Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment Y => Topic started 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 halfplane (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 $u(x,y)$:
$$\mathcal{F}^{1}_k (U(x,k)) = \mathcal{F}_k^{1}(\mathcal{F}_y(u(x,y))) \mapsto u(x,y) $$
$$ \implies u(x,y) = \mathcal{F}^{1}_k (\frac{2}{1+k^2} e^{ k x}) = \int_{\infty}^{\infty}(\frac{2}{1+k^2} e^{ k x}) e^{i k y} \frac{dk}{2 \pi} \phantom{\ } \blacksquare$$ \\ \\

In the last line, shouldn't the power to the second exponential be ikx instead of iky?

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

In the last line, shouldn't the power to the second exponential be ikx instead of iky?
Noit is Fourier integral for $k\mapsto y$ (remember FT was by $y$). Calvin, WTH, why are you not defending your solution?

In the last line, shouldn't the power to the second exponential be ikx instead of iky?
Noit 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.

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.

The question says we should do it with a Fourier transform, but why don't I just take
$$u(x,y)=e^{yix}$$? That does give me a zero Laplacian and satisfies the boundary condition, or am I wrong here?

The question says we should do it with a Fourier transform, but why don't I just take
$$u(x,y)=e^{yix}$$? 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).