Prove $F(r)=Cr^2\log(r)$ is a fundamental solution for $\Delta^2$ in $\mathbb R^2$

Question

Prove that for some $C$, $F(r)=Cr^2\log(r)$ is a fundamental solution for $\Delta^2$ in $\mathbb R^2$. Recall that

$$ \Delta^2u=u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta}$$

My answer is below.

Answer 1

Firstly, let $T_{F(r)}(\phi)$ be a regular distribution so that $\Delta^2 T_{F(r)}(\phi)= T_{F(r)}(\Delta^2\phi)$. Cutting out a ball of radius $\epsilon$, and letting the support of $\phi$ be contained in a ball of radius $R$,

$$ T_{F(r)}(\Delta^2\phi)=\int\int_{B(0,R)\setminus B(0,\epsilon)} F(r)\Delta^2\phi dA $$

Using Green's identity,

$$=\int\int_{B(0,R)\setminus B(0,\epsilon)}\Delta F(r)\Delta \phi dA+ \int_{r=R} F(r)\frac{\partial}{\partial r}\Delta \phi-\Delta \phi\frac{\partial}{\partial r}F(r) ds $$ $$ -\int_{r=\epsilon} F(r)\frac{\partial}{\partial r}\Delta \phi-\Delta \phi\frac{\partial}{\partial r}F(r) ds $$

Since $\phi\in C_0^\infty$, $\phi(R,\theta)=\phi_r(R,\theta)=(\Delta\phi)(R,\theta)=(\Delta\phi_r)(R,\theta)=0$, so this is

$$=\int\int_{B(0,R)\setminus B(0,\epsilon)}\Delta F(r)\Delta \phi dA-\int_{r=\epsilon} F(r)\Delta \phi_r-\Delta \phi\frac{\partial}{\partial r}F(r) ds $$

Using Green's Identity again,

$$=\int\int_{B(0,R)\setminus B(0,\epsilon)}\Delta^2 F(r)\phi dA+\int_{r=R}\Delta F(r) \phi_r -\phi\frac{\partial}{\partial r} \Delta F(r) ds $$ $$-\int_{r=\epsilon}\Delta F(r) \phi_r -\phi\frac{\partial}{\partial r} \Delta F(r) ds$$ $$ -\int_{r=\epsilon} F(r)\Delta \phi_r-\Delta \phi\frac{\partial}{\partial r}F(r) ds $$

Considering boundary terms again,

$$=\int^{2\pi}_0\int_{\epsilon}^R \Delta^2 F(r)\phi rdrd\theta-\int_{r=\epsilon}\Delta F(r) \phi_r -\phi\frac{\partial}{\partial r} \Delta F(r) ds$$ $$ -\int_{r=\epsilon} F(r)\Delta \phi_r-\Delta \phi\frac{\partial}{\partial r}F(r) ds $$

We see that for $r\neq 0$,

$$ \Delta^2 F(r) = C\Delta\left (\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} \right) r^2\log(r)$$

$$ = C\Delta\left (\frac{\partial}{\partial r}(2r\log(r)+r)+\frac{1}{r}(2r\log(r)+r) \right) $$

$$ = C\Delta\left (2\log(r)+2+1+2\log(r)+1) \right)=4C\Delta(\log(r)+1) $$ We note that $\Delta F(r)=4C(\log(r)+1) $.

$$ =4C\left (\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} \right)(\log(r)+1) $$

$$ =4C(-r^{-2}+r^{-2} )=0$$

So we have

$$=\int_{r=\epsilon}-\Delta F(r) \phi_r +\phi\frac{\partial}{\partial r} \Delta F(r)-F(r)\Delta \phi_r+\Delta \phi\frac{\partial}{\partial r}F(r) ds $$ or

$$=\int_{r=\epsilon}-4C(\log(r)+1) \phi_r +\phi\frac{\partial}{\partial r}(4C(\log(r)+1))-(Cr^2\log(r))\Delta \phi_r+\Delta \phi\frac{\partial}{\partial r}(Cr^2\log(r)) ds $$

$$=\int_{r=\epsilon}-4C(\log(r)+1) \phi_r +4C\phi r^{-1}-(Cr^2\log(r))\Delta \phi_r+C\Delta \phi(2r\log(r)+r) ds $$

$$=C\int^{2\pi}_0 \left( -4(\log(\epsilon))+1)\phi_r(\epsilon) +4\phi \epsilon^{-1}-(\epsilon^2\log(\epsilon))\Delta \phi_r+\Delta \phi(2\epsilon\log(\epsilon)+\epsilon) ds \right) \epsilon d\theta $$

$$=C\int^{2\pi}_0 \left( -4(\epsilon\log(\epsilon))+\epsilon)\phi_r(\epsilon) +4\phi -(\epsilon^3\log(\epsilon))\Delta \phi_r+\Delta \phi(2\epsilon^2\log(\epsilon)+\epsilon^2) ds \right) d\theta $$

Letting $\epsilon\to 0$, this becomes

$$=C\int^{2\pi}_0 4\phi(0,\theta) d\theta = 8\pi C \phi(0)$$

So we conclude that

$$\Delta^2 T_{F(r)}(\phi) = 8\pi C \phi(0) =8\pi C \delta(x) $$

and thus $F(r)$ is a fundamental solution.

Answer 2

Interesting. You know that $C\log(r)$ is a fundamental solution of $\Delta$. And, $$ \begin{align} \Delta( r^{2}\log(r) ) & = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial(r^{2}\log(r))}{\partial r}\right) \\ & =\frac{1}{r}\frac{\partial}{\partial r}\left(2r^{2}\log(r)+r^{2}\right) \\ & = \frac{1}{r}\left(4r\log(r)+2r+2r\right) \\ & = 4\log(r)+4. \end{align} $$ That should be enough to do the job, and it dictates a plan like the one you carried out.