Step size updating scheme adaptive embedded RK methods
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If I have a RK method $y$ of order $p$ and a RK method $z$ of order $p-1$ I have read I can estimate the local error as $r_{n+1} = y_{n+1} - z_{n+1}$. First of all I don't see how this estimates the local error defined as the error we make in a single step using correct previous values. What does a lower order method have anything to do with previous correct values? I can accept that this is the error we make using a lower order method instead of a higher order method, but what is the relevance of this?
Further, I read that the step size should be updated like
$$h_{n+1} = left( frac{TOL}{r_{n+1}}right)^{frac{1}{p}}h_n.$$
But I don't understand why this would work since we update our step size so that $r_{n+1} = TOL$, this would just result in $h_{n+1} = h_n$ after a while. What am I missing?
numerical-methods runge-kutta-methods
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
$endgroup$
|
show 2 more comments
$begingroup$
If I have a RK method $y$ of order $p$ and a RK method $z$ of order $p-1$ I have read I can estimate the local error as $r_{n+1} = y_{n+1} - z_{n+1}$. First of all I don't see how this estimates the local error defined as the error we make in a single step using correct previous values. What does a lower order method have anything to do with previous correct values? I can accept that this is the error we make using a lower order method instead of a higher order method, but what is the relevance of this?
Further, I read that the step size should be updated like
$$h_{n+1} = left( frac{TOL}{r_{n+1}}right)^{frac{1}{p}}h_n.$$
But I don't understand why this would work since we update our step size so that $r_{n+1} = TOL$, this would just result in $h_{n+1} = h_n$ after a while. What am I missing?
numerical-methods runge-kutta-methods
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
$endgroup$
$begingroup$
My understanding is that you usually compare to the result that a higher order RK method would give (after one step from the current start point), and use that as an estimate for the error of the lower order method. The idea of this is more or less the same as the idea of Richardson extrapolation.
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– Ian
Dec 28 '18 at 15:01
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@Ian If we have already calculated the higher order method, why would we need to use the lower order method or know the error of the lower order method?
$endgroup$
– Heuristics
Dec 28 '18 at 15:12
$begingroup$
You don't: you're using the lower order method as your "main approximation" and just using the higher order method to approximate the error you make along the way with the lower order method.
$endgroup$
– Ian
Dec 28 '18 at 15:37
$begingroup$
@Ian But to estimate the error we need to calculate the higher order method in the new point, so why not just use it instead of the lower order method, since the higher order method is better?
$endgroup$
– Heuristics
Dec 28 '18 at 16:03
$begingroup$
In fixed step size, you'd be right. In dynamic step size, you need something "better" to use as an error estimate so you know when to make the step size bigger or smaller. (Also, it's possible that the higher order method is more locally accurate but less stable, in which case you don't want to use it over and over.)
$endgroup$
– Ian
Dec 28 '18 at 16:28
|
show 2 more comments
$begingroup$
If I have a RK method $y$ of order $p$ and a RK method $z$ of order $p-1$ I have read I can estimate the local error as $r_{n+1} = y_{n+1} - z_{n+1}$. First of all I don't see how this estimates the local error defined as the error we make in a single step using correct previous values. What does a lower order method have anything to do with previous correct values? I can accept that this is the error we make using a lower order method instead of a higher order method, but what is the relevance of this?
Further, I read that the step size should be updated like
$$h_{n+1} = left( frac{TOL}{r_{n+1}}right)^{frac{1}{p}}h_n.$$
But I don't understand why this would work since we update our step size so that $r_{n+1} = TOL$, this would just result in $h_{n+1} = h_n$ after a while. What am I missing?
numerical-methods runge-kutta-methods
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
$endgroup$
If I have a RK method $y$ of order $p$ and a RK method $z$ of order $p-1$ I have read I can estimate the local error as $r_{n+1} = y_{n+1} - z_{n+1}$. First of all I don't see how this estimates the local error defined as the error we make in a single step using correct previous values. What does a lower order method have anything to do with previous correct values? I can accept that this is the error we make using a lower order method instead of a higher order method, but what is the relevance of this?
Further, I read that the step size should be updated like
$$h_{n+1} = left( frac{TOL}{r_{n+1}}right)^{frac{1}{p}}h_n.$$
But I don't understand why this would work since we update our step size so that $r_{n+1} = TOL$, this would just result in $h_{n+1} = h_n$ after a while. What am I missing?
numerical-methods runge-kutta-methods
numerical-methods runge-kutta-methods
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
asked Dec 28 '18 at 14:37
HeuristicsHeuristics
497211
497211
$begingroup$
My understanding is that you usually compare to the result that a higher order RK method would give (after one step from the current start point), and use that as an estimate for the error of the lower order method. The idea of this is more or less the same as the idea of Richardson extrapolation.
$endgroup$
– Ian
Dec 28 '18 at 15:01
$begingroup$
@Ian If we have already calculated the higher order method, why would we need to use the lower order method or know the error of the lower order method?
$endgroup$
– Heuristics
Dec 28 '18 at 15:12
$begingroup$
You don't: you're using the lower order method as your "main approximation" and just using the higher order method to approximate the error you make along the way with the lower order method.
$endgroup$
– Ian
Dec 28 '18 at 15:37
$begingroup$
@Ian But to estimate the error we need to calculate the higher order method in the new point, so why not just use it instead of the lower order method, since the higher order method is better?
