• Keine Ergebnisse gefunden

M I A Equal Distance Contours

N/A
N/A
Protected

Academic year: 2021

Aktie "M I A Equal Distance Contours"

Copied!
12
0
0

Wird geladen.... (Jetzt Volltext ansehen)

Volltext

(1)

Lecture 14

M I A Lecture 14

Geodesics and Curve Evolutions

Hamilton Jacobi Equations

Laplace-Beltrami Operator

1 2 3 4 5 6 7 8 9 10 11 12

(2)

Equal Distance Contours

M I A Equal Distance Contour

Given a source area K ⊂ S. We want to find a curve evolution s.t. the graph of α(·, t) is {p ∈ S : ds(p, K) = t}, the equal distance contour of distance t

Consider the general evolution

αt = N × −→

tα, α(u, 0) = α0(u)

Lemma: The curve β(t) := α(u, t)|u=u0 is a geodesic

1 2 3 4 5 6 7 8 9 10 11 12

(3)

Equal Distance Contours

M I A Equal Distance Contours

Proposition: The equal distance contour evolution of an initial curve u0 is given by

αt = N × −→

tα α(·,0) = u0(·) where −→

tα are the tangent unit vector of the equal distance contours α(·, t)

Given a source area K we can find the equal distance contours

{p ∈ S : ds(p, K) = t} choosing u0 with graph equal to the boundary of K

If source is a point choose K to be a small circle around the point

1 2 3 4 5 6 7 8 9 10 11 12

(4)

Equal Distance Contours

M I A 2D Projection

Implementing directly an evolution of a 3D curve is quite cumbersome. We are interested is the projection Π of this 3D curve in the xy plane.

Proposition: The projected equal distance contour evolution is given by

Ct = VN−→n c0 = ∂π(K) (1) where

VN =

D−→

n ,Π(N × −→ tα)

E

= s

(1 + q2)n21 + (1 + p2)n22 − 2pqn1n2 1 + p2 + q2 , with p = ∂z∂x, q = ∂z∂y and −→

n = (n1, n2)

This means that

VN = q

an21 + bn22 − cn1n2,

where a, b, c depend on the surface gradient and can be computed once at the start

1 2 3 4 5 6 7 8 9 10 11 12

(5)

Equal Distance Contours

M I A Level Sets Propagation

Lemma: Consider a function u : U ⊂ R2 → R whose level sets correspond to equal distance contours. Then

Π(N × −→

tα) = (ux(1 + q2) − pquy, uy(1 + p2) − pqux) q

(1 + p2 + q2)(u2x + u2y + (qux − pux)2)

and the 2D curve evolution C˜t = −Π(N × −→

tα) = − (ux(1 + q2) − pquy, uy(1 + p2) − pqux) q

(1 + p2 + q2)(u2x + u2y + (qux − pux)2)

(2)

can be used to compute geodesic paths Consequences:

The proposition in slide 4 follows using −→n = |∇u|∇u

We can first compute u and then solve this evolution to compute the paths of geodesics

1 2 3 4 5 6 7 8 9 10 11 12

(6)

Equal Distance Contours

M I A Hamilton Jacobi Equations

Computing the evolution of the equal distance contours corresponds to solving

|∇u|2 = VN2 = (1 + q2)u2x + (1 + p2)u2y − 2pquxuy 1 + p2 + q2 .

with boundary condition given by u = 0 at the source ∂K, where p = ∂z∂x, q = ∂z∂y

In other words the Hamilton Jacobi equation with hamiltonian H given by H(ux, uy) = (1 + q2)u2x + (1 + p2)u2y − 2pquxuy − (1 + p2 + q2) = 0.

1 2 3 4 5 6 7 8 9 10 11 12

(7)

Equal Distance Contours

M I A Hamilton Jacobi Equations

Summing up:

We compute u as the solution of the Hamilton Jacobi equation of Slide 6

Afterwards, we compute the path of a geodesic from a given point using the evolution (2) of Slide 5

1 2 3 4 5 6 7 8 9 10 11 12

(8)

Finding Minimal Paths Between Sets

M I A Finding minimal paths

Let A ⊂ S and MA(x, y) := dS((x, y, z(x, y)), A)

Lemma: All minimal paths between K, D ⊂ S are given by the set G := {(x, y, z(x, y)) : MK(x, y) + MD(x, y) = gm}

where gm = min(x,y)(MK + MD)

Let αK, αD denote distance contour evolutions starting from ∂K, ∂D respectively.

