An Elementary Note: Inverse & Implicit Vector Fields in the Complex Plane
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Transcript of An Elementary Note: Inverse & Implicit Vector Fields in the Complex Plane
An Elementary Note: Inverse & Implicit Vector Fields and Contours in the
Complex Plane
John Gill Spring 2015
Given a vector field : ( )f zF our task is to construct the inverse field 1: ( ) ( )g z f z−=G�� when
such a field exists. This is not easily accomplished if ( )g z is not available in closed form and
must be calculated implicitly. Additionally, Zeno contours of the inverse field should be
displayed as graphics. Implicit function analysis then follows, generalizing the inverse.
The contours in the original VF may be expressed as:
(1) , 1, , 1, 1,
: ( ( ) )k n k n k n k n k nz z f z zγ η− − −= + − or ( ): ( ) ( )
dzz t z f z z
dtϕ= = − , :0 1t →
Usually ,
1k n
nη = . For the inverse VF we have ( )( ), ( ) 0z f zζ ζ ζΦ = − = , to be solved for z for
each ζ by a simple gradient method (not requiring holomorphic functions). Then
(2) 1
, 1, , 1, 1,: ( ( ) )
k n k n k n k n k nzγ ζ ζ η ζ ζ−
− − −= + − or 1( ): ( ) ( )
dt g f
dt
ζζ ζ ζ ζ ζ−= − = −
Where { }, 1: lim
n
k n knzγ
=→∞ and { }1
, 1: lim
n
k n knγ ζ−
=→∞ are continuums of points in C .
We begin with a simple explicit example:
Example 1 : : and : ( )ze Ln zF G
In the next example the Liberty BASIC program I wrote uses an implicit evaluation procedure
that requires gradient searches to ascertain inverse function values at each grid point to plot
the vector field. Then the Zeno contour requires gradient searches, point to point, for 500
steps.
Example 2 : 3 1: ( ) and : ( ) ( )f z z iz g z f z−= + =F G
Example 3 : 1: ( ) and : ( ) ( )zf z z e g z f z−= + =F G
When the function is not univalent there can be an element of confusion in the program as the
inverse vectors and contour jump from one possibility to another. For analytic functions f , the
Jacobian 2
( )J f z′= . An inverse function exists wherever this is non-zero, (although inverse
function forms may change from one point neighborhood to another). In an intuitive sense the
reason is as follows: Suppose 1 2, ( )z z Nε α∈ . Then 1 2 1 2( ) ( ) ( ) ( )f z f z f z zα′− ≈ ⋅ − and
consequently 1 2 1 2 1 2( ) ( ) ( )f z f z f z z z zα δ′− ≈ ⋅ − > − so that in a small neighborhood of
α the function is univalent and the inverse is well-defined.
Example 4 : 2 11
: ( ) and : ( ) ( )f z z g z f zz
−= + =F G
Sometimes the implicit inverse is well behaved over a larger domain.
Example 5 :
1
: ( ) zf z z e= +F and 1: ( ) ( )g z f z−=G . Mapping of the unit circle under G :
Implicit Vector Fields and Contours
Suppose z and ζ are related by ( , ) 0zϕ ζ = and ( )z g ζ= in regions of C . Then the
vector field of ( )g ζ and associated Zeno contours can be graphed in those regions.
Example 6 : 2 1
( , ) 0z zz
ϕ ζ ζζ
= + − = , ( )z g ζ= . Mixed regions.
Example 7 :
1
( , ) 0zz e zζϕ ζ ζ+= + − = , ( )z g ζ=
Example 8 : ( , ) 0 , ( )zz e z z gζϕ ζ ζ ζ ζ+= + − = =
The topographical image over [-1,1] shows areas of questionable definition. For example,
(0)z g= is non-finite.
The Implicit Function Theorem tells us that a function ( )z g ζ= exists locally when 0z
ϕ∂≠
∂.
