20090921170009390_9-10장

20090921170009390_9-10장 - Homework 1...

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Unformatted text preview: Homework 1 § 9 . 8- § 9 . 9 1. Find the divergence and the curl of the vector function F ( x, y, z ) = ( x 2 + y 2 + z 2 )- 3 / 2 ( x i + y j + z k ) . Sol. Let F 1 = x ( x 2 + y 2 + z 2 ) 3 / 2 , F 2 = y ( x 2 + y 2 + z 2 ) 3 / 2 , F 3 = z ( x 2 + y 2 + z 2 ) 3 / 2 . Notice that ∂ ∂x F 1 = 1 ( x 2 + y 2 + z 2 ) 3 / 2- 3 x 2 ( x 2 + y 2 + z 2 ) 5 / 2 = y 2 + z 2- 2 x 2 ( x 2 + y 2 + z 2 ) 5 / 2 . Similarly, we have ∂ ∂y F 2 = x 2 + z 2- 2 y 2 ( x 2 + y 2 + z 2 ) 5 / 2 , ∂ ∂z F 3 = x 2 + y 2- 2 z 2 ( x 2 + y 2 + z 2 ) 5 / 2 . Thus div F = ∂ ∂x F 1 + ∂ ∂y F 2 + ∂ ∂z F 3 = y 2 + z 2- 2 x 2 ( x 2 + y 2 + z 2 ) 5 / 2 + x 2 + z 2- 2 y 2 ( x 2 + y 2 + z 2 ) 5 / 2 + x 2 + y 2- 2 z 2 ( x 2 + y 2 + z 2 ) 5 / 2 = 0 . Notice that ∂ ∂y F 3- ∂ ∂z F 2 =- 3 zy ( x 2 + y 2 + z 2 ) 5 / 2-- 3 yz ( x 2 + y 2 + z 2 ) 5 / 2 = 0 . Similarly, we have ∂ ∂x F 3- ∂ ∂z F 1 = 0 , ∂ ∂x F 2- ∂ ∂y F 1 = 0 . Thus curl F = fl fl fl fl fl fl i j k ∂ ∂x ∂ ∂y ∂ ∂z F 1 F 2 F 3 fl fl fl fl fl fl = ( ∂ ∂y F 3- ∂ ∂z F 2 ) i- ( ∂ ∂x F 3- ∂ ∂z F 1 ) j + ( ∂ ∂x F 2- ∂ ∂y F 1 ) k = 0 . Computational Science & Engineering (CSE) C. K. Ko Homework 2 2. The velocity vector v ( x, y, z ) of an incompressible fluid rotating in a cylindri- cal vessel is of the form v = w × r , where w is the constant rotation vector and r = [ x, y, z ]. Show that div v = 0. Sol. Let w = [ α, β, γ ]. Then div v = [ ∂ ∂x , ∂ ∂y , ∂ ∂z ] • ( w × r ) = [ ∂ ∂x , ∂ ∂y , ∂ ∂z ] • fl fl fl fl fl fl i j k α β γ x y z fl fl fl fl fl fl = [ ∂ ∂x , ∂ ∂y , ∂ ∂z ] • [ βz- γy, γx- αz, αy- βx ] = 0 . 3. Assuming sufficient differentiability, show that (a) div ( f ∇ g ) = f ∇ 2 g + ∇ f • ∇ g (b) div (curl v ) = 0 (c) curl( f v ) = (grad f ) × v + f curl v (d) div( u × v ) = v • curl u- u • curl v . Sol. (a) div ( f ∇ g ) = [ ∂ ∂x , ∂ ∂y , ∂ ∂z ] • [ fg x , fg y , fg z ] = ∂ ∂x ( fg x ) + ∂ ∂y ( fg y ) + ∂ ∂z ( fg z ) = ( f x g x + f y g y + f z g z ) + f ( g xx + g yy + g zz ) = ∇ f • ∇ g + f ∇ 2 g. (b) Let v = [ v 1 , v 2 , v 3 ]. By assumption, ( v 1 ) zy = ( v 1 ) yz , ( v 2 ) zx = ( v 2 ) xz , ( v 3 ) yx = ( v 3 ) xy . Hence div(curl v ) = [ ∂ ∂x , ∂ ∂y , ∂ ∂z ] • fl fl fl fl fl fl i j k ∂ ∂x ∂ ∂y ∂ ∂z v 1 v 2 v 3 fl fl fl fl fl fl = [( v 3 ) yx- ( v 2 ) zx ] + [( v 1 ) zy- ( v 3 ) xy ] + [( v 2 ) xz- ( v 1 ) yz ] = 0 . Computational Science & Engineering (CSE) C. K. Ko Homework 3 (c) Let v = [ v 1 , v 2 , v 3 ]. Then f v = [ fv 1 , fv 2 , fv 3 ] and curl ( f v ) = fl fl fl fl fl fl i j k ∂ ∂x ∂ ∂y ∂ ∂z fv 1 fv 2 fv 3 fl fl fl fl fl fl = [( fv 3 ) y- ( fv 2 ) z , ( fv 1 ) z- ( fv 3 ) x , ( fv 2 ) x- ( fv 1 ) y ] = f [( v 3 ) y- ( v 2 ) z , ( v 1 ) z- ( v 3 ) x , ( v 2 ) x- ( v 1 ) y ] +[ f y v 3- f z v 2 , f z v 1- f x v 3 , f x v 2- f y v 1 ] = f fl fl fl fl fl fl i j k ∂ ∂x ∂ ∂y ∂ ∂z v...
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This note was uploaded on 11/08/2011 for the course EE 111 taught by Professor Kim during the Spring '11 term at Korea University.

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20090921170009390_9-10장 - Homework 1...

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