stephen c. jardin princeton plasma physics...
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MHD Simulations for Fusion Applications
Lecture 3
The Galerkin Finite Element MethodStephen C. Jardin
Princeton Plasma Physics Laboratory
CEMRACS ‘10
Marseille, France
July 21, 2010
1
Outline of Remainder of Lectures
• Galerkin method• C1 finite elements• Examples• Split implicit time
differencing
Today Tomorrow• Split implicit time
differencing for MHD• Accuracy, spectral
pollution, and representation of the vector fields
• Projections of the split implicit equations
• Energy conserving subsets• 3D solver strategy• Examples
Weak Form of a Differential EquationConsider the ordinary differential equation on the domain [a,b]:
( ) ( ) ( ) ( ) (1)d dp x q x U x f xdx dx
⎡ ⎤⎛ ⎞− + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
We choose a function space H1, and multiply (1) by an arbitrary member of that function space, and integrate over the domain:
( ) ( ) ( ) ( ) ( ) ( ) (2 )b b
a a
d dV x p x q x U x dx V x f x adx dx
⎡ ⎤⎛ ⎞− + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫ ∫
( ) ( ) ( ) ( ) (2 )b b
a a
dV dUp x q x VU dx V x f x bdx dx
⎡ ⎤⎛ ⎞ + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫ ∫
Or, after integrating by parts (assuming the boundary term vanishes):
If we require that Eq. (2) hold for every function V(x) in the function space H1, it is called the weak form of Eq. (1). Note that only first derivatives occur in (2b), whereas 2nd derivatives occur in (1)
Galerkin Finite Element MethodThe Galerkin finite element method is a discretization of the weak form of the equation. To apply, we chose a finite dimensional subspace S of the infinite dimensional Hilbert space H1, and require that Eq. (2) hold for every function in S.
1( ) ( ) ( ) ( ) 1,
b bNji
j i j ij a a
dda p x q x dx x f x i Ndx dx
ϕϕ ϕ ϕ ϕ=
⎡ ⎤⎛ ⎞+ = =⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦∑ ∫ ∫
1( ) ( )
N
j jj
U x a xϕ=
= ∑
( ) ( )
( ) ( )
bji
ij i ja
b
i ia
ij j iM a
ddM p x q x dxdx dx
b x f x
b
dx
ϕϕ ϕ ϕ
ϕ
⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
=
=
∫
∫
known basis functions that span Saj are amplitudes to be solved for
Letting Equation (2b) becomes
Or,
In the finite element method, the basis functions are low order polynomials
( ) ( )iV x xϕ=
Simple Example: Linear Elements in 1D
1
( ) ( )N
j jj
U x a xϕ=
= ∑
( )i xϕ 1( )i xϕ +1( )i xϕ −
xixi-1 xi+10 L
1
0 0, (constants) ( )p p q q f f x= = =
, ati i j jx xϕ δ= =
linear: xi = ih
h=L/N
0 , 0
2 1 0 01 2 1 01
0 1 2 10 0 1 2
i j i jp dx K ph
ϕ ϕ
−⎡ ⎤⎢ ⎥− −⎢ ⎥∇ ∇ = =⎢ ⎥− −⎢ ⎥−⎣ ⎦
∫∫ i 0 , 0
4 1 0 01 4 1 00 1 4 160 0 1 4
i j i jhq v v dx A q
⎡ ⎤⎢ ⎥⎢ ⎥= =⎢ ⎥⎢ ⎥⎣ ⎦
∫∫“mass matrix”“stiffness matrix”
Leads to sparse, diagonally dominant matrices: Mi,j = Ki,j + Ai,j
i ib fdxϕ= ∫∫
ij j iM a b=
1( ) ( ) ( ) ( ) 1,
b bNji
j i j ij a a
dda p x q x dx x f x i Ndx dx
ϕϕ ϕ ϕ ϕ=
⎡ ⎤⎛ ⎞+ = =⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦∑ ∫ ∫
Linear elements only overlap with their nearest neighbors
Note: We were able to solve a 2nd order ODE with linear elements!
1( ) ( )
N
j jj
U x a xϕ=
= ∑
( )i xϕ 1( )i xϕ +1( )i xϕ −
xixi-1 xi+10 L
1
1( ) ( ) ( ) ( ) 1,
b bNji
j i j ij a a
dda p x q x dx x f x i Ndx dx
ϕϕ ϕ ϕ ϕ=
⎡ ⎤⎛ ⎞+ = =⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦∑ ∫ ∫
( ) ( ) ( ) ( )d dp x q x U x f xdx dx
⎡ ⎤⎛ ⎞− + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦( )
b
ia
dx xϕ∫
ij j iM a b=
Extension to Higher Order EquationsSuppose we have a 4th order equation:
4
4 ( ) ( )d U x f xdx
=
Form the weak form: 4
4
3
3
2 2
2 2
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
b b
i ia a
b bi
ia a
b bi
ia a
dx U x dx x f x dxdx
d d U dx x f x dxdx dx
d d U dx x f x dxdx dx
ϕ ϕ
ϕ ϕ
ϕ ϕ
=
− =
=
∫ ∫
∫ ∫
∫ ∫
(assumes the boundary terms vanish)
Application of the Galerkin method to a 4th order equation requires that the second derivative of the basis functions be defined everywhere.
