Toronto Math Forum
APM3462021S => APM346Tests & Quizzes => Test2 => Topic started by: aryakim on March 06, 2021, 10:07:36 AM

I had a question about problem $1$ of my Test$2$. (My version was called alternativeF, 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 $tx$ 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.

Solution is allowed to be discontinuous.