I'm trying to fuse multiple Bézier curves and get a C2 continuous curve. Every document or book that I read says that imposing:
$B_1(1) = B_2(0)$
$B_1'(1) = B_2'(0)$
$B_1''(1) = B_2''(0)$
But this is does not really means C2 continuity since the second derivative has only a point constraint, a peak can happen in the second derivative curve (which in turn will make curvature discontinuous). So where is the problem: is it my definition of continuity that is wrong or really some other constraint must be applied to achieve a C2 continuous bézier curve (and therefore a continuous curvature).
Let me give an example: using 5th Bézier curve to find a path between (0,0), (10,5) and (25,5). Using the method given in https://ieeexplore.ieee.org/document/5354805 without any optimization these are the results:
In the second image the blue plot is the 1st derivative and the red plot is the second derivative. One can see that in the red plot, in the connection point (x = 10) present is discontinuity, which is propagated in the curvature. This is consistent with the constraint, since the 2nd derivative curve does obey the point constraint, but nonetheless is discontinuous. This behavior is also observed considering a 3rd degree Bézier curve and $C_1$ continuity.
Ps.: curvature:
$\kappa(s) = \frac{Q'(s) \times Q''(s)}{\left||Q'(s)\right||^3} $
EDIT:
I mixed continuity with differentiability. The 2nd derivative is continuous by the $\epsilon-\delta$ definition, but is not differentiable due to different lateral limits. Therefore to obtain a continuous and differentiable curvature more than $C_2$ is necessary. Thanks.
When we say that the junction between two curves is $C_2$-continuous, we mean (by definition) that their first and second derivatives are the same at the junction point. Except in very unusual situations (like when the first derivative vector is zero), this means that the two curves will also have the same curvature at the junction.