## Area of the symmetric difference of two disks

This post goes back to high school: the area $\delta_d$ of the symmetric difference of two $d$-dimensional disks when one center is shifted a little bit. Let’s start with $d=1$. So we have two intervals $[-r,r]$ and $[x-r,x+r]$. It is easy to see the symmetric difference is of length $\delta_1(x)=2x$.

Then we move to $d=2$: two disks $L$ and $R$ of radius $r$ and center distance $x=2a. So the angle $\theta(x)$ satisfies $\cos\theta=\frac{a}{r}$.

The symmetric difference is the union of $R\backslash L$ and $L\backslash R$, which have the same area: $\displaystyle (\pi-\theta)r^2+2x\sqrt{r^2-x^2}-\theta r^2=2(\frac{\pi}{2}-\arccos\frac{x}{r})r^2+2x\sqrt{r^2-x^2}$. Note that the limit
$\displaystyle \lim_{x\to0}\frac{\text{area}(\triangle)}{2a}=\lim_{a\to0}2\left(\frac{r^2}{\sqrt{1-\frac{a^2}{r^2}}}\cdot\frac{1}{r}+\sqrt{r^2-a^2}\right)=4r$.
So $\delta_2(x)\sim 4rx$.

I didn’t try for $d\ge3$. Looks like it will start with a linear term $2d r^{d-1}x$.

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Now let ${\bf r}(t)=(a\cos t,\sin t)$ be an ellipse with $a>1$, and ${\bf r}'(t)=(-a\sin t,\cos t)$ be the tangent vector at ${\bf r}(t)$. Let $\omega$ be the angle from ${\bf j}=(0,1)$ to ${\bf r}'(t)$.
Let $s(t)=\int_0^t |{\bf r}'(u)|du$ be the arc-length parameter and $K(s)=|{\bf l}''(s)|$ be the curvature at ${\bf l}(s)={\bf r}(t(s))$. Alternatively we have $\displaystyle K(t)=\frac{a}{|{\bf r}'(t)|^{3}}$.

The following explains the geometric meaning of curvature:

$\displaystyle K(s)=\frac{d\omega}{ds}$, or equivalently, $K(s)\cdot ds=d\omega$. $(\star)$.

Proof. Viewed as functions of $t$, it is easy to see that $(\star)$ is equivalent to $K(t)\cdot \frac{ds}{dt}=\frac{d\omega}{dt}$.

Note that $\displaystyle \cos\omega=\frac{{\bf r}'(t)\cdot {\bf j}}{|{\bf r}'(t)|}=\frac{\cos t}{|{\bf r}'(t)|}$. Taking derivatives with respect to $t$, we get
$\displaystyle -\sin\omega\cdot\frac{d\omega}{dt}=-\frac{a^2\sin t}{|{\bf r}'(t)|^3}$. Then $(\star)$ is equivalent to

$\displaystyle \frac{a}{|{\bf r}'(t)|^{3}}\cdot |{\bf r}'(t)|=\frac{a^2\sin t}{\sin\omega\cdot |{\bf r}'(t)|^3}$, or
$\displaystyle \sin\omega\cdot |{\bf r}'(t)|=a\sin t$. Note that $\displaystyle \sin^2\omega=1-\cos^2\omega=1-\frac{\cos^2 t}{|{\bf r}'(t)|^2}$. Therefore $\displaystyle \sin^2\omega\cdot |{\bf r}'(t)|^2= |{\bf r}'(t)|^2-\cos^2 t=a^2\sin^2 t$, which completes the verification.

