The proof is by separation of variables, and is similar to Proposition 3.31. So we will not provide all the details, except pointing out one key point. For simplicity of notation we may assume $x_{1}=0$. After separation of variables we need to solve a PDE of the form on $\mathbb{R}^{m}$

(8.2)

$$-\Delta_{\mathbb{R}^{m}}u+\lambda u=\delta_{0^{m}},$$

where $\lambda$ is non-negative. When $\lambda=0$ a solution is given by the Green’s function on $\mathbb{R}^{m}$, so we only deal with the case $\lambda>0$. When $m=1$, this is the equation (3.357). When $m\geq 2$, we look for a radial solution $u=u(r)$, then (8.2) reduces to an ODE

(8.3)

$$-u^{\prime\prime}(r)-\frac{m-1}{r}\cdot u^{\prime}(r)+\lambda\cdot u(r)=0,r\in(0,\infty)$$

We make the transformation

(8.4)

$$f(r)\equiv u(r)\cdot r^{-\alpha},$$

where the exponent $\alpha$ is to be determined. Then $f(r)$ satisfies

(8.5)

$$r^{2}f^{\prime\prime}(r)+(2\alpha+m-1)rf^{\prime}(r)+\Big(\alpha(\alpha+m-2)-\lambda r^{2}\Big)f(r)=0.$$

Now let $2\alpha+m-1=1$, i.e. $\alpha=\frac{2-m}{2}$, and let $\sqrt{\lambda}\cdot r=s$, then we get the modified Bessel equation (c.f. (5.24))

(8.6)

$$s^{2}f^{\prime\prime}(s)+sf^{\prime}(s)-(\alpha^{2}+s^{2})f(s)=0.$$

Then we get a solution $u(r)=K_{\alpha}(\sqrt{\lambda}r)\cdot r^{\alpha}$, where $K_{\alpha}$ is the modified Bessel function defined by (A.5). So it follows that

(8.7)

$\displaystyle u(r)\sim\begin{cases}\varphi_{j}(p)\cdot r^{2-m},&m\geq 3,\\ \varphi_{j}(p)\cdot\log r,&m=2,\end{cases}$

as $r\to 0$, So in particular $u$ satisfies the distribution equation (8.2). Then we can define $G_{p}$ using a formal expansion, and the convergence and the asymptotic behavior follow from the uniform estimates on $K_{\alpha}(\sqrt{\lambda}r)$ for $r\geq 1$ in Proposition 5.5. ∎