### Author Topic: Web Bonus Problem to Week 8 (#3)  (Read 2575 times)

#### Victor Ivrii

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##### Web Bonus Problem to Week 8 (#3)
« on: November 01, 2015, 05:19:24 PM »

#### Xi Yue Wang

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##### Re: Web Bonus Problem to Week 8 (#3)
« Reply #1 on: November 09, 2015, 11:43:14 PM »
For the Robin boundary condition problem, we get transform $$\hat{u}(k,y) = A(k)e^{-|k|y} + B(k)e^{|k|y}$$
We discard $B(k)e^{|k|y}$, because it is unbounded.
The Robin boundary condition, $$(\hat{u}_y + \alpha\hat{u})|_{y=0} = \hat{f}(k)\\A(k) = -\frac{\hat{f}(k)}{(|k|-\alpha)}\\\hat{u}(k,y)= -\frac{\hat{f}(k)}{(|k| - \alpha)}e^{-|k|y}\\u(x,y) = \int_{-\infty}^{\infty}- \frac{\hat{f}(k)}{(|k| - \alpha)}e^{-|k|y+ikx} dk$$
For $\alpha<0$, the expression is a nice function (is correct from the physical point of view).
For $\alpha>0$, ?
For $\alpha = 0$, A(k) could be singular at $k = 0$, to avoid it we need $\int_{-\infty}^{\infty} f(x) dx = 0$.