Fundamental Solution of $\Delta^{2}$ in $\mathbb{R^{2}}$ we are to show that the fundamental solution of $\Delta^{2}$ in $\mathbb{R^{2}}$ is given by :
$$
V(x_{1},x_{2})= \frac{1}{8\pi}r^{2}\ln(r),\qquad r=|x-\xi|,
$$
and write the solution of $\Delta^{2}w=F(x_{1},x_{2})$.
My approach: Let $x=\left(x_{1}, x_{2}\right) .$ We have :
$$
V(x)=\frac{1}{8 \pi} r^{2} \log (r)
$$
In polar coordinates: (here, $V_{\theta \theta}=0$ )
$$
\begin{aligned}
\Delta V &=\frac{1}{r}\left(r V_{r}\right)_{r}=\frac{1}{r}\left(r\left(\frac{1}{8 \pi} r^{2} \log (r)\right)_{r}\right)_{r}=\frac{1}{8 \pi} \frac{1}{r}(r(2 r \log (r)+r))_{r} \\
&=\frac{1}{8 \pi} \frac{1}{r}\left(2 r^{2} \log (r)+r^{2}\right)_{r}=\frac{1}{8 \pi} \frac{1}{r}(4 r+4 r \log r) \\
&=\frac{1}{2 \pi}(1+\log r)
\end{aligned}
$$
The fundamental solution $V(x)$ for $\Delta^{2}$ is the distribution satisfying: $\triangle^{2} V(r)=\delta(r)$
$$
\begin{aligned}
\Delta^{2} V &=\Delta(\Delta V)=\Delta\left(\frac{1}{2 \pi}(1+\log r)\right)=\frac{1}{2 \pi} \Delta(1+\log r)=\frac{1}{2 \pi} \frac{1}{r}\left(r(1+\log r)_{r}\right)_{r} \\
&=\frac{1}{2 \pi} \frac{1}{r}\left(r \frac{1}{r}\right)_{r}=\frac{1}{2 \pi} \frac{1}{r}(1)_{r}=0 \quad \text { for } r \neq 0
\end{aligned}
$$
Therefore, we have that $\Delta^{2}V(r)=\delta(r)$. Hence, $V$ is the fundamental solution and the solution of:
$$
\Delta^{2} \omega=F(x)
$$
if given by:
$$
\omega(x)=\int_{\mathbb{R}^{2}} V(x-y) \Delta^{2} \omega(y) d y=\frac{1}{8 \pi} \int_{\mathbb{R}^{2}}|x-y|^{2} \log |x-y| F(y) d y
$$
Is this proof correct? if so, how could we establish an even more rigorous proof, since this proof is not rigorous enough?
 A: For a more rigorous proof (as you wish), we have that for $v \in C_{0}^{\infty}\left(\mathbb{R}^{n}\right),$ we want to show
$$
\int_{\mathbb{R}^{n}} K(|x|) \Delta^{2} v(x) d x=v(0)
$$
Suppose $v(x) \equiv 0$ for $|x| \geq R$ and let $\Omega=B_{R}(0) ;$ for small $\epsilon>0$ let
$$
\Omega_{\epsilon}=\Omega-B_{\epsilon}(0)
$$
$K(|x|)$ is biharmonic $\left(\Delta^{2} K(|x|)=0\right)$ in $\Omega_{\epsilon} .$ Consider Green's identity $\left(\partial \Omega_{\epsilon}=\partial \Omega \cup\right.$
$\left.\partial B_{\epsilon}(0)\right)$
$$
\begin{aligned}
\int_{\Omega_{\epsilon}} K(|x|) \Delta^{2} v d x &=\underbrace{\int_{\partial \Omega}\left(K(|x|) \frac{\partial \Delta v}{\partial n}-v \frac{\partial \Delta K(|x|)}{\partial n}\right) d s+\int_{\partial \Omega}\left(\Delta K(|x|) \frac{\partial v}{\partial n}-\Delta v \frac{\partial K(|x|)}{\partial n}\right) d s}_{\displaystyle=0,\;\;\;\text{since $v\equiv0$ for $x\geq R$}} \\
