Numerical integration of first order differential equation & fitting to experimental data












1












$begingroup$


We are working on phase transformation kinetics in Advanced High Strength steels and we are trying to fit our experimental data to the following differential for recristallization, by Kulakov et. al 2014. The variable model parameters are $b_0$, n, Q .



List item



We put this model into an Excel sheet and used both RK4 and Forward Euler, however this does not seem to yield the results.



Possible causes of problem:




  • Type-o's in the sheet. [Have been ruled out after 2 days I guess]

  • Incorrect use of constant $b_0$, I do not know what to do with the part withing square brackets of this variable: $1.1*10^{29}*[s^{-1.7}]$

  • Wrong application of numerical integration methods.


I used my textbook for the RK4 method and this little tutorial for the Forward Euler: http://bmi.bmt.tue.nl/sysbio/Education/Excel_Euler_simulation.pdf



If anyone can give me any tips ( or tell me how to upload my spreadsheet to let someone take a look ) please help me out. This is a rather big bump in the road towards finishing my bachelor thesis =]



Kind regards and thanks in advance!






ordinary-differential-equations numerical-methods data-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    1. Perhaps this is a silly question, but I have to ask it: does your experimental data correspond to a recrystallization? 2. Your DE there has no $t$ on the RHS, so far as I can see. (Is the $T$ in the equation for $b$ really upper-case? Or is it lower-case? Mathematics is case-sensitive!) That means it's theoretically separable. Have you tried integrating it that way? It actually looks doable.
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:24










  • $begingroup$
    Could you write your computations in some code or pseudo-code form into a code environment? It is not clear from the image what variables are connected, the same, ...
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:24






  • 1




    $begingroup$
    Is it $left(lnleft(dfrac{1}{1-X}right)right)^{(n-1)/n}$ or $lnleft(left(frac{1}{1-X}right)^{(n-1)/n}right)?$
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:41






  • 1




    $begingroup$
    @AdrianKeister : Looking at the original formula, it is more reasonable to assume that it is the first form.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:44








  • 1




    $begingroup$
    I think $s$ is the unit, seconds. This fits with $b^{1/n}$ having the unit $1/s$. In whatever way it makes sense to have fractional powers of units.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:56


















1












$begingroup$


We are working on phase transformation kinetics in Advanced High Strength steels and we are trying to fit our experimental data to the following differential for recristallization, by Kulakov et. al 2014. The variable model parameters are $b_0$, n, Q .



List item



We put this model into an Excel sheet and used both RK4 and Forward Euler, however this does not seem to yield the results.



Possible causes of problem:




  • Type-o's in the sheet. [Have been ruled out after 2 days I guess]

  • Incorrect use of constant $b_0$, I do not know what to do with the part withing square brackets of this variable: $1.1*10^{29}*[s^{-1.7}]$

  • Wrong application of numerical integration methods.


I used my textbook for the RK4 method and this little tutorial for the Forward Euler: http://bmi.bmt.tue.nl/sysbio/Education/Excel_Euler_simulation.pdf



If anyone can give me any tips ( or tell me how to upload my spreadsheet to let someone take a look ) please help me out. This is a rather big bump in the road towards finishing my bachelor thesis =]



Kind regards and thanks in advance!






ordinary-differential-equations numerical-methods data-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    1. Perhaps this is a silly question, but I have to ask it: does your experimental data correspond to a recrystallization? 2. Your DE there has no $t$ on the RHS, so far as I can see. (Is the $T$ in the equation for $b$ really upper-case? Or is it lower-case? Mathematics is case-sensitive!) That means it's theoretically separable. Have you tried integrating it that way? It actually looks doable.
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:24










  • $begingroup$
    Could you write your computations in some code or pseudo-code form into a code environment? It is not clear from the image what variables are connected, the same, ...
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:24






  • 1




    $begingroup$
    Is it $left(lnleft(dfrac{1}{1-X}right)right)^{(n-1)/n}$ or $lnleft(left(frac{1}{1-X}right)^{(n-1)/n}right)?$
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:41






  • 1




    $begingroup$
    @AdrianKeister : Looking at the original formula, it is more reasonable to assume that it is the first form.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:44








  • 1




    $begingroup$
    I think $s$ is the unit, seconds. This fits with $b^{1/n}$ having the unit $1/s$. In whatever way it makes sense to have fractional powers of units.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:56
















1












1








1





$begingroup$


We are working on phase transformation kinetics in Advanced High Strength steels and we are trying to fit our experimental data to the following differential for recristallization, by Kulakov et. al 2014. The variable model parameters are $b_0$, n, Q .



