Lecture 9: Contents
Finite Difference grids
Observation We have treated equations on rectangular or circular
grids.
7
1
Composite and Overlapping Grids
0.8
Interface variables. Block Structure.
Bilinear interpolation. Lagrange Interpolation Formula.
5
0.4
Orthogonal Grid Transformations.
Beziér curves.
0.2
Angle: θ
Coordinate: y
Grid Generation
6
0.6
0
4
3
-0.2
-0.4
2
-0.6
1
-0.8
-1
-1
0
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Radus: r
Coordinate: x
Question What to do with more complex domains?
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Example Consider the equation uxx + uyy = 0 in Ω with Dirichlet
boundary conditions u = f on ∂Ω.
Solution Divide Ω in two parts Ω1 and Ω2 .
(k)
(k)
Supply Ωk with aseparate grid {xi , yj )}, k = 1, 2.
(1)
(2)
Interior points are uh , uh and interior points on the interface w(12) .
Ω
What will the combined finite difference equations look like?
How to construct finite difference equations Ah u = b?
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Example Consider the equation uxx + uyy = 0 in Ω = Ω1 ∪ Ω2 , with
Dirichlet boundary conditions u = f on ∂Ω. Due to the specific
function f we want to use a smaller stepsize on Ω1 .
(1)
ui
(12)
wi−1
(12)
wi
(12)
wi+1
(2)
ui
Solution If the grid is a perfect match at the interface then
 (1)

 

(1,w)
(1)
Ah
0
Ah
uh
b(1)


