0
$\begingroup$

When using the Dual Reciprocity Boundary Element Method ( or any radial basis function method ) to solve a nonlinear differential equation it is necessary to approximate some derivatives of a potential field using radial basis functions.

If you have a potential field $\mathbf{u}_i$ with coordinates $\mathbf{x}_i=(x_i,y_i)$. The field can be approximated with radial basis functions such that $\mathbf{u}=\mathbf{F}\mathbf{\alpha}$. Where $\mathbf{F}$ is a matrix of radial basis functions based on the following: $$ f(\mathbf{x_i})=\Sigma_{j=1}^{N}\alpha_j]\phi(||\mathbf{x_i}-\mathbf{x_j}||)_2$$ $$ \mathbf{F}=\left[ \begin{matrix} \phi(||\mathbf{x_1}-\mathbf{x_1}||)_2 & \cdots & \phi(||\mathbf{x_1}-\mathbf{x_N}||)_2 \\ \vdots & \ddots & \vdots \\ \phi(||\mathbf{x_N}-\mathbf{x_1}||)_2 & \cdots & \phi(||\mathbf{x_N}-\mathbf{x_N}||)_2 \\ \end{matrix}\right] $$ $$ \mathbf{\alpha}=\left[\begin{matrix} \mathbf{\alpha_1} \\ \vdots \\ \mathbf{\alpha_N} \\ \end{matrix}\right]$$

For ATPS (Augmented Thin Plate Splines) the matrix of radial basis functions becomes: $$ \mathbf{x_i}=(x_i,y_i) $$ $$ r=\sqrt{ (x_i-x_j)^2 + (y_i+y_j)^2 }=||\mathbf{x_i}-\mathbf{x_j}|| $$ $$f(\mathbf{x_i})=\Sigma_{j=1}^{N}\alpha_j r^2 log(r) + \beta_1+\beta_2x_j+\beta_3y_j $$

$$\Sigma_{j=1}^N \alpha_j=\Sigma_{j=1}^N \alpha_j x_j=\Sigma_{j=1}^N \alpha_j y_j=0 $$

$$ \mathbf{P}= \left[ \begin{matrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_N \\ y_1 & y_2 & \cdots & y_N \\ \end{matrix} \right] $$ $$ \mathbf{F^*}= \left[ \begin{matrix} \mathbf{F} && \mathbf{P^T} \\ \mathbf{P} && \mathbf{0} \\ \end{matrix} \right] $$

$$ \left[ \begin{matrix} \mathbf{u} \\ \mathbf{0} \\ \end{matrix} \right]= \left[ \begin{matrix} \mathbf{F} && \mathbf{P^T} \\ \mathbf{P} && \mathbf{0} \\ \end{matrix} \right]\left[ \begin{matrix} \mathbf{\alpha} \\ \mathbf{\beta} \\ \end{matrix} \right]=\mathbf{F^*\alpha^*} $$

As $\alpha$ and $\beta$ are constants, a spatial derivative of $\mathbf{u}$ becomes a spatial derivative of the radial basis function series. Thus:

$$ \frac{\partial\mathbf{u}}{\partial x}=\frac{\partial\mathbf{F^*}}{\partial x}\mathbf{\alpha^*}=\frac{\partial\mathbf{F^*}}{\partial x}\mathbf{{F^*}^{-1}u} $$

similarly the second derivative would be $$ \frac{\partial^2\mathbf{u}}{\partial x^2}=\frac{\partial^2\mathbf{F^*}}{\partial x^2}\mathbf{{F^*}^{-1}u} $$

However when taking the second derivative of the ATPS function I obtain: $$\frac{\partial^2f(\mathbf{x_i})}{\partial x^2}=\Sigma_{j=1}^{N}\alpha_j \left[2log(r)+\frac{(y_i-y_j)^2}{r^2}+1 \right] $$

If you then take the limit of this as $\mathbf{x_i}$ approaches $\mathbf{x_j}$ you obtain $-\infty$ for the diagonals of the matrix $\frac{\partial^2\mathbf{F^*}}{\partial x^2}$.

Thus i do not understand how to obtain a second derivative approximation of a potential field u using Augmented Thin Plate Splines.

