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IMAGE RECONSTRUCTION OF ANISOTRO by fjzhangxiaoquan

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									       IMAGE RECONSTRUCTION OF ANISOTROPIC
              ELASTIC MODULI IN MRE

                     Jiah SONG 1 , Oh In KWON 2 and Jin Keun SEO 1

1) Department of Computational Science and Engineering, Yonsei University, Seoul, KOREA
2) Department of Mathematics, Konkuk University, Seoul, KOREA

                 Corresponding Author : : Jiah SONG, jasong@yonsei.ac.kr


                                         ABSTRACT
Magnetic Resonance Elastography(MRE) is a recent medical imaging modality to provide a
cross-sectional image of mechanical elastic properties of tissue using special phase encoding
sequence in MR system. Last two decades, MRE techniques have been made marked progress
in terms of reconstruction methods and measurement techniques, and now become successfully
commercialized as a routinely used clinical device for evaluation of liver disease. In MRE, we
apply a harmonically oscillating mechanical vibration using active and passive driver placed on
a surface of the object, and measure the displacement of the induced transverse acoustic strain
waves inside the body using MRI. The corresponding inverse problem is to recover shear mod-
ulus using measured displacement and elasticity equation. Most of research outcomes are based
on the assumption of isotropy and local homogeneity. However, several tissues such as mus-
cle show strong anisotropy. Hence, challenging issue in MRE is to identify Young’s modulus
and Poisson ratio as well as shear modulus that are different depending on directions. In this
work, we proposed a new MRE technique for recovering anisotropic elastic moduli under the
assumption of transversely isotropy. The proposed method decomposes the measured displace-
ment wave vector fields into curl-free and divergence-free components via Helmholtz-Hodge
decomposition and extract the three unknown elastic coefficients from a careful use of two
main components. Various numerical simulations show that the proposed method successfully
reconstruct transversely isotropic elastic moduli.


                MODEL PROBLEM AND NUMERICAL SIMULATION

Elastography is based on numerous finding that tissue stiffness is closely related to the velocity
of wave, and the shear modulus (or modulus of rigidity) varies over a wide range differentiating
various pathological states of tissues. Hence, the speed of the harmonic elastic wave provides
quantitative information for describing malignant tissues that are typically known to be much
stiffer than normal tissues.
     In elastography, we apply a time-harmonic excitation at a frequency ω in the range of 50
- 200 Hz through the surface of the object to be imaged. This produce harmonic mechanical
displacements inside an object. We can measure the resulting displacements using either ultra-
sound or MR system. The corresponding inverse problem is to identify elastic moduli from the
measured displacement vector, denoted by u, and the time-harmonic elasticity equation
                                          (                           )
                                            1
                                   ∇ · C : ( (∇u + ut )) + ρω 2 u = 0                                          (1)
                                            2
where ρ is density, ω the angular frequency, and C is the elastic 4th order tensor. In this work,
we assume that the tissue is transversely isotropic and incompressible so that C can be reduced
by 6 by 6 matrix form
                                                                
                C C12 C13 0
                11
                                               0        0        
                                                                
                        C11 C13 0             0        0        
                                                                
                                                                                 C11 = 4µ12 = 2C13 ,
                                                                
                                 C33 0        0        0        
                                                                
         C=                                                                                                  (2)
                                                                
                                       C44 0           0        
                                                                
                                                                              C33 = 4µ12 E3 , C55 = 2µ13
                       sym                   C44       0                                 E1
                                                                
                                                                
                                                     C11 −C12
                                                         2

By projecting the image domain to x-z plane, the equation (1) can be simplified into
                                                                                                 
                                         C11
                    C11 ∂x u1 +            ∂z u3 C2 ∂z u1
                                                    55
                                                                  +   C55
                                                                          ∂x u3           u
                                                                                         2 1
            ∇ ·                         2                            2
                                                                                   + ρω     = 0.           (3)
                        C55              C55
                         2
                            ∂z u1   +     2
                                             ∂x u3 C2 ∂x u1
                                                    11
                                                                  + C33 ∂z u3                    u3

