Toronto Math Forum
APM3462019 => APM346Lectures & Home Assignments => Home Assignment 3 => Topic started by: Wanying Zhang on February 01, 2019, 10:40:32 AM

Given the conditions:
$$g(x) = 0,
h(x) =
\begin{cases}
\text{1} & {x < 1}\\
\text{0} & {x \geq 1} \\
\end{cases}
$$
I know since both $g, h$ are even functions, as a result, $u$ is even with respect to $x$. But I don't quite understand how to get the conclusion that $u$ is odd with respect to $t$?

If you change $t\mapsto t$, equation does not change, and $u_{t=0}$ does not change, but $u_t_{t=0}$ acquires sign "$$". However, since $u_{t=0}=0$, if you replace in addition $u\mapsto u$, then nothing changes. Thus $u(x,t)=u(x,t)$.
If on the other hand, $h(x)=0$ then $u$ would be even with respect to $t$