Math 509: Advanced Analysis

Homework 9

Lawrence Tyler Rush

<me@tylerlogic.com>

April 9, 2015

http://coursework.tylerlogic.com/courses/upenn/math509/homework09

1 Problem 16 from Vector Calculus Notes

Let A : R³ → R³, B : R³ → R³, and f : R³ → R and denote

A(x,y,z) = (u(x,y,z),v(x,y,z),w(x,y,z))

and

B (x,y,z) = (r(x,y,z),s(x,y,z),t(x,y,z))

(a) Derive Rule (1)

▿(fg)	= (fg)_x + (fg)_y + (fg)_z
	= f_xg + fg_x + f_yg + fg_y + f_zg + fg_z
	= (f_x + f_y + f_z)g + f(g_x + g_y + g_z)
	= (▿f)g + f(▿g)

(b) Derive Rule (3)

▿⋅ (fA)	= ▿⋅ (fu,fv,fw)
	= (fu)_x + (fv)_y + (fw)_z
	= (f_xu + fu_x) + (f_yv + fv_y) + (f_zw + fw_z)
	= (f_xu + f_yv + f_zw) + (fu_x + fv_y + fw_z)
	= (f_x,f_y,f_z) ⋅ A + f(u_x + v_y + w_z)
	= (▿f) ⋅ A + f(▿⋅ A)

(c) Derive Rule (4)

▿⋅ (A × B)	= ▿⋅ (vt - ws,wr - ut,us - vr)
	= (vt - ws)_x + (wr - ut)_y(us - vr)_z
	= v_xt + vt_x - w_xs - ws_x + w_yr + wr_y - u_yt - ut_y + u_zs + us_z - v_zr - vr_z
	= (w_y - v_z)r + (u_z - w_x)s + (v_x - u_y)t + u(s_z - t_y) + v(t_x - r_z) + w(r_y - s_x)
	= (w_y - v_z,u_z - w_x,v_x - u_y) ⋅ B + A ⋅ (s_z - t_y,t_x - r_z,r_y - s_x)
	= (w_y - v_z,u_z - w_x,v_x - u_y) ⋅ B - A ⋅ (t_y - s_z,r_z - t_x,s_x - r_y)
	= (▿× A) ⋅ B - A ⋅ (▿× B)

(d) Derive Rule (5)

▿× (fA)	= ▿× (fu,fv,fw)
	= (fw)_y - (fv)_z,(fu)_z - (fw)_x,(fv)_x - (fu)_y
	= f_yw + fw_y - f_zv - fv_z,f_zu + fu_z - f_xw - fw_x,f_xv + fv_x - f_yu - fu_y
	= f_yw - f_zv,f_zu - f_xw,f_xv - f_yu + fw_y - fv_z,fu_z - fw_x,fv_x - fu_y
	= (f_x,f_y,f_z) × (u,v,w) + f(w_y - v_z,u_z - w_x,v_x - u_y)
	= (▿f) × (u,v,w) + f(▿× A)

2 Problem 17 from Vector Calculus Notes

3 Problem 18 from Vector Calculus Notes

Let

A = xi +2yj+ 3zk

and

B = 3yi- 2xj

(a) Check Product Rule 2

For the left hand side of rule 2, we have

▿(A ⋅ B)	= ▿(3xy - 4xy)
	= ▿(-xy)
	= (-y,-x,0)

and the right hand side we have

A × (▿× B) + B × (▿× A) + (A ⋅▿)B + (B ⋅▿)A	= A × (0,0,-5) + B × (0,0,0) + (A ⋅▿)B + (B ⋅▿)A
	= (-10y,5x,0) + (0,0,0) + (A ⋅▿)B + (B ⋅▿)A
	= (-10y,5x,0) + (A ⋅▿)B + (B ⋅▿)A
	= (-10y,5x,0) + B
	+ A
	= (-10y,5x,0) + (6y,-2x,0) + (3y,-4x,0)
	= (-y,-x,0)

which is the same as the left hand side.

(b) Check Product Rule 4

For the left hand side of rule 4, we have

▿⋅ (A × B)	= ▿⋅ (6xz,9yz,-2x² - 6y²)
	= 6z + 9z + 0
	= 3z

and the right hand side we have

(▿× A) ⋅ B - A ⋅ (▿× B)	= (0,0,0) ⋅ B - A ⋅ (0,0,-5)
	= -A ⋅ (0,0,-5)
	= -15z

which is the same as the left hand side.

(c) Check Product Rule 6

For the left hand side of rule 4, we have

▿× (A × B)	= ▿× (6xz,9yz,-2x² - 6y²)
	= (-21y,10x,0)

and the right hand side we have

(B ⋅▿)A - (A ⋅▿)B + A(▿⋅ B) - B(▿⋅ A)	= A -B
	+ A(▿⋅ B) - B(▿⋅ A)
	= (3y,-4x,0) - (6y,-2x,0) + A(▿⋅ B) - B(▿⋅ A)
	= (-3y,-2x,0) + A(▿⋅ B) - B(▿⋅ A)
	= (-3y,-2x,0) + A(0) - B(1 + 2 + 3)
	= (-3y,-2x,0) - 6B
	= (-3y,-2x,0) - (18y,-12x,0)
	= (-21y,10x,0)

which is the same as the left hand side.