Why is boundary information so significant? -- Stokes's theorem

Why is it that there are so many instances in analysis, both real and complex, in which the values of a function on the interior of some domain are completely determined by the values which it takes on the boundary?

I know that this has something to do with the general version of Stokes's theorem, but I'm not advanced enough to understand this yet -- does anyone have a (semi) intuitive explanation for this kind of phenomenon?

Solution 1:

Well it might be easier to start with the version of stokes theorem you probably know best, the fundamental theorem of calculus: $\int_a^b df = f(b) - f(a)$ (when applicable).

A sketch of (a) proof is that $\int_a^b df = \lim_{N \to \infty} \Sigma_1^N (x_i - x_{i-1})df(y_i)$, where $y_i \in [x_{i-1}, x_i]$. By the mean value theorem, we can choose $y_i \in [x_{i-1}, x_i]$ such that $f(x_i) -f(x_{i-1}) = (x_i - x_{i-1})df(y_i)$, and so in the sum you cancel a lot. But you can't continue this process by virtue past the edges, the buck stops there.

Somehow, in some of these things, you can think of the problem has being pushed off to the sides, because a little tiny "cell" and their interactions in the interior have certain nice properties that lets you do so, but the edge lacks this.

Another example is the residue theorem; nothing exciting happens except at the poles.

This might be a bit tangental, is in de Rham cohomology, contractible sets, the building blocks of manifolds are boring, but how they patch together is not. Interesting cohomology is a emergent phenomenon.

Solution 2:

Let's first present Stokes' theorem symbolically:

$$\int_{\partial\Omega}\omega=\int_\Omega d\omega$$

Intuitively, $\Omega$ is some 'domain', with boundary $\partial\Omega$, and $\omega$ is some 'function' with derivative $d\omega$. So the theorem is roughly stating "the integral of a function on a boundary is the integral of the derivative on the enclosed domain."

For more concreteness, and to see intuitively why this theorem is true, that is, why the boundary values are so important, let's consider two analogous examples from vector calculus.

Divergence Theorem

$$\iint_{\partial V}{\mathbf {F}}\cdot d\mathbf n=\iiint_V(\nabla\cdot\mathbf F)dV$$

Notice this takes the form of Stokes' theorem, where our domain $\Omega$ is now some volume $V$ in $\mathbb{R}^3$, our function $\omega$ is now a vector field $\mathbf F$, and our derivative $d$ is the divergence of a vector field. In terms of fluid flow, what does divergence measure? Imagine a small cube placed at a point $(x,y,x)$ in $V$. The divergence of $\mathbf F$ at this point is a scalar number measuring the tendency for fluid to move into or out of the cube. The right hand side of the Divergence theorem integrates this value throughout the volume $V$.

Now, if we fill the region $V$ with small cubes, it makes sense that boundaries of cubes touching each other on the interior will cancel out (fluid flow out of one is fluid flow into another), so all we need to calculate is the fluid that moves out across $\partial V$. This is exactly what the integral on the left hand side computes.

Kelvin-Stokes Theorem

$$\oint_{\partial S}{\mathbf {F}}\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot d\mathbf n$$

Notice this takes the form of Stokes' theorem, where our domain $\Omega$ is now some surface $S$ in $\mathbb{R}^3$ and our derivative $d$ is the curl of a vector field. In terms of fluid flow, what does curl measure? Imagine a small pinwheel placed at a point $(x,y,x)$ in $S$. The curl of $\mathbf F$ at this point is a vector measuring the tendency for the pinwheel to rotate. The vector has direction according to the right hand rule. The right hand side of the Kelvin-Stokes theorem integrates these vectors throughout the surface $S$.

Now, if we fill the surface $S$ with small pinwheels, it makes sense that boundaries of pinwheels touching each other on the interior will cancel out (a flow causing a pinwheel to rotate clockwise will contribute to a counterclockwise rotation of an adjacent pinwheel), so all we need to calculate is the fluid flow that moves around $\partial S$. This is exactly what the integral on the left hand side computes.