1211s3 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1211 Multivariable Calculus 2010-11 First Semester: Assignment 3 1. (a) By direct computation, we obtain: ∂f ∂x = 3 e 3 x +4 y sin(5 z ); ∂f ∂y = 4 e 3 x +4 y sin(5 z ); ∂f ∂z = 5 e 3 x +4 y cos(5 z ); ∂ 2 f ∂x 2 = 9 e 3 x +4 y sin(5 z ); ∂ 2 f ∂y 2 = 16 e 3 x +4 y sin(5 z ); ∂ 2 f ∂z 2 =- 25 e 3 x +4 y sin(5 z ) . It is direct to verify that ∂ 2 f ∂x 2 + ∂ 2 f ∂y 2 + ∂ 2 f ∂z 2 = 0. This indicates that f is harmonic in all of R 3 . (b) Let g ( x,y,z ) = f ( ax + by,cz ) = f ( u,v ), i.e., u = u ( x,y,z ) = ax + by, v = v ( x,y,z ) = cz . Applying chain rule, we have ∂g ∂x = ∂f ∂u ∂u ∂x + ∂f ∂v ∂v ∂x = ∂f ∂u a + ∂f ∂v 0 = a ∂f ∂u , ∂ 2 g ∂x 2 = ∂ ∂u a ∂f ∂u ∂u ∂x + ∂ ∂v a ∂f ∂u ∂v ∂x = a ∂ 2 f ∂u 2 a + a ∂ 2 f ∂v∂u 0 = a 2 ∂ 2 f ∂u 2 . Similarly, we have ∂ 2 g ∂y 2 = b 2 ∂ 2 f ∂u 2 . Also, ∂g ∂z = ∂f ∂u ∂u ∂z + ∂f ∂v ∂v ∂z = ∂u ∂z 0 + ∂f ∂v c = c ∂f ∂v , ∂ 2 g ∂z 2 = ∂ ∂u c ∂f ∂v ∂u ∂z + ∂ ∂v c ∂f ∂v ∂v ∂z = c ∂ 2 f ∂u∂v 0 + c ∂ 2 f ∂v 2 c = c 2 ∂ 2 f ∂v 2 . Since f is harmonic, it follows that ∂ 2 f ∂u 2 + ∂ 2 f ∂v 2 = 0. So if we want g to be harmonic, we need: 0 = ∂ 2 g ∂x 2 + ∂ 2 g ∂y 2 + ∂ 2 g ∂z 2 = ( a 2 + b 2 ) ∂ 2 f ∂u 2 + c 2 ∂ 2 f ∂v 2 = ( a 2 + b 2- c 2 ) ∂ 2 f ∂u 2 + c 2 ∂ 2 f ∂u 2 + ∂ 2 f ∂v 2 = ( a 2 + b 2- c 2 ) ∂ 2 f ∂u 2 , 1 which means either a 2 + b 2 = c 2 or ∂ 2 f ∂u 2 = 0. 2. ∂z ∂r = ∂z ∂x ∂x ∂r + ∂z ∂y ∂y ∂r = ∂z ∂x cos θ + ∂z ∂y sin θ, ∂z ∂θ = ∂z ∂x ∂x ∂θ + ∂z ∂y ∂y ∂θ =- r ∂z ∂x sin θ + r ∂z ∂y cos θ....
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1211s3 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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