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Elliptic Mesh Generation

An introduction to mesh generation

Part IV : elliptic meshing

Jean-Franois Remacle

Department of Civil Engineering,

Universit catholique de Louvain, Belgium

Jean-Franois Remacle Mesh Generation
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Curvilinear Meshes

Basic conceptA curvilinear mesh is build using a parametric space

Sponge analogy

Let us consider the real coordonates (x,y)and a parametric space (u,v).

The transformationx =x(u,v)and y =y(u,v)has to be build.

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Curvilinear Meshes

The transformation transforms a cartesian mesh in the parametric space into

a structured mesh in the real space with the following properties:

Two non intersecting lines in the parametric space do not cross in thereal space,

The mesh adapts to the boundary of on the real space,

The mesh has not to be orthogonal in the real space.

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Curvilinear Meshes






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Curvilinear Meshes







Elli ti M h G ti
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Curvilinear Meshes


a structured mesh in the real space with the following properties:Two non intersecting lines in the parametric space do not cross in the

real space,



The advantage of curvilinear meshes are numerous. They are usually high

quality/very regular meshes. Yet, the sponge analogy cannot be applied

easily to complex domains.


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Transfinite Interpolation

Let us consider a simply connected 2D manifold that is bounded by 4

curves ci(t), i=1. . . ,4.

We consider the four corners Pi,j that are the intersections if curves ciand cj.

We look for a transformation that maps the unit square in the

parametric space into the domain of interest.

We suppose that curves c1 and c3 will be the mapping of the vertical

segmentsv =0 andv =1.

We suppose that curvesc2 and

c4 will be the mapping of the horizontal

segmentsu =0 andu=1.


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Let us try the following intuitive formula

S(u,v) = (1 v)c1(u) +vc3(u) + (1 u)c4(v) +uc2(v)

Let us see if we get back the boundaries of the domain

S(u, 0) = c1(u) + (1 u)c4(0) +uc2(0)

S(u, 1) = c3(u) + (1 u)c4(1) +uc2(1)

S(0,v) = (1 v)c1(0) +vc3(0) +c4(v)

S(1,v) = (1 v)c1(1) +vc3(1) +c2(v)

Let us identify the four corners

c1(0) = c4(0) = P1,4 c1(1) = c2(0) = P1,2

c2(1) = c3(1) = P2,3 c3(0) = c4(1) = P3,4


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Taking into account the corners

S(u, 0) = c1(u) + (1 u)P1,4+uP1,2 = c1(u)

S(u, 1) = c3(u) + (1 u)P3,4+uP2,3 = c3(u)

S(0,v) = (1 v)P1,4+vP3,4+c4(v) = c4(v)

S(1,v) = (1 v)P1,2+vP2,3+c2(v) = c2(v)

Let us try an alternative formula

S(u,v) = (1 v)c1(u) +vc3(u) + (1 u)c2(v) +uc4(v) (1 u)(1 v)P1,4+uvP2,3+u(1 v)P1,2+ (1 u)vP3,4

.


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S(u,v) = (1 v)c1(u) +vc3(u) + (1 u)c2(v) +uc4(v) (1 u)(1 v)P1,4+uvP2,3+u(1 v)P1,2+ (1 u)vP3,4

.

Taking into account corners

S(u, 0) = c1(u) + (1 u)P1,4+uP1,2 (1 u)P1,4uP1,2= c1(u)

S(u, 1) = c3(u) + (1 u)P3,4+uP2,3 (1 u)P3,4uP2,3= c3(u)

S(0,v) = (1 v)

P1,4+v

P3,4+

c4(v) (1 v)P1,4v

P3,4=

c4(v)

S(1,v) = (1 v)P1,2+vP2,3+c2(v) (1 v)P1,2vP2,3= c2(v)

This formula is correct, it is called the transfinite interpolation formula.


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p


Using a transfinite interpolation does not guarantee that the mapping ispositive, i.e. the relation det |iSj| > 0 is not guaranteed, see some examples

withgmsh.


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Transfinite interpolation exists for 3-sides shapes,

S(u,v) = uc2(v) + (1 v)c1(u) +vc3(v) u(1 v)P1,2+uvP2,3

.

Transfinite interpolation exists for 3D shapes.


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Elliptic meshes

It is not always simple to devise an explicit formula for the mesh

mapping,

With the transfinite interpolation, mesh lines may cross each others,

leading to a wrong mesh

PDE based mesh have been developped in order to avoid those

difficulties

Two kind of approaches are possible, using either an hyperbolic PDE,

or using anellipticPDE. This last approach is now studied.


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Elliptic meshes : Thermal Analogy

Let us consider the same 4-sided domain as the one we used for

transfinite meshing,

We build mesh lines using a thermal analogy : mesh lines are

superimposed with the isolines of temperature.

Steady state heat equation has to be solved.

