Targeted Iterative Filtering Supplementary material SSVM 2013

Targeted Iterative Filtering

Supplementary material

SSVM 2013

Freddie ˚Astr¨om1,2_{, Michael Felsberg}1,2_,

George Baravdish3, and Claes Lundstr¨om2,4

1_{Computer Vision Laboratory, Link¨oping University, Sweden} 2_{Center for Medical Image Science and Visualization, Link¨oping University,}

Sweden

3_{Department of Science and Technology, Link¨oping University, Sweden} 4_{Sectra AB, Sweden}

{freddie.astrom,michael.felsberg, george.baravdish,claes.lundstrom}@liu.se

Contents

1 Necessary conditions for local minimum 84

2 Sufficient conditions for local minimum 86

3 Selection of mapping function 90

1

Necessary conditions for local minimum

Define the functional

E(u) = Z Ω (u − u0)2dx + λ Z Ω |∇m(u)|2dx (1) where m(u) ∈ C3(Ω) and u ∈ C2. The variational derivative of the regularization term of E(u) is computed using the Gˆateaux derivative

h∂R, vi = lim ε→0 Z Ω |∇m(u + εv)|2_{− |∇m(u)|}2 ε dx (2)

where v ∈ C1(Ω), v 6= 0 is an arbitrary test function.

Using the chain rule ∇m(u) = m0_{(u)∇u and rewriting the expression (2)}

yield h∂R, vi = lim ε→0 Z Ω |∇m(u + εv)|2_{− |∇m(u)|}2 ε dx (3) = lim ε→0 Z Ω |m0_{(u + εv)(∇u + ε∇v)|}2_{− |m}0_(u)∇u|2 ε dx (4) = lim ε→0 Z Ω

m0(u + εv)2(|∇u|2+ ε2∇ut∇v + ε2_|∇v|2_{) − m}0_(u)2_|∇u|2

ε dx (5) = lim ε→0 Z Ω |∇u|2_(m0_{(u + εv)}2_{− m}0_(u)2₎ ε +m 0_{(u + εv)}2_(2ε∇ut_{∇v + ε}2_|∇v|2₎ ε dx (6)

here use a first order Taylor expansion

m(u + εv) ≈ m(u) + εvm0(u) (7) m0(u + εv) ≈ m0(u) + εvm00(u) (8) then h∂R, vi = lim ε→0 Z Ω

|∇u|2_((m0_{(u) + εvm}00_(u))2_{− m}0_(u)2₎

ε +m 0_{(u + εv)}2_(2ε∇ut_{∇v + ε}2_|∇v|2₎ ε dx (9) = lim ε→0 Z Ω

|∇u|2_(m0_(u)2_{+ 2εvm}0_(u)m00_{(u) + ε}2_v2_m00_(u)2_{− m}0_(u)2₎

ε +m 0_{(u + εv)}2_(2ε∇ut_{∇v + ε}2_|∇v|2₎ ε dx (10) = Z Ω

by Green’s formula Z Ω (2m0(u)2∇ut)∇v dx = Z ∂Ω v(2m0(u)2∇u · n) dS − Z Ω

v div 2m0(u)2∇u
dx (12) and dS = bdy Ω, (bdy = boundary) we obtain

h∂R, vi = Z

Ω

2m0(u)m00(u)|∇u|2− 2 div m0_(u)2<sub>∇u

v dx (13)</sub>

Here observe that

div m0(u)2∇u
= 2m0(u)m00(u)|∇u|2+ m0(u)2∆u (14) which is equivalent to

2m0(u)m00(u)|∇u|2= div m0(u)2∇u
− m0_(u)2_{∆u (15)}

Inserting this into (13) gives that

h∂R, vi = Z

Ω

div m0(u)2∇u
− m0(u)2∆u − div 2m0(u)2∇u

v dx (16) =

Z

Ω

− div m0(u)2∇u
− m0(u)2∆u
v dx (17) Since v 6= 0, we obtain

− div m0(u)2∇u
− m0_(u)2_{∆u = 0 (18)}

where ∆ is the Laplacian operator. The u which minimizes the Euler-Lagrange equation is then obtained by solving the PDE



u − u0− λ(div m0_(u)2<sub>∇u
+ m</sub>0_(u)2_{∆u) = 0 in Ω}

m0(u)2∇u · n = 0 on ∂Ω (19) Since m0(u)2≥ 0 it is guaranteed that a solution of (19) exists.

