Toronto Math Forum
APM346-2016F => APM346--Lectures => Chapter 2 => Topic started by: ziyao hu on October 17, 2016, 06:15:10 PM
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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?
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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.
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So what's the meaning of "up to constants that will not affect u(x,t)"
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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).
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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