Author Topic: Problem 1  (Read 5895 times)

Calvin Arnott

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Problem 1
« 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 $u(x,y)$:
$$\mathcal{F}^{-1}_k (U(x,k)) = \mathcal{F}_k^{-1}(\mathcal{F}_y(u(x,y))) \mapsto u(x,y) $$
$$ \implies u(x,y) = \mathcal{F}^{-1}_k (\frac{2}{1+k^2} e^{- |k| x}) = \int_{-\infty}^{\infty}(\frac{2}{1+k^2} e^{- |k| x}) e^{i k y} \frac{dk}{2 \pi} \phantom{\ } \blacksquare$$ \\ \\
« Last Edit: November 11, 2012, 02:05:43 PM by Victor Ivrii »

Xuan Ju

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Re: Problem 1
« Reply #1 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?

Hanqing Liu

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Re: Problem 1
« Reply #2 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

Victor Ivrii

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Re: Problem 1
« Reply #3 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?

Zarak Mahmud

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Re: Problem 1
« Reply #4 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.

Calvin Arnott

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Re: Problem 1
« Reply #5 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.
« Last Edit: November 12, 2012, 12:29:44 PM by Calvin Arnott »

Thomas Nutz

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Re: Problem 1
« Reply #6 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?
« Last Edit: December 19, 2012, 02:28:48 PM by Victor Ivrii »

Victor Ivrii

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Re: Problem 1
« Reply #7 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).