2

Sufficient conditions for local minimum

Theorem 1. Let u0 _{be an observed image in a domain Ω ⊂ R}2, and denote by E(u) the functional

E(u) = Z Ω (u − u0)2dx + λ Z Ω |∇m(u)|2_dx

where u ∈ C2 _{and m(u) ∈ C}3_{. Let ε > 0 be arbitrary and consider the set}

Bε= n h, ∇h ∈ L2(Ω) : ||h||2_L2_(Ω)≤ ε 2 /CM, ||∇h||2L2_(Ω)≥ ε o where CM = max x∈Ω  [m0(u∗(x))m000(u∗(x)) − 3(m00(u∗(x)))2]|∇u∗(x)|2 .

Then u∗ is a local minimum of E(u) given by the solution of the E-L equation (19) if there exists ξ ∈ Ω such that

(m0(u∗(ξ)))2||∇h||2

L2(Ω)> ε

2 ₍₂₀₎

for every h ∈ Bε.

Proof. In order to find the sufficient condition for a minimum, define the regularization term in the functional as

J (u) = Z

Ω

|∇m(u)|2_{dx (21)}

Given a function ϕ ∈ C3

, a third order Taylor expansion at the point a ∈ R is given by ϕ(a) − ϕ(0) = aϕ0(0) +a 2 2 ϕ 00_{(0) +}a3 6 ϕ (3)_{(aθ) 0 < θ < 1. (22)}

Two types of minimums exist, global and local minimum:

• A local minimum is attained at 0 if it is a stationary point, that is ϕ0_{(0) = 0}

and ϕ00(x) ≥ 0 for every x in a neighborhood of 0.

• A global minimum is attained at 0 if it is a stationary point, that is ϕ0(0) = 0 and ϕ00(x) ≥ 0 for all x. Uniqueness of the point is given if the inequality is strict.

Let h ∈ C∞, then define ϕ(a) = J (u + ah), which determines the first variation δJ of J (u) as δJ = lim a→0 J (u + ah) − J (u) a = lima→0 ϕ(a) − ϕ(0) a = ϕ 0₍₀₎

In the same way the second variation δ2J follows. Since δJ is a linear functional in h and δ2J is a quadratic form in h define

L1(h) = δJ = ϕ0(0)

L2(h, h) = δ2J = ϕ00(0)

Given that ϕ is differentiable then so is J . If a = 1 in (22) then the Taylor expansion is given by

J (u(x) + h(x)) − J (u) = L1(h) + L2(h, h) + ||h||2ρ(h), (23)

where ρ(h) → 0, as h → 0 and we observe that

• L1(h) = 0 ⇔ ϕ0(0) = 0, that is u∗ is a stationary point to J iff. 0 is a

stationary point of ϕ.

• If L2(h, h) > 0 for every h in an neighborhood of u∗, then it follows from

(23) that u∗is a local minimum of J . Equivalent to this is that 0 is a local minimum to ϕ if ϕ00(x) > 0 is a neighborhood of 0.

• If L2(h, h) > 0 for every h, then it follows from (23) that u∗ is a global

minimum of J . Equivalent to this is that 0 is a global minimum to ϕ if ϕ00(x) > 0 for every x.

Applying the chain rule of differentiation, the functional J (u) is now ex-pressed as J (u) = Z Ω |∇m(u)|2_{dx =} Z Ω (m0(u))2|∇u|2_dx

then the derivatives of ϕ reads

ϕ(a) = J (u + ah) = Z

Ω

(m0(u + ah))2[|∇u|2+ 2a∇u · ∇h + a2|∇h|2_{] dx}

and

ϕ0(a) = 2 Z

Ω

m0(u + ah)m00(u + ah)h[|∇u|2+ 2a∇u · ∇h + a2|∇h|2_{] dx}

+2 Z

Ω

(m0(u + ah))2[∇u · ∇h + a|∇h|2] dx

and

ϕ00(a) = 2 Z

Ω

[(m00(u + ah))2+ m0(u + ah)m000(u + ah)]h2[|∇u|2

+ 2a∇u · ∇h + a2|∇h|2_{] dx}

+ 4 Z

Ω

m0(u + ah)m00(u + ah)h[∇u · ∇h + a|∇h|2] dx

+ 2 Z

Ω

2m0(u + ah)m00(u + ah)h[∇u · ∇h + a|∇h|2]