List item



We put this model into an Excel sheet and used both RK4 and Forward Euler, however this does not seem to yield the results.



Possible causes of problem:




  • Type-o's in the sheet. [Have been ruled out after 2 days I guess]

  • Incorrect use of constant $b_0$, I do not know what to do with the part withing square brackets of this variable: $1.1*10^{29}*[s^{-1.7}]$

  • Wrong application of numerical integration methods.


I used my textbook for the RK4 method and this little tutorial for the Forward Euler: http://bmi.bmt.tue.nl/sysbio/Education/Excel_Euler_simulation.pdf



If anyone can give me any tips ( or tell me how to upload my spreadsheet to let someone take a look ) please help me out. This is a rather big bump in the road towards finishing my bachelor thesis =]



Kind regards and thanks in advance!






ordinary-differential-equations numerical-methods data-analysis







share|cite|improve this question









$endgroup$




We are working on phase transformation kinetics in Advanced High Strength steels and we are trying to fit our experimental data to the following differential for recristallization, by Kulakov et. al 2014. The variable model parameters are $b_0$, n, Q .



List item



We put this model into an Excel sheet and used both RK4 and Forward Euler, however this does not seem to yield the results.



Possible causes of problem:




  • Type-o's in the sheet. [Have been ruled out after 2 days I guess]

  • Incorrect use of constant $b_0$, I do not know what to do with the part withing square brackets of this variable: $1.1*10^{29}*[s^{-1.7}]$

  • Wrong application of numerical integration methods.


I used my textbook for the RK4 method and this little tutorial for the Forward Euler: http://bmi.bmt.tue.nl/sysbio/Education/Excel_Euler_simulation.pdf



If anyone can give me any tips ( or tell me how to upload my spreadsheet to let someone take a look ) please help me out. This is a rather big bump in the road towards finishing my bachelor thesis =]



Kind regards and thanks in advance!






ordinary-differential-equations numerical-methods data-analysis




ordinary-differential-equations numerical-methods data-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 15:14









SimonisSimonis

233




233












  • $begingroup$
    1. Perhaps this is a silly question, but I have to ask it: does your experimental data correspond to a recrystallization? 2. Your DE there has no $t$ on the RHS, so far as I can see. (Is the $T$ in the equation for $b$ really upper-case? Or is it lower-case? Mathematics is case-sensitive!) That means it's theoretically separable. Have you tried integrating it that way? It actually looks doable.
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:24










  • $begingroup$
    Could you write your computations in some code or pseudo-code form into a code environment? It is not clear from the image what variables are connected, the same, ...
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:24






  • 1




    $begingroup$
    Is it $left(lnleft(dfrac{1}{1-X}right)right)^{(n-1)/n}$ or $lnleft(left(frac{1}{1-X}right)^{(n-1)/n}right)?$
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:41






  • 1




    $begingroup$
    @AdrianKeister : Looking at the original formula, it is more reasonable to assume that it is the first form.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:44








  • 1




    $begingroup$
    I think $s$ is the unit, seconds. This fits with $b^{1/n}$ having the unit $1/s$. In whatever way it makes sense to have fractional powers of units.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:56




















  • $begingroup$
    1. Perhaps this is a silly question, but I have to ask it: does your experimental data correspond to a recrystallization? 2. Your DE there has no $t$ on the RHS, so far as I can see. (Is the $T$ in the equation for $b$ really upper-case? Or is it lower-case? Mathematics is case-sensitive!) That means it's theoretically separable. Have you tried integrating it that way? It actually looks doable.
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:24










  • $begingroup$
    Could you write your computations in some code or pseudo-code form into a code environment? It is not clear from the image what variables are connected, the same, ...
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:24






