Fem Garlekine Method

8/8/2019 Fem Garlekine Method

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NBCR Summer Institute 2006:

Multi-Scale Cardiac Modeling withContinuity 6.3

Wednesday:Finite Element Discretization and

Anatomic Mesh Fitting

Andrew McCulloch and Fred Lionetti


2/67


3/67

The Finite Element Method

Solution is discretized using a finite number of functions

Piecewise polynomials (elements) Continuity across element boundaries ensured by

defining element parameters at nodes with associatedbasis functions,

12 13

14 15

21 22

23 24

FE equations are derived from the weak form of thegoverning equations

R = 0

Finite differences:

Finite elements:

R = 0


4/67

The Finite Element

Method Integrate governing equations in each element Assemble global system of equations by adding

contributions from each element

1 2

5 6

7 8

3 4

Element equationsk k k k k k k k

k k k k k k k k k k k k k k k k

k k k k k k k k

k k k k k k k k

k k k k k k k k

k k k k k k k k

k k k k k k k k

u

uu

u

u

u

u

u

11 12 13 14 15 16 17 18

21 22 23 24 25 26 27 28

31 32 33 34 35 36 37 38

41 42 43 44 45 46 47 48

51 52 53 54 55 56 57 58

61 62 63 64 65 66 67 68

71 72 73 74 75 76 77 78

81 82 83 84 85 86 87 88

1

2

3

4

5

6

7

8

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP

=

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP

f

ff

f

f

f

f

f

1

2

3

4

5

6

7

8

12 13

14 15

21 22

23 24

Global equations

L

N

MMMM

MMMMMMMMMMMMMMMMMMMMMM

MMMMMMMMMMMMMMMMMMM

O

Q

PPPP

PPPPPPPPPPPPPPPPPPPPPP

PPPPPPPPPPPPPPPPPPP

L

N

MMMM

MMMMMMMM

MMMMMMMMMMMMMM

MMMMMMMMMMMMMMMMMMM

O

Q

PPPP


PPPPPPPPPPPPPPPPPPP

L

N

MMMM

MMMMMMMMMMMMMMMMMMMMMM

MMMMMMMMMMMMMMMMMMM

O

Q

PPPP


PPPPPPPPPPPPPPP

=

PPPP


5/67

Consider the strong form of a linear partial differential

equation, e.g. 3-D Poissons equation with zero boundaryconditions:

0

),,(2

2

2

2

2

2

=

=

u

zyxfz

u

y

u

x

uOn region R

on boundary S

Strong Form Lu= f

Variational Principle, e.g. minimum potential energy

=Rv

vfLvu Vd)2(min

Weighted Residual (weak) form, e.g. virtual work

0Vd)( = R

wfLu

Integral Formulations


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0

),(2

2

2

2

=

=

u

yxf

y

u

x

uOn region S

on boundary C

Weak form

=

SSyxwfyxw

y

u

x

udddd

2

2

2

2

Integrate by parts

d d d d d dS C C S

u w u w u u x y w x w y f w x y x x y y y x

+ =

Where, u and w vanish at the boundary

0 0

Weak Form for 2-D Poissons Equation


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Choose a finite set of approximating (trial) functions,

i(x,y), i = 1, 2, , N

Allow approximations to uin the form

U(x,y) = U11 + U22 + U33 + + UNN(that can also satisfy the essential boundary conditions)

Solve N discrete equations for U1, U2, U3, , UN

( ) ij

jij

si

S

iNN

iNN

FUK

yxf

yxyy

Uy

Uxx

Ux

U

=

=

++

+

++

dd

dd...... 111

1

Galerkins Method for 2-D Poissons Equation


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yxfF

Kyxyyxx

K

Sii

jiS

jijiij

dd

dd

=

=

+

=

[K]U = F

[K] is the stiffness matrix and F is the load(RHS) vector

[K] is symmetric and positive definite

Galerkins Method for 2-D Poissons

Equation


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Galerkin is more general than Rayleigh-Ritz. If we add u/x, symmetry& the variational principle are lost, but Galerkin still works

Ifwis chosen as Dirac delta functions at N points, weighted residualsreduces to the collocation method

Ifwis chosen as the residual functions Lu-f, weighted residuals reduces

to the least squares method

By choosing wto be the approximating functions, Galerkins method

requires the error (residual) in the solution to be orthogonalto theapproximating space.

The integration by parts (Green-Gauss theorem) automatically

introduces the Neumann (natural) boundary conditions

The Dirichlet (essential) boundary conditions must be satisifed explicitly

when solving [K]U=F Since discretized integrals are sums, contributions from many elements

are assembledinto the global stiffness matrixby addition.

