If $\Delta u(x)=\delta(x)$ then $u(x)=C |x|^{2-d}$ via Fourier transform?

Question

Can the behaviour $$ \text{constant}\times|x|^{2-d} $$ be obtained for the solution of the distributional equation in $\mathbb R^d$, for $d\ge 3$, $$ \Delta u(x)=\delta(x) $$ via Fourier transform method?

Answer 1

The behaviour $|x|^{2-d}$ can be obtained also by the following arguments. Denoting by $\widehat{\cdot}$ the Fourier transform and by $$ f_\lambda(x)\equiv f(\lambda x), $$ then, by definition of the Fourier transform, \begin{align} \widehat {f_\lambda}(k)&=\int_{\mathbb R^n}e^{-ik\cdot x}f(\lambda x)dx\\ &=\int_{\mathbb R^n}e^{-ik\cdot x/\lambda}f(x)\lambda^{-d}dx\\ &=\lambda^{-d}\widehat f(k/\lambda). \end{align} Let $f(x)=F(|x|)$ be a radial function; if $M\in SO(n)$, then $f_M(x)\equiv f(Mx)=f(x)$ and hence, since $\hat f_M(k)=\widehat f(Mk)$, we have $\widehat f_M= \widehat f$, which in turn implies that also $\widehat f$ is radial. Now, suppose $f(x)=|x|^{-k}$, for $0<k<d$, then $$ f_\lambda(x) = \lambda^{-k}f(x) $$ and, by the above remark, $$ \widehat f (k/\lambda)= \lambda^{d}\widehat f_\lambda(k) = \lambda^{d-k}\hat f(k). $$ The only radial function $\hat f$ homogeneous of degree $-d+k$ is $$ \widehat f (k) = |k|^{k-d}. $$ In the present case, $k-d=-2$ so that $k=d-s$ and hence $f(x)=|x|^{2-d}$ up to constants. A nontrivial issue would be computing such a constant from the Fourier transform procedure: Fundamental solution to the Poisson equation by Fourier transform

Note that $f$ can be split as the sum of two pieces: let $B_a$ be the ball of radius $a$ centred at the origin, then \begin{align} f&\equiv u+v\\ &= \chi_{B_a}(x)\frac{1}{|x|^{d-2}}+\left(1-\chi_{B_a}(x)\right)\frac{1}{|x|^{d-2}}; \end{align} $u$ lies in $L^1(\mathbb R^d)$, since $0<d-2<d$, and $v$ lies in $L^2(\mathbb R^d)$, for $d\ge4$ (the case $d=3$ is easy to check explicitly). This ensures $\widehat u \in L^\infty(\mathbb R^d)$ and $\widehat v\in L^2(\mathbb R^d)$ and that $\widehat f$ needs indeed to be a function in $L^1_{\text{loc}}(\mathbb R^d)$ from an abstract point of view. So $f,\widehat f\in L^1_{\text{loc}}(\mathbb R^d)$.