# Solution to inhomogenous PDE

Given the equation $(1-\Delta)u=f$ for $f \in S(\mathbb{R}^n)$ (rapidly decreasing functions) we get by taking the Fourier transform that

$u = \left(\frac{1}{2\pi}\right)^{\frac{n}{2}}\mathcal{F}^{-1}\left(\frac{1}{1+|.|^2} \right)*f$

This is all well-defined in the sense of $S'(\mathbb{R}^n).$

Obviously, for $n \le 3$ the function $\frac{1}{1+|.|^2}$ is in $L^2$ so it is meaningful to write

$u$ as $u(x) = \int_{\mathbb{R}^n } K(x-y)f(y) dy$ for some integral kernel defined via the inverse Fourier transform, but is there also a way to make sense out of this integral kernel representation for $n>3$?- Or if not, how can I see that there is no such kernel in higher dimensions?

• Fourier transform is bounded from $L^p$ to $L^q$ where $1<p \leq 2$ and $q$ is the holder conjugate. This map is not necessarily surjective though for $p<2$. This then becomes a question of whether or not one can find an appropriate $L^p$ function with the fourier transform being your example... – Ali May 12 '16 at 22:25
• @Ali yes, this is also what I thought, see the linked question – Leopold May 12 '16 at 22:47

This kernel is called the Bessel potential. It is smooth away from $0$, and in your case, this kernel is locally integrable.
Alternatively, the self-adjoint operator $-\Delta$ on the domain $H^2\subseteq L^2$ has spectrum $[0,\infty)$ in any dimension. Since $-1$ is not in the spectrum, $(-\Delta+1)^{-1}$ is bounded on $L^2$ and thus for any $f\in L^2$, there is a unique $u\in H^2$ such that $(1-\Delta)u=f$.