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 Quote by fluidistic Ok thank you vela! This means that $\frac{\partial u }{\partial y} \big | _{y=0}=0$. Thus the PDE is equivalent to $\frac{\partial ^2 U_c (p,x)}{\partial x^2}-p^2 U_c (p,x)=0$. Since $p>0$, $U_c(p,x)=Ae^{px}+Be^{-px}$. Now I think it's time to take the inverse cosine transform.
$$U_c(x, p) = A(p)e^{px} + B(p)e^{-px}$$ You want a bounded solution as ##x \to \infty##, so you can toss the first term.
 Quote by fluidistic So this gives me $\mathbb{F} _c ^{-1} [U_c(p,x)]=u(x,y)=\frac{2}{\pi} \int _0 ^{\infty} U_c (p,x) \cos (py)dp$. Is this ok? $U_c(p,x)=Ae^{px}+Be^{-px}$. So that $u(x,y)=\frac{2}{\pi} \int _0 ^{\infty } (Ae^{px}+Be^{-px} ) \cos (py)dp$. 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.