Author Topic: HA #4 general  (Read 219 times)

ziyao hu

  • Newbie
  • *
  • Posts: 4
  • Karma: 0
    • View Profile
HA #4 general
« on: October 17, 2016, 06:15:10 PM »
I saw the solution professor posted last year, as the folloing picture.
Can we just assume constant = 0 and then put one at the end of the solution?

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: HA #4 general
« Reply #1 on: October 17, 2016, 10:42:24 PM »
Can we just assume constant = 0 and then put one at the end of the solution?
No since in the different places this constant enters differently.


ziyao hu

  • Newbie
  • *
  • Posts: 4
  • Karma: 0
    • View Profile
Re: HA #4 general
« Reply #2 on: October 18, 2016, 07:05:46 PM »
So what's the meaning of "up to constants that will not affect u(x,t)"

Shentao YANG

  • Full Member
  • ***
  • Posts: 24
  • Karma: 0
    • View Profile
Re: HA #4 general
« Reply #3 on: October 18, 2016, 09:13:21 PM »
I guess the meaning is: even if you take the constant as $0$ (which indeed should be $ \pm {c \over 2}$ for some constant $c$), $u$ function you get will still satisfy the defining equations in the problem, hence, be a solution.
And, in fact, I think you will always get the same solution to $u$ for any integration constant you put (since the constant in $\psi $ and $\phi $ will cancel each other).

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: HA #4 general
« Reply #4 on: October 18, 2016, 09:31:54 PM »
So what's the meaning of "up to constants that will not affect u(x,t)"

It should be in the context: $u(x,t)=\phi(x+ct)+\psi (x-ct)$; if we replace $\phi$ by $\phi+C$ and $\psi$ by $\psi-C$ then $u$ does not change