Two sides of the domain are adiabatical and a constant temperature isimposed on the two other sides.

!

" "

!




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We have to build up a PDE system that will lead to two families of

(orthogonal) mesh lines. Fo that purpose, two thermal problems will be

defined:1 The first family is constructed using the isolines of temperatures1 of

a thermal problem in which two opposite sides are adiabatical, i.e.

n1=0. The temperature is set to be constant on each of the two

other sides: 1=1 on the first side and1=0 on the other side. The

temperature field1 has to satisfy the Laplace equation inside thedomain :

21=0

2 The second family is constructed using the isolines of temperatures 2of the adjoint thermal problem in which adiabatic sides are replaced by

constant temperature sides and where constant temperature sides arereplaced by adiabatic sides. The co-temperature field2 satisfies the

Laplace equation as well :

22=0




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The two families of mesh lines will have the right properties, thanks to the

properties of the solutions of the Laplace equation:

The maximum principle : any solution of the Laplace equation has its

maximum on the boundaries (no local maxima is allowed),

The solution is unique,The solution belongs toH1, it is smooth,

Iso-contours cannot intersect,

Iso-contours stay inside the domain (obvious for a PDE, not for a

general mapping),




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The two families of mesh lines will have the right properties, thanks to the

properties of the solutions of the Laplace equation:

The maximum principle : any solution of the Laplace equation has its

maximum on the boundaries (no local maxima is allowed),

The solution is unique,

The solution belongs toH1, it is smooth,

Iso-contours cannot intersect,

Iso-contours stay inside the domain (obvious for a PDE, not for a

general mapping),




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Elliptic meshes : Formulation

The resulting mesh equations are two boundary value problems. We

have to find1 and2 solution of

2i=0 , i=1,2.

boundary conditions are

Boundary 1 2c1 1(x,y) =0 n2(x,y) =0

c2 n1(x,y) =0 2(x,y) =0

c3 1(x,y) =1 n2(x,y) =0c4 n1(x,y) =0 2(x,y) =1




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Problem : the equations have to be solved in the real space.

Disappointingly, solving such PDE requires... a mesh!




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This issue is addresded by the inversion of variablesx and y and 1 and

2,

The problem is solved in the space of temperatures 1 and2 and the

unknowns are the positions of the mesh in the real space x and y,

For a pair of temperatures(1,2), we look the point(x,y)in the

physical space where this temperature occurs.

Because of the properties of the Laplace operator, the exist such a

point(x,y)for any pair(1,2)for which 0 1 1 and for which

0 2 1

This is equivalent to the following change of variables

x=x1=x1(1,2) and y=x2=x2(1,2)




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2,







0 2 1


x=x1=x1(1,2) and y=x2=x2(1,2)



Elli i h F l i
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2,







0 2 1


x=x1=x1(1,2) and y=x2=x2(1,2)

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Elli ti h F l ti


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2,







0 2 1


x=x1=x1(1,2) and y=x2=x2(1,2)



Elli ti h F l ti


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We use the chain derivative rule to compute

xj=

2

j=1

j

j

xiand

2

xixk=

2

j=1

2

l=1

2

jl

l

xk

j

xi

In expanded form

2

x21= 2

21

1

x1

2

+ 2 2

12

1

x1

2

x1

+ 2

22

2

x1

2

.

2

x22= 2

21

1

x2

2+ 2

2

12

1

x2

2

x2

+ 2

22

2

x2

2.



Elli ti h F l ti
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The Laplace operator is written as

2 = 2

21

1

x1

2

+

1

x2

2

+ 2 2

12

1

x1

2

x1+1

x2

2

x2

+

+ 2

22

2

x1

2+

2

x2

2.

Therefore, in 2D, the PDEs that have to be solved can be written as

2x12

12x221

+ 2

2x1

122x2

12

+

2x12

22x222

=0.

with

=

1

x1

2+

1

x2

2, =

1

x1

2

x1+1

x2

2

x2, =

2

x1

2+

2

x2

2



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Note that

=J2

x11

2

+x12

2

=1

x1

2

+1

x2

2

,

= J2x11

x21

+x12

x22

=1

x1

2

x1+1

x2

2

x2,

=J2

x21

2+

x22

2=

2

x1

2+

2

x2

2,

with

J=det[jxi]



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The equations that have to be solved for the mesh problem are coupled and

non-linear! They can only be solved using numerical methods. There exists

several techniques to solve such systems. We are not going to detail all the

methods that are available (this is could be the subject of another course).

We are going to illustrate some basic principles. Two items will be detailed

hereThe numerical discretization of the mesh equations in the parametric

space(1,2),Their resolution.



