Local error per unit step












0












$begingroup$


The solution of the ODE $$ y' = f(t,y)$$ is being seeked.



Let $u_{m}$ be the numerical solution of a one step method and $y(t_m)$ its true solution. The local error $e_{loc} $ is then defined as $$e_{loc}(t_m):= y(t_m) - u_m$$
Related to this this definition, is the local error per unit step, defined as
$$e_{loc}(t+h)/h, quad h text{ being the step size.}$$
What is the deeper meaning of the defintion of the local error per unit step?



Any help would be greatly appreciated.






ordinary-differential-equations truncation-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE. I believe that when comparing various errors, they be "normalized" in some way so the comparison is reasonable (e.g., comparing an error over a length of $1$ compared to over a length of a billion doesn't usually provide much useful information). Dividing by $h$ means that the result is what the local error is roughly on average per unit change of $t$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 22:06












  • $begingroup$
    Could you add more context? One area where this is important is with adaptive step sizes. One criterion to control the step size is to hold the "local error per unit step" constant at the level of the provided tolerances, if $e_{loc}(t,h)=C(t)h^{p+1}+...$, one would strive to get $e_{loc}(t,h)/h=C(t)h^{p}=epsilon$ so that the global error computes approximately as $sum C(t_k)h_k^{p+1}=epsilonsum h_k=epsilon(t_{final}-t_{init})$.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:11










  • $begingroup$
    Thanks so much for the very quick answers so far. It is a very general context actually. It is used in a book for the analysis of the consistency of a general one step method for solving numerically ode's.
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:51










  • $begingroup$
    Unfortunately, the book is written in german. The author is Karl Strehmel, from his book "Numerik gewöhnlicher Differentialgleichungen" page 24. springer.com/de/book/9783834818478
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:53


















0












$begingroup$


The solution of the ODE $$ y' = f(t,y)$$ is being seeked.



Let $u_{m}$ be the numerical solution of a one step method and $y(t_m)$ its true solution. The local error $e_{loc} $ is then defined as $$e_{loc}(t_m):= y(t_m) - u_m$$
Related to this this definition, is the local error per unit step, defined as
$$e_{loc}(t+h)/h, quad h text{ being the step size.}$$
What is the deeper meaning of the defintion of the local error per unit step?



Any help would be greatly appreciated.






ordinary-differential-equations truncation-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE. I believe that when comparing various errors, they be "normalized" in some way so the comparison is reasonable (e.g., comparing an error over a length of $1$ compared to over a length of a billion doesn't usually provide much useful information). Dividing by $h$ means that the result is what the local error is roughly on average per unit change of $t$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 22:06












  • $begingroup$
    Could you add more context? One area where this is important is with adaptive step sizes. One criterion to control the step size is to hold the "local error per unit step" constant at the level of the provided tolerances, if $e_{loc}(t,h)=C(t)h^{p+1}+...$, one would strive to get $e_{loc}(t,h)/h=C(t)h^{p}=epsilon$ so that the global error computes approximately as $sum C(t_k)h_k^{p+1}=epsilonsum h_k=epsilon(t_{final}-t_{init})$.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:11










  • $begingroup$
    Thanks so much for the very quick answers so far. It is a very general context actually. It is used in a book for the analysis of the consistency of a general one step method for solving numerically ode's.
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:51










  • $begingroup$
    Unfortunately, the book is written in german. The author is Karl Strehmel, from his book "Numerik gewöhnlicher Differentialgleichungen" page 24. springer.com/de/book/9783834818478
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:53
















0












0








0





$begingroup$


The solution of the ODE $$ y' = f(t,y)$$ is being seeked.



Let $u_{m}$ be the numerical solution of a one step method and $y(t_m)$ its true solution. The local error $e_{loc} $ is then defined as $$e_{loc}(t_m):= y(t_m) - u_m$$
Related to this this definition, is the local error per unit step, defined as
$$e_{loc}(t+h)/h, quad h text{ being the step size.}$$
What is the deeper meaning of the defintion of the local error per unit step?