|cite|improve this question
$\endgroup$
  • $\begingroup$ I have an answer for you, but also a couple of questions. What is this called augmented TPS, not just TPS? $\endgroup$ – rych Jan 27 at 13:12
  • $\begingroup$ Now, something that will hopefully lead us to a solution. Do you know the value of $r^2 \log r$ at $r=0$? $\endgroup$ – rych Jan 27 at 13:14
  • $\begingroup$ It is augmented TPS because of the additional three polynomial terms after the $r^2log(r)$. The value of at $r=0$ you can take the limit as r goes to zero. $\lim_{r \to 0}[r^2log(r)]=\lim_{r \to 0}[\frac{log(r)}{\frac{1}{r^2}}]=\frac{-\infty}{\infty}$ then use L'hopital's rule to say that because the limit is $\frac{\infty}{\infty}$ you take the limit of the derivative. $\lim_{r \to 0}[\frac{log(r)}{\frac{1}{r^2}}]=\lim_{r \to 0}\frac{\partial}{\partial r}[\frac{log(r)}{\frac{1}{r^2}}]=\lim_{r \to 0}[\frac{\frac{1}{r}}{\frac{-2}{r^3}}]=\lim_{r \to 0}[\frac{-r^2}{2}]=0$ $\endgroup$ – N. Morgan Jan 28 at 2:12
  • $\begingroup$ The first spatial derivative of the matrix F also has zero diagonals because of the same limit as $r \to 0$. It is only the second spatial derivative that goes to $-\infty$. $\endgroup$ – N. Morgan Jan 28 at 2:17
1
$\begingroup$

It is worth remembering that approximation theory is a branch of numerical analysis. In numerical analysis when you're dealing with negligible terms, you don't look closely into how smooth those terms are: once they are sufficiently close to zero they are zero.

As you program TPS evaluation and want to avoid a runtime error you shouldn't attempt evaluating $r^2 \log r$ at $r=0$: although the limit value is zero, part of that expression is still $\log r$ which is simply undefined at $0$.

Instead, TPS is programmed as a piecewise function $\cases{r^2\log r, r>\varepsilon \\0}$.

So what you're trying to calculate is the second derivative of a constant $0$. It is zero.

Numerics aside, in pure analysis, we are not even allowed to differentiate a functions where its undefined; and if you make limit a part of definition, then you won't be allowed to just swap differentiation and limit at the singularity.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ How large should epsilon be? Should it be sized to only reduce the diagonal terms to zero? or should it potentially zero other parts of the matrix? $\endgroup$ – N. Morgan Jan 29 at 0:59
  • $\begingroup$ On a related note but not strictly following the original question... I sometimes observe a Runge's phenomenon effect on the edges of the area being interpolated and approximated. This effect tends to be more exaggerated in the derivatives. I added the polynomial smoother to the TPS (making it ATPS) in the hopes of addressing this overfit behavior. Do know any other methods to improve the edge predictions for the derivatives? (I am looking into using hermite interpolation instead for what has been labeled "mass conservative interpolation of velocity derivatives" in some papers/books) $\endgroup$ – N. Morgan Jan 29 at 1:03
  • $\begingroup$ Well, theoretically epsilon should be zero: you're removing the singularity at one single point $r=0$. But in practice, take $\varepsilon$ a smallest floating point number, for example. Actually, as an experiment, in your computer program, what happens if you try to evaluate $r\log r$ at $0$? $\endgroup$ – rych Jan 29 at 6:55
  • $\begingroup$ You cannot use TPS for data approximation without the linear polynomial in front, of course. See for example Table 3.1. in Armin Iske 2004 Multiresolution Methods in Scattered Data Modelling. $\endgroup$ – rych Jan 29 at 7:00
  • $\begingroup$ In the interpolation problem with RBF, such as TPS you have here, there are no "edges". However, for TPS, it's recommended to scale your data into a $[-0.5,0.5]^2$ interval first. $\endgroup$ – rych Jan 29 at 7:04
1
$\begingroup$

A "classic TPS spline $(\sigma)$ is a continuously differentiable function $\sigma \in C^1(\mathbb{R}^2)$". If you need to approximate second derivatives of the function then you have to use a twice continuously differentiable version of the Duchon spline.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Could you add a formula, or a reference link to a "twice continuously differentiable version of the Duchon spline" please? $\endgroup$ – rych Feb 12 at 6:17
  • 1
    $\begingroup$ Sure. Here we are: R. Arcangéli, Multidimensional Minimizing Splines, Springer, 2004 A. Bezhaev, V. Vasilenko, Variational Theory of Splines, Springer, 2001 J. Duchon, Splines minimizing rotation-invariant semi-norms in Sobolev spaces, Lect. Notes in Math., Vol. 571, Springer, Berlin, 1977 $\endgroup$ – NetTvor Feb 12 at 8:42
  • $\begingroup$ Also simple theory of something close to the Duchon splines could be found at normalsplines.blogspot.com/2019/01/… $\endgroup$ – NetTvor Feb 12 at 8:53
  • $\begingroup$ One month later and I've started reading the references you posted. Спасибо! A good chance to improve my approximation theory education. For now I'm a bit confused: say we reconstruct surface from a point cloud using TPS: isn't it smooth and curvature continuous despite TPS being only once differentiable at the centers? What's your opinion of my answer above? $\endgroup$ – rych Mar 13 at 10:16
  • 1
    $\begingroup$ TPS is infinitely differentiable everywhere but at the centers. BTW this book contains explicit formula for twice continuously differentiable TPS (D_m splines): Hebert Montegranario, Jairo Espinosa. "Variational Regularization of 3D Data Experiments with MATLAB". $\endgroup$ – NetTvor Apr 9 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.