What is the Jacobian of a Transformation with x=u/(u+v) and y=v/(u-v)?

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SUMMARY

The Jacobian of the transformation defined by x = u/(u+v) and y = v/(u-v) is calculated using the formula for the Jacobian determinant. The partial derivatives are computed as follows: ∂x/∂u = v/(u+v)², ∂x/∂v = -u/(u+v)², ∂y/∂u = -v/(u-v)², and ∂y/∂v = u/(u-v)². The resulting Jacobian determinant simplifies to 0, indicating that the transformation is not locally invertible at the given points.

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Homework Statement



Find the Jacobian of the transformation:

x=\frac{u}{u+v}, y=\frac{v}{u-v}

Homework Equations



Jacobian = \left|\stackrel{\frac{\partial x}{\partial u}}{\frac{\partial x}{\partial v}} \stackrel{\frac{\partial y}{\partial u}}{\frac{\partial y}{\partial v}}\right| =\left(\frac{\partial x}{\partial u}\right) \left(\frac{\partial y}{\partial v}\right) - \left(\frac{\partial x}{\partial v}\right) \left(\frac{\partial y}{\partial u}\right)

The Attempt at a Solution



Now, I got for my four partial derivatives...

\frac{\partial x}{\partial u} = \frac{v}{\left(u+v\right)^2}

\frac{\partial x}{\partial v} = - \frac{u}{\left(u+v\right)^2}

\frac{\partial y}{\partial u} = - \frac{v}{\left(u-v\right)^2}

\frac{\partial y}{\partial v} = \frac{u}{\left(u-v\right)^2}

So, multiplying these together gave me...

Jacobian = \frac{vu}{(u+v)^2 (u-v)^2} - \frac{uv}{(u+v)^2 - (u-v)^2} = 0

Am I supposed to get a Jacobian of 0?
 
Last edited:
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Fixed a bit of LaTeX.
 

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