The Ritz-Galerkin FEM finds the approximate solution that minimizes the

error in the energy

Comments on Galerkins Method


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1. Formulate the weighted residual (weak form)2. Integrate by parts (or Green-Gauss Theorem)

reduces derivative order of differential operator

naturally introduces derivative (Neumann) boundary

conditions, e.g. flux or traction. Hence called that

naturalboundary condition

3. Discretize the problem

discretize domain into subdomains (elements)

discretize dependent variables using finite

expansions of piecewise polynomial interpolating

functions (basis functions) weighted byparameters

defined at nodes

Steps in the Finite Element Method


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4. Derive Galerkin finite element equations substitute dependent variable approximation in

weighted residual integral

Choose weight functions to be interpolating

functions the Galerkin assumption (Galerkin,

1906)

5. Compute element stiffness matrices and RHS

integrate Galerkin equations over each element

subdomain

integrate right-hand side to obtain element load

vectors which also include any prescribed Neumann

boundary conditions

Steps in the Finite Element Method (contd)


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6. Assemble global stiffness matrix and load vector Addelement matrices and RHS vectors into global

system of equations

Structure of global matrix depends on node ordering

7. Apply essential (i.e. Dirichlet) boundary conditions at least one is required (essential) for a solution

prescribed values of dependent variables at specified

boundary nodes, e.g. prescribed displacements

eliminate corresponding rows and columns from

global stiffness matrix and transfer column effects of

prescribed values to Right Hand Side

the constraint reducedsystem



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8. Solve global equations

for unknown nodal dependent variables

using algorithms for Ax = b or Ax = x

9. Evaluate element solutions interpolate dependent variables

evaluate derivatives, e.g. fluxes

derived quantities, e.g. stresses or strain energy

graphical visualization;post-processing

10.Test for convergence

refine finite element mesh and repeat solution



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=

==

2

2d 2d

(1) 0

(4) 9

ux

u

u

solution

u x x( ) ( )= 1 2

1 2 3 4U1=0

2

4

6

8

x

u

U4=9

U3 =?

U2=?

Galerkin FEM: Simple 1-D Example


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0xd)( = R wfLu

2. Integrate by parts (or Green-Gauss Theorem)

0xd24

12

2

=

ww

dx

ud

0xd2

4

1

4

1 =+ wdxdu

wdx

dw

dx

du

1. Formulate the weighted residual (weak) form


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4 globalnodal parameters U1, U2, U3, U4

3 linear elements each with 2 elementnodal

parameters u1, u2.

Adjacent elements share global nodal parameters,e.g., global parameter U2 is element parameter u2 of

element 1 and u1 of element 2.

Two (linear) element interpolation functions for each

element, i(x), i = 1, 2

Allow element approximations to uin the form

u(x) = u1 1 + u2 2 = ui i i=1,2

3. Discretize the problem


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0 0.5 1

0

0.5

1

x

2 1

element basis functions

Element Basis Functions


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[ ] 0xd2

xd2xd2

4

1

4

3

3

2

2

1

=+

+

+

wdx

duwdx

dw

dx

du

wdx

dw

dx

duwdx

dw

dx

du

In each element, let

u(x) u1 1 + u2 2 = ui i (x)

and

w(x) i (x)

4. Derive Galerkin equations for each element


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( )i

jjij

2

1

2

1

2

2

1

1

fuk

d2dd

d

d

d

d

d

==

xxxxuxui

i

e.g. forElement 1 (no derivative boundary conditions):

[k] = [(kij)] is the element stiffness matrix

f = (fi) is the element load vector

4. Derive Galerkin equations for each element ( contd)


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x

x

xxd2f

kdk

ii

jiij

ji

=

=

=

[k]u = f

Element stiffness matrix, [k] and load(RHS) vector, f

12

:1Element

2

1

==

xx

=

=

=

11

11

d11

11

)1ele(

2

1

2

1)1ele(

[k]

[k]xx

xxx

5. Compute element stiffness matrices


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xd2f ii

=

1

2

:1Element

2

1

=

=

x

x

=

=

=

=

1

1

)21()44(

)14()48(

2

4d

22

24

)1ele(

2

22

1

2

1)1ele(

f

fxx

xxx

x

x

In this problem, each element is the same size and thus:

[k](ele1) = [k](ele2) = [k](ele3)

and:

f(ele1) = f(ele2) = f(ele3)

