Periodicity of a solution to a PDE 

I don't understand what is highlighted in green. 
Firstly, this is probably down to the ambiguity of the phrasing of the question, but at the points $(0,4)$ and $(2,4)$ in part i) $u=1$ and in part ii) $u=$sin($\pi x/2)$ so $u \neq 0$ on the entirety of $x=0$ & $x=2$?
More importantly if it were the case that:
$u=0$ on $x=0$ and $x=2$, i.e
$u(0,y)=u(2,y)=0$
does this not only mean that $u$ is periodic for the point $x=0$ rather than $\forall x$?
 A: The equation we get after separation of variables, $X''/X = c$, has two types of solutions:


*

*If $c = \lambda^2 > 0$ we have exponentially growing solutions $X = A\sinh(\lambda x) + B\cosh(\lambda x)$.

*If $c=-\lambda^2<0$ we have the periodic solutions $X = A\sin(\lambda x) + B\cos(\lambda x)$. 


The conditions $X(0)=X(2)=0$ implies $A=B=0$ for the first case so $X\equiv 0$ which cannot match the BC $u=1$ (or $\sin(\pi x/2)$) on the upper side. Therefore the second case - the periodic solution - applies for which $B=0$ and $\lambda = \pi k/2$ for integer $k$.
The full solution is as the author writes 
$$u = \sum B_k \sin(k\pi x/2)\sinh(k\pi y/2)$$
which is periodic in $x$ with period $4$. Since we are only interested in the solution in the region $0\leq x\leq 2$ there will only be $1/2$ period of the solution contained in this region so the word periodic might be a bit confusing here. Note that this means that $x=0$ and $x=2$ are not equivalent points - $x=0$ is equivalent to $x=4$. The way to think about it is that if you extend the solution to all $x$ then the resulting function will be periodic.
