the finite‐difference method · differential equation to solve conventional fdm (1 of 3) slide 20...
TRANSCRIPT
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Advanced Computation:
Computational Electromagnetics
The Finite‐Difference Method
Outline
• Finite‐Difference Approximations
• Finite‐Difference Method
• Numerical Boundary Conditions
• Matrix Operators
Slide 2
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Slide 3
Finite Difference Approximations
The Basic Finite‐Difference Approximation
Slide 4
1.5 2 1df f f
dx x
1f2f
df
dx
x
second‐order accuratefirst‐order derivative
This is the only finite‐difference approximation we will use in this course!
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Types of Finite‐Difference Approximations
Slide 5
Backward Finite‐Difference
1.5 2 1df f f
dx x
Central
Finite‐Difference
1 2 1df f f
dx x
Forward
Finite‐Difference
2 2 1df f f
dx x
The Generalized Finite‐Difference
Slide 6
n
ni
ii
d fa
xf
d i i
i
f aL f
The derivative of any order of a function at any position can be approximated as a linear sum of known points of that function.
In fact, any linear operation on the function can be approximated as a linear sum of known points of that function.
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Finite‐Difference “Atoms”
Slide 7
Any finite‐difference approximation can be summarized graphically as an “atom.”
1 1
2i i
i
f fdf x
dx
if 1if 1if
2
1 12 2
2i i ii
f f fdf x
dx
if 1if 1if
,i jf
1,i jf 1,i jf
, 1i jf
, 1i jf
1, , 1, , 1 , , 122 2
2 2i j i j i j i j i j i ji
f f f f f ff x
1
21
2
2
1
2
1
2
2
2
1
2
1
21
21
24
Slide 8
Finite‐Difference Method
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Overview of Our Approach to FDM
Slide 9
1. Identify and write the governing equation(s).
2
2a x f x b x f x c x f x g x
x x
2. Write the matrix form of this equation going term‐by‐term.
3. Put matrix equation in the standard form of [L][f] = [g].
2
2a x f x b x f x c x f x g x
x x
2x xA D f B D f C f g
2 where x xL f g L A D B D C 4. Solve [L][f] = [g].
1f L g
L = A*D2X + y*B*DX + C;
What’s the Catch?
Slide 10
2x xL A D B D C
How are these matrices constructed? What is their meaning?
A
2xD
B xD
C
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Functions Vs. Operations (1 of 2)
Slide 11
2
2f x f x f x ga b x x
xxx c
x
FunctionsThe only time functions appear in a differential equation is as the unknown or as the excitation.
unknown
excitation
f x
g x
OperationsEverything else in a differential equation is something that operates on a function.
2
2
, , point-by-point
multiplication on
, calculates derivatives of
scales entire
a x b x c x
f x
f xx x
f x
Functions Vs. Operations (2 of 2)
Slide 12
2x xA D B D Cf f f g
FunctionsFunctions are stored as column vectors.
OperationsOperations are always stored in square matrices. Any linear operation can be put into matrix form.
1 1
2 2
M M
f g
f gf g
f g
11 12 1
21 22 2
1 2
M
M
M M MM
l l l
l l lL
l l l
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Interpretation of Matrices
Slide 13
11 12 13 1a x a y a z b
21 22 23 2a x a y a z b 31 32 33 3a x a y a z b
11 12 13 1
21 22 23 2
31 32 33 3
a a a x b
a a a y b
a a a z b
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
Equation for x
Equation for y
Equation for z
EQUATION FOR… RELATION TO…
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
From a purely mathematical perspective, this interpretation does not make sense. This interpretation will be highly useful and insightful because of how we derive the equations.
Representing Functions on a Grid
Slide 14
A grid is constructed by dividing space into discrete
cells
Example physical
(continuous) 2D function
Function is known only at discrete points
Representation of what is
actually stored in memory
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Grid Cells
Slide 15
y
x
A function value is assigned to a specific point within the grid cell.
