Laplace's Equation on a Sphere - Semantic Scholar
Laplace’s Equation on a Sphere J. Robert Buchanan Department of Mathematics Millersville University P.O. Box 1002, Millersville, PA 17551–0302 [email protected] April 6, 2006
Problem Description Consider the heat equation on the sphere of radius 2. ut = k(uxx + uyy + uzz ) u(x, y, z, 0) = f (x, y, z) u(x, y, z, t) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4, t > 0
Find the steady state temperature distribution in the sphere. The temperature distribution on the surface of the sphere is shown in the plot below. y
2 1
0 -1 -2 2
1 z
0 -1 -2 -2 -1 0 1
x
2
1
Problem Solution At steady state, ut = 0 and thus we must solve Laplace’s equation on the sphere. 0 = uxx + uyy + uzz u(x, y, z) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4
Since the physical domain of the problem is a sphere, it will be to our advantage to convert the Laplacian and the boundary conditions to spherical coordinates. We will adopt the physical rather than mathematical convention of labeling the azimuthal angle φ and the latitudinal angle θ. Thus to convert from Cartesian to spherical coordinates we will make use of the following equations. x = ρ sin θ cos φ y = ρ sin θ sin φ z = ρ cos θ In spherical coordinates the Laplacian operator u xx + uyy + uzz = ∆u has the form ∂ ∂2u 1 ∂u 1 1 ∂ 2 ∂u , ρ + 2 sin θ + 2 2 ∆u = 2 ρ ∂ρ ∂ρ ρ sin θ ∂θ ∂θ ρ sin θ ∂φ2 where the coordinate ρ is the distance from the origin. The details of the conversion may be found in Appendix A. Since Laplace’s equations is linear and homogeneous, we may attempt to solve it using the method of separation of variables. To begin we will assume that u(ρ, θ, φ) = R(ρ)T (θ)P (φ) and substitute this expression in Laplace’s equation. This produces 1 1 ∂ 1 ∂ ρ2 R 0 T P + 2 sin θT 0 RP + 2 2 RT P 00 = 0 2 ρ ∂ρ ρ sin θ ∂θ ρ sin θ where the “prime” notation denotes the appropriate ordinary derivative. Multiplying both sides of 2 sin2 θ this equation by ρ RT P yields the equation sin θ ∂ P 00 sin2 θ ∂ ρ2 R 0 + sin θT 0 + = 0. R ∂ρ T ∂θ P Moving the functions depending only on φ to the right-hand side of the equations allows us to separate the φ and (ρ, θ) variables. sin θ ∂ P 00 sin2 θ ∂ ρ2 R 0 + sin θT 0 = − R ∂ρ T ∂θ P Since the left-hand side of the equation depends only on (ρ, θ) and the right-hand side depends only on φ, both sides of the equation must equal a constant.
2
φ-Dependent Factor of u Since the solution to Laplace’s equation must be 2π-periodic in φ the constant must be of the form n2 where n ∈ N ∪ {0}. Hence P 00 P P 00 + n2 P −
= n2 = 0
Pn (φ) = An cos nφ + Bn sin nφ. See [1, Chap. 3] for the classical method of solving a second-order, linear, homogeneous ordinary differential equation of the type shown above. Now we turn to the task of solving for the remaining two factors of the solution u to Laplace’s equation.
ρ-Dependent Factor of u The partial solution Pn (φ) found in the previous section imposes the following constraint on the solutions for ρ and θ. sin θ ∂ sin2 θ ∂ ρ2 R 0 + sin θT 0 = n2 , R ∂ρ T ∂θ
where n ∈ N ∪ {0}. As a next step we divide both sides of the equation by sin 2 θ to obtain ∂ 1 1 ∂ ρ2 R 0 + sin θT 0 = R ∂ρ T sin θ ∂θ 1 ∂ ρ2 R 0 = R ∂ρ
n2 sin2 θ n2 1 ∂ 0 2 − T sin θ ∂θ sin θT . sin θ
We see that the left-hand side of the equation depends only on ρ while the right-hand side depends only on θ. Thus both sides are equal to a constant. If the constant has the form m(m + 1) where m ∈ N ∪ {0} then the following ordinary differential equation is implied by the left-hand side of the last equation. 1 ∂ ρ2 R0 = m(m + 1) R ∂ρ 2 00 0 ρ R + 2ρR − m(m + 1)R = 0 The latter equation is known as Euler’s Equation (see [1, Sec. 5.5]). The solution to this equation has the form R(ρ) = Cρm + Dρ−m−1 . Since the solution to Laplace’s equation must be bounded at the origin (where ρ = 0) we must assume D = 0 and thus the ρ-dependent factor of the solution has the form R m (ρ) = ρm with m ∈ N ∪ {0}. Now we proceed to the task of finding the θ-dependent factor of the solution to Laplace’s equation.
