# Changes between Version 7 and Version 8 of ThinPlateShell

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Timestamp:
Dec 9, 2014, 5:30:04 PM (9 years ago)
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 v7 We consider a set of measurement points ${\bf r_i}, i = 1,2, \ldots,N$, and the corresponding set of measurements values for a quantity $f$ (a velocity component for PIV data). $f({\bf r_i})=f_i, i = 1,2, \ldots,N} \; [equ. 1]$ $f({\bf r_i})=f_i, i = 1,2, \ldots,N} \; [equ. 1]$ A pure interpolation function $f({\bf r})$ must exactly reach these values $f_i$ at the measurement points, while a  smoothing function should approach these values within a range smaller than the estimated error bar. Considering first the case of a pure interpolation, the function $f({\bf r})$ must minimise the 'elastic energy' or mean squared curvature $E = \int\left[\left(\frac{\partial^2 f}{\partial x^2}\right)^2 + 2\left(\frac{\partial^2 f}{\partial xy}\right)^2 + \left(\frac{\partial^2 f}{\partial y^2}\right)^2 \right] \textrm{d} x \, \textrm{d}y \;\textrm{[equ. 2]}$ $E = \int\left[\left(\frac{\partial^2 f}{\partial x^2}\right)^2 + 2\left(\frac{\partial^2 f}{\partial xy}\right)^2 + \left(\frac{\partial^2 f}{\partial y^2}\right)^2 \right] \textrm{d} x \, \textrm{d}y \;\textrm{[equ. 2]}$ with the constraints [->equ. 1]. These constraints can be also written as a domain integral with Dirac functions at measurement points: with the constraints [->equ. 1]. These constraints can be also written as a domain integral with Dirac functions at measurement points: ${\int f \delta({\bf r-r_i}) \textrm{d} x \, \textrm{d}y=f_i , i = 1,2, \ldots,N} \;\textrm{[equ. 3]}$ ${\int f \delta({\bf r-r_i}) \textrm{d} x \, \textrm{d}y=f_i , i = 1,2, \ldots,N} \;\textrm{[equ. 3]}$ This variational problem is solved by introducing a Lagrange parameter $8\pi S_i$ for each constraint (the factor $8\pi$ is put for further convenience). The variation of the energy can be rewritten by a double integration by part, leading to $\int \Delta^2 f \delta f dx dy$. The optimum function $f$ is thus solution of the  double Laplacian equation with pointwise source terms at locations $r_i$, The solution $\phi$ for a single source point is such that $\Delta^2 \phi=0$ everywhere outside the source. Using the axisymmetric form of the Laplacian $\Delta \phi=\frac{1}{r}\frac{d}{dr}(r\frac{d\phi}{dr})$, we can show that the general axisymmetric solution is $\phi=C_0 \log(r)+C_1 r^2 + C_2 r^2 \log (r)+C_3 \;\textrm{[equ. 5]}$$whose Laplacian writes$$\Delta\phi=\frac{1}{r}\frac{d}{dr}(r\frac{d\phi}{dr})=4 C_1+ 4 C_2 (1+\log(r))\;\textrm{[equ. 6]}$ $\phi=C_0 \log(r)+C_1 r^2 + C_2 r^2 \log (r)+C_3 \;\textrm{[equ. 5]}$ whose Laplacian writes $\Delta\phi=\frac{1}{r}\frac{d}{dr}(r\frac{d\phi}{dr})=4 C_1+ 4 C_2 (1+\log(r))\;\textrm{[equ. 6]}$ The coefficient $C_0$ must be zero to avoid divergence of $\phi$ at the origin. The constant $C_3$ can be included in a global constant $a_0$. The constant $C_1$ can be viewed as a change of unit for $r$ (replacing $r$ by $r/r_0$ with $\log(r_0)=-C_1/C_2)$), and can be set to 0 without loss of generality. The integral of $\Delta\phi$ over a small neighborhood of the source must be equal to $8\pi S_i$. By the divergence theorem, this is also equal to the flux of $\nabla \Delta \phi=4 C_2/r$ around the source, equal to $8\pi C_2$. This must be equal to $8\pi S_i$, which sets $C_2=S_i$. The elementary function is thus $\phi(r)=S_i r^2\log (r)$.^ The general form of $f$ is thus $\label{sol_gene} f({\bf r})=\sum S_i \phi({\bf|r-r_i}|)+a_0+a_1x+a_2y\;\textrm{[equ. 7]}$ $\label{sol_gene} f({\bf r})=\sum S_i \phi({\bf|r-r_i}|)+a_0+a_1x+a_2y\;\textrm{[equ. 7]}$ with $\phi(r)=r^2\log (r)$. The  values at the measurement points ${\bf r_j}$ are $f({\bf r_j})=\sum S_i \phi({\bf|r_j-r_i}|)+a_0+a_1x_j+a_2y_j;\;\textrm{[equ. 8]}$ $f({\bf r_j})=\sum S_i \phi({\bf|r_j-r_i}|)+a_0+a_1x_j+a_2y_j;\;\textrm{[equ. 8]}$ In other words, defining the vector $F=f({\bf r_j}) \{ j=1,...N \}, a_0, a_1,a_2$ and the matrix $M=\left[ \begin{array}{cc}^$ M=\left[  \begin{array}{cc}^ \phi(|r_j-r_i|) , 1,\;  x_j,\;  y_j \\ ... \;\;\;\;\;\;\;\;\;\;\;\;\; ... \;\;\; ... \;\;\; ...\\ The variational problem then leads to the equation $\rho\Delta^2 f= \sum (f_i-f) \delta ({\bf r-r_i})$ $\rho\Delta^2 f= \sum (f_i-f) \delta ({\bf r-r_i})$ The solution is still obtained as a sum of radial basis functions $\phi({\bf|r-r_i}|)$, whose weight $S_i$ now satisfies $M_\rho*S=F$, with the matrix $M_\rho=M+\rho I_{000}$ $M_\rho=M+\rho I_{000}$ where $I_{000}$ is the NxN identity matrix extended by three columns of 0.^ Spatial derivatives of the interpolated quantity $f$ can be obtained by direct differentiation of the result.  For any function $\phi(r)$, with radial distance $r=|{\bf r-r_i}|$, $r^2=X^2+Y^2$, $\partial_X \phi=(d\phi/dr) \partial_X r$, and $\partial_X r=X/r$. This yields $\partial_X \phi=X (2 \log(r) +1)$, so that, $\partial_x f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(x-x_i)+a_1;$ $\partial_y f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(y-y_i)+a_2;$ $\partial_x f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(x-x_i)+a_1;$ $\partial_y f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(y-y_i)+a_2;$ = Sub-domains =