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time, the solution looks like an exponential curve, but as time increases the growth
is reduced and the equilibrium point of s
D
1 acts as a barrier for the solution. The
monotonicity explained above is also observed.
The Error is O.t/
Let us look a bit closer at the error of the numerical solutions. We consider the
problem
s
0
.t /
D
s.t /.1
s.t //;
s.0/
D
0:2;
with t ranging from 0 to 5: We want to compare the analytical solution and the two
numerical solutions at time t
D
5: The analytical solution is
0:2
0:2
C
0:8e
5
s.5/
D
0:9737555;
see (2.52). In the table below we present the error for the numerical solution
generated by the explicit scheme, y
N
; and the explicit scheme,
z
N
:
jy
N
s.5/j
t
j
z
N
s.5/j
t
t
D
N
N
200
0:0250
0:0235
0:0234
400
0:0125
0:0235
0:0234
600
0:0083
0:0235
0:0234
800
0:0063
0:0235
0:0234
1; 000
0:0050
0:0234
0:0234
We observe from the table that both
j
y
N
s.5/
j j
z
N
s.5/
j
0:0234t
and thus we know that we can compute the solution as accurately as we want simply
by choosing a sufficiently small time step t:
2.3
Exercises
Exercise 2.1.
Consider the initial value problem
r
0
.t /
D
1
C
4r.t /;
(2.53)
r.0/
D
0: