I've come up with a somewhat simpler approach for presenting the solution for the EF geodesic equations.
Let r, v be the Eddington-Finklesein (EF) coordinates, which are presumed to be functions of proper time \lambda. Then let:
\dot{r} = \frac{dr}{d\lambda} \hspace{3cm} \dot{v} = \frac{dv}{d\lambda}
The Ingoing Eddington Finklestein metric (gemoetrized) is
http://en.wikipedia.org/w/index.php?title=Eddington–Finkelstein_coordinates&oldid=516198830
-(1-2m/r) dv^2 + 2\,dv\,dr
The Christoffel symbols are:
\Gamma^{v}{}_{vv} = \frac{m}{r^2}
\Gamma^{r}{}_{vv} = \frac{m(1-2m/r)}{r^2}
\Gamma^{r}{}_{vr} = \Gamma^{r}{}_{rv} = -\frac{m}{r^2}So we can write the geodesic equations as
<br />
\ddot{v} + \Gamma^{v}{}_{vv} \dot{v}^2 = \frac{d \dot{v}}{dr} \dot{r} + \frac{m}{r^2} \dot{v}^2 = 0<br />
<br />
\ddot{r} + \Gamma^{r}{}_{vv} \dot{v}^2 + 2 \,\Gamma^{r}{}_{vr} \dot{v} \dot{r} = \frac{d \dot{r}}{dr} \dot{r} + (1-2m/r)\frac{m}{r^2} \dot{v}^2 - 2 \frac{m}{r^2} \dot{r}\dot{v} = 0<br />
Note that I've used the chain rule to write
\ddot{v} = \frac{d^2 v}{d \lambda^2} = \frac{d \dot{v}}{d \lambda} = \frac{d \dot{v}}{dr}\frac{dr}{d\lambda} = \frac{d \dot{v}}{dr} \dot{r}
Then we can write the solution in infalling EF coordinates for m=2 as:
\dot{r} = - \sqrt{\frac{4}{r}}
\dot{v} = \frac{\sqrt{r}}{2+\sqrt{r}}<br />
And just plug them into the geodesic equations above to demonstrate that they are a solution.
Note that \dot{r} is negative, the first post had a sign error for the equivalent expression.
The math here is only mildly obnoxious compared to the previous expressions, though I've skipped over a lot of textbook stuff like computing the EF metric (the Wiki does that), and computing the Christoffel symbols for said metric.
[add]If you want r as a function of \lambda, it remains
r = \left(9 \lambda^2\right)^{\frac{1}{3}}
) Classically (i.e., without any quantum effects included), this *is* what GR says.
