Hypergeometric equation at z = infinity

[jncarter]

Homework Statement

Show that by letting $z = \zeta^-1$ and $u = \zeta^{\alpha}v(\zeta)$ that the differential equation,

$z(1-z)\frac{d^{2}u(z)}{d^{2}z}+{\gamma - (\alpha+\beta+1)z}\frac{du(z)}{dz}-\alpha \beta u(z) = 0$​

can be reduced to

$\zeta(1-\zeta)\frac{d^{2}v(\zeta)}{d\zeta^{2}} + {\alpha-\beta+1-(2\alpha-\gamma+2)\zeta}\frac{dv(\zeta)}{d\zeta}-\alpha(\alpha-\gamma+1)v(\zeta) = 0$​

and find the solutions of the first equation about z = ∞ in therms of hypergeometric functions of z.

Homework Equations

The first solution to the top equation about z=0 is given by
$u_{1} = 1 + \Sigma^{\infty}_{n=1}\frac{(\alpha)_{n}(\beta)_{n}}{(\gamma)_{n}n!}z^{n} = F(\alpha,\beta,\beta;z)$

The Attempt at a Solution

First, I'm trying to find expressions for all of the derivatives.
$z=\zeta^{-1} dz=-\zeta^{-2}d\zeta \frac{d\zeta}{dz}=-\zeta^{2}$
First question, is this the correct expression for d^{2z? I am concerned that I may have to use the chain rule in some manner that I am missing.
$\frac{d^{2}\zeta}{dz^{2}} = 2\zeta^{-3}$ or $d^{2}z = d(-\zeta^{-2}d\zeta) = 2 \zeta^{-3}d\zeta - \zeta^{-2}d^{2}\zeta$
which gives some funny expression for $\frac{d^{2}\zeta}{dz^{2}}$
$\frac{d}{dz}=\frac{d\zeta}{dz}\frac{d}{d\zeta} = -\zeta^{2}\frac{d}{d\zeta}$
$\frac{d^{2}}{dz^{2}} = \frac{d^{2}\zeta}{dz^{2}}\frac{d^{2}}{d\zeta^{2}} = \frac{\zeta^{3}}{2}\frac{d^{2}}{d\zeta^{2}}$
Now the next part can be done in a different order, but I have decided to go with a method that seems to be more careful.
$\frac{du}{dz} = \frac{d}{dz}(\zeta^{\alpha}v)$
$= \frac{d}{dz} \zeta^{\alpha} v+\zeta^{\alpha} \frac{dv}{dz}$
Now use $\frac{d}{dz}(\zeta^{\alpha}) = \frac{d}{dz}(z^{-\alpha}) = -\alpha \zeta^{\alpha - 1}$ to get
$\frac{du}{dz} = -\alpha \zeta^{\alpha - 1} v + \zeta^{\alpha}\frac{dv}{dz}$
$= -\alpha \zeta^{\alpha - 1} v - \zeta^{\alpha+2}\frac{dv}{d \zeta}$
Similarly,
$\frac{d^{2}u}{dz^2} = \frac{d^2}{dz^2}(\zeta^{\alpha})v + 2 \frac{d}{dz}(\zeta^{\alpha}) \frac{dv}{dz} + \zeta^{\alpha} \frac{d^2 v}{dz^2}$
$= - \alpha (\alpha - 1)\zeta^{\alpha -2} v +2 \alpha \zeta^{\alpha-1} \frac{dv}{d \zeta} +\frac{1}{2} \zeta^{\alpha +3} \frac{d^{2}v}{d \zeta^{2}}$
I've tried plugging this all in a couple of different times and keep getting weird answers. More specifically, I can't seem to get the $\zeta (1- \zeta)\frac{d^{2}v}{d\zeta^{2}}$. So I tried finding common factors that I could divide out of the equation. So far, it's been no good. This got me thinking that maybe I had the incorrect expression for the second derivatives. Please take a look at my work and point out any problems. I always find that a fresh pair of eyes is enlightening in these problems. Thanks for any help!}