3
θ-Dependent Factor of u The solutions found for R(ρ) and P (φ) have imposed two constraints on the differential equation for T (θ). It now has the form n2 ∂ 1 0 = m(m + 1), 2 − T sin θ ∂θ sin θT sin θ which is equivalent to the ordinary differential equation n2 cos θ 0 00 = 0. T + m(m + 1) − T + sin θ sin2 θ We can put this equation in a more convenient form for solving by making the change of variable w = cos θ. In this case, dT dθ d2 T dθ 2
dT and dw dT d2 T = − cos θ + sin2 θ 2 . dw dw = − sin θ
Substituting these expressions in the previous ordinary differential equation 2 dT n2 2 d T sin θ 2 − 2 cos θ + m(m + 1) − T dw dw sin2 θ dT n2 d2 T 2 + m(m + 1) − T (1 − cos θ) 2 − 2 cos θ dw dw 1 − cos2 θ 2 dT n2 2 d T (1 − w ) 2 − 2w + m(m + 1) − T dw dw 1 − w2
gives us = 0 = 0 = 0.
This last ordinary differential equation is known as the associated Legendre differential equation n (w) = T n (cos θ) with m ∈ N ∪ {0} and (see [3]). The solutions will be denoted by the functions T m m 0 ≤ n ≤ m (see [4]). Now we may summarize the product solution.
General Solution Combining the three factors of the product solution we have n n unm (ρ, θ, φ) = ρm Tm (cos θ)(Anm cos nφ + Bm sin nφ)
for m ∈ N ∪ {0} and 0 ≤ n ≤ m. Thus the series solution of Laplace’s equation on the sphere can be written as m ∞ X X n n ρm Tm (cos θ)(Anm cos nφ + Bm sin nφ). u(ρ, θ, φ) = m=0 n=0
n so that the boundary conditions can be The next step is to choose the coefficients A nm and Bm satisfied.
4
Boundary Conditions The boundary of the sphere is kept at a temperature distribution described by u(x, y, z) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4.
In spherical coordinates these boundary conditions are equivalent to u(2, θ, φ) = 6 + 6 cos2 θ − 2(1 − cos2 θ) cos 2φ for 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π. The details of this coordinate conversion are found in Appendix B. Rather than try to use the orthogonality property of the eigenfunctions, we will use the fact that the boundary condition is a finite sum of eigenfunctions and thus we merely need match the coefficients at the boundary. The first several associated Legendre functions are listed in the matrix below. T00 (w) = 1 p T10 (w) = w T11 (w) = − 1 − w2 p 1 T20 (w) = (3w2 − 1) T21 (w) = −3w 1 − w2 T22 (w) = 3(1 − w 2 ) 2 Thus by writing out the first several terms of the solution evaluated when ρ = 2 we get u(2, θ, φ) = =
m ∞ X X
n n 2m Tm (cos θ)(Anm cos nφ + Bm sin nφ)
m=0 n=0 A00 T00 (cos θ) + 2A01 T10 (cos θ) + 2T11 (cos θ)(A11 cos φ + B11 sin φ) + 4A02 T20 (cos θ) + 4T21 (cos θ)(A12 cos φ + B21 sin φ) + 4T22 (cos θ)(A22
cos 2φ + B22 sin 2φ)