With a stationary point at a = 0 then • ϕ(0) = J (u) =R Ω(m 0_(u))2_|∇u|2_dx • ϕ0_{(0) = L} 1(h) = 2 R Ω[m

0_(u)m00_(u)|∇u|2_{h + (m}0_(u))2_{∇u · ∇h] dx}

Thus a necessary condition of u∗ _{to be a minimum point of the functional J (u)}

is

ϕ0(0) = L1(h) = 2

Z

Ω

[m0(u∗)m00(u∗)|∇u∗|2_{h + (m}0_(u∗₎₎2_∇u∗_{· ∇h] dx = 0 (24)}

for every h in a neighborhood of u∗. This gives the E-L equation

m0(u∗)m00(u∗)|∇u∗|2h + (m0(u∗))2∇u∗· ∇h = 0 (25) for every h in a neighborhood of u∗.

Also given that ϕ00(0) = L2(h, h), then

1

2L2(h, h) = Z

Ω

[(m00(u))2+ m0(u)m000(u)]|∇u|2h2dx (26)

+ Z Ω 4m0(u)m00(u)[∇u · ∇h]h dx (27) + Z Ω (m0(u))2|∇h|2_{dx (28)}

According to the E-L equation the solution u∗ must satisfy

m0(u∗) 6= 0 (29)

otherwise the trivial solution J (u∗) = 0 is obtained. Then from the E-L equation it is given that

∇u∗· ∇h = − 1 m0_(u∗₎m

00_(u∗_)|∇u∗_|2_h

hence, the quadratic form in the stationary point u∗ reads

L2(h, h) = 2 Z Ω [m0(u∗)m000(u∗) − 3(m00(u∗))2]|∇u∗|2h2dx (30) +2 Z Ω (m0(u∗))2|∇h|2dx (31) Assume m ∈ C3 and u ∈ C1, then there is an upper bound CM > 0 such that

|[m0(u∗)m000(u∗) − 3(m00(u∗))2]|∇u∗|2| ≤ CM,

Let ε > 0 and Bε be a set defined by

Bε=

n

h, ∇h ∈ L2(Ω) : ||h||2_L2_(Ω)≤ ε2/CM, ||∇h||2L2_(Ω)≥ ε

Given that h ∈ Bε, then the first integral of L2(h, h) reads Z Ω [m0(u∗)m000(u∗) − 3(m00(u∗))2]|∇u∗|2_h2_{dx ≥ −C} M Z Ω h2dx ≥ −ε2. Since h ∈ Bε we have Z Ω (m0(u∗))2|∇h|2_{dx 6= 0 .}

By the mean value theorem of calculus there exists a ξ ∈ Ω such that m0(u∗(ξ)) 6= 0 and Z Ω (m0(u∗))2|∇h|2_{dx = m}0_(u∗_(ξ))2_||∇h||2 L2_(Ω) Hence L2(h, h) ≥ 2 Z Ω (m0(u∗))2|∇h|2_{dx − 2ε}2_{≥ 2(m}0_(u∗_(ξ)))2_||∇h||2 L2_(Ω)− 2ε2 > 2ε [(m0(u∗(ξ)))2− ε ] > 0 (32) since h ∈ Bε and we can always chose ε < (m0(u∗(ξ)))2 which is the sufficient

3

Selection of mapping function

In this work we are interested in a mapping function on the form

m(u) = (1 + e(u))−1

such that 0 ≤ u ≤ 1 and

e(u) = e(b−u)/a and

a = (u2− u1)/4 , 0 < a < 1/4

b = (u2+ u1)/2 , 0 < b < 1

where u1 and u2are two application dependent values and 0 < u1< u2< 1.