Whole Grid
A Single Grid Cell
, grid resolution parametersx y
Functions are Put Into Column Vectors
Slide 16
2‐D Systems
1f 5f 9f 13f
2f 6f 10f 14f
3f 7f 11f 15f
4f 8f 12f 16f
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
1
2
3
4
5
f
f
f
f
f
f
1‐D Systems
1f2f 3f 4f 5f
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Putting Functions into Column Vectors
Slide 17
1f 5f 9f 13f
2f 6f 10f 14f
3f 7f 11f 15f
4f 8f 12f 16f
1f
2f
3f
4f
5f
6f
7f
8f
9f
10f
11f
12f
13f
14f
15f
16f
1
2
3
4
5
6
7
8
9
10
11
12
13
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15
16
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
MATLAB ‘reshape’ commandf = F(:);F = reshape(f,Nx,Ny);
Locating Points in Column Vectors
Slide 18
1D Grids
Node located at nx m = nx
2D Grids
Node located at nx,ny m = (ny – 1)*Nx + nx
3D Grids
Node located at nx,ny,nz m = (nz – 1)*Nx*Ny + (ny – 1)*Nx + nx
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The Finite‐Difference Method
Slide 19
df xa x f x g x
dx
1 1
2
f k f ka k f k g k
x
L f g
1f L g
Conventional FDM
Tedious, but not difficult.
Very difficult and tedious.
Easy.
xD f a f g
x
L f g
L D a
Improved
FDM
Almost effortless.
Very easy and clean
Easy.
Most time consuming step.
Differential Equation to Solve
Conventional FDM (1 of 3)
Slide 20
Step 1 – Start with a differential equation to be solved.2
2
d f dfa bf c
dx dx
Step 2 – Approximate the derivatives with finite differences.
2
1 2 1 1 1
2
f k f k f k f k f ka k b k f k c k
Step 3 – Expand equation and collect common terms.
2 2 2
2 2 2
1 2 11 1 1 1
2 2
1 2 11 1
2 2
a k a kf k f k f k f k f k b k f k c k
a k a kf k b k f k f k c k
IMPORTANT RULE: Every term in a finite‐difference equation must exist at the same point.
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Conventional FDM (2 of 3)
Slide 21
Step 4 – The final equation is used to populate a matrix equation.
222
1 11
2 2
21b
a kkf k f k f k k
kc
a
1 1
2 2
3 3
4 4
5 5
1 1
f c
f c
f c
f c
f c
f N c N
f N c N
Filling in the matrix like this can be a very difficult and tedious task for more complicated differential equations or for systems of differential equations.
Conventional FDM (3 of 3)
Slide 22
Step 5 – The matrix equation is solved for the unknown function f(x)
L f c
Lf c
1
1
f L c
f L c
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Improved FDM
Slide 23
We want a very easy way to immediately write differential equations in matrix form. Starting with the same differential equation…
2
2
f fa bf c
x x
We will develop a procedure by which this will be directly written in matrix form without having to explicitly handle any finite‐differences.
2 1x x D f AD f Bf = c
2 1, , , and x xD A D Bare square matrices that perform linear operations on the vector f.
2
2f a f bf c
x x
2 1x x
L
L
D AD B f = c
1f L c
Locating Points in Matrices
Slide 24
1D m or n = nx2D m or n = (ny–1)*Nx + nx3D m or n = (nz–1)*Nx*Ny + (ny–1)*Nx + nx
Row mEquation for node (nx,ny,nz)
Column n
Relatio
n to
node (n
x,ny,nz)
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Slide 25
NumericalBoundary Conditions
2,1 1,1 1,2 1,11,12 2
3,1 2,1 1,1 2,2 2,
0,1 1,0
2,0
4,1
12,12 2
3,1 2,1 3,2 3,13,12 2
2,2 1,2 1,3 1,2 1,11,22 2
3,2 2,2 1
3,0
0,2
,2 2,3 2,2 2
2
2 2
2 2
2 2
2 2
2 2
f f f fg
x y
f f f f fg
x y
f f f fg
x
f f
y
f f f f fg
x y
f f f f f
f
f
f
x
f
f
,1
2,22
3,2 2,2 3,3 3,2 3,13,22 2
2,3 1,3 1,3 1,21,32 2
3,3 2,3 1,3 2,3 2,22,32 2
3,3 2,3 3,3 3,23,32
4,2
0,3 1,4
2,4
4,3 3
2
,4
2 2
2 2
2 2
2 2
gy
f f f f fg
x y
f f f fg
x y
f f f f fg
x y
f f f fg
x
f
y
f f
f
f f
0,1f
0,2f
0,3f
4,1f
4,2f
4,3f
1,0f 2,0f 3,0f
1,4f 2,4f 3,4f
The Problem at the Boundaries
Slide 26
Suppose it is desired to solve the following differential equation on a 33 grid.
