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# solh6 - Solution to homework Set 6 Math 716 1 Determine the...

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Solution to homework Set 6: Math 716 1. Determine the Green’s function G ( x , x 0 ) for Dirichlet condition for Laplace’s equation in 3-D dimensions in the hemispherical domain: Ω = { x : | x | < 1 , x 3 > 0 } Use this to determine an integral expression for u ( r, θ, φ ) in spherical polar-coordinates satisfying Δ u = 0 inΩ , with u = 0 for θ = 0 , and u (1 , θ, φ ) = ψ ( θ, φ ) for θ [0 , π 2 ] , φ [0 , 2 π ] Solution: Note that if x 0 = ( x 0 , 1 , x 0 , 2 , x 0 , 3 ) be the location of a unit positive charge inside the hemisphere, to satisfy the boundary condition at x 3 = 0, we need a negative unit charge at ¯ x 0 = ( x 0 , 1 , x 0 , 2 , - x 0 , 3 ). The image of positive charge x 0 outside the unit sphere is a negative unit charge at x 0 | x 0 | 2 , while the image of negative unit charge at ¯ x 0 outside the unit sphere is a positive unit charge at ¯ x 0 | ¯ x 0 | 2 . Combining, we have G ( x ; x 0 ) = - 1 4 π ( 1 | x - x 0 | - 1 | x - ¯ x 0 | - 1 | x - x 0 | x 0 | 2 | + 1 | x - ¯ x 0 | ¯ x 0 | 2 | ) We note that on | x | = 1, ∂n - 1 | x - x 0 | = x · ∇ 1 | x - x 0 | = x · ( x - x 0 ) | x - x 0 | 3 If x 0 = ( r 0 , θ 0 , φ 0 ), and on the spherical surface x = (1 , θ, φ ), then x · ( x - x 0 ) = 1 - r 0 cos Ψ, where Ψ is the angle between x and x 0 and is given by (using dot products): cos Ψ = sin θ 0 sin θ cos φ cos φ 0 + sin θ 0 sin θ sin φ sin φ 0 + cos θ 0 cos θ = sin θ 0 sin θ cos( φ - φ 0 ) + cos θ 0 cos θ Therefore, ∂n - 1 | x - x 0 | = 1 - r 0 cos Ψ (1 + r 2 0 - 2 r 0 cos Ψ) 3 / 2 We note that the image point ¯ x 0 has spherical coordinates ( r 0 , π - θ 0 , φ 0 ), and so using above formula makes an angle ˆ ψ with x , given by cos ˆ Ψ = sin θ 0 sin θ cos( φ - φ 0 ) - cos θ 0 cos θ Hence,

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