Jacobs University Bremen
Joachim Vogt
Earth and Planetary Physics
— Sample solutions —
Spring 2009
6
6.1
Aspects of geophysical potential theory
Upward continuation in cartesian coordinates
The Laplace equation is fulfilled if
∂2
∂2
0 = ∇ Φ =
+
Φ0 e−y/H cos(2πx/λ)
∂x2 ∂y 2
= −(2π/λ)2 + 1/H 2 Φ0 e−y/H cos(2πx/λ) = −(2π/λ)2 + 1/H 2 Φ
2
which yields −(2π/λ)2 + 1/H 2 = 0 and thus
λ
.
2π
H =
6.2
Planar harmonics in cartesian coordinates (Q)
Consider the complex function
(a) With z = x + iy, the function w2 = z 2 can be expressed in terms of x and y as follows:
w2 = (x + iy)2 = x2 − y 2 + 2ixy ,
hence the real part u2 = x2 − y 2 , and the imaginary part v2 = 2xy.
(b) We apply the two-dimensional Laplace operator to u2 and obtain
∇2 u2 =
∂2
∂2
+
∂x2 ∂y 2
2
x − y2 = 2 − 2 = 0 .
1
6.3
Differential operators in spherical coordinates (Q)
If the function Φ is given by
3
R
3 cos2 ϑ − 1
Φ(r, ϑ) = Φ0
,
r
2
then
∂Φ
∂r
∂Φ
∂ϑ
∂Φ
∂ϕ
= Φ0 R3 (−3) r−4
3 cos2 ϑ − 1
,
2
= Φ0 R3 r−3 3(− sin ϑ) cos ϑ ,
= 0.
The gradient in spherical coordinates can be expressed as
∇Φ =
∂Φ
1 ∂Φ ˆ
1 ∂Φ
ˆ
r+
ϑ+
ϕ
ˆ,
∂r
r ∂ϑ
r sin ϑ ∂ϕ
thus,
3 −4
B = −∇Φ = 3Φ0 R r
3 cos2 ϑ − 1
ˆ
ˆ
r + cos ϑ sin ϑ ϑ
2
which means that
Br = 3Φ0 R3 r−4
Bϑ = 3Φ0 R3 r−4
3 cos2 ϑ − 1
,
2
cos ϑ sin ϑ .
The divergence of B is given by
∇·B =
=
1 ∂
1
∂
1 ∂Bϕ
r2 Br +
(sin ϑBϑ ) +
2
r ∂r
r sin ϑ ∂ϑ
r sin ϑ ∂ϕ
2
3 cos ϑ − 1 ∂
1
3Φ0 R3
r−2
r2
2
∂r
| {z }
=−2r−3
+
∂
1
3Φ0 R3 r−4
sin2 ϑ cos ϑ
r sin ϑ
|∂ϑ
{z
}
=sin ϑ(3 cos2 ϑ−1)
3 −5
= 3Φ0 R r
6.4
2
−(3 cos ϑ − 1) + (3 cos2 ϑ − 1) = 0 .
Planar harmonics in polar coordinates
(a) Applying the first part (derivatives in r) of the Laplace operator to un = rn cos nϕ yields
1 ∂
∂un
1 ∂
∂(rn cos nϕ)
1 ∂
r
=
r
=
r(nrn−1 cos nϕ)
r ∂r
∂r
r ∂r
∂r
r ∂r
1 2 n−1
n2 un
=
n r
cos nϕ = n2 rn−2 cos nϕ =
.
r
r2
2
In a similar way, we find
1 ∂
r ∂r
∂vn
r
∂r
=
n2 vn
.
r2
Now we apply the second part (derivatives in ϕ) of the Laplace operator to yield
1 ∂ 2 un
r2 ∂ϕ2
=
=
1 ∂ 2 (rn cos nϕ)
1 ∂(−nrn sin nϕ)
=
r2
∂ϕ2
r2
∂ϕ
n2 un
1
2 n
2 n−2
−n
r
cos
nϕ
=
−n
r
cos
nϕ
=
−
,
r2
r2
and, analogously,
1 ∂ 2 vn
n2 vn
=
−
.
r2 ∂ϕ2
r2
Hence
2
1 ∂
=
r ∂r
∂vn
1 ∂ 2 vn
n2 vn n2 vn
r
+ 2
=
− 2 = 0.
