Changes between Version 6 and Version 7 of ThinPlateShell


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Timestamp:
Dec 9, 2014, 5:25:02 PM (6 years ago)
Author:
sommeria
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  • ThinPlateShell

    v6 v7  
    4949
    5050= Smoothing interpolation =
    51 To account for measurement noise, interpolation should involve smoothing, so that the interpolated data can deviate from the measurement at the measurement sites $r_i$. This can be performed by minimising a linear combination of the elastic energy and the quadratic deviation to the measurement values  $$                                                                                                                               E = \sum(|f({\bf  r_i})-f_i|^2)+ \rho\int\left[\left(\frac{\partial^2 f}{\partial x^2}\right)^2 + 2\left(\frac{\partial^2 f}{\partial xy}\right)^2 + \left(\frac{\partial^2 f}{\partial y^2}\right)^2 \right] \textrm{d} x \, \textrm{d}y   $$^
     51To account for measurement noise, interpolation should involve smoothing, so that the interpolated data can deviate from the measurement at the measurement sites $r_i$. This can be performed by minimising a linear combination of the elastic energy and the quadratic deviation to the measurement values 
    5252
    53 The variational problem then leads to the equation $$ \rho\Delta^2 f= \sum  (f_i-f) \delta ({\bf r-r_i}) $$ 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 $$ M_\rho=M+\rho I_{000} $$ where $I_{000}$ is the NxN identity matrix extended by three columns of 0.^
     53$ E = \sum(|f({\bf  r_i})-f_i|^2)+ \rho\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   $
     54
     55The variational problem then leads to the equation
     56
     57$ \rho\Delta^2 f= \sum  (f_i-f) \delta ({\bf r-r_i}) $
     58
     59 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
     60
     61$ M_\rho=M+\rho I_{000} $
     62
     63where $I_{000}$ is the NxN identity matrix extended by three columns of 0.^
    5464
    5565= Spatial derivatives =
    56 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,^
     66Spatial 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,
    5767
    58 $$\partial_x f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(x-x_i)+a_1;$$ $$\partial_y f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(y-y_i)+a_2;$$
     68$\partial_x f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(x-x_i)+a_1;$
     69$\partial_y f({\bf r})=\sum S_i (2 \log|{\bf r-r_i}|+1)(y-y_i)+a_2;$
    5970
    6071= Sub-domains =