Toronto Math Forum
APM346-2016F => APM346--Tests => Q5 => Topic started by: Roro Sihui Yap on November 03, 2016, 08:34:00 PM
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\begin{align}
&u_{xx}+u_{yy}=0,\qquad -\infty<x<\infty, 0<y<1, \\
&u|_{y=0}=f(x), \quad u_y|_{y=1}=g(x).
\end{align}
Taking fourier transform,
\begin{align}
&-k^2\hat{u} +\hat{u}_{yy}=0 \\
&\hat{u}|_{y=0}=\hat{f}(k) \quad \hat{u_y}|_{y=1}=\hat{g}(k)
\end{align}
From equation (3): $\hat{u}(k,y) = A(k)e^{-ky}+B(k)e^{ky}$
From equation (4):
$\hat{u}(k,0) = A(k) +B(k) = \hat{f}(k) $
$\hat{u_y}(k,1) = -kA(k)e^{-k} +kB(k)e^{k} = \hat{g}(k) $
Solving the two equations we get
$$A(k) = \frac{\hat{f}(k)ke^k - \hat{g}(k)}{2k\cosh(k)}$$
$$B(k) = \frac{\hat{f}(k)ke^{-k} + \hat{g}(k)}{2k\cosh(k)}$$
$$\hat{u}(k,y) = \frac{1}{2k\cosh(k)}\big[\hat{f}(k)ke^{k-ky} - \hat{g}(k)e^{-ky}+\hat{f}(k)ke^{-k+ky} + \hat{g}(k)e^{ky} \big]$$
$$\hat{u}(k,y) = \frac{1}{k\cosh(k)}\big[\hat{f}(k)k\cosh(k-ky) + \hat{g}(k)\sinh(ky) \big]$$
$$ u(x,y) = \int_{-\infty}^{\infty} \big[\frac{\hat{f}(k)\cosh(k-ky)}{\cosh(k)} + \frac{\hat{g}(k)\sinh(ky)}{k\cosh(k)} \big] e^{ikx} dk $$