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I know that a first-order linear constant coefficient PDE, such as $au_x + bu_y = 0$, can be transformed to an ODE by rotating the coordinate system so the $x'$ axis points to $(a,b)$ where the directional derivative vanishes.

As far as I know, a coordinate rotation is done by $x' = xcos\theta + y sin\theta$ and $y' = y' cos\theta - x' sin\theta$

However, in basic PDE textbooks this is done via:

$x' = ax + by$ and $y' = bx - ay$

My problem is that I do not understand how did we get this formulas. Although I guess that it should be quite simple, for example I know that the vector $(b,-a)$ is the orthogonal to $(a,b)$ and it should have a relationship with the problem, I do not get it.

Thanks!

DOMiguel
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The formulas with $a$ and $b$ not only rotate the coordinate system, they also change the scale of the axes by a factor of $\sqrt{a^2+b^2}$, and reverse the relative orientation of the axes (since the determinant of the coordinate change is negative). If you take $$ x'' = \frac{x'}{\sqrt{a^2+b^2}} = \frac{a}{\sqrt{a^2+b^2}} \, x + \frac{b}{\sqrt{a^2+b^2}} \, y $$ and $$ y'' = \frac{-y'}{\sqrt{a^2+b^2}} = -\frac{b}{\sqrt{a^2+b^2}} \, x + \frac{a}{\sqrt{a^2+b^2}} \, y , $$ then you can identify $a/\sqrt{a^2+b^2}$ with $\sin\theta$ and $b/\sqrt{a^2+b^2}$ with $\cos\theta$ to see that the $(x'',y'')$ coordinate system is rotated with respect to the $(x,y)$ system.

Hans Lundmark
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  • I am sorry but I still do not get the idea. First, how did you identify $a/\sqrt(a^2+b^2)$ with $sin\theta$? Second, I do not yet realize how to derive $x' = ax+by$ and $y' = bx-ay$? – DOMiguel Sep 25 '14 at 23:08
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    First question: $$\left( \frac{b}{\sqrt{a^2+b^2}},\frac{a}{\sqrt{a^2+b^2}} \right)$$ is a point on the unit circle, hence it's equal to $(\cos\theta,\sin\theta)$ for some $\theta$. But I don't think I quite understand the second question. I'm not deriving your those formulas from scratch, I'm just trying to show that the are really the same as the ones that you know (provided only that you change the direction of the $y'$ axis and change the scale on both the $x'$ axis and the $y'$ axis by a factor of $\sqrt{a^2+b^2}$). – Hans Lundmark Sep 26 '14 at 05:49