1, , 1, , 1 , , 1,2 2
2 2 i j i j i j i j i j i j
i j
f f f f f fg
x y
The terms in red exist outside of the grid.
How this is handled is called a boundary condition.
2 2
2 2
, ,,
f x y f x yg x y
x y
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2,1 1,1 1,2 1,11,12 2
3,1 2,1 1,1 2,2 2,12,12 2
3,1 2,1 3,2 3,13,12 2
2,2 1,2 1,3 1,2 1,11,22 2
3,2 2,2 1,2 2,3 2,2 2,12,22 2
3,2
02 2
2 2
2 2
2 2
2 2
2
0
0
0 0
0
0
f f f fg
x y
f f f f fg
x y
f f f fg
x y
f f f f fg
x y
f f f f f fg
x y
f
2,2 3,3 3,2 3,1
3,22 2
2,3 1,3 1,3 1,21,32 2
3,3 2,3 1,3 2,3 2,22,32 2
3,3 2,3 3,3 3,23,32 2
0 0
0
0 0
2
2 2
2 2
2 2
f f f fg
x y
f f f fg
x y
f f f f fg
x y
f f f fg
x y
0,1f
0,2f
0,3f
4,1f
4,2f
4,3f
1,0f 2,0f 3,0f
1,4f 2,4f 3,4f
Dirichlet Boundary Conditions (1 of 2)
Slide 27
The simplest boundary condition is to assume that all values of 𝑓 𝑥, 𝑦 outside of the grid are zero.
Dirichlet Boundary Conditions (2 of 2)
Slide 28
2
1
2
1
1
1
, , 0
2
, 0 ,
2
, , 0
2
, 0 ,
2
x
y
x
y
i i
x
i N i
x
i i
y
i i N
y
N
N
f y f x y
x
f y f x y
x
x
x
yf x f x y
y
f yx f x y
y
Dirichlet boundary conditions assume function values from outside of the grid are zero.
22 1
2 2
21
2 2
22 1
21
2
21
2 2
1, , 2 , 0
, 0 2 , ,
, , 2 , 0
, 0 2 , ,
x x
y y
x
y
i i i
x
i N i N i
x
i i i
y
i i N i
N
N
y
N
f y f x y f x y
x
f y f x y f x y
x
f x f x y f x y
y
f x f x y f x y
y
x
x
y
y
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Periodic Boundary Conditions (1 of 2)
Slide 29
If the function f(x,y) is periodic, then the values from outside of the grid can be mapped to a value from inside the grid at the other side.
3, 1, ,3 ,10, 4, ,0 ,4, , , and j j i ij j i if f ff f f ff
2,1 1,1 1,2 1,11,12 2
3,1 2,1 1,1 2,2 2,
3,1 1,3
2,3
1,1
12,12 2
3,1 2,1 3,2 3,13,12 2
2,2 1,2 1,3 1,2 1,11,22 2
3,2 2,2 1
3,3
3,2
,2 2,3 2,2 2
2
2 2
2 2
2 2
2 2
2 2
f f f fg
x y
f f f f fg
x y
f f f fg
x
f f
y
f f f f fg
x y
f f f f f
f
f
f
x
f
f
,1
2,22
3,2 2,2 3,3 3,2 3,13,22 2
2,3 1,3 1,3 1,21,32 2
3,3 2,3 1,3 2,3 2,22,32 2
3,3 2,3 3,3 3,23,32
1,2
3,3 1,1
2,1
1,3 3
2
,1
2 2
2 2
2 2
2 2
gy
f f f f fg
x y
f f f fg
x y
f f f f fg
x y
f f f fg
x
f
y
f f
f
f f
0,1f
0,2f
0,3f
4,1f
4,2f
4,3f
1,0f 2,0f 3,0f
1,4f 2,4f 3,4f
Periodic Boundary Conditions (2 of 2)
Slide 30
1
1
1
1
2
1
2
1
, ,,
2
, , ,
2
, ,,
2
, , ,
2
y
x
y
x
x
y
N
N
i ii
x
i i N i
x
i ii
y
i i i
N
N N
y
f x y f yf y
x
f y f y f x y
x
f x y f xf x
y
f x
xx
x x
yy
f x f xy y y
y
Periodic boundary conditions assume function values from outside of the grid can be taken from the opposite side of the grid.
