Background
The Poisson Integral
$$ \tilde h(r e^{i \theta}) = \frac{1}{2 \pi}\int_{-\pi}^\pi h(e^{i \phi}) P_r(\theta – \phi) d\phi$$
takes an arbitrary continuous complex-valued function $h(e^{i \theta})$ defined on the unit circle and extends it to a function $\tilde h(z)$ defined inside the unit disk such that $\tilde h(z)$ is harmonic ($\Delta \tilde h(z) = 0$) and has boundary values $h(z)$ on the unit circle.
$P_r(\theta)$ is the so-called Poisson Kernel and, among other representations,
$$P_r(\theta) = \Re \left( \frac{1 + r e^{i \theta}} {1 – r e^{i \theta}} \right) , \ \ \ \ r e^{i \theta} \in \mathbb{D}.$$
Question
Is there a different kernel function I could use which would make $\tilde h(z)$ biharmonic instead of just harmonic inside the unit disk? That is, I'd like to find something to replace $P_r(\theta)$ with so that $\Delta^2 \tilde h(z) = 0$.
Best Answer
After a tip from Conrad I was able to find a different paper (Tangential limit values of a biharmonic poisson integral in a disk) which gives the "Biharmonic Poisson Kernel" $\tilde P$ in terms of the regular Poisson Kernel:
$$ \tilde P_r(\theta) = \frac{1 - r^2}{2} (P_r(\theta) + P_r^2(\theta)) $$