  • 1




    $begingroup$
    Is it $left(lnleft(dfrac{1}{1-X}right)right)^{(n-1)/n}$ or $lnleft(left(frac{1}{1-X}right)^{(n-1)/n}right)?$
    $endgroup$
    – Adrian Keister
    Dec 14 '18 at 15:41






  • 1




    $begingroup$
    @AdrianKeister : Looking at the original formula, it is more reasonable to assume that it is the first form.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:44








  • 1




    $begingroup$
    I think $s$ is the unit, seconds. This fits with $b^{1/n}$ having the unit $1/s$. In whatever way it makes sense to have fractional powers of units.
    $endgroup$
    – LutzL
    Dec 14 '18 at 15:56


















$begingroup$
1. Perhaps this is a silly question, but I have to ask it: does your experimental data correspond to a recrystallization? 2. Your DE there has no $t$ on the RHS, so far as I can see. (Is the $T$ in the equation for $b$ really upper-case? Or is it lower-case? Mathematics is case-sensitive!) That means it's theoretically separable. Have you tried integrating it that way? It actually looks doable.
$endgroup$
– Adrian Keister
Dec 14 '18 at 15:24




$begingroup$
1. Perhaps this is a silly question, but I have to ask it: does your experimental data correspond to a recrystallization? 2. Your DE there has no $t$ on the RHS, so far as I can see. (Is the $T$ in the equation for $b$ really upper-case? Or is it lower-case? Mathematics is case-sensitive!) That means it's theoretically separable. Have you tried integrating it that way? It actually looks doable.
$endgroup$
– Adrian Keister
Dec 14 '18 at 15:24












$begingroup$
Could you write your computations in some code or pseudo-code form into a code environment? It is not clear from the image what variables are connected, the same, ...
$endgroup$
– LutzL
Dec 14 '18 at 15:24




$begingroup$
Could you write your computations in some code or pseudo-code form into a code environment? It is not clear from the image what variables are connected, the same, ...
$endgroup$
– LutzL
Dec 14 '18 at 15:24




1




1




$begingroup$
Is it $left(lnleft(dfrac{1}{1-X}right)right)^{(n-1)/n}$ or $lnleft(left(frac{1}{1-X}right)^{(n-1)/n}right)?$
$endgroup$
– Adrian Keister
Dec 14 '18 at 15:41




$begingroup$
Is it $left(lnleft(dfrac{1}{1-X}right)right)^{(n-1)/n}$ or $lnleft(left(frac{1}{1-X}right)^{(n-1)/n}right)?$
$endgroup$
– Adrian Keister
Dec 14 '18 at 15:41




1




1




$begingroup$
@AdrianKeister : Looking at the original formula, it is more reasonable to assume that it is the first form.
$endgroup$
– LutzL
Dec 14 '18 at 15:44






$begingroup$
@AdrianKeister : Looking at the original formula, it is more reasonable to assume that it is the first form.
$endgroup$
– LutzL
Dec 14 '18 at 15:44






1




1




$begingroup$
I think $s$ is the unit, seconds. This fits with $b^{1/n}$ having the unit $1/s$. In whatever way it makes sense to have fractional powers of units.
$endgroup$
– LutzL
Dec 14 '18 at 15:56






$begingroup$
I think $s$ is the unit, seconds. This fits with $b^{1/n}$ having the unit $1/s$. In whatever way it makes sense to have fractional powers of units.
$endgroup$
– LutzL
Dec 14 '18 at 15:56












1 Answer
1






active

oldest

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1












$begingroup$

Assuming the model is correct for your experimental data, it integrates directly:
begin{align*}
frac{df}{dt}&=left[n(1-f)lnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n} \
-frac{df}{f-1}&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n},dt \
-(ln|f-1|-ln|f_0-1|)&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
lnleft(frac{|f-1|}{|f_0-1|}right)&=-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
frac{|f-1|}{|f_0-1|}&=expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right} \
|f-1|&=|f_0-1|expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right}.
end{align*}



Other comments: your $n, b_0,$ and $Q$ are by no means independent. Depending on whether you're dealing with Ferrite-pearlite, Ferrite-bainite-pearlite, or Martensite, you'll have a relationship among those parameters that restricts the degrees of freedom considerably. It appears to me that you really only have one parameter: $Q$. So you can set up Excel's Solver routine to minimize the sum of squares of the errors between your experimental data and your model values by changing $Q$.






