Laplace s Equation on a Sphere

Transcription

1 Laplace s Equation on a Sphere Pierre Simon de Laplace Domenic Scorzetti Millersville University dascorze@marauder.millersville.edu April 5, 6 Problem Outline A spherical shell has an inner radius, and an outer radius. The temperatures of the inner and outer surfaces are given by: fi ( θ ) cosθ fo ( θ ) 5 + 3cosθ 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

2 a) Problem Statement Finding the steady-state temperature will require the use of the Laplacian operator: 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: u u () u + sinθ sin θ θ Since Laplace s equation is linear and homogeneous, I will begin solving by separation of variables. This assumes u (, θ ) R( ) T ( θ ) which is then substituted into (): θ ' R T + sin θ θ ( sinθt' ) R The primed variables indicate necessary regular derivatives. I continue the process by sin θ multiplying through by : RT -Dependent Portion of u sin θ R sinθ + T θ () R' ( sinθt ') To find this part, I begin by dividing equation () by sin θ and rearranging the terms: R T sinθ θ (3) R' ( sinθt' ) Since the left-hand side only depends on 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 + ) with

3 m N {} (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: Applying the product rule: ' R m( m + ) R R'' + R' m( m + ) R () R'' + R' m( m + ) R Equation () is Euler s Equation (see []) with a solution which looks like (5), for m N {} θ-dependent Portion of u (5) R( ) C m + D m Referring back to equation (3), I now isolate the θ component and set it equal to the same constant. T sinθ θ Once again, apply the product rule: T''sinθ + T'cosθ T sinθ (6) ( sinθt' ) m( m + ) [ ] m( m + ) T'' T'cosθ m( m + ) T T sinθ Multiply through by - and T: T'cosθ (7) T' ' + + ( m( m + )) T sinθ To further simplify the calculations, I put (7) into another form through a change of variables and by letting w cosθ (see []). This will require the chain rule: (8) dt dt dt sinθ d T d dt d dt sinθ 3

4 d T dt d T cosθ sinθ (9) Substitute (8) and (9) into (7): d T dt cos sin d T θ + θ d T dt cosθ dt sin cos + sin + ( ( + )) sin θ θ θ m m T θ d T dt dt sin θ cosθ cosθ + ( m( m + )) T d T dt sin θ cosθ + ( m( m + )) T Recall the trigonometric property sin θ + cos θ to make the following substitution: cos d T dt ( θ ) cosθ + ( m( m + )) T Also, recall that w cosθ : () d T dt ( w ) w + ( m( m + )) T Note that equation () is a Legendre polynomial which has solutions of the form: with m N {} (see [, Sec. 7..3]. () T m (w) m T (w) T (cosθ) m T (w) w T (cosθ) cosθ m T (w).5(3w -) T (cosθ).5(3cos θ -)

5 General Solution This is found by combining equations (5) and () inside an infinite series: () u (, θ) ( C m + D m ) (cosθ) m m m T m However, notice my original problem consists of only a finite sum of constants and cosines, so I can find an exact solution with and per equation (). Additionally, I only need to sum through m since there are no squared cosine terms to account for. u (, θ) fi( θ ) cosθ ( C + D ) T (cosθ) m m m m cosθ ( C + D ) T (cosθ ) + ( C + D ) T (cosθ) By substitution from the solution table above: (3) cosθ ( C + D ) + ( C + D ) cosθ u (, θ ) fo( θ) 5 + 3cosθ ( C m + D m ) T (cosθ ) m m m m 5 + 3cosθ ( C + D ) T (cosθ) + ( C + D ) T (cosθ ) 5 + 3cosθ ( C + D ) T (cosθ) + (C + D ) T (cosθ) By substitution from the solution table above: 5 + 3cos ( C + D ) + (C + D () θ ) cosθ 5

6 Using equations (3) and () I will match up coefficients and find C, D, C, and D and use these in my exact solution by plugging them into equation (). C + D 5 C + D C + D 3 C + D For clarity, the details of the coefficient calculations are in Appendix A. These calculations yield: C C D D - Here I will construct the final form of the steady-state temperature using equation () and my tables of Legendre solutions and coefficients: u(, θ ) ( C + D ) T (cosθ) + ( C + D ) T (cosθ ) u (, θ) ( + ) + ( )(cos θ) (5) u (, θ) + cosθ Equation (5) represents the steady-state temperature for the spherical shell. 6

7 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 to θ π and between and. 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

8 References [] Dr. Robert Buchanan, Laplace s Equation on a Sphere, < [] Richard Haberman, Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, th Edition, Pearson Prentice Hall, Upper Saddle River, NJ. Appendix A C D C 5 ( D ) + D 5 D D 5 D C D C 3 ( D ) + D 3 D + D 7 3 D 7 7 D D 8