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More generally let’s start with a plane curve ${\bf r}(x)=(x,y(x))$. Then $\displaystyle K(t)=\frac{|\ddot{y}|}{(1+\dot{y}^2)^{3/2}}$ (there should be some convention to choose the signature of the curvature. We set it as $\displaystyle \text{sgn}(\frac{d\omega}{ds})$),
$\displaystyle \cos\omega=\frac{{\bf r}'(x)\cdot {\bf j}}{|{\bf r}'(x)|}=\frac{\dot{y}}{\sqrt{1+\dot{y}^2}}$, and
$\displaystyle -\sin\omega\cdot\frac{d\omega}{dt}=\frac{\ddot{y}}{(1+\dot{y}^2)^{3/2}}$. Then we can check that $(\star)$ $\displaystyle K(s)=\frac{d\omega}{ds}$ also holds.

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Let $(M,\omega)$ be a symplectic manifold and $H:M\to\mathbb{R}$ be a smooth function, $X$ be the vector field satisfying $dH=\imath_X\omega$. Then $H$ is called the Hamiltonian function and $X$ the Hamiltonian vector field. It is well known that induced Hamiltonian flow $\phi_t$ preserves $\omega$.
Proof in Arnold’s book. To show $\phi^\ast_t\omega=\omega$, it suffices to show that $\int_{\phi_t D}\omega=\int_D \phi^\ast_t\omega=\int_D \omega$ for all small disks $D$ and all small $t>0$.

To this end let’s consider the track $D_\tau$ of $D$ under the flow $\phi_t$, $0\le t\le \tau$. It is easy to see $\partial D_\tau=\phi_\tau D-D+(\partial D)_\tau$. So we only need to show $\int_{C_\tau}\omega=0$ for any smooth circle $C:[0,1]\to M$, $C(0)=C(1)$.

Let $\xi=\frac{\partial}{\partial s}\phi_tC(s)$ and $\eta=\frac{\partial}{\partial t}\phi_tC(s)=X(\phi_tC(s))$. Then $\int_{C_\tau}\omega=\int_0^\tau\int_0^1\omega(\xi,\eta)ds dt=\int_0^\tau\int_0^1 dH_{\phi_tC(s)}(\xi)dsdt$ $=\int_0^\tau\int_{\phi_t C} dH dt=0$, since $C$ is closed. This ends the proof.

(From the same book) Let $(\mathbb{R}^{2n},\omega)$ be the standard symplectic space and $S$ be a symplectic isomorphism, that is, $\omega(S\xi,S\eta)=\omega(\xi,\eta)$ for all $\xi,\eta\in\mathbb{R}^{2n}$. The eigenvalues of $S$ are symmetric with respect to the real axis and with respect to the unit cirle. For example $\lbrace\lambda,\overline{\lambda},\frac{1}{\lambda},\frac{1}{\overline{\lambda}}\rbrace$ forms a quadruple of ev’s.

Note that the ev’s can leave $S^1$ only by colliding with another ev’s and from two pairs of ev’s on $S^1$ we obtain one quadruple. Conversely, the conjugate pairs $\lbrace\lambda,\frac{1}{\lambda}\rbrace,\lbrace\overline{\lambda},\frac{1}{\overline{\lambda}}\rbrace$ may collide and slide on $S^1$. In particular $S$ is (stably) Lyapunov-stable if all the ev’s are distinct and lie on the unit circle.

Krein noticed that not every collision will result in one of the previous bifurcations. In fact he defined the signs of complex ev’s, and proved that the complex ev’s with same sign will ‘go through one another’ after the collision and can’t leave $S^1$.

$\star$ Sign of a simple complex ev pair $\lambda,\overline{\lambda}\in S^1$. Let $\pi$ be the corresponding two-dimensional invariant plane. It can be shown that $\omega(\xi,S\xi)\neq0$ for all unit vectors $v\in\pi$ and hence have the same sign. Then $\lambda$ is said to be positive if $\omega(\xi,S\xi)>0$ (so well defined). Or equivalently, the quadratic form $\omega(\cdot,S\cdot)$ is positive-definite.

(Note that $\omega(\xi,S\xi)\to 0$ as $\lambda\in S^1\to \pm 1$. So the sign of $\lambda$ may be changed after crossing the real axis).

