# About the solutions of the Bessel differential equation and the modified Bessel differential equation

The modified Bessel differential equation can be obtained by replacing $x$ with $ix$ ($i$ is the imaginary unit) in the Bessel differential equation. If the general solution of the latter is

$$f(x) = A J_{n}(x) + BN_n(x)$$

While the general solution of the modified Bessel differential equation is

$$g(x) = CI_n(x) + DK_n(x)$$

Knowing the definitions of $I_n(x)$ and $K_n(x)$, I don't find a way to express $g(x)$ as $f(ix)$. That is, as far as I know,

$$f(ix) = A J_{n}(ix) + BN_n(ix) \neq g(x) = CI_n(x) + DK_n(x)$$

But if the modified Bessel differential equation is simply obtained by substituting the variable $x$ with $ix$ in the Bessel differential equation, why this can not be applied to the general solutions too?

In other words, why the change of variable can not immediately be "transferred" also to the solutions?

• For the Bessel functions and modified Bessel functions of the first kind, the relationship is simple : $$J_\nu(ix)=i^\nu I_\nu(x)$$. This is far to be the case for the relationship between the Bessel functions and modified Bessel functions of the second kind. All is the consequence of the respective definitions of those functions. – JJacquelin Dec 15 '15 at 21:38
• @jjacquelin The only difference between the two kinds of equations is the sign of the $x^2$ term. Is this sign capable of radically changing the solutions? (In particular, as you pointed out, the second kind of functions) – BowPark Dec 15 '15 at 21:43
• As I pointed out, this is the consequence of the definition of the modified Bessel function of first kind $Y_\nu(x)$ extended to non-integer $\nu$ : $$Y_\nu(x)=\frac{J_\nu(x)\cos(\nu\pi)-J_{-\nu}(x)}{\sin(\nu\pi)}$$ – JJacquelin Dec 15 '15 at 22:04

The behavior of solutions of $$r^2f''(r)+rf'(r)+(r^2-\nu^2)f(r) = 0$$ near $r=0$ is asymptotically the same as the solutions of $$r^2f''+rf'-\nu^2 f=0.$$ For $\nu > 0$, the solutions of the latter are $r^{\pm\nu}$. For $\nu=0$, the solutions are $1$ and $\ln(r)$. Because of this, solutions in the Complex plane inherit branch points at the origin, and the conventions used to deal with branch cuts are not uniquely determined. Solutions are locally holomorphic, but somewhere there must be a branch cut if you work in the Complex plane.
Example of Asymptotic for $\nu=0$: In this case, the solutions of $$r^2f''+rf'=0$$ are $1$ and $\ln r$. You can check that $$-\frac{1}{r^2}\left[r^2\frac{d^2}{dr^2}+r\frac{d}{dr}\right] \left(A+B\ln(r)+\int_{0}^{r}g(s)\ln(s)sds-\ln(r)\int_{0}^{r}g(s)sds\right)=g$$ (I may have the signs in front of the two integral terms swapped, but I'll let you work that out. It's a Green function solution.) Therefore a solution of $$\left[r^2\frac{d^2}{dr^2}+r\frac{d}{dr}\right]f+r^2f=0 \\ -\frac{1}{r^2}\left[r^2\frac{d^2}{dr^2}+r\frac{d}{dr}\right]f=f$$ must satisfy the following for some constants $A$ and $B$: $$A+B\ln(r)+\int_{0}^{r}f(s)\ln(s)sds-\ln(s)\int_{0}^{r}f(s)sds=f(r)$$ The integral terms must vanish at $r=0$ because the equation is in the limit circle case at $r=0$. This gives an asymptotic of $f\sim A+B\ln(r)$ near $r=0$. This is the simplest asymptotic, and this technique works for $0 \le \nu < 1$ because the equation is in the limit point case at $r=0$ for all such $\nu$, meaning that all eigenfunctions are in $L^2_{r}(0,1]$ (the weighted space $(f,g)_r=\int_{0}^{1}f(r)\overline{g(r)}rdr$.)
• @BowPark : I've given you an example of the simpler cases $0 \le \nu < 1$, where the equation is in the limit point case (this is Weyl's classification of singularities for singular Sturm-Liouville problems.) This new part has been appended to my solution above. – DisintegratingByParts Dec 16 '15 at 10:45
• Thanks for having integrated your answer. As before, I tried to follow your statements: they are a certainly valid explanation for why $J_{0}(r) \to 1$ and $N_0 (r) \to \ln(r)$ when $r \to 0$. But I think my question was slightly different: given the two equations $$r^2 \frac{\partial^2 f(r)}{\partial r^2} + r\frac{\partial f(r)}{\partial r} + (r^2 - n^2)f(r) = 0$$ and $$r^2 \frac{\partial^2 f(r)}{\partial r^2} + r\frac{\partial f(r)}{\partial r} - (r^2 + n^2)f(r) = 0$$, the only difference between them is in the $\pm r^2$ terms. Can this radically change the solutions? – BowPark Dec 18 '15 at 11:15
• The functions $J_n(e^{i\theta}x$ and $K_n(e^{i\theta}x)$ are linearly independent for $\theta=0$. What about for small $\theta > 0$? What can happen if $\theta$ varies across a branch cut of one of the solutions? And what are those branch cuts? – DisintegratingByParts Dec 18 '15 at 15:36