### Author Topic: integration constant in wave equation  (Read 8064 times)

#### Thomas Nutz

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##### integration constant in wave equation
« 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...
Thanks for your help!

#### Victor Ivrii

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##### Re: integration constant in wave equation
« Reply #1 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...
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
$$u(x,t)=\phi_1(x+ct)+\psi_1(x-ct)=\phi_2(x+ct)+\psi_2(x-ct) \qquad \forall x,t,$$
Then $\phi=\phi_1-\phi_2$ and $\psi=\psi_2-\psi_1$ satisfy
\phi (x+ct)=\psi (x-ct) \qquad \forall x,t
\label{V}

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 $\psi=-c_1$.
« Last Edit: September 24, 2012, 04:45:04 PM by Victor Ivrii »