0
$\begingroup$

Consider the following initial value problem $$\begin{cases} x'(t) = f(t,x(t)), \; t \in [t_0, T]\\ x(t_0) = \mu_0 \end{cases}$$ where $f \in \mathcal{C}\left([t_0, T] \times \mathbb{R}^n;\mathbb{R}^n \right)$ and is lipschitz on the second variable with Lipschitz's constant $L >0$. Given $n \in \mathbb{N}$, consider the net of nodes $t_i = t_0 + ih$ for $i = 0,1,\dots , N$ where $h = \frac{T-t_0}{N}$

Now consider the predictor-corrects method $P(EC)E$ where the predictor is the modified Euler Method $$x_{i+1} = x_i + hf\left( t_i + \frac{h}{2}, x_i + \frac{h}{2}f(t_i,x_i)\right)$$ and the corrector is the two step Adams- Moulton Method given by $$x_{i+2} = x_{i+1} + \frac{h}{12}\left( 5f_{i+2} + 8f_{i+1} - f_i \right)$$ where $f_{i+k} = f(t_{i+k}, x_{i+k})$

Show that the order of the resultant predictor-corrector method is $3$.

Now, I've already shown that the Adam-Mouton's Method has order $3$, and that Euler's modifies method has order two. Now there's a Theorem which barely says that

If I have two methods, both linear multistep methods of orders $p$ and $p^*$, the resultant predictor-corrector method taking as the predictor the method of order $p$ and the corrector the method of order $p^+$ has order $$\overline{p}=\min \{p,p^*\}$$

However, Euler's modified method is a Runge-Kutta method and therefore I don't think this theorem applies. My question is

How do I find the order of the method?

Thank you in advance for your help.

$\endgroup$
6
  • 1
    $\begingroup$ You could consider the method as a semi-implicit one-step method for $\Delta t=2h$ with tableau $$\begin{array}{c|ccc}0&&&\\\frac12&\frac12&&\\1&-\frac1{24}&\frac56&\frac{5}{24}\\\hline&-\frac1{24}&\frac56&\frac{5}{24}\end{array}$$ and check its order conditions. $\endgroup$
    – Lutz Lehmann
    Jun 12, 2017 at 14:05
  • $\begingroup$ I'm sorry but I don't know how did you get that tableau. For the Euler method I know how to get it, but I don't know much about semi-implicit methods... $\endgroup$
    – user313212
    Jun 12, 2017 at 16:49
  • $\begingroup$ It just means in this case that the equation for $k_3=Δt·f(t+Δt, x+...+\frac5{24}k3)$ depends non-linearly or implicitly on $k_3$ itself, which is natural since that is the part of the implicit corrector in the scheme. -- In this interpretation, your method is 2-periodic, $x_{2k+1}$ is always determined by the improved Euler method, $x_{2k+2}$ by the Adams- Moulton method. Is that correct? $\endgroup$
    – Lutz Lehmann
    Jun 12, 2017 at 16:55
  • $\begingroup$ I may have mixed up the halves and quarters. More explicitly, the interpretation as one-step method computes $k_1=Δt⋅f(t,x)$, $k_2=Δt⋅f(t+\frac14Δt, x+\frac14k_1)$, $x_{+1}=x+\frac12k_2$, $k_3= Δt⋅f(t+\frac12Δt, x+\frac12k_2)$, $x_{+2}=x_{+1}+\frac1{24}(5k_4+8k_3-k_1)$ where $k_4=Δt⋅f(t+Δt,x_{+2})$. Thus one gets a 4-stage method with tableau \begin{array}{c|cccc}0&&&&\\\frac14&\frac14&&&\\\frac12&&\frac12&\\1&-\frac1{24}&\frac12&\frac13&\frac{5}{24}\\\hline&-\frac1{24}&\frac12&\frac13&\frac{5}{24}\end{array} $\endgroup$
    – Lutz Lehmann
    Jun 12, 2017 at 17:10
  • $\begingroup$ Thank you. I'm trying to understand but I still don't see why is that. Maybe I'm not familiar with the notation you are using or something. Are you saying that my step $h$ should now be $2h$? Maybe I should review this whole predictor-corrector methods again and come back to this later. $\endgroup$
    – user313212
    Jun 12, 2017 at 17:30

1 Answer 1

Reset to default
2
$\begingroup$

Compare to the exact solution $x(t)$ of the ODE going through the initial point $(t_i,x_i)$.

The improved Euler step then relates to the exact ODE solution as $$x_{i+1}=x(t_{i+1})+O(h^3)$$ and thus also $$f_{i+1}=f(t_{i+1},x_{i+1})=x'(t_{i+1})+O(h^3).$$

The error of the Adams-Moulton step is $$ x(t_{i+2})=x(t_{i+1})+\frac{h}{12}(5x'(t_{i+2})+8x'(t_{i+1})-x'(t_i))+O(h^4) $$ Replacing $x'(t_{i+1})$ with $f_{i+1}$ only adds another $h·O(h^3)=O(h^4)$ error term. As we know that $x(t_{i+2})$ nearly solves the fixed-point equation for $x_{i+2}=X$, $$ X=C+\frac{5h}{12}f(t_{i+2},X),\qquad C=x_{i+1}+\frac{h}{12}(8f_{i+1}-f_i) $$ with an error $O(h^4)$, by the implicit function theorem or more basically the Banach fixed point theorem, the exact solution of the Adams-Moulton step also only has distance of $O(h^4)$ to $x(t_{i+2})$.

This local $O(h^4)$ error translates into a global $O(h^3)$ error, thus giving the order of the method as $3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.