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Follow up from this question Thin-Plate-Spline understanding and solution.

In the general case of $\mathbb{R}^N$ the following problem (interpolant which minimizes the Thin Plate Energy, specifically formula (1) in the document).

$$ \left\{ \begin{array}{ll} \text{min} \int_{\mathbb{R}^N} \sum_{|\alpha| = 2} {2 \choose \alpha} | D^{\alpha} u |^2 dx \\ \text{s.t.} \;\; u(x_i) = y_i & i = 1,\dots,n \end{array} \right. $$

After few tedious but easy calculations we can find that the solution $u$ has the form

$$ u(x) = \sum_{1 \leq i \leq n} \lambda_i \Phi(x - x_i) $$

where $\Phi(x)$ is the fundamental solution of the biharmonic equation $\Delta^2 u = 0$. Now no matter the choice of $N$ each $\Phi$ can have only one of the following expressions

$$ \Phi(x) = \begin{cases} A \ln r + B r^2 + Cr^2 \ln r + D & N = 2 \\ A \ln r + B r^2 + Cr^{-2} + D & N = 4 \\ A r^{4 - N} + B r^2 + C r^{2 - N} + D & \text{otherwise} \\ \end{cases} $$

Where the constants $A,B,C$ and $D$ have to be determined based on the interpolation problem, moreover $r = \sqrt{x_1^2 + \ldots x_N^2}$

The math matches the document I attached above. What I don't understand is how to deal with the singularities at $r = 0$ in practice.

For example in the paper Principal Warps for $N = 2$ it is only considered $U(r) = r^2 \ln r^2$, which has no singularity at the origin. This makes me think that a strategy might be to carefully choose a priory some coefficients $A,B,C$ and $D$ so that we don't have singular points.

Is this the reason why the Thin Plate Energy is normally used in interpolation problems more as a regularizer rather than the actual cost function when doing data fitting?

My personal understanding of regularizers (like Tikhonov_regularization) is because of practical numerical instabilities due to large linear systems becoming singular.

Is this understanding on how to use TPS correct?

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What I don't understand is how to deal with the singularities at $r=0$ in practice.

A quick answer is actually found later in that very ThinPlateSplines.pdf linked in your post: "The terms of $u(r)$ ... are discarded because they have no effect (in the distributional sense) in determining solutions to the minimization problem. In summary, the Green’s functions are" ... leading to the 2 cases of polyharmonic splines, simply $r^{4-n}\ln r$ or $r^{4-n}$ without disturbing yet irrelevant terms.

So, the solution to the biharmonic equation, the Euler-Lagrange for the given energy functional, has some terms that won't matter under that integral.

Unfortunately, for $n\geq4$, $r^{4-n}\ln r$ and $r^{4-n}$ themselves have a non-removable singularity "so the interpolating function is not defined at the sample points. This is clearly not desired." The Note after the eq. (29) continues with some close choices for the RBF that may no longer minimize this particular energy functional though.

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    $\begingroup$ So in practice there's no singularity (case $n =2,3$) because of the "distribution sense", right? $\endgroup$ Commented Jul 30 at 18:21
  • $\begingroup$ Yes, luckily(?) those terms don't contribute to (minimizing) the original elastic energy functional anyway, so we can choose their coefficients to be zero $\endgroup$
    – rych
    Commented Jul 31 at 10:05
  • $\begingroup$ And I guess the way to prove this is by solving the PDE with weak solution and show that those extra terms don't contribute to the solution $\endgroup$ Commented Jul 31 at 22:39

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