 

[jackmell]

Not sure I'm following your work. Here's what I'd do for the first one. If:

$$z=\frac{1}{\zeta}\rightarrow \zeta=\frac{1}{z}$$

$$u=\zeta^{-\alpha} v(\zeta)$$

then:

$$\begin{align} \frac{du}{dz}&=\frac{du}{d\zeta}\frac{d\zeta}{dz} \\ &=\left(\zeta^{-\alpha}v'-\alpha \zeta^{-\alpha-1}v\right)\left(-\frac{1}{z^2}\right)\\ &=\vdots \\ &=-\frac{1}{\zeta^2}\left(\zeta^{-\alpha}v'-\alpha\zeta^{-\alpha-1}v\right) \end{align}$$

 

[jncarter]

Oops, $u = \zeta^{\alpha}v$.
$\zeta = z^{-1} → d\zeta = -z^{-2}dz=-\zeta^{2} → \frac{d\zeta}{dz}=-\zeta^{2}$
Note the positive in the $\zeta]$, since z has a negative exponent, $\zeta$ has a positive one.
$\frac{du}{dz}=\frac{du}{d\zeta} \frac{d\zeta}{dz} = -\zeta^{2}\frac{d}{d\zeta}(\zeta^{\alpha})$
$\frac{d}{d\zeta}(\zeta^{\alpha}v) = \alpha \zeta^{\alpha -1}v+\zeta^{\alpha}\frac{dv}{d\zeta}$
Plus this into $\frac{du}{dz}$ and do some algebra:
$\frac{du}{dz} = -\alpha \zeta^{\alpha -1}v - \zeta^{\alpha+2}\frac{dv}{d\zeta}$
So I think I had a +1 where there should have been a +1 in the first term. I'm still concerned about the second derivatives. Will reply in a bit once I do some work...

 

[jncarter]

Okay, new attempt at the second derivative:
$\frac{d^2}{dz^2}(z^{-1})=2z^{-3}=2\zeta^{3}$
$\frac{d^2}{d\zeta^2}(\zeta^{\alpha}v) = v\frac{d^2}{d\zeta^2}(\zeta^{\alpha}) + \zeta^{\alpha} \frac{d^2v}{d\zeta^2}$
$\rightarrow = \alpha(\alpha-1)\zeta^{\alpha-2}v+\zeta^{\alpha}\frac{d^2v}{d\zeta^2}$
$\frac{d^2u}{dz^2} = \frac{d^2u}{d\zeta^2}\frac{d^2\zeta}{dz^2} = 2\zeta^3\frac{d^2}{d\zeta^2}(\zeta^{\alpha}v)$
$\rightarrow = 2\alpha(\alpha - 1)\zeta^{\alpha +1}v+2 \zeta^{\alpha + 3}\frac{d^2v}{d\zeta^2}$

Does that look about right?

 

[jackmell]

If:

$$\zeta=1/z\rightarrow \frac{d\zeta}{dz}=-\frac{1}{z^2}=-\zeta^2$$

$$u=\zeta^{\alpha}v$$

then I get:
$$\begin{align} \frac{du}{dz}&=\left(\zeta^{\alpha}v'+\alpha \zeta^{\alpha-1}v\right)\frac{d\zeta}{dz}\\ &=\vdots \\ &=\alpha\zeta^{\alpha+1}v-\zeta^{\alpha+2}v' \end{align}$$

and then one more time:

$$\begin{align} \frac{du}{dz}&=\alpha[\left(\zeta^{\alpha+1}v'+(\alpha+1)\zeta^{\alpha}v\right)(-\zeta^2)]-[\left(\zeta^{\alpha+2}v''+(\alpha+2)\zeta^{\alpha+1}v'\right)(-\zeta^2)\\ &=\alpha[-\zeta^{\alpha+3}v'-(\alpha+1)\zeta^{\alpha+2}v]+\zeta^{\alpha+2}v''+(\alpha+2)\zeta^{\alpha+2}v' \end{align}$$




. . . ok, I'm struggling with it. Not sure I have it exactly right.

