# Toronto Math Forum

## APM346-2015F => APM346--Home Assignments => HA7 => Topic started by: Victor Ivrii on November 01, 2015, 05:15:08 PM

Title: HA7-P6
Post by: Victor Ivrii on November 01, 2015, 05:15:08 PM
Problem 1

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.3.P.html#problem-5.3.P.1 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.3.P.html#problem-5.3.P.1)
Title: Re: HA7-P6
Post by: Xi Yue Wang on November 05, 2015, 12:31:17 PM
For part (a), let make partial Fourier transform with respect to $x\mapsto k, u(x,t) \mapsto \hat{u}(k,t)$, then we get $$\hat{u}_{yy} - k^2\hat{u} = 0\\\hat{u|}_{y=0} = \hat{f}(k)$$
Solving this ODE, then we have $$\hat{u}(k,y) = A(k)e^{-|k|y}+B(k)e^{|k|y}$$
And we discard the term $B(k)e^{|k|y}$, because it's unbounded.
So, we have $\hat{u} = A(k)e^{-|k|y}, A(k) = \hat{f}(k)$
Then we find the inverse Fourier transform of $e^{-|k|y}$, which is $\int_{-\infty}^{\infty} e^{-|k|y}e^{ikx} dk$. $$= \int_{-\infty}^{0} e^{(y+ix)k} dk +\int_{0}^{\infty} e^{(-y+ix)k} dk\\=\frac{1}{y-ix}+\frac{1}{y+ix}\\=\frac{2y}{x^2+y^2}$$
Then$$u(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(x')\frac{y}{(x-x')^2+y^2} dx'$$
Title: Re: HA7-P6
Post by: Victor Ivrii on November 05, 2015, 01:24:11 PM
The last thing was actually not needed for HA needs -- but it is "+"
Title: Re: HA7-P6
Post by: Yumeng Wang on November 05, 2015, 02:07:24 PM
Solution of part b is attached. Please correct me if I am wrong.
Title: Re: HA7-P6
Post by: Yeming Wen on November 06, 2015, 07:56:52 AM
part b.
First, we make a partial fourier transform $u(x,y) \mapsto \hat{u}(k,y)$.
Then $\hat{u}_{kk} = -k^2 \hat{u}$.
Thus we have $$\hat{u}_{yy}-k^2\hat{u}=0$$
Solve this ODE, we have $$\hat{u}=A(k)e^{-|k|y}+B(k)e^{|k|y}$$
Since the domain goes to infinity and we don't want the solution to escape to infinity, then $B(k)=0$.
Thus, $$\hat{u}=A(k)e^{-|k|y}$$
Notice $$\hat{u}_y=-|k|\cdot A(k)e^{-|k|y}$$
So we plug in the boundary condition, we get $$\hat{u}_y=-|k|\cdot A(k)=\hat{f}(k) \implies A(k)=-\frac{\hat{f}(k)}{|k|}$$
Thus $$\hat{u} = -\frac{\hat{f}(k)}{|k|}e^{-|k|y}$$
And we write $u$ in a fourier integral form as $$u=\int^{\infty}_{-\infty} -\frac{\hat{f}(k)}{|k|}e^{-|k|y} e^{ikx} dk$$
Title: Re: HA7-P6
Post by: Emily Deibert on November 15, 2015, 11:13:35 PM
For the solution to part a, at the very end, why did we put $(x-x')$? Thank you!
Title: Re: HA7-P6
Post by: Chi Ma on November 15, 2015, 11:44:54 PM
It's a convolution. We are applying Theorem 4a of Section 5.2.
Title: Re: HA7-P6
Post by: Emily Deibert on November 16, 2015, 08:22:12 AM
@Chi Ma, thank you!