Toronto Math Forum
APM3462012 => APM346 Math => Misc Math => Topic started by: Thomas Nutz on September 23, 2012, 12:10:16 PM

Good morning,
in the notes "Homogeneous 1D Wave equation" we get to
$$u(x,t)=\phi(x+ct)+\psi(xct)$$ as the general solution, but in the very last paragraph it is mentioned that we could add a constant to $\phi$ if we subtract that same constant from $\psi$, and that this constant would be the only arbitrariness of this solution. But why does this have to be the same constant?
I can add for instance 1434 to $psi$ and subtract 12i from $\phi$ and the sub of the two would still satisfy the PDE, as any derivative of 143412i is zero...
Thanks for your help!

Good morning,
in the notes "Homogeneous 1D Wave equation" we get to
$$
u(x,t)=\phi(x+ct)+\psi(xct)
$$
as the general solution, but in the very last paragraph it is mentioned that we could add a constant to $\phi$ if we subtract that same constant from $\psi$, and that this constant would be the only arbitrariness of this solution. But why does this have to be the same constant?
I can add for instance 1434 to $psi$ and subtract 12i from $\phi$ and the sub of the two would still satisfy the PDE, as any derivative of 143412i is zero...
Thanks for your help!
It is not an arbitrariness of the solution, but arbitrariness of the representation of the given solution in the given form. Really, consider the same solution
$$
u(x,t)=\phi_1(x+ct)+\psi_1(xct)=\phi_2(x+ct)+\psi_2(xct) \qquad \forall x,t,
$$
Then $\phi=\phi_1\phi_2$ and $\psi=\psi_2\psi_1$ satisfy
\begin{equation}\phi (x+ct)=\psi (xct) \qquad \forall x,t
\label{V}
\end{equation}
plugging $t=x/c$ we get$ \phi(2x)=\psi(0)$ and therefore $\phi(x)=C$ for all $x$. Thus $\phi=C$. Then (\ref{V}) implies that $\psi=c_1$.