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How to find the exact solution of a Sturm Liouville form, 2nd order ODE?
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I have a second order ordinary differential equation reducible to Sturm Liovelle form. The equation is given by
$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$
and boundary conditions are :
$f(0) =f_0; f(pm infty) = 0$
The equation can be reduced to the Sturm Liouville form as
$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $
(Mathematica gives solution in terms of Meijer functions). Now for a boundary condition $f(0)=0$ alone, we get a closed form analytical solution given by
$f(x) = c_1 x e^{frac{-x^2}{4}}left(I_0(frac{x^2}{4})+I_1(frac{x^2}{4})right)$ where $I_1$ and $I_2$ are modified Bessel functions.
This solution can be obtained by Frobenius method. My queries are as follows:
- Can the equation be solved analytically for any (or all) given sets of BC's?
- If a second order ODE can be reduced to a Strum Liouville form, is there a method for finding analytical solutions, perhaps limited to a certain class of problems?
- Mathematica gives the complete solution, with unspecified boundary conditions as a combination of the solution for boundary condition $f(0)=0$, and another involving Meijer function. I know that is possible to find a representation of the second solution of a second order linear ODE if one solution is available. But is it possible to find solutions of such equations in terms of Meijer functions directly?
ordinary-differential-equations laplace-transform bessel-functions sturm-liouville
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
$endgroup$
add a comment |
$begingroup$
I have a second order ordinary differential equation reducible to Sturm Liovelle form. The equation is given by
$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$
and boundary conditions are :
$f(0) =f_0; f(pm infty) = 0$
The equation can be reduced to the Sturm Liouville form as
$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $
(Mathematica gives solution in terms of Meijer functions). Now for a boundary condition $f(0)=0$ alone, we get a closed form analytical solution given by
$f(x) = c_1 x e^{frac{-x^2}{4}}left(I_0(frac{x^2}{4})+I_1(frac{x^2}{4})right)$ where $I_1$ and $I_2$ are modified Bessel functions.
This solution can be obtained by Frobenius method. My queries are as follows:
- Can the equation be solved analytically for any (or all) given sets of BC's?
- If a second order ODE can be reduced to a Strum Liouville form, is there a method for finding analytical solutions, perhaps limited to a certain class of problems?
- Mathematica gives the complete solution, with unspecified boundary conditions as a combination of the solution for boundary condition $f(0)=0$, and another involving Meijer function. I know that is possible to find a representation of the second solution of a second order linear ODE if one solution is available. But is it possible to find solutions of such equations in terms of Meijer functions directly?
ordinary-differential-equations laplace-transform bessel-functions sturm-liouville
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
$endgroup$
$begingroup$
NOTE : There was a mistake at the end of my answer. Now fixed.
$endgroup$
– JJacquelin
Jan 31 at 14:51
add a comment |
$begingroup$
I have a second order ordinary differential equation reducible to Sturm Liovelle form. The equation is given by
$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$
and boundary conditions are :
$f(0) =f_0; f(pm infty) = 0$
The equation can be reduced to the Sturm Liouville form as
$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $
(Mathematica gives solution in terms of Meijer functions). Now for a boundary condition $f(0)=0$ alone, we get a closed form analytical solution given by
$f(x) = c_1 x e^{frac{-x^2}{4}}left(I_0(frac{x^2}{4})+I_1(frac{x^2}{4})right)$ where $I_1$ and $I_2$ are modified Bessel functions.
This solution can be obtained by Frobenius method. My queries are as follows:
- Can the equation be solved analytically for any (or all) given sets of BC's?
- If a second order ODE can be reduced to a Strum Liouville form, is there a method for finding analytical solutions, perhaps limited to a certain class of problems?
- Mathematica gives the complete solution, with unspecified boundary conditions as a combination of the solution for boundary condition $f(0)=0$, and another involving Meijer function. I know that is possible to find a representation of the second solution of a second order linear ODE if one solution is available. But is it possible to find solutions of such equations in terms of Meijer functions directly?
ordinary-differential-equations laplace-transform bessel-functions sturm-liouville
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
$endgroup$
I have a second order ordinary differential equation reducible to Sturm Liovelle form. The equation is given by
$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$
and boundary conditions are :
$f(0) =f_0; f(pm infty) = 0$
The equation can be reduced to the Sturm Liouville form as
$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $
(Mathematica gives solution in terms of Meijer functions). Now for a boundary condition $f(0)=0$ alone, we get a closed form analytical solution given by
$f(x) = c_1 x e^{frac{-x^2}{4}}left(I_0(frac{x^2}{4})+I_1(frac{x^2}{4})right)$ where $I_1$ and $I_2$ are modified Bessel functions.