 

[jncarter]

If:

$$\zeta=1/z\rightarrow \frac{d\zeta}{dz}=-\frac{1}{z^2}=-\zeta^2$$

$$u=\zeta^{\alpha}v$$

then I get:
$$\begin{align} \frac{du}{dz}&=\left(\zeta^{\alpha}v'+\alpha \zeta^{\alpha-1}v\right)\frac{d\zeta}{dz}\\ &=\vdots \\ &=\alpha\zeta^{\alpha+1}v-\zeta^{\alpha+2}v' \end{align}$$

You lost a negative sign $\frac{du}{dz} = -\alpha \zeta^{\alpha+1}v- \zeta^{\alpha+2}v'$
I haven't checked the other part of your work yet.
My last post may also be wrong. I found this formula somewhere
$\frac{d^2}{dx^2}(uv) = \frac{d^2 u}{dx^2}v+2\frac{du}{dx}\frac{dv}{dx}+u\frac{d^2 v}{dx^2}$
where u=u(x) and v=v(x). I used this in my first post, where $\zeta^{\alpha}$ was the u in the last equation and dx becomes dζ.
With the correction to the powers I found
$\frac{d^2 u}{d\zeta^2}=2 \alpha( \alpha-1) \zeta^{\alpha+1}v+4 \alpha \zeta^{\alpha+2}v'+2 \zeta^{\alpha+3}v''$

 

[jackmell]

Homework Statement

Show that by letting $z = \zeta^-1$ and $u = \zeta^{\alpha}v(\zeta)$ that the differential equation,

$z(1-z)\frac{d^{2}u(z)}{d^{2}z}+{\gamma - (\alpha+\beta+1)z}\frac{du(z)}{dz}-\alpha \beta u(z) = 0$​

can be reduced to

$\zeta(1-\zeta)\frac{d^{2}v(\zeta)}{d\zeta^{2}} + {\alpha-\beta+1-(2\alpha-\gamma+2)\zeta}\frac{dv(\zeta)}{d\zeta}-\alpha(\alpha-\gamma+1)v(\zeta) = 0$​

I believe your latex caused the equations to be displayed incorrectly. I think they should be:

$$z(1-z)\frac{d^{2}u(z)}{d^{2}z}+\big[\gamma - (\alpha+\beta+1)z\big]\frac{du(z)}{dz}-\alpha \beta u(z) = 0$$

can be reduced to
$$\zeta(1-\zeta)\frac{d^{2}v(\zeta)}{d\zeta^{2}} + \big[\alpha-\beta+1-(2\alpha-\gamma+2)\zeta\big]\frac{dv(\zeta)}{d\zeta}-\alpha(\alpha-\gamma+1)v(\zeta) = 0$$

Once that's done then what we're doing above, minus the arithmetic errors, will yield those results I believe. I'm frustrated with all those greek letters in there so I used a,b, c and d and stuff and then just ran it through Mathematica (yeah I know, I'm not proud). When I do that I get for the second derivative (in Mathematica Latex):

$$a y^{2+a} v[y]+a^2 y^{2+a} v[y]+2 y^{3+a} v'[y]+2 a y^{3+a} v'[y]+y^{4+a} v''[y]$$

where a is alpha and y is zeta. Pretty sure that's correct now because when I substitute everything in the first equation above, I get the second one.. Also checked a reduced version numerically so I think we're good now.

 

[jncarter]

You're right about that second equation. That was my bad forgetting parentheses. I also got the chance to talk to my TA yesterday, and you're right about the second derivative. Thanks for all the help. There's just so much algebra with this type of problem, it's easy to lose track of things.