Biomedical Engineering Reference
In-Depth Information
θ
-scheme. Getting ahead of the spatial discretization in the next section, which
will lead to a set of linear differential equations, the time discretization is best
illustrated with a set of first-order linear differential equations:
d
dt
+
A
∼
=
f
∼
(
t
),
(15.11)
where
A
is a constant matrix and
f
∼
(
t
) a given column. To solve this time-dependent
problem the time domain is split into a finite number of time increments. Then,
attention is focussed on the increment with
t
n
<
t
<
t
n
+
1
=
t
n
+
t
. Assuming
that the solution
u
n
at time
t
n
is known, the unknown
u
n
+
1
at time
t
n
+
1
has to be
determined. The
θ
-scheme approximates these differential equations by
∼
n
+
1
−
∼
n
+
θ
A
∼
n
+
1
+
(1
−
θ
)
A
∼
n
=
θ
f
∼
n
+
1
+
(1
−
θ
)
f
∼
n
.
(15.12)
t
For
θ
=
0 this scheme reduces to the
Euler explicit
or
forward Euler
scheme,
while for
θ
=
1 the
Euler implicit
or
backward Euler
scheme results. Both of
these schemes are first-order accurate, i.e.
t
). This means that the accuracy
of the solution is linearly related to the size of the time step
O
(
t
. The accuracy
improves when the time step becomes smaller. For
θ
=
0.5 the
Crank-Nicholson
t
2
).
scheme results which is second-order accurate, i.e.
O
(
To illustrate the stability properties of the
θ
-method, consider a single variable
model problem:
du
dt
+
λ
u
=
f
,
(15.13)
with
λ>
0. This differential equation has the property that any perturbation to the
solution (for example induced by a perturbation of the initial value of
u
) decays
exponentially as a function of time. Assume that
u
satisfies the differential Eq.
(
15.13
) exactly, and let
u
be a perturbation of
u
, hence
u
=
u
+
u
. Consequently,
the perturbation
u
must obey
du
dt
+
λ
u
=
0.
(15.14)
If at
t
=
0, the perturbation equals
u
˜
=˜
u
0
, the solution to this equation is
e
−
λ
t
u
˜
=
u
0
.
˜
(15.15)
Clearly, if
0, the perturbation decays exponentially as a function of time.
Application of the
λ>
θ
-scheme to the single variable model problem (
15.13
) yields
u
n
+
1
−
u
n
+
θλ
u
n
+
1
+
(1
−
θ
)
λ
u
n
=
θ
f
n
+
1
+
(1
−
θ
)
f
n
.
(15.16)
t
