Implicit Euler scheme and stability

[wel]

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.

 

[pasmith]

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 <br /> y_{n+1} = g(y_n)<br /> 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 <br /> \epsilon_{n+1} = g&#039;(y^{*})\epsilon_n + O(\epsilon_n^2).<br />
Thus if g&#039;(y^{*}) \neq 0 then for |\epsilon_0| \ll 1 one has initially
<br /> \epsilon_n = (g&#039;(y^{*}))^n \epsilon_0<br /> and the fixed point will then be stable if 0 &lt; |g&#039;(y^{*})| &lt; 1 and unstable if |g&#039;(y^{*})| &gt; 1. If |g&#039;(y^{*})| = 0 or |g&#039;(y^{*})| = 1 then further investigation is necessary.

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

 

[wel]

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]

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
<br /> g(y) - y = hg(y)(1 - g(y))<br /> implicitly with respect to y as I suggested. This gives you g&#039;(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.

 

[wel]

You certainly don't want to do that. Instead differentiate
<br /> g(y) - y = hg(y)(1 - g(y))<br /> implicitly with respect to y as I suggested. This gives you g&#039;(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
<br /> g(y) - y = hg(y)(1 - g(y))<br />
<br /> g&#039;(y) =1-h(1 - g(y)-g(y))<br />
<br /> g&#039;(y) =1-h(1-2g(y))<br />
so
<br /> y =g(y)=0<br />
<br /> g&#039;(y) =1-h<br />

<br /> y =g(y)=1<br />
<br /> g&#039;(y) =1+h<br />
So what can i say with these results or did made any mistakes sir?

 

[pasmith]

So whay saying is that to differentiate this
<br /> g(y) - y = hg(y)(1 - g(y))<br />
<br /> g&#039;(y) =1-h(1 - g(y)-g(y))<br />

You should have <br /> g&#039;(y) - 1 = hg&#039;(y) - h(2g(y)g&#039;(y))<br /> so that <br /> g&#039;(y) = \frac{1}{1 - h + 2hg(y)}.<br />
Stability of the fixed points can therefore change depending on the value of h &gt; 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 <br /> \epsilon_{n+1} = A\epsilon_n<br /> is oscillatory if A &lt; 0: \epsilon_{n} = (-1)^n|A|^n \epsilon_0. Thus we require g&#039;(y) &gt; 0 at every fixed point. This again may impose constraints on the acceptable values of h.

 

[D H]

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).