Toronto Math Forum
APM3462018S => APM346Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 02:47:10 PM

Solve using (partial) Fourier transform with respect to $y$
\begin{align}
&\Delta u:=u_{xx}+u_{yy}=0, &&x>0,\label{71}\\
&u_x_{x=0}= h(y),\label{72}\\
&u\to 0 &&\text{as }x\to +\infty\label{73}
\end{align}
with $h(y)=\frac{4y}{(y^2+1)^2}$.
Hint. Fourier transform of $g(y)=\frac{2}{y^2+1}$ is $\hat{g}=e^{\eta}$ and $h(y)=g'(y)$.

We begin by transforming $$ \hat{u} = \hat{u}_{xx} \eta^2 \hat{u}_{\eta\eta} $$
and without having a singularity we only have one term. Then we know from the hint
$$ \hat{h}=i\eta e^{\eta} $$ and so
$$ \hat{u} = i\frac{\eta}{\eta} e^{\eta(x+1)} $$ of which we can inverse Fourier transform by the properties in the appendix
$$ u = \frac{2y}{y^2+(x+1)^2} $$
I missed the $ \frac{1}{\eta} $ term on the final and hope Dr. Ivrii is merciful.

By taking a partial fourier transform w.r.t. y, our Laplacian becomes $\hat{u}_{xx}k^2\hat{u}=0$ which resolves into $\hat{u}=A(k)e^{kx}+B(k)e^{kx}$. We can eliminate the first term, because it is unbounded on domain $x>0$, so $\hat{u}(x)=B(k)e^{kx}$. Then $\hat{u}_x=ke^{kx}B(k)$.
Before this can be solved, we must take the fourier transform of h(y). Using the hint that $h(y)=g'(y), \hat{g}=e^{k} \rightarrow \hat{h}=ike^{k}$.
Now: $\hat{u}_x(0,k)=kB(k)=ike^{k} \rightarrow B(k)=i sgn(k) e^{k} \rightarrow \hat{u}(x,k)=i sgn(k)e^{k(x+1)}$.
Let us take an inverse fourier transform back to (x,y) coordinates:
$u(x,y)=i\int_{\infty}^{\infty} sgn(k)e^{k(x+1)}e^{iky}dk = i \left ( \int_0^\infty e^{(iy(x+1))k}dk  \int_{\infty}^0 e^{(iy+(x+1))k}dk \right ) = i \left ( \frac{1}{iy(x+1)} + \frac{1}{iy+(x+1)} \right ) = \frac{2y}{y^2+(x+1)^2}$

Taking partial fourier transform gives us:
$\hat{u}_{xx}k^2\hat{u}=0$ giving us an ODE, which then can be solved: $\hat{u}=A(k)e^{kx}+B(k)e^{ky}$
The second term is discarded since we want a finite solution, otherwise it would violate boundary condition 3.
So now we take $\hat{u}_{x} = kA(k)e^{kx} = 0 = \hat{h}$
What is $\hat{h}$
We know that $h(y) = g'(y)$ and $g(y) = \frac{2}{y^2 + 1}$ with fourier transform of $e^{k}$. We then use a property of the fourier transform, where $\hat{g} = ik\hat{f}$ implies = $g = f'(x)$
Thus:
$\hat{h} = ik\hat{g}$
Thus
$ kA(k) = ik e^{k}$, rearranging gives us:
$A(k) = \frac{ike^{k}}{k} $
And $u(x,k) = \frac{ike^{k}}{k}e^{kx} $
We can put this in a fourier integral, and integrate to get u(x,y):
$u(x,y) = \int_{\infty}^{\infty} \frac{ike^{k}}{k}e^{kx} e^{iky} dk$
Split into two integrals, to simplify the process:
$u(x,y) = \int_{0}^{\infty} i e^{k(1+x  iy)} + \int_{\infty}^{0} ie^{k(1+x+iy)} = \frac{(i)((1+x  iy)  (1+x+iy))}{(1+x)^2 + y^2} = \frac{2y}{(1+x)^2 + y^2}$

To All,
I think you should have the minus sign.

Indeed, George and Tristan lost "$$" (but in Tristan's exam paper there was one).
How to calculate F.T. of $h(y)$? We observe that $h(y)'=\partial_y \phi(y)$ $\phi(y)=2(1+y^2)^{1}$ and therefore $\hat{h}(\eta)= i\eta \hat{\phi}(\eta)= i \eta e^{\eta}$.
More detailed final calculations:
\begin{align*}
u(x,y)=&\int_{\infty}^\infty \hat{u}(x,\eta)e^{i\eta y}\,d\eta =
i\int_{\infty}^\infty \sigma(y) e^{\etax+i\eta y}\,d\eta\\[3pt]
=&i\int_{\infty}^0 e^{\eta (1+x+yi)}\,d\eta + i\int_0^{\infty} e^{\eta (1+xyi)}\,d\eta \\[3pt]
=&\frac{i}{1+x+yi}+\frac{i}{1+xyi}= \frac{2y}{(1+x)^2+y^2}.
\end{align*}