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I can solve the following differential equation without any trouble using the method of Frobenius $$ x^2 y’’ - (2 + 3x) y = 0. $$ When I put the differential equation in Mathematica, it gives me the solutions in terms of modified Bessel functions of order 3 $$ y(x) = A \sqrt{x} I_3\big(2\sqrt{3x}\big) + B \sqrt{x} K_3\big(2\sqrt{3x}\big). $$ I cannot for the life of me see how to put the given ODE into the form of a modified Bessel equation. Can anyone point me in the right direction?

Some Added Information

Generally, the equation $$ \frac{d}{dx}\left(x^a \frac{dy}{dx}\right) + b x^c y = 0, $$ can be transformed to a Bessel equation with solution $$ y(x) = x^{\nu/\alpha} Z_\nu \left(\alpha\sqrt{|b|} x^{1/\alpha}\right), $$ where $Z_\nu$ is any Bessel function solution of the transformed equation, if we choose $$ \alpha = \frac{2}{c-a+2} \quad \text{and} \quad \nu = \frac{1-a}{c-a+2}, $$ Considering the equations for $\alpha$ and $\nu$, I can see from Mathematica’s solution that $\nu = 3$ and $\alpha = 2$, so I can solve them to find $a = -2$ and $c = -3$. Putting those into the ODE above, we get $$ \frac{d}{dx}\left(x^{-2} \frac{dy}{dx}\right) + b x^{-3} y = 0. $$ Expanding this out, I get $$ \frac{1}{x^2}y’’ - \frac{2}{x^3} y’ - \frac{b}{x^3} y = 0 \quad \text{or} \quad x^2 y’’ - 2x y’ - 3xy = 0. $$ This is my ODE if the $y’$ were instead a $y$. :-( So, I am stuck.

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  • $\begingroup$ I think you need a transformation first, as the modified vessel equation has solutions in linear combinations of modified vessel functions, not necessarily $\sqrt{x}I_3$ and $\sqrt{x}K_3$. The degree being 3 should also be a hint. $\endgroup$
    – DaveNine
    Dec 15, 2019 at 22:28
  • $\begingroup$ I’ve tried every transformation I can think of and it’s not happening. I will add some more to my question to show what I have tried. $\endgroup$ Dec 15, 2019 at 23:20

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Your equation is indeed a transformed modified Bessel equation. To see that, you need to take the modified Bessel equation $$ \xi^2 \frac{d^2\eta}{d\xi^2} + \xi \frac{d\eta}{d\xi} - (\xi^2+n^2)\eta=0 $$ and employ following transformation $$ \eta=\frac{y}{x^\alpha}, \quad \xi=\beta x^\gamma, $$ you should arrive at $$ \frac{d^2y}{dx^2} - \frac{(2\alpha-1)}{x}\frac{dy}{dx} - (\beta^2 \gamma^2 x^{2\gamma-2}+\frac{n^2\gamma^2-\alpha^2}{x^2})y=0. $$ To see the steps, I recommend to look into the book "Bowman: Introduction to Bessel functions, p.117", where it is done for standard Bessel equation.

Since $\eta$ satisfies the modified Bessel equation it follows that $\eta(\xi)=AK_n(\xi)+B I_n(\xi)$, $A,B$ being constants. Employing the transformations above, we have that the transformed Bessel equation has a general solution $$ y(x)=x^\alpha[A K_n(\beta x^\gamma)+B I_n(\beta x^\gamma)]. $$ Your specific equation is the case $\alpha=\frac12,\gamma=\frac12,\beta=2\sqrt{3},n=3$.

Note, that for non-integer $n$ we usually take the solution in a fom $$ y(x)=x^\alpha[A I_n(\beta x^\gamma)+B I_{-n}(\beta x^\gamma)]. $$

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The following Wolfram code

y[x_] := Sqrt[x] K[2 Sqrt[3 x]];
u = (x^2 D[y[x], {x,2}] - (2 + 3 x) y[x]) 8 Sqrt[3]/t;
u/. x -> (t^2/12) // PowerExpand // Simplify

evaluates to

-(9 + t^2) K[t] + t (K'[t] + t K''[t])

which goes a long way to establishing the Bessel connection.

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    $\begingroup$ That does it! Thank you. I will write up my solution soon when I have more time. $\endgroup$ Dec 16, 2019 at 2:48

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