(2)
(2,w)  
 0
  uh(2)  =  b(2) 
Ah
Ah
(w,2)
c(12)
w(12)
Ahw,1 Ah
Bw
(1)
(2)
Observation On the interface y = y0 = ym we have grid points
from both grids. Need one equation for each grid point.
The matrix has a block structure!
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Linear Interpolation
Example Let Ω be a rectangle [xi , xi+1 ] × [yi , yi+1 ]. Approximate
u(x, y) with error O(h2 ) in Ω
Observation Any bilinear polynomial on Ω is a linear combination of
1, x − xi , y − yi and (x − xi )(y − yi ). We need 4 interpolation points to
uniquely define the polynomial.
(12)
Example Discretize the equation ∆u = 0 at one point wi
interface. Use a O(h2 ) accurate approximation.
on the
Remark Bilinear interpolation on a grid will automatically create a
continuous function across interfaces.
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Overlapping grids
Example We have a rectangular domain with an elipse shaped hole.
Example Suppose we want O(h3 ) error in the interpolation? How to
organize the interpolation points?
Remark This is biquadratic interpolation. There is also bicubic
interpolation schemes, Trilinear, or Tricubic. All are easy to automate.
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Lagrange Interpolation
Example An object moves during a simulation.
Definition Let xj , j = 1, . . . , m be distinct points. The
Lagrange basis polynomial is
ℓi (x) = Πj6=i
x − xj
.
xi − xj
Remark We have ℓi (xi ) = 1 and ℓi (xj ) = 0 for i 6= j.
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2
Lemma The unique polynomial of degree m − 1 that
interpolates u(x) in the points xj , i = 1, 2, . . . , m is
1.45
1.5
1.4
1
1.35
0.5
uI (x) =
m
X
0
u(xi )ℓi (x).
1.3
-0.5
1.25
i=1
-1
1.2
-1.5
-2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
1.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
Example Find the unique polynomial interpolating the table
xi
ui
0.30
1.23
0.45
1.31
0.53
1.35
Results The Lagrange basis polynomials of degree 3 and the
interpolating polynomial. The error is O(h4 ).
0.67
1.42
Question How to generalize to 2D?
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Example We have m = 6 points (xi , yj ) and want to find the
interpolating polynomial uI (x, y).
Introduce basis functions ℓi such that ℓi (xj , yj ) = 1 if i = j and zero
otherwise. Then the interpolant is
uI (x) =
m
X
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4
3
2
2
0
1
-2
0
-4
-1
-6
1
-2
1
0.5
0.8
0.4
0.3
0.6
u(xi , xi )ℓi (x).
0.2
0.4
0.1
0
0.5
0.8
0.4
0.3
0.6
0.2
0.4
0.1
0
i=1
Two basis functions ℓ2 (x, y) and ℓ3 (x, y). We use m = 6 interpolation
points and basis functions 1, x, y, xy, x2 , y2 .
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Unstructured Grids
4
0.05
3.8
0.04
Example Complicated domains can be divided into triangles.
0.03
3.6
0.02
3.4
0.01
3.2
1
0
1
0.5
0.8
0.4
0.3
0.6
0.2
0.4
0.5
0.8
0.4
0.3
0.6
0.2
0.1
0
0.4
0.1
0
The interpolant uI (x, y) and the error |uI − u|, where
√
u = 3 + x y + cos(2x)y2 was interpolated using m = 6 points.
Remark Really simple and flexible method. The error terms are
obtained by comparing with a Taylor series expansion.
No structure in the linear system Ax = b means more memory needed.
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Orthogonal Coordinate Transformations
Definition A coordinate transformation
(x1 , x2 ) 7→ (ξ1 , ξ2 ) is orthogonal it two curves that are
orthogonal at (x∗1 , x∗2 ) are mapped onto curves that are
orthogonal at the intersection (ξ1∗ , ξ2∗ ).
Lemma Let T : (x1 , x2 ) 7→ (ξ1 , ξ2 ) be an orthogonal
transformation. Then
(a(x)ux1 )x1 + (b(x)ux2 )x2 = (ã(ξ)uξ1 )ξ1 + (b̃(ξ)uξ2 )ξ2 .
Two orthogonal grids. How can we construct these?
Remark A conformal mapping creates an orthogonal grid
transformation but can’t control the distribution of points on the
boundary.
Remark We do not introduce mixed derivatives.
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Beziér curves
Cubic Beziér curves
A cubic Beziér curve is defiend by four points and given by
Definition A Beziér curve (or surface) is given by a
number of control points.
s(t) = (1 − t)3 P1 + 3(1 − t)2 tP2 + 3(1 − t)t2 P3 + t3 P4 .
Exemple
P2
P2
P3
P2
P4
P1
P1
P1
P3
A linear Beziér curve is given by three points and a quadratic curve is
determined by three points. Beziér curves are approximating splines.
Lemma A cubic Beziér kurva s(x) interpolates the first
and last control point.
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Definition The convex hull formed by points {Pi }ni=1
consists of all points that can be written as convex linear
combinatiosn of {Pi }ni=1 .
Exemple what is the convex hull of the points
1
2
0
.
, and P3 =
, P2 =
P1 =
3
1
0
Theorem A Beziér curve lies within the convex hull of its
control points.
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Theorem For a Beziér curve with control points {Pi }ni=1
we have the tangent directions
s′ (0) = α(P2 − P1 ) and s′ (1) = α(Pn − Pn−1 ),
α > 0.
Remark It is easy to control the direction of the curve at its end
points.
Question How to make more complex curves?
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P2
P3
P7
Example Construct a curvilinear coordinate system for a region
P4
P5
P6
P1
Two cubic Beziér curves defined by the control points {P1 , P2 , P3 , P4 }
and {P4 , P5 , P6 , P7 }. Note that P4 is a common interpolation point.
Remark Algorithms for generating orthogonal grids is an active
research area. There are good methods for many types of geometry.
If P4 −P3 = P5 −P4 the curve has a continuous tangent.
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Grids Finite Differences use
• Structured Grids that are relatively simple.
• Combining Grids gives structured matrices.
The goal is to make solving the problem simpler and less resource
demanding.
Alternative The Finite Element method using unstructured grids.
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