3469 - Practical Numerical Simulations
Mike Peardon — mjp@maths.tcd.ie
http://www.maths.tcd.ie/~mjp/3469
School of Mathematics
Trinity College Dublin
Michaelmas Term 2015
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
1 / 14
Numerical Solution of PDEs - overview
Partial differential equations occur in many (maybe most)
numerical simulations of physical systems.
This is a HUGE topic; we’ll only cover the basic ideas.
Start by deciding which are dependent and independent
variables.
. . . then decide which method to use (just like for ODEs) by
considering the boundary data.
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
2 / 14
Classification of partial differential equations
Some second order PDEs can be classified as hyperbolic,
elliptic or parabolic.
The classification is made by looking at the discriminant. For
the family of 2nd order PDEs in u(x, y ),
A
∂ 2u
∂ 2u
∂u
∂ 2u
∂u
+
B
+D
+
C
+E
+ Fu = g
2
2
∂x
∂x∂x
∂y
∂x
∂y
, ∂u ) etc. , the discriminant is
with A = A(x, y , u, ∂u
∂x ∂y
d = B2 − 4AC
Condition
d>0
d=0
d<0
Classification
Example
2
2
hyperbolic
wave eqn: ∂∂tu2 − v 2 ∂∂xu2 =0
∂
parabolic
diffusion: ∂u
− ∂x
D(x, t) ∂u
=0
∂t
∂x
2
∂ u
∂2u
elliptic
Poisson’s eqn: ∂x 2 + ∂x 2 = ρ(x, y )
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
3 / 14
Boundary data for PDEs
As with numerical solutions to ODEs, the boundary data
usually decides the method to use.
Hyperbolic and parabolic equations are usually specified with
initial value (or Cauchy) data:
For example, u(x, t) is required for t > 0 and u(x, 0) = u0 (x)
is given (perhaps with other conditions true for all t > 0).
Elliptic eqations are usually specified with boundary value
data.
For example, u(x, y ) is required for (x, y ) ∈ A, and u is
∂u
specified on the boundary, ∂A (Dirichlet) or ∂u
and ∂y
are
∂x
specified on ∂A (von Neumann) or a mixture of the two
(Robin).
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
4 / 14
Boundary data for PDEs
t
u(dA) given
u(t,x0)=f0(t)
A
u(t,x1)=f1(t)
A
dA
Start solver here
and propagate in t
dA
u(0,x)=u0(x)
Construct solution globally
x
Hyperbolic or Parabolic
Mike Peardon — mjp@maths.tcd.ie (TCD)
u(x,y) in A required
Elliptic
3469
Michaelmas Term 2015
5 / 14
Overview of methods for solving PDEs
As with ODEs, the distict boundary data types lead to two
distinct algorithms for solving PDEs.
For parabolic and hyperbolic PDEs, a standard numerical
solver
Begins with the initial data, u(x, t0 ) in memory
Applies an update algorithm to compute the solution at a small
time-step later, u(x, t0 + h)
Repeats this stepping process, to compute
u(x, t0 + 2h), u(x, t0 + 3h), . . . .
For elliptic PDEs, the numerical solver
Begins with an initial guess, u0 (x, y ), ∀(x, y ) ∈ A.
Estimates how “wrong” the guess is, and improves it, giving a
better guess to the solution, u1 (x, y ).
Repeats this refinement process, until the solution is
sufficiently accurate.
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
6 / 14
Laplace equation on a grid
Find a numerical approximation for φ(x, y ) which solves the
Laplace equation
∇2 φ = 0
for φ(x, y ) inside the unit square, x, y ∈ [0, 1], given boundary
data on the edges of the square.
First step - discretise the unit square with a grid, with spacing
a = 1/n for some integer n.
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
7 / 14
Simple iterative schemes
Want to solve Ax = b where A is sparse.
Introduce any matrix P and re-write expression
Ax = b
Px = b + (P − A)x
Now promote x to be a sequence {x 0 , x 1 , x 2 , . . . } for some
initial guess x 0 , defined by the recursive iteration
Px (k+1) = b + (P − A)x (k) ,
k = 0, 1, 2, . . .