Differentiation up to third order of m gives

m(u) = (1 + e(u))−1 m0(u) = −(1 + e(u))−2e0(u)

m00(u) = 2(1 + e(u))−3(e0(u))2− (1 + e(u))−2e00(u) = (1 + e(u))−2 2(1 + e(u))−1(e0(u))2− e00(u)
m(3)(u) = −6(1 + e(u))−4(e0(u))3+ 2(1 + e(u))−32e0(u)e00(u)

+ 2(1 + e(u))−3e0(u)e00(u) − (1 + e(u))−2e(3)(u)

= (1 + e(u))−2 

− 6(1 + e(u))−2(e0(u))3

+ 6(1 + e(u))−1e0(u)e00(u) − e(3)(u) 

where

e(u) = exp((b − u)/a)

e0(u) = −1 ae(u) e00(u) = 1 a2e(u) e(3)(u) = −1 a3e(u)

Express the derivatives of m(u) in e(u), now dropping the parenthesis

m(u) = (1 + e)−1 m0(u) = 1 a(1 + e) −2_e m00(u) = 1 a2(1 + e) −2 _{2(1 + e)}−1_e2 − e
m(3)(u) = 1 a3(1 + e) −2 _{6(1 + e)}−2_e3 − 6(1 + e)−1e2+ e

Observe that m0(u) > 0 for all u, thus the condition (29) is satisfied.

For this choice of mapping function we show that the sufficient condition in (20) is satisfied, so that a local minimum exists. Then the lower bound of (m0(u∗))2is given by (m0(u∗))2= 1 a2 e2(b−u∗ )a (1 + eb−u∗a )4 ≥ 1 a2 e2(b−1)a (eba+ eab)4 ≥ 1 a2_16e2(b+1)_a (33)

thus the condition (32) reads

1

16a2_e2(b+1)_a > ε (34)

4

Noise transformation

Due to the nonlinear mapping function, m, it is of interest to investigate the transformation of the first and second statistical moments of the input signal.

We assume that the image signal can be described by a linear model

u0= u0+ η , (35)

where

η ∼ N (µ, σ2) (36) such that u0 _{is the observed signal, u}

0 is the noise signal and η is normally

distributed with mean µ and variance σ2_{. The Taylor expansion of m(u) reads}

m(u0) = ∞ X n=0 m(n)(u0+ η) (η − µ)n n! (37)

A second order approximation of the mean value is then

E[m(u0)] = E[m(u0+ η)] = E[m(u0+ µ + (η − µ))] ≈ E[m(u0+ µ) + m0(u0+ µ)(η − µ) + 1 2m 00_(u 0+ µ)(η − µ)2] = m(u0+ µ) + 1 2m 00_(u 0+ µ)σ2 (38)

The transformation of the variance is then

Var(m(u0)) = E[m(u0)2] − E[m(u0)]2

= E[m(u0+ η)2] − E[m(u0+ η)]2 (39)

Consider E[m(u0₎2_{] first}

E[m(u0)2] = E[m(u0+ η)2] = E[m(u0+ µ + (η − µ))2] ≈ E[(m(u0+ µ) + m0(u0+ µ)(η − µ))2] = E[m(u0+ µ)2+ 2m(u0+ µ)m0(u0+ µ)(η − µ) + m0(u0+ µ)2(η − µ)2)] = m(u0+ µ)2+ m0(u0+ µ)2σ2 (40)

Inserting (40) and (38) into (39) result in the variance

Var(m(u0)) = E[m(u0)2] − E[m(u0)]2 = m(u0+ µ)2+ m0(u0+ µ)2σ2 − {m(u0+ µ) + 1 2m 00_(u 0+ µ)σ2}2 = m(u0+ µ)2+ m0(u0+ µ)2σ2 − {m(u0+ µ)2+ m(u0+ µ)m00(u0+ µ)σ2+ 1 4m 00_(u 0+ µ)2σ4} = m0(u0+ µ)2σ2− m(u0+ µ)m00(u0+ µ)σ2− 1 4m 00_(u 0+ µ)2σ4 = m0(u0+ µ)2− m(u0+ µ)m00(u0+ µ)
σ2− 1 4m 00_(u 0+ µ)2σ4

An approximation to the transformed mean value and variance is

ˆ µm= m(u0+ µ) + 1 2m 00_(u 0+ µ)σ2 (41) ˆ σ_m2 = m0(u0+ µ)2− m(u0+ µ)m00(u0+ µ)
σ2 (42)