Any help would be greatly appreciated.






ordinary-differential-equations truncation-error







share|cite|improve this question











$endgroup$




The solution of the ODE $$ y' = f(t,y)$$ is being seeked.



Let $u_{m}$ be the numerical solution of a one step method and $y(t_m)$ its true solution. The local error $e_{loc} $ is then defined as $$e_{loc}(t_m):= y(t_m) - u_m$$
Related to this this definition, is the local error per unit step, defined as
$$e_{loc}(t+h)/h, quad h text{ being the step size.}$$
What is the deeper meaning of the defintion of the local error per unit step?



Any help would be greatly appreciated.






ordinary-differential-equations truncation-error




ordinary-differential-equations truncation-error






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 12:05









LutzL

57.1k42054




57.1k42054










asked Dec 30 '18 at 22:01









philo1707philo1707

1




1












  • $begingroup$
    Welcome to MSE. I believe that when comparing various errors, they be "normalized" in some way so the comparison is reasonable (e.g., comparing an error over a length of $1$ compared to over a length of a billion doesn't usually provide much useful information). Dividing by $h$ means that the result is what the local error is roughly on average per unit change of $t$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 22:06












  • $begingroup$
    Could you add more context? One area where this is important is with adaptive step sizes. One criterion to control the step size is to hold the "local error per unit step" constant at the level of the provided tolerances, if $e_{loc}(t,h)=C(t)h^{p+1}+...$, one would strive to get $e_{loc}(t,h)/h=C(t)h^{p}=epsilon$ so that the global error computes approximately as $sum C(t_k)h_k^{p+1}=epsilonsum h_k=epsilon(t_{final}-t_{init})$.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:11










  • $begingroup$
    Thanks so much for the very quick answers so far. It is a very general context actually. It is used in a book for the analysis of the consistency of a general one step method for solving numerically ode's.
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:51










  • $begingroup$
    Unfortunately, the book is written in german. The author is Karl Strehmel, from his book "Numerik gewöhnlicher Differentialgleichungen" page 24. springer.com/de/book/9783834818478
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:53




















  • $begingroup$
    Welcome to MSE. I believe that when comparing various errors, they be "normalized" in some way so the comparison is reasonable (e.g., comparing an error over a length of $1$ compared to over a length of a billion doesn't usually provide much useful information). Dividing by $h$ means that the result is what the local error is roughly on average per unit change of $t$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 22:06












  • $begingroup$
    Could you add more context? One area where this is important is with adaptive step sizes. One criterion to control the step size is to hold the "local error per unit step" constant at the level of the provided tolerances, if $e_{loc}(t,h)=C(t)h^{p+1}+...$, one would strive to get $e_{loc}(t,h)/h=C(t)h^{p}=epsilon$ so that the global error computes approximately as $sum C(t_k)h_k^{p+1}=epsilonsum h_k=epsilon(t_{final}-t_{init})$.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:11










  • $begingroup$
    Thanks so much for the very quick answers so far. It is a very general context actually. It is used in a book for the analysis of the consistency of a general one step method for solving numerically ode's.
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:51










  • $begingroup$
    Unfortunately, the book is written in german. The author is Karl Strehmel, from his book "Numerik gewöhnlicher Differentialgleichungen" page 24. springer.com/de/book/9783834818478
    $endgroup$
    – philo1707
    Dec 31 '18 at 15:53


















$begingroup$
Welcome to MSE. I believe that when comparing various errors, they be "normalized" in some way so the comparison is reasonable (e.g., comparing an error over a length of $1$ compared to over a length of a billion doesn't usually provide much useful information). Dividing by $h$ means that the result is what the local error is roughly on average per unit change of $t$.
$endgroup$
– John Omielan
Dec 30 '18 at 22:06






$begingroup$
Welcome to MSE. I believe that when comparing various errors, they be "normalized" in some way so the comparison is reasonable (e.g., comparing an error over a length of $1$ compared to over a length of a billion doesn't usually provide much useful information). Dividing by $h$ means that the result is what the local error is roughly on average per unit change of $t$.
$endgroup$
– John Omielan
Dec 30 '18 at 22:06














