Poisson Surface Reconstruction - II - IIT Bombay
Transcript of Poisson Surface Reconstruction - II - IIT Bombay
Poisson Surface Reconstruction - IISiddhartha Chaudhuri httpwwwcseiitbacin~cs749
Kazhdan Bolitho and Hoppe
Recap of differential operators (in 3D)
Gradient (of scalar-valued function)
In operator form Maps scalar field to vector field
nabla f=( part fpart xpart fpart ypart fpart z )
nabla=( partpart x partpart y partpart z )
Scalar fields (black high white low) and their gradients (blue arrows)
Wik
iped
ia
Recap of differential operators (in 3D)
Divergence (of vector-valued function)
Maps vector field to scalar field
Gam
asut
ra
nablasdotV=partV x
part x+
partV y
part y+
partV z
part z
Has divergence Divergence-free
Recap of differential operators (in 3D)
Curl (of vector-valued function)
Maps vector field to vector field
Gam
asut
ra
Curl-free Has curl
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Recap of differential operators (in 3D)
Laplacian (of scalar-valued function)
In operator form
Maps scalar field to scalar field
Gam
asut
ra
Δ=( part2
part x2 part2
part y2 part2
part z2 )
Δ f = nablasdotnabla f =
part2 fpart x2 +
part2 f
part y2+part2 f
part z2
Original function After applying Laplacian
Recap
The boundary of a shape is a level set of itsindicator function χ
The gradient nablaχ of χ is the normal field V at theboundary (after some smoothing which we wont go into here)
We can solve for χ by integrating the normal field hellip but in general we cant get an exact solution
since an arbitrary vector field need not be thegradient of a function (field needs to be curl-free)
So we find a least-squares fit minimizing ||nablaχ ndash V||2
Recap
So we find a least-squares fit minimizing ||nablaχ ndash V||2
This reduces to solving the Poisson Equation
We can discretize the system by representing thefunctions as vectors of values at sample points Gradient divergence and Laplacian operators become
matrices
Solving the resulting linear system gives a leastsquares fit at the sample positions
nablasdot(nabla χ )=nablasdotV hArr Δ χ=nablasdotV
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Recap of differential operators (in 3D)
Gradient (of scalar-valued function)
In operator form Maps scalar field to vector field
nabla f=( part fpart xpart fpart ypart fpart z )
nabla=( partpart x partpart y partpart z )
Scalar fields (black high white low) and their gradients (blue arrows)
Wik
iped
ia
Recap of differential operators (in 3D)
Divergence (of vector-valued function)
Maps vector field to scalar field
Gam
asut
ra
nablasdotV=partV x
part x+
partV y
part y+
partV z
part z
Has divergence Divergence-free
Recap of differential operators (in 3D)
Curl (of vector-valued function)
Maps vector field to vector field
Gam
asut
ra
Curl-free Has curl
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Recap of differential operators (in 3D)
Laplacian (of scalar-valued function)
In operator form
Maps scalar field to scalar field
Gam
asut
ra
Δ=( part2
part x2 part2
part y2 part2
part z2 )
Δ f = nablasdotnabla f =
part2 fpart x2 +
part2 f
part y2+part2 f
part z2
Original function After applying Laplacian
Recap
The boundary of a shape is a level set of itsindicator function χ
The gradient nablaχ of χ is the normal field V at theboundary (after some smoothing which we wont go into here)
We can solve for χ by integrating the normal field hellip but in general we cant get an exact solution
since an arbitrary vector field need not be thegradient of a function (field needs to be curl-free)
So we find a least-squares fit minimizing ||nablaχ ndash V||2
Recap
So we find a least-squares fit minimizing ||nablaχ ndash V||2
This reduces to solving the Poisson Equation
We can discretize the system by representing thefunctions as vectors of values at sample points Gradient divergence and Laplacian operators become
matrices
Solving the resulting linear system gives a leastsquares fit at the sample positions
nablasdot(nabla χ )=nablasdotV hArr Δ χ=nablasdotV
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Recap of differential operators (in 3D)
Divergence (of vector-valued function)
Maps vector field to scalar field
Gam
asut
ra
nablasdotV=partV x
part x+
partV y
part y+
partV z
part z
Has divergence Divergence-free
Recap of differential operators (in 3D)
Curl (of vector-valued function)
Maps vector field to vector field
Gam
asut
ra
Curl-free Has curl
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Recap of differential operators (in 3D)
Laplacian (of scalar-valued function)
In operator form
Maps scalar field to scalar field
Gam
asut
ra
Δ=( part2
part x2 part2
part y2 part2
part z2 )
Δ f = nablasdotnabla f =
part2 fpart x2 +
part2 f
part y2+part2 f
part z2
Original function After applying Laplacian
Recap
