Author Topic: P3  (Read 1495 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2553
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
P3
« on: February 15, 2018, 07:05:01 PM »
Find  continuous solution to
\begin{align}
&u_{tt}-4u_{xx}=0, &&   t>0, x>-t,\tag{1}\\
&u|_{t=0}=0, && x>0,\tag{2}\\
&u_t|_{t=0}=0, && x>0,\tag{3}\\
&u_x|_{x=-t}= \sin(t), &&t>0.\tag{4}
\end{align}

Ioana Nedelcu

  • Full Member
  • ***
  • Posts: 29
  • Karma: 3
    • View Profile
Re: P3
« Reply #1 on: February 15, 2018, 09:51:06 PM »
General solution for $ x >-t $: $ u = \psi(x+2t) + \phi(x-2t) $

Given initial conditions, we have $ \psi(x) = \phi(x) = 0 $ so $$ u = 0, x > 2t $$

Using the boundary conditions for $ -t < x < 2t $, ie where $\phi(x), x<0 $

$$ \psi'(t) + \phi'(-3t) = \sin(t) $$
From previous solution $$ \psi(x) = 0 \implies \psi'(x) = 0 $$
So $$ \phi(t) = -\cos(\frac{-t}{3}) + constant, t <0 $$
$$ \phi(t) = -\cos(\frac{t}{3}) + constant, t<0 $$
$$ \phi(x - 2t) = -\cos(\frac{(x-2t)}{3}) + constant $$

For $\phi$ to be constant at x = 2t, ie the same value from both sides of the characteristic equation:

$$ 0 = -\cos(0) + c \implies c = 1 $$

So $ u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t $

Jilong Bi

  • Jr. Member
  • **
  • Posts: 8
  • Karma: 9
    • View Profile
Re: P3
« Reply #2 on: February 15, 2018, 10:16:26 PM »
should $\phi(t) = 3cos(-\frac{1}{3}t) + constant $ ? And the answer for -t<x<2t is 3$cos(\frac{x-2t}{3}) -3$

Ioana Nedelcu

  • Full Member
  • ***
  • Posts: 29
  • Karma: 3
    • View Profile
Re: P3
« Reply #3 on: February 15, 2018, 11:09:40 PM »
Yeah you're right, looks like I forgot to write it with the chain rule

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2553
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: P3
« Reply #4 on: February 22, 2018, 06:47:10 AM »
OK