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What is $$\int_{K} e^{a \cdot x+ b \cdot y} \mu(x,y)$$ where $K$ is the Koch curve and $\mu(x,y)$ is a uniform measure look here.

Attempt: I can evaluate the integral numerically and I have derived a method to integrate $e^x$ over some cantor sets, look here. When I tried using that method to integrate the Koch Curve, I end up unable to express the integral in direct terms of its self. Here's a proof that integration can be done over the Koch Curve...

Information: I'd like a symbolic answer if its available, but infinite series/products for this integral are great too. If there's a reference that actually handles this specific function over fractals and derives a symbolic result, that's good to. Also feel free to change $K$ to any other (non-trivial of course ;) ) variant of the Koch curve if that makes it easier to compute. I warn only that because the goal is to integrate over any fractal rather than just one or two special examples, you shouldn't pick needlessly trivial examples...

Motivation: The derivation of this result allows for integration over a fractal, however the actual reason this is useful, is because of the usefulness of the exponential function. For instance, the concept of average temperature over a fractal is a very interesting concept. $e^x$ type functions allow for rudimentary temperature fields to be constructed and theoretically integrated over fractals. $e^x$ type functions are useful for many kinds of problems, but they seem to be difficult to integrate over fractals. In addition, developing a theory for integrals over fractals, requires a large library of results, and $e^x$ should definitely be included in that list of integrable functions.

Community
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Zach466920
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  • Interesting question, but what's the purpose? – user21820 Apr 28 '15 at 15:01
  • @user21820 I added my motivation. – Zach466920 Apr 28 '15 at 15:07
  • I see. Interesting! – user21820 Apr 28 '15 at 15:11
  • just for the non mathematicans, uniform measure is $dxdy$ right? – tired Apr 28 '15 at 15:14
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    @tired The short answer is yes. The long answer is no. The "measure" function just assigns mass to otherwise pathological functions like the above. For instance $\mu(K)$ is 1, which makes sense the measure of the whole object is unity. $\mu(K/3)$ is 1/4 however, since the object can be broken into 4 self-similar parts. You assign mass densities to the object using measure and integrate by adding up all of these densities...Note that regular integration of the above function leads to 0. – Zach466920 Apr 28 '15 at 15:18
  • Could you elaborate on how you define $\mu$ on $K$? – hot_queen Apr 28 '15 at 16:44
  • @hot_queen the mass is uniformly distributed across K (the koch curve) such that $\mu(K)$ equals 1 and $\mu(K/3)=1/4$. – Zach466920 Apr 28 '15 at 16:59
  • @Zach466920 What mass? $K$ isn't rectifiable so I don't see a natural measure on $K$. – hot_queen Apr 28 '15 at 17:10
  • @hot_queen What? The Koch curve at iteration n can be constructed from $4^n$ lines. Thus the measure of one iterate is $4^{-n}$. Just take the limit as n approaches infinity to get the desired measure. Its simple to see the summation of the measures is unity. Did you look at my link? It's an example of using this method. This is for the cantor set. – Zach466920 Apr 28 '15 at 17:16
  • ...you might be interested in this: http://math.stackexchange.com/questions/104210/do-integrals-over-fractals-exist – draks ... May 06 '15 at 14:32
  • I would be very surprised if there is any known closed form expression for this integral. For what it's worth (which might not be much), I computed the integral numerically for $a=b=1$ and found a value of $1.8800692$. The inverse symbolic calculator didn't find any particularly promising results. – Mark McClure May 07 '15 at 17:24
  • Looks like sort of a two-dimensional Laplace integral. Is that right? – Han de Bruijn May 08 '15 at 09:46
  • @HandeBruijn I wouldn't know to be honest... – Zach466920 May 08 '15 at 15:29
  • @MarkMcClure At this point, I'd also be interested in any infinite series/products that can represent this integral. – Zach466920 May 08 '15 at 15:31
  • @Zach466920: In your other question, where you integrate $x+y$, you use the fact that $K$ is the union of four smaller copies of itself. However, it seems to me that $K$ can also be expressed as the union of two smaller copies of itself: take $K$ and reflect it across the $x$-axis, then rotate it by $\frac{\pi}6$ and dilate by a factor of $\frac{\sqrt{3}}3$ to obtain the first copy. The other copy is then obtained from this one by reflecting it across the line $x=\frac12$. Perhaps this would simplify computations a bit? – Dejan Govc May 12 '15 at 16:20
  • Both my answers shall be withdrawn, because I think the question doesn't make sense: the integral equals an infinite sum of eponentials times an infinitesimal arc length. But the arc length of a Koch curve is infinite. Hence this "weighted" arc length is expected to be infinite as well. At least according to the "measure" standards in my universe. – Han de Bruijn May 13 '15 at 09:11
  • @HandeBruijn: The Koch curve is an object of dimension $\frac{\log(4)}{\log(3)}$, so it doesn't make sense to compute its "area" or "length" in the usual sense (just as trying to compute the length of a 2-dimensional object would give you an infinite answer at best). Instead, a fractal measure has to be used. – Dejan Govc May 13 '15 at 12:49
  • @DejanGovc: Simple question. What then is the "length" of the (0,0)-(1,0) Koch curve with help of that fractal measure of yours in that $\log(4)/\log(3)$ dimensional space of yours? Numbers please, no fancy stories. – Han de Bruijn May 15 '15 at 08:48
  • @HandeBruijn: it's $1$, by definition. To compute the "length" of its subsets, you may employ self-similarity, so for instance each of the four smaller building parts has a measure of $\frac14$. (Note that these are copies of the original curve, dilated by $\frac13$, since they are just rotated copies of the $(0,0)-(\frac13,0)$ Koch curve, so this agrees with the idea that dilating a set by a factor of $\alpha$ changes its $d$-dimensional measure by a factor of $\alpha^d$ -- in this case we have $(\frac13)^{\frac{\log 4}{\log 3}} = \frac14$.) – Dejan Govc May 15 '15 at 09:04

