How to solve this differential equation numerically in Python?












2












$begingroup$


I am trying to solve a differential equation in Python:
$$y'' + 2frac{y'}{x} + (1 - frac{e^{-x}}{x} - frac{l(l+1)}{x^2})y = 0$$
I have initial conditions at $x=0$ as:
$$y(0) = a$$
$$y'(0) = b$$
$a$ and $b$ are some known constants and they will be constrained by $l$. I tried using Euler forward method but solution was unstable. I tried Runge-Kutta 2nd order method but again the solution was unstable. What method should I use so that I will get a stable solution?






ordinary-differential-equations numerical-methods python







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The "industry standard" is Runge-Kutta 4. Have you tried that?
    $endgroup$
    – Arthur
    May 29 '18 at 10:02












  • $begingroup$
    @Arthur I haven't tried. I just want to ask, because I am new in this,, that does that work in such cases?
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 10:20










  • $begingroup$
    See dopri5 option at docs.scipy.org/doc/scipy-0.13.0/reference/generated/…
    $endgroup$
    – J.G.
    May 29 '18 at 10:32






  • 1




    $begingroup$
    You asked about the same differential equation in MSE question 2785298 but using Mathematica instead.
    $endgroup$
    – Somos
    May 29 '18 at 11:52










  • $begingroup$
    @Somos Yeah, and it worked on mathematica. But, I wanted to know that how would I implement it on python. Because mathematica doesn't show the method it used.
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 16:06


















2












$begingroup$


I am trying to solve a differential equation in Python:
$$y'' + 2frac{y'}{x} + (1 - frac{e^{-x}}{x} - frac{l(l+1)}{x^2})y = 0$$
I have initial conditions at $x=0$ as:
$$y(0) = a$$
$$y'(0) = b$$
$a$ and $b$ are some known constants and they will be constrained by $l$. I tried using Euler forward method but solution was unstable. I tried Runge-Kutta 2nd order method but again the solution was unstable. What method should I use so that I will get a stable solution?






ordinary-differential-equations numerical-methods python







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The "industry standard" is Runge-Kutta 4. Have you tried that?
    $endgroup$
    – Arthur
    May 29 '18 at 10:02












  • $begingroup$
    @Arthur I haven't tried. I just want to ask, because I am new in this,, that does that work in such cases?
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 10:20










  • $begingroup$
    See dopri5 option at docs.scipy.org/doc/scipy-0.13.0/reference/generated/…
    $endgroup$
    – J.G.
    May 29 '18 at 10:32






  • 1




    $begingroup$
    You asked about the same differential equation in MSE question 2785298 but using Mathematica instead.
    $endgroup$
    – Somos
    May 29 '18 at 11:52










  • $begingroup$
    @Somos Yeah, and it worked on mathematica. But, I wanted to know that how would I implement it on python. Because mathematica doesn't show the method it used.
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 16:06
















2












2








2


2



$begingroup$


I am trying to solve a differential equation in Python:
$$y'' + 2frac{y'}{x} + (1 - frac{e^{-x}}{x} - frac{l(l+1)}{x^2})y = 0$$
I have initial conditions at $x=0$ as:
$$y(0) = a$$
$$y'(0) = b$$
$a$ and $b$ are some known constants and they will be constrained by $l$. I tried using Euler forward method but solution was unstable. I tried Runge-Kutta 2nd order method but again the solution was unstable. What method should I use so that I will get a stable solution?






ordinary-differential-equations numerical-methods python







share|cite|improve this question









$endgroup$




I am trying to solve a differential equation in Python:
$$y'' + 2frac{y'}{x} + (1 - frac{e^{-x}}{x} - frac{l(l+1)}{x^2})y = 0$$
I have initial conditions at $x=0$ as:
$$y(0) = a$$
$$y'(0) = b$$
$a$ and $b$ are some known constants and they will be constrained by $l$. I tried using Euler forward method but solution was unstable. I tried Runge-Kutta 2nd order method but again the solution was unstable. What method should I use so that I will get a stable solution?






ordinary-differential-equations numerical-methods python




ordinary-differential-equations numerical-methods python






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 29 '18 at 9:57









Yaman SanghaviYaman Sanghavi

1876




1876








  • 1




    $begingroup$
    The "industry standard" is Runge-Kutta 4. Have you tried that?
    $endgroup$
    – Arthur
    May 29 '18 at 10:02












  • $begingroup$
    @Arthur I haven't tried. I just want to ask, because I am new in this,, that does that work in such cases?
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 10:20










  • $begingroup$
    See dopri5 option at docs.scipy.org/doc/scipy-0.13.0/reference/generated/…
    $endgroup$
    – J.G.
    May 29 '18 at 10:32






  • 1




    $begingroup$
    You asked about the same differential equation in MSE question 2785298 but using Mathematica instead.
    $endgroup$
    – Somos
    May 29 '18 at 11:52










  • $begingroup$
    @Somos Yeah, and it worked on mathematica. But, I wanted to know that how would I implement it on python. Because mathematica doesn't show the method it used.
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 16:06
















  • 1




    $begingroup$
    The "industry standard" is Runge-Kutta 4. Have you tried that?
    $endgroup$
    – Arthur
    May 29 '18 at 10:02












