Approximation of Gaussian Curvature

The sizing of the bottom topography is done using the Gaussian curvature of the surface using the GaussianCurvature class.

From wikipedia, the Gaussian curvature for a surface \(f(x,y) - z = F(x,y,z) = 0\) is

\[ \begin{equation} K=-\frac{1}{|\nabla F|^4} \left|\begin{array}{cc}H(F) & \nabla F^T \\\nabla F & 0\end{array}\right|=-\frac{1}{|\nabla F|^4}\left|\begin{array}{cccc}F_{xx} & F_{xy} & F_{xz} & F_x \\F_{xy} & F_{yy} & F_{yz} & F_y \\F_{xz} & F_{yz} & F_{zz} & F_z \\F_x & F_y & F_z & 0\end{array}\right| \end{equation} \]

where \(H\) is the Hessian matrix.

Filling in the derivatives,

\[ \begin{equation} K=-\frac{1}{|\nabla F|^4}\left|\begin{array}{cccc}f_{xx} & f_{xy} & 0 & f_x \\f_{xy} & f_{yy} & 0 & f_y \\0 & 0 & 0 & -1 \\f_x & f_y & -1 & 0\end{array}\right| \end{equation} \]

where \(|\nabla F|^4 = (f_x^2 + f_y^2 + 1)^2\).

Then

\[ \begin{equation}\begin{split} -{|\nabla F|^4}K &= \left|\begin{array}{cccc}f_{xx} & f_{xy} & 0 & f_x \\f_{xy} & f_{yy} & 0 & f_y \\0 & 0 & 0 & -1 \\f_x & f_y & -1 & 0\end{array}\right| \\[0.2cm] &= f_{xx}\left|\begin{array}{ccc}f_{yy} & 0 & f_y \\0 & 0 & -1 \\f_y & -1 & 0\end{array}\right| - f_{xy}\left|\begin{array}{ccc}f_{xy} & 0 & f_y \\0 & 0 & -1 \\f_x & -1 & 0\end{array}\right| -f_x\left|\begin{array}{ccc}f_{xy} & f_{yy} & 0 \\0 & 0 & 0 \\f_x & f_y & -1\end{array}\right| \\[0.2cm] &= f_{xx}f_{yy}-f^2_{xy} \end{split} \end{equation} \]

So, finally,

\[ \begin{equation} K = \frac{f_{xx}f_{yy}-f^2_{xy}}{ (f_x^2 + f_y^2 + 1)^2} \end{equation} \]

Example: For a sphere, \(z = \sqrt{1 - (x^2+y^2)}= f(x,y)\), \(K=1\).

To approximate \(K\), we compute a centered finite difference approximation to the first derivatives

\[ \begin{equation} \begin{gathered} f_x\approx \frac{f(x+h,y) - f(x-h,y)}{2h}\\[0.2cm] f_y\approx \frac{f(x,y+h) - f(x,y-h)}{2h}\\ \end{gathered} \end{equation} \]

and similarly for the second derivatives,

\[ \begin{equation} \begin{gathered} f_{xx} \approx \frac{f(x+h,y) -2f(x,y) + f(x-h,y)}{h^2}\\[0.2cm] f_{yy} \approx \frac{f(x,y+h) -2f(x,y) + f(x,y+h)}{h^2}\\[0.2cm] f_{xy} = f_{yx}\approx \frac{f(x+h,y+h) - f(x-h,y+h)- f(x+h,y-h) + f(x-h,y-h)}{4\Delta x\Delta y} \end{gathered} \end{equation} \]

The step size is taken to be \(h = 10^{-4}\).