Integration over arbitrary domains

In mathematical physics, we sometimes encounter situations where we have integrals of the forms:

$$\text{(case 1):}\ \ \ \ \int\limits_{D} f(x,y,z) dx dy dz =k$$ $$\text{(case 2):}\ \ \ \ \int\limits_{D} f(x,y,z) dx dy dz =\int\limits_{D} g(x,y,z) dx dy dz$$ where $D$ is an arbitrary domain (e.g. volume) of intergation over the $(x,y,z)$ variables and $k$ is some constant.

Because of $D$'s arbitrariness (i.e. can be taken small enough so that the integrand is taken outside integral sign), we usually proceed by taking $f,g$ outside the integral and ending up with the corresponding results: $$\text{(case 1):}\ \ \ \ f=k/V_{D}$$ $$\text{(case 2):}\ \ \ \ f=g$$

where $V_{D}$ is the volume $\int_{D}dxdydz$.

Questions:

(a) Is there any formal/rigorous theorems in math (e.g. in analysis) that specialise in such situations or explain its conditions in general? What branch/subbranch of math would cover such problems?

(b) If we are faced with a problem of case (2) above, for example, and we could find some kind of domains $D$ of a certain shape but arbitrary size (e.g. say $D$ as a sphere of arbitrary radius) for which both sides (integrals) are known to be equal (and hence we reach equality of integrands), will that be enough to conclude that equality of integrals will reduce to equality of integrands in general (or do we need to first show that it is also true for all other arbitrary shapes of $D$)?

(c) Finally, could the same discussion/conclusions of case (2) above be extended to the case of integrals over closed surfaces (via Divergence's theorem for instance?): $\oint\limits_{A} \boldsymbol{F}\cdot \boldsymbol{dA}=\oint\limits_{A} \boldsymbol{G}\cdot \boldsymbol{dA}$?

(Note - I have asked this question on math SE, but got no answers.)

• I think looking at how you transform between the integral and differential forms of Maxwell's equations could help with what you're asking – QtizedQ Aug 6 '17 at 21:08
• Case 1 cannot be constant (except zero) for arbitrary domains. Can you post a real life example for case 1? – lalala Aug 6 '17 at 21:32
• @lalala Please note that I have updated the question to read $f/V_{D}$ for case 1, because what is meant here is the volume of $D$, which I now denote as $V_{D}$. It should be clear now. – user135626 Aug 6 '17 at 21:49
• @lalala A real life example is a volume integral over a small enough volume to consider $f$ constant there, like density integral that gives a number $k$. – user135626 Aug 6 '17 at 21:51
• This question (v2) seems to be pure math. – Qmechanic Aug 6 '17 at 21:52

You may be interested in the book "Methods of Applied Mathematics" by Arbogast and Bona, available here as a pdf

Its quite a tough read, but of interest is proposition 1.39 on page 21, which can be restated for continuous functions that, if for all measurable $D \subset \Omega$, where $\Omega$ is the domain of our function (and a union of finite balls, so that we dont have any wierd measure zero bits), we have $$\int_D f dV = 0$$ Then $$f(x) = 0 \forall x \in \Omega$$

This does you for case 1 when the right hand side is zero. I'm not at all sure that the nonzero case is even true...

For case 2 $$\int_D f dV = \int_D g dV$$ $$\int_D f - g dV = 0$$ $$f-g=0$$ $$f=g$$

I hope that answers part a). For part b) the answer is no. If you have a series of concentric spheres for which the volume integrals are equal, then you can then conclude that the surface integrals over the spheres are equal, but the values may be distributed differently over those spherical surfaces.

For part c), if those surfaces are not necessarily closed, then a version the above theorum will apply and you have that $F=G$ (I dont have a proof of this, just intuition). If the surfaces must be closed then the theorum will not apply since we do not have the result for all domains, but we can use the divergence theorum and then the above theorum to obtain that $\nabla \cdot F=\nabla \cdot G$

Edit

Here I correct and expand on my answer to part b).

For a simply connected open set $D$ and a continuous function $f$ we have the multivariate mean value theorem for integrals, which states that there exists some $c \in D$ for which $$\int_D f(x) dV = f(c) |D|$$ where $|D|$ is the volume (specifically, the measure) of the open set.

Therefore, if we have an infinite sequence of simply connected open sets $D_0,D_1,...$ such that the sets all include some point $x_0$, and the sets 'tend towards' this point, specifically $$\lim_{n \rightarrow \infty} (\max_{x \in D_n} |x-x_0|) = 0$$ then for each domain we can specify a point $c_n$ for which $$\frac{1} {|D_n|} \int_{D_n} f(x) dV = f(c_n) =0$$ where the equality with zero is from the question. We have that $c_n \in D_n$ thus $c_n \rightarrow x_0$ as $n \rightarrow \infty$ and $f(x_0)=0$.

