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Linearizing an explicit differentiation scheme

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the following implicit scheme:
    [tex]y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n+1})+f(y_{n})][/tex]

    By linearization one can obtain an explicit scheme which is an approximation to this - with approximation error [tex]O(\Delta t^{3})[/tex]

    2. Relevant equations
    The solution is:
    [tex]y_{n+1}=y_{n}+\Delta t\left [1-\frac{1}{2}\Delta t f'(y_{n}) \right ]^{-1}f(y_{n})[/tex]

    And the notation: [tex]y'(x)=f(x,y(x))[/tex]

    3. The attempt at a solution
    I think I have to take the Taylor expansion of [tex]f(y_{n+1})[/tex]

    I get:
    [tex]f(y_{n+1})=f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})[/tex]

    Substituting in the main scheme:

    [tex]y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta ty'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})[/tex]

    This expression is, as far as I can see, not equal to the expression of the solution.

    Any help would be appreciated!
     
  2. jcsd
  3. Mar 29, 2012 #2

    SammyS

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    You are missing the "\right" code with several of your "]" symbols. (Corrected in Quote.)
     
  4. Mar 29, 2012 #3

    hunt_mat

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    Homework Helper

    This is what I did, it may help you it may not.
    [tex]
    \begin{array}{rcl}
    f(y_{n+1}) & = & f\left( y_{n}+\frac{\delta t}{2}X\right) \\
    & = & f(y_{n}+\frac{X\delta t}{2}f'(y_{n}) \\
    & = & f(y_{n})+ \frac{\delta t}{2}(f(y_{n+1})+f(y_{n}))f'(y_{n})
    \end{array}
    [/tex]
    where [itex]X=f(y_{n+1})+f(y_{n})[/itex] and we can solve for [itex]f(y_{n+1})[/itex], to obtain:
    [tex]
    f(y_{n+1})=\left( 1-\frac{\delta t}{2}f'(y_{n})\right)^{-1}\left( 2f(y_{n})+\frac{\delta t}{2}f'(y_{n})\right)
    [/tex]
    Substitute this into your equation.
     
    Last edited: Mar 29, 2012
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