### Author Topic: Halloween Challenge 8  (Read 173 times)

#### Victor Ivrii

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##### Halloween Challenge 8
« on: October 21, 2016, 12:07:50 PM »

#### XinYu Zheng

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##### Re: Halloween Challenge 8
« Reply #1 on: November 04, 2016, 03:05:56 PM »
(a) Make fourier transform in $x$ to obtain $u(x,y)\to\hat{u}(k,y)$. Then the ODE becomes
$$\begin{cases} -k^2\hat{u}+\hat{u}_{yy}=0\\ \hat{u}|_{y=0}=\hat{f}(k), \hat{u}_y|_{y=0}=\hat{g}(k) \end{cases}$$
This has solution $\hat{u}(k,y)=A(k)\sinh(|k|y)+B(k)\cosh(|k|y)$. Plugging in first boundary condition we have get $B(k)=\hat{f}(k)$. The second condition gives $\hat{g}(k)=|k|A(k)\implies A(k)=\hat{g}(k)/|k|$. Note that at $|k|=0$, we have a problem since $A(k)$ could be singular. So we must require as a necessary condition that $\hat{g}(0)=0$, which means $\int_{-\infty}^\infty g(x)\,dx=0$. Now we obtain solution via IFT:
$$u(x,y)=\int_{-\infty}^\infty \left(\frac{\hat{g}(k)}{|k|}\sinh(|k|y)+\hat{f}(k)\cosh(|k|y)\right)e^{ikx}\,dk$$

(b) Same story. We still have $A(k)=\hat{g}(k)/|k|$, but this time the first condition gives $|k|^2B(k)=\hat{f}(k)\implies B(k)=\hat{f}(k)/|k|^2$. Once again, at $|k|=0$ we have a problem, so we require as necessary conditions that $\hat{f}(0)=0$ and $\hat{g}(0)=0$, which is the same as $\int_{-\infty}^\infty g(x)\,dx=\int_{-\infty}^\infty f(x)\,dx=0$. Then the solution is given by the IFT:
$$u(x,y)=\int_{-\infty}^\infty \left(\frac{\hat{g}(k)}{|k|}\sinh(|k|y)+\frac{\hat{f}(k)}{|k|^2}\cosh(|k|y)\right)e^{ikx}\,dk$$