Thus, if you find a set of simply connected open domains of arbitrarily small size around every point, the above shows that $f=0$.

We can therefore state the following theorem:

If, for each point $x_0$ in an open domain $\Omega$, we have that, for a sequence of simply connected open subsets $D_n$, $n$ any natural number, $$\int_{D_n} f(x) dV =0$$ for some continuous function $f$, and $$\lim_{n \rightarrow \infty} (\max_{x \in D_n} |x-x_0|) = 0$$ then $$f=0$$

• Thanks for this. I agree with you on case 1, it should be k=0 or arbitrainess here is in doman shape only but fixed size. Regarding case 2, if we consider the limiting case of small sphere volume or surface, then shouldn't this lead us to equality at each point? And if we move the centre of concentric spheres to new points and take the limiting case again, wouldn't we gradually establish equality poitwise everywhere? The bigger question is: if this works in limiting case on spherical volumes, wouldn't that tell us something about cases of other shapes (e.g. small cubic volumes,etc) too? – user135626 Aug 7 '17 at 18:53
• Oh, yes. I assumed you ment that all the shapes would have the same centre, but that's not what you said. If the integral equality holds for every linear transform (translation, rotation, etc) of a particular measurable set, then it should hold for all measurable sets. – Eddy Aug 7 '17 at 18:58
• I really would like to make sure I understand you on this crucial point. So, first, would you please explain what you think the 'measurable' domain condition would mean in a real (physical) problem context? And, secondly, I am trying to use equality of integrals to reach equality of integrands. So, if I could prove equality of integrals for a particular family of domains $D$ (say $D$ set of particular concentric shapes & subset of all domains $\Omega$ ) at each point point, would that mean universal equality always for the integrads (even if later considered different shapes of domains)? – user135626 Aug 7 '17 at 19:17
• and does this theorem (integral equality holding for every linear transform of a particular measurable set means equality over all sets) havea name or reference you can refer me to? And would it be true only for the limiting case of small domains/set, or over any size? – user135626 Aug 7 '17 at 19:19
• Measurable sets include open sets, closed sets, and some more exotic sets of points. Any domain of physical interest will be measurable. I'll adf to the answer with more about part b) – Eddy Aug 7 '17 at 20:13

I am here posting a second answer, which will be a complete restructuring of the first, inclusion of discussion in the comments, and expansion of the scope of the answer. I will leave it to others to decide whether to keep or delete the other.

Firstly, it is important that "case 1" is only valid for $k=0$. "Case 2" is equivalent to it because

$$\int_D f dV = \int_D g dV$$ $$\int_D (f - g) dV = 0$$ then if "case 1" is true $$f-g=0$$ $$f=g$$ therefore "case 1" implies "case 2".

Part c) of the question is also equivalent to "case 1", by the divergence theorem (as noted in the question), if $A$ is the (piecewise smooth) boundary of the bounded open set $D$ then $$\oint_A f \cdot \hat{n} dA = \int_D \nabla \cdot f dV$$ thus if "case 1" is true and the and the integrals over all (or sufficiently many) surfaces are zero then $$\nabla \cdot f =0$$

However, if the surfaces are not necessarily closed then, whilst the above argument holds, a stronger case can be made and we will find that $f=0$. Note that this implies that $\nabla \cdot f =0$ and there is no contradiction.

Our first job, then, is to argue "case 1". This is a standard result in the field of functional analysis, see Methods of Applied Mathematics by Arbogast and Bona, available here as a pdf. Its quite a tough read, but of interest is proposition 1.39 on page 21, which can be restated for continuous functions.

Theorem 1 Let f be a continuous function whose domain is an open set $\Omega$. If for all measurable $D \subset \Omega$ we have $$\int_D f dV = 0$$ then $$f(x) = 0 \forall x \in \Omega$$

Note that 'measurable' just means that a Lebesgue volume integral can be defined over such a set, and so this is a very strong requirement on the function. In line with part b) of the question, we can construct a slightly more complicated theorem, which requires us to show the 'integral=0' condition on a smaller collection of sets.

