Implicit Euler Scheme and stability

[ra_forever8]

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.

 

[jvicens]

To determine the stability of the fixed points, we need to look at the behavior of $\epsilon^{n+1}$ as $n$ increases. If $\epsilon^{n+1}$ approaches 0 as $n$ increases, then the fixed point is stable. If $\epsilon^{n+1}$ increases or oscillates as $n$ increases, then the fixed point is unstable.

For the fixed point $y_n=0$, we can rewrite the equation as
\begin{equation}
\epsilon^{n+1}= \epsilon^n + h \epsilon^{n+1}(1-\epsilon^{n+1}) = \epsilon^n + h \epsilon^{n+1} - h \epsilon^{n+1} \epsilon^{n+1}
\end{equation}
Since $h$ is a small step size, we can assume $h\epsilon^{n+1}$ is small. Therefore, the term $h\epsilon^{n+1}\epsilon^{n+1}$ can be neglected. This gives us the equation
\begin{equation}
\epsilon^{n+1}= \epsilon^n + h \epsilon^{n+1}
\end{equation}
which can be rewritten as
\begin{equation}
\epsilon^{n+1}= (1+h)\epsilon^n
\end{equation}
This shows that as $n$ increases, $\epsilon^{n+1}$ will approach 0, meaning the fixed point $y_n=0$ is stable.

For the fixed point $y_n=1$, we can rewrite the equation as
\begin{equation}
\epsilon^{n+1}= \epsilon^n - h \epsilon^{n+1}(1+\epsilon^{n+1}) = \epsilon^n - h \epsilon^{n+1} - h \epsilon^{n+1} \epsilon^{n+1}
\end{equation}
Again, since $h$ is small, we can neglect the term $h\epsilon^{n+1}\epsilon^{n+1}$. This gives us the equation
\begin{equation}
\epsilon^{n+1}= \epsilon^n - h \epsilon^{n+1}
\end{equation}
which can be rewritten as
\begin{equation}
\epsilon^{n+1}= (1-h)\epsilon^n
\end{equation}