For a 2nd order equation, only the first derivative need be defined.
1
( ) ( )N
j jj
U x a xϕ=
= ∑
Integrate by parts twice:
Elements with higher continuityLinear elements are continuous, but only the function and first derivative are well defined. To represent a second derivative, both the function and first derivative must be continuous. They must be a subset of H2
j-1 j j+1j-1 j j+1
1 ( )j xϕ 2 ( )j xϕThe Hermite Cubic elements associate two basis functions with each node j, defined on the interval xj-1 < x < xj+1 as:
( ) ( )( )
21
22
1, 1
2, 2
( ) 1 2 1
( ) 1
( )
( )
jj
jj
y y y
y y y
x xx
hx x
x hh
ϕ
ϕ
ϕ ϕ
ϕ ϕ
= − +
= −
−⎛ ⎞= ⎜ ⎟
⎝ ⎠−⎛ ⎞
= ⎜ ⎟⎝ ⎠
These have the property that:1. In the interval xj < x < xj+1 , a complete
cubic can be represented by:
2. The first derivatives vanish at the element boundaries
3. The second derivatives are defined everywhere
1, 2, 1, 1 2, 1( ), ( ), ( ), ( ),j j j jx x x xϕ ϕ ϕ ϕ+ +
j-1 j j+1j-1 j j+1
1 ( )j xϕ 2 ( )j xϕ
Hermite Cubic Finite Elements
1, 2,1
( ) ( ) ( )N
j j j jj
U x a x b xϕ ϕ=
⎡ ⎤= +⎣ ⎦∑
In the interval [ j , j+1 ] , the function is written as:
( )( )
( )( )
1, 2, 1 1, 1 1 2, 1
21 1
31 1
2 30 1 2 3
( ) ( ) ( ) ( ) ( )
3 2 3
2 2
j j j j j j j j
j j j j j j
j j j j
U x a x b x a x b x
a b x a hb a hb x h
a hb a hb x h
c c x c x c x
ϕ ϕ ϕ ϕ+ + + +
+ +
+ +
= + + +
= + + − − + −
+ + − +
= + + +
Simple physical interpretation:
aj is value of function at node jbj is value of derivative at node j
0
12 2
123 2 3 2
13
1 0 0 00 1 0 0
3 / 2 / 3 / 1/2 / 1/ 2 / 1/
j
j
j
j
acbc
ac h h h hbc h h h h
+
+
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ − − −⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦ ⎣ ⎦
i
Complete cubic polynomial determined by values of function and derivative at endpoints
j j+1
1 ( )j xϕ
2 ( )j xϕ
Hermite Cubic Finite Elements-2
1, 2,1
( ) ( ) ( )N
j j j jj
U x a x b xϕ ϕ=
⎡ ⎤= +⎣ ⎦∑
In the interval [ j , j+1 ] , the function is written as:
( )( )
( )( )
1, 2, 1 1, 1 1 2, 1
21 1
31 1
2 30 1 2 3
( ) ( ) ( ) ( ) ( )
3 2 3
2 2
j j j j j j j j
j j j j j j
j j j j
U x a x b x a x b x
a b x a hb a hb x h
a hb a hb x h
c c x c x c x
ϕ ϕ ϕ ϕ+ + + +
+ +
+ +
= + + +
= + + − − + −
+ + − +
= + + +
1 1( )j xϕ +
2 1( )j xϕ +Complete cubic in each interval!
Linear elements
1st Hermite cubic
2nd Hermite cubic
( )xϕ
( )xϕ′
( )xϕ′′
j-1 j j+1j-1 j j+1
j-1 j j+1
not defined
Comparison of elements and derivatives
Types of Finite Elements in 2D
Shape Order Continuity
, 0
( , ) i kik
i k Mi k
x y C x yφ+ <
≥
= ∑rectangle
triangle
Order M if complete polynomial to that order
C0: continuous across element boundaries
C1: continuous first derivatives across element boundaries
DG: discontinuous Galerkin
Element Shape
Rectangle:
• natural for structured mesh
• allows tensor product forms
bi-linear: t=1, bi-quadratic: t=2, etc
•cannot refine locally without hanging nodes
Triangle:
• natural for unstructured mesh
• easier to fit complex shapes
• can easily refine locally0 ,
( , ) i kik
i k t
x y a x yφ≤ ≤
= ∑
Element OrderIf an element with typical size h contains a complete polynomial of order M, then the error will be at most of order hM+1
This follows directly from a local Taylor series expansion:
Thus, linear elements will be O(h2)quadratic elements will be O(h3)cubic elements will be O(h4)quartic elements will be O(h5)complete quintic elements will be O(h6)
0 0
10 0
0 0 ,
1( , ) ( ) ( ) ( )!( )!
