How to find Jacobian?

Question

I've learned to find Jacobian where x is equal to something and y is equal to something. Here I have the following problem: $u = x + 2y$ and $v = x - y$, in this case, how am I supposed to get $\partial(x,y) \over \partial(u,v)$? In other words, how to find $\partial x \over \partial u$?

Answer 1

While it is true that you are given $u$ and $v$ in terms of $x$ and $y$, these transformations are invertible, namely that $x$ and $y$ can also be expressed as functions of $u$ and $v$.

In this case, note that $u = x+2y$ and $v = x-y$. Multiply the second equation by two and add it to the first, and you get $3x =2v+u$, so that $x = \frac{2v+u}3$. Now, if we subtract the second equation from the first, then we get $3y = u-v$, so $y=\frac{u-v}3$.

Hence, we are in a position to calculate the Jacobian: $$ J = \begin{pmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}\end{pmatrix} = \begin{pmatrix} \dfrac{1}{3} & \dfrac{2}{3} \\ \dfrac{1}{3} & \dfrac{-1}{3}\end{pmatrix} $$

Answer 2

As an alternative:

$$ \begin{pmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}\end{pmatrix} = \begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial v}{\partial x} \\ \dfrac{\partial u}{\partial y} & \dfrac{\partial v}{\partial y}\end{pmatrix}^{-1} $$

$$ =\begin{pmatrix} 1 & 2 \\ 1 & -1\end{pmatrix}^{-1} $$

$$ =\begin{pmatrix} \dfrac13 & \dfrac23 \\ \dfrac13 & -\dfrac13\end{pmatrix} $$

Note This these types of problems become much harder once you no longer have linear expressions for $u$ and $v$.