What are the Boundary Conditions for Solving a Wave Equation?

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Homework Help Overview

The problem involves solving the wave equation \(\phi_{yy}-c^2 \phi_{xx} = 0\) with specified boundary conditions. The original poster discusses the general solution and the implications of applying the boundary conditions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply boundary conditions to the general solution but encounters difficulties in reconciling the conditions with the behavior of the functions involved. Some participants question the implications of the boundary conditions on the nature of the functions F and G.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the boundary conditions and the nature of the solution. Some guidance has been offered regarding the symmetry of the solution, which may address the original poster's concerns.

Contextual Notes

There is a mention of potential inconsistencies arising from the boundary conditions, and the original poster notes a detail regarding the symmetry of the solution that may affect their approach.

blalien
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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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