kM kl k l M
l k lk l x y
x y x x y y O hl k l x y
φφ − +−
= =
⎡ ⎤∂= − − +⎢ ⎥− ∂ ∂⎣ ⎦
∑∑
Element ContinuityTheorem: A finite element with continuity Ck-1 belongs to Hilbert space Hk, and hence can be used for differential operators with order up to 2k
Continuity Hilbert Space ApplicabilityC0 H1 second order equationsC1 H2 fourth order equations
This applicability is made possible by performing integration by parts in the Galerkin method, shifting derivatives from the unknown to the trial function
[ ]
( )2 2 2 2
recall:
( , ) ( , )
( , ) ( , )
i idomain domain
i idomain domain
v f x y dxdy f x y v dxdy
v f x y dxdy f x y v dxdy
φ φ
φ φ
∇ ∇ = − ∇ ∇
⎡ ⎤∇ ∇ = ∇ ∇⎣ ⎦
∫∫ ∫∫
∫∫ ∫∫
i i
Hk means that derivatives exist up to order k
The vast majority of the literature concerns C0
elements,.
Here we concentrate on C1 elements because we are interested in higher order equations
NOTE: requires the trial function have appropriate boundary conditions
Reduced Quintic 2D Triangular Finite Element: Q18
θP1(x1,y1)
P3(x3,y3)
P2(x2,y2)
origin
ξ
η
20
1
( , ) k km nk
k
aφ ξ η ξ η=
= ∑ k mk nk1 0 02 1 03 0 14 2 05 1 16 0 27 3 08 2 19 1 210 0 311 4 012 3 113 2 214 1 315 0 416 5 017 3 218 2 319 1 420 0 5For C1, require that the normal slope along the edges φn have only cubic variation:
5b4ca16 + (3b2c3 – 2b4c)a17 + (2bc4 – 3b3c2) a18 + (c5 – 4b2c3)a19 – 5bc4a20 = 0
5a4ca16 + (3a2c3 – 2a4c)a17 + (-2ac4 – 3a3c2) a18 + (c5 – 4a2c3)a19 – 5ac4a20 = 0
20 – 2 = 18 unknowns:
These are determined in terms of [ φ, φx, φy , φxx, φxy, φyy ] at P1,P2,P3
Implies C1 continuity at edges and C2 at nodes !
x
y
Specify the function and it’s first and second derivatives at the 3 nodes (φ, φx, φy, φxx, φxy, φyy)
Guarantees C0 continuity
Note: general quintic has 21 terms. 1 has been set to zero, and 2 others will be determined by constraints.
ξ,η are local orthogonal coordinates
No mk=4, nk=1 term
6 Unknowns at each node!
2 3 4 51
2 3 41
2 31
1
1
1
2
2
2
2
2
2
3
3
3
3
3
3
1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 2 0 0 3 0 0 0 4 0 0 0 0 5 0 0 0 00 0 1 0 0 0 0 0 0 0 0 0 0 0
00
b b b b bb b b b
b b bξ
η
ξξ
ξη
ηη
ξ
η
ξξ
ξη
ηη
ξ
η
ξξ
ξη
ηη
φφφφφφφφφφφφφφφφφφ
− − −⎡ ⎤⎢ ⎥ − −⎢ ⎥ − −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
2 3
2
2 3
2 3 4 5
2 3 4
2 3
2 3
2
0 0 00 0 0 2 0 0 6 0 0 0 12 0 0 0 0 20 0 0 0 00 0 0 0 1 0 0 2 0 0 0 3 0 0 0 0 0 0 0 00 0 0 0 0 2 0 0 2 0 0 0 2 0 0 0 2 0 0 01 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 2 0 0 3 0 0 0 4 0 0 0 0 5 0 0 0 00 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 2 0 0 6 0 0 0 12 0 0 0 0 20 0 0 0 00 0 0 0 1 0 0 2 0 0 0 3 0 0
b b bb b
b b ba a a a a
a a a aa a a
a a aa a
− −−
− −
2 3
2 3 4 5
2 3 4
2 3 4
2 3
2 3
2 3
4
0 0 0 0 0 00 0 0 0 0 2 0 0 2 0 0 0 2 0 0 0 2 0 0 01 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 2 0 0 0 3 0 0 0 0 4 0 0 0 0 50 0 0 2 0 0 0 2 0 0 0 0 2 0 0 0 0 2 0 00 0 0 0 1 0 0 0 2 0 0 0 0 3 0 0 0 0 4 00 0 0 0 0 2 0 0 0 6 0 0 0 0 12 0 0 0 0 20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5
a a ac c c c c
c c c cc c c c
c c cc c c
c c c
a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
172 3 4 5
4 184 3 2 2 3
192 3 4 54 4
204 3 2 2 3
3 2 52 3 43 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 52 3 4
aaaaaaaaaaaaaaaaa
a c ac c ac aca c a c a c ab c bc c ab c bcb c b c b c
⎡ ⎤⎡⎢ ⎥⎢⎢ ⎥
⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
− − + −⎢ ⎥⎢ ⎥⎢ ⎥− − − − ⎣⎢ ⎥⎣ ⎦
⎤⎥
⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎦
[ ] [ ][ ]T A′Φ =or,
20
1( , ) k km n
kk
aφ ξ η ξ η=
= ∑using the expansionCalculate the value of φ and it’s derivatives wrt local coords. ξ and η at the 3 vertex points
Combine with the 2 constraint equations:
[ ] [ ]1A T − ′⎡ ⎤= Φ⎣ ⎦
How to calculate the coefficients in terms of the unknowns:
How to calculate the coefficients in terms of the unknowns-2:The previous viewgraph calculated the coefficient matrix in terms of derivatives in the rotated coordinate system, which is different for each element. To get the coefficient matrix in terms of the vector containing the actual derivatives with respect the Cartesian coordinates (x,y), we have to apply a rotation matrix that depends on the triangle’s orientation.
