week9 - Week 9 Lectures, Math 716, Tanveer 1 1.1 Greens...

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Week 9 Lectures, Math 716, Tanveer 1 Green’s function as a distribution 1.1 Laplace Operator For the Poisson-Problem with homogeneous boundary condition: Δ u = f for x Ω , u = 0 on Ω (1) we know that u ( x 0 ) = Z Ω G ( x , x 0 ) f ( x ) d x (2) On the otherhand, if u is a test function with support inside Ω, we have from using corollary 4 of week 8 notes that u ( x 0 ) = Z Ω G ( x , x 0 u ( x ) d x = ( u, Δ G ( ., x 0 )) (3) Therefore, in the sense of distribution, Δ G ( x , x 0 ) = δ ( x - x 0 ) (4) Therefore, we view solution (2) as a principle of linear superposition. In the physical context ( n = 3), it means that the potential caused by charge density f in a domain Ω with boundary at zero potential is given by a linear superposition of point charge potentials satisfying the same boundary conditions, with a weighting proportional to the in±nitesimal charge f ( x ) d x present in a volume element d x at x . Further, note that G ( x , x 0 ) = G 0 ( | x - x 0 | ) + H ( x , x 0 ), where H is harmonic in x . It follows that Δ G 0 ( | x - x 0 | ) = δ ( x - x 0 ) (5) 1.2 Heat Equation Recall from last class that the source function: S = ± 1 4 πκt ² n/ 2 exp ³ - | x | 2 4 κt ´ (6) satis±es S t = κ Δ S for x R n , t > 0 , with S ( x , 0 + ) = δ ( x ) (7) It can be shown (exercise) that R ( x , t ) = S ( x - x 0 , t - t 0 ) for t > t 0 and R ( x , t ) = 0 for t < t 0 (8) satis±es R t - κ Δ R = δ ( x - x 0 ) δ ( t - t 0 ) (9) 1
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2 Wave equation 2.1 Solution in higher dimension through Spherical Means Assume u is a classical solution to the initial value problem for n -dimensional wave equation for n 2: u tt - Δ u = 0 for x R n for t > 0 with u ( x , 0) = φ ( x ) , u t ( x , 0) = ψ ( x ) (10) where φ C 2 and ψ C 1 . For t > 0, r > 0, we de±ne U ( x ; r, t ) to be the spherical average over the surface of an n -dimensional ball B ( x , r ) of radius r , centered at x , and denoted by U ( x ; r, t ) = 1 A r Z ∂B ( x ; r ) u ( y , t ) d y ≡ - Z ∂B ( x ; r ) u ( y , t ) d y (11) where A r is the surface area of an n dimensional ball of radius r . Note A r = ( n ) r n - 1 , where volume of the n -dimensional sphere is α ( n ) r n . It is to be noted that lim r 0 + U ( x ; r, t ) = u ( x , t ) from continuity of u . We can similarly de±ne G ( x ; r ) = - Z ∂B ( x ; r ) φ ( y , t ) d y (12) H ( x ; r ) = - Z ∂B ( x ; r ) ψ ( y , t ) d y (13) For ±xed x , we regard U as a function of r and t . We claim Lemma 1 For fxed x , U ( x ; r, t ) is a solution o± the initial value problem: U tt - U rr - n - 1 r U r = 0 for r > 0 , t > 0 and U ( x ; r, 0) = G ( x , r ) , U t ( x ; r, 0) = H ( x , r ) (14) Proof. For convenience, we depart from our usual convention and denote ‘surface area’ element on the n -dimensional ball as dS . Symbol dS y will denote surface area element in the variable y . We note that
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This note was uploaded on 07/26/2011 for the course MATH 716 taught by Professor Tanveer,s during the Spring '08 term at Ohio State.

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week9 - Week 9 Lectures, Math 716, Tanveer 1 1.1 Greens...

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