Finding fundamental solution to the biharmonic operator $\Delta^2=\Delta(\Delta)$?
Solution 1:
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!