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
1
1
1
RR R
R
2 2
2 2
2 2
1 0 0 0 0 00 cos sin 0 0 00 sin cos 0 0 00 0 0 cos 2sin cos sin0 0 0 sin cos cos sin sin cos0 0 0 sin 2sin cos cos
θ θθ θ
θ θ θ θθ θ θ θ θ θ
θ θ θ θ
⎡ ⎤⎢ ⎥⎢ ⎥
−⎢ ⎥= ⎢ ⎥
⎢ ⎥⎢ ⎥− −⎢ ⎥−⎣ ⎦
1R
2G = T RiRelates the coefficient matrix directly to the function and derivatives wrt (x,y) at the nodes
20x18 matrix consisting of the first 18 columns of T-1
Each triangle hasit's own 18 18
gij
× G
20
1
i im nj ij
i
v gξ η=
= ∑
The Trial Functions:20 20 18 18
1 1 1 1
i i i im n m ni ij j j j
i i j j
a g vφ ξ η ξ η= = = =
= = Φ = Φ∑ ∑∑ ∑
These are the trial functions. There are 18 for each triangle.
The 6 shown here correspond to one node, and vanish at the other nodes, along with their derivatives
Each of the six has value 1 for the function or one of it’s derivatives at the node, zero for the others.
Note that the function and it’s derivatives (through 2nd) play the role of the amplitudes: 6 at each node!
ai = gij Φj Function and derivatives wrt (x,y)
201 1
1
20 202 ( 2) ( ) ( ) ( 2)
1 1
( , ) k k k k
k l k l k l k l
m n m nk k k
k
m m n n m m n nk l kl kl k l k l
k l
a m n
a a K K m m n n
φ ξ η ξ η ξ ξ η η
φ ξ η ξ η
− −
=
+ − + + + −
= =
⎡ ⎤∇ = ∇ + ∇⎣ ⎦
∇ = = +
∑
∑∑
1 11
( ) ! !( , ) ( 2)!
( 2, ) ( , 2)
m mm n n
element
kl kl k l k l k l k l k l k lelement
a b m nd d cF m n m n
K K d d m m F m m n n n n F m m n n
ξ η ξ η
ξ η
+ ++
⎡ ⎤− −⎣ ⎦=≡ + +
≡ = + − + + + + −
∫∫
∫∫
Integrals involving powers of the local coordinates can be done analytically:
Thus, for example:
ai = gijΦj
All matrix elements easily calculated in terms of the ai (or gij)
20
1
i im nj ij
i
v gξ η=
= ∑
20 20 18 18
1 1 1 1
i i i im n m ni ij j j j
i i j ja g vφ ξ η ξ η
= = = =
= = Φ = Φ∑ ∑∑ ∑
Very compact representation compared to a popular C0 Element because all unknowns on nodes
6
66
Lagrange Cubic: C0, h4
Reduced Quintic: C1, h5
6
66
6
6 new unknowns: 2 new triangles
6/2 = 3 unknowns/ triangle
9 new unknowns: 2 new triangles
9/2 = 41/2 unknowns/ triangle
split
split
Comparison of reduced quintic to other popular triangular elements
Vertexnodes
Line nodes
Interior nodes
accuracyorder hp
UK/T continuity
linear element 3 0 0 2 ½ C0
Lagrange quadratic 3 3 0 3 2 C0
Lagrange cubic 3 6 1 4 4½ C0
Lagrange quartic 3 9 3 5 8 C0
reduced quintic 18 0 0 5 3 C1 *
N M
Linear Elements Reduced Quintic Elements21
LE aN
= 51
QE bM
=
15
2 / 5 2 / 5same ~error L QbE E M N Na
⎛ ⎞=> = => = ⎜ ⎟⎝ ⎠
. .. . .. . . .
. . . .. . .
. . . .. . . .
. . .. .
x xx x
x xx x x
x x xx x x
x xx x
x x
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
.. .
. .
. ..
x xx x
x x xx x
x xx x
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
N2
36N4/5
2N12N2/5
Win for N > 20 !
2 S∇ Φ =Second order equation:
N M
Linear Elements Reduced Quintic Elements21
LE aN
= 51
QE bM
=
15
2 / 5 2 / 5same ~error L QbE E M N Na
⎛ ⎞=> = => = ⎜ ⎟⎝ ⎠
. .. . .. . . .
. . . .. . .
. . . .. . . .
. . .. .
x xx x
x xx x x
x x xx x x
x xx x
x x
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
.. .
. .
. ..
x xx x
x x xx x
x xx x
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
4N2
36N4/5
4N12N2/5
Win for N > 6 !
