# Toronto Math Forum

## APM346-2015F => APM346--Home Assignments => HA7 => Topic started by: Victor Ivrii on November 01, 2015, 05:15:37 PM

Title: HA7-P7
Post by: Victor Ivrii on November 01, 2015, 05:15:37 PM
Problem 2

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.3.P.html#problem-5.3.P.2 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.3.P.html#problem-5.2.P.2)
Title: Re: HA7-P7
Post by: Xi Yue Wang 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$$\hat{u}(k,y) = \frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}\\u(x,y) = \int_{-\infty}^{\infty} [\frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}]e^{ikx} dk$$
Title: Re: HA7-P7
Post by: Yumeng Wang on November 05, 2015, 03:39:39 PM
This is my answer for part b. I am not sure whether I do calculation correctly.
Title: Re: HA7-P7
Post by: Xi Yue Wang on November 05, 2015, 03:53:46 PM
Yumeng, I think $e^{-|k|}+e^{|k|} = 2\cosh(|k|)$ not $-2\sinh(|k|)$
Title: Re: HA7-P7
Post by: Victor Ivrii on November 05, 2015, 03:54:21 PM
Yumeng, I think $e^{-|k|}+e^{|k|} = 2\cosh(|k|)$ not $-2\sinh(|k|)$
Indeed
Title: Re: HA7-P7
Post by: Yumeng Wang on November 05, 2015, 04:11:45 PM
Thanks. Hope I get correct answer this time
Title: Re: HA7-P7
Post by: Rong Wei 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.