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What does it mean for a curve to move in the normal direction?

Especially in the context of shape optimization and level set methods.

Does it mean that the boundary of a set is considered a curve and when this curve evolves through different shapes, then each point in the boundary is moving either to its inward normal or outward normal direction.

mavavilj
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  • "normal" means "perpendicular." However, no curve (in dimension > 1) is ever moving in the direction perpendicular to the direction it travels in... – Stella Biderman Mar 23 '17 at 15:46
  • @Stella Biderman Think to the movement of the boundary of an expanding drop of colored liquid. See my (very qualitative) answer. – Jean Marie Mar 23 '17 at 15:53

1 Answers1

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It means that you have to describe this "movement" through a family of implicitly defined functions under the form:

$$\tag{1}f_t(x,y)=0$$

where $t$ is a parameter that you can consider as a time, and that the gradient

$$(\partial f/\partial x,\partial f/\partial y)$$

indicates the direction of evolution of the curve ; the coefficient $k(x,y)$ of the normalized gradient you will attribute will give the local evolution speed (the intensity) of theses curves. Rather often, this speed is governed by a certain power of the local curvature, and the curves appear as solutions of differential equations or partial differential equations.

When you examine with another point of view, you can consider the family of curves defined by (1) as the level sets of a certain surface. This explains that "level sets" is a good keyword for the subject.

Jean Marie
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  • Yes, actually I'm studying level sets. However, I still don't understand how the curve or surface moves. And how can the gradient give just one direction, if the curve is a circle for example (and then all points of the circle would move to different directions). – mavavilj Mar 23 '17 at 15:53
  • The gradient of a circle $x^2+y^2=t^2$ is $(2x,2y)$, meaning that each point $(x,y)$ is flowing away with speed $(x,y)$, and normalized speed $(x,y)/\sqrt{x^2+y^2}$ ... – Jean Marie Mar 23 '17 at 15:56
  • So all points move with the same velocity? But into what direction? – mavavilj Mar 23 '17 at 15:58
  • In this case yes. In other cases no, if you multiply the normalized gradient (which is a vector) by a scalar quantity $k(x,y)$ which can be $>1$ or $=1$ or $<1$, you can mdulate the speed. – Jean Marie Mar 23 '17 at 16:17
  • I read from one source that actually the answer ought to be that the points of the boundary move to the direction of inward normal or outward normal with some velocity $v$. Could you clarify your answer to address this? – mavavilj Mar 24 '17 at 06:14
  • I will do it in a few hours. – Jean Marie Mar 24 '17 at 12:50