4 S∇ Φ =
2∇ Φ = Ψ2 S∇ Ψ =
Fourth order equation:
Anisotropic Diffusion
2BB S
t Bφ κ φ κ φ∂ ⎡ ⎤= ∇ ∇ + ∇ ∇ +⎢ ⎥∂ ⎣ ⎦
i i i
18
1j j
j
vφ=
= Φ∑20
( ) ( )
1
i im nl lj ij
i
v gξ η=
= ∑
Apply Galerkin method:
Typically κ|| >> κ⊥
( ) ( ) ( ) ( )2
( ) ( ) ( )2
l l l lj j j j
l l lj j j
BBv dxdz v v v S d dt B
v BB v v S d dB
φ κ φ κ φ ξ η
κφ κ φ ξ η
⎡ ⎤⎡ ⎤∂= ∇ ∇ + ∇ ∇ +⎢ ⎥⎢ ⎥∂ ⎢ ⎥⎣ ⎦⎣ ⎦
⎧ ⎫= − ∇ ∇ − ∇ ∇ +⎨ ⎬
⎩ ⎭
∫∫ ∫∫
∫∫
i i i
i i i
Now, ( ) ( ) ( ),ˆl l lj j jv B v z vψ ψ⎡ ⎤∇ = ∇ ×∇ = ⎣ ⎦i i
[ ]18
,1
( , ) , ,j j
j k kk
v d d v d d
R
ξ η φ ξ η ψ φ ψ ξ η
=
⎡ ⎤∇ ∇ = ⎣ ⎦
= Φ
∫∫ ∫∫
∑
BBi i
( )( )18 18 20 20 20 20
, , , , ,1 1 1 1 1 1
18 18
, , ,1 1
( 2, 2)
j k p j q i r k s l p q q p r s s ri l p q r s
p q r s p q r s i l
j k i l i li l
R g g g g m n m n m n m n
F m m m m n n n n
P
= = = = = =
= =
≡ − − ×
+ + + − + + + − Ψ Ψ
= Ψ Ψ
∑∑∑∑∑∑
∑∑
Evaluation of the Matrix Elements
( )( )20 20 20 20
, , , , , , ,1 1 1 1
( 2, 2)
j k i l p j q i r k s l p q q p r s s rp q r s
p q r s p q r s
P g g g g m n m n m n m n
F m m m m n n n n= = = =
≡ − −
× + + + − + + + −
∑∑∑∑
[ ]
ˆ
, x y y x
z
a b a b a b
a b a bξ η η ξ
ψ= × ∇
= −
= −
B
Anisotropic Diffusion
N..number of points per side
10 20 30 40 60
RM
S e
rror i
n st
eady
-sta
te
1e-6
1e-5
1e-4
1e-3
1e-2
1e-1κ|| = 10**4 κ|| = 10**5
κ|| = 10**6 κ|| = 10**7 κ|| = 10**8 power law
N-4
N-5
2BB S
t Bφ κ φ κ φ∂ ⎡ ⎤= ∇ ∇ + ∇ ∇ +⎢ ⎥∂ ⎣ ⎦
i i i
Solve to steady state
cos cosx ySL L
π πψ= =
Shows greater than N-5 convergence
2D Incompressible MHD
[ ]
2 2 2 4
2
, ,
,
t
t
φ φ φ ψ ψ μ φ
ψ ψ φ η ψ
∂ ⎡ ⎤ ⎡ ⎤∇ + ∇ − ∇ = ∇⎣ ⎦ ⎣ ⎦∂∂
+ = ∇∂
2 2 2 2 2
4 4
2 2
1 1
, ,
,
,
n n n
n n
n n n n
t t t t
t
t t t
t t
φ φ θδ φ φ θδ φ ψ θδ ψ ψ θδ ψ
μ φ θδ φ
ψ ψ θδ ψ φ θδ φ η ψ θδ ψ
φ φ ψ ψφ ψδ δ
+ +
⎡ ⎤ ⎡ ⎤∇ + ∇ + ∇ + − ∇ + ∇ +⎣ ⎦⎣ ⎦⎡ ⎤= ∇ + ∇⎣ ⎦
⎡ ⎤ ⎡ ⎤+ + + = ∇ + ∇⎣ ⎦⎣ ⎦
− −= =
18
1
n nj j
jvφ
=
= Φ∑18
1
n nj j
j
vψ=
= Ψ∑
20
1
i im nj ij
i
v gξ η=
= ∑Multiply equations by each trial function and integrate over space
“reduced MHD”
φ is stream function
ψ is poloidal flux
θ-centering….time centered about n+1/2 for θ=0.5
note:
Leads to the Matrix Implicit System
11 12 * *, , , , , ,
21 22 * *, , , , , ,
*, , , , ,
, , ,11 12,
21 22
[ ][ ]
[(
(1 ) ]
j j i j i j k k i j i j k k
j j i j k k i j i k j k i j
ni j i j k k i j k k n
i j k k ii jj j
j j
S S A t G B tGS S tK M t K A
A t G Gt G G
BD DD D
θδ μ θδθδ θδ η
δ θδ θ
θ μ
⎡ ⎤ ⎡ ⎤+ Φ + − Ψ=⎢ ⎥ ⎢ ⎥Ψ + Φ −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
⎧ ⎫− Φ − Φ⎪ ⎪ Ψ −⎨ ⎬+ −⎡ ⎤ ⎪ ⎪⎩ ⎭=⎢ ⎥
⎢ ⎥⎣ ⎦
*,
*1, , , 2*1
, , 2,
)
[ ( )( )
(1 ) ]
j k k
ni j i k j k kn
i j k k ki j
M t KtK
A
δ θδ θ
θ η
⎡ ⎤Ψ⎢ ⎥
⎢ ⎥⎢ ⎥
⎧ ⎫− Φ − Φ⎢ ⎥⎪ ⎪− Ψ + Ψ ⎨ ⎬⎢ ⎥− −⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦
11 12 1 11 12
21 22 1 21 22
n nj j j j j j
n nj j j j j j
S S D DS S D D
+
+
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤Φ Φ=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥Ψ Ψ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Solve each time step using SuperLU
For linear problem,only need to form LU decomposition once and do a back-substitution each time step.
