Finding fundamental solution to the biharmonic operator $\Delta^2=\Delta(\Delta)$? 
Show that when $n=2$, the function $u(x)=-\dfrac{1}{8\pi}\left\lvert x\right\rvert^2\log\left\lvert x\right\rvert$ is a fundamental solution to the biharmonic operator $\Delta^2=\Delta\left(\Delta\right)$. That is, show that $$\varphi(0)=\int_{\Bbb R^2}u(x)\,\Delta^2\varphi(x)\,dx$$ for all functions $\varphi$ smooth with compact support.

My attempt:
My first confusion: the question says it is equivalent to show $\varphi(0)=\int_{\Bbb R^2}u(x)\,\Delta^2\varphi(x)\,dx$. Why is that true? Considering the definition, I think it is $\;0=\int_{\Bbb R^2}u(x)\,\Delta^2\varphi(x)\,dx\;$ instead of $\varphi(0)$. Where am I wrong?
My second confusion is it seems I can directly compute $\Delta u=-\dfrac{1}{2\pi}\Big(1+
\log\left\lvert x\right\rvert\Big)$, then it is obvious $\,\Delta\left(\Delta u\right)=0$. Is that enough for proving this question?
The calculation of $\Delta u(x)$ is as follows:
$$u_{11}=-\dfrac{1}{8\pi}\left[2\ln\left(\sqrt{x_1^2+x_2^2}+\dfrac{2x_1^2}{x1^2+x_2^2}+1\right)\right]$$
and 
$$u_{22}=-\frac{1}{8\pi}\left[2\ln\left(\sqrt{x_1^2+x_2^2}+\dfrac{2x_2^2}{x1^2+x_2^2}+1\right)\right],$$
so 
$$\Delta u=u_{11}+u_{22}=-\frac{1}{2\pi}\Big(1+\log\left\lvert x\right\rvert\Big).$$
Could someone kindly help? Thanks!
 A: Hint: Note that $u(x)$ has a singularity at $0$ (the logarithm is not defined at $0$) and this must be dealt with. Split the integral over a ball $\mathbb{B}(0,\epsilon)$ centered at the origin and with radius $\epsilon$ and the rest of the space $\mathbb{R}^2\setminus \mathbb{B}(0,\epsilon)$, then analyse each integral separately. This can be written as
\begin{equation*}
\int_{\mathbb{R}^2}u(x)\Delta^2\varphi(x)dx=\int_{\mathbb{B}(0,\epsilon)}u(x)\Delta^2\varphi(x)dx+\int_{\mathbb{R}^2\setminus \mathbb{B}(0,\epsilon)}u(x)\Delta^2\varphi(x)dx.
\end{equation*}
Taking the modulus gives us the bound
\begin{equation*}
|\int_{\mathbb{B}(0,\epsilon)}-\frac{1}{8\pi}|x|^2\log|x|\Delta^2\varphi(x)dx| 
\leq c\sup_{x\in\mathbb{R}^2}|\Delta^2\varphi(x)||\int_{\mathbb{B}(0,\epsilon)}|x|^2\log|x|dx|.
\end{equation*}
Can you show this goes to zero as $\epsilon\to 0$ using polar coordinates? For the second integral, use the divergence theorem twice. Integrate over the boundary of the ball (denoted $\partial \mathbb{B}(0,\epsilon)$) and show everything goes to $0$.
Hope that helps!
