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Stability and Experimental Comparison of Prototypical Iterative Schemes for Total Variation Regularized Problems

Sören Bartels and Marijo Milicevic

Abstract

Various iterative methods are available for the approximate solution of non-smooth minimization problems. For a popular non-smooth minimization problem arising in image processing, we discuss the suitable application of three prototypical methods and their stability. The methods are compared experimentally with a focus on choice of stopping criteria, influence of rough initial data, step sizes as well as mesh sizes. An overview of existing algorithms is given.

MSC 2010: 65K15; 49M29

Funding source: Deutsche Forschungsgemeinschaft

Award Identifier / Grant number: SPP 1748: BA 2268/2-1

Funding statement: The authors acknowledge financial support by the Deutsche Forschungsgemeinschaft for the project “Finite Element Approximation of Functions of Bounded Variation and Application to Model of Damage, Fracture and Plasticity” (BA 2268/2-1) via the priority program “Reliable Simulation Techniques in Solid Mechanics, Development of Non-Standard Discretization Methods, Mechanical and Mathematical Analysis” (SPP 1748), and by the Sino-German Science Center (grant id 1228) on the occasion of the Chinese-German Workshop on Computational and Applied Mathematics in Augsburg 2015.

A Roots of Quartic Equations

Having (3.2) in mind we consider the problem of finding the roots of a quartic equation

(A.1)x4+ax3+bx2+cx+d=0

with a,b,c,d. A comprehensive discussion of this topic can be found, e.g., in [14]. With the substitution x=y-14a we equivalently obtain

(A.2)y4+b~y2+c~y+d~=0

with b~=-38a2+b, c~=18a3-ab2+c and d~=-3256a4+a2b16-ac4+d. Adding 14b~2 on both sides yields

(y2+12b~)2=14b~2-c~y-d~.

Our goal is to produce a perfect square on the right-hand side. The following procedure is due to Lodovico Ferrari. We introduce a new variable z, which is to be specified later, within the square on the left-hand side and obtain

(y2+12b~+12z)2=14b~2-c~y-d~+14z2+z(y2+12b~)
(A.3)=zy2-c~y+14z2+12b~z+14b~2-d~.

Now the right-hand side is a perfect square with respect to y if and only if the associated discriminant vanishes, i.e., if and only if

(A.4)0=4z(14z2+12b~z+14b~2-d~)-c~2
(A.5)=z3+2b~z2+(b~2-4d~)z-c~2.

The cubic polynomial in (A.5) is called cubic resolvent. We see that with z being any root of the cubic resolvent the term in (A.3) simplifies to a perfect square. Indeed, using (A.4), we have

(y2+12b~+12z)2=z(y2-c~zy+c~24z2)=z(y-c~2z)2.

We can take z to be, for instance, a real root of (A.5) and obtain

y2+12b~+12z=±z(y-c~2z),

so we have two quadratic algebraic equations in y and can compute the roots of (A.2). We obtain the roots {xi}1i4 of our original equation with the relation xi=yi-14a.

B Roots of Cubic Equations

In order to obtain the roots of (A.5) we briefly discuss how to compute the roots of a cubic algebraic equation. Given an equation of the form

(B.1)x3+ax2+bx+c=0

we set x=y-a3 and get

(B.2)y3+b~y+c~=0

with b~=b-a23 and c~=c-ab3+2a327. With a further substitution y=u+v, with – for the time being – arbitrary u and v, we get

y3=(u+v)3=(u+v)(u2+2uv+v2)=u3+v3+3uv(u+v)=u3+v3+3uvy
y3-3uvy-(u3+v3)=0.

Comparing coefficients yields

b~=-3uv,c~=-(u3+v3).

According to Vieta’s formulas, u3 and v3 are the roots of the quadratic algebraic equation

t2+c~t-b~327=0.

Hence we have

u=-c~2+c~24+b~3273,v=-c~2-c~24+b~3273.

The cube roots u and v have to be chosen such that uv=-b~3. Denoting by u1,v1 the cube roots defined by r3eiφ/3 for a complex number reiφ, the two other cube roots are given by u2=u1e2πi/3, u3=u1e4πi/3 and v2=v1e2πi/3, v3=v1e4πi/3, respectively. Since the product uv has to be real, we get the three feasible pairs (u1,v1), (u2,v3) and (u3,v2). Hence, the three roots of (B.2) are given by

y1=u1+v1,y2=u2+v3,y3=u3+v2.

The roots {xi}1i3 of (B.1) are then given by xi=yi-a3.

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Received: 2015-12-23
Revised: 2016-3-18
Accepted: 2016-3-20
Published Online: 2016-4-13
Published in Print: 2016-7-1

© 2016 by De Gruyter

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