1
$\begingroup$

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:

  1. Can the equation be solved analytically for any (or all) given sets of BC's?
  2. 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?
  3. 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?
$\endgroup$
  • $\begingroup$ NOTE : There was a mistake at the end of my answer. Now fixed. $\endgroup$ – JJacquelin Jan 31 at 14:51
1
$\begingroup$

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 : $$ x^2y''+\left((4b-1)x^3+(2a-1)x\right)y'+\left(2b(2b-1)x^4+a(4b-1)x^2+(a-1)^2 \right)y=0$$ By inspection, one see that the equation can be reduced to a Bessel form with particular values of $a$ and $b$. $$b=\frac14\quad\text{and}\quad a=1$$ leading to the ODE of Bessel kind : $$y''+\frac{1}{x}y'-\frac{x^2}{4}y=0$$ $$y=c_1I_0\left(\frac{x^2}{4}\right)+c_2K_0\left(\frac{x^2}{4}\right)$$ Modified Bessel functions of first and second kind and order $0$. $$f(x)=xe^{x^2/4}\left(c_1I_0\left(\frac{x^2}{4}\right)+c_2K_0\left(\frac{x^2}{4}\right) \right)$$

Condition $f(0)=f_0$ :

Using the series expansion of the Bessel functions around $0$ : $$xe^{x^2/4}I_0\left(x^2/4\right)=x-\frac{x^3}{4}+O(x^5) $$ $$xe^{x^2/4}K_0\left(x^2/4\right)=-2x\ln(x)+O(x) $$ Both tend to $0$ for $x\to 0$. As a consequence, $$\text{the problem has no real solution if } f_0\neq 0$$ If $f_0=0$ the above general solution satisfies the condition any $c1,c_2$.

\

Condition $f(\pm\infty)=0$ : $\qquad\color{red}{\text{Mistake corrected.}}$

Using the asymptotic expansion of the Bessel functions : $$xe^{x^2/4}I_0\left(x^2/4\right)\sim \sqrt{\frac{2}{\pi}}e^{x^2/2}$$ $$xe^{x^2/4}K_0\left(x^2/4\right)\sim \sqrt{2\pi}+O\left(x^{-2}\right)$$ The first tends to $\infty$ for $x\to\pm\infty$ which implies $c_1=0$

The second tends to $\sqrt{2\pi}$ for $x\to\pm\infty$ which implies $c_2=0$

\

Final result according to the specified conditions :

If $f_0\neq0$ 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/

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.