Theorem 2 If, for each point $x_0$ in an open domain $\Omega$, we have that, for a sequence of simply connected open subsets $D_n$, $n$ any natural number, $$\int_{D_n} f(x) dV =0$$ for some continuous function $f$, and $$\lim_{n \rightarrow \infty} (\max_{x \in D_n} |x-x_0|) = 0$$ then $$f(x) = 0 \forall x \in \Omega$$

Proof For a simply connected open set $D$ and a continuous function $f$ we have the multivariate mean value theorem for integrals, which states that there exists some $c \in D$ for which $$\int_D f(x) dV = f(c) |D|$$ where $|D|$ is the volume (specifically, the Lebesgue measure) of the open set. Therefore, if we have an infinite sequence of simply connected open sets $D_0,D_1,...$ such that the sets all include some point $x_0$, and the sets 'tend towards' this point, specifically $$\lim_{n \rightarrow \infty} (\max_{x \in D_n} |x-x_0|) = 0$$ then for each domain we can specify a point $c_n$ for which $$\frac{1} {|D_n|} \int_{D_n} f(x) dV = f(c_n) =0$$ where the equality with zero is because we have asserted that the integral is zero. We have that $c_n \in D_n$ thus $c_n \rightarrow x_0$ as $n \rightarrow \infty$ and $f(x_0)=0$. Thus, if you find a set of simply connected open domains of arbitrarily small size around every point, the above shows that $f=0$.

This covers question a) and also question b), it is fine to use open domains of a particular shape and arbitrary size, so long as they can be centred around any point and limited to zero size. We could also use different shapes at different sizes, so long as the conditions of the above are met.

At this point I will sidestep slightly from analysis, and address continuum mechanics, and the approximations of mathematical physics. In physical systems there is typically a split between the statistical macroscopic properties and the exact models that must be used to capture the small scale motion. The small scale motion may be the vibration of molecules in a body of water, or the bouncing of stones in an avalanche. We typically see that the statistical macroscopic variables accurately capture the large-scale behaviour, and the details of the microscopic are unimportant.

In such a case we typically have three scales of space-time. A large scale over which the dynamics of interest occurs so that the statistical variables are rapidly varying but approximately integral=0. A mid scale over which the statistical variables are slowly varying and approximately integral=0. A small scale at which the integral is no longer valid because the small scale behaviour dominates. Note that the statistical variables will be averages over a region of space-time of the same scale as the smallest regions at which the integral equation is valid.

What is typically then done is to define a macroscopic approximation to the physical system, where we claim that, since we have $$\int_{D_n} f(x) dV \approx 0$$ for a sequence of domains satisfying $$\lim_{n \rightarrow \infty} (\max_{x \in D_n} |x-x_0|) = L$$ where $L$ is the mid space-time scale (much smaller than the length scale of the dynamics), then we can approximate by $$f(x) = 0 \forall x \in \Omega$$ Notice how similar this is to Theorem 2. It is not something that can be 'proved' mathematically (or at least, not that I have ever seen...), rather an approximation that often works.

Finally, it is worth pointing out here that none of this discussion relied on $\Omega$ being a subset of three dimensional Euclidean space. The open domain of $f$ could be, say, a surface of a fluid, or any other 'boundary' of the physical domain which will be, in itself, a physical domain. This is how we can think of boundary conditions, down to a very small area (or length, etc) of a boundary domain, the equation integral=0 holds to a good approximation. Of course, to get any of the micro-scale behaviour the integral will have to have some thickness, but the length scale across the boundary will be much smaller than the length scale along the boundary (this is what makes it a boundary), so the domains of integration will be very thin disks. We can use these in Theorem 2 (ignoring the thin thickness and the fact that the limit can only be evaluated down to some small length scale) which yields $$f=0$$ on the 'boundary'.

Case 2 is evident, if it's satisfied for EVERY possible domain $D$.

As for case 1, it looks logical too, but if you're looking for a rigurous way, maybe you can do it using the Dirac Delta Function to re-write your domain. The properties of the Diract Delta function yield your result.

• Case 2 is wrong for arbitrary functions. Counterexample: Function f is zero everywhere but at the origin it is one. Then the integral over any domain is zero. Of course if you demand continuity then the statement is true. – lalala Aug 6 '17 at 21:28
• Okay, yes, I always suppose continuous and differentiable functions, I'll specify next time. – FGSUZ Aug 6 '17 at 21:41
• @FGSUZ The question is not whether case 2 is true if equality is met for "EVERY" possible domain. The question is rather whether we can infer equality over every possible domain by having knowledge that it is true for every domain of a particular shape (which is a subset of all possible domains). For example, if it holds true for concentric spherical domains of arbitrary radius, will it automatically also hold for domains of concetric cubic shape and arbirary size? – user135626 Aug 6 '17 at 22:00
• If you always suppose continous functions how come you mention the dirac delta function? :-) – lalala Aug 6 '17 at 22:13
• I answered to the questions (a) & (c). Before you say I forgot to say the branch of Maths, I'd say "calculus", so I didn't answer to (b) because I'm not sure enough. ||| @lalala Alright, I was thinking about $f$ and $g$, but the Dirac delta is the only exception I take, as USUAL in physics (monochromatic waves, plain wavefronts, quantum mechanics, and so on.). – FGSUZ Aug 7 '17 at 12:30