Toronto Math Forum
APM3462016F => APM346Lectures => Chapter 2 => Topic started by: ziyao hu 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?

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.

So what's the meaning of "up to constants that will not affect u(x,t)"

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).

So what's the meaning of "up to constants that will not affect u(x,t)"
It should be in the context: $u(x,t)=\phi(x+ct)+\psi (xct)$; if we replace $\phi$ by $\phi+C$ and $\psi$ by $\psiC$ then $u$ does not change