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Can we solve analytically (find a closed form solution) the second order ODE

$$x^{\alpha}y^{\prime\prime}=y,\quad x>0$$ where $\alpha\in\,]0,1[$. Consider the conditions $$y(0)=1,\quad y^{\prime}(0)=1$$ This equation appears in a fractional model that describes viscoelasticity properties in certain materials.

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  • $\begingroup$ Fourier transform? $\endgroup$
    – Rushabh Mehta
    Commented Nov 22, 2018 at 3:19
  • $\begingroup$ Could you precise the boundary conditions ? $\endgroup$
    – Claude Leibovici
    Commented Nov 22, 2018 at 3:23
  • $\begingroup$ @ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent. $\endgroup$
    – Medo
    Commented Nov 22, 2018 at 3:28
  • $\begingroup$ Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution. $\endgroup$
    – Medo
    Commented Nov 22, 2018 at 3:29
  • $\begingroup$ @ Mattos. That at best would give a special function. Can we go around that? $\endgroup$
    – Medo
    Commented Nov 22, 2018 at 3:32

2 Answers 2

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Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$ http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html

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Actually this is the solution of $$x^{1-\alpha}y^{\prime\prime}=y$$

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  • $\begingroup$ This does not seem to take into account the boundary conditions. $\endgroup$
    – Claude Leibovici
    Commented Nov 22, 2018 at 4:01
  • $\begingroup$ Thanks a lot for your answer. Can you get the solution for a general $\alpha$ ? $\endgroup$
    – Medo
    Commented Nov 22, 2018 at 7:34
  • $\begingroup$ What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $\frac{2x^{\frac{a+1}2}}{a+1}$ and a leading factor $\sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer. $\endgroup$
    – Claude Leibovici
    Commented Nov 22, 2018 at 7:44
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Too long for a comment.

As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.

Playing with the case where $\alpha=\frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them). $$\left( \begin{array}{ccc} m & n & y(x) \\ 1 & 2 & \, _0F_1\left(;\frac{1}{3};\frac{4 x^{3/2}}{9}\right)+x \, _0F_1\left(;\frac{5}{3};\frac{4 x^{3/2}}{9}\right)\\ 1 & 3 & \, _0F_1\left(;\frac{2}{5};\frac{9 x^{5/3}}{25}\right)+x \, _0F_1\left(;\frac{8}{5};\frac{9 x^{5/3}}{25}\right)\\ 1 & 4 & \, _0F_1\left(;\frac{3}{7};\frac{16 x^{7/4}}{49}\right)+x \, _0F_1\left(;\frac{11}{7};\frac{16 x^{7/4}}{49}\right)\\ 1 & 5 & \, _0F_1\left(;\frac{4}{9};\frac{25 x^{9/5}}{81}\right)+x \, _0F_1\left(;\frac{14}{9};\frac{25 x^{9/5}}{81}\right)\\ 1 & 6 & \, _0F_1\left(;\frac{5}{11};\frac{36 x^{11/6}}{121}\right)+x \, _0F_1\left(;\frac{17}{11};\frac{36 x^{11/6}}{121}\right)\\ 1 & 7 & \, _0F_1\left(;\frac{6}{13};\frac{49 x^{13/7}}{169}\right)+x \, _0F_1\left(;\frac{20}{13};\frac{49 x^{13/7}}{169}\right) \\ 2 & 3 & \, _0F_1\left(;\frac{1}{4};\frac{9 x^{4/3}}{16}\right)+x \, _0F_1\left(;\frac{7}{4};\frac{9 x^{4/3}}{16}\right)\\ 2 & 5 & \, _0F_1\left(;\frac{3}{8};\frac{25 x^{8/5}}{64}\right)+x \, _0F_1\left(;\frac{13}{8};\frac{25 x^{8/5}}{64}\right)\\ 2 & 7 & \, _0F_1\left(;\frac{5}{12};\frac{49 x^{12/7}}{144}\right)+x \, _0F_1\left(;\frac{19}{12};\frac{49 x^{12/7}}{144}\right)\\ 3 & 4 & \, _0F_1\left(;\frac{1}{5};\frac{16 x^{5/4}}{25}\right)+x \, _0F_1\left(;\frac{9}{5};\frac{16 x^{5/4}}{25}\right)\\ 3 & 5 & \, _0F_1\left(;\frac{2}{7};\frac{25 x^{7/5}}{49}\right)+x \, _0F_1\left(;\frac{12}{7};\frac{25 x^{7/5}}{49}\right)\\ 3 & 7 & \, _0F_1\left(;\frac{4}{11};\frac{49 x^{11/7}}{121}\right)+x \, _0F_1\left(;\frac{18}{11};\frac{49 x^{11/7}}{121}\right)\\ 3 & 8 & \, _0F_1\left(;\frac{5}{13};\frac{64 x^{13/8}}{169}\right)+x \, _0F_1\left(;\frac{21}{13};\frac{64 x^{13/8}}{169}\right)\\ 4 & 5 & \, _0F_1\left(;\frac{1}{6};\frac{25 x^{6/5}}{36}\right)+x \, _0F_1\left(;\frac{11}{6};\frac{25 x^{6/5}}{36}\right)\\ 4 & 7 & \, _0F_1\left(;\frac{3}{10};\frac{49 x^{10/7}}{100}\right)+x \, _0F_1\left(;\frac{17}{10};\frac{49 x^{10/7}}{100}\right)\\ 4 & 9 & \, _0F_1\left(;\frac{5}{14};\frac{81 x^{14/9}}{196}\right)+x \, _0F_1\left(;\frac{23}{14};\frac{81 x^{14/9}}{196}\right) \end{array} \right) $$

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