Polyharmonic spline

Remarks

Suppose the centres are unisolvent for $\Pi_{m-1}$, that is, the only $q \in \Pi_{m-1}$ vanishing at all $x_i$ is $0$. Then for any data $y_1, \dots, y_N$ there is exactly one polyharmonic spline with $s(x_i) = y_i$. This is radial basis function interpolation with the radial basis function $\varphi$. Up to a constant factor, $\varphi(\norm{x})$ is a fundamental solution of $\Delta^m$, where $\Delta$ is the Laplacian, so $s$ satisfies $\Delta^m s = 0$ away from the centres. The interpolant minimizes the seminorm $\sum_{\abs{\alpha} = m} \frac{m!}{\alpha!} \int_{\R^d} (\partial^\alpha f)^2$ among suitable functions with $f(x_i) = y_i$. Example: $d = 1$, $m = 2$ gives $\varphi(r) = r^3$ and $s(x) = \sum w_i \abs{x - x_i}^3 + a + bx$ with $\sum w_i = \sum w_i x_i = 0$; this is a natural cubic spline. Non-example: in the same setting, $s(x) = \abs{x}^3$ (one centre $0$, $w_1 = 1$) violates the side condition $\sum w_i = 0$, and indeed it is not linear outside the centres. For $d = 2$, $m = 2$ one gets $\varphi(r) = r^2 \log r$, the thin plate spline.