# Transforming Differential Equation to a Kummer's Equation

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!

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

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

• Thanks very much - I just saw this! I really appreciate everyone's help. Nov 28, 2012 at 17:54
• That really is beautiful! Nov 28, 2012 at 18:57

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.