+ ··· We can see by the absence of sin nφ terms in the spherical coordinate form of the boundary conditions that Bij = 0 for all i and j. Absent as well are any powers of cos θ larger than 2, thus A m n =0 for n > 2. Thus we have u(2, θ, φ) = A00 + 2A01 cos θ − 2A11 sin θ cos φ + 2A02 (3 cos2 θ − 1) − 12A12 cos θ sin θ cos φ + 12A22 (1 − cos2 θ) cos 2φ. Simplifying one step at a time we can set A 01 = A11 = A12 = 0. This will leave, u(2, θ, φ) = A00 + 2A02 (3 cos2 θ − 1) + 12A22 (1 − cos2 θ) cos 2φ = A00 − 2A02 + 6A02 cos2 θ + 12A22 (1 − cos2 θ) cos 2φ
and hence if we choose A02 = 1, A00 = 8, and A22 = − 61 we have satisfied the boundary conditions. Therefore the solution to Laplace’s equation on the sphere may be written as 1 1 u(ρ, θ, φ) = 8 + ρ2 (3 cos2 θ − 1) − ρ2 (1 − cos2 θ) cos 2φ. 2 2
5
Plots of Solution
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The following sequence of frames illustrates the φ-dependency of the solution for 0 ≤ φ ≤ steps of size ∆φ = π4 .
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The following sequence of frames illustrates the θ-dependency of the solution for 0 ≤ θ ≤ steps of size ∆θ = π8 .
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References [1] William E. Boyce and Richard C. DiPrima, Elementary Differential Equations and Boundary Value Problems, 7th edition, John Wiley & Sons, Inc., New York 2001. [2] Robert T. Smith and Roland B. Minton, Calculus, 2nd edition, McGraw-Hill Publishing Company, Boston, MA 2002. [3] Eric W. Weisstein, “Legendre Differential Equation”, From MathWorld–A Wolfram Web Resource, http://mathworld.wolfram.com/LegendreDifferentialEquation.html [4] Daniel Zwillinger (Ed.), CRC Standard Mathematical Tables and Formulae, CRC Press, Boca Raton, FL 1995.
A
Spherical Laplacian
The Laplacian operator in spherical coordinates can be derived by use of the chain rule for multivariable functions (see [2, Sec. 12.5]). Suppose that u ≡ u(ρ, θ, φ) then the first partial derivatives of u are uρ = u x xρ + u y yρ + u z zρ = ux sin θ cos φ + uy sin θ sin φ + uz cos θ uθ = u x xθ + u y yθ + u z zθ = ux ρ cos θ cos φ + uy ρ cos θ sin φ − uz ρ sin θ uφ = u x xφ + u y yφ + u z zφ = −ux ρ sin θ sin φ + uy ρ sin θ cos φ In a similar fashion we may find the three (out of the nine total) second partial derivatives. uρρ = uxx sin2 θ cos2 φ + 2uxy sin2 θ cos φ sin φ + 2uxz cos θ sin θ cos φ + uyy sin2 θ sin2 φ + 2uyz cos θ sin θ sin φ + uzz cos2 θ uθθ = uxx ρ2 cos2 θ cos2 φ + 2uxy ρ2 cos2 θ cos φ sin φ − 2uxz ρ2 cos θ sin θ cos φ + uyy ρ2 cos2 θ sin2 φ − 2uyz ρ2 cos θ sin θ sin φ + uzz ρ2 sin2 θ − ux ρ sin θ cos φ − uy ρ sin θ cos φ − uz ρ cos θ uφφ = uxx ρ2 sin2 θ sin2 φ − 2uxy ρ2 sin2 θ cos φ sin φ + uyy ρ2 sin2 θ cos2 φ − ux ρ sin θ cos φ − uy ρ sin θ sin φ Now we can verify that uxx + uyy + uzz
1 ∂ = 2 ρ ∂ρ
ρ
2 ∂u
∂ρ
∂ 1 + 2 ρ sin θ ∂θ
9
∂u sin θ ∂θ
+
∂2u 1 . ρ2 sin2 θ ∂φ2