Applying Helmholtz-Hodge decomposition, we obtain
                                                                                                       
                                                                                            ⊥
                                                                                   −∇φ1 + ∇ ψ1 
                                  C11
               C11 ∂x u1 +        2
                                      ∂z u3 C2 ∂z u1
                                             55
                                                            +   C55
                                                                 2
                                                                    ∂x u3   
                                                                           =                           .   (4)
                  C55
                   2
                      ∂z u1   +   C55
                                   2
                                      ∂x u3 C2 ∂x u1
                                             11
                                                            + C33 ∂z u3               −∇φ3 + ∇⊥ ψ3

Taking divergence operator to (4) leads to

                                        −∇2 φj = −ρω 2 uj                 (j = 1, 3)

which enable us to determine φ1 and φ3 . Neglecting ∇⊥ ψj (j = 1, 3) in the right side of (4),
                                 ∗     ∗     ∗
we can get principal components C11 , C33 , C55 which satisfy
                                                     ∗
                                   ∗               C11
                                  C11 ∂x u1 +       2
                                                       ∂z u3    = −∂x φ1
                                   ∗
                                  C55               ∗
                                                   C55
                                   2
                                      ∂z u1   +     2
                                                       ∂x u3   = (−∂z φ1 − ∂x φ3 )/2
                                   ∗
                                  C11            ∗
                                   2
                                      ∂x u1   + C33 ∂z u3 = −∂z φ3 .
                                                                      ∗    ∗    ∗
We can estimate ϕj (j = 1, 3) by substituting principal components C11 , C33 , C55 and taking
                                                              n+1   n+1   n+1
curl operator on the both side of (4). Finally we can update C11 , C33 , C55 by solving
                                          n+1
                                        C11
                    n+1
                   C11 ∂x u1 +             2
                                              ∂z u3 = −∂x φ1 − ∂y ψ1
                    n+1                  n+1
                   C55                  C55
                     2
                        ∂z u1     +       2
                                             ∂x u3 = (−∂z φ1 + ∂x ψ1             − ∂x φ3 − ∂y ψ3 )/2
                    n+1
                   C11
                     2
                        ∂x u1     + C33 ∂z u3 = −∂z φ3 + ∂x ψ3
                                     n+1
               Figure 1. Displacement data generated from two different places




                                               5     5         5
      Figure 2. The results of reconstruction C11 , C33 , and C55 with an iteration method




    Figure 3. The results of reconstruction C11 , C33 , and C55 without an iteration method
Fig 1 illustrates the measured displacement data obtained from two different positions. Fig 2 de-
scribes the reconstruction images of 5th iterations using the proposed algorithm. Additionally,
by using the symmetry, we can reconstruct them without iteration in the following way.

                               −∂z φ1 + ∂x ψ1 = −∂x φ3 − ∂z ψ3

The symmetry leads us to ∂x φ3 − ∂z φ1 = −∇ · Ψ where Ψ = [ψ1 ψ3 ]. We take gradient then
use the operator identity, ∇2 = ∇∇ · −∇ × ∇×, then

                                  −∇2 Ψ = ∇(∂x φ3 − ∂z φ1 )

assuming ∇×∇×Ψ ≈ 0. In this way, we can approximately determine curl-free and divergence-
free components so that we can reconstruct three unknown components without an iteration
method. Fig 3 depicts the results of C11 , C33 , and C55 without an iteration method.

                                       REFERENCES

1. Muthupillai R; Lomas, DJ; Rossman, PJ; Greenleaf, JF; Manduca, A; Ehman, RL,
   ”Magnetic resonance elastography by direct visualization of propagating acoustic strain
   waves”, Science 269, (5232): 1854-1857, 1995.
2. A.Manduca,         T.E.Oliphant,      M.A.Dresner,       J.L.Mahowald,        S.A.Kruse,
   E.Amromin, J.P.Felmlee, J.F.Greenleaf and R.L.Ehman,”Magnetic resonance elastography:
   non-invasive mapping of tissue elasticity”, Medical Image Analysis ,Vol 5:237-254, 2001.
3. T.H. Lee, C.Y. Ahn, O.I. Kwon and J.K. Seo , “A hybrid one-step inversion method for
   shear modulus imaging using time-harmonic vibrations”, Inverse Problems , Vol. 26, 2010.
4. Sebatian Papazoglou, Jurgen Braun, Uwe Hamhaber and Ingolf Sack , “Two-dimensional
   waveform analysis in MR elastography of skeletal muscles”, Physics In Medicine And
   Biology , Vol. 50, 2005.

								
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