Math 509: Advanced Analysis

Homework 8
Lawrence Tyler Rush
<me@tylerlogic.com>
April 6, 2015
http://coursework.tylerlogic.com/courses/upenn/math509/homework08

For the first five problems, let A,B,C,D ∈ R3 with A = (a1,a2,a3), B = (b1,b2,b3), C = (c1,c2,c3), and D = (d1,d2,d3).

1 Problem 1 from Vector Calculus Notes

First off, we have
B × C = (b2c3 - b3c2, b1c3 - b3c1, b1c2 - b2c1)
C × A = (c2a3 - c3a2, c1a3 - c3a1, c1a2 - c2a1)
A × B = (a2b3 - a3b2, a1b3 - a3b1, a1b2 - a2b1)
so that
A ⋅ (B × C) = a1(b2c3 - b3c2) + a2(b1c3 - b3c1) + a3(b1c2 - b2c1)
B ⋅ (C × A) = b1(c2a3 - c3a2) + b2(c1a3 - c3a1) + b3(c1a2 - c2a1)
= a1(b2c3 - b3c2) + a2(b1c3 - b3c1) + a3(b1c2 - b2c1)
C ⋅ (A × B) = c1(a2b3 - a3b2) + c2(a1b3 - a3b1) + c3(a1b2 - a2b1)
= a1(b2c3 - b3c2) + a2(b1c3 - b3c1) + a3(b1c2 - b2c1)

2 Problem 2 from Vector Calculus Notes

We have
A × (B × C) = (a2(b1c2 - b2c1) - a3(b1c3 - b3c1),
a1(b1c2 - b2c1) - a3(b2c3 - b3c2),
a1(b1c3 - b3c1) - a2(b2c3 - b3c2))
We also have
A ⋅C = a c + a c + a c 1 1 2 2 3 3

and

A ⋅B = a1b1 + a2b2 + a3b3

so that

B(A ⋅ C) - C(A ⋅ B) = ((a1c1 + a2c2 + a3c3)b1 - (a1b1 + a2b2 + a3b3)c1,
(a1c1 + a2c2 + a3c3)b2 - (a1b1 + a2b2 + a3b3)c2,
(a1c1 + a2c2 + a3c3)b3 - (a1b1 + a2b2 + a3b3)c3)
= (a1b1c1 + a2b1c2 + a3b1c3 - a1b1c1 - a2b2c1 - a3b3c1,
a1b2c1 + a2b2c2 + a3b2c3 - a1b1c2 - a2b2c2 - a3b3c2,
a1b3c1 + a2b3c2 + a3b3c3 - a1b1c3 - a2b2c3 - a3b3c3)
= (a2b1c2 + a3b1c3 - a2b2c1 - a3b3c1,
a1b2c1 + a3b2c3 - a1b1c2 - a3b3c2,
a1b3c1 + a2b3c2 - a1b1c3 - a2b2c3)
= (a2(b1c2 - b2c1) - a3(b1c3 - b3c1),
a1(b1c2 - b2c1) - a3(b2c3 - b3c2),
a1(b1c3 - b3c1) - a2(b2c3 - b3c2))
Hence A × (B × C) = B(A ⋅ C) - C(A ⋅ B).

3 Problem 3 from Vector Calculus Notes

Let A,B,C ∈ Rn. Then in light of the previous problem and the fact that the dot product is a commutative operation, we have
A × (B × C) + B × (C × A) + C × (A × B) = (B(A ⋅ C) - C(A ⋅ B)) + (C(B ⋅ A) - A(B ⋅ C)) + (A(C ⋅ B) - B(C ⋅ A) )
= (B(A ⋅ C) - B(C ⋅ A)) + (C(B ⋅ A) - C(A ⋅ B)) + (A(C ⋅ B) - A(B ⋅ C) )
= 0