Lemma: The tangential points of αK(u, t) and αD(˜u, t) for ˜t + t = gm generate the minimal paths from p1 to p2. i.e. lie on a constant parameter

u = u0(˜u = ˜u0) of the propagating curve αK(u, t)(αD(˜u,t))˜

1 2 3 4 5 6 7 8 9 10 11 12

(9)

Laplace-Beltrami Operator

M I A Riemannian Metric in Local Parametrisation

Let p ∈ S ⊂ RM and φ be a local parametrisation around p with φ(x) = p

Consider the basis of TpS, {∂i := ∂x∂φ

i(x) | 1 ≤ i ≤ N}

Consider the Riemmanian metric g on S induced by the Euclidean space of RM (The first fundamental form in the case of a surface: M = 3, N = 2).

Lemma: At TpS we have that

g(v, w) =

n

X

i,j=1

gi,jviwj

for all v = P

vii and w = wii. where

gi,j =

∂φ

∂xi, ∂φ

∂xj

= h∂i, ∂ji

1 2 3 4 5 6 7 8 9 10 11 12

(10)

Laplace-Beltrami Operator

M I A The Gradient Over a Surface

In RN, we have h∇f(x), vi = Df(x)(v), for all v ∈ RN

Lemma: For a function f : S → R :

i) Df(p)(v) := dtd f(γ(t))|t=0 for some γ : I → S, s.t. γ(0) = p, γ0(0) = v, is well defined (does not depend on γ)

ii) There exists a unique element ∇f ∈ TpS s.t.

g(∇f(p), v) = Df(p)(v)

for all v ∈ TpS. It is given by

∇f =

N

X

i=1

aii, with ai =

N

X

j=1

gi,j∂(f ◦ φ)

∂xj

with (gi,j)1≤i≤N the inverse of (gi,j)1≤i≤N and φ local parametrisation.

1 2 3 4 5 6 7 8 9 10 11 12

(11)

Laplace-Beltrami Operator

M I A The Divergence of a Vector Field

For a vector field in RN and a function f with compact support (zero outside a compact set)

Z

RN

f div(V )dx = − Z

RN

h∇f, V i dx (3)

If f is defined over S and φ : U → S are local coordinates Z

φ(U)

f dS = Z

U

fp

|g|dx

• Thus (3) transforms in the case of a surface into Z

S

f divφ(V )p

|detg|dx = − Z

S

g(∇φf, V )p

|detg|dx

for f with compact support, and we get

divφ(V ) = 1

p|detg|

N

X

i=1

∂xi(Vip

|detg|)

1 2 3 4 5 6 7 8 9 10 11 12

(12)

Laplace-Beltrami Operator

M I A Laplace-Beltrami Operator

The Laplace Beltrami operator of a function f : S → R in local coordinates φ : U → S, U ⊂ RN, is given by

φf := divφ(∇φf) = 1

p|detg|

N

X

i,j=1

∂xi

p|det g| gi,j ∂f

∂xj

1 2 3 4 5 6 7 8 9 10 11 12

Referenzen

ÄHNLICHE DOKUMENTE

We look at the com- plexity of set-chain matching based on the discrete Fr´echet distance, and although the original definition allowed points in the set to be reused in the path,

Introduction Preparations The Problem as an Inequality System A Center Curve Algorithm on the Line.. A better Way to guess Visiting

In addition to the striking progress in reducing female death rates before age 50, Figure 4 reveals systematic if less dramatic trends in reducing female

Repetitorium Impfen sowie eine Zusammenstellung der am häufigsten gestellten Fragen zur Corona- Impfung und den Antworten darauf, die Ihre Arbeit erleichtern sollen. Bleiben

I want you to go straight to Garrier's Cosmetic Factory. Eunice wants me to go straight back. She wants Channel 9 to fight against animal cruelty. The neighbours were nice. 'Leave

‘particles’. An intransitive verb doesn't have a direct object. • The plane took off. A) Complete the sentences with a suitable phrasal verb from this list. They had a problem

The point of this seminar is to understand how this works in the case of a 2D TQFT, where a beautiful classification in terms of algebra is possible.. The seminar follows the

Eines darf dabei jedoch nicht vergessen wer- den: Jeder Mensch kann letztendlich selber ent- scheiden, ob solche Unternehmen und ihre Stra- tegien unterstützenswert sind, indem