$endgroup$
– Heuristics
Dec 28 '18 at 16:03
$begingroup$
In fixed step size, you'd be right. In dynamic step size, you need something "better" to use as an error estimate so you know when to make the step size bigger or smaller. (Also, it's possible that the higher order method is more locally accurate but less stable, in which case you don't want to use it over and over.)
$endgroup$
– Ian
Dec 28 '18 at 16:28
|
show 2 more comments
$begingroup$
My understanding is that you usually compare to the result that a higher order RK method would give (after one step from the current start point), and use that as an estimate for the error of the lower order method. The idea of this is more or less the same as the idea of Richardson extrapolation.
$endgroup$
– Ian
Dec 28 '18 at 15:01
$begingroup$
@Ian If we have already calculated the higher order method, why would we need to use the lower order method or know the error of the lower order method?
$endgroup$
– Heuristics
Dec 28 '18 at 15:12
$begingroup$
You don't: you're using the lower order method as your "main approximation" and just using the higher order method to approximate the error you make along the way with the lower order method.
$endgroup$
– Ian
Dec 28 '18 at 15:37
$begingroup$
@Ian But to estimate the error we need to calculate the higher order method in the new point, so why not just use it instead of the lower order method, since the higher order method is better?
$endgroup$
– Heuristics
Dec 28 '18 at 16:03
$begingroup$
In fixed step size, you'd be right. In dynamic step size, you need something "better" to use as an error estimate so you know when to make the step size bigger or smaller. (Also, it's possible that the higher order method is more locally accurate but less stable, in which case you don't want to use it over and over.)
$endgroup$
– Ian
Dec 28 '18 at 16:28
$begingroup$
My understanding is that you usually compare to the result that a higher order RK method would give (after one step from the current start point), and use that as an estimate for the error of the lower order method. The idea of this is more or less the same as the idea of Richardson extrapolation.
$endgroup$
– Ian
Dec 28 '18 at 15:01
$begingroup$
My understanding is that you usually compare to the result that a higher order RK method would give (after one step from the current start point), and use that as an estimate for the error of the lower order method. The idea of this is more or less the same as the idea of Richardson extrapolation.
$endgroup$
– Ian
Dec 28 '18 at 15:01
$begingroup$
@Ian If we have already calculated the higher order method, why would we need to use the lower order method or know the error of the lower order method?
$endgroup$
– Heuristics
Dec 28 '18 at 15:12
$begingroup$
@Ian If we have already calculated the higher order method, why would we need to use the lower order method or know the error of the lower order method?
$endgroup$
– Heuristics
Dec 28 '18 at 15:12
$begingroup$
You don't: you're using the lower order method as your "main approximation" and just using the higher order method to approximate the error you make along the way with the lower order method.
$endgroup$
– Ian
Dec 28 '18 at 15:37
$begingroup$
You don't: you're using the lower order method as your "main approximation" and just using the higher order method to approximate the error you make along the way with the lower order method.
$endgroup$
– Ian
Dec 28 '18 at 15:37
$begingroup$
@Ian But to estimate the error we need to calculate the higher order method in the new point, so why not just use it instead of the lower order method, since the higher order method is better?
$endgroup$
– Heuristics
Dec 28 '18 at 16:03
$begingroup$
@Ian But to estimate the error we need to calculate the higher order method in the new point, so why not just use it instead of the lower order method, since the higher order method is better?
$endgroup$
– Heuristics
Dec 28 '18 at 16:03
$begingroup$
In fixed step size, you'd be right. In dynamic step size, you need something "better" to use as an error estimate so you know when to make the step size bigger or smaller. (Also, it's possible that the higher order method is more locally accurate but less stable, in which case you don't want to use it over and over.)
$endgroup$
– Ian
Dec 28 '18 at 16:28
$begingroup$
In fixed step size, you'd be right. In dynamic step size, you need something "better" to use as an error estimate so you know when to make the step size bigger or smaller. (Also, it's possible that the higher order method is more locally accurate but less stable, in which case you don't want to use it over and over.)
$endgroup$
– Ian
Dec 28 '18 at 16:28
|
show 2 more comments
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$begingroup$
My understanding is that you usually compare to the result that a higher order RK method would give (after one step from the current start point), and use that as an estimate for the error of the lower order method. The idea of this is more or less the same as the idea of Richardson extrapolation.
$endgroup$
– Ian
Dec 28 '18 at 15:01
$begingroup$
@Ian If we have already calculated the higher order method, why would we need to use the lower order method or know the error of the lower order method?
$endgroup$
– Heuristics
Dec 28 '18 at 15:12
$begingroup$
You don't: you're using the lower order method as your "main approximation" and just using the higher order method to approximate the error you make along the way with the lower order method.
$endgroup$
– Ian
Dec 28 '18 at 15:37
$begingroup$
@Ian But to estimate the error we need to calculate the higher order method in the new point, so why not just use it instead of the lower order method, since the higher order method is better?
$endgroup$
– Heuristics
Dec 28 '18 at 16:03
$begingroup$
In fixed step size, you'd be right. In dynamic step size, you need something "better" to use as an error estimate so you know when to make the step size bigger or smaller. (Also, it's possible that the higher order method is more locally accurate but less stable, in which case you don't want to use it over and over.)
$endgroup$
– Ian
Dec 28 '18 at 16:28