Toronto Math Forum
APM3462015F => APM346Home Assignments => HA7 => Topic started by: Victor Ivrii on November 01, 2015, 05:15:08 PM

Problem 1
http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter5/S5.3.P.html#problem5.3.P.1 (http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter5/S5.3.P.html#problem5.3.P.1)

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 $$\hat{u}(k,y) = A(k)e^{ky}+B(k)e^{ky}$$
And we discard the term $B(k)e^{ky}$, because it's unbounded.
So, we have $\hat{u} = A(k)e^{ky}, A(k) = \hat{f}(k)$
Then we find the inverse Fourier transform of $e^{ky}$, which is $\int_{\infty}^{\infty} e^{ky}e^{ikx} dk$. $$ = \int_{\infty}^{0} e^{(y+ix)k} dk +\int_{0}^{\infty} e^{(y+ix)k} dk\\=\frac{1}{yix}+\frac{1}{y+ix}\\=\frac{2y}{x^2+y^2}$$
Then$$ u(x,y) = \frac{1}{\pi}\int_{\infty}^{\infty} f(x')\frac{y}{(xx')^2+y^2} dx'$$

The last thing was actually not needed for HA needs  but it is "+"

Solution of part b is attached. Please correct me if I am wrong.

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 $$\hat{u}=A(k)e^{ky}+B(k)e^{ky}$$
Since the domain goes to infinity and we don't want the solution to escape to infinity, then $B(k)=0$.
Thus, $$\hat{u}=A(k)e^{ky}$$
Notice $$\hat{u}_y=k\cdot A(k)e^{ky}$$
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 $$\hat{u} = \frac{\hat{f}(k)}{k}e^{ky}$$
And we write $u$ in a fourier integral form as $$u=\int^{\infty}_{\infty} \frac{\hat{f}(k)}{k}e^{ky} e^{ikx} dk$$

For the solution to part a, at the very end, why did we put $(xx')$? Thank you!

It's a convolution. We are applying Theorem 4a of Section 5.2.

@Chi Ma, thank you!