Q. 60 - Calculus

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

Step-by-Step Solution

Answer

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

Step 1: 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,

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

Step 2: 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.