Path equation, radial trajectory not defined? I'm sorry if this is a rather silly question, but it keeps bothering me that I can't find a satisying answer. 
In Orbital Dynamics, we can describe the path taken by the body with the path equation: $$ \frac{d^2u}{d\theta^2} + u = - \frac{f(\frac{1}{u})}{L^2u^2}$$ $ u = \frac{1}{r(\theta)} $, and the force field $ F = mf(r)\hat{r} $.
However we can't solve this differential equation with the initial condition that the angular momentum is zero, i.e. the body is falling radially inwards, why isn't this orbit included in the differential equation?
 A: The starting point for the most general equation of motion in polar coordinates is to write $r = r(t)$ and $\theta = \theta(t)$ where $t$ is time.
In the special case of orbital motion, we effectively eliminate $t$ and write $r = r(\theta)$, and then introduce $u = 1/r$ to simplify the math.
But if the body is falling radially inwards, we have $\theta = \text{constant}$, so "$r = r(\theta)$" doesn't make any sense. The only thing it could mean is $r = \text{constant}$, which obviously doesn't represent an orbit.
A: Just like @alephzero already mentioned, the most general way of describing motion with polar coordinates is to write $r$ and $\theta$ as a function of time, i.e. $r(t)$ and $\theta(t)$. When deriving the path equation, we start from the acceleration formulas in polar coordinates and the fact that the body is in a central force field (inverse square in the case of orbital motion due to gravity), such that: 
$$ a_{\theta} = r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt}  = 0$$
$$ a_r = \frac{d^2r}{dt^2}+r(\frac{d\theta}{dt})^2 = F(r) $$
Our initial goal is to find two functions $r(t)$ and $\theta(t)$, that satisfy the above equations. However, this is hard to carry out analytically. Instead we make the assumption that $r$ is dependent on the variable $\theta$. We also know that the reduced angular momentum $L'$ is constant, because we have a central force field.
If $r$ is dependent on $\theta$, and the reduced angular momentum is zero, i.e. $r^2\frac{d\theta}{dt} = 0$, this implies that either $r(t) = 0 $ or $\frac{d\theta}{dt}=0$. In both cases the above equations for $a_r$ and $a_{\theta}$ vanish, so that we don't even have an equation to start with, we can't even speak of motion actually.
If $r$ is dependent on $\theta$, and $L'\not=0$ we can, after some manipulation of the above equations, eliminate the variable $t$, such that we try to find $r$ in function of the angle $\theta$, without wanting to know where the particle is at what time $t$. 
The final result is the path equation, which describes the path the object takes. The path equation makes the assumption that $r$ is dependent on $\theta$ and the reduced angular momentum $L'\not=0$. The radial trajectory is the case where the angular momentum is zero, and thus no solution of the path equation. 
