# PDE $y+k^2 \nabla^2 y = p(\mathbf{x})$ : no solution by Fourier transform

My partial differential equation of interest is as follows:

$y+k^2 \nabla^2 y = p(\mathbf{x})$

where $y$ and $p$ are functions of $\mathbf{x}$, which is a vector in $\mathbb{R}^D$, and the Laplacian operator $\nabla^2$ is defined as $\sum_{d=1}^{D} \frac{\partial^2}{\partial x_{d}^2}$. $k$ is a real, positive constant.

$p(\mathbf{x})$ may be an arbitrary function with good properties. (continuous, differentiable, ...) $p(\mathbf{x})$ is always positive and $p(\mathbf{x}) \to 0$ as $\mathbf{x} \to \infty$ and $\mathbf{x} \to -\infty$.

Boundary conditions are $y(\infty)=y(-\infty)=y'(\infty)=y'(-\infty)=0$.

A straightforward approach would be applying Fourier transform, but it would not yield a valid solution because of the positive sign of $k^2 \nabla^2 y$.

Does it have any solution? If so, how can it be derived?

• No, limits of $0$ at $\infty$ are not enough to ensure the Fourier transform exists. But I think the main problem is that in general this will have a bad singularity on $|\omega| = 1/k$. Apr 10, 2017 at 3:59
• @RobertIsrael Yes, the singularity hinders Fourier transform. Apr 10, 2017 at 14:09