- #1
woodssnoop
- 10
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Hello:
I am trying to determine the global error for the verlet integration algorithm: [itex]r_{i+1}=2r_{i}-r_{i-1}+\ddot{r}_{i}h^{2}[/itex]
It is easy to show that this approximation is locally 4th order, by way of a Talyor expansion. I have look around extensively for a proof of the global error but have been unsuccessful in finding a reasonable derivation anywhere (the articles I have found seem to make too many unjustified assumption and/or glaze over the dropping of terms; similar to the wikipedia page on verlet integration). (1) Has anyone else ran across a good derivation of the verlet integrators global error? My attempts at finding the global error for this expression has proved to be quite difficult.
The numerical testing I have done has shown 2nd order global error, but I can't prove it. (2) I know I have to show that [itex]d_{k} \leq M n^2 [/itex] for all [itex]k \leq n[/itex], where M is some constant, but how can I show this for an arbitrary M? In real analysis, you show an equation like this is true by showing that for any [itex]\epsilon[/itex] you can find a [itex]n[/itex] such that the expression holds, but I can't seem to frame my problem in those terms. I have been trying several different approaches, and my latest attempt is below.
My first step was to set up a difference equation for [itex]r_{i}[/itex] and [itex]r(h*i)[/itex] where [itex]r(h*i)[/itex] is the exact solution and [itex]r_{i}[/itex] is the approximate solution. After subtracting the two expression I get the difference formula:
[itex]d_{i+1}=2d_{i}-d_{i-1}-h^{2}(\ddot{r}_{i}-\ddot{r}(h*i)) - O(h^{4})[/itex]
using the mean value theorem:
[itex]d_{i+1}=2d_{i}-d_{i-1}-h^{2}(\ddot{\xi}_{i}d_{i}) - O(h^{4})[/itex] for [itex]\xi\in[r_{i},r(h*i)][/itex]
[itex]\Leftrightarrow d_{i+1}-(2-h^{2}(\ddot{\xi}_{i}))d_{i}+d_{i-1}=-O(h^{4})[/itex]
[itex]r[/itex] is differentiable on the interval [itex][0,h*n][/itex]. Which implies [itex]\ddot{\xi}_{i}[/itex] has a maximum. Letting [itex]\epsilon=h^{2}\ddot{\xi}_{i}[/itex] gives:
[itex]d_{i+1}-(2-\epsilon)d_{i}+d_{i-1}=-O(h^{4})[/itex]
condensing further:
[itex]d_{i+1}-xd_{i}+d_{i-1}=-O(h^{4})[/itex] where [itex]x=2-\epsilon[/itex].
This expression can be written out in matrix form (shown below is just a 12X12 example) where A=
Taking the inverse of A yields:
The rows of A have a closed form solution:
[itex]a(i)=\frac{(x-\sqrt{x^{2}-4})^{i}-(x-\sqrt{x^{2}-4})^{i}}{2^{i}\sqrt{x^{2}-4}}[/itex]
[itex]\Rightarrow d_{n}=\sum_{i=1}^{n}a(i)[/itex]
This two can be expressed in a better way:
[itex]d_{n}=\frac{(\frac{x-\sqrt{x^{2}-4}}{2})^{n}(\sqrt{x^{2}-4}+x-2)+(\frac{x+\sqrt{x^{2}-4}}{2})^{n}(\sqrt{x^{2}-4}-x+2)-2\sqrt{x^{2}-4}}{x-2}[/itex]
I am trying to show that [itex] d(n) \leq M n^2 [/itex], but I have been having trouble show that this true.
I am trying to determine the global error for the verlet integration algorithm: [itex]r_{i+1}=2r_{i}-r_{i-1}+\ddot{r}_{i}h^{2}[/itex]
It is easy to show that this approximation is locally 4th order, by way of a Talyor expansion. I have look around extensively for a proof of the global error but have been unsuccessful in finding a reasonable derivation anywhere (the articles I have found seem to make too many unjustified assumption and/or glaze over the dropping of terms; similar to the wikipedia page on verlet integration). (1) Has anyone else ran across a good derivation of the verlet integrators global error? My attempts at finding the global error for this expression has proved to be quite difficult.
The numerical testing I have done has shown 2nd order global error, but I can't prove it. (2) I know I have to show that [itex]d_{k} \leq M n^2 [/itex] for all [itex]k \leq n[/itex], where M is some constant, but how can I show this for an arbitrary M? In real analysis, you show an equation like this is true by showing that for any [itex]\epsilon[/itex] you can find a [itex]n[/itex] such that the expression holds, but I can't seem to frame my problem in those terms. I have been trying several different approaches, and my latest attempt is below.
My first step was to set up a difference equation for [itex]r_{i}[/itex] and [itex]r(h*i)[/itex] where [itex]r(h*i)[/itex] is the exact solution and [itex]r_{i}[/itex] is the approximate solution. After subtracting the two expression I get the difference formula:
[itex]d_{i+1}=2d_{i}-d_{i-1}-h^{2}(\ddot{r}_{i}-\ddot{r}(h*i)) - O(h^{4})[/itex]
using the mean value theorem:
[itex]d_{i+1}=2d_{i}-d_{i-1}-h^{2}(\ddot{\xi}_{i}d_{i}) - O(h^{4})[/itex] for [itex]\xi\in[r_{i},r(h*i)][/itex]
[itex]\Leftrightarrow d_{i+1}-(2-h^{2}(\ddot{\xi}_{i}))d_{i}+d_{i-1}=-O(h^{4})[/itex]
[itex]r[/itex] is differentiable on the interval [itex][0,h*n][/itex]. Which implies [itex]\ddot{\xi}_{i}[/itex] has a maximum. Letting [itex]\epsilon=h^{2}\ddot{\xi}_{i}[/itex] gives:
[itex]d_{i+1}-(2-\epsilon)d_{i}+d_{i-1}=-O(h^{4})[/itex]
condensing further:
[itex]d_{i+1}-xd_{i}+d_{i-1}=-O(h^{4})[/itex] where [itex]x=2-\epsilon[/itex].
This expression can be written out in matrix form (shown below is just a 12X12 example) where A=
Taking the inverse of A yields:
The rows of A have a closed form solution:
[itex]a(i)=\frac{(x-\sqrt{x^{2}-4})^{i}-(x-\sqrt{x^{2}-4})^{i}}{2^{i}\sqrt{x^{2}-4}}[/itex]
[itex]\Rightarrow d_{n}=\sum_{i=1}^{n}a(i)[/itex]
This two can be expressed in a better way:
[itex]d_{n}=\frac{(\frac{x-\sqrt{x^{2}-4}}{2})^{n}(\sqrt{x^{2}-4}+x-2)+(\frac{x+\sqrt{x^{2}-4}}{2})^{n}(\sqrt{x^{2}-4}-x+2)-2\sqrt{x^{2}-4}}{x-2}[/itex]
I am trying to show that [itex] d(n) \leq M n^2 [/itex], but I have been having trouble show that this true.