# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment Y => Topic started by: Calvin Arnott on November 11, 2012, 12:49:08 PM

Title: Problem 1
Post by: Calvin Arnott on November 11, 2012, 12:49:08 PM
I take the definition of the Fourier transform $\hat{f}$ for a function $f$ to be: $F\left(k\right) = \hat{f}\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx,$ with inverse Fourier transform $f\left(x\right) = \check{F}\left(x\right) = \int_{-\infty}^{\infty}F\left(k\right) e^{i k x}\frac{dk}{2 \pi}$.
Problem: Find a solution using the Fourier transform for the Laplace equation in the half-plane (it was a misprint in the problem. V.I.):
$$\Delta u = u_{xx} + u_{yy} = 0 : \{x > 0, -\infty < y < \infty\}$$
with the boundary conditions:
$$u(0,y) = e^{-|y|}, \phantom{\ } \max_{\{x,y\}} |u| < \infty$$
Answer: Now, $x > 0$ so we cannot perform a Fourier transform over $x$. We proceed instead by partially transforming with respect to $y \mapsto k$, $u(x,y) \mapsto \mathcal{F}_y (u)(x,k)$:
$$\text{Let: } U(x,k) = \mathcal{F}_y (u)(x,k) = \int_{-\infty}^{\infty} u(x,y) e^{-i k y} d y$$
We know that this transformation converges and exists as we have that $|u|$ is bounded by some constant $c < \infty$. Recall that the Fourier transform $\mathcal{F}$ is an operator, and that for a function $f(x)$ with transform $\mathcal{F}(f)(k)$, a property of the transform $\mathcal{F}$ is that for $g(x) = \partial_x f(x)$, $\mathcal{F}(g)(k) = i k \mathcal{F}(f)(k)$. Applying the partial transformation $\mathcal{F}_y$ to our PDE and BC then yields us:
$$\mathcal{F}_y((u_{xx} + u_{yy})(x,y) = (0)(x,y)) \mapsto \mathcal{F}_y(u_{xx})(x,k) + \mathcal{F}_y(u_{yy})(x,k) = \mathcal{F}_y(0)(x,k) = (0)(x,k)$$
$$= \mathcal{F}(\partial_{x}^2 u)(x,k) + \mathcal{F}(\partial_{y}^2 u)(x,k) = \partial_{x}^2 \mathcal{F}(u)(x,k) + (i k)^2 \mathcal{F}(u)(x,k) = 0$$
$$\text{And BC: } \mathcal{F}_y(u(0,y) = e^{-|y|}) \mapsto \mathcal{F}_y(u)(0,k) = \mathcal{F}_y (e^{-|y|})(k) = \frac{2}{1+k^2}$$
Where we differentiated $\mathcal{F}_y(\partial_{x}^2 u)(x,k)$ under the sign, and used that because $\mathcal{F}$ is an operator, we must have $\mathcal{F}(0) = 0$. Using $U(x,k) = \mathcal{F}_y (u)(x,k)$, we then have an ODE in $x$ with BC:
$$U_{xx} + (i k)^2 U = U_{xx} -k^2 U = 0$$
$$U(0,k) = \mathcal{F}_y (e^{-|y|})(k) = \frac{2}{1+k^2}$$
Using our previously derived transformation for $e^{-|y|}$. This has solution $U(x,k) = \Psi(k) e^{- |k| x} + \Phi(k) e^{+ |k| x}$, with $\{ \Psi(k),\Phi(k)\}$ arbitrary functions of $k$, where we take $|k|$ in place of $k$ to control the asymptotic behavior as $\{|k|,x\} \rightarrow \infty$. Since $x > 0$ we must exclude $e^{|k| x}$ in our solution since it grows without bound as $|k| \rightarrow \infty$ and so doesn't allow for the convergence of the inverse Fourier transformation. Then $U(x,k) = \Psi(k) e^{- |k| x}$. Plugging in our BC:
$$U(0,k) = \frac{2}{1+k^2} = (\Psi(k) e^{- |k| x})\bigr|_{x=0} = \Psi(k)$$
$$\implies \Psi(k) = \frac{2}{1+k^2} \implies U(x,k) = \frac{2}{1+k^2} e^{- |k| x}$$
Finally, we apply the inverse Fourier Transformation $\mathcal{F}^{-1}$ with respect to $k$ on $U(x,k)$ to solve in terms of $u(x,y)$:
$$\mathcal{F}^{-1}_k (U(x,k)) = \mathcal{F}_k^{-1}(\mathcal{F}_y(u(x,y))) \mapsto u(x,y)$$
$$\implies u(x,y) = \mathcal{F}^{-1}_k (\frac{2}{1+k^2} e^{- |k| x}) = \int_{-\infty}^{\infty}(\frac{2}{1+k^2} e^{- |k| x}) e^{i k y} \frac{dk}{2 \pi} \phantom{\ } \blacksquare$$ \\ \\
Title: Re: Problem 1
Post by: Xuan Ju on November 11, 2012, 07:13:53 PM
In the last line, shouldn't the power to the second exponential be ikx instead of iky?
Title: Re: Problem 1
Post by: Hanqing Liu on November 11, 2012, 11:39:44 PM
In the last line, shouldn't the power to the second exponential be ikx instead of iky?

Yes, I think it is a small typo
Title: Re: Problem 1
Post by: Victor Ivrii on November 12, 2012, 03:17:14 AM
In the last line, shouldn't the power to the second exponential be ikx instead of iky?

No--it is Fourier integral for $k\mapsto y$ (remember FT was by $y$). Calvin, WTH, why are you not defending your solution?
Title: Re: Problem 1
Post by: Zarak Mahmud on November 12, 2012, 11:45:59 AM
In the last line, shouldn't the power to the second exponential be ikx instead of iky?

No--it is Fourier integral for $k\mapsto y$ (remember FT was by $y$). Calvin, WTH, why are you not defending your solution?

He must be taking a long nap after typing it all up.
Title: Re: Problem 1
Post by: Calvin Arnott on November 12, 2012, 12:27:03 PM
Yeah, this method works by temporarily transforming the initial function $u(x,y)$ by $y \mapsto k$ and using the properties of the Fourier transform to get an ODE in the other variable $x$ of the function $U(x,k)$. Using that the Fourier transform is unique and invertible, after solving the ODE in $x$ we transform our ODE solution $U(x,k)$ back from $k \mapsto y$ to get a solution for the PDE in terms of $u(x,y)$.

If we had $e^{i k x}$ in the last line, our final transform would be an inverse Fourier transform mapping $k \mapsto x$ on the $U(x,k)$ solution, so our answer would be a single variable function $u(x,x)$. Since we wanted a solution to the initial PDE in both $(x,y)$ we instead apply the inverse transformation $k \mapsto y$, as is written in the initial solution.
Title: Re: Problem 1
Post by: Thomas Nutz on December 19, 2012, 02:19:38 PM
The question says we should do it with a Fourier transform, but why don't I just take
$$u(x,y)=e^{-|y|-ix}$$? That does give me a zero Laplacian and satisfies the boundary condition, or am I wrong here?
Title: Re: Problem 1
Post by: Victor Ivrii on December 19, 2012, 02:32:29 PM
The question says we should do it with a Fourier transform, but why don't I just take
$$u(x,y)=e^{-|y|-ix}$$? That does give me a zero Laplacian and satisfies the boundary condition, or am I wrong here?

For starters:  $e^{-|y|}$ is not twice differentiable (and your function definitely does not satisfy equation).