5. Compute element RHS matrices


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=

=

11001210

0121

0011

111111

1111

11

[K]

=

+

+

=

1

2

2

1

1

11

11

1

F

6.Assemble global stiffness matrix and load vector


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=

=

1

2

2

1

1100

1210

0121

0011

4

3

2

1

U

U

U

U

F[K]U

u U

u U

( )

( )

1 0

4 9

1

4

= =

= =That leaves global equations 2 and 3

+ =

+ =

2 3

2 3

0 2 2

2 9 2

U U

U U

7. Apply essential(i.e. Dirichlet) boundary conditions


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+ = = + = =

2 3 2

2 3 3

0 2 2 12 9 2 4U U U

U U UExact!

8. Solve global equations (constraint-reduced)


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Representing a One-Dimensional Field

Polynomials are convenient, differentiated and integrated readily For low degree polynomials this is satisfactory If the polynomial order is increased further to improve the accuracy,

it oscillates unacceptably Divide domain into subdomains and use low orderpiecewise

polynomials over each subdomain called elements

2 3Use a polynomial expression ( ) ...

and estimate the monomial coefficients a, b, c and d

u x a bx cx dx = + + + +


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Making Piecewise Polynomials Continuous

constrain the parameters to ensure continuity ofu

across the element boundaries

or better, replace the parameters a and b in the firstelement with parameters u1 and u2, which are the

values ofuat the two ends of that element:

where is a normalized measure of

distance along the curve

u u u( ) ( ) = +1 1 2 ( )0 1


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u = u(x)

+

+

++

+ ++

+ + ++

+

+

x

u

u = a + bx u = c + dx u = e + fx

++

+++ +

++ + +

+

+

+

x

u

u1

u2

u3

u4

0

1

u=(1- ) u1+ u2

element 1 element 2 element 3

nodes


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u

u1

u2

u=(1- )u1+u2

0 1

0 1

1

0

1

1

1 = 1-

2 =

Linear Lagrange Interpolation


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Global-Element Mapping

Associate the nodal quantity un with element node n Map the value U defined at global node onto local node n ofelement e by using a connectivity matrix (n, e),

u Un n e= ( , )

Thus, in the first element

u u u( ) ( ) ( ) = +1 1 2 2

withu1=U1 and u2=U2..

In the second element uis

interpolated by

u u u( ) ( ) ( ) = +1 1 2 2

Withu1=U2andu2=U3.


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We haveu ( ) but to defineu (x) we needx ( ).

Definexas an interpolation of nodal values, e.g.

u u

x x

n n

n

n n

n

( ) ( )

( ) ( )

=

=

Isoparametric Interpolation

u

x

u1

u2

x2

x1

1

1

u1

u2

u

x2x1x


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Quadratic Lagrange Basis Functions

Use three nodal parametersu1, u2 and u3

are the quadratic Lagrange basis functions.

( ) ( ) ( ) ( )

( ) ( )

( )( )

1 1 2 2 3 3

1

2

3

where

2 1 0.5

4 1

2 0.5

u u u u

= + +

=

= =

0 0.5 1.0

1.0

0 0.5 1.0

1.0

0 0.5 1.0

1.0

1 2

3


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Cubic Hermite Basis Functions

1

1

1

1

1

1

0

0

0

0

( ) 3211 231 +=

( ) ( ) 23212 =

( ) ( ) 221 1=

( ) ( )1222 =


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Scaling Factors

=0 =1 =0 =0s1 s2 s3

( ) ( )

( ) ( ) ( )

n e n,e

n en e n,e i

i i i(no sum on i)

=

=

U U

U U s

s

Global to local mapping:

Scaling Factors arc lengths

arc length


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Two-Dimensional

Tensor-Product Elements

1 1 2 1 1 1 2 1 2

2 1 2 2 1 1 2 1 2

3 1 2 1 1 2 2 2 2

4 1 2 2 1 2 2 1 2

( , ) ( ) ( ) (1 )(1 )

( , ) ( ) ( ) (1 )

( , ) ( ) ( ) (1 )

( , ) ( ) ( )

= =

= =

= =

= =

u( 1 , ) ( , ) ( , ) ( , ) ( , )2 1 1 2 1 2 1 2 2 3 1 2 3 4 1 2 4= + + +u u u uBilinear interpolation can be constructed

where

Bili T P d B i F i


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1

0 1 1

2

1

1

1

2

u

y

x

1

x=nxn

u=nun

y=nyn0

Bilinear Tensor-Product Basis Functions


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A Six-Noded

Quadratic-Linear Element

1

3

5

( , ) ( ) ( )