$\star$ Sign of a complex ev pair $\lambda,\overline{\lambda}\in S^1$ with multiplicity $k\ge2$. Let $\pi$ be the corresponding $2k$-dimensional invariant subspace. Then $\lambda$ is said to be positive definite sign if $\omega(\cdot,S\cdot)$ is positive-definite on $\pi$.

Then it follows that, $S$ is (stably) Lyapunov-stable if and only if all the ev’s lie on the unit circle and are of definite signs.

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Let $\omega$ be a 2-form on $\mathbb{R}^{2n+1}$, which can be represented by $\omega(\xi,\eta)=\langle A\xi,\eta\rangle$ for some asymmetric matrix $A$. Since $A^t=-A$, $\det(A)=\det(-A)=-\det(A)$ and hence $\det(A)=0$. So there always exists $\xi$ such that $\imath_\xi\omega=0$.

A 2-form $\omega$ on $\mathbb{R}^{2n+1}$ is said to be nonsingular if the $\xi$‘s with $\imath_\xi\omega=0$ form a 1-dimensional subspace $\langle\xi\rangle$. For example $\omega=dp\wedge dq-\alpha\wedge dt$ is nonsingular (for arbitrary 1-form $\alpha$).

Now let $M$ be a $(2n+1)$-dimensional manifold, $\alpha$ be a 1-form on $M$ such that $\omega=d\alpha$ is nonsingular. Then the zero-direction $V_x\langle\xi_x\rangle$ of $\omega$ is called the vortex direction of $\alpha$. The integral curves tangent to $V$ are called the vortex lines of $\alpha$. Pick arbitrary closed curve $\gamma\subset m$. The vortex lines passing $\gamma$ form a vortex tube. Then

Multidimensional Stokes Lemma. The integrals of $\alpha$ along any two curves are the same if they encircle the same vortex tube. That is, $\int_\gamma \alpha=\int_\beta \alpha$ if $\gamma-\beta=\partial \sigma$ for some piece $\sigma$ of the vortex tube.
Proof. $\int_\gamma \alpha-\int_\beta \alpha=\int_{\partial \sigma} \alpha=\int_\sigma d\alpha=\int_\sigma \omega=0$, since $\sigma$ contains the vortex direction.

Consider a Hamilton function $H=H(p,q,t)$ and 1-form $\alpha=pdq-Hdt$. Then $\omega=d\alpha=dp\wedge dq-H_p dp\wedge dt-H_q dq\wedge dt$ is nonsingular and corresponds to the matrix $A=\begin{pmatrix}0 & -I & H_p\\ I & 0 & H_q\\ -H_p & -H_q & 0\end{pmatrix}$. Now it is easy to see that $\xi=(-H_q,H_p,1)$ gives the vortex direction and the vortex line satisfies $\dot{p}=-H_q$, $\dot{q}=H_p$. In other words, the vortex lines coincide with the flow lines along the Hamiltonian vector field.

Consider a closed curve $\gamma$ consisting of simultaneous states (that is, lying in $\{t=t_0\}\simeq\mathbb{R}^{2n}$). Note that $\phi_{t_0,t_1}\gamma\subset \{t=t_1\}$. Along such curves $dt=0$ and hence $\displaystyle \int_\gamma p dq=\int_{\phi_{t_0,t_1}\gamma}pdq$. In particular Hamiltionian flow preserves the integral $\int pdq$ on closed curves in $\mathbb{R}^{2n}$.

Let $\omega=dp\wedge dq$. Then for any 2-disk $D\subset \mathbb{R}^{2n}$, $\int_D \omega=\int_\gamma pdq=\int_{\phi_{t_0,t_1}\gamma}pdq =\int_{\phi_{t_0,t_1}D}\omega$. So even the time-dependent Hamiltonian flow also preserves this 2-form $\omega$.

$latex$