Note that stream function and vorticity are solved together
Tilting of a Plasma Column
0 1
1
[2 / ( )] ( ) cos , 1( 1/ ) cos , 1
( ) 0
kJ k J kr rr r r
J k
θψ
θ<⎧
= ⎨ − >⎩=
Initial Condition:
Give small perturbation and evolve in time
Stream function and vorticity at final time
Flux (top) and current (bottom) at initial and final times
Tilting of a Plasma Column-cont
Converged (in time) growth rate the same for N=30,40 out to 6 decimal places
Boundary Conditions
x
y
x
y
xx
xy
yy
φφφφφφ
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥
Φ = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
If boundary aligned with x axis:
Homogeneous Dirichlet
Homogeneous Neumann
000
x
xx
φφφ
==
=
0
0y
xy
φ
φ
=
=
Non-Rectangular Boundary Conditions
At each boundary point, define a local normal and curvature:
ˆˆ ˆ /n n dt dlκ ≡ ⋅
We then define a new set of basis functions for that boundary node that are locally aligned with the boundary and curvature
Old basis functions
Ignoring curvature
with curvature
002 2
2 2
2 2
2 2
2 2
1 0 0 0 0 000 2 ( ) 20 0 00 0 0 2 20 0 0
x y y x y x xn
y x x y x y x y yt
x x y y xxnn
x y x y x y xynt
y x y x yytt
n n n n n nn n n n n n n n
n n n nn n n n n nn n n n
νμκ κ κ νμκ κ κ νμ
νμνμνμ
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎢ ⎥
− − − −⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥
⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ − −⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
i
34
3D C1 elements by combining Q18 triangles in (R,Z) Hermite Cubic representation in the toroidal angle φ
( ) ( ) ( )2 21 2( ) | | 1 2 | | 1 ; ( ) | | 1x x x x x xΦ = − + Φ = −
]
181 2
, 1 , 21
1 2, 1 1 , 1 2
( , , ) ( , ) ( / ) ( / )
( / 1) ( / 1)
j j k j kj
j k j k
U R Z R Z U h U h
U h U h
ϕ ν ϕ ϕ
ϕ ϕ=
+ +
⎡= Φ + Φ⎣
+ Φ − + Φ −
∑
Each toroidal plane has two Hermite cubic functions associated with it
Solution for each scalar function is represented in each triangular wedge as the product of Q18 and Hermite functions
All DOF are still located at nodes: => very efficient representation
35
2-Fluid MHD Equations:
( ) 2 2
2
2
2
( ) 0
( )
3 32 23 3
52
2 2
1
32
i
ee e e
GV
e H
ee
ii i i
n nt
t
nM pt
p p p J Qt
p
p hne
pp nne
p p V Qt
n
λ
μ
η
η
μ
Δ
Δ
∂+ ∇ • =
∂∂
= −∇ × = ∇ × = ∇×∂
∂+ • ∇ + ∇ = × − + ∇
∂
+ × = +
∂ ⎛ ⎞+
∇
× − ∇ − ∇
⎡ ⎤∇ − ∇⎢ ⎥⎣ ⎦∇ = − ∇ + + − ∇ +⎜ ⎟∂ ⎝ ⎠
∂ ⎛ ⎞+ ∇ = − ∇ + ∇ − ∇ −⎜ ⎟∂ ⎝ ⎠
V
B E B A J B
V V V J B V
E V
Π
J B JB J
V V q
V q
J
V
i i i
i i
i
i
i
2R
-esfl
istiuid
ve MHDterms
2 2 2R RU R χϕ ϕω −⊥= ∇ × ∇ + ∇ + ∇V
20 ˆlnR R Rf zψϕ ϕ= ∇ ×∇ + ∇ −A
8 scalar 3D variables:, , , , , , ,e if U n p pψ χ ω
No further approximations! Solve these as faithfully as possible in realistic geometry
ϕ
Z
R
36
t = 1
t = 16
t = 24
t = 32
t = 40
t = 8
“GEM” reconnection test problem1 for 2-fluid MHD
• Starts like resistive MHD
• Dramatic change in configuration for t > 20
In-plane current density contours at different times
1J. Birn, et al, J. Geophys. Res. 106 (2001) 3715
37
2-fluid reconnection requires high resolution for convergent results
• Note sudden transition where velocity abruptly increases
•These calculations used a hyperviscosity term in Ohm’s law proportional to h2 . . . required for a stable calculation
• Reconnected Flux at t=40 converges as 1/h5
38
vorticity field velocity divergence
out-of-plane current out-of-plane magnetic field
2F GEM Reconnection snapshot at time of maximum velocity t ~ 32 (2002 nodes)
Energy conservation to 1 in 103, flux conservation exact, symmetry preserved even though triangles were not arranged symmetrically.
