Understanding visually a change of variables (coordinate systems) in two dimensions.

Question

I understand visually what happens when we move from $(x,y)$ space to $(r,\theta)$ space, or vice versa.

But can someone explain to me this change of variables, $(\xi,\eta), \xi=x+y,\eta=x-y.$ Which points get map to where?

Thanks.

Answer 1

Ok, that's a good start. You're right about where the lines are getting mapped. Note that the $x$-axis being mapped to $\{\xi=\eta\}$ and $y$-axis being mapped to $\{\xi=-\eta\}$ already should suggest to you something fishy is going on, because the orientation seems to have been reversed, so some type of reflection must be taking place. Next, if $(x,y)=(1,0)$ then $(\xi,\eta)=(1,1)$, so something of length $1$ gets distorted to something of length $\sqrt{2}$; this suggests to us that pure rotation/reflection is out of the question. Consider the following three functions:

• $f_1:\Bbb{R}^2\to\Bbb{R}^2$, $f_1(x,y)=(y,x)$ ,
• $f_2:\Bbb{R}^2\to\Bbb{R}^2$, $f_2(a,b)=(\frac{a+b}{\sqrt{2}}, \frac{a-b}{\sqrt{2}})$,
• $f_3:\Bbb{R}^2\to\Bbb{R}^2$, $f_3(\alpha,\beta):=(\sqrt{2}\alpha,\sqrt{2}\beta)$.

Now, $f_1$ simply reflects across the line $\{x=y\}$, since we're just swapping the roles of $x$ and $y$. Next, $f_2$ takes a point $(a,b)$ and rotates it counter clockwise by $45^{\circ}$, or $\frac{\pi}{4}$ radians (the $\frac{1}{\sqrt{2}}$ is precisely the cosine and sine of $\frac{\pi}{4}$). Lastly, $f_3$ takes a point $(\alpha,\beta)$ and rescales by a factor of $\sqrt{2}$. Now consider the triple composition $f=f_3\circ f_2\circ f_1:\Bbb{R}^2\to\Bbb{R}^2$. If you carry out the computation, you'll see that \begin{align} f(x,y)=\cdots =(x+y,x-y) \end{align} Thus, the change of variables $\xi=x+y,\eta=x-y$ is simply the mapping $f$ I have just written down for you. So, it does three things: first reflects across the diagonal, then rotates counter clockwise by $\frac{\pi}{4}$, and finally rescales by $\sqrt{2}$.

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Since these are all linear transformations, we can also reword the argument above in terms of matrices (of the $f$'s with respect to the standard ordered basis of $\Bbb{R}^2$): \begin{align} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}&= \begin{pmatrix} \sqrt{2}&0\\ 0&\sqrt{2} \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \end{pmatrix}\cdot \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \end{align} The matrices, going right to left are the reflection across the diagonal, rotation counter clockwise by $\frac{\pi}{4}$ and scaling by $\sqrt{2}$. So, this is just another way of confirming what I already said above.

Answer 2

yes you just charge the coordinate system to the new axes $x=y ,;and x=-y$ example: $(t^2,t)$ black to, $(t^2+t,t^2-t)$green