- #1

wel

Gold Member

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\begin{equation} y_{n+1}-y_{n}= hf(t_{n+1},y_{n+1})

\end{equation}

when applied to the differential equation [itex]y'=y(1-y)[/itex] 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

[itex]y_{n+1}=y_{n}[/itex]

\begin{equation} y_{n}=y_{n}+hy_{n}-hy^2_{n}

\end{equation}

[itex]y_{n}=0[/itex] or [itex]1[/itex]

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

[itex]y_{n}= \alpha +\epsilon^n[/itex], [itex]y_{n+1}= \alpha +\epsilon^{n+1}[/itex],

\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 [itex]y_{n}=1=\alpha[/itex]

\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.