Laplacian in polar coordinates I am stuck with an exercise that requires me to find the Laplacian $\Delta u=(D_x^2u+D_y^2u)$ of a 2d-function $u$ in polar coordinates (in the standard Euclidean plane).
I found the following article on the net, and tried to follow its logic, but I could not understand two steps: http://www.sci.brooklyn.cuny.edu/~mate/misc/laplacian_polarcoord_higherdim.pdf
at first, the representation of $D_x$ and $D_y$ in terms of $r, \theta$, at the bottom of page 2: 
$D_y=\sin\theta D_r+\frac{\cos\theta}{r}D_\theta$ and
$D_x=\cos\theta D_r-\frac{\sin\theta}{r}D_\theta$ . When I draw a sketch of the plane with a circle and all the coordinates, I get that it should be $D_y=\sin\theta D_r+r\ \cos\theta D_\theta$, because the larger the radius is, the greater will be the impact of a change in $\theta$ on a change in $y$. What am I making wrong here?
And then, secondly, what looks like an easy multiplication, namely taking the square of the above terms (on the top of page 3 in the link):
$D_y^2=(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)$
$=\sin^2\theta D_r^2+\frac{2\sin\theta \cos\theta}{r}D_\theta D_r+\frac{\sin^2\theta}{r^2}D_\theta^2 \\ -\frac{\cos\theta}{r^2}D_\theta + \frac{\cos^2\theta}{r}D_r - \frac{\cos\theta \sin\theta}{r^2}D_\theta$
and similarly for $D_x$.
I don't see from where the last three summands come, what I see is: $(a+b)(a+b)=a^2+2ab+b^2+c+d+e$ and I cannot see the context of $c,d,e.$
It would be great if you could explain those issues to me!
 A: Here is another approach.
Since $x=r\cos(\theta)$ and $y=r\sin(\theta)$ we get the Jacobian
$$
\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)
\end{bmatrix}\tag{1}
$$
which inverted is
$$
\begin{align}
\frac{\partial(r,\theta)}{\partial(x,y)}
&=\frac1r\begin{bmatrix}r\cos(\theta)&-\sin(\theta)\\r\sin(\theta)&\cos(\theta)
\end{bmatrix}\\
&=\frac1r\begin{bmatrix}x&-y/r\\y&x/r
\end{bmatrix}\tag{2}
\end{align}
$$
Thus,
$$
\begin{align}
\frac{\partial}{\partial x}
&=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}\\
&=\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\tag{3}
\end{align}
$$
and
$$
\begin{align}
\frac{\partial}{\partial y}
&=\frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial y}\frac{\partial}{\partial \theta}\\
&=\sin(\theta)\frac{\partial}{\partial r}+\frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta}\tag{4}
\end{align}
$$
Applying $(3)$ twice yields,
$$
\begin{align}
\frac{\partial^2}{\partial x^2}&=
\cos^2(\theta)\frac{\partial^2}{\partial r^2}
-\frac{2\sin(\theta)\cos(\theta)}{r}\frac{\partial^2}{\partial r\partial \theta}
+\frac{\sin^2(\theta)}{r^2}\frac{\partial^2}{\partial \theta^2}\\
&+\frac{\sin^2(\theta)}{r}\frac{\partial}{\partial r}
+2\frac{\sin(\theta)\cos(\theta)}{r^2}\frac{\partial}{\partial \theta}\tag{5}
\end{align}
$$
and applying $(4)$ twice gives
$$
\begin{align}
\frac{\partial^2}{\partial y^2}&=
\sin^2(\theta)\frac{\partial^2}{\partial r^2}
+\frac{2\sin(\theta)\cos(\theta)}{r}\frac{\partial^2}{\partial r\partial \theta}
+\frac{\cos^2(\theta)}{r^2}\frac{\partial^2}{\partial \theta^2}\\
&+\frac{\cos^2(\theta)}{r}\frac{\partial}{\partial r}
-2\frac{\sin(\theta)\cos(\theta)}{r^2}\frac{\partial}{\partial \theta}\tag{6}
\end{align}
$$
Adding $(5)$ and $(6)$ produces
$$
\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\tag{7}
$$

Edit:
Judging from the second part of the question, a bit more detail regarding the composition of operators giving $(5)$ and $(6)$ might be useful.

As usual, there are four terms resulting from the distribution over addition:
$$
\begin{align}
&\left(\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)
\left(\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)\\
&=\left(\cos(\theta)\frac{\partial}{\partial r}\right)\left(\cos(\theta)\frac{\partial}{\partial r}\right)
-\left(\cos(\theta)\frac{\partial}{\partial r}\right)\left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)\\
&-\left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)\left(\cos(\theta)\frac{\partial}{\partial r}\right)
+\left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)
\left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)
\end{align}
$$
In the first term, there is no interaction between the first $\dfrac{\partial}{\partial r}$ and the second $\cos(\theta)$. Thus, the first term becomes
$$
\cos^2(\theta)\dfrac{\partial^2}{\partial r^2}
$$
However, in the second term, the $\dfrac{\partial}{\partial r}$ does interact with the $\dfrac{\sin(\theta)}{r}$. The product rule says the second term becomes 
$$
-\dfrac{\cos(\theta)\sin(\theta)}{r^2}\dfrac{\partial}{\partial \theta}+\dfrac{\cos(\theta)\sin(\theta)}{r}\dfrac{\partial^2}{\partial r\partial \theta}
$$
Similar things happen in the remaining two terms since $\dfrac{\partial}{\partial \theta}$ interacts with both $\cos(\theta)$ and $\dfrac{\sin(\theta)}{r}$.
A: $D_\theta$ and $D_y$ are not changes in $\theta$ and $y$, respectively, but changes with respect to these variables. Thus, precisely because, as you write, the greater the radius, the greater the impact of a change in $\theta$ on a change in $y$, it's also true that the greater the radius, the less $D_\theta$ contributes to $D_y$, since the changes in $\theta$ and $y$ are in the denominator of these derivatives, not the numerator.
On your second question, those terms arise because the differential operators in the first factor have to be applied to the entire result of applying the second factor to the function. The first three terms are what you get if you apply the differential operators in the first factor to the original function; the second three terms are what you get when you apply them to the functions in the second factor.
