Environmental Engineering Reference
In-Depth Information
to name the most important ones, and cellular automata, just to name a more
nowadays still exotic example. The specific branch of mathematics that deals
with this type of solutions is
numerics
- which by the way concerns more than
the solution of odes or pdes.
A first example of a numerical method was already given in Chap. 4.2. However
that worked for 1D only. In this chapter we show, how numerical methods can be
used to solve problems in higher dimensions. The pdepe solver was described in
Chap. 4.4. In fact there the MathWorks programmers have included sophisticated
knowledge from numerics, which cannot be outlined at this place. However, also
pdepe works for 1D spatial problems only. The MATLAB
®
finite element toolbox
may be used for higher dimensional spaces. However, in this topic it was intended
to use core MATLAB
only, and thus we show how here how to do it.
Starting point for a numerical solution is the differential equation itself. The
equality, given in the equation is in some way mimicked by a numerical algorithm.
The method of that mimicing makes the difference for most of the different
numerical methods mentioned above. In order to make the method powerful the
approximate mimicking of the differential equation is performed in small parts of
the model region and the boundary. I.e. the model region and/or the boundary have
to be sub-divided in many small parts, which may be called blocks, cells, elements
or volumes, depending on the method. This process of discretization is outlined in a
first introductory example.
®
21.1
Introductory Example
The differential equation
@c
@t
¼
l
c
(21.1)
is the simplest one to describe the decay or degradation of a chemical specie due to
whatever processes.
c
is the concentration and
the decay constant. This type of
problem was already introduced in Chap. 1 (with a positive coefficient it describes
exponential growth). There is a well-known analytical solution for that differential
equation for the case when the concentration is known at an initial time, i.e. for the
condition
l
cðt ¼
0
Þ¼c
0
(21.2)
with a concentration
c
0
. The solution is:
cðtÞ¼c
0
exp
ð
l
tÞ
(21.3)
