Find $J_f$ of a function $f:\Bbb R^2\to\Bbb R^2$.

Solution 1:

Inverse Function Theorem (I.V.F.) does say something about the inverse function. Instead of explicitly giving the inverse function itself, it tells us what the total differential (Jacobian) is in terms of the original function , which means the inverse is differentiable and you can obtain the total differential without even knowing the inverse.

Let $f\left(p\right)=q$ at which $\det J_f\left(p\right)\neq 0$. According to I.V.F, the Jacobian matrix is given by $$ J_g\left(x,y\right)=\left(J_f\left(g\left(x,y\right)\right)\right)^{-1}, $$ where $g$ is a local inverse defined on a neighborhood of $q$.

You don't have to know what exactly $g$ is if you simply want to compute $J_g\left(x,y\right)$. The point is you need to recognize to what value $g$ sends $\left(x,y\right)$. We illustrate it as follows.

Put $p_1=\left(1,0\right),p_2=\left(0,1\right),p_3=\left(-1,0\right),p_4=\left(0,-1\right)$. Note that $$f\left(p_1\right)=f\left(p_2 \right)=f\left(p_3\right)=f\left(p_4\right)=\left(1,0\right)$$ and $\det J_f\left(x,y\right)\neq 0$ at each of these four points, so according to the I.V.F., there exist four local inverses $g_i:V_i\to U_i$, where $V_i$ is an open neighborhood of $\left(1,0\right)$ while $U_i$ is an open neighborhood of $p_i$. Thus, notice that $$\left(J_f\left(x,y\right)\right)^{-1}=\frac{1}{4x^2-4y^2}\begin{bmatrix} 2x &-2y\\ -2y &2x \end{bmatrix},$$ so we obtain $$ J_{g_1}\left(1,0\right)=\left(J_f\left(p_1\right)\right)^{-1}=\begin{bmatrix} \frac{1}{2} &0\\ 0 &\frac{1}{2} \end{bmatrix}, $$ $$ J_{g_2}\left(1,0\right)=\left(J_f\left(p_2\right)\right)^{-1}=\begin{bmatrix} 0 &\frac{1}{2}\\ \frac{1}{2} &0 \end{bmatrix}, $$ $$ J_{g_3}\left(1,0\right)=\left(J_f\left(p_3\right)\right)^{-1}=\begin{bmatrix} -\frac{1}{2} &0\\ 0 & -\frac{1}{2} \end{bmatrix}, $$ $$ J_{g_4}\left(1,0\right)=\left(J_f\left(p_4\right)\right)^{-1}=\begin{bmatrix} 0 &-\frac{1}{2}\\ -\frac{1}{2} &0 \end{bmatrix}. $$ Does it make sense?

Edition. Thank Michael Hoppe for pointing out a mistake and I have revised this post. In addition, I supplement an example to make clear what I have tried to say.