Question on Jacobian with function composition and inverse functions

  • Thread starter mnb96
  • Start date
710
5
Hello,

let's suppose I have two functions [itex]\phi:U\rightarrow V[/itex], and [itex]T:V\rightarrow V[/itex] that are both diffeomorphisms having inverse.
Furthermore [itex]T[/itex] is linear.

I consider the function [itex]f(u) = (\phi^{-1}\circ T \circ \phi)(u)[/itex], where [itex]\circ[/itex] is the composition of functions.

Since [itex]T[/itex] is linear, we already know that the Jacobian determinant is constant: [itex]J_T(v)=\lambda[/itex].

What can we say about [itex]J_f(u)[/itex], the Jacobian of f ?
 

tiny-tim

Science Advisor
Homework Helper
25,790
249
710
5
you go first! :wink:
Ok, but you'll see I won't go very far :)
Let's try, though...

If I am not wrong the Jacobian obeys an analogous rule of the chain rule for derivatives, so we can write:
[tex]J_f(u) = J_{\phi^-1}(T(\phi(u))) \; J_{T}(\phi(u)) \; J_\phi(u)[/tex]

The Jacobian determinant of a product of Jacobians is given by the product of Jacobian determinants. We also know that the Jacobian determinant of T is constant, thus:

[tex]|J_f(u)| = \lambda \cdot |J_{\phi^-1}(T(\phi(u))) |\cdot|J_\phi(u)|[/tex]

...and here I get stuck.
I told you I wasn't going to go very far... :smile:
 

tiny-tim

Science Advisor
Homework Helper
25,790
249
well, that's quite a long way! :smile:

ok, you've proved that Jf is a multiple of Jφ-1Jφ

now what would you like to be able to say about Jφ-1Jφ ? :wink:
 
710
5
now what would you like to be able to say about Jφ-1Jφ ? :wink:
Basically, I would like to know if that term can be somehow simplified.
My guess is that we can't say much more than that because Jφ-1 is a function of [itex](T \circ \phi)(u)[/itex] while the term Jφ depends on u.
 
349
1
I think the difficulty here though is that the Jacobian of the inverse map is not being evaluated at the image point of the forward map phi. Rather it is being evaluated at T(phi(u)). I think that is why the OP is stuck. Am I missing a way to simplify this?
 
710
5
Hi Vargo,

the problem you mentioned in your post describes well why I get stuck.
Basically I can't see a way to simplify that expression, because the two Jacobians are evaluated at different points.
 

Related Threads for: Question on Jacobian with function composition and inverse functions

Replies
3
Views
468
  • Posted
Replies
10
Views
2K
  • Posted
Replies
1
Views
2K
Replies
1
Views
2K
Replies
19
Views
2K
  • Posted
Replies
18
Views
3K
Replies
3
Views
1K
  • Posted
Replies
3
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top