Question

Let $\Omega$and $\Omega \text{'}$be subsets of $R^2$. Use the results of Exercise 59 to prove that if a transformation $T:\Omega \to \Omega \text{'}$is invertible, and if both T and $T^{-1}$are differentiable, then role="math" $\frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (x,y)}=1$.

Short answer

It is proved that$\frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (x,y)}=1$.

Step-by-step solution

Given information

Consider a chain of functions is defined as,

$$\begin{gathered} x=x(u,v);y=y(u,v) \\ u=u(s,t);v=v(s,t) \end{gathered}$$

The chain rule of Jacobian is defined as,

role="math" $\frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (s,t)}=\frac{\partial (x,y)}{\partial (s,t)}$

Proof

Consider a transformation $T:(x,y)\to (u,v)and{T}^{-1}:(u,v)\to (x,y)$.

Use the above-given definition of the chain rule of Jacobians to derive the relation.

$\frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (x,y)}=\frac{\partial (x,y)}{\partial (x,y)}=1$

Hence, proved.