This solution can be obtained by Frobenius method. My queries are as follows:
- Can the equation be solved analytically for any (or all) given sets of BC's?
- If a second order ODE can be reduced to a Strum Liouville form, is there a method for finding analytical solutions, perhaps limited to a certain class of problems?
- Mathematica gives the complete solution, with unspecified boundary conditions as a combination of the solution for boundary condition $f(0)=0$, and another involving Meijer function. I know that is possible to find a representation of the second solution of a second order linear ODE if one solution is available. But is it possible to find solutions of such equations in terms of Meijer functions directly?
ordinary-differential-equations laplace-transform bessel-functions sturm-liouville
ordinary-differential-equations laplace-transform bessel-functions sturm-liouville
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
asked Jan 30 at 18:58
AyyappadasAyyappadas
638
638
$begingroup$
NOTE : There was a mistake at the end of my answer. Now fixed.
$endgroup$
– JJacquelin
Jan 31 at 14:51
add a comment |
$begingroup$
NOTE : There was a mistake at the end of my answer. Now fixed.
$endgroup$
– JJacquelin
Jan 31 at 14:51
$begingroup$
NOTE : There was a mistake at the end of my answer. Now fixed.
$endgroup$
– JJacquelin
Jan 31 at 14:51
$begingroup$
NOTE : There was a mistake at the end of my answer. Now fixed.
$endgroup$
– JJacquelin
Jan 31 at 14:51
add a comment |
1 Answer
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Sorry, I will not answer to so broad questions. My answer is limited to solve the ODE :
$$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$$
The Sturm Liouville form :
$$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $$
draw us to try a change of function on the form :
$$f(x)=x^ae^{b,x^2}y(x)$$
Of course, this is not a general method. It is a guess, hoping that the ODE will be transformed to a simpler form.
The transformation is an easy but boring calculus. Editing all the steps would be even more boring. So, going straightaway to the result :
$$ x^2y''+left((4b-1)x^3+(2a-1)xright)y'+left(2b(2b-1)x^4+a(4b-1)x^2+(a-1)^2 right)y=0$$
By inspection, one see that the equation can be reduced to a Bessel form with particular values of $a$ and $b$.
$$b=frac14quadtext{and}quad a=1$$
leading to the ODE of Bessel kind :
$$y''+frac{1}{x}y'-frac{x^2}{4}y=0$$
$$y=c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right)$$
Modified Bessel functions of first and second kind and order $0$.
$$f(x)=xe^{x^2/4}left(c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right) right)$$
Condition $f(0)=f_0$ :
Using the series expansion of the Bessel functions around $0$ :
$$xe^{x^2/4}I_0left(x^2/4right)=x-frac{x^3}{4}+O(x^5) $$
$$xe^{x^2/4}K_0left(x^2/4right)=-2xln(x)+O(x) $$
Both tend to $0$ for $xto 0$. As a consequence,
$$text{the problem has no real solution if } f_0neq 0$$
If $f_0=0$ the above general solution satisfies the condition any $c1,c_2$.
Condition $f(pminfty)=0$ : $qquadcolor{red}{text{Mistake corrected.}}$
Using the asymptotic expansion of the Bessel functions :
$$xe^{x^2/4}I_0left(x^2/4right)sim sqrt{frac{2}{pi}}e^{x^2/2}$$
$$xe^{x^2/4}K_0left(x^2/4right)sim sqrt{2pi}+Oleft(x^{-2}right)$$
The first tends to $infty$ for $xtopminfty$ which implies $c_1=0$
The second tends to $sqrt{2pi}$ for $xtopminfty$ which implies $c_2=0$
Final result according to the specified conditions :
If $f_0neq0$ no solution.
If $f_0=0$ the solution is trivial : $f(x)=0$.
Ref. :
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/01/04/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/02/01/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/01/04/01/02/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/02/01/01/
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
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add a comment |
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$begingroup$
Sorry, I will not answer to so broad questions. My answer is limited to solve the ODE :
$$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$$
The Sturm Liouville form :
$$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $$
draw us to try a change of function on the form :
$$f(x)=x^ae^{b,x^2}y(x)$$
Of course, this is not a general method. It is a guess, hoping that the ODE will be transformed to a simpler form.