  • $begingroup$
    @Arthur I haven't tried. I just want to ask, because I am new in this,, that does that work in such cases?
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 10:20










  • $begingroup$
    See dopri5 option at docs.scipy.org/doc/scipy-0.13.0/reference/generated/…
    $endgroup$
    – J.G.
    May 29 '18 at 10:32






  • 1




    $begingroup$
    You asked about the same differential equation in MSE question 2785298 but using Mathematica instead.
    $endgroup$
    – Somos
    May 29 '18 at 11:52










  • $begingroup$
    @Somos Yeah, and it worked on mathematica. But, I wanted to know that how would I implement it on python. Because mathematica doesn't show the method it used.
    $endgroup$
    – Yaman Sanghavi
    May 29 '18 at 16:06










1




1




$begingroup$
The "industry standard" is Runge-Kutta 4. Have you tried that?
$endgroup$
– Arthur
May 29 '18 at 10:02






$begingroup$
The "industry standard" is Runge-Kutta 4. Have you tried that?
$endgroup$
– Arthur
May 29 '18 at 10:02














$begingroup$
@Arthur I haven't tried. I just want to ask, because I am new in this,, that does that work in such cases?
$endgroup$
– Yaman Sanghavi
May 29 '18 at 10:20




$begingroup$
@Arthur I haven't tried. I just want to ask, because I am new in this,, that does that work in such cases?
$endgroup$
– Yaman Sanghavi
May 29 '18 at 10:20












$begingroup$
See dopri5 option at docs.scipy.org/doc/scipy-0.13.0/reference/generated/…
$endgroup$
– J.G.
May 29 '18 at 10:32




$begingroup$
See dopri5 option at docs.scipy.org/doc/scipy-0.13.0/reference/generated/…
$endgroup$
– J.G.
May 29 '18 at 10:32




1




1




$begingroup$
You asked about the same differential equation in MSE question 2785298 but using Mathematica instead.
$endgroup$
– Somos
May 29 '18 at 11:52




$begingroup$
You asked about the same differential equation in MSE question 2785298 but using Mathematica instead.
$endgroup$
– Somos
May 29 '18 at 11:52












$begingroup$
@Somos Yeah, and it worked on mathematica. But, I wanted to know that how would I implement it on python. Because mathematica doesn't show the method it used.
$endgroup$
– Yaman Sanghavi
May 29 '18 at 16:06






$begingroup$
@Somos Yeah, and it worked on mathematica. But, I wanted to know that how would I implement it on python. Because mathematica doesn't show the method it used.
$endgroup$
– Yaman Sanghavi
May 29 '18 at 16:06












1 Answer
1






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oldest

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3












$begingroup$

This equation is stiff for $xll1$ because the dominant term is the first derivative (you can check the coefficients w.r.t. the first one, which is one).



The Runge-Kutta methods are explicit and they are not suitable to solve these kind of stiff equations (however, for $xgg 1$ they fit perfectly).



You should go for implicit methods. Try the simplest one: backward Euler method.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is not only stiff, but downright singular at $x=0$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 14:39











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









3












$begingroup$

This equation is stiff for $xll1$ because the dominant term is the first derivative (you can check the coefficients w.r.t. the first one, which is one).



The Runge-Kutta methods are explicit and they are not suitable to solve these kind of stiff equations (however, for $xgg 1$ they fit perfectly).



You should go for implicit methods. Try the simplest one: backward Euler method.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is not only stiff, but downright singular at $x=0$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 14:39
















3












$begingroup$

This equation is stiff for $xll1$ because the dominant term is the first derivative (you can check the coefficients w.r.t. the first one, which is one).



The Runge-Kutta methods are explicit and they are not suitable to solve these kind of stiff equations (however, for $xgg 1$ they fit perfectly).



You should go for implicit methods. Try the simplest one: backward Euler method.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is not only stiff, but downright singular at $x=0$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 14:39














3












3








3





$begingroup$

This equation is stiff for $xll1$ because the dominant term is the first derivative (you can check the coefficients w.r.t. the first one, which is one).



The Runge-Kutta methods are explicit and they are not suitable to solve these kind of stiff equations (however, for $xgg 1$ they fit perfectly).



You should go for implicit methods. Try the simplest one: backward Euler method.






share|cite|improve this answer









$endgroup$



This equation is stiff for $xll1$ because the dominant term is the first derivative (you can check the coefficients w.r.t. the first one, which is one).



The Runge-Kutta methods are explicit and they are not suitable to solve these kind of stiff equations (however, for $xgg 1$ they fit perfectly).



You should go for implicit methods. Try the simplest one: backward Euler method.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered May 29 '18 at 10:28









HBRHBR

1,69359




1,69359












  • $begingroup$
    It is not only stiff, but downright singular at $x=0$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 14:39


















  • $begingroup$
    It is not only stiff, but downright singular at $x=0$.
    $endgroup$
    – LutzL
    Dec 10 '18 at 14:39
















$begingroup$
It is not only stiff, but downright singular at $x=0$.
$endgroup$
– LutzL
Dec 10 '18 at 14:39




$begingroup$
It is not only stiff, but downright singular at $x=0$.
$endgroup$
– LutzL
Dec 10 '18 at 14:39


















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