UNIVERSITY OF NOTRE DAME
Department of Civil Engineering
and Geological Sciences
CE 30125
November 24, 2010
J.J. Westerink
Due: November 30, 2010
Homework Set # 7 - Group Project 1
We will consider a pollutant advecting and diffusing down a river that is 500 m long. Assume that the pollutant
is distributed as a Gaussian distribution with a standard deviation equal to
m and an initial location equal
to
x
0
=100 m. We must solve the advection-diffusion
equation to solve this problem.
i.c.’s
b.c.’s
The exact solution to this problem is:
We want to find out what the distribution looks like after 250 seconds from the initial time. Let us solve this
problem using a second order accurate central difference approximation in space and a first order accurate
forward approximation in time (Explicit in time, central in space solution), a first order backward approxima-
tion in time (Implicit in time, central in space solution) and a second order accurate Crank-Nicolson approxi-
mation in time (Central in time at the half step, central in space solution). We will apply the exact solution to
compute the initial and boundary conditions.
Corbitt has been kind enough to code this up for us in Matlab in ADDIF.m
The code input parameters are grid
size
, time step
, the standard deviation of the Gaussian distribution
, the current speed
, the diffusion
coefficient
, the implicit fraction
(equal to 0 for and explicit scheme, 0.5 for a Crank-Nicolson scheme,
and 1.0 for a fully implicit scheme).
Note that the code also computes the Courant number,
, a
measure of the how far information propagates across a space step during a time step; the Peclet number,
, a measure of the ratio of advection to diffusion; the simulation time on your computer; the
L
2
discrete error norm,
;
and the
error norm,
σ
5
=
∂
C
∂
t
------
U
∂
C
∂
x
+
D
∂
2
C
∂
x
2
---------
=
0
x
500
≤≤
Cxt
,
()
xx
0
–
2
2
σ
2
×
--------------------
–
⎝⎠
⎜⎟
⎛⎞
exp
=
0
x
500
C
0
t
,
σ
σ
2
2
Dt
×
×
+
-------------------------------------
0
x
0
–
Ut
×
–
2
2
σ
2
4
×
+
×
--------------------------------------
–
exp
×
=
C
500
t
,
σ
σ
2
2
×
×
+
500
x
0
–
×
–
2
2
σ
2
4
×
+
×
--------------------------------------------
–
exp
×
=
,
σ