Author Topic: Halloween Challenge 2  (Read 197 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii

Roro Sihui Yap

  • Full Member
  • ***
  • Posts: 30
  • Karma: 16
    • View Profile
Re: Halloween Challenge 2
« Reply #1 on: October 21, 2016, 05:47:44 PM »
10a)
$$u(x,t) = f(x+t) + g(x-t) $$

From $u|_{t=0} = 0$, $\ \ f(x) + g(x) = 0$ $\rightarrow f(x)= -g(x)$
From $u|_{x=0} = 0$, $\ \ f(t) + g(-t) = 0 \ \ $
Using $f(x)= -g(x), \ $we get $\ -g(t) + g(-t) =0 $ $\rightarrow g(-t)= g(t)$

From $u|_{x=a} = 0$, $\ \ f(a+t) + g(a -t) = 0$ $\rightarrow  g(t-a) = g(t+a)$ $\rightarrow  g(x) = g(x +2a)\label{A}$
From $u|_{t=b} = 0$, $\ \ f(x+b) + g(x-b) = 0$ $\rightarrow g(x-b) = g(x+b)$ $\rightarrow g(x) = g(x+2b)\label{B}$

$g$ is a periodic function, suppose g(x) is not constant function, let the period of $g$ be $p$, the smallest number such that $g(x) = g(x+np)$ where $n$ is any integer and $p \ne 0$,

Then, $2a = k_1p$ and $2b = k_2p$ where $k_1,k_2$ are integers
$$\frac{2a}{k_1} = \frac{2b}{k_2} \implies \frac{a}{b} = \frac{k_1}{k_2}$$

Contradiction, we assumed a and b are not commensurable. Thus $g(x)$ must be a constant function, suppose $g(x) = k$, then $f(x) = -k$ where k is any number

$$u(x,t) = f(x+t) + g(x-t) = k - k = 0$$
The only solution is the trivial solution
« Last Edit: October 21, 2016, 05:51:09 PM by Roro Sihui Yap »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Halloween Challenge 2
« Reply #2 on: October 30, 2016, 08:02:31 AM »
10b) is a simple substitution