∂r
r ∂ϕ2
r2
r
and also
∇ vn
∂un
r
∂r
1 ∂ 2 un
n2 un n2 un
=
− 2 = 0,
r2 ∂ϕ2
r2
r
∇ un
1 ∂
=
r ∂r
2
+
(b) The elementary functions un or vn tend to zero in the limit r → ∞ for n < 0. Since
sin(−2ϕ) = −2 sin ϕ cos ϕ
we can write
fR (ϕ) = sin ϕ cos ϕ = −
sin(−2ϕ)
R2
. = −
v−2 (r = R, ϕ) .
2
2
Hence the function
f (r, ϕ) = −
R2
R2 sin(−2ϕ)
R2 sin 2ϕ
v−2 (r, ϕ) = −
=
2
2r2
2r2
goes to zero as r tends to infinity, solves the Laplace equation in the exterior of the circle
r = R, and satisfies the boundary condition f (r = R, ϕ) = fR (ϕ).
(c) The elementary functions un or vn are regular at the origin for n ≥ 0. Since
cos 2ϕ = cos2 ϕ − sin2 ϕ = 2 cos2 ϕ − 1 ,
we can write
fR (ϕ) = cos2 ϕ =
1
1
cos 2ϕ + 1
=
u2 (r = R, ϕ) + u0 (r = R, ϕ) .
2
2
2R
2
Hence the function
f (r, ϕ) =
1
1
r2 cos 2ϕ + R2
u
(r,
ϕ)
+
u
(r,
ϕ)
=
2
0
2R2
2
2R2
is regular at the origin, solves the Laplace equation in the interior of the circle r = R, and
satisfies the boundary condition f (r = R, ϕ) = fR (ϕ).
3
6.5
Planar harmonics in cartesian coordinates (E)
(a) For n = 0 and n = 1 we obtain w0 = 1 and w1 = x + iy and
u0 = 1 ,
v0 = 0 ,
u1 = x ,
v1 = y .
Since (x + iy)2 = x2 − y 2 + i2xy, we get for n = 2:
u2 = x2 − y 2 ,
v2 = 2xy .
Let z¯ = x − iy denote the complex conjugate of z = x + iy. Since
z¯
z¯
x − iy
1
=
=
= 2
,
2
z
z z¯
|z|
x + y2
w−1 =
we obtain for n = −1:
x
,
+ y2
y
= − 2
.
x + y2
u−1 =
v−1
x2
Finally, n = −2 gives
w−2 =
1
z¯2
x2 − y 2 − i2xy
=
=
z2
|z|4
(x2 + y 2 )2
and thus
x2 − y 2
,
(x2 + y 2 )2
xy
= −2 2
.
(x + y 2 )2
u−2 =
v−2
(b) The function u2 = x2 − y 2 yields the correct values at the boundary y = 0. Since v2 (x, y =
0) = 0, it can be added to u2 and the boundary condition is still satisfied. This means
that, both, u2 and also u2 + v2 are harmonic functions in the half-plane y ≥ 0 that satisfy
the boundary condition at y = 0.
6.6
Surface spherical harmonics
Recursion formula for Legendre polynomials:
(n + 1)Pn+1 − (2n + 1)µPn + nPn−1 = 0 .
4
(a) Application of the recursion formula and differentiation yields
P4 (µ) =
dP4
dµ
=
The normalization factor p14 =
r
P41 (µ)
=
1
35µ4 − 30µ2 + 3
and
8
1
5
35µ3 − 15µ =
7µ3 − 3µ .
2
2
p
1/10, and the associated Legendre function
5
(1 − µ2 )1/2 7µ3 − 3µ =
8
r
5
(1 − µ2 )1/2 7µ µ +
8
r !
r !
3
3
µ−
.
7
7
The surface spherical harmonic with A14 = 1 und B41 = 0 is given by
r
Y41 (ϑ, λ)
=
5
sin ϑ 7 cos3 ϑ − 3 cos ϑ cos λ =
8
The zero lines are λ = 90◦ , ϑ = 90◦ und ϑ = ± arccos
be found at sin λ = 0 (i.e., λ = 0◦ or λ = 180◦ ) and
r
51
sin 2ϑ (7 cos 2ϑ + 1) cos λ .
84
p
3/7 ' 90◦ ± 40.89◦ . Extrema can
∂ϑ 1
1 5
P4 = √
14 cos2 2ϑ + cos 2ϑ − 7 = 0 .
∂t
10 4
The corresponding ϑ values are the solution of a quadratic equation, one obtains
ϑext
1
= arccos
2
!
√
−1 ± 393
,
28
therefore, ϑext = 23.88◦ , 69.02◦ , 110.98◦ , 156.12◦ . The surface spherical harmonic is antisymmetric with respect to the equator ϑ = 90◦ and symmetric with respect to the zero
meridian λ = 0◦ . A few contour lines are shown in the figure below.
5
6