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Thread: Euler's implicit formula for diff. eqs. - Convergence

  1. #1
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    Euler's implicit formula for diff. eqs. - Convergence

    Hi good people, I'm doing some old exam questions and here's one I don't quite get.

    We have the ordinary diff. eq,

    $\displaystyle y'(t) = e^{y(t)}$


    where $\displaystyle y(0)=1$ and $\displaystyle t\in [0,1]$.

    If we want to solve this by using Euler's implicit formula, we need solve an equation by say fixed-point iteration in each step. How small must the step size $\displaystyle h$ be in order for the fixed-point iteration to converge on each step.

    First of all, the problem does not say what kind of implicit formula I need to use. I get to the formula in the following manner,

    $\displaystyle \int^{t_{i+1}}_{t_i} y'(t)dt = \int^{t_{i+1}}_{t_i}f(t,y(t))dt$,

    and so,

    $\displaystyle y(t_{i+1}) = y(t_i) + \int^{t_{i+1}}_{t_i}f(t,y(t))dt$.

    To get an implicit formula from this I could simply approximate the integral by,

    $\displaystyle f(t_{i+1},y(t_{i+1}))(t_{i+1}-t_i) = e^{y(t_{i+1})}h$,

    or I could approximate it by the Trapezoidal rule.. Let's assume I do the former such that,

    $\displaystyle y(t_{i+1}) = y(t_i) + e^{y(t_{i+1})}h.$

    Now I need to use fixed-point iteration to solve this equation. If I let my first guess at $\displaystyle y(t_{i+1})$ be $\displaystyle y(t_i)$ which I believe is a reasonable guess, then I could write this thing in the familiar way;

    $\displaystyle x_{n+1} = x_n + e^{x_n}h = g(x_n)$.

    The sequence defined by the expression above converges if $\displaystyle 0\leq|g'(x)|<1$ for all $\displaystyle x$ in the interval. But $\displaystyle |g'(x)|=|1+e^xh|<1$ and so $\displaystyle h<0$ but that doesn't make much sense does it? I don't think the step size must be zero Hope someone can help me out here.

    Thanks.
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  2. #2
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    This DE is solvable exactly...

    $\displaystyle \displaystyle \frac{dy}{dt} = e^{y}$

    $\displaystyle \displaystyle e^{-y}\,\frac{dy}{dt} = 1$

    $\displaystyle \displaystyle \int{e^{-y}\,\frac{dy}{dt}\,dt} = \int{1\,dt}$

    $\displaystyle \displaystyle \int{e^{-y}\,dy} = t + C_1$

    $\displaystyle \displaystyle -e^{-y} + C_2 = t + C_1$

    $\displaystyle \displaystyle -e^{-y} = t + C$ where $\displaystyle \displaystyle C = C_1 - C_2$

    $\displaystyle \displaystyle e^{-y} = -t - C$

    $\displaystyle \displaystyle -y = \ln{(-t - C)}$

    $\displaystyle \displaystyle y = -\ln{(-t-C)}$
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  3. #3
    Member Mollier's Avatar
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    Ah, I did not notice that. Thanks.
    But it should still be solvable by Euler's implicit formula and so my argument must have some kind of mistake I think..
    Anyone see where I went wrong?
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