# Toronto Math Forum

## APM346--2019 => APM346--Lectures & Home Assignments => Home Assignment 3 => Topic started by: Wanying Zhang on February 01, 2019, 10:40:32 AM

Title: Problem2(17) even or odd?
Post 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 $t$?
Title: Re: Problem2(17) even or odd?
Post by: Victor Ivrii on February 01, 2019, 04:56:18 PM
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 $t$