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I need the opposite direction, I was not quite seeing it. I then decided to try and find a transformation from square to triangle in a hope to invert it.

I am trying to find an invertible transformation between the unit square $x,y \in[0,1] \times [0,1]$ and the triangle $T$ bounded by the coordinate axes and the line $y = -\frac{b}{a}x + b$, ($\xi,\eta \in T$) .

My way of thinking: If we fix an $x$ then run though $y$, we want the $\eta $ ordinate to end at $b$ and the $\xi$ ordinate to end at 0. $$\xi = ax(1-y)$$ $$\eta = by$$

After a few diagrams I am confident this is correct. I am going to try to use it to simplify an integral so I would like to be sure that this transformation is good before moving on.

EDIT

Putting this in a different way. I would like an invertible transformation between the unit square defined above and a right-angled triangle bounded by the lines $y = 0$, $x = 0$ and $$y = -\frac{b}{a}x + b$$

This triangle will have vertices $(0,0),(0,b)$ and $(a,0)$.

From the constructive comments, I might have to have the restriction that the area of the triangle and the square must be equal. It is not important that I am mapping from the unit square it just seemed the most simple to begin with.

The transformation that I found is not invertible either. If we consider the determinant of the Jacobian of the transformation we find that

$$|J| = ab(1-y)$$ which is $0$ when $y = 1$. So I have failed in that respect.

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    1. We know nothing about your T. 2) By mapping a square to a triangle, do you mean that they are of equal area? – 2017-02-25
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    What do you mean: " We know nothing about your triangle?" I thought it would be clear from "the triangle bounded by the coordinate axes and the line...". Ah I did not consider whether or not they need to have equal area. – 2017-02-25
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    Well you can work out on your own that you have constrained a valid triangle, now what about defining the type of transformation $T$? Is it a linear transformation, a continuous one, invertible? – 2017-02-25
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    The "T" is referring to that T in $"\xi,\eta \in T"$. – 2017-02-25
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    That can not be determined from what you have written in the question. – 2017-02-25
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    This cannot be invertible: all the square side with $y=1$ is mapped to a single point. It should be fine if you map only the interior points. – 2017-02-25
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    Yes, you are right! Could you suggest another approach ? – 2017-02-25

2 Answers 2

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What about something like this?

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  • 0
    This looks good. – 2017-02-25
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    I will have to take some time to think about how I am going to implement this. – 2017-02-25
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    @HMPARTICLE: essentially, split both the square and the triangle in four triangular parts, and map them accordingly to a dilation along a suitable axis. – 2017-02-25
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