This post goes back to high school: the area of the symmetric difference of two -dimensional disks when one center is shifted a little bit. Let’s start with . So we have two intervals and . It is easy to see the symmetric difference is of length .

Then we move to : two disks and of radius and center distance . So the angle satisfies .

The symmetric difference is the union of and , which have the same area: . Note that the limit

.

So .

I didn’t try for . Looks like it will start with a linear term .

—————–

Now let be an ellipse with , and be the tangent vector at . Let be the angle from to .

Let be the arc-length parameter and be the curvature at . Alternatively we have .

The following explains the geometric meaning of curvature:

, or equivalently, . .

Proof. Viewed as functions of , it is easy to see that is equivalent to .

Note that . Taking derivatives with respect to , we get

. Then is equivalent to

, or

. Note that . Therefore , which completes the verification.

—————–

More generally let’s start with a plane curve . Then (there should be some convention to choose the signature of the curvature. We set it as ),

, and

. Then we can check that also holds.

———————–

———————

Let be a symplectic manifold and be a smooth function, be the vector field satisfying . Then is called the Hamiltonian function and the Hamiltonian vector field. It is well known that induced Hamiltonian flow preserves .

Proof in Arnold’s book. To show , it suffices to show that for all small disks and all small .

To this end let’s consider the track of under the flow , . It is easy to see . So we only need to show for any smooth circle , .

Let and . Then , since is closed. This ends the proof.

(From the same book) Let be the standard symplectic space and be a symplectic isomorphism, that is, for all . The eigenvalues of are symmetric with respect to the real axis and with respect to the unit cirle. For example forms a quadruple of ev’s.

Note that the ev’s can leave only by colliding with another ev’s and from two pairs of ev’s on we obtain one quadruple. Conversely, the conjugate pairs may collide and slide on . In particular 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 .

Sign of a simple complex ev pair . Let be the corresponding two-dimensional invariant plane. It can be shown that for all unit vectors and hence have the same sign. Then is said to be positive if (so well defined). Or equivalently, the quadratic form is positive-definite.

(Note that as . So the sign of may be changed after crossing the real axis).

Sign of a complex ev pair with multiplicity . Let be the corresponding -dimensional invariant subspace. Then is said to be positive definite sign if is positive-definite on .

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

————————

Let be a 2-form on , which can be represented by for some asymmetric matrix . Since , and hence . So there always exists such that .

A 2-form on is said to be nonsingular if the ‘s with form a 1-dimensional subspace . For example is nonsingular (for arbitrary 1-form ).

Now let be a -dimensional manifold, be a 1-form on such that is nonsingular. Then the zero-direction of is called the vortex direction of . The integral curves tangent to are called the vortex lines of . Pick arbitrary closed curve . The vortex lines passing form a vortex tube. Then

Multidimensional Stokes Lemma. The integrals of along any two curves are the same if they encircle the same vortex tube. That is, if for some piece of the vortex tube.

Proof. , since contains the vortex direction.

Consider a Hamilton function and 1-form . Then is nonsingular and corresponds to the matrix . Now it is easy to see that gives the vortex direction and the vortex line satisfies , . In other words, the vortex lines coincide with the flow lines along the Hamiltonian vector field.

Consider a closed curve consisting of simultaneous states (that is, lying in ). Note that . Along such curves and hence . In particular Hamiltionian flow preserves the integral on closed curves in .

Let . Then for any 2-disk , . So even the time-dependent Hamiltonian flow also preserves this 2-form .

$latex $