Here is a sketch of a proof of that theorem:
Given that ( , )zϕ ζ is analytic in both variables and 0 0( , ) 0zϕ ζ = . The question arises: Is there
1 0z z≠ such that 1 0( , ) 0zϕ ζ = or is ( )z z ζ= a proper function there? Let’s assume
0 0( , ) 0z zϕ ζ ≠ . Define an auxiliary function 0 0
( , )( , ):
( , )z
zF z z
z
ϕ ζζ
ϕ ζ= − . Then
( , ) ( , ) 0F z z zζ ϕ ζ= ⇔ = . Now 0 0( , ) 0zF z ζ = , and this means that there are
ε -neighborhoods about 0 0 and z ζ in which ( , ) 1zF z ζ ρ< < . Therefore
( , ) ( , ) ( , )z
sF z F F s ds
ξ
ζ ξ ζ ζ− = ⋅∫ and hence ( , ) ( , )F z F zζ ξ ζ ρ ξ− < ⋅ − , demonstrating that F
is a contraction mapping on the first variable, which in turn (using the Banach Fixed Point
Theorem) shows there is a unique α such that 0( , )F α ζ α= . I.e., 0( , ) 0ϕ α ζ = , so that
0 0( )z z ζ α= = is indeed unique. This reasoning extends to a small neighborhood about 0ζ .
In Example 8 0z
ϕ∂≠
∂ takes the form 0ze ζ ζ+ + ≠ . Solving the system
0
0
z
z
e z
e
ζ
ζ
ζ ζ
ζ
+
+
+ − =
+ = to
find an exceptional point, gives 1.557 0iζ ≈ − + .
Example 9 : 2( , ) 0zz e zϕ ζ ζ= + − = , ( )z g ζ=
Implicit Functions and Attractors
Suppose a function ( )z z ζ= is defined implicitly by ( , ) 0zϕ ζ = .
(1) Choose and fix 0ζ ζ= . Thus 0 0
( ) ?z z ζ= = and 0 0( , ) 0zϕ ζ = .
(2) Define a vector field (VF) by 0 0( , ) ( , )F z z zζ ϕ ζ= +
(3) Observe that 0 0( , ) ( , ) 0F z z zζ ϕ ζ= ⇔ = and therefore 0z is a fixed point of F .
(4) If 0 0( , ) 1zF z ζ < then 0z is an attracting fixed point of F .
(5) Form the Zeno contour ( ) ( ), 1, 1, 0 1, 1, 0 1,( , ) ( , )
k n k n n k n k n n k n k nz z z z F z zη ϕ ζ η ζ− − − − −= + = + −
(6) It has been shown that , 0
as n nz z n→ → ∞ for starting points close to the attractor.
(7) Write * * *1
0 0 0 02( , ) ( , ) and ( , ) ( , )F z F z z F z zζ ζ ϕ ζ ζ= = − .
(8) Then ( ) ( )* *
, 1, 1, 0 1, 1, 0 1,( , ) ( , )
k n k n n k n k n n k n k nz z z z F z zη ϕ ζ η ζ− − − − −= + = + − and
, 0 0 0( ) ( ( ), )
n nz α ζ ϕ α ζ ζ→ = , the fixed point of the implicit function.
As a consequence, it is possible sometimes to determine ( )z ζ for particular values of ζ as
the end point – an attractor - of a Zeno contour.
Example 10 : 2
0
1( , ) ( 1) 1 ,
2z z z iϕ ζ ζ ζ ζ= + − + = ,
0 0( ) ? if ( , ) 0z zζ ϕ ζ= =
The Zeno contour has been boosted by employing 10 nη rather than nη . ( ) .414 .414z i i≈ +
Example 11 : ( )ζ
ζ ζ
Φ = − =
, 010
zz Cos z . Then ( )
ζζ ζ
=
,
10
zF z Cos and the Zeno
contour terminates at ζ= ( )z z for ζ in a neighborhood of the origin and initial values of z
near the fixed points. For example, + ≈ +(1 3 ) .1573 2.5197z i i
Example 12 : 1 14 2
( , ) 0 , 1zz e z iϕ ζ ζ ζ= − − + = = . (1) 1.97 1.58 , (1) .909 1.021z i iα≈ + ≈ −
Contours defined Implicitly
Instead of ( , )zϕ ζ , consider an ordinary contour defined implicitly:
Example 13 : 3 2( , ) 2 4 =0, :0 1 , ( ) ?z t tz t z i t z tΦ = − + → = for various values of t .
