Could there be a linear algebra way for converting a PDE to another coordinate system?

Question

I tried to convert a partial differential equation from rectangular to polar coordinates in 2D.
As one of the steps required, I expressed the first derivatives of each of the rectangular coordinates in terms of the polar coordinates by using the chain rule, and I got:

$\frac{\partial w}{\partial x} \;=\; \frac{\partial w}{\partial r}\!\cdot\!\color{red}{\cos{\theta}} \;+\; \frac{\partial w}{\partial \theta}\!\cdot\!\frac{\color{red}{-\sin{\theta}}}{\color{blue}{r}}$

$\frac{\partial w}{\partial y} \;=\; \frac{\partial w}{\partial r}\!\cdot\!\color{red}{\sin{\theta}} \;+\; \frac{\partial w}{\partial \theta}\!\cdot\!\frac{\color{red}{\cos{\theta}}}{\color{blue}{r}}$

Now this looks oddly familiar! The coefficients I marked with red colour look awful lot like the elements of a transformation matrix that rotates something by the angle $\theta$ ! :o

$\color{red}{\begin{bmatrix}\cos{\theta}&-\sin{\theta}\\\sin{\theta}&\cos{\theta}\end{bmatrix}}$

But the $\theta$ coordinate is additionally divided by $\color{blue}{r}$ to compensate, similarly to how we divide the arc length by the radius to get the angle (the corresponding arc length on a unit circle).

My question is then:
Is there any linear algebra trick underlying this that would show this relation more explicitly? (Of the sorts of matrix-by-vector multiplication involving the partial derivatives.)

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Aside:
In case there were are any doubts about the correctness of my calculations, here they are:

First, we have the following relations between the coordinates that will be used:

$\begin{cases} x = r\cdot\cos\theta \\[2ex] y = r\cdot\sin\theta\end{cases}$       $\begin{cases} r = \sqrt{x^2 + y^2} = \color{green}{(x^2 + y^2)^{\frac{1}{2}}} \\[2ex] \theta = \color{green}{\arctan \left( \frac{y}{x} \right)} \end{cases}$

Now, by the chain rule, we got:

$\begin{cases} \frac{\partial w}{\partial x} = \frac{\partial w}{\partial r} \color{red}{\frac{\partial r}{\partial x}} + \frac{\partial w}{\partial\theta} \color{red}{\frac{\partial\theta}{\partial x}} \\[2ex] \frac{\partial w}{\partial y} = \frac{\partial w}{\partial r} \color{red}{\frac{\partial r}{\partial y}} + \frac{\partial w}{\partial\theta} \color{red}{\frac{\partial\theta}{\partial y}} \end{cases}$

Now let's calculate the $\color{red}{\frac{\partial r}{\partial x}}$ and $\color{red}{\frac{\partial r}{\partial y}}$ that are used in these equations:

$\color{red}{\frac{\partial r}{\partial x}} = \frac{\partial}{\partial x}\left[ \color{green}{(x^2 + y^2)^{\frac{1}{2}}} \right] = (2x)\!\cdot\!\frac{1}{2}\left(x^2 + y^2\right)^{-\frac{1}{2}} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r} = \frac{r\cdot\cos\theta}{r} = \color{red}{\cos\theta}$

$\color{red}{\frac{\partial r}{\partial y}} = \frac{\partial}{\partial y}\left[ \color{green}{(x^2 + y^2)^{\frac{1}{2}}} \right] = (2y)\!\cdot\!\frac{1}{2}\left(x^2 + y^2\right)^{-\frac{1}{2}} = \frac{y}{\sqrt{x^2 + y^2}} = \frac{y}{r} = \frac{r\cdot\sin\theta}{r} = \color{red}{\sin\theta}$

and also the other two, $\color{red}{\frac{\partial\theta}{\partial x}}$ and $\color{red}{\frac{\partial\theta}{\partial y}}$:

$\color{red}{\frac{\partial\theta}{\partial x}} = \frac{\partial}{\partial x}\left[ \color{green}{\arctan \left( \frac{y}{x} \right)} \right] = \frac{1}{\left(\frac{y}{x}\right)^2 \;+\; 1} \cdot \frac{-y}{x^2} = \frac{-y}{\left( \frac{y^2}{x^2} \;+\; 1 \right)\cdot x^2} = \frac{-y}{y^2 \;+\; x^2} = \frac{-y}{r^2} = \frac{-r\cdot\sin\theta}{r^2} = \frac{\color{red}{-\sin\theta}}{\color{blue}{r}}$

$\color{red}{\frac{\partial\theta}{\partial y}} = \frac{\partial}{\partial y}\left[ \color{green}{\arctan \left( \frac{y}{x} \right)} \right] = \frac{1}{\left(\frac{y}{x}\right)^2 \;+\; 1} \cdot \frac{1}{x} = \frac{1}{\left( \frac{y^2}{x^2} \;+\; 1 \right)\cdot x} = \frac{1}{\frac{y^2}{x} \;+\; x} = \frac{1}{\frac{y^2 \;+\; x^2}{x}} = \frac{x}{x^2 \;+\; y^2} = \frac{x}{r^2} = \frac{r\cdot\cos\theta}{r^2} = \frac{\color{red}{\cos\theta}}{\color{blue}{r}}$

If you still think that there is some error with these $\color{blue}{r}$s, kindly point them out.

Also it would be nice if someone corrected the formatting, because some symbols turned out to be very small and I don't know how to fix it.

Answer 1

Yes, there is!

Directional derivatives (or rather the corresponding differential operators) behave very much like vectors, so much that if you ever get to learning about differential geometry and manifolds, you might see some sources defining a "tangent vector" to be a linear differential operator.

Anyway, to keep this more basic, the partial derivative $\partial/\partial x$ tells you what happens to the function as you move (with unit speed) in the direction of the unit vector $\mathbf{e}_x=(1,0)$. Similarly, $\partial/\partial r$ should tell you what happens to the function as you move in the direction of the unit vector $\mathbf{e}_r=(\cos\theta,\sin\theta)$.

But $\partial/\partial \theta$ is a little bit trickier; it tells you what happens to the function as you vary $\theta$ with unit speed, which actually means moving in the $xy$ plane with speed $r$ (the further out you are, the larger will the effect of varying $\theta$ be). So $\partial/\partial \theta$ corresponds to the vector $r \, \mathbf{e}_\theta = (-r \sin\theta,r \cos\theta)$ in the $\theta$ direction. Or put differently, it's the scaled operator $(1/r) \partial/\partial \theta$ that corresponds to the unit vector $\mathbf{e}_\theta = (-\sin\theta, \cos\theta)$.

With these correspondences, the relations between the differential operators are exactly as for the corresponding vectors: $$ \mathbf{e}_r=(\cos\theta,\sin\theta) = \cos\theta \, \mathbf{e}_x + \sin\theta \, \mathbf{e}_y $$ gives $$ \frac{\partial}{\partial r}= \cos\theta \, \frac{\partial}{\partial x} + \sin\theta \, \frac{\partial}{\partial y} $$ and $$ \mathbf{e}_\theta=(-\sin\theta,\cos\theta) = -\sin\theta \, \mathbf{e}_x + \cos\theta \, \mathbf{e}_y $$ gives $$ \frac{1}{r} \frac{\partial}{\partial \theta}= -\sin\theta \, \frac{\partial}{\partial x} + \cos\theta \, \frac{\partial}{\partial y} . $$ Or with matrix notation: $$ \begin{pmatrix} \frac{\partial}{\partial r} \\ \frac{1}{r} \frac{\partial}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{pmatrix} , $$ which gives (if we invert the matrix) $$ \begin{pmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \frac{\partial}{\partial r} \\ \frac{1}{r} \frac{\partial}{\partial \theta} \end{pmatrix} , $$ as you found using the chain rule.