Review of accuracy analysis
Euler:
Local error = O(h2)
Global error = O(h)
Runge-Kutta Order 4:
Local error = O(h5)
Global error = O(h4)
But there’s more to worry about:
stability and convergence
Stability & Convergence
Stability: Suppose we perturb initial
condition by ε. Then 1) effect  0 as
ε  0, and 2) effect grows only
polynomially fast as h  0.
Convergence: Solution of
discrete problem  solution of
continuous problem as h  0.
Stability & Convergence
For Ordinary Differential Equations
(ODEs), Stability ↔ Convergence
But for Partial Differential Equations
(PDEs), where there are more than one
variable--- time and space, for example--Stability and convergence are not
equivalent. We require an additional
condition.
Lax’s Theorem
Consistent: A finite-difference
scheme is consistent if the local
truncation error  0 as the grid size
 0. (Not always true for PDEs, as
we shall see.)
Lax’s Theorem: If a finite-difference
scheme for an initial-value PDE is
consistent, then
Stability ↔ Convergence
PDEs
Partial differential equations are at the
very heart of many sciences, and
provide our best understanding of the
way the world works. Some examples:
Quantum mechanics; propagation of waves
of all kinds; elasticity; diffusion of particles,
population, prices, information; spread of
heat; electrostatic field; magnetic fields,
fluid flow; etc., etc., etc.
To see the power…
Suppose you work for the government
and your job is to worry about the
possibility of terrorist nuclear weapons.
What is the critical mass of U92235 (“25”)?
The following material was classified, but
is now public: see The Los Alamos
Primer: The first lectures on how to build
an atomic bomb, R. Serber, Univ. of Calif.
Press, Berkeley, 1992. QC773 .A1S47
Simplest model of neutron diffusion
Laplace operator:
2
2
2

f

f

f
2
 f ( x, y , z )  2  2  2
x
y
z
In spherical coordinates:
1  2 f
 f  2 (r
)  terms in  ,
r r
r
2
So for spherically symmetric systems:
1  2 f
 f  2 (r
)
r r
r
2
Consider a sphere of “25”
Let N(t,x,y,z) be the number of
neutrons in a tiny cube and
consider the net growth of N at
any given point in space and any
particular time:
N
 1
2
 D N 
N
t

Rate of change of
neutron flux
Diffusion influx
fission
Consider a sphere of “25”
where


= mean time between fissions
= avg. no. of neutrons
produced per fission
D = diffusion constant
Separation of variables: an
important technique
 t / 
N  N1 ( x, y, z )e
where   = effective neutron number
Leads to
  1  
 N1  
N1
D
2
Separation of variables: an
important technique
For sphere of radius R, can check solution
sin r / R
N1 
r
With the boundary condition
    1 
 D
2
R
2
So critical mass is determined by
 D
R 
 1
2
2
Answers
For Uranium:
 2 D  220
  2.3
Rcritical  13.5 cm
Masscritical  200 kg
More accurate boundary condition gives
56 kg, and thick U tamper gives 15 kg
Little Boy