(1) 1.18z i≈ (.5) 1.834z i≈ (.2) 2.67z i≈
For each value of t a Zeno contour is drawn to an attractor. The actual contour ( )z t is not
shown (and is embedded in a time dependent vector field (TDVF) not graphed).
Example 14 : 2( , ) 0tzz t e z itΦ = + − = , :0 1t → . To derive the TDVF differentiate the equation
and solve for dz
dt. Then ( , ) ( , )
dzz t F z t z
dtϕ= = − , where
(1 ( 1)) 2( , )
1
tz
tz
z e t tiF z t
te
+ − +=
+ .
Here the contour is determined directly from the original implicit format, rather than
constructing it as a Zeno contour using ( , )dz
z tdt
ϕ= .
Example 15 : 2( , ) 0 , 1zz t ze t C C i−Φ = − + = = − ,
2( , ) ( , )
1
zdz tez t F z t z
dt zϕ= = = −
−
(0) 1.34 5.25z i≈ + . The Zeno contour has been boosted using 10n
η .
Implicit Generating Functions
Normally, a Zeno contour is defined as follows: Let , ,
( ) ( )k n k n
g z z zη ϕ= + where z S∈ and
,( )
k ng z S∈ for a convex set S in the complex plane. Require
,lim 0
k nn
η→∞
= , where (usually)
1,2,...,k n= . Set 1, 1,
( ) ( )n n
G z g z= , ( ), , 1,( ) ( )k n k n k nG z g G z−= and
,( ) ( )
n n nG z G z= with
( ) lim ( )nn
G z G z→∞
= , when that limit exists. The Zeno contour is a graph of this iteration. The word
Zeno denotes the infinite number of actions required in a finite time period if ,k nη describes a
partition of the time interval [0,1]. Normally, ( ) ( )z f z zϕ = − for a vector field ( )=F f z , and
( , ) ( , )ϕ = −z t f z t z for a time-dependent vector field , in which case , ,
( ) ( , )η ϕ= + ⋅ kk n k n n
g z z z .
An implicit generator , , ,
( ) ( , ( ))k n k n k ng z z z g zη ϕ= + produces a contour
( ), 1, , 1, , 1, , 1, , 1,: ( , ) ( , )
k n k n k n k n k n k n k n k n k n k nz z z z z f z z zγ η ϕ η− − − − −= + = + −
Example 16 : ,
, 2
,
( )1( )
1 ( )
k n
k n
k n
z g zg z z
n g z
⋅ = + +
A result virtually indistinguishable from
2
1,
, 1, 2 2
1,
1 , ( )
1 1
k n
k n k n
k n
z zz z f z z
n z z
−
−
−
= + = + + +
. . . . for good reason: 2
*
2( , ) ( )
1
dz zz z z
dt zϕ ϕ= = =
+
Implicit Iteration
Suppose z and ζ are related by ( , ) 0zϕ ζ = and ( )z g ζ= in regions of C . How does
one iterate ( )z g ζ= ? In theory, iteration would look like this:
0 0 0 0 1 1 2( ) ( ) ( )z z z z z z z zζ ζ→ = → = → = →� or 1( , ) 0k kz zϕ − = , 1,2,k = �
A simple application is seen in the following example:
Example 17 : Suppose we wish to find a root of the equation 4 32 2 0α α− + = . Observe that
the implicit iteration scheme 2
1
1
1 10
2n n
n n
Z ZZ Z
−
−
+ − = , should it converge, will provide a root.
Toward this end set 21 1
( , ) 02
z zz
ϕ ζ ζζ
= + − = and begin iteration with 1 iζ = + .
The image is magnified, with the point 1+i in green, and the 250th
iterate in red. There are a
fair number of iterate points scattered in a roughly circular pattern (not shown) around the
initial value until finally the iterates fall into the spiral as shown. 500 iterates lead to the same
point, with a bit more accuracy.
Time Dependent Implicit Vector Fields
From Example 9,
Example 18 : 2( , ; ) 0ztz t e ztϕ ζ ζ= + − = , :0 1t →