22 1
2 2
21
2 2
22 1
2 2
21
2
1
1
1
1
2
, 2 , ,,
, , 2 , ,
, 2 , ,,
, , 2 , ,
x x
y y
x
x
y
y
Ni i ii
x
i i N i N i
x
i i ii
y
i i i N i N
y
N
N
N
f x y f x y f yf y
x
f y f y f x y f x y
x
f x y f x y f xf x
y
f x f x f x
xx
x x
y f x y
y
yy
y y
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Neumann Boundary Conditions
Slide 31
We use Neumann boundary conditions for non‐periodic functions or functions that are not zero at the boundary. Here the function continues linearly outside of the grid.
Spatially variant grating that is not periodic and not zero at the boundaries.
Here is a 1D function with Neumann boundaries
Neumann BC’s for 1D Function
Slide 32
The finite‐difference approximation for a 1D function is
1 1
2i i
i x
f fdf
dx
At i=1, we have a problem…
2 0
1 2i x
f fdf
dx
This term doesn’t exist!
2 1
1i x
f fdf
dx
At i=Nx, we have another problem…
1 1
2x x
x
N N
i N x
f fdf
dx
This term doesn’t exist! 1x x
x
N N
i N x
f fdf
dx
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What About the 2nd‐Order Derivatives for the Neumann Boundary Condition?
Slide 33
In order for the function to continue in a straight line, the second‐order derivate should be set to zero at the boundary.
IMPORTANT: This is NOT Dirichlet boundary conditions. A Dirichlet BC sets the function itself to zero outside of the grid, not the derivative. Here, the 2nd‐order derivative is set to zero.
2 22 1 0
2 2 2
1 1
0xi i
f f fd f d f
dx dx
2 21 1
2 2 2 0x x x
x x
N N N
xi N i N
f f fd f d f
dx dx
Neumann Boundary Conditions for 2D Functions
Slide 34
1 2 1
1
1 2 1
1
, , ,
, , ,
, , ,
, , ,
x x x
y y y
N N
i i i
x
i i i
x
i i i
y
N
Ni Ni
y
N i
f y f y f y
x
f y f y f y
x
f x f x f x
y
f x
x x x
x x x
y y y
y f x f x
y
y y
Neumann boundary conditions are used when a function should be continuous at the boundary. That is, the first‐order derivative is continuous and the second‐order derivative is zero.
2
2
2
2
2
2
2
2
1
1
,0
,0
,0
,0
x
y
i
i
i
N
Ni
f y
x
f y
x
f x
y
x
x
yf
y
x
y
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Slide 35
Matrix Operators
Origin of Matrix Operators
Slide 36
We start with a governing equation.
2
2
,,
d f x yg x y
dx
We construct a grid to store the functions.
2,1 1,11,12
3,1 2,1 1,12,12
3,1 2,13,12
2,2 1,21,22
3,2 2,2 1,22,22
3,2 2,23,22
2,3 1,31,32
3,3 2,3 1,32,32
3,3 2,33,32
2
2
2
2
2
2
2
2
2
0
0
0
0
0
0
f fg
xf f f
gxf f
gx
f fg
xf f f
gxf f
gx
f fg
xf f f
gxf f
gx
We approximate the governing equation with finite‐differences and then write the finite‐difference equation at each point the grid.
1,1 1,1
2,1 2,1
3,1 3,1
1,2
2,22
3,2
1,3
2,3
3,3
2 1 0 0 0 0 0 0 0
1 2 1 0 0 0 0 0 0
0 1 2 0 0 0 0 0
0 0 2 1 0 0 0 01
0 0 0 1 2 1 0 0 0
0 0 0 0 1 2 0 0
0 0 0 0 0 2 1 0
0 0 0 0 0 0 1 2 1
0 0 0 0 0 0 0 1
0
0
0
2
0
f g
f g
f g
f
fx
f
f
f
f
1,2
2,2
3,2
1,3
2,3
3,3
g
g
g
g
g
g
We collect the large set of equations into a single matrix equation.