mid
plan
e
39
Midplane Current density collapses to the width of 1-3 triangular elements
t=32 time of previous contour plot( note sudden collapse at t=23+)
40
Midplane electric field before and after transitiont=20
X-Midplane
10 15 20 25 30 35-0.01
0.00
0.01
0.02
0.03
0.04
t=30
X-Midplane
10 15 20 25 30 35-0.2
0.0
0.2
0.4
Total Electric FieldResistivity J x BV x BHyper-resistivity
( ) 2 21ˆ e HJ B p hJVEne
B Jz η λ×−⎡ ⎤= − ∇+ +⎢ ⎥⎣ ⎦× − ∇iReconnection rate:
41
Blowup showing electric field after transition
X-Midplane
20 21 22 23 24 25-0.2
0.0
0.2
0.4
Total Electric FieldResistivity J x BV x BHyper-resistivity
( ) 2 21ˆ e HJ B p hJVEne
B Jz η λ×−⎡ ⎤= − ∇+ +⎢ ⎥⎣ ⎦× − ∇i
one triangle
Hyper-resistivity coefficient must be large enough that current density collapse is limited to 1-2 triangles: reason for factor h2
t=30
X-Midplane
10 15 20 25 30 350.2
0.0
0.2
0.4
43
2 22
2 2 0
u vcu ut x c
v u t xct x
∂ ∂ ⎫= ⎪ ∂ ∂⎪∂ ∂ − =⎬∂ ∂ ∂ ∂⎪=⎪∂ ∂ ⎭
Consider a simple 1-D Hyperbolic System of Equations (Wave Equation)
44
2 22
2 2 0
u vcu ut x c
v u t xct x
∂ ∂ ⎫= ⎪ ∂ ∂⎪∂ ∂ − =⎬∂ ∂ ∂ ∂⎪=⎪∂ ∂ ⎭
1 1 11/2 1/2 1/2 1/2
1 1 11/2 1/2 1 1
(1 )
(1 )
n n n n n nj j j j j j
n n n n n nj j j j j j
u u v v v vc
t x x
v v u u u uc
t x x
θ θδ δ δ
θ θδ δ δ
+ + ++ − + −
+ + ++ + + +
⎡ ⎤⎛ ⎞ ⎛ ⎞− − −= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞ ⎛ ⎞− − −= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
Consider a simple 1-D Hyperbolic System of Equations (Wave Equation)
Implicit Centered Difference:
Stable for θ ≥ ½ 2nd order for θ = ½
45
1 1 11/2 1/2 1/2 1/2
1 1 11/2 1/2 1 1
(1 )
(1 )
n n n n n nj j j j j j
n n n n n nj j j j j j
u u v v v vc
t x x
v v u u u uc
t x x
θ θδ δ δ
θ θδ δ δ
+ + ++ − + −
+ + ++ + + +
⎡ ⎤⎛ ⎞ ⎛ ⎞− − −= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞ ⎛ ⎞− − −= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
( ) ( )
1 1 11 1 1 1 1/2 1/21 2 2
2 2
1 1 11/2 1/2 1 1
2 2( ) (1 )
(1 )
n n n n n n n nj j j j j j j jn n
j j
n n n n n nj j j j j j
u u u u u u v vu u tc tc
x x x
tcv v u u u ux
δ θ θ θ δδ δ δ
δ θ θδ
+ + ++ − + − + −+
+ + ++ + + +
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + −= + + − +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
⎡ ⎤= + − + − −⎣ ⎦
Consider a simple 1-D Hyperbolic System of Equations (Wave Equation)
Substitute from second equation into first:
These two equations are completely equivalent to those on the previous page, but can be solved sequentially!
Only first involves Matrix Inversion … Diagonally Dominant
46
2 2
2 2 2
2 2 2
2 2 2
2 2 2
2 2
1 211 21
1 211 21
1 211 21
1 11 1
1 11 1
1 11 1
S SSS S SS S
S S SS SS S SS S
S S SS SS SS S
S SS S
S SS S
S SS
⎡ ⎤+ −−⎡ ⎤⎢ ⎥⎢ ⎥ − + −− ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ − + −−⎢ ⎥⎢ ⎥ − + −− ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ − + −−⎢ ⎥⎢ ⎥
− +− ⎢ ⎥⎣ ⎦⎢ ⎥ →⎢ ⎥− ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
2N
N
N
Substitution takes us from having to invert a 2N x 2N system that has large off-diagonal elements to sequentially inverting a N x N system that is diagonally dominant + the identity matrix.
