fundamental solution of biharmonic equation of dimension 2

Question

How can I get the fundamental solution of biharmonic equation of dimension 2,$\Delta^2u=0,$ with the help of fundamental solution of Laplace equation?

Indeed I just to calculate the following integration:

$$\int_{0}^{+\infty}\int^{2\pi}_{0}r\ln{r}\ln{\sqrt{R^2+r^2-2 R\, r\cos{(\theta-\theta_0)}}\,\mathrm{d}\theta \mathrm{d}r},$$

where $R$ and $\theta_0$ are positive constants.

OK, my difficulty is just to calculate the integration above. And the answer is $Cr^2\ln{r}$, and $C$ is a constant of $R$. — Daniel, Jun 08 '13 at 13:32
I have never heard of double Laplace equation, where did you see this term? — Shuhao Cao, Jun 09 '13 at 00:09
Also it sounds like from your comment that the upper limit for the polar radius should be $r$, not $+\infty$. — Shuhao Cao, Jun 09 '13 at 00:19
@ShuhaoCao Also known as the biharmonic equation $\Delta^2 u=0$. — ˈjuː.zɚ79365, Jun 09 '13 at 03:04

score 3 · Answer 1 · edited Jun 10 '13 at 19:07

The integral $$\int^{2\pi}_{0}\ln{\sqrt{R^2+r^2-2 R\, r\cos{(\theta-\theta_0)}}}\,\mathrm{d}\theta \tag1$$ does not depend on $\theta_0$, as you can find by changing the variable to $\theta-\theta_0$ and using the periodicity of $\cos$. What's more, it does not depend on $r$ either (as long as $r<R$). Indeed, the function under the integral is harmonic in the disk $|x|<R$, being the fundamental solution of $\Delta$ with the pole at $(R,\theta_0)$. Therefore, its average over circle of radius $r$ is equal to the value at $0$.
$$\int^{2\pi}_{0}\ln{\sqrt{R^2+r^2-2 R\, r\cos{(\theta-\theta_0)}}}\,\mathrm{d}\theta = 2\pi \ln R \tag2$$

Just for completeness: there are other ways to get the findamental solution for the bi-Laplacian.

Use the fact that every biharmonic function is of the form $r^2 g + h $ where $g,h$ are harmonic.
Or, write down $\Delta^2$ in polar coordinates and solve the ODE.

fundamental solution of biharmonic equation of dimension 2

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