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Subject: Re: rotationally symmetry of laplacian



"news.cse.ohio-state.edu" wrote in message
news:e1k5ko$f9d$1@news1.cse.ohio-state.edu...

> In the context of edge-detection using lapacian operators, I came across
> the statement that the lapacian is rotationally invariant. I am unable to
> see why this is true. Any pointers or ideas?

Consider a C2-function F(x,y); then Laplacian(F) = Fxx + Fyy
where Fx = dF/dx, Fy = dF/dy, Fxx = d^2 F/dx^2,
Fxy = d^2 F/dx dy, and Fyy = d^2 F/dy^2 (all partial derivatives).
Define u = c*x - s*y and v = s*x + c*y, where c = cos(A) and
s = sin(A). This is a rotational change of variables. Define
G(u,v) = F(x,y). Use the chain rule
Fx = Gu*du/dx + Gv*dv/dx = c*Gu +s*Gv
Fy = Gu*du/dy + Gv*dv/dy = -s*Gu + c*Gv
Differentiate again,
Fxx = c^2*Guu + 2*s*c*Guv + s^2*Gvv
Fyy = s^2*Guu - 2*s*c*Guv + c^2*Gvv
Add these together,
Fxx + Fyy = Guu + Gvv
so Laplacian(F) = Laplacian(G).

--
Dave Eberly
http://www.geometrictools.com



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