Answer to Question #134097 in Real Analysis for Senzo

Question #134097
Let F : R² → R² be defined by
F(x,y)= (x²-y²,2xy)

(a) Using the inverse function theorem, determine the points in R²
at which F has a local inverse
(b) Determined whether F has an inverse defined on all of R²
2. Is such an inverse of F unique?
1
Expert's answer
2020-09-22T13:51:25-0400

(a) We remind the Inverse function theorem:

Theorem.

Let "f:\\mathbb{R}^n\\rightarrow\\mathbb{R}^n" be continuously differentiable on some open set containing a, and suppose that "det Jf(a)\\neq0" , where "Jf(a)" denotes a Jacobi matrix at the point "a." Then, there is an open set "V" containing "a" and an open set "W" containing "f(a)" such that "f:V\\rightarrow W" has a continuous inverse "f^{-1}:W\\rightarrow V" , which is differentiable for all "y\\in W" .


1.(a). We compute "JF(x,y)=\\begin{pmatrix}\n \\frac{\\partial F_1}{\\partial x} & \\frac{\\partial F_1}{\\partial y} \\\\\n \\frac{\\partial F_2}{\\partial x} & \\frac{\\partial F_2}{\\partial y}\n\\end{pmatrix}=\\begin{pmatrix}\n 2x & -2y \\\\\n 2y & 2x\n\\end{pmatrix}" .

"det JF(x,y)=4x^2+4y^2" . The latter is greater than 0 for all "x,y" except of the point (0,0). Thus, the inverse function exists for all points (any neighborhood of (0,0) contains points, where function has a continuous inverse)

(b). "F(x,y)" has an inverse for all points

2. The inverse is not unique. E.g, we can take F(-2,-1)=F(2,1)=(3,4). More generally, F(-x,-y)=F(x,y)



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