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fluidistic
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#9
Feb11-12, 10:52 AM
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P: 3,173
Ok thank you vela!
This means that [itex]\frac{\partial u }{\partial y} \big | _{y=0}=0[/itex].
Thus the PDE is equivalent to [itex]\frac{\partial ^2 U_c (p,x)}{\partial x^2}-p^2 U_c (p,x)=0[/itex]. Since [itex]p>0[/itex], [itex]U_c(p,x)=Ae^{px}+Be^{-px}[/itex]. Now I think it's time to take the inverse cosine transform.