2 Answers2

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not an answer yet, just some thoughts.

Say our Koch curve $K$ starts at $(0,0)$, ends at $(1,0)$ and the midpoint is at $(1/2, 1/(2\sqrt{3}\;))$. Mark seems to have used this, since his computation with $a=b=1$ agrees with mine.

Self-similarity is described by two maps of the plane to itself: $$ L(x,y) = \left(\frac{x}{2}+\frac{y}{2\sqrt{3}},\frac{x}{2\sqrt{3}}-\frac{y}{2}\right), \\ R(x,y) = \left(\frac{x}{2}-\frac{y}{2\sqrt{3}}+\frac{1}{2},-\frac{x}{2\sqrt{3}}-\frac{y}{2}+\frac{1}{2\sqrt{3}}\right), $$ So $L(K)$ is the left half and $R(K)$ is the right half. Set $K$ is the unique nonempty compact set with $K = L(K) \cup R(K)$. Map $L$ shrinks by factor $1/\sqrt{3}$, reflects in the $x$-axis, rotates by $\pi/6$, and fixes the point $(0,1)$. Map $R$ shrinks by factor $1/\sqrt{3}$, reflects in the $x$-axis, rotates by $-\pi/6$ and fixes the point $(1,0)$.

The measure $\mu$ on $K$ is made up of two parts, which are images of $\mu$ under $L, R$, respectively, with half the measure. That is, for integrable $f$ we have $$ \int_K f\,d\mu = \int_{L(K)} f\,d\mu+\int_{R(K)} f\,d\mu = \frac{1}{2}\int_K f\circ L\,d\mu + \frac{1}{2}\int_K f\circ R\,d\mu $$

Now if we write $$ q(a,b) := \int_K e^{ax+by}d\mu(x,y) $$ the self-similarity shows $$ q(a,b) = \frac{1}{2}q\left(\frac{a}{2}+\frac{b}{2\sqrt{3}}, \frac{a}{2\sqrt{3}}-\frac{b}{2}\right)+\frac{1}{2}\exp\left(\frac{a}{2}+\frac{b}{2\sqrt{3}}\right)q\left(\frac{1}{2}-\frac{b}{2\sqrt{3}},-\frac{a}{2\sqrt{3}}-\frac{b}{2}\right) $$

We could use this recursively to evaluate $q(a,b)$ numerically. At each iteration, the point $(a,b)$ where $q$ should be evaluated moves closer to the origin by factor $1/\sqrt{3}$. We stop when we are "close enough" to $(0,0)$, since we know $q(0,0)=1$. But, of course, at each iteration the number of exponentials we have to evaluate doubles, so it is a slow method.

GEdgar
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  • You don't answer all my questions, but you make serious headway. I have no problem giving you the bounty. You clearly put a lot of thought into this. Good job! – Zach466920 May 13 '15 at 14:55
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This "answer" is in response to your comment that you'd be interested in seeing series/product solutions. As I'm sure you know, it's not difficult (in principle) to compute the integral of $x^p$ or $y^p$ with respect to a self-similar measure. (I have Mathematica code that automates the procedure.) Thus, we can get an approximation by simply writing $$ e^{ax+by}=e^{ax}e^{by}, $$ replacing the exponential expressions with a finite sum approximation, and then integrating. The result is: $$ \left(1+\frac{a}{2}+\frac{19a^2}{120}+\frac{3 a^3}{80}+\frac{92983 a^4}{13023360}+\frac{5935 a^5}{5209344}+\frac{618497323 a^6}{3948161817600}+\cdots\right)\times \\ \left(1+\frac{b}{6 \sqrt{3}}+\frac{b^2}{120}+\frac{b^3}{1008\sqrt{3}}+\frac{83b^4}{2604672}+\frac{601 b^5}{234420480 \sqrt{3}}+\frac{2095657 b^6}{35533456358400}+\cdots\right) $$ Unfortunately, I see no significant simplification beyond this. In particular, I am not able to find closed form expressions for the integrals of the power functions - only exact expressions for specific integers.

Mark McClure
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    I don't think the integral of the product will be the product of the integrals... Putting $a=b=1$ in your product is $1.8829$ and not the $1.88006$ that you computed for that case. My computation agrees with your $1.88006$. So for a series solution, we will also need to compute integrals of $x^py^q$. – GEdgar May 12 '15 at 15:54