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Math Help - Trapezoid method, fixed-point iteration

  1. #1
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    Trapezoid method, fixed-point iteration

    Hi,

    problem:

    The trapezoid method for solving ordinary differential equations is,

    y(x_{i+1}) = y(x_i) + h \frac{f(y(x_{i+1})) + f(y(x_i))}{2}.

    Since this method is implicit, we must solve for y(x_{i+1}) in every step. If we use fixed-point iteration x=g(x), we will have convergence if |g'(x)|<1 for  x \in[y(x_i), y(x_{i+1})].

    (a) Give the expression for g.

    (b) Give a condition on h such that the fixed-point iterations converges at each step.

    attempt:

    (a)

    g(x) = y(x_i) + h \frac{f(x) + f(y(x_i))}{2}

    For each iteration i, we have to run fixed-point iteration enough times to make sure we get the wanted accuracy.
    We take a guess at x=y_{i+1}, calculate g(x) and see how big |x-g(x)| is. If it's close enough to 0, we stop the fixed-point iteration.

    (b)

    I'm not really sure what to make of this question...
    Fixed-point methods converge if |g'(x)|<1.

    <br />
\begin{aligned}<br />
&|g'(x)| = \frac{h}{2}\Big[f'(x) + f'(y(x_i))\Big] < 1&\\<br />
&\Rightarrow h < \frac{2}{\Big[f'(x) + f'(y(x_i))\Big]} &<br />
\end{aligned}<br />

    I do not see much sense in my last expression.
    Thanks!
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  2. #2
    A Plied Mathematician
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    I think your answer for part (a) is correct. However, for part (b), everything except plain ol' x is treated like a constant when you take the derivative, right? I mean, you've already found y(x_{i}), so when you're trying to find y(x_{i+1}), the previous value doesn't change. I think this may simplify your expression a bit.
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  3. #3
    Member Mollier's Avatar
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    Hi Ackbeet, thanks for replying.

    To kick of fixed-point iteration I guess a value for x = y_{i+1} so that I can calculate f(x). But this means that I know f'(x). I now calculate g(x) and use it as the new y_{i+1}.
    To me it looks like I always know that value of f'(y_{i+1}).

    But then again, if I think of x just as some unknown I guess I end up with,

    h < \frac{2}{f'(x)}

    This bound still does not make much sense to me though...
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  4. #4
    A Plied Mathematician
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    Well, the fraction on the RHS there, namely, 2/f'(x), is a bound on the step size for your method. What if you always knew that

    h<\dfrac{2}{\max|f'(x)|}?

    Then you would always have a contraction mapping, and the iteration would converge (by the Banach fixed-point theorem). Make sense?
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  5. #5
    Member Mollier's Avatar
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    That does make sense. Thank you!
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  6. #6
    A Plied Mathematician
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    You're welcome. Have a good one!
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