We start with the right-hand side. 1 ∂ ∂ ∂2u 1 ∂u 1 2 ∂u ρ + sin θ + 2 2 2 2 ρ ∂ρ ∂ρ ρ sin θ ∂θ ∂θ ρ sin θ ∂φ2 1 cos θ 1 2 uθ + 2 2 uφφ = uρρ + uρ + 2 uθθ + 2 ρ ρ ρ sin θ ρ sin θ = uxx sin2 θ cos2 φ + 2uxy sin2 θ cos φ sin φ + 2uxz cos θ sin θ cos φ + uyy sin2 θ sin2 φ 2 + 2uyz cos θ sin θ sin φ + uzz cos2 θ + (ux sin θ cos φ + uy sin θ sin φ + uz cos θ) ρ 2 2 2 + uxx cos θ cos φ + 2uxy cos θ cos φ sin φ − 2uxz cos θ sin θ cos φ + uyy cos2 θ sin2 φ 1 − 2uyz cos θ sin θ sin φ + uzz sin2 θ − (ux sin θ cos φ + uy sin θ cos φ + uz cos θ) ρ cos θ + 2 (ux ρ cos θ cos φ + uy ρ cos θ sin φ − uz ρ sin θ) + uxx sin2 φ ρ sin θ 1 (ux cos φ + uy sin φ) − 2uxy cos φ sin φ + uyy cos2 φ − ρ sin θ = uxx sin2 θ + cos2 θ cos2 φ + sin2 φ + uyy sin2 θ + cos2 θ sin2 φ + cos2 φ + uzz cos2 θ + sin2 θ + 2uxy sin2 θ + cos2 θ cos φ sin φ − cos φ sin φ + 2uxz (cos θ sin θ cos φ − cos θ sin θ cos φ)
1 cos2 θ (ux cos φ + uy sin φ) − uz cos θ ρ sin θ ρ 1 1 + (ux sin θ cos φ + uy sin θ sin φ + uz cos θ) − (ux cos φ + uy sin φ) ρ ρ sin θ 1 − sin2 θ 1 = uxx + uyy + uzz + (ux cos φ + uy sin φ) + (ux sin θ cos φ + uy sin θ sin φ) ρ sin θ ρ 1 − (ux cos φ + uy sin φ) ρ sin θ 1 sin θ (ux cos φ + uy sin φ) + (ux sin θ cos φ + uy sin θ sin φ) = uxx + uyy + uzz − ρ ρ = uxx + uyy + uzz + 2uyz (cos θ sin θ sin φ − cos θ sin θ sin φ) +
10
B
Spherical Boundary Conditions u(x, y, z) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4
u(2, θ, φ) = (2 sin θ cos φ)2 + 2(2 sin θ sin φ)2 + 3(2 cos θ)2 = 4 sin2 θ cos2 φ + 2 sin2 θ sin2 φ + 3 cos2 θ
= 4 [1 − cos2 θ] cos2 φ + 2[1 − cos2 θ] sin2 φ + 3 cos2 θ = 4 [3 − cos2 φ − 2 sin2 φ] cos2 θ + 2 sin2 φ + cos2 φ = 4 [2 − sin2 φ] cos2 θ + 1 + sin2 φ 1 − cos 2φ 1 − cos 2φ 2 = 4 [2 − ] cos θ + 1 + 2 2
= 2(3 + cos 2φ) cos 2 θ + 2(3 − cos 2φ) = 6 + 6 cos2 θ − 2(1 − cos2 θ) cos 2φ
11
Problem Description Consider the heat equation on the sphere of radius 2. ut = k(uxx + uyy + uzz ) u(x, y, z, 0) = f (x, y, z) u(x, y, z, t) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4, t > 0
Find the steady state temperature distribution in the sphere. The temperature distribution on the surface of the sphere is shown in the plot below. y
2 1
0 -1 -2 2
1 z
0 -1 -2 -2 -1 0 1
x
2
1
Problem Solution At steady state, ut = 0 and thus we must solve Laplace’s equation on the sphere. 0 = uxx + uyy + uzz u(x, y, z) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4
Since the physical domain of the problem is a sphere, it will be to our advantage to convert the Laplacian and the boundary conditions to spherical coordinates. We will adopt the physical rather than mathematical convention of labeling the azimuthal angle φ and the latitudinal angle θ. Thus to convert from Cartesian to spherical coordinates we will make use of the following equations. x = ρ sin θ cos φ y = ρ sin θ sin φ z = ρ cos θ In spherical coordinates the Laplacian