The transformation is an easy but boring calculus. Editing all the steps would be even more boring. So, going straightaway to the result :
$$ x^2y''+left((4b-1)x^3+(2a-1)xright)y'+left(2b(2b-1)x^4+a(4b-1)x^2+(a-1)^2 right)y=0$$
By inspection, one see that the equation can be reduced to a Bessel form with particular values of $a$ and $b$.
$$b=frac14quadtext{and}quad a=1$$
leading to the ODE of Bessel kind :
$$y''+frac{1}{x}y'-frac{x^2}{4}y=0$$
$$y=c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right)$$
Modified Bessel functions of first and second kind and order $0$.
$$f(x)=xe^{x^2/4}left(c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right) right)$$
Condition $f(0)=f_0$ :
Using the series expansion of the Bessel functions around $0$ :
$$xe^{x^2/4}I_0left(x^2/4right)=x-frac{x^3}{4}+O(x^5) $$
$$xe^{x^2/4}K_0left(x^2/4right)=-2xln(x)+O(x) $$
Both tend to $0$ for $xto 0$. As a consequence,
$$text{the problem has no real solution if } f_0neq 0$$
If $f_0=0$ the above general solution satisfies the condition any $c1,c_2$.
Condition $f(pminfty)=0$ : $qquadcolor{red}{text{Mistake corrected.}}$
Using the asymptotic expansion of the Bessel functions :
$$xe^{x^2/4}I_0left(x^2/4right)sim sqrt{frac{2}{pi}}e^{x^2/2}$$
$$xe^{x^2/4}K_0left(x^2/4right)sim sqrt{2pi}+Oleft(x^{-2}right)$$
The first tends to $infty$ for $xtopminfty$ which implies $c_1=0$
The second tends to $sqrt{2pi}$ for $xtopminfty$ which implies $c_2=0$
Final result according to the specified conditions :
If $f_0neq0$ no solution.
If $f_0=0$ the solution is trivial : $f(x)=0$.
Ref. :
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/01/04/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/02/01/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/01/04/01/02/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/02/01/01/
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
$endgroup$
add a comment |
$begingroup$
Sorry, I will not answer to so broad questions. My answer is limited to solve the ODE :
$$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$$
The Sturm Liouville form :
$$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $$
draw us to try a change of function on the form :
$$f(x)=x^ae^{b,x^2}y(x)$$
Of course, this is not a general method. It is a guess, hoping that the ODE will be transformed to a simpler form.
The transformation is an easy but boring calculus. Editing all the steps would be even more boring. So, going straightaway to the result :
$$ x^2y''+left((4b-1)x^3+(2a-1)xright)y'+left(2b(2b-1)x^4+a(4b-1)x^2+(a-1)^2 right)y=0$$
By inspection, one see that the equation can be reduced to a Bessel form with particular values of $a$ and $b$.
$$b=frac14quadtext{and}quad a=1$$
leading to the ODE of Bessel kind :
$$y''+frac{1}{x}y'-frac{x^2}{4}y=0$$
$$y=c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right)$$
Modified Bessel functions of first and second kind and order $0$.
$$f(x)=xe^{x^2/4}left(c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right) right)$$
Condition $f(0)=f_0$ :
Using the series expansion of the Bessel functions around $0$ :
$$xe^{x^2/4}I_0left(x^2/4right)=x-frac{x^3}{4}+O(x^5) $$
$$xe^{x^2/4}K_0left(x^2/4right)=-2xln(x)+O(x) $$
Both tend to $0$ for $xto 0$. As a consequence,
$$text{the problem has no real solution if } f_0neq 0$$
If $f_0=0$ the above general solution satisfies the condition any $c1,c_2$.
Condition $f(pminfty)=0$ : $qquadcolor{red}{text{Mistake corrected.}}$
Using the asymptotic expansion of the Bessel functions :
$$xe^{x^2/4}I_0left(x^2/4right)sim sqrt{frac{2}{pi}}e^{x^2/2}$$
$$xe^{x^2/4}K_0left(x^2/4right)sim sqrt{2pi}+Oleft(x^{-2}right)$$
The first tends to $infty$ for $xtopminfty$ which implies $c_1=0$
The second tends to $sqrt{2pi}$ for $xtopminfty$ which implies $c_2=0$
Final result according to the specified conditions :
If $f_0neq0$ no solution.
If $f_0=0$ the solution is trivial : $f(x)=0$.
Ref. :
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/01/04/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/02/01/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/01/04/01/02/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/02/01/01/
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
$endgroup$
add a comment |
$begingroup$
Sorry, I will not answer to so broad questions. My answer is limited to solve the ODE :
$$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$$
The Sturm Liouville form :
$$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $$
draw us to try a change of function on the form :
$$f(x)=x^ae^{b,x^2}y(x)$$
Of course, this is not a general method. It is a guess, hoping that the ODE will be transformed to a simpler form.