Mathematically equivalent same answers!
Matrices to be inverted
1c tSx
θ δδ
=
Coupled system Un-coupled system
47
2 2 2 1 2 2
1 11/2 1/2 1/2 1/2
1 1 (1 )
(1 )
n n nx j x j x j
n n n nj j x j x j
S u S u S v
v v S u u
θ δ θ θ δ δ
θδ θ δ
+
+ ++ + + +
⎡ ⎤ ⎡ ⎤− = + − +⎣ ⎦ ⎣ ⎦⎡ ⎤= + + −⎣ ⎦
2 1 1 1 11 1
1/2 1/2
2n n n nx j j j j
n n nx j j j
u u u u
v v v
tcSx
δ
δ
δδ
+ + + ++ −
+ −
≡ − +
≡ −
≡
( ) ( )
1 1 11 1 1 1 1/2 1/21 2 2
2 2
1 1 11/2 1/2 1 1
2 2( ) (1 )
(1 )
n n n n n n n nj j j j j j j jn n
j j
n n n n n nj j j j j j
u u u u u u v vu u tc tc
x x x
tcv v u u u ux
δ θ θ θ δδ δ δ
δ θ θδ
+ + ++ − + − + −+
+ + ++ + + +
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + −= + + − +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
⎡ ⎤= + − + − −⎣ ⎦
Rewrite using standard finite-difference notation:
Operator to invert
48
u vct xv uct x
∂ ∂=
∂ ∂∂ ∂
=∂ ∂
n
n
u vc v tt x tv uc u tt x t
θδ
θδ
∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦
An alternate derivation:
Expand RHS in Taylor series in time to time-center
49
u vct xv uct x
∂ ∂=
∂ ∂∂ ∂
=∂ ∂
n
n
u vc v tt x tv uc u tt x t
θδ
θδ
∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦
n n
n
u uc v t c u tt x x t
v uc u tt x t
θδ θδ
θδ
⎡ ⎤∂ ∂ ⎛ ∂ ∂ ⎞⎡ ⎤= + +⎢ ⎥⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦⎝ ⎠⎣ ⎦∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦
An alternate derivation:
Expand RHS in Taylor series in time to time-center
Substitute from second equation into first
50
2 22 2 2 1 2 2
2 2
1 1
1 ( ) 1 (1 )( )
(1 )
n n n
n n n n
t c u t c u tc vx x x
v v tc u ux x
θ δ θ θ δ δ
δ θ θ
+
+ +
⎡ ⎤ ⎡ ⎤∂ ∂ ∂− = + − +⎢ ⎥ ⎢ ⎥∂ ∂ ∂⎣ ⎦ ⎣ ⎦
∂ ∂⎡ ⎤= + + −⎢ ⎥∂ ∂⎣ ⎦
u vct xv uct x
∂ ∂=
∂ ∂∂ ∂
=∂ ∂
n
n
u vc v tt x tv uc u tt x t
θδ
θδ
∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦
n n
n
u uc v t c u tt x x t
v uc u tt x t
θδ θδ
θδ
⎡ ⎤∂ ∂ ⎛ ∂ ∂ ⎞⎡ ⎤= + +⎢ ⎥⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦⎝ ⎠⎣ ⎦∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦
An alternate derivation:
Expand RHS in Taylor series in time to time-center
Substitute from second equation into first
Use standard centered difference in time:
51
u vct xv uct x
∂ ∂=
∂ ∂∂ ∂
=∂ ∂
n
n
u vc v tt x tv uc u tt x t
θδ
θδ
∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦
n n
n
u uc v t c u tt x x t
v uc u tt x t
θδ θδ
θδ
⎡ ⎤∂ ∂ ⎛ ∂ ∂ ⎞⎡ ⎤= + +⎢ ⎥⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦⎝ ⎠⎣ ⎦∂ ∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂ ∂⎣ ⎦
2 22 2 2 1 2 2
2 2
1 1
1 ( ) 1 (1 )( )
(1 )
n n n
n n n n
t c u t c u tc vx x x
v v tc u ux x
θ δ θ θ δ δ
δ θ θ
+
+ +
⎡ ⎤ ⎡ ⎤∂ ∂ ∂− = + − +⎢ ⎥ ⎢ ⎥∂ ∂ ∂⎣ ⎦ ⎣ ⎦
∂ ∂⎡ ⎤= + + −⎢ ⎥∂ ∂⎣ ⎦
An alternate derivation:
Expand RHS in Taylor series in time to time-center
Substitute from second equation into first
Use standard centered difference in time:
This is the same operator as before when centered spatial differences used
Summary• Finite elements with C1 continuity can be used with differential equations up
to 4th order.
• Hermite cubic element is an attractive C1 element in 1D– 2 DOF at each node
• Q18 is an attractive C1 element in 2D– All DOF defined at nodes very compact compared to other elements
– Shown to be very accurate
– Curved domains can be accommodated with normal vector and curvature
– High accuracy has been verified in demanding tests
• 3D Element can be made as the tensor product of Q18 and Hermite cubic
• Fits naturally into implicit method for MHD in tokamak (next lecture)