operator u xx + uyy + uzz = ∆u has the form ∂ ∂2u 1 ∂u 1 1 ∂ 2 ∂u , ρ + 2 sin θ + 2 2 ∆u = 2 ρ ∂ρ ∂ρ ρ sin θ ∂θ ∂θ ρ sin θ ∂φ2 where the coordinate ρ is the distance from the origin. The details of the conversion may be found in Appendix A. Since Laplace’s equations is linear and homogeneous, we may attempt to solve it using the method of separation of variables. To begin we will assume that u(ρ, θ, φ) = R(ρ)T (θ)P (φ) and substitute this expression in Laplace’s equation. This produces 1 1 ∂ 1 ∂ ρ2 R 0 T P + 2 sin θT 0 RP + 2 2 RT P 00 = 0 2 ρ ∂ρ ρ sin θ ∂θ ρ sin θ where the “prime” notation denotes the appropriate ordinary derivative. Multiplying both sides of 2 sin2 θ this equation by ρ RT P yields the equation sin θ ∂ P 00 sin2 θ ∂ ρ2 R 0 + sin θT 0 + = 0. R ∂ρ T ∂θ P Moving the functions depending only on φ to the right-hand side of the equations allows us to separate the φ and (ρ, θ) variables. sin θ ∂ P 00 sin2 θ ∂ ρ2 R 0 + sin θT 0 = − R ∂ρ T ∂θ P Since the left-hand side of the equation depends only on (ρ, θ) and the right-hand side depends only on φ, both sides of the equation must equal a constant.
2
φ-Dependent Factor of u Since the solution to Laplace’s equation must be 2π-periodic in φ the constant must be of the form n2 where n ∈ N ∪ {0}. Hence P 00 P P 00 + n2 P −
= n2 = 0
Pn (φ) = An cos nφ + Bn sin nφ. See [1, Chap. 3] for the classical method of solving a second-order, linear, homogeneous ordinary differential equation of the type shown above. Now we turn to the task of solving for the remaining two factors of the solution u to Laplace’s equation.
ρ-Dependent Factor of u The partial solution Pn (φ) found in the previous section imposes the following constraint on the solutions for ρ and θ. sin θ ∂ sin2 θ ∂ ρ2 R 0 + sin θT 0 = n2 , R ∂ρ T ∂θ
where n ∈ N ∪ {0}. As a next step we divide both sides of the equation by sin 2 θ to obtain ∂ 1 1 ∂ ρ2 R 0 + sin θT 0 = R ∂ρ T sin θ ∂θ 1 ∂ ρ2 R 0 = R ∂ρ
n2 sin2 θ n2 1 ∂ 0 2 − T sin θ ∂θ sin θT . sin θ
We see that the left-hand side of the equation depends only on ρ while the right-hand side depends only on θ. Thus both sides are equal to a constant. If the constant has the form m(m + 1) where m ∈ N ∪ {0} then the following ordinary differential equation is implied by the left-hand side of the last equation. 1 ∂ ρ2 R0 = m(m + 1) R ∂ρ 2 00 0 ρ R + 2ρR − m(m + 1)R = 0 The latter equation is known as Euler’s Equation (see [1, Sec. 5.5]). The solution to this equation has the form R(ρ) = Cρm + Dρ−m−1 . Since the solution to Laplace’s equation must be bounded at the origin (where ρ = 0) we must assume D = 0 and thus the ρ-dependent factor of the solution has the form R m (ρ) = ρm with m ∈ N ∪ {0}. Now we proceed to the task of finding the θ-dependent factor of the solution to Laplace’s equation.