4 Problem 4 from Vector Calculus Notes

Since
A × (B × C) = (a2(b1c2 - b2c1) - a3(b1c3 - b3c1),
a1(b1c2 - b2c1) - a3(b2c3 - b3c2),
a1(b1c3 - b3c1) - a2(b2c3 - b3c2))
= (a2b1c2 - a2b2c1 - a3b1c3 + a3b3c1,
a1b1c2 - a1b2c1 - a3b2c3 + a3b3c2,
a1b1c3 - a1b3c1 - a2b2c3 + a2b3c2)
and
(A × B) × C = (a2b3 - a3b2, a1b3 - a3b1, a1b2 - a2b1) × C
= (c2(a1b2 - a2b1) - c3(a1b3 - a3b1),
c1(a1b2 - a2b1) - c3(a2b3 - a3b2),
c1(a1b3 - a3b1) - c2(a2b3 - a3b2))
= (a1b2c2 - a2b1c2 - a1b3c3 + a3b1c3,
a1b2c1 - a2b1c1 - a2b3c3 + a3b2c3,
a1b3c1 - a3b1c1 - a2b3c2 + a3b2c2)
Hence we have that A × (B × C) = (A × B) × C whenever all of
a2b1c2 - a2b2c1 - a3b1c3 + a3b3c1 = a1b2c2 - a2b1c2 - a1b3c3 + a3b1c3
a1b1c2 - a1b2c1 - a3b2c3 + a3b3c2 = a1b2c1 - a2b1c1 - a2b3c3 + a3b2c3,
a1b1c3 - a1b3c1 - a2b2c3 + a2b3c2 = a1b3c1 - a3b1c1 - a2b3c2 + a3b2c2
are satisfied.

5 Problem 5 from Vector Calculus Notes

6 Problem 9 from Vector Calculus Notes

Let r = (x,y,z) be an element in R3 and put r = |r| and ˆr = r∕r.

(a) Show ▿(r2) = 2r

Since
2 2 (∘ -----------)2 2 2 2 r = |r| = x2 + y2 + z2 = x + y + z

so that

▿ (r2) = (2x,2y,2z) = 2(x,y,z) = 2r

(b) Show ▿(1∕r) = -ˆr ∕r2

Since r is a function of x, y, and z, then
▿( ) 1 r = ▿(r-1)
= ( dr dr dr) - r-2dx-,- r-2dy,- r-2dz
= -r-2( ) dr-, dr, dr dx dy dz
= -r-2( ) -1 -1 1- 2r(2x),2r(2y),2r(2z)
= -r-2( 1) 2r(2x,2y,2z)
= -r-2( ) 1 r(x,y,z)
= -r-2( ) 1 rr
= -r-2(r) r
= -ˆr- r2

7 Problem 10 from Vector Calculus Notes

8 Problem 11 from Vector Calculus Notes

9 Problem 12 from Vector Calculus Notes

Define a vector field V in R3 by
V (x,y,z) = (- y,x,1)

Then for any φt in the group of one-parameter diffeomorphisms generated by this vector field must have

d dtφt(x,y,z) = V (φt(x,y,z))

By defining φt(x,y,z) = (x*(t,x,y,z),y*(t,x,y,z),z*(t,x,y,z)), the above equation yields

d- dt(x*(t,x,y,z),y*(t,x,y,z),z*(t,x,y,z)) = V ((x*(t,x,y,z),y*(t,x,y,z),z*(t,x,y,z)))
= (-y*(t,x,y,z),x*(t,x,y,z),1)
further implying that
d- dtx* = -y*
d- dty* = x*
d- dtz* = 1
Hence the group {φt} of one-parameter diffeomorphisms generated by V is the group of φt such that
φt(x,y,z) = (ft(x,y,z) cost, ft(x,y,z)sint, t)

where ft is an arbitrary function on R3 but fixed for each t.

Furthermore,

( ) cost∂∂ftx- cost∂∂fyt cost∂∂fzt Jφt = |( sin t∂∂ftx- sint∂∂fyt sint∂∂fzt |) 0 0 0

which has determinant zero. Hence d- dt|t=0 detJφt is zero, and thus ▿⋅ V is also zero, indicating the {φt} are volume preserving.

10 Problem 13 from Vector Calculus Notes

11 Problem 14 from Vector Calculus Notes

Let
2 2 -1 V = (x + y ) (- yi+ xj)

in other words

( - y x ) V (x,y,z) = x2 +-y2,x2-+-y2,0

Therefore

curlV = ( ∂ ∂ x ∂ - y ∂ ∂ x ∂ - y ) ∂y0- ∂z-x2 +-y2,∂z-x2 +-y2-- ∂x0,∂x-x2 +-y2-- ∂yx2-+-y2
= ( ( 2 2 ) ( 2 2 ) ) 0,0, -y2-+-x22- - -y2-+-x22- (x + y) (x + y)
= (0,0,0)
as desired.

12 Problem 15 from Vector Calculus Notes