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
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