The boundary of a shape is a level set of itsindicator function χ
The gradient nablaχ of χ is the normal field V at theboundary (after some smoothing which we wont go into here)
We can solve for χ by integrating the normal field hellip but in general we cant get an exact solution
since an arbitrary vector field need not be thegradient of a function (field needs to be curl-free)
So we find a least-squares fit minimizing ||nablaχ ndash V||2
Recap
So we find a least-squares fit minimizing ||nablaχ ndash V||2
This reduces to solving the Poisson Equation
We can discretize the system by representing thefunctions as vectors of values at sample points Gradient divergence and Laplacian operators become
matrices
Solving the resulting linear system gives a leastsquares fit at the sample positions
nablasdot(nabla χ )=nablasdotV hArr Δ χ=nablasdotV
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Recap of differential operators (in 3D)
Curl (of vector-valued function)
Maps vector field to vector field
Gam
asut
ra
Curl-free Has curl
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Recap of differential operators (in 3D)
Laplacian (of scalar-valued function)
In operator form
Maps scalar field to scalar field
Gam
asut
ra
Δ=( part2
part x2 part2
part y2 part2
part z2 )
Δ f = nablasdotnabla f =
part2 fpart x2 +
part2 f
part y2+part2 f
part z2
Original function After applying Laplacian
Recap
The boundary of a shape is a level set of itsindicator function χ
The gradient nablaχ of χ is the normal field V at theboundary (after some smoothing which we wont go into here)
We can solve for χ by integrating the normal field hellip but in general we cant get an exact solution
since an arbitrary vector field need not be thegradient of a function (field needs to be curl-free)
So we find a least-squares fit minimizing ||nablaχ ndash V||2
Recap
So we find a least-squares fit minimizing ||nablaχ ndash V||2
This reduces to solving the Poisson Equation
We can discretize the system by representing thefunctions as vectors of values at sample points Gradient divergence and Laplacian operators become
matrices
Solving the resulting linear system gives a leastsquares fit at the sample positions
nablasdot(nabla χ )=nablasdotV hArr Δ χ=nablasdotV
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Recap of differential operators (in 3D)
Laplacian (of scalar-valued function)
In operator form
Maps scalar field to scalar field
Gam
asut
ra
Δ=( part2
part x2 part2
part y2 part2
part z2 )
Δ f = nablasdotnabla f =
part2 fpart x2 +
part2 f
part y2+part2 f
part z2
Original function After applying Laplacian
Recap
The boundary of a shape is a level set of itsindicator function χ
The gradient nablaχ of χ is the normal field V at theboundary (after some smoothing which we wont go into here)
We can solve for χ by integrating the normal field hellip but in general we cant get an exact solution
since an arbitrary vector field need not be thegradient of a function (field needs to be curl-free)
So we find a least-squares fit minimizing ||nablaχ ndash V||2
Recap
So we find a least-squares fit minimizing ||nablaχ ndash V||2
This reduces to solving the Poisson Equation
We can discretize the system by representing thefunctions as vectors of values at sample points Gradient divergence and Laplacian operators become
matrices
Solving the resulting linear system gives a leastsquares fit at the sample positions
nablasdot(nabla χ )=nablasdotV hArr Δ χ=nablasdotV
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Recap
The boundary of a shape is a level set of itsindicator function χ
The gradient nablaχ of χ is the normal field V at theboundary (after some smoothing which we wont go into here)
We can solve for χ by integrating the normal field hellip but in general we cant get an exact solution
since an arbitrary vector field need not be thegradient of a function (field needs to be curl-free)
So we find a least-squares fit minimizing ||nablaχ ndash V||2
Recap
So we find a least-squares fit minimizing ||nablaχ ndash V||2
This reduces to solving the Poisson Equation
We can discretize the system by representing thefunctions as vectors of values at sample points Gradient divergence and Laplacian operators become
matrices
Solving the resulting linear system gives a leastsquares fit at the sample positions
nablasdot(nabla χ )=nablasdotV hArr Δ χ=nablasdotV
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Recap
So we find a least-squares fit minimizing ||nablaχ ndash V||2
This reduces to solving the Poisson Equation
We can discretize the system by representing thefunctions as vectors of values at sample points Gradient divergence and Laplacian operators become
matrices
Solving the resulting linear system gives a leastsquares fit at the sample positions
