I had a question about problem $1$ of my Test$2$. (My version was called alternative-F, night section).There was something confusing about this problem that I realized during the test.

Here is the problem:

\begin{equation}

\nonumber

\left\{ \begin{aligned}

& u_{tt}-u_{xx}=0, &&0< t < \pi, 0 < x < \pi, &(1.1) \\\

&u|_{t=0}= 2\cos (x), && 0< x < \pi, &(1.2)\\

&u_t|_{t=0}= 0, && 0< x < \pi, &(1.3)\\

&u|_{x=0}= u|_{x= \pi}=0, && 0< t < \pi. &(1.4)

\end{aligned}

\right.

\end{equation}

So, on the $t-x$ diagram, the lines $t = 0$ and $x = \pi$ intersect at ($t=0,x=\pi$), which will be a boundary point of the region where $0<t<x<\pi$.

If this point (i.e. ($t=0,x=\pi$)) on the diagram is approached by the line $t = 0$, equation $(1.2)$ is used to conclude that the value of $u(x,t)$ approaches $-2$.

On the other hand, if the point is approached from the line $x = \pi$, $u(x,t)$ should become zero, which is in contradiction with the other boundary condition. It seems this would make it impossible to incorporate conditions $(1.2)$ and $(1.4$) at the same time. I was hoping someone could clear my confusion regarding this problem.