$begingroup$
Could you add more context? One area where this is important is with adaptive step sizes. One criterion to control the step size is to hold the "local error per unit step" constant at the level of the provided tolerances, if $e_{loc}(t,h)=C(t)h^{p+1}+...$, one would strive to get $e_{loc}(t,h)/h=C(t)h^{p}=epsilon$ so that the global error computes approximately as $sum C(t_k)h_k^{p+1}=epsilonsum h_k=epsilon(t_{final}-t_{init})$.
$endgroup$
– LutzL
Dec 31 '18 at 12:11




$begingroup$
Could you add more context? One area where this is important is with adaptive step sizes. One criterion to control the step size is to hold the "local error per unit step" constant at the level of the provided tolerances, if $e_{loc}(t,h)=C(t)h^{p+1}+...$, one would strive to get $e_{loc}(t,h)/h=C(t)h^{p}=epsilon$ so that the global error computes approximately as $sum C(t_k)h_k^{p+1}=epsilonsum h_k=epsilon(t_{final}-t_{init})$.
$endgroup$
– LutzL
Dec 31 '18 at 12:11












$begingroup$
Thanks so much for the very quick answers so far. It is a very general context actually. It is used in a book for the analysis of the consistency of a general one step method for solving numerically ode's.
$endgroup$
– philo1707
Dec 31 '18 at 15:51




$begingroup$
Thanks so much for the very quick answers so far. It is a very general context actually. It is used in a book for the analysis of the consistency of a general one step method for solving numerically ode's.
$endgroup$
– philo1707
Dec 31 '18 at 15:51












$begingroup$
Unfortunately, the book is written in german. The author is Karl Strehmel, from his book "Numerik gewöhnlicher Differentialgleichungen" page 24. springer.com/de/book/9783834818478
$endgroup$
– philo1707
Dec 31 '18 at 15:53






$begingroup$
Unfortunately, the book is written in german. The author is Karl Strehmel, from his book "Numerik gewöhnlicher Differentialgleichungen" page 24. springer.com/de/book/9783834818478
$endgroup$
– philo1707
Dec 31 '18 at 15:53












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The local error measures how much error you make taking one step of length $h$. If you cut the step size in half you need to take twice as many steps to cover a given distance, so unless the local error is cut in half as well you are doing worse to take the shorter steps. Dividing by the step size normalizes this, so the local error per unit step is an approximation of the error you make while you advance the independent variable by $1$ regardless of the step size you take.






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    $begingroup$

    The local error measures how much error you make taking one step of length $h$. If you cut the step size in half you need to take twice as many steps to cover a given distance, so unless the local error is cut in half as well you are doing worse to take the shorter steps. Dividing by the step size normalizes this, so the local error per unit step is an approximation of the error you make while you advance the independent variable by $1$ regardless of the step size you take.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The local error measures how much error you make taking one step of length $h$. If you cut the step size in half you need to take twice as many steps to cover a given distance, so unless the local error is cut in half as well you are doing worse to take the shorter steps. Dividing by the step size normalizes this, so the local error per unit step is an approximation of the error you make while you advance the independent variable by $1$ regardless of the step size you take.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The local error measures how much error you make taking one step of length $h$. If you cut the step size in half you need to take twice as many steps to cover a given distance, so unless the local error is cut in half as well you are doing worse to take the shorter steps. Dividing by the step size normalizes this, so the local error per unit step is an approximation of the error you make while you advance the independent variable by $1$ regardless of the step size you take.






        share|cite|improve this answer









        $endgroup$



        The local error measures how much error you make taking one step of length $h$. If you cut the step size in half you need to take twice as many steps to cover a given distance, so unless the local error is cut in half as well you are doing worse to take the shorter steps. Dividing by the step size normalizes this, so the local error per unit step is an approximation of the error you make while you advance the independent variable by $1$ regardless of the step size you take.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 0:08









        Ross MillikanRoss Millikan

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