I have a question regarding an alternate form of the Bessel equation and how that alternate form translates to the modified Bessel equation and its solution. The modified form is from: http://mathworld.wolfram.com/BesselDifferentialEquation.html

and looks like this: $$ \frac{d^2 y}{dx^2}+\frac{1-2\alpha}{x}\frac{dy}{dx}+\left(\beta^2\gamma^2x^{2\gamma-2}+\frac{\alpha^2-n^2\gamma^2}{x^2}\right)y=0 $$ and has the following solutions: $$ y= \begin{cases} x^\alpha\left[AJ_n(\beta x^\gamma)+BY_n(\beta x^\gamma)\right] &\text{ for integer }n \\ \\ x^\alpha\left[AJ_n(\beta x^\gamma)+BJ_{-n}(\beta x^\gamma)\right] &\text{ for noninteger }n \end{cases} $$ For the moment, I am only concerned with the case where $\gamma=1$ and $\alpha=n$ so the equation simplifies to: $$ \frac{d^2 y}{dx^2}+\frac{1-2\alpha}{x}\frac{dy}{dx}+\beta^2y=0 $$ with the following solutions: $$ y= \begin{cases} x^\alpha\left[AJ_\alpha(\beta x)+BY_\alpha(\beta x)\right] &\text{ for integer }n \\ \\ x^\alpha\left[AJ_\alpha(\beta x)+BJ_{-\alpha}(\beta x)\right] &\text{ for noninteger }n \end{cases} $$ What I'd like to know is if instead of being the unmodified Bessel equation, the equation was in the form of the modified Bessel equation as shown below, what would the solutions be? $$ \frac{d^2 y}{dx^2}+\frac{1-2\alpha}{x}\frac{dy}{dx}-\beta^2y=0 $$ A quick search on wolframalpha tells me they might look like this: $$ y= \begin{cases} x^\alpha\left[AJ_\alpha(-i\beta x)+BY_\alpha(-i\beta x)\right] &\text{ for integer }n \\ \\ x^\alpha\left[AJ_\alpha(-i\beta x)+BJ_{-\alpha}(-i\beta x)\right] &\text{ for noninteger }n \end{cases} $$ Could these then be translated to the modified Bessel functions so: $$ y= \begin{cases} x^\alpha\left[CI_\alpha(\beta x)+DK_\alpha(\beta x)\right] &\text{ for integer }n \\ \\ x^\alpha\left[CI_\alpha(\beta x)+DI_{-\alpha}(\beta x)\right] &\text{ for noninteger }n \end{cases} $$

Any insights you could provide would be greatly appreciated, thanks!


$x^\alpha I_\alpha(\beta x)$, $x^\alpha K_\alpha(\beta x)$ and $x^\alpha I_{-\alpha}(\beta x)$ are indeed solutions to your "modified" differential equation. Of course, there are only two linearly independent solutions on, say, $(0,\infty)$: if $\alpha$ is a non-integer, we have $$ K_\alpha(x) = \frac{\pi}{2 \sin(\pi \alpha)} (I_{-\alpha}(x) - I_{\alpha}(x))$$ while if $\alpha$ is an integer, $I_{-\alpha}(x) = I_\alpha(x)$. So for non-integer $\alpha$ you can use any two of $x^\alpha I_\alpha(\beta x)$, $x^\alpha K_\alpha(\beta x)$ and $x^\alpha I_{-\alpha}(\beta x)$ as basic solutions, while for integer $\alpha$ you want to use $x^\alpha I_\alpha(\beta x)$ and $x^\alpha K_\alpha(\beta x)$.

  • $\begingroup$ Thank you for the insight! $\endgroup$ – Travis Jan 23 at 14:32

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