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?

  • $\begingroup$ Stoke's and its variants relate the integral over the domain to the integral over the boundary. Is that what you're asking about, or are you asking about values in the domain? $\endgroup$ May 17, 2013 at 18:30
  • 3
    $\begingroup$ Someone needs to address the nature of elliptic operators here. I believe what the OP is after is recovering interior values from boundary values alone. $\endgroup$ May 17, 2013 at 18:56

6 Answers 6


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.


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.

  • $\begingroup$ Does this idea apply to all the possible types of derivative is Stoke's Theorem? -- for all of them, do the volume elements, for instance, cancel out inside the domain and not on the boundary? $\endgroup$
    – user27182
    May 18, 2013 at 23:55
  • $\begingroup$ @user27182: Yes, roughly this is the idea behind Stokes' theorem. I've given two concrete examples to see the pattern, but there should be similar explanations when applying Stokes' theorem in any setting, not just the two I've described. $\endgroup$
    – Jared
    May 19, 2013 at 0:42

When a function has zero derivative within a volume, its values are determined by sources outside the volume, which are measured only by values of the field on the boundary.

Got that? Here's the mathematical explanation.

There are many functions $f$ that obey the basic differential equation $\nabla F = 0$. We could be talking about a scalar field, or if we extend the idea of the "gradient" to act on vector fields, then this condition is that the field is both divergenceless and curlless. Let's not fret on how the gradient can be extended in this fashion--that's a lot of tensor stuff that we don't really need to worry about. Let's just keep in mind that this is the condition being imposed.

We often say that equations of the type $\nabla F = j$ represent a field $F$ that permeates space with a source $j$. Let's understand first how that source $j$ determines the field.

To do that, we first need the Green's function for $\nabla$. We'll call it $G$, and it obeys $\nabla G = \delta$, the Dirac delta function.

This is where the generalized Stokes theorem comes in. It tells us that

$$\oint_{\partial M} G(r-r') \, dS' \, F(r') =- \int_M \delta(r-r') \, dV' \, F(r') + (-1)^{n-1} \int_M G(r-r') \, dV' \, j(r')$$

Or, equivalently,

$$F(r) = (-1)^{n-1} \oint_{\partial M} G(r-r') \, \hat n |dS'| \, F(r') + \int_M G(r-r') j(r') \, |dV'|$$

It tells that, at any point $r$, the function $F(r)$ is determined by the values on a boundary plus the values of its derivative throughout the volume.

Moreover, only when that derivative is zero is the function determined entirely by its values on the boundary.

In physics, this is common in electromagnetism problems: there may be no currents in a given volume, but there is some source outside the volume instead. These produce values of the field on the boundary, which in turn affect values within the volume.

Edit: In short, if you know a field's derivative everywhere, you should be able to reconstruct the field itself. Stokes' theorem allows us to divide that information into two categories: sources outside a given volume, and sources inside. Information from sources outside is entirely captured by the surface integral; information from sources inside has to be computed through the volume integral.

Functions entirely determined by their surface values are those that, in particular, have no sources within the volume.

  • $\begingroup$ Thank you. How does this extend to more general derivatives, though? I have seen many instantiations of this type of thing, but I don't see what connects them in a general, heuristic sense. $\endgroup$
    – user27182
    May 17, 2013 at 19:04
  • $\begingroup$ What do you mean by "more general derivatives"? Do you mean, for instance, fields that obey more complicated differential equations? Or something else? $\endgroup$
    – Muphrid
    May 17, 2013 at 19:05
  • $\begingroup$ @Murphid. You used the example of $\nabla$, but the theorem is true for many types of derivative -- are there green's functions for all of these types of derivatives too? Or did you not mean grad when you wrote $\nabla$? $\endgroup$
    – user27182
    May 18, 2013 at 23:53
  • $\begingroup$ There are Green's functions for other differential operators, but Stokes' theorem relies on using $\nabla F$ (which I replaced by $j$) specifically. Hence, if $\nabla F$ is not specified (or not easily found), it is more difficult to reconstruct $F$. $\endgroup$
    – Muphrid
    May 19, 2013 at 1:02

You have to be a bit carefull with that comparison. In complex analysis, you really have (for holomorphic functions) that the values on the boundary completely determine the values within. My interpretation of why this works there is that being differentiable is a much stronger property for complex functions than for real ones. For complex functions, it implies being representable (locally) by a power series, and by one which converges in a way that allows component wise integration and differentiation. So what this comes down to is that once you know the results of evaluating the power series on all points on the boundary, you can determine the individual coefficients and the function is thus uniquely determined.

Stoke's theorem, on the other hand, speaks about the relationship of an integral over a boundary vs. the integral of some kind of derivative over the whole volume. You can view this as the (pretty vast) generalization of one-dimensional anti-derivatives. If $F$ is the anti-derivative of $f$, i.e. $\frac{dF}{dx} = f$, then $ \int_a^b f(x) \,dx = F(b) - F(a)$. You the view the left-hand side as an integral over the "volume" [a,b] and the right-hand-side as an integral over the boundary "{a,b}".

  • 1
    $\begingroup$ Complex analysis works this way because the holomorphic functions in CA have zero derivative with respect to the vector derivative $\nabla$--they are "sourceless". Meromorphic functions have point sources. In this way, saying that holomorphic functions are differentiable is actually kinda misleading: the analogues of these functions in 3d and beyond have no divergence or curl. $\endgroup$
    – Muphrid
    May 17, 2013 at 19:00

If you want to compute rate at which something is growing inside a surface (say, ppm of some molecule in a fluid, or electrons, etc) then it is useful to know the rate at which those particles are flowing through the boundary of that surface.

Consider, for example, electric current. The current to and from the boundary of a surface indicates how much the charge inside the surface is building or diminishing.


This is because many phenomena in nature can be described by well-defined fields. When we look at the boundaries of some surface enclosing the fields, it tells us everything we need to know.

Take a conservative field, such as an electric or gravitational field. If we want to know how much energy needs to be added or removed to move something from one spot in the field to another, we do not have to look at the path. We just look at the initial and final location, and the potential energy equation for the field will give us the answer.

The endpoints of a path are boundaries in a 1D space, but the idea extends to surfaces bounded by curves and so on.

This is not so strange. For instance, to subtract two numbers, like 42 - 17, we do not have to look at all the numbers in between, like 37, right? 37 cannot "do" anything that will wreck the subtraction which is determined only by the values at the boundary of the interval.


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