*})| = 0$ or $|g'(y^{*})| = 1$ then further investigation is necessary.

Here you have $$y_{n+1} - y_n = hy_{n+1}(1 - y_{n+1})$$ so that $$g(y_n) - y_n = hg(y_n)(1 - g(y_n))$$ and it may be easier to differentiate implicitly with respect to $y_n$ than to solve for $g(y_n)$ and then differentiate.

Gold Member
so from there
\begin{equation}
g(y_{n})-y_{n}=hg(y_{n})(1-g(y_{n}))
\end{equation}
\begin{equation}
g(y*+\epsilon_{n})-y*+\epsilon_{n}=hg(y*+\epsilon_{n})(1-g(y*+\epsilon_{n})
\end{equation}
i guess do the taylor expaninsion but you get nothing to determine the stability

pasmith
Homework Helper
so from there
\begin{equation}
g(y_{n})-y_{n}=hg(y_{n})(1-g(y_{n}))
\end{equation}
\begin{equation}
g(y*+\epsilon_{n})-y*+\epsilon_{n}=hg(y*+\epsilon_{n})(1-g(y*+\epsilon_{n})
\end{equation}
i guess do the taylor expaninsion but you get nothing to determine the stability
You certainly don't want to do that. Instead differentiate
$$g(y) - y = hg(y)(1 - g(y))$$ implicitly with respect to $y$ as I suggested. This gives you $g'(y)$ as a function of $h$ and $g(y)$ and you can then set $y = g(y) = 0$ or $y = g(y) = 1$ as appropriate.

Gold Member
You certainly don't want to do that. Instead differentiate
$$g(y) - y = hg(y)(1 - g(y))$$ implicitly with respect to $y$ as I suggested. This gives you $g'(y)$ as a function of $h$ and $g(y)$ and you can then set $y = g(y) = 0$ or $y = g(y) = 1$ as appropriate.
So whay saying is that to differentiate this
$$g(y) - y = hg(y)(1 - g(y))$$
$$g'(y) =1-h(1 - g(y)-g(y))$$
$$g'(y) =1-h(1-2g(y))$$
so
$$y =g(y)=0$$
$$g'(y) =1-h$$

$$y =g(y)=1$$
$$g'(y) =1+h$$
So what can i say with these results or did made any mistakes sir?

Last edited:
pasmith
Homework Helper
So whay saying is that to differentiate this
$$g(y) - y = hg(y)(1 - g(y))$$
$$g'(y) =1-h(1 - g(y)-g(y))$$
You should have $$g'(y) - 1 = hg'(y) - h(2g(y)g'(y))$$ so that $$g'(y) = \frac{1}{1 - h + 2hg(y)}.$$
Stability of the fixed points can therefore change depending on the value of $h > 0$. You will want to bear in mind the following:
• We want the fixed points of the iteration to share the same stability as the corresponding fixed points of the original ODE. This may impose constraints on the acceptable values of $h$, which is not a parameter occurring in the original ODE.
• The ODE whose solution you are approximating is $$\dot y = y(1- y),$$ which is first-order, one-dimensional and autonomous, so its solutions are constant or strictly monotonic. But the solution of $$\epsilon_{n+1} = A\epsilon_n$$ is oscillatory if $A < 0$: $\epsilon_{n} = (-1)^n|A|^n \epsilon_0$. Thus we require $g'(y) > 0$ at every fixed point. This again may impose constraints on the acceptable values of $h$.

• 1 person
D H
Staff Emeritus
I don't what to say or do after that to determine the stability.
Break the problem down into parts.

When you are using implicit Euler you have to solve for ## y_{k+1} ## given a previous value ## y_k ##. Your goal is to solve ## y_{k+1} -y_k = hy_{k+1}(1-y_{k+1}) ##. Apparently you are asked to use a fixed point iteration scheme to do this part of the job. Let ## y_{k+1,0} \equiv y_k ##. Now iteratively solve for ## y_{k+1,n+1} = y_k + hy_{k+1,n}(1-y_{k+1,n}) ##.

Question #1: Does the sequence ## \{y_{k+1,0}, y_{k+1,1}, y_{k+1,2}, \cdots \} ## converge to some value ##y_{k+1} ##? If it doesn't, it's obviously not stable.

Question #2: If it does converge, does the sequence ## \{ y_k,y_{k+1},y_{k+2},\cdots \} ## converge toward the fixed point? If it doesn't it's also not stable (but not so obviously).

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