1.12 Jacobians

If \(F(u,v)\) and \(G(u,v)\) are differentiable in a region, then Jacobian determinant (or Jacobian) of \(F\) and \(G\) with respect to \(u\) and \(v\) is the second order functional determinant defined by \[ \frac {\partial \big (F,G\big )}{\partial \big (u,v\big )} = \begin {vmatrix} \dfrac {\partial F}{\partial u} & \dfrac {\partial F}{\partial v}\\\\ \dfrac {\partial G}{\partial u} & \dfrac {\partial G}{\partial v}\\ \end {vmatrix} \]

\(|||_y\), the third order determinant \[ \frac {\partial \big (F,G, H\big )}{\partial \big (u,v,w\big )} = \begin {vmatrix} \dfrac {\partial F}{\partial u} & \dfrac {\partial F}{\partial v} & \dfrac {\partial F}{\partial w}\\\\ \dfrac {\partial G}{\partial u} & \dfrac {\partial G}{\partial v} & \dfrac {\partial G}{\partial w}\\\\ \dfrac {\partial H}{\partial u} & \dfrac {\partial H}{\partial v} & \dfrac {\partial H}{\partial w}\\ \end {vmatrix} \] is called the determinant of \(F\), \(G\), \(H\) with respect to \(u,v,w\).

Properties of Jacobians

1.: If \(u\) and \(v\) are functions of \(x\) and \(y\) then \[\frac {\partial \big (u,v\big )}{\partial \big (x,y\big )}\times \frac {\partial \big (x,y\big )}{\partial \big (u,v\big )}=1\]
2.: If \(x\) and \(y\) are functions of \(u\) and \(v\) while \(u\) and \(v\) are functions of \(r\) and \(s\), then \[\frac {\partial \big (x,y\big )}{\partial \big (r,s\big )} = \frac {\partial \big (x,y\big )}{\partial \big (u,v\big )}\times \frac {\partial \big (u,v\big )}{\partial \big (r,s\big )}\hspace {0.3cm}\text {Chain Rule}\]

Example 1.12.1. If \(x = uv\) and \(y =\dfrac {u + v}{u -v}\) find \[\frac {\partial \big (u,v\big )}{\partial \big (x,y\big )}= \begin {vmatrix} \dfrac {\partial u}{\partial x} & \dfrac {\partial u}{\partial y}\\\\ \dfrac {\partial v}{\partial x} & \dfrac {\partial v}{\partial y}\\ \end {vmatrix} \]

Solution. but \(\dfrac {\partial u}{\partial x}, \dfrac {\partial u}{\partial y}, \dfrac {\partial v}{\partial x}\) and \(\dfrac {\partial v}{\partial y}\) are comparatively difficult than \(\dfrac {\partial x}{\partial u}, \dfrac {\partial x}{\partial v}, \dfrac {\partial y}{\partial u}\) and \(\dfrac {\partial y}{\partial v}\).

So we find \(\frac {\partial \big (x,y\big )}{\partial \big (u,v\big )}\). \begin {align*} \frac {\partial \big (x,y\big )}{\partial \big (u,v\big )} & =\begin {vmatrix} \dfrac {\partial x}{\partial u} & \dfrac {\partial x}{\partial v}\\\\ \dfrac {\partial y}{\partial u} & \dfrac {\partial y}{\partial v}\\ \end {vmatrix}= \begin {vmatrix} v & u\\\\ \dfrac {-2v}{(u-v)^2} & \dfrac {2u}{(u-v)^2}\\ \end {vmatrix}\\\\ & = \frac {4uv}{(u -v)^2} \end {align*}

but \(\dfrac {\partial \big (u,v\big )}{\partial \big (x,y\big )}\times \dfrac {\partial \big (x,y\big )}{\partial \big (u,v\big )}=1\) \[\therefore \hspace {0.5cm}\frac {\partial \big (u,v\big )}{\partial \big (x,y\big )} = \frac {(u -v)^2}{4uv}\]

Exercise 1.12.2. If \(xyz =h,\hspace {0.3cm} v = x^2 + y^2 + z^2,\hspace {0.3cm} w = x+y+z,\) find \(\dfrac {\partial \big (x,y,z\big )}{\partial \big (h,v,w\big )}\)

Example 1.12.3. Find the Jacobian \(\dfrac {\partial \big (u,v\big )}{\partial \big (r,\theta \big )}\) where \(u = x^2 - y^2,\hspace {0.3cm} v = 2xy\) and \(x = r\cos \theta ,\hspace {0.3cm} y = r\sin \theta \).

Solution. \begin {align*} \frac {\partial \big (u,v\big )}{\partial \big (r,\theta \big )} & = \frac {\partial \big (u,v\big )}{\partial \big (x,y\big )}\times \frac {\partial \big (x,y\big )}{\partial \big (r,\theta \big )}\\\\ & = \begin {vmatrix} \dfrac {\partial u}{\partial x} & \dfrac {\partial u}{\partial y}\\\\ \dfrac {\partial v}{\partial x} & \dfrac {\partial v}{\partial y}\\ \end {vmatrix} \times \begin {vmatrix} \dfrac {\partial x}{\partial r} & \dfrac {\partial x}{\partial \theta }\\\\ \dfrac {\partial y}{\partial r} & \dfrac {\partial y}{\partial \theta }\\ \end {vmatrix}\\\\ & = \begin {vmatrix} 2x & -2y\\ 2y & 2x\\ \end {vmatrix}\times \begin {vmatrix} \cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \\ \end {vmatrix}\\\\ & = \big (4x^2 + 4y^2\big )\cdot r\\ & = 4r^2\cdot r\\ & = 4r^3\\\\ \end {align*}

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