( , ) ( )

( , ) ( )

1 2 1 1 2

1 2 1 1 2

1 2 1 1 2

2 11

21

21

2

1

21

4 1

= FHIK

= FHIK FH

IK

=

2

4

6

( , ) ( )( )

( , )

( , )

1 2 1 1 2

1 2 1 1 2

1 2 1 1 2

4 1 1

2 11

2

2 12

=

= FHIK

= FH IK

b g

1

21.0

1.000

0.5


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Three-dimensional Linear Basis Functions

e.g. trilinear element has eight nodes with basis functions:

( ) ( ) == 21 ;1

1

2

3

4

5

6

7

8

1

2

3

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ).,,

;,,

;,,

;,,

;,,

;,,

;,,

;,,

3222123218

3222113217

3221123216

3321113215

3122123214

3122113213

3121123212

3121113211

=

=

==

=

=

=

=

( )=

=8

1

321 ,,i

ii uu


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1

2

3

5

6

7

1

2

3In each node we define:

221

3

32

2

31

2

321

2

21

,,

,,,,,

uuu

uuuu

u

Tri-Cubic Basis Functions


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( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

+

+

+

+

+

+

+++= =

321

3

321

8

32

2

321

7

31

2

321

6

3

321

5

21

2

321

4

8

1 2

3213

1

3212

3211

321

,,,,

,,,,,,

,,,,,,,,

ii

ii

ii

ii

ii

i

ii

iiii

uu

uuu

uuuu

( ) ( ) ( ) ( )

8,...2,1,;2,1,,,,,

;,, 321321

==

=

jirqnmlk

r

q

n

m

l

k

j

i

Tri-Cubic Basis Functions (Contd)


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Scaling Factors

=0 =1 =0 =0s1 s2 s3

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

n e n,e

n en e n,e i

i i i

n e n e n en,e2 i j

i j i j i j

n e n,e3

i j k i j k

(no sum on i)

(no sum on i,j)

=

=

=

=

U U

U U s

s

U U s ss s

U U

s s s

( ) ( ) ( )n e n e n ei j k

i j k(no sum on i,j,k)

s s s

Global to local mapping:

Scaling Factors arc lengths

arc length


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Coordinate Systems

Rectangular Cartesian global reference

coordinate system Orthogonal curvilinear coordinate

system to describe geometry and

deformation Curvilinear local finite element

coordinates

Locally orthonormal body coordinatesdefine material symmetry and

structure, related to the

finite element coordinates by a rotation

about the -normal axis through

the "fiber angle" ,

1 2 3{ }Y ,Y ,Y

1 2 3{ }, ,

1 2 3{ }, ,

1 2 3{ }X ,X ,X

1 2( , ) 1

From Costa et al, J Biomech Eng 1996;118:452-463

C iliA) Rectangular Cartesian Coordinates: { A}=(X Y Z)


42/67

Curvilinear

World

Coordinates

R

Y1

Y2

Y3

A) Rectangular Cartesian Coordinates: { A} (X,Y,Z)

C) Spherical Polar Coordinates: { A}=(R,,)

Y1 = R cos cos

Y2 = R cos sinY3 = R sin

B) Cylindrical Polar Coordinates: { A}=(R,,Z)

Y1 = R cosY2 = R sinY3 = Z

Y3=ZY2=Y

Y1=X

Y1

R

Y2

Y3=Z

+

Y2

Y1

Y3

d=focus

= d cosh cos= d sinh sin cos= d sinh sin sin

Y2

Y1

Y3

D) Prolate Spheroidal Coordinates ( ,, )


43/67

Fiber/Sheet Coordinates

State of

B d I di C di t

Covariant

B V t

Covariant

M t i T


44/67

Coordinate

System

Notations

* Represents a Lagrangian description of the deformation from B to B .

Body Indices Coordinates Base Vectors Metric Tensors

A) rectangular Cartesian reference coordinates

B R,S RY Re R,S

B r,s Ry re r,s

B) curvilinear world coordinates

B A,B A ( )

R

A RA

Y =

G e ( )

R R

AB A B

Y YG

=

B , ( )

r

ry

=

g e

( )r ry y

g

=

C) normalized finite element coordinates (Lagrangian)

B K,L K ( ) ( )K AK

=

G G ( ) ( )A B

KL ABK LG G

=

B *( ) ( )K K

=

g g *( ) ( )KL K L

g g

=

D) locally orthonormal body/fiber coordinates

B I,J IX ( ) ( )

K

xI KIX

= G G ( ) ( )

K Lx

IJ KL IJI J

G GX X

= =

B i,j ix ( ) ( )

Kx

i Kix

=

g g ( ) ( )

K Lx

ij KL iji jg g

x x

= =

B *( ) ( )

Kx

I KI

x

X

=

g g *( ) ( )

K Lx

IJ KLI J

x xg g

X X

=


45/67

Fitting with Linear Lagrange 1-D Elements

Two linear Lagrange elementsfit the data witha root-mean-squared-error (RMSE) of 0.614892.

