Let $T:\R^2 \to \R^2$ be a rotation, and $u$ be any function. Then $$ \lap (u \circ T) = (\lap u) \circ T $$ This is also true in higher dimensions.
Given that the Laplacian has this rotational symmetry, one might expect a nice cancellation / compact formula for the Laplacian in polar coordinates. We try this next.
Let $\hat x$, $\hat y$ be the unit vectors in the $x$ and $y$ direction respectively, and let $\hat r$ and $\hat \theta$ be the unit vectors in the $r$ and $\theta$ direction respectively. Explicitly, $$ \hat x = \begin{pmatrix}1\cr 0\end{pmatrix},\quad \hat y = \begin{pmatrix}0\cr1\end{pmatrix},\quad \hat r = \frac{1}{r}\begin{pmatrix}x\cr y\end{pmatrix},\quad \hat \theta = \frac{1}{r}\begin{pmatrix}-y\cr x\end{pmatrix}. $$
First we compute that $$ \grad u = \partial_x u \hat x + \partial_y u \hat y = \partial_r u \hat r + \frac{1}{r} \partial_\theta u \hat \theta. $$ Now we compute $$ \lap u = \dv \grad u = (\grad \partial_r u) \cdot \hat r + \partial_r u (\dv \hat r) + (\grad \frac{1}{r} \partial_\theta u) \cdot \hat \theta + 0 = \partial_r^2 u + \partial_r u (\dv \hat r) + \frac{1}{r^2} \partial_\theta^2 u. $$
To finish the calculation, we only need to compute $\dv \hat r$. We do this as follows $$ \dv \hat r = \dv \left( \frac{1}{r} \begin{pmatrix} x \cr y \end{pmatrix} \right) = \frac{2}{r} + \grad\left( \frac{1}{r} \right) \cdot \begin{pmatrix} x \cr y \end{pmatrix} = \frac{2}{r} - \frac{1}{r} = \frac{1}{r}. $$ Substituting back gives $$ \lap u = \partial_r^2 u + \frac{1}{r} \partial_r u + \frac{1}{r^2} \partial_\theta^2 u. $$
Here’s a superscript without math: a^b, a_b. Here they are with math:
One backslash: \( a^b, a_b \).
Two backslashes: \( a^b, a_b \).
Three backslashes: \( a^b, a_b \).
{x} produces {x} in normal mode.
In math mode it will mess things up!
However, smileys need whitespace before and after to be recognized.
So while $ {x} $ will mess things up, ${x}$ will be OK and produce ${x}$.
The code
$$ a * b + b * c \qquad a_b + b_c, \qquad a^{b + c} $$
produces $$ a * b + b * c \qquad a_b + b_c, \qquad a^{b + c} $$
Here is a labeled equation: \begin{equation}x+1\over\sqrt{1-x^2}\label{ref1}\end{equation} with a reference to ref1: \ref{ref1}, and another numbered one with no label: $$x+1\over\sqrt{1-x^2}$$ This one uses \nonumber: \begin{equation}x+1\over\sqrt{1-x^2}\nonumber\end{equation}
Here’s one using the equation environment: \begin{equation} x+1\over\sqrt{1-x^2} \end{equation} and one with equation* environment: \begin{equation*} x+1\over\sqrt{1-x^2} \end{equation*}
This is a forward reference [\ref{ref2}] and another \eqref{ref2} for the following equation: \begin{equation}x+1\over\sqrt{1-x^2}\label{ref2}\end{equation} More math: $$x+1\over\sqrt{1-x^2}$$ Here is a ref inside math: \(\ref{ref2}+1\) and text after it.
\begin{align} x& = y_1-y_2+y_3-y_5+y_8-\dots && \text{by \eqref{ref1}}\cr & = y’\circ y^* && \text{(by \eqref{ref3})}\cr & = y(0) y’ && \text {by Axiom 1.} \end{align}
Here’s a bad ref [\ref{ref4}] to a nonexistent label.
An alignment: \begin{align} a&=b\label{ref3}\cr &=c+d \end{align} and a starred one: \begin{align*} a&=b\cr &=c+d \end{align*}
This tests some of my predefined macros.
$$ a < b, \quad a \leq b \quad a \geq b \quad a > b. $$
Inline limits: $\dlim_{x \to a} \frac{1}{x}$, $\dlimto{a} \frac{1}{x}$, $\dmax_{0, 1}$.
Displayed \esssup_{x \in \R} produces
$$
\esssup_{x \in \R}
$$