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    1 Answer
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    active

    oldest

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    1












    $begingroup$

    Assuming the model is correct for your experimental data, it integrates directly:
    begin{align*}
    frac{df}{dt}&=left[n(1-f)lnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n} \
    -frac{df}{f-1}&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n},dt \
    -(ln|f-1|-ln|f_0-1|)&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
    lnleft(frac{|f-1|}{|f_0-1|}right)&=-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
    frac{|f-1|}{|f_0-1|}&=expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right} \
    |f-1|&=|f_0-1|expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right}.
    end{align*}



    Other comments: your $n, b_0,$ and $Q$ are by no means independent. Depending on whether you're dealing with Ferrite-pearlite, Ferrite-bainite-pearlite, or Martensite, you'll have a relationship among those parameters that restricts the degrees of freedom considerably. It appears to me that you really only have one parameter: $Q$. So you can set up Excel's Solver routine to minimize the sum of squares of the errors between your experimental data and your model values by changing $Q$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Assuming the model is correct for your experimental data, it integrates directly:
      begin{align*}
      frac{df}{dt}&=left[n(1-f)lnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n} \
      -frac{df}{f-1}&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n},dt \
      -(ln|f-1|-ln|f_0-1|)&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
      lnleft(frac{|f-1|}{|f_0-1|}right)&=-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
      frac{|f-1|}{|f_0-1|}&=expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right} \
      |f-1|&=|f_0-1|expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right}.
      end{align*}



      Other comments: your $n, b_0,$ and $Q$ are by no means independent. Depending on whether you're dealing with Ferrite-pearlite, Ferrite-bainite-pearlite, or Martensite, you'll have a relationship among those parameters that restricts the degrees of freedom considerably. It appears to me that you really only have one parameter: $Q$. So you can set up Excel's Solver routine to minimize the sum of squares of the errors between your experimental data and your model values by changing $Q$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Assuming the model is correct for your experimental data, it integrates directly:
        begin{align*}
        frac{df}{dt}&=left[n(1-f)lnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n} \
        -frac{df}{f-1}&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n},dt \
        -(ln|f-1|-ln|f_0-1|)&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
        lnleft(frac{|f-1|}{|f_0-1|}right)&=-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
        frac{|f-1|}{|f_0-1|}&=expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right} \
        |f-1|&=|f_0-1|expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right}.
        end{align*}



        Other comments: your $n, b_0,$ and $Q$ are by no means independent. Depending on whether you're dealing with Ferrite-pearlite, Ferrite-bainite-pearlite, or Martensite, you'll have a relationship among those parameters that restricts the degrees of freedom considerably. It appears to me that you really only have one parameter: $Q$. So you can set up Excel's Solver routine to minimize the sum of squares of the errors between your experimental data and your model values by changing $Q$.






        share|cite|improve this answer











        $endgroup$



        Assuming the model is correct for your experimental data, it integrates directly:
        begin{align*}
        frac{df}{dt}&=left[n(1-f)lnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n} \
        -frac{df}{f-1}&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n},dt \
        -(ln|f-1|-ln|f_0-1|)&=left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
        lnleft(frac{|f-1|}{|f_0-1|}right)&=-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0) \
        frac{|f-1|}{|f_0-1|}&=expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right} \
        |f-1|&=|f_0-1|expleft{-left[nlnleft(frac{1}{1-X}right)^{(n-1)/n}right]b^{1/n}(t-t_0)right}.
        end{align*}



        Other comments: your $n, b_0,$ and $Q$ are by no means independent. Depending on whether you're dealing with Ferrite-pearlite, Ferrite-bainite-pearlite, or Martensite, you'll have a relationship among those parameters that restricts the degrees of freedom considerably. It appears to me that you really only have one parameter: $Q$. So you can set up Excel's Solver routine to minimize the sum of squares of the errors between your experimental data and your model values by changing $Q$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 16:26

























        answered Dec 14 '18 at 15:55









        Adrian KeisterAdrian Keister

        5,27371933




        5,27371933






























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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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