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Implicit Euler scheme and stability

  1. Apr 13, 2014 #1

    wel

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    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 [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.
     
  2. jcsd
  3. Apr 14, 2014 #2

    pasmith

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    Homework Helper

    I think you mean [itex]y_n = \alpha + \epsilon_n[/itex], etc.

    If [tex]
    y_{n+1} = g(y_n)
    [/tex] and [itex]y^{*} = g(y^{*})[/itex] is a fixed point, then one can set [itex]y_n = y^{*} + \epsilon_n[/itex] and expand [itex]g[/itex] in a taylor series to obtain [tex]
    \epsilon_{n+1} = g'(y^{*})\epsilon_n + O(\epsilon_n^2).
    [/tex]
    Thus if [itex]g'(y^{*}) \neq 0[/itex] then for [itex]|\epsilon_0| \ll 1[/itex] one has initially
    [tex]
    \epsilon_n = (g'(y^{*}))^n \epsilon_0
    [/tex] and the fixed point will then be stable if [itex]0 < |g'(y^{*})| < 1[/itex] and unstable if [itex]|g'(y^{*})| > 1[/itex]. If [itex]|g'(y^{*})| = 0[/itex] or [itex]|g'(y^{*})| = 1[/itex] then further investigation is necessary.

    Here you have [tex]
    y_{n+1} - y_n = hy_{n+1}(1 - y_{n+1})
    [/tex] so that [tex]
    g(y_n) - y_n = hg(y_n)(1 - g(y_n))
    [/tex] and it may be easier to differentiate implicitly with respect to [itex]y_n[/itex] than to solve for [itex]g(y_n)[/itex] and then differentiate.
     
  4. Apr 14, 2014 #3

    wel

    User Avatar
    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
     
  5. Apr 14, 2014 #4

    pasmith

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    Homework Helper

    You certainly don't want to do that. Instead differentiate
    [tex]
    g(y) - y = hg(y)(1 - g(y))
    [/tex] implicitly with respect to [itex]y[/itex] as I suggested. This gives you [itex]g'(y)[/itex] as a function of [itex]h[/itex] and [itex]g(y)[/itex] and you can then set [itex]y = g(y) = 0[/itex] or [itex]y = g(y) = 1[/itex] as appropriate.
     
  6. Apr 15, 2014 #5

    wel

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    Gold Member

    So whay saying is that to differentiate this
    [tex]
    g(y) - y = hg(y)(1 - g(y))
    [/tex]
    [tex]
    g'(y) =1-h(1 - g(y)-g(y))
    [/tex]
    [tex]
    g'(y) =1-h(1-2g(y))
    [/tex]
    so
    [tex]
    y =g(y)=0
    [/tex]
    [tex]
    g'(y) =1-h
    [/tex]

    [tex]
    y =g(y)=1
    [/tex]
    [tex]
    g'(y) =1+h
    [/tex]
    So what can i say with these results or did made any mistakes sir?
     
    Last edited: Apr 15, 2014
  7. Apr 15, 2014 #6

    pasmith

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    Homework Helper

    You should have [tex]
    g'(y) - 1 = hg'(y) - h(2g(y)g'(y))
    [/tex] so that [tex]
    g'(y) = \frac{1}{1 - h + 2hg(y)}.
    [/tex]
    Stability of the fixed points can therefore change depending on the value of [itex]h > 0[/itex]. 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 [itex]h[/itex], which is not a parameter occurring in the original ODE.
    • The ODE whose solution you are approximating is [tex]\dot y = y(1- y),[/tex] which is first-order, one-dimensional and autonomous, so its solutions are constant or strictly monotonic. But the solution of [tex]
      \epsilon_{n+1} = A\epsilon_n
      [/tex] is oscillatory if [itex]A < 0[/itex]: [itex]\epsilon_{n} = (-1)^n|A|^n \epsilon_0[/itex]. Thus we require [itex]g'(y) > 0[/itex] at every fixed point. This again may impose constraints on the acceptable values of [itex]h[/itex].
     
  8. Apr 15, 2014 #7

    D H

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    Staff Emeritus
    Science Advisor

    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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