Result of twice refining the mesh (yielding 8

elements) andre-fitting: RMSE = 0.0930764

Least Squares Fitting


46/67

( ) ( )

( )( ) ( )( )

( )( ) ( )( ) ( )( )

2

1,

2 2

1 2

1 2

2 2 22 2 2

3 4 52 2

1 2 1 2

d

d d d

d D

d d d d

d d d d d d

F

d

=

= +

+ + + +

X X X

X X X X

X X X X X X

The least squares fit minimizes the objective function:

dX

( )dX

i

d

where is measured coordinate or field variable;

aresmoothingweights

is the interpolated value at

Least Squares Fitting

d are weights applied to the data points

( ) ( ) ( )i id dN

d j j N

N

jN

j

0 a linear system of equations for nodal parameters

X X

FX

X

= =

=

X


47/67

Fitting a Coronary Vascular Tree with Quadratic Lagrange

1-D Elements


48/67

anesthetized & ventilated NewZealand White rabbit

heart arrested in diastole,

excised

pulmonary vessels removed,aorta cannulated

heart suspended in Ringers

lactate, perfused in unloaded

state with buffered formalin at

80 mm Hg for 4 minutes

heart cast in polyvinylsiloxane

plunger

tube

knife

heart cast in rubber

Rabbit Ventricular Anatomy


49/67

plunger

knife

Rabbit Ventricular Anatomy

BASE

APEX


50/67


51/67

2

1

data point

projects onto

surface at

( d, d, )

Bicubic Hermite

isoparametric interpolation

( 1, 2) = { ii1i=

4

( 1, 2) +

i

1i

2( 1, 2) +

i

2i

3( 1, 2) +

2 i

1 2i

4( 1, 2)}

1

x = d cosh cos y = d sinh sin cos

z = d sinh sin sin


52/67


53/67


54/67


55/67

endo >0

epi


56/67

8,351 geometric

points 14,368 fiber angles

36 elements 552 geometric DOF

RMSE = 0.55 mm 184 Fiber angle DOF

RMSE = 19

Anatomic Model

Vetter & McCulloch Prog Biophys & Mol Biol69(2/3):157 (1998)


57/67

Strain Analysis

X2, longitudinal

X1, circumferential

X 3, radial

Xc , crossfiber

X f , fiber

X r , radial

( )T122 2 1

2

=

d d

i i kij i j i j i j

j k j

ij i j

x xF

X X

s S E dX dX

= = =

=

F e e e e e e

E F F I


58/67

A/P View Lateral View

Reconstructed

3D Coordinates

Transform


59/67

Baseline 2 minutes ischemia

End-Systolic Circumferential Strain

0.04

0.00

-0.04

-0.07

( ) ( )x2

F = +X X X


60/67

( ) ( )

( )( ) ( )( )

( )( ) ( )( ) ( )( )

x

x x

x x xx

1,

2 2

1 2

2 2 22 2 2

2

2 2

1 2 1 2

d

d d d

d D

d d d d

d d d d d d

F

dx

=

= +

+ +

+ +

X X X

X X X X

X X X X X X

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

RMS

Fitting

Error(mm

)

10-7 10-6 10-5 10-4 10-3 10-2 10-1 100

Smoothing Weight

10-2 0 032


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0.018

0.02

0.022

0.024

0.026

0.03

0.028

10-6

10-5 10-4 10-3 10-2

10-6

10-5

10-4

10-3

10 0.032

Fiber Strain Cross-fiber StrainMyocardial Blood Flow


62/67

y

Contr o

l

LAD

Occlu

sion

-0.05 0.00 0.050.0 1.5 3.0mL/min/


63/67

3 th t


64/67

SEP

TAL

LATE

RAL

3months post-surgeryPre-surgery


65/67

Base Bead

Apex Bead3 Columns ofradiopaque beads


66/67

C

L

R

Undeformed

state

Deformed

state


67/67

Three-Dimensional Strain Analysis

Documents

Fem Garlekine Method