I consider the following PDE: \begin{equation} c b x^2 h''(x) + b x h'(x) - x h(x) = 0, \text{ for } x>0 \end{equation} where $b>0$ and $c>0$.
Firstly, I want to know what is the general form of the solution to this equation.
Then I want to present the solutions for the particular values of $c$, e.g. $c=\frac{1}{2}$, $c=1$, $c=\frac{3}{2}$.
Using WolframAlpha, I obtained the general solution of the form: \begin{equation} h(x) = c^{\frac{1}{2}(\frac{1}{c} - 1)} k_1 r^{\frac{1}{2}(\frac{1}{c} - 1)} x^{\frac{c-1}{2c}} Γ\left(\frac{1}{c}\right) I_{\frac{1}{c} - 1}\left(\frac{2\sqrt{x}}{\sqrt{c}\sqrt{r}}\right) + (-1)^{\frac{c - 1}{c}} c^{\frac{1}{2}(\frac{1}{c} - 1)} k_2 r^{\frac{1}{2}(\frac{1}{c}-1)} x^{\frac{c-1}{2c}} Γ\left(2 - \frac{1}{c}\right) I_{\frac{c-1}{c}}\left(\frac{\sqrt{x}}{\sqrt{c}\sqrt{r}}\right), \end{equation} where $I_n(x)$ is the modified Bessel function of the first kind.
Then, I've used WolframAlpha again to substitute the particular values for $c$ and check if the form of this solution is for sure the correct one.
What I' ve observed:
- For $c\in(\frac{1}{2}, 1)$ - the solution agrees with the above formula.
- For $c\in(0, \frac{1}{2}]\cup\{1\}$ - the solution is not a sum of two modified Bessel functions of the first kind as above, but it is a sum of the modified Bessel function of the first kind and modified Bessel function of the second kind.
- For $c>1$ - the solution agrees with the above formula, but I've noted that for $c=2$ it is a sum of hyperbolic sine and cosine.
To sum up:
- Am I right that there is no one formula of solution for each $c\in(0, \infty)$?
- Could anyone give me any reference where I will find such PDE with all possible forms of solutions?
- Is any straightforward way do derive the solution (not using WolframAlpha)?
$$h''(x) + \frac{1}{cx} h'(x) - \frac{1}{bcx} h(x) = 0$$ See Eqs.(6-7) in: https://mathworld.wolfram.com/BesselDifferentialEquation.html
$1-2\alpha=\frac{1}{c}\quad;\quad 2\gamma-2=-1\quad;\quad \alpha^2-n^2\gamma^2=0\quad;\quad \beta^2\gamma^2=-\frac{1}{bc}$ $$\begin{cases} \alpha=\frac{c-1}{2c}\\ \gamma=\frac12\\ \beta=\pm \frac{2\:i}{\sqrt{bc}}\\ n=1-\frac{1}{c} \end{cases}$$ NOTE : Since $\beta$ is complex, one have to change the Bessel function into modified bessel functions. $$y(x)=A\:x^{\frac{c-1}{2c}}I_{1-\frac{1}{c}}\left(\frac{2}{\sqrt{bc}}x^{1/2}\right)+B\:x^{\frac{c-1}{2c}}I_{-1+\frac{1}{c}}\left(\frac{2}{\sqrt{bc}}x^{1/2}\right)\quad\text{case of non-integer}\quad 1-\frac{1}{c}.$$ $A,B$ are constant.
If you want find this solution by yourself directly from the original ODE, change of variables $$\begin{cases} X=\frac{2}{\sqrt{bc}}x^{1/2}\\ Y=x^{\frac{1-c}{2c}}y(x) \end{cases}$$ This will transform the original ODE into a standard modified Bessel ODE : https://mathworld.wolfram.com/ModifiedBesselDifferentialEquation.html
For particular cases of reduction of Bessel functions to elementary functions, see the properties of Bessel functions of order $\pm 1/2$.