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Recently I received an exercise that I do not quite understand. It is as follows.

Let $$ y'' + 20y' + 19y = 0, \qquad y(0) = 1 \quad y'(0) = -10 \quad t \geq 0 $$

be the relevant IVP. Solve it using 3rd order Adams-Moulton method, i.e.

$$y_{j+2} - y_{j+1} = \frac{h}{12}(5f(t_{j+2},y_{j+2})+8f(t_{j+1}, y_{j+1})-f(t_j,y_j)).$$

Estimate the value $y_{j+2}$ via fixed-point iteration. What is the maximum size of $h$ s.t. the fixed-point iteration still converges?

First of all, how am I supposed to solve it? There is no specified range for approximations. Or should I consider taking the limit $h \to 0$ for the method above?

Second, how do I estimate $y_{j+2}$ via fixed-point iteration and determine the maximum step size? I know for example Banach's fixed-point theorem but I can't seem to figure out its application to multi step methods.

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The equation as it is is a fixed point iteration for $y_{j+2}$. What is its Lipschitz constant considering that the Lipschitz constant of $f$ can be bounded by $20$?

For the initial point of the iteration, you could just extrapolate from the known values of $y_j, y_{j+1}$. This could be enhanced by also using the (previously already computed) value of $f(t_{j+1},y_{j+1})$.


Update: The directly read off fixed point equation is $$ y_{j+2}=T(y_{j+2})=C+\frac{5h}{12}f(t_{j+2},y_{j+2}). $$ $T$ has thus Lipschitz constant $L_T=\frac{5}{12}L_fh$ if $L_f$ is the $y$-Lipschitz constant of $f$. Now one has to ensure that $L_T<1$ or $L_fh<2.4$ if the associated fixed-point iteration is to converge.

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  • $\begingroup$ Honestly, LutzL, I have no idea what the Lipschitz constant might be in this particular multistep setting. How would you determine this here? I know that in the general case you just determine it by $$\vert\vert f(x) - f(y)\vert\vert \leq L\vert\vert x-y\vert\vert$$ $\endgroup$ – Taufi Jan 15 '17 at 20:18
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    $\begingroup$ added one more step. Now you need to define what you want to use as $f$ and what the related constant is. $\endgroup$ – Dr. Lutz Lehmann Jan 15 '17 at 21:14
  • $\begingroup$ So $C = y_{j+1} + (\frac{8h}{12}f(t_{j+1},y_{j+1})-\frac{h}{12}f(t_j,y_j))$? How you come up with $L_jh < 2.4$? $\endgroup$ – Taufi Jan 16 '17 at 13:07
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    $\begingroup$ Yes. And $\frac{12}5=\frac{24}{10}=2.4$. Which one gets from transforming $L_T=\frac{5}{12}L_f\,h<1$. $\endgroup$ – Dr. Lutz Lehmann Jan 16 '17 at 13:21
  • $\begingroup$ Thank you, LutzL. I marked it as an accepted answer. The only thing left is what to use as $f(t_{j+2},y_{j+2})$. Is it just the ODE rearranged such that $y'' = -20 y' - 19 y$? $\endgroup$ – Taufi Jan 16 '17 at 14:13

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