An Elementary Note (2) on Implicit functions and Vector Fields - Images

John Gill Summer 2015

Implicit functions of the form ( ), 0 , :0 1z t t = can be solved explicitly in the following way: differentiate 0

d

dt

= and solve for ( , ) ( , )

dzz t F z t z

dt= = , with vector field ( , )F z t .

Set 0t = to find the starting point for a Zeno contour

( )1, 1, 1,: , , kk n k n n k n nz z z n = + Then for each value of t there is a corresponding point ( )z t on the contour.

Example 1 : ( ) 0zz e t+ = , 2( ) 2 1t t it = + + , (0) 1 (0) 0z = =

2( )

( , ) ( , )1z

dz t iz t F z t z

dt e += = =

+. The green vector field is ( , )F z t . The -contour is red

(vector field red) and the z -contour is green. Each point ( )t on the red contour has a corresponding point ( )z t on the green contour that is its implicit function value. Here

1 ,

1

, 1,

(1 ): ,

1 k n

kn

k n k n n z

iz z n

e

+ = + +

For instance, to find (.5)z start at (0) 0z = and stop at /2,

.1798 .4642n nz i + for n

It is not necessary for the z and the t to be separated:

Example 2 : 2( 2 1) 0zzt e t ti+ + + =

(.5) .278 .571z i +

Example 3 : 2( ) ( ) 0Sin zt z t it+ + =

(1) .47 .52z i +

Example 4 : 2 2( ) 0 , ( ) 1zzte z t t t it = = + + . The TDVF for ( )t is in red.

Non-holomorphic functions:

Example 5 : ( ) ( ) 2( ) 0 , ( ) 1 2y xx e i e y t t t ti + + = = + +

2( ) 2( 1)

1 1

y x

x y x y

dz dx dy t e tei i

dt dt dt e e

+ +

+ = = + = +

+ + , (0) .669 .512z i +

(1) .786 .194z i +

Implicit Time-dependent Vector Fields and Contours

Consider implicit functions of three variables, written ( , ; ) 0z t = where is independent and :0 1t , so that ( )z z t= . The -vector field is time-dependent and requires implicit computations of vector clusters at each point . Once the field has been graphically displayed, a (Zeno) contour beginning at an arbitrary

0 must be computed point-by-point , each step

requiring the evaluation of the implicit function:

, 1, 1,

: k n k n n k n

z z Z

= + where ( )11, 1,, ; 0kk n k n nZ z = , and 0, 0nz =

Example 6 : ( ) , ; 0ze zt z t = = , 05 Re 5 , -5 Im 5 , 3 3i =

There are imperfections in the implicit value search program (jumping from branch to branch).

Example 7 : ( ), ; 0ztz t e z = = , 06 Re 6 , -5 Im 5 , 5.2 3.4i = +

The implicit value search program is relatively stable in quadrant I.

Topographical Images of Implicit Functions

The Zeno process can be used to paint images of implicit functions much more efficiently than a

gradient algorithm that must search discretely for each value. Consider the following

expression: ( ) 0ze z t+ = where ( ) , :0 1 , :0 1t t i t = + . The region outlined by is the unit square and an algorithm that starts at 0, 0t = = and advances with incremental steps of , say .005, and at each step generates a Zeno contour as :0 1t

incrementally , calculating the modulus of ( )z t at each step and coloring the appropriate pixel

in the unit square , will produce an image in the plane that represents the topography of the

implicit function ( )z t over the region bounded by the square, or rectangle in general. The one

major drawback is the primitive nature of the implicit search routine I devised, which goes

haywire for periodic functions.

Example 8 : ( ) 0ze z t+ = , ( ) , : 2 4 , : 2 2t t i t = +

Example 9 : 2 ( ) 0z t = (square root of z) ( ) , :0 2 , :0 2t t i t = +

Example 10 : 31100

( ) 0z iz t = , ( ) , : 2 2 , : 2 2t t i t = +

Suppose the implicit function takes a non-holomorphic form:

( ( ), ( )) ( ( ), ( )) ( ) 0U x t y t iV x t y t t+ = . Then ( , )dz dx dyi z tdt dt dt

= + =

Example 11 : ( ) ( ) ( ) 0x ye y i e x t+ + = , ( ) , :0 2.5 , :0 2.5t t i t = + . 1 1

, 1 1 1 1

y y

x y x y x y x y

dx e dy ei

dt e dt e e e

+ + + += = = +

+ + + +

Example 12 : ( ) ( )sin 2 ( ) 0x y i x y t + = , ( ) , : 2 2 , : 2 2t t i t = +

Example 13 : ( ) ( )2 ( ) 0x ye e i x y t + = , ( ) , : 2 2 , : 2 2t t i t = +