If the sequence converges, then it converges to the solution
to Ax = b.
A good choice for P is invertible with P −1 easy to apply
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
8 / 14
Simple iterative schemes (2)
Sequence is
Px (k+1) = b + (P − A)x (k)
Define d k to be the difference between successive entries in
the sequence.
d (k) = x (k+1) − x (k)
k = 0, 1, 2, . . .
Then sequence for d becomes
Pd (k+1) = (P − A)d (k)
k = 0, 1, 2, . . .
so if P is invertible can solve for this recursion
k = 0, 1, 2, . . .
d (k) = (I − P −1 A)k d (0)
Sequence converges if d → 0 so if
Eigenvalues of this matrix inside unit disk: |λ(I − P −1 A)| < 1
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
9 / 14
The Jacobi iteration
Perhaps simplest choice for P is the diagonal elements of A.
P = D. If none of these are zero, this is easy to invert. This
simple iteration is called...
Jacobi iteration: P = D
x (k+1) = D −1 b − (I − D −1 A)x (k)
For the Laplace equation, the iteration can be written
1 (k)
(k)
(k)
(k)
+
φ
+
φ
+
φ
φ(k+1)
=
φ
x,y
x+1,y
x,y −1
x,y +1
4 x−1,y
The new value of the field at a site is given by the average of
the four neighbours in the old iteration.
For Dirichlet boundaries, Laplace solver always converges
(for simple rectangular geometries, this can be seen using
Fourier analysis).
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
10 / 14
The Gauss-Seidel iteration
With iterative sequence
Px (k+1) = b + (P − A)x (k)
and writing A = D + L + U with D the diagonal part of A, and
L, U the upper/lower components, choose
P =D+L
and the iteration is now
Gauss-Seidel iteration:
x (k+1) = (D + L)−1 b − (D + L)−1 U x (k)
Applying the inverse of P = D + L is easily done using
forward substitution.
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
11 / 14
Gauss-Seidel iteration and Laplace’s equation
What does this look like for Laplace’s equation?
(D + L)x (k+1) = b − U x (k)
To define which hopping terms appear in L and U means
specifying a site ordering. With the “natural” site order, GS
becomes
1 (k+1)
(k)
(k)
φ x−1,y + φ(k+1)
x,y −1 + φ x+1,y + φ x,y +1
4
This has a simple interpretation as an algorithm; loop over
sites in order and over-write φ with the average of the four
nearest neighbours.
φ(k+1)
=
x,y
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
12 / 14
Successive over-relaxation
Gauss-Seidel has better convergence than Jacobi, but simple
trick improves further. Introduce parameter ω and write
1
P = D+L
ω
this gives the iteration
Successive over-relaxation
(D + ωL) x (k+1) = ωb − (ωU + (ω − 1)D) x (k)
Idea is to move further along current update direction.
For Laplace problem, the iteration becomes
ω (k+1)
(k)
(k+1)
(k)
(k+1)
φx,y =
φx−1,y + φx+1,y + φx,y −1 + φx,y +1 + (1 − ω)φ(k)
x,y
4
ω = 1 is the Gauss-Seidel iteration, 0 < ω < 2 converges for
large grid. ω close to 2 best.
Mike Peardon — mjp@maths.tcd.ie (TCD)
3469
Michaelmas Term 2015
13 / 14
Von Neumann boundary data
Von Neumann boundary
condition constrains
derivative of φ normal to the
boundary.
δφ.n
How can this be
implemented for our mesh?
Rectangular domain: with a
simple mesh. At one edge,
can represent derivative
perpendicular using another
Taylor approx.
n
For all points on the
boundary, if n is a vector
normal to the surface, then
∂φ · n is given along the
surface.
Mike Peardon — mjp@maths.tcd.ie (TCD)
φ0,y
3469
φ1,y
φ2,y
Michaelmas Term 2015
14 / 14