Radial Yukawa source power fails the homogeneous equation

For any twice differentiable radial function $w(x)=q(r)$ on $r>0$, ${\mathrm{\partial }}_{j}w=q^{\mathrm{\prime }}(r)x_j\mathrm{/}r$ and ${\sum }_{j=1}^3{\mathrm{\partial }}_j^{2}w=q^{\mathrm{\prime }\mathrm{\prime }}(r){\sum }_{j=1}^3\frac{x_j^2}{r^2}+q^{\mathrm{\prime }}(r){\sum }_{j=1}^3(\frac{1}{r}-\frac{x_j^2}{r^3})=q^{\mathrm{\prime }\mathrm{\prime }}(r)+\frac{2}{r}{q}^{\mathrm{\prime }}(r)\mathrm{.}$ Thus the radial Laplacian formula follows directly from the chain rule on ${\mathbb{R}}^3\setminus \{0\}$.

For $q(r)=e^{-\mu r}{r}^{-2}$, differentiation gives $q^{\mathrm{\prime }}(r)=e^{-\mu r}(-\frac{\mu }{r^2}-\frac{2}{r^3}),q^{\mathrm{\prime }\mathrm{\prime }}(r)=e^{-\mu r}(\frac{{\mu }^2}{r^2}+\frac{4\mu }{r^3}+\frac{6}{r^4})\mathrm{.}$ Substituting into step 1.1 yields $(\mathrm{\Delta }-{\mu }^2)u_{\mu }(x)=2e^{-\mu r}(\frac{\mu }{r^3}+\frac{1}{r^4})>0(r>0),$ including $\mu =0$. This contradicts [A1] at every point off the origin.

For comparison, set $v_{\mu }(x)=e^{-\mu r}\mathrm{/}r$ for $r>0$. Direct differentiation gives $v_{\mu }^{\mathrm{\prime }}(r)=e^{-\mu r}(-\frac{\mu }{r}-\frac{1}{r^2}),v_{\mu }^{\mathrm{\prime }\mathrm{\prime }}(r)=e^{-\mu r}(\frac{{\mu }^2}{r}+\frac{2\mu }{r^2}+\frac{2}{r^3})\mathrm{.}$ Using step 1.1, the terms $2\mu \mathrm{/}{r}^2+2\mathrm{/}{r}^3$ cancel against $2v_{\mu }^{\mathrm{\prime }}\mathrm{/}r$, leaving $\mathrm{\Delta }{v}_{\mu }={\mu }^2{v}_{\mu }$ on $r>0$. Thus the $r^{-1}$ power passes the off-origin homogeneous check, but this does not prove its full distributional Green-function normalization.