Author Topic: HA7-P7  (Read 1928 times)

Victor Ivrii

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Xi Yue Wang

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Re: HA7-P7
« Reply #1 on: November 05, 2015, 02:28:48 PM »
For part (a), similar to problem 6, we get $$\hat{u}(k,y) = A(k)e^{-|k|y}+B(k)e^{|k|y}$$
Because we have two boundary conditions $$\hat{u|}_{y=0} = \hat{f}(k)\\\hat{u|}_{y=1} = \hat{g}(k)$$
We cannot discard anything. Then,$$A(k) + B(k) = \hat{f}(k)\\A(k)e^{-|k|}+B(k)e^{|k|} = \hat{g}(k)$$
Then we get, $$A(k) = \frac{e^{|k|}\hat{f}(k)}{2\sinh(|k|)} - \frac{\hat{g}(k)}{2\sinh(|k|)}\\B(k)=-\frac{e^{-|k|}\hat{f}(k)}{2\sinh(|k|)}+\frac{\hat{g}(k)}{2\sinh(|k|)}$$
Hence, we get$$\hat{u}(k,y) = \frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}\\u(x,y) = \int_{-\infty}^{\infty} [\frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}]e^{ikx} dk$$

Yumeng Wang

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Re: HA7-P7
« Reply #2 on: November 05, 2015, 03:39:39 PM »
This is my answer for part b. I am not sure whether I do calculation correctly.

Xi Yue Wang

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Re: HA7-P7
« Reply #3 on: November 05, 2015, 03:53:46 PM »
Yumeng, I think $e^{-|k|}+e^{|k|} = 2\cosh(|k|) $ not $ -2\sinh(|k|)$

Victor Ivrii

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Re: HA7-P7
« Reply #4 on: November 05, 2015, 03:54:21 PM »
Yumeng, I think $e^{-|k|}+e^{|k|} = 2\cosh(|k|) $ not $ -2\sinh(|k|)$
Indeed

Yumeng Wang

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Re: HA7-P7
« Reply #5 on: November 05, 2015, 04:11:45 PM »
Thanks. Hope I get correct answer this time 

Rong Wei

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Re: HA7-P7
« Reply #6 on: November 05, 2015, 05:08:01 PM »
hmmmm...... I think the final answer from Yumeng is still not correct.
should be something about cosh(|y|(1-y)) exists.
Furthermore,  e−|k|y+e|k|y=2cosh(|k|y) instead of 2cosh(|k|)