nablasdot(nabla χ )=nablasdotV hArr Δ χ=nablasdotV
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Why cant we solve it exactly Over a non-loop 1D range (which we studied closely)
this isnt very useful ndash the gradient is invertible byintegration and we can solve the system exactly We can also do this in the discrete setting ndash the
corresponding operator matrix is invertible
But in 2 and higher dimensions the gradient is notinvertible and neither is its operator matrix Gradient maps scalar field to vector field intuitively ldquolower-
dimensionalrdquo to ldquohigher-dimensionalrdquo In 1D scalars and vectors are the same
ddx d χ
dx=V
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
A vector field (over a simply-connected region) is thegradient of a scalar function if and only if it iscurl‑free (has no circulation about any point) In other words we can solve nablaχ = V (over a simply-
connected region) if and only if nablatimesV = 0
If the region is not simply-connected even this maynot be enough
nablatimesV=( partpart x partpart y partpart z )times(V x V y V z)
Curl-free Not curl-free
Gam
asut
ra
Non-invertibility of k-D continuous operators
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Non-invertibility of k-D discrete operators
nabla(k-D discrete
gradient)
χ
= V
n columns
kn rows
n rows
kn rows
(1 row for eachcoordinate ofeach point)
Overdetermined
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
nablasdot(k-D discretedivergence)
g=V
n rows
kn columns
kn rows
(1 row for eachcoordinate ofeach point)
n rows
Non-invertibility of k-D discrete operators
Underdetermined
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Thought for the Day 1
What about the Laplacian Is it invertible
Is this over- or under-determined
nablasdotnabla(k-D discreteLaplacian)
g=n columns
n rows
n rows χn rows
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
What we have so far
Transform continuous variational problems todiscrete linear algebra problems
Solve in a least squares sense since the problemis overdetermined in higher dimensions
BUT the results are also discrete the values ofthe function f at the sampled points Solution A different type of discretization
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Galerkin Approximation
Restrict the solution space F to weighted sums ofbasis functions ie F = sumi wi Bi for some set offunctions B1 B2 Bm
Why Allows us to discretize the problem in termsof the m-D vector of weights
We will choose functions that are locally supported hellip ie each fi is non-zero only around some local region
of space This keeps the resulting linear system sparse
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Basis Functions with Local Support
A finite element model Discretize space into cells then define a basis
function centered around each cell
Instead of values at points we now have values locally around points
Wik
iver
sity
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Basis Functions with Local Support
A potential grid of cells
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Basis Functions with Local Support
A single basis function centered at a gridcell but overlapping adjacent cells
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Basis Functions with Local Support
A potential grid of cellsProblem Not enough detail where its needed (boundary) too much
detail where its not (empty space or interior)
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Basis Functions with Local Support
A hierarchical adaptive grid (octree)
Puts resolution where it matters One basis function per octree cell
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Projecting to the Finite Basis
Assume we want to reconstruct the function overrange Ω (eg [0 1] in 1D or [0 1]3 in 3D)
The original Poisson problem is Δχ = nablaV
BUT since weve now restricted our solutions tothe space spanned by Bi this equation may nothave an exact solution Solution Least squares to the rescue again
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Projecting to the Finite Basis
Solve Δχ = nablaV for χ isin F
To find the best solution within the space spannedby the basis we minimize the sum of squaredprojections onto the basis functions
where measures theprojection of function f onto basis function Bi
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
⟨ f Bi⟩=intΩf (x )Bi (x )dσ
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap
Projecting to the Finite Basis
Minimize
(skipping some algebra) This amounts tominimizing ||Lw ndash v||2 where
sumi=1
m
⟨Δ χminusnablasdotV Bi ⟩Ω2
= sumi=1
m
|⟨ Δ χ Bi ⟩minus ⟨nablasdotV Bi ⟩Ω|2
Lij = ⟨ part2Bipart x2 B j⟩+⟨ part2Bipart y2 B j⟩+⟨ part2Bi
part z2 B j ⟩v i = ⟨nablasdotV Bi ⟩
w1
w2
wm
w =
Mostly zero since mostpairs of basis functions
dont overlap