Toronto Math Forum
APM3462015F => APM346Home Assignments => HA7 => Topic started by: Victor Ivrii on November 01, 2015, 05:15:37 PM

Problem 2
http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter5/S5.3.P.html#problem5.3.P.2 (http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter5/S5.3.P.html#problem5.2.P.2)

For part (a), similar to problem 6, we get $$\hat{u}(k,y) = A(k)e^{ky}+B(k)e^{ky}$$
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$$\hat{u}(k,y) = \frac{\sinh(k(1y))\hat{f}(k)}{\sinh(k)}+\frac{\sinh(ky)\hat{g}(k)}{\sinh(k)}\\u(x,y) = \int_{\infty}^{\infty} [\frac{\sinh(k(1y))\hat{f}(k)}{\sinh(k)}+\frac{\sinh(ky)\hat{g}(k)}{\sinh(k)}]e^{ikx} dk$$

This is my answer for part b. I am not sure whether I do calculation correctly.

Yumeng, I think $e^{k}+e^{k} = 2\cosh(k) $ not $ 2\sinh(k)$

Yumeng, I think $e^{k}+e^{k} = 2\cosh(k) $ not $ 2\sinh(k)$
Indeed

Thanks. Hope I get correct answer this time

hmmmm...... I think the final answer from Yumeng is still not correct.
should be something about cosh(y(1y)) exists.
Furthermore, eâˆ’ky+eky=2cosh(ky) instead of 2cosh(k)