This matrix calculates the derivative of f(x,y) and puts the answer in g(x,y). This is a matrix operator.
2xD
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Other Matrix Operators
Slide 37
A square matrix can always be constructed to perform any linear operation on a function that is stored in a column vector.
1
2
3 ?
x
x
N
f
fd
f x fdx
f
D
D f
FFT f x Ff
f x dx f
g x f x Gf
h x f x Hf
Point‐by‐Point Multiplication (1 of 2)
Slide 38
Since we are storing our “functions” in vector form, how do we perform a point‐by‐point multiplication using a square matrix?
f xb x Bf
1x 2x 3x 4x 5x 6x
1f
2f3f 4f 5f 6f
1b2b 3b 4b 5b 6b
1 11 1
2 2
3 3
4 4
2 2
3 3
4
5 5
4
5 5
6 66 6
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
f f
f f
f
b b
b b
b b
b b
f
f f
f f
f f
b b
b b
f fB B
?
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Point‐by‐Point Multiplication (2 of 2)
Slide 39
Since we are storing our “functions” in vector form, how do we perform a point‐by‐point multiplication using a square matrix?
f xb x Bf
1x 2x 3x 4x 5x 6x
1f
2f3f 4f 5f 6f
1b2b 3b 4b 5b 6b
1 11 1
2 2
3 3
4 4
2 2
3 3
4
5 5
4
5 5
6 66 6
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
f f
f f
f
b b
b b
b b
b b
f
f f
f f
f f
b b
b b
f fB B
First‐Order Partial Derivative (1 of 2)
Slide 40
How do we construct a square matrix so that when it premultiplies a vector, we get a vector containing the first‐order partial derivative?
1 xf xx
D f
1x 2x 3x 4x 5x 6x
1f
2f3f 4f 5f 6f
1 1
2 0
3 1
4
1
2
23
4
6
7
5
6
5 3
4
5
2
2
2
2
2
0 1 0 0 0 0
1 0 1 0 0 0
0 1 0 1 0 01
0 0 1 0 1 02
0 0 0 1 0 1
0 0 0 0 1 0 2
x
x
x
x
x
x
x
x
x
f f
f f
f f
f f
f
f
f
f
f
f f
f ff
D fDf
?
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First‐Order Partial Derivative (2 of 2)
Slide 41
1 xf xx
D f
1x 2x 3x 4x 5x 6x
1f
2f3f 4f 5f 6f
1 1
2 0
3 1
4
1
2
23
4
6
7
5
6
5 3
4
5
2
2
2
2
2
0 1 0 0 0 0
1 0 1 0 0 0
0 1 0 1 0 01
0 0 1 0 1 02
0 0 0 1 0 1
0 0 0 0 1 0 2
x
x
x
x
x
x
x
x
x
f f
f f
f f
f f
f
f
f
f
f
f f
f ff
D fDf
How do we construct a square matrix so that when it premultiplies a vector, we get a vector containing the first‐order partial derivative?
Second‐Order Partial Derivative (1 of 2)
Slide 42
How do we construct a square matrix so that when it premultiplies a vector, we get a vector containing the second‐order partial derivative?
2
2
2 xx
f x
D f
1x 2x 3x 4x 5x 6x
1f
2f3f 4f 5f 6f
2
22 1 0
23 2 1
24 3 2
25 4 3
26 5 4
27 6
1
2
3
4
5
2
6 5
2 1 0 0 0 0
1 2 1 0 0 0
0 1 2 1 0 01
0 0 1 2 1 0
0 0 0 1 2 1
0 0 0 0
2
2
2
2
2
2 21
x
x
x
xx
x
x
x
f f f
f f f
f f f
f f f
f f f
f
f
f
f
f
f
f f f
fD
2x
fD
?
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8/24/2019
22
Second‐Order Partial Derivative (2 of 2)
Slide 43
2
2
2 xx
f x
D f
1x 2x 3x 4x 5x 6x
1f
2f3f 4f 5f 6f
2
22 1 0
23 2 1
24 3 2
25 4 3
26 5 4
27 6
1
2
3
4
5
2
6 5
2 1 0 0 0 0
1 2 1 0 0 0
0 1 2 1 0 01
0 0 1 2 1 0
0 0 0 1 2 1
0 0 0 0
2
2
2
2
2
2 21
x
x
x
xx
x
x
x
f f f
f f f
f f f
f f f
f f f
f
f
f
f
f
f
f f f
fD
2x
fD
How do we construct a square matrix so that when it premultiplies a vector, we get a vector containing the second‐order partial derivative?
