Is this an inconsistent example from Griffiths electrodynamics? 
In Griffiths electrodynamics example 3.3 we've to find potential in a region subject to the conditions
(i) $\quad V=0$ when $y=0$
(ii) $V=0$ when $y=a$
(iii) $V=V_{0}(y)$ when $x=0$
(iv) $V \rightarrow 0$ as $x \rightarrow \infty$

And the example goes on to solve it.
My question:  Any point on the z axis has coordinates $x=y=0$. Now according to (i) $V=0$ but according to(iii)  $V=V_{0}(y)$ which is a contradiction.
Now if this example is therefore poorly constructed how can one find a solution to an inconsistent problem, which Griffiths goes on to do? And if there is really no contradiction please explain that as well.
 A: Firstly, this is a problem in 2D, so there is no $z$-axis. Secondly, you're right. Those boundary conditions are contradictory at $(0,0)$ and $(0,a)$.  However, I wouldn't say that the example is poorly constructed; it just sweeps certain technical issues like this under the rug in an effort to get to the physics with as little formalism as possible.  After all, if you ignore the corners of the region then everything works fine.
The problem is that we are imposing discontinuous boundary conditions on a function which is apparently twice-differentiable (and therefore continuous).  This should be troubling.  Even if we say that $\nabla^2 V = 0$ only on the interior of the region, it's not at all obvious that a lovely and well-behaved harmonic function could suddenly become discontinuous on the boundary.
As it turns out, we can have this behavior, but it comes with a few technical issues which are centered around the points of discontinuity - in this case, the corners $(0,0)$ and $(0,a)$.  The main takeaway is that the best we can do is to find a function $V$ where $\nabla^2 V = 0$ on the interior of the region, and such that for all $x\in (0,\infty)$, $\lim_{y\rightarrow 0} V(x,y) = \lim_{y\rightarrow a}V(x,y) = 0$ and for all $y\in(0,a), \lim_{x\rightarrow \infty}V(x,y)=0$ and $\lim_{x\rightarrow 0} V(x,y)= V_0(y)$.
Such a solution will (by necessity) be discontinuous at the points of discontinuity of the boundary values.  In this sense, it doesn't matter whether you choose the boundary value at $(0,0)$ to be $0$ or $V_0(0)$, since $V(0,0)$ will differ from both of those.  In fact, I believe that if you take Griffiths' solution, set $x=y\equiv r$, and take the limit as $r\rightarrow 0$, you should find that the solution approaches $V_0(0)/2$.
At the end of the day you can add formalism to understand exactly how to handle discontinuous boundary conditions, but the end result in cases like this is simply to "ignore the corners," which is precisely what Griffiths does in the first place.  It's an excellent question to raise, but in the context of the level of his book I would say that it's not an unreasonable omission on his part.
