Why radial coordinate of a particle must decrease continuously once it is inside the Schwarzschild radius? Suppose we are inside the Schwarzschild radius of a black hole and throw a ball radially outward. It is said that the ball has no possibility to increase its radial coordinate. It must continuously decrease its radial coordinate and reach finally to the center. 
I don't understand why this is so. The metric inside the Schwarzschild radius for radial motion is:
$$ds^2=-c^2(2r_s/r-1)dt^2+(2r_s/r-1)^{-1}dr^2$$
I do understand that the ball must follow timelike world-line, i.e., we must have $ds^2>0$, whatever way the ball moves. And for this to happen it is necessary that $dr\ne0$. But the metric does not appear to put any restriction on whether $dr$ shall be positive or negative, because $dr$ appears there as a squared term.
So why the ball can't move with positive $dr$, i.e, radially outward?
 A: You are quite correct that the metric alone cannot tell you which way the ball is moving. That's because the metric describes both the black hole and the white hole. But what we can work out from the metric is that once inside the event horizon the velocity $dr/dt$ can never change sign. That means if you cross the horizon headed inwards into a black hole, i.e. with a negative value of $dr/dt$, the radial velocity can never increase to zero and become positive so you can never head outwards again. The same argument applies to white holes. In this case a ball inside the horizon and headed outwards can never come to a halt and fall inwards again.
To show this rigorously is involved, but if you'll forgive a bit of arm waving we can make a reasonable non-rigorous argument. The metric for a radial trajectory $(d\theta = d\phi = 0)$ is:
$$ c^2d\tau^2 = c^2\left(1 - \frac{r_s}{r}\right)dt^2 - \frac{dr^2}{1 - r_s/r} $$
For the direction of the velocity to change we require that the object be momentarily stationary, i.e. $dr=0$, but this gives us:
$$ c^2d\tau^2 = c^2\left(1 - \frac{r_s}{r}\right)dt^2 $$
The problem is that once inside the horizon $1 - r_s/r$ is negative and this will give us a negative value for $d\tau^2$. Since this is impossible the conclusion is that $dr$ can never be zero i.e. the ball can never change the direction of the radial velocity.
A: You are correct that although the metric can tell you whether or not a given line is timelike, it cannot on its own determine which is the future-pointing direction and which the past-pointing direction. This is determined by other considerations, such as the initial conditions. We can tell which is the future-pointing direction within the horizon by continuity from outside the horizon. Once it is determined for one worldline, then it is also determined for adjacent worldlines and hence for the whole spacetime, under ordinary conditions. However I don't know whether, from a mathematical point of view, there might be exceptions to this in special cases, such as spacetime configurations with closed timelike lines or naked singularities, but there is good reason to believe those don't occur in the physical world.
A: The metric is in fact
$$ds^2=-c^2(r_s/r-1)dt^2+(r_s/r-1)^{-1}dr^2$$
Or writing in terms of proper time
$$d \tau^2 = -(r_s/r-1)dt^2+(r_s/r-1)^{-1}dr^2$$
When $r<r_s$, then since $d\tau > 0$ for an object with mass and $dt^2 \geq 0$ then $dr$ cannot equal zero.
This means that $dr/dt$ cannot change sign once $r<r_s$. Thus once moving into a black hole ($dr <0$), then the radial coordinate cannot increase.
A: To illustrate the points of the previous answers, you can think about flat minkowskian metric and just equivalently ask why an object can't move backwards in time, because  the line element $ds^2$ only depends on $dt$ squared! the answer is that the equations of motion themselves are symmetric in time, and the direction we call 'future' or 'past' in time depends on other considerations, such as the direction in which entropy increases. 
so similarly, you can draw a timelike curve with one end inside the event horizon and the other end outside it. But in order for it to be consistent with our usual definition of time direction, you will have to conclude that it describes an object falling into the black hole, because we can only cross the event horizon going inwards and not vice versa.
(what you can't do is draw a timelike curve in which $r$ both increases and decreases inside the event horizon. that would not be timelike)  
