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We will look for the Green's function for R2In particular, we need to find a corrector function hx for each x 2 R2 , such that ∆yhx(y) = 0 y 2 R2 hx(y) = Φ(y ¡x) y 2 @R2 Fix x 2 R2We know ∆yΦ(y ¡ x) = 0 for all y 6= xTherefore, if we choose z =2 Ω, then ∆yΦ(y ¡ z) = 0 for all y 2 Ω Now, if we choose z = z(x) appropriately, z =2 Ω, such that Φ(y ¡ z) = Φ(y ¡ x) for y 2Z Z R f(x,y)dxdy = Z Z R g(u,v) ∂(x,y) ∂(u,v) dudv , where g(u,v) is obtained from f(x,y) by substitution, using the equations (3) We will derive the formula (5) for the new area element in the next section;N B V C Z S K J S S S S O I S Y T R I I S S G H B D A S I B S S S S B I MerryNoel Chamberlain, MA, Teacher of Students with Visual Impairments NAME_____ T = BLUE T = BLUE T T T B J T J T N B D S J T J T T T K M J V T R B T J B T N V X T B Z E MerryNoel Chamberlain, MA, Teacher of Students with Visual Impairments
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(Последнее обновление ноябрь 19) H e Z k b f _ g _ g b y h e Z k b _ B g n h j f Z p b y, h l h j m x u h b j Z _ fG ∂ vy ∂ v Curl ∇ × v = (x x ∂ vz − ∂ vy ) xˆ (∂ v − ∂ vz ) yˆ ( − )zˆ ∂ y ∂ z ∂ z ∂ x ∂ x ∂ y 2 2 2∂ t ∂ t Laplacian ∇2t = ∂ t 2 ∂ x 2 ∂ y ∂ z 2 Spherical ∂ t Gradient ∇t = r ˆ 1 ∂ t θˆ 1 ∂ t φ ∂ r r ∂θ r sinθ ∂φ G 1 ∂ 2 1 ∂ 1 ∂ vφ Divergence ∇ ⋅If g(0) ≠ 0 then put y = 0 in the equation and you get for all other x, g(x) = 2 First solution g(x) = 2 Now if g(0) = 0 there are other possibilities Choose x = y = 2 and then either g(2) = 0 or g(2) = 2 If g(2) = 0, then either g(1) = 0 or g(1) = 3 For g(1) = 0, you get g(n) = 0 for all n ∈ Z
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Mar 01, 17 · $\begingroup$ You may want to rethink how you're asking questions I've seen (and answered) several of your questions It may be good if you could think deeper about the problem be careful about writing "junk" or random facts;Department of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the US
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