Showing posts with label stokes theorem. Show all posts
Showing posts with label stokes theorem. Show all posts

The correct multi-variate mean-value theorem (no inequality)

You may notice the similarity between the mean-value theorem and the fundamental theorem of calculus. Indeed:

\[f(b) - f(a) = \int_a^b {f'(x)\,dx} \]\[f(b) - f(a) = f'(c)(b - a) \,\, (\exists\, c\,\, \rm{s.t.})\]
And naturally so: the fundamental theorem of calculus tells us that the boundary term $f(b) - f(a)$ is naturally related to $f'(x)$ on the interior -- specifically it's equal to its sum, and the mean-value theorem talks about the mean, which is proportional to the sum.

One may wonder: if Stokes' theorems (Navier-Stokes, Divergence, etc.) are the generalization of the fundamental theorem of calculus: can we make a "Stokes' theorem" version of the mean-value-theorem?

Actually, we can do better: the relationship between the mean-value-theorem and the fundamental theorem of calculus can be "suppressed" by equating the above two equations to reveal the key, new, general insight provided by the mean value theorem, which is that a function achieves its average value on a compact domain:

\[\exists\, c,\,\,  g(c) = \frac{1}{{b - a}}\int_a^b {g(x)\,dx} \]
Where we replace $f'$ with $g$. This theorem can be generalized easily to higher dimensions:

\[\exists\, c,\,\, g(c) = \frac{1}{{\left| R \right|}}\int_R {g(x)\,dx} \]
Equating with various Stokes theorems will then get you appropriate generalizations.

Why does this not work for vector-valued functions? What does this tell you about the topology of $\mathbb{R}$ vs $\mathbb{R}^n$? What is the "best" generalization you can make to vector-valued functions?

Contour Integration I: Cauchy and Morera's Integral Theorems

We're interested to find out if there exists an integral form of the Cauchy-Riemann equations. On one hand, this sounds absurd -- this is asking if there's an "integral form" of complex differentiability. On the other, the Cauchy-Riemann equations are just partial differential equations.

The standard relationship between differential and integral formulations of things is Stoke's theorem -- the theorem that tells you that adding things on a lot of tiny curves gives you a thing on a big curve. So let's see what a complex integral on a tiny square looks like.
Observe that the integral on AB is (using the midpoint as the partition tag) is $\varepsilon$ times the midpoint of $f(A)f(B)$, while the integral on CD is $-\varepsilon$ times the midpoint of $f(C)f(D)$. The sum of these is $\varepsilon$ times the line connecting these midpoints (the red arrow in the diagram below). Similarly, the sum of the other two parts of the integral is $i$ times the blue arrow in the diagram below.
Because a holomorphic function preserves squares and their orientation, these cancel out, and the integral gives zero. One can then use Green's theorem to show that the integral of a holomorphic (on $D$) function $f$ on the closed curve $\partial D$ is zero. (If you wanted to be completely formal, the equivalent would be to just apply Green's theorem and note that the local integral is zero, which is what the geometry above shows).

$$\oint_\gamma f(z)dz=0$$
Alternatively, one may write, for a simply-connected region $D$: if $f$ is holomorphic on $D$, the integral of $f$ on all closed curves contained in $D$ is zero. This is known as Cauchy's Integral theorem (or the Cauchy-Goursat theorem).

One also immediately sees that the converse holds -- if the function weren't holomorphic, the blue arrow would not be a right-angle rotation of the red one, and you could construct closed curves on which this cancellation doesn't occur. This converse -- if the integral of a continuous function $f$ on all closed curves contained in an open region $D$ are zero, then $f$ is holomorphic -- is called Morera's Integral theorem.

(The "openness" requirement in Morera's theorem is important because we want to ensure the integral is an actual global property -- that it's across some amount of "space".)

Think about how surprising this is for a moment.
  • Cauchy's theorem tells us that for a simply-connected region, existence of a derivative implies existence of a primitive. 
  • Morera's theorem tells us that for a continuous function, existence of a primitive implies existence of a derivative.
Morera's theorem does not show that a holomorphic function is infinitely-differentiable. Do you see why?

Trace, Laplacian, the Heat equation, divergence theorem

The aim of this article is to help build an intuition for the trace of a matrix, "the sum of the elements on the diagonal" -- the basic idea is that the trace is an "average" of some sort, an average of the action of an operator or a quadratic form. We'll make this idea clearer with an example from classical physics: the heat equation.



Consider an $n$-dimensional space with some temperature distribution $T(\vec{x},t)$. We wish to set up a differential equation for this function.

In the case that $n = 1$, this differential equation is exceedingly easy to write down, considering the difference $(T(x+dx)-T(x))-(T(x)-T(x-dx))$ as the double-derivative upon division by $dx^2$. More rigorously, what we're doing here is applying a localised version of the fundamental theorem of calculus. I.e. we're writing down:

$$\begin{align}
\lim_{\Delta x \to 0} \frac{1}{\Delta x}(T'(x + \Delta x) - T'(x)) &= \lim_{\Delta x \to 0} \frac{1}{{\Delta x}}\int_x^{\Delta x} {T''(x)dx}  \\
& = T''(x)
\end{align}
$$
More generally, we may consider the $n$-dimensional case.

Analogously to before, one may try to look at temperature flows in each direction -- here, we have an integral, done on the boundary of an infinitesimal region $V$ (this symbol will also represent the volume of the region):

$$ \frac{{\partial T}}{{\partial t}} = \lim_{V \to 0} \frac{\alpha }{V}\int_{\partial V} {\hat u\,dS \cdot \vec \nabla T} $$
At this point, one may apply the divergence theorem, converting this to:

$$\frac{{\partial T}}{{\partial t}} = \mathop {\lim }\limits_{V \to 0} \frac{\alpha }{V}\int\limits_V {\vec \nabla  \cdot \vec \nabla T\;dV}  = \alpha{\left| {\vec \nabla } \right|^2}T$$
In this sense, the divergence theorem is analogous to the fundamental theorem of calculus for manifolds with boundaries that are more than one-dimensional (see the bottom of the page for a link to a formalisation/an abstraction based on this analogy). But there are more ways to intuitively understand this. Note how the Laplacian is the trace of the Hessian matrix (note: we use $\vec{\nabla}^2$ to refer to the Hessian and $\left|\vec\nabla\right|^2$ to refer to the Laplacian):

$${\left| {\vec \nabla } \right|^2}T = {\mathop{\rm tr}} \left({\vec{\nabla} ^2}T\right)$$
The trace of a matrix is fundamentally linked to some notion of averaging -- the simplest interpretation of this is that it is the mean of the eigenvalues. But more relevant to our situation, it can be shown that the trace of a matrix is the expected value of the quadratic form defined by the matrix on the unit sphere -- or on a general sphere $S$:

$${\mathop{\rm tr}} A = \frac{1}{S}\int_S {\frac{{\Delta {x^T}A\,\Delta x}}{{\Delta {x^T}\Delta x}}\,dS} $$
One may check that taking the limit as $\Delta x \to 0$, substituting $\nabla^2$ for the operator and writing ${\overrightarrow \nabla ^2}f\,d\vec x = \overrightarrow \nabla  f$, one gets the original "average of directional derivatives" expression.

Can you interpret the other coefficients of the characteristic polynomial in terms of statistical ideas?


Further reading:
  • Using the "infinitesimal region" idea to define divergence, curl and Laplacian rigorously: Khan Academy
  • An abstraction based on the "analogy" between FTC, Divergence Theorem, Navier-Stokes Theorem, etc. Stokes' theorem (Wikipedia)