Transforming Differential Equation to a Kummer's Equation

Question

I'm trying to transform an equation of the form

$$ yw^{\prime\prime}(y) - [b - ay] w^\prime(y) - [d + ey]w(y) = 0 $$

into the form of a Kummer's or confluent hypergeometric differential equation:

$$ y w^{\prime\prime}(y) + [f - y] w^\prime(y) + g w(y) = 0 $$

I know it may have something to do with merging two of the singularities of the original equation, and maybe doing something with $y$, making it $\frac{y}{b}$ and taking b to infinity, but I don't know and can't find the details for this process, and for my equation specifically. There's not too much difference between the two but just enough so that I can't get it.

Thanks in advance for any help!

Answer 1

In fact this is not the business about the singularities but just the business about the variable transformation.

Let $w=e^{ny}w_1$ ,

Then $\dfrac{dw}{dy}=e^{ny}\dfrac{dw_1}{dy}+ne^{ny}w_1$

$\dfrac{d^2w}{dy^2}=e^{ny}\dfrac{d^2w_1}{dy^2}+ne^{ny}\dfrac{dw_1}{dy}+ne^{ny}\dfrac{dw_1}{dy}+n^2e^{ny}w_1=e^{ny}\dfrac{d^2w_1}{dy^2}+2ne^{ny}\dfrac{dw_1}{dy}+n^2e^{ny}w_1$

$\therefore y\left(e^{ny}\dfrac{d^2w_1}{dy^2}+2ne^{ny}\dfrac{dw_1}{dy}+n^2e^{ny}w_1\right)-(b-ay)\left(e^{ny}\dfrac{dw_1}{dy}+ne^{ny}w_1\right)-(d+ey)e^{ny}w_1=0$

$y\left(\dfrac{d^2w_1}{dy^2}+2n\dfrac{dw_1}{dy}+n^2w_1\right)-(b-ay)\left(\dfrac{dw_1}{dy}+nw_1\right)-(d+ey)w_1=0$

$y\dfrac{d^2w_1}{dy^2}+2ny\dfrac{dw_1}{dy}+n^2yw_1-b\dfrac{dw_1}{dy}-bnw_1+ay\dfrac{dw_1}{dy}+anyw_1-dw_1-eyw_1=0$

$y\dfrac{d^2w_1}{dy^2}+(-b+(2n+a)y)\dfrac{dw_1}{dy}+(-(bn+d)+(n^2+an-e)y)w_1=0$

Choose $n^2+an-e=0$ , i.e. $n=\dfrac{-a\pm\sqrt{a^2+4e}}{2}$ , the ODE becomes

$y\dfrac{d^2w_1}{dy^2}+(-b\pm\sqrt{a^2+4e}~y)\dfrac{dw_1}{dy}+\dfrac{ab-2d\mp b\sqrt{a^2+4e}}{2}w_1=0$

Let $y_1=ky$ ,

Then $\dfrac{dw_1}{dy}=\dfrac{dw_1}{dy_1}\dfrac{dy_1}{dy}=k\dfrac{dw_1}{dy_1}$

$\dfrac{d^2w_1}{dy^2}=\dfrac{d}{dy}\left(k\dfrac{dw_1}{dy_1}\right)=\dfrac{d}{dy_1}\left(k\dfrac{dw_1}{dy_1}\right)\dfrac{dy_1}{dy}=k\dfrac{d^2w_1}{dy_1^2}k=k^2\dfrac{d^2w_1}{dy_1^2}$

$\therefore\dfrac{y_1}{k}k^2\dfrac{d^2w_1}{dy_1^2}+\left(-b\pm\sqrt{a^2+4e}\dfrac{y_1}{k}\right)k\dfrac{dw_1}{dy_1}+\dfrac{ab-2d\mp b\sqrt{a^2+4e}}{2}w_1=0$

$ky_1\dfrac{d^2w_1}{dy_1^2}+\left(-b\pm\dfrac{\sqrt{a^2+4e}}{k}y_1\right)k\dfrac{dw_1}{dy_1}+\dfrac{ab-2d\mp b\sqrt{a^2+4e}}{2}w_1=0$

$y_1\dfrac{d^2w_1}{dy_1^2}+\left(-b\pm\dfrac{\sqrt{a^2+4e}}{k}y_1\right)\dfrac{dw_1}{dy_1}+\dfrac{ab-2d\mp b\sqrt{a^2+4e}}{2k}w_1=0$

Choose $\pm\dfrac{\sqrt{a^2+4e}}{k}=-1$ , i.e. $k=\mp\sqrt{a^2+4e}$ , the ODE becomes

$y_1\dfrac{d^2w_1}{dy_1^2}+(-b-y_1)\dfrac{dw_1}{dy_1}\mp\dfrac{ab-2d\mp b\sqrt{a^2+4e}}{2\sqrt{a^2+4e}}w_1=0$

Answer 2

The differential equation $$ y \cdot w^{\prime\prime}(y) + (a y - b) \cdot w^{\prime}(y) - (d+ e y) \cdot w(y) = 0 \tag{1} $$ has two singular points, namely $y=0$ and $y=\infty$. The origin is regular singular point with exponents $0$ and $b-1$. Infinity is irregular singular point of $(1)$, i.e. its solutions have essential singularity at infinity. Asymptotically the fundamental solution reads: $$ w(y) = c_+ \exp\left(-\lambda_+ y\right) y^{\mu_+} \left(1 + \mathcal{o}(1)\right) + c_- \exp\left(-\lambda_- y\right) y^{\mu_-} \left(1 + \mathcal{o}(1)\right) $$ where $2\lambda_{\pm}= b \pm \sqrt{b^2 + 4 e}$, and $\mu_\pm = \pm\dfrac{ a \lambda_\pm - d}{\lambda_+ - \lambda_-}$.

The second equation also has two singular points, $y=0$ and $y=\infty$. It can be obtained by eq. $(1)$ by setting $a=-1$, $b=-f$, $d=-g$ and $e=0$. Thus, the second differential equation also has the origin as its regular and infinity as its irregular singular points.

Since both equation are of the same type, they can not be obtained from one another by confluent limit, which takes a differential equation with three regular singular points (like Gauss's hypergeometric differential equation you were most certainly given), and merges two of those singularities.