# Toronto Math Forum

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

Title: integration constant in wave equation
Post by: Thomas Nutz on September 23, 2012, 12:10:16 PM
Good morning,
in the notes "Homogeneous 1D Wave equation" we get to
$$u(x,t)=\phi(x+ct)+\psi(x-ct)$$ 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 1434-12i is zero...
Title: Re: integration constant in wave equation
Post by: Victor Ivrii on September 23, 2012, 02:33:46 PM
Good morning,
in the notes "Homogeneous 1D Wave equation" we get to
$$u(x,t)=\phi(x+ct)+\psi(x-ct)$$
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 1434-12i is zero...
$$u(x,t)=\phi_1(x+ct)+\psi_1(x-ct)=\phi_2(x+ct)+\psi_2(x-ct) \qquad \forall x,t,$$
Then $\phi=\phi_1-\phi_2$ and $\psi=\psi_2-\psi_1$ satisfy
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 $\psi=-c_1$.