Solving Poisson equation with Fourier transform.

Question

Is possible solving the poisson equation with Fourier transform?

For example, with application of the Fourier transform, the problem \begin{align} (1-\partial_x^2)u=f \end{align} with $f\in L^2(\mathbb{R})$ is converted

\begin{align} (1+\xi^2)\widehat{u}=\widehat{f} \end{align} then the solution is \begin{align} u_f=\mathcal{F}^{-1}\left( \frac{1}{1+\xi^2}\widehat{f}\right) \end{align}

Now, \begin{align} -\partial_x^2u=f \end{align}

\begin{align} u_f=\mathcal{F}^{-1}\left( \frac{1}{\xi^2}\widehat{f}\right) \end{align} but, there is a singularity in the origin. How can I get solving this trouble? any reference? Thanks!

Answer 1

Finally, something that I can answer. This was a problem I ran into when implementing the Vorticity-Streamfunction Algorithm for 2D Fluid Dynamics. $$ \nabla^2 \left[ \psi(x, y) \right] = - \omega(x, y) $$ $$ \frac{\partial^2}{\partial x^2} \left[ \psi(x, y) \right] + \frac{\partial^2}{\partial y^2} \left[ \psi(x, y) \right] = - \omega(x, y) $$ By discretizing $\psi(x, y)$ to $\psi(x_i, y_j) = \psi_{ij}$: $$ \psi_{ij} = \text{ifft2} \left[ \Psi_{pq} \right] = \text{ifft2} \left[ \frac{ - \Omega_{pq}}{- k_p^2 - k_q^2} \right] $$

Now, you would be correct that there is a singularity at $p=q=0$, but my question to you is, does that matter? In my case, because I would then be taking derivatives of $\psi_{ij}$, rather than using the actual value, it didn't. And so, if we just force the following: $$ \Psi_{00} = \frac{-\Omega_{00}}{-k_0^2 - k_0^2} = 0 \neq \infty $$ Our recovered $\psi_r$ will be the analytical $\psi_a$ minus the average $\bar{\psi}$, as the $\Psi_{00}$ value is the 'DC' value. If the average is zero anyways, then $\psi_r = \psi_a$. Hopefully that helps. Let me know if I can clarify anything.

Answer 2

The problem is that you go from $-\partial_x^2u=f$ to the conclusion that $$u_f=\mathcal{F}^{-1}\left( \frac{1}{\xi^2}\widehat{f}\right)$$ What does $\frac 1 {\xi^2 }f$ really mean? It is most likely only defined for $\xi\neq 0$.

A more rigorous approach to this is to use the theory of distribution. If $u$ is a distribution that satisfies $\xi \hat u=0$ then $\hat u$ is a multiple of the Dirac delta distribution. Likewise, if $\xi^2 \hat u=0$ then $\hat u$ is a linear combination of $\delta$ and its derivative $\delta^\prime$.

Suppose now that $u_0$ and $u_1$ are two solutions of $\partial_x u = f$. Then $u=u_1-u_0$ verifies $-\partial^2_x u = 0$. Consequently, $$\hat u_1=\hat u_0 + c_1\delta + c_2\delta^\prime$$ and $$u_1 = u_0 + c_1 + c_2 i x$$ (which you could derive directly without the Fourier transform).

Another way to think about this without distributions is to consider that in some sense $$\frac 1 {\xi^2} = \lim_{\varepsilon\rightarrow 0} \frac 1 {\varepsilon^2 + \xi^2}$$ Thus a solution $u$ can be thought of as $$u=\lim_{\varepsilon\rightarrow 0} \mathcal F^{-1}\left(\frac 1 {\varepsilon^2 + \xi^2}\hat f\right)$$

The inverse Fourier transform of $$\xi\mapsto \frac 1 {\varepsilon ^2 + \xi^2}$$ is something like $x\mapsto c_\varepsilon e^{-\varepsilon|x|}$.

Thus a solution $u$ will can also be expressed in the time/space domain as $$u = \lim_{\varepsilon\rightarrow 0} c_\varepsilon \left( e^{-\varepsilon|x|} \ast f\right )$$ This is comes down to approaching the Dirac delta as the limit of $c_\varepsilon e^{-\varepsilon|x|}$ as $\varepsilon \rightarrow 0$. Something known as approximation of the identity.

Of course, the exchange of limits and Fourier transforms is not super rigorous, but the rigorous way is to use Distributions.