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If $\left(x_{0},y_{0}\right)\in\left(0,\infty\right)\times\left(0,\infty\right)$, then it easy to get $\left(u_{0},v_{0}\right)\in\left(0,\infty\right)\times\left(0,1\right)$ though the parametric equations $$u=x+y\quad\textrm{and}\quad v=\frac{x}{x+y}.$$

So I tried running the below code:

Region[ParametricRegion[{{x + y, x/(x + y)}, 0 < x && 0 < y}, {x, y}],Axes -> True, AxesOrigin -> {0, 0}]

but I got the below region:

enter image description here

Why the region does not coincide with $\left\{\left(u,v\right):0<u \wedge 0<v<1 \right\}$? Could you please help me diagnose this issue?

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1 Answer 1

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  • Since we can only plot a mapping from a finite region to another finite region. And by default, we map a finite rectangle to another region which the boundary may not be a rectangle.
ani = Manipulate[
  GraphicsRow[{ParametricPlot[{x, y}, {x, 0, c}, {y, 0, c}, 
     Mesh -> 10, PlotStyle -> Yellow, MeshStyle -> Cyan, 
     BoundaryStyle -> Red, PlotRange -> Automatic], 
    ParametricPlot[{x + y, x/(x + y)}, {x, 0, c}, {y, 0, c}, 
     Mesh -> 10, PlotStyle -> Yellow, MeshStyle -> Cyan, 
     BoundaryStyle -> Red, PlotRange -> {{0, 6}, {0, 1}}, 
     PerformanceGoal -> "Quality"]}, ImageSize -> Full], {c, 1, 8}]

enter image description here

  • If we want the region be a rectangle,we need to limit the range domain be a triangle. ImplicitRegion[{x > 0, y > 0, x + y <= c}, {x, y}]
c = 10; 
GraphicsRow[{ParametricPlot[{x, y}, {x, y} ∈ 
    ImplicitRegion[{x > 0, y > 0, x + y <= c}, {x, y}], Mesh -> 10, 
   MeshFunctions -> {#3 + #4 &}, PlotStyle -> Yellow, 
   MeshStyle -> Cyan, BoundaryStyle -> Red, PlotRange -> Automatic], 
  ParametricPlot[{x + y, x/(x + y)}, {x, y} ∈ 
    ImplicitRegion[{x > 0, y > 0, x + y <= c}, {x, y}], Mesh -> 10, 
   MeshFunctions -> {#3 + #4 &}, PlotStyle -> Yellow, 
   MeshStyle -> Cyan, BoundaryStyle -> Red, PlotRange -> All]}, 
 ImageSize -> Full]

enter image description here

Appendix

FunctionBijective[{{x + y, x/(x + y)}, x > 0 && y > 0, 
  u > 0 && 0 < v < 1}, {x, y}, {u, v}]

True

FunctionRange[{{x + y, x/(x + y)}, x > 0, y > 0, x + y > 0}, {x, 
   y}, {u, v}, Reals] // Simplify

u > 0 && v > 0 && v < 1

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1
  • 1
    $\begingroup$ Thanks for your help.Yes,indeed,this transformation $\psi :\mathbb{R^2}\rightarrow\mathbb{R^2}$ $$\psi(x,y)=(x+y,\frac{x}{x+y})$$ is one-to-one on a unit square $\mathcal{R}=\{(x,y):0<x<1,0<y<1\}$ and maps the region $\mathcal{R}$ in the $xy$-plane onto the non-retangle $\mathcal{R^*}=\{(u,v):0<uv<1,0<u(1-v)<1\}$ in the $uv$-plane. $\endgroup$
    – AplehKevin
    Jan 29 at 7:36

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