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vela
#11
Feb12-12, 10:45 PM
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Quote Quote by fluidistic View Post
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.
Remember that the "constants" can still depend on p. That is,
$$U_c(x, p) = A(p)e^{px} + B(p)e^{-px}$$ You want a bounded solution as ##x \to \infty##, so you can toss the first term.

Quote Quote by fluidistic View Post
So this gives me [itex]\mathbb{F} _c ^{-1} [U_c(p,x)]=u(x,y)=\frac{2}{\pi} \int _0 ^{\infty} U_c (p,x) \cos (py)dp[/itex]. Is this ok?
[itex]U_c(p,x)=Ae^{px}+Be^{-px}[/itex]. So that [itex]u(x,y)=\frac{2}{\pi} \int _0 ^{\infty } (Ae^{px}+Be^{-px} ) \cos (py)dp[/itex]. This doesn't look a correct answer to me though, let alone how to simplify it and calculate A and B from the boundary conditions.
Before you take the inverse transform, you want to incorporate the boundary condition for x=0 by doing essentially what was done on pages 242 and 243 in Mathews and Walker to determine B(p).