plugin-Math255-Pset14

plugin-Math255-Pset14 - 0 centered at , and by V , the...

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Winter 2010 Math 255 Problem Set 14 Due Friday 16 April Section 17.8: 6, 12, 14, 16, 20 Section 17.9: 2, 6, 18, 26, 30 Challenge Problem : This is taken with modification from: Edwards Advanced Calculus of Several Variables Marsden and Tromba Vector Calculus Let f be a harmonic function (that is 2 f = 0), on the open set U R 3 . If B is the 3-dimensional ball in U with center p and radius a , (i.e. x B if | x - p | ≤ a ), by following the steps below or another method of your choice, prove that f ( p ) = 1 4 πa 2 Z δB f d A. That is, the value of f at the center of the ball is the average of its values on the boundary. Remark This can be used to prove a strengthened version of the maximum principle for harmonic functions. Question for thought What does this tell you about how solutions to the heat equation 2 T = 0 behave? a) Without loss of generality, assume that p is the origin. Define g : R n / { 0 } → R 1
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Winter 2010 Math 255 by g ( x ) = 1 r , r = | x | . Denote by S ± a small sphere of radius ± >
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Unformatted text preview: 0 centered at , and by V , the region between S ± and S , as shown in Fig. 1. Figure 1: A figure showing the domain of integration. 1. Show g ( x ) is harmonic. 2. Apply the formula Z V f ∇ 2 g-g ∇ 2 f d V = Z ∂v ± f ∂g ∂n-g ∂f ∂n ² d A to obtain Z S ± f ∂g ∂n-g ∂f ∂n ² d A = Z S ± ± f ∂g ∂n-g ∂f ∂n ² d A ? 3. Observe that Z S ± f ∂g ∂n-g ∂f ∂n ² d A =-1 a 2 Z S f d A-1 a Z S ∂f ∂n d A and similarly for the right hand side of ? with a re-placed by ± because ∂g ∂n =-1 r 2 . 2 Winter 2010 Math 255 4. Use the divergence theorem to show that Z S ∂f ∂n d A = Z S ± ∂f ∂n d A = 0 so equation ? reduces to-1 a 2 Z S f d A =-1 ± 2 Z S ± f d A ? ?. 5. Now obtain the average value property for f by tak-ing the limit in ?? as ± → 0. 3...
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This note was uploaded on 12/31/2011 for the course MATH 255 taught by Professor Jackwaddell during the Fall '08 term at University of Michigan.

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plugin-Math255-Pset14 - 0 centered at , and by V , the...

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