### Author Topic: Q5  (Read 1904 times)

#### Roro Sihui Yap

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##### Q5
« on: November 03, 2016, 08:34:00 PM »
\begin{align}
\end{align}

Taking fourier transform,
\begin{align}
&-k^2\hat{u} +\hat{u}_{yy}=0 \\
\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
$$A(k) = \frac{\hat{f}(k)ke^k - \hat{g}(k)}{2k\cosh(k)}$$
$$B(k) = \frac{\hat{f}(k)ke^{-k} + \hat{g}(k)}{2k\cosh(k)}$$

$$\hat{u}(k,y) = \frac{1}{2k\cosh(k)}\big[\hat{f}(k)ke^{k-ky} - \hat{g}(k)e^{-ky}+\hat{f}(k)ke^{-k+ky} + \hat{g}(k)e^{ky} \big]$$
$$\hat{u}(k,y) = \frac{1}{k\cosh(k)}\big[\hat{f}(k)k\cosh(k-ky) + \hat{g}(k)\sinh(ky) \big]$$

$$u(x,y) = \int_{-\infty}^{\infty} \big[\frac{\hat{f}(k)\cosh(k-ky)}{\cosh(k)} + \frac{\hat{g}(k)\sinh(ky)}{k\cosh(k)} \big] e^{ikx} dk$$
« Last Edit: November 03, 2016, 08:57:01 PM by Roro Sihui Yap »