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We assume that the mesh in the parametric space is a mnfinite

difference mesh. The mesh is uniform in the sense that the spacings and

in both directions are constant.

f

fi+1,j fi1,j2

f

fi,j+1 fi,j12

2f

2

fi+1,j 2fi,j +fi1,j2

2f

2

fi,j+1 2fi,j +fi,j12

2f

fi+1,j+1 fi+1,j1 fi1,j+1+fi1,j1

4





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We recall the mesh equations

2x1212x22

1

+ 2 2x112

2x212

+ 2x122

2x22

2

=0.Their discrete version can be written, for any vertex (i,j)of the mesh in the

parametric space, as

[xi+1,j 2xi,j+xi1,j ]+ [xi,j+1 2xi,j+ xi,j1]

2 [xi+1,j+1 xi1,j+1 xi+1,j1+xi1,j1] = 0

[yi+1,j 2yi,j+ yi

1,j] +

[yi,j+1 2yi,j+yi,j

1]2 [yi+1,j+1 yi1,j+1 yi+1,j1+yi1,j1] = 0



Elliptic meshes : Resolution
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With the numerical evaluation of coefficients , and

= (xi,j+1 xi,j1)

2 + (yi,j+1 yi,j1)2

(2)2

= (xi+1,j xi1,j)

2 + (yi+1,j yi1,j)2

(2)2

= (xi+1,j xi1,j)(xi,j+1 xi,j1)

(4)2

+(yi+1,j yi1,j)(yi,j+1 yi,j1)

(4)2



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Applying the formulation to the mnnodes of the finite difference mesh,

we obtain a coupled system of non linear equations. In fact, , and

are all functions of the unknowns, i.e. the position of the nodes in the realspace. Several methods are available for solving such a system. They can be

classified in two categories:

1 direct methods,

2 iterative methods.

The choice of the method is essentially based on the following criterion

The CPU time cost and the memory cost,

The ability of the method to converge to the solution,

The relative ease of programming !



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Direct methods are not the choice that is usually made for building elliptic

meshes. Direct methods have the advantage of ensuring the convergence in

a finite number of step (yet this not true for non-linear systems). Theirprincipal drawback is their memory signature and their (relative)

complexity.

Iterative methods have the following advantages:

1 They are usually easier to code,

2 They are well adapted to structured grids,

The following smoothers are often used for building elliptic meshes

Point Jacobi iteration,

Block Jacobi iteration (the block may be a line or a column of nodes),

ADIs, i.e. alternated direction implicit !



Elliptic meshes : Point Jacobi Iteration
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Elliptic meshes : Point Jacobi Iteration

We look for the values of the variables x and y, notedx andy at node

(i,j)verifying

xi+1,j 2x

i,j+ x+

i1,j

+

xi,j+1 2x

i,j+ x+

i,j1

2xi+1,j+1 xi1,j+1 x

+i+1,j1+x

+i1,j1

= 0

yi+1,j 2y

i,j+y+

i1,j

+

yi,j+1 2y

i,j+ y+

i,j1

2yi+1,j+1 yi1,j+1 y

+

i+1,j1+y+

i1,j1

= 0

Here,x andy are for the values ofx and y at the previous sweep and x

andy are the values at the present sweep. The linearization of the

coefficients consist in using previous values ofx and y in , and .



Elliptic meshes : Over relaxation
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Elliptic meshes : Over relaxation

We know thatx and y are updated by doing

x=x+ (xx) and y=y+ (yy).

Over relaxing the system consist in doing

x+ =x+(xx) and y+ =y+(yy).

with 1. Current values are computed as

xi,j=xi,j+Cxi,j/ and y

i,j=yi,j+Cyi,j/.

with the corrections

Cxi,j=x+

i,jxi,j and Cyi,j=y+

i,jyi,j.

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Elliptic meshes : Boundary conditions


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p y

We do not want to move the boundary nodes. The way to implement it is

very simple : do not apply any corrections to those nodes

Cx1, j = Cxm, j = 0

Cy1, j = Cym, j = 0

Those are Dirichlet boundary conditions.

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Elliptic meshes : Line Jacobi Iteration


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p

Iterative process can be accelerated using a more implicit method. We

cound compute one line at a time

valeurs courantes

valeurs ltape n + 1

valeurs ltape n

j+1

i-1 i i+1

j

j-1

The resulting system is tridiagonal and ad hoc linear solvers can be used for

accelerating the process.



Elliptic meshes : Line Jacobi Iteration
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p

The system of equations can be assembled solved using the following finite

difference formula.

xi+1,j 2x

i,j+ x

i1,j

+

xi,j+1 2x

i,j+ x+i,j1

2xi+1,j+1 xi1,j+1 x

+i+1,j1+x

+i1,j1

= 0

yi+1,j 2y

i,j+ y

i1,j

+

yi,j+1 2y

i,j+ y+i,j1

2yi+1,j+1 yi1,j+1 y

+

i+1

,j1

+y+i1

,j1

= 0

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