Toronto Math Forum
APM346-2015F => APM346--Home Assignments => HA7 => Topic started by: Victor Ivrii on November 01, 2015, 05:15:37 PM
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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)
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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$$\hat{u}(k,y) = \frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}\\u(x,y) = \int_{-\infty}^{\infty} [\frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}]e^{ikx} dk$$
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This is my answer for part b. I am not sure whether I do calculation correctly.
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Yumeng, I think $e^{-|k|}+e^{|k|} = 2\cosh(|k|) $ not $ -2\sinh(|k|)$
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Yumeng, I think $e^{-|k|}+e^{|k|} = 2\cosh(|k|) $ not $ -2\sinh(|k|)$
Indeed
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Thanks. Hope I get correct answer this time
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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|)