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Laplace’s Equation on a Sphere

Pierre Simon de Laplace

Domenic Scorzetti

Millersville University dascorze@marauder.millersville.edu

April 25, 2006

Problem Outline A spherical shell has an inner radius ρ = 1, and an outer radius ρ = 2. The temperatures of the inner and outer surfaces are given by:

θθ cos20100)( −=fi θθ cos3050)( +=fo

a) State the boundary value problem for the steady-state temperature within the shell. b) Find the steady-state temperature inside the shell. c) Plot the steady-state temperature inside the shell for several slices parallel to the shell’s equator. Problem Cross-Section

ρ = 1

ρ = 2

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a) Problem Statement Finding the steady-state temperature will require the use of the Laplacian operator:

0=∆u

As described above, the solution will be for the region bounded by the two surfaces. Also, note the functional dependence only on θ (in this case used as the angle from the sphere’s north pole) and the absence of φ (azimuthal angle). b) Problem Solution The spherical Laplacian with only a θ component looks like this:

(1) ⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂+

∂∂

∂∂=∆

θθ

θθρρρ

ρρuuu sin

sin212

21

Since Laplace’s equation is linear and homogeneous, I will begin solving by separation of variables. This assumes )()(),( θρθρ TRu = which is then substituted into (1):

( ) 0'sinsin21'2

21 =

∂∂+

∂∂

⎟⎠⎞⎜

⎝⎛ RTTR θ

θθρρ

ρρ

The primed variables indicate necessary regular derivatives. I continue the process by

multiplying through by RT

θρ 22 sin :

(2) ( ) 0'sinsin'22sin =∂∂+

∂∂

⎟⎠⎞⎜

⎝⎛ T

TR

Rθ

θθρ

ρθ

ρ-Dependent Portion of u To find this part, I begin by dividing equation (2) by sin2θ and rearranging the terms:

(3) ( )'sinsin

1'21 TT

RR

θθθ

ρρ ∂

∂−=∂∂

⎟⎠⎞⎜

⎝⎛

Since the left-hand side only depends on ρ and the right-hand side depends solely on θ, then both sides must equal a constant. For convenience, say this constant has the form m(m + 1) with

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}0{∪∈ Nm (meaning m is a natural number and can also be zero). Under these assumptions, I am left with an ordinary differential equation for the LHS:

)1('21 +=∂∂

⎟⎠⎞⎜

⎝⎛ mmR

Rρ

ρ

Applying the product rule:

)1('2''21 +=+ ⎥⎦⎤

⎢⎣⎡ mmRR

Rρρ

(4) 0)1('2''2 =+−+ RmmRR ρρ

Equation (4) is Euler’s Equation (see [1]) with a solution which looks like (5), for }0{∪∈ Nm

(5) 1)( −−+= mDmCR ρρρ

θ-Dependent Portion of u Referring back to equation (3), I now isolate the θ component and set it equal to the same constant.

(6) ( ) )1('sinsin

1 +=∂∂− mmT

Tθ

θθ

Once again, apply the product rule:

[ ] )1(cos'sin''sin

1 +=+− mmTTT

θθθ

0)1(sincos''' =+−−− mm

TT

TT

θθ

Multiply through by -1 and T:

(7) 0))1((sincos''' =+++ TmmTTθθ

To further simplify the calculations, I put (7) into another form through a change of variables and by letting θcos=w (see [1]). This will require the chain rule:

(8) dwdT

ddw

dwdT

ddT θ

θθsin−==

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ −==dwdT

dd

ddT

dd

dTd θ

θθθθsin2

2

4

θθθ

θ ddw

dwTd

dwdT

dTd

22

sincos22

−−=

(9) 222sincos2

2

dwTd

dwdT

dTd θθ

θ+−=

Substitute (8) and (9) into (7):

0))1((sinsincoscos2

22sin =++−+− ⎟⎟⎠

⎞⎜⎜⎝

⎛ TmmdwdT

dwdT

dwTd θ

θθθθ

0))1((coscos222sin =++−− Tmm

dwdT

dwdT

dwTd θθθ

0))1((cos2222sin =++− Tmm

dwdT

dwTd θθ

Recall the trigonometric property 1cossin 22 =+ θθ to make the following substitution:

0))1((cos222

)2cos1( =++−− TmmdwdT

dwTd θθ

Also, recall that θcos=w :

(10) 0))1((222

)21( =++−− TmmdwdTw

dwTdw

Note that equation (10) is a Legendre polynomial which has solutions of the form:

(11) Tm(w)

with }0{∪∈ Nm (see [2, Sec. 7.10.3].

m = 0 T0(w) = 1 T0(cosθ) = 1 m = 1 T1(w) = w T1(cosθ) = cosθ m = 2 T2(w) = 0.5(3w2-1) T2(cosθ) = 0.5(3cos2θ -1)

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General Solution This is found by combining equations (5) and (11) inside an infinite series:

(12) )(cos0

)1(),( θρρθρ ∑∞

=−−+=

m mTmmDm

mCu

However, notice my original problem consists of only a finite sum of constants and cosines, so I can find an exact solution with ρ = 1 and ρ = 2 per equation (12). Additionally, I only need to sum through m = 1 since there are no squared cosine terms to account for. ρ = 1

)(cos0

)(cos20100)(),1( θθθθ ∑∞

=+=−==

m mTmDmCfiu

)(cos1)11()(cos0)00(cos20100 θθθ TDCTDC +++=−

By substitution from the solution table above:

(13) θθ cos)11()00(cos20100 DCDC +++=−

ρ = 2

)(cos0

)122(cos3050)(),2( θθθθ ∑∞

=−−+=+==

m mTmmDm

mCfou

)(cos1)1121121()(cos0)120

020(cos3050 θθθ TDCTDC −−++−+=+

)(cos1)141

12()(cos0)021

0(cos3050 θθθ TDCTDC +++=+

By substitution from the solution table above:

(14) θθ cos)141

12()021

0(cos3050 DCDC +++=+

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Using equations (13) and (14) I will match up coefficients and find C0, D0, C1, and D1 and use these in my exact solution by plugging them into equation (12).

00100 DC += 1120 DC +=−

021

050 DC += 141

1230 DC +=

For clarity, the details of the coefficient calculations are in Appendix A. These calculations yield:

C0 = 0 C1 = 20 D0 = 100 D1 = -40

Here I will construct the final form of the steady-state temperature using equation (12) and my tables of Legendre solutions and coefficients:

)(cos1)111

11()(cos0)1

00

0(),( θρρθρρθρ TDCTDCu −−++−+=

))(cos240120()11000(),( θρρρθρ −−+−+=u

(15) θρ

ρρ

θρ cos24020100),(

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+=u

Equation (15) represents the steady-state temperature for the spherical shell.

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c) Plot the Solution For my solution plot, I used the fact that the problem is symmetrical with respect to φ the azimuthal direction. I then used Mathematica’s DensityPlot function for slices parallel to the equator and through the region of interest, from θ = 0 to θ = π and between ρ = 1 and ρ = 2. In the schematic diagram below, the shaded red box depicts one slice and the red outlined rectangle shows the orientation of the output plot. You can think of the plot being the blue area bent straight instead of curved.

The color scheme shows cooler light greens and yellows with warmer dark blue and violet tones. Notice the temperature is relatively uniform around the north pole and becomes increasingly varied between the surfaces approaching the south pole. Lastly, note the hottest temperature distribution near the inner surface and the coolest temperatures near the outer surface and south pole.

0 0.5 1 1.5 2 2.5 31

1.2

1.4

1.6

1.8

2

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References [1] Dr. Robert Buchanan, “Laplace’s Equation on a Sphere”, <http://banach.millersville.edu/~bob/math467/sampleproject.pdf> [2] Richard Haberman, Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 4th Edition, Pearson Prentice Hall, Upper Saddle River, NJ 2004. Appendix A

00 100 DC −= 00 =C ______________________________

( ) 00 2110050 DD +−=

02110050 D−=

5021

0 =D

1000 =D _______________________________

11 20 DC −−= 201 =C _______________________________

( ) 11 4120230 DD +−−=

11 4124030 DD +−−=

1474030 D−−=

14770 D−=

401 −=D _______________________________