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Product of Jacobians Proof

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data

    The pair of variables (x, y) are each functions of the pair of variables (u, v) and vice versa.
    Consider the Jacobians A=d(x,y)/d(u,v) and B=d(u,v)/d(x,y). Show using the chain rule that the product AB of these two matrices equals the unit matrix I.

    2. Relevant equations



    3. The attempt at a solution

    I wrote out the two Jacobians and tried to multiply them but I got the following:
    (dx/du)(du/dx)+(dx/dv)(dv/dx) (dx/du)(du/dy)+(dx/dv)(dv/dy)
    (dy/du)(du/dx)+(dy/dv)(dv/dx) (dy/du)(du/dy)+(dy/dv)(dv/dy)

    = 2 2dy/dx
    2dy/dx 2

    Where did I go wrong/ how do I continue this question?
     
  2. jcsd
  3. Jan 27, 2012 #2
    Your product of A and B is correct, but I don't quite see how you got the result after that. Can you explain "= 2 2dy/dx " ?
    2dy/dx 2

    Perhaps you mean that those are what you get in the diagonals? I'm not sure.
     
  4. Jan 27, 2012 #3
    I see what you were getting at. You're close enough to the solution, re-work the part where you simplified the partial derivatives. Check your differentials on the off-diagonal.

    Edit: I removed the part about a constant multiplying the identity. Indeed B should be the inverse mapping that takes (u,v) back to (x,y), or the other way if you wish.
     
    Last edited: Jan 27, 2012
  5. Jan 27, 2012 #4
    Sorry, I wasn't sure how to format matrices so it's a bit unclear.
    For the first row, first column: (dx/du)(du/dx)+(dx/dv)(dv/dx)=1+1=2
    First row, second column: (dx/du)(du/dy)+(dx/dv)(dv/dy)= (dx/dy)+(dx/dy)=2(dx/dy) as the u's and v's cancel out.
    Second row, first column:(dy/du)(du/dx)+(dy/dv)(dv/dx)=(dy/dx)+(dy/dx)=2(dy/dx) as the u's and v's cancel out.
    Second row, second column:(dy/du)(du/dy)+(dy/dv)(dv/dy)=1+1=2

    I'm not sure where I've made a mistake...

    Thanks so much for helping! :-)
     
  6. Jan 29, 2012 #5
    I understand the terms on the main diagonal now, by manipulating the total derivative, but I'm still confused about the other two terms.

    Any help would be great.
     
  7. Jan 29, 2012 #6

    Deveno

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    Science Advisor

    if y isn't a function of x at all, and x isn't a function of y at all, what can you say about ∂y/∂x and ∂x/∂y? think of the simplest example:

    x(u,v) = u
    y(u,v) = v

    (so that u(x,y) = x, v(x,y) = y).
     
  8. Jan 29, 2012 #7
    If y and x don't depend on each other, would the partial derivatives be zero?
     
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