Multiple View Geometry: Exercise Sheet 10
Prof. Dr. Florian Bernard, Florian Hofherr, Tarun Yenamandra
Computer Vision Group, TU Munich
Link Zoom Room , Password: 307238
Exercise: July 17th, 2019
Part I: Theory
1. Variational Calculus and Euler-Lagrange
(a) Under the assumption that h vanishes at the boundary of Ω, prove that
dE(u)
du
=
∂L(u,∇u)
∂u
− div
(
∂L(u,∇u)
∂(∇u)
)
.
We can expand L(u+ �h,∇u+ �∇h) in terms of �:
L(u+ �h,∇u+ �∇h) = L(u,∇u) + �h
∂L
∂u
+ �∇h
∂L
∂(∇u)
+O(�2)
Inserting into δE(u)
δu
∣∣∣
h
gives
δE(u)
δu
∣∣∣∣
h
= lim
�→0
1
�
∫
Ω
(
�h(x)
∂L
∂u
∣∣∣∣
u(x)
+ �∇h(x)
∂L
∂(∇u)
∣∣∣∣
∇u(x)
+O(�2)
)
dx
The � in the first two terms cancels, and in the O(�2) it will go to zero for �→ 0. Integra-
tion by parts of the second term yields∫
Ω
∇h(x)
∂L
∂(∇u)
dx =
∫
∂Ω
h(x)
∂L
∂(∇u)
ds−
∫
Ω
h(x)div
(
∂L
∂(∇u)
)
dx =
= −
∫
Ω
h(x)div
(
∂L
∂(∇u)
)
dx .
Thus,
δE(u)
δu
∣∣∣∣
h
=
∫
Ω
h(x)
(
∂L
∂u
− div
(
∂L
∂(∇u)
))
dx ⇒ claim .
(b) Which condition must hold true for a minimizer u0 of E(u) …
– … in general?
dE(u)
du
= 0 ⇒
∂L(u,∇u)
∂u
= div
(
∂L(u,∇u)
∂(∇u)
)
.
– … if L(u,∇u) = L(u)?
∂L(u)
∂u
= 0 .
1
https://tum-conf.zoom.us/s/62772800235?pwd=SUpZN2QrV0JpeXJyR2R1TWx5cHEwdz09
– … if L(u,∇u) = L(∇u)?
div
(
∂L(∇u)
∂(∇u)
)
= 0 .
2. Multiview Reconstruction as Shape Optimization
(a) Write down the Euler-Lagrange equation for the given energy E(u).
The E-L equations are
dE(u)
du
= −div
(
∂L(∇u)
∂(∇u)
)
= 0 with L(∇u) = ρ|∇u|
Taking the derivative w.r.t. ∇u gives
0 = −div
(
ρ(x)
∇u(x)
|∇u(x)|
)
.
It is also possible (but not neccessarily required) to expand this further using the product
rule for divergence:
div
(
ρ(x)
∇u(x)
|∇u(x)|
)
= ∇ρ(x)>
∇u(x)
|∇u(x)|
+ ρ(x)div
(
∇u(x)
|∇u(x)|
)
(b) Write down one gradient descent iteration for E(u).
u(k+1)(x) = u(k)(x)−τ
dE(u)
du
= u(k)(x)+τ
(
∇ρ(x)>
∇u(x)
|∇u(x)|
+ ρ(x)div
(
∇u(x)
|∇u(x)|
))
2