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Using the Fourier transform, $\mathscr{F}\{\cdot\} = \hat{f}(\cdot)$, may I transform the $u_{xy}$ where $u=u(x,y,t)$, $x,y \in (-\infty, \infty)$ and $t \in (0,1]$? The initial condition is $u(x,y,t)=\delta(x) \delta(y)$ for $t=0$.

I tried the following:

$ \begin{align} \mathscr{F}\{u_{xy}\}=\hat{f}(\xi_1, \xi_2)& =\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{\partial^{2}{u}} {\partial{x}\partial{y}} e^{-2 \pi i (\xi_1 x + \xi_2 y)} dx dy \tag{1} \\ & = \int_{\mathbb{R}} \frac{\partial}{\partial{y}} e^{-2 \pi i \xi_2 y} \left(\int_{\mathbb{R}} \frac{\partial{u}}{\partial{x}} e^{-2 \pi i \xi_1 x} dx\right) dy \tag{2} \\ & = 2 \pi i \xi_1 \hat{f}(\xi_1) \int_{\mathbb{R}} -2 \pi i \xi_2 e^{-2 \pi i \xi_2 y} dy \tag{3} \\ & = 4 \pi^{2} \xi_1 \hat{f}(\xi_1) \xi_2 \delta(\xi_2) \tag{4} \end{align} $

where in (2) I used Leibniz rule, in (3) differentiated the integrand w.r.t. $y$ and in (4) used the Fourier transform of $1$ referenced here.

If this is so, then I can take the inverse of the transform to get the original function.

Edit: The PDE is the following:

$ \begin{cases} u_{t} = u_{xx} + 2 \rho u_{xy} + u_{yy}, \ & \text{for} \ t \in (0,1],\\ u(x,y,t) = \delta(x) \delta(y), \ & \text{for} \ t = 0 \end{cases} $

where $\rho \in (-1, 1)$.

Thank you.

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  • $\begingroup$ In your integral, integrate by parts to see what the result comes to be. I often find this the best way to determine whether there are any issues in the transform without appealing to measure theoretic ideas. $\endgroup$ – Mattos Dec 17 '15 at 2:49
  • $\begingroup$ @Mattos I showed my work of the Fourier transform. I'm not sure if it is correct. Thanks. $\endgroup$ – AnonymousJ Dec 17 '15 at 3:36
  • $\begingroup$ What is the PDE? $\endgroup$ – Julián Aguirre Dec 17 '15 at 15:13
  • $\begingroup$ @JuliánAguirre I edited and typed in the PDE. Thanks. $\endgroup$ – AnonymousJ Dec 17 '15 at 15:37
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Let $\widehat u(\xi,\eta,t)$ be the Fourier transform (in the spatial variables) of $u(x,y,t)$. Each time you derivate $u$ with respect to $x$ you multiply $\widehat u$ by $-2\,\pi\,\xi\,i$; each time you derivate $u$ with respect to $y$ you multiply $\widehat u$ by $-2\,\pi\,\eta\,i$. Taking Fourier transforms in the equation we get then $$ \frac{d\widehat u}{dt}=-4\,\pi^2\bigl(\xi^2+2\,\rho\,\xi\,\eta+\eta^2\bigr)\widehat u. $$ This is first order linear ODE, which must be complemented with the initial condition $$ \widehat u(\xi,\eta,0)=\mathcal{F}(\delta(x)\,\delta(y))=1. $$

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  • $\begingroup$ Thank you very much. I worked it out and checks. Thus, since we have the Fourier transform of the above PDE, we can take the inverse Fourier transform to find $u(x,y,t)$. $\endgroup$ – AnonymousJ Dec 18 '15 at 5:45
  • $\begingroup$ Yes, that's the way to procede. $\endgroup$ – Julián Aguirre Dec 18 '15 at 10:27

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