Geography Reference
In-Depth Information
interval is divided into J subintervals of length δx, then ψ(x) can be approxi-
mated by a set of J
+
1 values as
j
=
ψ(jδx), which are the values of the field
at the J
L/J .
Provided that δx is sufficiently small compared to the scale on which ψ varies,
the J
+
1 grid points given by x
=
jδx,j
=
0, 1, 2,...,J, where δx
=
1 grid point values should provide good approximations to ψ(x) and its
derivatives.
The limits of accuracy of a finite difference representation of a continuous vari-
able can be assessed by noting that the field can also be approximated by a finite
Fourier series expansion:
+
a
m
cos
2πmx
L
J/2
a
0
2
+
2πmx
L
ψ (x)
=
+
b
m
sin
(13.1)
m
=
1
1
coefficients in (13.1). That is, it is possible to determine a
0
plus a
m
and b
m
for
wave numbers m
The available J
+
1 values of
j
are just sufficient to determine the J
+
1, 2, 3, ..., J/2. The shortest wavelength component in
(13.1) has wavelength L/m
=
2δx. Thus, the shortest wave that can be
resolved by finite differencing has a wavelength twice that of the grid increment.
Accurate representation of derivatives is only possible, however, for wavelengths
much greater than 2δx.
We now consider how the grid point variable
j
can be used to construct a finite
difference approximation to a differential equation. That is, we wish to represent
derivatives such as dψ/dx and d
2
ψ/dx
2
in terms of the finite difference fields.
We first consider the Taylor series expansions about the point x
0
:
=
2L/J
=
O
(δx)
4
(13.2)
ψ
(x
0
)
(δx)
2
2
ψ
(x
0
)
(δx)
3
6
ψ
(x
0
) δx
ψ (x
0
+
δx)
=
ψ (x
0
)
+
+
+
+
O
(δx)
4
(13.3)
where the primes indicate differentiation with respect to x and O[(δx)
4
] means
that terms with order of magnitude (δx)
4
or less are neglected.
Subtracting (13.3) from (13.2) and solving for ψ
ψ
(x
0
)
(δx)
2
2
ψ
(x
0
)
(δx)
3
6
ψ
(x
0
) δx
ψ (x
0
−
δx)
=
ψ (x
0
)
−
+
−
+
(x) give a finite difference
expression for the first derivative of the form
O
(δx)
2
ψ
(x
0
)
=
[ψ (x
0
+
δx)
−
ψ (x
0
−
δx)] / (2δx)
+
(13.4)
while adding the same two expressions and solving for ψ
(x) give a finite differ-
ence expression for the second derivative of the form
O
(δx)
2
(13.5)
ψ
(x
0
)
δx)] / (δx)
2
=
[ψ (x
0
+
δx)
−
2ψ (x
0
)
+
ψ (x
0
−
+