3
θ-Dependent Factor of u The solutions found for R(ρ) and P (φ) have imposed two constraints on the differential equation for T (θ). It now has the form n2 ∂ 1 0 = m(m + 1), 2 − T sin θ ∂θ sin θT sin θ which is equivalent to the ordinary differential equation n2 cos θ 0 00 = 0. T + m(m + 1) − T + sin θ sin2 θ We can put this equation in a more convenient form for solving by making the change of variable w = cos θ. In this case, dT dθ d2 T dθ 2
dT and dw dT d2 T = − cos θ + sin2 θ 2 . dw dw = − sin θ
Substituting these expressions in the previous ordinary differential equation 2 dT n2 2 d T sin θ 2 − 2 cos θ + m(m + 1) − T dw dw sin2 θ dT n2 d2 T 2 + m(m + 1) − T (1 − cos θ) 2 − 2 cos θ dw dw 1 − cos2 θ 2 dT n2 2 d T (1 − w ) 2 − 2w + m(m + 1) − T dw dw 1 − w2
gives us = 0 = 0 = 0.
This last ordinary differential equation is known as the associated Legendre differential equation n (w) = T n (cos θ) with m ∈ N ∪ {0} and (see [3]). The solutions will be denoted by the functions T m m 0 ≤ n ≤ m (see [4]). Now we may summarize the product solution.
General Solution Combining the three factors of the product solution we have n n unm (ρ, θ, φ) = ρm Tm (cos θ)(Anm cos nφ + Bm sin nφ)
for m ∈ N ∪ {0} and 0 ≤ n ≤ m. Thus the series solution of Laplace’s equation on the sphere can be written as m ∞ X X n n ρm Tm (cos θ)(Anm cos nφ + Bm sin nφ). u(ρ, θ, φ) = m=0 n=0
n so that the boundary conditions can be The next step is to choose the coefficients A nm and Bm satisfied.
4
Boundary Conditions The boundary of the sphere is kept at a temperature distribution described by u(x, y, z) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4.
In spherical coordinates these boundary conditions are equivalent to u(2, θ, φ) = 6 + 6 cos2 θ − 2(1 − cos2 θ) cos 2φ for 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π. The details of this coordinate conversion are found in Appendix B. Rather than try to use the orthogonality property of the eigenfunctions, we will use the fact that the boundary condition is a finite sum of eigenfunctions and thus we merely need match the coefficients at the boundary. The first several associated Legendre functions are listed in the matrix below. T00 (w) = 1 p T10 (w) = w T11 (w) = − 1 − w2 p 1 T20 (w) = (3w2 − 1) T21 (w) = −3w 1 − w2 T22 (w) = 3(1 − w 2 ) 2 Thus by writing out the first several terms of the solution evaluated when ρ = 2 we get u(2, θ, φ) = =
m ∞ X X
n n 2m Tm (cos θ)(Anm cos nφ + Bm sin nφ)
m=0 n=0 A00 T00 (cos θ) + 2A01 T10 (cos θ) + 2T11 (cos θ)(A11 cos φ + B11 sin φ) + 4A02 T20 (cos θ) + 4T21 (cos θ)(A12 cos φ + B21 sin φ) + 4T22 (cos θ)(A22
cos 2φ + B22 sin 2φ)
+ ··· We can see by the absence of sin nφ terms in the spherical coordinate form of the boundary conditions that Bij = 0 for all i and j. Absent as well are any powers of cos θ larger than 2, thus A m n =0 for n > 2. Thus we have u(2, θ, φ) = A00 + 2A01 cos θ − 2A11 sin θ cos φ + 2A02 (3 cos2 θ − 1) − 12A12 cos θ sin θ cos φ + 12A22 (1 − cos2 θ) cos 2φ. Simplifying one step at a time we can set A 01 = A11 = A12 = 0. This will leave, u(2, θ, φ) = A00 + 2A02 (3 cos2 θ − 1) + 12A22 (1 − cos2 θ) cos 2φ = A00 − 2A02 + 6A02 cos2 θ + 12A22 (1 − cos2 θ) cos 2φ
and hence if we choose A02 = 1, A00 = 8, and A22 = − 61 we have satisfied the boundary conditions. Therefore the solution to Laplace’s equation on the sphere may be written as 1 1 u(ρ, θ, φ) = 8 + ρ2 (3 cos2 θ − 1) − ρ2 (1 − cos2 θ) cos 2φ. 2 2
5
Plots of Solution
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2
3
3
2.5
2.5
2
2
1.5
1
0.5
0.5
0
0.5
1 Ρ
1.5
0
2
3
3
2.5
2.5
2
2
1.5
1
0.5
0.5
0
0.5
1 Ρ
1.5
0
2
7
1 Ρ
1.5
2
0
0.5
1 Ρ
1.5
2
0
0.5
1 Ρ
1.5
2
0
0.5
1 Ρ
1.5
2
1.5
1
0
0.5
1.5
1
0
0
1.5
1
Θ
Θ
0
3
0
Θ
1.5
1
0
Θ
Θ
3
Θ
Θ
The following sequence of frames illustrates the φ-dependency of the solution for 0 ≤ φ ≤ steps of size ∆φ = π4 .
