Author Topic: Problem2(17) even or odd?  (Read 312 times)

Wanying Zhang

  • Full Member
  • ***
  • Posts: 21
  • Karma: 6
    • View Profile
Problem2(17) even or odd?
« 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 $t$?

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2466
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem2(17) even or odd?
« Reply #1 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 $t$