&+\int_{\partial B_{\epsilon}(0)}\left(K(|x|) \frac{\partial \Delta v}{\partial n}-v \frac{\partial \Delta K(|x|)}{\partial n}\right) d s+\int_{\partial B_{\epsilon}(0)}\left(\Delta K(|x|) \frac{\partial v}{\partial n}-\Delta v \frac{\partial K(|x|)}{\partial n}\right) d s \\
\lim _{\epsilon \rightarrow 0}\left[\int_{\Omega_{\epsilon}} K(|x|) \Delta^{2} v d x\right] &=\int_{\Omega} K(|x|) \Delta v^{2} d x
\end{aligned}
$$
On $\partial B_{\epsilon}(0), \quad K(|x|)=K(\epsilon) .$ Thus :\begin{aligned}
\left|\int_{\partial B_{\varepsilon}(0)} K(|x|) \frac{\partial \Delta v}{\partial n} d S\right| &=|K(\epsilon)| \int_{\partial B_{\epsilon}(0)}\left|\frac{\partial \Delta v}{\partial n}\right| d S \leq|K(\epsilon)| \omega_{n} \epsilon \max _{x \in \bar{\Omega}}|\nabla(\Delta v)| \\
&=\left|\frac{1}{8 \pi} \epsilon^{2} \log (\epsilon)\right| \omega_{n} \epsilon \max _{x \in \bar{\Omega}}|\nabla(\Delta v)| \rightarrow 0, \quad \text { as } \epsilon \rightarrow 0 \\
\int_{\partial B_{\epsilon}(0)} v(x) \frac{\partial \Delta K(|x|)}{\partial n} d S=& \int_{\partial B_{\epsilon}(0)}-\frac{1}{2 \pi \epsilon} v(x) d S \\
&=\int_{\partial B_{\epsilon}(0)}-\frac{1}{2 \pi \epsilon} v(0) d S+\int_{\partial B_{\epsilon}(0)}-\frac{1}{2 \pi t}[v(x)-v(0)] d S \\
&=-\frac{1}{2 \pi \epsilon} v(0) 2 \pi \epsilon-\underbrace{\max _{x \in \partial B_{\epsilon}(0)}|v(x)-v(0)|}_{\rightarrow 0,(v \text { is continuous})}=-v(0) .
\end{aligned}
\begin{aligned}
\left|\int_{\partial B_{\epsilon}(0)} \Delta K(|x|) \frac{\partial v}{\partial n} d S\right|=&|\Delta K(\epsilon)| \int_{\partial B_{\epsilon}(0)}\left|\frac{\partial v}{\partial n}\right| d S \leq\left|\frac{1}{2 \pi}(1+\log \epsilon)\right| 2 \pi \epsilon \max _{x \in \bar{\Omega}}|\nabla v| \rightarrow 0, \quad \text { as } \epsilon \rightarrow 0 \\
\int_{\partial B_{\epsilon}(0)} \Delta v \frac{\partial K(|x|)}{\partial n} d S &=\int_{\partial B_{\epsilon}(0)}\left(-\frac{1}{4 \pi} \epsilon \log \epsilon-\frac{1}{8 \pi} \epsilon\right) \Delta v(x) d S \\
& \leq \frac{\epsilon}{4 \pi}\left|\log \epsilon+\frac{1}{2}\right| \cdot 2 \pi \epsilon \max _{x \in \partial B_{\epsilon}(0)}|\Delta v| \rightarrow 0, \text { as } \epsilon \rightarrow 0
\end{aligned}
$$
\therefore \quad \int_{\Omega} K(|x|) \Delta^{2} v d x=\lim _{\epsilon \rightarrow 0} \int_{\Omega_{\epsilon}} K(|x|) \Delta^{2} v d x=v(0) $$
[Note] : To prevent any misunderstanding, note that in $K(|x|)=K(c)$, for $|x|=\epsilon$, we have :
$$
\begin{aligned}
K(|x|) &=K(\epsilon)=\frac{1}{8 \pi} \epsilon^{2} \log \epsilon, & \Delta K=\frac{1}{2 \pi}(1+\log \epsilon) \\
\frac{\partial K(|x|)}{\partial n} &=-\frac{\partial K(\epsilon)}{\partial r}=-\frac{1}{4 \pi} \epsilon \log \epsilon-\frac{1}{8 \pi} \epsilon, & \frac{\partial \Delta K}{\partial n}=-\frac{\partial \Delta K}{\partial r}=-\frac{1}{2 \pi \epsilon}
\end{aligned}
$$