7π 4
in
6
5
5
4
4
3
2
1
1
0.5
1 Ρ
1.5
0
2
5
5
4
4 Φ
6
3
2
1
1
0
0.5
1 Ρ
1.5
0
2
6
6
5
5
4
4
3
2
1
1
0
0.5
1 Ρ
1.5
0
2
6
6
5
5
4
4
3
2
1
1
0
0.5
1 Ρ
1.5
0
2
8
1 Ρ
1.5
2
0
0.5
1 Ρ
1.5
2
0
0.5
1 Ρ
1.5
2
0
0.5
1 Ρ
1.5
2
3
2
0
0.5
3
2
0
0
3
2
Φ
Φ
0
6
0
Φ
3
2
0
Φ
Φ
6
Φ
Φ
The following sequence of frames illustrates the θ-dependency of the solution for 0 ≤ θ ≤ steps of size ∆θ = π8 .
7π 8
in
References [1] William E. Boyce and Richard C. DiPrima, Elementary Differential Equations and Boundary Value Problems, 7th edition, John Wiley & Sons, Inc., New York 2001. [2] Robert T. Smith and Roland B. Minton, Calculus, 2nd edition, McGraw-Hill Publishing Company, Boston, MA 2002. [3] Eric W. Weisstein, “Legendre Differential Equation”, From MathWorld–A Wolfram Web Resource, http://mathworld.wolfram.com/LegendreDifferentialEquation.html [4] Daniel Zwillinger (Ed.), CRC Standard Mathematical Tables and Formulae, CRC Press, Boca Raton, FL 1995.
A
Spherical Laplacian
The Laplacian operator in spherical coordinates can be derived by use of the chain rule for multivariable functions (see [2, Sec. 12.5]). Suppose that u ≡ u(ρ, θ, φ) then the first partial derivatives of u are uρ = u x xρ + u y yρ + u z zρ = ux sin θ cos φ + uy sin θ sin φ + uz cos θ uθ = u x xθ + u y yθ + u z zθ = ux ρ cos θ cos φ + uy ρ cos θ sin φ − uz ρ sin θ uφ = u x xφ + u y yφ + u z zφ = −ux ρ sin θ sin φ + uy ρ sin θ cos φ In a similar fashion we may find the three (out of the nine total) second partial derivatives. uρρ = uxx sin2 θ cos2 φ + 2uxy sin2 θ cos φ sin φ + 2uxz cos θ sin θ cos φ + uyy sin2 θ sin2 φ + 2uyz cos θ sin θ sin φ + uzz cos2 θ uθθ = uxx ρ2 cos2 θ cos2 φ + 2uxy ρ2 cos2 θ cos φ sin φ − 2uxz ρ2 cos θ sin θ cos φ + uyy ρ2 cos2 θ sin2 φ − 2uyz ρ2 cos θ sin θ sin φ + uzz ρ2 sin2 θ − ux ρ sin θ cos φ − uy ρ sin θ cos φ − uz ρ cos θ uφφ = uxx ρ2 sin2 θ sin2 φ − 2uxy ρ2 sin2 θ cos φ sin φ + uyy ρ2 sin2 θ cos2 φ − ux ρ sin θ cos φ − uy ρ sin θ sin φ Now we can verify that uxx + uyy + uzz