What About 2D Grids (1 of 2)?
Slide 44
Two‐dimensional grids are a little more difficult
2
2
2 , xx
f x y
D f1f 2f 3f
4f 5f 6f
7f 8f 9f
x
y
2
22 1 ???
23 2 1
2??? 3 2
25 4 ?
1
??
6 52
2
3
4
5
6
7
8
9
2
2
0 2
2 1
1
0 2
2 1
1 2
2
2
11
1 2 1
1 2
2 1
1
0
0
1 2
1 2
x
x
x
x
x
x
f
f
f
f
f
f
f f f
f f f
f f f
f f
f
f
f f
f
f
f
fD
2
24
2??? 6 5
28 7 ???
29 8 7
2??? 9 8
2
2
2
2
x
x
x
x
x
x
f f f
f f f
f f f
f f f
fD
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8/24/2019
23
What About 2D Grids (2 of 2)?
Slide 45
2
24 1 ???
25
1
2
3
4
2 ???
6
7
2
8
9
3
5
6
2 1
2 1
2 1
1 2 11
1 2 1
1 2 1
1 2
1 2
1
20 0
20 0 0
20 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 2
y
y
y
y
f
f
f
f
f
f
f
f
f
f f f
f f f
f f f
D f
2
2???
27 4 1
28 5 2
29 6 3
2??? 7 4
2??? 8 5
2??? 9 6
2
2
2
2
2
2
y
y
y
y
y
y
y
y
f f f
f f f
f f f
f f f
f f f
f f f
fD
Two‐dimensional grids are a little more difficult
2
2
2 , xx
f x y
D f1f 2f 3f
4f 5f 6f
7f 8f 9f
x
y
Derivative Operators with Dirichlet Boundary Conditions
Slide 46
2
2
0 0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 01
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0 0
2 1
1 2 1
1 2
2 1
1 2 1
1 2
2
0
0
0
1
1 2 1
1
0
2
xx
D
2
2
2 1
2 1
2 1
1 2 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 01
0 0 0 0 0 0
0 0 0 0 0 0
1 2 1
1 2 1
1 2
1 2
1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 20 0 0
yy
D
Both of these matrices only have numbers along three of their diagonals.
This is called tridiagonal.
This suggests a fast way to construct these matrices.
Dx(2) has some corrections shown in blue along two of
its diagonals.
These matrices contain mostly zeros.
These are called a sparse matrices.
See MATLAB sparse() command.
Also see MATLAB spdiags() command.
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8/24/2019
24
Why Do We Need Separate Matrix Operators for First‐ and Second‐Order Derivatives?
Slide 47
It is known that2 2
2 2
x x x y y y
Can we just calculate D(2) from D(1)?
? ?2 1 1 2 1 1 x x x y y y D D D D D D
Yes, but this does not make efficient use of the grid. For a 5‐point, 1D grid, we have
1 1
2
1 0 1 0 0
0 2 0 1 01
1 0 2 0 12
0 1 0 2 0
0 0 1 0 1
x x
x
D D
This is not as accurate because it calculates the derivative with poorer grid resolution than is available.
2
2
2 1 0 0 0
1 2 1 0 01
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
xx
D
This matrix operator makes optimal use of the available grid resolution.
2D Derivative Operators for 1D Grids
Slide 48
When Nx=1 and Ny>10 0
0
0 0
x
D
When Nx>1 and Ny=1
is standard for 1D grid
zero matrixx
y
D
D Z
zero matrix
is standard for 1D gridx
y
D Z
D
0 0
0
0 0
y
D
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USE SPARSE MATRICES!!!!!!!
Slide 49
WARNING !!The derivative operators will be EXTREMELY large matrices.
For a small grid that is just 100200 points:
Total Number of Points: 20,000Size of Derivate Operators: 20,000 20,000Total Elements in Matrices: 400,000,000Memory to Store One Full Matrix: 6 GbMemory to Store One Sparse Matrix: 1 Mb
NEVER AT ANY POINT should you use FULL MATRICES in the finite‐difference method.
Not even for intermediate steps. NEVER!
49