[Math] Transforming Differential Equation to a Kummer’s Equation

hypergeometric functionordinary differential equationstransformation

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!

Best Answer

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

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

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