Solving a wave equation

  • Thread starter blalien
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  • #1
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Homework Statement


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

Homework Equations


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

The Attempt at a Solution


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

So [tex]F'[/tex] and [tex]G'[/tex] both vanish. Then how do we apply the first boundary condition?
 

Answers and Replies

  • #2
fzero
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The first boundary condition is (in your notation)

[tex]F'(\omega) - G'(\omega) = f'(\omega).[/tex]

As long as G is neither odd nor even, there's no inconsistency.
 
  • #3
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The problem is that the two boundary conditions
[tex]F(\omega)+G(-\omega)=0[/tex] and
[tex]F'(\omega)+G'(-\omega)=0[/tex]
imply that [tex]F(\omega)[/tex] and [tex]G(\omega)[/tex] are constant. This does create an inconsistency.
 
  • #4
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I missed a detail that the solution is symmetric with y. That solves the problem. Thanks anyway.
 
Last edited:

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