[Differential Geometry of Curves] Regular Closed Curve function

[cheersdup]

Let m : [0,L] → ℝ2 be a positively oriented C1 regular Jordan curve parametrized with arc length. Consider the function F : [a,b] x [a,b] → ℝ defined by F(u,v) = (1/2) ||m(u) - m(v)||2

Define a local diameter of m as the line segment between two points p = m(u) and q = m(v) such that:
The open line segment (p,q) is included in the interior of m and
F has a local maximum at (u,v)

(1) Compute the first derivatives of F, expressing them as functions of (m(v) - m(u)) and the unit tangent Tm

(2) Prove that if the segment [p,q] is a local diameter then it is normal to m at both points p and q

(3) Assume that (u,v) is such that the open segment (m(u),m(v)) is interior to m, and that it is normal to the curve at both extremities
(i) Prove that Tm(u) and Tm(v) are collinear. For the rest of the question admit without proof that Tm(v) = -Tm(u)
(ii) Prove that the Hessian of second derivatives of F at (u,v) can be put in the form
H = ( 1- k(u)||m(u)-m(v)|| ...1
......1... 1-k(v)||m(u)-m(v)|| )

Homework Equations

Given above

The Attempt at a Solution

I can get part 1, but for part 2 I'm not sure how to prove that a local diameter is normal to p and q. I believe it's something to do with the distance between the tangent lines is should be perpendicular?

Part 3 is quite beyond me I'm not sure how to tackle that one yet

 

[mighty2000]

.

For part 2, to prove that the local diameter is normal to m at both points p and q, we can use the definition of a normal vector: a vector that is perpendicular to a given vector. In this case, the given vector is the tangent vector Tm at points p and q. Since the local diameter is a line segment between p and q, and the tangent vector is a vector tangent to the curve at those points, the local diameter must be perpendicular to the curve at those points. This can also be seen geometrically as the tangent vector and the local diameter form a right angle at those points.

For part 3, we can start by using the definition of collinearity: three or more points are collinear if they all lie on the same line. In this case, the points Tm(u), Tm(v), and -Tm(u) all lie on the tangent line at point p, and similarly for the points Tm(u), Tm(v), and -Tm(v) at point q. Since the tangent line is unique at each point, this means that Tm(u), Tm(v), and -Tm(u) are collinear at point p, and Tm(u), Tm(v), and -Tm(v) are collinear at point q. This is because the local diameter is normal to the curve at both points, and therefore the tangent vector Tm is perpendicular to the local diameter at those points.

For part 3(ii), we can use the definition of the Hessian matrix, which is the matrix of second partial derivatives of a function. In this case, the function is F(u,v) and the Hessian matrix is defined as H = [∂^2F/∂u^2 ∂^2F/∂u∂v; ∂^2F/∂u∂v ∂^2F/∂v^2]. Since we are given that Tm(v) = -Tm(u), we can simplify the Hessian matrix to H = [∂^2F/∂u^2 -∂^2F/∂u∂v; -∂^2F/∂u∂v ∂^2F/∂v^2]. We can then substitute in the given values for k(u) and k(v) to get the final form of H =