Toronto Math Forum

APM346-2018S => APM346--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 02:47:10 PM

Title: FE-P7
Post by: Victor Ivrii on April 11, 2018, 02:47:10 PM
Solve using (partial) Fourier transform with respect to $y$
&\Delta u:=u_{xx}+u_{yy}=0, &&x>0,\label{7-1}\\
&u_x|_{x=0}= h(y),\label{7-2}\\
&u\to 0 &&\text{as   }x\to +\infty\label{7-3}
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 $h(y)=-g'(y)$.
Title: Re: FE-P7
Post by: Andrew Hardy on April 11, 2018, 06:18:47 PM
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.
Title: Re: FE-P7
Post by: George Lu on April 11, 2018, 06:33:56 PM
By taking a partial fourier transform w.r.t. y, our Laplacian becomes $\hat{u}_{xx}-k^2\hat{u}=0$ which resolves into $\hat{u}=A(k)e^{|k|x}+B(k)e^{-|k|x}$. We can eliminate the first term, because it is unbounded on domain $x>0$, so $\hat{u}(x)=B(k)e^{-|k|x}$. Then $\hat{u}_x=|k|e^{-|k|x}B(k)$.

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: $\hat{u}_x(0,k)=|k|B(k)=-ike^{-|k|} \rightarrow B(k)=-i sgn(k) e^{-|k|} \rightarrow \hat{u}(x,k)=-i sgn(k)e^{-|k|(x+1)}$.

Let us take an inverse fourier transform back to (x,y) coordinates:

$u(x,y)=-i\int_{-\infty}^{\infty} sgn(k)e^{-|k|(x+1)}e^{iky}dk = -i \left ( \int_0^\infty e^{(iy-(x+1))k}dk - \int_{-\infty}^0 e^{(iy+(x+1))k}dk \right ) = -i \left ( \frac{1}{iy-(x+1)} + \frac{1}{iy+(x+1)} \right ) = \frac{2y}{y^2+(x+1)^2}$
Title: Re: FE-P7
Post by: Tristan Fraser on April 11, 2018, 06:34:09 PM
Taking partial fourier transform gives us:   
 $\hat{u}_{xx}-k^2\hat{u}=0$ giving us an ODE, which then can be solved: $\hat{u}=A(k)e^{-|k|x}+B(k)e^{|k|y}$
The second term is discarded since we want a finite solution, otherwise it would violate boundary condition 3.

So now we take $\hat{u}_{x} = -|k|A(k)e^{-|k|x} = 0 =  \hat{h}$

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 = $g = f'(x)$

$\hat{h} =  -ik\hat{g}$

$ -|k|A(k) = -ik e^{-|k|}$, rearranging gives us:

$A(k) = \frac{ike^{-|k|}}{|k|} $

And $u(x,k) = \frac{ike^{-|k|}}{|k|}e^{-|k|x}  $

We can put this in a fourier integral, and integrate to get u(x,y):

$u(x,y) = \int_{-\infty}^{\infty} \frac{ike^{-|k|}}{|k|}e^{-|k|x} e^{iky} dk$
Split into two integrals, to simplify the process:

$u(x,y) = \int_{0}^{\infty} i e^{-k(1+x - iy)} + \int_{-\infty}^{0} -ie^{k(1+x+iy)} = \frac{(i)((1+x - iy) - (1+x+iy))}{(1+x)^2 + y^2} = \frac{2y}{(1+x)^2 + y^2}$
Title: Re: FE-P7
Post by: Jingxuan Zhang on April 11, 2018, 07:21:38 PM
To All,

I think you should have the minus sign.
Title: Re: FE-P7
Post by: Victor Ivrii on April 14, 2018, 09:48:43 AM
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:
u(x,y)=&\int_{-\infty}^\infty \hat{u}(x,\eta)e^{i\eta y}\,d\eta =
i\int_{-\infty}^\infty \sigma(y) e^{-|\eta|x+i\eta y}\,d\eta\\[3pt]
=&-i\int_{-\infty}^0 e^{\eta (1+x+yi)}\,d\eta + i\int_0^{-\infty} e^{-\eta (1+x-yi)}\,d\eta \\[3pt]
=&-\frac{i}{1+x+yi}+\frac{i}{1+x-yi}= -\frac{2y}{(1+x)^2+y^2}.