Elementary commentary and graphics of implicit functions and vector fields.
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An Elementary Note: Inverse & Implicit Vector Fields and Contours in the Complex Plane John Gill Spring 2015 Given a vector field : () fz F our task is to construct the inverse field 1 :() () gz f z - = G when such a field exists. This is not easily accomplished if () gz 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, : (( ) ) kn k n kn k n k n z z fz z γ η - - - = + - or ( ): () () dz zt z fz z dt ϕ = = - , :0 1 t → Usually , 1 kn n η = . For the inverse VF we have ( ) ( ), () 0 z fz ζ ζ ζ Φ = - = , to be solved for z for each ζ by a simple gradient method (not requiring holomorphic functions). Then (2) 1 , 1, , 1, 1, : (( ) ) kn k n kn k n k n z γ ζ ζ η ζ ζ - - - - = + - or 1 ( ): ( ) ( ) d t g f dt ζ ζ ζ ζ ζ ζ - = - = - Where { } , 1 : lim n kn k n z γ = →∞ and { } 1 , 1 : lim n kn k n γ ζ - = →∞ are continuums of points in C . We begin with a simple explicit example: Example 1 : : and : () z e Ln z F G
Transcript
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 .