1 ∂ = 2 ρ ∂ρ
ρ
2 ∂u
∂ρ
∂ 1 + 2 ρ sin θ ∂θ
9
∂u sin θ ∂θ
+
∂2u 1 . ρ2 sin2 θ ∂φ2
We start with the right-hand side. 1 ∂ ∂ ∂2u 1 ∂u 1 2 ∂u ρ + sin θ + 2 2 2 2 ρ ∂ρ ∂ρ ρ sin θ ∂θ ∂θ ρ sin θ ∂φ2 1 cos θ 1 2 uθ + 2 2 uφφ = uρρ + uρ + 2 uθθ + 2 ρ ρ ρ sin θ ρ sin θ = uxx sin2 θ cos2 φ + 2uxy sin2 θ cos φ sin φ + 2uxz cos θ sin θ cos φ + uyy sin2 θ sin2 φ 2 + 2uyz cos θ sin θ sin φ + uzz cos2 θ + (ux sin θ cos φ + uy sin θ sin φ + uz cos θ) ρ 2 2 2 + uxx cos θ cos φ + 2uxy cos θ cos φ sin φ − 2uxz cos θ sin θ cos φ + uyy cos2 θ sin2 φ 1 − 2uyz cos θ sin θ sin φ + uzz sin2 θ − (ux sin θ cos φ + uy sin θ cos φ + uz cos θ) ρ cos θ + 2 (ux ρ cos θ cos φ + uy ρ cos θ sin φ − uz ρ sin θ) + uxx sin2 φ ρ sin θ 1 (ux cos φ + uy sin φ) − 2uxy cos φ sin φ + uyy cos2 φ − ρ sin θ = uxx sin2 θ + cos2 θ cos2 φ + sin2 φ + uyy sin2 θ + cos2 θ sin2 φ + cos2 φ + uzz cos2 θ + sin2 θ + 2uxy sin2 θ + cos2 θ cos φ sin φ − cos φ sin φ + 2uxz (cos θ sin θ cos φ − cos θ sin θ cos φ)
1 cos2 θ (ux cos φ + uy sin φ) − uz cos θ ρ sin θ ρ 1 1 + (ux sin θ cos φ + uy sin θ sin φ + uz cos θ) − (ux cos φ + uy sin φ) ρ ρ sin θ 1 − sin2 θ 1 = uxx + uyy + uzz + (ux cos φ + uy sin φ) + (ux sin θ cos φ + uy sin θ sin φ) ρ sin θ ρ 1 − (ux cos φ + uy sin φ) ρ sin θ 1 sin θ (ux cos φ + uy sin φ) + (ux sin θ cos φ + uy sin θ sin φ) = uxx + uyy + uzz − ρ ρ = uxx + uyy + uzz + 2uyz (cos θ sin θ sin φ − cos θ sin θ sin φ) +
10
B
Spherical Boundary Conditions u(x, y, z) = x2 + 2y 2 + 3z 2
for x2 + y 2 + z 2 = 4
u(2, θ, φ) = (2 sin θ cos φ)2 + 2(2 sin θ sin φ)2 + 3(2 cos θ)2 = 4 sin2 θ cos2 φ + 2 sin2 θ sin2 φ + 3 cos2 θ
= 4 [1 − cos2 θ] cos2 φ + 2[1 − cos2 θ] sin2 φ + 3 cos2 θ = 4 [3 − cos2 φ − 2 sin2 φ] cos2 θ + 2 sin2 φ + cos2 φ = 4 [2 − sin2 φ] cos2 θ + 1 + sin2 φ 1 − cos 2φ 1 − cos 2φ 2 = 4 [2 − ] cos θ + 1 + 2 2
= 2(3 + cos 2φ) cos 2 θ + 2(3 − cos 2φ) = 6 + 6 cos2 θ − 2(1 − cos2 θ) cos 2φ
11
