What are the Boundary Conditions for Solving a Wave Equation?

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SUMMARY

The discussion focuses on solving the wave equation \(\phi_{yy}-c^2 \phi_{xx} = 0\) with specific boundary conditions. The general solution is given by \(\phi(x,y)=F(x+cy)+G(x-cy)\). The application of boundary conditions leads to two equations: \(F(\omega)+G(-\omega)=0\) and \(F'(\omega)+G'(-\omega)=0\). The inconsistency arises when both functions \(F\) and \(G\) are constant, which is resolved by recognizing the symmetry of the solution with respect to \(y\).

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Homework Statement


The problem is to solve
\phi_{yy}-c^2 \phi_{xx} = 0
\phi_y (x,0) = f'(x), x>0
\phi_x (0,y) = \phi(0,y) = 0, y>0 or y<0

Homework Equations


The solution, before applying boundary conditions is obviously
\phi(x,y)=F(x+c y)+G(x-cy)

The Attempt at a Solution


I start with the general solution
\phi(x,y)=F(x+c y)+G(x-cy)
and apply the two vanishing boundary conditions
\phi(0,y)=F(c y)+G(-cy)=0 or
1) F(\omega)+G(-\omega)=0
\phi_x(0,y)=F'(c y)+G'(-cy)=0 or
2) F'(\omega)+G'(-\omega)=0
Take the derivative of equation 1:
F'(\omega)-G'(-\omega)=0

So F' and G' both vanish. Then how do we apply the first boundary condition?
 
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The first boundary condition is (in your notation)

F'(\omega) - G'(\omega) = f'(\omega).

As long as G is neither odd nor even, there's no inconsistency.
 
The problem is that the two boundary conditions
F(\omega)+G(-\omega)=0 and
F'(\omega)+G'(-\omega)=0
imply that F(\omega) and G(\omega) are constant. This does create an inconsistency.
 
I missed a detail that the solution is symmetric with y. That solves the problem. Thanks anyway.
 
Last edited:

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