The transformation is an easy but boring calculus. Editing all the steps would be even more boring. So, going straightaway to the result :
$$ x^2y''+left((4b-1)x^3+(2a-1)xright)y'+left(2b(2b-1)x^4+a(4b-1)x^2+(a-1)^2 right)y=0$$
By inspection, one see that the equation can be reduced to a Bessel form with particular values of $a$ and $b$.
$$b=frac14quadtext{and}quad a=1$$
leading to the ODE of Bessel kind :
$$y''+frac{1}{x}y'-frac{x^2}{4}y=0$$
$$y=c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right)$$
Modified Bessel functions of first and second kind and order $0$.
$$f(x)=xe^{x^2/4}left(c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right) right)$$
Condition $f(0)=f_0$ :
Using the series expansion of the Bessel functions around $0$ :
$$xe^{x^2/4}I_0left(x^2/4right)=x-frac{x^3}{4}+O(x^5) $$
$$xe^{x^2/4}K_0left(x^2/4right)=-2xln(x)+O(x) $$
Both tend to $0$ for $xto 0$. As a consequence,
$$text{the problem has no real solution if } f_0neq 0$$
If $f_0=0$ the above general solution satisfies the condition any $c1,c_2$.
Condition $f(pminfty)=0$ : $qquadcolor{red}{text{Mistake corrected.}}$
Using the asymptotic expansion of the Bessel functions :
$$xe^{x^2/4}I_0left(x^2/4right)sim sqrt{frac{2}{pi}}e^{x^2/2}$$
$$xe^{x^2/4}K_0left(x^2/4right)sim sqrt{2pi}+Oleft(x^{-2}right)$$
The first tends to $infty$ for $xtopminfty$ which implies $c_1=0$
The second tends to $sqrt{2pi}$ for $xtopminfty$ which implies $c_2=0$
Final result according to the specified conditions :
If $f_0neq0$ no solution.
If $f_0=0$ the solution is trivial : $f(x)=0$.
Ref. :
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/01/04/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/02/01/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/01/04/01/02/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/02/01/01/
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
$endgroup$
Sorry, I will not answer to so broad questions. My answer is limited to solve the ODE :
$$frac{f(x)}{x^2} - (frac{1}{x}+x)f'(x) + f''(x) =0$$
The Sturm Liouville form :
$$frac{d}{dx} (x e^{frac{x^2}{2}}f'(x))=frac{e^{frac{x^2}{2}}}{x}f(x) $$
draw us to try a change of function on the form :
$$f(x)=x^ae^{b,x^2}y(x)$$
Of course, this is not a general method. It is a guess, hoping that the ODE will be transformed to a simpler form.
The transformation is an easy but boring calculus. Editing all the steps would be even more boring. So, going straightaway to the result :
$$ x^2y''+left((4b-1)x^3+(2a-1)xright)y'+left(2b(2b-1)x^4+a(4b-1)x^2+(a-1)^2 right)y=0$$
By inspection, one see that the equation can be reduced to a Bessel form with particular values of $a$ and $b$.
$$b=frac14quadtext{and}quad a=1$$
leading to the ODE of Bessel kind :
$$y''+frac{1}{x}y'-frac{x^2}{4}y=0$$
$$y=c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right)$$
Modified Bessel functions of first and second kind and order $0$.
$$f(x)=xe^{x^2/4}left(c_1I_0left(frac{x^2}{4}right)+c_2K_0left(frac{x^2}{4}right) right)$$
Condition $f(0)=f_0$ :
Using the series expansion of the Bessel functions around $0$ :
$$xe^{x^2/4}I_0left(x^2/4right)=x-frac{x^3}{4}+O(x^5) $$
$$xe^{x^2/4}K_0left(x^2/4right)=-2xln(x)+O(x) $$
Both tend to $0$ for $xto 0$. As a consequence,
$$text{the problem has no real solution if } f_0neq 0$$
If $f_0=0$ the above general solution satisfies the condition any $c1,c_2$.
Condition $f(pminfty)=0$ : $qquadcolor{red}{text{Mistake corrected.}}$
Using the asymptotic expansion of the Bessel functions :
$$xe^{x^2/4}I_0left(x^2/4right)sim sqrt{frac{2}{pi}}e^{x^2/2}$$
$$xe^{x^2/4}K_0left(x^2/4right)sim sqrt{2pi}+Oleft(x^{-2}right)$$
The first tends to $infty$ for $xtopminfty$ which implies $c_1=0$
The second tends to $sqrt{2pi}$ for $xtopminfty$ which implies $c_2=0$
Final result according to the specified conditions :
If $f_0neq0$ no solution.
If $f_0=0$ the solution is trivial : $f(x)=0$.
Ref. :
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/01/04/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/02/01/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/01/04/01/02/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/02/01/01/
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
edited Jan 31 at 14:50
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
answered Jan 31 at 9:34
JJacquelinJJacquelin
45.4k21856
45.4k21856
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$begingroup$
NOTE : There was a mistake at the end of my answer. Now fixed.
$endgroup$
– JJacquelin
Jan 31 at 14:51